I
«
Presented to
TKe Library) of
TKe Ontario College of
Education
TKe University of Toronto
by
James Amoss
A TREATISE
ON
MENSURATION
FOR THE USE OF SCHOOLS.
THINTED AND PUBLISHED BT DIRECTION OF THB
DUBLIN:
ALEXANDER THOM, PRINTER AND PUBLISHER,
87 & 88, ABBEY-STREET.
Sold bt Longmans, Gkeen & Co. ; Simpkin, Marshall & Co. ;
Hamilton, Adams & Co. ; Trubneb & Co. ; \V. Kent & Co. ;
F. Warne & Co. ; Groombridge & Sons ; and J. C. Tacey, London.
John Heywood, and Abel Heywood & Son, Manchester.
Philip, Son, & Nephew, Liverpool.
Oliver & Boyd, and John Menzies & Co., Edinburgh.
George Robertson, Melbourne, Acsxr.\lia.
And all Booksellers.
1874.
OUflLIN :
ALBXANCER THOM, 87 & 88, ABBEY-STREET,
PRINTER TO THE COMMISSIONERS Of NATION-^*- •WLCAXiON.
PREFACE TO SECOND EDITION.
To tbid Edition there is em Appendix, printed in a separate
form, for the u&o of teachers, containing the leading proper-
ties of the Conic Sections, and the Demonsti'atious of the
liules of Mensuration. These were, in the First Edition,
mterspersed through the work, partly interwoven with the
cext, and partly in the shape of notes. It is hoped that the
present arrangement will better suit the convenience of botb
teachers and pupils. Several other alterations have been
made, which, it is hoped, will be found to be improvements.
Teachers should direct their pupils to learn only such
portions of the work as may be necessary for their intended
occupations : for moiit pupils, the first and tocoud nections,
(lud a few problems in the fourth and sixth will be quitt
suilicieut.
1 •">
CONTENTS.
SECTION 1.
Pkactical Geomhtht
I
SECTION U.
Mensuration or ScPEaFicies —
Cross Mnltijili cation ..... 26
To find the area of a square .... 27
To find the area of a rectangle .... 28
To find the area of a rhombun .... 29
To find the araa of a trinngle .... 29
Hanng the three •ides of a triangle, to find its area . 30
To find the area of an equilateral triangle . . 32
Given the area and altitude of a triangle, to find the base 32
Given the area of a triangle and it« base, to find rts altitude 33
Given any two sides of a right-angled triangle, to find the
third side, emd ihence iti area ... 33
Given the base and perpendicular of a right-angled tri-
angle, to find the perpendicular let fall on the hypolhe-
ntise from the right angle, Ac. ... 34
To find the area of a traperinm .... 36
To find the area of a trapezium inscribed in a circle . 36
To find the area of a trapezoid .... 37
To find the area of an irregular polygon . 37
To find the area of a regular polygon ... 39
Given the diameter of a circle, to find the circumference 41
To find the length of an arc of a circle ... 43
To find the area of a circle ... 43
Given the diameter of a circle, to find the side of a sqnaro
equal in area to the circle . . . M
•VI
UOTTETfTS.
-Mensuratiow ot SuPERFiciEs^continned).
Given tie circnmference of a circle, to find the side of a
Bqrare equal in area to the circle . . 45
Given the diameter, to find the side of an inscribed square i6
Given the area of a circle, to find the aide of an inscribed
square ....... 4fi
Given the side of a square, to find the diameter of tbe
circumscribed circle . . . . . 46
Given the side of a square, to find the circnmference of
the oircnmscribed circle .... 46
Given the side of a square, to find the diameter of a circle
equal in area to the square .... 46
Given the side of a square, to find the circumference of a
circle, whose area is equal to the square whose side is
given ....... 47
To find the area of a sector of a circle ... 47
To find the area of a pegment of a circle . . 48
To find the area of a zone of a circle ... 61
To find the area of a circular ring ... 63
To find the area of a part of a ring, or of the segment of
a sector ...... 53
To fir d the area of a lune . . . . t>4 •
To measure long irregular figures ... 66
Exercises in Mensuration of Superficies ... 67
SECTION m.
loNir Sections —
The Ellipsis
Th" Parabola
The Hyperbola
61
68
73
SECTION IV.
\Ie.nsuration of Solids —
Definition* .
To find the solidity of a cnbe
Of a pamllelopipedou
'J'apri«m
Of* cylinder
To tind the conttmt of a solid, formed by e. plane paseiug
parallri to the axis of a cylinder
80
83
85
86
i7
coNTKJrrs.
vu
MivsDiLATiftiT or SoLioi — (cootuiued).
Vb^
To find the tolidity «f a p^ruaid • • i
b-
Of a cone .... •
^f{
Of the fruBtTim of » pyramid . . , ,
h'-
Of the frudtnm of h coae . . . ,
(''■t
Of a wedge ......
Q)
Of a pri-'moid . . . . •
9-2
Of a cjlindroid . . . . <
9f
Of a uphere . . . . .
If.
Of the segment of a gphere . . . ,
9b
Of the froitnni of a spher**
96
Of a circular npindle
Of the middle Irngtcm of a circnlar spindle .
98
99
Of a spheroid ....
100
Of the middle T^ae of a spheroid
IfnJ
Of a parabolic conoid
103
Of the rhiBtnm of a parabolic conoid
103
Of a pcirabolic gpindie
104
Of the mi<idle frustum of a parabolic spindle
105
Of a hyperbolic conoid
106
Of a firoatnm of a hvperbolio conoid
Of a frustum of im elliptical Fpindle
107
107
Of a drcul&r ring ....
108
SBCnON ▼.
Im FiTB Reodlar Bopies —
To €nd the §olid contents of the regnlar bodies
To find their »npt.rficial contents
111
115
SBCnON VI,
''iBFACBs OF Solid*
To tind the Horface of a prirm
IB
Of a pyramid
120
Of a cone
121
<X a frustum of a pyramid
123
Of a fructnm of a cone
123
Of a wedire
124
Of the fru«t ~Ti of a wedge
124
Of a globe
125
Of a segment or xone of a snhent
128
Of a cylinder . .
120
Of a dxcnlar cylindm
127
Vlll
CONTESTS.
SECTION vn.
Meksdeatios of Timb£r awd ov Artiftcbbs' Work-
Description of tbe carpent<*r's rule ,
Use of the sliding rule
► -
P«ge
128
130
Timber measure
132
Carpenters and joiners' w(>rk
Bricklayers' work
Of walling
145
163
154
Of chimneys
Masons' work
156
167
Plasterers' work ,
169
Plumbers' wctk
160
Painters' work
163
Glaziers' work
163
Vaulted and arched roofe
165
SECTION
vni.
Specific Gravity —
A table of specific gravities
To find the tonnage of ships
Floating bodies
173
176
183
186
SECTION li.
VVeightfl and dimensions of halls and shells
Piling of balls and shells
Ueterminiug distances by sound*
IbS
193
197
SECTION \
iiAVQlNO —
Of the gauging rule . . . . 199
Verie's sliding rule . . . . 201
A table of luuUipliers and divisors «Dd >{auf<*^P^''^*'' ^f"
squares and cirrles ..... 20f
The g;<uguig or diagonal rod .... 214
imaging . , 225
OONTKJn*.
It
SECTION n.
SCBVEYISO
Tun
Form of a field book ..... 233
To meaiiure laud with the chain onlr . , . 23S
To siiney a finld by means of the thoodclite . . 243
To sarvey a field with crooked hedges . . . 245
To !<un-ey any piece of land from two statioat; . . 247
To surrey a large estate .... 243
To durvey a town or city . , . , 25 1
To compote the content of any gnrrey . , 252
VflSCELLANEOOS PROBLEMS . . . . • 255
A Table op tbb Areas or rar. SiawcMTS or a CiRCLBf
whosa di&meter if i . . 26<u
MENSURATION.
SECTION I.
PRACTICAL GEOMETRY.
DEFIXmOKS.
1. GEOireTRT teaches and demonstrates the properties
of all kinds of magpaitudes, or extension ; as solids, surfaces,
lines, and angles.
2. Geometry is di\'i(3ed into tVo parts, theoretical and
practical. TheoreticAl Geometry treats of the various pro-
perties of extension abstractedly ; and Practical Geometry
apjiliea these theoretical properties to the various purposes
of lil'e. ^^'hen length and breadth only are considered, the
icience ■which treats of them is called Plane Geometry ; but
when length, breadth, and thickness ore considered, the
science which treats of them is called Solid Geometry.
3. A Solid ia a figTire. or a bo<ly, having
throe dimensions, vii. length, breadth, and
thickness ; se A-
A
Th« houDdaries of a solid are surfaces nr sn^)erf cies.
PRACTICAL- GEOMETRf.
4. A Superficies, or surface, has lenglh
and breadth only ; as B. [_
B
B
The boundaries of a superficies are lines.
5. A Line is lenjjth without
breadth, and is formed by the mo- ^•"
tion of a point; as CB.
The extremities of a line are points.
6. A Straight or Right Line is the shortest distance be-
tween two points, and lies evenly between these two points.
7. A Point is that which has no parts or magnitude ; it is
indivisible ; it has no length, breadth, or thickness. If it
had len^h, it would then be a line ; were it possessed of
length and breaxltn, it would be a superficies ; and had it
len^h, breadth, and thickness, it would be a solid. Hence
a point is void of length, breadth, and thickness, and only
marks the position of their origin or termination in every
instance, or of the direction of a line.
8. A Plane rectilineal Angle is
the inclination of two right lines,
which meet in a point, but are not
in the same direction ; as S.
9. One angle is said to be less
than another, when the lines which
form that angle are nearer to each
other than those which form the
other, measuring at equal distances
from the poiuts in which the lines,
meet. TaJce Bn, Bm, Elr, and E>-,
equal to one another ; then if m r»
be greater than x n, the angle ABC
is greater than the angle FED. By
conceiving the point A to move to- £
wards C, till m n becomes equal to
X n, the angles at B and E would
then be equal ; or by conceiving the
point F to recede from D, till « n becomes equal to m n, then
the angles at li ami E would be CQuaL
PRACnCAJi OEOilETRI.
Hence it appears that the nearer the extremities of the
lines forming an angle approach each other, while the point
at which they meet remains fixed, the less the angle ; and
the farther the extreme points recede from each other, the
vertical point remaining fixed, as before, the greater the
angle.
ane figure contained
1 U. A Circle is a pi
by one line called the circuuiierence, which
13 every where equally distant from a point
rtithin it, called its centre, as o: and an axe
of a circle is any part of its circumference ;
AB.
as
11.
of
angle
The magnitude ot an
does not consist In the
length of the lines which form
it: the angle CBG is less than
the angle ABE, though the
lines CB, GB are longer than
AB, EB.
12. WTien an angle is expressed by three letters, as ABE,
the middle letter always stands at the angular point, and the
other two any where along the sides; thus the angle ABE
is formed by AB and BE. The angle ABG by AB and GB.
13. In equal circles, angles have the same ratio to each
other aa the arcs on wliich they stand, (33. VI.) Hence
also, in the same, or equal circles, the angles vary as the
arcs on which tliey stand; and therefore the arcs may be
issumed as proper measmes of angles. Every angle then is
aieasurod by an arc of a circle, described about the angular
point as a centre; tlius the angle ABE is measured by the
are AE; the angle ABG by the arc AF.
14. The circumference of every circle is generally divided
int-o 3G0 equal parts, called degrees; and every degree
into 60 equal parts, called minutes ; and each minute into
60 equal parts, called seconds. The angles are measured
by the niunber of degrees contained in the arcs wliich sub-
to nd them ; thus, if the arc AE contain 40 degrees, or the
nintli part of the circumference, the aiijle .\BE is said t*.
measure 40 deifree*.
PRACTICAL OEOMETBY.
15. When a stralglit line
HO, standing on another AB,
makes the angle HOA equal
to the angle HOB ; each of
these angles is called a right
angle ; and the Une HO i»
said to be a perpendicular to
AB. The meaaure of the
angle HOA i» 90 degrees, A O B
or the fourth part of 360 degrees. Hence a nght angle ig
90 degrees.
16. An acute angle b less than a right angle ; as AOG,
or GOH.
17. An obtuse angle is greater than a right angle ; aa
GOB.
18. A plane Triangle is the snace en-
closed by three straight lines, and has three
angles ; as A.
19. A right angled Triangle ia
that which has one of its angles
right; aa ABC. The aide BC, oj)-
posite the right angle is called the
hypothenuae ; the side AC is called
the perpendicular ; and the side AB
is called the base.
20. An obtune angled Triangle has one
of its angles obtuse ; as the triangle B,
which haa the obtuse aogle A.
21. An acute angled Triangle has all ita three anglef
acute, as b fijjare A, imu*«w^i '.^ 'ti'iiiiibou i !^.
¥RA.CTICA1> UEOiCBTRY.
22, An equilateral Ti-iaiigU has ita three /
/ .
sidea equil, and also its three angles ;
as C.
k
L
\
23. An isosceles Triangle is that which has
two of its sides equal ; as D. / X)
24. A scalene Triangle is that which
has all its sides unequal ; as £.
25. A quadrUateral fi^jure is a space included by foui
straight lines. If its four angles be right, it is called »
rectangular parallelogram.
26. A Parallelogram \a a plane figiire bounded by fouj
straight lines, the opposite ones being parallel ; that is, i/
produced ever so far, would never meet.
27. A Square is a four-«ided figure,
having all its sidea equal, and all its angles
right angles ; as H.
28. An Oblong, or rectangle, L» a
right angled parallelogram, whose length
exceeds its breadth ; as I.
29. A Rhombus IS a paralldo-
grum having all its sides equal, but
Its angles not right angles ; as K*
H
l:
\.
PRACTICAL GEOMETRY.
30. A Rhomboid is a parailelo-
gram having its opposite sides equal,
but its angles ore not right angles, \^ M
and its length exceeds its breadth ;
as M.
31. A Trapezium is a figure included
bj four straight hnes, no two of which / J^
are parallel to each other ; as N.
A line connecting any two of its opposite angles, is called
a diagonal.
32. A Trapezoid is a four-sided figiu-e / \
having two of its opposite sides parallel ; / F i
as F. /_ \
33. Multilateral Figures, or Polygons, are those which
have more than four sides. They receive particular names
from the number" of their sides. Thus, a Pentagon has five
sides ; a Hexagon has six sides ; a Heptagon, seven ; ac
Octagon, eight; a Nonagon, nine; a Decagon, ten; aa
Undecniron, eleven ; and a Dodecagon has twelve sides.
If all the sides and angles of each figure he equal, it is
called a regular polygon ; but if either or both be unequal,
an irregular polygon.
34. The Diameter of a circle is a straight line passing
through the centre, and terminated both ways by the circum-
ference ; thu« AB is the diameter of the circle. The du-jueter
divides the ciryXe into two equal parts,
each of which \3 eddied a semicircle ;
the diameter also divides the circum-
ference into two equal parts, each
containing 180 degrees. Any line
drawn from the centre to the cir-
cumference is called the radius, as
AO, OB, or OS. If OS be dra%TO
from the centre perpendicidar to AB, ^
it divides the semicircle into two equal parts. AGS and BOS,
each of which is called a quadrant, or one-fourth of the circle ;
and the arcs AS and BS cont&iu each 90 degrees, and the)
!>r»> said to be the measure of the aneles AGS and BOS.
PRACTICAL UEO\lETRI.
35. A Sector of u circle is a part of the circle com-
preliended iinder two Radii, not
forming one line, and the part
of the circ\unference between them.
From this dellnitiou it appears that
a sector may be either greater or less
than a seiui circle ; thus AOB is a
sector, and is less than a semicircle ;
and the remaining part of the circle
is a sector also, but is greater than
a semicircle.
36. A Chord of an arc is a straight line joining its estre-
Tiities, and is less than the diameter ; T S is the chord of
Jie arc T H S, or of the arc T A B S.
37. A Segment of a circle is that part of the circle con-
tained between the chord and the circumference, and may be
either greater or less than a semicircle; thus TSHT and
TABST are segments, the latter being greater than a sem«
circle and the former less.
38. Concentric circles are those hav-
ing the same centre, and the space in-
cluded between their circumferences is
called a ring ; as FE.
PROBLEM I.
To bisect a given straight line AB ; that is, to divide it
into two equal parts.
From the centres A and B, with
any radius, greater than half the
given line A B, describe two arcs
intersecting each other a1 O and S,
then the Une joining O "S> will bisect
A B
PRACnCAl. GKOMETRV.
PROBLEM II.
Through a given point x to draw a straight Une C D
parallel to a given straight line A B.
lu A B take any point *,
and with the centre s and ra-
dius s a? describe the arc o x ;
w'ith .r as a centre and the same
radius * x, describe the arc sy.
Lay the extent u x taken with the compasses from * to ^ ;
through X y draw CD, which will be parallel to AB.
PROBLEM IIL
To draw a straight line CD parallel to AB, and at a give^-
distance F from it.
In A B take any
two points X, /';
And from the two
points as centres
tvith the extent F
Uiken with the con>-
p;tsses, describe two
area s, r ; then
/
A-
.X
^
B
F-
draw a line C D
touching' these arcs at r and *, and it will be at the giver,
ilistance from A B, and parallel to it.
PROBLEM IV.
To divide a straight line AB into any number oj^ equal parts.
Draw A K making any angle with A B ; and through B
draw B T parallel to A K ; take any part A E and repeat
it as often as there are
parts to be in A B, and
from the point B on
the line B T, take B I,
I S, S V, and V T
equal to the parts taken
on the fine A K ; then
join AT, EV, GS, HI,
and KB, which wil"' di-
vide the line AB into
tho number of equal parts required, as AC, CD, DF, IB.
PBACTICAL GEOiDKl'BY.
PROBLEM V.
Prcm a given point P in a straight line A B to erect a
perpendicular.
1. When the given point is in, or near the middle of the lin'
On each side of the point P take
equal portions, P X, Py"; and from
the centres, x, f, with any radius
greater than P ar, describe two arcs,
cutting' each other at D ; then the
line joining D P will be perpendicu-
lar to A B.
Or thus:
From the centre P, TTith auj
radius P », describe an arc
n X y; set off the distance, P n
from n to X, and from a; to y ;
then from the points x and y
with the same or anv other
radius, describe two arcs in-
tersecting each other at D ;
then the line joining the points
D and P will be perpendicular
to A B.
D
/
\
•^ P y
E
2. W/ien the poin- it at the end of the line.
From any centre q out of
the line, and with the distance
^ B as radius, describe a circle,
cutting A B in /) ; draw p q O;
and the line joining the pointi
O, B, will be perpendio.nlflj* to
A B.
10
PlUCnCAL GBOmSTBT.
Or thus:
Set one leg of the compasses
on B, and vrith any extent B p
describe an arc p z ; Bet off the
hame extent from p to q ; join
p q ; from y as a centre, with the
extent p q as radius, describe an
arc r ; produce p q to r, and the
line joining r B will be perpen-
dicular to A B.
/.■
/
r
/N
X
B
PROBLEM VI.
From a given point D to let fall a perpendicular upon a
given line AB.
I. TThen the point is nearly oppc-nte the middle of the
given line.
From the centre D, with
any radius, describe an arc
X y, cutting A B in jr and w,
from X and y as centres, and
with the same distance a*
radius, describe two arcs cut-
tinsr each other at S ; then
the line joining D and S will
be perpendicular to x\ B.
A
Od
D
B
\
/
PRACTICAL. GEOMETUT.
n
2. When the point is nearly opposite the und of the given lin?,
and when the eiven line cannot be conveniently produced.
D
Draw any line D jr, which
bisect in o ; from o as a centre
with the radius o x describe an
arc cutting A 13 in j/ ; then the
line joining D y will be per-
pendicular to A B.
'J
PROBLEM VII.
To draw a perpendicular, frnm
any angle of a triangle ABC,
to its opposite side.
Bisect either of the sides con-
taining the angle from which the
perpendicular is to be drawn, as
B C in the point r ; then with the
radius r C, and from the centre
r, describe an arc cutting A B
(or A B produced 11 necessary, as
in the second figiire,) in the point
P ; the Une joining C P will be
perpendicular to A B, or to A B
produced.
PROBLEM VIIL
Upon a given right line A Hi to describe an equi
triangle.
From the centres A and B, with
the given line A B as ra^lius, de-
scribe two arcs cutting each other at
C ; then the lines d^a^^^l from the
point C to the points A and B will
form, with the given line A B, an
equilateral triangle, ad A B C,
'lalrral
12
PRACTICAL GEOMETRY.
PROBLEM IX,
fo make a triangle tchose sides shall be equal to thre> givet*
right lines AB, AD, and BD, any two of tchich are
greater tha7i the third.
From the centre A wita the ex-
tent A D, descrihe an arc, and
from the centre B with the radius
B D describe another arc cutting
the former at D ; then join D A,
D B, and the sides of the triangle
A B D will be respectiyely equal to
the three given right lines.
PROBLEM X.
Two sides AB and BC of a right
angled triangle being given, to
find the hi/pothenuse.
Place B C at right angles to
A B ; draw A C, and it will be
the hypothenuse required.
B
PROBLEM XL
The hypothenuse A B, and one tide A C, of a right angled
triangle being given, to find the otlier tide.
Bisect A B in a; ; with the
centre x, and x A as radius,
describe an arc , and with A
as a centre, and A C as radius,
describe another arc cutting the
former at C ; tlien join A C and
C B ; and A 13 C will be a right
angled triangle, and B C the
enquired &ida.
PRACTICAL OEOMKTHY,
13
angle
PROBLEM XIL
To bisect a given angle; that is, to divide it into two e^ual
parts.
Let A C B be the angle to be
bisected.
From C as a centre, with
any radius C jt, describe the
arc X y ; from the points x and
J/ as centres, with the same ra-
dius, describe two arcs cutting
each other at O ; join O C,
and it wUl bisect the
ACB.
PROBLEM XIIL
At a given point A in a given right line A B, to make an
angle equal to the givun angle C.
in
From the centre C with any ra-
flius C ijy describe an arc x y ; and
from the centre A, with the same ra-
dius, describe another arc, on which
take the distance m n equal to r it ;
then a line drawn from A throu<rh in
will nuike the angle m A « equal \a-\
the angle x C y.
PROBLEM XIV.
To make an angle containing any proposed number of
dtgrees.
1. When the required angle is less tha-t a (piadrant, as 40
degrees.
Take in the compasses the extent of 60 degrees from
the line of chords, marked cho. on the scale ; and with thii
chord of Go degrees as radius, and
the centre A, describe an arc x y ;
take from the line of chords 40 de-
grees, which set oiY from n to tn ;
from A draw a line tlu-ough m ; and
the angle m A n will coutaia *kO
detjrei^. i^
i
14
PRACnCAli GEOiCETRT.
2. When the required angle is greater than a quadrant^
as 120 degrees.
From the centre o, with the
chord of 60 degrees as radius,
describe the eemicircle yxn B ;
set off the chord of 90 degrees
from B to n, and the remain- a
ing 30 degrees from n to x \ " ^
join o X ; and the angle Boa: will contain 1 20 degrees ; or
subtract 120 from 180 degrees, and set off the remainder
(60 degrees) taken from the line of chords from ytox: then
join X 0, and Box will contain 120 degrees as before.
PROBLEM XV
4w angle being given, tu find, by a scats of chords, hiw
many degrees it contains.
From the vertex A as centre,
with the chord of 60 degrees as
radius, describe an arc x y ; take
the extent x y with the compasses,
and setting one foot at the begin-
ning of the line of chords, the other leg will reach to the
number of degrees which the angle contains : but if the
extent x y should reach beyond the scale, find the number
of degrees in x y, which deducted from 1 80, will leave the
degrees in the angle Box. See figinre to the second case of
the last Problem.
PROBLEM XVL
Upon a given right line A B, to construct a sqi/nrf.
With the distance A B as
radius, and A as a centre,
describe the arc EDB ; and
with the distance A B as ra-
dius, and B as a centre, de-
scribe the arc AFC, cutting
the former in x ; make x E
equal to a: B ; join E B ;
make x C and x D each
eijual to A F. or F a;; then join AD, DC, CB, and A D C 13
"^ be the required square.
rBJLCnCAl, OKO.M^XHX.
u
Or thutt
Draw B C at right angles to
A B, and equal to it ; then from
the centres A and C, •wntx the
radius A B and C B, describe two
arcs cutting each other at D ; join
D A and D C, which will complete
tlie square.
PROBLEM XVII.
To make a rectanfpUar parallelogram of a given length and
breadth.
lyet A B be the length, and
B C the breadth.
Erect B C at right angles to
A B ; through C and A draw
C D and A D, parallel to A B
and B C.
PROBLEM XVin.
To find the centre oj" a given circle.
^1 c
Draw any two chords A C,
C B ; from the points A, C, B,
as centres, with any radius greater
tlian half the lines, describe four
arcs cutting in r x, and y v,
draw r x and y v, and produce
them till they meet in O, which
will be the centre.
16
FaACTlC\- GEOilETRl
PROBLEM XIX.
(Jj}on a given right line A B, to describe a rhoinhus having
an angle equal to a given angle A.
A<r
^ B
Make the angle CAB equal to tlie angle at A ; make
A C equal to A B ; then from C and B as centres, valh. the
ra<3iufl A B describe two arcs crossing each other at D ;
join D C and D B, which wiU complete the rhombus.
PROBLEM XX,
To find a mean proportional between two given right linei
A B anr/ B C.
Plac« A B and B C in one
straight line ; bisect A C in
'J ; from as a centre, with
A o or C as radius, describe
ii semicircle ASC ; erect the
perpendicular BS, and it will
be a mean proportional be-
tween A B and B C ; that is,
AB: BS: : BS : BC.
PROBLEM XXL
To divide a given right line AB into two such parts ^ as shaU
be to each other as x o to o f.
From the point A draw
AS equal to x o, and pro-
duce it till F S becomes
equal to o f\ join F B,
and draw S T parallel to
F B; then will A T :
T B : :x ox of.
JC-
O
— t-
-^
PHACriCAL OKOilETRT.
17
PROBLEM XXII.
To find a third proportional to two given right lines AB, AS.
Place A B and A S bo as to
make any angle at A : from the
centre A, with the distance A S
describe the arc S D ; then draw
D X parallel to BS, and A jc wilJ
be the third proportional required ;
that is,AB:AS: :AS:A:r.
PROBLEM XXIIL
Tu Jind a fourth proportional to three given right lines.
A B, A C, and A D.
Place the right lines A B
and A C 80 as to niake any
lugle at A ; on A B eet off
A. D ; join B C ; and draw
D S parallel to it ; then A S
will be the fourth propor-
tional required, vi«, A B :
AC: : A D : A S.
PROBLEM XXIV.
In a given circle to insci'ihe a square.
Draw any two diameters AC, DB
It right angles to each other ; then
join their extremities, and the GgTire
A B C D will be a square inscribed
in the given circle.
If a line be drawn from the centre
circle.
octag
i
it
PRAUTICAL, GJiOiDiTKt.
PROBLEM XXV.
To make a regular polygon on a aiven right Unp, A B.
Diride 360 degrees by the num-
ber of sides contained in the po-
lygon ; deduct the quotient from
ISO degrees, and the remainder will
be the number of degrees in each
angle of the jx)lygon. At the points
A and B make the angles o A B
and B A each equal to half the
angle of the polygon ; then from o
as a centre, and with o A or o B
as radius, describe a circle, in which place A B continually.*
Or tL
lus:
Take the given line A B fiom the scale of equal parts,
and multiply the ntmaber of equal jiarts in it by the nimi'
ber in the tliird column of the following table, answering
to the given number of sides ; the product will give the
number of equal parts in the radius A o, or o B, v.hich
taken from the scale of equal parts in the compasses, will
give the radius, with which describe a circle, and place
in it the line A B continually, as shown in the first
method.f
* b«« Appendix, Demon^tntion i. f Kies Appendix, Demonstration 2.
PRACTICAX. OEOMETRT.
19
TABLE I.
When the aide of the polygon t$ 1,
So. or
Name of tbe
Radliii of toe circunucrlb-
1
Angle OAB, or ;
SidM.
Polygon.
tag rlrclw.
OBA.
3
Tricjon
•5773.'^0?
30
4
Tetragoii
•TWlOoS
45
5
Pentagon
•8506506
54
6
Hexagon
1, Si(fe^=>radius.
60
7
Heptagon
1 -1523825
64^
8
Oetjigon
1-3065(330
67.i
9
Nonagon
l-4GUyO-22
70
10
Decagon
l-61SfJ340
72
11
Undecagon
1-7747329
73/,
12
Dodecagon
1-9318516
75
PROBLEM XXVI.
/w a gicen circle to inscribe any regular polygon ; or, to
divide the circumference of a given circle into aiiy numbr^
of equal parts. OC
Divide the diameter AB
into as many equal parts as
the figure hfis sides ; erect
the perpendicular o x, from
the centre n ; divide the ra-
dius oy into four equal parts,
and set oflP three of these
parts from y to X ; draw a
line from r to the second di-
vision z, of the diameter A B,
and produce it to cut the
circumference at C ; join AC,
and it will be the side of the
required polygon.^
**w Arf^ndix. Demonstration 3.
\
20
PRACTICAL GEOMETRY.
PROBLEM XXVII.
To draw a straight line equal to any given are, of a circie,
A B.
Divide the chord A B into four ^ ^.^^^ -p
equal parts ; and set off one of these /^ ..^--'^x^
parts from B to D ; then join D C, a / ^-^-""''^ ^ -p
and it will be equal to the len^h of t^- -"
half the given arc nearly.*
Or thus:
From the extremity of the arc AB,
whose length is required to be foimd,
draw h. m, passing through the cen-
tre ; divide o n into four equal parts,
and set off three of these parts from n
to T» ; draw m B, and produce it to
meet A C drawn at right angles to
A m ; then will A C be nearly equal
in length to the arc A B.f
PROBLEM XXVin.
To viake a square equal in area to a given circle.
First divide the diameter A R
into fourteen equal parts, and
set off eleven of them from A
to S ; from S erect the perpen-
dicular S C and join A C, the
square of which will be very
nearly eqnnl to the area of the
given circle. J
• See Appendix, DemonBtration 4. t See Appendia. Demonstration 5
{ See Appendix. OeinoiLstration 0-
PRArnCAL 0T5OMETRT.
21
PROBLEM XXIX.
To construct a diagonal tcale.
Draw an indefinite straight line ; set off any dietance
A E according to the intended length of the scale ; re-
peat A E any number of times, EG, Q B, &c. ; draw
2 D parallel to A B at any
convenient distance ; then
Iraw the perpendiculars A C,
EF, GH, BD, &c. Divide
\ E and A C each into ten
equal parts ; through 1, 2,
3, &c, draw lines parallel to
A B, and through x, y, &c.
draw X Y, y 2., ho,, aa in the
Minexed figure.
The prinicpal use of this ^
scale is, to lay down any
line from a given measure ;
or to measure any line and
?ompare it with others. —
'Aliatever number C F re-
presents, F Z will be the
tenth of it, and the sub-
ill visions in the vertical direc-
tion FE will be each one-hun-
dredth part. Thus, if C F
be a unit, the small divisions
in C F, viz. F Z, &c. will be
lOths, and the divisions in the
altitude will be the 100th
parts of a unit. If C F be
ten, the small divisions FZ,
&c. will be units, and th<)8e
in the vertical line, tenths ;
if C F be a hundred, the
others will be tens and units.*
* H*-p An»«i)'hT. I *MtpiTirtT»ti(m 7.
S3
PRACTICAL GEOMETRT.
To take any number off the scale, as suppose 2j^, that
*8, 2*38 : place one foot of the compasses at D, and extend
'Jbe other to the division marked 3 ; then move the com
passes upward, keeping one foot on the line D B, and the
ither on the line 3 *, till you arrive at the eighth interval,
marked 88, and the extent on the compasses vnH be that
required. This, however, may express 2"3S, 23*8, or 233,
according to the magnitude of the assimied unit.
XoU. If C F -were divided into 12 equal parts, each division would
fje 1 inch, and earh vertical division 1-lOth of an inch, by making
"'■ F one foot.
PROBLEM XXX.
lo reduce a reclilinear Jigure to a similar one upon eithei
a smaller or a larger scale.
Take any point
P in the figure
A. B C D E, and
from this assumed
point draw lines to
all the angles of
the figure ; upon
one of which P A ^
take P a agreeably
to the proposed
scale ; then draw
a b parallel to AB,
b c to B C, &c.
then shall the figure a b c d e \)e similar to the original one,
and upon the required scale. Or measure all the sideh and
diagonals of the figure by a scale, and lay down the same
measures respectively from another scale, in the required
proportion.
When the figure is complex, the reduction to a different
scale is best accomplished by means of the Eidograph, an
instrument invented by Professor Wallace, or bjr means of
the improved Peutoifia^k.
niACTiCAL a;i.OJuj£TBV.
23
PROBLEM XXXL
To divide a circle into any number of equal parts, hainru
their perimeters equal also.
Divide the diameter A B
into the required number of
equal parts, at the points C,
D, £, &c. ; then on one side
describe the semicircles 1, 2,
3, 4, &c. and on the other side
of the diameter describe the
semicircles 7, 8, 9, 10, &c. on
the diameters B F, BE, B D,
B C, &c. ; so shall the parts
I 11, 2 10, 3 y, 4 8, &c. be
o^ual both in area and perimeter — I..hblie'8 GEOitETRY.
24
HKACTIt'ii, OTBOitETKY.*^
MENSURATION OF SUPERFICIES.
SECTION II.
The area of any plane figiire is the space contained withis
its boundaries, and is estimated by the number of square
miles, square yards, square feet, &c. which it contains.
I.
Long Measvre.
12 Inches
3 Feet
6 Feet
l(j{ Feet Eng.
5^ Yards
40 Percbt-a .
8 Furloiig* ,
1 Foot
1 Yard.
1 Fathom.
I Pole or
Perch.
1 FnrloDg.
1 Mile.
II.
Square Mfosure,
144 Inches
9 Feet .
36 Feet . .
272} Feet Eng.
301 Vards
1600 Perches .
64 Fnrlongs
. 1 Fool.
. 1 Yard.
. 1 Fa;hom.
I 51 Pole or
S "I Perch.
1 Furlong.
. 1 Mile.
In Ireland 21 feet make 1 pole or perch, and 7 yard-s
therefore will make a pole or perch. There are other
vneasures used, for which see Arithmetical Tables.
Land is generally measured by a Chaiuy of 4 poles, or 22
yards ; it consists of 100 links, each link being "22 of a yard.
See Section XI. Surveying.
Duodecimals are calculations by feet, inches, and parts,
which decrease by twelves : hence they take their name.
Multiplication of feet, inches, and parts, is sometimes
called Cross Multiplication, from the factors being multiplied
crosswise. It is lised in finding the contents of work done
by artificers, where the diniensions are taken in feet, inches,
and parts.
icbnsuratiox of surerficiks, 26
Rule.
I. Write the multiplier under the multiplicand in
such a manner, that feet shall be under feet, inches under
inches, &c.
II. Multiply each term of the multiplicand by the number
if feet in the multiplier, proceeding from ng'ht to left ;
carry 1 for every 12, in each product, and set down the
remainder uader the term multiplied.
III. Next multiply the terms of the multiplicand by
the number imder the denomination inches, in the multiplier ;
carry 1 for every 12, as before, but set dowTi each remainder
jne place farther to the rig'ht than if multiplying' by a nuin-
I er under the denomination feet.
IV. In like manner proceed with the number in the
oiultipher under the denomination parts or lines, remem-
oeriiig to set do^^-n each remainder one place farther to the
right than if multiplying by a number under the denomin-
ation inches. And so on with numbers of inferior denomin-
ations.
V. Add the partial products thus placed, and their gum
w^ill be the whole product.
IN CU0S8 MULTIPLICATION IT IS USUAL TO SAY,
Feet multiplied by feet, give feet.
Feet by inches, give inches.
Feet by parts, give parts.
Inches by inches, give parts.
Inches by parts, give tliirds.
Inches by thirtls, give fourths.
Parts by parts, give fom-ths.
Parts by thirds, give fiftlis.
Parts by fourths, give sixths, &c.*
I — — — ^^— ^■—
* In multipiicatioo, the multiplier mast always b« • number o(
times ; to talk of multiplying feet by feet, ic. ii absurd, for ■what notion
cuQ b« formed of 7 feet taken a times ? However, sincr Che above
•jasily lugj^esti the correct meaning, and ii « concise method of expreuing
the i-ulei. It has been thought proper to retsin it. i}«e Appendix, Demou-
'trction 8.
o2
26 MENSURATION OF SUPERFICIES.
1. Multiply 7 feet 9 inches by 3 feet 6 inches.
P. L
7 . 9
3 . 6
23 . 3
3 . 10 . 6
27 . 1 . 6 Ans.
p. I. p. F. I.
2. Multiply 240 . 10 . 8 by 9 . 4 .
9.4.6
p.
2168 .0.0
80 . 3 .6.8
10 . .5.4
2258 .4.0. OAns.
I. p. F. I. p. F I.
•0 ith.
3. Multiply 8 .
5 . by 4 .
7. Ans. 38 .
6.
11.
4. Multiply 9 .
8 . by 7 .
6. — 72 .
6.
5. Multiply 7 .
6 . by 5 .
9. 43 .
1 .
6.
6. Multiply 4 .
7 . by 3 .
10. 17 .
6.
10.
7. Multiply 7 .
5 .9by 3.
5.3. 25 .
8.
6.2.3.
8. Multiply 10 .
4 . 5 by 7 .
8.6.— 79 .
11 .
0.6.6.
9. Multiply 75 .
7 . by 9 .
8 . 0. — 730 .
7.
8.
10. Multiply 57 .
9 . by 9 .
5 . 0. 543 .
9.
9.
11. Multiply 75 .
9.0byl7.
7 . 0. 1331 .
11 .
3.
12. Multiply321 .
7 . 3 by 9 .
3 . 6. —2988 .
2.
10.4.6.
13. Multiply 4 .
7 . 8 by 9 .
6. 44.
0.
10.
14. Mtdtiply 39 .
10.7bYl8.
8.4.— 745 .
4 .
10.2.4.
.Vj/f J— All these can be soked by tJie methxxl of aliqnot parts, th.M :—
MENSURATION OP GUPERFICIFfi.
27
F. ' ' F.
16. Multiply 368 . 7 . 5 by 137 8.4
137 . 8 . 4
6'= i
2' = i
4"= )
6' = l
2576
1104
368
. . 184 .
3.8.6
61 .
5 . 2 . 10
10 .
2 . 10 . 5 .
8
68 .
6
11 .
5
3 .
9 . 8
.
11 . 5
A
ns. 50756 .
7 . 10 . 9 .
8
1
PROBLEM I.
Tojind the area of a square.
Role. Multiply the length
of the side by itself, and the
product will be the area.*
1. Let the side of the square
A.BCD be 6 : what ia its area?
Ans. 6 X 6 = 36, the area,
2. Wliut is the area of a square
whose side is 15 chains?
Ans. 225.
'3. What is the area of a square
whose side ia 7 feet 9 inches?
Ans. 60iV»
6
'
3)
C
4. What ia the area of a square whose side is 4769 links ?
Ans. 22743361.
* See Appendix, DemonstratioD 1.
i
as
MENSUEATION OF SUPERFICIES.
PROBLEM II.
To find the area of a rectangle.
Rule. Multiply the length of the rectangle by its breadth
and the product will be the area.*
A.
12
B
,
1 _
£
C
1. Let the sides of the rectangle A B C D be 12 and 9.
what is its area? -4»*. 12 x 9= 108, the area.
2. \Vhat is the superficial content of a plank, whose length
is 5 feet 6 inches, and breadth 7 feet 8 inches?
Ans. 42 feet 2 inches.
3. What is the area of a field whose bourularies form a
rectiingle, its length being 176 links and brea<ith 154 links V
Ans. "27104 of an acre.
4. What ifl the superficial content of a floor, whose length
IB 40 feet 6 inches, and breadth 28 feet 9 inches?
Ans. 1164 feet, 4 inches, 6 part*.
* See Appendix, DemoMtration 8.
MENSURATION OF SUPE&FICEES.
29
PROBLEM III.
To^nd the area of a rhombus.
Rule. Multiply the j^
lenj^-th by the perpen-
dicular breadth, and the
product will be the
area.*
1. What is the area
of a rhomb ufl, whose
aide is 16 feet, and per-
pendicular breadth 10
feet? Ans. 16 x 10
= 160 feet, the area.
2. What is the content of a field in the form of a rhombus,
whose length i« l-\j chains, and perpendicidar height 57
chains ?
Ans. 43-32 chains.
3. VMiat is the area of a rhombus, whose side is 7 feet
6 inches, and perpendicular height 3 feet 4 inches ?
Ans. 25 feet.
4. V\Tiat is the area of a rhombus whose length is 3 yards,
and perpendicular height 2 feet 3 inches ?
Ans. 20 feet 3 inches.
PROBLEM IV.
To find the area of a triangle.
Rule. Midtiply the base by the perpendicular height, and
llvide the product by two for the area.^
1. The base of a triangle is 76 "5 feet, and perpendiculai
92*2 feet ; what is its area ?
Ans. 76-5 x92-2-^2 = 3526-65 square feet, the area.
* See Appendix, Demonstration 9.
t See Appendix, Demonatruiior ^.
30
MENSUBA'j'iON OF SUPERFICIES.
2. The base cf a triangle is 72-7 yards, and the perpen
dicular height of 36*5 yards ; what is its area ?
Ans. 1326-775 yards.
3. Tlie base of a triangular field is 1276 links; and
perpendicular 976 links ; how many acres in it ?
Ans. 6 acres 36*3008 perches.
4. The base of a triangle measures 15 feet 6 inches, and
the perpendicular 12 feet 7 inches ; what is its area ?
Ans. 97 feet 6;^ inches.
PROBLEM V.
Earing the three sides of ant/ triangle given, tojindits area
RxjLK T. From half the sum of tlie three sides subtract
each side separately, then multiply the half Bxaa and the
tluee remainders together, and the square root of the last
product will be the area of the triangle.*
Rule II. Divide the difference between the squares oi
two sides of the triangle by the tliird side ; to half this third
side add half the quotient, and deduct the square of this sum
from the square of the greater side, the remainder will be
the square of the perpendicular, the square root of which,
mvdtiplied by half the base, wUl give the area of the
triangle.f
* 8e« Appendix, Demonstration 11.
t Se« Appendix, D«monftration 12.
MEN8URAT10:i OV SUPERFICTEB. 31
1. Given the side A B = 9-2, B C = 75, and A C = 5-6;
c\3quired the area of the triangle ?
9-2
7-5
5-5
Sum 22-2
I Sum 11-1— 9-2=l-9) then ^ (11-1 X 1-9 X 3-6 X 5-6)
^ 425-1744=20-619 the area
Rule I.
11-1— 9-2=l-9)
IM— 7-5=3-6[- =
ll-l— 5-5 = 5-6)by
A^ain, 9-2'— 7-5 ^=84-64— 56-25 = 28-3«); then 28-39
-i-5 5 = 5" 161818, quotient.
Now (5'l6l818-f.2) + (5-5-J-2) =^ 2580909 + 2-75 =
5-3309 = half quot. plu3 half third side: then 84-64 —
28-41849481=56-22150519, and y' 56-22150519 = 7*498
= perpendicular; then 7-498 X 2-75 = 20-619 the area as
before.
2. WTiat ia tbe area of a triangle whose sides are 50, 40,
and 30 ? Ans. 600.
3. The sides of a triangular 'Md are 4900, 5025, and 2569
links ; how many acres does it contain ?
Ans. 61 acres, 1 rood, 39-68 perches.
4. What is the area of an isosceles triangle, whose base is
20, and each of its equal sides 15? Ans. 111-803.
5. How many acres are there in a triangle, whose three
sides are 380, 420, and 765 yards ?
Ans. 9 acres 38 poles.
6. How many square yards in a triangle, whose three
sides are 13, 14, and 15 feet ? Ans. 93 square yards.
7. How many acres, &c. in a triangle, whose three sidee
are 49, 50-25, and 25-69 chains?
Ant. fcl acres, 1 rood, 39-68 perchec
h(3 MENSURATION OP SUPERFICIES.
PROBLEM VI.
To find the area of an equilateral triangle.
Rule. Square the side, and from this square deduct its
fourth part ; then multiply the remainder by the fourth part
of the square of the side, and the square root of the product
will give the area.* Or multiply ^ by -v^ 3 for the area.t
1. Each side of a triangtdar field, ABC, measures 4 pferches,
what is its area ?
4^ = 16, then 16^4 = 4 and 16 - 4 = 12 : thee
12 X ^4p = 12 X 4rr48, and V48 = 6-928, the area.
2. How many acres in a field of a triangular form, each
of whose sides measm*es 70 perches ?
Ann. 13 acres, 1 rood, 1 perch.
3. Tlie perimeter of an equilateral triangle is 27 yards,
what is its area ? Ans. 35-074.
Xote. When the triangle is isosceles, the perpendicular is equal to the
iquare root of the difference between the squares of either of the equal sidesi,
uid half the base.
PROBLEM Vn.
Given the area and altitude of a triangle, toji^id the base.
Rule. Divide the area by the C
altitude or perdendicular, and double
the quotient will give the base.J
1. Given the area of a ti-iangle
= 12 yards, and altitude = 4 ; what
is its base?
Ans. 12-^4 = 3 ; then 3x2 = 6
yards, the base, AB. A jT
2. A surveyor having lost his field book, and requiring
^ See Appendix, Demonstration 13.
+ See Appendix, Demonstration 14.
X See Appendix. DomonstratioQ IS,
MBKSUIUTIO:* OF SUPERFICIES. 33
the base of u tnang^lar field, whose content he kiiew from
recollection was 14 acres, and altitudft 7 yards, how much ia
the base? Ans. 19360 yards.
PROBLEM VIII.
Given the area of a triangle, and its base, tojind its altitude.
Rule. Divide the area by the g^ven base, and doable
the quotient will give the perpendicular.
The reason of this rule is manifest, from tfcfl lart.
1 . Given the area of a triangle = 1 2, iud its base = 6 ;
what is its perpendicular height?
Ans. 12-1-6 = 2 ; then 2 x 2 = 4 the altitude.
PROBLEM IX.
Given any two sides of a right angled triangle, to find the
third side, and thence its area.
RUL.E.
I. To the square of the perpendicular add the square
'jf the base, and the square root of the sum wHl give the
hypothenuse.
IL The square root of the difference of the squares of
the hypothenuse, and either side will give the other.
III. Or multiply the sum of the h^'pothenuse, and either
side, by their difference ; and the square root of the product
will g^ve the other.*
1. Given the base A C 3, the per-
pendicular C B 4 ; required the hypo-
thenuse A B ?
3- + 4- = 25 ; then ^ 25 = 5,
the hypothenuse AB.
2. Given A B 5, A C 3 ; required
CB?
5* - 3* = 16; then ^/ 16 - 4
the side B C; or, (5-|-3) X G*^ -3)= /
8x2=16; then ^ 16 = 4, us before. A^
3. Given A B 5, B C 4 ; required AC?
5* - 4* = 9 then -y/ 9 = 3 the side A C ; or (5 -f- 4)
X (5-4) = 9x1=9; then v^ 9 = 3, as before. And
3 X 4-f-2=6 the area of the triangle.
* Sae Appeadiji. I >«n)/in8tniDaD 16.
34 jiKNsuKAnux oy buperficies.
4. The wall of a building on the brink of a riyer is 1 20
feet, and the breadth of the river is .^0 yards ; what is the
length of the chord in feet that will *reach from the top of
the building across the river? A)}s. 241*86 feet.
5. A ladder 60 feet long, wiU reach to a window 40 feet
from the flags on one side of a street, and by timiing the
ladder over to the other side of the street, it wUl reach a
window 50 feet from the flags ; required the breadth of the
street? Ans. 77-8815 {eet.
6. The roof of a house, the side walls of which are the
game height, forms a right angle at the top, the length of
one rafter being 1 feet, and its opposite one 1 4 feet ; whaJ
is the breadth of the house ? Ans. 17*204.
PROBLEM X.
Giveii, the base and perpendicular of aright angled triangle, to
Jind tlie perpendicular let fall on the hypothenuse frcm tho
right angle; and also the segments into which the hypo-
thenuse is divided by this perpendicular.
Rule. Find the hypothenuse by Prob.
IX. Then divide the square of the greater
side by the hypothenuse, and the quotient
will give the greater segment, which de-
ducted from the entire wiU give the less.
Having found the segments, multiply them
together, and the square root of the pro-
duct wiU give the perpendicular.*
1. Given ACS yards, and C B 4 yards; required the
segments B D, D A, and the perpendicidar D C.
3* + 4* = 25 : then ^ 25 = 5 = A B.
4*-^5 = 16-^5=3-2=: B D ; then 5—3-2=1 •8= A D.
Again, 3-2 X 1'8 = 5-7G ; then -/ 5-76 = 2-4 = D C.
2. The roof of a house whose side walls are each 30 feel
high, forms a right angle at the top ; now if one of the
rafters be 10 feet long, and its opposite yoke fellow 12,
required the breadth of the building, the length of the prop
set upright to support the ridge of the roof, and the part of
the floor at which it must be placed ?
Ans. Breadth of the building 15-6204 feet, greater segment
* S<« Appendix^ Demonstntioa 17.
MENSURATION OF 8UPEBFICIZS. 35
9*2186 feet, lesser segment 6'4018 feet, and length of the prop
37-68 feet.
PROBLEM XI.
Tojind the area of a trapezium.
Rule. Divide the trapezium into two triangles, by joining
two of its opposite angles ; find the area of each triangle, and
the sum of both areas wiU give the area of the trapezium.
Or,
Draw two perpendiculars from the opposite angles to the
diagonal ; then multiply the sum of these perpendiculars b>
the diagonal, and halt the product vriU give the area.*
1. In the trapezium ABCD, the diagonal AC is 100 yards,
the perpendicular DE 35, and BF 30 ; what is its area ?
D E = 35 ,
B F = 30 ■"
65
100
2)6500
3250 the area.
2. What is the area of a field, whose south side b 2740
links, east side 3575 links, north side 3755 links, west side
4105 linkfi, and the diagonal from south-west to north-east
1835 links? Am. 123 acres 11-8633 perches.
3. In the trapeziimi ABCD, the side A D is 15, DC 13,
C B 1 4, and A B 1 2 ; also the diagonal A C 1 6 ; what is its
area? Ans. 172-5247.
4. In the trapezium ABCD, there are given A B 220
yards, DC 265 yards, and A C 378 yards; also A F 100
yards, and E C 70 yards ; what is its area ?
Ans. 85342*2885 yards =17 acres, 2 roods, 21 perches.
5. In the trapezium ABCD, there are given A B 2'20
yards, D C 266 yards, B F 195-959 yards, D E 255*5875
yards; also F E 208 yards ; required the area of the trnjie-
rimn ? Ans. 85342-2885 yartis.
* Se« Appendix, DemoiutrMiuu 12.
36 MEXSURAIION Oe SUFERFICTES.
6. Suppose ia the trapezium A B C D, on account of ob-
stacles, I can only measure A B, D C, B F, D E, and F D,
which are respectively 22 yards, 26 yards, 19 yards, 25 yards,
and 32 yards ; required the ai-ea ?
Ans. 840-55 squaie yards.
PROBLEM XII.
ToJ'md the area of a trapezium inscribed in a circle, or of
any one whose opposite angles are together equal to two
right angles.
Rule. Add the four sides together, and take half the sum,
from this half sum deduct each side separately ; and the
square root of the product of the four remainders will give
the area of the trapezium.*
1. WTiat is the area of a four-sided field, whose opposite
angles are together equal to two right angles, the length of the
four sides being as follows, viz. A B 12'o, AD 17, DC
17-5, and B C 8 yards?
i^'5
17
17-5
8
2)55
27-5
27-5 27-5 27-5
12-5 17 17-5
15 X 10-5 X 10 X 19-5 = 307 12-50; then v/
307 12-50= 175-25, the area in yards.
2. There is a trapezium whose opposite angles are together
equal to two right angles ; the sides are an follows, viz. A H
25, A D 34, D C 35, and B C 16 ; required its area?
Ans, 700-99.
"Uat- Appeiutix. DemoiutraMan 19.
liKNSURATION OP SUPERFICIKB. "*»
PROBLEM XIII.
To find the area of a trapezoid.
Rule. Multiply half the sum of the two parallel sides bj
the perpemlicular distance betvppen tliem, aud the product
wriU give the area,*
1. Let A B C D be a i^ [) (;
trapezoid, the side AB = 40,
DC = 25, CP= 18; re-
quired the area?
40
25
£ m
z\ z-1'
65^2=32-5 X 18 = 585 area.
2. VV'hat is the area or a trapezoid, whose parallel sides
art 750 and 1225 links, and the perpendicular height 1540
links? ' Ans. 15 acres 33*2 perches.
3. Wliat is the area of a trapezoid whose parallel sides are
4 feet 6 inches, aud 8 feet 3 inches ; and the perpcndiculai
haight 5 feet 8 inches? Ans. 36 feet 1^ inches.
4. What is the area of a trapezoid whose parallel sides est-
1476 and 2073 yards, and perpendicular height 976 yards?
Ans. 220 acres, 3 roods, 25 perches, 7 yards Irish.
PROBLEM XIV.
To find the area of an irregular polygon.
Rule. Divide the 6gure into triangles and trapeziums,
And find the area of each separately, by Problem IV. or XI.
Add these areas together, and the sum will be the area of the
polygon.f
I . WTiat is the area of the irregular polygon ABCDEFGA,
the following lines being given i
* S«e Appendix, Demoiutntioii 20.
t la fioding the aipa of an irregular figure, draw a line through th«i
extreme angles of the tigure, on -which let fall perpendicular* from all tb«
other arijles of the polyj;on, which will divide it into triaaglei aod tntp*-
soidi* ; then find the area of theM hv Problonu I V. and XIlL
38
.yIESSURATION of supeefictes.
A0= 9
GB = 29
Cn =11
G C = 28-4
¥ X = 14-5
C 1/ = 13
FD = 35
Ez = 7-4
A0= 9
C» = 11
2)20 sum
10 half
29 diag. Q B
290 = area of A B C G A.
Cy = 13
Ez = 7-4
2)20-4 sum
357-0 area of F C D E F.
F ar = 14-5
1 G C = 14-2
205-9 area of G F C.
290 = area of A B C G A*
357 = area of F C D E F
205-9= prea of GFC
/.n». 852-9 = area of A B C D E F O A.
IIEXSURATIOW OP SUPERFICIES.
39
2. In a five-sided field G C D E F G there b G C = 28
perches, Fx = 1 4 perches, Cy = 13 perches, zE = 1 perches,
and FD = 35 perches ; required ita area, ?
Ans. 3 acres, 1 rood, 26 perches.
3. In the annexed figiire, there are given in perches,
AX = 15
A P = 17
F R = 10
X R = 8
PS = 14
ET = 12
R T = 14
SD = 12
B P = 20
TD = 6
GX = 5
C S = 14
Required the area?
Ans. 4 acres,
3 roods, ly^ pe
PROBLEM XV.
Tojfind the area of a regular polygon.
Rule I. Add all the sides together and multiply half the
8um by the perpendicular drawn from the centre of the
polygon to the middle of one of the sides, and the product
will give the area. This perpendicular is the radius of the
inscribed circle.
Rule II. Multiply the square of the side of the polygon
by the number standing opposite to its name in the following
table, under the word area, and the product will give the area
of the polygon.
Rule III. Multiply the side of the polygon by the number
;tanding opposite to its najne iii the column zi the following
40
ilEXSURATION OP SUPERFICIES.
table, headed " Radius of inscribed Circle," and tlie product
will be the perpendicular from the centre of the polygon to
the middle of one of its sides ; then multiply half the sum
of the sides by this perpendicular, and the product snll give
the area.*
TABLE II.
When the side of the polygon is 1.
I Sides.
No. of Radius of Inscrib-
ed Circle.
Aret of Polygon.
3
4
5
6
7
8
9
10
11
12
0-2886751
0-5000000
0-6881910
0-8660254
1-0382617
1-2071068
1-3737387
1-5388418
1-7028437
1-8660254
0-4330127 =|f tan. 30"=^^/^
1-0000000 = I tan. 45'^= 1 x 1
1-7204774 =1 tan. 54''=|^ (l+f>/5)
2-5980762 =|f tan.60" = f>/3
3-6339124 ='|tan.64°f
4-8284-271 =f tan.67"i^=2x(l+v/2)
6-1818242 = I tan. 70"
7.6942088 =1^' tan. 72°=!^/ (5+2^5) '
9-3656404 = 'J tan. 7^5"-/^
11-1961524 =!V tan.75' = 3x(2+v'3)
Note. The radiua of the circumscrihed circle, when the side of th«
polygon IB 1, may be seen in Table I.
The expressions in the fourth column may be seen in Triponometrt/, tc
which the pupil is referred for a full investigation of them. The tangents ol
the angle O a C in the heptagon, nonagon, and undecagon, are extremely
difficult to be found without a table of tangents.
1. The side of a pentagon is 20 yards, and the perpendi-
cular from the centre to the middle of one of the sides is
13-76382; required the area?
By Rule I. 20x 5 x 13-76382 -j- 2 = 1376-382 h-2 =
688 191. Ans.
By Rule II. 20 x 20 X 1*720477 = 688-19, the area aa
before.
2. The side of a hexagon is 14, and the perpendicular from
the centre 12-1243556; required the area? Ans. 509-2229352.
3. The side of an octagon is 5-7, required its area ?
Am. 156-875596479.
* S«o Appendix. DeniuDStration 21.
MENSURATION OF SUPERFICIES.
41
c^^'i^.
4. The 9id6 of a hepta<roai8 19*38 yards, what is its areat
Ans 1364-84.
5. The side of an octagon is 1 feet, what is its area ?
Ans. 482-84271.
6. The side of a notdgon is 50 inches, what is its area ?
Ans. 15454*5605.
7. The side of an undecagon is 20, what is its area ?
Ans. 374G-25616.
8. The side of a dodecagon is 40 yards, what is its area ?
Ans. 17913-84384.
PROBLEM XVI.
'■liven the diameter of o circle, to Jind the circumference;
or the circumference to Jind the diameter, and thence the
area.
Rule.*
I. Say as 7 : 22 : : the given dia-
neter : circuxoference.
Or, as 1 13 : 355 : : the diameter :
■he clrcuraference.
Or, as 1:3-416:: the diameter :
the circmnference.
II. Say as 22 : 7 : : the given cir-
cumference : the diameter.
Or, as 355 : 113 :: the circumfe-
rence : the diameter.
Or, as3-14l6: I:: the circumference : the diameter.
1. The diameter of a circle is 15, what is its circumfe-
rence ?
7 : 22 : : 15 : 22 x 15 -h 7 = 330 -i- 7 = 47-142857.
Or, 113 : 355 :: 15 : 355 X 15 4- 113 = 5325 -r- 113
= 47-124.
Or, 1 : 3-1416:: \5 : 3-1416 x 15 = 47-124.
2. The circumlerence of a circle is 80, what is its dia-
meter ?
22 : 7 : : 80 : 7 X 80 -^ 22 = 25-45.
355 : 113 :: 80 : 113 X 80 -1. 355 = 25-4647.
3-1416 : 1 : : 80 : 80 -t- 3-1416 = 25-4647.
•S«e App«ndiz, r>»muaMf»uon 22.
D 2
42
MEN9UKATI0N OP SUPERFICEE8.
3. What Is the circumference of a circle whose diameter
IB 10? Ans. 31-4285.
4. WTiat is the diameter of a circle whose circumference
b 50 ? • Ans. 15-909.
5. The diameter of the earth is 7958 miles, what is it3
circumference? Ans. 25000-8528 miles.
6. The circumference of the earth being 25000'8528
miles, what is its diameter ? Ans. 7958 miles.
PROBLEM XVII.
Tojind the length of an arc of a circle.
Rule I. Multiply the radius of the circle by the number
of degrees in the given arc, and that product by -01745329,
and the last product wiU be the length of the arc*
Rule II. From eight times the chord of half the arc,
subtract the chord of the whole arc, one-third of the re-
mainder will give the length of the arc, nearly .|
1. If the arc A B contain 30 degrees, the radius being
9 feet, what is the length of the arc ?
30x9 = 270, and 270 x -01745329 = 4-7124. Ans.
2. If the chord AD of half the arc ADB be 20 feet, and the
abord A B of the wbole arc 38; what is the length of the arc?
20 X 8 - 38 = 1 22 ; then 1 22 ^3 = 40§ feet. Ans.
• See Appendix, J>einon8tration 23.
f Sc« Appendix, DemoiutrafioD 34.
ILENSUaATlOa OF SUPJiRKlCLBS. 43
3. The chord of an arc is 6 feet, and the chord of half
the arc is 3^ ; required the length of the whole arc ?
Ans. 7^.
4. The chord of the whole arc is 40, and the versed sine*
or height of the segment 15 ; what is the length of the arc?
Ans. 53^.
5. The chord A B of the whole arc is 48'74, and the
chord A D of half the arc 30"25 ; required the length of the
arc?
6. A B = 30, D P = 8 ; required the length of the arc ?
Ans. 35^.
PROBLEM XVIII.
Tojind the area of a circle.
Rule I. Multiply half the circumference by half the
diameter, for the area.f
Rule II. Multiply the square of the diameter by "7854,
for the area.J
Rule 111. Multiply the square of the circumference by
•07958.§
Rule IV. Aa 14 to 11, so is the square of the diameter
to the area.
Rule V. As S'^ to 7, so is the square of the circum-
ference to the area.
1. To find the area of a circle whose diameter is 100 and
p'jcumference 3 14" 16.
By Rule I.
314 16
100
By Rule II.
•7854
100-= 10000
By Rule III.
98696-0 sq. clr.
•07958
4)31416
Area 7854
Area 7854
7S54^ Area,
• By " versed sine," in works on mensurKtion, is not meant the trigooo-
metrical verged line of the whole arc, be* ai half the arc.
"t* See Appendix, Demonstration 'J'2.
J See Appendix, Demonrtration iJ2.
§ See Appendix, Demonstration 26.
44 MEirSURATIOlT OF SUPERFICIES.
By Rule IV. Bv Rule V.
/00» = 10000 98696-5 sq. cir.
11 7
2)110000 8)690875-5
7)55000 11)86359-4
Area 7857 7350-85
2. \Miat is the area of a circle whose diameter is 7 ?
Jns. 38^ nearly,
3. How many square yards are in a circle whose diameter
b l^yard? Ans. 1-069.
4. The sm-veying wheel turns twice in the length of 16^
feet ; in going round a circular bowling green it turns exactly
200 times ; how many acres, roods, and perches in it?
Ans. 4 acres, 3 roods, 35-8 perches.
5. The circumference of a fish-pond is 56 chains, what
is its area? Ans. 2-l9-5G'-iSS.
6. AVTiat is the area of a quadrant, the radius being 100?
A71S. 7854.
7. Required the length of a cord fastened to a stake at
oue eud, and to a cow's horns at the other, so as to allow
her to feed on an acre of grass and no more ?
Ans. 39^ yards.
8. The circumference of a circle is 91, what is its area?
Arts. 65 9 -00 198.
9. The diameter of a circle is 15 perches, what is its
area? A7U. 176-715.
10. What is the area of the semicircle of which 20 if
the radius ? Ans. 628-32.
PROBLEM XIX.
Given the diameter of a circle to find the side of a square
equal in area to the circle.
Rule. Multiply the diameter by '8862269, and the pro-
duct will be the side of a square equal in area to the circle.*
* !3ee Appendix, Demonstration '26.
MEUSURATION OF SUPERl'ICtES.
4£
1. If the diameter of a circle be 100, what is the side of
& square equal in area to the circle ? Ana, 88*62269.
2. The diameter of a circular fish-pond is 200 feet, what
IB the side of a square fish-pond equal in area to the circular
one? Ans. 177-24538.
PROBLEM XX.
Given the circumference of a circle, to Jind the side of a
square equal in area to the circle.
Rule. Multiply the circumference by '2820948, and the
product will be the side of the square.*
1. The circumfereuce of a circle is 100, what is the side
of a sijuare equal in area to the circle? Ans. 28*20948.
2. The circumference of a roimd fish-pond is 200 yards,
what is the side of a square fish-pond eq'uiJ ii. area to the
round one? ^n*. 56"41896.
PROBLEM XXL
Given the diameter, to Jind the side of the inscribed square.
Rule. Midtiply the diameter by
"707 1 068, and the product will give
the side of the inscribed square. f
1. The diameter of a circle is
1 00, what is the side of the in- a
scribed square? Ans. 70*71068.
2. The diameter of a circle is
200, what is the side of the in-
scribed square? ^n*. 14r42136.
PROBLEM XXIL
Given tha aiea of a circle, to Jind tlm side of the inscribed
square.
Rule. Midti[/Iy the area by •6366197, and extract the
square root of the product, which will give the side of the
xiscribed square. i
* Se« Appendix, I)«monBtration 27.
i* 8oe Appendix, Demonjtration 28.
£Se« AppftDflir. Domonstration 29
*6 MEN8UEATI0N OP SUPERFICIES,
1. The area of a circle is 100, what is the side of the
uiscnbed square ? Ans. 7-97884.
_ 2. The area of a circle is 200, what is the side of the
inscribed square ?
200 X -6366197 = 127-3239400; then ^ 127-3239400
= 11-2837. Ans. ^
PROBLEM XXIII.
Given the side of a square, to find the diameter of the
circumscribed circle.
Rule. Multiply the side of the square by 1-4142136, and
the product will give the diameter of the circumscribed
circle.*
1. Kthe side of a square be 10, what is the diameter ol
the circumscribed circle ? A>is. 14-142136.
2. If the side of a square be 20, find the diameter of the
circumscribed circle ? A71S. 28-284272.
PROBLEM XXIV.
Given the side of a square, to find the circumference of tht
circumscribed circle.
Rule. Multiply the side of the square by 4-4428934,
and the product will be the circumscribed circle. f
1. If the side of a square be 100, what is the circum-
ference of the circumscribed circle ? Ans. 444-28934.
2. If the side of the square be 30, what ia the clrcumfo-
rence of the circumscribed circle? Ans. 133-28G802.
PROBLEM XXV.
Given the side of a square, to find the diarneter of a circU
equal in area to the square.
Rule. Multiply the side of the square by M 283791,
and the product will be the diameter of a circle equal in
area to the square whose side is given.J
• See Appendix, Demonstration 30.
+ See Appendix, Demonstration 31.
X S«e Appendix, Demonstration 32.
aDBUeURATION OP SUFERFICIES.
47
1. If tLo side of a square be 100, what is the diameter ol
the circle whose area is equal to the square whose side is
100? ^ns. 112-83791.
2. What is the diameter of a circle equal in area to a
square whose side is 200 ? Aiis. 225-67582.
PROBLEM XXVI.
Given the side of a square, to Jind the circumference of a
circle whose area is equal to the square whose side is given.
Rule. Multiply the side of the square by 3-5449076, and
the product will g^ve the circumference of a circle equal
in area to the given square.*
1. What is the circumference of a circle, whose area
may be equal to a square whose side is 100 ?
Ans. 354-49076.
2. Find the circumference of a circle equal in area to a
Bquare whose side is 300? Ans 106347228.
PROBLEM XXVIL
To find the area of a sector of a circle.
Rule L Multiply half the length of the arc by the radius
of the circle, and the product is the area of the sector.f
Rule. IL As 360 is to the degrees in the arc of the sector,
so is the area of the whole circle to the area of the sector.J
1. Let AC BO be a sector less than a
semicircle whose radius A O is 20 feet,
and chord AB 30 feet, what is the area?
First, -/ (AO^ - AD^) = ^ 400
-225)= 13-228= OD; then OC-OD
= 20 - 13-228 = 6-772 = C D.
Again, V (A D^+ C D*) = -/ 225
+ 45-859984) = 16-4578 = A C, the
chord of half the arc.
• See Appendix, Demonrtrntion 3S.
+ See Appendix, Demonrtraticin 54.
X Sm Appendix, DemoostnUi'ra fS,
48
llEKSD RATION OF SUPERFICIES.
Hence, by Problem XVII. the arc A B is 33-8874 ; then
X 20 = 338-874, the area required.
2. Let A E F B O A be a sector
greater than a semicircle, whose ra-
dius A O is 20, the chord E B, 38,
and chord B F of half E F B 23 ;
required the area ?
23 = chord B F
8
184
38 = chord B E
3)146
48'6GG &c. = arc B F E
20
973^ area.
3. What is the area of a sector whose arc contains 18
dt-grees, the diameter being 3 feet?
•7854
9
Then 360 : 18 *. : 7-0686 : the area of the sector;
Or, 20 : 1 :: 7-0686 : -35343. Am.
4. What is the area of a sector whose arc contains 147
degrees 29 minutes, and radius 25 ? Ans. 804*3986.
5. WTiat is the area of a sector whose arc contains 18
degrees, the radius being 3 feet? Ayis. 1-4 137 2.
PROBLEM XXVm.
To find the area of the segment of a circle.
Rule I. Find the area of the sector having the same arc
with the segment, by the la«i problem ; find also the area
MKNSTTRATION OF SUPERFICIES.
49
oi the triangle, formed by the chord of the segment and tht
two radii of the sector. Then add these two area* together,
when the segment is greater than a semicircle, but find theii
difference when it is less tlian a semicircle, the result will
eA-idently be the answer.
1 . \^'^lat is the area of the segment
A C B D A, its chord A B being 24,
and ra<lius A E or E C 20?
-/ (A E* — A D-) = ^ (400 -
1 14) = 16 = DE; EC — ED=20
— 16 = 4 = CD; v/(AD* + DC-)
= v/(l'l-* + l'')= 12-64911= AC;
= 25-7309=
we A C B,
And 12-8654 = half arc
20 = radius
12
16
AD
DE
257-308 = area of sector EBCA. 192 =areaof A ABE
192 = area of a A B E
65-308 = area of segment A B C A.
2. Let A G F B A be a segment greater than a semi-
circle, there are given the chord A B 20-5, F D 17" 17,
A F 20, F G 11-5, and A E, 11-64, required the area of the
segment ?
(FGx8)-AF^(Ul)<8)_-^0 ^ „^ ^^ ,^^^^ „,
o o
the arc A G F (Problem XVII.); then 24 x 11-64 =
279-36, area of sector A E B F G A (Problem XXVTI).
Again, F D^ E F= 17-17 — 11-64 = 5-53 = E D ; then
A^ X E D _ 20-5 X 5 -53
2 ~ 2
= 56-6825 the area of the tri-
angle ABE, which being added to the area of the sector
beForc found will give the area of the segment, Tiz. 279-36
+ 56-6825 = 3360426 the area ol' the segment A Q F B A.
50 MENSURATION OP SUPERFICIES.
Rule II. To two-thirds of the product of the chord and
versed sine of the segment, add the cube of the versed sine
divided by twice the chord, and the sum will give the area
of the segment, nearly.
WTien the segment is greater than a semicircle, find the
area of the remaining segment, and deduct it from the area
of the whole circle, the remainder wiU give the area of the
segment.*
3. \Miat is the area of the segment A C B, less than a
semicircle, its chord being 18 9, and height or versed sine
D C 2-4?
A B X D C = 18-9 X 2-4 = 45-36, and f A B x D C
2-4^
= 1 X 45-36 = 30-24; then ^ — TgTg = -30571; hence 30-24
-«- -36571 = 30-60571 the area.
Note. If two chords of a circle cut one aoother, the rectangle contained by
the segments of one of them is equal to the rectangle contained by the segmentfi
of the other. This ia the 35th Proposition of Book III. of EucUd.
4. Required the area of the segment A G F B whose height
F D is 20, and chord A B 20 ?
^^ = ^= 10 = A D, and AD^= 100; but A D' = F D
A D^ 100_
xDC.*. CD — pj-v:= "20 — o.
The area of the segment A C B is, by the last case,
69-7916 ; and the area of the whole circle, by Problem XVIIl.
is 490-875; then 490-875 —69-7916 = 421-0334 = area of
the segment A G F B.
5. What is the area of the segment A G F B, greater than
a semicircle, whose chord A B is 12, and versed sine 18 ?
Ans. 297-81034.
* 8e« Appendix, DemooBtnition dfi.
MEKSURATION OF SUPERFICIES. 51
Rule III. Dh-ide the height of the segment by the
diameter of the circle, to three places of decimals. Find the
quotient in the column Height of the Table at the end of
the practical part of this treatise, and take out the corres-
ponding Area Seg., which multiply by the square of the dia-
meter, and the product ■will be the area of the segment
required.*
\ote I. If the quotient of the height by the diameter be greater than -5
tubtract it from 1, and find the Area Seg. corresponding to the remainder,
which subtract from •7854 for the correct Area Seg.
Xot* 11. If the Quotient of the height by the diameter does not terminate in
three figures, find the Area See. corresponding to the first three decimal figures
of the quotient, subtract it from the next greater Area Seg. multiply the
remainder by the fractional part of the quotient, and add the product to the
area segment first, taken out of the table. When great accuracy is not required,
the fractional part may be omitted.
6. Let the diameter be 20, and the versed sine 2, required
the area of the segment ?
3*^ = •!, to which answers •040875
Scjuare of diameter, 400
16"35 area.
7. What is the area of a segment, whose diameter is 52,
and versed sine 2 ?
j^ =: 'OJS^^ which is the tabular versed sine. Then to
•038 answers •0097<>3, and the difference between tliis area
and the next is '000385, which multiplied by j^ gives •000177
which added to "0097(33 gives •009940, which is the area cor-
responding to the versed sine -0381^5. Then 52- X 009940
= 26*87776 is the area required.
PROBLEM XXIX.
To Jind the arpa rf a zone, or the space included hif two
parallel chords and the arcs contained between them.
Rule. Join tlie extremities of the parallel chords towards
•.he same parts, and these connecting lines will cut off two
• See AppendiiL. Demonstration 37.
52
MENSUEATION OF SUFERnCIES.
equal eegments, the areas of whicli added to the area of the
trapezoid then formed will g^ve the area of the zone.
1. Suppose the greater
chord A B = 30, the less
C D 20, and the perpen-
dicular distance D ar = 25,
required the area of the
zone A B D C.
i(AB-CD) = a:B = i
(30 -20) = 5 : thenWa^D'-
+ x'B'')= DB=.-v/(25* +
5') = 25-49. A B - Bx
= A J? = 30 - 5 = 25, and
(Axx Bx) -h Ox = ¥x =
(25 X 5) -r 25 = 5. D a: + F ar = D F = 25 + 5 = 30 ;
lv/(CD« fDF*) = ^CF=G^=|A/ (20« + 30^) =
18-027, the radius of the circle ;(DBxAa;) + 2Dar=: Qy*
~ (25-49 X 25) + (2 X 25) = 12-745 ; Oz-Qif=zy =
18-027 - ■ 12-745 = 5-282, the height of the segment A z C.
36'05)5-28(-l46, the tabular area segment answering to which
is -071033, then -071033 X (36-05)^ = 92-315 = the area of
the segment A z C.
^ f A B + C D) X D ar = -1(30 + 20) X 25 = 625 the area
of tlie trapezoid A B D C : then 625 + 92-315 x 2 =
80 9 '63 = the area of the zone.
2. Let the chord A B be 48, the chord C D 30, the chord
A C 15-8114 ; what is the area of the zone A B D C ?
Ans. The diameter C F = 50, height of the segment
A z C = 1-2829, area by the table of segments = 13-595.
Area of the zone A B D C = 534-19 ?
3. Let A B = 20, C D = 15, and their distance = 17^;
requii-ed the area ? Ans. 395-4369.
4. Let A B = 95, C D = 60, and their distance = 26 ;
reqrored the area? Ans. 2136-7527.
" Soe Appendix, Demonrtratiou 88.
MBMSURATION OF SUPERFlCXaS.
53
PROBLEM XXX.
Tojind the area of a circular ring, or of the space included
between two concentric circles.
Rule. Multiply the sum of the two diameters by their
difl'erence, and the product arising by '7854 for the area of
the ring.*
1. The diameter A B is 30,
and C D 20 ; what is the area
of the ring XX?
30
20
50 sum
10 difference
500
•7854
392-7000 area of the ring XX.
2. \Miat is the area of the circular ring, when the dia-
meters are 40 and 30? Ans. 549*78.
3. \VTiat 18 the area of a circular ring, when the dia-
meters are 50 and 45 ? Ans. 373-065.
PROBLEM XXXI.
Tojind the area of a part of a ring, or of the segment oj
a sector.
Rule. Multiply half the sum of the bounding arcs by
their distance asunder, and the product will give the
area.f
* See Appendix, Demonstration 39.
f See Appendix. Demonstration H).
64
MENSURATION OF SUPERFICIES
1 . Let A B be 50, and a b 30, and the
distance a A 1 ; what is the area of the
space a b B A?
Ans
50 + 30
X 10=400.
2. Let A B = 60, a 6 = 40, and a A
^ 2 ; required the area of the space
o 6 B A? Ans. 100.
3. Let A B = 25, o 6 = 15, and a A
= 6 ; required the area of the segment
of the sector? Ans. 120.
A
PROBLEM XXXIL
Tojind the area of a lune, or the space included between
the intersecting arcs of two eccentric circles.
Rule. Find the areas of both segments which form the
lune, and deduct the less from the greater ; the remainder
will evidently be the area required.
1. Let the chord A B =
40, EC= 12, and ED = 4;
what is the area of the lune
A D B C A ?
By note, page 50, (A E^
-j- E C) -f E C = diameter
of the circle of which A C B is an arc ; and (A E* •+- E D1
-H E D = the diameter of the circle of which A D B is an
arc; hence (20* + 12) + 12 = 45-3; and (20» + 4) + 4
=: 104 ; are the two diameters.
12 + 45-3 = -264.
104 = -038.
The Area Seg. answering to "264 is •165780, and
(45-3)' X -165780 = 340-1954802 = area of the segment
A E B C A.
The Area Seg. anflwerlng to '038 Lb '009763, and
IIENSURATION OF SUPERFICIES. 56
(104)' X -009763 = 105-596608 = area of the segment
AEBDA ; then 340-1954802 - 105-59G608 = 234-5988722
= the area of the lune,
2. Let the chord A B be 40, and the heights of the seg-
ments E C and E D 1 5 and 2 ; required the area of tht
lune? ^n*. 388-5.
PROBLEM XXXin.
TO MIlASDllE LONG IRREGULAR FIGURES.
When irregular JigureSt not reducible to any known Jigurt,
present themselves, their contents are best found by the
method of eqiii-distant ordinates.
Rule. Talce the breadths in several places, at equal dis-
tances, and divide the sum of the first and last of them
by 2 for the arithmetical mean between those two. Add
together this mean and all the other breadths, omitting
the first and last, and divide tlieir sum by the mmiber of
parts so added, the quotient will give the mean breadth of
the whole, which being multiplied by tlie given length will
give the area of the figure, very nearly.
It is not necessary sometimes to take the breadths at
equal distances, but to compute each trapezoid separately,
and the sum of all the separate areas thus found will giM'
the area of the entire nearly.
Or, add all the breadths together and divide by the
number of them for a mean breadth, which being multiplied
by the length, as before, will give the area nearly.
1. Let the ordinate A D be 9-2, i/7, c^ 9, <f ^ 10, B C 8-{?
»ud the length A li 20 ; re«iuired the art»a ?
B
56
MENSUBATION OP SUPERFICIES.
9-2 AD
8-8 B C
2)18
9 mean bnsadth of first and last.
7 6/
9cg
\0 dh
4)35 sum
8*75 mean breadth of ail.
30
262 50 area of the whole figure.
2. The length of an irregrdar figure is 39 yards, and its
breadths, in fiye equi-di«tant places are 4*8, 5*2, 4*1, 7 '3,
and 7'2 ; what is its area ? Ans. 220-3o square yards.
3. The length of an irregular figure h .SO yards, and its
breadtJis, at seven equi-distant places, are 5*5, 6*2, 7'3, 6,
7'5, 7, and 8*8 ; what is its area? Ans. 342'9150 sq. yards.
4. The length of an irreg^ar figure being 37*6, and the
breadths, at idne equi-distant places, 0, 44, 6*5, 7*6, 5*4,
8, 5-2, 6-5, 6-1 ; what is the ai-sa? Ant. 218-315.
61
EXERCISES.
1. Find the area of a square whoa© side i« 35"25 chains.
^ 124 acres, 1 rood, 1 perch.
2. Find the area of a rectangular board, whose length
is 12^ feet, and breadth, 9 inches. Ans. 9| feet.
3. The side* of three squares being 4, 5, and 6 feet, what
is the length of the aide of a square which is equal to all
three ? Ans. 8-7749 feet.
4. Required the area of a rhomboid whose length is 1051
chains, and breadth, 4*28 chains ?
Ans. 4 acres, I rood, 39 perches.
5. There is a triangle whose base ifl 12'6 chains, and alti-
tude 6*4, chains, what is its area ? Ans. 40'32.
6. Find the area of a triangle whose sides are 30, 40, and
50 yards. Ans. (iOO square yards.
7. There is a triangidar corn-field whose sides are 150,
2(^0, and 260 yards, determine the number of acres contained
ii) the field, and the expense of reaping the com at 99» 6d.
per acre.
Ans. Content of the field, 3 acres 15 perches; expense
jf reaping, £ 1 9s. 5d.
8. ^^^lat must the base of a triangle be to contain 36
jiquare feet, whose vertex ia to be 9 feet from the base ?
An$. 8 fetrt.
e2
58
EXERCISES
9. What must be the altitude of a triangle equal in arta
to the last, -whose base is 12 feet ? Ans. 6 feet.
10. The height of a precipice standing close by the side
of a rirer is ] 03 feet, and a line of 320 feet will reach from
the top of it to the opposite bank ; required the breadth of
the river? Ans. 302-97 feet.
1 1 . A ladder 1 2^ feet in length stands upright against a
fpall, how far must the bottom of it be pulled out from th«
wall so as to lower the top 6 inches ?
Ans. 3^ feet.
12. A person wishing to mea-
sure the distance from a point
A, at one side of q canal, to an
object O, at the other, and hav-
ing no iustnunent but a book,
placed a comer of it on the point
A, and directed an edge of it,
us in the figure, in a straight line
with the object O, and drew the
straight lines A B, AC; bo then
placed the book so that a comer
of it rested on the point B, at
the distance of eight times its
length from the point A, and di-
rected an edge of it, as before,
to the object O, and drew the
straight line B C which met A C
at the distance of three times the
length of the book from A ; how
many times the length of the book
is the object O from the points
A and B ?
Ans. 21^ and 22-78 times.
13. What is the area of a trapezium whose diagonal is
70*5 feet, and the two perpendiculars 26-5 and 30'2 feet ?
Ans. 1998-675 square feet.
»lKN8CRA.nON OF SrPERnCDift. -^9
14. WTiut 18 the area of a trapezium whose diagonal 13
108 feet 6 inches, and the perpendiculars 56 feet 3 inches,
md 60 feet 9 inches? Ans. 6347 feet 36 incho^
15. ^Vhat is the area of a trapezoid whose two parallel
sides are 75 and 122 links, and the perpendicular distanc«^
154 links? Ans. 15169 square links.
16. A field in the form of a trapezoi^p whose parallel sides
are 6340 and 4380 yards, and the perpendicular distance
between them 121 yards, lets for £207 Ns. per annum:
w hat is that per acre ? Ayis. £1 lis.
17. Two opposite angles of a four-sided field are toge-
ther equal to two right angles, and the sides axe 24, 26,
28, and 30 yards ; what is its area ?
Ans. 723'99 square yards, nearly.
18. Required the area of a figure similar to that annexed
to the first question under Problem XIV., whose dlmen-
iions are double of those there giyen ?
Ans. .3411^6
19. What is the side of an equilateral triangle equal in
area to a square, whose side is 10 feet?
Ans. 15* 196 feet, nearly.
20. Required the area of a regulur nonagon, one of whosf
iides is 8 feet, and the perpendicular from the centre =
10'99 feet? Ans. 395'64 square feet.
21. Required the area of a regirlar decagon, one of whose
sides is 20*5 yards ? Ans. 3233-491125 square yards.
22. A wheel of a car turns round 4400 times in a distance
of 10 miles ; what is its diameter ?
Ans. 3*8 19708 feet,
23. If the diameter of a circle be 9 feet, what is the length
of the circumferenoe ? Ans. 2S| feet, nearly.
f50 EXERCISES IN MENSURATION OF SUPERFICIES.
24. Required the length of an arc of 60° ; the radius of
the circle being 14 feet? Ans. 14-660772 feet.
25. The chord of an arc is 30 feet and the height is b
feet, what is the length of the arc? Ans. 35^ feet, nearly.
26. The diameter of a circle is 200, what is the area of
the quadrant ? Ans. 7854.
27. The diameters of two concentric circles are 15 and
1 0, what is the area of the ring formed by those circles ?
Ans. 98-175.
28. The circumference of a circle is 628*32 yards, what
is the radius of a concentric circle of half the area 2
Ans. 70-71.
29. What is the side of a square equal in area to the circle
whose diameter is 3 V Ans. 2-6o86807.
30. The two parallel chords of a zone are 16 and 12 and
their perpendicular distance i& 2, what is the area of the
zone ? Ans. 28-376.
31. The length of a chord is 15, and the heights of two
segments of circles on the same side of it are 7 and 4 ; what
'8 the area of the lune formed by those segments ?
Ans. 38, nearly.
32. The base and perpendicular of a right-angled triangle
rxe each I, what is the area of a circle having the hypo-
thenuse for its diameter 'i Ans. l-o70S.
33. If the area of a circle be 100, what is the area of the
inscribed 8qu."u-e? Ans. 63-66.
61
CONIC SECTIONS.
SECTION III.
OF THE ELLIPSIS.*
PROBLEM I.
The transverse and conjugate diameters of an ellipsis beiiig
given, to find the area.
Rule. Multiply the transverse and conjugate diameters
together, and the product arising by '7854, and the result
will be the area.f
1. Let the transverse axis be 35, and the conjugate axia
25 ; required the area ?
35 X 25 X •7854 = 687-225. Ans.
2. The longer diameter of an ellipse is 70, and the shorter
50 ; what is the area? Ans. 2748-9.
3. Wbat is the area of an ellipse whose longer axis is 80f
and shorter axis is 60 ? Ans. 3769-92.
4. WTiat is the area of an ellipse, whose diameters are
50 and 45? Ans. 1767-15.
PROBLEM IL
To find the area of an elliptical ring.
Rule. Find the area of each ellipse separately, and
their difference wiU be the area of the ring.
* For deflnitiona of the ellipsii (or, m it li frequectly irrittaa, ellipite)
hod the other Conie Sections, tee Appendix, Properties of tho Conk
8c>rtious.
-t* See Appendix. DemooBtntloc 41.
62
CONIC SECriOXS.
Or, From the product of the two diameters of the greater
ellipse deduct the product of the two diameters of the less, and
multiply the remainder by '7854 for the area of the ring.*
1. The transverse diameter A B is 70, and the conjugate
C D 50 ; and the transverse diameter E F of another ellipse
having the same centre 0, is 35, and the conjugate G H is
25 ; required the area of the elliptical space between their
circumferences ?
70 X 50 X -7854 = 2748-9 ; and 35 X 25 x -7854
= 687-225 ; then 2748-9 - 687-225 = 2061-675 = arer
of the elliptical ring.
70 X 60 = 3500
35 X 25 = 875
2625 X -7854 = 206 1 -675 = area.
2. The transverse and conjugate vliameters of an ellipse
are 60 and 40, and of another 30 and 10 ; required the area
of the space between their circumferences? Ans. 1649'34.
3. A gentleman has an ellijitical flower garden, whose
greater diameter is 30, and less 24 feet ; and has ordered a
gravel walk to be made round it of 5 feet 6 inches in width ;
required the area of the walk ? Ans. 371-4942 feet.
PROBLEM III.
Given the height of an elliptical segment, trhose base is pa-
rallel to either of the axes of the ellipse, and the two axes
of the ellipse, to find the area.
Rule. Divide the height of the segment by that diameter
of which it 18 a part, to throe places of decimals, find the
• See A^petidjjc, Demonstraliou 42.
MENeURATlOU.
63
quotleut in the column Height of the Table referred to in page
51, and take out the correspondent Area Seg. Multiply the
Area Seg. thus found and both the axes of the ellipsis together,
and the result will g^ve the area required.*
1. Required the area of an elliptical segment R A Q,
Q D
whose height A P is 20 ; the transverse axis A B being 70,
and the conjugate axis C D 50 ?
20 -I- 70 = '285^ = the tabular versed sine, the corres-
ponding segment answering to which, is '185166 ; then
•18a 166 X 70 X 50 = 648-081, the area.
2. \Miat 13 the area of an elliptical segment cut off by a
chord parallel to the shorter axis, the height of the segment
being 10, and the two diameters 35 and 25 ?
Ans. 162-0202.
3. \Miat is the area of an elliptical segment cut off by a
chord parallel to the longer axis, the height of the segment
being 10, and the two diameters 40 and 30 ?
Atis. 275-0064.
4. What is the area of an elliptical segment cut off by 8
chord parallel to the shorter diameter, the height being 10.
and the two diameters 70 and 50 ? A7if. 240-884.
* Sea Appendix, D«moni^tion 43.
'54 lOOTC SECTIONS.
PROBLEM IV.
To find tke circumference of an ellipse, by having the tioo
diameters given.
Rule. Multiply the sum of the two diameters by 1 '5708,
aud the product will g^ve the circimiference nearly ; that is,
puttinff t for tlie transverse, c for the conjugate, and p for
3 1416 ; the cire»unference will be {t-\-c)x\ p.*
1. Let the transverse axis be 24, and the conjugate 18
required the area ?
(24 + 18) X 1-5708 = 42 X 1-5708 = 65-9736 is the
circumference nearly.
2. Required the circumference of an ellipse whose trans-
verse axis is 30, and conjugate 20? Ans. 78-54.
3. Required the circumference of an ellipse whose dia-
meters are 60 and 40 ? Ans. 157-08.
4. ^Vhat is the circumference of an ellipse whose dia-
meters are 6 and 4 ? Ans. 15-708.
5. What is the circumference of an ellipse whose dia-
meters are 3 and 2? Ans. 7-854.
PROBLEM V.
To find the length of any arc of an ellipse.
Rule. Find the length of the circular arc x y, inter-
cepted by O C, O B, and whose radius is half the sum of
O C, O B : and it will be equal to the eUipticaJ arc B C
nearly.f
Note. The ne&rer the axes of the ellipse approach towards eqaalitj,
the more exact the result of the operation by this Rule ; and the lets the
elliptical arc, the nearer its exact length will approach the arc x y.
• See Appendix, DemonBtration 4-1.
•¥ S»e Appendix. Dernuiutration 4n.
MENSUnAnON.
65
1. Let the axis A D be 24, C K 18, and O T 3 ; re-
quired the length of the axe B C ?
Here we have T D = 9, and A T = 15 : then from the
property of the ellipsis, we liave A O* I O C* : : A T x
T D : T B- = '--^^J' = 5^-3^, and O B =
12 X 12
16
v^ (O T' + T B*) = v/ (9 +^^6^^) = 9-21616, tht.
radius of the circle of which G B is an eu-c ; but O C is the
rudius of the circle of which C V is an arc ; therefore the ra-
dius of the circle of wliich a: ^ is an arc, is ■^ O C + ^ O H
= 9-10808. But by 'Ingonometry* HB^-0B = 3-^
9-21616 = -325515, is the sine of the angle COB,
or arc x y, to the ra<iius 1, answering to 1 8*9963 degree*.
Tlerefore, by Problem XVII. Rule I, the length of the &ii
r.y is -01745 x 18-9968 x 9-10808 = 3-0192, which is also
equal to the length of the elliptical arc C B, nearly.
2. Given A D 30, C K 20, and O T 5 ; required th^
leng h of the arc B C ? Ans. 5-03917786255.
3. Given A D 40, C D 30, and O T 5 ; required th
length of the arc B C? Ans. 5-033i<80786.
*It may b« done without Trigonometry, by first finding the length of th«
arc G a by Role IL Prob. XVII. SJec. 2. then O vl : O Y : : G B :
1 Z
glj CONIC SECTIONS.
PROBLEM VI.
Given the diameter and abscissas, to find the ordinate.
Rule. Say, as the transverse Is to the conjugate, so u
the square root of the rectangle of the two abscissas, to
the ordinate.*
1. In the ellipse A C D K, the transverse diameter A D
is 100, the conjugate diameter C K 80, and the abscissa
D T 10 ; required the length of the ordinate T B ?
100 : 80 : : -v/ (90 X lO) : T B = 24. (See the last figure.)
2. Let the transverse axis be 35, the conjugate 25, and
the abscissa 7 ; required the ordinate? Ans. 10.
3. Given the tvro diameters 70 smd 60, and the abscissa
10 ; required the ordinate? A7is. 20"9956.
PROBLEM Vn.
{liven tne transverse axis, conjugate and ordinate, to find
the abscissas.
Rule. As the conjugate is to the transverse diameter, so
is the square root of the difference of the squares of the
ordinate and semi- conjugate, to the distance between the
ordirat: ztA centre. Then this distance being nr^'.xl to,
and subtracted from, the semi-diameter, will give the two
abscissas.!
1. Let the diameters be 35 and 25, and the ordinate 10 ;
required the abscissas ?
„ ^ „ , 35 35 /r25i' ,/\ 35+21 „„
By the Rule 2- ±25 V([ 2 J -10)=— g— =28
and 7, the two abscissas.
2. Let the diameters be 120 and 40, and the ordinate
16; required the abscissas? Ans. 96 and 24.
• S«« Appendix, Demonstration 46.
f Sm> Appendix, DemonatraUon 47.
ICBNgURATION.
67
PROBLEM VIII.
Given tTib conjugate axis, ordinate, and abscissas^ to ^find the
I transverse axis.
Rule. Find the square root of the difTerence of the
squares of the semi-conjugate axis and the ordinate, which
add to, or subtract froui, the semi-conjugate, according a^
the k'ss abscissa or greater is given.
Then say, as the square of tlie ordinate is to the rectangle
of the conjugate, and the abscissa, so is the sum or difference
foimd above to the transverse required.*
1. Let the ordinate be 10, and the less abscissa 7 ; what L<>
the diameter, allowing the conjugate to be 25 ?
10* : 25 X 7 : : 20 : 35 the transverse required.
2. Let the ordinate be 10, the greater abscissa 28, and
the conjugate 25 ; required the transverse diameter?
Ans. .35.
PROBLEM IX.
Given the transverse axis, ordinate^ and abscissa, to find (he
conjugate.
Rule. The square root of the product of the two ab-
scissas is to the ordinate, as the transverse axis is to the
conjugate. I
1. Let the transverse axis be 35, the ordinate 10, and
the abscissas 28 and 7 ; required the conjugate ?
35 X 10 35 X 10
v' (28 X 7) : 10 : : 35 :
the conjugate.
v(28x7)
14
':5,
* Se« Appendix, Demonstration +H.
tbee Appendix, Demougtr»tiuQ ■19.
fl8
CO?nC 8ECTIOS3.
2. Let the transverse diameter be 120, the ordinate 16,
aud the abocissas 24 and 96 ; required the conjugate ?
An8, 40.
OF THE PARABOLA.
PROBLEM X.
CUven the base and height of a parabola, to find its area.
N'ote. Any doable ordinate, A B, to the axis
of a parabola may be called its base, and the
ibscissa O D, to that ordinate its height.
Rule. Multiply the base by the
height, and | of the product \vill be q
the area,* ^i I
1. Required the area of a parabola, whose height is 6 and
baae 12 ?
6 X 12 X § = 48 the area.
2. What is the area of a parabola, whose base is 24, and
height 4 ? Ana. 64.
3. What is the area of a parabola, whose base is 1 2, and
height 2 ? Am.lQ.
* See Appendix, Demonstration M.
MENSUBATION. tj^
PROBLEM XI.
To Jind the area of the zone of a parabola, or the tjiact
between two parallel double ordmate$.
Rule I. When the two double ordinates, their distance,
and the altitude of tlie whole parabola are given ; find the
irea of the whole parabola, and find also the area of the
apper segment, their diflFerence will be the area of the zone.
II. When the two double ordinates and their distance are
given ; to the sum of the sqxiares of the two double ordinates,
add their product, divide the sum by the sum of the two
double ordinates, miJtiply the quotient by ^ of the altitude
of the f one, and the product wiU be the ai */f the zone.*
1. Given A B 20, S T 12, and D ar 8 ; what \a the area
of the zone A S T B, the altitude D O being ll'-5 ?
(20 X 12-5) X f = 166| = area of the parabola ABO,
and (12-5 - 8) X 12 = 54, and 54 X f = 36 ; hence 166|
— 36 = 130| the area.
III. When the altitude of the whole parabola is not g^ven.
2. Suppose the double ordinate A B = 10, the double ordi-
nate S T = 6, and their distance D « = 4 ; what is the area
of the zone A S T B ?
AOL±^_^Ojil = .2 J , ,,,„ 12 i X 4 X } = 3.2|,
the area as before.
3. Let the double ordinate A B =: 30, C P = 25, and
their distance D Q = 6 ; required the area of the zone
\BPC? Ana.lQS^^.
PROBLEM XIL
To Jind th-e length of tht> curve, or arc of a parabola, cut oj^
by a double ordinate to the axis.
RXTLB.
I. Divide the doubb ordinate bj the paremeter, and call
the quotient q.
* Se« Appendix. DemonstratioD fil.
70 COMO «iSCTIOH§.
IL Add 1 to the square of the quotient q, and call the
square root of the sum s.
III. To the product of q and s, add the hyperbolic loga-
rithm of their sum, then the last sum multiplied by half tlie
parameter, will give the length of the whole curve on both
sides of the axis.
Putting c for the curve, q for the quotient of the double
ordinate divided by the parameter, s for v/ (1 + q') and a
for half the parameter ; then
c = a X {^ s + hyp. log. of (q + «.)]*
Note. The common logarithm of any nnmher multiplied by 2'302585093
gives the hyperbolic logarithm of the same number.
1. Wliat is the length of the curve of a parabola, cut ofl
by a double ordinate to the axis, whose length is 1 2, the
abscissa being 2 ?
X = 2 and t/ = 6 ; then a = ^—^ = '^ =
9, and y = ^ = S = §, also * = V- (1 + /) = v- (1 + ^)
= x/ (l^) = ^ x/ (13) = 1-2018504 = s. Then f -f
l •2018504 = 1*8685 17, whose common logarithm is
•271497, which being multiplied by 2-302585093, pro-
duces '6251449 for its hyperboUc logarithm ; and also § X
1-2018504 = -8012336 ;' the sum of these two is 1-4263785,
therefore 9 X 1-4263785 = 12-8374065, is the length of the
curve required.
Rule II. Put y equal to the ordinate, and q equal the
quotient arising from the division of the double ordinate by
the parameter, or from the division of double the abscissa
by the ordinate; then the length of the double curve will be
expressed by the infinite series.
• S«e App«nd)x, Demoustnt^.tm fi3L
mensuratios. 7J
Note. Thif series wlU converge no longer than till j = 1. For when q it
^raX*x than 1, the series will diverge.
Let the last example be resumed, in which the abscissa
is 2, and the ordinate 6.
Hence, 2x2 + 6 = § = y; then employing | instead
of q in the la^st series, we get
,; X (I + fl -1^+3X24^^)= 12-637 the length
of the curve as before.
Rule III. To the square of the ordinate, add \ of the
square of the abscissa, and the square root of the sum will
be the length of the single curre, the double of which wilJ
be the length <>f the double ciuttc, nearly.*
A'ote. The two first rules are not recommended in practice. — The practical
ap;ilicatinn of thii is much bimpler, and ia therefore to be eaiployed in pre-
ference to either.
Retaining the same example, in wliich x = 1, and y = 6,
we shall «ret »• - s' (v* + | ar') = v/ (36 -f '/) = 6-1291,
dnd C = r.i-N582, nearly.
2. Required the length of the parabolic cm-iire, whose ab-
scissa is 3, and ordinate 8 ? Ans. 17*435.
PROBLEM Xin.
Given am/ two abscissas and the ordijiate to one of them, to
Jind the cvrresponditig ordinate to the second abscissa.
Rule. Say, as the abscissa, whose ordinate is given, is
to the square of the given ordinate, so Ls the other given
abscissa to the square of its corresponding ordinat«.f
1. If the absci-ssa j: O = 10, and the onlinate t S == 8,
what is the ordinate A D, whose abscissa D O is 20 ?
ar O : a; S* : : D O : A D-, viz. 10 : 64 : : 20 : 128, the
•jquare root of which is 11 "3 13, &c., = A D.
• .See Appendix, Demonftration .53.
t See Appendix, DaBooustntioQ 6^
72 CONIC SECTIONS.
2. If 6 be the ordinate corresponding to the abscise 9,
required the ordinate corresponding to the abscissa 16 ?
Ans. 8.
PROBJf.EM XIV.
Given tivo ordinat^s^ and the abscissa corresponding to one
of them, tojind the abscissa corresponding' *o the other.
Rule. Say, as the square of the ordinate whose abscissa
is given, is to the given abscissa, so is the square of the
other ordinate to its corresponding abscissa.*
1. Given S a: = 6, a; O = 9, Pid A D = 8 ; required
the abscissa O D ? 36 : 9 : : 64 : 16 = O D.
2. Given S j? =: 8, j? O = 10, and A D = 9 ; required
D? Ans. 12-656.
PROBLEM XV.
Given two ordinates perpendicular to the a^cia and their dit-
tancCf to find the corresponding abscissas.
Rule. Say, as the diflPerence of the squares of the ordi-
nates is to their distance, so is the square of either of them
to the corresponding abscissa.f
1 . Given S a: = 6, A D = 8, and ar D = 7 ; required
t he abscissas ^
(64 - 36) : 7 : : 64
28 : 7 : : 64 : 16 = O D, and
28 : 7 : : 36 : 9 = O a?.
2. Given S * = 3, A D = 4, and * D = 2 ; required
the abscissa* ? A ns. 4^ and 2^.
• See Appendix, Deiuonstiation 54.
+ Sen Appendix, Demonetration 56.
aiENBUKiin0I7. 73
OF THE HYPERBOLA.
PROBLEM XVI.
Given the transverse and conjugate diameters, and ant
abscissa, tojind the corresponding ordinate.
Rule. As the transverse is to the conjugate, so is the
mean proportional between the abscifisas to the ordinate.*
1. If the transverse be 24, the conjugate 21, and the
less abscissa ADS; rcrjuired the ordinate ?
Note. The less abscissa added to the transverse gives the greater.
24 : 21 :: ^ (32 x 8) : '^±^^p^) =, u the
ordinate.
2. If the transverse aiis of an hyperbola be 120, the less
abscissa 40 the conjugate 72 ; required the ordinate ?
Ans. 48.
3. The transverse axis being 60, the conjugate 36, and
the less abscissa 20 ; what is the ordinate ? Ans. 24.
• See Appendix. I><niiont)tr»tiun 56.
-^
74 CONIC SECTIONr.
PROBLEM XVII.
GitiTfi tht> transvene, cotijugate, and ordinate, to find the
abscissa.
KULE. To the square of half the conjugate, add the
square of the ordinate, and extract the square root of the
sum. Then say,
As the conjugate is to the transverse, so is that square
root to half the sum of the abscissas.
Then to this half sum, add half the transverse, for the
greater abscissa; and from the half sum take half the trans-
verse for the less abscissa.*
1. If the transverse be 24, and the conjugate 21 ;
required the abscissas to the ordinate 14 ?
10*5 = -^ conjugate 14 = ordinate
10-5 14
110-25 196
196
306'25 the square root of which is 17*5 ; then 21 :
24 :: 17-5 : 20 = half sum, 20 + 12 = 32 the greater
abscissa, and 20 — 12 = 8 the less abscissa.
2. The transverse is 120, the ordinate 48, and the con-
jugate 72 ; required the abscissas ? Ans. 40 and l60.
PROBLEM XVIIl.
Given the conjugate, ordinate, and abscissas, to find the
trayisverse.
Rule. To or from the square root of the sum of the squarei
of the ordinate and semi-conjugate, add or substract the semi-
coniugate, according as the less or greater abscissa is used ;
■ S«t) AyytfXidix, Demonaratioii />7.
MKNRORaTIOK. 1C>
then, aa the square of the ordinate is to the product of the
abscissa and conjugate, so is the sum or difference, above
found, to the transverse.*
1. Let the conjugate be 21, the less abscissa 8, and iti-
ordinate 24 ; required the transverse ?
21 X 8 X a/ (14' +^^ ) + 10^
14» ~
= 3 X v^ (3' + 4') + 3) = 3 X (5 + 3) = 24 the trans-
verse.
2. The conjugate axis is 72, the less abscissa 40, the
ordinate 48 ; recjuired the transverse ? Ans. 120.
3. The conjugate i& 36, the less abscissa 20, and it*
ordinate 24 ; required the transverse ? A/is. GO.
PROBLEM XIX.
Given the abscissa, ordinate, and transverse diatneter, to
Jind the conjugate.
Rule. As the mean proportional between the abscissas
is to the ordinate, so is the transverse to its conjugate. f
1. What is the conjugate to the transverse 24, the less
abscissa being 8, and its ordinate 14 ?
24 X 14 „, ^, . ^
— r,^ ON = •^l the conmgate.
a/ (32 X 8) - ^
2. The transverse diameter is 60, the ordinate 24, and
the less abscissa 20 ; what is the conjugate ? Ans. 36.
PROBLEM XX.
Oiven any two abscissas, X, x, and their ordinate s, Y, if, to
Jind the transverse to which they belong.
Rule. Multiply each abscissa by the square of the or-
dinate belonging to the other ; roultiplv also the square of
each abscissa by the square of the other's ordinate ; then
• H«e Appendix, ri«tDonitration 58.
^ S«e Appeni^x, I')«moastration 59.
76
CONIO BECnOi^.
divide the difference of the latter products by the differenct
of the former ; and the quotient will be the transverse dia-
meter to which the ordinates belon":.*
1. If two abscissas be 1 and 8, and their corresponding
ordinates 4| and 14, required the transverse to which thev
belong ?
8- X 4g X 4^ — 1* X 14^ _ 35 X 35 — 14 X 14
4g - 14 X 14 — 35 X 4|
Here
1 X 14» -
5 X 5 — 2 X 2
2 x2 — 5 X i^ 7
PROBLEM XXL
8 X 4| X
21 X 8
::;;; 24, the tTansvcrse.
To find the area of a space A N O B, hounded on one side
by thu curve of a hyperbola, by means of equi-distant
ordinates.
Let A N be divided into any given number of equal parts,
AC, C E, EG, &c., and let perpendicular ordinates A B,
CD, E F, &c., be erected, and let these ordinates be ter-
minated by any hyperbolic curve B D F, &c. ; and let A =
AB + NO, B = CD + GH+LM, &c., and C = E F
+ I K, &;c. ; then the common distance A C, of the ordi-
nates, being multiplied by the sum arising from the addition
of A, 4 B, and 2 C, and one-third of the product taken will
>r o
be the area, very nearly. That is, -^ + ^ ^ -f 2 C ^ j) _
the area, putting D = A C.f
• See Appendix, Demonstration 60.
t See Appendix, Demonstration 61.
MENSURATION. 7*1
1. Gftrcv '^e lengths of 9 equi-distant ordinates, viz., 14,
15, 16, 17, J 8, '2t>, 22, 23, 25 feet, and the common dis-
tance 2 feet ; roquirec the area ? Ans. 300| feet.
2. GiTen the lenfrths of 3 eqm-distant ordinates, viz., A B
= 5, C D = 7, and E F = 8, also the lenq^h of the hase
A E 10 ; what is the area of the ^gwre At V E ?
Ans. 681 feet.
3. If the length of the asymptote of a hyperbola be 1 , and
there be 11 equi-distant ordinates between it and the curve,
the common distance of the ordinates will then be ,^o> ''"'3
from the nature of the curve their lengths will be \^, \^,
TT> Tj> i4> i5> i6> nt i»> ie> 10 > wuai 13 tue area oi uie
curbed fiffure ? -4n*. •39315021.
'o'
This formida will answer for finding the area of all curve*
by using the sections perpendicular to the axis. Th<
greater the number of ordinates employed, the more accurate
the result; but in real practice three or five are in most
cases sufficient.
PROBLEM XXIl.
To Jind the length of' any arc of an hypurhola beginning ai
the vertex,
RULE.
I. To 19 times the square of the transverse, add 21 times
the square of the conjugate ; also to 9 times the square of
tlie transverse add, as before, 21 times the square of the
conjugate, and multiply each of these sums by the abscissa.
II. To each of these two products, thus found, acid 15
times the product of the transverse and the square of the
conjugate.
HI. Tben, as the less of these results is to the greater,
so ijj the ordinate to the length of the curve, nearly.*
■ See App«ndix, Damonitntioa 62.
78 CONIC SECTIONS.
1. In tie hjperbola B A C, the transverse diameter is 80
ihe conjugate 60, the ordinate B D 10, and the abscissa
A D 2 ; required the length of the arc B A C ? (Fig. p. 73.)
Here 2 (19 X 80* -»- 21 X 60^) = 2 (121600 + 75600) =
394400. ^
And 2 (9 X 80» + 21 X 60^ = 2 (57600 + 75600^ =
2G6400. > \ ^ ]
Whence 15 x 80 X 60' + 394400 = 4320000 -f
314400 = 4714400.
And 15 X 80 X 60» + 266400 = 4320000 + 266400
= 4586400.
Then 4586400 : 4714400 :: 10 : ^^^^'^'^QQ _ in-o-n
4586400 ~ '"-'^
= A B.
Hence A B C = 10-279 X 2 = 20-558.
2. In the hyperbola B A C, the transverse diameter is
80, the conjugate 60, the ordinate B D 10, and the abscissa
VD 2-1637 ; required the length of the arc A B ?
Ans. 10-3005.
PROBLEM XXIII.
Given the transverse aTi.9 of a hf/perb'^la, tha conjugate,
and the abscissa, to find the area.
RCI^E.
I. To the product of tlie transverse and abscissa, add il
of the square of the abscissa, and multiply the square root
of the sum by 21.
il. Add 4 tiroes the square root of the product of the
transverse and abscissa, to the preceding product, and divide
the sum by 75.
III. Divide 4 times the product of the conjugate and ab-
aniNavTHATioN. Jj
flcusa by the transverse ; this quotient, multiplied by the
former quotient, will give the area of the hyperbola, nearly.*
1. In the hyperbola B A C, ( see fig-ure, page 73) the trans-
verse axis is 30, the conjugate 18, and the abscissa A D is
10 ; wliat is the area ?
Here 21 a/ (30 X 10 + f X lO') = 21 a/ (300 +
71-4L>857) = 21 v' (371-42857) = 21 x 19272 =
•( 1-
4 a/ (30 X 10) + 404-712 4 x 17*3206 -f 404-712
And .-5 — = ^3
69-282 + 404-712 _ 473994 _ g.^^^^g
~ IS 7.')
WTience ^ X 6-3199 = 24 x 6-3199 =
15 1*6776, the area required.
2. ^V^lat ifl the area of an hyperbola whose abscissa is 25,
Xhe transverse and conjugate being 50 and 30 ?
Ans. 805-0909.
3. The transverse axis is 100, the conjugate 60, and ab-
x:is8a 60; required the area? Ans. 322-3633584.
* hee Appecdiz, L>«mon<t3»tiou 63.
au
MENSUKATION OF SOLIDS.
SECTION IV.
DEFINITIONS.
1. A solid is that which has length, breadth, and thick-
oess.
2. The solid content of any body is the number of cubio
inches, feet, yards, &c., it contains.
3. A cube is a solid, having six equal sides
at right angles to one another.
4. A prism is a solid
whose ends are plane fig^es,
which are parallel, equal, and
similar. Its sides are paral-
lelograms.
It is called a triangular prism, when its ends are triangles ;
a square prism, when its ends are squares ; a pentagonal
prism, when its ends are pentagons ; and so on.
5. A parallelopipedon is a
solid having sli rectangular
sides, every opposite pair of
which are equal and parallel.
81
aXEXaURATION Ob' SOLIDS.
6. A cylinder is a round solid,
having circular ends, and may be
conceived to be described by tlie re-
volution of a rectan«rle about one of
its sides, which remains fixed.
7. A pjrainid is a solid, having a plane figTjre
for its base ; and whose sides are triangles mei-t-
ing in a point, called the vertex.
Pyramids have their names from their bases,
like prisms.
When the base Is a triangle, the solid is
called a triangular pyramid ; when the base is a
square, it is called a square pyramid ; and so on.
8. A co7ie is a round pyramid, having a
•ircle for its beise.
9. A sphere is a round solid, which may
be conceived to be formed by the revolution
of a semicircle about its diameter which
remains fijxed.
10. The axis of a solid is a line joining the middle ot
botli ends.
1 1 . When the axis is perpendicular to the base, the solid is
called a right prism or p^Tamid, othenvise it is oblique.
12. The height or altitude of a solid, is a line tlrawn from
its vertex, perpendicular to its base, and is equal to the axis of
a right prism or pyramid ; but in an oblique one, the altitude
is the perpendicular of a right-angled triangle, whose hypo-
thenuse is the axis.
1 3. When the base is a regular figure, it is called a regular
prism or pyramid ; but when the base is an irregular figure,
the solid on it is called irregular.
14. The segment of any solid, is a part cut off from tl^e toi
by a plane parallel to its base.
62
JiCENSURATION OF 60LrD3.
15. A frustum is the part remaining at the bottom, aft^n
the segment is cut off.
16. A zone of a sphere is a part intercepted between two
planes, which are parallel to e-ach other.
17. A circular spindle is a solid gene-
rated by the revolution of a segment of a
circle about its chord, which remains fixed.
18. A wedge is a solid, having a rectan-
gular base, and two of its opposite sides
meeting in an edge.
19. A prismoid is a solid, having
for its two ends two right-angled pa-
rallelograms, parallel to each other, and
its upright sides are four trapezoids.
20. A tpheroid is a solid, generated
by the rotation of a semi-ellipsis about
one of its axis, which remains fixed. fi
When the ellipsis revolves round the ^
transverse axis, the figure is called a \
prolate, or oblong spheroid ; but when ~*— — —
the ellipsis revolves round the shorter axis, the figure is called
an oblate spheroid.
21. An elliptical spindle is a solid,
generated by the rotation of a segment
of an ellipsis about its chord.
\
22. A parabolic conoid, or paraboloid, is a
solid generated by the rotation of a aemi-
p&rabula about its axin.
MBarSTTRATION OF 80LID8.
83
23. An ungula or hoof, is a part cut off a soHd by a plane
t>bliquc to the base.
PROBLEM L
To Jind the solidity of a cuoe.
Rule. Multiply the side of a cube by itself, and that
product again by the side, for the solidity required.*
1. If the side of a cube be 4 inches, required its solidity ?
E
/V
^
//J
/ /
! ^
1/
D
P
C
B
Here, 4x4 = 16, the number of cubes of 1 inch deep
in the square E F G D, and as the entire solid consists of
four such dimensions, its oontent is 16 X 4 = 64 cubic
inches.
2. Wliat is the solidity of a cubical piece of marble, each
side being 5 feet 7 inches ? Ans. 174 feet, nearly.
3. A cellar is to be dug, whose length, breadth, and depth,
are each 12 feet 3 inches; rtquired the number of solid feet
in it ? Ans. 1838 feet 3 inches, nearly.
• S«>« Appendix, I>txconjur»tkn< ^.
Hi IfSNSrRATION OP SOLID*.
PROBLEM II.
To find the solidity of a parallelopipedon.
Rule. Multiply continually the len^h, breadth, and depth
together for the soUdity.*
1. \\Tiat is the solidity of tlie parallelopipedon A B C D
E F G, the length A B being 10 feet, the breadth A G 4
feet, and thickness ADS feet ?
ABxAGxAD= 10x4x5 = 200 feet.
2. A piece of timber is 26 feet long, 10 Inches broad, and
8 inches deep ; required its solid content ? Ans. 14^ feet.
3. A piece of timber is 1 inches square at the ends and 40
feet long ; required its content ? Ans. 27^ feet.
4. A piece of timber 15 inches square at each end, and
18 feet long, is to be measured; required its content, and
how far from the end must it be cut across, so that the piece
cut off may contain 1 solid foot ?
Ans. The solidity is 28-125 feet; and 7-68 b length will
make one foot..
5. W}^t length of a piece of square timber will make one
solid foot,' being 2 feet 9 inches deep, and 1 foot 7 inches
broad?
Ana. 2*756 inches in length wiE make one solid foot.
' Bee ApjKOidijr.. IXaaiojKtTatioii 64
MENBURATION OF SOLIDS.
85
PROBLEM III.
To ^find the solidity of a prism.
Role. Multiply the area of the base by the perpendiciiiar
aeight, and the product will be the solidity.*
1. AMiat is the solidity of a prism, A B C F I E, whose
base C A is a pentagon, e«^h side oJ which being f/JS, and
heig'lit 15 feet ?
When the side of a j>entagoi. '« 1, its area is ^%^0477
(Table II.); therefore 1-720477 X 3-75» = 24-J942 =
the area of the base in square feet; li «nce 24* 1942 X ^5 =
362*9 13 solid feet, the content.
2. WTiat is the solidity of a square pnstu, vhose length is
5^ feet, and each side of its base 1^ foot?
Ans. 9^ bOijr' feet.
3. What is the solidity of a prism, whose base t? an equi-
lateral triangle, each side being 4 feet, and height 10 feet?
Ans. 69-282 feet.
4. WTiat quantity of water will a prismatic ressel contain,
its base being a square, each side of which is 3 feet, and
heigrht 7 feet? Ans. 63 feet.
* 8«e Apptudix. Democstntion 64.
bS
MENSURATIO-V OP SOUl»6.
PROBLEM IV;
To Jind the solidity of a cylinder.
Rule. Multiply the area of the base by its height, and the
product will be the solid content.*
1. What is the capacity of a right cylinder A B G C,
whose height, and the circumference of its base, are each
20 feet?
20
First— ---—^ — the diameter, half of which multiplied by
half the circumference will give the area of the base (Prob.
XVIII. Sec. II.), that is, 10 X ~^^^. = ^irr= tbe
area of the end; then ^^ X 20 = 636-61828, the content.
•7854
2. V\'Tiat is the content of the oblique cylinder A B F E,
the circumference of whose base ia 20 feet, and altitude A C
20 feet?
25 25
As before, the area of the base is ^r^ ^ 5 then .7=^72
X 20 = 636'61828, the solid content, as before.
3. The length of a cylindrical piece of timber b 18 feet,
and its circumference 96 inches ; how many solid feet in it ?
Ans. 91-676 feet.
4. Three cubic feet are to be cut off a rolling stone 44
inches in circumference ; what distance from the end must
the section be made? Ans. 33'64 inches.
• See Appondn*. OfwuDiutratJoB ♦U
MJSKSDRATION O*" oOLIDS. 87
PROBLEM V.
To Jind the content of a solid J'urmed bif a plane passing
parallel to the ajcis of a cylinder.
Rule. Hnd by Prob. XXVIII., Sec. II., the area of the
base, which, multiplied by the height, will give the solidity.*
1. In the cylinder A B G C, whose diameter is 3, and
height 20 feet ; let a plane L N pass parallel to the axis,
and 1 foot from it ; what is the solidity of each of the two
prisms into which the cylinder is divided ? — (See the last
tigure).
|-^ = (| — l)-f-3 = | = g= -ItiGI the tabular ve.-sed
sine, to which, in the Table of Circular Segments, corres-
ponds the aiea ...... '08004117
which taken from ..... '785398 lb
leaves the other segment .... '6993569^
Then 3' = 9 which X '08604117 = 7'7437053 = seg.
D CN.
Also 9 X -69935699 = 6-29421291 = seg. D G N.
Hence 20 X 7'7437053 = 15'4874 = the slice L K A C
N D; and 20 X 6-29421699 = 125-88434 = the slice
L K B G N D.
2. Suppose the right cylinder, whose length is 20 feet, and
diameter 50 feet, is cut by a plane parallel to, and at the
distance of, 2 1 -75 feet from its aiis ; required the solidity
of the smaller slice? Ans. 1 082-95 feet.
PROBLEM VL
To Jind the solidity nf a pyramid.
Rule. Midtiply the area of the base by the one-third of
the height, and the product will be the solidity. f
* S«« Appendix, Demooiitration 64.
t So« A£p«ndix« Demonstration 65.
88 AIEXSUIIA-TIOK OF SOLIWS.
1. What ia tlie solidity of a square pyramid, each side oi
its base being 4 feet, and height 12 feet ?
4x4 = 16 the area of the base :
Then 16 X V = 64 feet, the solidity.
2. Each side of the base of a triangxdar pyramid is 3, jind
height 30 ; required its solidity? Ans. 38-971 17-
3. The spire of a church is an octagonal pyramid, each side
at the base being 5 feet 10 inches, and its perpendicular height
45 feet; also each side of the cavity, or hollowpart, at the
base is 4 feet 11 inches, and its perpendicular height 41 feet;
it is required to know how many solid yards of stone the spure
contains. .
Ans. 32-19738 yards.
4. The height of a hexagonal pyramid is 45 feet, each side
of the hexagon of the base being 10 : required its soUdity ?
Ans. 3897-1143.
PROBLEM VII.
To find the solidity of a cone.
Rule. Multiply the area of the base by one-third of the
height, and the product will be the solidity.*
1. The diameter of the base of a cone is 10 feet, and its
perpendicular height 42 feet ; what is its solidity ?
10' = 100 X -7854 = 78-54; then 78-54 X V =
1099-56 feet.
2. The diameter of the base of a cone is 12 feet, and its
perpendicular height 100 ; required its solidity?
Ans. 3769-92 feet.
3. The spire of a church of a conical form measures
37-G992 feet round its base ; what is its soUdity, its perpen-
dicular height being 100 feet ? Ans. 3769-92.
4. How many cubic yards in an upright cone, the circum-
ference of the base being 70 feet, and the slant height 30 ?
Ans. 134-09.
* See Appendix, DemonstnitioD 6^.
tTENSUUATlON OP SOLIDS.
89
5. How many cubic feet in an oblwjue cone, xne greatest
slant height being 20 feet, the least 1(5, and the diameter of
the base 8 feet ? Ans. 254 -656588 feet.
PROBLEM VIII.
To find the solidity of the frustum of a pyrainid.
RnUB. Add the areas of two ends and the mean propor-
^onal between them together; then multiply the sum by oue-
ihird of the perpendicular height, and the product will give
the solidity.*
1. In a square pyramid, let A O = 7, P D = 5, aud the
height O Q = 6 ; the solidity of the frustum is required.
7 X 7 = 49 = the area of the base.
5 X 5 = 25 = the aiea of the section S D.
7 X 5 — 35 = the mean proportional between 49 and 25.
Therefore,
49 + 35 + 25
X 6 = 218 = the content of
the frustum.
2. What is the content of a pentagonal frustum, whose
height is 5 feet, each side of the base 1 foot 6 inches, and each
side of the less end G inches ?
Alts. y"3I925 cubic feet.
3. WTiat is tlie content of a hexagonal frustum, whose
height is 6 feet, and the side of the greater end 18 inches, and
of tin- less 1 2 inches ? A ns. 24 -08 1 72 4 .
4. How many cubic feet in a squared piece of timber, the
areas of the two ends being 504 and 372 inches, and ite
length 31^ feet? Am: 95'447 feet.
•iji»e App«ndix, U«tauiuitratioB 67.
90 MEXSURATION OP SOLIDS.
5. WTiat is the solidity of a squared piece of timber, its
length being 18 feet, each side of the greater base 18 inches,
and each side of the small end 12 inches?
Alls. 28-5.
PROBLEM IX.
To find the solidity of the frustum of a cone.
Rule. Add the two ends, and the mean proportional
between them together, then multiply one-third of the smn
by the perpendicular height, and the product will be the
content.*
1 . How many solid feet in a tapering round piece of timber,
whose length is 26 feet, and the diameters of the ends 22 and
1 8 inches respectively ?
Here 22* x -7854 = 380-134 inches, the area of the greater
end, and
18^ X -7854 = 254-47 inches = the area of the less end,
1^380-134 X 254-47) h. = 311-018 = the mean proportional
between the areas of the ends ; then by the rule
254-47 + 380-134 + 311-018 ^ 26 x 12 = 98345 cubic
inches = 56-9 cubic feet, the answer
2. How many cubic feet in a round piece of timber, the
diameter of the greater end being 18 inches, and that of the
less 9 inches, and length 14*25 feet?
Ans. 14-68943 feet.
3. \\niat is the solid content of the frustum of a cone,
whose height is 1 foot 8 inches, and the diameters of the
ends 2 feet 4 inches, and 1 foot 8 inches ?
Ans. 5-284.
* S«e Appendix, Detsoneiration 00.
MENSURATION OF SOLIDS.
91
PROBLEM X.
To find the solidity of a wedge.
Rule I. Add the three parallel edges together, and mul-
tiplv one-third of the sum by the area of that section of tha
weage which is perpendiciilar to these three edges, and th«
product will give the content.*
S'ote. When the quadrangalar rides are parallelograms, the wedge is »
triangular prism, having for its base the triangle B O C ; when the auadranglee
are rectangular, A O ia the height of the prism, and the area of the triangie
B O C multiplied by A O ■will ).'ive its content ; when the triangle B O C
13 isosceles and perpendicular to the jilano A C, the wedge \» of the common
kind ; C G is its edge, and A K B O iu back.
RtJLE II. To twice the length of the base, add the length
of the edge, raiiltiply the sum by the breadth of the base, and
the product by the height of the wedge, and one-sixth of
the last product will be the solidity, that is, (2 L -|- /) X
^ 6 A, by putting L = R B, the length of the base, I =
G C, the length of the edge, i = A R, the breadth of the
base, h = the perpendicular height of the wedge.f
1. Let A O = 4, G C = 3, R B = 2^, the perpendicular
D T = 12, and p the perpendicular distance of B R from
the plane of the face A C = 3^ feet; required the solid
content ?
l^-Jjt^l X 12 X ^ = 66t cubic feet.
• See Appendi-T, DemoDstration 6^.
t 6m Appeuiiix, Demynstrntion 70.
92
MENSURATION OP SOLIDS.
2. The perpendicular height from the point T to the middle
of the back A B is 24*8, the length of the edge C G HO
inches, the base R B 70 inches, and its breadth A R 30
inches ; required the soHdity ?
Ans. 31000 cubic inclies.
3. Ho'w many cubic inches in a wedge whose altitude is 14
inches, its edge 21 inches, the length of its base 32 inches,
and its breadth 4^ inches ?
Ans. 892*5 cubic inches.
PROBLEM XI.
To Jind the solidity of a pinsmoidy which is the frustum
of a wedge»
Rule, By either of the foregoing rules, find the solidity of
two wedges whose bases are the two ends of the frustum, and
height the distance between them, and the sum of both will
be the solidity of the prismoid or frustum.*
1. In the prismoid A B P Q, there is given R B = 18,
A O = 27, P D = 21, S Q = 24, B O = 12, D Q = 4,
and B I = 30 ; what is its solidity ?
18 -I- 27 + 21
= 3960 = the content of the
greater wedge, and x n — ^^ 1440, the
content of the other; then 3960 + 1440 = 5400, the con-
«nt of the frustiuiL,
* See Appendix, Deinocstr»yoti 71.
JaKJrSURATION OP SOLIDS. 93
2, What is the solidity of a piece of wood in the form of a
prismoid, whose ends are rectantrles, the length and breadth
of one being 1 foot 2 inches and 1 foot respectively, and the
corresponding sides of the other 6 and 4 inches respectively ;
the perpendicular height being 30^ feet?
Ans. 18*074 cubic feet.
Note. The following rule will answer for any prismoid, of whatever figure
each end may be.
RcLR. If the bases be dissimilar rectangles, take two corresponding dimen-
sions, and moltiplj each by the snm of double the other dimension of the aamfa
end, and the dimension of the other end corresponding to this last dimension;
then multiply the sum of the prodactt by the height, and ODe-sixth of the laal
product will be the solidity. "
PROBLEM XII.
I'o Jind the $olidity of a cylindroid ; or the frustum cj
ajh elliptical cone.
RTTLE.
I. To the longer diameter of the greater end, add half the
longer diameter of the less end, and multiply tlie sum by
the shorter diameter of the greater end.
II. To the longer diameter of the less end, add half the
longer diameter of the greater end, and multiply the sum bv
the shorter diameter of the less end.
III. Add the two preceding products together, and multi-
ply the sum by •2618 (one-third of '7854) and then by the
height ; the last product wiU be the solidity.f
I. Let A B C D be a cylindroid, the base of which is
an ellipsis, whose two diameters are 40 and 20 inches, the
• See Appendix, Demonstration 72.
t See Appendix, Demonstration 73.
94
MENSURATION OP SOLIDS.
top a circle, whose diameter is 30 inches ; what is it« solidity,
allowing the height to be 10 feet ?
(A B + 4 C D) X G H = (40 + 15) X 20
(C D + I A B) X m P = (30 + 20) X 30
1100
1500
sum = 2600
Then (2600 X '2618 X 10) 6806-8, which, divided by 144,
gives 47'27 feet, the answer.
2. The transverse diameter of the greater base of a cylin-
droid is 13, and conjugate 8 ; the transverse diameter of the
less base 10, and conjugate 5*2; what is the solidity of the
cylindroid, its height being 12?
Ans. 721-93968.
3. The transverse diameter at the top of the cylindroid
is 12 inches, and conjugate 7; the longer diameter at the
bottom is 14 inches, and shorter 12, and its height 10 feet;
required its solidity ? ^n^. 6-78 feet.
PROBLEM XIIL
To Jind the solidity of a sphere.
Rule I. Multiply the cube of the diameter by '5236, and
the product will be the content.
Rule II. Multiply the diameter by the circumference of the
Inhere, and the product multiplied by one-sixth part of the
lameter will be the soUdity.*
* See Appendix, DemosBtr&tioii 74.
bUSMSCZLATlON OF &OLID6. 95
1. Suppose the earth to be a per-
fect sphere, and its diameter 7957|
miles, how many solid miles does it
contain?
79571 X 3-1416 = the circum-
ference of the earth (Prob. XVI., Sec.
II.); then
7957| X 3-1416 x 7957| = 198943750 = the surface of
the sphere ; then
19S943750 x 79571 X i = 263857437760 miles, the solidity
bj Rule II.
Aj,'ain, -5236 X rf» =-5236 X (7957J)' = 263858149120
nules, the solidity by Rule I., which gives the result too great
on account of taking '5236 a little too great.
2. ^^^lat is the solidity of a sphere, whose diameter is 24
inches? Ans. 7238"2464 cubic inches.
3. What is the solid content of the earth, allowing its
circumference to be 25000 miles ?
Ans. 263858149120 miles.
4. Required the solidity of a globe whose diameter is 3o
feet? Ans. 14137-2.
PROBLEM XIV.
To Jind the solidity of the segment of a sphere.
Rule I. From three times the diameter of the sphere
deduct twice the height of the segment ; multiply the
remainder by the square of the height, and that product
by '5236 ; the last product will be the solidity.*
Rule II, To three times the square of the radius of the
-.gment's base add the square of its height; multiply this
sum by the height, and the product by '5236 ; the last result
will be the solidity.
• S«« Appendix, D^moniitntioo IS,
Beg
96
MENSURATION OF SOLIDS.
1 . \VTiat Is the solidity of each of
the frigid zones, the diameter of the
earth being 7957f miles, and half
the breadth, or arc of the meridian
intercepted between the polar circle
and the pole 23^ degrees ; that is,
A D = 23^ degrees, supposing A B
to represent the polar circle.
By Rule I.
As 1 (= tabular radius) : 3978^ (= radius of the earth)
: : -0829399 (= tabular versed sine of 23^ degrees) :
330*0074946, the versed sine, or height of the segment.
Then -5236 A* = (3 rf — 2 A) = -5236 x 330-0074946'
X 23213-2350108 = 1323(179710, the soUd content.
By Rule II.
As 1 : 3978| :: '3987491 (= the tabular sine of 23^
degrees) : 1586-57282526, the radius of the base.
Then -5236 h x (3 r* + ^-) = -5236 X 3300074946
X 7660544-936 = 1323680299-69, the solid\>
2. Let A B D O be the segment of the sphere whose
solidity is required. The diameter A B of the base is 16
inches, and the height O D 4 inches.
Ans. 435-6352 cubic inches.
3. Required the solidity of the segment of a sphere, whose
diameter is 20 feet, and the height of the segment 5 feet ?
Ans. 654-5 feet.
PROBLEM XV.
To Jind the solidity of the frustum or zone, of a sphere.
RTILE.
I. To the sum of the squares of the radii of the two
ends, add ^ of the square of their distance, or of the height
of the zone ; this sum multiplied by the height of the
zone, and the product again by 1-5708, will be the solidity.
MKNSURATION OP SOLIDB,
.^7
II. For the middle lone of a sphere. To the squaie of
the diameter of the end add two-thirds of the square of the
height; multiply this sum by the height, and then by '7304,
the last result will be the soUdity.
Or. From the square of the diameter of the sphere,
deduct one-third of the square of the height of the middle
zone; miJtiply the reraninder by the height, and then by
•7854, the last residt will be the solidity.*
1. Reqtiired the solidity of the
fni-^tmn of s sphere, the diameter
of whose greater end is 4 feet, the
diameter of the less end 3 feet, an«i
the height 2^ feet ?
(o: ^.1.52^^ X2-5-) X 1-5708
X 2-5 = 8^ X 3-927 = 32-725, the
soliditv of the frustum.
2. ^^^lat is the solidity of the temperate zone, its breadth
being 43 degrees, the radius of the top being 1586'57282526,
and the radius of the base 3648'8675053S, and height
20f)2-2Go5 ?
(3648-86750538^+ 1586-57282526' + ^ X 2062-2655') x
2062-2655 X 1-5708 = 17249136 x 2062-2955 x 1-5708 =
55877778668, the solidity of each temperate zone.
3. Required the solidity of the torrid loue, v^lach exicndj
23^ degrees on each side of the equator, the diameter being
79571 miles, and the height 3173*14565052?
(7957-75' — ^ X 3173-14565052') x 3173-14565052 x
•7854 = 149455081137, the answer.
4. \Miat is the solidity of the middle zone of a sph^e, whose
top and bottom diamtters are each 3 inches, and heiglit 4
inches? Ana. 61-7848.
5. WTiat is the solid content of a «one, whose greater
diameter is 20 feet, less diameter 15 feet, and the height
U) feet? Ans. 189-58.
* See ADpendix. I)<>moiistntioD 76.
98
^IBNSURATION OP SOLIDS.
6. How many solid feet in a zone, whose greater diameter
is 12 feet, and less diameter 10; the height being 2 ?
Ans. 195-8264.
PROBLEM XVI.
To find the solidity of a circular spindle.
Rule. Find the distance of the chord of the generating
circular segment from the centre of the circle, and also the
area of this segment.
Then, from one-third of the cube of half the length of the
spLadle or half chord of the segment, subtract the product of
the central distance, and half the area of the segment ; the
remainder, multiplied by 12-5664, will give the solidity.*
1. Let the axis A C of V B
a circular spindle be 40
inches, and its greater di-
ameter B L 30 inches ; what
is its solidity ?
20^ -^ 15 = 26§, then
26f -f- 15 = 41f, the dia-
meter of the circle. Again,
4 1 2 3Q
—^ — = 5f, the central
distance.
Now 15 ..f. 4 If = -36, the
area segment corresponding
to which is -254550, which
multiplied by the square of 4 If, produces 441-92708 the
area of the generating segment ABC, the half of which is
220-96354.
Lastly, (20» -i- 3) — (5^x220-96354) = 1377-71268, and
this multiplied by 12-5664 produces 17312-88862 cubic inches,
the solidity required.
* S«« Appendix, Demonstratioii 77.
IffENSCRATION OF sOr-fDS. 99
2. The asla of a circular spindle is 48, and tTie middle
leter 36 ; required the solidity of the spindle ?
Ans. 29916-6714.
PROBLEM XVII.
To find the snlidity of the middle frustum of a circular
spindle.
RULE.
I. Find the distance of the centre of the middle frustum,
'xom the centre of the circle.
II. Find the area of a segment of a circle, the chord of
which is equal to the length of the frustum, and height hali
',he difference between its greatest and least diameters ; to
which add the rectangle of the length of the frustum anc^
lialf its least diameter ; the result will be the generating
surface.
III. From the stjuare of the radius subtract the square of
the central distance, the square root of the remainder wriii
give half the length of the spindle.
IV. From the square of half the length of the spindle
take one-third of the square of half the length of the middle
frustum, and multiply the remainder by the said half length.
V. Multiply the central distance by the generating sm^ace,
and subtract this product from the preceding ; the remainder,
multiplied by 6*283'2, will give the solidity.*
1 . Required the solidity of the middle frustum of a circular
spindle, the length D E being 40, the greatest diameter Q F
32, and the least diameter P S 24 ?
First, 20- + 4 = 100, and 100 + 4 = 104, the diameter
of the circle.
Again, 52 — 16 = 36, the central distance. Also, ^ (32
— 24) = 4, and 4 + 104 = '038 j^, the area segment cor
* S«« Appeadiz. DcmoniitratioD 7((
100 MEXBURATION OF SOLIDS.
responding to which is '009940, which, multiplied by the
square of 104, produces 107'51104, the area of P L Q; and
40 X 12 = 480 the area of the rectangle P D E L.
Hence 107-51104 + 480 = 587-51104 the area of the
generating surface P D L E.
Next y/ (52^—36^) = -/ (1408) = 8 ^ (22) = B
half the length of the spindle ;
And (1408 _ H^ X 20 = 25493^.
Then 36 X 587'51104 = 21150-39744, and
(25493^ — 21150-39744) x 6-2832 = 27287-5347, the
"equired solidity.
2. WTiat is the solidity cf the middle frustum P S R L oi
a circular spindle, whose middle diameter F Q is 36, the
diameter P S of the end 16, and its length D E 40 ?
Ans. 29257-2904.
PROBLEM XVIIl.
' To find the solidity of a spheroid.
Rule. Multiply the square of the revolving axis by the
6xed axis, and this product again by "5236 for the solidity.*
1. \Vhat is the solidity of a prolate ^-^frrr^
spheroid whose longer axis A B is 55 ^' •— «■—
indies, and shorter axis C D 33 ?
Here33'x55x -5236 =31361-022
cubic inches, the answer.
2. WTiat is the solidity of an oblate spheroid, whose longer
axis is 100 feet, and shorter axis 6 ?
A71S. 31416 cubic feet.
3. WTiat is the solidity of a prolate spheroid, whose axes are
*^and 50? ^,,,. 41888.
4. W^hat is the solidity of an oblate spheroid, whose axes
^20 and 10? ^ ^J. 2094-4.
• Soe Appendix, Domonstr&tioii 79.
AUSXSUaATIOS OF SOLIDS.
101
PROBLEM XIX.
To find the solidity of the segment of a spheroid, the base
of the segment being parallel to the retolving ajcis of the
spheroid.
CASE I.
Rule. From three times the fixed axis, deduct twice the
height of the segment, multiply the remainder by the square
of the height, and that product by '5230.
Then say, as the square of the fixed axis is to the square of
the revoK-ing axis, so is the product found above to the solidity
of the spheroidal segiuent.*
!. What is the content of the segment of a prolate
spheroid, the height O C being 5, the fixed axis 50, aud
the revolving axis 30 ? — See last figure.
50 X 3 — 5 X 2 = 150 — 10 = 140 ; then
140 X 5- = 3500, aud 3500 X '5236 = 18326 ; then
25 : 9 : : 1832-6 : 659-736, the answer.
CASE n.
When the base is elliptical, or perpendicular to the revolving!
axis.
Rule. From three times the revolving axis, take double the
height ; multiply that difl'erence by the square of the height,
and the product again by '5236.
Then as the revohnng axis is to the fixed axis, so is the last
product to the content.f
2. \Miat is the content of the segment
of a spheroid, whose fixed axis is 50,
revdviug uiis 30, and height 6 ?
30 X3 — 2x6 = 90— 12 = 78;
Then 78 X 6' = 2808 ; and 2808 X '5236
= 1470-2688:
Then 30 : 50: : 1470-268& : 2450-448, the
answer.
D
a!— _J
^
F
B
• Se« Appendix, Demonstration 80.
T tie« App«ndix, Demosutoation 8 1.
102
MENSURATION OP SOLIDS.
3. In a prolate spheroid, the transverse or fixed axis is 100,
the conjugate or revolving axis is 60, and the height of the
segment, whose base is parallel to the revolving axis, is 10;
required the solidity?
Ans. 5277-888.
4. K the axes of a prolate spheroid be 10 and 6, required
the content of the segment, whose height is 1, its base being
parallel to the revolving axis ? Ans. 5"277888.
PROBLEM XX.
To find the solidity of the middle zone of a spheroid, the
diameter of the ends being perpendicular to the fixed axis,
the middle diameter, and that of either end being given,
together with the length of the zone.
Rule. To twice the square of the middle diameter, add tha
iquare of the diameter of the end ; multiply the sum by the
length of the zone, and the product again by '2618 for the
solidity.*
1. What is the solidity of the
middle zone of an oblate spheroid,
the middle diameter being i 00, the
diameter of the end 80, and the
length 36?
100^ X 2 + 80' = 26400 ; then
26400 X 36 = 950400, and 950400
X -2618 = 248814-72, the answer.
2. \Miat is the solidity of the
liddle frustum of a spheroid, the greater diameter being 30,
he diameter of the end 18, and the length 40 ?
Ans. 22242-528.
■ Se« Appendix, Dcmonstratioa 82.
UKN8URAT10N OF 80LTD6.
103
PROBLEM XXI.
To find the loliditi/ of a parabolic conoid.
Rule. Multiply the square of the diameter of its base by
•3927, and that product by the height ; the last product will
be the solidity.*
1 . \Miat ifi the solidity of the parabolic
conoid, whose height is 10 feet, and the
diameter of its base 1 feet ?
10- X -3927 = 39-27 ; then 39-27 X
10 = 392'7, the solidity required.
2. What is the solidity of a parabolic
conoid, whose height is 30, and the dia-
meter of its base 40 ? ^V
Ans. 18849-6.
3. What is the content of the parabolic conoid, whose alti-
tude is 40, and the diameter of its base 12?
Ans. 2261-952.
4. Required the solidity of a parabolic conoid, whose height
is 30, and the diameter of its base 8 ?
Ans. 753-984.
PROBLEM XXIL
To find the solidity of the frustum of a parabolic conoid.
Rule. Multiply the sum of the squares of the diameters
vi the two ends by the height, and that product by '3927 ;
the last product will be the solidity .f
• See Appendix, Demonstration 83.
f bee Appendix, DemomtrKtion Bi.
II
104
ME!fSUBATIOX oV SOLIDS.
1. The greater diameter of the
frustum 18 10, and the less diameter
5 ; what is the solidity, the length
Wngl2?
l(f=: 100
5«= 25
125. Then 125 X 12 = 1500, and
1500 X -3927 = 589-05, the solidity.
2. The greater diameter of the fi-ustmn of a parabolic
conoid is 20, the less 10, and the height 12; what is the
V)lidity? ^n*. 2357-4.
3. The greater diameter of the frustum of a parabolic
conoid is 30, the less 10, and the height 50 ; required the
solidity? Ans. 19635.
4. The greater diameter of the frustmn of a parabolic
conoid is 15, the less 12, and the height 8 ; required the
•oUdity? ^715. 1159-8408.
PROBLEM XXIII.
To find the tolidity of a parabolic spindle.
Rule. Multiply the square of the middle diameter by
■7854, and that product by the length; then fj of this
product will be the solidity.*
1 . The middle diameter C D, of (^
a parabolic spindle is 1 feet, and ^ \ -.^^^^
the length A B is 40 ; required its ''^^-^II~~~1 '~~^0
solidity ? Y)
W X -7854 X 40 = 3141-6 feet.
Then 1^ X 3141-6 = 1675-52 feet, the answer.
• Se« Appendix, Demonftration 85.
STEXSURATION OP SOLIDg.
105
2. The middle diameter (/ D, of a parabolic spindle ii 12
's-ei, and the length A B is 30 ; required the solidity ?
Ans. 1809-5616.
3. The middle diameter of a parabolic spindle is 3 feet,
ad the length 9 feet ; required its solidity ?
Ans. 33-92928.
4. The middle diameter of a parabolic spindle is 6 feet, and
he length 1 ; required its solidity ?
Ans. 150-7968.
5. The middle diameter of a parabolic spindle is 30 feet,
ind the length 50 ; required its solidity ?
Ans. 18849-6.
PROBLEM XXIV.
To Jind the solidity of the middle frustum of a parabolic
spindle.
Rule. To double the sqiiare of the middle diameter, add
the square of the diameter of the end ; and from the sum
mbtract -j^ of the square of the difference between these
diameters ; the remainder multiplied by the length, and that
product by '2618, will be the sohdity.*
T C
1. In a parabolic spindle, the
aiiddle diameter of the middle frus-
tum is 16, the least diameter 12, ^
ftnd the length 20 ; re<|uired the
BoUdity of the frustmn ?
Here 2 X 16' + 12' — ^^ X 4» = 512 + 144 — 6-4 =
>49-6; hence 649-6 X 20 x -2618= 3401-3056, the solidity.
* S«e Appeodix, Dv>mon5U«tioD 80.
H 2
106
MENSURAriON OP SOLIDS.
2. The Diing diameter of a cast is 30 inches, the hea&
diameter 20 inches, and the lengih 40 Inches ; required its
content in ale gallons, allowing 282 cubic inches to be equal
to one gallon? Aiis. 80'211 gallons.
3. The bung diameter of a cask is 40 Inches, the hearl
diameter 30 inches, and the lengtli 60 ; how many Avine gal-
lons does it contain, 231 cubic inches being equal to one
gallon? Ans. 276*08 gallons.
PROBLEM XXV.
To Jind the soliditj/ of a hyperbolic conoid.
Rule. To double the height of the solid add three times
tbe transverse axis, multiply the sum by the square of the
radius of the base, and that product by the height, and thi«
£
last product by '5236 ; the result di-
vided by the sum of the height and
transverse axis, will give the solidity.*
1. Required the solidity of an hyper-
lolic conoid, whose height V m is 50,
Jie diameter AH 103-923048, and the
transverse axis V E 100 ?
Here (2 x 50 + 3 x 100) X C"^'^^^""'^)'. =400x2700=
1080000; ^i}2^$^$^^><::^=imm,fk.s.mij.
2. WTiat is the content of an hyperboloid, whose altitude
is 10, the radiujB of its base 12, and the transverse 30?
Ans. 2073-451 151369.
'See Appendix. Demoiuitniinon 87.
MENSURATION OP BOLIDb. 107
PROBLEM XXVL
To find the solidity of the frustum of an hyperholoid, or
hyperbolic conoid.
Rule. To four times the square of the middle diameter,
add tlie Bum of the equares of the greatest and least diame-
ters ; multiply the result by the altitude, and that product by
•1309, for the soUdity.*
1. Required the solidity of the
frustimi A C E H D B of an hyper-
bolic conoid, whose greatest diameter
A. B is 96, least diameter E H 54,
middle diameter C D 76-4264352,
and the altitude m n 25 ?
Here 4 C D* -f A B* -f E H- =
(5841 X 4) + 9216 + 2916 =
35496, and 35496 x 25 x '1309 = .^
1 16 160*66, the answer.
2. What is the solidity of an hyperboloidal cask, its bunv,
diameter being 32 inches, its head diameter 24, and the
diameter in the middle between the bung and bead § -v/ 310,
and its length 40 inches ?
Ayis. 24998-69994216 inches.
PROBLEM XXVn.
To find the solidity of a frustum of an elliptical spindle,
or any other solid formed by the revolution of a conic
section about an axis.
^ Rule. Add together the squares of the greatest and least
diameters, and the square of double the diameter in the
" S«« Appendix, Demonstration 88.
108
MENSURATION OP SOLIDS.
middle between the two ; multiply the sum by the lengthy
and the last product by '1309 for the solidity.*
1. What is the content of the middle frustum C D I H of
any spindle, the length O P being 40, the greatest, or middle
"iameter E F 32, the least, or diameter at either end C D 24,
od the diameter G K 30-157568?
Here 32^ + (2 X 30-157568)' + 24« = 5237-89 sum;
Then 5237-89 X 40 = 209515-6, and
209515-6 X -1309 = 27425-7, the answer.
2. What is the content of the segment of any spindle, the
length being 20, the greatest diameter 1 0, the least diameter at
either end 5, and the diameter in the middle between these 8 ?
Ans. 997-458.
PROBLEM XXVIII.
To find the solidity of a circular ring.
Rule. To the thickness of the ring add the inner diameter ;
multiply the sum by the square of the thickness, and the product
by 2-4674, for the solidity.-j"
1 . The thickness of a cylindrical
ring is 2 inches, and the diameter
C D 5 inches ; required its soUdity ?
(2 -f 5) X 4 = 28 ; then 28 x
2-4674 = 69-0872 cubic inches,
the answer.
2. Required the solidity of an
iron ring whose axis forms the cir-
* See Appendix, ^^"""""tnition 89.
t See Appendix, UemottBtration 90,
ilEN'SURATJOX OF SOLIDS. 109
cumference of a circle ; the diameter of a section of the ring
2 inches, and the inner diameter, from side to side, 1 8 inches ?
Ans. 197*3925 cuhic inches.
3. The thickness of a eylindrical ring is 7 inches, and the
inner diameter 20 inches ; required its solidity ?
Ann. 3264-3702.
4. WTiat is the solidity of a circtdar ring, whose thickness
if 2 inches, and lU diameter 1 2 inches ?
Ant. 138'1744 cubic inchea.
110
THE FIVE REGULAR BODIES.
SECTION V.
DEFINITIONS,
A regular body is a solid contained under a certain number
of similar and equal plane figures.
Only five regular bodies can possibly be formed. Because
it is proved in Solid Geometry that only three kinds of equi-
lateral and equi-angular plane figures, joined together can
make a solid angle.
1. The tetraedron, or equi-lateral
pyramid, is a solid having four trian-
G^ular faces.*
2. The hexaedron, or cube, is a
V)lid having six square faces.
■«
re-
3. The octaedron is a
gular 8ohd having eight tri
angular faces
• If figures siinllar to those annexed to the definitions, be drawn on pasta*
board, and cut out, by cutting through the bounding lines, and if the other linet
be cut half through, and then the parts be turned up and glued together, the
bodies defined will be formed.
THE REGULAJl BODIES. Ill
4. The dodecaedron has twelve pentagonal faces.
S. The icosaedron has twenty triangular faces.
PROBLEM I.
To find the solidity of a tetraedron.
Rule I. Multiply ^ of the cube of the lineal side by
the square root of 2, and the product will be the solidity.
Rule II. Multiply the cube of the length of a side of the
body by the tabular solidity, and the product will be the solidity
of the tody.* This nile is general for all the regular bodies.
1. If the side of each face of a fj
tetraedron be 1 ; required its soli-
dity?
Here ^l^ X 1' X ^/ 2 = ^J,- x
V^ 2 = -117851 13, the solidity.
2. The side of a tetraedron is
12; what is its solidity?
Ant. 203-6467.
* 8«« App«ndix. Demosatration 91.
112
MENSURAllON OP SOLIDS.
PROBLEM II.
To find the solidity of a hesaedron, or a cube.
Rule. Cube the side for its solidity.*
1. If the linear side of a hexaedron be 3, what is it»
content ?
Ans. 3 X 3 X 3 = 27.
PROBLEM IIL
To jftnd the solidity of an octaedron.
Rule. Multiply the cube of the side by the square root ol
2, and ^ of the product will be the content.|
I. What is the solidity of an octaedron, when the linear
side la 1 ?
2 X
^ = ^
^/ 2 = F,
1' X v"
4714045.
2. What is the solidity of the
octaedron, whose linear side is 2?
Ans. 3-7712.
PROBLEM IV.
To find the solidity of a dodecaedron.
Rule. To 21 times the square root of 5 add 47, and divide
the sum by 40 ; multiply the root of the quotient by 5 times
the cube of the lineal side, and the product will be the
solidity.^
• See Appendix, Demoiutration 64.
t See Appendix, Demourtration 92.
X See Appendix, Demonntrbtion iA
TILE UEGULAR B00xl:.-S.
lis
1. If the lineal side of the dodecaedron be 1, what Ls iu
solidity ?
47 V 21 */ 5
Here A = 1 , consequently 5 A' v' — ^ ^
f-66311896, the content.
2. The side of a regular dodecaedron is 12 inches ; how
many cubic inches does it contain ?
Ans. 13241-8694592.
PROBLEM V.
To Jind the soliditi/ of an icosaedron.
Rule. To 7 add three times the square root of 5, taktj
ha4f the sum, multiply the square root of this half sum by ^
of the cube of the lineal side, and the product will be the
solidity.*
*8ee AppendU, i>MuvuKnaion 9A.
114
MENSURATION OF SOLIDS.
1 . UTiat is the solidity of an icosaedron, whose lineal side
b 1?
Let the side be denoted by A. Then A = 1, and coa
sequently
I A" V ^ + ^^^ ^ I V '^ + ^^^ = 2-18169499,
the coDtent.
2. What is the solidity of an icosaedron, whose lineal sidt
is 12 feet ? Ans. 3769-9689 feet.
Note. ITie following table may be collected from the examples ^ven in the
foregoing rules, each of -which has been demonstrated under its particular head.
It has also been demonstrated that the cube of the lineal side of any regular
Bolid multiplied by the tabular number corresponding to the figure, will give its
content. It is particularly recommended to the pupil to employ the general
rule given in Problem I. whenever the content of any of the five regular bodies
is required.
TABLE III.
Showing the solidity of the Jive regular bodies, the length
of a side in each being 1.
No. of
sidei.
Names.
SoUdity,
4
6
8
20
12
Tetraedron
Hexaedron
Octaedron
Icosaedron
Dodecaedron
•1178511
1-0000000
•4714045
2-1816950
7-6631189
I
6DRFACBS OP THE REGULAR BODIES. 116
PROBLEM VI.
7'ojind the surface* of a tetraedron.
Rule I. Multiply the square of the linear side by the
Bquai-e root of 3, aud the product will be the whole surface.f
Rule II. Multiply the square of the length of a side of the
body, by the tabular area corresponding to the figure, and the
product will be the surface of the body. This is a general
rule for finding the surfaces of the regular bodies.
1. If the side of a tetraedron be 1, what is its surface?
Here, V X v" 3 = x/ 3 = 1-7320508 = the whole
MM-face.
'2. The side of a tetrae«lrou is 12 ; what is its surface ?
^n*. 249-4153152.
PROBLEM Vn.
To fnd the surface of a hexaedron, or cube.
Rule. Square the side and multiply it by 6, and the product
will be the surface.}
1 . If the side be 1 , what is the surface of a hexaedron ?
1 * X 6 = 6 the whole surface.
2. If the side be 4, what ia the surface of a hexaedron ?
Ans. 9&.
* Though the next' section treats exclusively of the surfaces of solids, and
would therefore sieeiii to be the proper place for this problem and the following
ones In this section, vet it has been thought more convenient to place together
the rul«8 both for findinc; the solidities and surfaces of those curious bodies.
f See Appendix, Demonstration 95.
i See Appendix, Demonstration 96.
lib SURFACES OF THE REGULAK BODIES. I
PROBLEM VIII.
To find the surface of an octaedron.
Rule. Multiply the square of the side by the square root
of 3, and double the product will be the surface.*
1. If the side of an octaedron be 1, what is its siuface ?
2xl'i/3 = 2v'3 = 3-4641016 = the whole surface.
2. If the side of an octaedron be 12, what is its super
6cies? ^n*. 498-8306304.
3. If the side of an octaedron be 4, what is its surface ?
.ans. 55-4256256.
PROBLEM IX.
To find the superficies of a dodecaedroii.
Rule. To 1 add | of the root of 5 ; multiply the root of
the sum by 15 times the square of the lineal side, and tlie
f)roduct will be the surface.f
1. If the lineal side be 1, what is the surface of a regular
dodecaedron ?
Here P X 15 a/ (1 + i v^ 5) = 15 v (1 + | v^ 5) =:
20-645728S07, the surface.
2. \\Tiai Is the surface of a dodecaedron, whose lineal side
is 2 ? Ans. 82-58292.
• S«e AppendiT, Demonstration 91 .
+ S«e Appendix, D«moiuti»tioD 9S,
SURFACES OF TITb liEUULAB BODIKa.
117
PROBLEM X.
To find the superjicies of an icosaedron.
Rule. Multiply five times the square of the lineal side by
the square root of 3, and the product will be the surface.*
1. The side of an icosaedron is 1, what is its surface?
Sxl'Xv/SinS'v/S- 8-66025403.
2. What is the surface of an icosaedron whose side is 2 ?
Ans. 34-641.
3. What is the siirface of an icosaedron whose side is 3 ?
Am. 77-9423.
Not«. In finding the gnperficial content of the regnlar bodies, it is par.
ticularlj recomni ended to employ the gener&l ml« given in Problem VI.
in practice, in preference to anj other, llie p&rtictilar rules ^ven for each
solid are introduced merely to find the tabular numbers by which the pupil it
to ■work.
From the examples given in the preceding rules, in which the lineal side d
sach regular solid is 1, the following tabular numbers may be collected.
TABLE IV.
Showing the mrfacet of th« Jive regular bodies, when the
linear side is 1.
Num)>er
of iiaes.
Naowm.
iiurface.
4
6
8
12
20
Tetraedron . 1 -7320508
Hexaedron . 6-0000000
Octaedron . \ 3-4641016
Dodecaedron . 20-6457288
Icosaedron . : 8-6602640
* See Appendix, Demoostration ^9.
118
SUEFACES OF SOLIDS.
SECTION VL
PROBLEM I.
To find the surface of a prison.
Rtji:e. Multiply the perimeter of the end of the solid bj
ita length, to the product axid the area of the two ends, a^iq
the sum will be the surface.*
K
U I
\
7E
/
C D
'^•e 4pp«&dix. DemoiuftraidoQ ICO.
MESIMURATION Of SOUDS. ^ ^ ^
1. If the side H I of the pentagon be 25 feet, and height
ID 10, what is its surface ?
25 X 5 = 1 25, the perimeter ;
Then 125 X 10 = 1250 = the upright surface ;
25' X 1-720477 = 1075-298125 = the area of one end;
And 1075-298125 X 2 = 2150-596250 = the area of both
ends ;
Then 2150-596250 + 1250 = 340U-5962c = the entire
surface.
2. If the side of a cubical piece of timber be 3 feet 6
inches, what is the upright surface and whole superticiai
content ?
jdng ^ ^^ ^^^ upright surface.
^73 feet 6 in. whole superficial content.
3. If a stone in the form of a parallelopipedon be 12 feet
9 vnciies long, 2 feet 3 inches deep, and 4 feet 8 inches broad,
what is the upright surface and whole superticiai content?
J (176 feet 4 in. (> sec. upright surface.
' \ 197 feet 4 in. 6 sec. whole sup. content.
PROBLEM II.
To Jind the surface of' a pyramid.
Rule. Multiply the slant height by half the circuitife-
rence of the base, and the product will be the sm-face of the
sides, to which add the area of the base for the whole
gurface.*
Xott. Thu slant height of a pyramid is the perpendicular distance from the
vertex to the middle of one of tho sides, and the perpendicular height is »
straight line drawa from the vert«x to the middle of the base.
• See Appendix, DtiuonBtration IW.
120
MENSURATION OP SOLIDS.
B D
1. The slant height of a trlangvilar pyramid is 10 feet, and
each side of the base is 1 ; what is its surface ?
Half circumference = 5-
Slant height
= 10
Upright surface = 15
Ar'ia of the base = '433013
The entire surface = 1 o '4 330 13
2. The perpendicular height of a heptagonal pyramid is
13'5 feet, and each side of the base 15 inches ; required its
surface ? A ns. 65 '0 1 28 feet.
PROBLEM III.
To find the surface rf a cone.
Rule. Multiply the slant height by half the circumference
of the base, and the product, with the area of the base, will
be the whole surface.*
iSee Appendix. Demonstration 1U2.
SCUKACKS Of SOLIDS. 121
1. Wliat is the surface of a cone whose side is 20, ami the
circumference of its base 9 ?
Here 20 X ,^ = 90 = the convex surface.
9- X -07958 ='G-445i<8 = the area of the bcise.
Then 90 + 6-44598 = 9(i-44598 = the whole surface.
2. The perpendicular height of a cone is 10-5 feet, and
ihe circumference of its base is 9 feet ; what is its super-
ficies ? ^ «A\ 5 4 • 1 3.3() feet.
PROBLEM IV.
To find the superficies of the J'ruslurn nf a right, vfgulai
pyramid.
Rule. ^\dd the perimeters of the two ends together, and
multiply half the sura by the slant height, the product will
be the upright surface ; to which add the areas of both ends,
and the sum will be the whole surface.*
1. \Vliat is the superficies of the frustum of a ^^ ^- ju
square pyramid, each side of the greater base /jri \
A. B being 10 inches, and each side of the less
hase C D 4 inches, and slant height 20 inches ?
Here 10 X 4 = 40 the perimeter of the greater
base.
And 4x4 = 1 6 the perimeter of the less end.
Sum 50, the half of which is 28.
Then 28 X 20 = 560 = the u[>right surface.
10 X 10 = 100 = the area of the greater base.
4x4= 16 = the area of the less end.
Hence 500 + 100 + 16 = 676 = the whole surface.
2. What is the superficies of the frustum of an octagonat
pyramid, each side or the greater base being 9 inches, each
side of the less base 5 inches, and the height 10-5 feet?
A71S. 52-59 feet.
See Appendix, Demonstration 103.
r 2
122 MENSURATION OP SOLIDS.
PROBLEM V.
To Jind the superjicies of the frustum of a cone.
Rule. Add the perimeters of both ends together, and mul-
tiply half the sum by the slant height, to which add the areas
of both ends, for the whole superficies.* ^
1. If the diameters of the two ends C D and A B are 7
and 3, and the slant height D B 9, what is the whole surface
of the frustum A B C D ?
■^ + 1 X 31416 X 9 = 141-372, the convex surface.
7 X 7 X -7854 = 38-4846, the area of the base C D.
3 X 3 X -7854 = 7-0686, the area of the end A B.
Then 141-372 + 45-5532 = 186-9252 = the whole
siirface of the frustum.
2. What is the superficies of the frustiun of a cone, whose
greater diameter is 18 inches, and less diameter 9 inches,
and the slant height 171-0592 inches?
Ans. 7572-981.
• S«eAppec3ix, Demonstration 104.
SUHFACES OF SOLIDS.
123
PROBLEM VI.
To Jind the stiperjicies of a wedge.
Ri'LE. Find the area of the back, which is a right-ang;lc(]
parallelogram ; find the areas of both ends, which are tri-
angles ; and also of both sides, which are trapezoids ; all
these areas added together will evidently be the whole
surface.*
1. The back of a wedge is 10 inches A H
long, and 2 inches broad, each of its
taced is 10 inches from the edge to the
back ; required its whole surface ?
10 X 2 = 20 = the area of the back.
10 X 10 X 2 = 200 the areas of both
faces,
v/ (A E- — Ea:-) = v/ (100 — 1) = - ^ -^ ,■
9-949 = A x; then
9-i)49 X 2 = 19-898 = areas of both
ends.
Hence 200 + 20 + 19-898 = 239-898 z= the whol^ sur-
face of the wedjre.
2. The back of a wedge is 20 inches long, and 2 incheh
broad; each of its faces is 10 inches from tlie back to the
edge; what is its whole surface ? Ans. 459-898.
PROBLEM Vn.
To Jind the area of the frustum of a wedge.
Rule. Eind the areas of the back and top sections ; A
the two faces ; and of the two ends ; the sums of all the
separate results will evidently be the whole sui-face.
• See Appendix Demonstraticn 105.
124
ilJS>)Sr«ATlOX OF SOLIDS.
1. The length and breadth of the
back are 10 and 2 inches, the lengtli
and breadth of the upper section are
10 and 1 inches, the length of the
edge from the back to the upper sec-
tion is 1 inches ; required the whole
surface ?
10 X 2 = 20 = the area of the back.
10x1 = 10 = the area of the upper
section.
10 X 10 X 2 = 200 = the areas of both faces.
2 — I , , ,
~^— = 1 = -0, and V (IC'J — -25) = 9-98 = B y.
Then (2 + 1) X 9'98 = 29-94 = areas of both ends.
Hence 20 + 10 + 200 + 29-94 = 259-94 inches, the ans^ver.
2. The length and breadth of the back are 10 and 4, the
length and breadth of the upper section are 5 and 2, and
the length of each of tlie faces is 20 ; required the whole
superficies? ^»*. 470-78.
PROBLEM VIIL
To find the surface of a globe or sphei-j-
Rule. Multiply the diameter of the sphere by its circum-
ference, and the product will be its convex surface.*
1. Wliat \s> the surface of a globe, whose diameter is 24
mches ?
24 X 3-MH) = 75-3984, the circumference :
75-3984 X 24 = 1809-5616 inches, the answer.
2. WTiat is the surface of the earth, its diameter being
7957|, and circumference 25000 miles ?
Ans. 198943750 square miles.
• See Appendix, UemoMtration IC7.
SURFACES OF SOLIDS. 125
PROBLEM IX.
Tc find the convex surface of any segment, or zone of a
sphere.
Rule. JIultiply the circumference of the whole sphere
by the height of the segment, or zone, and the product will
be the convex surface.*
1. If the diameter of the earth be 7970 miles, th«» height
of the frigid zone will be 2D2'36r2S3 miles, wXiaX is its
surface ?
Here 7970 X 3* 1416 = the circumference; then
7970 X 3-1416 X 252-361283 = 6318761-107182216
miles.
2. If the diameter of the earth be 7970 miles, the height
of the temperate zone will be 2143'623553o miles; what i;*
its surface ? Ans. 53673229-812734532 miles.
3. If the diameter of the earth be 7970 miles, the height
of the torrid zone will be 3178*030327 miles; what is ita
sm-faee ? Ans. 79573277-600166504 miles.
XuU. By ad'ling the surfaces of both frigid zones and botb temperate
tones, to the surface of the torrid zone, the sum 19y537259*44, ig the surface
9f the earth in square miles.
4. Tlie diameter of a sphere is 3, the height of the seg-
ment 1 ; what is its convex surface? Ans. 9-4248.
5. The circumference of a sphere is 33, the height of the
segment is 4 ; what is its convex surface ? Ans. 1 32.
PROBLEM X.
To find the surface of a cyliiuler.
Rule. Multiply the circumference by the length, and the
product will be the convex surface ; to which add the area
of the t\ro ends, and the stmi will be the surface of the entire
solid.f
* ^pe Appendix, Demonstration I OS.
T "^(.-e Ai>Pend:x Deiuonnration loy.
126 MEXSURATIOX OF SOLIDS.
1 . What is the entire surface of a cylinder, whose length is
10 feet, and its diameter 5 feet ?
3-1416
5
15-7080, then io-708 X 10 = loT-OS the convex
surface.
5 X 5 X '785-i — the area of the base ; then
S X 5 X 6 X -7854 = 50 X -7854 = 39-2700 = the area
of both bases ; then
157-08 4- 3'J-27 = 196-35, the answer.
' Kequired the superficial content of a cylinder, whose
•li.inieter is 21-5 inches, and height 16 feet. Ans. 95-1 ft.
3. \Miat is the surface of a cylinder whose diameter is
2075 inches, and its length 55 inches ? Ans. 29*595 feet.
PROBLEM XL
Til Jind (lie superjicies of a circular cylinder.
Rule. Add the inner diameter to the thickness of tht
ling, multiply the sum by the thickness, and that product by
9'8696 for the superficies.*
1. The thickness AC of a cvlindrical rvAfr is 2 inches,
«io mner diameter C D 5 inches ; required its superficial
content.
lioro ^2 + 5) X 2 = 14; then 14 X 9-86J;6 = 138-1744
square inches.
* Jiee Appendix, "Demonstration 1 10.
SUKFACE8 OF THE REQCLAR B0DIE9. i- '
PROBLEM XII.
To find the surface of a parallelopipedon.
Rule. Find the area of the sides and ends, and their suiii
v ill be the surface.
1 . ^Miat is the surface of a parallelopipedon, whose length
10 feet, breadth 4, and depth 2 ? Ans. 136 feet.
10. X 4 = 40 = the area of one face.
10 X 4 = 40 = the area of its opposite face.
10 X 2 = 20 = the area of one face.
10 X 2 = 20 = the area of its opposite face.
4x2= 8 = the area of one end.
4x2= 8 = the area of its opposite end.
1 jti = the surface of the whole sobcL
2. The length of a parallelopipedon is 5, breadth 4, anc
depth 3; what is its siirface? A^ns. 94.
12S
SECTION VII.
DEiSCRIPTlOS OF THE CARPEXTER'S RULE.
This instrument is sometimes called the sliding rule, and
is used in measurins: timber, and artificers* works. Bv it
dimensions are aken, and contents computed.
It consists of two equal pieces of box-wood, eacb one foot
iong, connected by a folding joint.
One face of the rule is divided into inches and half-quar-
ters or eighths. On the same side or face are several plane
Bcales divided by diagonal lines into twelfths^ these are
i;hiefly. used in planning dimensions which are taken in fee;
and inches. The edge of the rule is di\'ided decimally ; that
s, each foot is divided into 10 equal parts, and each of thos«
again into 10 equal parts. By means of this last scale,
dimensions are taken in ft;et, tenths, and hundredths ; and
then multiplied as common decimal numbers.
In one of these equal pieces, there is a slider on which are
marked the two letters B, C ; on the same face are marked
the letters A, D. The same numbers serve for both these
two middle lines, the one being above the numbers, and the
other below.
Three of these lines, viz., A, B, C, are called double Ihies,
as they proceed from 1 to 10 twice over. These three lines
are exactly alike both in division and numbers, and are
numbered from the left hand towards the right, 1, 2, 3, 4, 5,
(>, 7, 8, 9 to 1 , which stands in the middle ; the numbers then
go on, 2, 3, 4, 5, 6, 7, 8, 9 to 10, which stands at the right-
hand end of the rule.
These four lines are logarithmic ones ; the lower line D,
is n single one, proceeding from 4 to 40, and is called the
girt line, from its use in finding the content of timber.
Upon it are also marked W G at 17-15, AG at 18-95,
SLIDZ AND RULE. 129
uul I G at 15"8. These are the wine, ale, and imp<»riaJ
ju.'i^e points.
^u this face is a table of the vaUie of a lond. - ' ■ o
eet, of timber, at all prices from G jM?nce to 'J ■ r
bot.
To ascertain the values of tlie fij^tircs on the ri:I«.', wiiioh
lave no determinate val-ie of their owni, but dejHiul »{H>ti
he value set on the unit at the left hand of that part of th«
ule marked 1, 2, 3, &c. ; if the first unit be railed 1. tl.o
I in the middle wiW be 10, the other figures tlmt follow
ivill be 20, 30, 40, &c., and the 10 at the right-hand end
,vill be 100. If the left-liand unit be calletl 10, the I in
:he middle yviW be 100, and the following fibres will Iw
200, 300, 400, 500, &c. ; and the 10 at the nght-hfind cud
,vill be 1000. If the 1 at the left-hand end be call (hI 1<'»
he middle 1 v-ill be 1000, and the following figures will U-
JOOO, 3000, 4tK)0, &c., and the 10 at the right-hand will
)e 10,000. From this it appears that the values of all Ow
igures depend upon the value set on the first unit.
The use of th.^ double line A, H, is to fintl a fourth pro-
jortional, and also to find the areas of plane figures.
The use of the several lines described here is l)Cflt Icnnud
n practice.
If the rule be mifolded, and the slider moved out of the
;rove, the back part of it will be «een divided like the edge
)f the rule, all measuring 3 fcot in length.
Some rules have oUier scales and tables delineated unon
hem; such as a table of board measure, one of tiioUr
neasure, another for showing what length for any In." •'-
vill make a square foot. There L* al<w a line showK-g
twgth for anjr tliickness will make a siJid foot.
130
THE USE OF THE SLIDING RULE.
PROBLEM :.
7u> multinlij numbers together.
\
Set 1 on B to the multiplier on A ; then against the mul-
tiplicand on B, stands the product on A.
1, Multiply 12 and 18 together.
Set 1 on B, to ] 2 on A ; then against 1 8 on B, stands
the product 216 on A.
2. Multiply 36 by 22.
Set 1 on B, to 36 on A ; then as 22 on B goes beyond
the rule, look for 2*2 on B, and against it on A stands 79*2 ;
but as the real multiplier was divided by 10, the producJ
79"2 must be multiplied by 10, wliich is effected by taking
away the decimal point, leavhig the product 792.
PROBLEM 11.
To divide one number by another.
Set the divisor on A, to 1 on B ; then against the dividend
on A, stands the quotient on B.
1. Divide 11 into 330.
Set the divisor 11 on A, to 1 on B ; then against tht
dividend 330 on A, stands the quotient 30 on B.
2. Divide 7680 by 24.
Set 24 on A, to 1 on B ; then because 7680 goes beyonc
the rule on A, look for 768 (the tenth of 7680), on A, anc
against it stands 32 on B ; but as the tenth of the divident
was taken that the number should fall within the compass o
the scale A, the quotient 32 must be multiplied by 10, whicl
gives 320 for the answer.
SLIDING UU I,E. ]31
PROBLEM III.
To square any number.
Set 1 upon C, to 10 upon D ; then if you call the 10 upon
D 1, the 1 on C will be 10; il' you call the 10 on D, 10,
then the 1 on C will be 100 ; if you call the 10 on D, 100,
then the 1 on C will be 1000 ; this being understood, you wil
observe that against every number on D, stands its square
on C.
1. WTiat are the squares of 25, 30, 12, and 20 ?
Proceeding according to the above directions, 625 stands
against 25, 900 against 30, 144 against 12, 400 against 20.
PROBLEM IV.
To extract the square root of a number.
Set 1 or 100, &c., on C, to 1 or 10, &c., on D ; then against
tfvery number found on C, stands its root on D.
1. WTiatare the square roots of 529, and 1600?
Proceeding according to the above directions, opposite 529
stands 23 ; opposite 1600 stands 40, and so on.
PROBLEM V.
To Jind a mean proportional between two numbers as 9
and 25.
Set the number 9 on C, to the same 9 on D ; then against
25 on C, stands 1 5 on D, the required mean proportional.
The reason of this may be seen from the proportion, viz.,
9 : 15 :: 15 : 25.
1. What is the mean proportional between 29 and 430 ?
Set one number 29 on C, to the same on D ; then agaiu5t
the other number 430 on C, stands 112 on D, which i^ the
aiean proportional, nearly.
132 SLIDING RULE.
PROBLEM VI.
To find a third proportional to two numbers, as 21 and 32
Set the first number 21, on B, to the second, 32, on A ;
then against the second, 32, on B, stands 48-8 on A, wliich is
the required third proportional.
FROBLEM VII.
To find a fourth proportional to three given numbers.
Set the first term on B, to the second on A ; then against
the third term on B, stands the fourth on A.
If either of the middle numbers fall beyond tne line, take
one-tenth part of that number, and increase the fourth num-
ber found, ten times.
I. Find a fourth proportional to 12, 28, and 114.
Set the fust term, 12, on B, to the second term, 28, on
A; then against the third term 114 on B, stands 266 on A.,
Tkhich is the answer.
TIMBER MEASURE.
PROBLEM L
Tu find the sujterficial content of a hoard or plank. .
Rule. Multiply the length by the breadth, and the product
will be the area.
Xote. When the pliink is broader at one end than at the other, add bolt
end} together, and take hall the sum for a mean breadth.
TliinER MEA8UKK. 133
BY THE CAllPENTEU'S RULE.
Set 12 on B, to tlie breadth in inches on A ; then against
the length in feet, on B, will be found the superficieson A,
in feet.
1, If a board be 12 feet 6 inches long, uud 2 feet 3 inches
bri)ail, how many feet are contained in it?
12.6 12-5
2 . 3 2-2.J
25 .0
3. 1 .G
cmipexter's
625
250
250
28 . 1 G Ans
UY THE
28 125 ^'/*.
RULE.
As 12 on B : 27 on A : : 12-5 on B : 28-125 on A.
2. Wliat is the value of a board whose lengtli is 8 feet 6
inches, and breadth 1 foot 3 inches, at 5rf. per foot ?
Ajis. 4s. 5d.
3. What is the value of a board whose leny;tli is 12 feei
9 inches, and breadth 1 foot 3 inches, at 5d. per foot ?
Ans. Cs. 7^d.
4. What is the value of a plauk whose breadth at one end
is 2 feet, and at the other end 4 feet, at Gd. per foot, the
length being 12 feet ? Ans. ld».
5. How many square feet in a boarJ, whose breadth at
one end is 15 inches, and at the other 17 inches, the li-ngtb
lieing G feet ? Ans. b.
6. How many square feet in a plank, whose length ia 2(t
feet, and mean breadth 3 feet 3 inches? Ans, CO.
PROBLEM II.
To find the solid content of squared or four-sided timber.
Rule. Take half the sum of the breadth «ujd depth ii
134 TIMBER MEASURE.
the middle, (that is, the quarter girt,) square this half sum,"
and multiply it by the length for the solid content.*
BY TUE carpenter's RULE.
As 12 on D : length on C : : quarter girt on D : the solid
content on C.
1. If a piece of squared timber be 3 feet 9 inches broad,
2 feet 7 inches deep, and 20 feet long ; how many solid feet
are contained therein ?
3.9
2 .7
2)6,4
3 . 2 quarter girt.
3. 2
9.6
6.4
10.0.4 square of the quarter girt.
20 length of the piece.
200 .6.8 solid content.
BY THE carpenters' RULE.
As 12 on D : 20 on C : : 38 on D : 200^ on C.
2. A squared piece of timber is fifteen inches broad, 15
inches deep, and 1 8 feet long ; how many solid feet does it
contain ?
Ans. 28|- feet, which is the accurate content, as the
breadth and depth are equal.
3. What is the solid content of a piece of timber, whose
breadth is 16 inches, depth 12 inches, and length 12 feet?
Ans. 16 feet.
* This rule, which is genenilly employed in practice, is far from being correct
(^ hen the breadth and depth differ matcrlallj from each other, and the timbe:
Joe* no! taper.
TIMDKK MEASUUE. 136
Rule II. Multiply the breadth in tne middle oy tlit
depth in the middle, and that product by the length, for tin
solidity.*
4. The length of a piece of timber is IS feet G inches ;
the breadths at the greater and less end 1 foot G inches, and
1 foot 3 inches, and the thickness at the greater and less end
1 foot 3 inches, and 1 foot ; what is the solid content ?
1-5 1-25
1-25 1
2)2-75 2)2-1
or*
r375 mean breadth. 1-125 mean depth.
1-125 mean depth.
1-375 mean breadth.
1-546875
18-5 length.
28-6171875 solid content.
BY THE SLIDING KLLE.
B A B A
As 1 : 13Jf : : IGJ^ : 223 the mean square.
C D' C" D
As 1 : 1 : : 223 : 14-9 quarter girt.
CD DC
As 18^ : 12 : : 14-9 : 2S-6 the content.
Note. \N'Tien the niece to be me.i«ured tapers regnilarly frorr one end
to tbe otber, eitber take tbe mean breadth and depth in the middle, or take
the dimensions at both ends, and half their sum for the mean dimention.
rbis, however, though very easy in practice, is but k very imperfect approxi-
mation.
^^^ieu the piece to be measured does not taper regularlr, but is thick Id
some parts and small in others, in this case take several dimensions ; add them
• Tliis rule is correct vs-hen the timber does not taper ; bnt when tbe
timber tapers considerably, and tlie breadth and denth are nearly equal, the
role is very erroneous. The measurer, therefore, ougfit to considiT tlie ihap«
of the timber he is about to measure before he applies either 0/ «be abovr
rules* K
]o6 TIMBER MEASrRE.
all together, and <ii v-ae n-oir sum bj tne number oi (Tiinensions bo case/,
and use the quotient as the mean dimeusion.
RcLE III. Multiply the sum of the breadths of the two
ends by the sum of the depths, to which add the product of
the breadth and depth of each end ; one-sixth of this siim.
multiplied by the length, will g-ive the exact sotidity of any
piece of squared timber tapering regularly.*
5. How maz-y feet in a tree, whose ends are rectangles,
the length and breadth of one being 14 and 12 inches, and
the corresponding dimensions of the other 6 and 4 inches;
oJso the length 30^ feet ?
U 12 12 X 14 = 163
6 4 6 X 4 = 24
_ _ 20 X 16 = 320
20 16
512 square inches
= ^ " squa^'e feet.
32 2
Then ^ X -— X ;50^ =18-. feet, the solidity.
G. How manv solid inches in a mahofjan- plank, the lenjrth
and breadth of one end being 91^ and 55 inches, the length
and breadth of the other end 41 and 29^^ inches, and the
length of the plank 47^ inches ?
Ans. 1 26340-59375 cubic inches.
PROBLEM III.
Given the breadth of a rectangular plank in inches, to Jind
how much in length will make afoot, or any other required
(juantitii.
Kli-e. Di\-ide 144, or the area to be cut off, by the breadth
in inches, and the quotient will be the length in inches.
• Thii rule is correct, being that given for finding the soliJity of tho pris-
moid — which sea.
L«'t n und b he Uie br.-adlhs of the two ends, D and d the denfh*,
uDd L the length : j (B D 4- (B -f i) X (D -f- (f) -f J «/) X L =
th»> true sohdiiy, aa in the rule for the prismoij.
TIMBER MEASURE.
137
The Carpenter's rule is furnislicd with a scale which an-
swers the purpose of this rule. It is called a table oi* buaxd
measure, and is in the following form :
1
5
-sT'
6
Inches.
Feet.
"12
6
4
3
1 '^
'2
1
1
1
2
3
1 4
5
" 6
7
8
Breadth.
If the breadth be 1 inch, the length standing against it is
12 feet ; if the breadth be 2 mclies, the length standing
against it is fi feet ; if the breadtli be 5 inches, the length is
2 feet 5 inches, &c.
When the breadth goes beyond the limits of the table on
the rule, it must be shut, and then you are to look for the
breadth in the line of board measiure, which runs along the
rule from the table of board measure, and over against it
nn the opposite side, in tho scale of inches, will be fimnd
the length reauired. For example, if the breadth be 9
inches, you will find the length against it to be 1 6 inches ;
if the breadth be 1 1 inches, the length will be found to be a
little above 13 inches.
1. If a board be 6 inches broad, what length of it will make
a; square foot ? Ans. 2 feet.
2. If a board be 8 inches broad, what length of it wil)
make 4 square feet ? Am. 6 feet.
3. If a board be 16 inches broad, what length of it will
make 7 square feet? Ans. 5^ feet.
When the board b broader at one end than at the other,
proceed according to the following
Rule. To the square of the product of the length, and
narrow end, add twice the continual product of these quan-
tities, viz. the length, the difference Wtween the breadtlis of
the ends, and the area of the part required to be cut otY,
extract the square root of the sura ; from the result deduct
tlie product of the length and narrow end, and divide the
remainder by the difference between the breadths of the
onds.*
• S«e Appendix, DamoottrmticD lli^
K 2
138
TIMBER iLEASUUE.
If it were required to cut off 60 square inches from the
smaller end of a board, A D being' 3 inches, C E 6 inclies.
and A B 20 Inches.
nC
Here A x =
1
2 j^^ (v^{(AB X AD)='+4BC xAB
X 60} - A B X A D) = ^(v^ {(20x3)- + 6x20x60]
— 20 X 3) n:: 14-64, the length required.
PROBLEM IV.
To Jind how much in length will make a solid foot, or any
oilier required quantity, of squared timber, of equal dimen-
sions from end to end.
Rule. Divide 1728, the solid inches in a foot or the
solidity to be cut off, by the area of the end in inches, and
the quotient will be the end in inches.
1. If a piece of timber be 10 inches square, how much in
length will make a solid foot ?
10 X 10 = 100 the area of the end; then 172rf h- 100
= 17-28 Ans.
2. If a piece of timber be 20 inches broad, and 10 inches
detY> ho^' niuch of it will make a solid foot ?
Ans. 8^5 inches.
2. If a piece of timber be 9 inchos broad, and 6 inches
deep, how much of it will make 3 solid feet? Ans, 8 ft.
TlMnER MEA&UKE.
139
Ou some Larpenter's rules, there is a taMo to answer the
purpose of the hist rule; it is calleJ a Table of Timber, aiul
IS ill the followin<ir form :
II
3
9
liichi'«J.
144
3(3
16
9
5
4
iJ|
2
1
Feet.
1
2
3|
4
5
6
7
6
9
Side of s(|uarc.
PROBLEM V.
Tojina, the solidity of' round or un squared timber.
Rile I. Gird the piece of timber to be measured round the
middle with a striuj^, take one-fourth part of the girth, aiid
scjuare it, and multiply this sijuare by the length fur the solidity.
BY THE SLIDING RULE.
As the length on C : 12 or lU on D :: quarter girt, in
I2ths or lOths ou D : content ou C.
Note. When the tree is very irregular, divide it into several leugthi and liml
the solidity of each j>art separately; or add all the girts togeuu'i, and divid*
the sum by the number of tLem.
1. Let the length of a piece of round timber be 9 feet 6
inches, and its mean quarter girt 42 inches; what is ita
content ?
3*5 quarter girt.
3-5
12-25
9-5 length.
116375 content.
3 .
6
6
juaiter girt
10
1 .
6
9
12
9
.3
6 length.
110.
6
3
. 1
. i'>
i 1(3 . 4 . 6 conl*iiit.
:-io
TIMBER MEASUKE.
BT THE SLIDES'G RULE.
As 9-5 on C : 10 onD
Or 9-5 : 12 :
: 35 on D
42
ll6i.
116^ on
Rule II. Multiply the area corresponding to the quarter
girt in inches, by the length of the piece in feet, and the
product will be the solidity.
Xote. It may sometime* happen that tie quarter girt exceeds the limits
■3'. the table : in this ca»e, take t'jtl^ of it, and four timea the content thus found
irill give the required content.
A TABLE FOR MEASURING TIMBER.
(
1 Quarter
; Girt.
Area.
Quarter
Girt.
A/ea.
Quarter
Gin.
1 1
j Area. 1
Inchei.
Fut.
Incl>£i.
Fett.
Inchttt.
Feet.
6
•250
12
1-000
18
2-250
6i
•272
\2\
1-042
181
2-376
6i
•294
124
1-085
19
2-506
61
•317
12}
1129
194
2640
7
•340
13
1-174
20
2-777
7i
•364
131
1-219
204
2-9 J 7
rt
•390
i3t
1-265
21
3-06-2
rj
•417
131
1-313
214
3-209
8
•444
14
1-361
22
3-362
8J
•472
141
1-410
221
3-5 i6
H
•501
14i
1-460
• 23
3-673
M
•531
141
1-511
231
3-835
9
•662
15
1-562
24
4-000
n
•594
151
1-615
241
4-168
yi
•626
loi
^6«8
25
4-340
91
•659
16}
1-722
251
4-516
10
•694
16
1-777
26
4-694
lOJ
•731)
161
1-833
264
4-876 *
lot
•766
161
1-890
27
5-062
101
•&03
It'l
1-948
271
6-Sl2
1 11
•840
17
2-006
28
i5-444
' m
•«78
171
2-066
281
5-640
11*
•918
171
2-12G
29
5-ts40
111
•969
171
2-187
291
6-044
TIMBER MEASCRE. 141
2. If a piece ot round timber be 10 feet long, and the
quarter girt 12^ inches; required the solidity. Ans. 10-85.
To find the solid content bj this table, look for the quarter
i:i;irt 12^, in the column marked, Quarter Girt, and in adjoin-
ing column marked, Ai-ea, will be found 1 "OSS, which mul-
tiplied bv the length, iufeet, will give 10-85 feet for the solid
content.
3. A piece of round timber is 20 feet long, and the quarter
girt 14:^; how many feet are contained therein ?
^7jv. 2S-2 feet.
4. How many solid feet are contained in a tree +0 feet IijiicTi
its quarter girt being 9 inches ? Ans. 22'4y.
5. How many solid feet in a tree 32 feet long, its quartci
girt being 8 inches ? • Arts. 14-208.
G. How many solid feet in a tree 8^ feet long, its quartef
girt being 7i inches? Ans. 3-315 feet.
7. Required the content of a tree, whose length is 4(.'
feet, and quarter girt 27^ inches ? Ans. 210-08 feet.
8. What is the content of a tree, whose length is 30 feef
<j Inches, and quarter girt 27-^ inches ? Aris. 160*186 feet.
9. Required the content of a piece of timber, whose length
is 25 feet i) inches, and quarter girt 12^ inches ?
Aim. 29-071 feet.
10. What is the solid content of a piece of timber, whose
length is 12 feet, and quarter girt 13^ inches ?
Ans. 15-18 feet.
1 1 . What is the solid content of a piece of timber, whose
(luarter girt is 14^ inches, and length 38 feet ?
Ans. 57-418 feet.
Wh«i tlie pquaTB of ihfc quarter girt is multiplied by the
length, the prvMiuct gifts a lesult nearly one foiu-th less than
the true quantity in the tree. This rule, however, is invariably
practised by tiir.Lei merchants, and is not likely to be abo
lished. When the tree is in the form of a cylinder, its con
tent ought to be found by Prob. l\'. Sec. 1\\, whicii give?
the content greater than that found by the last rule, nearly ir
the proportion of 1 I tc 1 1. Notwithstanding that tlie trui
142 TIMBER MEASTTRE.
content is not found by means of the square of tlie quarter girt,
vet some allowance ought to be made to the purchaser on ac-
count of the waste in squaring the wood so as to be fit for use.
If the cylindrical tree be reckoned no more than what the in-
scribed square will amount to, the last rule, wliich is said to give
Loo little, gives too much. When the tree is uot perfectly
circular, the quarter girt is always too great, and therefore the
couteut, on that account, will be too great.
Doctor Hutton recommends the follo\ving rule, wliich
will give the content extremely near the truth :
Rdle. Multiply the square of one-fifth of the girt, or cir-
cumference, by twice tlie length, and the product wUl be the
content.
RY tue sliding rule.
As double the length on C : 1 2 or 1 on D : : ^ of the
irirt, in 12ths or lOths on D : content on C.
12. Required the content of a tree, its length being 9 feel
6 inches, and its mean girt 1 4 feet.
ft. in. p.
14-^5 = 2-8 = 2. 9. 7 = ^ of the girt ; then
ft. in.
2-S 9.6 2.9.7
2-8 2 2.9.7
o
7-84 19 . 5.7
19 2.1.2.3
■ 1.7.7.1
148-96 content.
9 . 11 . 10 . 1
19
148 . 9 . 8.11.7 content.
C D D C
As 19 : 10 :: 28 : 149 content by the Sliding Rule.
Or 19 : 12 :: 33-6 : 149 content without it.
Dr. GuEGORT recommends the following rules given by
Mr. Andrews :
TIMBER AiEASl'llE. 143
Let L denote the Icn^li of the tree in feet ano declnials,
unci G the mean girt in inches.
Rule I. Making no allowance for bark.
— = cubic feet, customary ; and — —
2304 ^ 1807
content.
= cubic feet, customary ; and — — =: cubic feet, tnic
^ 1807
Rule IT. Allowing ^ for bark.
L G- L G* .
— = cubic feet, customary : -rz-rr^ = cubic feet, true con-
30uy • 2360
tent.
Rule III. Allowing -^g for baik.
=z cubic feet, customary ; ;r-7., = cubic feet, true
2845 • 2231 '
"ontent.
Rule IV. Allowing ^ for bark'.
L C" L G"
" = cubic feet, customary; J- — = cubic feet, true
2742 ' 2150
content.
VNHiat is the solid content of a tree, whose circumference-,
or girt, is GO inches, and length 40 feet ?
Bif Rule I,
= 62i cubic feet, customary.
2304 ^
40 x CQ- _ yfj3 j^j J. customary.
1807 * ^
Bj, liulc 11.
__^ = 47-85 cubic feet, customary.
3009
^^Z^^^' r= Gl cubic feet, true content.
23GO
144 • artificers' work.
By Rule III.
40_X^ _ 5Q.gj ^^^.g f-gg^ customary.
2S45 ^
— 1^ = 64 "04 cubic feet, true content.
2231
40 X 6 0"
2742
40 X 60-
2150
By Rule IV.
= 52*47 cubic feet, customary.
= 66*97 cubic feet, true content.
WTien the two ends are very unequal, calculate its content
by the rule given for finding' the solidity of the frustum of a
cone, and deduct the usual allowance from the result.
\Mien it is required to find the accurate content of an irre-
gular body, not reducible to any figure of which we have alreadv
treated, provide a cylindrical or prismatic vessel, capable of
contaiiwng the solid to be measured ; put the solid into the
vessel, and pour in water to cover it, marking the height to
which the water reaches. Then take out the sohd, and observe
how much the water has descended in consequence of its re-
moval ; calculate the capacity of the part of the vessel thus left
dry, and it will evidently be equal to the solidity of the body
whose content is required.
ARTIFICERS' WORK.
Artificers compute their works by several difTerent mea-
sui'es :
Glazing and masonry by the foot.
Plastering, painting, paving, &c., by the yard of 9 square
feet.
Partitioning, roofing, tiling, flooring, &.c., by the square
of 100 square feet.
Brick work is computed, either by the yard of 9 square
iVtt, or by the perch, or square rood, containing 272^ square
feet, or 30| square yards ; 272;^ and 30;^ being the squares
ot 161 feet and 5^ yards respectively.
145
CARPENTERS AND JOINERS' WORK.
1. OF FLOORING.
To measure joists, multiplj the breadth, dvpth, and length
together for the content.*
If a floor be 50 feet 4 inches long, and 22 feet 6 inche*
broad ; how many squares of flooring are in that room ?
50-333 50 . 4
22-5 22 . G
251665 1107.4
100666 25 . 2
100666
luu) 11,32 . 6
100)1132-4925
11-325
11-3249 squares.
A)is. 11 squares 32^ feet.
2. If a floor be 51 feet 6 inches long, and 40 feet 9 inched
broad ; how many squares are contained in that floor ?
Ajis. 20986 squares.
3. If a floor be 36 feet 3 inches long, and 16 feet G inche»
broad, how many squares are contained in that floor ?
jins. 5 squares 98^ feet.
4. If a floor be 86 feet 11 inches long, and 21 feet 'J
inches broad ; how many squares jue contained in it ?
An.i. 18-3972.
5. In a uaJvcd floor the girder is 1 foot 2 inches deep, 1
foot l»road, and 22 feet long; there are 9 bridgings, the
scanthug of each (riz. breadth and depth,) being 3 inches, by
6 inches, and length 22 feet; 9 biudiijg joists, the leiij^ of
* Joists receive various names from their position ; inch as fjirJen, bimlinf-
/dists, trimming-joists, ci'nituon-joii.t.«, ceiiiiig-joistj, Ac. V' - — -'rr» and
joists of flooring are desiimed to bear considerable weight, ! i b« l«t
Into the vrali at each end about two-'h-rds of the thickoesaof ut« >.....
146 CARPENTERS aXU JOINEIiS' WORK.
each being 10 feet, and scantlings 8 mches by 4 inches ; the
ceiling joists are 25 in number, each 7 feet long, and their
scantlings 4 inches by 3 inches ; what is the solidity of the
whole ? Ans. 85 feet.
6. \\Tiat would the flooring of a house three stories high
come to, at £5 per square ; the hoiise measures 30 feet
long, and 20 broad ; there are seven fire-places,* two oi
which measure, each 6 feet by 4 feet, two others, each 6
feet by 5 feet 6 inches ; two, each of 5 feet 6 inches b\
4 feet ; and the seventh 5 feet by 4 ; the well-hole for the
nau-s is 10 feet by 8 ? Ans. £69 25.
OF PARTITIOXING.
Partitions are measured by squares of 100 feet, as floor-
ing ; their dimensions are taken by measuring from wall to
wall, and from floor to floor ; then multiply the length and
height for the content in feet, which bring to squares by
dinding 100, as in flooring. When doors and windows are
not included by agreement, deductions must be made for
their amount.f
1. A partition measiures 173 feet 10 inches in length, and
10 feet 7 inches in height; required the number of squares
m it ? Ans. 18'3972 squares.
2. A partition between two rooms measures 80 feet in
length, and 50 feet 6 inches in height ; how many squares
in it ? A7ts. 40|^ squares.
3. If a partition measure 10 feet 6 inches in length, and
1 feet 9 inches in height ; how many squares in it ?
Ans. 1 square 12^ feet.
4. \Miat is the number of squares in a partition, whose
length is 50 feet 6 inches, and heiglit 12 feet 9 inches ?
Ans. 6 squares, 43 feet, 10| inches.
• Fire-nlace«, &c. are of course to be deducted.
+ Tiie best and strongest paHitions arc those made with feamod work.
The king-posts are nuaBureJ as roofing, the reit as flooring.
CABrENTERS A.NU J01>KliS* WOUR. 147
In roefiiig, the lenglh of the rafters is equal to the length
of a string stretched from the ridge down the rafter till it
meets the top of the wall.
To find the content, multiply this length hv the breadth
and depth of the rafters, and the result will 6e the content
of one rafter ; and that multiplied by the number of them will
give the content of all the ral'ters,*
1. If a house within the walls be 42 feet G inches long,
and 20 feet 3 inches broad ; how many squares of roofiny
in that house ?
ft. ft. in
42-5 4J . 6
20-25 20.3
2125
850
8500
860-625 flat
430-3125
100)1290-9375
840
6^
10 .
1
3j
10 .
t
860 .
8
flat
430 .
4
100)
1291
12-91 squares. 12 : 91
2. What cost the roofing of a house at 1 l.v. per square ;
ihe length within the walls being 50 feet 9 inches, and the
breadth 30 feet ; the roof being of a true pi;cli ?
Atis. i:i2 11j. 2^irf.
• Workmen generally take the flat and half the flat of any houM, taken
within the walls, to be' the measiire of the roof of the tame Iioiim. Thit,
however, is only when the roof is vi a true pitch, lie u»ual pitches are the
common, or true pitches, in which the rafters are three-fourtln of the hreaiith
of the huildinij ; the CJothic pit.h is when the lenjflh of the princ.pal raften
>s equal to the breadth of the building : tlio pediment pitch u when the per-
pendicular height it two-ninths of tlie breadth.
When the covering of the building is to be riain tiles or slates, the roof
IS tenerallv of a tnie or common pitch ; fbe Gothic pitch is used when t)ie
eovenng is' of pantiles; the pediment p:t«h is OMd -vhea the rt>of is to be
aovurfd with leaiL
148
CARPENTERS AST) JODTERS WORK.
3. What number of squares are contained in a house,
whose length within the walls is 40 feet, and breadth 18
feet ; the roof being common pitch ?
Ans. 10 squares and 80 feet.
4. How many squares in the roof of a. building, the length
of the house being 60 feet, and the length of the rafter 14
feet 6 inches? Am. 1 7 squares and 40 f&et.
5. How many squares in a building, whose length is 50
feet, and the length of the rafter 15 feet ?
Ans. 15 squares.
6. How many squares in the roof of a building, whosf
length is 37 feet, the length of the rafter being 13 feet ?
Ans. 9 squares and 62 feet.
7. How many squares In the roof of a building, whosn
.ength is 70 feet 6 inches, the length of the rafter being 14
feet 6 inches ? Ans. 20 squares and 44^ feet.
8. How many squares in the roof of a building, whose
length is 50 feet, and the length of a string reaching across
the ridge from eave to eave being 30 feet ?
Ans. 15 squares.
Xote. All the timbers employed in roofing are measured like those nsed iii
flooring, except ■where there is a necessity of cutting out parallel pieces equal
to, or-exceeding 2M iuches broad and 2 feet long. In this case the amount ol
the jiieces so cut out must be deducted from the content of the whole piece «
found from its greatest scantlinge. When the pieces cut out do sot amount to
the above dimeijsions, they are considersd as useless, and thereiore no deduction
•f to be made for them.*
• In the above figure, K is called tlie kint-post, and in meastuiiu; the.
pieces cut out of it, the shortest len^h i.» to'bf taken. T B is caiied th*
caRfkntrrs and join bus* wonK. 140
10. Let the tie-beam T B be 36 feet long, 9 inchf*
broad, and 1 fool 2 inches thick ; the king^-post K 1 1 f i . t
f^ Inches hig-h, 1 foot br<xid nt the U.tti.m, nnd 5 inches
thick ; ont of this post are sawn two eqnul pieces from the
side8, each 7 feet long and 3 inches brojul. The braces
li B, are 7 feet 6 inches long, and 5 inches bv 5 inches
square ; the rafters R R are 19 feet long, 5 inches broad,
and 10 inches deep ; the struts S S are 3 feet <> inches lonir,
4 inches broad, and 5 inches deep ; vrhat is the measurement
for workmanship an d als*^ ^^o- materials ?
St. in. p.
31 . 6.0 solidity of the tie-beam T R.
4.9.6 solidity of the king-post K.
2.7.3 solidity of the braces B B.
13 . 2.4 solidity of the rut'ters R R.
11.8 solidity of the struts S S.
53 . 0.9 solidity for workmansliip.
1.5.6 solidity cut from the king-post.
61 7.3 solidity for materials.
OF WAIXSCOTTINO.
Wainscotting is measured by the yard square, which Is 9
square feet.
In taking the dimensions, the string is nia<le to ply c! -
:ver the cornice, swelling panels, moulding, &c. i
neight of the room from the tioor to the ceiling being thus
taken, is one dimension, and the compass of the room tnkt n
all round the floor is the second dimeusiou.
tie-b«im, ivhich prevents the rafters ft R from pressine cnt the wall. The
braces B B serve to »tr«ntrthen the rofters ; the Bttuu S 3 mrv- lor a, fimilw
purpose. B«diJes rtrengtlieniii? the rafter?, the hraoes »nd Mnilj irrwe u>
bintl the roof together. W» «q bead-room Is required, the nften u« hr»«e^
»iini>!y bv R R. '
150 CARPEXTERS A_VD JOINERS* WORK..
Doors, win dows, shutters, &c., where both their sides are
planed, are considered as work and half ; therefore in mea-
suring the roo m, they need not be deducted ; but the super-
ficial content of the whole room found as if there were no
door, window, &c., then the contents of the doors and windows
must be found, and half thereof added to the content of the
whole room.
When there are no shutters, the content of the windows
must be deducted ; chimneys, window-seats,' check-boards,
sopheta-boar ds, linings, &c., must be measured by them-
<«elves.
Windows are sometimes valued at so much per window,
and sometimes by the superficial foot. The dimensions of a
window are taken in feet and inches, from the under side of
the sill to the upper side of the top-rail; and from the outside
to outside of the jambs.
When the doors are paneLed on both sides, take double
the measure for the workmanship.
For the surrounding architrave, girt round it and inside
ihe jambs, for one dimension, and add the length of the
jambs to the length of the cap-piece, (taking the breadth ol'
the opening for the length,) for the other dimension.
Weather-boarding is measured by the yard square, and
sometimes by the square.
Frame-doors are meisured by the foot, or sometimes by
the yard square.
Staircases are measured by the foot superficial. The dimen-
sions are taken with a string passing over the riser and
tread for one dimension, and the length of the step for the
other. By the length of the step is meant the ler.gth of the
front and the retnma at the two aitds.
For the baliwtrade, take the whole length of the upper
part of the hand-rail, and girt it over its end till it meet the
top of the newel-post, for one dimension ; and twice the
length of the baluster upon the landing, v\-ith the girt of thfl
liand-rail, for the other dimension.
Modilliau cornices, coves, &c., are generally measured by
the foot superficial.
CARPENTERS AJSD JOISEJIS' WOUH. 151
Beads, stops, astragals, copings, fillets, boxings to window*,
skirting-boards, and water-trunks, art* paid ior by lim.il
measure.
Frontispieces are measured by the foot superficial, and the
architrave, frieze, and cornice, are measured separately."
To find the contents of the foregoing work, multiply the
two corresponding dimensions together for the superficial
content.
1. A room, or wainscot, being girt downwards over the
moiddings, measures 12 ft. G in. and l.)0 ft. in. in compass
how many yards does that room contain ?
ft. in. ft.
1.30 . 9 130-75
12 . 6 12-5
1560 65375
65 . 4 . 6 26150
6.0.0 13075
3.0.0
9)1634 .4.6
181-5 Ans.
9)1634-375 ft.
181 yards, 5 ft.
* Baluster is a small column or pillar, u5«<l for balustrades.
Balustrade is a row of balusters, joLaed by a rail ; lerving for a re»t to lh«
irms, or a» an inclosure to balconiee, stairaises, altars, A:c.
Cornice is the third and upp'-nnost part o{ the eutablatur* of a column, or
the uppermost ornament of any wainscutting, &c.
lifad is a ro'ind moulding carved like beads in necklaces. Tb»re is aUo
a kind of plain bead, often ."^et on the edire of each fascia of an arcbitraTc,
oil the uiiper ed^e of skirting -hoards, on the linine-board of a door-case, &c
Architrave is that part of a column that l>ears unrae'liately on the capital.
It is supposed to represent the priiioiml beam in timber buildio^'s, «■
Rhich it 18 sometimes called the m:c(ter-piece or reason-piece. In chimneys
it is called the mantel-pieco. Architrave doors are those which have an
irchitrave on the jumb? and over the doors. Architrave windows of timbej
»re usually raised out of the solid timber, and sometimes the mouldings urr
itruck and laid on.
Astragal is a small round moulding, encompassing the top of the shaft ol
> column, like a ring or bracelet. The sluft tcrminaIeK at the lop wl h
■in astragal, and at bottom with » fillet, which is this place is cail«J •».».
L
152 BRIOKLAYEKS' WORK.
2. If the wainscot of a room be 15 ft. 6 in. high, and the
compass of the room 142 ft. 6 in. ; how many yards are con-
tained in it ? Ans. 245-i^, yards.
3. If the window shutters about a room be 60 ft. 6 in.
broad, and 6 ft. 4 in. high ; how many yards are contained
therein, at work and a half ? Ans. 63§^ yards.
4. A rectangrdar room measvires 129 feet 6 inches round,
and is to be wainscotted at 3s. 6d. per square yard ; after
due allowance for girt of corruce, &c., it is 16 feet 3 inches
liigh ; the door is 7 feet by 3 feet 9 inches ; the window
shutters, two pair, are 7 feet 3 inches by 4 feet 6 inches ;
the cheek-boards round them come 15 inches below the
shutters, and are 1 4 inches in breadth ; the lining-boards
round the doorway are 16 inches broad : the door and win-
dow-shutters being worked on both sides, are reckoned as
work and half, and paid for accordingly; the chimney 3 feet
9 inches by 3 feet, not being enclosed, is to be deducted from
the superficial content of the room. The estimate of the
charge is required. Ans. £43 4*. 6|rf.
5. The height of a room, taking in the cornice and mould-
ings, is 12 feet 6 inches, and the whole compass 83 feet b'
inches ; the three window-shutters are each 7 feet 8 inches
by 3 feet 6 inches, and the door 7 feet by 3 feet 6 inches; the
door and shutter, being worked on both sides, are reckoned
work and a half. Required the estimate, at 6s. per square
yard. Ans. £36 12*. 2^^.
OF BRICKLAYERS' WORK.
OP TILING OR SLATING.
Tiling and slating are measured by the square of 100
feet. There is no material difference between the method
employed for finding the estimate of roofing and tiling :
bricklaysrs iometlmes require double measure for hips and
vallevs.
BRICKLATEIiS' WORa. 15^
NVTien gutters are ulIowcJ double measure, the usuiJ niutle
is, to measure the lecj^h along the riJge tile, und atid it to
the conteuti of the roof : this makes aa allowance of one foot
ill breadth along the hips or valleys. Double measure ia
usually allowed for the eaves, so much as the projector id
over the plate, wliich is generally Is or 2U inches.
When sky-lig^hts and chimney-shafts are not large, uo
allowance is to be made for them ; but wheu they uro larg«;,
their amount is to be deducted.
1, There is a roof covered with tiles, whose depth on
both sides (with the usual allowance at the eaves) is 30 feet 6
inches, and the length 42 feet ; how many squares of tiline
are contained therein ?
ft. in. ft.
30 . 6 30-5
42 42
1260 610
21 1220
100)1281 100)12,81U
12 . 81 12 squares 31 feet.
2. There is a roof covered with tiles, whose depth on
ooth sides (with the usual allowante at the caves, is 40 feet
9 inches, and the length 47 feet G inches ; required the num-
ber of squares contamed therrin ?
Ans. 19 squares 35^ feet.
3. What will the slating of a h.)use cost at £\ 5». 6r/.
per square; the length being 43 fevt 10 inches, and the
breadth 27 feet 5 inches, on the Hat ; the eaves proji-cti:./
16 inches on each side — true pitch V Ans. f'H 9*. 6^'/.
4. What is the content of a slated roi^f. the length bcink
45 feet 9 inches, and the wli,,le girt 34 fet< 3 incliob '
/!»»•. 1 74* 1 04 vanl*.
154 EEICKLAYERS' WORK.
OF WALLING.
Brick-work is estimated at tlie rate of a brick anJ a half
thick ; so that if a wall be more or less than the standard
thickness, it must be reduced to it : thus, multiply the super-
ficial content of the wall by the number of half bricks in tbc
thickness, and divide the product by 3.
The superficial content is found by multiplying the lengthy
by the height. Bricklayers estimate their work by the rod
of 16^ feet, or 272^ square feet. Sometimes 18 feet are
allowed to the rod, that is, 324 square feet ; sometimes the
work is measured by the rod of 21 feet long, and three feet
high, that is, 63 square feet ; in this case, no regard is paid
to the thickness of the wall in measuring, but the price is
rcirulated according to the thickness.
When a piece of brickwork is to be measured, the first
thing to be done is to ascertain which of the above measures
is to be employed: then, having multiplied the length and
breadth together (the dimensions being feet) the product is '
to be divided by the proper divisor, namely, 272 25, 324,
or 63, according to the measure of the rod, and the quotient
will bo the measure in square rods of that measure.
To measure any arched way, arched window, or door, &c.,
the heiirht of the window or door from the crown or middle
of the arch, to the bottom or sill, is to be taken, and like-
wise from the bottom or sill to the spring of the arch, that
is, where the arch begins to turn. Then to the latter height
add twice the former, and multiply the sum by the breadth
of the window, door, <tc., and one-third of the product will
be the area sufficiently near the truth fur jjractice.
1. If a wall be 72 feet 6 inches long, and 19 feet 3 inches
high, and 5 bricks and a half tliick, how many rods of
brick-^nrk are contained therein, when reduced to the
standard?
Xote. The ftan(t/ird means a wall a bnVk and a half thick; therefore, to
reduce any wall to the standard, multiply the superficial content of it by the
Buuiber of half bricks in iti thicknees, and divide by 3.
BRICKLAITEUS WOKK.. ,55
ft. in.
72.6
19. 3
648
72
18. 1 .
6
9. '5.
1395 .7.6
11
3)15351 . 10 . 6
272)51 17(18 rods.
2397
68)221(3 quarters.
17 feet.
A'ote. That 6806 is the fourth part of 27226, *nd 6« u on«»-ft>urtb
»f T,2.
In reducing- feet into rods, it is usual to divide '272, re-
jecting the decimal -25. By this method, the ausacr found
above is about 4^ feet too much.
2. How many rods of str'^dard brick- work are in a wall
whose length is 67 feet 3 intnes, and height 24 feet G inches ;
the wall being 2^ bricks tliick ?
Ans. 8*58fi6 rods.
3. The end wall of a house is 23 feet 10 inches long, anil
55 feet 8 inches high to the eaves ; 20 feet high is 2^ bricks
thick, another 20 feet high is 2 bricks thick, and the re-
maining 15 feet 8 inches is 1^ bricks thick, above which i? a
triangular gable one brick thick, which rises 42 courses oi
bricks, of which every 4 courses make a foot. What is the
whole content in standard measure ?
Ans. 253'G2 vutUa.
156 BRICKT.ATERS' WORK.
OP CUIMNEYS.
Ulicn a chimney stands by itself, without any party-wall
being' adjoined, take the girt in the middle for the length,
and the height of the story for the breadth ; the thickness
is to be the same as the depth of the jambs ; if the chimney
he built upright from the mantel-piece to the ceiling, no
deduction is to be made for the vacancy between the floor
(or hearth) and mantel-tree, on account of the gatherings of
the breast and wings, to make room for the hearth in the
next story.
When the chimney-back forms a party-wall, and is mea-
sured by itself, then the depth of the two jambs is to be
measured, and the length of the breast for a length, and the
height of the story for the breadth ; the thickness is the
same as the depth of the jambs. That part of the chimney
which appears above the roof, called the chimney-shaft, is
measiu-ed by girding it round the middle for the length, and
the heia:ht is taken for the breadth.
In consideration of plastering and scaffolding, the thick-
ness is generally reckoned half a brick more than it reallj
is ; and in some places double measure is allowed on account
f)f extra trouble.
1. Let the dimensions of a chimney, having a double
funnel towards the top, and a double shaft, be as follows,
viz., in the parlour the breast and two jambs measure 18 feet
9 inches, and the height of the room is 12 feet 6 inches; in
the first floor, the breast and two jambs girt 1 4 feet 6 inches,
and the height 9 feet ; in the second floor, the breast and
the jambs girt 10 feet 3 inches, and the height is 7 feet ;
above the roof, the comp<iss of the shaft is 13 feet 9 inches,
nnd its height 6 feet G inches ; lastly, the length of the
middle partition, which parts the funnel, is 12 feet, and its
thickness 1 foot 3 inches; how many rods of brick-work, stan-
dard moa5tire, are contained in the chimney, double measure
being allowed, and thf-^kiiess \^ bricks ?
MAflONS' WORK. J 57
ft. In. ft. in. p.
1st. 18 . 9 5th. 1.3.0
12.6 12
3rcl.
4th.
•2ml. 14. 6 541 .0 . sura
2
234
.4 .6
ft.
in.
14.
6
9
130.
6
ft.
in.
10.
3
7
71 .
9
ft.
in.
13
n
.9
82. 6
6. 10.6
89 . 4.6
225 .0 15 .0 partition.
9.4. 234 ."1.6 parlour.
130 . 6 . hrst floor.
71 . 9 • second lloor.
89 . 4 . 6 shaft.
272)1082.0. double.
68)»66(3 rods 3 quarters.
62 feet.
Ans. 3 rod-s 3 quarters, and 62 feci.
MASONS' WORK.
To masonry belong all sorts of stone-work. Tlio work
is sometimes measured bj the foot solid, sometime* by the
foot in lenglh, and sometimes by the foot superficial. >!a.sons
in taking dimensions, girt all their mouldings, in th»- same
manner a* joiners.
150 KASONS' WOKK..
\V'ali«, colun.ns, blocks of stone or marMe, &c., are mea-
sured by the solid foot, and pavements, slabs, chimney-
pieces, &c., by tVip square »oot.
In estimating for the workmanship, square measure is
generally used, but for the materials, solid measure.
In the solid measure, the length, breadth, and thickness,
are multiplied together.
In tlie superficial meastire, there must be taken the length
und breadth of every part of the projection, which is seen
without the general upright face of the building.
1. If a wall be 82 feet 9 inches long, 20 feet 3 inches
high, and 2 feet 3 inches thick ; how many solid feet are
contained in that wall ?
ft. in. ft.
82 . 9 82-75
20 . 3 20-25
1640 41375
3 = ^ 20 . 8 . 3 16550
G =1 10 . 3 . 165500
3=i 5.0.0
1675 .8.3
2. 3
3
3351 .4.6
= ^ 418. 11 . Of
1675-6875
2-25
83784375
33513750
3.^513750
3770 . 3 . 6J 3770-296875 Ans.
2. rf a wall be 120 feet 4 inches long, and 30 feet 8 inches
high ; how many superficial feet are contained therein ?
A>i.i. 3690| feet.
3. If a wall be 112 feet 3 inches long, and 16 feet 6 inches
liigh ; how many superficial rods of 63 square feet are con-
tained therein ? Ans. 29 rods 25 feet.
4. What is the value of a marble slab at 8s. per foot, the
length being 5 feet 7 inches, and breadth 1 foot 10 inches ?
Ans. £4 l.». 10^.
169
PLASTERERS' V OHK.
Plasterers' work Is of two kinds, vi z., ceiling-, wtiich is pl< '
tering- upou laths; and rrnd^ring, \* hich is plasteiine upo^
walls. These are measured separate! % .
The content is sometime!" estimated by the foot, someti '.c.
by the yard, and sometimes uy the S(j uare of lOO feet. En
riched mouldings are calculated by the running foot or yanj
Deductions are made for chimneys, doors, windows, &«.
In phistering timber partitions, where several of the largo
braces and other large timbers proj« t from the plastering, a
fifth is usually deducted,
\Miitening and colouring are ivieasured in the same mannr-r
as plastering. In timbered partitions, one-tourth, or oik"
fifth of the whole area is usually added, to compensate f<i»
*he trouble of colouring the sides of the quai'ors and braces,
In arches, the girt ronnd them is malflpH'^d by the length
for Uie superficial content.
*. If a ceiling be 40 feet 3 inches lon^, and 16 feet 9
inches broad, how many square yards contained therein?
h. in. ' ft.
40.3 46 -.iS
16.9 '0-7o
■ ^* '■■■
•20125
28170
24150
4025
640
6
=
i
20
. 1
.6
3
=r
i
10
.
.9
3
'
4
.
.
9)674 .
o
3
9)'-74-187'i
Ans. 74 yards 8 feet. Ans. 74-9097 yaids.
2. The length of a room is 14 feet 5 inches, breadth 1.1
feet 2 inches, and height 9 feet 3 inches to the under fi<le ot
the cornice,* which projects 5 inches from the wall, on the
upper part next the ceiling; required the quantity of ren
• Cornices, festoons, &c., are rnt on after 'he room i* plastered mm) arc oot
)f coursca taken into account l-i tii* plasterer.
lUO
PLUMBERS' WORK.
ilering and plastering, there being no deduction but for one
door, wliich is 7 feet by 4 ?
Ans. 53 yards 5 feet of rendering, 18 yards 5 feet of ceiling.
3. The circular Taulted roof of a church measures 105 feet
6 inches in the arch, and 275 feet 5 inches in length ; what
vj-ill the plastering come to at Is. per yard?
Am: £161 8s. 5|rf.
4. The length of a room is 18 feet 6 inches, the breadth
12 feet 3 inches, and height 10 feet 6 inches ; to how mucb
amount the coiling and rendering, the form r at 8d. and the
latter at 3d. per yard ; allowing for the door of 7 feet by 3
♦'e«t 8, and a fire-place of 5 feet square ?
Ans. £1 13^. 36^.
PLUMBERS' WORK.
Plumbers' work is rated by the pound, or by the hundred
weight of 1 121bs. Sheet lead, used m roofing, guttering, &c.,
weighs from 6 to 12 pounds per square foot, according to the
thickness ; and leaden pipes vary in weight per yard, accord-
ing to the diameters of the bore.
The following table shows the weight of a square foot of
sheet lead, according to its thickness ; and the common
weight of a yard of leaden pipe according to the diameter of
its bore.
Thickness
Pounds to a
Bore of
Pounds ;/er
of lead.
square foot.
leaden pipes.
yard.
Inch.
^
5-899
0;;
10
6-554
1
12
1
7-373
11
16
I
8-427
18
r,
9-831
ll
21
^
11-797
24
plumbers' work. lg|
1. A piece of sheet lead measures 20 feet G inohin, in
len£rth, and 7 feet [) inches in breadth; what is Ita wei -lit at
S^lbs. to the square foot? '^
ft. in. ft
20. 6 20'.';
7. 9-
7-75
143 . 6 1025
15 . 4.6 1435
158. 10 .6
1435
158-875
1271-000
39-719
• cwt. qrs. Ib«.
112)1310-719(11 . 2 . 2J?, neuilj.
112
I'JO
1!2
2S)78(2
56
22
2. What weight of load ^^i "'* ^^ '"ch thick will cover a
flat, 15 feet 6 inches long, and 10 feet 3 inches broad, tlie
lead weighing Gibs, to the square' f<x)t ?
Ans. 8 c%Tt., 2 qrs., l^lb.
3. ^^^lat will be the expense of covering ami gntterinij
1 roof with lead, at 18s. per cwt.; the k-iigtli of the roof
being 43 feet, and the girt over it 32 feet ; the g\Ut«"nTi^
being 57 feet in length, and 2 feet in breadth, allowing 1
square foot of lead to weigh 8i^li'8. ?
An3.X]04 \5s. 3^'/.
iL2 PAIXTERS' WORK.
4. What will be the expense of 130 jards of leaden pipe
of an inch and half bore, at 4d. per lb., admitting each yard
losvelgh ISlbsV? • ^ns. £39.
PAINTERS' WORK.
Paintei s' work is computed in square yards. Every pan
19 measup d where the colour lies, and the measuring line i?
forced in' all the mouldings and comers. Double measure
is allowed for carved mouluings, &c.
^Vindows are done at so much a- piece. Sash-frames at
a certain price per dozen ; sky-lights, window-bars, case-
ments, &c., are charged at a certain price per piece.
To measure balustrades, take the length of the hand-rail
for one dimension, and twice the height of the baluster upon
the landing, added to the girt of the hand-rail, for the other
dimension.
No general rule can be given for measuring trellis-work ;
but, however, double the area of one side is often taken for
the measure of both sides.
1. If a room be painted, whose heiqht (being girt over
the moulding) is 16 feet 4 inches, and the compass of the
room 120 feet 9 inches ; how many yards of painting in it ?
ft.
120-75
16-3
ft.
in
120
.9
16
.4
1920
4 - ^ 40
. 3
6 = iL 8
.0
3 = 1 4
9U972 . A
Ayis. 219 yards 1 foot.
36225
72450
12075
9)1968-225
4ni. 218-691 yard*
GLAZIERS' WOUR. !(,.(
2. A gentleman had a room to be painted, it« length
being 24 feet 6 inches, breadth \6 feet 3 inches, and lei^lit
12 feet 9 inches; also the size of the door 7 feet bv 3 feet
6 inches, and the size of the window-shutters to eacli of the
windows, there being two, is 7 feei 9 inches by 3 feet
inches ; but the breaks of the windows themselves are b feet
6 inches high, and 1 foot three inches deep ; what will be the
expense of giving it three coats, at 2rf. per yard each ; the
size of the fire-place to be deducted, being 5 feet hy 5 feet
6 inches ? An.s. £6 3s. lO^rf.
3. The length of a room is 20 feet, its breadth 14 teet C\
inches, and height 10 feet 4 inches; how many yards of
painting in it, deducting a fire-place of 4 feet by 4 feet 4
inches, and two window-shutters each 6 feet by 3 feet 'J
Inches? A.-m. 73^^ yardi.
GLAZIERS' WORK.
Glaziers take their dimensions either in feet, inches, and
^arts ; or feet, tenths, and hundredths. They compute then
work in square feet.
Windows are sometmies meaBured by taking the dimen-
sions of one pane, and multiplying its superficies by lh«j
number of panes. But generally they take ine length and
breadth of the whole frame for the glazing. Circular win-
dows are measured as if they were square, taking for their
dimeusions their greatest length and breadth.
I. If s pane of glass be 3 feet G iaches and 9 parts long,
and 1 foot 3 inches and 3 parts broad ; how many feet of
glass in that pane ?
3 . 6 . 9 ■i-SG
1.3.3 I -'77
3 . G . 9 2492
10. 8. 3 2492
10. 8.'d 712
356
Ans. 4 . 6 . 3.11.3
Jh.v. 4-54G12 fe*^.
164 PAVERS' WOltK..
2. If there be 10 panes of glass, each 4 feet 8 inches S
parts long, and 1 foot 4 inches and 3 parts broad ; how maii^
feet of glass are contained in the 10 panes ? Ans. 64-0407.
3. There are 20 panes of glass, each 3 feet 6 inches 9
parts long, and 1 foot 3 inches and 3 parts broad ; how manj
feet of glass are in the 20 panes ? Ans. 90-9224 ft.
4. If a window be 7 feet 6 inches high, and 3 feet 4
inches broad ; how many square feet of glass contained
therein ? Ans. 25.
5. How many feet in an ellij)tical fan-light of 14 feet 6
inches in length, and 4 feet 9 inches in breadth ?
Ans. 68 feet 10 inches.
6. VVTiat will the glazing of a triangular sky-light ci»me
to at 20d. ; the base being 12 feet 6 inches, and the perpen-
dicular height 6 feet 9 inches ? Ans. £3 10*. 3^d.
PAVERS' WORK.
Pavers' work is computed by the square yard ; and the
content is found by multiplying the length by the breadth.
1. What will be paid for paving a foot-path, at 4*. thi
yard, the length being 40 feet 6 inches, and the breadth
feet
ft.
40-5
7-25
3 inches ?
ft. in.
40. 6
7. 3
283. 6
10 . 1 .
6
Ans. 293 . 7
6
2025
810
2835
Ans. 293-625 feet.
2. WTiat wdl be the expense of paving a rectangrdur court
«anl, whose length is 62 feet 7 inches, and breadth 44 feet
5 inches ; and in which there is a foot-path, whose whole
VAULTED AXD AUCUED ROOFS. 166
length 13 62 feet 7 inches, and breadth 5 feet G inchet, this
at 3*. per yard, and the rest at 2«. Gd. per yard ?
Alls. £3d Hi. :iu.
3. What is the expense of paving a court, at 3jf. 2a. por
yard ; tlie length being 27 feet 1 inches, and the bro-odih
1 4 feet 9 inches ? An^i. £7 As. 5\d.
4. What will the paving of a walk round a circular bowl-
ing-green come to, at 2*. 4d. per yard, the diameter of the
bowling-greuii being 40 feet, and the breadtli of the walk
5 feet? A us. £<J 3*. 3^'j(/.
5. How many yards of paving in an ellipticaJ walk 1 feet
broad, the longer diameter ocing GO feet, and shorter 50 ?
Atui. 62'31'J7 yardi.
VAULTED AND ARCHED HOOFS.
Arched roofs are either domes, vaults, saloons, or groinfl.
Domes are formed of arches springing from a circula",
*r polygonal base, and meeting in a point directly i>vfr th.
centre of that base.
Saloons are made by arches connecting tlie side walls of
a
building to a flat roof or ceiling.
Groins are made by the intersection of vaulted roofs with
each other.
Vaulted roofs are sometimes circular, sometimes clliptlciJ,
and sometimes Gothic.
Circular roofs are those oi ^hich the arch is a part of tlie
cLrcumfereuce of the circle.
Elliptical roofs are those of which tlie arcli is a }»art I'l
the circmiii'erence of an ellipsis.
Gothic roofs are made by the meeting of two e«iual circulur
arches, exactly above the span of the arch.
Groins are generally measured like a parallelopijMHbm, aiiJ
the content is found by multiplying tht length and brvuultij
of the base by the height.
166
VAULTED AXD ARCnEB KOOF3.
Sometimes ooe-tenth is deducted from the solidity thus
fomid, and the remainder is reckoned as the solidity of the
eacuity.
PROBLEM I.
To find the solidity of a circular, elliptical^ or Gothic
vaulted rooj'.
Rule. Find the area of one end, by one of the foregoing
rules, and multiply the area of the end by the length of tiie
roof, or vault, and the product wUl be the content.
iVo/e. When the arch is a segment of a circle, the area is foand by Prob,
XXVIIl. Sec. II. When the arch is a segment of an eilipsis, multiplv the
span by the height, and that product by •71554 for the area of the end. When
it is a Gothic arch, find the area of an isosceles triangle, whose base is equal *'o
the span of the arch, and its sides eqaal to the two chords of the circular tee-
uient of the arch ; then add the areas of the two segments to the area of the
triangle, and the sum will give the area of the eni.
Nrr-
O-
m
K
l> \
:ml'-LL L.„.
1. What is the content of a concavity of a semi-circular
vaulted roof, the span being 30 feet, and the length of the
vault 150 feet ?
30 X 30 = 900; then 900 x '7854 = 70G-86, henc^
70(i-86 -4- 2 = 353-43 the area of the end;
then 353-43 x 150 = o30l4-5 the content.
VAULTED AND AnClTED U00F8. JJiy
2. \Muit is the solid content of the vacuity A O V. l\ <^
Gothic vault, whose spun A B is (JO feet, the chord H O, or
A O, of each arcii 60 feet ; the distance of each arch from
the middle of the chords as D E = 12 feet, and the lenplh
of the vault 40 feet?
In this example, the triangle A H O e'jul-lateral. and its
area is ^ A B" ^/ 3 = 'JOO ^3 = 15'>7. Again, j (li O x
■^ '^^ + ^ = 5 (00 X 12) + j;-,^= 4943=.ro.
of segment O E B, and 404j X 2 = 9S8) the areas of the
two segments O E B and O il A ; then 15j7 + 1^88)) x
40 = 101832 the solidity required.
Let M N K L represent a perpendicular section of a
vaulted roof (Gothic). '1 lie span A B is ('A) feet, the thick-
ness of the wall M A, or B L, at the spring of the arch = 4
IVc't, the thickness O P at the crown of the arch = 3, una
tlie length of the roof = 4(t feet, the chord A O or O B =
6 feet, and the versed sine D E = 12 feet; re(piire<l tliu
solidity of the materials of the arch.
First, v^ (A O- - A C-'j = v/ (60^ - AO') = ryl'dC =
S O the height of the vacuity t>f the arch, and S O + O P =
Gl-96 + 3 = 54-<(G= S P; 'again, A B + M A -f B L =:
60 -f 4 + 4 = 68 = M L, and M L x !5 P = the area
of the rectangle M N K L ; hence M L X S P x 40 —
101832, (the solidity of the vacuity A O B by tlie l.ust
Problem), gives the solidity of the materials ; that is,
68 X 54-96 X 40 - 101832 = 476,59-2 feet, th« solidity
required.
.We. NVliPii the arch A O H is an clliplic.il tecrui.t, it< »r<a nr. t:-
plied by tho length of the roof gives the solidity of the racuity, ftii<l .M 1.
nuiltii.lied hy S P, and the product by the Icn^h of the wrh, gives th«
."itlidity of the cubic figure whose end is M N K 1.; »nrl the di(T<Trnre of
the two solidities is tho solidity of the mixed solid w! ' ii is A .M .N K I.
BF. OH A. The materials of a bridge may be >■ . after the «anie
m.inner, by adding the solidities of T, '1', and of tho tiAltUmenU, to th«
solidity as found in this Problem.
3. Piequired the capacity of the vacuity of an elliptical
vault, whose span is 30 feet, and height 15 feet, the length
of the vault hemv: 90 feet, Ans. 3180.'*-7 fevt.
168 VAULTED XSD .UICIIED ROOKS.
PROBLEM II.
To find the concave nr convex surface of a circutctr^
ellipticdl, or Gothic vaulted roofi
Rule. Multiply the len^h of the arch hv the lecgtii of
the vault, and. the product will be the superficies.
Aote. To find the length of the arch, make a line ply close to it, quite acrost
from side to side.
1 . WTiat is the surface of a vaulted roof, the len^h of tlit
3rch being 45 fe'it, aad the length of the vault 140 feet ?
140 X 45 = 6300 square feet.
2. Required the svu^'ace of a vaulted roof, the length of tht
arch being 40 feet 6 inches, and the length of the vault 100
feet ? A)is. 4050 feet.
3. WTiat 13 the surface of a vaulted roof, the length of th*-
Tch being 40-5 feet, and the length of the vault 60 feet ?
A?is. 2430 feet.
PROBLEM in.
To find the solidity of a dovie, having the height and th*-
dimensions of its base given.
Rile. Multi[ily the area of the base by the height, and
two-thirds of the product will give the solid content,*
* This rule is correct only in one case, namely, ■when the dome is half a sphere,
and ill this case the heiijht is equal to the radius of the circular base. It is a
Rell-known property toat the solidity of a sphere is two-thirds of that of a
cylinder having the same base aiul height. liut the soliditv of a cylinder if
found by muitiplj-ing the area of its base by the heieht. iJence the reason ot
the rule when applied to this particular case. No L'eneral rule can be given
to answer every case, as some domes are circular, some elliptical, gome poly-
roual, &c. ; they are of various heights, and their sides of different curvature.
VV'hen the height of the dome is equal to tlie raiiius of its base, ^the curved
sides bein? circulAr or elliptical quadrants.) or to half the mean proportional
between t'le two axes of jts ellipticii ba^e. tli« aiiove niJe will answer pretty
well ; but with anv other dimen>ions it ought not to be umhL
VAULTED AND ARCHED ROOFS. 1^9
1. What is the solid content of a dome, in tlio form of a
hemisphere, the diameter of the circular base being 4U feet ?
402 X -78.54 =1 1256-Gl = the area of ba«c.
§ (1250-64 X 20) = § (25132-8) = IU755-2, answer.
2. What is the solid content of nn octaq^onaPdome, each
side of its base being 20 feet, and the hciglit 21 feet?
Ans. 2703yi917 cubic feet.
3. Required the solidity of the stone-work of an elliptical
dome, the two diameters of its base being 40 and 30 fcef.
tlie height 17'32 feet, and the stone-work in every part 4
feet thick. Ans. 9479'08GS4S cubic feet.
PROBLEM IV.
To Jind the superficial contrnt of a dome, the height and
diynensiuns of itx base being given.
Udle. Multiply the square of the diameter of the base
by 1'570S, and the product will be the superficial content.*
For an elliptical dome, multiply the two diameters of its
base together, and the product resulting by I '5708 for the
superficial content, sufficiently correct for practical purposes.
1. The diameter of the base of a circular dome Is 20 fott,
and its height 10 feet ; required its concave superficies.''
20^ X 1-570S = 628-32 feet, the imswer.
2. The two diameters of an elliptical dome are 40 and 30
feet, and its height I7'32 feet; required the concave surface?
Ans. 1684*9(3 square feet.
3. What is the superficies of a hexagonal spherical dome,
each side of the base being 10 feet ? Ans. 6 1 90 152.
• Tliis nile is correct only when the dome i« circu!»r,«ncl iti b«'gbtt)o«l to
tlie radius o(,the base.- See Appendix, Dcnioustrnion, 11-.
u 2
170
VAULTED AND ARCHED ROOFS.
PROBLEM V.
To find the solid content of a saloon.
Rule. Multiply the area of a transverse section by the
compass or ca rcuniference of the solid part of the 8aloo«i,
taken round the middle part. Subtract this product from the
whole vacuity of the room, supposing the waJls to go upright
from the spring of the arch to the flat ceiling, and the differ-
ence will be the answer, as will appear evident from the fol-
lowing example.
I. WTiat is the solid content of a saloon with a circulai
quadrantal arch of 2 feet radius, springing over a rectan-
gular room of 20 feet long and 16 feet wide.
cVE
2' X -7854 = 3-1416 = area of the quadrant C D A F.
2x2+2 = 2= area of the triangle CDF; then 3-1416
— 2 = 1'1416 = area of the segment D A F. Now,
2x2 = 4 = area of the rectangle C D E F ; then 4 —
3-14 16 = -8.584 = area of the section D E F A D.
v/ (2^ + 2'-) = ^ 8 = 2-8284271. 2 x 16 -f 2 x 20 =
72 = the compass within the walk. ^ (2-S284271 — 2)
= -4142130= ES and 2-8284271 : -4142136 :: 2 : -2928932
= E t/ : hence 72 = (-2928932 X 8) = 69-6568.544 = the
circumference of the middle of the snlid part of the saloon ;
VAULTED AND ARCHED ROOFS. 171
therefore G0-G5GS.54-t x -8584 = 59-7031 1381G9G = tho
content of tlio Bolid part of the saluou.
20 X IG = 320 the area of the room floor, md 320 x 2
= 040 = the solidity of tlie upi)cr part of the room ; then
640 — 59-79344 = 580-20GJG feet, the eoliJity of the
aloon.
2. If the height D E of the ealoon he 3-2 feet, the chore!
D F = 4-5 feet, and its versed sine = 9 inches ; what is the
solid content of the solid part, the mean compass being 60
feet?
Ans. 13S-2G4S9 feet
PROBLEM VI.
To find the superficies of a saloon.
Rule. Find its breadth by applyin:^ a string close to it
across the surface ; find also its length by measuring along
the middle of it, quite round the room ; then multiply these
two dimensions together for the superficial content.
1. The girt across the face of the saloon is 5 feet, and its
mean compass 100 feet ; what is its superficial content?
100 X 5 = 500, the answer.
2. The girt across the face of the saloon is 12 feet, and
its mean compass 98: required its surface ?
An3. 11 7C feet.
172
SECTION VIII.
SPECIFIC GRAVITY.
1 . The specific gravity of a body Is the relation which the
weight of a given magnitude of that body has to the weight
of an equal uiagnitude of a body of another kind.
In this sense a body ia said to be specifically heavier than
another, when under the same bulk it weighs more than that
other. On the contrary, a body is said to be specifically fighter
than another, when under the same bulk it weighs less than
that other. Thus, if there be two equal spheres, each one
loot or one inch in diameter, the one of lead and the othei
of wood, then since the leaden sphere is found to be heavier
than the wooden one, it is said to be specifically, or in specie,
heavier, and the wooden sphere specifically fighter.
2. l£ two bodies be equal in bulk, their specific gravities
are to each other as their weights, or as their densities.
3. If two bodies be of the same specific gravity or density,
their absolute weights will be as their magnitudes or bulks.
4. If two bodies be of the same weight, the specific
gravities will be reciprocally as their bulks.
5. The specific gravities of all bodies are in a ratio com-
[)Ounded of the direct ratio of their weights, and the reci-
procal ratio of their magnitude. Hence, again, the specific
gravities are as the densities.
6. The absolute weights or gravities of bodies are in the
compound ratio of their specific gravities and magnitudes oi
bulks.
aPKCIFlC OIUVITT. 173
7. T'ae ma{»TiituJes of bodies are directly- &£ their weight*,
ind reciprocally as their specific gravities.
8. A body specifically hea^-ier than a fluid, lose« as much of
its weight, when immersed in it, as i^ eijual to the weight
of a quantity of the fluid of the same bulk or magnitude ;
if the body be of ecjual density with the fluid, it loses all it/
weight, and requires no force but the lluid to (sustain it. I(
it be hea>'ier, its weight in the fluid will be only the diflerenci'
between its own weight and the weight of the same bulk of
tlie fluid : and therefore it will require a force equiU to thi»
diffeience to sustain it. But if the body iramerseo be lightei
than the fluid, it will require a force equal to the difTerenc*
between its own weii^ht and that of the same bulk of ih*
fluid, to keep it from ri^iiug iu the fluid.
9. In comparing the weights of bodies, it is nece^A^r^ u.
consider some oue as the standard with which all other
bodies may be compared. Rain water is geni-rally taken a>
the standard, it being found to be nearly alike in all places.
A cubic foot of rain water is found, by re[H?atod experi-
ments, to weigh 6'i^ pounds avoirdupois, or 1000 ounces, and
a cubic foot containing 17-S cubic inclu's, it follows that a
uubic inch weighs •03G I < !8Ll8 1 4 8 of a pound. Therefore if tht
.pecific gravity of any body be niultiplled by 'O-SolfJ^B^l 1>,
;he product will be the weight of a cubic inch of that bod)
in pounas avoirdupois ; and if this weight be multiplied b»
175, and the product be divided by 1-44, the quotient will
be the weight of a cubic inch in iMmndj* troy, 144 pouixi*
avoirdupois being exactly equal to 175 poumis troy.
10. Since the specific grantios of bodies are aa their abso-
lute gravities under the siune bulk; the specific graTilT of a
fluid will be to the specific gravity of any body imuioried iu
it, as the part of the weight lost bv the solid is to the wboU
weight. Hence the speciric gravities of ditierynt fluid* ar«
uB the weights lost by the »*uie solid immer»cd lii thoin.
174 SPECIFIC GuAMTY.
PROBLEM I.
To fnd t/ie s}tecifjc p-avity of a body.
Case I. IVhen the body is heavier than watjr.
Weigh the body first in water, and afterwards in the opeo
air ; the difference will give the weight lost in water ; then
sav, aa the weight lost in water is to the absohite weight of the
body, so is the specific gravity of water to the specific gra-
Tity of the body.
Case II. When the bodi) is lighter than water.
Fix another body to it, so heavy as that both may sink in
*-ater to"-ether, as a compound mass. Weigh the compound
mass and the heavier body separately, both in the water and
open air, and find how much each loses in water, by taking
its wel"-ht in water from its weight in the open air. ^ Then
eay, as the difference of these remainders is to the weight ol
Uie lighter body iu air, so is the specific gravity of wate?
;o the specific granty of the lighter body.
Case III. For a jiuid of any kind.
Weigh a body of known specific gravity both in the fiuid
and open air, and find the loss of weight, by subtracting the
weight in water from the weight out of it. Then say, as
llie whole, or absolute weight, is to the loss of weight, so
is the specific gravity of the sohd to the specific gravity of
the fluid.
The usuiJ way of fiudiug the specific gravities of bodie?
18 the following, viz. —
On one arm of a balance suspend a globe of lead by a fine
thread, and to the other arm of the balance fasten an equal
weight sufficient to balance it in the open air ; immerse the
globe into the fluid, and observe what weight balances it
then, by which the lost weight is ascertained, which is pro-
pt>rtional to the specific gravity.
SPECmc URAVITT, 1 75
Immerse the globe successively in all tlie fluids whose nr>»-
portional specific gravity you require, observing the weight
lost in each ; then these weights lost in each will be the pro-
portions of the fluids sought.
£xa>/ij)les. — Case I,
1. A piece of platina weighed 8.'M88<1 pounds out of water,
and in water only 7y'5717 poimds; what is its specillc gravity,
that of water being 1000 ?
8;M88fi — 79 5717 = 3-6169 pounds, which is the w.i-ht
lost in water: then 3-61 (iO : 83-1SH6 :: U)()0 : L'3(>U(i tin-
specific gravity, or the weight of a cubic foot of metal in
ounces,
2. A piece of stone weighed lOlbs. in the open air, but \i«
water only 6|lb3. ; what is its specific gravity ?
Ans, 3077.
Examples. — C<ue II.
3. If a piece of elm weigh l^lbs. in the open air, and thai
A piece of copper, which weighs 18lbs. in open air, and l()!b»
in water, is ailixt-d to it, and that the compound wi-iglis Q\>>.
u) water ; required the specific gra\-ity of the elm ?
Copper. Compound-
is in air. 33
16 in water. 6
2 loss. -7
As '.ij: 15:: 1000:600, ti «
specific gravity of the elm.
4. A piece of cork weighs 2Ulbs. in open air, and a piece I'l
granite oeing affixed to it, which weiglis TJOlbs. in iiir, aid
only 80lbs. in water, the compound mass weighs ir)|lbs. ip
water; required the specific gravity of tlie cork. Am. 2-\[i.
lixamjilft. — Coxa III.
5. A piece ot cast-ixon weighed 259' 1 ounce* in a lluid, am
298' I ounces out of it; requir«>d the specific gravity of tb
i7o
SPECIFIC GRAVnr.
Buid, allowing the specific gravity of the cast-iron to be
7645.
298-1 — 259-1 = 39 loss of weight in the iron; then
29S-1 : 39 :: 7645 : lOOO, the specific gravity of the fluids
ghowing the fluid to be water.*
6. A piece of lignum vitae weighed 4'2| ounces in a fluid,
and 166f ounces out of it ; what is the specific gravity of
the fluid, that of the ligniun vitae being 1333?
Ayis. 991 is the specific gravity of the fluid, which shows
it to be liquid turpentine or Burgundy vane.
TABLE OF SPECIFIC GRAVITIES.
Spec. grav.
wi, cub. in
oz.
Platins
19500 .
11-286
Do. hammered
20336 .
11-777
Cast zinc
7190 .
4-161
Ca-'it iron
7207 .
4-165
Caj<t tin
7291 .
4-219
Bar iron
7788 .
4-507
Hard steel
7816 .
4-523
Cast bra««
8395 .
4-858
Cast copper
8788 .
6-085
Pure cast silver
10474 .
6-061
Cast lead
11352
6-569
Mercury
13568 ,
7-872
Pure cast gold
19-258 .
11-145
Amber
1078 . w
t. cub. ft.
l:irick .
2000 .
l'i:.-00
Sniphnr
2033 .
12706
Cast nichel
7807 .
4513
Cast cobalt
7811 .
4520
Paring; stones
2416 .
151-00
Comiiion sioLio
2620 .
157-60
Flint and spar
2594 .
162-12
Green glass
2642
White glass
2892
Pebble
2rt64
166-60
* In this manner may the Bpecles of a fioid or «oIid b« urert^ned, hj means
of its specific gravity, and the above table. 'Dm table baa boen taken from
(■irtyory's vork lor practicaJ iu«u.
SPECIFIC GIIA\1TT.
it;
Slate .
Pearl
Alabaster
Marble .
Chalk .
Limestone
Wax
Tallow .
Campbor
Beer*' wax
Honey .
Bone of an ox
Ivory
Air at the oarth'» sarfar*
Liquid tiirpenaue
Olive oil
Burgundy wine
Distilled water
Sea water
Milk
Beer
Cork
Poplar
Larch
Elm, and Went India ti
Maboganj
Cedar .
Pitch pine
Pear tret<
Walnut
Elder tree
Beech .
Cherry tree
Maple and Ri){:i tit
Ash and Dani7.
Apple tree
Aider .
Oak, Canadian
Box, French
Lof^wood
Oak, English
Oak, 60 year* old
Ebony
Ligaaa vitar
tfrC. Ifrn*
»l. cnb. ft
lU.
267J .
167(>o
2684
27311
274J .
171-3H
27b< .
. 174-0«1
3179 .
193(>!1
897
945
9»9
965
1466
1659
1832
14
991
915
991
1000
1028
1030
1-034
240 .
j6-oe
383 .
23-94
644 .
3400
666 .
M-7i
660 .
J5-00
69G .
-17-05
660 .
4 1 -25
661 .
i\A\
671 .
il'Ji
695 .
*3-44
696 .
» f..O
7:6 .
;* '.si
760 .
46-1^7
7M .
47 50
793 .
iHi6
800 .
•M <■!*•)
872 .
. .)
913 .
1
9l:< .
'ti
970 .
1 *^ >
1170 .
."-MJ
1331
■ 1 U
1333 .
^,(31
178 SPECIFIC GRAViTT.
PROBLEM II.
77>e specific g^ avity of a body, and its weight, being given, to
find its solidity.
Rule. Say, as the tabular specific gravity of the body is
to its weight, in ounces avoirdupois, so is 1 cubic foot to the
content.
1. Wliat IB the solidity of a block of marble that weighs
'0 tons, its specific gravity being 2742 ?
First, 10 tons = 200 hundreds = 22400 pounds = 358400
ounces ; then
2742 : 358400 : : 1
1
2742)358400(1 30 ,V^,
2742
8420
8226
<^^1940. n,o
"'^2742^*'^'^
2. How many cubic inches in an irregrdar block of marble
which weighs 112 pounds, allowing its specific gravity to be
2520? Ans. 1228|§.^g cubic inches.
3. How many cubic inches of gtmpowder are there in 1
pound weight, its specific gravity being 1745 ?
Ans. 15|, nearly.
4. How many cubic feet are there in a ton weight of dry
oak, its Bpecific gravity being 925 ? Ans. 38 [^f^.
PROBLEM in.
The linear dimensions, or magnitude of a body, being given,
and also its i^jecific gravity, to find its weight.
Rule. One cubic foot Lh to the solidity of the body, as
SPECIFIC GUaVITY. ,-o
I I J
the tabular specific graWtj of the body is to the weight ii
avoirdupois ounces.
1. What is the weight of a pioic of <lrv oaJv, in the forni o\
a parallelopipedon, whose leng-th is sC Luchos, br.;i.lili IS
inches, and depth 12 ?
56 X 1>< X 12 = 12090 cubic inches, the solid content.
Then 1723 : I'iOWG :: 932 : 65'24 ounces = 4U7^ poui.d-,,
the weig'ht required.
2. What is the wein;ht of a block of drv oak, which mea-
sures 10 feet long, 3 feet brond, and 2^ feet deep ; its specific
gravity being 9'25 ? A ns. 4:536 { jSlbs.
3. What is the weight of a block of marble, whose length
is 63 feet, and its breadth and tliickness, each 1 2 feet ?
Atis. <394iVij tons.
PROBLEM IV.
Tojind the guanlities of two inffredientsin a given compound.
Rule. Take the difference of every pair of the three spe-
cific gravities, viz., of the compound and each ingredient ; an
multiply the difference of every two by the third.
Then as the greater product is to the whole weight of the
compound, so is each of the other pro<luct.s to the weights of
tiie two ingredients.*
1. A couiposition of ITilbs. being made of tin and copper,
A^hose specific gravity is found to be 8784 ; what is the «|uan-
tity of each ingredient, the specific gravity of tin being 73_'('.
and of copper 9000 '(
9000 9000 8784
7320 6784 7320
1680 21G 1464 dilT.
8784 7320 9(100
14757120 1581120 13176000. Then
1475-1'>0 • 112 •• i'^^l"^'^"^^: lOOlbs. copjM-r.
1475<l-0 , 112 .. I 1581100. I2lbs. tin.
• For the reason of this role. •*» Alli?i»»i"n ToUl it tb« fvcoDtJ hotk •(
Antbnietic. pabliabed by tue ("omniienotiprm.
ISO SPECIFIC GRAVITY.
2. Fliero, king of Sicily, furnished a goldsmith with a
quantity of gold, to make a crown. When it came home, he
Ruspt'cted that the goldsmith had used a greater quantity of
silver than was necessary in the composition; and applied to
r,he famous mathematician, Archimedes, a Syracusian, to dis-
cover the fraud, without defacing the crown.
To ascertain the quantity of gold and silver in the crown,
he procured a mass of gold and another of silver, each ex-
actly of the same weight with the crown ; justly considering
that if the crown were of pure gold, it would be of equal bulk,
and therefore displace an equal quantity of water with the
gold en mass ; and if of silver, it would be of equal bulk and
displace an equal quantity of water with the silver mass ; but
if of a mixture of the two, it would displace an intermediatfl
quantity of water.
Now suppose that each of the three weighed 100 ounces ;
arid that on immersing them severally in water, there wer«
displaced 5 ounces of water by the golden mass, 9 oimces bj
silver mass, and 6 ounces by the crown; what quantity ofgoi<i
and silver did the crown contain ?
J (75 oimces of jjold.
( 25 oimces of silver.
Note. Questions relating to specific gravities mav be WTonght hj the ralef
of Alligation in Arithinetic. as -well as by any Algebraic process that migtii
Le employed.
PROBLEM V.
Tojind how many inches ajloatmg body will sink tn ajiuid.
KuLE. Find, by Problem III. the weight of the floating body
from its solidity and specific gravity, and that will be the
weight of the fluid which it will displace.
Then say, as the specific gravity ot the fluid is to 1728 cubic
inches, »o is the weight of the body, in omices, to the cubic
inches immersed. The depth will be f.'Und from tJas a^veo
dimen sioof-
SPECIFIC flRAVlTT. 181
1 . Suppobt, a piece of drv oak, in the form of a parallelo-
pipedou, whose length is 56 "inches, breadth 18, and depth \2,
is to be floated upon common smooth witer. on ita broadest
side; how manv incht, will it sink, its spt'cific trravitT bei- '
932? ' b J
By Problem L I., the weight of the piece of oak is 65" 1
ounces, ■which by the preliminarj part or this section, ia tise
Weight of water displaced.
Then 1000 : 1728 : : 6524 : 1 ^273-472 cubic inches of oak
immersed. Therefore, 1! 273-472 -»- (56 X IS) = 1 1-1^4
inches the depth it will sink.
To find how far it will sink, allowing it to float on its nar-
rower side, 11273-472 -h (56 x 12) = 16-776 inches.
2. How many inches will a cubic foot of dry oak sink in
rommon water, allowing the specific gravity of the oak to b«
(70? Ans. 11-64.
PROBLEM VI.
io Jind what weight may he attached to a ftnatinfr body, so
that it may be just covered with a given jhnd.
Rule. Multiply the cubic feet in the body by the difTtTt-ni^
between its specific gravity, and that of the fluid, n- ! •
product will be the weight in ouuces avoirdupois, juit ^
to immerse it in the fluid.
1. \Miat weight must be atuiohed to a piece of dry <>..k,
56 inches ong, 18 inches broad, and 12 inches di-t-p. r.i
keep it from rising above the surface of a frcsh-wuUT !ak.' :
the specific gravity of the water being I()(>0, ami that of th*-
oak 932 ?
Here 56 X 18 x 12 = 12096 ct;bic inchM.
Then 12096 + 1728 = 7 feet.
Theu (1000 — 932) x 7 - 6« v T - 4*^0 nac.'*« = ^H
(lounfli- 12 ourP**^
182 BPECiriO GRAVITy.
2. What weight fixed to a piece of dry oak, 9 inches Ions',
6 inches broad, and 3 inches deep, will keep it from rising
above tlie surface of conmaon water, the specific gravity of
water being 1000, and that of the oak 970 ?
Ans. 2\^ ounces.
3. A sailor had half an anker of brandy, the specific gra-
vity of the liquor was 927, the cask was oak, and contained
2 1 6 cubic inches, and its specific gravity was 932; to secure
his prize from the custom-house officers, he &xed just as
much lead to the cask as would keep it imder water, and then
threw it into the sea ; what weight of lead was necessary for
'lis purpose ?
Ans. The cask of brandy contained 1371 cubic inches,
tlie weight of sea-water of an equal bulk was bl7*204ytj
ounces, the cask weighed 116*5 ounces, the brandy
6I9'(J09375, both together weighed 736*19375 ounces. The
diiference between the specific gravity of lead and sea-watei
is to this remainder, as the specific gravity of lead to iti
weight in ounces, which will be found to be 89'09-i95 oimces,
or 5 pounds 9 ounces.
PROBLEM VII.
I'o find the solidity of a body, lighter than a jluid, which
will be .su//icifnt to prevent a body mvch heavier than the
Jluid, from sinking.
Rule. Find the solidity of the body to be floated ; from
its weiglit and specific gravity, by Problem II. Find also the
weight of an equal bulk of the tluld by Problem III. Then
sav, as the difTert-nce between the specific gravity of the fluid,
and that of the body lighter than the fluid, is to the differ-
ence between the weight of the body to be floated and the
woiglit of an equal bulk of the fluid, so is 1728 to the solidity
at' the lighter body in cubic inches.
I. flow manv solid foet of yellow fir, whose specific gravity
is G57, will be suflicient to keep a brass cannon, weighing 56
TONNAGE ((F iJlUl'8, Jf^^J
cwt.., afloat at sea, the specific gravity of bras* being 83'J^)
and of sea-water 1 030 ?
First, 56 cwt. = 1 00352 ounces, weight of the body to b.
floated.
Then, 8396 : 100352 :: 1728 : 20653G75 cubic indies ir.
the cannon.
And, 1728 : 20653-G75 :: 1030 : lli31U-U2S9, the weight of
sea-water equal in bulk to that of the cannon.
Hence, 1030 — 657 : 100352 — 123101)289 :: 1728 :
407868*5545 cubic inches = 236*036 feet, the answer.
2. The specific gravity of k-ad is 1 1325, of cork, 2-l(t, and
of sea-water 1030; now it is required to know how iuan\
cubic inches of cork will be suflicient to keep 49| pounds of
lead afloat at sea? Aus. 1570*84 cubic inches.
TO FIND THE TONNAGE ur cjHIPS.
Ist. VESSELS AOllOCND.
li^ the Parliametitary Rule.
PROBLEM VIll.
For a sliip or vessel, tlie length Is to be measured on ■
^.traight line along the rabbet of the keel, from a perpen-
dicular, let fall from the back of the niaju post, at the height
m1 the •.uug-tiausoiu, to a porpendicuhir at tlie hei^'lit of tlu*
jpper deck, (but the middle deck of three-decked ships,) fn-iii
the forepart of the stem ; then from the length bctwc^^ti tJ.. •
perpendiculars subtract three-fiftlis of the extreme brcndtli for
tlie rake of the stem, and 2^ inches for every foot t th«-
height of the wing-traiu^om above the louer part of the nibljet
of the keel, for the rake abaft ; luid the remainder will br
the length of the keel for tonnage.
J84 TOis'NAOE OF 6HI1B.
The nisia breadth is to be taken from the outside of the
outside plvuk, in the broadest part of ihe ship, either above
or belc vv the wales, deducting therefi jm all that it exceeds
the thickness of the plank of the bottom, which shall be
iiccountixd the main breadth; so that the moulding breadth,
or the breadth of the frame, will then be less than tho inwi>
breadth, so found, bj double the thickiiebs of the plank »■?
the bottom.
Then multiply the length of the keel for tonnage, by the
main bre/xdth, so taken, and the product by half the breadth,
then divide the whole by 94, and the 4U0tient ^-iH give the
tonnage.
In cutters and brigs, where the rake of the stem-post ex-
ceeds 24 inches to every foot in height, the actual rake is
generally subtracted Lofitead of the 2^ inches to every foot,
H.S before mentioned.
1 . Let US suppose the length from the fore-part of the stem,
at the height of the upper deck, to the afler-p&rt of the stem-
post, at the height of the wing-transom, to be 155 feet 8
mches, the breath from out to outside 40 fe«;t 6 inches,
and the height of the wing-tmnsom 21 feet 10 inchefl, what
Is the tonnage ?
ft.
40*6 bread '.h.
deduct 3
40-3
3
5)12U-9
24-l|^ = 24-15
21'10 height of wing-transom.
2^ multiply
4-55 -I- 24-15 = 2tt-70
155-66 — 28-70 = 126-96 = length.
126-96 X 40-25 -1-20-125 , ^ ,
j-TT -= 1094, the aabTTar
TOXKAOE Uf dUiPS. 185
2. Suppose the length of the keel to be 50*5 feet, brw-dth
of the miofihip-beam 20 feet; required the tonnage?
Au». 107-4.
3. If the lenf^th of the keel be 100 fe»»t, and the bre>..':>)
of the beam 30 feet; what is the touuage ? .Am. 47i>.
2nd. — VESSELS AFLOAT.
Drop a pliunb-Une over the btera of the ship, and mea-
sure the dLitauoe between such line and the after-part of the
bteru-post, at the load wuter-mark : in a piirallel direction
with the water, to a perpendicular point unniediatelv over
tlie load water-murk, at the fore-part of tlie maiu-utcm, sub-
tracting from such measurement the above di.s!;ince, the
remainder will be the slup's extreme Idugth ; fron; which i.«
to be deducted three inches for every foot of the loat! draught
of water for the rake abaft, and also tliree-fifths of the iihip'*
breadth for the rake forward, the remainder shall be esteemnJ
the just length of tlie keel to find the tonnage ; and U;»
breadth shall be taken from outside to outside of the plaiil,
in the broadest part of the ship, either above or below the
aiain-wales, exclusive uf all manner of sheathing or douUin;,'
that mav bo wrought upon tlie sides of the ship ; then mul-
tiply the' length of the keel, taken as before directed, by the
breadth, as before taught, and that prodiict l.y half iIm? said
breadth, and dividing the product by 94, the quotient in tlit
tonnage.
3rd. — STKAil VESSELS.
The length ehall be taken on a stiuight line, along the
rabbet of the keel, from the back of the main-»t4'ru-posi
to a perpendicular line from the fore-part of the m i
under the bow-sprit; fruui which deducting the h „ ♦
the engine-room, and subtracting three-fifths of the brca»lth,
the remainder shall be euteemed the ju*t lengtli of tlie keel
to find the touuage ; and the breadth Hhull l>e tiikm from t^w
out^iide of the ouuside plank in the broade.sl pLu'e of iIk
ihip or vessdi, l)e it either above or below the uuuu- ralci
A 2
18b FLOATi-VO BODIES.
exclusively of all manner of doubling planks that may bo
wrought upon the sides of the sliip or vessel ; then multiply
the length and breadth so found together, and that product
by half the same breadth, and dividing by 94, the quotient
will be the toimnge, according to which all such vessels shall
be measured,
A'ote. Under certain enaltie« nothuij but the fuel can be stowed in th«
engine-room.
Some divide the last product by 100, to find the tonnage o<
king's ships and by 95, to find that of merchaxit's ships.
FLOATING BODIES.
1. The buoyancy ol casks, or the load which they will
Oi<Xiy without sinking, may be estimated by reckoning lOlbs.
GVulrdupois to the ale gallon, or 8^ lbs. to the wine gallon.
2. The buoyancy of pantoons may be estimated at about
I alf a hundred weight, or 561bs. for each cubic foot. There-
lore a pantoon which contained 96 cubic feet, would sustain
48 himdred weight before it would sink.
N. B. — Tliis is an approximation, in wbich the difference between .j*^ and ^,
Hz., ^], of the whole weight is allowed for that of the pantoon itself.
3. The principles of buoyancy are very ingeniously applied
in the self-acting dood-gate, wliich, in the case of common
sluices to a mill-dam, prevents inundation when a sudden
Hood occurs. By meana of the same principle, it is that a
hollow ball attach €>d to a metallic lever of about a foot long,
i"" made to rise with the liquid in a water-cask, and thus ta
close the cock and stop the supply from the pipe, just before
t^^e time when the water would otherwise run over the top oi
tie vebbfl.
The property of buoyancy has also been successfully era-
plcved in rai.sing ships which had sunk under water, and in
^)uUing up old piles in a river when the tide ebbs and flows,
tk Ivso barere is broutrht over a pile as the water begins to
FLOATING BODIES. I «7
rise; a strong chain which has been previously fixet.1 tu tne
pile by a ring, &c., is made to girJ the hurgo, ami ia thi-n
firmly fastened ; then, 83 tlie tide risen, the barge rises also,
;ind bv means of its buoyant force draws up the pile with it.
In a case which actually occurred, a barge of 50 feet long,
12 feet wide, 6 deep, and drawing two feet water, was em-
ployed. Then 50 X 12 X (6 — 2) x ^ =» ^U X 12 x IG
^ 192 X 7f = 1344 + 27? = 1371? cwt, = C6J tons,
nearly, which is the measure of the forte with which the
'oarfe acted in pulling up the pile.
i6S
SECTION IX,
HEIGHT AND DIMENSIONS OF BALLS
AND SHELLS.
Tne foregoing problems furnish rules for finding the
vpeight and dimensions of balls and shells. But they may
ho found much easier by means of the experimental weight
of a ball of a given size, and from the well-knovm geome-
trical property, that similar solids are as the cubes of their
diameters.
PROBLEM I.
To find the tcpight of an iron hall from its diameteif.
Rule. Nine times the cube of the diameter being divided
by 64, will express the required weight in pounds.*
1. The diameter of an iron shot is 5 inches ; required its
weight?
5x5x5= 1 25 = cube of the ball's diameter.
Then 125 x 9 + 64 = 17Hlbs-» ^^^ answer.
2. The diameter of an iron shot being 3 inches ; required
its weight ? Ans. S'Slbs.
3. The diameter of an iron shot is 5-54 inches ; what v
Hi wpight? ^^- 24lbs.
■b«« App«D(Ux, L>*nion8tr&t-ioB 113.
iTETOnT Ay-D DnrENSTON-8 OP BALIA, TTC. 180
PROBLEM II.
To /iii^ the weight of a UaHen ball, by having ita diameter
gwi-n,
RULB. Multiply the cube of ita diameter by 2, and dirid*-
the product by y, and the quotient w-iil give the weiifbt in
pounds.*
1. ^V^lat ifl the 'veight of • leaden ball of 5 inchen dia-
meter ?
5 X 5 X 5 = 125 cube of ball'ii .iiameter.
Then, 125 X 2 + 9 = 250 + 9 = 27yb8., answer.
2. What i» tho weight of a leaden ball, whose rllameteT 1»
6-6 inches ? Am. hSS'^^XH.
3. \Miat \a the weight of a leaden hall, whose diameter is
3-5 inches? .I'u. 9-53lbs.
4. ^\^lat It the weight of a leaden ball, whose dianiPtcr l»
finches? J»w. I^IU.
PROBLEM in.
Havmg the weight of an iron ball, to determine its diameter.
Rule. ^luitiply the weight by 7^, then take the cube root
of the product tor the diameter.f
1. \Miat is the diameter of on iron ball, who«e weight i»
421b8. ?
42 X 7 i = 29Sf
Then, v' 29S — 6'€>85 inchea, the answer.
2. Required the diameter of an iron ball, whoM weight if
24lb8. ? Ans. 5-54 Loch. a.
* 8«« App«i>dix. DsTDoiLitrmtion 114.
f'inu role IS r>b7ioui< from Pmhieci I.. haUut t^ eonvarM XiMnat
190 WEIGHT ANT) DIMENSIONS
3. WTiat is the diameter of an iron ball, whose weight i?
3-81bs. ? Ans. 3 inches.
PROBLEM IV.
Having the weight of a leaden bally to determine its diameter.
Rule. Multiply the weight by 9, and divide the product
by 2 ; and the cube root of the quotient will express the
diameter.*
1. What ifl the diameter of a leaden ball, whose weight i*
64lb3.
64 X 9 = 576.
Then, 576 -I- 2 = 288.
Hence, \/ 288 = 6'6 inches, the answer.
2. Required the diameter of a leaden ball, whose weic;ht is
^7^1bs. ? Ans. 5 inches.
3. What ifl the diameter of a leaden ball, whose weight if
63-8«81bs. ? Ans. 6-6 inches.
PROBLEM V.
Having given the external and internal diameter of an iron
shell, to Jind xts weight.
Rule. Find the difference between the cubes of the two
diameters, and multiply it by 9 ; divide the product by 64
and the quotient will express the weight in pounds. f
1. Wliat is the weight of an 18-inch iron bomb-shell,
whose mean thickness is 1^ inches?
18 — 2^ =: 15^ = internal diameter.
Then, 18' = 5832 the cube of external diameter.
(15-5)' = 3723*875 the cube of internal diameter.
And, 5832 _ 3723-875 = 2108-125 = difference of cubes.
Hence, 2108-125 x 9 + 64 = 296-45lb3., the answer.
• 'Riis rule is manifon from Problem III., being it* conTers*.
4- See App«Ddiz, I>«moQ«tr»tiou 11&.
OF BAIiLS AND 8,TKJ.LIL 19*
-• What is the weight of a 9-inch iron borab-shcll, whoM
men- thickness is 1^ inch Atis. 7'2\\\\m.
"^ What is the weifrht of «n imn boinh-shfll, w'
ternal diameter is 'JS mches, aud inU'iual diameter 7
Ans. 84jllbs.
PROBLEM VI.
Tofindhnui much potcder ttiUjill a shell oj" given dimeujiou*
Rule. Divide the cube of the internal diameter in inche^
by 57*3, and the quotient will express the answer.*
1. WTiat quantity of powder will fill a shell, whusM; iutcr-
aal diameter is 10 Inches ?
First, 10 X 10 X 10 = 1000 = cube of eter.
57*3)1000(17-45lb8., answer.
573
4270
4011
•2590
22<J2
29S0
2865
115, &c.
S'ote. In some recent works, Uie cnbf of the diMnetrr u diT■:d^d by .19 J"l
for the weight of powder in pound*.
2. How many pounds of gunpowder are required to fill ■
hollow shell, whose internal diameter is 13 imlies?
Jus. 371bs., according to the iii)le.
3. Required the number of pounds of powder that will fil!
ft shell, whose intemnl diameter is 7 inches ?
Atis. Gibs, by the rule in the tciU
• Sm Ap)wndui DamoiutnHoa 116.
id'£ WEIGHT aXD dimensions OF BALLS, BTO.
PROBLEM VII.
Tojind how much powder will Jill a rectangular box of givtn
dimeninons.
Rule. Multiply the length, breadth, and depth together iri
hiches, and the last result by '0322, and the last product will
give the weight in pounds.*
1. How many pounds of powder will fill a rectangular box.
whose length is 16 inches, breadth 12 inches, and depth 6
inches ?
16 X 12 X 6 = 1152 = content of the box.
Then, 1152 X -0322 = 37*0944, the answer.
2. How many pounds of powder will fill a rectangular
Dox, whose length is 10 inches, breadth 5 inches, and depth
2 inches ? Ans. 3-22lb8.
3. How many pounds of powder will fiU a rectangular
box, whose length is 5 inches, breadth 2 inches, and depth
10 inches? An$. 3-22lbs.
PROBLEM VIIL
Having the length and diameter of a cyiindt'r, to determine
how many pounds of gunpowder will ^ fill it.
Rule. Multiply the square of the diameter by the length
md divide the product by 40, for the weight in poxmds.f
1. The diameter of a hollow cylinder is 10 inches, and the
length 14 inches; how many pounds will it hold?
10x10 = 100 = square of diameter.
Then, 100 x 14 = 1400.
Hence, 1 400 + 40 = 351b8., the answer.
* ^e Appendix, DemonstratioD 117.
i" S«« Appendix, Uemooit'atioa 118.
PILIN'O OF BALKS AND SUKLLjI. I9J
2. The diameter of a hollow cylinder is 5 inches, and :f<«
len^h 40 inches ; how much powder will it hold ?
Ans. S.Slhii.
3. The diameter of a hollow cylinder is 5 inches, and th^
length 12 inches; how many pouudu will it hold?
PROBLEM IX.
Tr> Jind what portinn of a cylind'-r will ha occupied I", n
given qnantitif of powder^ the diameter of the cylinder
being given.
Rule. Multiply the given weif^ht of powder by 40, and
divide the product by the squiu-e of the diameter of the
cylinder, and the quotient will Uv ibe pounds require*!.*
1. The diameter of a hollow cylinder is 10 inches; how
omch of it will hold SOlbs. of powder?
50 X 40 = 2000.
Then, 2000 + 100 = 2«J inches, the answer.
2. How much of a cylinder of 14 inche* diameter will hol.i
iOlbs. of powder? Ans. 2'05 inches.
3. How much of a cylinder, 12 inches in diamofer, will huld
I44lb8. of powder? Ans, 40 incn
PILING OF BALLS AND SHELLS.
Iron-shot and shells are usually piled in horiiont'J cour««,
either in a pyramidical or in a wedp-e-Iike form ; the oa5«
being either an equi-lutcral triangle, a square, or a rectangle,
• See Ap:'«i.d:x. IVmonitrafioa 'l"*.
lyi PILING OF BALLS AXD BHKLLS.
Those piles whose bases are trianglaa or squares, terminate
in one ball at the top : but piles whose bases are rectangles
terminate in a single row of balls.
In triangular and square piles, the number of horizontal
rows or courses, is always equal to the number of balls ir
one side of the bottom row.
And in rectangular piles, the number of rows b equal tn
the number of baUs in the breadth of the bottom.
Also the number in the top row or edge, is one more than
the diflFerence between the length and breadth of the bottom
row.
PROBLEM I.
To Jind the number of balk in a rectangular pile.
Rule. Multiply the munber in one side of the bottom
row, by that number increased by 1, and the result by the
same number increased by 2 ; then the one-sixth of the last
product will give the number of balls required.*
1. Required the number of shot in a complete triangulai
pile, one of whose sides contains 22 balls ?
22 = the number in one side of base.
23 = the number -f- 1.
66
44
506
24 = the number + 2,
2024
1012
6)12144
2024 = the number of shot in the pile.
• S>>M A.ppeD(lix. DemondtratioD 120.
PILING OlT BAi.LS AND SUbLUi. 195
2. Required the number of shot iu a coraplete truuig^uiu
pile, one side of whose base contains 15 bolls r
Ant. (iSO balls.
3. Reijuired the number of l.alls in a trian^uhu" pilr^ each
side of the base containing 3U bails ? Anj. 4y(iO.
PROBLEM II,
To /tnd the numbtsr of balls in a square pile.
Rule. Multiply continuaUj together the number iu on«
side of the bottom course, that number increased bj 1, and
double the sajne number increased by 1 ; then one-sixth of
tlie last product will be the answer.*
1. How many balls are in a square pile of 30 rows ?
30 =: number in one side.
31 = uimiber in one side + I.
930
61 = twice the number in one side 4- 1.
6)56730
9455 answer.
2. Required the nambcr of shot iu a complete square pile,
one side of whose base contains 19 ? Aru. 'J470.
3. How many shot iu a finished square pile, when a side of
the base contains 2 1 shot ? ^ nj. 33 1 1 ■
PUUliLEM III.
To find the number of shot in a finuhtid rectangular piU
Rule. Add 1 to three times the number of shot contaiucd
in the length of the base, subu-act the number of shot in th«
■ ijM ▲piMiadiA' iJvaoaaUM.oL I'iU
ly6 PILING OF BAIXP AXD chtLLS.
breadth of the base, muitiply tne remamdei by the said
number increased by 1, and this result again by the number
in the breadth ; then one-sixth of the last result will give
the number of shot in the rectangular pile.*
1. Required the number of shot in a finished rectangular
pile, the length of the base containing 59, and its breadth
containing 20 balls
59 = the niunber of shot in the length.
3
177; then 177 + I = 178, and 178 — 20= 158.
158 X 21 = 3318, and 3318 x 20 = 66360. Hence
66360 -^ 6 = 11060, the answer.
2. How many balls are in a rectangular complete pile, th"
length of the bottom couise being 46, aud its breadth 15 ?
Am. 4960.
PROBLEM IV.
To determine the number of balls contained in a pile which
is not JiJiished, the highest course being complete, and the
nu?/iber of baUs in each side thereof being given.
Rule. Find the number of shot which would be contaiued
iu the pile if it were complete. Find also the nmnbor in
that complete pile, each side of whose base contains one shot
fewer than the corresponding side of the uppermost course
uf the uufmished pile, aud the difference between these results
will e>'idently give the umnber of balls in the unfinished
pile.
1. How many shot axe there in an unfinished triangular
pile, a side of whose base contains 23, and a side of the
uppermost course 7 shot?
* boo Appendix, DemoDstration l'J2L
DISTERMLNINO DISTANCES BY 80CIil>. ]y^
23 = number of balls iii the ba«j.
24 = number of bulls in the bm>e -f- 1.
552
25
6)13800
2300 = number of the pile when complete,
6
4'i
8
6)336
56 number of bailfl in the imag-iiiarjr oile.
Therefore, 2300 — 56 = 2244, the tlitwa.
2, How many balls In an incomplete square pile, the »'ui*
of the bajse being 24, and of the to;- 8 ? Am. 4760.
3. How many balla ore there in the incomjilete rectaugiilar
pile of 12 courses, the length and breadth of the ba&e being
lOand^U? Ant. GliC.
DETERMINING DISTANCES BY SOUND.
The velocity of Bound, or the bpace throut' V'oh it i*
f^ropaguted in a given tiiut-, has been rery (. . ly i-sU-
uiated by philosophers who have written uu thia cuhject.
We shall, nowever, take it to be 1142 feet in k Mooui.
From re{)eatcd experiments it hu been aAcertainc«i ihAt
fiound moves uniformly, or, to i«{>eak more philoaophiciJiy,
il'.at the pulses of air vthich excite it move uniformly. The
velocity of sound is the same with that of the rrial wave*,
and does not vary much whoiber it go with tht wiijd or
against it. By the wind, no doubt, a certain c{iiaiiL;l^ jl air
198 uETEBMl>ri>G DlaT-NCiia lik SOUSL.
is carried from one place to ato her, and the souna is some-
what accelerated while its wavos move through that part of
the air, if their direction be the same as that of the wind.
But as the velocity of sound vastly swifter t. an the wind,
the acceleration it will thereby r ceive i^ !jut inconsiderable,
being at most but ^g of the wholi velocity.
The chief effect perceptible from the wind is, that it in-
creases aaid diminishes the space through which sound is
Eropagated. The utmost distance at which sound has been
eard is about 200 miles. It is said that the unassisted
human voice has been heard from Old to New Gibraltar, a
distance of about 12 miles. Dr. Derham, placing cannon at
different distances, and causing thet to be fired off, observed
the intervals between the flash and report, by means of which
he found the velocity of sound to be as above stated,
] . Having observed the fla^h of a cannon, I noticed by
my watch that 5 seconds elapsed previous to my hearing thf
report ; determine my distance from the gun,
'll42
5
5710 feet, tne answer.
2. Being at sea I saw the Qash of a cannon, and counted
8 seconds between the flash and tht report ; required the
distance? An$. 1^ mile.
IW
SECTION 3L
GAUGING.
Gauging' is the art of measxiring the capacities of reMels,
*uch as casks, vata, &c.
The business of gauging- is general! j performeU by meaui
of two instruments, uaiuelj, the gauging or sliding rule, and
the gauging or diagonal rod.
1. O*- THE OAUOLXO RULE. — LE>J>nETTER's.
By this instnunent is computed the contents of c-aiks, &c.,
after the dimensions have been taken. It is a sipuire rult-,
having various logarithmic lines on its four faces, and three
sliding pieces capable of being moved through grot)Vfs in
which they fit, in three of these faces.
On the first face are delineated three lints, namely, two
marked A B, on which multiplication and division are per-
formed; and the third mjirked M D, signifu'S malt deptli,
and serves to Erauire malt. The middle one H is on the
slider, and is a kind of double line, being markiil ut both
edges of the slider, for applying it to both the lines \ &nd
M D. These three hues are all of the same radius, or din-
tance from 1 to lU, each containing twice the lfn;;th of the
radius. A and B are numbered and placed txiicily alike,
each commencing at 1, which maybe either 1, or 10, 1(.K>,
&c., or 1, or -01, or UOI, Sic. Whatever the I at tJie be-
ginning is estimated at, the middle division 10, will be 10
times as much, ancj the last division 100 times om mttcb.
But 1 on the line M D Ls opposite 2J20, or more eiaclly
2*J18'2 on the other lines, which nun\ber 2218"2 denotes
riie cubic inch in an imperial malt bushel ; and its divisions
ivumbered retrogade tc those uf A and B. Ou tlieao two
o
zoo eAUGiNo.
lines are also several other marks and letters ; thus on tl-e
fine A are M B, or sometimes only B, for malt bushel, at the
number 2218'2, and A for ale, at 282, the cubic inches in
an old ale gallon ; and on the line B, is W, for wine, at 231,
the cubic inches in an old wine gallon.
These marks are now usually omitted upon the rule, since
the late uew act of parliament for uniformity of weights and
measures, and G for gallon is put at 277*274 the inches
b an imperial gallon,* whether of ale, wine, or spirits.
On many siiomg rules are also found s i, for square in-
scribed at "707, the side of a square inscribed in a circle,
whose diameter is 1 ; a e, for square equal at "886, the side
of a square wliich is equal to the same circle ; and c for cir-
cumference, at 3' 14 16, the circumference of the same circle.
On the second face, or that opposite the first, are a slidej
ind four lines marked D, C, D, E, at one end, and root
square, root cube at the other end ; the lines C and D con-
taining, respectively, the squares and cubes of the opposit*
Qumbers on the lines D, D ; the radius of D being double to
ihat of A, B, C, and triple to that of E ; therefore whateve
the first 1 on D denotes, the first on C is its square, and th
first on E its cube ; that is, if D begin with 1, C and E wiE
begin with 1 ; but if D begin with 1 0, C will begin with 1 00,
and E with 1000 ; and so on.
On the line C are marked o r at '0796, for the area of the
• Until 5 George IV., in which a uniform system of -weights and measnres
was establiflhed under the denomination of imperial weights and measdkf-s
thert were, amongst other sources of inconvenience, different measure*, though
of the same name, for ale and -wine. A gallon of ale contained 282 cubic
incbea, and a gallon of wine 231 ; a bushel of malt contained 2150-42 cubic
Inches. • . v • u
To red uce old measure into new, say, as the number of cubic inches m the
^vperial standard is to the number of cnbic inches in the old standard, so is the
number of gallons or bushels, &c., old mftasure, to the number of gallons, &c..
I<«nal measure.
When great accuracy is not requijred, old vrine gallons ma.j be reduced
or
n. ami dividing the prodca* by 3VL
CAlGINr.. 201
circle whose circumference is 1 ; nml o d, at •7?54 for ilu-
axea of the circle whoso diameter ia 1.
On the line D are maxkeci G S, for pallon tniiare, at
16-65, and G R for jrallou round at 18'769; alw M S for
malt square at 47*01)7, and M K for malt round at /iS-M I.
Thwe are the respective gaiif^e-points for t,'all.mii nnd
bushels. The first 16"r)5 is the side of a 8<juare, which at nn
inch depth holds a gallon; the second 1 8*789, the diameter
of a circle, which at an inch depth holds a gallon ; the third
47*097 tlte side of a sqiuire, which at an inch depth holds a
bushel ; the fourth, 53*144, the diameter of a circle, which
at an inch depth holds a bushel.
On the third face ar« throe lines : one on a slider marked
N ; and two on the stock, marked S S and S L, for seginent
standing and segment lying, which serve uUaging, stiiiidintj
and lying ca-sks.
And on the fourth side, or opposite face, are a scale i»f
inches, and three other scales, marked spheroid, or 1st varietv,
2nd variety, 3rd variety ; the scjile for the fourth or conic
variety, being on the inside of the slider in the third faoe.
The use of these lines is, to find the mean diameter of C4L.-k<i.
On the inside of the two first sliders, besides all thos*-
already described, bse two other lines, being continued from
one slider to the other.
The one of these is a scale of inch^*, from 'J^ to 3<i, and
the other is a scale of ale gallons, between the corresjMjndiii;;
number 435 an 36 1 ; wluch form a table, to show, in ale
gallons, the contonta of all cylinders whose diameter* are from
12^ to 36 inche«, their common alutude being 1 inch,
VKRIE's SLIDDiO RLLK.
This niie is in the form of a paraiielopipedon, and i» gene-
rally made of lx)i.
1. The line marked A, on the face of this rule, is calivd
Gunter's line, and is u umbered, I, 2, 3, 4, 5, 6, 7, 8, 9i 10.
At 22 18' 19*2 Is fixed a bnws nln. marked IM, H. Bignifvliig
the cubic inches in «ui imperi;u bushel; at •J77'274 is tmJ
201 GAUGINy.
another brass pin, marked IM, G, denoting the number ol
cubic inches in an imperial gallon.
2. The line marked B is on the slide, and is divided
t^actly Hke that marked A. There is another sUde B, on
the opposite side, which is used along vrith this. The slide
on the first face is called the second radius, and that on
the opposite face, the ^rst radius. The two brass ends,
when placed together, make a double radius, numbered
from the left-hand towards the right. At 277*274, on the
second radiu-s, is a fixed brass pin, marked IM, G, denoting
the cubic inches in an imperial gallon ; at 314 is fixed
onother brass pin, marked C, signifying the circimiference
if a circle whose diameter is 1. These lines are used and
-ead exactly as the Hnes A and B, on the carpenter's rule,
which have been already described.
3. The back of one slide or radius, marked B, haa
the dimensions for imperial gallons, and bushels, green
starch, dry starch, hard soap hot, hard soap cold, greeii
Boft soap, white soft soap, flint prlass, &c., &c., as in table,
page 206.
The back ot the other slide or radius, marked B, contain*
the gauge-points corresponding to these divisors, where S
denotes squares, and C circles,
4. The line VT D on the rule, denoting malt depth, is a
line of numbers commencing at 2218-192, and is numbered
from the left to the right-hand 2, 10, 9, 8, 7, 6, 5, 4, 3.
This rule is ussd in malt gauging.
5. The two slides B, just described, are always used
rogether, either Vith the line A. M D, or the hue D, which
is on the opposite face of the rule to that already described-
The line D is numbered from the left-hand towards the
right, 1, 2, 3, .'-51, to 32, which is at the right-hand end; it
is then continued from the left-hand end of the other edge of
the rule, 32, 4, 5, 6, 7, 8, 9, 10. At 16-651 is a brass piu
G S, signify" Jig a gauge square, being the square gauge-
point for imperial gallons. At 18-789 is fixed a brass pin,
marked G R, denoting gauge round, or circular gauge-point
ff»r imperial gallons. At 47'0')7, M S signifies malt square,
the fcquar« i^auge-point for malt bushels. At 53-1 14, M R
GAUUING. 20J
denotes mutt round, the round or circulai gau^e-point for
malt bushels. The line D on this rule is of the same naturr
as the line marked D on the carpenter's rule, whieh ha* been
already described. The line A and the two slides ii, are
used together, tor perfomuug nmltiiilicatiun, division, simple
proportion, &c. ; and the line 1), and the same slides li, arc
used together for extractini^ the square and i-u!>« rooLs.
6. The other two slides belong^ing to this rule are marked
C, and are divided in the same manner, and uaed tt>ireiheE
like the slides B.
The back of the first slide or radius, marked C, is divided,
next the edg-e, into inches, and numbered from the left-liand
towards the rip-ht 1, 2, 3, 4, 5, &c., and these inches are
ag'ain sub-divided into ten equal parts. 1 he second line ij
marked spheroid, and is ntmibered from the left-hand towardj
the right 1, 'J, 3, 4, 5, 6, 7, H. The third lino is marked second
rariety, and is numbered 1, 2, 3, 4, 5, G. These lines ar«
used, with the scale of inches, for ficUnjj a mean diameter.
The back of the second slide or radius, marked C, lui»
several factors for reducing goods of one denomination to
others of equivalent values. Thus | X. to VI. 6. | siguiiiea
that to reduce strong beer at S,v. j)er barr«l. to small beer at
Is. Ad. you are to multiply by U. | V 1. to X. 1 7. | signilies tluit
to reduce small beer at 1*. 4rf. per barrel to strong beer at 8j.
per barrel, you are to niultij)ly by •17. | C 4 ^ to X. 27. |
signifies that *J7 is the multiplier for reducing cider at 1». ; er
barrel to another at 8.v., &c.
7. The two slides C, ]\xM. described, are always u>ed toge-
ther, with the lines on the rule marked Scg. St., or S S, »fg-
ments standijig ; and S«'g. L ^ or S L, segment* lying ; for
ullaging casks. The former of these lines is nuinberi«i 1, 2, 3,
4, 5, 6, 7, 8, which sUnds at the right hand end ; it then gott
on from the left-hand on the other edge 8, 9, 10, &c., to lUO.
the latter is numbered in the same manner 1, 2. 3, 4, which
stands at the right hand end ; it then goes «>n frum the left-
u&aA ou the other edcre, 4, 5 tif 7. &c., to 100.
204 GAUGINO.
PROBLEM 1.
To find the several multipliers, divisors, and gauge-points,
belonging to the several measures now used.
MULTIPLLERS FOR SQUARES.
As 211'21^ solid inches are contained in one imperial
g'allon, and 2218*192 solid inches in an imperial bushel; then
it is obvious that if 1 be divided by 277'274, and 2218-192,
respectively, the quotients will be the multipliers for imperial
g-aiions and bushels respectively.
Hence the method of finding the foUovring multipliers is
obvioua : —
277-274) 1-00000(-0036065 multiplier for imperial gaUons.
22l8-l92)100000(-0004508 multiplier for imperial bushels.
Now it is manifest that if the solid inches contained in any
vessel be multipUed by the first of these multipliers, the pro-
duct will be tne imperial gallons that vessel vn\\ contain ;
and if multiplied by the other, the product •will be the in*-
perial busheU.
MULTIPLIERS AND DIVISORS FOR CIRCLES.
It has been shown that when the diameter of a circle is 1 ,
the area of that circle is -785398, &c., or '7854, nearly; then
by dividing the solid capacity of any figure by '7854, thf
uuotient will be the proper divisor for the square of the
diameter of a circular figure. Then to reduce the area at
one inch deep into gaUons, divide '7854, or "785398, &c., by
277-274, and 2218192, and the quotients \\-ill give the multi-
pliers for imperial gaUons and bushels respectively; and -7854
divided into 277-274 and 2218-192, will give the divisors for
the imperial gallons and bushels.
277-274)-785398(-002832 multipUer for imperial gallons.
•2218-l92)-785,^98(-00354 multipUer for imperial bushehi.
•785398)277-274(350-0362 divisor for imperial gaUons.
•7b53y8)22l8-l92( 2824-2897 divisor for inipurial bushels.
oauoi.no. Wft
The gaupe points are found by extractinj^j the stjuare ro.i!
of the divisors.
OAUOE-POIM8 FOB SQCARKS.
V lll-TA = 16-n5l imiKfriul f,'allon«.
v/ 2218-192 = 47-0U7 iniperifU biiaheU.
GADGB-P01>T8 FOR CIRCLES.
\/ 353-0362 = 18-789 imperial pallooB.
\/ 2824-2897 = 53-144 imperial buiheU.
In thi^ nviQiier the numbers in the following table were
c'ulculat«d.
S06
GAUGING.
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jVote. It very often happenB in the practic* of ganging, that 'when the oan
eiven uomber is get to the gaage-point on the sliding mle, the other given num-
ber will tali off the mle; hence in many cases it will be necessary to find
a second, or new gauge-point. The second eunge-points are the square roots of
ten tiroes the divMon in the above table. Thus, for squares, the new gauge-
point for imperial g;allons is 52(].% for bushels 14803 ; and for circles, the new
gang^e-poirt for gallons is 69*42, for malt bushels 168'05
oAComo. 20^
PROBLEM II.
To find the area, in imperial gallunt, of any rectilineal plan*
figure.
Rule. By the rules given in Mensuration of Su(>erfici«'!i,
find the area of the ^^pae in inches, wliich being div-ided bv
277*274, or multiplied by 'OO-SGOGS, will gire the area in
gallons.*
1. Suppose a back or cooler in the fomi of a panillehv
gram to be 100 inches In lenjrtli, and 40 in breadth; re-
quired the area in imperial gallonB.
^100_^ X 40 ^ 4000 the area in inches, which divided by
277*274 the quotient I4*42f) = the number of imj>eriaJ
gallons; or if we multiply 40UU by -OOJGuoj, tlie product
14*42G is the number of imperial gallons as before.
BY TUB SLICING KULE.
On A On H On A On B
As 277*274 : 40 : : 100 : U*4, nearly.
2. If the side of a square be 40 Inches, what is thf area io
imperial gallons ? Ana. S'll galion».
3. If the side of a rhombus be 40 inches, and its perpcn-
dicular breadth 37 inches ; required its area in wine giiJli>ns.
Ans. 6*4 I.
4. What is the area of a square cooler, in imperial gallons,
the side being 144 inches? Ans. 74*765.
6. Allowing the side of a hexagon to be 64 indies, and the
perpendicular from the centre to the middle of one of the sidef
• Tbe awM of plane SgnrM, in gmuging, ar» expres»«d In j[«J'om, or horft*-'*
For there ^^ill be <ii ni»nj lolid inches in kaj ▼•m«1 of on« inch irf^, %.- {^-r*
are guperficial inchei iu lU bue. What ii c»ll»<l in ?»u^in«f &
in in reality • njlid of oue inch deep, which mniunliau bj the i.. . •
the whole content in (alloni or ouiuel*.
208 UAUUUiG.
55 42 inches ; requii'e<l its area in imperial gallons and malt
bushels ?
- ( 38'38 imperial gallon*.
'{ 4-6 malt bu&heLs.
PROBLEM III
The diameter of a circular vessel being given in inches, to
Jind its area in imperial gallons, 6fc.
Rule. Multiply the square of the diameter by '002832 ;
or divide the square of the diameter by 353'036, the product
or quotient will give the area in imperial gallons.
When it is required to find the area in any other deno-
mination than imperial gaUons, use the proper multiplier or
divisor for the required denomination, as given in the table,
page 206.
1 . The diameter of a circular vessel is 32*6 inches ; re-
Quired the area in imperial gallons?
(32-6)* = 1062-76. Then,
1062-76 X -002832 = 30 1 gallons.
Or, 1062-76 + 353-036 = 3-Oi.
BT THE SLIDIXO BTJLE.
As 18-78 is the circular gauge-point for impcriaJ gallons,
say.
On D On B On D On B
As 18-78 : 1 :: 32-6 : 3
2. If the diameter of a circular vessel be 10 inches, what
is the area in imperial gallons ? Ans. -283.
3. Suppose the diameter of a circular vessel is 30 inches,
what is its area in imperial gallons ? Ans. 2-548-
4. What is the area in imperial gallons of a round vessel,
wQose diameter is 21 iuches ? Ajis. 1-631,
OAUOINO. 2(>0
PROBLEM IV.
Given the tiantverse and conjugate diameter of an elltptnaJ
vetxel, to Jind its area in imperial measure.
Rule. Multiply the product of the two diamctprs bt
•002832 ; or divide the product of the two diiuiu-ters bV
353'036 ; the product or quotient wili (^ve the imperial gal-
ious required.
When any other denomination is required, the pro[)er mul-
tiplier or divisor in the table is to be emplojcd.
1. Suppose the longer diameter of an elliptical vessel u 10,
and the shorter diameter 6, required tin- area in hIu aii.l wine
fCiillons.
Here, 10 X 6 = 60.
Then, 60 X -002832 = -17 of a gaUon.
2. The transverse or longer diameter of an elliptical ves««)
.6 20, and the conjugate or shorter diameter 10 inches ; what
i« the area in imperial measure ? Ans. '^GG of a gallon.
On A On B On A On H
As 353 : 20 : : 10 : -566 of a giiilon.
3. Suppose the transverse diameter of an elliptical vessel it
70 inches, conjugate oO inches ; required ita art-a in iuipcritti
gaUous and malt bushels ?
J 5 'J-914 gallons.
^""•7 1-24 mallbuiheU.
Sate. Aa vessels ar« seldom or nrver m»J« truly elliptic*]. 1 ■»J1t
ovals, the area found b*' the »t>o%e rule is not curr«cl, »xt€ «mk1
be a truly mathenu-itical ellipsis ; ■wlion the vessel ii of %a oral form, tXie m««
is best tuund by the method of equi-dittant ordinate*
Let A B C D be the oval vessel whose area u required,
and let A B and C D be the transverse and ciinju{rat« <ii»-
meters, at right angles to each other, the former lieintf IO'J'8
210
11 IB
Inches. Divide this transverse (102*8) bj some even number
which will leave a small remainder, the quotient will be tha
distance of the ordinates ; which distance may be laid off ou
both sides of the conjugate diameter a number of times equa
to half the even niimber by which the transverse was divided ;
tlien with chalk and a parallel ruler, draw the ordinates
through the points 1, 2, 3, 4, &;c. Then, by Problem XXI.,
Sec. III., the area may be found, which being multiplied of
di%dded by the proper tabular numbers, will give the area ia
gallons, &c. Or,
Ist. Add together the first and last ordinates.
2nd. Add together the even ordinates, that is, the 2, 4, 6,
8, 10, &c,, and multiply the sum by 4.
3rd. Add together the odd ordinates, except the first and
Ijist ; that is, add the ordinates 3, 5, 7, 9, &c., and multiply
the sum by 2.
4 th. Multiply the sum of the extreme ordinates by their
distance from the curve.
5th. Add the three first found sums together, and multiply
the sura by the common distance of the ordinates, and to the
product add the fourth found sum, and divide the total by 3,
and the quotient resulting by 277'274, or 221S'192, for the
area in imperial gallons, or malt bushels, respectively.
OAUOINO
ail
First, 102 8 + 10 = 10 the distance of the ordinate!
Asunder, and the remainder 2'8 is double the dintance of the
extreme ordinates from the <nirve ; that is, 1*1 = A 1, or
B 11.
Now let us suppose the lenjrths of the ordinatrs to ht
20, 40-2, 57, 66-6, 73, 75, 73, (iG-fi, 57, 40-2, 20, respcctivclT
beginning at 1, and proceeding to 1 1 .
^^'- ? 1 1 = 20
40 inches, sum of the first and lust.
2iid. <^
r 2 =
4 =
6 =
8 =
10 =
66
4(1-2
75-0
6Cv6
40-2
;^rd. <^
289-6 X 4 = 1154 4
r 3 = 57
5 = 73
7 = 73
9 = 57
260 X 2 = 520
Tlitii, 40 + 1 154-4 + 520 = 1714-4 sum of fin-t thn-e mr
10
17144
56
3)1720*'
5733-3; then,
5733-3 + 277-274 = 2(t-64 .'allonv
5733-3 -}- 2218-192 = 2-58 nuUl bushel
2!2 GAUGING.
VVTien the vessel is not circular, or elliptical, it b l)est to
measure the equi-distant ordinates, which though ever fXi
unequal, will, by proceeding as above, serve to find the area
of the base. Whenever the vessel is an irregular ciu-ved
figure, the area should be invariablj found by the method
of equi-distant ordinates, as the true result cannot be foimd
bv anv other method.
4. What is the area, in imperial measure, of an ellipse,
whose transverse axis is 24, and conjugate 18 ?
Avs. 1-2234 gallons.
PROBLEM V.
To find the content of a prism, in tinpm ial gallons.
Rule. Find the area of the base, by Problem II., in
Gauging, which being multiplied by the depth witliin, wilJ
^ive the content in gallons.
Or, find the ^•olid content by mensuration, and divide that
content by 277'274 for imperial gallons.
A vessel, whose base is a right-angled parallelogram, is
•*9'3 inches in length, the breadth 36-5 inches, and the depth
42'6 inches ; required its content in imperial gallons ?
Here, 49-3 X 3G-5 X 42-6 = 7^6.56-57.
Then, 76656-57 ^ 277-274 = 276-465 gallons.
Ard 76656-57 -f- 2218-192 = 34-558 malt bushels
BY TILE SLIDLNG RULE.
On B On D On B
41y3 : 49-3 : : 36-5 : 42-42.
On D On B On D
16-65 \ . g .. . f 27-6 gaUonj.
2 Each side of the sq'aare ba«e of a teswl Li 20 inchf<»
and its depth 10 inches, what is the content in oid nJe
gallons? J n«. 14-29 gallon*.
3. The side of a vessel in the form of a rhomhus ia 20
inches, breadth 15 inches, and depth 10 inches; required the
content in old ale gallons? Am. 1 0'G3tt gallons.
4. ^Tiat IB the content, in old wine gallons of a TCMe! in
the form of a rhomK>id, whose longest side is 20 irich<-«,
breadth from side to side 8 inchen, and depth 10 inches ?
Ant. b-88 wine gallons-
PROBLEM VI.
To Jind the content of any te*>el, tchose etiih are tquarr*
or rtctangies, of any dtinentioui.
Rule. Multiply the sum of the lengths of the two ends,
by the siun of their breadths, to which add the areas of the
two ends ; this sum, ni;Utip!ied by one-sixth of the depth, wdl
give the solidity in cubic inches ; then diride by 'lll-'llA, or
22 18" 192 for the content in imperial gallons, or malt bushels.
1. Suppose the top and bottom of a vessel are parallelo*
grams, the length of the top is 40 inches, and **' breadth
30 inches ; the length of the bottom is 30 inches, and »t.«
breadth 20 ; and the depth GO Inches ; required the conienu
in imperial gallons ?
40 -f 30 = 70 sum of the lengths.
30 + 20 = 50 sum of the bread tlis.
3500 product.
40 X 30 = 1200 area of the greater b:i*«.
30 X 20 = 600 area of the leaser base.
5300
10 one-sixth '*f the depth.
53000 solidity »o cubic inchea.
Than. .^^(HK) + 2"7-'.;74 = lill-'-Wi.
214 GAUGING.
BY THE SLIDING RULE.
Find a mean proportional (a/ (40 X 30) = 34"64,) be-
tween the length and breadth at the top, and a mean propor-
tional ( v/ (30 X 20) = 24-49,) between the length and
breadth at the bottom; the sum of these is 59' 1^, twice a
mean proportional between the length and breadth in the
middle. Then,
On D Oq B On D On B
l6-fi5 • 60 . . ^ 94.4q • Zm l^"'^ 191-146 imperial gal-
2. Suppose the top and bottom of a vessel are parallelo-
grama, the length of the top is 100 inches, and its breadth 70
inches ; the length of the bottom 80, and its breadth 56,
and the depth 42 inches ; what is its content in imperial
gallons ? Ans. 862'59 imperial gallons.
THE GAUGING OR DIAGONAL ROD.
The diagonal rod is a square rule, ha\ing four faces, and
IS generally 4 feet long. It folds together by joints. Tliis
instrument is employed both for gauging and measmring
casks, and computing their contents ; and that from one
dimension only, namely, the diagonal of the cask, or the
length from the middle of the bung-hole to the meeting of
the cask with the stave opposite the bung ; being the longest
line that can be drawn from the middle of the bung-hole to
any part within the cadk.
On one face of the riile is a scale of inches for measuring
this diagonal ; to which are placed the areas, in ale gallons,
of circles to the corresponding diameters, in like manner as
the lines on the under sides of the three slides in the sliding
rule.
On the opposite face, there are two scales of ale and wine
gallons, expressing the contents of casks having the corres-
Dondent diagonals.
All tho other lices on the instrument are similar to ihoMs i»i
the sliding rule, and are used in the i»ame manuer.
Exampli:. The diao;onal, or distance between the middle
of the bung-hole to the most distiint part of the ca«k, a>
found by the diagonal rod, is 31*4 inches: what is tho content
in gallons?
To 34'4 inches correspond, on the rnd, QOj ule gallons,
or HI w-iue gallons, 9'2$ imperial gallons, the cuiitcnt re-
quired ?
Xote. Tbe coatenta shown bj the ruil answer to the most common fum ol
'Aski, and fall in between tbu 'Jnd and 3rd variutiei following.
OF CASKS AS DIVIDED IXTO VARFETIES,
Casks are usually divided into four varieties, which arp
easily distinguished by the curvature of their sides.
1 . The middle frustum of a spheroid belongs to tho firs'
vailety.
'J. The middle frustum of a parabolic spindle belongs U
llic second variety.
il. The two equal frustums of a paraboloid Ix-Ioug to tl»f
thiid variety.
4. And the two equal fruatums of a cone belong to th»
fourth variety.
If the content of any of these be found in inches by tlulr
proper rules, and this divided by '277-'274, or 'J"Jlb-J, llif
quutient will be the content in imperial gallon5, or bu.>.!iel».
respectively.
PROBLEM VII.
Tojinrf the content of a tvwc/ in the form of the fi-ustutr,
uf a conti.
Rule. To three times the product of the two dlaiuelcf>
odd tbe square of their dilierei.t^) : :ualii['ly the «um bj on»
r
216 GATJGINO.
third of the depth, and divide the product bj 353-0362 foi
mperial gallons, and by 2824'289 for malt bushels.
1 . VvTiat is the content of a cone's frustum, whose oreater
diameter is 20 inches, leajst diameter 15 inches, and depth 21
inches ?
20 X 15 X 3 = 900
20 — 15 = 5 & 5' = 25
925 X 7 = 6475. Then,
353-0362)6475(1 8-34 imperial gaUona.
294-12)6475(22-01 wine gallons.
2. The greater diameter of a conical frusttmi is 38 inches,
the less diameter 20-2 and depth 21 Laches ; what is the
content in old ale gallons? Ajis. 51 "07 gallons.
PROBLEM VIII.
To find the content of the frustum of a square pyramid.
Rule. To three times the product of the top and bottom
eides, add the square of their difi'erence, multiply their sum
by one-third of the depth, and divide the product by 232
and 231, for old ale and wine gallons, respectively; and by
•^77*274, for imperial gallons.
1. Suppose the greater base is 20 inches, the less base 15
inches, and depth 21 inches; recjuiied the content in old wine
measure ?
20 X 15 X 3 = 900
2U— 15 =5
Then, 5 X 5 = 25
925 X 7 + 231 = 27'8 gallons.
Sot«. Tb« oont«iit of the fhutum of ft pyramid is foand jnst like that of
« cuue, witli tii« exctj ti'n of the tabular divisor, or multiplier. tL« eoae r^
^uiriug the circa Lar factor, and the pyramid the square one.
OADOINO. J17
PROBLEM IX.
To Jind the content of a globe.
Rule. Multiply the diameter of the globe by its circui;>
ference, and the resulting product by one- sixth of the dia-
meter; then the List product multiplie<l or divided by tht
circular factor, ^^-ill gire the content in gallons.
1 . Let the diameter be 34 inches, what is its content ?
.34 X M X 34 X •5i3G = 20579-5744.
Then, '20o7'J-5744 -h 282 = 72-y772 old ale gallon*.
And, 20o79-57l4 -h 231 = 8908 old wine gallons.
RCLK n. Or cube the diameter of the globe, which mJil-
tjply by -001 888 (5 of •002ti32) for the content in impiriJ
gallons.
34^ = 39304; then 39304 X -001888 = 74-2 imperial gallons.
2. ^V^lat is the content of a globe in old ale and wiue
measure, the diameter being 20 inches?
J 14-848 old ale gallon*.
^"^' \ 18-1-28 old wine gallons.
3. Required the content of a globular vessel, whose dia-
Qieter is 100 inches?
Ans. 1888 J imperial gallon*.
PROBLEM X.
To /i)id the content nf the segmrnt of a sphere, at th« rising
crown of a copper stilt, fyc.
Rule. M'.isnre the diameter, or chord of tJie »ci»^ment,
and the altitude just in the middle. Multiply the Mjuar* of
half the diameter by 3; to the product add the »«niarv »f the
;iltitude; multiply this sum by the altitude, ami the priHluol
again by -00185^5, or •0022G(j, for old ole or wine nica.<iiu«,
respectively, and by -001 8bS for inipcrinl gallon*.
218
GAUGING.
1. The diameter of the crown of a copper still is 27*0 its
depth 9 "2 ; required its content ?
Hers, 27-6 ^ 2 = 13'8.
Then, 13-8 xlS'B X 3 = 571-32
9-2 X 9-2= 84-64
655-96 sum.
9-2 depth.
imperial gallons.
6034-832 X -001888 = 13-39
PROBLEM XI.
To gougt; a copper having either a concave or convex bottom ;
or what is called a falling bottom, or rising crotvn.
Rule. If the side of the vessel be straight with a falling'
botti>m, find the content of the segment Cj/ D, by Prob. X.;
find also the content of the upper part A B D C, by Prob.
\ 1 1. ; the sum of l*oth will give the content of the copper.
\Vlien the copper h.if= a rising crown, find tlie content of
A B C D, by Prob. Vll., from which deduct the content of
the segment C a* D, and the remainder will be the content of
the vessel A B D a: C.
OAUOUiO.
319
PROBLEM \II.
I'll gauj^e a vessel whose side is curved from top U bottorn.
Take the diameters at equal distances of 2, 2, 4, or '»
ndies, according' us the case m&y require; if the side of
fhe vessel be considerably curved, the number of diametci>
iiat will be required will be cousidi-rable ; the less the cur-
vatui e of the side, the less the cumber of diameters tluii
will oe required.
To gauge the vessel, or copper, A IJ D C, fasten a pieo
'i pack-thread at A and B, as A F B; then with some con
venient instrument find the distance a C of the dcej)*it |>nri
of the cupper, which let us suppose to be 47 iuches.
By means of the same instrument meoAure the di»tanc.>
F from the top of the crown to F the midille of A IJ; «liich
let us suppose to be 42 inches, this deducted from a C, 47,
will leave o (= o G) the height of the crown.
To Jind the diainfter C D, of the bottom of the croim.
Measure the top diameter A B, which suppose to b« 99
inches: then hold a thread, so that a plummet attached to
the end thereof, may hang just over C, and measure A a =
B F) each of wliich let us ^iiuit to be 17*5 inches; mUd
220
GAUGING.
these together, and deduct their siim (35) from 99, and the
remainder (64) vrill evideutly be equal to C D, the diameter
at the bottom of the crown. Measure the diameter in o n,
Nvhich touches the top of the crown, which suppose is 65
inches.
Now, as this copper is not considerably curved, the dia-
meters may be taken in the middle of every 6 inches of the
depth, which suppose to be as in the second column of the
following table ; to each diameter find the area in imperial
gallons, by Prob. III., which write in the third column ; find
also the content of every 6 inches, corresponding to these
diameters, whj*vh write in the fourth cx)luinn of the table ;
lastly, find the tontent of the crown by Prob. X., and sub-
tract it from the content of A B D G C, the remainder
will give the capacity of the copper.
Or thus, C D being 64 inches, the area answering to it
is li'6022, this multiplied by half the altitude of the crown,
viz., by 2-5, gives 29"0055 gallons, the content of the crowu.
The content of the part »i n D C is 53-9222 gallons, froti
which the content of the croTvn being deducted, the remaiudei:
(-9-9167 gallons) is the quantity of liquor which covers tb
erown.
Parw of
the depLh.
Diameter*.
Area*.
Coolant of every
6 inches.
6
6
6
6
6
6
6
95-3
90-1
So-
so-
75-2
70-5
66'
25-7257
22-9948
20-4653
1 18-1284
16-0183
14-0786
12-3387
154-3542
137-9688
122-7918
108-7704
96-1098
84-4716
74-0322
The sum
To cover c
» <
;rowu
778-1 9^8
29-9167
The whole content . 808-4155
UAUGI^O. T2y
PROBLEM XIII.
To Jind the content nf any clnt« cafft.
Whatever be the form of the casK, the following dimtuiionj
must be taken ; that is,
The bung diameter, )
The head diameter, ^ witliin.
The length of the cask, )
On account of the tlifficultvv in ascertaining the figure of
the cask, it is not, in manj cases, e&aj lo find the exAct
contenta of casks.
In taking the dimensiorw of a cask, it ij etwential that iha
bung-hole be in the middle of the cask, and also, that th«
bung-stave, and the stave opposite tc» it, are both reg^ar and
even within.
It is likewise essential that the bea'is of casks are equal
and truly circular ; and if so, the distance l-^rtween the inside
of the chimb to the outside of the oppo«»ite stare, will bo th»»
head diameter within the cask, nearlj.
From the variety in the forms of casks, no penoral nJ*
eould be given to answer everj form ; two cisks may hare
equal head diameters, equal bung diumeters, ami equal lengths,
\nd yet their contents may be vfo^y unequal.
PROBLEM XIV.
To Jind the content of a cask of th4 firtt rnriety.
Rclt:. To the square of the head diameter a<ld double
the square of the bung diameter, and multinly tlie lum b\
the length of the cask. Then multiply the lukt jiroduct by
•0009^, or divide by 1059*1, the pr-^luct or quotient will bt
the content in injperial gallons.
) . What is the content of a sphe-
ro dal cask, whose length is 40 inches,
bung diameter 32 in<-!)t>';. and head
diameter 24 inclu'.> ?
^22 QAUGIJSO,
24 X 24 = 5"'6
32 X 32 = 1024
2
2048
576
2624 X 40 = 104960
•0009|
944640
34987
11662
99' 1289 Imperial gallons.
BY THE GArOING RULE.
Set 40 on C, to the G R 18-79 ou D, against
24 on D, stands 64 -99 on C,
32 on D, staudii 116-2 on C,
I- llG-2
3)297-39
99-13 gallons.
2, What is the content of a spheroidal cask, whose leiigtl
IS 2 inches, bung diameter 16 inches, and head diameter"l'2
inches ?
j^^ f 12-3G old ale gaUons.
' ( 14-869 old wine gallons.
Tojind the content of a cask by the mean diameter.
Rule. Multiply the difJerence of the head and bung
diameters by -68 for the first variety ; by -62 for the second
variety ; by '55 for the third ; and by -5 'for the fourth, when
the' difference between the head and bung diameters is less
than 6 inches ; but when the difference between these exceeds
b inches, multiply that difference by '7 for the first variety,*
by -64 for tlie socoml ; by -57 for tlie tliinl ; nml bv -5'-' for
the fourth. AdJ tliis product to the hca.l diameter,' nnd the
sum will be a mean diameter. Suuare this mean diameter,
ttnd multiply the square by the ItMit^th of the ca»k ; thia
product multiplied, or divided, by tiie proj>cr multiplier oi
divisor, will give the content.
By resuming the last example but one, we have
Bung diameter 32 2[)'6 mean dianieier
Head diameter 24 2[)-6
8 87^' 16 square
•7 40 lengtli.
5-6 359-5);i504G-40
24
mean diameter 29 6
97'6 g^ous.
In the same manner the cotjtcut for the second variety will
be 94-46 ale gallons ; for the third variety 90"87 ale gallons ;
and for the fourth variety 83'34 gallons.
PROBLEM XV.
To find the content of a cask of the seeoud varirty.
Rule. To the square of the head diameter u.ld doubl*
the square of the bung diameter, an<l from the sum deduct
two-fifths ot the square of the dilVerenco of the diuuKieni;
multiply the remainder by the length, and the pr.jducl
again by -0009^ for th* «»iitent in imperial gallons.
1. What ijT the content of a ciisk,
whose length is 40 inches, bung dia-
meter 32 inches, and head diameter
24 inches?
224
GADGDfO.
32 — 24 = 8 ; then 8* = 64, and J of 64 = 25-6
24* = 576, and 32* = 1024, then 1024 X 2 = 2048
2048 + 576 = 2624, and 2624 _ 25'6 = 2598-4
40
) 03936 X -00091 = 98-1617 gallons.
103936
PROBLEM XVI.
To Jind the content of a cask of the third variety.
Rule. To the square of the bung diameter add the square
of the head diameter ; multiply the sum by th^ length,
and the last product by -001416 for the answer in imperial
gallons.
Let us resume the last example : thus
32* = 1024
24*= 576
1600 X 40 = 64000
•001416
90*624 imperial gallons.
PROBLEM XVIL
To find ihs content of a catk of the fourth variety.
RuL^. Add Uiti sq'iare of the difference of the diameters
to 3 time* the »q'.:£Cs of their sum ; multiply the sum by the
length, and the \mt product by -000236 for the content in
gallona.
Resuming still the last example, 32 ^.— !^^«c^h^
-f- 24 = 56, and 56* X 3 = 3136 x ^"^"^^^W^^
3 = 9408, and 8* = 64, then 9408 fM
-f 64 = 9472; then 9472 x 40 = \ -.:j
378880, BDd 378880 x '000236 = ''^-.,
69'H668 imperial gallons.
OACOINO. i'-iA
PROBLEM XVIII.
To Jind the content of any catk by Doctor IluHotii gtmrr^i
rule.
Rule. Add into one vun^ 39 time« the sqaare of tbr
biing diameter, 25 times the sqiiAre of the heiid diameter,
and 20 times the product of the two diameters; then iiiultiplT
ilie sum by the leugth, and the product again bj •00u3 1 \ foi
the content in gallons.
1. What is tJie content of a caiik, who«« length i^ .
inches, and the buiig and head diameters 3*2 and 2 1 ?
32^ = 1024 24» = 576 32 x 24 = TCS
39 25 2fi
39936
14400
19968
40
14400
74304 X
= 29721 GO
•00031^
93-4579 gallons.
ULLAQING.
19968
PFxOBLEM XIX.
To ullarrc a lying cask.
This is the finding what quantity of liquor u oontaincil ib
a cask when portly empty.
To ullage a lying cjwk, the wet an.l dry inches n.'i«» b^
known, ^ al»o the content of tlir ci-n'^ and bung dianiclcr.
22^ GAUGING.
Rule. Take the wet inches, and divide them by the bung
diameter; find the quotient in the column of versed sines,
in the table at the end of the practi-
cal part of this book, and take out
its corresponding segment ; multiply
this segment by the whole content
of the cask, and the product aris-
ing by 1^ for the ullage required,
nearly.
1. Find the ullage for 8 wet inches, the bung diametei
being 32 inches, and the content 92 ale gallons ?
32)8('25, whose tabular segment is -153546.
Then, -153546 X 92 = 14-1262.S2.
And, 14-126232 x H = 1 7-65779 gallons.
PROBLEM XX.
To ullage a standing cask.
Rule. Add together the square of the diameter at the
surface of the liquor, the square of the diameter of the nearest
end, and the square of double the diameter taken in the
middle between the other two; multiply the sum by the length
between the surface and nearest end, and the product arising
by -000472 for the gallons in the less part of the cask whether
empty or filled.
I. Wliat is the ullage for 10 wet inches, the three dia-
meters being 24, 27, and 29 inches ?
(2 X
24^
292
27)^
= 576
= 841
= 2916
4333
10
43330
•000472
86660
303310
173320
43330 2'j-45170galion«.
OAUOUiU.
m
PROBLEM XXI.
To Jind the content of an ungula, or hoof, of the frui'mn
of a cone.
Rule, For the less hoof, multiply the product of the le&a
diameter and height, by the product of the ^Tcatcr diamcttT
multiplied by a mean proportional between both diameters,
less the snuare of the less diameter, and tiiis la^t divided by
three times the circular factor multiplied by the dilTcrenr«
of the diameters, gives the content of the less hoof.
1. CD = 30, AB= 40, C(f-
20, required the content of the less A
hoof.
40 X 30 = 1200, and ^ 1200
= 34 '6 mean.
.SO X 20 = 600, 1st product.
'0 X 34-6 = l;^'^ 1, 2nd product.
30 X 30 = 1*1..
494 remainder.
484 X <)00 = 290400
40 — 30 = 10, then .1nf» x 3 X 10 = 10:7<sund
21)0400 ^ 10770 = 2f;-96 gallons.
Rule. For the greater hoof multiiily the product of the
greater diameter and the height ot the fru.*tum, by the
square of the greater diameter made less by the pro^hict o<
the less diameter multiplied by a mean j)roj»ortional iM-twcen
those diameters ; this remainder, divided by three tiroes the
circular divisor multiplitd by the difference of the diameter*,
gives the content of the greater hoof.
Resuming the last example, we have
40 X 41) = KJUO
20 X 40 = 800 Ist product.
40 X 30 = 1200, and y/ 1200 = 34r)
34-6 X 30 — l(t3"^ 2nd producU
40 — 3U = 10.
228 OADGLNO.
Then 1600— 1038 = 562
800
359 X 3 X 10 = 10770)449600 last product
41-74 old ale irallons.
PROBLEM XXII.
To gauge a still.
Fill the still with water, and draw it off in another vessel
of some regTilar form, whoae content is easily computed.
rhia is by far the most accurate method that can be em-
ployed.
Or gauge the shoulder by itself, and gauge the body b}
taking a greater number of diameters at near and equal dis-
tances throughout, first covering the bottom, if there be anj
cavity, with water, the quantity jf which is known.
229
SECTION XL
LAND SURVEYING.
Land surveying is that art which enables u£ to pire » tTT3«
Slan or represent^ition of any field or parcel of land, and to
etennine the superficial content thereof.
In measuring land, the area or superficial content i^ alway>i
expressed in acres, or in acres, roods, and perchea ; each acre
wntaining 4 roods, and each rood 4U perclica.
Land i» measored with a chain, called Guntor'i c' ' '
I poles or 22 yards in length, which coususts of li ■ . ,
links, each link being y^ of a yard long, or ,6^ of a f«K>i.
)r 7*92 inches. 10 sqmuro chaina, or 10 chains in ' '>
iind 1 in breadth, make an acre ; or 4810 squjire y.i. , '
square poles, or 100,000 souare links make an acre. The
length of lines measured with a chain, are geiifraliy wt down
in links as integers; every chain being 100 links in lei'.'''i
Therefore, after the content is found, it will b« iu ><;
links, and as 100,000 square links make an acre, it wiii i-v
necessary to cut off fivo of the figures on the r' '^ ' hand for
decuuals, and the rest will be acres. The t' ^ arc re-
duced to roods by multiplying by 4, and cutting otV fiv*- figures
as before for aecinuJs, which decimal part is rr' ' '
fierches by multiplying by 40, and cuttiug olT Bto
rom the product. As an example :
Suppose the length of a rectangular piece of groand to
be 792 links, and its brea/lth 3''5 ; re<pured the namb«r of
acres, roods, and perches it contains ?
230 LAND SURVEYING
792 3-04920
3S5 4
3960 -19680
fi.^S6 40
2376
7-87200
304920
A. R. P.
Ans. 3. 0. 7.
The statute perch is 5^ yards, but the Irish plantation
perch is 7 yards ; hence the length of a plantation link i?
10-08 inches.
PROBLEM I.
To measure a line or distance on the ground, two persom
are emploijed ; the foremost, for the sake of distinction,
is called the leader, and the hindermost, the follower.
Ten small arrows or rods, to stick in the ground at the
end of each chain, are provided; also some station-stages,
nr long poles with coloured flags, to set up in the direction of
the line to be measured, if there do not appear some marks
naturally in that direction.
The leader takes the 10 arrows in one hand, and one end
uf the chain by the ring, in the other ; the follower stands
at the beginning of the line, holding the ring at the end of
the chain in his hand, while the leader drags forward the
chain by the other end of it, till it is stretched straight, and
the leader directed by the follower, by moving liis hand, to
the riglit or left, till the follower see him exactly in a line
with the mark or direction to be measured to ; then both of
them holding the chain level and stretched, the leader
sticks an arrow upright in the ground, as a mark for tho
follower to come to, and advances another chain forward,
being directed in his position by the follower standing a
ihe arrow, as before, as also by himself, now and at every
succeeding chain's length, by mo>'ing himself from side to
side, till the follower and back-mark be in a rllract line
LAND 8UKVKY1N0. 211
Huviiifr then stretched the chain, and Btuofc iTown an arrow,
8.S betore, the follower takes up tlie arrow, und ' '
proceed till the 10 arrows are employed, or in tin-
the follower, mid the leader without an arrow, is nrrivi-d nt
the end of the eleventh ehain-liii;rth. Tj-.e follower tlio:.
sends or brings the 10 arrows to the Itiulir, who puts our of
them do^vn at the end of his chain, and advances with hin
chain, as before. And thus the arrows are clmn;,'c<l from
one to the other at every 10 chains' hn^nh, till tl.e whoU
line is finished, if it exceed 10 chains; and the aun;bcr of
changes shows how many times Id chains the liue (^>ntuin^,
to which the follower adds the arrows he hi)lds in his liand.
and the number of links of another chain over to tlu' mark
or end of the line. Thus, if tlic whole line measure 30
diains 45 links, or 3G45 links, the arrows ha>x' boon cliunpc*!
three times, the follower will have 5 arrows in Ms hand, tlie
leatler 4, and it will be 45 links from the last arrow, to be
■^tiken up by the follower, tn the end of the line.
In works on Surveying, it u usual to describe the ».... u.^
instruments used in the arL The pupil, however, will U*st
learn the use of these instruments when actually cngay;cd ia
the practice. The chief instruments employed are t' ' '.n,
the plain table, the theo<h)lite, the cross, the circui. .r,
the olfset sUiff, the perambulator, used in measuring road>,
and other great distances.
Levels, with telescopic or other sights, arc used to !'.• <1 tiic
/evels between two or more places, or how much one place is
higher or lower than the other.
Besides all these, various scales are used in protracting and
meiLsuring on paper; such as f)!uiic scalers, line of chords,
protractor, compas,ses, reducing hcidis, parullel and jn-rfH-n-
dicular rulers, &c.
TUB FIELU UOOK.
In surveying with the plain table, a field book ut uol
required, as every thing is drawn ou the table immediately
whrn it Is measured. But when the theodolite, or any • ''
instrument is us*:^}, some sort of a held IxKik is u.««(l i .
232 ULNT) SURVEYING.
to register all that is done relative to the survey in hand.
This book every one contrives and rules as he thinks fit. It
is, however, iisually divided into three columns. The middle
column contains the different distances on the chain-line,
angle«, bearings, &c., and the columns on the right and left
are for the offsets on the right and left, which are set against
their corresponding distances in the middle coliunn ; as also
for such remarks as may occur, and may be proper to note
in drawing the plan ; such as houses, ponds, castles, churches,
rivers, trees, &c., &c.
But in smaller surveys, an excellent way of setting down
the work is, to draw by the eye, on a piece of paper, a figure
resombling that which is to be measured ; and then write
the dimensions, as they are fotmd, against the corresponding
parts of the figure. This method may be practised even in
larger surveys, and is far superior to any other at present
practised. A specimen of this plan will be seen further on.
LAKO bURVETINO.
S33
FORM OP THE FIELD-BOOK.
Olhetf ind rvnurki on th«
left.
SUUont,
BMringi, and
OiMADM*.
1
OSVu tad rvaarU mi Ika
a 1
104' 25'
00
Cross a hedee 24
, a brook 30
67
Brown's bam.
120
734
1
1
954
Tree.
1
736
67 stile.
82
62*25'
00
House comer 61
40
67
j Foot-path 15
84
95
467
44
1
14 Spring.
1
976
D 6
54" 17'
62
30 PoDd
124
Clayton's hedge 24
630
7G7
•
767
30 Stil«.
•
305
760
^*'
«2
234 LAND SDRVEYINO.
In this form of a field-book D 1 is the first station, where
the angle or bearing is 1 04° 25'. On the left, at 67 links in
the distance or principal line, is an offset of 24 ; and at 120
an offset of 30 to a brook on the right; at 67 Brown's barn
is situated ; at 954 is an offset of 20 to a tree, and at 736 an
offset to a stile.
And so on for the other stations.
A line is drawn under the work, at the end of every station,
to prevent confusion.
PROBLEM II.
To take angles and bearings.
Let it be required to take the bear-
•j3gs of the two objects B, C, from the
station A.
In this problem it is required to
■neasure the angle at A, formed by
two lines, passing from the station A, through two objects
B and C.
1. By measurement with the chain, 3fc,
Measure with the chain any distance along the two lines
A B, AC, as A 6, A c ; then measure the distance be; and
this being done, transfer the tliree sides of the triangle Abe
to paper, on which measure the angle c A b, as in Problem
XV., Practical Geometrj.
2. With the m.agnetic needle and compass.
Turn the instrument, or compass, so that the north end
of the needle may point to the flower-de-luce. Then direct
the sights to a mark at B, noting the degrees cut by the
needle. Next direct the sights to another mark at C, noting
the degrees cut by the needle as before. Then their sura or
difference, as the case may be, will give the number of degrees
in the angle CAB.
LAND srilVKVIXO. o<«<
3. IVith the theodolite, c^r.
Direct the fixed sights along tlie line A 15, hv tn"" ■' tlie
rnstrument ubout till you see the mark 11 throiigfi t! Iitii,
ami in that position srrow the instnuncnt fast. Then turn the
moveable index about till, through its sight?, jou see tliv
other mark C. Then the degrees cut by the index, on the
graduated limb or ring of the ir strumcnt, show the nuiubei
of degrees in the nnglt- C A R.
A. fl'ith the plain table.
Having covered the table with paner, and fixeil it on it*
stand, plant it at the station A, and fix a fine pin, or a point
of the compass in a pr(.M>ei' point of the paper, to repn-sont
the station A. Close by the side of this pin, lav the tiduciol
edge of the index, and turn it about, still toiiciiing the pin,
tiC one object 15 can be seen through the sights ; then by th«
fiducial edge of the index draw a line. Hy a sintilur procesi
draw another line in the direction of the object C. And it xr
done.
PHOBLKM HI.
To measure the aJTsi ts.
Let A i r fl e f'g be a crooked hedge, river, or brook, &«.,
and A G a biise line.
Begin at the point A, and measure towards G ; and nheu
you come opposite any of the comers bed, &c., which i*
/ah.
'.A^^) SURVEf ING.
aicertained by means of the cross-staff, measure the offsets
B 6, C c, D rf, fcc, with the chain, and register the dimen-
sion, as in the annexed field-book.
FIELD BOOK.
91
785 = A G.
■
57
634
98
510
70
340
84
220
62
45
D A go North.
Offsets
Base line A G, or
Offsets
Left.
D Station.
Right.
To lay dotvn the plan.
Draw the line A G of an indefinite length ; then by u
diagonal scale, set off A B equal to 45 links ; at B erect the
perpendicular B b equal to 62 Hnks taken from the same
scale. Next set off A C equal to 220 links, or 2 chains 20
links, and at C erect the perpendicular C c, equal to 84
hnks ; in the same way set off A D equal to 340 links, or 3
chains 40 links, and at D erect the perpendicular D d equal
to 70 links. Proceed in a similar manner with the remaining
offsets, and straight lines joining the points A, b, c, d, e, &c.,
will complete the figure.
To Jind the content.
Some authors direct to add up all the perpendiculars B 6,
C c, &c., and divide their sum by the nmnber of them, then
multiply the quotient by the length A G. This method,
however, should never be used, except when the offjets B
C c, &c., are equally tlistaut from each other.
L.AND SUKVEYIXO.
237
VN'hen the ofTscts are not tsjually distant frrtm ^^ch other,
which indeed is generally tlie ca:>e, this methixl is erruncoiu ;
therefore the foUomn^ method ouglit to be employed.
Find the content ot the space A U 6 as a triangle, by Pro-
blem v., Section II. Find the contents of the figures 15 Crb,
C D d c, &c., OS trapezoids, by I'roblem XIII., Section II.,
the sum of all these separate results will be »ho content of
the figure A G gfe d c b A..
The actual calculation is as follows :
CALCULATION.
AB
B^
= 46
r= 62
A c = 220
AB= 46
1 r
AD = 340[ab = 610 AF = 634
AC =z 220 AD := 340 AB = 5J0
CD = 12o'db =r 170 tr = 124
AO = 7^'
Af = (i:n
or ^ 1^1
r/ = 67
ag CD 91
Snm 14H 1
r« = 161 1
1
!
90
270
2790
Bcz= 175
B^= 62
cc=: 84
Sam 146
Bc= 176
ce = 84 Drf = ;o E* = 9S
Ddz= 70 r.e = 98 r/= 67
Snm 154| Sum 168 Sam 165
CD = 120 Di =9 170 tr = 124
I
»rod. 25660
18480 38660 19220
2334 s
These respective product^s are cndcntly double the true
(iontents of the respective figures A H b, 15 C < /, (' I) t/c,
iJk:., that ifi,
2790 = double area of A B o.
25550 = double area of U C c b.
18480 = double area of C D d c.
28500 = double an-a of D E «f d.
1 9220 = double area of E V f e.
22:J 18 = double ar-a of F G gf.
2) 11 69 « 8 = double area of the whole in square lixJu.
58474 = area in square links.
•58474 = area in acres = Oa.. 2r., l3-55S4r.
238
LA..\D SUKVEVING.
2. Required the plan and content of part of a field, from
the fbli owing field-book : —
AC 45
62 C li
A d 220
84 d %
A e 340
70 e k
A/510
88 f I
A ^634
51 g m
AB785
91 B n
Ans. Oa., 2ii., 12p.
C cl e F 9
PROBLEM IV.
To measure a field of a triangnlar form.
1. B^ the chain.
Set up marks at the three cor-
ners A, B, C, and measure witli
the chain, the distance A 1), D being
the point at whi(;h a perpendicular
doniitted from C, would meet the
line A B ; measure also the distance
D B ; hence you have the measure
of A B. Next measure the perpen-
•licular D C ; tlien from the two
dimensions A B and D C, the content may be found bv
IVoblem IV., Section II. ^
r.ANO STTRVETINO.
'tKf
Ut A D = 794, A B = 1321, D C = 8'J6 ink*.
I. Til X H2G + 2 = 54.X.73 links.
Then. 545.')T3 + lOdOOO = 5-4.'j.'i7:' ncr A-
45573 X 4 = l-8*J-21i2 roods.
82292 X 40 = 32-9 H580 perches.
Hriice the answer Is 5a., 1 rt., 33p., nearly.
L. Wliat Is tlie area of n tnaii;^ular field, whose bam- ii
I2'2.j chains, and height 8*5 chains?
Alts. 5a., Or., 33r.
2. lit/ taking one or more of the atigUt.
I^leasure two sides A B, A C, and the anplto A. included
between them ; then half the continual prtKhiot of the two
sides, and the natural sine of the lontaim-d aiiL^Ii' will ^ivr
the area.*
Or, me.'isiire the two angKs A and 1>, and the ndjuei'iit
side A B, from whieli the fig-nre n.av be planned, ai.d the
>erj)endieular C D found, whi;h perpendicular being multi-
plied by half the base A I^ will give the area. Or by snca-
>uriiig the three sides of th« ♦nangle, its area may be found
l>y Pioblem V., Section II.
PKOBLKM \
1. iiy the chatn.
To surret/ a four-tidfa Jitld.
Meastire the dijxgonal A C, and, as before dirrcto<l. mo.n-
3Ure the perpendiculars D I" and 1? I'; then the area of each
See Appendix. n«nionjU«tioa II.
240
LAND SURVEYING.
of the triangles A B C, A D C may be found, as in the Last
problem, and both areas being added together, will give the
content of the foxir-sided figure A B C D.
1. Let A C = 592, D E = 210, B F = 306 links.
592 X 210 = 124320 double area of A D C.
592 X 306 = 181152 double area of A B C.
2)305472 double area of A B C D.
1-52736 = area of A B C D.
4
2-10944
40
4-37760
Hence 1a., 2r., 4p., the answer.
2. By taking one or more of trui angles.
Measure the diagonal A C, also the sides A D and A b.
^Jext measure the angles D A C and BAG: then the area
of each of the triangles ABC and ADC maj be found bj
case 2, last problem.
2. Required the plan and content of a field by the follow-
ing field-book :
riELD-BOOK.
342
1360 = A B.
1190
600
a D go East.
1
625
Offsets
Left.
Station n , Offsets
or base line. Right.
An$. 6a., 2r., 12p.
L.OCD snnvEYixo.
24 1
3. How many acres are there in a four-«itle<l field, wIm^.
diac^onal is 475 chains, and the two perpcndiculnn fall:: .'
on it, from its opposite angles, 2-25 niul 3-^) chain*.
t'lyeh? Ans. Ia., l!L- 1. ..,.
PROBLEM VI.
To survey a Jield of many sides by the chain only.
Let ABCDEFGbe the field wnose content is rc<juire*l.
Set up marks at the corners of the field, if there be uotio
there naturally. Consider how the field may be bent dirided
into trapeziums and triangles; measure them senaratoly, a*
in the two last problems; and the sum of all the separate
results will give the aro:i of the whole field.
In this way of measuring with the chain, the field thouta
be divided into trnpeziums and triangles by dm-
nals from comer to comer, so that all the ptij «
may be within the figure.
The last figure is divided into two trancxiums A H C O,
G D E F, and the triangle G C I). In the fip<t tr
measure the diaLronal A C, nud the two pcriH-ndicul... ■ ■ • *
and B n. In the triangle G C D, mea.«;ure the ha»e G C,
an<l the perpendicular f) g. Fmally, measure the diagonal
242
LiJST) SURVKfENG.
F D, and the two perpeudictilars G o and E p. Having
drawn a rough figure resemhling the field, set all these mea-
sures against the corresponding parts of the figure. Or «et
them down thus :
A m 135
A n 4J5
AC 550
1 130
m G
180 n B
C&44o}230^I^
Fo 206^ ,^^ ^ p
Vp 288 \^Zl%
Fa520j ^^P^
CALCULATION.
130 + 180 = 310, 550 + 2 = 275'
275 X 310 = 85250 = A B C G.
440 X 230 + 2 = 50G00 ="
C G D.
120 + 80 = 200, 520 ^ 2 = 2Go,
2G0 X 200 = 52000 = D E F G.
1-87850 =ABCDEFG.
4
3-51400
40
20-56000
1a., 3r., 20-5Gp., answer.
Other methods will naturally present themselves to an
ingenious practitioner who has read the preceding part of
this work, or who has heen previously actjuainted with the
principles of Mathematics. Every surveyor ought to be well
acquainted with Plane Geometry at least. This, with a
knowledge of Trigonometry, would be sufficient for the pur-
pose of most surveyors.
The content of the last hgure may be found by measuring
the sides A B, B C, C D, D E, E F, F G, G A ; and the
diagonals A C, C G, G D, D F, by which the figure is
divided into triangles, the content of each of which may be
foimd by Problem V.. Section II.
LASD SCKVETtjro.
243
t2. Required the plan and content of a field o an .r,...,!...
Jormfrom the fullowinjr '
FIELD BOOK.
I 900 = EG
268 550
D E,goS.W.
1
280
1 1100 = HE
790
350
D H, go East.
1
1
410
1
1
, ^— —
140
1180= CH
710
350
D C, go S VV.
280
1
200
900 = AC.
430
300
D A,goS.E.
450
Offsets
Left.
1
Stations, a , or
Base Lines.
—
Ofistta
Right.
AiiJ. IOa., In., 24r>4p.
PKOnLKM VII.
To surri'j/ a Jirld with the theoduiitt, !fc.
1. I't'-iii <iitr ji'.i.ii '>r atation.
VtHien all the oxiglea can be seen from ouc putuL, u luip
pose C.
244
UkKD SURVEYINQ.
Having placed the instrument, at C, turn
it about till, through the fixed sights, the
mark B may be seen. Fixing the instru- .
ment in thia position, turn the moveable ■*
index about, till the mark A is seen through
the sights, and note the degrees on the in-
strument. In the same mauoer, turn the
index successively to the angles E and D, taking care to
note the degrees cut off at each ; by which you have all the
angles, viz., B C A, B C E, B C D. Now, having obtained
the angles, measure the lines C B, C A, C E, C D ; entering
the respective measures against the corresponding part of a
rough figure, drawn to resemble the figure.
2. By going round the Jield.
Set up marks at B, C, D, &c. Place the instrument at
the point A, and turn it about till the fixed index be ia the
direction A B, and then screw it fast : turn the moveable
mdex in the direction A F, and the degrees cut off will U
the angle A ; next measure A B, and planting the instrument
at B, measm-e, as before, the angle B ; measure the line
B C, and the angle C : and so proceed round the figure,
always meaaiuing the side aa you go along, as also the
angles.
The 32nd Proposition of the Ist Book of Euclid affords an
easy method of proving the vork : thus, add all the internal
LANS BUK>'EY1X0. f4j
angles, A, B, C, &c., ot" the figure together, ami tlu ir ium
raiLit be equal K) twice as iirniiy right angles as the fi^-ure
has sides, wautinj^ four right angle*. But when the figure
hiis a re-enterant angle aa F, meaaure the ext*nud an^'Ie,
vvhich is less than two right-angles, and deduct it from foul
riglit-angles, or 260 degrees, the remainder will give the
internal angle, (if such it nmy be caJltxl,) which is greater
than ISO degrees.
When the field is surveyed from one station, at in tht
first case shown above, the content of the figure is found as
in the second case of Prob. IV., since we hare two sides and
the angle included between them in each triangle of ihf
Ggpire.
PROBLEM VIII.
To tur vey a Jield with crooked hedjfu.
Measure the length.s and positions of lines runnicg a.,
.^ear the sides of the field as you c&n ; and, in proceixling
along these lines, measure the offsets to the different cumen,
as before taught, and join the ends of the offsotn ; those
connecting lines will represent the required figure. When
the plane table is used, the plan will be truly represented on
the paiier which covers it. But when the survey is made
with the theodolite, or other instnunent, the different nua-
sures are to be noted in the field-botik, from which th«
sides and angles are laid down on a r.ii.p, after returning
from the field.
In survcnng the piece A B C D K F G H I K L .M, set up
marks at « E F j:. Begin at the station i, and measure the
lines J E, E ^, p J-, X *, as also their iK>«ition*, or the
angles E * i, * E F, K f x, and p t i ; aiKi in going along
the four-sided fit,'ure i E /> J, measure the offwrts at a, h, J,
g, k, I, m, as before tauglit. By means of tJie figure u K /' x,
and of the offsets, the groimd is easily planned.
When the principal lines are talin within the figurr, as
the above case, the contents of the eiterior jKirtion*
246
LAND S-(JRVETING.
5 C B A, C D E, &c., must be added to the area of the
quadralateral s x ¥ E. But when the principal lines are
tiiken outside the fij^ure, the portions included bet^Yeen them
and the boundaries of the field are to be deducted from th«i
content of the quadralateral, and the remainder will give tb*
'rue content of the field.
a B
When there are obstructions within the figure, such as
wrood, water, hills, &c., measure the lengths and positions of
the four-sided figure abed, taking care to measui-c the
offset>! from the dlffereat comers as .you £:o aloni^.
LASO SUUVKTINO.
247
PROHLEM IX-
To aurwy any yiece of land by twu # k/iomj.
Choose two stations, from wLIch all the ■ - of Um
groTUid can be net-n, if fxjhsible ; measure u... _. a.c* b*>
tweeu the stationa ; at each sUition tal^e tJ»e anj^le* (unaad
by every ohject, from the station line, or Ji.-.taiicc. Thaa
the Blation line, auJ thtse diilereut un^'lcu \h:iu^^ laid down
from a regular scale, and the exterual poinU of inienectiou
connected, the connecting linea will give the b«>ui:dArT,
The two stations may be taken within tlie boiif.riii, in one
ol the sides, or without the Injiindj of the ground to I*
surveyed-
Let m and n be two stations, from which all the mark*
A, H, C, Sic, can bo seen, phmt the ixutrumenl at fi, and by
it, measure tlio angles A m n, IJ tn n, C m n^ Lo. Next mea-
sure m u, and planting the iiistrument at is, measure the angles
A fJ m, H n m, C n rn, &c. Tiiese obMTratloui 1 ".liiuetl.
the lines joining the pouits of eitorual iutcrscH :. ... .A give
itrue map of the ground- The Uiothod of Gnding the ouO
tcBt will bo shown further on-
The principal ubj©<:t8 on the gr Kind may b« d riinw it ^ «»
a
V48 LAND SUKVEIINO.
the map, by measuring- the angles at each station, which
every object makes with the station line m n. When all the
objects to be surveyed cannot be seen from two stations,
then three or four may be used, or as many as may be found
necessary ; taking care to measure the distance from one
station to another ; placing the instrument at CA'ery station,
and observing the angles formed by all the visible objects
with tlie respective station line ; then the intersection of the
lines forming these respective angles, will give the positions
of all the remarkable objects thus observed.
In this manner may very extensive surveys be taken; and
the positions of hills, rivers, coasts, &c., ascertained.
PROBLEM X.
To survey a large estate.
The following method of survejang a large estate was first
given by Emerson, in his siu-veying, page 47. It has been
followed bj Uutton and Keith.
Wh en the estate is very large, and contains a great num-
ber of fields, it cannot be accurately surveyed and planned
by measuring each field separately, and then adding all the
Sf'para te results together ; nor by taking all the angles, and
measuring the boundaries that enclose it. For in these
cases the small errors will be so multi^jlied as to render it
very much distorted.
1 . WaHc over the estate two or three times, in order to
got a p erfect idea of its figure. And to helj) your memory,
nake a rough draft of it on paper, inserting the names of the
<liderent fields within it, and noting down the principal
I'n'pcts.
2. Choose two or more elevated places in the estate, i(n
vour stations, from wi^'oh you can see all the principal parti
of it and let tl.ese stations be as far distant frona aacli other
I
LAMJ BURVLYINU. 949
a.' possible, a* the fewer stations tou huTe to comiiiacd Ui«
whole, the mure eTuct the work will be.
Ill selecting Ibe iHationi, care sbould be titkon that the
lines which coimfci them may nin along the boimtlaries of
the estate, or gomp of the h««dges, to wliich ofiWu majr be
taken when necossurj.
3. Take such angles, Wlween the 8ta»ion», u yju ihiak
necessary, aid measure tlie distance from btatiou to •tittioI^
aJvvays in a right line; those things must be done till you
get as many lines and angles &a are Bullicient for deli-r-
mining all the st/ition poinui. In niea«uring anv of thcM
st;ition distances, niiirk accurately where these line* meet
with any hedges, ditcht-s, romis, lanes, paths, rivulets, tc,
and where any remarkable object is placed, by mea.-tiring ita
distance from the station line; and where a {leriK-ndicular
from it cuts that line ; and always mind, in any of thf»e ob-
servations, that you be iu a right line, which you may easily
know by taking a back-sight and fore-sight, along the station
line. In going along any main station line, take (>!T*etji to
the ends of all hedged, and to a:vy pond, ho'i<c. mill, bridge,
fcc, omitting nothing that is remarkable All these things
i>ust be noted down; for these are the data by which ti»e
places of such objects are to be dtttrniincd on the plan.
Be careful to set up marks at the intersections of all
hedges with the station lino, that you nuty know wbero t«
measure from when vou come to survey the {mrticulor lieKU
that are crossed bv this line.
These fields must be measured as soon as you hate eonv-
pleted your station-line whilst they ore frrsh in your m©
mory. In this manner all the sUition lines must be meA«ur«J,
and the situations of all adjarent objects dctemiinrd It will
be proper to lay do\>n the work on paper e»cry r.:ght, that
yoa may see how you go on.
4. With respect to the internal parts of the estate, thry
n>'ist be determined by new station lines ; for, after the
imiin stations are determined, and every thing adjuinioir to
tJii'ui, then the estrte must be ■uUiirided into two or ihrte
250 LAXD SURVEYTSO.
parts by new station lines : taking the inner stations at proper
places, where you can have the best view. Measure these
station lines as you did the first, and all their intersections
with hedges, ditches, roads, &c., also take offsets to the
bends of hedges, and to such objects as appear near these
lines. Tht-n proceed to siirvey the adjoining fields, by taking
the angles which the sides make \\nth the station line at the
intersections, and measuring the distances to each corner
from these ir+ersections ; for, every station line wiU be a
basis to all future operations, the situation of every object
being entirely dependent ontnem; and therefore they should
lie taken of as great length as possible; and it is best for
them to run along some of the hedges or boundaries of one
or more fields, or to pass througii some of their angles.
All things being determined for these stations, you must
take more inner stations, and continue to divide and sub-
divide, till at la^t you come to single fields; repeating the
same work for the inner stations as for the outer ones, till
the whole is finished. The oftener you close yoiu- work,
and the fewer lines you make use of, the less you will I e
liable to error.
5. An estate may be so situated that the whole cannot be
sm-veyed together, because one part of the estate may not
be seen from another. In this case you may divide it into
three or four parts, and survey these parts separately, as if
dhey were lands, belonging to different persons, and at last
ioiu them t^)gether.
6. As it is necessary to protract or lay down the work as
you proceed in it, you must have a scale of due length to do
it by. To get such a scale, me;isure the whole length of
the estate in chains ; then consider how many inches long
the map is to be ; and from these you will know bow many
diains you must have in an inch ; then make your scale
accordingly or choose one already made.
7. The trees in every hedge-row may be placed in their
pioper situation, which is soon done by the plane table ; but
may be done by the eye without an instrument ; and bfliujf
LA.VU SLIINTSTIXU. 151
thus taken hy puoss in a rongh draft, ihojr will be rsort
enough, bciii;^ only to l(x>k at ; except it Iw cuch bm are at
any rcniarkabio plnccs, ns at tbe eniU of hc«I;;««, at •tllr«,
gates, &c., and these must be nic<u>urt>4l or taken with fK*-
plane tuble, or some other instrument. Hut all thi5 ni-ol
not be done till the draft is finished. And ubM-nre, in all
hedges, what side the gutter or ditch is on, and to nhuni tb«
fence belongs.
PUOHLKM XI.
To sm-rr^ a li-un or cj." /.
To survey a town or city, it will I* proper to hare an
instrument for taking angles, such ns a tluiHloIite or plane
table; the latter is a very convenient instrument, \- * •■ 'h*
minute parts may be drawn upon it on the sjKit. .\ ul
oO feet long, divided into 50 links, will be nioro convenient
than the common surveying chain, and nn off?. • • ''■ of lO
tV-et long will be very useful. IJegin at the i; ■•( two
or more of the principal streets, through which you can hav«
tlie longest prospects, to get the longi>t st;,' ' " ' • re
having fixed tlie instruments, draw l!ne^ < : ■ g
tiiese streets, using two men as marks, or ikiIc* $et in wnodro
pedestals, or perhaps some remarkable pfaet'S in the Iioum-*
at the farther ends, as windows, door«, corners, &c. Mea-
sure these lines with the chain, taking otTscts with the •taff,
at all corners of streets, l>endiiig8, or winding*, and to aU
-emarkable objects, as churche**, niarketjs hn"-, •■■".•?»,
eminent buildings, &c. Then remove the i to
another station, along one of these lines, and there n-iieal
rhe same process as before. And »o continue until the whole
is finished.
Thus, fix the instrument at A, end draw line* in the direc-
tions of all the streets meeting there ; tlicu tneajure A C,
noting the street" nt T. At the second itation (', draw th*
directions of flU the streits meeting there : mea»tirr from C
to D, noting the place of the »tii'« t K, ut you ;.4.%* bj it, A«
252
LAND SUH\'^riNG.
B/>1
'^^r^^r'^^'=^-^
the third station D, take the direction of all the streets
nieetiiig' there, and measure D S, noting the cross street at T.
Proceed in like manner through all the principal streets, aftei
ft'kich proceed to the smaller and intermediate streets ; and
.ast of all to the lanes, alleys, courts, yards, and every other
place which it may be thought proper to represent in th#
plan.
PROBLEM XII.
To compute the content of anif survey.
I. In small and separate pieces, the method generally
employed is, to compute their contents from the measures of
the lines taken in surveying them, without drawing any
correct map of them : rules for this purpose have been given
in the preceding part of the work. But in large pieces, and
whole estates, corigifiting of a great number of fields, the
usual method is, to make an tmfinished but correct plan of
the whole, and from this plan, the boundaries of which
include the whole estate, compute the contents quite inde-
pendent of the measures of the Unes and angles that were
taken in Burveying. Di>'ide the plan of the survey into
fri&ngles and traxx'Ziuma, by drawing new lines through it :
LA.ND SLKVETINU. 263
me&xure all the bases and j>iTj)uii<iiculuj(i t-f all iho»« nev»
figures, by means of tl»e scale from whlcli lli.j plan waj
di'awn, antl from these dimensions compute th« coutcutii.
whether triangles, or tra|H-/.iuni3, bj the pro{)cr rules f.>r
finding the area.s of such figures.
The cnief dlfTicultj in comjiutlng consists in fin<ling tl.<-
con-tenta of land bounded hv curved or vorj irregular line*,
or Id reducing such crooked sides ur boundaries to straieiit
lines, that shall enclose an etjuaJ area with thoM cruoke<l
"sides, and eo obtain the area of the curved figxxre b^ means of
the right-lined one, which in general will Ik* u tru]>ezium.
The reduction of crooked sides to straiglit ones is ea«ilj
p)erionned, thus :
Apply a horse-hair or silk threa»l acroes the crooked •id*-*
in such a manner, that the small parts cut otT from the croi>iv< .1
figure by it, may be equal to those taken in. A little pruo
tice will enable you to exclude exaitly as much m jon
include ; then, with a pencil, draw a line along tlie thread,
or horse-hair. Do the same by the otlier sides of the figure,
and you will thus have the figure reduced to a 8truJght-»idwi
figure equal to the curved one: the content of wliicu, being
computed na before directed, will b« the content of tlw
cur\'ed figure proposed.
The best way of using the thread or horv-hair it, to •trin»»
a small slender bow with it, cither of whtdubooe or wirr,
which will keep it stretched.
If it were required to find the contenU or the follovr--.-
CTijtiked-sided figure; draw the four dotte*! str .light ..
A B, B C, C I), and D A, excluding as much from up
sunev as is taken in hy the straight lines; br which tlje
crooked figure is reducecl to n right-lined one, l>oth f.jual in
area. Then draw the diagonal U 1), which b«'ing njcasiirrJ
by a projier scile, and nniliiplifd by half tho nuu of the i>rr-
pendiculars let fall from A and C upon If I), MUc.wurtJ »wi
the same scale '■ill pve the area required-
254
L.1ND SUR\'EYtNO.
3:?^
B
Many other methods might have been given for computing
the contents of a survey, but they are omitted, the above
being, perhaps, the most expeditious.
255
MISCELLANEOCS PROl'.LEMS.
1. The tlirce sides of a triangle are 12, 20, and 28 , wlim
IS the area? Ann, 60 v^ 3.
2. Find the difiiTciice between the or-ja of a tr
sides are 'A, 4, and 5 feet; and tbe urea of an >,...». . .1
triangle having an eijual perimeter ?
Aiif. "U'iS of a »t]uare foot.
3. There is a segment of a sphere, tlu- diameter ot v'
base is 24 inches, and its altitude It) inches; re<|tjinii .:
iolidity ? Ans. 27S5'S5'2 inche?..
4. There is a hushel iu the form nf a cvlinder, whiw*- depth
is 8 inches, and breadth \>^l indies; required tod
the breadth of another cvlintrrica! vessel of the same c\.^ \
is the former, whose deiith la only 7^ inches?
Ans. 1 9' 107 inches
5. A ladder 40 feet l>>nq;, may be so planteil, that it -' .
reach a window 33 leet from the gr«)und on one »ide of ■
stri-et; and by ou\y turning it over, without moving the
out of its plac^, it will ilo the same bv a window 21 fevl lu^u
on the other side ; what is the breadth of the strr«-i ?
Am. 5l> fevt 7 j inches.
6. lu turning a one-horse chaise within a ring of a certain
diameter, it was observed that the outer whcid m
t\niis while the inner made but one; the wheel* w
4 feet high ; and supposing tin m fixed b.i the jtatu:
256 wibcellaneous problems.
tance of 5 feet asunder on the axle-tree, wnat vras the
circumference of the track described bv the outer wheel ?
Ans. 63 feet, nearly.
7. A cable wliich is 3 feet long, and 9 inches in compass,
weighs 221b? ; what will a fathom of that cable weigh, which
measures a foot about? Auo: 78|lbs.
8. How many solid cubes, a side of which equals 4
inches, may be cut out of a large cube,, vhose side is 8
Tiches? Ans. 8.
9. Determine the areas of an equilateral triangle, a square,
a hexagoa, the perimeter of each being 4U feet ?
Ans. 76-9b0035— 100— 115-47.
U). A person wants a cylindrical vessel 3 feet deep, that
shall contain twice as much as another cylindrical vessel
whose diameter is 3^ feet, and altitude 5 feet; find tht-
diameter of the required vessel ? ^n*. G'3y feet.
11. Three persons having bought a conical sugar-loaf,
tvish to di s-ide it into three equal parts by sections paralW
to the base ; it is re(juired to find the altitude of each person
si>are, the uUitude of the loaf being 20 inches ?
Afis. Altitude of the upper part = 13"867, of the middlt
part = 3'604, of the lower part 2*528 iiiches.
12. There is a frustum of a p}Tainid, whose bases are
regular octagons; each side of the greater base is 21 inches,
and each side of the less base 9 inches, and its perpendlculai
length 15 feet, how many solid feet are contained in it?
Ans. 119-2 feet.
13. Rpqiiiring to find the height of a May-pole, I pro-
cured a staffs feet in length, and placing it in the sunsliine,
perpendicular to the liorizon, I found its shadow to be 4*1
feet. Next I measured the shadow of the May-pole, whioli
I found to be fi5 fc'Ct; from this data the height of the polt
is requlre'^ ' -^ Ans. 79*26 feet.
14. Givc-iA tyyiiy sides of an obtuse-angled triangle, which
are 20 aud 40 poles ; required the third side, that the tri«
aiigle may con tain just an acre of lanfl ?
Ans. 58-876 or 23-099.
MISCELLANEOL'i. rUOULtMS. * 25/
15. A cirtnlar fi.sh-ponJ is to bo made in a panlcn, fiaf
shall take up just hult" uu acre ; what imi.st he tlic Kn-th tii
the chord that strikes the circle? Ans. '27 ^ vard*.
16. A gentleman has a garden lOO (cet lonjf, and 80 fi-ct
l)road ; now a f,'ravcl walk is to ho made of an c<ju.d width all
rouml it; what must the hriinlih of the walk In-, to take uii
just half the ground? Ana. l2*Jtmj fccu
17. A silver cup, in form of a frustum of a cnno, who**
top diameter is 3 inches, its b ittom diamotcr 4, a!>d it# alti-
tude 6 inches, being filled with liouor, a person drank out if
it till he could see the middle of tl»e bottom ; it is required i«
Hnd how much he drank? Am. • 1 5-121 ale gallons.
IS. I have a right cone, which cost me £5 l.li. 7r/., at
lOs. a cubic foot, the diameter of its base being t<>
as 5 to 8 ; ami would have its convex surface diN .
same ratio, by a plane parallel to the base ; tJio «p[»fr part to
be the greater; required the slant luigh; «)f each piUt ?
J j 3'950<i486, the slant heiglit of tlie upi>tr part.
( 1 •08546 12, the tlaut height of the under pan.
19. HoMT many acres of the earth's surface may be »e«n
from the top of a steeple whose height is 4U0 feet, t'l •
being suj)posed to be a perfect sphere, whose circuudt r. : . . .;
25000 miles ? Anji. 1 2 1 'JOyb ! -338267 1 1 2 acre*.
20. Two boys meeting at a farm-house, had a tankard nl
milk set down to them ; the (me being very thirst;
he could see the centre of the Ujttom of the ta;
other drank the rest. Now, if we supinwc that ll»e ludk
cost 4^(/., and the tankard measiued 4 iiiches diai:
top and bottom, and G inches in depth; it is i - ■
know what each boy had to pay, projK>rl»»>nable to the tiuU)-
tlty of miilk he drank.
\ 14-IS02S15 far-''- ' •' ' •'■ '"• '.
^"*-{ 3SU)71S5 Hu . iuL
21. If the linear side of a certain cube, bo incrr»»cd ont
uich, the surface of the cube will he incrvased 24 G **\MMn
inches : determine the side of the cube.
Am. 20 inch**.
258 MISCELLANEOUS PKOBLEMft.
22. If from a piece of tin, in the form of a sector of a
circle, whose radius is 30 inches, and the length of its arc
3G inches, be cut another sector whose radius is 20 inches;
and if then the remaining frustum be rolled up so as to fonr
the frustum of a cone; it is required to find its content,
supposing one-eighth of an inch to be allowed off its slant
height for the bottom, and the same allowance of the circum-
ference, of both top and bottom, for what the sides fold over
each other, in order to their being soldered together ?
Ans. 685*3203 cubic inches.
23. Three men bought a grinding-stone of 40 inches dia-
meter, which cost 20*., of which svun the first man paid 9*.,
the second 6s., and the third 5*., how much of the stone must
each man grind down, proportionably to the money he paid ?
A71S. The first man must grind down 5*1C7G03 inches oi
the radius; the second 4*832397 inches, and the third 10
inches.
24. There is a frustum of a cone, whose so'id content is 20
eet, and its length 12 feet ; the greater diameter is to the
ess as 5 to 2 ; what •»•'» the diameters ?
. (2*02012 feet.
^"*- \ -80804 feet.
25. A tarraer borrowed of his neighbour part of a hiiy-
dck, which measured 6 feet in length, breadth, and thick-
ness ; at the next hay-time he paid back two equal cubiciil
pieces, each side of which was 4 feet. lias the debt been
discliarged ? Ans. No ; 8S cubic feet are due.
26. There is a bowl in form of the segment of an oblong
spheroid, whose axes are to each other in the proportion of
3 to 4, the depth of the bowl one-fourth of the whole trans-
verse axis, and the diameter of its top 20 inches ; it is re-
quired to determine what number of glasses a company of
10 persons would have in the contents of it, when filled,
using a conical glass, whose depth is 2 inches, and the dia-
meter of its top an inch and a half?
Alls. 1 14*0444976 glasses each,
27. If a cubical foot of brass were to be drawn into wire,
MISCELLAXEOUS I'KOULEMS. 26^
of ^(^ of ail incli in diiinirtcr ; it is ri-<jtilri'«l to determine th#
longth of the said wirt-, allowing no losa in the metal ?
Am. !>'>^ miles.
28. How many slh)t ar<» tlicre in un ti- " ' •
p.'lc, the k'liglh 'ind hriadlh of whose ha.»e i ... ',
and the length and breadth of the highest courno beinp 2-1
and 6? An». \~'Mii\.
29. How many shot are tht-re in an c- '"■ ■ ' i « t . .
pile of 12 coiurses ; lonirth and breadth of i
40 and 10 shot resfK-ctively ? Aus. r»,t»o »hot.
30. Of what diameter must the bore of a canin';! bo in-'t
for a ball of 24 j)ound.s weigiit, so that the diamtter of tlt«
bore may be ^\^ of a-i inch more tlian that of the Imll ?
Aus. 5 '75 Toys inches.
31. What is the content of a tree, whose length i« 17^
feet, and which girts in five ditVerenl places m» fullown, viz..
in the first place U 43 feet, in the second 7 "92, in the thirJ
615, in the fourth 4-74, and the fifth 316?
An^. 4 2.') 195.
32. What three numbers will express the pruporlions »uK-
disting between the solidity of a sphere, that of the circum-
scribing cylinder, and circumscribing eijuilateml cone?
Aus. 4, G, 9.
33. Given the side of an eoailnteral triangle U>, it U rv-
uuired to 6nd the radii of its circumscribiug circle ?
Ans, o I 4 JO.
34. Given the perpendicular of a plane triangle 300, t)i«
sum of the two sides 1 1 50, and the ditVerencc of •
ment of the base 495; required the bjuoe and tlie ^,^v^.
Alts. 1(45, .175, and 780.
35. A side wall of a house is 30 feet high, oiid the oi>-
posite one 40, the roof forms a right angle, iit '
lengths of the riiftcrs are 10 feet and 12; the
ehorttr is placed on the higher wall, aiid rice rrrtd; retj-
the length of the upright, which supi^rtj the ndge ol ti.«
rwof, and the breatlth of the house?
Aus. 41-80.3, length ..f ..[light, and 12 feel ih* brwdlk
of tlie house.
260
A TABLE
OP THE AREAS OF THE SEGMENTS OF A CIRCLE,
Who!:e diameter is 1, and supposed to be divided into 1000 equal parti.
He'?ht.
Area Seg.
Height.
Area Seg. 1
Height.
Area Seg. Height.
Area Seg.
•001
•000042
•038
•009763
•075
•026761
112
•048262
•002
•000 119
•039
•010148
•076
•027289
113
•04Sb94
•003
•00(1219
•040
■010537
•077
•027821
114
•049528
•0114
.000;i37
•041
•010931
•078
•028356
115
•050165
•006
•000470
•042
•011330
•079
•028894
116
•050804
•OOG
•000618
•043
•011734
•080
•029435
117
•051446
•007
•000779
•044
•012142
•081
•029979 •
118
•052090
•oos
•000951
•045
•01-2554
•082
•030526
119
•052736
•009
•001135
•04P
•012971
•083
•031076
120
•053385
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