MENSURATION
AND
PRACTICAL GEOMETRY;
CONTAINING
TABLES OF WEIGHTS AND MEASURES, VULGAR AND
DECIMAL FRACTIONS,
MENSUKATION OF AEEAS, LIKES,
SURFACES, AND SOLIDS,
LENGTHS OF CIRCULAR ARCS, AREAS OF SEGMENTS AND
ZONES OF A CIRCLE, BOARD AND TIMBER MEASURE,
CENTRES OF GRAVITY, &c, &c.
TO WHICH 18 APPENDED A
TREATISE ON THE CARPENTER'S SLIDE-RULE AND
GAUGING.
By CHAS. H. HASWELL,
CIVIL AND MAEINE ENGINEER.
SECOND EDITION.
OF THE ^>X\
SITY )
y
NEW YORK:
HARPER & BROTHERS, PUBLISHERS,
FRANKLIN SQUARE.
18 6 3."
<Q.A4<
W
W
a
Entered, according to Act of Congress, in the year one thousand eight hundred
and sixty-three, by
HARPER & BROTHERS,
In the Clerk's Office of the District Court of the Southern District of New York.
TO
B. H. BARTOL, ESQ,,
A TRIBUTE TO AN EARLY AND ESTEEMED FRIEND OF
THE AUTHOE.
New Yoek, January* 1863.
103910
P K E F A C E.
The following work is designed for the use of Stu-
dents, Mechanics, and Engineers, and the purpose of the
author has been to present a full set of rules whereby
may be readily determined the Lines, Areas, Surfaces,
Solidities or Volumes, and Centres of Gravity of va-
rious Regular and Irregular Figures.
In the progress of the work, it was essayed to pre-
sent a familiar rule for the surface, volume, and centre
of gravity of every figure, but, in consequence of the im-
practicability of reducing the rules in some cases to a
formula at all consistent with the design of the work,
such cases as presented this difficulty were abandoned :
as it occurs, however, the number of them has been
very essentially reduced by the aid and advice of Pro-
fessors A. E. Church, U. S. Military Academy, West
Point, and G. B. Docharty, N. Y. Free Academy.
In consequence of the great number of works on
Mensuration that have been submitted to the public,
the inference is a fair one that the author, in this case,
has not presented any novelties whereby he may antici-
pate any particular notice or attention ; he trusts, how-
ever, that a reference to the result of his labors will
show that, in the essential points of the extent of the
figures submitted, as well in their number as variety of
VI PREFACE.
section, and in the introduction of rules for determining
their centres of gravity, he has submitted some features
of so new and useful a purpose as to entitle him to the
attention of those upon whom he confidently relies for
patronage.
New York, April 1st, 1858.
CONTENTS,
Page
Explanation op Characters 9
MEASURES AND WEIGHTS.
Measures of Length 11
Measures of Surface 12
Measures of Capacity 14
Measures of Solidity 15
Measures of Weight 15
Miscellaneous Measures 16
Measure of Value IT
Ancient Measures IT
VULGAR FRACTIONS. 19
Reduction of Vulgar Fractions 20
Addition of Vulgar Fractions 25
Subtraction of Vulgar Fractions 26
Multiplication of Vulgar Fractions. . 26
Division of Vulgar Fractions 2T
Rule of Tliree in Vulgar Fractions . . 2T
DECIMALS. 28
Addition of Decimals 28
Subtraction of Decimals 29
Multiplication of Decimals 29
Division of Decimals 31
Reduction of Decimals 32
Rule of Three in Decimals S5
DUODECIMALS 36
Involution 3S
Evolution 39
To extract the Square Root 39
Square Roots of Vulgar Fractions. . . 40
To extract the Cube Root, etc 41
Properties of Numbers 44
GEOMETRY
Definitions 45
MENSURATION OF AREAS, LINES,
AND SURFACES.
Parallelograms (Square, Rectangle,
Rhombus, Rhomboid, and Gno-
mon) 4T-49
Page
Triangles 50
Trapeziums and Trapezoids 5T
Polygons 59
Regular Polygons 59
Irregular Figures 6T
Circle 68
Useful Factors Tl
Circular Arc T2
Section of a Circle 81
Segment of a Circle 84
Sphere 88
Segment of a Sphere 89
Spherical Zone 90
Spheroids or Ellipsoids 91
Circular Zone 95
Cylinder 93
Prisms 99
Wedges 100
Trismoids 101
Ungulas 102
Lune 10T
Cycloid 108
Rings, Circular 109
Cylindrical 110
Cones 112
Pyramids 114
Helix (Screw). 116
Spirals HT
Spindles 120
Circular Spindle 120
Elliptic, Parabolic, and Hyperbolic . . 124
Ellipsoid, Paraboloid, and Hyper-
boloid op Revolution 124
Ellipsoid 125
Paraboloid 126
Any Figure op Revolution 12T
Vlll
CONTENTS.
Page
Lengths of Circular Arcs 130
Areas of Segments of a Circle 134
Areas of Zones of a Circle 139
Promiscuous Examples 144
MENSURATION OF SOLIDS.
Cube , 155
Parallelopipedon 156
Frisms, Prismoids, and Wedges . . . 156
Regular Bodies 160
Tetrahedron 161
Hexahedron, Octahedron 162
Dodecahedron, Icosahedron 163
Regular Bodies 164
Cylinder 165
Cone 166
Pyramid 168
Spherical Pyramid 171
Cylindrical Ungulas 1T1
Sphere 176
Segment of a Sphere 177
Spherical Zone 178
Spheroids (Ellipsoids) 179
Segments of Spheroids 180
Frustrums of Spheroids '. . 1S2
Cylindrical Ring . 184
Links, Elongated or Elliptical 185
Spherical Sector 187
Spindles 188
Circular Spindle 188
Frustrum or Zone of Circular Spindle 189
Segment of Circular Spindle 191
Cycloidal Spindle 192
Elliptic Spindle : 193
Parabolic Spindle 196
Hyperbolic Spindle 200
Ellipsoid, Paraboloid, and Hyper-
boloid of Revolution (Conoids). . 202
Ellipsoid of Revolution (Spheroid) . . 203
Pag«
Paraboloid of Revolution 203
Hyperboloid of Revolution 206
Any Figure of Revolution .... 209
Irregular Body .'. 210
Promiscuous Examples 211
CONIC SECTIONS.
Definitions 228
Conoids 233
Ellipse 234
Parabola 240
Parabolic Curve 24?
Area of Parabola 244
Hyperbola 246
Arc of Hyperbola 250
Conic Ungulas 253
Regular Polygons 258
Table of Units for Elements 261
Mensuration of Regular Bodies or
Polyhedrons 262
Tables of Units for Elements 267
Orophoids :
Domes, Arched and Vaulted Roofs. . . 268
Saloons 269
Board and Timber Measure 276
APPENDIX.
Mensuration of Surfaces 278
Centres of Gravity of Surfaces . . 281
Mensuration of Solids 285
Solids of Revolution 286
Centres of Gravity of Solids 292
Sjjherical Triangles, Pyramids, Epi-
cycloids, Heli-coids, &c, &c 296
Carpenters' Slide-Rule 297
Gauging 307
Ullaging of Casks 319
EXPLANATIONS OF CHARACTERS
Used in the following Calculations, etc., etc.
Equal to, as 12 inches=l foot, or 8x8 = 16x4.
Plus, or more, signifies addition; as 4 + 6+5 = 15.
Minus, or less, signifies subtraction; as 15 — 5 = 10.
Multiplied by, or into, signifies multiplication ; as
8x9=72. axd, a.d, or ad, also signify that a
is to be multiplied by d.
Divided by, signifies division; as 72-j-9=8.
Is to, I
o • I signifies Proportion, as 2 : 4::8 : 16; that
' f is, as 2 is to 4, so is 8 to 16.
To, J
V Radical sign, which, prefixed to any number, sig-
nifies that the square root of that number is re-
quired; as V9, or Vaxb. The degree of the
root is indicated by the number placed over the
sign, which is termed the index of the root, or
radial; as ^, y , etc.
', *, added or set superior to a number, signifies that
that number is to be squared, cubed, etc. ; thus 43
means that 4 is to be multiplied by 4 ; 43, that it
is to be cubed, as 43 is =4 X 4 X 4 = 64. The pow-
er, or number of times a number is to be multi-
plied by itself, is shown by the number added, as
3 3 4 5 pfp
The vinculum, or bar, signifies that the numbers are
to be taken together; as 8 — 2 + 6=12, or
3x5+3=24.
Decimal point, signifies, when prefixed to a number,
that that number has a unit (1) for its denom-
inator; as .1 is •jJtf, .155 is i^j-, etc.
co Difference, signifies, when placed between two
quantities, that their difference is to be taken.
X EXPLANATIONS OF CHARACTERS.
Signify Degrees, minutes, seconds, and thirds.
/_ Signifies angle.
± Signifies perpendicular.
A Signifies triangle,
D Signifies square, as □ inches.
> T Signify inequality, or greater than, and are pnt be-
tween two quantities ; as a T b reads a greater
than b.
< L Signify the reverse ; as a lb reads a less than b.
Signifies therefore or hence.
Signifies because.
( ) [.] Parentheses and brackets signify that all the fig-
ures within them are to be operated upon as
if they were only one ; thus (3 + 2) X 5 = 25 ;
[3 +2] X 5 =25.
p or it Is used to express the ratio of the circumference of
a circle to its diameter = 3.1415926, etc.
A A' A" A'" Signify A, A prime, A second, A third, etc.
± tf Signify that the formula is to be adapted to two
distinct cases.
a~ ', a-2, a~3, etc. Denote inverse powers of a, and are equal to
ill
S* a* a* etC*
sin.^a, cos.~*a, etc. Signify the arc or angle, the sine or cosine, etc., of
which is a, the arc or angle being expressed in
terms of the radius, as the unit, unless other-
wise stated. If A0 denotes the arc or angle in
•7rA°
degrees, in terms of the radius it is A=— — .
If sin. A.— a, ,'.sin. ~~ 1a=A, etc.
, , etc. Set superior to a number, signify the square or
cube root, etc., of the number; as 22 signifies
the square root of 2.
I # ?-
a, 3, a, etc. Set superior to a number, signify the square or cube
root, etc., of the 4th power, etc., etc.
*-*\ 3*6, etc. Set superior to a number, signify the tenth root of
the 17th power, etc., etc.
'** OF THE ^4fa
UNIVERSITY
HASWELL'S MENSURATION,
MEASURES AND WEIGHTS
Used in this Work,
MEASURES OF LENGTH.
LINEAL.
12 inches = 1 foot.
3 feet = 1 yard.
5 J yards = 1 rod or pole.
40 rods —1 furlong.
8 furlongs =1 mile.
1 degree =69.77 statute
miles.
1 geographical mile=2046.58 yards or 6139.74 feet.
CIRCLES.
60 thirds =1 second. i 60 minutes =1 degree.
60 seconds =1 minute. | 360 degrees =1 circle.
1 day is .002739 of a year.
1 minute is .000694 of a year.
6 points =1 line.
12 lines =1 inch.
1 palm =3 inches.
MISCELLANEOUS.
1 hand =4 inches.
1 span ss 9 inches.
6 feet =1 fathom.
1 yard =.000568 of a mile.
1 foot =.000199 "
1 inch =.0000158 "
Gunter's Chain is 4 poles or 22 yards in length, and has
100 equal links of .666 of a foot, or 7.92 inches.
80 chains=l mile.
12
MEASURES AND WEIGHTS.
Foreign.
Great Britain.
Imperial yard
=39.1393 imperial inches.
Mile
= 1760 U. S. yards.
France.
Metre
=39.37079 inches, or 3.2809 feet
Foot (old system)
= 12.7925 inches.
Common league
=4264.16 U. S. yards.
Austria.
Foot
= 12.445 inches.*
Mile
=8296.66 U. S. yards.
China.
Foot, builder's
= 12.71 inches.
" mathematic
= 13.12 "
" surveyor's
= 12.58 "
" tradesman's
= 13.32 "
Li
=629 U. S. yards.
Copenhagen.
Foot
= 12.35 inches.
Genoa.
Foot
=9.72 "
Hamburgh.
Foot
= 11.29 "
Mile
=8244 U. S. yards.
Lisbon.
Foot
= 12.96 inches.
Mexico.
Foot
= 11.1284 inches.
Common league
=4636.83 U. S. yards.
Prussia.
Foot
= 12.361 inches.
Mile
=8468 U. S. yards.
Rome."
Foot
= 11.60 inches.
Mile
=2025 U. S. yards.
Russia.
Foot
=21.1874 inches.*
Versta
= 1167 U. S. yards.
Spain.
Foot
= 11.1284 inches.* ■ '
Judicial league
=4636.83 U. S. yards.
Sweden.
Foot
= 11.6865 inches.*
Mile
= 11700 U. S. yards.
Turkey.
Pick
= 17.905 inches.
Berri
=1826 U. S. yards.
MEASURES OF SURFACE.
144 inches =1 foot.
9 feet =1 yard
272ifeet ">
SQUARE.
40 rods =1 rood.
4 roods =1 acre.
640 acres = 1 mile.
SOi yards)
— 1 rod or pole.
* U. S. Ordnance Manual, 1850.
WEIGHTS AND MEASURES.
13
10 square chains
4840 " yards
160 " poles
100000 " links
69.5701 yards square
220x198 feet
208.710321 feet square
235.5041 feet diameter, or
43560 square feet
► = 1 acre.
MISCELLANEOUS.
24 sheets^l quire. 20 quires = 1 ream.
Drawing Paper.
Cap 13 xl6
Demy 19Jxl5J
Medium 22 xl8
Koyal 24 xl9
Super-royal ... 27 X 19
Imperial 29 x21J
Elephant 27|x22i
in.
u
Columbier 33|
Atlas 33
Doub. Elephant 40
Theorem ..... 34
Antiquarian. . . 52
Emperor 40
Uncle Sam .... 48
X
23 in
x
26 "
X
26 "
X
28 "
X
31 "
X
60 "
Xl20
Foreign. ,
France. Old System,
New System, 1 are=
Amsterdam.
Berlin.
Hamburgh.
Portugal.
Prussia.
Rome.
Russia.
Spain.
Switzerland.
1 square inch = 1.13587 U. S. inch.
: 100 square metres = 1 19.603 square yards.
Morgen = 9722 " "
* great = 6786 " "
" =11545 " "
Geira = 6970 " "
Morgen = 3053 " "
Pezza == 3158 " "
Desiatina =13066.6 " "
Fanegada = 5500 " . "
Taux = 7855 " "
Square Foot in U. S. Square Inches.
Amsterdam 124.255
Antwerp 126.337
Berlin 148.603
Bologna 224.700
Bremen 129.504
Cologne 117.288
Dantzic 127.690
Denmark 152.670
14
MEASURES AND WEIGHTS.
Dresden 124.099
France 163.558
Geneva 369.024
Hamburgh 127.441
Leipsic 123.432
Lisbon 167.547
Milan 243.984
Rhineland 152.670
Riga 116.425
Rome 137.358
Spain 123.832
Sweden 136.515
Venice 187.182
Vienna 155.002
MEASURES OF CAPACITY
LIQUID AND DRY.
7.21875 cub. i
4 gills
2 pints
1 gill-
1 pint.
1 quart.
4 quarts = 1 gallon.
2 gallons =1 peck.
4 pecks ss 1 bushel.
MISCELLANEOUS.
1 chaldron— 36 bushels, or 57.244 cubic feet, when heaped
in the form of a cone.
Note. — The standard U. S. bushel is the Winchester (British), and it
measures 2150.42 cubic inches, and contains 543391.89 troy grains, or
77.627413 pounds avoirdupois of distilled water at its maximum density.
Its dimensions are 18£ inches in diameter inside, 19£ inches outside,
and 8 inches deep. When heaped, the cone must not be less than 6
inches high, and it contains 2986.4765 cubic inches.
The standard U. S. gallon=231 cubic inches, and contains 58372.1754
troy grains (8.3389 pounds avoirdupois) of distilled water at its maxi-
mum density (39°.83).
Foreign.
Great Britain.
France/
Amsterdam.
Antwerp.
Bremen.
Constantinople.
The imperial gallon measures 277.274 cubic inches.
Imperial bushel 2218.192 cubic inches, and when
neaped in the form of a true cone (6 inches high)
it contains 2815.4872 cubic inches.
1 chaldron =58.68 cubic feet, and weighs 3136 lbs.
Old System. 1 Pinte=0.931 litre, or 56.817 cubic
inches.
New System. 1 Litre =61.027 U. S. inches.
Anker 2331 cub. in. Mudde 6786 cub. in.
Stoop 168 " Viertel 4705 "
Stubgens 194.5 " Scheffel 4339 "
Almud 319 " Kislos 2023 "
MEASURES AND WEIGHTS.
15
Copenhagen.
Anker
2335 cub. in.
Toende
8489 cub. in.
Genoa.
Pinte
90.5 "
Mina
7366 "
Lisbon.
Almudi
1010
Alqueire
827 "
Rome.
Boccali
80
Quarti
4226 "
Russia.
Vedro
750.58 "
Chet.wert 12800 "
Spain.
Quartillos
30.5 "
Catrize
41269 "
■
Arroba
4.2455 gall.
Fanega
1.593 bush.
Sweden.
Kann
160 cub. in.
Tunnar
8940 cub. in.
Tripoli.
Mattari
1376 "
" Caffiri
19780 "
Vienna.
Eimer
3443 «
Metzen
3753 "
MEASURES OF SOLIDITY.
CUBIC.
1728 inches=l foot. | 27 feet = l yard.
1 foot =7.4806 gallons.
128 feet=:l cord.
24.75 " =1 perch.
Foreign.
France. Stere (1 cubic metre) =6 102 7.1 U. S. inches, or
35.3166 cubic feet.
Note. — For the solid measures of other foreign countries, take the
cube of the measures given in the preceding tables.
MEASUBES OF WEIGHT.
AVOIRDUPOIS.
1G drams =1 ounce.
1G ounces =1 pound.
112 pounds =1 cwt.
20 cwt. =1 ton.
TROY WEIGHT.
24 grains r=l pennyweight.
20 penny weights = 1 ounce.
12 ounces =1 pound.
16
MEASURES AND WEIGHTS.
The pound, ounce, and grain are the same in apothecaries'
and troy weights.
7000 troy grains = 1
175 « pounds=144
175 " ounces =192
lb. avoirdupois,
lbs. "
oz. "
oz. "
.8228 lb. "
The standard U. S. pound contains 7000 troy grains, or
27.7015 cubic inches of distilled water at its maximum
density.
437.5 troy grains = 1
1 " pound =
Foreign.
Great Britain.
Pound avoi
irdupois
=27.7274 cubic
: inches of (
tilled water at the temperature <
of 62°. Hen
22.815689 cubic inches weigh 1 troy pound.
France.
1 Gramme
=
15.43316
troy grains.
Alexandria.
1 Rottoli
5=
.9346
pounds avoirdupois.
Amsterdam.
1 Pound
=
1.0893
a
(i
Austria.
1 "
BS
1.2351
u
H
Bengal.
1 Seer
55
1.8667
"
M
Bremen.
1 Pound
BS
1.0997
u
M
Cairo.
1 Rottoli
=
.9523
ii
U
China.
1 Catty
=
1.3253
«
11
Constantinople.
1 Oke
BS
2.8129
(I
(I
Copenhagen.
1 Pound
BT
1.1014
u
U
Corsica.
1 "
SB
.7591
II
u
Genoa.
1 " (heavy) ==
1.0768
U
a
Japan.
1 Catty
BS
1.3000
a
(<
Prussia.
1 Pound
BE
1.0333
u
II
Rome. I
1 "
as
.7479
U
ii
Russia.
1 "
S3
.9020
II
ii
Spain.
1 " •
=
1.0152
(4
it
Sweden.
1 "
SB
.9376
II
ii
Tripoli.
1 Rottoli
BS
1.1200
II
it
Venice.
1 Pound (heavy) =
1.0555
M
it
MISCELLANEOUS.
1 cubic foot of anthracite coal from 50 to 55 lbs.
1 cubic foot of bituminous coal from 45 to 55 lbs.
1 cubic foot of Cumberland coalm 53 lbs.
MEASURES AND WEIGHTS.
17
1 cubic foot of charcoal = 18.5 lbs. (hard wood.)
1 cubic foot of charcoal = 18. " (pine wood.)
1 cord Virginia pine =2700 "
1 cord Southern pine =3300 "
1 stone = 14 "
Coals are usually purchased at the conventional rate of 28
bushels (5 pecks) to a ton =43.56 cubic feet.
MEASURES OF VALUE.
1 eagle =258 troy grains.
1 dollar=412.5 « "
1 cent =168 " "
The standard of gold and silver is 900 parts of pure metal
and 100 of alloy in 1000 parts of coin.
A digit
A palm
A span
ANCIENT MEASURES.
MEASURES OF LENGTH.
Scripture.
0.912
3.648
10.944
A cubit =1
A fathom =7
9.888
3.552
Feet
=0
Grecian.
A digit =0 0.7554^- A stadium:
A pous (foot)=l 0.0875 A mile
A cubit =1 1.5984|
A Greek or Olympic foot= 12.108 inches.
A Pythic or natural foot = 9.768 "
Feet.
604
4835
Inches.
4.5
Jewish.
A cubit
A Sabbath day's
journey
= 1.824
=3648.
A mile = 7296
A day's journey =175104
(or 33 miles 864 feet).
18
MEASURES AND WEIGHTS.
Roman.
A digit :
An uncia (inch):
A pes (foot)
Arabian foot
Babylonian foot
Egyptian "
Inches.
.72575
.967
11.604
Feet. Inches.
A cubit = 1 5.406
A passus= 4 10.02
A mile =4835
Miscellaneous.
Feet.
= 1.095
= 1.140
= 1.421
Feet.
Hebrew foot =1.212
" cubit =1.817
" sacred cubit= 2.002
FRACTIONS. 19
VULGAE FRACTIONS.
A Fraction, or broken number, is one or more parts of
a Unit.
Illustration. — 12 inches are 1 foot.
Here 1 foot is the unit, and 12 inches its parts ; 3 inches,
therefore, are the one fourth of a foot, for three is the quarter
or fourth of 12.
A Vulgar Fraction is a fraction expressed by two numbers
placed one above the other, with a line between them, as 50
cents is the \ of a dollar.
The upper number is called the Numerator, because it shows
the number of parts used.
The lower number is called the Denominator, because it de-
nominates, or gives name to the fraction.
The Terms of a fraction express both numerator and de-
nominator ; as 6 and 9 are the terms of f .
A Proper fraction has the numerator equal to, or less than
the denominator ; as \, -J, &c.
An Improper fraction is the reverse of a proper one ; as
f , 4, &c.
A Mixed fraction is a compound of a whole number and a
fraction ; as 5-J , &c.
A Compound fraction is the fraction of a fraction ; as \ of
f,4 of-!, &c.
A Complex fraction is one that has a fraction for its numer-
i 3—
ator or denominator, or both ; as i, or %, or J, or S, &c.
A Fraction denotes division, and its value is equal to the quotient
obtained by dividing the numerator by the denominator ; thus ±£- is
equal to 3, -^ is equal to 44, and % is equal to -^.
20 FRACTIONS.
REDUCTION OF VULGAR FRACTIONS.
To find the greatest Number that will divide two or
more Numbers without a Remainder,
Rule. — Divide the greater number by the less ; then divide
the divisor by the remainder ; and so on, dividing always the
last divisor by the last remainder, until nothing remains.
When there are more than two numbers, find the greatest
common measure of two of them, and then that for this com-
mon measure and the remaining number.
Example. — What is the greatest common measure of 1908
and 936? 936)1908(2
1872
36)936(26
72
216
216
Hence 36 is the greatest common measure.
Ex. 2. What is the greatest common measure of 246 and
372 ? Ans. 6.
Ex. 3. What is the greatest common measure of 1728, 864,
and 3456 *
864)3456(4
3456
864 is the greatest common measure of 3456 and 864.
864)864(1
864
Hence 864 is the greatest common measure of the three num-
bers.
Ex. 4. What is the greatest common measure of 216 and
288? AnsM2.
To find the least common Multiple of two or more
Numbers.,
Rule. — Divide by any number that will divide two or more
FRACTIONS. 21
of the given numbers without a remainder, and set the quo-
tients with the undivided numbers in a line beneath.
Divide the second line as before, and so on, until there are no
two numbers that can be divided ; then the continued product
of the divisors and quotients will give the multiple required.
Example. — What is the least common multiple of 40, 50,
and 25?
5)40 . 50 . 25
5) 8.10. 5
2) 8. 2. 1
4. 1. 1
Then 5x5x2 x 4=200, Ans.
To reduce Fractions to their lowest 'Terms.
Rule. — Divide the terms by any number that will divide
them without a remainder, or by their greatest common meas-
ure at once.
Example. — Reduce -£§ $ of a foot to its lowest terms.
l™ + 10=™ + 8=^+3=%, or 9 inches.
. Ex. 2. Reduce -|ff to its lowest terms. Ans. § .
To reduce a Mixed Fraction to its equivalent Improper
Fraction.
Note. — Mixed and improper fractions are the same; thus 5£=-2J-.
Rule. — Multiply the whole number by the denominator of
the fraction, and \g the product add the numerator, then set
that sum above the denominator.
Example. — Reduce 23J to a fraction.
23x6 + 2 140 ,i
-— ■ — = — — = the answer.
o 6
Ex. 2. Reduce 20$ inches to a fraction. Ans. ±%2-.
Ex. 3. Reduce 5-J to a fraction. Ans. $£-.
Ex. 4. Reduce 183^5T to a fraction. Ans. 3|j9.
Ex. 5. Reduce 125£ to a fraction. Ans. £JI*
FRACTIONS.
To reduce an Improper Fraction to its equivalent
Whole or Mixed Number.
Rule. — Divide the numerator by the denominator, and the
quotient will be the whole or mixed number required.
Example. — Reduce ^ to its equivalent number.
^ or 12-^-3=4, the answer.
Ex. 2. Reduce ^^ to its equivalent number. Ans. 8.
To reduce a Whole Number to an equivalent Fraction
having a given Denominator.
Rule. — Multiply the whole number by the given denomi-
nator, and set the product over the said denominator.
Example. — Reduce 8 to a fraction whose denominator shall
be 9.
8 x 9=72 ; then ^-—the answer.
Ex. 2. Reduce 12 to a fraction whose denominator shall
be 13. Ans. J#,
To reduce a Compound Fraction to an equivalent
Simple one.
Rule. — Multiply all the numerators together for a numer-
ator, and all the denominators together for a denominator.
N0tE# — When there are terms that are common, they may be omitted.
Example. — Reduce \ of f of § to a simple fraction.
ix|x#=A=i, Am.
Or, - x -X-=-» ty canceling the 2's and 3's.
Ex. 2. Reduce -| of f to a simple fraction.
ix|=f, Ans.
Ex. 3. Reduce £ of % to a simple fraction. Ans. Jf.
Ex. 4. Reduce •§ of f of -f of £ of -^ to a simple fraction.
Ans. *%.
Ex. 5. Reduce 2, and -| of £ to a fraction. ^4tw. §#.
Ex. 6. Reduce 2^ and $ of § to a fraction. -4ws. £.
FRACTIONS. 23
To reduce Fractions of different Denominations to
equivalent ones having a common Denominator.
JJule. — Multiply each numerator by all the denominators
except its own for the new numerators, and multiply all the
denominators together for a common denominator.
Note. — In this, as in all other operations, whole numbers, mixed, or com-
pound fractions, mustjirst be reduced to the form of simple fractions.
Example. — Keduce -J, -§-, and f to a common denominator.
1x3x4 = 12^
2x2x4 = 16
3x2x3 = 18
12 16 18 An<!
Til -2TJ Tfj JinS'
2x3x4 = 24^
The operation may be performed mentally ; thus,
Reduce -J-, -§> !> an(^ "i to a common denominator.
1 1 3 12 6 6 n71fJ 5 2 0
"5" — "§"* H — ~W~' "F — ~S> ana ~2 — -W
Ex. 2. Reduce 2, and -§ of -J to a common denominator.
A„o 60 12 25
Ex. 3. Reduce -§ , 2f , and 4 to a common denominator.
Ana 25 78 120
XX /to. ^-y-, -g-g, -Jo-*
Ex. 4. Reduce 4, ^, 14, and 54 to a common denominator.
-2~> 4> A"2"> *"" WB~
J.,, 24 36 72 256
AB* 4¥? T¥J 4~g-> _4~r*
Note 1. When the denominators of two given fractions have a com-
mon measure, divide them by it ; then multiply the terms of each given
fraction by the quotient arising from the other denominator.
Ex. 5. Reduce ■£$ and -^ to a common denominator.
^andif=^and^,Ans.
Note 2. When the less denominator of two fractions exactly divides
the greater, multiply the terms of that which has the less denominator
by the quotient.
Ex. 6. Reduce -f- and -^ to a common denominator.
5
f and 1?=-^ and ^, Ans.
To reduce Complex Fractions to Simple ones.
Rule. — Reduce the two parts both to simple fractions, then
multiply the numerator of each by the denominator of the other.
24 FRACTIONS.
22
Example. — Simplify the complex fraction jj.
2|=f 8x 5=40_fl A
Ex. 2. Simplify the complex fraction f.
|=4 f==fc Am-
8f
Ex. 3. Simplify the complex fraction jf. ^4w5. -Jf*
7b ^/mtZ £/i<? Value of a Fraction in parts of a whole
Number.
Rule. — Multiply the whole number by the numerator, and
divide by the denominator ; then, if any thing remains, mul-
tiply it by the parts in the next inferior denomination, and
divide by the denominator as before, and so on as far as nec-
essary ; so shall the quotients placed in order be the value of
the fractions required.
Example. — What is the value of J of ■§• of $9 ?
J of § =£, and f of £=J^-=$3, Ans.
Ex. 2. Reduce J of a pound to avoirdupois ounces.
3
1
4) 3(0 lbs.
16 ounces in a lb.
4)48
12 ounces, Ans.
Ex. 3. Reduce -^ of a day to hours.
i^X^=li=7£> hours, Ans.
Ex. 4. Reduce % of a pound troy to ounces and pennyweights.
4
12
5)48
ounces 9 3
20
5)60
pennyweights 12=9 oz., 12 dwts., Ans.
FRACTIONS. 25
Ex. 5. What is the value of $ of an acre ?
.Arcs. 3 roods, 20 poles.
Ex. 6. What is the value of f of $4 83 ?
Ans. $1 93J.
2b reduce a Fraction from one Denomination to
another.
Rule. — Multiply the number of parts in the next less de-
nomination by the numerator if the reduction is to he to a less
name, but multiply by the denominator if to a greater.
Example. — Reduce J of a dollar to the fraction of a cent.
Ex. 2. Reduce \ of an avoirdupois pound to the fraction
of an ounce.
is/ 16 16 8 An*
Ex. 3. Reduce -f- of a cwt. to the fraction of a lb.
2 VH2 224 32 A„Q
Ex. 4. Reduce § of f of a mile to the fraction of a foot.
2 nf 3 6 w 6280 31680 2640 A^o
~s 0I ¥— T2 x — i — — — ra — — — i — J ^ns.
Ex. 5. Reduce J of a square foot to the fraction of an inch.
Ans. -2j£.
ADDITION OF VULGAR FRACTIONS.
Rule. — If the fractions have a common denominator, add
all the numerators together, and place their sum over the de-
nominator.
Note. — If the fractions have not a common denominator, they must be
reduced to one, and compound and complex must be reduced to simple frac-
tions.
Example. — Add \ and J together.
i+4=|=l, Ans.
Ex. 2. Add J of J of tV to 2f of f .
Then, W+«=-,ftW+MS*=l«*4=144fc Ans'
Ex. 3. Add | and $ together. Ans. 1^%.
Ex. 4. Add I, 7fc and } of | together. Ans. 8|.
B
26 FRACTIONS.
/
Ex. 5. Add T of an eagle, -J of a dollar, and -j^ of a cent
together. Ans. 165f|f.
SUBTRACTION OF VULGAR FRACTIONS.
Rule. — Prepare the fractions, when necessary, the same as
for other operations, then subtract the one numerator from
the other, and set the remainder over the common denom-
inator.
Example. — What is the difference between -f and ^ ?
|-i=|» Ans.
Ex. 2. Subtract -§ from -§.
6x9=54\
3x8 = 24 Uff-ff=ff, Ana.
8x9 = 72j
Ex. 3. Subtract -^ from ^-. Ans. T^-.
Ex. 4. Subtract $ of -^ from f of 5 J- of 1. ,4ns. -|f$£.
MULTIPLICATION OF VULGAR FRACTIONS.
Rule. — Prepare the fractions, when necessary, as previous-
ly required ; multiply all the numerators together for a new
numerator, and all the denominators together for a new de-
nominator.
Example. — What is the product of -§ and § ?
ix%=&=ih,Ans.
Ex. 2. What is the product of 6 and % of 5 ?
6xf of5=6x-^=-$f=20, Ans.
Ex. 3. What is the product of § , 3£, 5, and f of f ?
3
-x^x^x(fof^=^xf=¥=41,^.
2 4
Ex. 4. What is the product of 5, f, f of % and 4£?
FRACTIONS. 27
DIVISION OF VULGAR FRACTIONS.
Rule. — Prepare the fractions, when necessary, as previous-
ly required ; then divide the numerator by the numerator, and
the denominator by the denominator, if they will exactly di-
vide ; but if not, invert the terms of the divisor, and multiply
the dividend by it, as in multiplication.
Example. — Divide -%§. by f .
25 . 5 5 12 Anli
^ ■ . Tj- — tj- — -"-"3"? JaLilo.
Ex. 2. Divide j by ^.
5 . 2 5V15 7 5 A3 A„Q
Ex. 3. Divide -^ by f . ^rcs. ^.
Ex. 4. Divide § by 2. <£*& ^
Ex. 5. Divide f of J by $ of 7-f. jitaft yf,-.
RULE OF THREE IN VULGAR FRACTIONS.
Rule. — Prepare the fractions, when necessary, as previous-
ly required ; invert the first term, and multiply it and the sec-
ond and third terms continually together ; the product will be
the result required.
Example. — If f of a barrel cost -§ of a dollar, what will
-^ of a barrel cost ?
f:f::-rV— ^X^X^=i=$0 33+, Ans.
2
2
Ex. 2. What will 3f ounces of silver cost at -^ of a pound
sterling per ounce?
Q 3 27 19 «f 20 380 38 19
9
- X -y- X — =: shillings.
,8 8 8
fci|ixi^-2^i:=256||)«,or£l Is. ±\d.,Ans.
Ex. 3. What part of a ship is worth $60,120, when J of
her cost $17,535 1 flfo+itf-^ Am.
28 DECIMALS.
DECIMALS.
A Decimal is a fraction which has for its denominator a
unit with as many ciphers annexed as the numerator has
places ; it is usually expressed by setting down the numerator
only, with a point on the left of it. Thus, -^ is .4, ■££$ is .85,
tottdit is -OO^, and towW is -00125. When there is a de-
ficiency of figures in the numerator, ciphers are prefixed to
make up as many places as there are ciphers in the denomi-
nator.
Mixed Numbers consist of a whole number and a fraction,
as, 3.25, which is the same as o.y2^, or -f-§-§.
Ciphers on the right hand make no alteration in the value
of a decimal, for .4, .40, .400 are all of the same value, each
being equal to -^j-. •
ADDITION OF DECIMALS.
Rule. — Set the numbers under each other, according to
the value of their places, as in whole numbers, in which po-
sition the decimal points will stand directly under each other.
Then, beginning at the right hand, add up all the columns as
in whole numbers, and place the point directly below all the
other points.
Example. — Add together 25.125, 56.19, 1.875, and
293.7325.
25.125
56.19
1.875
293.7325
376.9225 the Sum.
Ex. 2. Add together 27.62, .358, 17.3, .007, and 173.1.
Sum 218.385.
Ex. 3. Add together .001, .09, and .909. Sum 1.000.
Ex. 4. Add together 87.5, 56.25, 37.5, and 43.75.
Sum 225.
DECIMALS. 29
SUBTEACTION OF DECIMALS.
Rule. — Set the numbers under each other, as in addition ;
then subtract as in whole numbers, and point off the decimals
as in the last rule.
Example.— Subtract 15.150 from 89.1759.
89.1759-
15.150
74.0259, the Rem.
Ex. 2. Subtract 96.50 from 100. Rem. 3.50.
Ex. 3. Subtract 3.1416 from 4.5236. Rem. 1.3820.
Ex. 4. Subtract 14.56789 from 2486.173.
Rem. 2471.60511.
MULTIPLICATION OF DECIMALS.
Rule. — Set the factors, and multiply them together the
same as if they were whole numbers ; then point off in the
product just as many places of decimals as there are decimals
in both the factors. But if there are not so many figures in
the product as there are decimal places required, supply the
deficiency by prefixing ciphers.
Example. — Multiply 1.56 by .75.
1.56 ,
.75
780
1092
1.1700, the Product.
Ex. 2. Multiply 79.25 by .460. Product 36.455.
Ex. 3. Multiply 79.347 by 23.15.
Product 1836.88305.
Ex. 4. Multiply .385746 by .00464.
Product .00178986144.
30 DECIMALS.
BY CONTRACTION.
To contract the Operation so as to retain only as many Decimal
places in the Product as may be thought necessary.
Rule. — Set the unit's place of the multiplier under the
figure of the multiplicand whose place is the same as is to be
retained for the last in the product, and dispose of the rest of
the figures in the contrary order to what they are usually
placed in. Then, in multiplying, reject all the figures that
are more to the right hand than each multiplying figure, and
set down the products, so that their right-hand figures may
fall in a column directly below each other; and observe to
increase the first figure in every line with what would have
arisen from the figures omitted; thus, add 1 for every result
from 5 to 14, 2 from 15 to 24, 3 from 25 to 34, 4 from 35
to 44, &c, &c., and the sum of. all the lines will be the prod-
uct as required.
Example.— Multiply 13.57493 by 46.20517, and retain
only four places of decimals in the product.
13.574 93
71 502.64
54 299 72
8 144 96 + 2 for 18
27150+2
" 18
6 79+4
" 35
14+1
" 5
9 + 2
" 21
627.23 20
6 is the unit of the multiplier, and 9 is the figure of the multi-
plicand whose place is the same as is to be retained for the last in
the product
Ex. 2. Multiply 27.14986 by 92.41035, and retain only
five places of decimals. Product 2508.92806.
Ex. 3. Multiply 480.14936 by 2.72416, and retain only
four places of decimals. Product 1308.0035.
DECIMALS. 31
Ex. 4. Multiply 325.701428 by .7218393, and retain only
three places of decimals. Product 235.103.
Ex. 5. Multiply 81.4632 by 7.24G51, retaining only three
places of decmials. Product 590.324.
DIVISION OF DECIMALS.
Rule. — Divide as in whole numbers, and point off in the
quotient as many places for decimals as the decimal places in
the dividend exceed those in the divisor ; but if there are not
so many places, supply the deficiency by prefixing ciphers.
Example. — Divide 53.00 by 6.75.
6.75)53.00(7.85 +
47 25
5 750
5 400
3500
3375
125
Here 2 ciphers were annexed to carry out the division.
Ex. 2. Divide 45.5 by 2100. Quotient .0216 + .
Ex. 3. Divide 12 by .7854. Quotient 15.278.
Ex. 4. Divide .061 by 79000.
Quotient .00000077215+.
Ex. 5. Divide 2.7182818 by 3.1415927.
fc Quotient .865256—.
Ex. 6. Divide .00128 by 8.192. Quotient .000156.
BY CONTRACTION.
Rule. — Take only as many figures of the divisor as will be
equal to the number of figures, both integers and decimals, to
be in the quotient, and find how many times they may be con-
tained in the first figures of the dividend, as usual.
Let each remainder be a new dividend ; and for every such
dividend leave out one figure more on the right-hand side of
the divisor, carrying for the figures cut off as in Contraction
of Multiplication.
32 DECIMALS.
Note. — When there are not so many figures in the divisor as are required
to be in the quotient, continue the first operation until the number of figures in
the divisor be equal to those remaining to be found in the quotient, after which
begin the contraction.
Example.— Divide 2508.92806 by 92.41035, retaining only
four places of decimals in the quotient.
92.4103|5)2508.928|06(27.1498
1848 207+1
660 721
646 872+2
13 849
9 241
4 608
3 696
912
832+4
80
74+2
6
Ex. 2. Divide 4109.2351 by 230.409, retaining only four
decimals in the quotient. Quotient 17.8345.
Ex. 3. Divide 37.10438 by 5713.96, retaining only five
decimals in the quotient. Quotient .00649.
Ex. 4. Divide 913.08 by 2137.2, retaining only three deci-
mals in the quotient. Quotient .427.
REDUCTION OF DECIMALS.
To reduce a Vulgar Fraction to its equivalent Decimal.
Rule. — Divide the numerator by the denominator, as in
division of decimals, annexing ciphers to the numerator as
far as necessary, and the quotient will be the decimal re-
quired.
. Example. — Reduce \ to a decimal.
5)^0
.8, Quotient.
DECIMALS. 33
Ex. 2. Keduce -^j- to a decimal.
700)35.00(.05, Quotient
35 00
Ex. 3. Reduce 4 to a decimal. Quotient .625.
Ex. 4. Reduce 44 to a decimal. Quotient .9375.
Ex. 5. Reduce -j-f-g- to a decimal. Quotient .03125.
To find the Value of a Decimal in Terms of an in-
ferior Denomination.
Rule. — Multiply the decimal by the number of parts in the
next lower denomination, and cut off as many places for a re-
mainder to the right hand as there are places in the given
decimal.
Multiply that remainder by the parts in the next lower
denomination, again cutting off for a remainder, and so on
through all the denominations of the decimal.
Then the several denominations pointed off on the left hand
will give the result required.
Example. — What is the value of .875 dollar?
.875
100
87.500 cents,
10
5.000 milk.
Ans. 87 cents, 5 mills.
Ex.2.
What is the content of .140 cubic foot in inches?
.140
1728 cubic inches in a cubic foot
241.920
Ans. 241.920 cubic inches.
Ex.3.
What is the value of .00129 of a foot?
Ans. .01548 inches.
Ex.4.
What is the value of 1.075 ton in pounds?
Ans. 2408.
Ex.5.
Reduce .0125 lb. troy to pennyweights.
Ans. 3 dwts.
B2
34 DECIMALS.
Ex. 6. Reduce .95 mile to its equivalent decimals in its
lower denominations.
.95
8 furlongs,
TM
40 rods.
24.00 Ans. 7 furlongs and 24 rods.
Ex. 7. Reduce .05 mile to its equivalent decimals.
Ans. 0 furlongs and 16 rods.
Ex. 8. Reduce ^ of a mile to its equivalent decimals.
Ans. § furlongs and 16 rods.
Ex. 9. Reduce ^ of a cubic yard to its equivalent decimals.
Ans. 2.9999 feet+.
Ex. 10. Reduce J of a degree to its equivalent decimals.
Ans. 19 minutes and 59.999 seconds -f.
To reduce a Decimal to its equivalent in a higher De-
nomination.
Rule. — Divide by the number of parts in the next higher
denomination, continuing the operation as far as required.
Example. — Reduce 1 inch to the decimal of a foot.
12
1.00000
.08333, &c, Ans.
Ex. 2. Reduce 14 minutes to the decimal of a day.
60
24
14.00000
.23333
.00972, &c, Ans.
Ex. 3. Reduce 14" 12'" to the decimal of a minute.
14" 12/7/
60
60
60
852.'
X4.2y
.23666', &c, Ans.
Note. — When there are several numbers, to be reduced all to the decimal
of the highest.
DECIMALS. 35
Reduce them all to the lowest denomination, and proceed
as for one denomination.
Ex. 4. Reduce 5 feet 10 inches and 3 barleycorns to the
decimal of a yard.
Feet. Inches. Be.
5 10 3
12
70
3
213.
71.
5.9166
1.9722, &c, Ans.
Ex. 5. Reduce 1 dwt. to the decimal of a pound troy.
Ans. .004166+ lb.
Ex. 6. Reduce 1 yard to the decimal of a mile.
Ans. .000568+ mile.
Ex. 7. Reduce 8 feet 6 inches to the decimal of a mile.
Ans. .0016098 mile.
Ex. 8. Reduce 4J miles to the decimal of 80 miles.
Ans. .05625.
Ex. 9. Reduce 14', 18", and 36'" to the decimal of a de-
gree. • Ans. .2385 degree.
Ex. 10. Reduce 17 yards, 1 foot, and 5.98848 inches to the
decimal of a mile. Ans. .009943 mile.
RULE OF THREE IN DECIMALS.
Rule. — Prepare the terms by reducing vulgar fractions to
decimals, compound numbers to decimals of the highest de-
nominations, and the first and third terms to the same denom-
ination ; then proceed as in whole numbers.
Example. — If ^ a ton of iron cost f of a dollar, what will
.625 of a ton cost?
i = .5 \ .5:. 75::. 625
:=.75j
|=.75j .625
.5).46875
.9375, Ans.
36 DUODECIMALS.
Ex. 2. If -§ of a yard cost § of a dollar, what will ^ of a
yard cost? Am. .3333+ dollar.
Ex. 3. If t^t of a mile cost $15.75, what will ^ of a fur-
long cost? Ans. $4 05.
DUODECIMALS.
In Duodecimals, or Cross Multiplication, the dimensions are
taken in feet, inches, and twelfths of an inch.
Rule. — Set down the dimensions to be multiplied together,
one under the other, so that feet may stand under feet, inches
under inches, &c.
Multiply each term of the multiplicand, beginning at the
lowest, by the feet in the multiplier, and set the result of each
directly under its corresponding term, carrying 1 for every 12
from 1 term to the other.
In like manner, multiply all the multiplicand by the inches
of the multiplier, and then by the twelfth parts, setting the
result of each term one place farther to the right hand for
every multiplier. The sum of the products is the result re-
quired.
Example. — Multiply 1 foot 3 inches by 1 foot 1 inch.
Feet. Inches.
1 3
1 1
3
14 3
Proof. — 1 foot 3 inches is 15 inches, 1 foot 1 inch is 13
inches; and 15x13 = 195 square inches. Now the above
product reads 1 foot, 4 inches, and 3 twelfths of an inch, and
1 foot =144 square inches.
4 inches =48 "
3 twelfths = 3 "
195 "
which is the product required.
DUODECIMALS. 37
Ex. 2. How many square feet, inches, &c, are there in a
platform 35 feet 4^ inches long, and 12 feet 3 J inches wide?
Feet.
35
Inches.
4
Twelfths.
6
12
3
4
424
6
8
10
1
6
11
9
6
0
434 3 11 0 0.
Or 434 feet, 3 inches, and 11 twelfths.
Ex. 3. Multiply 20 feet 6£ inches by 40 feet 6 inches.
Ans. 831 feet, 11 mcAes, 3 twelfths, which is equal to 831
square feet and 135 square inches.
By decimals, 40 ft. 6 in. =40.5
20 ft. 6 J in. = 20541666, &c.
831.937499 square feet.
144
134.999856 square inches.
Table showing the value of Duodecimals in Square
Feet and Decimals of an Inch.
1 Foot
1 Inch
1 Twelfth
^ of 1 Twelfth
-jL of -j^ of 1 Twelfth
Sq. Feet. Sq. Inches.
1 or 144.
. 1 « 1
i « 08W
17 2tJ .VOOOc
1 " OOfi<V
20 73 6 .UUOJ^
Application of this Table.
What number of square inches are there in a floor 100J
feet broad, and 25 feet, 6 inches, and 6 twelfths long %
100^ feet = 100.5 /jig.
~25 « (25xl44)=3600. iwctes.
6kk (6x12) = 72. "
6 fegeffifo =; 6. "
25 feet 6 wcfos 6 twelfths =3678. inches.
AOPTHE "^>
38 INVOLUTION.
12)3678. inches.
12) 306.5
2544166 feet*
As the 3678 are square inches, it is necessary to divide by 144 to produce
square feet, and the operation is more readily performed by dividing twice by
12 (12 X 12 = 144) than by 144 in one division.
Then 25.54166
100.5
t feet 2566.936830
12
inches 11.241960
12
twelfths 2.903520
Or, 2566 /art.
11x12 = 132. inches.
2 = 2. "
.9 (.9-^12)= .75 "
2566 feet, 134.75 inches.
INVOLUTION.
Involution is the multiplying any number into itself a cer-
tain number of times. The products obtained are called
Powers. The number is called the Root, or first power.
When a number is multiplied by itself once, the product is
the square of that number ; twice, the cube ; three times, the
biquadrate, &c. Thus, of the number 5,
5 is the Root, or 1st power.
5x5=25 " Square, or 2d power, and is ex-
pressed 52.
5x5x5 = 125 " Cube, or 3d power, and is expressed 53.
5x5x5x5 = 625 " Biquadrate, or 4th power, and is ex-
pressed 54.
The little figure denotes the power, and is called the Index
or Exponent. -
EVOLUTION. 39
Example. — What is the cube of 9 ?
9x9x9=729,^715.
Ex. 2. What is the cube of 1 1 Ans. |£.
Ex. 3. What is the 4th power of 1.5 ? Ans. 5.0625.
Ex. 4. What is the square of 4.16 ? Am. 17.3056.
Ex. 5. What is the square off? Ans. J.
Ex. 6. What is the third power of f 1 Ans. fjf .
Ex. 7. What is the fourth power of .025 !
Ans. .000000390625.
Ex. 8. What is the fifth power of 5 ! Ans. 3125.
Ex. 9. What is the fifth power of .05 !
Ans. .0000003125.
EVOLUTION.
Evolution is finding the Root of any number.
The sign -y/ placed before any number indicates the square
root of that number is required or shown.
The same character expresses any other root by placing the
index above it.
Thus, <t/25=5, 4+2 = ^36.
^27=3, and {/6± = 4.
Roots which only approximate are called Surd Roots, but
those which are exact are called Rational Roots.
TO EXTRACT THE SQUARE ROOT.
Rule. — Point off the given number from the place of units
into periods of two figures each by setting a point over the
place of units, another over that of hundreds, and so on over
every second figure, both to the left hand in integers and to
the right hand in decimals.
Find the greatest square in the left-hand period, and place
its root in the quotient ; subtract the square number from the
left-hand period, and to the remainder bring down the next
period for a new dividend.
Double the root already found for a divisor ; find how many
40
EVOLUTION.
times this incomplete divisor is contained in the dividend ex-
clusive of the right-hand figure ; with the consideration that
the result, or root, is to be the units' figure of the complete di-
visor, place the result in the quotient, and at the right hand
of the divisor.
Multiply this divisor by the last quotient figure, and sub-
tract the product from the dividend; bring down the next
period, and proceed as before.
Example. — What is the square root of 2 ?
2.000000(1.414, &c, Ans.
1
24
4
100
96
281
1
400
281
2824
4
2828
11900
11296
604
Ex. 2. What is the square root of 144 ?
144(12, Ans.
1
22
044
44
Ex. 3. What is the square root of 17.3056 !
17.3056(4.16, Ans.
16
130
81
826
4956
4956
Ex. 4. What is the square root of .000729 1 Ans. .027.
SQUARE ROOTS OF VULGAR FRACTIONS.
Rule. — Reduce the fractions to their lowest terms, and
that fraction to a decimal, then proceed as in whole numbers
and decimals.
EVOLUTION. 41
Note. — When the terms of the fractions are squares, take the root of each
and set one above the other ; as, the square root of^ is %.
Example. — What is the square root of ^^
-^=.41666666, &c
6 .4i666666(.6454, Ans.
6 36
124
4
566
496
1285
5
12904
4
7066
6425
64166
51616
12908 12550
Ex. 2. What is the square root of -^ I
Ans. 0.8660254.
Ex. 3. What is the square root of 17-f ?
171=17x84-3=-^^ and i§£=17.875,
which is the integer and decimal from which the root is to be
extracted. . Ans. 4.1683, &c.
TO EXTRACT THE CUBE ROOT
By the ordinary Method.
Point off the given number from the place of units into
periods of three figures each, by setting a point over the place
of units, and another also over every third figure from thence
to the left hand in integers, and to the right hand in decimals.
Find the greatest cube in the left-hand period, and set its
root in the quotient ; subtract the cube number from the left-
hand period, and to the remainder bring down the second pe-
riod for*a new dividend.
To three times the square of the root just found add three
times the root itself, setting this one figure place more to the
right than the former ; add these products together, and call
their sum the divisor. Divide the new dividend, less the last
figure to the right hand, by the divisor for the next figure of
42
EVOLUTION.
the root, which annex to the former, calling this last figure e,
and the part of the root before found a.
Add the following three products together, viz., thrice a
square multiplied by e, thrice a multiplied by e square, and e
cube, setting each of them one figure place more to the right
than the former, and call their sum the subtrahend; which
must not exceed the dividend, but if it does, then make the
last figure e less, and repeat the operation for finding the sub-
trahend till it be less than the dividend. •
From the dividend take the subtrahend, and to the remain-
der join the next period of the given number for a new divi-
dend, to which form a new divisor from the whole root already
formed, and from thence another figure of the root as already
directed, and so on until the extraction is complete.
Example. — Extract the cube root of 48228.544.
3x32:
3x3 :
divisor
:27
: 09
279
3x32x6 =162 "
3x3 x62= 324
63 = 216,
3x362r=3888
3x36 = 108
divisor
3X362x4
3x36
48228.544(36.4, root.
21228 dividend.
19656 subtrahend.
38988
15552
X42= 1728
43= 64
1572544 dividend.
1572544 subtrahend.
000 remainder.
Ex. 2. Extract the cube root of 46656. Ans. 36.
Ex. 3. Extract the cube root of 46656000. Ans. 360.
Ex. 4. Extract the cube root of 2. Ans. 1.259921.
Ex. 5. Extract the cube root of -j-^. Ans. .25.
TO EXTRACT THE CUBE ROOT.
RUle. — From a table of Roots take the nearest cube to the
given number, and call it the assumed cube.
EVOLUTION. 43
Then, as the given number added to twice the assumed cube
is to the assumed cube added to twice the given number, so is
the root of the assumed cube to the required cube root, nearly.
By using, in like manner, the root thus found as an as-
sumed cube, and proceeding as above, another root will be
found still nearer, and in the same manner as far as may be
necessary.
Example. — What is the cube root of 10517.9 ?
Nearest cube 10648, root 22.
10648. 10517.9
2_ 2
21296 21035.8
10517.9 10648.
31813.9 : 31683.8:: 22 : 21.9+, Ans.
Tafind the fourth Root of a Number.
Rule. — Extract the square root twice.
Example.— What is the 4th root of 625 ? Ans. 5.
To find the sixth Root of a Number.
Rule. — Take the cube root of its square root.
Example.— What is the ■{/ of 441?
V441 = 21, and^21 = 2.758923, Ans.
To find the eighth Root of a Number.
Rule. — Extract the square root thrice.
Example.— What is the 8th root of 390625 1 Ans. 5.
To extract any Root whatever.
Let P represent the number,
n " the index of the power,
A ' " the assumed power, r its root.
R " the required root of P.
Then, as the sum of n-f-1 x A and n— 1 xP is to the sum
of n-f-1 xP and n— 1 x A, so is the assumed root r to the
required root R.
44 PROPERTIES OP NUMBERS.
Example. — What is the cube root of 1500 ?
The nearest cube is 1331, root 11.
P=1500, n=3, A=1331, r=ll;
then,
«+lxA=5324, 7i+lxP=6000
7z-lxF-3000, yz-lxA-2662
8324 : 8662 :: 11 : 11.446+, Ans.
PEOPERTIES OF NUMBERS.
1. A Prime Number is that which can only be measured
(divided without a remainder) by 1 or unity.
2. A Composite Number is that which can be measured by
some number greater than unity.
3. A Perfect Number is that which is equal to the sum of
all its divisors or aliquot parts; as 6=-§, •§-, -§.
4. If the sum of the digits constituting any number be di-
visible by 3 or 9, the whole is divisible by them.
5. A square number can not terminate with an odd num-
ber of ciphers.
6. No square number can terminate with two equal digits,
except two ciphers or two fours.
7. No number the last digit of which is 2, 3, 7, or 8, is a
square number.
GEOMETRY. 45
GEOMETRY.
Definition^,
For the Definitions of the Surfaces of Figures, Solids, Lines,
&c, &c, see Mensuration of Areas, Lines, and Surfaces, Men-
suration of Solids and Conic Sections.
A Point has position, but not magnitude.
A Line is length without breadth, and is either Right, Curved, or
Mixed. \
A Right Line is the shortest distance between two points.
A Mixed Line is composed of a right and a curved line.
A Superficies has length and breadth only, and is plane or curved.
A Solid has length, breadth, and thickness.
An Angle is the opening of two lines having different directions, and
is either Right, Acute, or Obtuse.
A Right Angle is made by a line perpendicular to another falling
upon it.
An Acute Angle is less than a right angle.
An Obtuse Angk is greater than a right angle.
An Arc is any part of the circumference of a circle.
A Chord is a right line joining the extremities of an arc.
The Radius of a circle is a line drawn from the centre to the circum-
ference.
A Semicircle is half a circle.
A Quadrant is a quarter of a circle.
A Secant is a line that cuts a circle, lying partly within and partly
without it.
A Cosecant is the secant of the complement of an arc.
A Sine of an arc is a line running from one extremity of an arc per-
pendicular to a diameter passing through the other extremity, and the
sine of an angle is the sine of the arc that measures that angle. ,
The Versed Sine of an arc or angle is the part of the diameter inter-
cepted between the sine and the arc.
The Cosine of an arc or angle is the part of the diameter intercepted
between the sine and the centre.
A Tangent is a right line that touches a circle without cutting it.
A Cotangent is the tangent of the complement of the arc.
The Circumference of every circle is supposed to be divided into 360
equal parts, called Degrees ; each degree into 60 Minutes, and each min-
ute into 60 Seconds, and so on.
The Complement of an angle is what remains after subtracting the
angle from 90 degrees.
The Supplement of an angle is what remains after subtracting the
angle from 180 degrees.
Note. — A Triangle is also called a Trigon, and a Square a Tetragon.
46
GEOMETRY.
To exemplify these definitions, let A c b, in the following diagram,
be an assumed arc of a circle described with the radius A B.
B k, the Cosine of the arc A c b.
A g, the Tangent of do.
C B 6, the Complement, and b B E,
the Supplement of the arc Ac b.
C g, the Cotangent of the arc, writ-
ten cot.
B g, the Cosecant of the arc, written
cosec.
m C, the Cover sed sine of the arc,
or, by convention, of the angle
A B b : written coversin.
A c &, an Arc of the circle A C E D.
A 6, the Chord of that arc.
e D a, a Segment of the circle.
A B, the Radius.
A B b, a Sector.
A D E B, a Semicircle.
C B E, a Quadrant.
A e a E, a Zone.
n o h, a Lune.
B g, the Secant of the arc A c b.
b k, the Sine of do.
A k, the Versed Sine of do.
The Vertex of a figure is its top or upper point. In conic sections it
is the point through which the generating line of the conical surface
always passes.
The Altitude, or height of a figure, is a perpendicular let fall from its
vertex to the opposite side, called the base.
The Measure of an angle is an arc of a circle contained between the
two lines that form the angle, and is estimated by the number of de-
gi-ees in the arc. ' •
A Segment is a part cut off by a plane, parallel to the base.
A Frustrum is the part remaining after the segment is cut off.
The Perimeter of a figure is the sum of all its sides.
A Problem is something proposed to be done.
A Postulate is something required.
A Theorem is something proposed to be demonstrated.
A Lemma is something premised, to render what follows more easy.
A Corollary is a truth consequent upon a preceding demonstration.
A Scholium is a remark upon something going before it.
MENSURATION OF AREAS, LINES, AND SURFACES.
47
MENSURATION OF AREAS, LINES, AND SUR-
FACES.
OF FOUR-SIDED FIGURES.
!
Parallelograms.
Definition. Quadrilaterals having their opposite sides parallel.
To ascertain tlie Area of a Square, a Rectangle, a Rhombus, or a
Rhomboid.
Rule. — Multiply the length by the breadth or height, and
the product will be the area.
Or, length x breadth — area.
Note. — The surface of any Quadrilateral is equal to half the product
of the diagonals X the sine of their angle.
Or, abxb c=area.
Or, Ixb—area, I representing the length and b the breadth.
Square.
Definition. A plane superficies having equal sides and angles.
Fig. 1. a
b
X
c
Example. — The sides a b, b c,fig. 1, are 5 feet 6 inches
(5.5) ; what is the area.
5.5x5.5 = 30.25 square feet.
Centre of Gravity. Is in its geometrical centre,* o.
* The geometrical centre of any Parallelogram, regular Polygon,
Circle, Ellipse, &c, &c, is the point of intersection of two or more of
their respective diagonals, radii, or diameters.-
48
MENSURATION OF AREAS, LINES, AND SURFACES.
The side of a Square is equal to the square root of its area.
Example. — What is the side of a square when the area of
it is 1024 feet? Ans. 32 feet.
Rectangle.
Definition. A plane Superficies with parallel sides and equal
angles.
Fig. 2.
Example. — The side a &, Jig. 2, is 5 feet, and b c 1 feet
3 inches (7.25); what is the area?
5 x 7.25=36.25 square feet.
The length of a Rectangle is equal to the area divided by its breadth.
Example. — What is the length of a rectangle, the area
being 2048 feet and the breadth 32 ! Ans. 64 feet.
Centre of Gravity. Is in its geometrical centre, o.
Rhombus (Lozenge).
Definition. A plane superficies with equal sides, but its angles
not right angles.
Fig.
Example. — The height a b, fig. 3, is 5 feet 9 inches (5.75),
the length a c, 7 feet ; what is the area ?
5.75x7=40.25 square feet.
Note. — The opposite angles of a Rhombus are equal.
Centre of Gravity. Is in its geometrical centre, o.
MENSURATION OF AREAS, LINES, AND SURFACES. 49
Rhomboid.
Definition. A plane superficies with parallel sides, but its angles
not right angles.
Fig. 4. a c
b
Example. — The breadth a b, Jig. 4, is 5 feet, and the length
a Cj 6 feet ; what is the area 1
5x6=30 square feet.
The opposite angles of a Rhomboid are equal.
Centre of Gravity. Is in its geometrical centre, o.
Gnomon.
Definition. The space included between the lines forming two
parallelograms, of which the smaller is inscribed within the larger,
so that the angles of each are common to both.
To ascertain the area of a Gnomon.
. Rule. — Find the areas of the two parallelograms, and sub-
tract the less from the greater; the difference will give the
area required.
Or, a—a/=:area.
Example. — The dimensions of a gnomon are 10 by 10 and
6 by 6 inches ; what is its area ?
10x10=100. 6x6=36.
Then, 100— 36 = 64 ^difference of areas of the parallelograms
— the area required. . .
Ex. 2. The dimensions of two concentric parallelograms
are 15x8 and 12x6 feet ; what is the area of the gnomon ?
Ans. 48 feet.
Centre of Gravity. Is in its geometrical centre.
C
50 MENSURATION OF AREAS, LINES, AND SURFACES.
Triangles.
Definition. Plane superficies having three sides and angles,
and are designated
Right angled, when one of the angles is a right angle.
Acute angled, when all its angles are less than a right angle.
Obtuse angled, when one angle is greater than a right angle.
Equilateral, when the sides are equal.
Isosceles, when two of the sides are equal.
Scalene, when all the sides are unequal.
Notes. — Equiangular triangles are similar, or have their like sides
proportional.
The Hypothenuse is that side of a right-angled triangle which is op-
posite to the right angle.
The perpendicular height of a triangle is equal to twice its area di-
vided hy its base.
To ascertain the Area of a Triangle, Figs. 5, 6, 7.
Rule. — Multiply the base a b, by the height c d, and half
the product is the area.
„ a bxc d
Or, = = area.
area, b representing the base, and h the height.
c Fig. 6. c Fi9- 7-
) a d bad b
Example. — The base a b,fig. 5, is 4 feet, and the height
c b, 6 feet ; what is the area %
4x6=24, and 24^-2 = 12 square feet.
Ex. 2. The base a b, fig. 6, is 5 feet 6 inches, and the height
c d, 5 feet ; what is the area ?
5 feet 6 inches =5.5.
5.5 x5=27.5, and 27.5-^2 = 13.75 square feet.
MENSURATION OF AREAS, LINES, AND SURFACES. 51
V
Ex. 3. The base a b, fig. 7, is 6 feet 3 inches, and the height
c d is 5 feet 8 inches ; what is the area ?
6 feet 3 inches=6.25, and 5 feet 8 inches=5.666.
Then, 6.25 x 5.666=35.413, and 35^13 = 17.706 square feet.
Ex. 4. The base a b, fig. 7, is 18 feet 4 inches, and the
height c d, 10 feet 11 inches; what is the area?
18 ft. 4 in. = 18.333, and 10 ft. 11 in. = 10.916.
200 12S
18.333 x 10.916 = 200.123, and — ^ = 100.0625 feet.
Ex. 5. What is the area of a triangle when the base of it is
20 feet and the height 10.25 ? Ans. 102.5 feet.
Ex. 6. When the height of a triangle is 16.75 feet and the
base 6.24, what is its area? Ans. 52.26 feet.
Ex. 7. What is the area of a right-angled triangle when its
two shortest sides are 26 and 28 feet? Ans. 364 feet.
Ex. 8. The base of an equilateral triangle is 10 feet, what
is its area ?
10-7-2=5, or the length of the base of two equal right-angled
triangles, of which the length of the remaining sides of the triangle
is iiie hypothenuse.
Therefore, 102— 52 = 75, and y75 = 8.66, the height of each
of the right-angled triangles.
Hence, 8.66x5=43.3 square feet.
Ex. 9. The equal sides of an isosceles triangle are 50 feet,
and its base 28 ; how many square yards does it contain ?
Ans. 74f yards.
To ascertain the area of a Triangle by the length of its Sides
{Figs. 6 and 7).
Eule. — From half the sum of the three sides subtract each
side separately ; then multiply the half sum and the three re-
mainders continually together, and the square root of the prod-
uct is the area.
Or, Vs (s—a)x{s—b)x(s—c)=area, a, b, c being the sides,
. a-{-b-{-c
and s= — .
52 MENSURATION OF AREAS, LINES, AND SURFACES.
When all the Sides are equal.
Rule. — Square the length of a side, and multiply one fourth
of the product by 1.732, and this product will give the area
required.
S2
Or, — x 1.732 =area.
Example. — The sides of a triangle are 30, 40, and 50;
what is the area in square feet ?
30+40 + 50 120 r(. . . . ;,, .,
! ! = ==60, or half sum of the sides.
2 2
60—30=30^
60 — 40 = 20 y remainders.
60-50 = 10j
Whence 30x20x10x60=360000, and ^360000=600
square feet.
Ex. 2. How many acres are there in a triangle, the sides
of which are 49, 50.25, and 26 chains? Ans. 62.1875.
Ex. 3. What is the area of a triangle, the three sides being
26, 28, and 30 feet? Ans. 336 feet.
Ex. 4. What is the area of a triangle when its sides are
each 50 feet?
502
— — =z§2o =one fourth of the square of the length of a side,
and 625 x 1.732 = 1082.5 feet.
Ex. 5. What is the area of a right-angled triangle when
its sides are 50, 50, and 70.7107 feet? Ans. 1250 feet.
Ex. 6. What is the area of a triangle in square yards
when its sides are 500, 400, and 300 feet?
Ans. 6666.66 yards.
To ascertain the length of one side of a right-angled Triangle, the
length of the other two sides being given (Fig 5).
When the two legs are given, To ascertain the hypothenuse.
Rule. — Add together the squares of the two legs a b, b c,
and extract the square root of the sum.
MENSURATION OP AREAS, LINES, AND SURFACES. 53
Or, V« b2-\-b c2 = hypothenuse.
Or, Vb2 + h2, b representing the base, and h the height.
Example. — The base a b is 30 inches, and the height b c
40 ; what is the length of the hypothenuse %
302+402 — 2500, and -/2500=50 inches.
Ex. 2. The base of a right-angled triangle is 38.5 feet, and
the perpendicular 18 feet; what is the hypothenuse?
Ans. 42.5 feet.
Ex. 3. The height of a church steeple is 103 feet, and the
length of its shadow is 320 feet ; what is the distance from
the point of the shadow to the top of the steeple ?
Arts. 336.17 feet.
Ex. 4. The base of a triangle is 14 feet, and the height of
it 48 ; what is the length of its hypothenuse ?
Ans. 50 feet.
Ex. 5. The sides of a triangle are 6 yards, 1 foot, and 11.4
inches ; what is the length of its hypothenuse ?
Ans. 9 yards, 1 foot, and .2135 in.
To ascertain the other Leg {Fig. 5), the Hypothenuse and one of
the Legs being given.
Rule. — Subtract the square of the given leg from the square
of the hypothenuse, and the square root of the remainder is
the length of the leg required.
Or, ■jhyp.2- -Lj^
Or Ja c2 Sab2 = b c-
Or, ya c - |5 c2=a h
Example. — The base of a triangle is 30 feet, and the hy-
pothenuse 50 ; what is the height of it?
502-302z=2500-900, and 2500-900 = 1000.
Then -/1600 = 40/ee*.
Ex. 2. The hypothenuse of a triangle is 50 feet, and the
perpendicular 40 ; what is the base ?
502-402 = 2500-1600, and 2500-1600 = 900.
Then -/900= 30/^.
54
MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 3. It is required to find the length of a ladder, the
lower end placed 15 feet from the face of a wall, and the up-
per end resting on it at a height of 26 feet.
Ans. 30.017 feet.
Ex. 4. A pole* 50 feet in length, being placed in a street,
reached the sill of a window 30 feet from the ground on one side,
and being turned over without removing the foot from its lo-
cation, it reached a window on the opposite side 45^ feet high ;
what was the breadth of the street ? Ans. 60.734 feet.
Ex. 5. A ladder is to be placed so as to reach the top of a
wall 33.75 feet high, and the foot of it can not be set nearer
to the base of the wall than 18 feet ; what must be the length
of the ladder ! Ans. 38.25 feet.
Ex. 6. The base of an isosceles triangle is 25 feet, and the
sides of it are 32.5 feet; what is its perpendicular height?
Ans. 30 feet.
Ex. 7. If a ladder 100 feet in length was set upright against
a vertical wall, and then set out at its foot 10 feet from the
wall, how far would the top of the ladder fall?
Ans. .50125 inches.
Ex. 8. The width of the wall plates of a house is 48 feet,
and the height of the ridge is 10 feet ; what must be the length
of the rafters ? Ans. 2Q> feet.
When any two of the dimensions of a triangle and one of the correspond-
ing dimensions of a similar figure are given, and it is required to find the
other corresponding dimensions of the last figure.
Let A B C, a b c, be two similar triangles, Figs. 8 and 9.
C - Fig. 8.
Fig. 9.
B Aba.
nmAB:B C ::a b : b c, or a b : b c\\ A B : B C.
Note. — The same proportion holds with respect to the similar lineal
parts of any other similar figures, whether plane or solid.
MENSURATION OF AREAS, LINES, AND SURFACES. 55
Example. — The shadow of a cone 4 feet in length, set ver-
tical, was 5 feet ; at the same time, the shadow of a tree was
found to be 83 feet ; what was the height of the tree, both
shadows being on level ground ?
ab: k::AB:BC
5 : 4:: 83 : 66|
4
5)332
66$ /ee*.
Ex. 2. The side of a square is 5 feet, its diagonal 7.071
feet ; what will be the side of a square, the diagonal of which
is 4 feet? Ans. 2.828 feet.
Ex. 3. The length of the shadow of a spire is 151.5 feet,
while the shadow of a post 8 feet high is 6 feet ; what is the
height of the spire ? Ans. 202 feet.
To ascertain the length of a Side when the Hypothenuse of a
right-angled Triangle of equal sides alone is given.
Rule. — Divide the hypothenuse by 1.414213, and the quo-
tient will give the length of a side.
Or, - ; , * ; —the length of a side.
' 1.414213 * J
Example. — The hypothenuse of a right-angled triangle is
300 feet; what is the length of its sides?
300-^ 1.414213 = 212.1320 feet.
Ex. 2. The diagonal of a square is 28.28426 feet ; what is
the length of a side of it V Ans. 20 feet.
To ascertain the Perpendicular or Height of a Triangle when the
+Base and Area alone are given.
Rule. — Divide twice the area of the triangle by its base,
and the result is the length of the perpendicular.
Or,— = h.
Example. — The area of a triangle is 10 feet, and the length
of its base 5 ; what is its perpendicular 1
10x2 = 20, and 20-r-5 = 4/<W.
56 MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 2. The base of an isosceles triangle is 25 feet, and* its
area 375 feet; what is the height of its perpendicular?
Ans. 30 feet.
To ascertain the Perpendicular or Height of a Triangle when the
two Sides and the Base are given (Fig. 10).
Rule. — As the base is to the sum of the sides, so is the dif-
ference of the sides to the difference of the divisions of the
base. Half this difference being added to or subtracted from
half the base will give the two divisions thereof. Hence, as
the sides and their opposite division of the base constitute a
right-angled triangle, the perpendicular thereof is readily found
by preceding rules.
L. b c+c axb cCOC a . _ _
Or, 5 —b dcod a.
b a
Or, — —a d, whence -y/a c2—ad2—dc.
2 a b '
Fig. 10. c
Example. — The three sides of a triangle, a, b, c, Fig. 10,
are 9.928, 8, and 5 feet ; what is the length of the perpen-
dicular on the longest side ?
As 9.928 : 8 +5 : : 8 co5 : 3.928, the difference of the divisions of
the 'base.
Then 3.928-^2 = 1.964, which, added to -^p. =4.964 +
1.964 = 6.928, the length of the longest division of the base.
Hence we have a right-angled triangle with its base 6.928, and
its hypothenuse 8 ; consequently, its remaining side or perpendicular
is y'(82-6.9282)=4/^. -
Ex. 2. The three sides of a triangle are 42, 40, and 26 feet ;
what is the height thereof? Ans. 24 feet.
MENSURATION OF AREAS, LINES, AND SURFACES.
57
Centres of Gravity. On a line drawn from any angle to the
middle of the opposite side, at two thirds of the distance from
the angle, as at o,fig. 5.
TRAPEZIUMS AND TRAPEZOIDS.
Trapeziwn.
Definition. Quadrilaterals having unequal sides.
To ascertain the Area of a Trapezium (Fig. 11).
Rule. — Multiply the diagonal a c by the sum of the two
perpendiculars falling upon it from the opposite angles, and
half the product is the area.
~ acxb-t-d
Or, =area.
Fig. 11.
Example. — The diagonal a c,fig. 11, is 125 feet, and the
perpendiculars b and d, 50 and 37 feet ; what is the area?
125x50 + 37 = 10875, and 10875 -f- 2 =5437.5 square feet.
Ex. 2. What is the area of a plot of ground, the diagonal
being 12.5 poles, and the perpendiculars 50 and 37 poles?
Ans. 543.75 poles.
Ex. 3. What is the area of a trapezium, the diagonal being
42 feet, one perpendicular 18 feet, and the other 16?
Ans. 714 feet.
When the two opposite angles are supplements to each other, that
is, when a trapezium can be inscribed in a circle, the sum of its op-
posite angles being equal to two right angles, or 180°.
Rule. — From half the sum of the four sides subtract each
C 2
58 MENSURATION OF AREAS, LINES, AND SURFACES.
side severally ; then multiply the four remainders continually
together, and the square root of the product will be the area.
Example. — In a trapezium the sides are 15, 13, 14, and
12, and the diagonal 16 inches; required its area, its opposite
angles being supplements to each other.
15 + 13 + 14+12=54, and ^=27.
27 27 27 27
15 13 14 12
12x14x13x15 = 32760, and -^32760 = 180.997, Am.
Centre of Gravity. Draw the two diagonals, and find the
centres of gravity of each of the four triangles thus formed ;
join each opposite pair of their centres, and the intersection
of the two lines fs the centre of gravity.
Trapezoid.
Definition. A Quadrilateral with only one pair of opposite
To ascertain the Area of a Trapezoid (Fig. 12).
Rule. — Multiply the sum of the parallel sides a b, d c, by
a h, the perpendicular distance between them, and half the
product is the area.
a b-\-d cxa h
Or,
2
Or, - —area, s+/ representing the sides.
b
c h d
Example. — The parallel sides are 100 and 132 feet, and
the distance between them 62.5 feet ; what is the area !
100 + 132x62.5 = 14500, and 14500-4-2=7250 square feet.
MENSURATION OF AREAS, LINES, AND SURFACES. 59
Ex. 2. Required the area in feet, the distance between the
parallel sides being 12.5 feet, and the sides being 20 and 26.4
yards. Ans. 870 feet.
Ex. 3. How many square yards are there in a trapezoid
the breadth of which is 65 feet, and the two sides 28 and
38.5 feet? Ans. 240.1388.
Ex. 4. What is the area of a trapezoid the height of which
is 54.25 feet, and the sides 28 feet 1£ inches and 30 feet 4^
inches'? ' Ans. 1586.8125.
Centre of Gravity. On a line, e, joining the middle points
of the parallel sides a b, d c, the distance from
e (cd+2ab\
dc=-x( — -j- — r)
3 \c d-\-a b )
POLYGONS.
Definition. Plane figures having more than four sides, and
are either regular or irregular, according as their sides or angles
are equal or unequal, and they are named from the number of
their sides and angles. Thus,
A Trig on (triangle) has 3 sides.
An Octagon lias 8 sides.
Tetragon (square) 4 "
A Nonagon 9 "
Pentagon 5 "
Decagon 10 "
Hexagon 6 "
An Undecagon 11 "
Heptagon 7 "
A Dodecagon 12 "
fyc, fyc, fyc.
All of which being composed of isosceles triangles, they may be sim-
ilarly measured.
Regular Polygons.
To ascertain the Area of a regular Polygon (Fig. 13).
Rule. — Multiply the length of a side, a b, by the perpen-
dicular distance to the centre e c, and the product multiplied
by the number of sides and divided by 2 will be the area.
Or, —area, where n represents the number of sides.
60 MENSURATION OF AREAS, LINES, AND SURFACES.
Fig. 13. a
Example. — "What is the area of a pentagon, the side a b
being 5 feet, and the distance c e 4.25 feet?
5 x 4.25 x5 (n) = 106.25 =product of length of a side, the dis-
tance to the centre, and the number of sides.
Then 106.25^2 = 53.125 feet, the result required.
Ex. 2. What is the area of a hexagon when its side is
14.6 and its perpendicular 12.64 feet?
A ns. 553.632 square feet.
Ex. 3. What is the area of an octagon when its sides are
4.9705 and its perpendicular 6 feet? Ans. 119.292 feet.
To ascertain the Area of a Regular Polygon when the length of a
Side only is given.
Eule. — Multiply the square of the side by the multiplier
opposite to the name of the polygon in the following table,
and the product will be the area.
A.
B.
c.
D.
No. of
Sides.
Name of Polygon.
Area.
Radius of
Circumscribed
Circle.
Length of
the Side.
Radius of
Circumscribing
Circle.
Radius of
Inscribed
Circle.
3
Trigon
0.4330
2.
1.7320
.5773
.2887
4
Tetragon
1.
1.414
1.4142
.7071
.5
5
Pentagon
1.7205
1.238
1.1756
.8506
.6882
6
Hexagon
2.5981
1.156
1.
1.
.8660
7
Heptagon
3.6339
1.110
.8677
1.1524
1.0383
8
Octagon
4.8284
1.083
.7653
1.3066
1.2071
9
Nonagon
6.1818
1.064
.6840
1.4619
1.3737
10
Decagon
• 7.6942
1.051
.6180
1.6180
1.5388
11
Undecagon
9.3656
1.042
.5634
1.7747
1.7028
12
Dodecagon
11.1962
1.037
.5176
1.9319
1.8660
Example. — What is the area of a square when the length
of its sides is 7.0710678 inches?
7.07106782=50, and 50x1. =50 inches, Ans.
MENSURATION OF AREAS, LINES, AND SURFACES. Gl
Ex. 2. "What is the area of a hexagon when the length of
its sides is 5 inches'? Ans. 64.9525 inches.
Ex. 3. Kequired the area of an octagon, the length of its
sides being 3.8265 inches. Ans. 70.698 inches.
I
To ascertain the Radius of a Circle that contains a given Polygon
when the length of a Perpendicular from the Centre alone is
given.
Rule. — Multiply the distance from the centre to a side of
the polygon by the unit in column A, and the product will be
the radius of a circle that will circumscribe the figure.
Example. — What is the radius of a circle that contains a
hexagon, the distance to the centre being 4.33 inches'?
4.33x1.156=5 inches.
Ex. 2. What is the radius of a circle that contains an oc-
tagon, the distance from a side to the centre being 4.6168
feet? Ans. 5 feet.
Ex. 3. What is the radius of a circle that contains a square,
the distance from a side to the centre being 3.5355 feet?
Ans. 5 feet
To ascertain the length of a Side of a Polygon that is contained in
a given Circle when the radius of the Circle is given.
Rule. — Multiply the radius of the circle by the unit in
column B, and the product will be the length of the side of
the figure which the circle will contain.
Example.- — What is the length of the side of a pentagon
contained in a circle 8.5 feet in diameter ?
-^- = 4.25 radius, and 4.25 x 1.1756 = 5 feet.
Z
Ex. 2. What is the length of the side of a hexagon when
the diameter of the circumscribing circle is 20 feet?
Ans. 10 feet.
Ex. 3. What is the length of the side of an octagon when
the diameter of the circle is 10 feet? Ans. 3.8265 feet.
62 MENSURATION OF AREAS, LINES, AND SURFACES.
To ascertain the Radius of a Circumscribing Circle ivhen the
length of a Side is given.
Rule. — Multiply the length of a side of the polygon by the
unit in column C, and the product will give the radius of the
circumscribing circle.
Example. — What is the radius of a circle that will contain
a hexagon, a side being 5 inches ?
5x1=5 inches, Ans.
Ex. 2. What is the radius of a circle that will contain a
pentagon, a side of it being 5.878 inches. Ans. 5 inches.
Ex. 3. What is the radius of a circle that will contain an
octagon, when each of its sides is 7.653 feet ?
Ans. 10 feet.
Ex. 4. Each side of a pentagon is required to be 9 feet ;
what are the radii of the circumscribing and inscribed circles ?
9 x. 8506 = 7. 6554 = 7 feet 7.8648 inches, radius of circum-
scribing circle.
9 x. 6882=6.1938-6 feet 2.3256 inches, radius of inscribed
circle.
To ascertain the Radius of a Circle that can be inscribed in a given
Polygon when the length of a Side is given.
Rule. — Multiply the length of a side of the polygon by the
unit in column D, and the product is the radius of an in-
scribed circle.
Example. — What is the radius of the circle that is bound-
ed by a hexagon, its sides being 5 inches ?
5 x. 866 = 4.33 inches, Ans.
Ex. 2. The sides of an octagon are 8 inches ; what is the
radius of the inscribed circle? Ans. 9.657 inches.
Ex. 3. The sides of a pentagon are 5.878 inches ; what is
the radius of its inscribed circle? Ans. 4.045 inches.
To ascertain the Length of a Side and Radius of a regular Poly-
gon when the Area alone is given.
Rule. — Multiply the square root of the area of the polygon
MENSURATION OF AREAS, LINES, AND SURFACES.
63
by the multiplier in column E of the following table for the
length of the side ; by the multiplier in column G of the same
table for the radius of the circumscribing circle, and by the
multiplier in column H, also in the same table, for the radius
of the inscribed circle or perpendicular.
No. of
Sides.
Name of Polygon.
E.
Length of
the Side.
G.
Radius of
Circumscribing
Circle.
H.
Radius of
Inscribed
Circle
Angle.
Angle of
Polygon
3
4
5
6
Trigon
Tetragon
Pentagon
Hexagon
1.5197
1.
.7624
.6204
.8774
.7071
.6485
.6204
.4387
.5
.5247
.5373
120°
90
72
60
60°
90
108
120
7
8
9
10
Heptagon
Octagon
Nonagon
Decagon
.5246
.4551
.4022
.3605
.6045
.5946
.5880
.5833
.5446
.5493
.5525
.5548
5125'
45
40
36
128f
185
140
144
11
12
Undecagon
Dodecagon
.3268
.2989
.5799
.5774
.5564
.5577
32 43'
30
147^
150
Example. — The area of a square (tetragon) is 16 inches;
what is the length of its side ?
-y/16z=4, and 4 x 1=4 inches.
Ex. 2. The area of an octagon is 70.698 yards; what is
the length of the diameter of its circumscribing circle ?
Ans. 10 yards.
Ex. 3. The area of a square is 50 inches ; what is the
length of the radius of its inscribed circle?
Ans. 3.5355 inches.
Ex. 4. The area of a hexagon is 64.9525 inches; what is
the length of its sides'? Ans. 5 inches.
Ex. 5. The area of a decagon is 144 inches ; what are the
lengths of its sides, and of the radii of its* circumscribing and
inscribed circles'?
•v/144 = 12, and 12 x. 3605=4.326 inches,}
y!44 = 12, and 12 x.5833 = 6.9996 " *>Ans.
T/lU = 12,and 12 x. 5548=6.6576 " J
Additional uses of the foregoing Table.
The sixth and seventh columns of the table will greatly fa
cilitate the construction of these figures with the aid of a sec-
64
MENSURATION OF AREAS, LINES, AND SURFACES.
tor. Thus, if it is required to describe an octagon, opposite
to it, in column sixth, is 45 ; then, with the chord of 60 on the
sector as radius, describe a circle, taking the length 45 on the
same line of the" sector; mark this distance off on the circum-
ference, which, being repeated around the circle, will give the
points of the sides.
The seventh column gives the angle which any two adjoin-
ing sides of the respective figures make with each other.
REGULAR BODIES.
To ascertain the Surface or Linear Edge of any Regular Solid
Body*
Rule. — Multiply the square of the linear edge, or the radius
of the circumscribed or inscribed circle, by the units in the
following table, under the head of the dimension used, and
the product will be the surface or edge required.
Number
Radius of
Radius of
of sides.
Names of figures.
Surface.
circum. circle.
inscribed circle.
4
Tetrahedron
1.73205
1.63294
4.89898
6
Hexahedron
6.
1.15470
2.
8
Octahedron
3.46410
1.41421
2.44949
12
Dodecahedron
20.G4578
.71364
.89806
20
Icosahedron
8.66025
1.05146
1.32317
Example. — What is the surface of a hexahedron or cube
having sides of 5 inches ?
52 x 6 = 25 x 6 = 150 inches, Ans.
¥%, 2. What is the linear edge of a hexahedron circum-
scribed by a circle of 10 feet radius? Ans. 11.547 feet.
Centre of Gravity. In all regular polygons and bodies it is
at their geometrical centre.
Irregular Polygons.
Definition. Figures with unequal sides.
To ascertain the Area of an Irregular Polygon {Fig. 14).
Rule. — Draw diagonals to divide the figures into triangles
* See Appendix (p. 258-61) for additional rules both for Polygons
and Regular Bodies.
MENSURATION OF AREAS, LINES, AND SURFACES.
65
and quadrilaterals : find the areas of these separately, and the
sum of the whole is the area.*
Note or- ^b ascertain the area of mixed or compound figures, or such as
are composed of rectilineal and curvilineal figures together, compute the areas
of the several figures of which the whole is composed, then add them together,
and the sum will be the area of the compound figure : and in this manner
may any irregular surface or field of land be measured, by dividing it into
trapeziums and triangles, and computing the area of each separately.
a
Fig. U.
Example. — What is the content of Fig. 14?
e £=125 inches be— 35.7 inches ae=25 inches
ag= 20 " ec=130 " and d i-20 "
ebxa-g=125x20 =2500 -^-2 = 1250 ") Ans.
ebxb c=125 x 35.7=4462.5-^-2 = 2231.25 I 4781.25
ecxd {=130x20 =2600 ^2 = 1300 Jj ins.
Ex. 2. Required the area of the irregular figure abode
fga, Fig. 15. e
#£=30.5 Fi2'15'
gd=29
fd=2A.S
an-11.2
* Polygons containing one or more re-entering angles are called
Re-entering, as Fig. 15. The term re-entering is opposed to salient.
It is a property of a salient polygon that no right line can be drawn ex-
ternal to it that will cut its perimeter in more than two points, while in
a re-entering polygon such line may cut it in more than two points.
66 MENSURATION OF AREAS, LINES, AND SURFACES.
a ni° ° *9 b=U'2J~6 'x30.5 = 8.6x30.5 = 262.3=arga of
the trapezium a b c g.
'f ltC & *9 ^=11t6'6 X 29=8.8 x29=255.2=area of the
trapezium g c df.
f dxe P 24.8 x4
*== — — z=99. 2=49. 6=area of the triangle f d e.
Then 262.3 + 255.2 + 49.6 = 567.1 = area of the figure required.
Ex. 3. Required the area of an irregular polygon.
e 6=100 feet e c=110 feet
ag— 18 " b c— 12 "
a e= 45 " d i= 15 "
Ans. 2385 feet.
Ex. 4. In a pentangular field, beginning at the south side
and running toward the east, the first side is 2735 links, the
second 3115, the third 2370, the fourth 2925, and the fifth
2220 ; also the diagonal from the first to the third is 3800
links, and that from the third to the fifth 4010 ; what is the
area of the field? Ans. 117 ac. 2 ro. 39 po.
Note. — As this figure consists of three triangles, all of the sides of
which are given, by calculating their areas according to the rule, p. 51,
the sum will be the area of the whole figure accurately, without draw-
ing perpendiculars from the angles to the diagonals.
The same thing may also be done in most other cases of this kind.
When any part of the Figure is bounded by a Curve, the Area
may be found as follows :
Rule. — Erect any number of perpendiculars upon the base,
at equal distances, and find their lengths.
Add the lengths of the perpendiculars thus found together,
and their sum divided by their number will give the mean
breadth. Multiply the mean breadth by the length of the
base, and it will give the area of that part of the figure re-
quired.
MENSURATION OF AREAS, LINES, AND SURFACES.
G7
To ascertain the Area of a long Irregular Figure {Fig. 16).
Rule. — Take the breadth at several places and at equal
distances apart ; add them together, and divide their sum by
the number of breadths for the mean breadth ; multiply this
by the length of the figure, and the product is the area.
Example. — What is the area o£fig. 16 in square feet?
a c=50 inches, 3 = 54 inches,
1=52 " hd-= 60 "
2=48 " cd=150 "
50+52+48 + 54+60 = 264, and 264^-5=52.8.
Then 52.8 x 150 = 7920, which-h- 144 = 55 square feet, Ans.
To ascertain the Centre of Gravity of any. Plane Figure.
Rule. — Divide it into triangles, and find the centre of gravity of
each ; connect two centres together, and find their common centre ;
then connect this common centre and the centre of a third, and find
the common centre, and so on, always connecting the last found com-
mon centre to another centre till the whole are included, and the last
common centre will be that which is required.
Illustration. Where is the centre of gravity of Fig. 17 ?
Fig. 17. a Fig. 19.
Fig. 18.
68 MENSURATION OP AREAS, LINES, AND SURFACES.
Fig. 20. c Fig. 21.
The point • represents the centre of gravity of each triangle, a b c,
Fig. 18, of which the figure is composed.
Connect two centres of gravity, as in Fig. 19 ; join their common cen-
tre to the centre of gravity of the next triangle, as in Fig. 20 ; join their
common centre o to the centre of the remaining triangle, Fig. 21, and
the common centre of this last connection is the required centre of
gravity of the figure.
CIRCLE.
Definition. A plane figure bounded by a true curve, called its
Circumference or Periphery.
The Diameter of a Circle is a right line drawn through its
centre, bounded by its periphery.
The Radius of a Circle is a right line drawn from the centre
of it to its circumference.
The Circumference of a Circle is assumed to be divided into
360 equal parts, called degrees; each degree is divided into 60
parts, called minutes ; each minute into 60 parts, called seconds;
and each second into 60 parts, called thirds, and so on.
To ascertain the Circumference of a Circle {Fig. 22).
Rule. — Multiply the diameter a b by 3.1416,* and the
product is the circumference.
* The exact proportion of the diameter of a circle to its circumfer-
ence has never yet been ascertained. Nor can a square or any other
right-lined figure be found that shall be equal to a given circle. This
is the celebrated problem called the squaring of the circle, which has
exercised the abilities of mathematicians for ages, and been the occa-
MENSURATION OF AREAS, LINES, AND SURFACES. 69
Or, as 7 is to 22, so is the diameter to the circumference.
Or, as 113 is to 355, so is the diameter to the circumference.
sion of so many disputes. Several persons of eminence, have, at dif-
ferent times, pretended that they had discovered the exact quadrature ;
but their errors have soon been detected, and it is now conceded as a
thing impracticable of attainment.
Though the relation between the diameter and circumference can not
be accurately expressed, it may yet be approximated to great exactness.
In this manner was the problem solved by Archimedes, about two thou-
sand years ago, who discovered the proportion to be nearly as 7 to 22.
This he effected by showing that the perimeter of a circumscribed reg-
ular polygon of 192 sides is to the diameter in a less ratio than that
of 3^- to 1, and that the perimeter of an inscribed polygon of 96 sides
is to the diameter in a greater ratio than that of 3 -ff- to 1, and from
thence inferred the ratio above mentioned, as may be seen in his book
De Dimensione Circuit.
The proportion of Vieta and Metius is that of 113 to 355, which is
something more than the former. This is a commodious proportion ;
for,- being reduced into decimals, it is correct as far as the sixth figure
inclusively. It was derived from the pretended quadrature of a M. Van
Eick, which first gave rise to the discovery.
But the first who ascertained this ratio to any great degree of ex-
actness was Van Ceulen, a Dutchman, in his book De Circulo et Ad-
scriptis. He found that if the diameter of a circle was 1, the circum-
ference would be 3.141592653589793238462643383279502884 nearly;
which is exactly true to 36 places of decimals, and was effected by the
continual bisection of an arc of a circle, a method so extremely trouble-
some and laborious that it must have cost him incredible pains. It is
said to have been thought so curious a performance, that the numbers
were cut on his tomb-stone in St. Peter's Church-yard at Leyden.
This last number has since been confirmed and extended to double the
number of places by the late ingenious Mr. Abraham Sharp, of Little
Horton, near Bedford, in Yorkshire.
But since the invention of Eluxions, and the Summation of Infinite
Series, there have been several methods discovered for doing the same
thing with much more ease and expedition. The late Mr. John Machin,
Professor of Astronomy in Gresham College, has by these means given
a quadrature of the circle which is true to 100 places of decimals ; and
M. de Lagny, M. Euler, &c, have carried it still farther. All of which
proportions are so Extremely near the truth, that, except the ratio
could be completely obtained, we need not wish for a greater degree
of accuracy. — Bonntcastle.
\
70 MENSURATION OF AREAS, LINES, AND SURFACES.
Fig. 22.
Example. — The diameter of a circle, fig. 22, is 1.25 inches ;
what is its circumference ?
1.25x3.1416 = 3.927 inches.
Ex. 2. The diameter of a circle is 17 feet; what is its cir-
cumference? Ans. 53. 4072 feet.
To ascertain the Diameter of a Circle (Fig. 22).
Divide the circumference by 3.1416, and the quotient is the
diameter.
Or, as 22 is to 7, so is the circumference to the diameter.
To ascertain the Area of a Circle (Fig. 22).
Rule. — Multiply the square of the diameter by .7854, and
the product is the area.
Or, multiply the squai*e of the circumference by .07958.
Or, multiply half the circumference by half the diameter.
Or, multiply the square of the radius by 3.1416.
Or, p r2=area, where p represents the ratio of the circumfer-
ence to the diameter, and r the radius.
Example. — The diameter of a circle is 8 inches ; what is
the area of it ?
82 or 8 x 8 = 64, and 64 x .7854=50.2656 inches.
Ex. 2. What is the area of a circle, the radius being 39J
yards? . Ans. 4839.8311 yards.
Ex. 3. What is the area of a circle in feet, the diameter
being 15j poles? Ans. 3214.6587 /«?*.
Ex. 4. What is the circumference of a circle, the area of
which is an acre? Ans. 246 yds., 1 ft., 10^ in.
Centre of Gravity. Is in its geometrical centre. Semicircle.,
4 r
k — —distance from centre.
'6.p J
MENSURATION OF AREAS, LINES, AND SURFACES.
71
USEFUL FACTORS.
In which p represents the Circumference of a Circle the Diameter
of which is 1.
Then
p = 3.1415926535897932384626 +
2p = 6.283185307179 +
4 /> = 12.566370614359 +
±p= 1.570796326794 +
ip= 0.785398163397 +
!>■?
4.188790
ip=
.523598
ip=
.392699
•fsP =
.261799
sh>P=
.008726
l_
P
2
P
.318309
.636619
4_
1.273239
.079577
1.772453
.886226
3.544907
F97884
.564189
I
4~"/>
Vp —
Wp=
2Vp=
p
p
— =114.591559
P
fp=t 2.094395
5= 1.909859
P
36/> = 113.097335
In which the Diameter of a Circle is 10.
1. Chord of the arc of the semicircle =10.
2. Chord of half the arc of the semicircle == 7.071067
3. Versed sine of the arc of the semicircle = 5.
4. Versed sine of half the arc of the semicircle = 1.464466
5. Chord of half the arc of the half of the arc of the
semicircle - 3.82683
6. Half the chord of the chord of half the arc = 3.535533
7. Length of arc of semicircle = 15.7075*63
8. Length of half the arc ,of the semicircle == 7.853981
Square of the chord of half the arc of the semicircle (2.) =50.
Square root of versed sine of half the arc (4.) = 1.210151
Square of versed sine of half the arc (4.) = 2.144664
Square of the chord of half the arc of half the arc of the
semicircle (5.) =14.64467
Square of half the chord of the chord of half the arc (6.) =12.5
In all the following calculations, p is taken at 3.1416, ip at
.7854, \p at .5236, and whenever the decimal figure next to
the one last taken exceeds 5, one is added. Thus, 3.14159
for four places of decimals is set down 3.1416.
72 MENSURATION OF AKEAS, LINES, AND SURFACES.
Circular Arc.
Definition. A part of the Circumference of a Circle.
Fig. 23. c
The Sine of an Arc is a line running from one extremity of an arc
perpendicular to a diameter joining the other extremity, as a d,Jig. 23.
The Sine of an Angle is the sine of the arc that measures that angle.
The Versed Sine of an Arc or Angle is the part of the diameter inter-
cepted between the sine and the arc, as d b. It is frequently written
versed sine of half the arc.
The Complement of an Arc or Angle is what,remains after subtracting
the angle from 90°, as a o c, Fig. 23.
The Supplement of an Arc or Angle is what remains after subtracting
the angle from 180°, as a o e, Fig. 23.
TJie Coversed Sine is the sine of the complement of the arc, as c g,
Fig. 23.
Note. — For other illustrations of these definitions, see figure,
page 46.
To ascertain the Length of an Arc of a Circle {Fig. 24) when the
Number of Degrees and the Radius are given.
Rule 1. As 180 is to the number of degrees in the arc,
so is 3.1416 the radius to its length.
Rule 2. Multiply the radius, o a, of the circle by
.01745329, and the product by the degrees in the arc.
If the length is required for minutes, multiply the radius
by .000290889; if for seconds, multiply by .000004848.
(See Table of length of Arcs, page 130.)
MENSURATION OF AREAS, LINES, AND SURFACES. 73
Fig. 24. c
Example. — The number of degrees in an arc, oab, are 90,
and the radius, o b, 5 inches ; what is the length of the arc?
180 : 90 :: 5 x 3.1416 : 7.854 inches, Ans.
Ex. 2. The radius of an arc is 10, and the measure of its
angle 44° 30' 30" ; what is the length of the arc 1
10 x. 01745329 = . 1745329, which x 44=7.679447£, the
length for 44°.
10 x. 000290889 = .00290889, whichx 30 = .0872667, the
length for 30r.
10 x. 000004848 = .00004848, wMx 30 = .00 14544, the
length for 30".
Then, 7.67944761
.0872667 1=7.7681687, the length required.
.0014544 J
Or, reduce the minutes and seconds to the decimal of a degree,
and multiply by it.
See Rule, page 34. 30' 30//=.5083, and .1745329 from
above x 44.5083 = 7.768163.
Ex. 3. The degrees in the arc of a circle are 90, and the
diameter of the circle is 20 feet ; what is the length of the
arc? Ans. 15.708.
Ex. 4. The degrees in the arc of a circle are 32° 38' 42",
and the radius of it is 25 inches ; what is the length of the arc %
32o 38' 42"=32.645 degrees.
Ans. 14.2441 inches.
Ex. 5. The degrees in the arc of a circle are 147° 21' 18",
and the radius of its circle is 25 feet ; what is the length of
the arc ? Ans. 64.2959 feet.
D
74 MENSURATION OF AREAS, LINES, AND SURFACES.
When the Chord of half the Arc and the Chord of the Arc are
given.
Rule. — From 8 times the chord, a b, of half the arc, Fig.
25, subtract the chord, a c, of the arc, and one third of the
remainder will be the length nearly.
k 8 c-c ,
Or, — - — , c representing the chord of half the arc, and c
the chord of the arc.
Fig. 25. b
Example. — The chord of half the arc, a b, is 30 inches,
and the chord of the arc 48 ; what is the length of the arc ?
30 x 8 = 240 = 8 times the chord of half the arc.
240-48 = 192, and 192-^-3=64 inches, Ans.
Ex. 2. The chord of half the arc is 60 feet, and the chord
of the arc 96 ; what is the length of the arc?
Ans. 128 feet.
Ex. 3. The chord of half the arc is 17.67765 feet, and the
chord of the arc 25 ; what is the length of the arc ?
Ans. 38.80706*/^.
Ex. 4. The chord of half the arc is 3.8268 inches, and the
chord of the arc 7.071 ; what is the length of the arc1?
Ans. 7.8478-f
When the Chord of the Arc and the Versed Sine of the Arc are
given.
Rule. — Multiply the square root of the sum of the square
of the chord, a c, Fig. 25, and four times the square of the
* The exact length is 39.27 feet.
+ The exact length is 7.854 inches.
MENSURATION OF AREAS, LINES, AND SURFACES. 75
versed sine, b r (equal to twice the chord of half the arc), by-
ten times the square of the versed sine ; divide this product
by the sum of fifteen times the square of the chord and thirty-
three times the square of the versed sine ; then add this quo-
tient to twice the chord, a b, of half the arc,* and the sum
will be the length of the arc very nearly.
v c +4uaXl0u2
Or, — — — — + \/c2+4: v2} c representing the chord, and v the
LO C ~v OoV
versed sine.
To ascertain the Versed Sine when the Chord of Half the Arc
and Chord of the Arc are given.
Rule. — From the square of the chord, a b, of half the arc, sub-
tract the square of half the chord of the arc a c (—a r2), and the
square root of the remainder is the versed sine (jb r).
Or Vc'z — (c-r-'2,y = versed sine.
To ascertain the Length of the Chord of Half the Arc.
Rule 1. Divide the square root of the sum of the square of the
chord of the arc and four times the square of the versed sine by two,
and the result will be the length. .
. Or, -\/c2-\-4: v2-r-2=chord of half the arc.
Rule 2. From the sum of the squares of half the chord of the arc
and the versed sine, take the square root, and the result will be the
length.
7(i)v=
Or,\J [-) -\-v2=zchord of half the arc.
Rule 3. Multiply the diameter by the versed sine, and the square
root of their product will be the length.
Or, -\/dXv=chord of half the arc.
To ascertain the Versed Sine when the Chord of Half the Arc
and the Diameter are given.
Rule. — Divide the square of the chord of half the arc by the
diameter, and the quotient will be the versed sine.
Or, (c'2-7-d)=zversed sine.
* The square root of the sum of the square of the chord and four times
the square of the versed sine is equal to twice the chord of half the arc.
76 MENSURATION OF AREAS, LINES, AND SURFACES.
When the Chord of the Arc and the Diameter are given.
Rule. — From the square of the diameter subtract the square of
the chord and extract the square root of the remainder ; subtract this
root from the diameter, and half the remainder is the versed sine.
Or, r —versed sine.
m
When the Versed Sine is greater than a Semidiameter.
Proceed as before, but add the square root of the remainder (of the
squares of the diameter and chord) to the diameter, and half the sum
is the versed sine.
Or, ~ =versed sine.
To ascertain the Diameter.
Rule 1 . Divide the square of the chord of half the arc by the
versed sine, and the result will be the diameter.
c'2
Or, — =diameter.
v
Rule 2. Add the square of half the chord of the arc to the square
of the versed sine ; divide this sum by the versed sine, and the re-
sult will give the diameter.
(I)V
Or, ^.diameter.
v
To ascertain the Chord of the Arc when the Chord of Half the
Arc and the Versed Sine are given.
Rule. — From the square of the chord of half the arc subtract the
square of the versed sine, and twice the square root of the remainder
will give the chord.
Or, VV2— v2)X2— chord of the arc.
Example. — The chord of half the arc is 60 inches, and the versed
sine 36 ; what is the length of the chord of the arc 1
602— 362=2304, and -/2304 X 2=96, Ans.
When the Diameter and Versed Sine are given.
Rule. — Multiply the versed sine by 2, and subtract the product
from the diameter ; then subtract the square of the remainder from
MENSURATION OF AREAS, LINES, AND SURFACES. 77
the square of the diameter, and the square root of the remainder will
give the chord.
Or, V(oX2— dY— d2—c.
If the diameter and chord of half the arc only are given, find the
versed sine, as per rule, p. 75, then proceed as above.
Example. — The diameter of a circle is 100 feet, and the versed
sine of half the arc is 36 ; what is the length of the chord of the arc 1
36 X 2— 100=28, then 28s— 1002=9216, and -^9216=96, Ans.
Example. — The chord of an arc is 80 inches, and its versed
sine 30 ; what is the length of the arc?
802 = 6400 —square of the chord.
302 = 900= square of the versed sine.
-y/(6400-f-900x4) — 100 =zsquare root of the square of the
chord and four times the square of the versed sine, which is,
twice the chord of half the arc.
Then, 100 x302 x 10 = 900000 —product of ten times the square
of the versed sine and the root above obtained.
802x 15 = 96000«^15 times the square of the chord.
302x33 = 29700 = 33 times the square of the versed sine.
125700
. .100x302xl0 900000 ,._ ___
l25700~ ^ 125700 ? ' '
or twice the chord of half the a? c = 107. 1599 feet.
Ex. 2. The chord of an arc is 7.07107 inches, and the
versed sine 1.46447 ; what is the length of the arc?
7.07 1072 =50 —the square of the chord.
1.464472 x 4= 8.5787 =4 times the square of the versed sine.
V '58.57 87 = 7. 6536 = twice the chord of half the
arc.
1.464472 X 10 = 21.4467 = 10 times the square of the versed sine.
7.07l072x 15 = 750. =15 times the square of the chord.
1.464472 X 33 = 70.7742 =33 times the square of the versed sine.
820.7742
Then> 7^S^P^+7-Q5d6=7'S53Q inches-
78 MENSURATION OF AREAS, LINES, AND SURFACES. ■
Ex. 3. The chord of an arc is 96 feet, and the versed sine
36 ; what is the. length of the arc? Ans. 128.5918 feet.
Ex. 4. The chord of an arc is 40 inches, and the versed sine
15 ; what is the length of the arc? Ans. 53.58 inches.
Ex. 5. The chord of an arc is 48 inches, and the versed sine
18 ; what is the length of the arc? Ans. 64.2959 inches.
Ex. 6. The chord of an arc is 60 inches, and .its versed
sine 10 ; what is the length of the arc?
Ans. 64.3493 inches.
Ex. 7. The versed sine of an arc is 2.5658, and the chord
31.6228 inches; what is the length of the arc?
Ans. 32.1747 inches.
Ex. 8. The chord of an arc is 7.071, the chord of half the
arc is 3.8268, and the diameter of the circle 10 inches; what
are the lengths of the versed sine and arc ?
By Note, page 75.
3.82682— 14.6444 =square of half the chord of the arc.
7.071^-2=3.5356, and 3.53562= 12.5 ^square of half the
chord of the arc. ,
Then, -^14.6444 — 12.5 = 1 AG ±4:= versed sine.
Or, by preceding rule, page 75,
3.82682 14.6444 , AnAA
— — — == — — — = 1Ao4:4:= versed sine.
Length of arc by rule, p. 74, Example 2, p. 77, 7.8536 inches.
Ex. 9. The chord of an arc is 96, and the versed sine 36
inches ; what is the chord of half the arc ? Ans. GO inches.
Ex. 10. The chord of an arc is 70.7107, and the versed
sine 14.6447 inches; what are the lengths of
The arc? Ans. 78.54 inches.
The chord of half the arc ? Ans. 38.268 "
And the diameter of the circle? Ans. 100. "
When the Diameter and Versed Sine are given.
Eule — .Multiply twice the chord of half the arc, a b, Fig.
25, by 10 times the versed sine, b r; divide the product by
27 times the versed sine subtracted from 60 times the diam-
MENSURATION OF AREAS, LINES, AND SURFACES. 79
eter, and add the quotient to twice the chord of half the arc ;
the sum will be the length of the arc very nearly.
c x 2 x 10 v
Or, — —z (- 2c', d representing the diameter.
Example. — The diameter of a circle is 100 feet, and the
versed sine of the arc 25 ; what is the length of the arc 1
V25 x 100 =50= chord of half the arc.
50 x 2 x25xl0=25000 = fance the chord of half the arc by 10
times the versed sine.
100x60—25x27=5325 = 27 times the versed sine subtracted
from 60 times the diameter.
25000
Then =4.6948, and 4.6948 + 50x2 = 104.6948 feet.
Ex. 2-. The diameter of a circle is 10 inches, and the versed
sine 1.46447 ; what is the length of the arc"?
V 1.46447 X 10 = 3.8268=c/wrd of half the arc.
3.8268 x 2 xl0xl.46447 = 112.0847 = ^'ce the chord of half
the arc by 10 times the versed sine.
10 x 60 — 1.46447 x 27=560.459 = 27 times the versed sine sub-
tracted from 60 times the diameter.
Then 112.0847 -f-560.459 = . 19998, and .19998 added to
3.8268 x 2 {twice the chord of half the arc) = 7.8536 inches.
Ex. 3. The diameter of a circle is 10 inches, and the versed
sine 1.46447 ; what is the length of the arc ?
Ans. 7.854 inches.
Ex. 4. The diameter of a circle is 41.66 feet, and the versed
sine 15 ; what is the length of the arc?
Ans. 53.5799 feet.
Ex. 5. The diameter of a circle is 200 feet, and the versed
sine of the arc 72 ; what is the length of the arc?
Ans. 257.1837 feet.
Ex. 6. The diameter of a circle is 80 feet, and the chord
of half the arc is 16.018 ; what is the versed sine and what is
the length of the arc !
Ans. Versed sine, 3.2072; Length of arc, 32.2539 feet.
80 MENSURATION OF AREAS, LINES, AND SURFACES.
Centre of Gravity. Multiply the radius of the circle by the
chord of the arc, and divide the product by the length of the
arc ; the quotient is the distance of it from the centre of the
circle.
r x c
Or, — — =zdista?ice from the centre of the circle.
EXAMPLES UNDER THE SEVERAL RULES.
1. The degrees in the arc of a sector are 30° 38/ 42", and
the radius of the circle 50 ; what is the length of the arc ?
Ans. 26.7429.
2. The chord of an arc is 70.71067, and the chord of half
the arc 38.268 ; what i&the length of the arc?
Ans. 78.536.
3. The chord of an arc is 48, and the versed sine 18 ; what
is the length of the arc? Ans. 64.2959.
4. The chord of an arc is 50, the radius 40, and the versed
sine 8.775 ; what is the length of the arc ? Ans. 54.0096.
*5. The diameter of a circle is 10, and the versed sine 5 ;
what is the chord of the arc, the chord of half the arc; and
the length of the arc ?
f Chord of the arc, 10.
Ans. -j Chord of half the arc, 7.07107.
\ Length of the arc, 15.708.
6. The diameter of a circle is 100, and the chord of half
the arc 60 ; what is the versed sine? Ans. 36.
7. The diameter of a circle is 100, and the chord of the
arc 60 ; what is the versed sine, the chord of half the arc,
and the length of the arc ?
f Versed sine, 10.
Ans.l Chord of half the arc, 31.6228.
I Length of the arc, 64.3493.
8. The chord of an arc is 96, and the versed sine 36 ; what
are the lengths of the chord of half the aro, the diameter, and
the arc? ( Chord of half the arc, 60.
Ans-i Diameter of the circle, 100.
I Length of the arc, 128.5918.
MENSURATION OP AREAS, LINES, AND SURFACES. 81
Proportions of the Circle, its Equal, Inscribed, and Circumscribed
Squares.
CIRCLE.
1. Diameter X.8862) 0.-, ^ ,
. , )™~ >■ = Side of an equal square.
2. Circumference X. 2821) ^ H *
3. Diameter X .7071"
4. Circumference x .2251 f = Side of the inscribed square.
5. Area X .9003,
SQUARE.
6. A Side X 1.4142 ^Diameter of its circumscribing circle.
7. " x 4.443 = Circumference of its circumscribing circle.
8. " X 1.128 ^Diameter * - , . ,
°' it ' , y of an equal circle.
9. " X 3.545 = Circumference) *
10. Square inches X 1.273 = Round inches.
Note. — The square described within a circle is one half the area of
one described without it.
Sector of a Circle.
Definition. A part of a Circle bounded by an arc and two
radii.
To ascertain the Area of a Sector of a Circle when the Degrees
in the Arc are given {Fig. 26).
Kule. — As 360 is to the number of degrees in the sector,
so is the area of the circle of which the sector is a part to the
area of the sector.
, •»
Or, —area, d representing the degrees in the arc and a
ooO
the area of the circle.
Fig. 26. b
a sZ^Z. .7^T>s. c
82 MENSURATION OP AREAS, LINES, AND SURFACES.
Example. — The radius of a circle, o a, is 5 inches, and the
number oidegrees of the sector is 22° 30' ; what is the area?
Area of a circle of 5 inches radius is 78.54 inches.
Then, as 360° : 22° 30' : : 78.54 : 4.90875, Ans.
Ex. 2. The degrees in the arc of a sector are 147° 21/ 18",
and the area of the circle is 1963.5 feet; what is the area of
the sector?
To reduce 147° 21' 18" to a decimal.
21 18
60 #
60)1278
60)213
.355
Then .355 + 147 = 147.355.
Ans. 803.6987/*?*.
Ex. 3. The degrees in the arc of a sector are 32° 38' 42",
and the area of the circle is 1963.5 inches; what is the area
of the sector? Ans. 178.0512 inches.
Ex. 4. The degrees in the arc of a sector are 90°, and the
area of the circle is 981.75 inches; what is the area of the
sector? • Ans. 245.4375 inches.
Ex. 5. The degrees in. the arc of a sector are 90°, the versed
sine 1.46446 feet, and the chord of half the arc is 3.8268 ;
what is its area ?
3.82682 = 14.6446, the square of the chord of half the arc.
14.6446 -r- 1.46446 = 10 = the square of the chord of half the arc
-^-the versed sine, which is the diameter.
102 x .7854 = 78.54, the area of the whole circle.
Then, as 360° : 90° :: 78.54 : 19.635 feet
Note. — Divide the area by .7854, and the square root of the quotient
is the diameter of the circle.
Illustration. The area of a circle is 176.715 ; what is its diameter?
176.715-^.7854=225, and ^225 = 15, Ans.
When the Length of the Arc, fyc, are given {Fig. 26).
Rule. — Multiply the length of the arc, a c b, by half the
length of the radius, a o, and the product is the area.
MENSURATION OF AREAS, LINES, AND SURFACES. 83
T
Or, ay,-— area, a representing the arc, and r the radius.
z
Example. — The length of the arc of a sector is 7.854 inches,
and a radius of it is 5 ; what is its area ?
5
7.854x^ = 7.854x2.5 = 19.635 inches.
Ex. 2. The length of the arc of a sector is 10.472 inches,
and a radius 5 ; what is its area? Ans. 26.18 inches.
Ex. 3. The length of the arc of a sector is 14.19 inches, the
diameter of the circle being 100 ; what is its area ?
Ans. 354.75 inches.
Ex. 4. The radius of a circle is 25 feet, and the versed sine,
b r, of the arc of a sector is 18 ; what is the area of the sector?
By Eule, page 78, the length of the arc is 64.2959.
Ans. 803.69875 feet.
Note. — If the diameter or a radius is not given, see Rules, page 7G.
Ex. 5. What is the area of a sector when the versed sine
of its arc is 15, and the chord 40 inches'?
40-1-2 =400= square of half the chord of the arc.
152= 225= square of the versed sine.
lien, - — ~- =41.666, the diameter, and — '- — =20.83
15 L
the radius.
20.833
Length of arc by Mule, page 78, 53.58, and 53.58 x
23
558.116 inches.
Ex. 6. The radius of a circle is 50 inches, and the versed
sine of the arc of a sector of it is 25 ; what is its area ?
Ans. 2617.37 inches.
Ex. 7. The diameter of a circle is 100 feet, the versed sine
of the arc of a sector is 36, and the chord of half the arc is
60 ; what is the area of the sector? Ans. 3214.795 feet.
Ex. 8. The length of the arc of a sector is 104.6948 inches,
the chord of the arc is 86.6024, and the versed sine of it is
25 ; what is the area of the sector ?
Ans. 2617.37 inches.
84 MENSURATION OF AKEAS, LINES, AND SURFACES.
Centre of Gravity. Multiply twice the chord of the arc by
the radius of the sector, and divide their product by three
times the length of the arc ; the quotient is the distance from
the centre of the circle.
2 c r
Or, ———^distance from centre of circle ; r representing radi-
us, and I the length of the arc.
Example. — Where is the centre of gravity of the sector
given in example 5 ?
40 X 2 = 80 = twice the chord of the arc.
80 x 20.833 = 1666. 64 =product of twice the chord and the
radius.
53.58x3 = 160.74=^ree times the length of the arc.
Then, 1666.64—160.74= 10.369, the distance from the centre
of the circle.
Segment of a Circle.
Definition. A part of a circle hounded by an arc and a chord.
To ascertain the Area of a Segment of a Circle, Fig. 27, when
the Chord and Versed Sine of the Arc, and Radius or Diam-
eter of the Circle are given.
When the Segment is less than a Semicircle, as ab c, Fig. 27.
Rule. — Find the area of the sector having the same arc as
the segment ; then find the area of the triangle formed by the
chord of the segment and the radii of the sector, and the dif-
ference of these areas will be the area required.
Note. — Subtract the versed sine from the radius; multiply the re-
mainder by one half of the chord of the arc, and the product will be
the area of the triangle.
Or, a— a' —area of segment; a representing area of the sector,
and a' the area of the triangle.
When the Segment is greater than a Semicircle, as a e c, Fig. 27.
Rule. — Find, by the preceding rule, the area of the lesser
portion of the circle, a b c ; subtract it from the area of the
whole circle, and the remainder is the area required.
MENSURATION OF AREAS, LINES, AND SURFACES. 85
Or, c—c'=area of segment; c representing area of circle, and
c' area of the lesser portion.
(See Table of Areas, page 134.)
Fig. 27. b
Example. — The chord, a c, Fig. 27, is 14.142, the diameter,
b e, is 20, and the versed sine, b d, is 2.929 inches ; what is the
area of the segment ?
14.142
By Rule, page 75, — - — = 7.071 = half the chord of the arc.
m
V7.0712+2.9292=7.654 = ^e square root of the sum of the
squares of half the chord of the arc and versed sine, which is
the chord a b of half the arc a b c.
By Rule, page 78.
7.654 x 2 x 10 x 2.929 = 448.371 = twice the chord of half the
arc by 10 times the versed sine.
20 x 60 — 2.929 X 27 = 1120.917 = 00 times the diameter sub-
tracted from 27 times the versed sine.
Then, 448.37 1-r- 1120,917 =.400, and .400 added to 7.654x2
{twice the chord of half the arc) = 15.708 inches, the length of
the arc. By Rule, p. 82, 15.708 x —= 78.54 = the arc multi-
plied by half the length of radius, which is the area of the sector.
10 — 2.929 = 7.071 = ^6 versed sine subtracted from a radius,
14.142
which is the height of the triangle a o c, and 7.071 x — ~ —
= 50 = araz of the triangle.
Consequently, 78.54—50 = 28.54 inches.
86 MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 2. The chord of the arc of a segment is 86.6024, the
versed sine 25, and the radius 50 feet; what is the area of
the segment ? Ans. 1534.84: feet.
Ex. 3. The chord of the arc of a segment is 28 feet, the
diameter of the circle 100, and the versed sine of the arc 2 ;
what is the area of the segment? Ans. 37.4852 feet.
Ex. 4. The diameter of a circle is 50 feet, the chord of the
arc of a segment of it is 30, and its versed sine 5 ; what is
the length of the arc, and what the area of the segment ?
Ans. Arc, 32.17 46 feet; Segment, 102.183 feet.
Ex. 5. The chord of a segment is 56 inches, its versed sine
4, and the radius of the circle 200 inches ; what is its area1?*
Ans. 149.941 inches.
When the Chords of the Arc, and of the half of the Arc, and
the Versed Sine are given.
Rule. — To the chord, a c, of the whole arc, add the chord,
a b, of half the arc and one third of it more ; multiply this
sum by the versed sine, b d, and this product, multiplied by
.40426, will give the area nearly.
c'
Or, c-\-c -\-— xv x .40426= a?*ea nearly,
o
Example. — The chord of a segment is 28 feet, the chord
of half the arc is 15, and the versed sine 6 ; what is the area
of the segment ?
28 + 15-j- — =4:8 = the chord of the arc added to the chord of
o
half the arc and ^ of it more.
48 x 6 = 288— product of above sum and the versed sine.
Then, 288 x .40426 = 116.427 feet, the area required.
Ex. 2. The chord of a segment is 40 feet, its versed sine
10, and the chord of half the arc is 22.36 ; what is its area?
Ans. 282.226 feet.
Ex. 3. What is the area of a segment, its half chord being
14, the chord of half its arc 14.142, and its versed sine 2 yards?
Ans. 37.884 yards.
MENSURATION OF AREAS, LINES, AND SURFACES. 87
Ex. 4. The chord of a segment is 150 feet, the chord of
half the arc is 106.066, and the versed sine 75 ; what is the
area? Ans. 8835.739 feet*
When the Chord of the Arc (or Segment) and the Versed Sine
only are given.
Rule. — Find the chord of half the arc, and proceed as be-
fore.
Example. — The chord of the arc of a segment is 28 yards,
and its versed sine 2 ; what is the length of the chord of half
the arc, and what the area of the segment in feet ?
28
—=14, and 142-f-22=200=swm of square of half the chord
and the versed sine.
y200 = 14. 142 13 =square root of preceding sum = chord of
half the arc.
Area of segment 113.6525 feet.
Ex. 2. The chord of a segment is 48 inches, its versed sine
32, and the diameter of the circle 50 ; what is its area?
The versed sine being greater than half the diameter, the segment
is consequently greater than a semicircle.
Area of circle, 502 x .7854=1963.5
Area of lesser portion, versed sine 18.= 640.3478
Ans. 1323.1522 inches.
Ex. 3. The chord of an arc is 86.6024, and its versed sine
25 feet ; what is the area of the segment in feet and inches?
Ans. 1549 feet, .156 inches.
Centre of Gravity. Divide the cube of the chord of the seg-
ment by twelve times the area, and the quotient is the distance
from the centre of the circle.
c3
Or, — — — d, when c represents the chord of the segment, a the
area, and d the distance from the centre of the circle.
Example. — The chord of a segment is 14.14213, its radius
10, and its area 28.53 ; where is its centre of gravity?
* The exact area is 8835.729.
88 MENSURATION OF AREAS, LINES, AND SURFACES.
14.142133 = 2828.426 = cwfo of the chord.
28.53 x 12 = 342.36 = 12 times the area.
Then, 2828.426 -4- 342. 36 =8.261 from the centre of the circle,
and 10 — 8. 26 1 = 1.7 39 from the base of the segment.
Sphere.
Definition. A figure, the surface of which is at a uniform
distance from the centre.
To ascertain the Convex Surface of a Sphere (Fig. 28).
Fig. 28. c
Rule. — Multiply the diameter, a b, by the circumference,
abed, and the product will give the surface required.
Or, dxc= surf ace, d representing diameter, and c the circum-
ference.
Or, 4 p r2=surface* Or, p d2 — surface.
Example. — What is the surface of a sphere of 10 inches
diameter u?
10x31.416 = 314.16 inches.
Ex. 2. The diameter of a sphere is 17 inches; what is the
surface of it in feet? Ans. 6.305 square feet.
Ex. 3. If the circumference of a sphere is 50.2656 inches,
what is its surface in feet? Ans. 5. 585 feet.
Centre of Gravity. Is in its geometrical centre.
* p or 7T represents in this, and in all cases where it is used, the ra-
tio of the circumference of a circle to its diameter, or 3.1416.
f
MENSURATION OF AREAS, LINES, AND SURFACES. 89
Segment of a Sphere.
Definition. A section of a sphere.
To ascertain the Surface of a Segment of a Sphere, Fig. 29.
Rule. — Multiply the height, b o, by the circumference of
the sphere, and the product, added to the area of the base,
a o c, is the surface required.
Or, hxc-\-b=z surf ace, when h represents the height, c the cir-
cumference of the sphere, and b area of base.
Or, 2prh=zconvex surface alone.
Fig. 29. b
Example. — The height, b o, of a segment, a be, is 36 inches,
and the diameter, b e, of the sphere, 100 ; what is the convex
surface, and what the whole surface %
36 x 100 x 3.1416 = 11309.76 —height of segment multiplied by
the circumference of the sphere.
Then, to ascertain the area of the base. The diameter and versed
sine being given ; the diameter of the base of the segment, being
equal to the chord of the arc, is, by rule, page 76,
36x2-100=28.
V282-1002 = 96.
962x. 7854 = 7238.2464 ^convex surface, and 7238.2464 +
11309.76 = 18548.0064 —convex surface added to area of
base^the whole surface.
Note. — When the convex surface of a figure alone is required, the
area or areas of the base or ends must be omitted.
90 MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 2. The height of a segment is 10, and the diameter of the
sphere 100 feet ; what is the surface"? Ans. 5969.04 feet.
Ex. 3. The diameter of a sphere is 200 inches, and the
height of a segment of it is 1 foot 8 inches ; what is its surface
in feet ! Ans. 162.4733 feet.
When the diameter of the Base of the Segment and the Height
of it are alone given.
Rule. — Add the square of half the diameter of the base to
the square of the height ; divide this sum by the height, and
the result will give the diameter of the sphere.
2
Or, d -^-2 -{-h2-i-h— diameter.
Example. — The semi-diameter of the base of a segment of
a sphere is 48 feet, and the height of it is 36 ; what is the
surface of the segment in square yards I
Ans. 2060.8896 yards.
Centre of Gravity of Convex Surface. At the middle of its
height.
Spherical Zone {or Frustrum of a Sphere).
Definition. The part of a sphere included between two paral-
lel chords.
To ascertain the Surface of a Spherical Zone, Fig. 30.
Fig. 30, *
\
\
Rule. — Multiply the height, c g, by the circumference of
the sphere, and the product added to the area of the two ends
is the surface required.
Or, h x c -f a -f a' = surface.
Or, 2 p x r x'h— convex surface.
MENSURATION OF AREAS, LINES, AND SURFACES. 91
Example. — The diameter of a sphere, a b, from which a
segment is cut, is 25 inches, and the height of it, c g, is 8 ;
what is its convex surface?
25x3.1416x8 = 628.32 = height x circumference of sphere =
convex surface.
Ex. 2. The height of a zone is 36 inches, and the radius
of the sphere is 50 inches ; what is its convex surface ?
Ans. 11309.76 inches.
When the Diameter of the Sphere is not given.
Multiply the mean length of the two chords by half their difference,
divide this product by the breadth of the zone, and to the quotient add
the breadth. To the square of this sum add the square of the lesser
chord, and the square root of their sum will be the diameter of the
circle.
Ex. 3. The greater and lesser chords of a segment of a
sphere are 96 and 60, and the height of the segment is 26 ;
what is its convex surface and what its surface j
Convex surface, 8168.160") \
Surface, ' 18233.846 j
Centre of Gravity. At the middle of its height.
Spheroids or Ellipsoids.
Definition. Figures generated by the revolution of a semi-
ellipse about one of its diameters.
When the revolution is about the transverse diameter they are
Prolate, and when it is about the conjugate they are Oblate.
To ascertain the Surface' of a Spheroid {Fig. 31).
When the Spheroid is Prolate.
Rule. — Square the diameters, a b and c d, and multiply the
square root of half their sum by 3.1416, and this product by
the conjugate diameter.
a"2 4- d"2.
Or, <y/ — - x3.1416xd=: surface, d representing conjugate
diameter.
92 MENSURATION OF AREAS, LINES, AND SURFACES.
Fig. 31. c
Example. — A prolate spheroid has diameters of 10 and 14
inches ; what is its surface ?
102-{-142=296 = sum of squares of diameters.
296 + 2 = 148, and ^/ 148 = 12.165 5= square root of half the
sum of the squares of the diameters. -^
12.1655 x 3.1416 x 10=382.191 =/?rato of root above ob-
tained X 3.1416, and that product by the conjugate diameter.
Ex. 2. A prolate spheroid has diameters of 16 and 22
inches; what is its surface? Ans. 966.879 inches.
When the Spheroid is Oblate.
Rule. — Square the diameters, a b and c d, and multiply
the square root of half their sum by 3.1416, and this product
by the transverse diameter.
d2-\-d/2 , ,
Or, -y/ - — X 3.1416 x d' =z surf ace, d' representing trans-
verse diameter.
Example. — An oblate spheroid has diameters of 14 and
10 inches; what is its surface?
142-f-102=296 = sum of squares of diameters.
296 + 2 = 148, and ^148 = 12. 1655= square root of half the
sum of the squares of the diameters.
12.1655 x 3.1416 x 14 = 535.0679=proto of root above ob-
tained x 3.1416, and that product by the transverse diameter.
Ex. 2. An oblate spheroid has diameters of 22 and 16
inches; what is its surface? Ans. 1329.4585 inches.
Centre of Gravity. Is in their geometrical centres.
Note. — For centre of gravity of semi-spheroids, see Appendix, p. 281.
MENSURATION OF AREAS, LINES, AND SURFACES. 93
To ascertain the Convex Surface of a Segment ofa Spheroid (Fig. 32),
Rule. — Square the diameters, and take the square root of
half their sum. Then, as the diameter from which the seg-
ment is cut, is to this root, so is the height of the segment to
the proportionate height required.
Multiply the product of the other diameter and 3.1416 by
the proportionate height of the segment, and this last product
will give the surface required.
Or, 5 ^ x cT or d x 3. 14 1 6 = surface.
a or a
Fig. 32. c
Example. — The height, a o, of a segment, e f, of a prolate
spheroid, Fig. 32, is 4 inches, the diameters being 10 and 14
inches ; what is the convex surface of it ?
Square root of half the sum of the squares of the diameters, as
by previous examples, page 92, 12.1655.
Then, 14 : 12.1655 ::4 : 3A7 58 = height of segment, proportion-
ate to the mean of the diameters.
10 X 3.1416 X 3.4758 = 109.1957 = remaining diameter x
3.1416, and again by proportionate height of segment.
Ex. 2. The height of a segment of a prolate spheroid, Fig.
32, is 6 inches, the diameters being 15 and 21 inches; what
is the convex surface of it? Ans. 252.9663 inches.
94
MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 3. The height^ c o, of a segment, e f of an oblate
spheroid, Fig. 33, is 5 inches, the diameters being 14 and 10 ;
what is the convex surface? Ans. 267.5339 inches.
To ascertain the Convex Surface of a Frustrum or Zone of a
Spheroid {Fig. 34).
Rule. — Proceed as by previous rule, page 93, for the surface
of a segment, and obtain the proportionate height of the frus-
trum. Then multiply the product of the diameter parallel to
the base of the frustrum and 3.1416 by the proportionate height
of the frustrum, and it will give the surface required.
Or, d or oV x 3.1416 x h=surface.
Fig. Si. c
Example. — The middle frustrum, o <?, of a prolate spheroid,
Fig. 34, is 6 inches, the diameters of the spheroid being 10
and 14 inches ; what is its convex surface ?
Mean diameter, as per example, page 92, is 12.1655.
'Diameter parallel to base of frustrum is 10.
As 14 : 12.J.655 :: 6 : 5. ,2138= proportionate height of frustrum.
10x3.1416 x5.2138 = 163.7967=swr/ace.
Fig. 35. c
Ex. 2. The middle frustrum of an oblate spheroid, o e9 Fig.
35, is 2 inches in height, the diameters of the spheroid, as in
MENSURATION OF AREAS, LINES, AND SURFACES. 95
the preceding examples, being 10 and 14 ; what is its convex
surface? Aris. 107-0136 inches.
Centre of Gravity. Zone. Is in its geometrical centre.
Segment and Frustrum of Spheroid. See Appendix, p. 28 1.
Circular Zone.
Definition. A part of a circle included between two parallel
chords.
To ascertain the Area of a Circular^ Zone (Fig. 36).
Rule. — To the area of the trapezoid, a b c d, or of the par-
illelogram, a h c g, as the case may be, add the area of the
segments, a b, c d, or a h, c g, and the sum is the area.
Or, subtract the areas of the segments a i c, h k g, from the
area of the circle.
Or, a-{-a'z=zS, a representing area of trapezoid, or parallelo-
gram, and a' area of segments.
(See Table of Areas of Zones, page 130.)
Fig. 36. i
When the Diameter of the Circle i$ not given.
Multiply the mean length of the two chords by half their difference ;
divide this product by the breadth of the zone, and to the quotient add
the breadth.
To the square of this sum add the square of the lesser chord, and
the square root of their sum will be the diameter of the circle.
Example. — The greater chord, b d, is 96 inches, the lesser,
a c, is 60, and the breadth of the zone, a e, is 26.; what is its
area?
96 MENSURATION OF AREAS, LINES, AND SURFACES.
=78= mean length of chords. =18= half their
difference.
78x18
54:=product of chords and their difference, divided by
*o
the breadth of the zone.
54 + 26 = 80=swm of above quotient and breadth of zone.
802 + 602 = 10000 =; sum of square of above sum and lesser chord.
Then, -y/1 0000 = 100 = diameter required.
— ±— ^=78, and 78 X 2Q>-202S-area of trapezoid.
To ascertain the Area of the Segments.
It is necessary, first, to ascertain the chord of their arcs ; second, the
versed sine of their arcs.
To ascertain the Chord. The breadth of the zone is the perpendic-
ular, a e, of the triangle, of which either chord, a b, c d, is the hypothe-
nuse. Further, half the difference of the chords a c and b d of the zone
is the length of the base, b e, of this triangle.
Hence, having the base and the perpendicular, the hypothenuse or
chord of the arc of the segment is readily found.
Thus, 2Q=breadtk of the zone or perpendicular of triangle.
— =18 = length of base of triangles.
Then, 18, + 262 = 1000, and V 1000 =31. 6223 = chord of arc of segments
a b, c d.
To ascertain the Versed Sine. From the square of the radius subtract
the square of half the chord, and the square root of the remainder sub-
tracted from the radius is the versed sine.
Thus, 100-^-2=50, and 502= 2500= square of radius.
31.0228-^2 = 15.8114, and 15.81142=250=sgware of half the chord.
2500-250=2250, and V 2250=47.4342 =square root of the difference
of the squares of the radius and half the chord.
Then, 50-47.4342 = 2. 5658 = versed sine.
Having obtained the chord of the arcs (31.6228), their
versed sines (2.5658), and the diameter of the circle (100),
then, by rule, page 75,
VlOOx 2.5658 = 16.0181 =chord of half the arc.
And by rule, page 78, to ascertain the length of an arc,
MENSURATION OF AREAS, LINES, AND SURFACES. 97
16.0181 x 2 x 10 x 2.5658=821.9848=ta;ice the chord of half
the arc by 10 times the versed sine.
100 x 60 — 2.5658 x 27=5830.7234 = 27 times the versed sine
subtracted from 60 times the diameter.
821.984:8+5930.7234: = .l385=quotientofaboveproductand
remainder, and .1385 + 32.0362 (16.0181 x 2) =32.1747
= length of the arc.
32. 1747x50 Tl — 804.3675 =the product of the length of the
arc and half the radius of the circle =area of sector.
And 804.3675-31-6228^47-4342=54.3664=amz ofthetri-
angle subtracted from the area of the sector— area of each seg-
ment.
54.3564x2 = 108.7 328= area of segments.
Area of trapezoid = 2028.
2136.7328 —area of zone.
Ex. 2. The greater chord is 24, the lesser 15, and the
breadth of the zone 6.5 inches ; what is its area ?
(Length of arc 8.0437.) Ans. 133.5467 inches.
Ex. 3. The lesser chord is 96, the greater 100, and the
breadth of the zone 14 inches ; what is the area of it %
(Length of arc 14.189.) Ans. 1381.4851 inches.
Ex. 4. The chords of a zone are 96 inches, and its height
28 ; what is its area 1
(Length of arc 28.379.) Ans. 2762.95 inches.
Centre of Gravity. Find the centres of gravity of the trap-
ezoid and the segments comprising the zone; draw a line
(equally dividing the zone) perpendicular to the chords ; con-
nect the two centres of the segments by a line cutting the per-
pendicular to the chords ; then will the centre of gravity of
the figure be on the perpendicular, toward the lesser chord,
at such proportionate distance of the difference between the
centres of gravity of the trapezoid and line connecting the cen-
tres of the segments as the area of the two segments bears to
the area of the trapezoid.
E
98
MENSURATION OF AREAS, LINES, AND SURFACES.
Cylinder.
Definition. A figure formed by the revolution of a right-
angled parallelogram around one of its sides.
To ascertain tlie Surface of a Cylinder {Fig. 37).
Eule. — Multiply the length, a b, by the circumference, and
the product added to the area of the two ends will be the sur-
face required.
Or, lxc-\-2 a=zs, where a represents area of end.
Note. — "When the internal surface alone is wanted, the areas of the
ends are to be omitted.
Fig, 37.
Example. — The diameter of a cylinder, b c, is 30 inches,
and its length, a b, 50 inches ; what is its surface?
30 x 3.1416 = 94.2480 inches = circumference.
94.248 x50=4712A=area of body.
And 302x- 7854 = 706.86 = area of one end.
706.86 X 2 = 1413.72 =amz of both ends.
Then, 4712.4 + 1413.72 = 6126.12=swr/ace required.
Ex. 2. The diameter of a cylinder is 100 inches, and its
length 12 feet; what is its surface*? Ans. 423.243 feet.
Ex. 3. The diameter of a hollow cylinder is 36 inches, and
its length 10 feet; what is its internal surface?
Ans. 94.248 feet.
Centre of Gravity. Is in its geometrical centre.
MENSURATION OF AREAS, LINES, AND SURFACES.
99
Prisms.
Definition. Figures the sides of which are parallelograms, and
the ends equal and parallel.
Note. — When the ends are triangles, they are called triangular
prisms; when they are square, they are called square or right prisms ;
when they are pentagons, pentagonal prisms, &c, &c.
To ascertain the Surface of a Prism (Figs. 38 and 39).
Kule. — Find the areas of the ends and sides as by the rules
for the mensuration of squares, triangles, &c., and add them
together ; the sum will be the surface of the figure.
Or, 2 a-{-a/—s, where a represents the area of the ends, and
a/ the area of the sides.
Fig 38. a b Fig. 39. b
Example. — The side a b, Fig. 38, of a square prism is 12
inches, and the length, b c, 30; what is the surface %
12 x 12 = 144=area of one end.
144 x 2 = 288 —area of both ends.
12 x 30 — 360 —area of one side.
360x4=1440=araz of four sydes.
Then, 2884-1440 = 1728 inches, the surface required.
Ex. 2. What is the surface of a triangular prism, the sides
ab,bc, and c a, Fig. 39, being each 12 inches, and the length,
c d, 30 inches'?
12-^2 = 6, and VQ2 — 122= 10. 3923= width of prism.
Hence, 10.3923 x 1 2 -j- 2 = 62.3538^= area of each end.
100 MENSURATION OF AREAS, LINES, AND SURFACES.
12 x 30 = 360= area of one side.
62.3538 x 2 = 124.7076 —area of ends,
Then, 360 x 3 = 1080 = area of sides,
and 124.7076 + 1080=1204.7076 inches— surface required.
Ex. 3. What is the surface of a rhomboidal prism, the depth
of it being 5 feet 9 inches, the width 7 feet, and the length 10
feet? Ans. A02.5 feet.
Centre of Gravity. When the ends are parallelograms, it is
in their geometrical centre.
When the ends are triangles, trapeziums, etc., it is in the mid-
dle of their length at the same distance from the base as that
of the triangle or trapezoid which is a section of them.
Wedge.
Definition. A wedge is a prolate triangular prism, and its
surface is found by the rule for that of a right prism.
Fig. 40. e
b c
Example. — The back of a wedge, abed, Fig. 40, is 20 by
2 inches, and its end, ef, 20 by 2 inches ; what is its surface?
2
202+2-f- 1 —401 =sum of the squares of half the base, a f, and
the height, ef, of the triangle, efa.
y^401 = 20.025 —square root of above sum = length of e a.
Then, 20.025 x 20 x 2=801 —area of sides.
And 20 x 2=4,0— area of bach, and 20x2-4-2 x 2=40=area
of ends.
Hence, 8014-40 + 40 = 881= surface required.
Centre of Gravity. See rule for prisms.
MENSURATION OF AREAS, LINES, AND SURFACES. 101
Prismoids.
Definition. Figures alike to a prism, but having only one pair
of thdr sides parallel.
To ascertain the Surface of a Prismoid {Fig. 41).
Rule. — Find the area of the ends and sides as by the rules
for squares, triangles, &c., and add them together.
Fig. 41. a b
g h
Example. — The ends of a prismoid, efgh and abed, Fig.
41, are 10 and 8 inches square, and its slant height 25 ; what
is its surface ?
10 x 10 = 100=arai of base.
8 x 8 = 64 = area of top.
— i-x25 = 225, and 225 x4=1000=ara* of sides.
It
Then, 100 + 64 + 1000= \\^— surface required.
Ex. 2. The ends of a prismoid are 15 and 12 inches square,
and its slant height 40 ; what is its surface %
Ans. 2529 inches. .
Ex. 3. The ends of a prismoid are 12x16 and 14x18 inches,
and its vertical height is 33 ; what is its surface %
Ans. 2424 inches.
Centre of Gravity. Is at the same distance from its base
as that of the trapezoid or trapezium which is a section of it.
102 MENSURATION OF AREAS, LINES, AND SURFACES.
Ungulas.
Definition. Cylindrical ungulas are frustrums of cylinders.
Conical ungulas* are frustrums of cones.
To ascertain the Curved Surface of an Ungula, Figs. 42, 43,
44, 45, and 46.
1. When the Section is parallel to the Axis of the Cylinder, Fig. 42.
Rule. — Multiply the length, a b c, of the arc line of one
end by the height, b d, and the product will be the curved
surface required.
Or, cxh=zs, where c represents length of arc line.
Fig. 42. a
Example. — The diameter of a cylinder from which an un-
gula is cut is 10 inches, its length 50, and the versed sine or
depth of the ungula is 5 inches ; what is the curved surface
of it?
10-4-2=5 = radius of cylinder.
Hence the radius and versed sine are equal; the arc line, therefore,
of the ungula is one half the circumference of the cylinder,
which w 31.416-^2 = 15.708, v
and 15.708x50=785.400 inches. •
Ex. 2. The base line of the section of a cylindrical ungula
is 48 inches, the height or versed sine of the arc is 20, and
* For mensuration of conical ungulas, see Conic Sections, p. 253.
MENSURATION OF AREAS, LINES, AND SURFACES.
103
the length of the ungula is 20.5 feet ; what is its curved sur-
face? Ans. 109.8388 feet.
2. When the Section passes obliquely through the opposite Sides of
the Cylinder, Fig. 43.
Rule. — Multiply the circumference of the base of the cylin-
der by half the sum of the greatest and least heights, d b and
e a, of the ungula, and the product will give the curved surface
required.
Fig. 43.
Example. — The diameter of a cylindrical ungula is 10
inches, and the greater and less heights are 25 and 15 inches ;
what is its surface ?
10 diameter = 31.416 circumference.
25 + 15=40, and 40-=- 2 = 20.
Hence, 31.416 x 20=62.8320 inches.
Ex. 2. The circumference of an ungula is 60.75 inches, and
the mean height of it 13 feet ; what is its surface?
Ans. 65.8125 feet.
3. When the Section passes through the Base of the Cylinder and one
of its Sides, and the Versed Sine does not exceed the Sine, Fig. 44.
Rule. — Multiply the sine, a d, of half the arc, d g, of the
base, d g f, by the diameter, e g, of the cylinder, and from this
product subtract the product* of the arc and cosine, a o. Mul-
* When the cosine is 0, this product is 0.
104 MENSURATION OF AREAS, LINES, AND SURFACES.
tiply the difference thus found by the quotient of the height,
g c, divided by the versed sine, a g, and the product will be
the curved surface required.
Fig. 44.
Example. — The sine, a d, of half the arc of the base of an
ungula is 5, the diameter of the cylinder is 10, and the height
of the ungula 10 inches ; what is the curved surface ?
5 x 10 = 50= sine of half the arc hy the diameter.
Length of arc, the versed sine and radius being equal, under ride,
page 78 = 15.708.
Again, as the versed sine and the radius are equal, the cosine is 0.
Hence, when the cosine is 0, the product is 0. 50 — 0 = 50 = the
difference of the product before obtained and the product of t/ie
arc and the cosine.
50xlO-f-5 = 50x2 = 100=*Ae difference multiplied by the height
divided by the versed sine, which is the surface required.
Ex. 2. The sine of half the arc of the base of an ungula is 12
inches, the versed sine is 9, the diameter of the cylinder is 25,
and the height of the ungula is 18 ; what is its curved surface l?
12 x 25 = 300 =product of sine and diameter.
Arc of base of ungula, by rule, p. 77, the versed sine being 9, is
32*14795.
Then, 32.14795 x 12.5-9 = 112.51782,
and 300-112.51782 = 187.48218,- which, multiplied by 18-f-9
= 374,96436 inches.
MENSURATION OF AREAS, LINES, AND SURFACES.
105
4. When the Section passes through the Base of the Cylinder, and
the Versed Sine exceeds the Sine, a g, Fig. 45.
Rule. — Multiply the sine of half the arc of the base by the
diameter of the cylinder, and to this product add the product
of the arc and the excess of the versed sine over the sine of
the base.
Multiply the sum thus found by the quotient of the height
divided by the versed sine, and the product will be the curved
surface required.
Fig. 45.
/"" "\
i 1
Example. — The sine, a d, of half the arc of an ungula is
12 inches, the versed sine, a g, is 16, the height, eg, 16, and
the diameter of the cylinder 25 inches ; what is the curved
surface ?
12 x2o=30Q=sine of half the arc by the diameter of the cyl-
inder.
Length of arc of base, by rule, p. 74:=a?v of d b f— circumfer-
ence of base— 46.392.
Then 46.392 x 16 — 12.5 = 162.372 ^product of arc and the ex-
cess of the versed sine over the sine.
300-|-162.372=462.372 = *Ae sum of the above products.
16 -i-16 = l= quotient of height divided by the versed sine.
462.372x1 = 462.372 inches=the sum of the products and the
height divided by the versed sine = the curved surface required.
E2
106 MENSURATION OF AREAS, LINES, AND SURFACES.
Ex. 2. The sine of half the arc of the base of an ungula is
0, the diameter of the cylinder- is 10 inches, and the height of
the ungula is 20 inches; what is its curved surface?
Note. — The sine of the arc being 0, the versed sine is equal to the
diameter (10), and the sine of the base is 10-^2=5.
0x 10 = 0 = product of sine of half the arc and diameter of the
cylinder.
0-f(31.416 {length of arc) x 10oo5)= 157.08 — the sum of the
product above obtained and the product of the arc and the ex-
cess of the versed sine over the sine.
157.08 x 20-r-10=314.16 = ^e above sumxthe height-i-the
versed sine=the result required.
5. When the Section passes obliquely through both Ends of the
Cylinder, abed. Fig. 46.
Rule. — Conceive the section to be continued till it meets
the side of the cylinder produced ; then, as the difference of
the versed sines of the arcs of the two ends of the ungula is
to the versed sine of the arc of the less end, so is the height
of the cylinder to the part of the side produced.
Find the surface of each of the ungulas thus found by the
rules 3 and 4, and their difference will be the curved surface
required.
Fig.M. k
Example. — The versed sines a e, d o, and sines i Tc, g r, of
MENSURATION OP AREAS, LINES, AND SURFACES. 107
the arcs of the two ends of an ungula, Fig. 46, are respectively
5 and 2.5, and 5.and 4.25 ; the height of the ungula within the
cylinder, cut from one having 10 inches diameter, is 5 inches ;
what is the height of the ungula produced beyond the cylinder?
5cv>2.5 = 2.5, and 2.5 : 2.5 :: 5 : 5= height of ungula produced
beyond the cylinder,
Ex. 2. The versed sines of the base and arc of an ungula
cut from a cylinder of 6 inches diameter are 6 and 2 inches,
and its height within the cylinder is 4 inches ; what is the
distance it extends above or beyond the cylinder !
Ans. 2 inches.
Lune.
Definition. The space between the intersecting" arcs of two
eccentric circles.
To ascertain the Area of a Lune, Fig. 47.
Kule. — Find the areas of the two segments from which the
lune is formed, and their difference will be the area required.
Or, s—s'=a, wlien s and s/ represents the areas of the segments.
Fig. 47. d
Example. — The length of the chord a c is 20, the height
e d is 3, and e b 2 inches ; what is the area of the lune ?
By Rule 2, p. 76, the diameters of the circles of which the lune is
formed are thus found:
lor ad c, ^ —=25.
5
102+22 ffo
For a e c, - — 52.
108 MENSURATION OF AREAS, LINES, AND SURFACES.
Then, by rule for the areas of segments of a circle, page 87,
the area of a d c is 70.5577 in.
. " a e c 27.1638 in.
their difference 43.3939 in., the area of the lune required.
Ex. 2. The chord of a lune is 40, and the heights of the
segments 10 and 4 inches; what is its content?
Ans. 173.5752 inches.
' Ex. 3. The chord of a lune is 6 feet 8 inches, and the
heights of the arcs 1.666 feet and 8 inches; what is its area?
Ans. 694.2996 inches.
Ex. 4. The chord of a lune is 86.6024 inches, and the
heights of the segments 25 and 15 inches; what is its area?
Ans. 653.3551.
Centre of Gravity. On a line connecting the centres of
gravity of the two arcs at a point proportionate to the respect-
ive areas of the arcs.
Note. — If semicircles be described on the three sides of a right-
angled triangle as diameters, two lunes will be formed, and their united
areas will be equal to that of the triangle.
Cycloid.
Definition. A curve generated by the revolution of a circle on
a plane.
To ascertain the Area of a Cycloid, Fig. 48.
Rule. — Multiply the area of the generating circle a b c by
3, and the product will give the area required.
Or, a x^— area.
Fig. 48.
MENSURATION OP AREAS, LINES, AND SURFACES. 109
Example. — The generating circle of a cycloid has an area
of 115.45 inches ; what is the area of the cycloid1?
115.45 x 3 = 346.35 inches.
Ex. 2. The area of a circle describing a cycloid is 1.625
feet; what is the area of the cycloid in inches?
Ans. 702 inches.
Ex. 3. The diameter of a circle describing a cycloid is
66.5 feet ; what is the area of the cyloid in inches ?
Ans. 1500434.064 inches.
To ascertain the Length of a Cycloidal Curve, Fig. 48.
Rule. — Multiply the diameter of the generating circle by
4, and the product will give the length of the curve.
Or, d x 4 = length of curve.
Example. — The diameter of the generating circle of a cy-
cloid, Fig. 48, is 8 inches ; what is the length of the curve dsc?
8 x 4= 32 '= product of diameter and 4 = the length required.
Ex. 2. The diameter of the generating circle is 20 inches ;
what is the length of the cycloidal curve %
Ans. 80 inches.
Centre of Gravity. At a distance from the centre, n, of
the chord, d c, of the curve d s c— -| of the radius of the gen-
erating circle.
Note. — The curve of a cycloid is the line of swiftest descent ; that is,
a body will fall through the arc of this curve, from one point to another,
in less time than through any other path.
RINGS.
Circular Rings.
Definition. The space betiveen two concentric circles.
To ascertain the Sectional Area of a Circular Ring, Fig. 49.
Rule. — From the area of the greater circle, a b, subtract
that of the less, c d, and the difference will be the area of the
ring.
Or, a— a' =: area.
110 MENSURATION OF AREAS, LINES, AND SURFACES.
Fig. 49.
Example. — The diameters of the circles forming a ring
are each 10 and.,15 inches ; what is the area of the ring?
Area of 15 = 176.7146
" 10= 78.5400
98.1746 inches.
Ex. 2. The diameters of a circular ring are 10.75 and 18.25
inches; what is its area? Ans. 171.82 inches.
Centre of Gravity. Is in its geometrical centre.
Definition.
Cylindrical Rings.
A ring formed by the curvature of a cylinder.
To ascertain the Convex Surface of a Cylindrical Hing, Fig. 50.
Kule. — To the thickness of the ring, a b, add the inner
diameter, be; multiply this sum by the thickness and the
product, by 9.8696, and it will give the surface required.
Or, d+d' xdx9.S696=surface.
Fig. 50.
Example. — The thickness of a cylindrical ring, a b, is 2
inches, and the inner diameter, b c, is 18 ; what is the surface
of it?
MENSURATION OF AREAS, LINES, AND SURFACES.
Ill
2 + 18 — 20 = thickness of ring added to the inner diameter.
20X 2 X 9. 8696 = 394.784 = Z/*e sum above obtained X the thick-
ness of the ring, and that product by 9.8696, the result re-
quired.
Ex. 2. The thickness of a ring of metal of 20 inches diam-
eter (internal) is 2 inches ; what is the surface of it 1
Ans. 434.2624 inches.
Link.
Definition. An elongated ring.
To ascertain the Convex Surface of a Link, Figures 51.
Rule. — Multiply the circumference of a section of the body,
a b, of the link by the length of its. axis, and the product .will
give the surface required.
Or, c x l=swface.
Note. — To ascertain the Circumference or Length of the Axes.
When the Ring is elongated. To the less diameter add its thickness,
multiply the sum by 3.1416 ; multiply the difference of the diameters by
2, and the sum of these products will give the result required.
When the Ring is elliptical. Square the diameters of the axes of the
ring, and multiply the square root of half their sum by 3.141G; the
product will give the length of the body of the ring.
Figs. 51.
Example. — The link of a chain is 1 inch in diameter of
body, a b, and its inner diameters, b c and ef, are 12.5 and 2.5
inches ; what is its circumference.
2.5 + 1x3.1416 — 10.995Q = length of axis of ends.
12.5 — 2.5x2x2 = 15 = length of sides of body.
Then, 10.9956 + 15 = 25.9956 = length of axis of link, which,
X 3.1416 (cir. of 1 in.) ^ 1.6678 inches.
Centres of Gravity. Are in their geometrical centres.
112 MENSURATION OP AREAS, LINES, AND SURFACES.
Cones.
Definition. A figure described by the revolution of a right-
angled triangle about one of its legs.
For Sections of a Cone, see Conic Sections, page 228.
To ascertain the Surface of a Cone, Fig. 52.
Kule. — Multiply the perimeter, or circumference of the
base, by the slant height, or side of the cone, and half the
product added to the area of the base will be the surface.
Or, — — - —surface.
Fig. 52. „
Example. — The diameter, a b, of the base of a cone is 3 feet,
and the slant height, a c, 15 ; what is the surface of the cone ?
Perimeter of 3 fcet= 9.4248, and 9'424^ X 15^70.686-
m
surface of side of cone.
Area of 3 feet=7. 068, and 70.686 +7.068 = 77.754 ^surface
required.
Ex. 2. The diameter of the base of a cone is 6.25 inches,
and the slant height 18.75 ; what is the surface of it?
Ans. 214.757 inches.
Ex. 3. The diameter of the base of a cone is 20 inches, and
the slant height 14.142 ; what is the surface of the cone?
Ans. 758.445 inches.
MENSURATION OF AREAS, LINES, AND SURFACES. 113
To ascertain the Surface of the Frustrum of a Cone, Fig. 53.
Rule. — Multiply the sura of the perimeters of. the two ends
by the slant height of the frustrum, and half the product add-
ed to the areas of the two ends will be the surface required.
_ p+p'xh . . , ,
Or, - — ~ \-a-\-a — surface.
&
Fig. 53.
Example. — The frustrum, abed, Fig. 53, has a slant
height of 26 inches, and the circumferences of its ends are
15.7 and 22. inches respectively; what is its surface?
15.7 + 22. x 26^-2— 490. 1 =su?face of sides.
(i*7\2 /22\2
HiTe) *-7854+(04l6) X .7854=58.12=™/^,
Then, 490.1 + 58.12 = 548.22 ^surface.
Ex. 2. What is the surface of the frustrum of a cone, the
diameters, of the ends, a c and b d, being 4 and 8 feet, and the
length of the slant sides 20 feet ! Ans. 439.824 feet.
Ex. 3. What is the surface of the frustrum of a cone, the
diameter of the ends being 6.66 and 10 feet, and the length
of the slant side 3.73 feet ?
Ans. 210.989.
Centres of Gravity. Cone or Frustrum. — At the same dis-
tance from the base as in that of the triangle or parallelo-
gram, which is a right section of them.
114 MENSURATION OF AREAS, LINES, AND SURFACES.
Pyramids.
Definition. A figure, the base of which has three or more
sides, and the sides of which are plane triangles.
To ascertain the Surface of a Pyramid, Fig. 54.
Eule. — Multiply the perimeter of the base by the slant
height, and half the product added to the area of the base
will be the surface.
r\ Pxh i *
Or, — - — \- a— surface.
Fig. 55.
be b
Example. — The side of a quadrangular pyramid, a b, Fig. 54,
is 12 inches, and its slant height, c e, 40 ; what is its surface?
12x4=48
48x40
perimeter of base.
960 =" area of sides.
Then 12 X 12 + 960 = 1104=swr/ace.
Ex. 2. The sides of a hexagonal pyramid are 12.5 inches,
and its slant height 62.5 ; what is its surface?
Ans. 2749.703 inches.
Ex. 3. The sides, a b, b c, of an oblong quadrangular pyr-
amid, Fig. 55, are 15 and 17.5 inches, and its slant height,
d e, 36 ; what is its surface? Ans. 1432.5 inches.
Ex. 4. The sides of an octagonal pyramid are 4 feet 2
inches, and its slant height 6 feet 9 inches ; what is its sur-
face? Ans. 196.326 square feet.
MENSURATION OF AREAS, LINES, AND SURFACES. 115
Ex. 5. What is the surface of a pentagonal pyramid, its
slant height being 12 feet, and each side of its base 2 feet?
Ans. 66.882.
To ascertain the Surface of the Frustrum of a Pyramid, Fig. 56.
Rule. — Multiply the sum of the perimeters of the two ends
by the slant height or side, and half the product added to the
area of the ends will be the surface.
_ p+p'xh , „
Or, - — ^r \-a+a = surface.
Fig. 56. d c
ExAarPLE. — The sides, a b, c d, Fig. 56, of a quadrangular
pyramid are 10 and 9 inches, and its slant height, e o, 20 ;
what is its surface 1
10x4=40
9x4=36
7Q=sum of perimeters.
76X20 = 1520, and -—— =7 60= area of sides.
z
10x10=100, and 9x9=81.
Then 100 + 81+760 = 941, the surface.
Ex. 2. The ends of a frustrum of a quadrangular pyrami
are 15 and 9 inches, and its slant height 40 ; what is its sur-
face? . Ans. 2226 inches.
Ex. 3. The ends of a frustrum of a triangular pyramid are
20 and 10 inches, and its slant height 50 ; what is its surface ?
Ans. 2466.5 inches.
Ex. 4. The sides of a frustrum of a hexagonal pyramid are
15 and 25 inches, and its slant height 20 ; what is its sur-
face? . Ans. 32.002 square feet.
116 MENSURATION OF AREAS, LINES, AND SURFACES.
Centres of Gravity. Pyramid or Frustrum. — At the same
distance from the base as in that of the triangle or parallelo-
gram, which is a right section of them..
Helix {Screw).
Definition. A line generated by the progressive rotation of a
point around an axis and equidistant from its centre.
To ascertain the Length of a Helix, Fig. 57.
Rule. — To the square of the circumference described by
the generating point, add the square of the distance advanced
in one revolution, and the square root of their sum multiplied
by the number of revolutions of the generating point will give
the length of the line required.
Or, ^{c2-\-h2)xn— length of line, n representing the number
of revolutions.
Fig. 57.
Example. — What is the length of a helical line running
3.5 times around a cylinder of 22 inches in circumference and
advancing 16 inches in each revolution1?
222 + 162 = 740 =sum of squares of circumference and of the dis-
tance advanced.
-y/740 X 3.5 = 95.21 —square root of above sum x number of rev-
olutions == length of line required.
Ex. 2. What is the length of the helical line described by
a point in a screw in one revolution at a radius from its axis
MENSURATION OF AREAS, LINES, AND SURFACES. 117
of 11.3 inches, the progression of the line or pitch of the
screw being 17 inches ? Ans. 73.
Ex. 3. What is the length of the helical line described by
a point on the periphery of a screw of 10 feet in diameter,
having a pitch of 20 feet? Ans. 37.242.
Centre of Gravity. Is in its geometrical centre.
Spirals.
Definition. -Lines generated by the progressive rotation of a
point around a fixed axis.
A Plane Spiral is when the point rotates around a central
point.
A Conical Spiral is when the point rotates around an axis
or a cone.
To ascertain the Length of a Plane Spiral Line, Fig. 58.
Rule. — Add together the greater and less diameters,* di-
vide their sum by two, multiply the quotient by 3.1416, again
by the number of revolutions, and the product will give the
length of the line required.
Or,. when the circumferences are given, take their mean
length, multiply it by the number of revolutions, and the prod-
uct will give the length required.
Or, — — - X 3.1416 Xn= length of line, n representing the
number of revolutions.
Fig. 58.
* When the spiral is other than a line, measure the diameters of it
from the middle of the material composing it.
118 MENSURATION OP AREAS, LINES, AND SURFACES.
Example. — The less and greater diameters of a plane spiral
spring, as a b, c d, Fig. 58, are 2 and 20 inches, and the num-
ber of revolutions 10 ; what is the length of it ?
2 + 20
— - — =11— swm of diameters -f- 2.
z
llX3.U16 = 34:.5576 = above quotient X 3.1416.
34.5576 X 10 = 345.576 =above product X number of revolu-
tions z^the length of line.
Ex. 2. The greater and less diameters of a plane spiral are
4 and 30 inches, and the number of revolutions 5 ; what is
the length of it % Ans. 267.036 inches.
To ascertain the Length of a Conical Spiral, Fig. 59.
Rule.-1- Add together the greater and less diameters ;* di-
vide their sum by two, and multiply its quotient by 3.1416.
To the square of the product of this circumference and the
number of revolutions of the spiral, add the square of the
height of its axis, and the square root of the sum will be the
length required.
Or, V\— g— X3.1416xrc + A = length of spiral.
Fig. 59. c
Example. — The greater and less diameters of a conical
spiral, Fig. 59, are 20 and 2 inches, its height, c d, 10, and the
number of revolutions 10; what is the length of it?
* See Note to Rule, page 117.
MENSURATION OP AREAS, LINES, AND SURFACES. 119
20+2_i-2 = ll x3.1416=34.5576=swm of diameters +2 and
X3.1416.
34.5576X10=345.576, and 345.5762= 11 9422.77 =square
of the product of the circumference and number of revolutions.
Vll9422.77 + 102=345.72 = ^ square root of the sum of the
above product and the square of the height of the spiral— the
result required.
Ex. 2. The greater and less diameters of a conical spiral are
1.5 and 8.75 feet, its height 6 feet, and the number of its revo-
lutions is 5 ; what is the length of it ? Ans. 80.725 inches.
Ex. 3. The greater and less diameters of a conical spiral
are 3 and 9 feet, its height 12.5 feet, and the number of its
revolutions 10; what is the length of it? Ans. 188.91.
Centres of Gravity. Plane Spiral. — It is in its geometrical
centre.
Conical Spiral. — It is at a distance from the base J of the
line joining the vertex and centre of gravity of the base.
Note.^ — This rule is applicable to winding engines where
it is required to ascertain the length of a rope, its thickness,
the number of revolutions, diameter of drum, etc., etc.
Illustration. — The diameter over the roll of a flat rope upon
the drum of a winding engine shaft is 134.5 inches, the diam-
eter of the drum is 94.5 inches, and the number of revolutions
20 ; what is the length of the rope and what is its thickness?
Ans. Length of the rope, 7194.247 inches.
Area of 134.5 m. = 14208.049
" " 94.5 = 7013.802
7194.247
Then, 7194.247 -^ 20 = 359.712 =area of rope~-revolutions=
area of each thickness. #
134.5-94.5-^2 + 94.5 = 114.5 and 114.5x3.1416 = 359.712
= circumference of the mean diameter of each thickness.
Hence, 359.712 -f-359.7l2 = l inch—the width or thickness of
the rope.
* For Rules to ascertain the elements of Winding Engines, see Has-
well's Engineers' and Mechanics' Pocket-book, p. 263-4.
120 MENSURATION OF AREAS, LINES, AND SURFACES.
. SPINDLES.
Definition. Figures generated by the revolution of a plane
area, ivhen the curve is revolved about a chord perpendicular to
its axis, or about its double ordinate, and they are designated by
the name of the arc or curve from which they are generated, as
Circular, Elliptic, Parabolic, etc., etc.
Circular Swindle,
To ascertain the Convex Surface of a Circular Spindle, Fig. 60.
Rule. — Multiply the length, / c, by the radius, o c, of the
revolving arc ; multiply this arc, fac,by the central distance,
o e, or distance between the centre of the spindle and centre
of the revolving arc ; subtract this product from the former,
double the remainder, multiply it by 3.1416, and the product
will be the surface required.
Or, lxr—(aX -y/r2 — I-) )X2 p; a representing the length
of the arc, c the chord, and p 3.1416.
Fig. 60. a
Example. — What is the surface of a circular spindle, Fig.
60, the length of \t,fe, being 14.142 inches, the radius of its
arc, o c, 10, and the central distance, o e, 7.071 ?
14.142x10 = 141.42 = length x radius.
Length of arc, by rules, p. 76, 78 = 15.708.
15.708 x 7.071 = 1 1 1.0713 — length of arc X central distance.
141.42 — 111.0713=30.3487=^mice of products.
30.3487 x 2 = 60.6974 x 3.1416 = 190.687 = the remainder
doubled x 3.1416, which is the result required.
MENSURATION OF AREAS, LINES, AND SURFACES. 121
Ex. 2. The length of a circular spindle is 28.284 feet, the
radius of its arc 20, and the distance between the centre of
the spindle and the centre of the revolving arc is 14.142 ;
what is the surface of it * Ans. 762.7484 feet.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Convex Surface of a Zone of a Circular Spin-
dle, Fig. 61.
Rule. — Multiply the length, i c, by the radius, o «, of the
revolving arc ; multiply the arc, d a b, by the central distance
o e; subtract this product from the former, double the re-
mainder, multiply it by 3.1416, and the product will be the
surface required.
Or, lxr—(ax yV2— ( -J ) x2p, I representing t/ie length oj
the zone.
•v/
O
Example. — What is the convex surface of the zone of a
circular spindle, Fig. 61, the length of it being 7.653 inches,
the radius of its arc 10, the central distance 7.071, and the
length of its side or arc, d b, 7.854 inches ?
7.653 X 10=76.53=lengthxradius.
7.854:X7.071=55.5S56 = length of ' arcX central distance.
76.53— 55.5356= 20.9944 =^rerace of products.
20.9944x2=41.9888x3.1416 = 131.912 = ^6 remainder
doubled X 3.1416, which is the result required.
- Ex. 2. The zone of a circular spindle is 23 inches in length,
F
122 MENSURATION OF AREAS, LINES, AND SURFACES.
the radius of its arc 30, its central distance 21.2, and the
length of its side 23.56 ; What is its convex surface ?
Ans. 1197.2255 inches.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Convex Surface of a Segment of a Circular
Spindle, Fig. 62.
Kule. — Multiply the length, i c, by the radius of the re-
volving arc, o a; multiply the arc by the central distance, o e ;
subtract this product from the former, double the remainder,
multiply it by 3.1416, and the product will be the surface
required.
Or, lxr—(ax \/r2— I - ) ) X 2p, I representing the length of
the segment.
o
Example. — What is the convex surface of a segment of a
circular spindle, Fig. 62, the length of it being 3.2495 inches,
the radius of its arc 10, the central distance 7-071, and the
length of its side, id, 3.927 inches?
3.2495 x l0=S2A95 = length X radius.
3.927 X 7.071 =27.7678= length of arc x central distance.
32.495 — 27.7678 =4.7272 = difference of products.
4.7272x2 = 9.4544x3.1416=29.702, which is the result re-
quired.
Ex. 2. The segment of a circular spindle is 14.142 feet in
length, the radius of its arc is 20, and the distance between
the plane of the segment, i e, and the centre of the revolving
arc, o e, Fig. 62, is 14.142, and the length of its side, t d, is
15.708 ; what is its convex surface 1 Ans. 381.3745 feet.
MENSURATION OF AREAS, LINES, AND SURFACES. 123
For Surface of a Circular Spindle, Zone, or Segment
General Formula. S = 2(lr—ac)p, I representing length of
spindle, segment, or zone, a the length of its revolving arc, r the
radius of the generating circle, and c the central distance.
Illustration. — The length of a circular spindle is 14.142
inches, th5 length of its revolving arc is 15.708, the radius
of its generating circle is 10, and the distance of its centre
from the centre of the circle from which it is generated is
7.071 ; what is its surface?
2 x (14.142 x 10-15.708 x 7.071) X 3.1416 = 190.6869 = re-
sult required.
Centre of Gravity. See Appendix, page 283.
Note. — The surface of the frustrum of a spindle is obtained by the
division of the surface of a zone.
Cycloidal Spindle.
To ascertain the Convex Surface of a Cycloidal Spindle, Fig. 63.
Rule. — Multiply the area of the generating circle by 64,
and divide it by 3 ; the quotient will give the surface re-
quired.
_ «X64
\)r, — - — —surface.
Fig. 63.
Example. — The area of the generating circle, a be, of a
cycloidal spindle, d e, is 32 inches ; what is the surface of the
spindle ?
124 MENSURATION OF AREAS, LINES, AND SURFACES.
32 x 64 = 2048 =area of circle x 64.
2048 -r-3 = 682.667 —above product-^- 3 —surface required.
Ex. 2. The diameter of the generating circle of a cycloidal
spindle is 20.375 inches ; what is the surface of the spindle?
Ans.- 6955.7483 in.
Ex. 3. The diameter of the generating circle of a cycloidal
spindle is 14.5 inches ; what is the surface of the spindle?
Ans. 3522.773 in.
Ex. 4. The radius of the generating circle of a cycloidal
spindle is 8.5 inches; what is the surface of the spindle in
square feet ! Ans. 33.633 feet.
Centre of Gravity. Is in its geometrical centre.
Note. — The area of a cycloidal spindle is twice the area of the cy-
cloid, to ascertain which, see rule, page 108.
Elliptic, Parabolic, and Hyperbolic Spindles.
The rules to ascertain the surfaces of either an Elliptic, Parabolic, or
Hyperbolic Spindle, or of zones or segments of them, are of a character
to preclude their being given in such a form as would be consistent with
the design of this work ; hence they are omitted here.
See Appendix, p. 278, 279.
Ellipsoid, Paraboloid, or Hyperboloid of Revolution.
Definition. Figures alike to a cone generated by the revolu-
tion of a conic section around its axis.
Note. — These figures are usually known as Conoids.
When they are generated by the revolution of an ellipse,
they are called ellipsoids, and when by a parabola, parabo-
loids, &c, &c, &c.
The revolution of an arc of a conic section around the axis of the
curve will give a segment of a conoid.
MENSURATION OF AREAS, LINES, AND SURFACES. 125
Ellipsoid.*
To ascertain the Convex Surface of an Ellipsoid, Fig. 64.
Kule. — Add together the square of the base a b and four
times the square of the height c d ; multiply the square root
of half their sum by 3.1416, and this product by the radius
of the base. The product will give the surface required.
Or, \J — x 3.1416 X r=zsurface,h representing the height
of the ellipsoid.
Fig. 64. s*r±.
Example. — The base a b of the ellipsoid, Fig. 64, is 10
inches, and the height c d 7 ; what is its surface ?
102-f72 x 4 = 296— sum of the square of the base and 4 times
the square of the height.
296-f-2 = 148, and ^148 = 12.1655 —square root of half
the above sum.
12.1655 x 3.1416 x-^r = 191. 0957 =product of root above
obtained x 3.1416, and that product by 'the radius of the base =
the surface required.
Ex. 2. The base a b of an ellipsoid is 14 inches, and the
height c d 5 ; what is the convex surface of it ?
Ans. 267.534 in.
To ascertain the Convex Surface of a Segment, Frusirum, or Zone
of an Ellipsoid.
See rules for the convex surface of a segment, frustrum, or
zone of an ellipsoid, p. 93-95.
* An ellipsoid is a semi-spheroid. (See p. 91-94.)
126 MENSURATION OF AREAS, LINES, AND SURFACES.
Paraboloid.
To ascertain the Convex Surface of a Paraboloid, Fig. 65.
Rule. — From the cube of the square root of the sum of four
times the square of the height, b d, and the square, of the ra-
dius of the base, d a, subtract the cube of the radius of the
base ; multiply the remainder by the quotient of 3.1416 times
the radius of the base divided by six times the square of the
height, and the product will give the surface required.
rxp
Or, [(-\/4A2-f r2)3— r3] x ~ — ^= convex surface.
d
Example. — The axis b d of a paraboloid, Fig. 65, is 40
inches, the radius a d of its base is 18 inches; what is its
convex surface ?
402x4z=6400r=4 times the square of the height.
6400-t-182 = 6724=:sm« of the above product and the square
of the radius of the base.
(V6724)3-183=:545536=^ remainder' of the cube of the
radius of the base subtracted from the cube% of the square root of
the preceding sum.
3.1416 x 18-^(6 x402)=.0058905=r^e quotient of 3.1416
times the radius of the base +-6 times the square of the height.
Hence 545536 x.0058905z=3213.48=Me product of the
above remainder and the preceding quotient =the result required.
Ex. 2. The axis b d of a paraboloid is 20 inches, and the
diameter of its base a c is 60 inches ; what is its convex sur-
face? . Ans. 3848.46 in.
MENSURATION OF AREAS, LINES, AND SURFACES. 127
Ex. 3. The axis of a parabolic conoid is 18 inches, the ra-
dius of its base 40 inches ; what is its convex surface <?
Ans. 5937.16 in.
Centre of Gravity.- See Appendix, p. 284.
Any Figure of Revolution.
To ascertain the Convex Surface of any Figure of Revolution,
Figs. 66, 67, and 68.
Rule. — Multiply the length of the generating line by the
circumference described by its centre of gravity, and the prod-
uct will give the surface required.
Or, lx%rxp— surface ; r representing radius of centre of
gravity.
Fig. 66. e
, a
-«f
Example. — If the generating line a c of the cylinder a c df
10 inches in diameter, Fig. 66, is 10, then the centre of grav-
ity of it will be in b, the radius of which is b r—5.
Hence 10x5 x 2 x3. 1416 = 314.16 = ^e convex surface of
the cylinder.
Again, If the generating line is c a e g, and it is (e a=z5,
a c=10, and c <7=5)=20, then the centre of gravity o will
be in the middle of the line joining the centres of gravity of
the triangles e a c and a c #=3.75.
Hence 20 x 3775x2 X 3.141 6 = 471.24 =zthe entire surface
of the cylinder.
C Convex surface as above 314.16
Proof ■< Area of each end, 102x-7854 = 78.54,
( and 78.54x2= 157.08
471.24
128 MENSURATION OF AREAS, LINES, AND SURFACES.
Fig. 67. a
Ex. 2. If the generating elements of a cone, Fig. 67, are
a dz=\0,d c=10, and a c the generating line = 14.142, the
centre of gravity of which is in o, and o r=z5.
Then, 14.142 x5 x2 x 3.1416=444.285=^6 convex surf ace
2
of the cone, and 10x2 x. 7854=314. \Q> — area of base.
Hence, 444.285-j-314.16 = 758.445 = w/zo/6 surface of cone.
Again. If the generating line is a c d— 24.142, then the
centre of gravity will be in n, in the middle of the line, join-
ing the angle of the generating line and the base a d at r=5.
Hence, 24.142 x 5~x2 X 3.1416 =758.445 —whole surf ace of
cone.
Fig. 68
Ex. 3. If the generating elements of a sphere, Fig. 68, are
a c=10, a b c will be 15.708, the centre of gravity of which
is in o, and by rule, page 80, o r=3.183.
Hence 15.708x3.183 x2x 3.1416 =314.16 -the surface of
tlie sphere.
To ascertain the Area of an Irregular Figure.
Rule. — Take a uniform piece of board or pasteboard,
weigh it, cut out the figure of which the area is required, and
weigh it ; then, as the weight of the board or pasteboard is to
the entire surface, so is the weight of the figure to its surface.
MENSURATION OP AREAS, LINES, AND SURFACES. 129
CAPILLARY TUBE.
To ascertain the Diameter of a Capillary Tube.
Rule. — Weigh the tube when empty, and again when filled
with mercury ; Subtract the one weight from the other ; re-
duce the difference to troy grains, and divide it by the length
of the tube in inches. Extract the square root of this quo-
tient, and multiply it by .0192245, and the product will be
the diameter of the tube in inches.
/ w
Or,*/ — x. 0192245 —diameter; w representing difference
in weights in Troy grains, and I the length of the tube.
Example. — The difference in the weights of a capillary
tube when empty and when filled with mercury is 90 grains,
and the length of the tube is 10 inches ; what is the diameter
of it?
90-t-lO = 9 =weight of mercury -+■ length of tube; ^9 — 3, and
3 x .0192245 =.0576735 = the square root of the above quotient
X .0192245 = diameter of tube required.
Proof. — The weight of a cubic inch of mercury is 3442.75
Troy grains, and the diameter of a circular inch of equal area
to a square inch is 1.128 (p. 81).
If, then, 3442.75 grams occupy 1 cubic inch, 90 grains will
require .0261419 cubic inch, which, -^ 10 for the height of the
tube, gives .00261419 inch for the area of the section of the tube.
Then v'.00261419 = .051129=SKfeo/Me square of a column
of mercury of this area.
Hence .051129 x 1.128, which is the ratio between the side of
a square and the diameter of a circle of equal area — .057 '67 '35.
F2
130 MENSURATION OF AREAS, LINES, AND SURFACES.
LENGTHS OF CIRCULAR ARCS.
Table of the Lengths of Circular Arcs, the Diameter of a Circle
being Unity, and assumed to be divided into 1000 equal Parts.
Height.
Length.
Height.
Length.
Height.
Length.
Height.
Length.
.100
1.0265
.134
1.0472
.168
1.0737
.202
1.1055
.101
1.0270
.135
1.0479
.169
1.0745
.203
1.1065
.102
1.0275
.136
1.0486
.170
1.0754
.204
1.1075
.103
1.0281
.137
1.0493
.171
1.0762
:'205
1.1085
.104
1.0286
.138
1.0500
.172 1.0771 .206
1.109G
.105
1.0291
.139
1.0508
.173
1.0780
.207
1.1106
.106
1.0297
.140
1.0515
.174
1/0789
.208
1.1117
.107
1.0303
.141
1.0522
.175
1.0798
.209
1.1127
.108
1.0308
.142
1.0529
-.176
1.0807
.210
1.1137
.109
1.0314
.143
1.0537
.177
1.0816
.211 1.1148
.110
1.0320
.144
1.0544
.178
1.0825
.212
1.1158
.111
1.0325-
.145
1.0552
.179
1.0834
.213
1.1169
.112
1.0331
.146
1.0559
.180
1.0843
.214 1.1180
.113
1.0337
.147
1.0567
.181
1.0852 .215;1.1190
.114
1.0343
.148
1.0574
.182
1.0861
.21611.1201
.115
1.0349
.149
1.0582
.183
1.0870
.217 1 1.1212
.116 1 1.0355
.150
1.0590
.184
1.0880
.218
1.1223
.11711.0361
.151
1.0597
.185
1.0889
.219
1.1233
.118 1.0367
.152
1.0605
.186
1.0898
.220 1.1245
.119 1.0373
.153
1.0613
.187
1.0908
.221 |1.1256
.120 ; 1.0380
.154
1.0621
.188
1.0917
.222
1.1266
.12111.0386
.155
1.0629
.189 1.0927
.223
1.1277
.122 1.0392
.156
1.0637
.190 1.0936
.224
1.1289
.123
1.0399
.157
1.0645
.191
1.0946
.225
1.1300
.124
1.0405
.158
1.0653
.192
1.0956
.226
1.1311
.125
1.0412
.159
1.0661
.193
1.0965
.227
1.1322
.126
1.0418
.160
1.0669
.194
1.0975
.228 11.1333
.127
1.0425
.161
1.0678
.195
1.0985
.229 1.1344
.128
1.0431
.162
1.0686
.196
1.0995
.230; 1.1356
.129
1.0438
.163
1.0694
.197
1.10051.231 ! 1.1367
.130
1.0445
.164
.165
1*0703
.198
1.1015S. 232 1.1379
.131
1.0452
1.0711
.199
1.10251.233 1.1390
.132
1.0458
.166
1.0719
.200
1.1035 .234 1.1402
.133
1.0465
.167
1.0728
.201
1.10451
.235
1.1414
MENSURATION OF AREAS, LINES, AND SURFACES. 131
Table of Circular Arcs — Continued.
Height.
Length. 1
Height.
Length.
Height.
Length.
Height
Length.
.236
1.1425
.274
1.1897
.312
1.2422
.350
1.3000
.237
1.1436
.275
1.1908
.313
1.2436
.351
1.3016
.238
1.1448
.276
1.1921
.314
1.2451
.352
1.3032
.239
1.1460
.277
1.1934
.315
1.2465
.353
1.3047
.240
1.1471
.278
1.1948
.316
1.2480
.354
1.3063
.241
1.1483
.279
1.1961
.317
1.2495
.355
1.3079
.242
1.1495
.280
1.1974
.318
1.2510
.356
1.3095
.243
1.1507
.281
1.1989
.319
1.2524
.357
1.3112
.244
1.1519
.282
1.2001
.320
1.2539
.358
1.3128
.245
1.1531
.283
1.2015
.321
1.2554
.359
1.3144
.246
1.1543
.284
1.2028
.322
1.2569
.360
1.3160
.247
1.1555
.285
1.2042
.323
1.2584
.361
1.3176
.248
1.1567
.286
1.2056
.324
1.2599
.362
1.3192
.249
1.1579
.287
1.2070
.325
1.2614
.363
1.3209
.250
1.1591
.288
1.2083
.326
1.2629
.364
1.3225
.251
1.1603
.289
1.2097
.327
1.2644
.365
1.3241
.252
1.1616
.290
1.2120
.328
1.2659
.366
1.3258
.253
1.1628
.291
1.2124
.329 1.2674
.367
1.3274
.254
1.1640
.292
1:2138
.330
1.2689
.368
1.3291
.255
1.1653
.293
1.2152
.331
1.2704
.369
1.3307
.256
1.1665
.294
1.2166
.332
1.2720
.370 1.3323
.257
1.1677
.295
1.2179
.333
1.2735
.371 i 1.3340
.258
1.1690
.296
1.2193
.334
1.2750
.372 ! 1.3356
.259 1.1702
.297
1.2206
.335
1.2766
.373 1.3373
.260 1.1715
.298
1.2220
.336
1.2781
.374 1.3390
.261
1.1728
.299
1.2235
.337
.338
1.2796
.375 1.3406
.262
1.1740
.300
1.2250
1.2812
.376 11.3423
.263
1.1753
.301
1.2264
.339
1.2827
.377 i 1.3440
.264
1.1766
.302
1.2278
.340
1.2843
.378
1.3456
.265
1.1778
.303
1.2292
.341
1.2858
.379
1.3473
.266
1.1791
.304
1.2306
.342
1.2874
.380
1.3490
.267
1.1804
.305
1.2321
.343
1.2890
.381
1.3507
.268
1.1816
.306
1.2335
.344
1.2905
.382
1.3524
.269
1.1829
.307
1.2349
.345
1.2921
.383
1.3541
.270
1.1843
.308
1.2364
.346
1.2937
.384
1.3558
.271
1.1856
.309
1.2378
.347
1.2952
.385
1.3574
.272
1.1869
.310
1.2393
.348
1.2968
.386
1.3591
.273
1.1882
.311
1.2407
.349
1.2984
.387
1.3608
132 MENSURATION OF AREAS, LINES, AND SURFACES.
Table of
Circular Arcs —
■Continued.
Height. | Length.
Height. | Length.
Height. | Length. Height. | Length.
.388 1.3625
.417
1.4132
.445
1.4644
.473
1.5176
.389
1.3643
.418
1.4150
.446
1.4663
.474
1.5196
.390
1.3660
.419
1.4168
.447
1.4682
.475
1.5215
.391
1.3677
.420
1.4186
.448
1.4700
.476
1.5235
.392
1.3694
.421
1.4204
.449
1.4719
.477
1.5254
.393 1.3711
.422
1.4222
.450
1.4738
.478
1.5274
.394; 1.3728
.423
1.4240
.451
1.4757
.479
1.5293
.395 1.3746
.424
1.4258
.452
1.4775
.480
1.5313
.396
1.3763
.425
1.4276
.453
1.4794
.481
1.5332
.397
1.3780
.426
1.4295
.454
1.4813
.482
1.5352
.398
1.3797
.427
1.4313
.455
1.4832
.483
1.5371
.399
1.3815
.428
1.4331
.456
1.4851
.484
1.5391
.400
1.3832
.429
1.4349
.457
1.4870
.485
1.5411
.401
1.3850
.430
1.4367
.458
1.4889
.486
1.5430
.402
1.3867
.431
1.4386
.459
1.4908
.487
1.5450
.403
1.3885
.432
1.4404
.460
1.4927
.488
1.5470
.404
1.3902
.433
1.4422
.461
1.4946
.489
1.5489
.405
1.3920
.434
1.4441
.462
1.4965
.490
1.5509
.406
1.3937
.435
1.4459
.463
1.4984
.491
•1.5529
.407
1.3955
.436
1.4477
.464
1.5003
.492
1.5549
.408
1.3972
.437
1.4496
.465
1.5022
.493
1.5569
.409
1.3990
.438
1.4514
.466
1.5042
.494
1.5585
.410
1.4008
.439
1.4533
.467
1.5061
.495
1.5608
.411
1.4025
.440
1.4551
.468
1.5080
.496
1.5628
.412
1.4043
.441
1.4570
.469
1.5099
.497
1.5648
.413
1.4061
.442
1.4588
.470
1.5119
.498
1.5668
.414
1.4079
.443
1.4607
.471
1.5138
.499
1.5688
.415
1.4097
.444
1.4626
.472
1.5157
.500
1.5708
.416
1.4115
To find the Length of an Arc of a Circle by the foregoing Table.
Rule. — Divide the height by the base, find the quotient in
the column of heights, and take the length of that height from
the next right-hand column. Multiply the length thus ob-
tained by the base of the arc, and the product will be the
length of the arc required.
MENSURATION OF AREAS, LINES, AND SURFACES. 133
Example. — What is the length of an arc of a circle, the
span or base being 100 feet, and the height 25 feet?
25 -f- 100 = .25, and .25, per table, =1.1591, which, multiplied
by 100, =1 15,9100 /ee*.
Note. — When, in the division of a height by the base, the quo-
tient has a remainder after the third place of decimals, and great
accuracy is required,
Take the length for the first three figures, subtract it from
the next following length, multiply the remainder by the said
fraction, and add the product to the first length ; the sum will
be the length for the whole quotient.
Example. — What is the length of an arc of a circle, the
base of which is 35 feet, and the height or versed sine 8 feet ?
8^35=.2285714; the tabular length for .228 = 1. 1333, and
for .229 = 1.1344, the difference between which is .0011. Then,
.5714 x .0011 =.00062854. i /
Hence, .228 = 1.1333
.0005714= .00062854
1.13392854, the sum by which the base
of the arc is to be multiplied; and 1.13392854 x 35 = 39 feet
.6874989, which, x 12 for inches =8. 25, making the length of the
arc 39 feet 8.25 inches.
134 MENSURATION OF AREAS, LINES, AND SURFACES.
AREAS OF SEGMENTS OF A CIRCLE.
Table of the Areas of the Segments of a Circle, the Diameter of
which is Unity, and assumed to be divided into 1000 equal
Parts.
Versed
Bine.
Seg. Area.
Versed
sine.
Seg. Area.
Versed
sine.
Seg. Area.
Versed
sine.
Seg. Area.
.001
.00004
.034
.00827
.067
.02265
.100
.04087
..002
.00012
.035
.00864
.068
.02315
.101
.04148
.003
.00022
.036
.00901
.069
.02336
.102
.04208
.004
.00034
.037
.00938
.070
.02417
.1(53
.04269
.005
.00047
.038
.00976
.071
.02468
.104
.04310
.006
.00062
.039
.01015
.072
.02519
.105
.04391
.007
.00078
.040
.01054
.073
.02571
.106
.04452
.008
.00095
.041
.01093
.074
.02624
.107
.04514
.009
.00113
.042
.01133
.075
.02676
.108
.04575
.010
.00133
.043
.01173
.076
.02729
.109
.04638
.011
.00153
.044
.01214
.077
.02782
.110
.04700
.012
.00175
.045
.01255
.078
.02835
.111
.04763
.013
.00197
.046
.01297
.079
.02889
.112
.04826
.014
.00220
.047
.01339
.080
.02943
.113
.04889
.015
.00244
.048
.01382
.081
.02997
.114
.04953
.016
.90268
.049
.01425
.082
.03052
.115
.05016
.017
.00294
.050
.01468
.083
.03107
.116
.05080
.018
.00320
.051
.01512
.084
.03162
.117
.05145
.019
.00347
.052
.01556
.085
.03218
.118
.05209
.020
.00375
.053
.01601
.086
..03274
.119
.05274
.021
.00403
.054
.01646
.087
.03330
.120
.05338
.022
.00432
.055
•01691
.088
.089
.03387
.121
.05404
.023
.00462
.056
.01737
.03444
.122
.05469
.024
.00492
.057
.01783
.090
.03501
.123
.05534
.025
.00523
.058
.01830
.091
.03558
.124
.05600
.026
.00555
.059
.01877
.092
.03616
.125
.05666
.027
.00587
.060
.01924
.093
.03674
.126
.05733
.028
.00619
.061
.01972
.094
.03732
.127
.05799
.029
.00653
.062
.02020
.095
.03790
.128
.05866
.030
.00686
.063
•02068
.096
.03849
.129
.05933
.031
.00721
.064
.02117
-097
.03908
.130
.06000
.032
.00756
.065
.02165
.098
.03968
.131
.06067.
.033
.00791
.066
.02215
.099
•04027
.132
.06135
MENSURATION OF AREAS, LINES, AND SURFACES. 135
Table of Areas of Segments of a Circle — Continued.
Versed
sine.
Seg. Area.
Versed
sine.
Seg. Area,
1 Versed
1 sine.
Seg. Area.
Versed
sine.
Seg. Area.
.133
.06203
.171
.08929
.209
.11908
.247
.15095
.134
.06271
.172
.09004
.210
.11990
.248
.15182
.135-
.06339
.173
.09080
.211
.12071
.249
.15268
.136
.06407
.174
.09155
.212
.12153
.250
.15355
.137
.06476
.175
.09231
.213
.12235
.251
.15441
.138
.06545
.176
.09307
.214
.12317
.252
.15528
.139
.06614
.177
.09384
.215
.12399
1.253
.15615
.140
.06683
.178
.09460
.216
.12481
.254
.15702
.141
.06753
.179
.09537
.217
.12563
.255
.15789
.142
.06822
.180
.09613
.218
.12646
.256
.15876
.143
.06892
.181
.09690
.219
.12728
.257
.15964
.144
.06962
.182
.09767
.220
.12811
.258
.16051
.145
.07033
.183
.09845
.221
.12894
.259
.16139
.146
.07103
.184
.09922
.222
.12977
.260
.16226 *
.147
.07174
.185
. 10000
.223
.13060
.261
.16314
.148
.07245
.186
.10077
.224
.13144
.262
. 16402
.149
.07316
.187
.10155
.225
.13227
.263
.16490
.150
.07387
.188
.10233
.226
.13311
.264
.16578
.151
.07459
.189
.10312
.227
.13394
.265
.16666
.152
.07531
.190
.10390
.228
.13478
.266
.16755
.153
.07603
.191
.10468
.229
.13562
.267
.16844
.154
.07675
.192
. 10547
.230
.13646
.268
.16931
.155
.07747
.193
.10626
.231
.13731
.269
.17020
.156
.07820
.194
.10705
.232
.13815
.270
.17109
.157
.07892
.195
.10784
.233
.13900
.271
.17197
.158
.07965
.196
.10864
.234
.13984
.272
.17287
.159
.08038
.197
.10943
.235
.14069
.273
.17376
.160
.08111
.198
.11023
.236
.14154
.274
.17465
.161
.08185
.199
.11102
.237
.14239
.275
.17554
.162
.08258
.200
.11182
.238
.14324
.276
.17643
.163
.08332
.201
.11262
.239
.14409
.277
.17733 4
.164
.08406
.202
.11343
.240
.14494
.278
.17822
.165
.08480
.203
.11423
.241
.14580
.279
.17912
.166
.08554
.204
.11503
.242
.14665
.280
.18002
.167
.08629
.205
.11584
.243
.14751
.281
.18092
.168
.08704
.206
.11665
.244
.14837
.282
.18182
.169
,08779
.207
.11746
.245
.14923
.283
.18272
.170
.08853
.208
.11827 .246 .15009
THE ^\
.284
.18361
I U
NIVFl
3-21TV
136 MENSURATION OF AREAS, LINES, AND SURFACES.
Table of Areas of Segments of a Circle — Continued.
Versed
sine.
Seg. Area.
Versed
sine.
Seg. Area
Versed
sine.
.285
.18452
.323
.21947
.361
.286
.18542
.324
.22040
.36?
.287
.18633
.325
.22134
.363
.288
.18723
.326
.22228'
.364
.289
.18814
.327
.22321
.365
.290
.18905
.328
.22415
.366
.291
.18995
.329
.22509
.367
.292
.19086
.330
.22603
.368
.293
.19177
.331
.22697
.369
.294
.19268
.332
.22791
.370
.295
.19360
.333
.22886
.371
.296
.19451
.334
.22980
.372
.297
.19542
.335
.23074
.373
..298
.19634
.336
.23169
.374
.299
.19725
.337
.23263
.375
.300
.19817
.338
.23358
.376
.301
.19908
.339
.23453
.377
.302
.20000
.340
.23547
.378
.303
.20092
.341
.23642
.379
.304
.20184
.342
.23737
.380
.305
.20276
.343
.23832
.381
.306
.20368
.344
.23927
.382
.307
.20460
.345
.24022
.383
.308
.20553
.346
.24117
.384
.309
.20645
.347
.24212
.385
.310
.20738
.348
.24307
.386
.311
.20830
.349
.24403
.387
.312
.20923
.350
.24498
.388
.313
.21015
.351
.24593
.389
.314
.21108
.352
.24689
.390
.315
.21201
.353
.24784
.391
.316
.21294
.354
.24880
.392
.317
.21387
.355
.24976
.393
.318
.21480
.356
.25071
.394
.319
.21573
.357
.25167
.395
.320
.21667
.358
.25263
.396
.321
.21760
.359
.25359
.397
.322
.21853
.360
.25455
.398
Seg. Area
Versed
sine.
Seg. Area.
.25551
.399
.29239
.25647
.400
.29337
.25743
.401
.29435
.25839
.402
.29533
.25936
.403
.29631
.26032
.404
.29729
.26128
.405
.29827
.26225
.406
.29925
.26321
.407
.30024
.26418
.408
.30122
.26514
.409
.30220
.26611
.410
.30319
.26708
.411
.30417
.26804
.412
.30515
.26901
.413
.30614
.26998
.414
.30712
.27095
.415
.30811
.27192
.416
.30909
.27289
.417
.31008
.27386
.418
.31107
.27483
.419
.31205
.27580
.420
.31304
.27677
.421
.31403
.27775
.422
.31502
.27872
.423
.31600
.27969
.424
.31699
.28067
.425
.31798
.28164
.426
.31897
.28262
.427
.31996
.28359
.428
.32095
.28457
.429
.32194
.28554
.430
.32293
.28652
.431
.32391
.28750
.432
.32490
.28848
.433
.32590
.28945
.434
.32689
.29043
.435
.32788
.29141
.436
.32887
MENSURATION OF AREAS, LINES, AND SURFACES. 137
Table, of Areas c
)f Segments of a
Circle — (
Continued.
Versed Versed Versed
-, , Versed
sine.
Seg. Area.
sine.
Seg. Area.
sine.
Seg. Area.
sine.
Seg. Area.
.437
.32987
.453
.34577
.469
.36172
.485
.37770
.438
.33086
.454
.34676
.470
.36272
.486
.37870
.439
.33185
.455
.34776
.471
.36371
.487
.37970
.440
.33284
.456
.34875
.472
.36471
.488
.38070
.441
.33384
.457
.34975
.473
.36571
.489
.38170
.442
.33483
.458
.35075
.474
.36671
.490
.38270
.443
.33582
.459
.35174
.475
.36781
.491
.38370
.444
.33682
.460
.35274
.476
.36871
.492
.38470
.445
.33781
.461
.35374
.477
.36971
.493
.38570
.446
.33880
.462
.35474
.478
.37071
.494
.38670
.447
.33980
.463
.35573
.479
.37170
.495
.38770
.448
.34079
.464
.35673
.480
.37276
.496
.38870
.449
.34179
.465
.35773
.481
.37370
.497
.38970
.450
.34278
.466
.35872
.482
.37470
.498
.39070
.451
.34378
.467
.35972
.483
.37570
.499
.39170
.452
.34477
.468
.36072
.484
.37670
.500
.39270
To find the Area of a Segment of a Circle by the above Table.
Rule. — Divide the height or versed sine by the diameter of
the circle, and find the quotient in the column of versed sines.
Take the area noted in the next column, and multiply it by
the square of the diameter, and it will give the area required.
Example. — Required the area of a segment, its height be-
ing lO, and the diameter of the circle 50 feet.
10 + 50 = .2, and .2, per table, =.11182; then, .11182 x502
=279.55feet.
Note. — When, in the division of a height by the base, the quo-
tient has a remainder after the third place of decimals, and great
accuracy is required,
Take the length for the first three figures, subtract it from
the next following length, multiply the remainder by the said
fraction, and add the product to the first length ; the sum will
be the length for the whole quotient.
138 MENSURATION OF AREAS, LINES, AND SURFACES.
Example. — What is the area of a, segment of a circle, the
diameter of which is 10 feet, and the height of it 1.575 feet?
1.575-f- 10=. 1575 ; the tabular area for .157=.07892, and
for .15.8 = .07964, the difference between which is .00072.
Then, .5 x .00072 = .000360.
Hence, .157 =.07892
.0005 = .00036
.07928, the sum by which the square
of the diameter of the circle is to be multiplied; and .07928 x 102
= 7.928 /£<*.
MENSURATION OF AREAS, LINES, AND SURFACES. 139
AREAS OF THE ZONES OF A CIRCLE.
Table of the Areas of the Zones of a Circle, the Diameter of
which is Unity, and assumed to be divided into 1000 equal
Parts.
Height.
Area.
Height.
Area.
Height.
Area,
Height.
Area.
.001
.00100
.034
.03397
.067
.06680
.100
.09933
.002
.00200
.035
.03497
.068
.06780
.101
.10031
.003
.00300
.036
.03597
.069
.06878
.102
.10129
.004
.00400
.037
.03697
.070
.06977
.103
.10227
.005
.00500
.038
.03796
.071
.07076
.104
.10325
.006
.00600
.039
.03896
.072
.07175
.105
.10422
.007
.00700
.040
.03996
.073
.07274
.106
.10520
.008
.00800
.041
.04095
.074
.07373
.107
.10618
.009
.00900
.042
.04195
.075
.07472
.108
.10715
.010
.01000
.043
.04295
.076
.07570
.109
.10813
.011
.01100
.044
.04394
.077
.07669
.110
.10911
.012
.01200
.045
.04494
.078
.07768
.111
.11008
.013
.01300
.046
.04593
.079
.07867
.112
.11106
.014
.01400
.047
.04693
.080
.07966
.113
.11203
.015
.01500
.048
.04793
.081
.08064
.114
.11300
.016
.01600
.049
.04892
..082
.08163
.115
.11398
.017
.01700
.050
.04992
.083
.08262
.116
.11495
.018
.01800
.051
.05091
.084
.08360
.117
.11592
.019
.01900
.052
.05190
.085
.08459
.118
.11690
.020
.02000
.053
.05290
.086
.08557
.119
.11787
.021
.02100
.054
.05389
.087
.08656
.120
.11884
.022
.02200
.055
.05489
.088
.08754
.121
.11981
.023
.02300
.056
.05588
.089
.08853
.122
.12078
.024
.02400
.057
.05688
.090
.08951
.123
.12175
.025
.02500
.058
.05787
.091
.09050
.124
.12272
.026
.02599
.059
.05886
.092
.09148
.125
.12369
.027
.02699
.060
.05986
.093
.09246
.126
.12469
.028
.02799
.061
.06085
.094
.09344
.127
.12562
.029
.02898
.062
.06184
.095
.09443
.128
.12659
.030
.02998
.063
.06283
.096
.09540
.129
.12755
.031
.03098
.064
.06382
.097
.09639
.130
.12852
.032
.03198
.065
.06482
.098
.09737
.131
.12949
.033
.03298
.066
.06580
.099
.09835
.132
. 13045
140 MENSURATION OP AREAS, LINES, AND SURFACES.
Table of the Zones of a Circle — Continued.
Height.
Area.
Height.
Area.
Height
Area.
Height.
Area.
.133
.13141
.171
.16761
.209
.20274
.247
.2365
.134
.13238
.172
.16855
.210
.20365
.248
.2374
.135
.13334
.173
.16948
.211
.20456
.249
.2382
.136
.13430
.174
.17042
.212
.20546
.250
.2391
.137
.13527
.175
.17136
.213
.20637
.251
.2400
.138
.13623
.176
.17230
.214
.20727
.252
.2408
.139
.13719
.177
.17323
.215
.20818
.253
.2417
.140
.13815
.178
.17417
.216
.20908
.254
.2426
.141
.13911
.179
.17510
.217
.20998
.255
.2434
.142'
.14007
.180
.17603
.218
.21088
.256
.2443
.143
.14103
.181
.17697
.219
.21178
.257
.2451
.144
.14198
.182
.17790
.220
.21268
.258
.2460
.145
.14294
.183
.17883
.221
.21358
.259
.2469
.146
.14390
.184
.17976
.222
.21447
.260
.2477
.147
. 14485
.185
.18069
.223
.21537
.261
.2486
.148
.14581
.186
.18162
.224
.21626
.262
.2494
.149
.14677
.187
.18254
.225
.21716
.263
.2502
.150
.14772
.188
.18347
.226
.21805
.264
.2511
.151
.14867
.189
.18440
.227
.21894
.265
.2520
.152
.14962
.190
.18532
.228
.21983
.266
.2528
.153
.15058
.191
.18625
.229
.22072
.267
.2537
.154
.15153
.192
.18717
.230
.22161
.268
.2545
.155
.15248
.193
.18809
.231
.22250
.269
.2553
.156
.15343
.194
.18902
.232
.22335
.270
.2562
.157
.15438
.195
.18994
.233
.22427
.271
.2570
.158
.15533
.196
.19086
.234
.22515
.272
.2579
.159
.15628
.197
.19178
.235
.22604
.273
.2587
.160
.15723
.198
.19270
.236
.22692
.274
.2595
.161
.15817
.199
.19361
.237
.22780
.275
.2604
.162
.15912
.200
.19453
.238
.22868
.276
.2612
.163
.16006
.201
.19545
.239
.22956
.277
.2620
.164
.16101
.202
.19636
.240
.23044
.278
.2629
.165
.16195
.203
.19728
.241
.23131
.279
.2637
.166
.16290
.204
.19819
.242
.23219
.280
.2645
.167
.16384
.205
.19910
.243
.23306
.281
.2654
.168
.16478
.206
.20001
.244
.23394
.282
.2662
.169
.16572
.207
.20092
.245
.23481
.283
.2670
.170
.16667
.208
.20183
.246
.23568
.284
.2678
MENSURATION OF AREAS, LINES, AND SURFACES.
141
Table
of the Zones oj
a Cm
lie — Continued.
Height.] Area. | Height.
Area.
Height.
Area. 1
Height.
Area.
.285
.26871
.323
.29886
.361
.32656
.399
.35122
.286
.26953
.324
.29962
.362
.32725
.400
.35182
.287
.27035
.325
.30039
.363
.32794
.401
.35242
.288
.27117
.326
.30114
.364
.32862
.402
.35302
.289
.27199
.327
.30190
.365
.32931
.403
.35361
.290
.27280'
.328
.30266
.366
.32999
.404
.35420
.291
.27362
.329
.30341
.367
.33067
.405
.35479
.292
.27443
.330
.30416
.368
.33135
.406
.35538
.293
.27524
.331
.30491
.369
.33203
.407
.35596
.294
.27605
.332
.30566
.370
.33270
.408
.35654
.295
.27686
.333
.30641
.371
.33337
.409
.35711
.296
.27766
.334
.30715
.372
.33404
.410
.35769
.297
.27847
.335
.30790
.373
.33471
.411
.35826
.298
.27927
.336
.30864
.374
.33537
.412
.35883
.299
.28007
.337
.30938
.375
.33604
.413
.35939
.300
.28088
.338
.31012
.376
.33670
.414
.35995
.301
.28167
.339
.31085
.377
.33735
.415
.36051
.302
.28247
.340
.31159
.378
.33801
.416
.36107
.303
.28327
.341
.31232
.379
.33866
.417
.36162
.304
.28406
.342
.31305
.380
.33931
.418
.36217
.305
.28486
.343
.31378
.381
.33996
.419
.36272
.300
.28565
.344
.31450
.382
.34061
.420
.36326
.307
.28644
.345
.31523
.383
.3^125
.421
.36380
.308
.28723
.346
.§1595
.384
.34190
.422
.36434
.309
.28801
.347
.31667
.385
.34253
.423
.36488
.310
.28880
.348
.31739
.386
.34317
.424
.36541
.311
.28958
.349
.31811
.387
.34380
.425
.36594
.312
.29036
.350
.31882
.388
.34444
.426
.36646
.313
.29115
.351
.31954
.389
.34507
.427
.36698
.314
.29192
.352
.32025
.390
.34569
.428
.36750
.315
.29270
.353
.32096
.391
.34632
.429
.36802
.316
.29348
.354
.32167
.392
.34694
.430
.36853
.317
.29425
.355
.32237
.393
.34756
.431
.36904
.318
.29502
.356
.32307
.394
.34818
.432
.36954
.319
.29580
.357
.32377
.395
.34879
.433
.37005
.320
.29656
.358
.32447
.396
•3494p
.434
.37054
.321
.29733
.35Q
.32517
.397
.35001
.435
.37104
.322
.29810
.360
.32587
.398
.35062
.436
.37153
142 MENSURATION OF AREAS, LINES, AND SURFACES.
Tabh
i of ike Zones of a Circle — Continued.
Height.] Area.
Height.
Area.
Height.
Area. Height.
Area.
.437
.37202
.453
.37931
.469
.38549
.485
.39026
.438
.37250
.454
.37973
.470
.38583
.486
.39050
.439
.37298
.455
.38014
.471
.38617
.487
.39073
.440
.37346
.456
.38056
.472
.38650
.488
.39095
.441
.37393
.457
'.38096
.473
.38683
.489
.39117
.442
.37440
.458
.38137
.474
.38715
.490
.39137
.443
.37487
.459
.38177
.475
.38747
.491
..39156
.444
.37533
.460
.38216
.476
.38778
.492
.39175
.445
.37579
.461
.38255
.477
.38808
.493
.39192
.446
.37624
.462
.38294
.478
.38838
.494
.39208
.447
.37669
.463
.38332
.479
.38867
.495
.39223
.448
.37714
.464
.38369
.480
.38895
.496
.39236
.449
.37758
.465
.38406
.481
.38923
.497
.39248
.450
.37802
.466
.38443
.482
.38950
.498
.39258
.451
.37845
.467
.38479
.483
.38976
.499
.39266
.452
.37888
.468
.38514
.484
.39001
.500
.39270
To find the Area of a Zone by the above Table.
Rule 1. — When the zone is less than a semicircle, divide the
height by the diameter, and find the quotient in the column
of heights. Take out the area opposite to it in the next col-
umn on the right hand, and multiply it by the square of the
longest chord ; the product will be the area of the zone.
Example. — Required the area of a zone, the diameter of
which is 50, and its height 15 ?
15^50 = . 300; and .300, as per table, =.28088.
Hence, .28088 x 502= 702.2, the area of the zone.
Rule 2. — When the zone is greater than a semicircle, take the
height on each side of the diameter of the circle, and find, by
Rule 1, their respective areas; add the areas of these two
portions together, and the sum will be the area of the zone.
Example. — Required the area of a zone, the diameter of
the circle being 50, and the height of the zone on each side of
the diameter of the circle 20 and 15 respectively.
MENSURATION OF AREAS, LINES, AND SURFACES. 143
20-r-50=.400; A00, as per table, = .35182 ; and .35182 x
502=879.55.
15-^50=.300; .300, asper table, =.28088 ; and .28088 x
502= 702/2.
Hence, 879.55 + 702.2 = 1581.65, the result required.
Note. — -When, in the division of a height by the chord, the
quotient has a remainder after the third place of decimals, and
great accuracy is required,
Take the area for the first three figures, subtract it from
the next following area, multiply the remainder by the said
fraction, and add the product to the first area ; the sum will
be the area for the whole quotient.
Example. — What is the area of a zone of a circle, the
greater chord being 100 feet, and the breadth of it 14 feet 3
inches %
14 feet 3 inches=U.25, and 14.25 -f- 100 = .1425 ; the tab-
ular- length for .142 = .14007 ', and for .143=. 14103, the differ-
ence between which is .00096.
Then, .5 x .00096 = .000480.
Hence, .142 =.14007,' .
.0005 =.00048
.14055, the sum by which the square
of the greater chord is to be multiplied; and .14055 xl002=:
1405.5 feet.
^
144 MENSURATION OF AREAS, LINES, AND SURFACES.
PEOMISCUOUS EXAMPLES.
1. If a load of wood is 8 feet long, 3 feet 10 inches wide,
and 6 feet 6 inches high, what are its contents ?
Ans. 1.72 cords.
2. Add | of a ton to -j^ of a cwt. Ans. 12.329 cwt.
3. What are the contents of a board 25 feet long and 3 feet
wide? Ans. 75 feet.
4. What is the difference between the contents of two floors ;
one being 37 feet long and 27 feet wide, and the other 40 feet
long and 20 feet wide? Ans. 199 feet.
5. How many yards of paper that is 30 inches wide will it
require to cover the wall of a room that is 15^ feet long, 11^
feet wide, and 7 J feet high ? Ans. 55.2833 yards.
6. If i of a post stands in the mud, J in the water, and 10
feet above the water, what is the length of the post ?
Ans. 18.182 feet.
7. What fraction is -that to which if f- of -J be added the
sum will be 1 ? Ans. •§-§•
8. If the earth make one complete revolution in 23 hours
5£ minutes 3 seconds, in what time does it move one degree ?
Ans. 3 mm. 59.3417 seconds.
9. From a plank 26 inches broad a square yard and a half
is to be sawed off; what distance from the end must the line
be struck ? Ans. 6. 23 feet.
10. What is the side of a triangle that may be inscribed in
a circle, the circumference of which is 1000 feet?
Ans. 275.6556 feet.
11. How large a square field can be made in a circle of
100 rods in diameter?
Ans. 22 rods 2 yards 2 feet 4.5 inches.
12. A rectangular field is 12 rods 2 yards 2 feet and 3
inches in length, by 9 rods and 1 yard in breadth ; what is
its area in square yards? Ans. 3471.875 yards. .
MENSURATION OF AREAS, LINES, AND SURFACES. 145
13. The sides, of a triangular plot of ground are 24, 36,
and 48 feet ; what is its area in square feet ?
Ans. 418.282 feet.
14. In turning a chaise within a ring of a certain diameter,
the outer wheel made two turns while the inner wheel made
but one, the wheels being four feet in diameter, and five feet
asunder on the axle-tree ; what was the circumference of the
track described by the outer wheel? Ans. 62.832 feet.
15. The ball on the top of a church is 6 feet in diameter ;
what did the gilding of it cost at 8 cents per square inch ?
Ans. 1302.884 dollars.
16. A roof of a house is 24 feet 8 inches by 14.5 feet, and
is to be covered with lead weighing 8 lbs. per foot; what
will be the weight of the lead required ?
Ans. 2861.324 lbs.
17. The area of an equilateral triangle, whose base falls on
the diameter, and its vertex in the middle of the arc of a semi-
circle, is equal to 100 ; what is the diameter of the semicircle?
Ans. 26.32148.
18. The distance of the centres of two circles, the diameters
of which are each 50, is equal to 30 j what is the area of the
space inclosed between the two circles by arcs of their circum-
ferences? Ans. 559.115.*
19. In the latitude of London, the distance around the
earth, measured on that parallel, is about 15,550 miles ; now,
as the earth revolves in 23 hours and 56 minutes, at what
rate per hour does the city of London move from west to
east? Ans. 649.7214 miles per hour.
20. A father left his son an estate, ^ of which he ran
through in 8 months ; f- of the remainder lasted him 12 months
longer, wheit he had barely $820 left ; what sum did his fa-
ther leave him ? Ans. $1913.34.
21. There is a segment of a circle the chord of which is 60
feet, its versed sine 10 feet ; what will be the versed sine of
that segment of the same circle when the chord is 90 feet?
Ans. 28.2055.
* By Table of Areas of the Segments of a Circle, p. 134.
G
146 MENSURATION OF AREAS, LINES, AND SURFACES.
22. If a line 144 feet long will reach from the top of a
fort to a point on the opposite side of a river that is 64 feet
wide, what is the height of the fort above that point 1
Ans. 128.99 feet.
23. A certain room is 20 feet long, 16 feet wide, and 12 feet
high ; how long must a line be to extend from one of the lower
corners to an opposite upper corner'? Ans. 28.2843 feet.
24. Two ships sail from the same por.t ; one sails due north
128 miles, the other due east 72 miles; how far are the ships
from each other? Ans. 146.86+ miles.
25. There are two columns in the ruins of Persepolis left
standing upright ; one is 70 feet above the plain, and the oth-
er 50 ; in a straight line between these stands an ancient
statue 5 feet in height, the head of which is 100 feet from the
summit of the higher, and 80 feet from the top of the lower
column ; required the distance between the tops of the two
columns'? * Ans. 143.543 feet.
26. The height of a tree growing in the centre of a circular
island 100 feet in diameter is 160 feet, and a line extending
from the top of it to the further shore is 400 feet ; what is
the breadth of the stream, assuming the land on each side of
the water to be level % Ans. 316.6065 feet.
27. A ladder 70 feet long is so placed as to reach a win-
dow 40 feet from the ground on one side of a street, and with-
out removing it at the foot, will reach a window 30 feet high
on the other side ; what is the breadth of the street ?
Ans. 120.6911/^.
28. If a tree stand on a horizontal plane 80 feet in height,
at what height from the ground must it be cut off so that the
top of it may fall on a point 40 feet from the bottom of the
tree, the end where it was cut off resting on the%tump ?
Ans. 30 feet.
29. Four men, A, B, C, D, bought a grindstone, the diam-
eter of which was 4 feet ; they agreed that A should grind
off his share first, and tnat each man should have it alternately
until he had worn off his share ; how much will each man grind
off? Ans. A 3.215 + , B 3.81 + , C 4.97 + , D 12 inches.
MENSURATION OF AREAS, LINES, AND SURFACES. 147
30. The classification of a school is as follows, viz., -^ of
the boys are taught geometry, -| grammar, -^5- arithmetic, ^
writing, and 9 reading; what is the number in each branch'?
j, (5 geometry, 30 grammar, 24 arithmetic,
' \ 12 writing, and 9 reading.
31. A certain general has an army of 141,376 men. How
many must he place in rank and file to form them into a
square? Ans. 376.
32. If *he area of a circle be 1760 yards, how many feet
must the side of a square measure to contain that quantity ?
Ans. 125.8571 jeet.
33. If the diameter of a round stick of timber be 24 inches,
how large a square stick may be hewn from it ?
Ans. 16.97 inches.
34. To set out an orchard of 2400 mulberry trees so that
the length shall be to the breadth as 3 to 2, and the distance
of each tree one from the other 7 yards, how many trees must
there be in the length of the orchard, and how many in its
breadth, and how many square yards of ground do they stand
on ? C Trees in length, 60.
Ans. < Trees in breadth, 40.
{Square yards, 1 1 7,600.
35. Suppose the expense of paving a semicircular plot of
ground, at 30 cents per square foot, amounted to $25.63,
what is the diameter of it1? Ans. 14.75 feet.
36. Two sides of an obtuse-angled triangle are 20 and
40 poles ; what must be the length of the third side, that
the triangle may contaiiFJust an acre ?
Ans. 58.876 ^ofes.
37- If two sides of an obtuse-angled triangle, the area of
which is =60 x -/3, are 12 and 20, what is the third side?
Ans. 28.
38. If an area of 63 feet is cut off from a triangle, the
three sides of which are 13, 20, and 21 feet, by a line paral-
lel to the longest side or base of the triangle, what are the
lengths of the sides of the triangle which will include that
area?
148 MENSURATION OF AREAS, LINES, AND SURFACES.
Operation. — A triangle of the above dimensions has an area of 126.
See Rule, p. 51.
Then, as 126 {—area of triangle) : J-f^ ( — area of required triangle)'.'.
hyp.2 (400) : hyp/2 (200)=square of hyp. of required triangle; and -y/200
= 14.142 =square root of square of hyp.— hyp.' of required triangle.
Hence, as hyp. (20) : base (21) : : hyp.' (14.142) : base of required trian-
gle (14.849).
Consequently, 14.849 is the base, 14.142 is the hyp., and by Rule, p.
53, V 14.1422 -14.8492 =4.527, the length of the remaining side.
39. Seven men bought a grindstone of 60 inches in diam-
eter, each paying \ part of the cost;, what part of the diam-
eter can each grind down for his share ?
(The 1st, 4.4508 ; 2d, 4.8400 ; 3d, 5.3535 ;
Ans. < 4th, 6.0765 ; 5th, 7.2079 ; 6th, 9.3935 ;
( • and the 7th, 22.6778.
This problem may be thus constructed, Fig. 69 :
On the radius, a c, describe a semicircle ; also divide a c into as many
equal parts, c d, d e, ef &c, as there are shares, and erect the perpen-
diculars d I, e m,fn, &c, meeting the semicircle described on a c in /,
m, n, o, p, q.
Then, with the centre c and radii c I, c in, c n, &c, describe
circles, and the diameter which each is to grind down will be thus
shown.
Fig. 69.
For, the square of the chords or radii c I, cm, en, &c, arc as the co-
sines c d,c e, cf <fec.
40. A gentleman has a garden 100 feet long and 80 feet
broad, and a gravel walk is to be made of an equal width
MENSURATION OF AREAS, LINES, AND SURFACES. 149
half around it ; what must be the width of the walk so that
it will take up just half the gr.ound ?*
Operation. — 100 X 80= 8000 =area of garden.
100 + 80
=90=half length of half the sides of garden.
2
8000
902 — - =4100=araz of walk if the garden was a square.
Hence, V 4100 =64.0312, and 64.0312 - 90=25.9688= width of the
walk.
41. In the midst of a meadow, well covered with grass,
It just took an acre to tether an ass ;
How long was a line, that, reaching all round,
Restricted his grazing to an acre of ground ?
Ans. 39.2507 yards.
42. A maltster has a kiln that is 16 feet 6 inches square;
it is necessary to pull it down, and build a new one that will
dry three times as much at a time as the old one did ; what
must be the length of its side % Ans. 28.58 feet.
43. In a round garden containing 75 square rods, how
large a square garden can be laid out 1
Ans. 47.7475 square rods.
44. If a circular garden contain 75 square rods, what
must be the side of a square field that would inclose it ?
Ans. 9.772 rods.
45. There is a circular field 25 rods in diameter ; what is
the difference of the areas of the inscribed and circumscribed
squares, and how much do they differ from the areas of the
field?
f312.5 rods, the difference of the squares; 134.1261
Ans. •< rods, the circumscribed square more than the area;
C 178.374 rods, inscribed square less than the area.
* This problem may be constructed thus:
Let abed represent the garden ; make c e =
c b, and with the centre d and radius d e de-
scribe the semicircle g ef. Make b i=$b g,
b l=\bf, and complete the rectangle i b I h,
and the result is obtained.
Fig. 70.
SCAe
150 MENSURATION OF AREAS, LINES, AND SURFACES.
46. Two persons start from the same place ; one goes south
4 miles per hour, the other west 5 miles per hour; how far
apart are they in 9 hours? A ns. 57.6281 miles.
47. The four sides of a field, Fig. 71, the diagonals of which
are equal to each other, are 25, 35, 31, and 19 poles; what
is its area ?
Construction. — In this question, the sums of the squares of the oppo-
site sides of the trapezium being equal (d c2 (19) -\-a b2 (35) = 1586, and
d a2 (25) +c b2 (31) = 1586), the figure may be constructed as follows :
Fig. 71.
Draw a b and a e at right angles, and each equal to the longest of the
given sides (35) ; join b e, and from the points e and b, with radii equal
to the two remaining opposite sides (25 and 31) respectively, describe
arcs intersecting in c on the farther side of b e; join a c, and draw bf
at right angles to it. With the centre c, and radius equal to the re-
maining side (19), describe an arc cutting 6/produced in d. Join a d
and c d; then will a b c d be the figure required.
Hence, in the two triangles a b d and e a c, we have the two sides b a,
a d equal the two a e, e c, each to each ; and the angles a b d and e a c
equal (each being the complement of b a/), and e c and a d similarly
situated; wherefore b d—a c.
Calculation. — On b e let fall the perpendiculars c g and a h.
Now b e*=a b2+a e2=352X2 ; b e = -/352x2=35\/2=49.4975.
a h=bh=\b e=24.7487.
,, c , Ka , bc2+be2-ce> 312+2~x35"2-25a 2
By formula, p. 56, b g=
:28.1428.
26 e 2x49.4975 98.995
g h=bg-b A=28.1428-24.7487=3.3941.
c^ = -/6c2-65r3=-/312-28.14282 = -/ 168.9828=12.9993.
By similar triangles (see Note, p. 50), a h+c g (37.748) :g h (3.3941)
: : a h (24.7487); h i=2.2253.
ai = Va h2 + h i2 = -/24.7487a + 2.22532 = V 612.4981 + 4.9520 =
V 617.4501 =24:8485.
MENSURATION OP AREAS, LINES, AND SURFACES. 151
Again, by similar triangles, h i (2.2253) : h g (3.3941): \a i (24.8485)
: a c= 37.8997 =6 d ; now, by Rule, p. 57, the area of the trapezium* abed
acxbf+fd acXbd ac2 37.89972 1QQ_ . :
= si — ^— = =—-= — 718.1936 poles=4: ac. 1 ro.
2 2 2 2 r
38 po. 5 yds. 7.7076 feet, Ans.
48. A messenger traveling 8 miles an hour was sent to
Mexico with dispatches for the army; after he had gone 51
miles, another was sent with countermanding orders who
could go 19 miles as quick as the former could 16 ; how long
will it take the latter to overtake the former, and how far
must he travel ?
Operation. — If the first messenger travels 8 miles in an hour, and the
second 19 while the first travels 16, the second travels 1.5 miles an hour
Pr-*)
faster than the first.
Then, 51-f-1.5=34 hours, and 3ixd. 5 =323 =number of miles traveled
by second messenger when he overtook the first.
51
Hence, 34+— -=40.375, and 40.375 X 8 =323 miles = ihe distance reach-
8 •
ed by first messenger.
49. The hour and minute hands of a watch are exactly to-
gether at 12 o'clock ; when are they next together?
Operation. — The velocities of the two hands of a watch are to each
other as 12 to 1 ; therefore the difference of velocities is 12 — 1 = 11.
Then, as 11 : i " * \ I : : : 1 : i J »■ *> "" 2?ft«C- ?***
\ 12 X 2 ) \2h.l0m. 54-jSj- sec, 2d tune.
50. A person being asked the hour of the day, replied,
The time past noon is equal to f of the time till midnight ;
what was the time? Ans. 20 minutes past 5.
Operation.— If the time required is | of the time to midnight, then the
whole time from noon to midnight (12 hours or 720 minutes) is divided
into-fd+f+1).
Hence, if -|=720, £=80 minutes, and ^=320 minutes, or 5 hours and
20 minutes, the Ans.
51. A person being asked the time of day, replied that |
of the time from noon was equal to -^T of the time to mid-
night; what was the time? Ans. 40 minutes past 4.
* a c and b d being equal.
152 MENSURATION OF AREAS, LINES, AND SURFACES.
52. What is the radius of a circular acre?
Operation. — Side. of a square (p. 81) X 1.128= diameter of an equal
circle.
The side of a square acre (p. 13) is 208.710321, which, X 1.128 =
235.5, and 235.5^2 to obtain radius=ll7. 75 feet.
53. The time of the day is between 4 and 5, and the hour
and minute hands are exactly together; what is the time?
Operation. — The speed of the hands is as 1 to 11.
4 hours X 60=240, and 240-f-ll=21 min. and 49^ sec, which, added
to 4, =4 hours 21 min. and 49^- sec.
54. A person being asked what o'clock it was, replied that
it was between 5 and 6 ; but, to be more particular, the min-
ute-hand was as far beyond the 6 as the hour-hand wanted of
being to the 6 ; that is, that the hour and minute hands made
equal acute angles with a line passing from the 12 through
the 6 ; required the time.
Operation. — 5 hours— 300 minutes, and 6 hours = 360 minutes.
Then, 300 4- the time by the hour-hand past 5=360— the time by the
minute-hand past 6.
As the relative speed of the hour and minute hands is as 1 to 12,
300 + 1,=360-12.
Q£f) 300
Consequently, ~\%—^i 36^§=the space between the hour
1 ~rl2
and minute hands, which, -J- 2 to obtain the half space (each side of the
6), gives 2 min. 18-^g- sec, which, added to 5 hours and 30 min., =5 hours
32 min. and 18^- sec, the time required.
4
55. Two persons, A and B, start at the same time to meet
each other when apart 100 miles ; after 7 hours they meet,
when it appears that A had ridden \\ miles per hour faster
than B ; at what rate per hour did each ride %
Ans. A 7.893, B 6.392 miles per hour.
56. Swift can travel 7 miles in -§ of an hour, but Slow
can travel only 5 miles in -^ of an hour ; »both started from
one point at the same time to walk a distance of 12 miles;
how much sooner will Swift arrive than Slow ?
Ans. 12.467 seconds.
MENSURATION OP AREAS, LINES, AND SURFACES. 153
57. At a certain time between two and three o'clock, the
minute-hand of a clock was between three and four ; within
an hour after, the hour and minute hands had exactly changed
places with each other ; what was the precise time when the
hands were in the first position ?
Ans. 2 h. 15 m. 56^- sec.
58. If a traveler were to leave New Haven at 8 o'clock on
a morning, and walk toward Albany at the rate of 3 miles an
hour, and another traveler were to set out from Albany at
4 o'clock in the evening, and walk toward New Haven at the
rate of 4 miles an hour, whereabout on the road would they
meet, supposing the distance to be 130 miles ?
Ans. 69.4286 miles from New Haven.
59. A thief, escaping from an officer, has 40 miles the
start, and travels at the rate of 5 miles an hour ; the officer
in pursuit travels at the rate of 7 miles an hour ; how far
must he travel before he overtakes the thief?
Ans. 20 hours, and 140 miles.
60. If 12 oxen graze 3^ acres of grass in 4 weeks, and 21
oxen 10 acres^in 9 weeks, how many oxen would it require to
graze 24 acres in 18 weeks, the grass to be growing ?•
Operation. — Each ox grazes a certain quantity in each week, which
we suppose to be 100 pounds, and of the whole quantify grazed in each
case, a part must have grown during the time of grazing.
Then, by the first condition,
1 2 X 4 X 100—4800 lbs. = whole quantity on 3£ acres for 4 weeks.
4800-7-3^=1410 lbs.=whole quantity on 1 acre for 4 weeks.
By the second condition,
21 X 9 X 100=18900 lbs. =whole quantity on 10 acres for 9 weeks.
18900-7-10 = 1890 lbs.=whole quantity on 1 acre for 9 weeks.
1890 — 1440=450 lbs. = the quantity grown on an acre for 9— 4= 5 weeks.
450-f-9—4=90 lbs. = the quantity which grows on each acre for 1 week.
90x3^x4 = 1200 lbs. = quantity grown on 3^ acres for 4 weeks.
4800—1200=3600 lbs. = original quantity of grass on 3^ acres.
3600-7-3^=1080 lbs. ^original quantity on 1 acre.
And by the last condition,
24 X 1080=25920 lbs. ^original quantity on 24 acres.
24 X 90 X 18 = 38880 lbs. =quantity which grows on 24 acres in 18 weeks.
G2
154 MENSURATION OF AREAS, LINES, AND SURFACES.
25920+38880=64800 lbs.=whole quantity on 24 acres for 18 weeks.
64800-r-18=3600 lbs.— quantity to be grazed from 24 acres each week.
3600-r-100 =36 =m«?i&er of oxen required to graze the whole.
61. A tract of land, exactly square, is inclosed by a three-
railed fence ; the length of each rail is 15 feet, and the num-
ber of rails in the fence is equal to the number of acres in-
closed ; required the area of this tract in acres, and the length
of its side in feet.
Operation. — If the tract of land was inclosed by one rail, then, 15-r-
(4x3) =1.2 5 feet, the length of its side.
Then, if 43,560 square feet make an acre, as 1.252 : 43,560: :1 rail:
27878.4, the number of rails in the fence, or the number of acres in the
tract; and (27878.4 X 15)-r-(4 X 3) =34,848, the length of the side in
feet.
62. What is the radius of a circular acre %
Operation. — Side of a square X 1.128 = diameter of an equal circle.
By table, p. 13, 208.710321 =the side of a square acre.
Then, 208.710321x1.128=235.50, which, -r-2 (for radius), =117.75
feet,
63. There is an island 20 miles in circumference, and three
men start together to travel the same way about it ; A goes
2 miles per hour, B 4 miles per hour, and C 6 miles per hour ;
in what time will they come together again ?
Ans. 10 hours.
64. A hare starts 12 rods before a hound, but is not per-
ceived by him till she has been off 1^ minutes ; she runs at
the rate of 36 rods a minute, and the dog, on view of her,
makes after her at the rate of 40 rods a minute ; how long
will the course hold, and what distance will the dog run ?
Ans. 14 J minutes, and he will run 570 rods.
MENSURATION OF SOLIDS.
155
MENSURATION OF SOLIDS.
OP CUBES AND PAEALLELOPIPEDONS.
Cube.
Definition. A solid contained by six equal square sides.
% To ascertain the Contents of a Cube, Fig. 72
Eule. — Multiply a side of the cube by itself, and that prod-
uct again by a side, and this last product will give the con-
tents required.
Or, s^S, s representing the length of a side, and S the con-
tents.
Fig. 72.
X
~~[\
m
"\
ap
Example. — The side a b of the cube, Fig. 72, is 12 inches;
what are the contents of it f
12.X 12 X 12 = 1728 inches.
Ex. 2. The side of a cube is 15 inches ; what are its con-
tents in feet and inches %
^725.1.953125/^, or 1 foot and H^Voo inches.
Ex. 3. The sides of a cube are 12.5 feet; what are its
contents in cubic feet and yards ?
A C1953.125 cubic feet.
t 72.338 cubic yards.
Centre of Gravity. Is in its geometrical centre.
156
MENSURATION OF SOLIDS.
Parallelopipedon.
Definition. A solid contained by six quadrilateral sides, every
opposite two of which are equal and parallel.
To ascertain the Contents of a Parallelopipedon, Fig. 73.
Rule. — Multiply the length by the breadth, and that prod-
uct again by the depth, and this last product will give the 9
contents required.
Or, lxbxd=S.
Fig. 73.
Example. — The length a b, Fig. 73, is 15, the breadth c d
12, and the depth c b is 11 inches; what are the contents?
15 X 12 X 11 = 1980 inches.
Ex. 2. The length of a parallelopipedon is 15 feet, and
each side of it is 21 inches ; what are its contents?
Ans. 45.9375 feet.
Ex. 3. The dimensions of a parallelopipedon are 20 feet in
length, 11.5 in breadth, and 7 in depth ; what are its contents
in feet? Ans. 1610 feet.
Centre of Gravity. Is in its geometrical centre.
PEISMS, PRISMOIDS, AND WEDGES.
Prisms.
Definition. Solids, the ends of which are equal, similar, and
parallel planes, and the sides of which are parallelograms.
Note. — When the ends of a prism are triangles, it is called a trian-
gular prism ; when rhomboids, a rhomboidal prism ; when squares, a
square prism; when rectangles, a rectangular prism, &c.
MENSURATION OP SOLIDS.
157
To ascertain the Contents of a Prism, Figs. 74 and 75.
Rule. — Multiply the area of the base d efby the height,
and the product will give the contents required.
Or, axh=S. r
Fig. 74. Fig. 75.
Example. — A triangular prism, ahcdef Fig. 75, has sides
of 2.5 feet, and a length c d of 10 feet ; what are its contents ?
(By Eule, p. 52) -^-=1.5625, which, X 1.732^2.70625 =area of end
a be, and 2.70625 X 10 = 27.0625 =feet.
Ex. 2. A side of the end of a triangular prism is 18 inches,
and the length of the prism is 9 feet ; what are its contents !
Ans. 8.7 682 feet.
Ex. 3. What is the solidity of a prism 15 feet in length,
the ends of which are hexagonal, with sides 16 inches in
length f
(By Eule, p. 60) 162X 2.5981 = 665.1136 = square of a side multiplied
by the tabular number for the area of a hexagon; then, 665.1136x15 =
91)76.704, the contents required.
Ex. 4. The sides of an octagonal prism are 3 feet, and its
height 6.75 feet; what are its contents in feet and yards?
* A ("293.3388/^.
t 10.8644 yards.
" Centre of Gravity. For rule, see Mensuration of Areas,
Lines, and Surfaces, p. 100. «.
158
MENSURATION OF SOLIDS.
Prismoids.
Definition. Figures alike to a prism, but having only one pair
of their sides parallel.
Note. — Prismoids, alike to prisms, derive their designation from the
figure of their ends, as triangular, square, rectangular, pentagonal, &c.
To ascertain the Contents of a Prismoid, Fig. 76.
Rule. — To the sum of the areas of the two ends, abed,
efgh, add four times the area of the middle section at i k
parallel to them ; multiply this sum by -J- of the height, and
the product will give the contents required.*
Or, a-\-a/-\-4mxh-r-6 = Si a and a' representing areas of
ends, and m area of middle section.
Or, (bxa-{-4mxn-\-dxc)xh^-6=zS, a b, c d representing
dimensions of ends, and m n of the middle section.
Note. — The length and breadth of the middle section are respective-
ly equal to half the sum of the lengths and breadths of the two ends.
Fig. 76. a b
g h
Example. — What are the contents of a rectangular pris-
moid, Fig. 76, the lengths and breadths of the two ends being
7 and 6, and 3 and 2 inches, and the height 15 feet?
7x6+3x2=42+6=48=sttm of the areas of the two ends.
7+3-i-2=10-7-2 =5 =length of the middle section.
6+2-f-2=8-7-2=4 = breadth of the middle section.
5x4 X 4 = 80 =four times the" area of the middle section.
Then, 48 + 80X
15ft.
~6~
128 X 30=3840 cubic inches.
* This is a general rule, and applies equally to figures of proportion-
ate or dissimilar ends. v
MENSURATION OF SOLIDS. 159
Ex. 2. What is the capacity of a prismoid, the ends of which
are respectively 6 by 8 and 9 by 12 inches, and the height of
it is 5 feet?* Ana. 2.6389 feet.
Ex. 3. What are the contents of a prismoid when the ends
of it are respectively 40.75 by 27.5 inches and 20.5 by 14.75
inches, and the length of it is 23.625 inches ?
Ans. 9.1392 cubic feet
Centre of Gravity. For rule, see Mensuration of Areas,
Lines, and Surfaces, p. 101.
Wedge.
Definition. A prolate triangular prism'.
To ascertain the Contents of a Wedge, Fig. 77.
Rule. — To the length of the edge, e g, add twice the length
of the back ; multiply this sum by the perpendicular height,
ef, and then by the breadth of the back, and i of the product
will be the solidity required.
Or, (J+77x2xAx&)-f-6=S. * '
Fig. 77.
Example. — The back of a wedge, a I, d c, is 20 by 2 inches,
and its height, ef 20 inches; what are its contents'?
20+20 X 2 = 60=length of the edge added to twice the length of the back.
60 X 20 X 2 =2400 —above sum multiplied by the height, and that product
by the breadth of the bach.
2400-f-G=400=£ of above product = the contents required.
* An excavation or embankment of a road, when terminated by par-
allel cross sections, is a rectangular prismoid.
160 MENSURATION OF SOUDS-
Ex. 2. The back of a wedge is 15 inches by 3 broad, the
edge of it 15 inches in length, and the height 30; what are
its contents in inches? Ans. 675 inches.
Ex. 3. The back of a wedge is 64 inches by 9 broad, the
length of the edge is 42 inches, and the height is 2 feet 4
inches ; how many cubic feet are contained in it ?
Ans. 4.1319 cubic feet
Ex. 4. The height of a wedge is 15 inches, the edge 7, and
the base 9 by 3 J ; what are its contents'?
Ans. 218.75 cubic inches.
Centre of Gravity. For rule, see Mensuration of Areas,
Lines, and Surfaces, p. 100.
Note. — "When a wedge is a true prism, as represented by Fig. 77,
the contents of it are equal to the area of an end multiplied by its
length.
regular bodies (Polyhedrons).
Definition. A regular body is a solid contained under a cer-
tain number of similar and equal plane faces* all of which are
equal regular polygons.
Note 1. The whole number of regular bodies which can possibly be
formed is five.
Note 2. A sphere may always be inscribed within, and may always
be circumscribed about a regular body or polyhedron, which will have a
common centre.
1. The Tetrahedron, or Pyramid, Fig. 78, which has four
triangular faces.
2. The Hexahedron, or Cube, Fig. 79, which has six square
faces.
S.^The Octahedron, Fig. 80, which has eight triangular faces.
4. The Dodecahedron, Fig. 81, which has twelve pentagonal
faces.
5. The Icosahedron, Fig. 82, ivhich has twenty triangular faces.
* The angle of the adjacent faces of a polygon is called the diedral
angle.
MENSURATION OF SOLIDS.
161
If the following figures are made of pasteboard, and the lines be so
cut that the parts may be turned up and secured together, they will rep-
resent the five bodies.
Fig. 78.
g. 71).
Fig. 81.
Fig. 80.
Fig. 82.
Tetrahedron.
To ascertain the Contents of a Tetrahedron, Fig. 83.
Rule. — Multiply -^ of the cube of the linear side by the
square root of 2 (1.414213), and the product will be the con-
tents required.
I3
Or, —x -v/2z=S, I representing the length of a side.
Fig. 83.
Example. — The linear side of a tetrahedron, a be, Fig. 83,
is 4 ; what are its contents ?
43 G4
jnX V2=— X lAU = 7.5U3=result required.
162 ^ MENSURATION OP SOLIDS.
Ex. 2. Required the contents of a tetrahedron, the side of
which is 6. Ans. 25.452.
Centre of Gravity. Is in the common centre of the centres
of gravity of the triangles made by a section through the cen-
tre of each side of the figures.
Hexahedron,*
To ascertain the Contents of a Hexahedron {Cube), see Fig. 72,
and Rule, p. 155.
Octahedron.
To ascertain the Contents of an Octahedron, Fig. 84.
Rule.— Multiply •£- of the cube of the linear side by the
square root of. 2 (1.414213), and the product will be the con-
tents required.
Or, Z/xV2=S.
Fig. 84
Example. — What are the contents of the octahedron abed,
Fig. 84, the linear side of which is 4 1 '
43 64
— X V 2 =— - x 1.414 =30.1649 -result required.
o o
Ex. 2. Required the contents of an octahedron, the side of
which is 8? Ans. 241.3226.
Centre of Gravity. Is in its geometrical centre.
* A hexahedron and a cube are identical figures, being solids having
the same number of similar and equal plane faces.
MENSURATION OF SOLIDS. . . 163
Dodecahedron.
To ascertain the Contents of a Dodecahedron, Fig. 85.
Rule. — To 21 times the square root of 5 add 47, and di-
vide the sum by 40 ; then, the square root of the quotient be-
ing multiplied by 5 times the cube of the linear side, will give
the contents required.
A/5x 21 + 47 *
Qr'V 40 X*3*g=S.
d
Example. — The linear side of the dodecahedron abed e*is
3 ; what are its contents 1
/•/5X21+47 __ _ / 2.23606X21 +47 J^
y/ — X27X5=W — X 135= 206.901 = re-
sult required.
Ex. 2. The linear side of a dodecahedron is 1 ; what is the
capacity of it? 4ns. 7.6631.
Centre of Gravity. Is in its geometrical centre.
Icosahedron.
To ascertain the Contents of an Icosahedron, Fig. 86.
Eule. — To 3 times the square root of 5 add 7, and divide
the sum by 2 ; then, the square root of this quotient being
multiplied by -§ of the cube of the linear side, will give the
contents required.
V5x3+7 Px5
°rV"
=S.
164 MENSURATION OP SOLIDS.
Fig. 86. d
Example. — The linear side of the icosahedron ah c d e f
is 3 ; what are its contents %
/-/5X3 + 7 33X5 /2.23606X3 + 7 27x5 ,
y/ T2 X~6-=V 2~§ X-g—v' 6.85409X22.5
=58.9056 ^result required.
Ex. 2. Required the contents of an icosahedron, the linear
side of which is 1 * Ans. 2. 1817.
Centre of Gravity. Is in its geometrical centre.
REGULAR BODIES.
To ascertain the Contents of any regular Solid Body.
• When the Linear Edge is given.
Rule. — Multiply the cube of the linear edge by the mul-
tiplier in column A in the table on the following page, and
the product will be the contents required.
Example. — What is the capacity of a hexahedron having
sides of 3 inches ?
33 X 1=27 =tabular volume multiplied by cube of edge =contents required.
When the radius of the Circumscribing Sphere is given.
Rule. — Cube the radius of the circumscribing sphere, and
multiply it by the multiplier opposite to the figure in column B.
Example. — The radius of the circumscribing sphere of a
hexahedron is 1.732 inches; what is the volume of it?
1.7323 X 1.5S06=product of cube of radius and tabular multiplier =8 =
result required.
MENSURATION OF SOLIDS.
165
When the radius of the Inscribed Sphere is given.
Rule. — Cube the radius, and multiply it by the multiplier
opposite to the figure in column C.
Example. — The radius of the inscribed sphere of a hexahe-
dron is 1 inch ; what is its volume ?
13 x 8 —product of cube of radius and tabular multiplier =8 =result re-
quired.
No. of
sides.
Figures.
A,
By linear
B.
By radius
of circum.
sphere.
C.
By radius of
inscribing
sphere.
Angle between
two adjacent
faces.
4
6
8
12
20
Tetrahedron
Hexahedron
Octahedron
Dodecahedron
Icosahedron
0.11785
1.
0.47140
7.66312
2.18169
.51320
1.53960
1.33333
2.78516
2.53615
13.85641
70° 31' 42'
90
109 28 18
5.55029116 33 54
5.05406 138 11 23
8.
6.92820
Note. — For further rules to ascertain the elements of polyhedrons,
see Appendix, p. 262.
Centre of Gravity. Is in their geometrical centre.
Cylinder.
Definition. A figure formed by the revolution of a right-an-
gled parallelogram around one of its sides.
To ascertain the Contents of a Cylinder, Fig. 87.
Rule. — Multiply the area of the base by the height, and
the product will give the contents required.
Or, axh=zS.
Fig. 87.
166
MENSURATION OF SOLIDS.
Example. — The diameter of a cylinder, b c, is 3 feet, and
its length, a £, 7 feet ; what are its contents ?
Area of 3 feet = 7.068. Then, 7.068 X 7=49.476 =result required.
Ex. 2. "What are the contents of a cylinder, the height of
which is 5 feet, and the diameter 2 feet ?
Ans. 15.708 feet.
Ex. 3. The circumference of the base of a cylindrical col-
umn is 20.42 feet, and the height of the column is 9.695 feet;
what is its volume'? Ans. 320.515 feet.
Centre of Gravity. Is in its geometrical centre.
Cone.
Definition. A figure described by the revolution of a right-
angled triangle about one of its legs.
To ascertain the Contents of a Cone, Fig. 88.
Eule. — Multiply the area of the base by the perpendicular
height, and one third of the product will be the contents re-
quired.
axh
Or,
3
:S.
Fig. 88.
Example. — The diameter, a b, of the base of a cone is 15
inches, and the perpendicular height, c d, 32.5 inches ; what
are the contents of the cone ?
MENSURATION OF SOLIDS. 167
Area of 15 inches =176.7146.
Then, 176,7UQ6X32-5 = 1914.4082 cubic inches.
Ex. 2. The diameter of the base of a cone is 20, and its
height 24 inches ; what are its contents in cubic inches %
Ans. 2513.28 inches.
Ex. 3. What are the contents of a cone when the diameter of
its base is 1.5 feet, and its height 15 feet ? Ans. 8.8358 feet.
Ex. 4. The diameter of the base of a cone is 12.732 feet,
and the height of it is 50 feet ; what is its volume?
Ans. 2121.9386 feet.
Centre of Gravity. It is at a distance from the base J of
the line joining the vertex and centre of gravity of the base.
To ascertain the Contents of the Frustrum of a Cone, Fig. 89.
Rule.— Add together the squares of the diameters of the
greater and less ends and the product of the two diameters ;
multiply their sum by .7854, and this product by the height ;
then divide this last product by three, and the quotient will
give the contents required.
Or, add together the squares of the circumferences of the great-
er and less ends and the product of the two circumferences ; mul-
tiply their sum by .07958, and this product by the height; then,
divide this last product" by three, and the quotient will give the con-
tents required.
r\ 79 i j/9 . 3 — T/ -7854 x A ~
Or, d2+d2+dxd'x ^ =S.
o
Fig. 89.
168 MENSURATION OF SOLIDS.
Example.-*— What are the contents of the frustrum of a
cone, the diameters of the greater and less ends, bd,ac, being
respectively 5 and 3 feet, ami the perpendicular height, e o,
9 feet?
52 + 32+5x3=49=^e sum of the squares of the diameters and the
product of the diameters.
49 X .7854 =38.4846 =the above sum by .7854.
— '■ — =1 15.4538 — the last product X the height and divided by
three, which is the result required.
Ex. 2. What are the contents of the frustrum of a cone,
the diameters of the ends being respectively 2 ahd 4 feet, and
the height 9 feet? Ans. 65.9736 feet.
Ex. 3. The frustrum of a cone is 12 inches in height, and
has diameters of 7 and 9^ inches ; what are the contents of
it? Ans. 646.3842 inches.
Centre of Gravity. It is at a distance from the base £ of
(R-f-r)2-l-2R2
the smaller end— £ height x-m , ' — =-} R and r radii of
(R-j-r)2— Rr "
the greater and less ends.
Pyramid.
Definition. A figure, the base of which has three or more
sides, and the sides of which are plane triangles.
Note. — The volume of a pyramid is equal to one third of that of a
prism having equal bases and altitude.
To ascertain the Contents of a Pyramid, Fig. 90.
Rule. — Multiply the area of the base by the perpendicular
height, and one third of the product will be the contents re-
quired.
Or, — =&
MENSURATION OF SOLIDS.
Fig. 90.
169
Example. — What are the contents of a hexagonal pyramid,
a b c, Fig. 90, a side, a b} being 40 feet, and its height, c e,
60 feet?
40ax 2.5981 (tabular multiplier, p. 60)=4156.96=area of base.
A 1 R.Ct CiCt \s f*f\
'— =83139.2 —one third of the area of the base X the height=
the contents required.
Ex. 2. The height of a quadrangular pyramid is 67 feet,
and the width of its base is 16.5 feet; what are its contents
in cubic feet? Ans. 6080.25.
Ex. 3. What are the contents of a pentagonal pyramid, its
height being 12 feet, and each of its sides 2 feet?
Am. 27.528 feet.
Centre of Gravity. It is at a distance from the base i of
the line joining the vertex and centre of gravity of the base.
To ascertain the Contents of the Frustrum of a Pyramid, Fig. 91.
Rule. — Add together the squares of the sides of the great-
er and less ends and the product of these two sides ; multiply
the sum by the tabular multiplier for areas in Table, p. 60,
and this product by the height ; then, divide the last prod-
uct by three, and the quotient will give the contents re-
quired.
Or, (ss+s'z+s x/) X tab. mult. x^=S} s and sr representing
o
the lengths of the sides.
If
170
MENSURATION OF SOLIDS.
Note. — When the areas of the ends are known, or can be obtained
without reference to a tabular multiplier, use the following.
a-f a/4-yaxa/XQ=S, a and a' representing areas of the
o
ends.
Fig. 91.
Example. — What are the contents of the frustrum of a hex-
agonal pyramid, Fig. 91, the lengths of the sides of the great-
er and less ends, a b, c d, being respectively 3.75 and 2.5 feet,
and its perpendicular height, e o, 7.5 feet %
3.752+2.52=20.3125=sww of the squares of sides of greater and less
ends.
20.3125 +3.75 X 2.5 =29.6875 =above sum added to the product of the
two sides.
29.6875 X 2.5981 X 7.5 =578.48 —the last sum X tab. mult., and again by
the height, which, -+3 = 192.82/eef.
Ex. 2. The frustrum of a hexagonal pyramid has sides of
4 and 3 feet, and a height of 9 feet ; what are its contents !
Ans. 294.3891 feet.
When the Ends of a Pyramid are not those of a Regular Polygon,
or when the Areas of the Ends are given.
Rule. — Add together the areas of the two ends and the
square root of their product ; multiply the sum by the height,
and one third of the product will be the contents.
Or, a+a/+"V/axa/Xg=S.
MENSURATION OF SOLIDS. 171
* Example. — What are the contents of an irregular-sided
frustrum of a pyramid, the areas of the two ends being 22
and 88 inches, and the length 20 inches ?
22 + 88 = 110= sum of areas of ends.
22x88 = 1936, and V 1936= 44 = square root of product of areas.
— =1026.66=one third of sum of above sum and product X
o
the height —feet.
Ex. 2. The areas of the ends of an irregular-sided frustrum
of a pyramid are 81 and 100 inches, and the length 25 inches;
what are its contents'? Arts. 2258.33 inches.
Centre of Gravity. It is at a distance from the centre of
the smaller end=i heiqhtx\n — r^ — oc— 5 R <*nd r radii °f
(R-f-r)2— Rr
the greater and less ends.
/Spherical Pyramid.
A Spherical Pyramid is that part of a sphere included within
three or more adjoining plane surfaces meeting at the centre of the
sphere. The spherical polygon defined by these plane surfaces of
the pyramid is called the base, and the lateral faces are sectors of
circles.
To ascertain the Elements of Spherical Pyramids, see Docharty
and Hackley's Geometry.
Cylindrical Ungulas.
Definition. Cylindrical ungulas are frustrums of cylinders.
Conical ungulas are frustrums of cones*
To ascertain the Contents of a Cylindrical Ungula, Fig. 92.
1 . When the section is parallel to the axis of the cylinder.
Rule — Multiply the area of the base by the height of the
cylinder, and the product will be the contents required.
Or, axh=S.
* For Mensuration of Conical Ungulas, see Conic Sections, p. 253.
172
MENSURATION OF SOLIDS.
Fig. 92.
Example. — The area of the base, def Fig. 92, of a cylindric-
al ungula is 15.5 ins., and its height 20 ; what are its contents?
15.5 X 20 =310 =product of area and height =result required.
Ex. 2. The area of the base of a cylindrical ungula is
168.25 inches, and the height of it 22 ; what are its contents
in cubic feet? Ans. 2.148 cubic feet.
2. When the section passes obliquely through the opposite sides
of the cylinder, Fig. 93.
Rule. — Multiply the area of the base of the cylinder by
half the sum of the greatest and least lengths of the ungula,
and the product will be the contents required.
Or, axl+T'+2=S.
Fig. 93.
MENSURATION OP SOLIDS.
173
Example. — The area of the base d efof a cylindrical un-
gula is 25 inches, and the greater and less heights of it, a d,
b e, are 15 and 17 inches; what are its contents?
25 X- —4:00=product of half the sum of the heights and the area
of the base=result required.
Ex. 2. The area of the base of a cylindrical ungula is 75.8
inches, and the greater and less heights of it are 4.25 and 5.65
feet ; what are its contents in cubic feet %
Ans. 2.6056 cubic feet.
3. When the section passes though the base of the cylinder and
one of its sides, and the versed sine does not exceed the sine, Fig. 94.
Eule. — From two thirds of the cube of the sine, a d, of the
arc, dcfof the base, subtract the product of the area of the
base and the cosine,* a e, of the half arc.
Multiply the difference thus found by the quotient arising
from the height divided by the versed sine, and the product
will give the contents required.
s32 h
Or, -5 a xcX — =S, v s representing the versed sine.
o v s
Fiff.di. ,
Example. — The sine a d of half the arc of the base of an
ungula, Fig. 94, is 5, the diameter of the cylinder is 10, and
the height of the ungula 10 ; what are the contents of it ?
* When the cosine is 0, the product is 0.
174
MENSURATION OF SOLIDS.
§• of 53 = 83.333 = taw thirds of the cube of the sine.
As the versed sine and radius of the base are equal, the cosine is 0.
Hence, area of base X cosine =0.
83.333 -OX-^ =166.666 ^difference of $ of cube of the sine and the
product of area of base and the cosine, Xthe height-— the versed sine=the
contents required.
Ex. 2. The sine of half the arc of the base of an ungula is
12 inches, the diameter of the cylinder is 25, and the height
of the ungula 18 ; what are its contents ?
Ans. 1190.34375 inches.
For rules to ascertain the area of base, see pp. 85-87.
4. When the section passes through the base of the cylinder,
and the versed sine exceeds the sine, Fig, 95.
Rule. — To two thirds of the cube of the sine of half the
arc of the base, add the product of the area of the base and
the excess of the versed sine over the sine of the base.
Multiply the sum thus found by the quotient arising from
the height divided by the versed sine, and the product will be
the contents required.
^ 2 s3, h
Or, -j-+ax(vscvs)X — =S.
Fig. 95.
Example. — The sine a d of half the arc of an ungula, Fig. 95,
is 12 inches, the versed sine a g is 16, the height c g 20, and
the diameter of the cylinder 25 inches ; what are the contents?
MENSURATION OF SOLIDS.
175
§■ of 123 = ll52 = two thirds of cube of sine of the arc of the base.
Area of base (see Rules, p. 84-134) =331.78.
1152+(331.78xl6-12.5)=2313.23=swm of % of the cube of the sine
of the base, and product of area of base, and difference between the versed
sine and sine of the base.
2313.23 X 20-T-16 = 2891.5375 =product of above sum and the height, di-
vided by the versed sine=result required.
5. When the section passes obliquely through both ends of the
cylinder, Fig. 96.
Rule. — Conceive the section to be continued till it meets
the side of the cylinder produced ; then, as the difference of
the versed sines of the arcs of the two ends of the ungula is to
the versed sine of the arc of the less end, so is the height of the
cylinder to the part of the side produced.
Find the contents of each of the ungulas by rules 3 and 4,
and their difference will be the contents required.
v' X h
Or, — — -p = h\ v and v' representing the versed sines of the
arcs of the two ends, h the height of the cylinder, and li the height
of the part produced.
Fig. 96. k/
Example. — The versed sines, a e,do, and sines, % h, g r, of
the arcs of the two ends of an ungula, Fig. 96, are assumed to
be respectively 8.5 and 25, and 11.5 and 0 inches, the length
of the ungula within the cylinder, cut from one having 25
inches diameter is 20 inches ; what is the height of the un-
176 MENSURATION OF SOLIDS.
gula produced beyond the cylinder, and what the contents of
the ungula?
25<X)8.5 : 8.5 : : 20 : 10.303 =height of ungula produced beyond the cylinder.
Lower ungula, the sine, g r, being 0, the versed sine = the diameter.
Base of ungula being a circle of 25 inches diameter, area=490.874.
The versed sine and diameter of the base being equal (25), the sine is 0.
490.874 X 25 =6135.925 =/>rodWtf of area of base and excess of versed
sine over the sine of the base.
30.303-4-25 = 1.2121 —quotient of heights-versed sine.
Then, 6135.925X1. 2121 =7 437. 35 i7=product of above product and
quotient=the residt required.
Definition. A solid, the surface of which is at a uniform dis-
tance from the centre.
To ascertain the Contents of a Sphere, Fig. 97.
Rule. — Multiply the cube of the diameter by .5236, and
the product will be the contents required.
Or, d3x.5236=S, d representing the diameter.
7-7.7. 97. c
d
Example. — What are the contents of a sphere, Fig. 97, its
diameter, a b, being 10 inches?
103=1000> and 1000 x .5236=523.6 cubic inches.
Ex. 2. The diameter of a sphere is 17 inches ; what are its
contents'? Ans. 1.4887 cubic feet.
Ex. 3. What are the contents of a globe 10.5 feet in di-
ameter? Ans. 606.132 cubic feet.
Centre of Gravity. Is in its geometrical centre.
Note.— .5236=f of 3.1416.
MENSURATION OF SOLIDS. 177
Segment of a Sphere.
Definition* A section of a sphere.
To ascertain the Contents of a Segment of a Sphere, Fig. 98.
Rule 1. To three times the square of the radius of its base
add the square of its height ; multiply this sum by the height,
and the product, multiplied by .5236, will give the contents
required.
Or, (3r2-f/*2)x^X.5236=S.
2. From three times the diameter of the sphere subtract
twice the height of the segment ; multiply this remainder by
the square of the height, and the product, multiplied by .5236,
will give the contents required.
Or, 3 d-2 hxh2x.523Q=S.
0
Fig. 98.
Example. — The segment of a sphere, Fig. 98, has a radius,
a o, of 7 inches for its base, and a height, b o, of 4 inches ;
what are its contents ?
72 X 3 +4? = 163= the sum of three times the square of the radius and the
square of the height.
1G3 X 4 X .5236 =341 .3872 =the above sum multiplied by the height, and
by .5236=inches.
Ex. 2. The radius of a spherical segment is 48 inches,
and the height 12 inches; what are its contents?
Ans. 44334.2592 cubic inches.
Ex. 3. The height of a spherical segment is 2 inches, and
the diameter of the sphere 6 inches ; what are its contents?
Ans. 29.322 cubic inches.
112
178 MENSURATION OF SOLIDS.
Centre of Gravity. Distance from the centre, 3.1416 v2lr— -J
—s, v being the versed sine, s the contents of the segment, and r
the radius of the sphere.
=•5-* — TTjh, h representing the height
or versed sine of the segment.
Of a Hemisphere. Distance from the centre -§ r.
Spherical Zone {or Frustrum of a Sphere).
Definition. The part of a sphere included between two paral-
lel chords.
To ascertain the Contents of a Spherical Zone, Fig. 99.
Rule. — To the sum of the squares of the radii d c and/ g
of the two ends, df eh, add one third of the square of the
height, c g, of the zone ; multiply this sum by the height, and
again by 1.5708, and it will give the contents required.
h2
Or, r2+r/2+-XAxl.5708=S.
Fig. 99.
/
Example. — What are the contents of a spherical zone, Fig.
99, the greater and less diameters, / h and d e, being 20 and
15 inches, and the distance between them, or height of the
zone, being 10 inches.
102 + 7.52 = 1 56.25 =sum of the squares of the radii of the two ends.
102
156.25+—=: 189.58 —the above sum added to one third of the square
o
of the height.
189.58 X 10 X 1.5708 =2977.9226 —the, last sum multiplied by the height
and again by 1.5708 =feet.
MENSURATION OF SOLIDS.
179
Ex. 2. A zone of a sphere has the radii of its ends each 6
inches, and its height is 8 inches ; what are its contents ?
Ans. 1172.86 cubic inches.
Ex. 3. What are the contents of the zone of a sphere, the
radii of its ends being 10 and 12 inches, and the height of it
4 inches? Ans. 1566.6112 cubic inches.
Centre of Gravity. Right Zone. Is in its geometrical centre.
Of a Frustrum. — — =d, representing the distance from the
lor — 4A
vertex of the frustrum.
Spheroids (Ellipsoids).
Definition. Solids generated by the revolution of a semi-ellipse
about one of its diameters.
When the revolution is about the transverse diameter, they are
Prolate, and when about the conjugate, they are Oblate.
To ascertain the Contents of a Spheroid, Figs. 100 and 101.
Rule. — Multiply the square of the revolving axis, c d, by
the fixed axis, a b, and this product by .5236, and it will
give the contents required.
Or a2 x at X -5236 = S, a and a' representing the axes.
4
Or, - 3.1±16 xr2 xr'=S, r and ?%/ representing the semi-axes.
o
Note. — The contents of a spheroid are equal to two thirds of a cylin-
der that will circumscribe it.
Fig. 100.
Fig. 101.
it?
Example. — In a prolate spheroid, Fig. 100, the fixed axis
a b is 14, and the revolving axis c d 10 ; what are its con-
tents?
102X 14 = 1400 ^product of square of revolving axis and fixed axis.
1400 X .5236 =733.04 = above product by .523S=result required.
180 MENSURATION OF SOLIDS.
Ex. 2. The axes of a prolate spheroid are 100 and 60
inches ; what are its contents'? Ans. 188496 inches.
Ex. 3. The axes of an oblate spheroid, Fig. 101, are 10 and
14 inches ; what are its contents'? Ans. 1026.256 inches.
Ex* 4. What are the contents of an oblate spheroid, its
transverse axis, c d, being 24, and its conjugate, a b, 18 inches !
Ans. 5428.685 inches.
Ex. 5. What are the contents of an oblate spheroid, the axes
of which are 50 and 30 inches t Ans. 22.7257 cubic feet.
Centre of Gravity. Is in their geometrical centre.
Segments of Spheroids.
To ascertain the Contents of the Segment of a Spheroid.
When the base, e f, is Circular, or parallel to the revolving
cms, as c d, Figs. 102 and 102*.
Rule. — Multiply the fixed axis, a b, by three, the height of
the segment, a o, by two, and subtract the one product from
the other ; multiply the remainder by the square of the height
of the segment, and the product by .5236.
Then, as the square of the fixed axis is to the square of the
revolving axis, so is the last product to the contents of the
segment.
Or, j— '• =S, a and a/ representing
the fixed and revolving axes.
Fig. 102. c Fig. 102*. a
Example. — In a prolate spheroid, Fig. 102, the fixed or
transverse axis, a b, is 100, and the revolving or conjugate,
c d, GO; what are the contents of a segment of it, its height,
a o, being 10 inches ?
MENSURATION OF SOLIDS.
181
100x3 — 10X2 =280=twice the height of the segment subtracted from
three times the fixed axis.
280 X 102X. 5236 = 14660. 8 =product of above remainder, the square of
the height, and .5236.
Then, 1002: 602:: H660.8: 5277.888 =the result required.
Ex. 2. The height of a segment of a prolate spheroid, Fig.
102, is 5 inches ; what are its contents, the transverse axis
being 4 feet 2 inches, and the conjugate 2.5 feet?
Ans. 659.736 cubic inches.
Ex. 3. The height of a segment of an oblate spheroid, Fig.
102*, is 10 inches, the transverse diameter being 100, and
the conjugate 60; what are its contents?
Ans. 23271.111 cubic inches.
When the base, e ft is Elliptical, or perpendicular to the revolv-
ing axis, as c d, Figs. 103 and 103*.
Eule. — Multiply the revolving axis, c d, by three, the height
of the segment, c o, by two, and subtract the one from the
other ; multiply the remainder by the square of the height
of the segment, and the product by .5236.
Then, as the revolving axis is to the fixed axis, so is the
last product to the contents.
Or, * ~— - ' = S ; a representing the fixed
a
and af the revolving axes.
Fig. 103. c Fig. 103*.
Example. — The diameters, c d and a b, of an oblate sphe-
roid, Fig. 103*, are 100 and 60 inches, and the height of a
segment, c o, thereof is 12 inches ; what are its contents ?
182
MENSURATION OF SOLIDS.
100x3— 12x2 =276= twice the height of the segment subtracted from
three times the revolving axis.
276 X 122 X .5236 -20809.9584 -product of above remainder, the square
of the height, and .5236.
Then, 100: 60:: 20809.9584 : 12485.975 = ^6 result required.
Ex. 2. The segment of a prolate spheroid, Fig. 103, is 20
inches in height, the revolving diameter of the spheroid being
10 feet, and the fixed axis 16 feet 8 inches; what are its con-
tents in cubic inches. Ans. 111701.333 inches.
Ex. 3. The segment of an oblate spheroid, Fig. 103* is 2
feet, and the axes of the spheroid are 200 and 120 inches ;
what are its contents in cubic feet ? Ans. 57 .8054 feet.
Frustra of Spheroids.
To ascertain the Contents of the Middle Frustrum of a Spheroid.
When the ends, e f and g h, are circular, or parallel to the
revolving axis, as c d, Figs. 104 and 104*.
Rule. — To twice the square of the revolving axis, c d, add
the square of the diameter of either end, eforg h; multiply
this sum by the length, { o, of the frustrum, and the product
again by .2618, and it will give the contents required.
Or, 2a/2+d2x/x.2618=S.
Fig. 104.
Fig. 104
Example. — The middle frustrum, i o, of a prolate sphe-
roid, Fig. 104, is 36 inches in length, the diameters of it
being, in the middle, c d, 50 inches, and at its ends, e f and
g h, 40 ; what are its contents ?
502X 2 +402 =6600 =sum of twice the square of the middle diameter
added to the square of the diameter of the ends.
6600x36x.2618=62203.68=^ro(fMc/ of the above sum, the length of
the frustrum, and .2618 = ^e result required.
MENSURATION OF SOLIDS.
183
Ex. 2. What are the contents of the middle frustrum of an
oblate spheroid, Fig. 104*, the transverse diameter being 100
inches, the diameters of the ends of the frustrum each. 80
inches, and the length of it 3 feet 4 inches'?
Ans. 276460.8 cubic inches.
Ex. 3. The middle frustrum of a prolate spheroid, Fig. 104,
is 80 inches in length, the diameter of it being, in the middle,
60 inches, and at its ends 38.5 inches; what are its contents
in cubic feet? Ans. 105.232 cubic feet.
Ex. 4. The diameter of the middle frustrum of a prolate
spheroid, Fig. 104, is 100 inches, that of the end of the frus-
trum is 60 inches, and the length of it is 8 feet ; what are
its contents'? Ans. 593134.08 cubic inches.
When the ends, e f and g h, are elliptical, or perpendicular to
the revolving axis, as c d, Figs. 1 05 and 1 05*.
Eule.— To twice the product of the transverse and conju-
gate diameters of the middle section, a b, add the product of
the transverse and conjugate of either end, ef or g h; multi-
ply this sum by the length, o i, of the frustrum, and the prod-
uct by .2618, and it will give the contents required.
Or, a/xax2-fO<?x/x.2618=S.
Fig. 105. c Fig. 105*. a
d b
Example. — In the middle frustrum of a prolate spheroid,
Fig. 105, the diameters of its middle section are 50 and 30
inches, and of its ends 40 and 24 inches ; what are its con-
tents, its length being 18 inches?
50 X 30 X 2 =3000 =twice the product of the transverse and conjugate di-
ameters.
3000+40 X 24 = 3960 =sum of the above product and the product of the
transverse and conjugate diameters of the ends.
3960 X 18 X. 2618 = 18661. 104=^e preceding product into the length
.2618 — the contents required.
184 MENSURATION OF SOLIDS.
Ex. 2. The diameters of the middle frustrum of an oblate
spheroid, Fig. 105* are 100 and 60 inches in the middle, and
60 and 40 inches at the ends, and the length of it is 6 feet 8
inches ; what are its contents ? Arts. 301593.6 cubic inches.
Ex. 3. The middle frustrum of a prolate spheroid is 3 feet
4 inches in height, and has diameters of 8 feet 4 inches and
5 feet in its middle, and of 6 feet 8 inches and 4 feet at its
ends ; what are its contents in cubic feet ?
Ans. 95.993 cubic feet.
Centres of Gravity. Frustrum. See Appendix, p. 282.
Middle Frustra, or Zones. Is in their geometrical centre.
Semi-spheroids. — Distance from the centre of the spheroid.
Prolate, fa. Oblate. f 5.
(a-4-d)2 (b + d')2
Segments. — Prolate. § — -. Oblate, f — — — r, b repre-
senting the semi-conjugate axis, a the semi-transverse, and d, d/ the
distances of the base of the segments from the centre of the spheroid.
Cylindrical Ring.
Definition. A ring formed by the curvature of a cylinder.
To ascertain the Contents of a Cylindrical Ring, Fig. 106.
Rule. — To the diameter of the body of the ring, a b, add
the inner diameter of the ring, be; multiply the sum by the
square of the diameter of the body, the product by 2.4674, and
it will give the contents required.
Or, d-\-d' xd7 x 2.4674 =rS, d and d/ representing the diam-
eter of the body and inner diameter.
Or, a x £=S, a representing area of section of body, and I the
length of the axis of the body.
Fig. 106. ^~ZT^\
MENSURATION OF SOLIDS. 185
Example. — What are the contents of an anchor-ring, the
diameter of the metal being 3 inches, and the inner diameter
of the ring 8 inches %
, 3 + 8x3*=9d=p?'oduct of sum of diameters and the square of diameter
of body of ring.
99 x2A674:=24:4:.2726=above product X 2.4674 = *//e contents required.
Ex. 2. The diameter of the body of a cylindrical ring is 2
inches, and the inner diameter of the ring is 12 inches; what
are its contents'? Ans. 138.1744 cubic inches.
Ex. 3. The dimensions of a cylindrical ring are 14 inches
in diameter of body and 16 inches in the ring ; what are its
contents'? Ans. 14508.312 cubic inches.
Centre of Gravity. Is in its geometrical centre. .
LINKS.
Definition. Elongated or elliptical rings.
Elongated or Elliptical Links.
To ascertain the Contents of an Elongated or Elliptical Link,
Figs. 107 and 108.
Rule. — Multiply the area of a section of the body, a b,
of the link by its length, or the circumference of its axis, and
the product will give the contents required.
Or, a x 1= S, a representing area of section of body, and I the
length of the axis of the body.
Note. — By rule, p. Ill, the circumference or length of the axis
of an elongated link=the sum, of 3.1416 times the sum of the less
diameter added to the thickness of the ring, and the product of twice
the remainder of the less diameter subtracted from the greater.
Note. — By rule, p. Ill, the circumference or length of the axis of
an elliptical ring = the square root, of half the sum of the diameters
squared x 3.1 4 1G.
18G
MENSURATION OF SOLIDS.
Fig. 107. Fig. 108
Example. — The elongated link of a chain, Fig. 107, is 1
inch in diameter of body, and its inner diameters, b c and e f,
are 10 and 2.5 inches ; what are its contents?
Area of 1 inch = .7854.
2.5 + 1x3.1416 = 10.9956=3.1416 times the swn of the less diameter
and thickness of the ring = length of axis of ends.
10— 2.5x2 = 15 = twice the remainder, of the less diameter subtracted
from the greater =length of sides of body.
Then, 10.9956 + 15 =25.9956 = length of axis of link.
Hence, .7854x25.9956=20.417=area of link X its length=result re-
quired.
Ex. 2. The elongated link of a chain cable is 1.5 inches
in diameter of body, and its inner diameters are 3.5 and 4.5
inches; what are its contents? Ans. 38.8584 inches.
Ex. 3. The elliptical link of a chain, Fig. 108, is 1 inch in
diameter, a b, of body, and its inner diameters, b c and e f,
are 10 and 2.5 inches; what are its contents ?
v/
2.5 + 1 +10 + 1 =133. 25= diameters of axes squared.
133.25
X 3. 1416=25. 6iS=square root of diameters squared X 3.141 6
= circumference of axis of ring.
Area of 1 inch =.7854.
Then, 25.643x.7854=20.14=reswft required.
Ex. 4. An elliptical link has diameters of 11.5 and 5.5
inches, its diameter of body being 2.5 inches; what are its
contents? Ans. 185.8272 inches.
Centres of Gravity. Is in their geometrical centres.
MENSURATION OP SOLIDS. 187
Spherical Sector.
Definition. A figure generated by the revolution of a sector
of a circle about a straight line drawn through the vertex of the
sector as an axis.
Note. — The arc of the sector generates the surface of a zone termed
the hase of the sector of a sphere, and the radii generates the surfaces
of two cones, having a vertex in common with the sector at the centre
of the sphere.
To ascertain the Contents of a Spherical Sector, Fig. 109.
Rule. — Multiply the surface of the zone, which is the base of
the sector, by one third of the radius of the sphere, and the
product will give the contents required.
r
Or, ax-z=S,a representing the area of the base.
Fig. 109.
si ■ u^-
Example. — What are the contents of a spherical sector,
Fig. 109, generated by the sector c a h, the height of the zone
abed being, a o, 12 inches, and the radius, g h, of the sphere
15 inches?
12 X 94.248 = 1130.976 =height of zone X circumference of sphere=sur-
face of zone (see p. 90).
1130.976 X ^=5654.88 =product of surf ace of zone and £ of diameter
(—$ of radius) = result required.
Ex. 2. The diameter of a sphere is 10 inches, and the height
of the zone or base of a sector is 5 inches ; what are its con-
tents? Ans. 261.5 inches.
Centre of Gravity. Distance from the centre =j(r— JA).
Distance from the vertex = — - — , h=zthe height of the zone.
o
188 MENSURATION OF SOLIDS.
Note. — The surface of a spherical sector=the sum of the areas of
the zone and the two cones.
SPINDLES.
Definition. Figures generated by the revolution of a plane
area bounded by a curve, when the curve is revolved about a chord
perpendicular to its axis, or about its double ordinate, and they are
designated by the name of the arc from which they are generated,
as Circular, Elliptic, Parabolic, etc.
Circular Spindle.
To ascertain the Contents of a Circular Spindle, Fig. 110.
Rule. — Multiply the central distance, o e, by half the area
of the revolving segment, a c ef Subtract the product from
one third of the cube of half the length,/ e; then multiply the
remainder by 12.5664, and the product will give the contents
required.
a (1—2)3
Or, c x r ~- X 12.5664 = S, I representing the length of
the spindle, and a the area of the revolving segment.
Fig. 110. a
Example. — What are the contents of a circular spindle,
when the central distance, o e, Fig. 110, is 7.071067 inches,
the length,/*?, 14.14213, and the radius, o c, 10 inches?
Note. — The area of the revolving segment, /e, being=the side of the
square that can be inscribed in a circle of 20, is 202x .7854 — 14.14213*
-H4=28.54.
MENSURATION OF SOLIDS. 189
7.071067 Xl4.27=100.9041=cen*ra/ distanceX half area of revolving
segment.
7 071673
100.9041 '—— = 16.947= remainder of above product and i of
o
cube of half the length.
16.947 X 12.5664=212.9628 =result required.
Ex. 2. The central distance of a circular spindle is 3 inches,
the area of the revolving segment is 11.5 inches, and the length
of the spindle 8 inches ; what are the contents of it ?
Ans. 51.3086 inches.
Ex. 3. The length of a circular spindle is 24 inches, and its
diameter 18 ; what are the contents of it?
Ans. 3739.585 inches.
The chord of the arc, 24, and the versed sine C1^), 9, being given, the
diameter of the circle is found by rules p. 75, 76.
v©
2
+92 = 15= chord of half the arc.
152
Hence -—■ = 25 = diameter.
Area of revolving segments, per table, p. 136, 137=159.094.
Centre of Gravity. Is in its geometrical centre.
Frustrum or Zone of a Circular Spindle.*
To ascertain the Contents of a Frustrum or Zone of a Circular
Spindle, Fig. 111.
Rule. — From the square of half the length, / g, of the
whole spindle, take | of the square of half the length, t c, of the
frustrum, and multiply the remainder by the said half length
of the frustrum ; multiply the central distance, o c, by the re-
volving area which generates the frustrum; subtract this
product from the former, and the remainder, multiplied by
6.2832, will give the contents required.
Note. — The revolving area of the frustrum can be obtained by di-
viding its plane into a segment of a circle and a parallelogram.
* The middle frustrum of a circular spindle is one of the various
forms of casks.
190 MENSURATION OF SOLIDS.
2
Or, 7^2-l-^x~(cxa)xG.2832 = S, I and f repre-
senting the lengt/is of the spindle and of the frustrum, a area of
the revolving section of frustrum, and c the central distance.
Example. — The length of the middle frustrum of a circu-
lar spindle, I c, is 6 inches,, the length of the spindle, fg,i$ 8
inches, the central distance, o e, is 3 inches, and the area of the
revolving or generating segment is 10 inches ; what are the
contents of the frustrum 1
(8+2)*-^— J- = 13x3 =39 =product of half the length of the frus-
3
trurn and the remainder of% the square of half the length of the frustrum
subtracted from the square of half the length of the spindle.
39— 3 X 10= 9 =product of the central distance and the area of the seg-
ment subtracted from preceding product.
9 X 6.2832 =56.5488 = last product X 6.2832 =result required.
Ex. 2. The length of a circular spindle is 1.333 feet, the
length of the middle frustrum of it is 1 foot, the central dis-
tance is 6 inches, and the area of the revolving segment is 40
inches ; what are the contents of the frustrum *?
Ans. 452.3904 cubic inches.
Ex. 3. The length of the middle frustrum of a circular spin-
dle is 12 inches, the length of the spindle being 24, the central
distance is 3.5, and the area of the revolving segment is 96
inches ; what are the contents of the frustrum ?
Ans. 2865.1392 cubic inches.
Centre of Gravity. Is in its geometrical centre.
MENSURATION OF SOLIDS. 191
Segment of a Circular Spindle.
To ascertain the Contents of a Segment of a Circular Spindle,
Fig. 112.
Rule. — Subtract the length of the segment, i c, from the
half length, t e, of the spindle ; double the remainder, and as-
certain the contents of a middle frustrum of this length.
Subtract the result from the contents of the whole spindle,
and half the remainder will give the contents of the segment
required.*
Or, C—c-r- 2 = S, C and c representing the contents of the spin-
dle and middle frustrum.
Fig. 112.
\ \ /
o
Example. — The length of a circular spindle, i a, Fig. 112,
is 14.14213, the central distance, o e, is 7.07107, the radius
of the arc, o a, is 10, and the length of the segment, t c, is
3.53553 inches; what are its contents'?
— '— 3. 53553 X 2 =7. 071 07 =doub!e the remainder, of the length
of the segment subtracted from half the length of the spindle = length of the,
middle frustrum .
Note. — The area of the revolving or generating segment of the whole
spindle is 28.54 inches, and that of the middle frustrum is 19.25.
The contents of the whole spindle is 212.9628 cubic in.
middle frustrum is 162.8982 " "
Hence 50.0646^2 =
25.0323= Me contents required.
* This rule is applicable to the segment of any spindle or any conoid,
the volume of the figure and frustrum being first obtained.
192
MENSURATION OF SOLIDS.
Centre of Gravity.*
^(r3+a2)(^-Za)-i(^-^)+|a[a3-(^-P)f]
(r>+a*X9-l)-^-t3)-2aS
of spindle.
When g=b o.
l—o c.
r— radius of circle, a d.
a=a o = V r2— #2.
S=area of the semi-zone, /
efff h. /
= distance from centre
Cycloidal Spindle.^
To ascertain the Contents of a Cycloidal Spindle, Fig. 113.
Rule. — Multiply the product of the square of twice the di-
ameter of the generating circle, a b c, and 3.927 by its circum-
ference, and this product divided by 8 will give the contents
required.
Or,
2d x 3.927x^x3.1416
8
S, d representing the diameter
of the circle, a half width of the spindle.
Fig. 113.
\
/c
Example. — The diameter of the generating circle, a b c, of
a cycloid, Fig. 113, is 10 inches ; what are the contents of the
spindle, d e ?
* By Professor A. E. Church, U. S. M. A.
t The contents of a cycloidal spindle are equal to § of its circumscrib-
ing cylinder.
MENSURATION OF SOLIDS. 193
10X2X3.927 = 1570. 8 ^product of twice the diameter squared and
3.927.
1570.8 X 10X3. 1416-f-8 =6168.5316 cubic inches =product of the pre-
ceding product and the circumference divided by 8 = result required.
Ex. 2. The diameter of the generating circle of a cycloid is
5G.5 inches ; what are the contents of its spindle in cubic feet ?
Ans. 64:3.9292 cubic feet.
Ex. 3. The diameter of the generating circle of a cycloid is
6 feet ; what are the contents of its spindle in cubic feet ?
Ans. 1332.4028 cubic feet
Elliptic /Spindle.
To ascertain the Contents of an Elliptic Spindle, Fig. 114.
Rule. — To the square of its diameter, c d, add the square
of twice the diameter ef at J of its length ; multiply the sum
by the length, a b, the product by .1309, and it will give the
contents required.
Note. — For all such solids, this rule is exact when the body is form-
ed by a conic section, or a part of it, revolving about the axis of the
section, and will always be very near when the figure revolves about
another line.
Or, d2-\-2d'2 x^X«1309=:S, d and <t representing the diam-
eters as above.
Fig. 114.
Example. — The length of an elliptic spindle, a b, Fig. 114,
is 75 inches, its diameter c d 35, and the diameter efat £ of
its length 25 ; what are its contents?
I
194
MENSURATION OF SOLIDS.
35s +25 X 2 =3725= sum of squares of diameter of spindle and of twice
its diameter at £ of its length.
3725 X 75 =279375 =above sum X the length of the spindle.
Then 279375 X. 1309 =36570.1875 =result required.
Ex. 2. The length of an elliptic spindle is 100 inches, its
diameter 20, and the diameter at J of its length 15 ; what are
its contents? Ans. 17017 cubic inches.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Contents of the Middle Frustrum or a Zone of
an Elliptic Spindle, Fig. 115.
Rule. — Add together the squares of the greatest and least
diameters, a b, c d, and the square of double the diameter g h
in the middle between the two ; multiply the sum by the
length, e f, the product by .1309, and it will give the con-
tents required.*
Or, d2+<r2+(2xcO2x/x.l309=S, d,d', and d" repre-
senting the different diameters.
Fig. 115.
Example. — The greatest and least diameters, a b and c d, of
the frustrum of an elliptic spindle, Fig. 115, are 68 and 50 inches,
its middle diameter, g h, 60, and its length, ef 75 ; what are its
contents !
G82 + 502 + G0x2=21524=swm of squares of greatest and least diam-
eters and of double the middle diameter.
21524 X 75 X. 1309=211311. 87=product of above sum, the length, and
. 1 309 = the result required.
* For all such solids, this rule is exact when the body is formed by
a conic section, or a part of it, revolving about the axis of the section,
and will always be very near when the figure revolves about another
line.
MENSURATION OF SOLIDS. 195
Ex. 2. The greatest and least diameters of the zone of an
elliptic spindle are 20 and 5 inches, its middle diameter 16,
and its length 42 ; what are its contents ?
Ans. 7966.312 cubic inches.
Ex. 3. The greatest, middle, and least diameters of the zone
of an elliptic spindle are 25, 23.5, and 20 inches, and its length
42.5; what are its contents'? Ans. 17991.55 cubic inches.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Contents of a Segment of an Elliptic Spindle,
Fig. 116.
Rule. — Add together the square of the diameter of the
base, c d, of the segment, and the square of double the diam-
eter g h, in the middle between the base and vertex ; multiply
the sum by the length, o e, of the segment, the product by
.1309, and it will give the contents required.*
Or, rf2+2 d//2x/x.l309 = S, d and d" representing the di-
ameters.
Fig. 116.
-->
d —
Example. — The diameters, c d and g h, of the segment of
an elliptic spindle, Fig. 116, are 20 and 12 inches, and the
length, o e, is 16 inches ; what are its contents?
20a + 12x2 = 976=£ttfrc of squares of diameter at base and of double the
diameter in the middle.
976 X 16 X. 1309 =2044.134 =product of above sum, the length of the
segment, and .1309 = <^e result required.
Ex. 2. The diameters of the segment of an elliptic spindle
are 25 and 20 inches, and the length of it is 42.5 inches ;
what are its contents'? Ans. 12378.231 cubic inches.
Centre of Gravity. At two thirds of the length, measuring
from the end.
* See note at bottom of page 194.
196 MENSURATION OF SOLIDS.
Parabolic Spindle.
To ascertain the Contents of a Parabolic Spindle, Fig. 117.
Rule 1. — Multiply the square of the diameter, a b, by the
length, c d, the product by .41888,* and it will give the con-
tents required.
Or, d2xl XA1888=S.
Rule 2. — To the square of its diameter add the square of
twice the diameter at \ of its length ; multiply the sum by
the length, the product by .1309, and it will give the contents
required.!
Or, d2+2d'2 x lx .1309 = S, d and d/ representing the diam-
eters as above.
Fig. 117. c
Example. — The diameter of a parabolic spindle, a b, Fig.
117, is 40 inches, and its length, c d, 20 ; what are its con-
tents ?
402 X 20 =32000= square of diameter X the length.
Then 32000 X. 41888 = 13404. 16 =a&ore product X A1888 =result re-
quired.
Again, If the middle diameter at one fourth of its length is 29. G5,
then, by Rule 2,
40-+29.65x2x20X.1309 = 13394.97=resuftre£i«Vee/.
Ex. 2. The length of a parabolic spindle is 15.75 feet, and
its diameter is 3 feet ; what are its contents ?
Ans. 59.376 cubic feet.
* T8sof.7854.
f See note at bottom of page 194.
MENSURATION OF SOLIDS. 197
Ex. 3. The length of a parabolic spindle is 40 feet, and
its diameter is 80 feet ; what are its contents in cubic feet t
Ans. 107233.28 cubic feet. ■
Centre of Gravity. Is in its geometrical centre.
To ascertain the Contents of the Middle Frustrum of a Parabolic
Spindle, Fig. 118.
Rule 1. — Add together 8 times the square of the greatest
diameter, a b, 3 times the square of the least diameter, e f, and
4 times the product of these two diameters ; multiply the sum
by the length, c d, the product by .05236, and it will give the
contents required.
Or, d2x84-^^3+^xd7x4x^X.05236 = S.
Rule 2. — Add together the squares"of the greatest and least
diameters, a b, ef, and the square of double the diameter in
the middle between the two ; multiply the sum by the length,
c d, the product by .1309, and it will give the contents re-
quired.
Or, cZ2 + d/2+(2O2xJx.l309=S, d" representing the di-
ameter between the two.
Fig. 118.
c **«•«.
•— -
Example. — The middle frustrum of a parabolic spindle,
Fig. 118, has diameters, a b and ef of 40 and 30 inches, and
its length, c d, is 10 inches ; what are its contents'?
402 X 8+302 X3-H0x30x4=20300=^e sum of 8 times the square of
ihc greatest diameter, 3 times the square of the least diameter, and A; times
the product of these diameters.
20300 X 10 X .05236 = 10G29.08=resuft required.
198 MENSURATION OP SOLIDS.
Ex. 2. The middle frustrum of a parabolic spindle has di-
ameters of 20 and 15 feet, and its length 5 feet ; what are
its contents? Ans. 1328.635 cubic feet.
Ex. 3. The middle frustrum of a parabolic spindle has di-
ameters of 80 and 56.5 feet, and its length is 20- feet; what
are its contents'? Ans. 82578.789 cubic feet.
Centre cf Gravity. Is in its geometrical centre.
To ascertain the Contents of a Segment of a Parabolic Spindle,
Fig. 119.
Rule. — Add together the square of the diameter of the
base, e f, of the segment, and the square of double the diam-
eter, g h, in the middle between the base and vertex ; multiply
the sum by the length, c d, of the segment, the product by
.1309, and it will give the contents required.
Or, d2+2 dr2x/X.1309=S.
Example. — The segment of a parabolic spindle, Pig. 119,
has diameters, e f and g h, of 15 and 8.75 inches, and the
length, c d, is 2.5 inches ; what are its contents ?
152 + 8.75x 2=531. 25=sum of square of base and of double the diam-
eter in the middle of the segment.
531.25 X 2.5 X.130S =173.852 =product of above sum, length of seg-
ment, and .1309 =the result required.
Ex. 2. The segment of a parabolic spindle has diameters
of 30 and 20 inches, and the length of it is 30 inches ; what
are its contents in cubic feet? Ans. 5.6814 cubic feet.
MENSURATION OF SOLIDS. 199
Ex. 3. The segment of a parabolic spindle has diameters
of 56.5 and 40 feet, and the length of it is 10 feet ; what
are its contents in cubic feet ?
Ans. 12556.255 cubic feet
Ex. 4. The segment of a parabolic spindle has diameters
of 28 and 20 feet, and the length of it is 20 feet ; what are
its contents? Ans. 6241.312 cubic feet.
Ex. 5. The segment of a parabolic spindle has diameters
of 42 and 30 feet, and the length of it is 60 feet ; what are
its contents'? Ans. 42128.856 cubic feet.
Centre of Gravity*
b6 d6
u "ap (bi-di)+2a2p2(b2-d2)
— — —distance from centre
h J^J^L^-d^+latp^b-d)
5 5 o
of spindle, a representing semi-diameter of spindle, b the half
length, d the distance of the base of the segment from the centre
b2
of the spindle, andp one half parameter, which is equal to — .
Illustration of rules, p. 196, 197, and 198 :
Solidity of spindle (Ex. 3, p. 197) by rule 2,
80 feet in diameter by 40 feet in length,
the diameter at £ of its length being 56.5
feet = 100368.884 cubicfeet.
Solidity of middle frustrum (Ex. 3, p. 198) by rule 2,
20 feet in length, the diameter at £ of its length
being 69.25 feet \ = 75331.641
Solidity of segment (Ex. 3, p. 199) 10 feet in length
= 12556.255, which x 2, for two end segments = 25112.510
100444.151
Solidity of spindle as above = 100368.884
Difference, arising from the impracticability of ob-
taining the middle diameters of the frustrum and
segment, from a figure of so minute a scale as
that of the example taken for illustration = 75.267
* By Professor A. E. Church, U. S. M. A.
200 MENSURATION OF SOLIDS.
Hyperbolic Spindle.
To ascertain the Contents of a Hyperbolic Spindle, Fig. 120.
Rule. — To the square of its diameter, c d, add the square
of double the diameter, e f at ^ of its length ; multiply the
sum by the length, a b, the product by .1309, and it will give
the contents required*
Or, d2-f-2d/2x£x.l309:=S, d and d' representing the diam-
eters as above.
Fig. 120. a
b
Example. — The length, a b, Fig. 120, of a hyperbolic spin-
dle is 100 inches, and its diameters, c d and ef are 150 and
1503 + 110 X 2 X 100 = 7090000 =product of the sum of the squares of the
greatest diameter and of twice the diameter at £ of the length of the spin-
dle and the length.
Then, 7090000 X. 1309 =928081 = the result required.
Ex. 2. The length of a hyperbolic spindle is 120 inches,
and its diameters in the middle and at J of its length are 100
and 80 inches ; what are its contents %
Ans. 323.614 cubic feet.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Contents of the Middle Frustrum of a Hyper-
bolic Spindle, Fig. 121.
Rule. — Add together the squares of the greatest and least
diameters, a b,c d, and the square of double the diameter, g h,
in the middle between the two ; multiply this sum by the
* See note at bottom of page 194.
MENSURATION OF SOLIDS. 201
length, e f the product by .1309, and it will give the contents
required.*
Or, rf2+^2+(2x^//)2x/x.l309 = S.
Fig. 121. ....- ;
^b
m
^ 1. — ..-^
Example. — The diameters a b and c d, of the middle frus-
trum of a hyperbolic spindle, Fig. 121, are 150 and 110 inches,
the diameter g h, 140 inches, and the length, e f, 50; what
are its contents I
1502 + 1102 + 14:0x2 = 113000=swm of squares of greatest and least di-
ameters and of double the middle diameter.
113000 X 50 X .1309 = 739585 =product of above sum X the length X .1309
=7-esult required.
Ex. 2. The diameters of the middle frustrum of a hy-
perbolic spindle are 16 and 10 inches, the diameter at J
of its length is 13.5, and the length of it is l6 ; what are
its contents in cubic inches'? Ans. 1420.265 cubic inches.
Ex. 3. The diameters of the middle frustrum of a hyper-
bolic spindle are 16 and 12 feet, the diameter at J of its length
14.5, and the length of it 20 feet ; what are its contents ?
Ans. 3248.938 cubic feet.
Centre of Gravity. Is in its geometrical centre.
To ascertain the Contents of a Segment of a Hyperbolic Spindle,
Fig. 122.
Rule. — Add together the square of the diameter of the
base, e f, of the segment, and the square of double the diam-
eter, g h, in the middle between the base and vertex ; multiply
the sum by the length, a b, of the segment, the product by
.1309, and it will give the contents required.
* See note at bottom of page 194.
12
202 MENSURATION OF SOLIDS.
Or, d2+<f/2x*X.1309 = S, d and d" representing the diam-
eters.
Fig. 122. a
Example. — The segment of a hyperbolic spindle, Fig. 122,
has diameters, e f and g h, of 110 and 65 inches, and its
length, a b, 25 ; what are its contents ?
1102 + 65x2=29000=sttm of squares of diameter of base and of double
the middle diameter.
29000 X 25 X .1309 =94902.5 =product of above sum X the length X .1309
=result required.
Ex. 2. The segment of a hyperbolic spindle has diameters
of 8 inches at its base and 6 at half its length, its length be-
ing 10 ; what are its contents?
Ans. 272.272 cubic inches.
Ex. 3. The segment of a hyperbolic spindle has diameters
of 50 and 31 inches at its base and half its length, its length
being 25 ; what are its contents ?
Ans. 20760.74 cubic inches.
Centre of Gravity. At J of the height, measured from the
vertex.
ellipsoid, paraboloid, and hyperboloid of revolution*
(conoids).
Definition. Figures like to a cone, described by the revolu-
tion of a conic section around and at a right angle to the plane of
their fixed axes.
* These figures have been known as conoids. For the definition of
a conoid, see Conic Sections, page
MENSURATION OP SOLIDS.
203
Ellipsoid of Revolution {Spheroid).
An ellipsoid of revolution is a semi-spheroid. (See p.
180-184.)
Paraboloid of Revolution*
To ascertain ' the Contents of a Paraboloid of Revolution,
Fig. 123.
Rule. — Multiply the area of the base, a b, by half the alti-
tude, d c, and the product will give the contents required.
Note. — This rule will hold for any segment of the paraboloid, whether
the base be perpendicular or oblique to the axis of the solid.
or, «x|==a
Fig. 123.
Example. — The diameter, a b, of the base of a paraboloid
of revolution, Fig. 123, is 20 inches, and its height, d c, 20
inches ; what are its contents ?
Area of 20 inches diameter of base =314.16.
20
314.16 X — =3141. 6=area of base X half the height=result required.
Ex. 2. The diameter of the base of a paraboloid of revolu-
tion is 11.5 inches, and its height 7 ; what are its contents?
Ans. 363.5411 cubic inches.
* The contents of a paraboloid of revolution are=i of its circum-
scribing cylinder.
204 MENSURATION OP SOLIDS.
Ex. 3. The diameter of the base of a paraboloid of revolu-
tion is 11 feet 3 inches, and its height 8 feet; what are its
contents in cubic feet? Ans. 397.608 cubic feet.
Centre of Gravity. At two thirds of the height, measured
from the vertex.
Frustrum of a Paraboloid of Revolution.
To ascertain the Contents of a Frustrum of a Paraboloid of Hev->
olution, Fig. 124.
Rule. — Multiply the sum of the squares of the diameters,
a b and d c, by the height, e f, of the frustrum, this product
by .3927, and it will give the contents nearly.
Or, d2-fd/2x^X.3927=S.
Fig. 124.
Example. — The diameters, a b and d c, of the base and
vertex of the frustrum of a paraboloid of revolution, Fig. 124,
are 20 and 11.5 inches, and its height, ef 12.6; what are
its contents ?
202 + 11.52=5B2.25=sum of squares of the diameters.
532.25 X 12.6 X. 3927=2033.5837 ^product of above sum, the height,
and .3927=^6 result- required.
Ex. 2. The diameters of the frustrum of a paraboloid of
revolution are 30 and 58 inches, and the height 18 ; what are
its contents in cubic feet? Ans. 17.4424 cubic feet.
Ex. 3. The diameters of the frustrum of a paraboloid of
revolution are 48 and 60 inches, and the height 18 ; what are
its contents in cubic feet? Ans. 24.151 cubic feet.
2R2-f-r2
Centre of Gravity. From the vertex JA 2 , R and r
representing radii of base and diameter, and h the height.
MENSURATION OF SOLIDS. 205
Segment of a Paraboloid of Revolution.
To ascertain the Contents of the Segment of a Paraboloid of
Revolution, Fig. 125. *
Rule. — Multiply the area of the base, a b, by half the alti-
tude, e f and the product will give the contents required.
Note. — This rule will hold for any segment of the paraboloid, whether
the base be perpendicular or oblique to the axis of the solid.
^ h ^
Ortax^=S.
Fig. 125.
Example. — The diameter, a b, of the base of a segment of
a paraboloid of revolution, Fig. 125, is 11.5 inches, and its
height, e /, is 7.4 ; what are its contents ?
■*
Area of 11.5 inches diameter of base = 103.869.
7.4
103.869 X— =384.315 —arm of base X half the height=the result re-
quired. ...
Ex. 2. The diameter of the base of a segment of a parabo-
loid of revolution is 30 inches, and its height 25 ; what are
its contents in cubic inches ?
Ans. 8835.73 cubic inches.
Ex. 3. The diameter of the base of a segment of a parabo-
loid of revolution is 48 inches, and the height 15 ; what are
its contents in cubic feet? Ans. 7.854 cubic feet.
Centre of Gravity. At two thirds of the height, measured
from the vertex.
206 MENSURATION OF SOLIDS.
Hyperboloid of Revolution.
To ascertain the Contents of a Hyperboloid of Revolution, Fig.
126.
Rule. — To the square of the radius of the base, a b, add
the square of the middle diameter, c d; multiply this sum by
the height, ef the product by .5236, and it will give the con-
tents required.
Or, r2-(-cZx/«X.5236r=S, d representing the middle diam-
eter.
Fig. 126,
Example. — The base, a b, of a hyperboloid of revolution,
Fig. 126, is 80 inches, the middle diameter, c d, 66, and the
height, ef 60 ; what are its contents ?
8(M-2 + 662=5956=s«m of square of radius of base and middle diam-
eter.
5956 X 50=297800 X .5236 = 155928. 08 =product of above sum, the
height, and .52S6=result required.
Ex. 2. The base of a hyperboloid of revolution is 20 inches,
its middle diameter is 16, and its height 20 ; what are its con-
tents ! Ans. 3728.032 cubic inches.
Ex. 3. The base of a hyperboloid of revolution is 104
inches, the diameter at half its height is 68, and the height of
it is 50 inches; what are its contents in cubic feet?
Ans. 111.0226 cubic feet
Centre of Gravity. Distance from the vertex, x h,
b representing the base, and h the height of the figure.
MENSURATION OF SOLIDS. 207
Frustrum of a Hyperboloid of Revolution,
To ascertain the Contents of the Frustrum of a Hyperboloid of
Revolution, Fig. 127-
Rule. — Add together the squares of the greatest and least
semi-diameters, e b and * d, and the square of the whole di-
ameter, g h, in the middle of the two ; multiply this sum by
the height, i e, the product by .5236, and it will give the con-
tents required.
Or, d2+d'2-{-d/'2xhx.5236=S, d, d', and d" represent-
ing the several diameters.
Fig. 127. f
Example. — The frustrum of a hyperboloid of revolution,
Fig. 127, is in height, e i, 50 inches, the diameters of the great-
er and less ends, a b and c d, are 110 and 42, and that of the
middle diameter, g h, is 80 ; what are the contents %
110-^2=55 and 42-f-2=21. Hence, 552+212+802=9866=sw7w of
the squares of the semi-diameters of the ends and that of the whole diame-
ter in the middle.
9866 X 50=493300 X .5236=258291.88 cubic inches.
Ex. 2. The height of a frustrum of a hyperboloid of revolu-
tion is 1 foot, the greatest and least diameters 10 and 6 inch-
es, and the middle diameter 8.5 inches; what are its con-
tents in cubic feet? Ans. .38633 cubic foot.
Ex. 3. The height of the frustrum of a hyperboloid of rev-
olution is 10 inches, the radii of the ends 21 and 1 inches, and
the middle diameter 25 inches ; what are its contents in cubic
feet? Ans. 3.2331 cubic feet.
208 MENSURATION OF SOLIDS.
Centre of Gravity.
_ (d+d>)(2a2-d/a4-d2) » , . *-« .
x ^ , ,,, -7 — ^ -^distance from centre of the base of
* 3a2— d't+d'd+d2 J J J
the figure, a representing the semi-transverse axis, or distance from
centre of the curve to vertex of figure, f; d and d/ the distances
from the centre of the curve to the centre of the less and greater
diameter of the frustrum.
Segment of a Hyperboloid of Revolution.
To ascertain the Contents of the Segment of a Hyperboloid of
Revolution, Fig. 128.
Eule. — To the square of the radius of the base, a e, add
the square of the middle diameter, c d ; multiply this sum by
the height, e f, the product by .5236, and it will give the con-
tents required.
Or, r2 -\-d2 X h x .5236 == S, r representing radius of base. •
Fig. 128. f
Example. — The radius, a e, of the base of a segment of a
hyperboloid of revolution, Fig. 128, is 21 incnes, its middle di-
ameter, c d, is 30, and its height, e f, 15 ; what are its con-
tents 1
21 2+302X 15=20115=<Ae product of the sum of the squares of the ra-
dius of the base and the middle diameter multiplied by the height.
20115 X .5236 = 10532.214=resM& required.
Ex. 2. The radius of the base of a segment of a hyperboloid
of revolution is 55 inches, its middle diameter 70, and its
height 65 ; what are its contents?
Ans. 269719.45 cubic inches.
Centre of Gravity. (See rule for hyperboloid, page 206.)
MENSURATION OF SOLIDS.
209
ANY FIGURE OF REVOLUTION.
To ascertain the Contents of any Figure of Revolution, Fig. 129.
Rule. — Multiply the area of the generating surface by the
circumference described by its centre of gravity.
Or, ax%rxp=contents, r representing radius of centre of
gravity.
Fig. 129.
HH
Example. — If the generating surface, a b c d, of the cylin-
der, b e df, Fig. 129, is 5 inches in width and 10 in height,
then will a b=5 and b d = 10, and the centre of gravity will
be in o, the radius of which is r o— 5-r-2 = 2.5.
Hence, 10x5 = 50 =area of generating surface.
50 X 2.5 X 2 X 3.1416 =785.4 =area X circumference of its centre of grav-
ity = the contents of the cylinder.
Proof. Volume of a cylinder 10 inches in diameter and 10 inches in
height. 10s X. 7851 =78.54, and 78.54x10=785.4.
' Fig. 130.
Ex. 2. If the generating surface of a cone, Fig. 130, is a e
= 10, d e=5, then will a d— 11.18, and the area of the tri-
angle = 10 x 5-^2 = 25, the centre of gravity of which is in o,
and o r, by rule, page 57, =1.666.
Hence, 25 X 1.606x2 X3. 1416 =261.8 =area of generating surfaceX
circumference of its centre of gravity • = the contents of the cone.
210 MENSURATION OP SOLIDS.
Fig. 131.
Ex. 3. If the generating surface of a sphere, Fig. 181, is
/102x.7854\
a b c, and ac=10, ab c will be I - J =39.27, the cen-
tre of gravity of which is in o, and by rule, page 87, or— 2.122.
Hence, 39.27 X 2.122 x 2 X 3.1416 =523.6 =area of generating surface
X circumference of its centre of gravity — the contents of the sphere.
To ascertain the Contents of an Irregular Body.
Rule. — Weigh it both in and out of fresh water, and note
the difference in pounds; then, as 62.5* is to this difference,
so is 1728f to the number of cubic inches in the body.
Or, divide the difference in pounds by 62.5, and the quo-
tient will give the volume in cubic feet.
Note. — If salt water is to be used, the ascertained weight of a cubic
foot of it, or 64, is to be used for 62.5.
Example. — An irregular shaped body weighs 15 pounds in
water, and 30 out ; what is its volume in cubic inches ?
30 — 15 = 15= difference of weights in and out of water.
62.5 : 15 : : 1728 : 414.72 =volume of the body in cubic inches.
Or, 15^62.5 = .24, and .24 X 1728 =414.72 =volume of the body.
Ex. 2. An irregular body weighs 187.5 pounds in water,
and 250 out ; what is its volume? Ans. 1728 cubic inches.
Ex. 3. The difference in weights of a bronze gun in and out
of water is 625 pounds ; what is its volume in cubic feet?
Ans. 10 cubic feet
* The weight of a cubic foot of fresh water.
• f The number of inches in a cubic foot.
MENSURATION 'OF SOLIDS. 211
PROMISCUOUS EXAMPLES.
1. If a stone measures 4 feet 6 inches long, 2 feet 9 inches
broad, and 5 feet 4 inches deep, how many cubic feet does it
contain? Ans. 66 feet.
2. The dimensions of a bushel measure are 18^ inches in
diameter, and 8 inches deep ; what should be the dimensions
of a measure of like form that would contain 8 bushels'?
Ans. 37 inches in diameter, and 16 inches deep.
3. If a box, of plank 3.5 inches thick, is 4 feet 9 inches in
length, 3 feet 7 inches in breadth, and 2 feet 11 inches in
height, how many square feet did it require to make the box,
how many cubic feet does it contain, and how many does it
measure ?
Ans. 70.208 square feet, 29.167, and 49.644 cubic feet.
4. A well 40 feet in depth is to be lined, of which the di-
ameter is 6.5 feet, the thickness of the wall is to be 1.5 feet,
leaving the inner diameter of the well 3.5 feet ; how many cubic
feet of stone will be required? Ans. 942.478 cubic feet.
5. How many bricks, exclusive of mortar, 8 inches long,
4 inches wide, and 2 inches thick, will it take to build a wall
40 feet long, 20 feet high, and 2 feet thick !
Ans. 43,200 bricks.
6. How many bricks, exclusive of mortar, will it take to
build the walls of a house which is 80 feet long, 40 feet wide, and
25 feet high, the walls to be 12 inches thick, the bricks being
8 inches long, 4 broad, and 2 thick? Ans. 159,300 bricks.
7. How many bricks will it require to construct the walls
of a house 64 feet long, 32 feet wide, and 28 feet high ; the
walls are to be 1 foot 4 inches thick, and there are to be three
doors 7 feet 4 inches high, and 3 feet 8 inches wide ; also 14
windows 3 feet wide and 6 feet high, and 16 windows 2 feet
8 inches wide and 5 feet 8 inches high : each brick is to be 8
inches long, 4 inches wide, and 2 inches thick ?
Ans. 171,990 bricks.
212 MENSURATION OF SOLIDS. »
8. If a garden 100 feet long and 80 feet wide is to be in-
closed with a ditch 4 feet wide, how deep must it be dug that
the soil taken from it may raise the surface one foot ?
Aiis. 5.319 feet.
9. If a man dig a small square cellar, which will measure
6 feet each way, in one day, how long would it take him to
dig a similar one that measured 10 feet each way?
Ans. 4.629 days.
10. If a lead pipe f of an inch in diameter will fill a cis-
tern in 3 hours, what should be its diameter to fill it in 2
hours'? Ans. .918 inch.
11. What are the contents of a stick of round timber 20
feet long, the diameter at the larger end being 12 inches,
and at the smaller end 6 inches ?
Ans. 10.9126 cubic feet.
12. Required the volume of Bunker Hill Monument, the
height of which is 220 feet, by 30 feet square at its base, and
15 feet at its* vertex. Ans. 115500 cubic feet.
13. What are the contents of a spherical segment 3 feet in
height, cut from a sphere 10 feet in diameter?
Ans. 113.0976 cubic feet.
14. The largest of the Egyptian pyramids is square at its
base, and measures 693 feet on a side ; its height is 500 feet.
Supposing it to come to a point at its vertex, what would be
its contents, and how many miles in length of wall, 5 feet in
height and 2 feet thick, would it make ?
An (80041500 cubic feet.
•■ (,1515.9375 miles in length.
15. What are the contents of a sphere, the diameter of
which is 20 inches'? Ans. 4188.8 cubic inches.
16. What is the weight of an iron spherical shell 5 inches
in diameter, the thickness of the metal being 1 inch, estimat-
ing a cubic inch of iron to weigh .25 of a pound ?
Ans. 12.8282 jtounds.
17. How many cubic feet of water are there in a pond that
measures 200 acres, and is 20 feet deep !
Am. 174240000 cubic feet
MENSURATION OF SOLIDS. 213
18. If rain was to fall to the depth of 3 inches on a surface
of 20000 square acres, what would be the number of hogs-
heads of water fallen, assuming each hogshead to contain 100
gallons, and each gallon 231 cubic inches?
A < 16292571 hhds. 42 galls. 3 qts. 0 pts. 3.43 gills,
US' 1 or 217800000 cubic feet.
19. The ditch of a fortification is 1000 feet long, 9 feet
deep, 20 feet broad at bottom, and 22 at top ; how much wa-
ter will fill the ditch, allowing 231 cubic inches to make a
gallon? Ans. 1413818.1819 gallons.
20. What must be the height of a bin that will contain
600 bushels, its length being 8 feet and breadth 41*
Ans. 23.333 feet.
Note. — As a bushel contains very nearly one fourth more than a
cubic foot, the dimensions of a bin, etc., for any required number of
bushels, may be readily found by adding one fourth to the number of
bushels ; the result will give the number of cubic feet the bin will con-
tain, or that may be required. Therefore, when two dimensions of a bin,
etc., are given, divide the number of cubic feet by their product, and the
quotient will be the other dimension.
In the above example, then,
600-^-4 = 150, and 600 + 150 = 750 = ^Ae number of' cubic feet the bin is
to contain.
Then 750-r-8 X 4 =£3. 4375 feet = the height of the bin required.
21. The length of a bin is 4 feet, its breadth 5 feet 6 inch-
es ; what must its height be, by the above rule, that it may
contain 272 bushels? Ans. 15 feet 5.454 inches.
22. There are 1000 men besieged in a town with provis-
ions for 5 weeks, allowing each man 16 ounces a day ; if they
are re-enforced by 500 more, and no relief can be received till
the end of 8 weeks, how many ounces must be given daily to
each man ? Ans. 6.G6 ounces.
23. An officer drew up his company in a square, the num-
ber in each rank being equal ; on being re-enforced with three
times his first number of men, he placed them all in the same
form, and then the number in each rank was just double what
it was at first ; he was again re-enforced with three times his
* The standard United States bushel is 2150.42 cubic inches.
214 MENSURATION OF SOLIDS.
whole number of men, and, after placing them all in the same
form as at first, his number in each rank was 40 men ; how
many men had he at first? Am. 100 men.
24. The volume of a sphere is 381.7044 cubic inches ; re-
quired its radius. Ans. 4.5 inches.
25. If an iron wire TL of an inch in diameter will sustain a
weight of 450 pounds, what weight might be sustained by a
wire an inch in diameter? Ans. 45000 pounds.
26. The edge of a cube is 36 inches ; what is the volume
of a sphere that may be inscribed within it ?
Ans. 24429.0816 cubic inches.
27. In the walls of Balbeck, in Turkey, there are three
stones laid end to end, now in sight, one of which is 63 feet
long, 12 feet thick, and 12 feet broad ; what is the weight,
supposing its specific gravity to be 3 times that of water?
Ans. 759.375 tons.
28. If two men carry a burden of 200 pounds suspend-
ed near the middle of a pole, the ends of, which rest on
their shoulders, how much of the load is borne by each man,
it hanging 6 inches from the middle, and the whole length
of the pole being 4 feet ?
Ans. 125 pounds, and 75 pounds.
29. A joist is 8 \ inches deep and 3^ broad; what will be
the depth of a beam of twice the contents of the joist that is
4f inches broad? Ans. 12.526 inches.
30. Bunker Hill Monument is 30 feet square at its base,
15 feet square at its top, and its height is 220 feet; from the
bottom to the top, through its centre, is a cylindrical opening
15 feet in diameter at the bottom and 11 feet at the top;
how many cubic feet are there in the monument ?
Ans. 86068.444 cubic feet.
31. If a bell 4 inches in height, 3 inches in diameter (ex-
ternal), and i of an inch in thickness, weigh 2 pounds, what
should be the dimensions of a bell, of like proportions, that
would weigh 2000 pounds %
, j 3 feet 4 inches high, 2 feet 6 inches diameter,
' \ and 2£ inches thick.
MENSURATION OF SOLIDS. 215
32. If a round column 7 inches in diameter has a capacity
of 4 cubic feet, of what diameter is a column of equal length
that contains 10 times as much ?
Note. — The contents of cylinders, prisms, parallelopipedons, etc., of
equal altitudes, are to each other as the squares of their diameters or
like sides. The same rule is applicable to frustrums of a cone or pyr-
amid when the altitude is the same and the ends proportional.
Hence, as 4 : 40, or as 1 : 10 : : 72 : 490 = the square of the required di-
ameter, and V 490 =22. 1359, the diameter required.
33. A frustrum of a cone is 12 inches in height, and the
diameters of the greater and smaller ends 5 and 3 inches re-
spectively. Required the diameter of a frustrum of the same
altitude that will contain 3848.46 cubic inches, and have its
diameters in the same proportion as the smaller one.
Aixs. The greater diameter 25, and less diameter 15 inches.
34. There is a fish the head of which weighs 15 pounds,
his tail weighs as much as his head and half as much as his
body, and his body weighs as much as his head and tail. Re-
quired the weight of the fish. Ans. 72 pounds.
35. A certain grocer has only 5 weights; with these he
can weigh any quantity by pounds from 1 to 121 pounds.
Required the weights. Ans. 1, 3, 9, 27, and 81 pounds.
36. A gentleman has a bowling-green 300 feet long and
200 feet broad, which he desires to raise one foot higher by
means of earth to be taken from a ditch that is to go around
it; to what depth must the ditch be dug, supposing its
breadth to be 8 feet? Ans. 7 feet 3.21 inches.
37. Required a cylindrical vessel 3 feet in depth that shall
hold twice as much as a vessel 28 inches deep and 46 inches
in diameter ; what must be its diameter *?
Ans. 57.37 inches.
38. A cubic foot of brass is to be drawn into a wire of ^_
of an inch in diameter ; what will be the length of the wire,
assuming there is to be no loss of metal in the operation I
Ans. 97784.5684 yards.
39. One end of a pile of wood is perpendicular to the hori-
zon, the other is an inclined plane ; the length of the pile at
the bottom is 64 feet, at the top 50 feet, in height 12 feet, and
216 MENSURATION OF SOLIDS.
the length of the wood 5 feet ; required the number of cords
it contains? Ans. 26 cords 92 feet.
40. If a vessel of war, with her ordnance, rigging, and ap-
pointments, is depressed so as to displace 50000 cubic feet of
water, what is the weight of the vessel"?*
Ans. 1428.571 tons.
41. The monument erected in Babylon by Queen Semira-
mis at her husband Ninus's tomb is said to have been one
block of solid marblef in the form of a square pyramid, the
sides of the base being 20 feet, and the height of the monu-
ment 150 feet ; if this monument had been sunk in the Eu-
phrates, what weight would it have required to raise its apex
to the surface of the water ?{ Ans. 2450000 pounds.
42. If the pyramid described in the last example were di-
vided into three equal parts by planes parallel to its base,
what would be the length of each part, measured from the
top? Ans. 104.0042, 27.0329, and 18.9629 feet.
43. There is a mill-hopper in the form of a square pyra-
mid, the contents of which are 13.5 feet, the side of its great-
er end and depth are in the proportion . of 1 to 1.5 ; but one
foot has to be cut off its length to make a passage for the
grain from the hopper to -the mill-stone. Required its con-
tents in corn measure"? Ans. 10.446 bushels.
44. A crucible is in the form of a conic frustrum ; the bot-
tom of it is 2 inches in diameter, the top 3, and the depth
6.7365 ; this crucible is filled with melted metal, of which it
is required to make a sphere ; what is the diameter of the
mold ? Ans. 4 inches.
45. Suppose a cistern has two pipes, and that one can fill
it in 8^ hours, the other in 4| ; in what time can both fill it
together'? Ans. 3 hours 2 min. 49.5 sec.
46. A certain cistern has three pipes ; the first will empty
it in 20 minutes, the second in 40 minutes, and the third in
75 minutes ; in what time would they all empty it ?
Ans. 11 min. 19 sec. 15 thirds.
* A cubic foot of sea water weighs 64 pounds.
t The weight of a cubic foot of marble is assumed to be 185 pounds.
% The weight of a cubic foot of fresh water is 62.5 pounds.
MENSURATION OF ^SOLIDS. 217
47. A reservoir of water has two supply cocks ; the first
will fill it in 40 minutes, and the second in 50 ; it has also a
discharging cock, by which it may be emptied, when full, in
25 minutes. Now if all the cocks are opened at once, and
the water runs uniformly, how long before the cistern will be
filled'? Ans. 3 hours 20 minutes.
Operation. If it will fill once in 40 minutes by one cock, and also in
50 minutes by the other cock, it will fill 2.25 times in 50 minutes by
both cocks.
Then, 2.25 times : 50 minutes'. '. 1 time : 22.222 minutes — the time in which
it will Jill by both cocks.
Now, asdt will empty in 25 minutes, the time gained in filling over
emptying will be as 25 to 22.222, or 9 to 8.
Consequently, it will fill 9 times while it empties 8 times ; and as
the time of filling is 22.222 minutes, 22.222x9 = 199.998 minutes = the
time required.
48. A cistern containing 60 gallons of water has three un-
equal cocks for discharging it ; the largest will empty it in one
hour, the second in two hours, and the third in three ; in what
time will the cistern be emptied if they all run together %
Ans. 32 min. 43 sec. 26 thirds.
Operation.
60 galls, by smallest cock in 3 hours =-%°-= 20 galls, in 1 hour.
60 " "second cock in 2 hours=4|>=30 " 1 "
60 " " largest cock in 1 hour =-S,G=60 " 1 "
Then, 20+30+60=110 gallons in 1 hour.
Hence, 110 gallons:! hour'.: 60 gallons : .5454 hours =32 minutes, 43
seconds, and 26 thirds—the result required.
49. A reservoir has three pipes; the first can fill it in 12
days, the second in 11 days, and the third can empty it in 14
days ; in what time will it be filled if they are all running to-
gether? Ans. 9 days 17 hours 24 min.
50. If a pipe 1J inches in diameter will fill a cistern in 50
minutes, how long would it require a pipe that is 2 inches in
diameter to fill the same cistern ? Ans. 28 min. 7.5 sec.
51. If a pipe 6 inches in diameter will draw off a certain
quantity of water in 4 hours, in what time would it take 3
pipes of four inches in diameter to draw off twice the quan-
tity % Ans. 6 hours.
K
218 MENSURATION OF SOLIDS.
52. A water tub contains 147 gallons; the supply pipe
gives 14 gallons in 9 minutes ; the tap discharges 40 gallons
in 31 minutes ; now, supposing the tap, the tub being empty,
to be carelessly left open, and the water to be turned on at 2
o'clock in the morning ; a servant at 5, finding the water run-
ning, shuts the tap. Required the time in which the tub will
be filled after this discovery.
Ans. 6 hours 3 min. 48.7 sec.
53. If the diameter of the earth is 7930 miles, and that
of the moon 2160, required the ratio of their surfaces and
their solidities, assuming them to be spheres.
Note. — The surfaces of all similar solids are to each other as the
squares of their like dimensions, such as diameters, circumferences,
linear sides, etc., etc. ; and their solidities are as the cubes of those
dimensions.
Operation. Hence, the surface of the moon : the surface of the earth
Also, the solidity of the moon : solidity of the earth :: 21603 : 79303,
21fi03 1
54. A sugar-loaf is to be divided equally among three per-
sons by sections parallel to the base ; it is required to find the
height of each person's share, assuming the loaf to be a cone
the height of which is 20 inches.
203
Operation. By similar cones, f3: 1::203: — =2666. 667 =</*e cube of
o
the height of the upper section; hence, -^2666.667 = 13.867, the upper part.
ov 203
Also, 3:2: :203: — —-=5333.333, and ^5333.333-13.867=3.604,
the middle part; consequently, the lower part will be 20—13.867+3.604
= 2.529 inches.
* The ratio of one quantity to another may be obtained by dividing
the antecedent by the consequent.
t This proportion, as well as all others of the kind, may be ex-
20
pressed thus : v' 3 : y/ 1 : : 20 : -^-^ = the height of the upper section ; and
V o
in some instances this ii the most convenient method.
MENSURATION OF SOLIDS. 219
55. A ship has a leak by which she would fill and sink in
15 hours, but by means of her pumps she can be pumped out,
if full, in 16 hours ; now, if the pumps are worked from the
time the leak begins, how long before the ship will sink ?
Ans. 240 hours. ■*
Operation. She will fill -^ in an hour ; then, if -fa is pumped out,
the water gains -j^— ■£$= ^io °f tne sniP Per hour.
56. Three men bought a piece of tapering timber, which
was the frustrum of a square pyramid ; one side of the base
was 3 feet, one side of the top 1 footj and the length 18 feet ;
what is the length of each man's piece, assuming they are to
divide equally ?
Operation. By similar triangles, 3 — 1 : 18: :1 : 9 = the length of the
piece cut off from the end, by which the piece is deficient of being a
pyramid.
Then, by rules, pages 168, 169, the contents of the piece of timber (a
frustrum of a pyramid) and the piece cut off (a pyramid) are 78 and 3
cubic feet. Also, 78-^-3 = 26 — the contents of each person' s share, which,
a(Jded to the contents of the piece cut off=26 + 3=29=*fo contents of
the first division (including the piece cut off), and 26 X 2 +3 =55 = the con-
tents of the second division (including the piece cut off).
Now, by similar pyramids, 3:29::93 (length of piece cut off):70<!t7,
and V 7047 = 19.172; hence, 19.172 -9 = 10.172 = the length of the first
division of the frustrum.
Again, 29 : 55: : 19.1723 (length of first division) : 13365, and v7 13365
=23.731; hence, 23.731 -19. 172 =4.559 =the length of the second di-
vision of the frustrum.
Hence,
Length of timber... 18.000
" first division 10.172
" second division 4.559 14.731
" third division 3.269
57. Assuming the earth to be a sphere, and a quarter of
its radius 1000 miles, what is the area of its surface, its vol-
ume, and weight, the mean density of it being 353.75 pounds
per cubic foot ?
i $ Surface, 201062400 square miles.
US' {.Volume, 268083200000 cubic miles.
220 MENSURATION OF SOLIDS.
58. The sides of the base of an irregular tetrahedron are
21, 20, and 13 feet, and its height 9 ; what is its volume?
By rules, pages 56 and 60, the area of the base of the figure is 12
feet.
Ans. 756 cubic feet
59. A regular tetrahedron contains 1.8414 cubic feet; re-
quired its side and surface.
Ans. Side, 30 inches ; surface, 1558.845 square inches.
60. Required the volume of the frustrum of a triangular
pyramid, the base of which has a side of 9 inches, its vertex
4, and its lateral edge 5.
Operation. It is required to ascertain the perpendicular height or
length of the frustrum.
1. 92-(9-^2)2 = 60.75, and V 60. 7 5 =7. 7942 = perpendicular of trian-
gle of base.
2. By rule, p. 62, to ascertain centre of base (an equilateral triangle),
9 X. 5773 — 5.1957= radius of circumscribing circle of base, or centre of
base.
3. The base and vertex of the frustrum have sides of 9 and 4, and the
slant height is 5 ; hence, 9co4: 5 '.'A: 4:= slant height of end of pyramid
if continued ; and 5 + 4=9 =height of whole pyramid. •
4. If 5.196 =radius of base of pyramid, and 9=its slant height, then,
V (5A962-9*)=7M8=height of ichole pyramid; and 9: 5:: 7.348 : 4.082
= height of frustrum of pyramid.
Then, by rule, p. 169, 92+42+9 x4 X .433 X 4.082-r-3=78.3595 = ?e-
sult required.
6 1 . The sides of the base of an irregular tetrahedron are
12, 15, and 17 inches, and its height 9 ; required its volume.
Ans. 263.248 cubic inches.
62. How large a cube may be inscribed in a sphere 40
inches in diameter? Ans. 23.094 inches.
63. How many cubic inches are contained in a cube that
may be inscribed in a sphere 20 inches in diameter ?
Ans. 1480.2936 cubic inches.
By rule, page 61.
64. If a stone is put into a vessel of 14 cubic feet in capac-
ity, and it then requires but 2.5 quarts of water to fill it, what
is the volume of the stone?
Ans. 13 feet 1560 inches.
MENSURATION OF SOLIDS. 221
65. A cone, the diameter of which is 12 inches and altitude
10, being put into a vessel filled with rain water, with its base
upward, was depressed to a point where the area of its sec-
tion, parallel to the base, was eighty inches ; required the
weight of the cone.
As 12 : 10:: ■/ (80-4-. 7854) : 8.41 inches=depth of the cone in water;
then, 80x8.41x^=224.266, cubic inches, the volume of the part im-
mersed. Consequently, as 1728 : 224.266 : : 1000* : 129.784 ounces.
66. Into a vessel filled with rain water, suppose there be put
a cone of dry wood, having a volume of one cubic foot, with
the lesser end downward, and its axis perpendicular to the
surface of the water, and if a plane passing through the cen-
tre of gravity of the piece, parallel to its base, coinciding with
the water's surface, is found to rest in equilibrium, required
the quantity of water that will run over from the vessel, and
the specific gravity of the cone.
Operation. — The centre of gravity of a cone is distant from the ver-
tex f of its axis ; hence, 43 : 33 .* : 1728 : 729 =cubic inches immersed in the
water.
^Consequently, the quantity of water run over will be 729 cubic inches,
and the specific gravity of the water will be to that of the cone as 729
to 1728, or as 27 to 64, Ans.
67. A right cone cost $1363, at $120 per cubic foot, the
diameter of its base being to its altitude as 5 to 8. It is re-
quired to have its convex surface divided in the same ratio by
a plane parallel to the base, the upper part to be the greater ;
what is the slant height of each part ?
1363—120 = 1 1.358= the volume of the cone in feet; and 53X.7854x
•§•=52.36 =volume of a cone similar to it, the altitude of which is 8.
Again, the surface of similar solids being as the squares of their like
dimensions, \/(5 +8) : V8 : : \/(2.52 + 82) the side of the said similar cone
: V&ffi = the slant height of the upper part of this cone when its surface is
divided in the ratio proposed.
Consequently, V 70^-V ^-=V ^-V ^-=the slant height of the
under part of it.
Then, as similar solids are to each other as the cubes of their like di-
mensions, ^52.36: / 11.358: :V^-:S.9506 = the length of the slant
height of the upper part.
* Weight of a cubic foot of fresh water in ounces.
222 - MENSURATION OF SOLIDS.
Again, ^52.36 : f 11.358 : : V&&-V ^f* : 1.0855 = *Ae length of the
slant height of the lower part.
68. An elliptic inclosure has diameters of £30 and 612
links within its wall, which is 14 inches thick ; required the
area it incloses and covers.
a jit incloses 4 acres and 6 poles, and
' \ covers 1760.5 square feet.
69. A block of marble, in the form of a square pyramid,
weighs 18 tons, the perpendicular height being twice the di-
agonal of the base ; required its dimensions and volume in
cubic feet.
A cubic inch of the marble is assumed to weigh 1.6 ounces avoir-
dupois.
A (Side, 4.35325 feet.
nS' t 233.334 cubic feet.
70. A pail containing 2.1215 cubic feet is 12 inches in
depth ; what are its top and bottom diameters, they being in
the proportion of 5 to 3 ?
Ans. 14.64 and 24.4 inches.
71. If a sphere, the diameter of which is 4 inches, is de-
pressed in a conical glass full of water, the diameter of which
is 5 inches and altitude 6, it is required to know the quantity
of water which will run over? Ans. 26.272 inches.
By Construction, Fig. 132.
Draw a section of the glass, as A B C, and of the sphere, d ef.
Fig. 132. e
MENSURATION OF SOLIDS.
223
Then will AB=5; Ci = 6; Bt' = 54-2 = 2.5; A C and B C =
V(C i2+B i2)=6.5; and o/=4-j-2 (half diameter of sphere) =2.
The triangles B»C and o f C are similar, for they have the com-
mon angle C, and the right angles i andy; hence, their remaining an-
gles are equal.
Therefore, B i : B Cl'.of-.o C ; that is, as 2.5 : 6.5 : : 2 : 5.2 = the depth
of the glass from where the centre of the sphere rests.
Consequently, 5.2 — 6 = .8, and 2 + .8 =2. 8 inches of the sphere immersed,
and 4 — 2.8 = 1.2 inches of it above the glass.
Hence, the volume of the segment immersed (hy rule 2, p. 177) de-
ducted from that of the sphere =26.2722 cubic i?iches = the result re-
quired.
72. If a sphere is depressed in a conical glass full of water,
the diameter of which is 5 inches and altitude 6, so that its
upper edge is in a line with the rim of the glass, while its side
rests upon the side of the glass ; required the quantity of wa-
ter that will overflow from the glass.
By Construction, Fig. 133.
Draw a section of the glass, as A B C, and of the sphere, as d ef.
Fig. 133.
Then will C c=6, B C=f =2.5, and, by preceding example, B C
=6.5.
Hence, o e and of are equal, and the line B o, bisecting B e o f, is
a hypothenuse common to both triangles, B e o, Hfo.
Consequently, B c and B f are equal to 2.5, and B C— ~B f=C f=
6.5-2.5 = 4.
/. 6 (C e) : 2.5 (e B) : : 4 (C/) : 1.667 (/o), and 1.667 X 2=3.334 inches,
the diameter of the sj>here, the volume of which (by rule, p. 176) = 19.387
inches = the quantity of water that will over/low from the glass.
224 MENSURATION OF SOLIDS.
73. If a sphere, the diameter of which is 4 inches, is de-
pressed in a conical glass -L full of water, the diameter of
which is 5 inches and altitude 6, it is required how much of
the vertical axis of the sphere is immersed in water.
By Construction, Fig. 134.
Draw a section of the glass, as A B C, and of the sphere, d ef
Fig. 134. __i_
Let m n be the original level of the water, and n r the level when the
sphere is immersed.
Then will the cone n C r=cone m C n+the volume of the segment of
the sphere d sf=^ of cone A B C-\-the volume of the segment d sf.
A C = VCi2(6)+Ai2(5+2)=6.5 = length of slant side.
As A i (2.5) : A C (6.5) : : of radius of sphere (2) : C o {o.2)=dis-
tancefrom the centre of sphere at rest and the bottom of the glass.
C s=C o-o s=5.2-2=3.2.
Contents of cone (by rule, p. 166)=39.27, •£ of which =7.854.
Put x=s u—the immersed part of the axis of the sphere, and C w=C s
+s u=S.2+x.
Then, as similar solids are to each other as the cubes of their like di-
mensions,
63: (3.2 +:r)3:: 39.27: cone nCr; .'.cone n C r^^^x 39.27.
216
Segment dsf (by rule 2, p. 177)=(4x3t-2a;)Xa:2X.5236.
(3 2+a:)3
Since Cone n C r=cone m C n+segment d sf, .'.— X 39.27=
7.854+(4 x 3-2z) x x2 x .5236.
And 25 X (3.2 +x)3=5 X 216 + 4 X 62 x (6 -x) X x2.
Cube 3.2+x and 25 (3.23+3x3.22x+3x3.2xs+x3)=5x63+4x6»
(6-x)Xx2.
MENSURATION OF SOLIDS. 225
Actually multiplying the terms in the first member by 25,*
163
— +3 • 162x+3 • 5 • 16x2+25x3=5 • 63+4 • 62 (6-x)x'j
5
But 4 • 62 (6-x)x*=4: • 63o:2-4 ■ 62*3.
lfi3
/.—-+3 • 162o;+3 • 5 ■ 1 6a;5 -f 25a:3 =5 ' 63+4 • 63x2-4 • 62x3.
5
Multiplying by 5 throughout,
163+3 • 5 • 162x+3 • 52 • 16a;2 +5V =52 63+4 ■ 5 • 63x2-4 • 5 ■ 62x3.
By transposition,
(53-4 • 5 • 62)*3+(3 ; 52 • 1G-4 • 5 • 63)x3 + 3 • 5 • 162o;=52 • 63-163.
But, S= + 4 ■ 5 • G'=^p=169 = f ±jjj X5 = 13»x5.
3-5-16-4-5-6»=-3120=-2^,^,^ = 16=3-5-13-16
6 o lo
=3120.
52 • 63-163 = 1304=-^=163=8 ■ 163.
3.5.162=8-^,^,^, 16=3-5. 162=3840.
3 5 16
.'.5 • 132o;3-3 • 5 • 13 • 16x3+3 -5 • 163ar=8 ■ 163.
.-.5 (132*3-3 • 13 • 16x2+3 • 162x)=8 ■ 163.
Or, 132o:3-3 ■ 13 • 16x2+3 • 16'a; = — - — .
5
Multiplying both members by 13,
iSV-8 • 132 • 16x2+3 • 13 • 162*=8'13/163=^=3390.
Subtract 163 from both numbers
lS3x* - 3 • 132 ' 16a:2 + i
8-13-163-5-163 -3528
133x3 - 3 • 132 • 16a:2 + 3 ■ 13 • 162a: - 163 = 16» =
705-6.
5 5
The first member is now a perfect cube, the root of which is
13a; - 16 = -^ -705.6=8.90265.
13x= 16 - 8.90265 =7.09735.
* = — = .54595 inch.
13
Proof. C s=3.2/and s w=.54595.
Hence, C «=3.2 + .54595=3.74595.
* By Professor G. B. Docharty, New York.
f See Explanation of Characters, page 10, for use of a period be-
tween two factors.
K2
226 MENSURATION OF SOLIDS.
Then, 63 : 39.27 1 : 3.745953 : 9.5563 =volume of water in cone, nCr, from
which is to be deducted the volume of segment d sf and the remainder
should be equal to \ of 39.27 = 7.854.
Thus, volume of segment (by rule 2, p. 177) = 1.7023, and 9.5563
— 1.7023 = 7.85 4= result required.
74. A lady having three daughters had a farm of 450.758
acres, in a circular form, with her dwelling-house in the cen-
tre. Being desirous of having her daughters near her, she
gave to them three equal parcels of land as large as could be
made in three equal circles within the periphery of her farm,
one to each, with a dwelling-house in the centre of each ; that
is, there were to be three equal circles as large as could be
drawn within the periphery of the farm ; required the diam-
eters of the farm and of the three parcels.
V450.758 x43560-r-.7854=5000/ee*=dww»efer of farm.
By Construction, Fig. 135.
Draw the given circle, with o as its centre, and divide its periphery
into three equal parts, as at A B C ; connect A o, B o, and C o, and
assume d ef&s the centres of the required circles.
As the three circles required touch one another and the given circle,
the points, as A, B, and C, the centres, d ef, of the required circles,
and o, are necessarily in right lines. Connect d, ^ &ndf.
Then, as d ejfand e of axe isosceles triangles, the angle d and the
base e/are bisected at right angles in i by the line d i, and e o, in like
manner, bisects the angle e.
The triangles, e di, e o i, are equiangular:
Hence, ed:ei=($ed)::e o:oi=($e o); :.e o=2o i.
MENSURATION OF SOLIDS. 227
Put B e=x=(e i), R=B o=2500=radius of given circle = ( j ;
„ i . R— *
then, e o=K—x. :.^eo=oi= — - —
eoa=et9+o»8,or(R-a;)a=ara+
Or, R2-2Rx+x2=a;2 +
2
(R-*)2
4
R2-2R*+a:2
4
Or, 4R2-8Ra;=R2-2Ra:+a;2.
Transposing the formulae,
x2 + 6Rar=3R2:'x2 + 6Rx+9R2=3R2+9R2=12R?.
x+3R=±V12R2: x=±-/l2R1'-3R
x s= ±Ra/ 1 2 — 3R : x = ±R( \/ 12 — 3) = radius of required circles.
.'.Radius being=-/12— 3=3.4641— 3 = .4641, wfo'c^ is a constant
multiplier for all like problems.
Consequently, 2500 x .4641 = 1160.25=A d, B e, and Cf=product of
radius of circle and multiplier ■=radii of each of the circles required.
75. The weight of a quantity of silt in 30 cubic inches of
salt water is 4.21 grains, assuming the weights of silt and
salt water to be respectively 125 and 64 pounds per cubic
foot ; what is the volume of the silt compared to that of the
water? Am. 1. to 3608.307.
228 CONIC SECTIONS.
CONIC SECTIONS.
Definition. Plane figures generated by the cutting of a cone.
A Cone is a figure described by the revolution of a right-
angled triangle about one of its legs.
c
The axis (of a cone) is the line about which the triangle re-
volves, as C o.
The base is the circle which is described by the revolving
base of the triangle, as B o.
Notes. — If a cone is cut by a plane through the vertex and base, the
section will be a triangle, as A C B.
If a cone is cut by a plane parallel to its base, the section will be a
circle.
An Ellipse is a figure generated by an oblique plane cutting
a cone, as a b c d. v
The transverse axis or diameter (of an ellipse) is the longest
right line that can be drawn in it, as a b.
The conjugate axis or diameter is a line drawn through the
centre of the ellipse perpendicular to the transverse axis, as c d.
CONIC SECTIONS. 229
A Parabola is a figure generated by a plane cutting a cone
parallel to its side, as a b c.
The axis (of a parabola) is a right line drawn from the ver-
tex to the middle of the base, as b o.
Note. — A parabola has no conjugate diameter.
An Hyperbola is a figure generated by a plane cutting a
cone at any angle with the base greater than that of the side
©f the cone, as a b c.
ft--
« jg
The transverse axis or diameter, o b (of an hyperbola), is that
part of the axis e b, which, if continued, as at o, would join
an opposite cone, ofr.
The conjugate axis or diameter is a right line drawn through
the centre, g, of the transverse axis, and perpendicular to it.
The straight line through the foci is the indefinite trans-
verse axis ; that part of it between the vertices of the curves,
as o bf is the definite transverse axis. Its middle point, g, is
the centre of the curve.
230
CONIC SECTIONS.
The eccentricity of an hyperbola is the ratio obtained by di-
viding the distance from the centre to either focus by the semi-
transverse axis.
The asymjrtotes of an hyperbola are two right lines to which
the curve continually approaches, touches at an infinite dis-
tance, but does not pass ; they are prolongations of the diag-
onals of the rectangle constructed on the extremes of the axes.
Two hyperbolas are conjugate when the transverse axis of
the one is the conjugate of the other, and contrariwise.
An Ordinate is a right line from any point of a curve to
either of the diameters, as a e and do; ab and dfare double
ordinates.
c d
e i
—\
i o
An abscissa is that part of the diameter which is contained
between the vertex and an ordinate, as c e, g o.
The parameter of any diameter is equal to four times the
distance from the focus to the vertex of the curve ; the param-
eter of the axis is the least possible, and is termed the param-
eter of the curve.
The parameter of the curve of a conic section is equal to
the chord of the curve drawn through the focus perpendicular
to the axis.
The parameter of the transverse axis is the least, and is
termed the parameter of the curve.
The parameter of a conic section and the foci are sufficient
elements for the construction of the curve.
In the Parabola the parameter of any diameter is a third proportional
to the abscissa and ordinate of any point of the curve, the abscissa and
ordinate being referred to that diameter and the tangent at its vertex.
CONIC SECTIONS.
231
In the Ellipse and Hyperbola, the parameter of any diameter is a third
proportional to the diameter and its conjugate.
Note. — To determine the Parameter of an Ellipse or Hyperbola.
Kule. Divide the product of the conjugate diameter, multiplied by it-
self, by the transverse, and the quotient is equal to the parameter.
In the annexed figures of an Ellipse and Hyperbola, the transverse and
conjugate diameters, ab, c d, are each 30 and 20.
TJien, 30 : 20 : : 20 : 13.333 ^parameter.
Hence, the parameter of the curve =ef a double ordinate passing
through the focus s.
In a Parabola. The abscissa a b, and ordinate c b, are also equal
to 30 and 20.
a
c b
Hence, the parameter of the curve =e/.
232
CONIC SECTIONS.
A Focus is a point on the principal axis where the double
ordinate to the axis, through the point, is equal to the pa-
rameter, ase/ in the preceding figures.
It may be determined arithmetically thus : Divide the square of the
ordinate by four times the abscissa, and the quotient will give the focal
distances a s and s in the preceding figures.
The Directrix of a conic section is a straight line, such that
the ratio obtained by dividing the distance from any point of
the curve to it by the distance from the same point to the fo-
cus shall be constant.
It is always perpendicular to the principal axis ; and if the
curve is given, it is constructed as follows :
Let ABC represent the curve or curves, e f their axis,
and s the focus.
Through s draw s n or n' perpendicular to the axis till it
meets the curve in n or n' ; at n, n' draw the tangents weor
n' e% cutting the axis at e and e' ; through e, ef draw g h and
g' h' perpendicular to the axis, and they will be the directrices
of two conic sections.
If d s, drawn from any point, as d, > (is less than) d d, the
curve is an ellipse ; if equal to each other, it is the curve of a
parabola ; and < (if greater), as d/ s, d/ d/, it is the curve of
an hyperbola.
Ellipsoid, Paraboloid, and Hyperboloid of Revolution. Fig-
ures generated by the revolution of an ellipse, parabola, etc.,
around their axes. (See p. 124 and 202.)
CONIC SECTIONS.
233
A Conoid is a warped surface generated by a right line be-
ing moved in such a manner that it will touch a straight line
and curve, and continue parallel to a given plane. The straight
line and curve are called directrices, the plane a plane direc-
trix, and the moving line the generatrix.
Z\
j7
Thus, let a b c be a circle in a horizontal plane, and d d'
the projection of a right line perpendicular to a vertical plane,
d e ; if right lines, d a, r s, r' b, r" s, and df c, be moved so
as to touch the circle and right line d d/, and be constantly-
parallel to the plane r b, it will generate the conoid d b s n.
Note. — All the figures which can possibly be formed by the cutting
of a cone are mentioned in these definitions, and are the five following,
viz., a triangle, a circle, an ellipse, a parabola, and an hyperbola; but the
last three only are termed the conic sections.
234
CONIC SECTIONS.
ELLIPSE.
To describe an Ellipse.
The Transverse and Conjugate Diameters being given, Fig. 1.
Rule 1. — Draw the transverse and conjugate diameters,
A B, C D, bisecting each other perpendicularly in o.
Make A e equal to D C ; divide e B into three equal parts ;
set off two of those parts from o to e and from o to c ; then
with the distance c e make the two equilateral triangles c b e
and q d e.
These angles are the centres, and the sides being contin-
ued are the lines of direction for the several arcs of the ellipse
A C B D.
Fig. 1.
J
Note. — Mechanics are oftentimes required to work an architrave,
etc., about windows, of this form ; they may, by the help of the four
centres c, d, e, b, and the lines of direction h d, b f d g,b i, describe
another ellipse around the former, and at any distance required.
Rule 2. — Draw the line C D equal in length to the trans-
verse diameter ; also, E F equal in length to the conjugate di-
ameter, and at right angles with C D.
Take the distance C o or o D, and with it, from the points
E and F, intersect the diameter C D at h and f which points
are the foci.
Secure a string at h and f of such a length that it may just
reach to E or F.
Introduce a pencil, and bearing upon the string, carry it
around the centre o, and it will describe the ellipse required.*
* It is a property of the ellipse that the sum of two lines drawn from
the foci to meet in any point in the curve is equal to the transverse di-
CONIC SECTIONS.
235
Fig. 2.
The Transverse Diameter alone being given, Fig. 3.
Rule. — Let A B be the given length.
Divide it into three equal parts, as A s i b. Then, Vith
the radius A s, describe A / o i n c, and from i the circle
~B dn s o e ; then with n f and o c describe f e and c d, and
the required ellipse is made.
Fig. 3.
When any three of the four following Terms of an Ellipse are
given, viz., the Transverse and Conjugate Diameters, an Or-
dinate, and its Abscissa, to find the remaining Term.
To ascertain the Ordinate, the Transverse and Conjugate Diam-
eters and the Abscissa being given, Fig. 4.
Rule. — As the transverse diameter is to the conjugate, so
is the square root of the product of the two abscissas to the
ordinate which divides them.
Or, -x Vax(t— «0— °9 t ^presenting the transverse diam-
eter, c the conjugate, a' the less abscissa, and o the ordinate.
ameter, and from this the correctness of the above construction is ev-
ident.
236
Fig. 4.
CONIC SECTIONS.
C
Example. — The transverse diameter, A B, of an ellipse,
Fig. 4, is 25 inches, the conjugate, C D, is 16, and the ab-
scissa. A i, 7 ; what is the length of the ordinate i el
25 — 7 = 18= second abscissa.
Vl x 18 = 11.225 = square root of the abscissce.
Hence, 25 : 16:: 11.225 : 7.184 inches, the length of the ordinate re-
quired.
Ex. 2. The transverse diameter or axis of an ellipse is 100
inches, the conjugate 60, one abscissa 20, and the other 80 ;
what is the length of the ordinate? Ans. 24 inches.
To ascertain the Abscissa?, the Transverse and Conjugate Diameters
and the Ordinate being given, Fig. 4.
Rule. — As the conjugate diameter is to the transverse, so
is the square root of the difference of the squares of the or-
dinate and semi-conjugate to the distance between the ordi-
nate and centre, and this distance being added to, or subtract-
ed from the semi-transverse, will give the abscissas required.
~] x representing the dis-
' I tame obtained, and a
' the greater and less
-X=a>\ ab
abscissa?.
Example. — The transverse diameter, A B, of an ellipse,
Fig. 4, is 25 inches, the conjugate, C D, 16, and the ordinate
t e 7.184 ; what is the abscissa « B ?
Vj. 184* — 82 =3.519943 —square root of difference of squares of semi-
conjugate and ordinate.
Hence, as 16 : 25 : : 3.52 : 5.5 ^distance between ordinate and centre.
CONIC SECTIONS. 237
■ ok _.
Then, y = 12.5, and 12.5 +5.5 = 18 =B z, |
\ abscissas required.
-^ = 12.5, and 12.5-5.5= 7=A if, f
Ex. 2. The transverse diameter, A B, of an ellipse is 50
inches, the conjugate, C D, 32, and the ordinate i e 14.368 ;
what are the lengths of the abscissae ?
Ans. 11 and 39 inches.
To ascertain the Transverse Diameter, the Conjugate, Ordinate,
and Abscissa being given, Fig. 4.
Rule. — To or from the semi-conjugate, according as the
greater or less abscissa is used, adVjl or subtract the square root
of the difference of the squares of the ordinate and semi-con-
jugate.
Then, as this sum or difference is to the abscissa, so is the
conjugate to the transverse. '*
axe
Or,c-f-2 +
CH-2-
W*ffl
Example. — The conjugate diameter, C D, of an ellipse,
Fig. 4, is 16 inches, the ordinate i e is 7.184, and the abscissas
B i, i A are 18 and 7 ; what is the length of the transverse
diameter ?
^/7.1842— ( — J =3.52 =square root of difference of squares of ordi-
nate and semi-conjugate.
y + 3.52:18::16:25,l
-g > =transverse diameter required.
— -3.52: 7:: 16: 25, J
Ex. 2. The conjugate diameter of an ellipse is 60 inches,
the ordinate 24, and the abscissa 20 ; what is the length of
the transverse diameter? Ans. 100 inches.
238 CONIC SECTIONS.
To ascertain the Conjugate Diameter, the Transverse, Ordinate,
and Abscissa being given, Fig. 4.
Rule. — As the square root of the product of the abscissae
is to the ordinate, so is the transverse diameter to the con-
jugate.
Or, oxt-i- Vci X a' =c.
Example. — The transverse diameter, A B, of an ellipse,
Fig. 4, is 25 inches, the ordinate i e 7.184, and the abscissae
B i and i A 18 and 7 ; what is the length of the conjugate di-
ameter %
viSxl = 11.225= square root of product of abscissae.
11.225 : 7.184: :25 : 16=conjugate diameter required.
Ex. 2. The transverse diameter of an ellipse is 100 inches,
the ordinate 24, and the abscissae 20 and 80 ; what is the
length of the conjugate diameter'? Ans. 60 inches.
To ascertain the Circumference of an Ellipse, Fig. 4.
Rule. — Multiply the square root of half the sum of the
squares of the two diameters by 3.1416, and this product will
give the circumference nearly.
/d2-\-d'2
Or, \/ — x 3 . 1 4 1 6 = circumference.
Example. — The transverse and conjugate diameters, A B
and C D, of an ellipse, Fig. 4, are 24 and 20 inches ; what is
its circumference I
°4? + 202
- — = 488, and V 488 =22.09 =square root of half the sum of the
squares of the diameters.
Hence, 22.09 X3.1416=69.398=^e above rootx3.14:W=the result re-
quired.
Ex. 2. The diameters of an ellipse are 30 and 20 inches ;
what is its circumference? Ans. 80.0951 inches.
CONIC SECTIONS.
239
To ascertain the Area of an Ellipse, Fig. 4.
Rule. — Multiply the diameters together, the product by
.7854, and the result will give the area required.
Or, multiply one diameter by .7854, and the product by
the other.
Or, dx^X -7854= arm.
Example. — The transverse diameter of an ellipse, A B,
Fig. 4, is 12 inches, and its conjugate, C D, 9 ; what is its
area?
12 X 9 X.7854:= 84. 8232 =product of diameters and .7 854:= result re-
quired.
Ex. 2. The diameters of an ellipse are 70 and 50 feet ;
what is its area*? Ans. 2748.9 feet.
Centre of Gravity. Is in its geometrical centre.
/Segment of an Ellipse.
To ascertain the Area of a Segment of an Ellipse when its base is
parallel to either axis, as e if Fig. 5. /
Rule. — Divide the height of1 the segment b i by the diam-
eter or axis, a b, of which it is a part, and find in the table of
areas of segments of a circle, p. 134-138, a segment having
the same versed sine as this quotient ; then multiply the area
of the segment thus found and the two axes of the ellipse to-
gether, and the product will give the area required.
Gr, h-^-dx tab. area x d . d/z=area.
Fig. 5.
240 CONIC SECTIONS.
Example. — The height, b i, Fig. 5, is 5 inches, and the
axes of the ellipse are 30 and 20 ; what is the area of the
segment ?
^-=.1666 = tabular versed sine, the area of which (p. 135) is .08554.
Hence, .08554 X 30 x20=51.324=*Ae area required.
Ex. 2. The height of a segment at right angles or perpen-
dicular to the transverse diameter of an ellipse is 6.25 inches,
and the diameters are 16 and 25 % what is the area of the
segment? Ans. 61.42 inches.
Ex. 3. The height of a segment of an ellipse at right angles
to the conjugate diameter is 25 inches, and the diameters are
50 and 70 ; what is the area of the segment ?
Ans. 1374.415 inches.
The area of an elliptic segment may also be found by the following rule:
Ascertain the segment of the circle described upon the same axis to
which the base of the segment is perpendicular.
Then, as this axis is to the other axis, so is the circular segment to
the elliptical segment.
Illustration. — In the above example, the axis to which the base of
the segment is perpendicular is the conjugate, 50, and the height of the
segment 25. Also, the area of the segment is one half of that of a circle
of 50 ^a^er=1963;4954^981.7472.
Hence, 50: 70: : 981.75 : 1374.45 =area of elliptic segment.
PARABOLA.
1. To describe a Parabola, the Base and Height being given, Fig. 6.
Operation. — Draw an isosceles triangle, as A B D, the
base of which shall be equal to, and its height, B c, twice that
of the proposed parabola.
Divide each side, A B, D B, into any number of equal
parts; then draw lines, 1 1, 2 2, 3 3, etc., and their intersec-
tion will define the curve of a parabola.
Fig. 6.
CONIC SECTIONS.
15,
241
2. To describe a Parabola, any Ordinate to the Axis and its Ab-
scissa being given, Fig. 7.
Operation. — Bisect the ordinate, as A o in r; join B r,
and draw r s perpendicular to it, meeting the axis continued to s.
Draw B c and B n each equal to o s, and n will be the fo-
cus of the curve.
Take any number of points, 1 1, etc., on the axis, through
which draw the double ordinates 2 12, etc., of an indefinite
length. Then, with the radii cn,cl, etc., and focal centre n,
describe arcs cutting the corresponding ordinates in the points
2 2, etc., and the curve ABC, drawn through all the points
of intersection, will define the parabola required.
Note. — The line 2 n 2, passing through the focus n, is the parameter.
Fig. 7.
B.
242
CONIC SECTIONS.
To ascertain either Ordinate or Abscissa of a Parabola, the other
Ordinate and the Abscissa?, or the other Abscissa and the Or-
dinates being given, Fig. 8.
Rule. — As either abscissa is to the square of its ordinate,
so is the other abscissa to the square of its ordinate.
Or, 1. = o
—if*.
3. — j^-=a.
2.
4.
o zxa
a
—a'
Or, as the square root of any abscissa is to its ordinate, so is
the square root of any other abscissa to its ordinate.
„ ox Va' /
Hence, -. — =zo .
ya
Fig. 8.
|
Example. — The abscissa a b, of the parabola, Fig. 8, is 9,
its ordinate, b c, 6 ; what is the ordinate d e, the abscissa of
which, a d, is 16?
Hence, 9 : 62 : : 1 6 : 64, and V 64 = 8 = length of ordinate required
Or, V 9 : 6 : : V 16 : 8 =ordinate as before.
Ex. 2. The less abscissa of a parabola is 16, its ordinate
10 ; what is the ordinate, the abscissa of which is 36 ?
Ans. 15.
Ex. 3. The abscissae of a parabola are 9 and 16, and their
I corresponding ordinates 6 and 8 ; any three of these being
taken, it is required to find the fourth.
CONIC SECTIONS. ' 243
1. 62x16^— ^64, and t/64: =8= ordinate.
j y
2. ZJll __Ar_36, and v'36 = 6=Wiwafe.
3. • — — — =— — = 9z=less abscissa.
82 64
. 82x9 576 f. _ .
4. — — - =— -=16=abscissa.
6"1 ob
Parabolic Curve.
To ascertain, the Length of the Curve of a Parabola cut off by a
Double Ordinate.
Rule. — To the square of the ordinate add ^ of the square
of the abscissa, and the square root of this sum, multiplied by
two, will give the length of the curve nearly. .
4a2\
Or, -y/{o2-\ — — J x2 = length of curve.
Example. — The ordinate, d e, Fig. 8, is 8, and its abscis-
sa, a d} 16 ; what is the length of the curve/ a e f
4x 16a
82 + — ^ — =105. 333= sum of square of the ordinate and % of the
o
square of the abscissa, and V 405.333= 20.133, which X 2 =10.267= length
required.
Note. — This rule can be used only when the abscissa does not ex-
ceed half the ordinate. The length of the curve in other cases is to be
found by means of hyperbolic logarithms, as shown by writers on
fluxions.
Ex. 2. The ordinate is 16 inches, and its abscissa 32;
what is the length of the curve? Ans. 80.533 inches.
Ex. 3. The abscissa is 20, and its ordinate 12 inches;
what is the length of the curve ? Ans. 52.05 inches.
Ex. 4. The abscissa is 5 feet, and its ordinate 3 ; what is
the length of the curve? Ans. 13.014 feet.
244 * CONIC SECTIONS.
Parabola.
To ascertain the Area of a Parabola, Fig. 9.
Rule. — Multiply the base by the height and J of the prod-
uct will give the area required.* ^
Or, § bxh=zarea.
Fig. 9. b
Example. — What is the area of the parabola a b c, Fig. 9,
the height, b e, being 16 inches, and the base, or double ordi-
nate, a c, 16 ?
16 X 16 = 256 =product of base and height, and § of 256 = 170.667 =
area required.
Ex. 2. The height of an abscissa of a parabola is 32 inches,
and the base or double ordinate is 16; what is its area?
Ans. 341.333 square inches.
Ex. 3. The height of a parabola is 50 inches, and its base
24 ; what is its area ? Ans. 800 square inches.
To ascertain the Area of a Frustrwn of a Parabola, Fig. 10.
Rule. — Multiply the difference of the cubes of the two
ends of the frustrum, a b, c d, by twice its altitude, e o, and
divide the product by three times the difference of the squares
of the ends.
d3(^>d/3 x%h
Or, -= -q — —=area, d and d/ representing the lengths of
the base and lesser end.
* Corollary. — A parabola is § of its circumscribing parallelogram.
CONIC SECTIONS. 245
Fig. 10.
Example. — The ends of a frustrum of a parabola, a b and
c d, Fig. 10, are 10 and 6 inches, and the height, e o, is 10
inches ; what is its area ?
103co63X 10 x 2 = 15680= difference of cubes of the ends X twice the
height.
15680-f-102co 62 X 3=81.667 =preceding product-r-3 times the differ-
ence of the squares of the ends — area required.
Ex. 2. The base and lesser end of a frustrum of a parabola
are 30 and 24 inches, and the height 22 inches ; what is its
area*? Arts. 596.444 inches.
Ex. 3. The base and upper end of a frustrum of a parabo-
la are 30 and 20 inches, and the height 20 inches ; what is
the area"? Ans. 506.667 inches.
Ex. 4. The ends of the frustrum of a parabola are 5 and
4 feet, and its height 2.5 feet ; what is its area ?
Ans. 1 1.296 feet.
Ex. 5. The ends of a frustrum of a parabola are 8.5 and 7
feet, and its height 6 feet ; what is its area *?
Ans. 46.645 feet.
Note. — Any parabolic frustrum is equal to a parabola of the same
altitude, the base of which is equal to the base of the frustrum, in-
creased by a third proportional to the sum of the two ends and the
lesser end.
Illustration. — In Example 1, the base and end are 10 and 6.
Then, 10 + 6 : 6: '.6 : 2. 25 = third proportional to the sum of the two ends
and the lesser end.
Hence, 10 + 2.25 = 12.25=swn of length of base of parabola and third
proportional, the area of which, the height being 10, is=81.667.
246
CONIC SECTIONS.
HYPERBOLA.
To describe an Hyperbola, the Transverse and Conjugate Diam-
eters being given, Fig. 11.
Fig. 11.
/
er/_
C
//£ s
A
i\ a
Bj/1
^ s
1
\
0
«v iX/
f n n n n
s
y
Let A B represent the transverse diameter, and C D the
conjugate.
Draw C e parallel to A B, and e r parallel to CD; draw
o e, and with the radius o e, with o as a centre, describe the
circle F e f r, cutting the transverse axis produced in F and
/; then will F and / be the foci of the figure.
In o B produced take any number of points, n, n, etc.,
and from F and f as centres, with A n and B n as radii, de-
scribe arcs cutting each other in s, s, etc. Through s, s, etc.,
draw the curve s s s B s s s, and it will be the hyperbola required.
Note. — If straight lines, as o e y and o r y} are drawn from the cen-
tre o through the extremities e r, they will be the asymptotes of the
hyperbola, the property of which is to approach continually to the curve,
and yet never to touch it.
When the Foci and the Conjugate Axis are given.
Let F and /be the foci, and C D the conjugate axis, as in the pre-
ceding figure.
Through C draw g C e parallel to F and/; then, with o as a centre
and oFas a radius, describe an arc cutting g C e at g and e ; from
these points let fall perpendiculars upon the line connecting F and/ and
the part intercepted between them, as A B, will be the transverse axis.
CONIC SECTIONS.
247
To ascertain the Ordinate of an Hyperbola, the Transverse and
Conjugate Diameters and the Abscissa being given, Fig. 12.
Rule. — As the transverse diameter is to the conjugate, so
is the square root of the product of the abscissae to the ordi-
nate required.
Or,
c X Va X a*
~~t
■.ordinate.
Note. — 1. In hyperbolas, the less abscissa, added to the axis (the
transverse diameter), gives the greater.
2. The difference of two lines drawn from the foci of any hyperbola
to any point in the curve is equal to its transverse diameter.
Fig. 12.
Example. — The hyperbola, a b c, Fig. 12, has a transverse
diameter, a t, of 120 inches, a conjugate, df of 72, and the
abscissa, a e, is 40 ; what is the length of the ordinate e c?
40 + 1 20 = 1 60 =sum of less abscissa and axis=gveater abscissa.
120 : 72: : -/(40 X 160) : ±8=ordinate required.
Ex. 2. The transverse diameter of an hyperbola is 25, the
conjugate 15, and the less abscissa 6 inches; what is the
length of the ordinate? Ans. 8.1829 inches.
Ex. 3. The transverse diameter of an hyperbola is 5 feet,
the conjugate 3, and the less abscissa 1.8 feet ; what is the
length of the ordinate ? Ans. 2 feet.
Ex. 4. The transverse diameter of an hyperbola is 5 feet,
the conjugate 3, and the greater abscissa 8.667 feet; what is
the length of the ordinate ? Ans. 3.67 feet.
248 CONIC SECTIONS.
To ascertain the Abscissa, the Transverse and Conjugate Diameters
and the Ordinate being given, Fig. 12.
Eule. — As the conjugate diameter is to the transverse, so
is the square root of the sum of the squares of the ordinate
and semi-conjugate to the distance between the ordinate and
the centre, or half the sum of the abscissas.
- Then, the sum of this distance and the semi-transverse will
give the greater abscissa, and their difference the less abscissa.
*-v/o2 + (c~2)2 a+a' . 7/. 7 „ , , .
Or, s= — <r — =z half the sum of the abscissa.
a-\-a' t n a-\-a' t
Example. — The transverse diameter, a t, of an hyperbola,
Fig. 12, is 120 inches, the conjugate, d f, 72, and the ordi-
nate, e c, 48 ;■ what are the lengths of the abscissas, t e and ae?
72 : 120: : V48a+(72-f-2)a = 60 : 100 —half the sum of the abscissce.
100 +(120-^2) = 160 =above sum added to the semi-transverse = the
greater abscissa, and
100 — (120-i-2)=40=aJore sum subtracted from the semi-transverse^
the less abscissa.
Ex. 2. The transverse and conjugate diameters of an hy-
perbola are 25 and 15 inches, and the ordinate 8.1829 ; what
are the lengths of the abscissae? Ans. 6 and 31 inches.
To ascertain the Conjugate Diameter, the Transverse Diameter,
the Abscissce, and Ordinate being given, Fig. 12.
Rule. — As the square root of the product of the abscissas
is to the ordinate, so is the transverse diameter to the conju-
gate.
o X t
Or, —77 77 = conjugate diameter.
V(axa )
Example. — The transverse diameter, a b, of an hyperbo-
la, Fig. 12, is 120, the ordinate, e c, 48, and the abscissas,
t e and a e, 160 and 40 ; what is the length of the conju-
gate^/?^ ;.
V 40X160=80 ; 48 : 1 120 : 12=conjugate required.
CONIC SECTIONS. 249
Ex. 2. The transverse diameter of an hyperbola is 25
inches, the ordinate 8.1829, and the abscissas 6 and 31;
what is the length of the conjugate? Ans. 15 inches.
To ascertain the Transverse Diameter, the Conjugate, the Ordi-
nate, and an Abscissa being given, Fig. 12.
Eule. — Add the square of the ordinate to the square of the
semi-conjugate, and extract the square root of their sum.
Take the sum or difference of the semi-conjugate and this
root, according as the greater or less abscissa is used.^
Then, as 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 diameter required.
Or, (a or a x c x ( Vo2-\-(c-i-2)2 d-c-±-2))-~o2 = transverse
diameter.
Example. — The conjugate diameter, df, of an hyperbola,
Fig. 12, is 72 inches, the ordinate, e c, 48, and the less ab-
scissa, a e, 40 ; what is the length of the transverse diam-
eter, a t f
V ±8* + {7 2-r-2)2=Q0= square root of the squares of the ordinate and
semi-conjugate.
60 + 72-*-2=96=sm??i of above root and the semi-conjugate (the less
abscissa being used).
40 X 72 =2880 =product of abscissa and conjugate.
482 : 2880 : : 96 : 120 = transverse diameter.
Ex. 2. The conjugate diameter of an hyperbola is 15
inches, the ordinate 8.1829, and the greater abscissa 31 ;
what is the length of the transverse diameter ?
Ans. 25 inches.
Ex. 3. The conjugate diameter of an hyperbola is 6 feet 8
inches, the ordinate 4 feet, and the less abscissa 3 feet 4
inches*, what is the length of the transverse diameter?
Ans. 142.336 feet.
* When the greater abscissa is used, the difference is taken, and
contrariwise.
L 2
250 CONIC SECTIONS.
To ascertain the Length of any Arc of an Hyperbola, commencing
at the vertex, Fig. 13.
Rule. — To 19 times the transverse diameter add 21 times
the parameter of the axis, and multiply the sum by the quo-
tient of the less abscissa divided by the transverse diameter.
To 9 times the transverse diameter add 21 times the pa-
rameter, and multiply the sum by the quotient of the less ab-
scissa divided by the transverse diameter.
To each of the products thus found add 15 times the pa-
rameter, and divide the former by the latter ; then this quo-
tient, being multiplied by the ordinate, will give the length of
the arc nearly.
tx 19 + 21 xp X-+ 15 xp
Or, xo—arc nearly.
tx 9+21Xi?Xy-fl5xp
Fig. 13.
Example. — In the hyperbola a b c, Fig. 13, the transverse
diameter is 120, the conjugate 80, the ordinate e c 48, and the
less abscissa, a e, 40 ; required the length of the arc b a c?
120 : 80 : : 80 : 53.3333 ^parameter.
40
120 x 19+53.3333 X 21 X—= 1133.3333 =proe&c« of the sum of Id
times the transverse and 21 times the parameter, by the quotient of the less
abscissa and transverse.
Note. — As the transverse diameter is to the conjugate, so is the con-
jugate to the parameter. (See rule, p. 231.)
CONIC SECTIONS. 251
40
120x9+53.3333x21 X-— = 733. 3333=product of the sum of 9 times
the transverse and 21 times the parameter, by the quotient of the less abscis-
sa and transverse.
1133.3333 +53.3333 xl5+-(733.3333+53.3333x 15) = 1.2622 = quo-
tient of former product and 15 times the parameter flatter product and 15
times the parameter.
1.2622 X 48 = 60.5856 =above quotient X the ordinate = the length re-
quired.
Ex. 2. The transverse diameter of an hyperbola is 60, the
conjugate 36, the ordinate 24, and the less abscissa 20
inches ; required the length of the curve.
Ans. 31.324 feet.
Ex. 3. The transverse diameter of an hyperbola is 25, the
conjugate 15, and the less abscissa 6 ; required the length of
the curve. Ans.lO.2Hl.
Ex. 4. The transverse diameter of an hyperbola is 50, the
conjugate 15, and the greater abscissa 31 ; what is the length
of the curve? Ans.5.9395.
To ascertain the Area of an Hyperbola, the Transverse, Conju-
gate, and less Abscissa being given, Fig. 13.
Rule. — To the product of the transverse diameter and less
abscissa add 4 of the square of this abscissa, and multiply the
square root of the sum by 21.
Add 4 times the square root of the product of the trans-
verse diameter and less abscissa to the product last found, and
divide the sum by 75.
Divide 4 times the product of the conjugate diameter and
less abscissa by the transverse diameter, and this last quo-
tient, multiplied by the former, will give the area required
nearly.
n V*Xa/+-X2x21+-(\/*Xa/x4) cxa'x*
Or, n X ——=area.
Example. — The transverse diameter of an hyperbola, Fig.
13, is 60 inches, the conjugate 36, and the less abscissa or
height, a e, 20 ; what is the area of the figure ?
252 CONIC SECTIONS.
60x20+f of 202 = 1485.7143=swm of the product of the transverse
and abscissa and -f- of the square of the abscissa.
V 1485.7143 X 21 = 809.424=21 times the square root of the above sum.
V 60 X 20 X 4 + 809. 424=947.988 = sum of above result and the square
root of 4 times the product of the transverse and abscissa.
947.988-^75 = 12.6398 ^quotient of above result-r-75.
36 x 20 x 4
— X 12.6398 = 606.7 104:=product of 4 times the product of the
conjugate and abscissa-^-the transverse and the above quotient = area re-
quired.
Ex. 2. The axes of an hyperbola are 100 and 60, and the
less abscissa 50 inches ; what is its area ?
Ans. 22.3636 square feet.
Ex. 3. Required the area of an hyperbola, the greater ab-
scissa being 6.25 feet, the ordinate 2.165, and the transverse
diameter 4.167. Ans. 9.6008 square feet.
CONIC SECTIONS. 253
CONICAL UNGULAS.
To ascertain the Convex Surface of a Conic Ungula, Fig. 14.
Fig. 14.
c / T~y% 6
1. When the Section passes through the opposite Extremities of the Ends
of the Frustrum, as a e.
Rule. — Let d represent diameter of greater end, d' diameter of less end,
and h the perpendicular height of the frustrum, as o r.
Then, '^^VW+(d^dyx d* (.d+d^xdd0 -convex surface of
elliptic ungula, a eh. -v
Example. — The diameters ab,ce, of the frustrum of the cone, Fig.
14, are 10 and 5 inches, and the height, o r, 10; what is the convex
surface of the elliptic ungula a eh?
.7854 .7854
d-d' 10-5
.15708, and .15708 X V 4 h* + (d-d')2 = .15708 X
V(d+dr+d<f)
•/(4x 100+10-5 =3.2383, and 3.2383 xda- v a J =3.2383
X100~
required.
X 100 (10+5 X10X5)=32383 x iQM7=zl52mS2=convexsurface
To ascertain the Contents of a Conic Ungula, Fig. 14.
Rule. — Let n=.7854.
,™ d2—d'Vdd' ndh „ . „. . 7 ,
lnen, X — — = contents of the greater elliptic ungula a e b.
dVdd'-d'* nd'h '* ... . .
-j — -s; — X— - — ^contents of the less elliptic ungula ace.
(d%-d'%y nh „_,
3 — X — = difference of these ungnlas.
254 CONIC SECTIONS.
Example. — The diameters, height, and section of the frustrum being
as in the preceding example, what are its contents ?
102-5\/10x5 100-5X7.071 100-35.355 __ ,.,
ttt-t = - = = 12.920, which X
10 — 5 5 5
<1?=78^4_26 and 12.929x26.18=338.4812=co»ten<*
3
required.
To ascertain the Convex Surface of a Conic Ungula, Fig. 15.
Fig. 15.
2. When the Section cuts off" part of the Base, and makes the Angle erb
less than the angle cab.
Note.— When a Segment greater than a Semicircle is to he found,
Subtract the quotient of ita veraed sine, divided by the diameter of the circle, from
unity. Then take the tabular segment (p. 134-137) which corresponds to the re-
mainder from .78539, the whole tabular area, and the remainder will be the tabular
segment corresponding to the segment required.
RULE._ _Ly 4 h*+(d-dy x (tab. seg. ibs~x lx(^,}"ar
d—d' ..-.,. d2 d'—ar
X V-p X area of segment of circle a b, the height of which is
d X — — — J == convex surface of elliptic ungula iebs.
Example. — The diameters ab, ce, of the frustrum of the cone, Fig.
15, are 10 and 5 inches, and the height o n 10; what is the convex
surface of the elliptic ungula, the base b r being 6 inches ?
^-^=—^=.2, and .2 X V±h2 + (d-dy = .2 xV(ix 100+To^53)
= 4.1231.
Tab. seg. i b s=-fo=.6, and.G — 1 = .4, and tab. seg. of .4 X 102=29.337,
which -78.539 =49.202.
d'*_ 25 ^X(<J+Q-ar_^X(10 + 5)-4_3.5
da~100* d'-ar ~ 5-4 " 1
vtLL_=v««2Am.
d—ar 5 — 4 1
CONIC SECTIONS.
255
Area of seg. cir. a b . height=dx — — — = 10x — — =2, and 2-MO
= .2, tab. seg. =.11823, which X 102=1 1.1823.
Then, 4. 1231 X (49.202-^x3.5 X 2.4495 X 11.1823) = 104.0464 =
convex surface required.
To ascertain the Contents of a Conic Ungula, Fig. 15.
Rule. — Let S=tabular segment qfbr, the versed sine being-^-a b ; s=
tabular segment, the versed sine being b r — (d—d')-r-d'.
b r br
Then^SXcP-sXd^X
br \ Ik
, J X , 7= contents of
—d—d J d—d
br — d—d' br-
the ungula ieb s.
Example. — The dimensions of the frustrum and ungula being the
same as in the preceding example, what are its contents ?
S=-r=— , and ttl^.G .'. 1 — .6 = .4, the tabular segment of which is
(page 135) .29337, and .78539 -.29337 =.49202.
6—5
s=br—(d—d'-7-d')= = .2, the tabular segment of which is .11182.
S X d3 = . 49202 X102 =492.02.
br br
Xd'3X
IV
-=.11182 X 53 X 6 X 2.4495=205.427,
br—d—d' br—d—d'
which-492.02=286.593.
_|& 3J533 _ ^ and 2g6 593 x >6667 = 19L0715 = resuh n_
d—d 5
quired.
To ascertain the Convex Surface of a Conic Ungula, Fig. 16.
Fin. 10.
C k J — + d
3. When the Section is parallel to one of the Sides of the Frustrum, as d r.
-rl-pV4:h>+(d-dy X area of seg. i b «- f| {d-d) X Vd'x(d-d'))
= convex surface of parabolic ungula idb s.
256 CONIC SECTIONS.
Example. — The diameters ab, cd, of the frustrum of the cone, Fig.
16, are 10 and 5 inches, and the height or 10; what is the convex sur-
face of the ungula ?
-1— =— — = .2, and .2x V 4 h*+(d-dy = V ±x 100 + (10-5)2 =
d—d 10 — 5
.2X20.6155=4.1231.
Area ofseg. i b s.-Q.(d-d')X Vd'X(d-d'j) =39.27- (? (10-5) X
V5x(10-5))=39.27-16.667 = 22.603.
Then, 4.1231 x22.603=93.1944=conuex surface required.
To ascertain the Contents of a Conic Ungula, Fig. 16.
Rule. — Let A=area of base i b s.
Then, ( - — - — - d'V (d— d')X d')\ X-h ^contents required.
Example. — The diameters of a frustrum of a cone, Fig. 16, a b and
c d, are 10 and 5 inches, and the height o r 10; what are the contents
of the ungula i d b s?
A= 102 X .7854-^2 =39.27.
Axd 39.27x10 392.70
d-d' 10-5
78.54.
-d'V{d-d')xd'=^5V{iO-b)x 5 = 6.667X5=33.335, and 33.335
o o
■78.54=45.205, which X- 10=45.205 X 3.334 = 150.7135 ^contents re-
quired.
To ascertain the Convex Surface of a Conic Ungula, Fig. 17.
4. When the Section cuts off part of the Base, and makes the Angle drb
greater than the Angle cab.
-±-j, X V ±h* + (d-dy X (dr. seg. i b s - ^ X o r - ^{d - d')
d~d V * br-d=d'
V ) = convex surface of hyperbolic ungula idbs.
br—d'—d/
CONIC SECTIONS. 257
Example. — The dimensions of the frustrum being the same as in the
preceding example, and the height o n and base b r of the ungula be-
ing 10 and 2.5 inches, what are its contents ?
d-d'
— xV4:h2-\-(d-dy = .2x 20.6155=4.1231.
2.5
And (cir. seg. i b s=— = .25, tab. seg. = .153546 X 103 = 15.3546.
cT2_25 br-$(d-d') br _2.5— frx (10-5) _ 6.25
d~100' br-(d-d'y br-(d-d')~ 2.5-(10-5) ~~275~
V si =i.
V 2.5 -(5 -10)
ok 6 2^
Then, 4.1231 X (15.346--— X-^ X 1) = 15.346 - .625 = 14.721,
100 Z.o
which X 4.1231 = 60.6962 =convex surface required.
To ascertain the Contents of a Conic Ungula, Fig. 17.
Let the area of the hyperbolic section i d s=A, and the area of the cir-
cular seg. ib s=a.
Then, _/* ,x(aXd— Ax — -, = contents of the hyperbolic ungula
d — h dr
i d b s.
Note.— The transverse diameter of the hyp. seg.= _ , the conjugate =
br
dV-; r, — r— i an<i tne abscissa = dr, from which its area may be found by rule,
a — a — or
page 251.
Example. — The diameters a b, c d, of the frustrum of the cone, Fig.
17, are 10 and 5 inches, the heights o n and dr, of an hyperbolic un-
gula are 10, and the base b r 2.5 inches ; what are its contents ?
A (by rule, page 251)=53.675, a = 15.355.
Then, -i^ = ?'S 3* = 3.334, and («x<2-Ax-^— -^=(15.355 X 10
. a — n 10 — 10 dr J
= 153.55-53.675=99.875, which X ~ ' =24.9688, which X3.SS4 =
83.246 = contents required.
258 MENSURATION OF REGULAR POLYGONS.
REGULAR POLYGONS.
(Appendix to Rules, see Table, page 261.)
TO ASCERTAIN THE ELEMENTS OP ANY REGULAR POLYGON.
To ascertain the JRadius of the Inscribed or Circumscribing Circles
of a Polygon.
1. When the Length of a Side is given.
Rule. — Multiply the length of the side by the units in col-
umns A and B of the following table under the head of the
element required.
Example. — The length of the side of a square (tetragon)
is 2 inches ; what are the radii of its inscribed and circum-
scribing circles?
2x .5 = 1. =product of length of side and tabular multipliers^ radius of
its inscribed circle.
2 X. 70711 = 1.41422 =product of length of side and tabular mulliplier
= radius of its circumscribing circle.
Ex. 2. The length of the side of a hexagon is 2 inches ;
what is the radius of its inscribed circle ?
Ans. 1.73204 inches.
2. When the Area is given.
Rule. — Extract the square root of the area, and multiply
it by the units in columns C and D of the following table un-
der the head of the element required.
Example. — The area of a square is 4 inches; what are
the radii of its inscribed and circumscribing circles ?
V 4 X .5 = 1 . =radius of its inscribed circle.
■/4X .70711 = 1.41422 = radius of its circumscribing circle.
Ex. 2. The area of a hexagon is 64.952 inches; what is
the radius of its circumscribing circle? Ans. 5 inches.
MENSURATION OP REGULAR POLYGONS. 259
3. When the Radius of the Circumscribing Circle is given.
Rule. — Multiply the radius given by the unit in the col-
umn E opposite to the figure for which the radius is required.
Example. — The radius of the circumscribing circle of a
square is 2 inches ; what is the radius of its inscribed circle %
2 X. 70711 = 1.414:22 =radius of its inscribed circle,
Ex. 2. The radius of the circumscribing circle of a hexa-
gon is 5 inches ; what is the radius of its inscribed circle %
Ans. 4.3301 inches.
4. When the Radius of the Inscribed Circle is given.
Rule. — Multiply the radius given by the unit in the col-
umn F opposite to the figure for which the radius is required.
Example. — The radius of the inscribed circle of a square
is 1.41421 inches; what is the radius of its circumscribing
circle 1
1.41421 X 1.41421 =2 =radius of its circumscribing circle.
Ex. 2. The radius of the inscribed circle of a hexagon is
.866025 inches; what is the radius of its circumscribing cir-
cle? Ans. 1 inch.
To ascertain the Area.
1. When the Radii of the Inscribed or Circumscribing Circles
are given.
Rule. — Square the radius given, and multiply it by the
unit in the columns G and H of the following table under the
head of the element given.
Example. — The radius of the inscribed circle of a square
is 2 inches ; what is its area ?
22X 4= 16 =area.
Ex. 2. The radius of the circumscribing circle of a hexa-
gon is 5 inches; what is its area? Ans. 64.952 inches.
260 MENSURATION OP REGULAR POLYGONS.
3. When the Length of the Side is given.
. Eule. — Square the length of the side, and multiply it by
the unit in column I.
ExAMPLE.-^-The length of the side of a square is 4 inches ;
what is its area?
42Xl = 16 =area.
Ex. 2. The length of the side of a hexagon is 5 inches ;
what is its area*? Ans. 64.952 inches.
To ascertain the Length of a Side.
1. When the Radii of the Inscribed or Circumscribing Circles
are given.
Rule. — Multiply the radius given by the unit in the col-
umns K and L of the following table under the head of the
radius given.
Example. — The radius of the inscribed circle of a square
is 2 inches ; what is the length of a side %
2x2=4 = length of side.
Ex. 2. The radius of the circumscribing circle of a hexa-
gon is 5 inches ; what is the length of a side ?
Ans. 5 inches.
2. When the Area is given.
Rule. — Extract the square root of the area, and multiply
it by the unit in column M of the following table.
Example. — The area of a square is 16 inches ; what is the
length of a side ?
V 16 X 1=4= length of side.
Ex. 2. The area of a hexagon is 64.952 inches ; what is
the length of a side? Ans. 5 inches.
Note. — For other rules to ascertain the surface or linear edge of a
polygon, see page 64.
MENSURATION OF REGULAR POLYGONS.
261
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262 MENSURATION OF POLYHEDRONS.
MENSURATION OF REGULAR BODIES, OR
POLYHEDRONS.
To ascertain the Elements of any Regular Body, Figs. 72, 83, 84,
85, and 86, p. 72, 161-164.
To ascertain the Radius of a Sphere that will Circumscribe a given
Regular Body, and the Radius of one also that may be Inscribed
within it.
1. When the Linear Edge is given.
Rule. — Multiply the linear edge by the multiplier opposite
to the body in the columns A and B in the following table,
under the head of the element required.
Example. — The linear edge of a hexahedron or cube is 2
inches ; what are the radii of the circumscribing and inscribed
spheres ?
2 X. 86602 = 1. 73204 =product of edge and tabular multiplier = radius
of circumscribing sphere.
2 X .5 = l=product of edge and tabular multiplier — radius of inscribed
sphere.
Ex. 2. The linear edge of a hexahedron is 10 inches;
what are the radii of its inscribed and circumscribing spheres ?
Ans. 5 and 8.6602 inches,
2. When the Surface is given.
Rule. — Multiply the square root of the surface by the mul-
tiplier opposite to the body in the columns C and D in the
following table, under the head of the element required.
Example. — The surface of a hexahedron is 25 inches ;
what are the radii of the circumscribing and inscribed spheres'?
V25 X .35355 = 1.76775 = radius of circumscribing sphere.
V2S x .20412 = 1. 0206 = radius of inscribed sphere.
MENSURATION OF POLYHEDRONS. 263
Ex. 2. The surface of a hexahedron is 24 inches ; what is
the radius of its inscribed sphere ? Arts. 1 inch.
3. When the Volume is given.
Rule. — Multiply the cube root of the volume by the mul-
tiplier opposite to the body in the columns E and F in the
following table under the head of the element required.
Example. — The volume of a hexahedron is 8 inches ; what
are the radii of its circumscribing and inscribed spheres !
V 8 X .86602 = 1.73204:= radius of circumscribing sphere,
y/ 8 X .5 =1 = radius of inscribed sphere.
4. When one of the Radii of the Circumscribing or Inscribed
Sphere alone is required^ the other being given.
Rule. — Multiply the given radius by the multiplier oppo-
site to the body in the columns G and H in the following ta-
ble, under the head of the other radius.
Example. — The radius of the circumscribing sphere of a
hexahedron is 1 inch; what is the radius of its inscribed
sphere ?
1 X .57735 =.57735 = radius required.
Ex. 2. The radius of the inscribed sphere of a hexahedron
is 2 inches ; what is the radius of its circumscribing sphere %
Ans. 3.4641 inches.
To ascertain the Linear Edge of a Polyhedron.
li When the Radius of the Circumscribing or Inscribed Sphere
is given.
Rule. — Multiply the radius given by the multiplier oppo-
site to the body in the columns I and K in the following table.
Example. — The radius of the circumscribing sphere of a
hexahedron is 1 inch ; what is its linear edge 1
1 X 1.1547 = 1. 1547 =edge required.
264 MENSURATION OF POLYHEDRONS.
Ex. 2. The radius of the inscribed sphere of a hexahedron
is .57735 inch ; what is its linear edge ?
Ans. 1.1547 inch.
2. When the Surface is given.
Rule. — Multiply the square root of the surface by the mul-
tiplier opposite to the body in the column L in the following
table.
Example. — The surface of a hexahedron is 6 inches ; what
is its linear edge ?
V 6 X .40825 = 1 =linear edge.
Ex. 2. The surface of a hexahedron is 24 inches ; what is
its linear edge? Ans. 2 inches.
3. When the Volume is given.
Rule. — Multiply the cube root of the volume by the mul-
tiplier opposite to the body in the column M in the following
table.
Example. — The volume of a hexahedron is .3535 inch;
what is its linear edge ?
v7 .3535 X 1 = .7071=Unear edge.
Ex. 2. The volume of a hexahedron is 8 inches ; what is
its linear edge"? Ans. 2 inches.
To ascertain the Surface of a Polyhedron.
1. When the Radius of the Circumscribing Sphere is given.
Rule. — Multiply the square of the radius by the multi-
plier opposite to the body in the column N in the following
table.
Example. — The radius of the circumscribing sphere of a
hexahedron is .866025 inch ; what is its surface?
.86602a X 8=6= surface required.
MENSURATION OF POLYHEDRONS. 265
2. When the Radius of the Inscribed Sphere is given.
Rule. — Multiply the square of the radius by the multi-
plier opposite to the body in the column O in the following
table.
Example. — The radius of the inscribed sphere of a hexa-
hedron is .5 inch ; what is its surface ?
.52X 24 = 6= surface required.
3. When the Linear Edge is given.
Rule. — Multiply the square of the edge by the multiplier
opposite to the body in the column P in the following table.
Example. — The linear edge of a hexahedron is 1 inch ;
what is its surface ?
I2 X 6 = 6 = surface required.
4. When the Volume is given.
Rule. — Extract the cube root of the volume, and multiply
the square of the root by the multiplier opposite to the body
in column Q in the following table.
Example. — The volume of a hexahedron is 8 inches ; what
is its surface ?
^8 = 2, and 22 X 6 = 24 = surface required.
Ex. 2. The volume of a hexahedron is 353.5533 inches;
what is its surface? Ahs. 300 inches.
To ascertain the Volume of a Polyhedron.
1. When the Linear Edge is given.
Rule. — Cube the linear edge, and multiply it by the multi-
plier opposite to the body in column R in the following table.
Example. — The linear edge of a hexahedron is 2 inches ;
what is its volume ?
23 Xl = 8=volume required.
M
266 MENSURATION OF POLYHEDRONS.
2. When the Radius of the Circumscribing Sphere is given.
Rule. — Multiply the cube of the radius given by the mul-
tiplier opposite to the body in the column S in the following
table.
Example. — The radius of the circumscribing sphere of a
hexahedron is .5 inch ; what is the volume of the figure"?
53 X 1.5396 =.1925 cubic inch.
Ex. 2. The radius of the circumscribing sphere of a hexa-
hedron is 1.73205 inch; what is the volume of the figure?
Ans. 8 cubic inches.
3. When the Radius of the Inscribed Sphere is given.
Rule. — Multiply the cube of the radius given by the mul-
tiplier opposite to the body in the column T in the following
table.
Example. — The radius of the inscribed sphere of a hexa-
hedron is .5 inch ; what is its volume ?
.53x8 = 1 cubic inch.
Ex. 2. The radius of the inscribed sphere of a hexahedron
is 3.535 inches ; what is its volume?
Ans. 353.3932 cubic inches.
4. When the Surface is given.
Rule. — Cube the surface given, extract the square root,
and multiply the root by the multiplier opposite to the body
in the column U in the following table.
Example. — The surface of a hexahedron is 6 inches ; what
is its volume ?
VQ3X. 06804=1 cubic inch.
Ex. 2. The surface of a hexahedron is 24 inches ; what is
its volume? Ans. 8 inches.
Ex. 3. The surface of an octahedron is 125 inches; what
is its volume? Ans. 102.1743 cubic inches.
MENSURATION OP POLYHEDRONS.
267
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268 orophoids {Domes, Arched and Vaulted Roofs, etc.).
OROPHOIDS* {Domes, Arched and Vaulted Roofs, etc.).
Definition. Figures generated by a curved line running from
the periphery or the angles alone of their base to a common cen-
tre at the top. When the curve runs from the periphery alone,
and their point of connection is in the centre, they are Regular ;
when the point of connection is eccentric, they are Oblique ; ichen
curves run both from the angles and some intermediate point, or
there is any combination of elements in their generation, they are
Compound or Irregular.
Orophoids or Arched Roofs are either Domes, Saloons,
Vaults, or Groins.
A Dome is formed by arches springing from a circular or
polygonal base, and meeting in the centre at the vertex.
A Saloon (frustrum of an orophoid) is formed by arches
springing from a circular or polygonal base, and connecting
with a flat roof or ceiling in the middle.
A Vault is formed by arches springing from two opposite
bases alone, and meeting in a line at the vertex.
A Groin] is formed by the intersection of one vault with an-
other at any angle.
Orophoids are of the following forms :
1. Circular Dome, Saloon, or Vault, when generated by an
arc of a circle.
2. Elliptical Dome, Saloon, or Vault, when generated by an
arc of an ellipse.
* From opodog, a roof. The absence of any generic term to denote
this class of figures has induced the adoption of the above term.
t The curved surface between two adjacent groins is termed the sec-
troid.
orophoids (Domes, Arched and Vaulted Roofs, etc.). 269
3. Gothic Dome, Saloon, or Vault, when generated by two
circular or elliptical arcs meeting in the centre of the arch.
4. Curvilinear Dome, Saloon, or Vault, when generated by
an irregular curve.
To ascertain the External or Internal Surface of a Spherical
Dome.*
Rule. — Multiply the area of the base by 2, and the prod-
uct will give the surface required.
Or, a x 2 —surface.
Example. — The external diameter of a hemispherical dome
is 20 feet ; what is the surface of it ?
202X .7854 =314. 16 =area of base.
314.16 x2=628.S2 = twice the area of the base=surface required.
Ex. 2. The sides of a quadrangular spherical dome are 20
feet; what is the surface of it? Ans. 800 square feet.
Ex. 3. The internal sides of a hexagonal spherical dome
are 10 feet ; what is the surface of if?
Ans. 519.62 square feet.
To ascertain the Surface of a Side of a Polygonal Spherical
Dome.
Rule. — Multiply the base by its length, divide by 1.5708,
and the quotient will give the surface required.
#)n —surface-
' 1.5708 y /
Example. — The side of a quadrangular spherical dome is
20 feet ; what is the surface of it ?
* An elliptical dome or vault is either a semi-spheroid (ellipsoid) or
a segment of a spheroid ; for rules to ascertain the surface of these, see
p. 91-93, and 125.
A parabolic dome or vault is a paraboloid, or a segment of one ; for
rules to ascertain the surface of which, see page 126.
A Gothic dome or vault is a semi-circular spindle, or a segment of
one; for rules to ascertain the surface of which, see p. 12CF-122.
270 orophoids (Domes, Arched and Vaulted Hoofs, etc.).
2(K-2 X 3.1416=31.416-^2 = 15.708 = radius of dome X 3.1416-^2 =
perimeter or length of side.
20 X 15.708=314.16=/>rocto of length of base and length of side.
314.16 -M.5708 =200 = above product -f- 1.5708= surface of side re-
quired.
Ex. 2. The side of a hexagonal spherical dome is 10 feet ;
what is the surface of it ? Ans. 86.6025 feet.
To ascertain the Volume of a Spherical Dome*
Rule. — Multiply the area of its base by § of the height,
and the product will give the volume required.
Or, ax%h=volume.
Example. — The diameter of the base of a spherical dome
is 20 feet ; what is its volume?
202x. 7854=314. 16=«rai of base.
314.16x| of -^ = 2094.4 =areaxf of height =volwne required.
Ex. 2. The side of a hexagonal spherical dome is 10 feet ;
what is its volume? Ans. 1732.0506 cubic feet.
To ascertain the External or Internal Surface of a Saloon.
Rule. — To the area of the ceiling add the surface of its
sides, and the sum will give the surface required.
Or, a+s=surface.
Example. — A saloon roof has a quadrangular arch of 2
feet radius springing over a rectangular base of 20 by 16
feet ; what is its surface ?
20— 2 -{-2 = 16= length of ceiling.
16—2+2 = 12 =width of ceiling.
16 X 12 = 192 =area of ceiling.
— __ •
* An elliptic dome is either a semi-spheroid (ellipsoid) or a segment
of a spheroid. For rules to ascertain the volume of them, see p.
179-181.
Parabolic and hyperbolic domes are paraboloids and hyperboloids of
revolution. For rules to ascertain the volumes of them, see p. 203-206.
A Gothic dome is a semi-circular spindle or a segment of one. For
rules to ascertain the volume of them, see p. 188-191.
OROPHOIds (Domes, Arched and Vaulted Roofs, etc.). 271
2+2 X 3. 1416 = 12. 5664= circumference of arch having a radius of 2
feet.
12.5664-f-4=3.1416 = /e«s*A of arch at sides.
3.1416 X 16 + 16 + 12 + 12 = 175.9296 = /en#*A of arch X length of sides
and ends.
2+2 X 4
z=4:^=sum of sides of mitred ends of sides of saloon-^-4:=side of
4
a quadrangular dome.
4x4x2=32=area of base X 2 = surf ace of quadrangular spherical
dome. ,
192 + 175.9296 + 32=399.9296=s«r/ace of saloon required.
Ex. 2. A saloon roof has a hexagonal arch of 4.33 feet ra-
dius springing over an oblong hexagonal base having sides of
20 feet and ends of 10 feet ; what is its surface?
Ans. 556.5832 square feet.
To ascertain the Volume of a Saloon.
Rule. — Multiply the square of twice the height of the arc
by .7854, divide the product by 4, and multiply the quotient
by the length of the sides of the ceiling.
Subtract the side of the ceiling from the side of the saloon,
ascertain the area of a like figure having a side equal to this
remainder (see rule, p. 60), and multiply this area by two thirds
of the height of ceiling.
Multiply the area of the ceiling by the height of it; and
this product, added to the preceding, will give the volume re-
quired.
2/i2x.7854
Or, x l-\- %a'-\-a x h= volume.
Example. — The sides of a quadrangular circular saloon
roof are 20 feet, and the height of the ceiling 2 feet ; what is
the volume of it ? •
Note. — When the Saloon is a Circle.
To ascertain the Volume of the Ring projecting beyond the Diameter of
the Ceiling.
Ascertain the centre of gravity of a section of the ring, and multi-
ply the area of this section by the circumference described by its centre
of gravity. (See Example 2, page 272.)
272 okophoids (Domes, Arched and Vaulted Roofs, etc.).
2
2 X 2 X .785i-7-4:=3.l4:lG=07ie fourth of product of the square of twice
the height of the arc and .7854.
3.1416 X (20-2 + 2) X 4 =201.0624 =pn>ofac* of preceding quotient and
the length of the sides of the ceiling.
2+2—20=16, and 16 — 20=1= diameter of ceiling subtracted from
diameter of saloon.
42X 1 = 16 =product of square of side and tabular multiplier = area of
figure. ■"_
16 X§ of 2 =21.333 =product of area of base and § of height of ceiling.
162X 2 =512 =product of area of ceiling and its height.
512+21.333+201.0624 = 734.3954=sm7;j of above products =volume
required.
Ex. 2. A circular room 40 feet diameter, and 25 feet high
to the ceiling, is covered with a saloon having a circular arch
of 5 feet radius ; required the contents of the room in cubic
feet. Ans. 307 '7 9 .453 feet.
Operation.— Area of 40 X 25-5 = 1256.64 X 20=25132.8 = volume of
body of room.
Area of flat portion of ceiling, 40-5 + 5 = 706.86x25 co 20 = 3534.3
= volume of body of roof .
Area of quadrant of a circle having a radius of 5= — - — = 19.635.
Volume of quadrantal ring of 5 feet height and base =area thereof
X the circumference described by its centre of gravity. (See p. 209.)
Hence (by rule, p. 84), the centre of gravity of the ring having the sec-
tion of a sector of a circle of 5 feet radius is 3.001 feet from the angle of it.
Then (by rule, p. 55), the hypothenuse of the right angle (3) being alone
given, the length of the side (that is, the distance of the centre of gravity
of the sector from the vertical side ofit)=2.122. THxerefore, 40s— 5 + 5
+2.122 X 2 =34.244 =diameter of circle described by centre of gravity of
quadrantal ring.
Consequently, circumference of 34.244 X 19.635 =2112. 353 =volume of
quadrantal ring.
Volume of body of room, 25132.8
" " roof, 3534.3
" ring, 2112.353
30779.453 cubic feet = volume required.
Ex. 3. A quadrangular building having sides of 40 and 30
feet is covered with a saloon 25 feet in height from the floor,
having an arch of 5 feet radius ; what is the volume of the
saloon? Ans. 29296.83 cubic feet.
orophoids (Domes, Arched and Vaulted Hoofs, etc.). 273
To ascertain the Surface of a Vault.
Eule. — Multiply the length of the arch by the length of
the vault, and the product will give the surface required.
Or, px l=zsurface,p representing the perimeter of the arch.
Example. — What is the concave or internal surface of a
circular vault, the width of it being 40 feet and the length 8(^|
40 X 3.1416-^2 X 80 =5026.56 =product of length of arch and length of
vault=result required.
Ex. 2. The width of an elliptic arched vault is 18 feet, its
height 12, and its length 50 ; what is its internal surface*?
Ans. 1666.085 square feet.
To ascertain the Volume of a Vault.
Eule.— Multiply the area of a section of the vault by its
length, and the product will give the volume required.
Or, a x I— volume.
Example. — The width of a semi-circular arched vault is
10 feet, and its length 60 ; what is its volume?
102 X .7854-^2 =39.27 =area of semicircle of 10 feet span.
39.27 X 60 =235 6. 2 =product of area and length=volume required.
Ex. 2. The width of a semi-elliptic arched vault is 25 feet,
its height 17.5, and its length 40 feet ; what are its contents'?
Ans. 13744.5 cubic feet.
To ascertain the Internal ffurface of a Circular Groin.
Eule. — Multiply the area of the base by 1.1416, and the
product will give the surface required.
Or, a x 1.1416 ^surface.
Note. — The exact surface or volume of a groin is obtained by subtract-
ing from the sum of tfce surfaces or volumes of the two vaults composing
the groin, the surface or volume of the quadrantal arch formed by them.
Thus, in Example 1, the surface of a circular groin of 12 feet base is
as follows :
Surface of vault, 12 X 3.1416 -f- 2 = 18.8496, which X 12 and 2 =
452. 3904= sur/ ace of the two vaults.
122 X 2 =288 =surface of the quadrantal arch.
Hence, 452.3904-288= 164.3904 =surface.
M2
274 orophoids (Domes, Arched and Vaulted Roofs, etc.).
Example. — What is the surface of a circular groin, a side
of its square being 12 feet.
122X 1. 1416 ^1G4. 390 i= product of area of base and 1.14 16 ^surface
required.
Ex. 2. What is the surface of a circular groin, a side of
its square being 8 feet? Ans. 73.0624 feet.
•■%'
Note. — This rule may be observed in elliptical groins, as the error
or difference is too small to be regarded in ordinary practice.
To ascertain the Volume of a Circular or an Elliptical Groin.
Rule. — Multiply the area of the base by the height, and
the product again by .904, and it will give the volume re-
quired.
Or, axhx .904c=:volwne.
Example. — What is the volume of the vacuity or space
formed by a circular groin, one side of its square being 10
feet?
1 02X 5 X. 904= 452 =product of area of base, the height and .904 =
volume required.
Ex. 2. What is the volume of the vacuity formed by an
elliptic groin, one side of its square being 24 feet, and its
height 9 feet? Ans. 4686.336 cubic feet.
To ascertain the Internal Surface of a Triangular Groin.
Rule. — Ascertain the length of a side of the arch, multi-
ply it by twice the width of the vault, and the product will
give the surface required.*
Or, lxbx2=^surface.
Example. — The width of a triangular groin is 12 feet, and
its height 12 ; what is its internal surface?
-/(122 + 12H-2 ) = 13AlG4:=length of one side of arch.
13.4164x12x2=321. 9936=product of length of side and twice the
base=result required.
* See note, page 273.
orophoids (Domes, Arched and Vaulted Roofs, etc.). 275
To ascertain the Volume of the Materials that form the Groin.
Rule. — Multiply the area of the base by the height, inclu-
ding the work to the top of the groin, and from this product
subtract the volume of the vacuity ; the result will give the
volume required.
A General Rule for the Measurement of the Contents of Arches
is thus :
From the volume of the whole, considered as a solid, from
the springing of the arch to the outside of it, deduct the vacu-
ity contained between the said springing and the under side
of it, and the remainder will give the contents of the solid
part.
In measuring works where there are many groins in a
range, the cylindrical pieces between the groins, and on their
sides, must be computed separately.
When the upper sides of orophoids, whether vaults or groins, are
built up solid, above the haunches, to the height of the crown, it is
evident that the product of the area of the base and the height will
be the whole 'contents. And for the volume of the vacuity to be de-
ducted, take the area of its base, computing its mean height according
to its figure.
276 . BOARD AND TIMBER MEASURE.
BOARD AND TIMBER MEASURE.
To ascertain the Surface of a Board or Plank.
Rule. — Multiply the length by the breadth, and the prod-
uct will give the surface required.
Or, lxb=surface.
Note. — When the piece is tapering, add the breadths of the two ends
together, and take half the sum for the mean breadth.
Example. — The length of a plank is 16 feet, and its breadth
15 inches; what is its surface ?.
16 X 1.25(15) =20=product of length and breadth = surface required.
Ex. 2. The length of a plank is 25 feet, and its breadth
14 inches ; what is its surface? Am., 29.167 square feet.
Ex. 3. The length of a plank is 18 feet, and its widths at
the ends are 17 and 19 inches; what is its surface?
Ans. 27 square feet.
To ascertain the Contents of Squared Timber.
Rule. — Multiply the breadth by the thickness, and this
product by the length, and it will give the contents required.
Or, bxtxl=contents.
Example. — The length of a piece of square timber is 20
feet, its sides at its less end are 15 inches, and at its greater
end 19 ; what are its contents'?
19 + 15^2 = 17, and 172X20-M44 = 40.1388 ^product of square of
mean side and the kngth-^-lH to produce feet — contents required.
Ex. 2. The ends of a piece of timber are 18 and 22 inches
square, and the length of it is 22.5 feet ; what are its con-
tents? Ans. 62.5 cubic feet.
Note. — 1. If the piece tapers regularly from one end to the other, the
breadth and thickness, taken in the middle, will be the mean breadth
and thickness.
2. If the piece does not taper regularly, but is thicker in some places
than in others, take several different dimensions, and their sum, divided
by the number of them, will give the mean dimensions.
BOARD AND TIMBER MEASURE. 277
To ascertain the Contents of Round or Unsquared Timber.
Eule. — Multiply the square of one fifth of the girth by
twice the length, and the product will give the contents
nearly.* 2
Or, c-=-5 x 2l=z contents.
Example. — The diameter of a round piece of timber is
23^ inches, and its length 18 feet ; what are its contents'?
75 (circumference of 23-|)-f-5 = 15, and 152=225, and 225x18x2
=8100, which-j-144 =56.25 cubic feet. f
Ex. 2. Xhe circumference of a round piece of timber is 168
inches, and its length 15 ; what are its contents %
Ans. 235.2 cubic feet.
* The ordinary rule is to square one fourth of the girth, and multi-
ply it by -the length.
In order to show the fallacy of taking one fourth of the girth for the
side of a mean square, take the following example :
A diameter of 13.5 inches will give an area of 143.13. Hence, a
piece of round timber of this diameter will have nearly a square foot of
area of section. j.
The circumference for a diameter of 13.5=42.41, and 42.41-7-4 =
10.6, and 10.62 = 112.36 =area of section of timber.
By the above rule, 42.41-7-5=8.48, and 8.482 = 71.91, which X 2 =
143.82 =area of section of timber.
t When feet are multiplied by inches, -=-144 to obtain cubic feet.
278
APPENDIX TO MENSURATION OF SURFACES.
MENSURATION OF SURFACES.
To ascertain the Length of an' Elliptic Curve which is less than
half of the entire Figure, Fig. 1.
Fig. 1. >
Geometrically. — Let the curve of which the length is re-
quired be a b c.
Extend the versed sine b d to meet the centre of the curve in e.
Draw the line c e, and from e, with the distance e b, de-
scribe b h ; bisect c h in a, and from e, with the radius e i, de-
scribe k i, and it is equal to half the arc ab c.
To ascertain the Length when the Curve is greater than half the
entire Figure.
Rule. — Find by the above problem the curve of the less
portion of the figure, and subtract it from the circumference
of the ellipse, and the remainder will be the length of the
curve required.
Parabolic Spindle.
To ascertain the Surface of a Parabolic Spindle* Fig. 2.
Fig. 2.
* By Professor G. B. Docharty, New York Free Academy.
APPENDIX TO MENSURATION OF SURFACES. 279
Let o T>=Pf A o—q,
a o = p r =x, and
ap =z or —y.
Then, from the nature of the curve,
p r2 : A o2 :: Dr:Dc;
that is, x2:q2:\ p—y.p.
p x2 = q2 p—q2 y /
q2 y~p(q2— X2);
And, by differentiating,
_ 2pxdx _ _ 4»2a;2cZa:2
<*2/= *--= — ; .\dy2=-+—- .
q2 Q
Let d s represent the differential of the surface ; then,
ds=2 ny\/dx2-\-dy2.
Substitute for y its value, ^ (q2—x2), and for dy2 its value,
2
ldx2
, we have
9
4:p2x2dx2
~* , we have
rt4 '
^JL(tf.-Xi)\J dx
^ A 27TPS 1 iVi 4 A4 + 4j02a.2
Or,ds= — f- (q2— x2) dxy ±— — |
= <^^(q2-x2)dx(q*+4p2x2f
=^dx(q^+4p2x2f-?-^x2(q^4p2x2fdx;
.\s=^jqodx(q^4p2x2f-^^CqoX2dx(q^4p2x2f
=rP {Jfe4+^2^)i+41°g-^+^^T4^^)-^
Wp2q2-q* q^\
32 p* g'2pS
280 APPENDIX TO MENSURATION OP SURFACES.
Which is the formula for the surface of one half of the spin-
dle, orADc.
The logarithms indicated in the preceding and subsequent formula
are hyperbolical or Naperian.
If a table of them is not at hand, one of common logarithms may
be used instead, by dividing the common logarithm by .4343, and
using the quotient when the logarithm is indicated to be used in the
formula.
To ascertain the Surface of a Frustrum of a Circular Spindle,
Fig. 3'.
L i I
a d
Let R represent the radius of the circle, an arc of which
generates the spindle, as ad; D the distance of the centre
of the spindle from the centre of the circle, asao; d and d'
the distances of the two bases from the centre of the spindle,
as o c, os; h—d/^pd, the altitude, c s, of the frustrum, or the
distance between the bases.
d d/
Find z and zf from the formulae sin. z——, sin. zf ——.
AC xC
Find Z from the formula Z— — .
Find the surface from the formula S=2 7rR(A— D Z).
In the formulae for h and z, the upper sign is used when both
bases are on the same side of the centre of the spindle ; the
lower sign when the centre is between them. Z is the arc
which generates the frustrum expressed in units of the radius.
Corollary. — The entire surface of the spindle may be found
from the formula S = 2 7r R (I— D z), I being the length of the
spindle, and z being determined from sin. £*=— multiplied
by ttt^, and D from D^Rcos. \z.
APPENDIX TO MENSURATION OF SURFACES.
281
CENTRES OF GRAVITY OF SURFACES.
Semi- Spheroid or Ellipsoid, Fig, 4.
Fig. 4. g
Prolate.
f»3 y 3
Distance from C— ^ , in which C d, £fc radius of the
auxiliary circle, =zr=-
a-
-., a and b representing the semi-
Va2-b2
transverse and conjugate axes, y=.ef, and S=ara« of segment of
plane C g ef.
Oblate. = — _= 5 2-V ,
|(&vV2 + &2-r/2log.r'+r/2log.(\/r/2 + 62 + &))
£2
in which r'= _, 5 awd a representing the semi-conjugate
Va2 — b2
and transverse axes.
Segment of Semi- Spheroid or Ellipsoid, Figs, 5 and 6.
Fig. 5.
Prolate.
A
282 APPENDIX TO MENSURATION OP SURFACES.
3
Distance from C:
(r2_/2)7_(ra_ 2V-8T y3-y'3 .
ivhich y=g s,y' '—e o, and S—area of plane s g e o, /=Cs.
Oblate.
/
Distance from C —
r'* + b^-(r'* + P)%
(V
§(bVr' * + bi+r'2log.Vr'* + b* + b-l^r'2 + l2-r'*log.(Vr,* + l2+l))
b2
in which r'= — , l=Cs.
Va2-b*
Surface of a Frustrum of a Circular Sjpindle,
Fig. 7, p. 283.
r2 r'2
The distance from the centre of the spindle =—— — =r — -, r
z{h — D.z)
and r' being the radii of the two bases, e and s; h the distance
between the two bases ; D the distance of the centre of the spindle
from the centre of the circle, as a o ; z the generating arc, ex-
pressed in units of the radius.
'Fig. 7,
APPENDIX TO MENSURATION OF SURFACES. 283
Surface of a Segment of a Circular Spindle, as b c,
Fig. 8.
Fig. 8.
t
/
/
g h a d
r2
Distance from centre of spindle = r— ; — =r — r, r representing
2(h— D.z)
the radius of the base ; the other symbols the same as given on
page 282.
Note. — This last formula is essentially the same as the follow-
ing.
Segment of a Circular Spindle.
Distance from centre =-
-\q -1A~]
2 \_g — I — a (sin. sin. - J
_1 9 -1 1
sin. - and sin. - denoting the arcs, the sines of which are
r r
7
respectively - and - ; g=fo, l=zoc, r— radius of circle— ad, a
z=:ao=z Vr2—g2, and b — radius of end circle of segment
Paraboloid of Revolution.
(See Fig. 65, on page 126.)
tv * c 1 CJp2 + J2)*(8ft2-2pa) + 2jp6
Distance from vertex——— xr T — '— - - — * '^ r ,
b2
a — altitude, b — radius of base, and p = — .
284
APPENDIX TO MENSURATION OF SOLIDS.
MENSURATION OF SOLIDS.
Cycloidal Swindle.
To ascertain the Contents of the Frustrum of a Cycloidal Spin-
dle, b eco, Fig. 9.
Fig. 9.
e
~>N
L
f
iiiiiiiii!!
1>
|((2r- Z)^(2p+5^r+15Z^r2-15r3ver.sin. ^+15^r3))
dc
—contents, I representing ef, r——, and p, as before, 3.1416,
m
~H . /
ver. sin. — , symbol for the arc, the v. s. of which =—.
To ascertain the Contents of the Segment of a Cycloidal Spindle,
a be, Fig. 9.
(l5 r3 ver. sin. '--(2 r-$ (2 P + 5 fir + 15 /*r2))
P
6
tents.
con-
\ %
APPENDIX TO MENSURATION OF SURFACES. 285
Frustra of Spheroids, or Ellipsoids of Revolution,
e c df Figs. 104 and 104* page 182.
Distance from centre of spheroid.
_ J4 3d(2a*-d*) „tJ 4 3d(2P-o?)
Prolate^ g ^^ ; 0»A=i___J; arepresent-
ing the semi-transverse axis, h the semi-conjugate, and d the height
of the frustrum.
286
APPENDIX TO MENSURATION OF SOLIDS.
SOLIDS OF REVOLUTION.
To ascertain the Volume of a Solid of Revolution* Fig. 7.
Fig. 7.
Let A X, the axis of x, be the axis of revolution, and c m d
the generating curve, A n=:r, m n=y, the co-ordinates of any
point of the curve, and let the solid be terminated by planes
perpendicular to the axis, cutting it at o and s.
Let Ao=a, and As=b, the abscissae of these points; o c
— r, and s d=r', the radii of the two bases. The origin, A,
may be taken at any convenient point on A X.
The general formula,! when A X is the axis of revolution, is
V=zpfy2 d xy in which p — 3. 1416 ; f is the symbol of integra-
tion, and d that of the differential.
If, in the expression for V, y2 or d x be eliminated by means
of the equation of the generating curve, and the integration be
effected between the limits x=a and x— b, or y=r and y=r',
the value of V is determined.
Corollary. — If A Y, or the axis of y, is the axis of revolu-
tion, then,
Y=pfx2dy,
which differs from the preceding simply by the interchange
of the letters x and y.
* By Professor J. H. C. Coffin, U. S. N.
f This formula is thus read : The volume is equal to p times the in-
tegral of y- multiplied by the differential of x.
APPENDIX TO MENSURATION OF SOLIDS.
287
Example. To ascertain the Volume of a Cylinder, Fig. 8.
Fig. 8. A
c
r
0
m
i\
r'
s
The generating line, c m d, is a right line parallel to the
axis, AX; its equation is y=r=r/.
Whence V z=pr2fdx; and integrating between x=a and
x = b,
V=pr2(b—a); or, letting h= b— a(=o s), the altitude of
the cylinder,
V—pr2h; or, since pr2 = the area of the base,
The volume of a cylinder is equal to the area of its base multi-
plied by its altitude.
Example 2. To ascertain the Volume of the Frustrum of a
Cone with a Circular Base.
The generating curve is a straight line inclined to the axis.
Let h=:b— a( — o s), the altitude of the frustrum. The equa-
tion of the generating line is (the origin being at the vertex),
y = — r — x, whence dx=— dy, and
h r — r
V a= / y2dy; and integrating between y = r and y — r ',
ph(r/3~r2) .
Y=z 3(r/-^~; orreducing>
V=^hp(r2-±-r'2-\-rr'); i. e. (since pr2 and pr2 are the
two bases, and p r r' is a mean proportional between the two),
The volume of such a conical frustrum is equal to the sum of the
two bases and their mean proportional, multiplied by one third of
the altitude.
288
APPENDIX TO MENSURATION OF SOLIDS.
Or the following rule may be used :
Add together the squares of the radii of the two bases and the
product of those radii, and multiply the sum by one third of the
altitude and the number 3.1416.
Consequently, if the volume of the cone itself be required,
r'—o, and V— \hpr2 ; that is,
The volume of a cone is equal to its base multiplied by one
third of its altitude.
Example 3. To ascertain the Volume of a Spherical Segment,
Fig. 9.
Fig. 9. A
\
The generating curve, c m d, is an arc of a circle, and its
equation, if the origin be taken at the centre of the sphere, C, is
y2=~R2 — x2 (R being the radius of the sphere); whence,
V =pf(R2 —x2) d x; and integrating between x—a andx—b,
V=p[R2(b — a)— i(b3— a3)] ; or, letting h=b—a, the al-
titude of the segment, and substituting for R2 its value,
W2+r'2)+\{a2 + b2),
V=hp fa (r2 + r/2) + i h2~\ ; that is,
The volume of a spherical segment is equal to half the sum of
its bases -\- one sixth the area of the circle, the radius of which is
equal to the altitude, multiplied by the altitude.
Or the following rule may be used :
Add together the square of the altitude and three times the
squares of the radii of the two bases, and multiply the sum by one
sixth of the altitude and by the number 3.1416.
APPENDIX TO MENSURATION OP SOLIDS. 289
Corollary 1. If one of the bases=0, V=£ hp (3 r2 + h2) ;
that is,
The volume of a spherical segment with a single base is equal to
the sum of the area of a circle, the radius of which is the alti-
tude, and three times the base, multiplied by one sixth the alti-
tude. >
2. If both bases =0, the segment becomes the entire sphere,
and
V=^p h3 ; or, since A=2E, the diameter,
V=fpR2x^=f^R3; that is,
The volume of a sphere is equal to two thirds of a cylinder leav-
ing the same diameter and altitude.
Or, It is equal to four thirds of the cube of its radius, multi-
plied by 3.1416.
2. If both bases =0, the segment becomes the entire sphere,
and
V=^p h3 ; or, since ^2E=D, the diameter,
V=:%p~R2xh=%pTL3=ipT>3; that is,
T/ie volume of a sphere is equal to two thirds of a cylinder hav-
ing the same diameter and altitude.
Or, It is equal to four thirds of the cube of its radius, multi-
plied by 3.1416.
Or, It is equal to one sixth of the cube of its diameter, multiplied
by 3.1416.
Example 4. To ascertain the Volume of a Segment of a Pro-
late Spheroid.
The generating curve, c m d, is an arc of an ellipse ; and
if the origin be taken at the centre, C, the equation is
B2
y2=.-— (A2— x2), in which
A
A and B are respectively the semi-transverse and semi-con-
jugate axes of the ellipse. Whence,
B2 r
Y=zp— I (A2— x2)d x; and integrating between x= a
and x=zb,
N
290 APPENDIX TO MENSURATION OF SOLIDS.
B2
Y=p — [A2(b-a)-i(P-a3y];or,lettiiagh=b-a(=os),
.A.
the altitude of the segment, substituting for A2 its value,
A2
2 "r2 (r2+r/2)+i («2+&2)> an<* reducing,
B2 h2~\
Y=hp[^(r2+r^)+ — .-J; that is,
The volume of such a segment is equal to half the sum of the bases
4- one sixth of the area of the circle the radius of which is the alti-
/B'2\
tude, multiplied by (-Tg J> the square of the ratio of the axes of
the generating ellipse, multiplied by the altitude.
Or the following rule may be used :
Multiply the squares of the altitude and semi-conjugate axis to-
gether ; divide by the square of the semi-transverse axis ; add to-
gether the quotient and three times the squares of the radii of the
two bases ; and multiply the sum by one sixth of the altitude and
by the number 3.1416.
Corollary 1. If one of the bases =0, the expression for the
[r2 B2 h2~\
"2+A2'"6_r
2. If both bases =:0, the segment is the entire spheroid, and
the altitude A=2A; and
V=;?B2x^; or,
The volume of such a spheroid is equal to -| the volume of a
cylinder of the same altitude, the base of which is equal to the mid-
dle section of the spheroid.
Example 5. To ascertain the Solidity of a Segment of an Ob-
late Spheroid.
A2
The equation is y2=— (B2— x2) ; whence
A2 C
V—p^ l(B2—x2)dx; and integrating between x—a
and x—b, and reducing, as in Example 4,
v=AKiO-2+>-'2)+p.J}
APPENDIX TO MENSURATION OF SOLIDS. 291
Example 6. To ascertain the Volume of a Segment of an Hy-
perboloid of Revolution.
B2
The equation is y2z=—(x2— A2) ; whence
A.
B2 C •
V=p-r^ I (x2—A2)dx; and integrating and reducing, as
in Examples 4 and 5,
B2 h2~\
V=Ai»[i(r»+r")-5; -gj-
Example 7. To ascertain the Volume of a Segment of a Par-
aboloid.
The equation is y2 = 2 ~Px, in which 2 P=the parameter.
V=2 Fpfxdx; or, integrating between x=.a and x—b,
V^~Pp(b2 — a2); or, since r2=2P«, and r/2=2PJ, and
h~b— a,
Y=^hp(r2+r'2).
Another method of determining the volume of a solid of
revolution is to ascertain the area of the generating surface
and the distance of its centre of gravity from the axis of rev-
olution.
Let A=zthe area osdc (Fig. 7).
g—the distance of its centre of gravity, G, from the axis
AX; then +,
Y=z2pgxA; that is,
The volume is equal to' the area of the generating surface mul-
tiplied by the circumference of the circle described by its centre of
gravity.
292 APPENDIX TO MENSURATION OF SOLIDS.
CENTRES OF GRAVITY.
To ascertain the Centre of Gravity of a Solid of Revolution.
The centre of gravity is upon the axis of revolution, and
it is necessary to determine only its distance from some par-
ticular point, as, for example, the vertex, or the intersection
of the axis by a base, or the origin of co-ordinates, A, Fig. 7.
As before, Let AX, the axis ofx, be the axis of revolution, etc.
(See page 286.)
Let G/ be the centre of gravity of the solid, and AGc'=g.
The general formula is ■ _
pfy2xdx fy2xdx
9=z V = fy2dx'
If y or x and d x be eliminated by means of the equation
of the curve which generates the surface, and the integration
be effected between the limits x=a and x=b, or y=r and
y—rf, the distance of the centre of gravity, G', from the ori-
gin of co-ordinates is determined.
Example. To ascertain the Centre of Gravity of a Cylinder
with a Circular Base. (See Fig. 8, page 287.)
The equation is y=r=r/ ; whence
r 2 I'x d x
<7= — J- — - — , and integrating between the limits x=a and
x=b,
b2—a2
g=^-rr -—^(b-\-a) ; or, if the origin be taken at o,
g=^b = half the altitude.
Example 2. To ascertain the Centre of Gravity of the Frus-
trum of a Cone with a Circular Base. (See Example 2, p. 287.)
The equation is y— — ; — x; whence x=—z y, and d x
h r —r"
=-7 dy; and
APPENDIX TO MENSURATION OP SOLIDS. 293
fij3dy
and integrating,
r'—r' fy2dy
3 h (V/4 r4)
g- — —, /3 ; and reducing,
* 4(r — r)(r3— H)
3ft(r/a + r2)(r/+r)
0- 4(r^3_r3)
IPtoce, To ascertain the distance of the centre of gravity from
the pertex of the cone,
Multiply together the sum of the radii of the two bases, the
sum of their squares, and f the altitude, and divide the product by
the difference of the cubes of those radii. .
Corollary 1. If the distance from the greater base (or 5G')
be required,
r/4_4/r3 + 3r4 r/24-2rr' + 3r2
_x; (r/+r)a + 2f8
-4/i'(r/+r)2_rr/'
2. For the entire cone, r=0, and
<7=-f ^, or g'=\h; that is,
T^e distance of the centre of gravity from the vertex is equal to
2- the altitude ; or, from the base, is equal to J the altitude.
Example 3. To ascertain the Centre of Gravity of a Spher-
ical Segment. (See Example 3, page 288.)
Equation of curve, y2=~R2— x2 ; whence xdx= —ydy; and
pf—y3dy r'4 — r4 y/4_r4
9=~ ~~V =4^(r/2+r2)+^2] = 2%/2 + r2+^2];
Hence, To ascertain the distance of the centre of gravity of the
segment from the centre of the sphere,
Take the difference of the Aih powers of the radii of the bases
as a dividend; and for the divisor, multiply the sum of the squares
of the radii and ^ the square of the altitude by twice the altitude ;
the quotient is the distance required.
The centre of gravity is between the centre of the sphere
and the lesser base.
294 APPENDIX TO MENSUBATION OP SOLIDS.
Corollary. For a segment with a single base, r=0 ; and
,•* (R—lh)2
<7— — fT3— "^5 — TT~> ^e distance from the centre of
£Ae sphere. "
Example 4. To ascertain the Centre of Gravity of a Frustrum
of a Prolate or Oblate Spheroid.
The distance from the centre of the spheroid is
A2
(r'*—rA)
n(d±d/)(2A2-d/2-d2) p
T*2 J/>_j/,7 3T- i0r a
••5
2A(r 2+r2+^-— A2)
prolate spheroid.
B2
A2V ; ^d±:d')(2B2-d"2-d2) r
9~9i,<'^_ 2^iAV~* 3W-d'* + d'd-d* t0ran
oblate spheroid, A cmd B representing semi-transverse axes, d
and a" respectively the distances from the centre of the spheroid to
the base and end of the frustrum. If both these are on the same
side of the centre, the upper sign is used, but if they are on differ-
ent sides, the lower sign is used.
Example 5. To ascertain the Cenfre of Gravity of a Segment
of a Prolate or Oblate Spheroid.
(j± ± h)2
<7=^-r — t"T~j f°r a Pr°late spheroid.
g=-^5 — TT"? tor an oblate spheroid.
-fc> Tf tl
Frustrum of Hyjperboloid of Revolution.
^. * „ 11M Ad'+d){d'2+d2-2a2)
Distance trom.centre of hyperboloid=:f — ^ — -,- — 2 ,
a— semi-transverse axis, d— distance from centre to base of seg*
ment.
APPENDIX TO MENSURATION OF SOLIDS. 295
Segment of Hyperboloid of Revolution, Fig. 12, p. 247.
Distance from centre of hyperboloid (point of intersection
of the diameters t a and d f)=% — -, a and das before,
and d/ —distance from centre to base of frustrum.
Frustrum of a Paraboloid of Revolution.
Distance from vertex of paraboloid =§ —, — , d
representing height of paraboloid, and d/ the distance between the
frustrum and vertex.
296 APPENDIX TO MENSURATION.
For the mensuration of Spherical Triangles and Pyra-
mids, Spirals, Epi-cycles, Epi-cycloids, Cardioids, Heli-
coids, Peli-coids, etc., etc., see Loomis's Analytical Geometry
and Calculus; Davies and Peck's Dictionary of Mathematics;
Docharty's Geometry; Hackletfs Geometry.
For a Glossary and Explanation of Geometrical Figures,
see Davies and Peck, and the Library of Useful Knowledge,
vols. i. and ii.
carpenters' slide-rule. 297
CARPENTERS' SLIDE-RULE, OR GUNTER'S LINE.
This instrument is commonly called a Sliding Rule. It is
constructed of two pieces of box- wood or ivory, of a foot in
length each, connected together by a joint, which enables them
to be folded up lengthwise upon an edge.
On one side, the whole rule is divided upon its outer edges
into inches and eighths, for the purpose of taking dimensions,
and upon its inner edges there are several plane scales, each
divided into twelve parts, which are designed for the reduction
of dimensions for the purposes of drawing.
On the other side there is a metallic slide, and four lines
marked A, B, C, and D, the two middle, B and C, being
upon the slide.
Three of these lines, A, B, and C, are doubled in their di-
mensions ; that is, they proceed from 1 to 10 twice ; and the
fourth line, D, is a single one, running from 1 to 10, and is
called the line of roots*
The use of the double lines, A and B, is for operations in
arithmetic, and ascertaining the areas of plane figures.
Upon the other part of this face there is a table of gauge
points for the ascertaining and measurement of the different
elements and substances there given, under the respective col-
umns of Square, Circular, and Globe ; and the gauge point to
be selected for operation is determined by the denomination
in which the dimensions are given ; thus, if in three dimen-
sions, and all in feet, that under F.F.F. is to be taken ; if one
dimension is in feet and the others in inches, that under F.I.I.
is to be taken ; and if all are in inches alone, that under I.I.I,
is to be taken. Again, if of two denominations, or of one
only, the gauge point is to be taken from under F.I., 1. 1., or
F. and I., as the case may be.
* Upon some rules, the fourth line, D, is divided from 4 to 40, in
which case it is termed a girth-line, and is then used in the measure-
ment of timber.
N2
298 carpenters' slide-rule.
The divisions on the first, second, and third lines are mark-
ed alike, each beginning at 1, which may be 1, 10, 100, or
1000, etc., or .1, .01, .001, etc. ; but, whatever it is assumed
to be, the second or middle line of these divisions must be
taken at 10 times as many as the first, and the third line
must be taken at 10 times as many as the second.
Numeration is the first operation to be acquired upon this
instrument, for when that is understood, all other operations
will become quite easy. It is first to be observed that the
values of the divisions upon the rule are all arbitrary, and the
value set upon them must be such as will meet the require-
ments of the question, which must be determined when a
question is proposed.
The principal divisions indicated by figures at 1, 2, 3, 4,
and so on to 10, are termed primes, and the next divisions
tenths, and these again are, or may be, subdivided into hund-
redth and thousandth parts.
If the 1 (next to the joint) represents one tenth, then will
the middle 1 be a unit, or a whole number, and the other
figures toward the right hand are likewise whole numbers,
from the middle 1 to 10 at the end; but if the first 1 rep-
resents a unit, then the middle 1 will be 10, and the 10 at
the right hand 100 ; if the first 1 represents 10, the middle 1
will be 100, and that at the right hand 1000; always in-
creasing in a tenfold proportion, according to the value set
upon the first 1. The figures between 1 and 10 are designated
after the same manner ; that is, if 1 at the beginning is one
tenth, 2 will be two tenths, and the next 2 toward the right
hand 2 units ; but if 1 at the beginning is 1 unit, then 2 will
be 2 units, and the other 2 will be 20 ; if the first 1 is called
10, then 2 will be 20, and the next 2 is 200, etc. When the
1 at the left hand is taken 88. 1, or one tenth, the line is read
in the following order: 1, 2, 3, 4, 5, 6, 7, 8, 9 tenths, unity
or 1 ; 2, 3, 4, 5, .6, 7, 8, 9, 10; when a higher value is set
on them, they will read thus, beginning next the joint, 10, 20,
30, 40, 50, 60, 70, 80, 90, 100, 200, 300, and so on to 1000.
carpenters' slide-rule. 299
To multiply two Numbers.
Set 1 on A (1st line) to either of the given numbers on B
(2d line), then at the other number on A (1st line) will be
found the product on B (2d line).
Example. — Multiply 12 by 25.
Under 1 on A put 12 on B, and under 25 on A is 300 on B.
Ex: 2. Multiply 64 by 15. Ans. 960.
Note. — If the third terra runs beyond the end of the line, look for it
on the first radius or other part of the line, and increase it ten times.
To divide one Number by another.
Set 1 on A to the divisor on B, then at the dividend on B
will be found the quotient on A.
Example. — Divide 300 by 25.
Under 1 on A put 25 on B, and at 300 on B is 12 on A.
Note. — If the dividend runs beyond the end of the line, diminish it
10 or 100 times, as may be required, to make it fall upon A, and in-
crease the quotient accordingly.
To ascertain a Fourth Proportional.
Set the first term upon A to the second on B, then at the
third on A is the fourth on B.
Example. — What is the fourth proportional to 8, 20,
and 30 "?
Under 8 on A put 20 on B, then at 30 on A is 75 on B.
To ascertain a Mean Proportional.
Set the first term on C to the same term on D, then at the
second on C is the mean on D.
Example. — What is the mean proportional between 20
and 80 !
Under 20 on C set 20 on D, and at 80 on C is 40 on D.
'■j
300 carpenters' slide-rule.
Bute of Three Direct.
In the Rule of Three Direct, there are three numbers given
to find a fourth, that shall have the same proportion to the
third as the second has to the first.
The operation is, as the first term upon A is to the second
upon B, so is the third term upon A to the fourth upon B.
Or, bring the first term upon B to the second upon A ;
then against the third upon B is the result upon A.
Example. — If a man can walk 20 miles in 5 hours, how
long will he require to walk 125 miles at the same rate"?
Over 20 upon B put 5 upon A, and against 125 upon B is 31.25, the
result, upon A.
Rule of Three Inverse.
In this rule, there are three numbers given to find a fourth,
that shall have the same proportion to the second as the first
has to the third.
Note. — If more requires more, or less requires less, the question he-+
longs to the rule of three Direct ; but if more requires less, or less re-
quires more, it then belongs to the rule of three Inverse.
Example. — If 6 men can do a certain piece of work in 8
days, how many will it require to perform the same in 3 days ?
Note. — In inverse proportion, thex slide is to be inverted (by with-
drawing it and introducing the opposite end of it) ; then the question
will be operated in the same way as in direct proportion.
Invert the slide in the groove, and over 8 upon C set 6 upon A;
then at 3 upon C is 16, the result, upon A.
Vulgar and Decimal Fractions.
To reduce a Vulgar Fraction to its equivalent Decimal Expression.
The operation is, as the denominator upon A is to 1 upon
B, so is the numerator upon A to the decimal required upon B.
Example. — Reduce i to a decimal.
Set 1 upon B to 4 upon A; then at 1 upon A is .25., the result,
upon B.
carpenters' slide-rule. 301
To extract the Square Root.
When the lines C and D are equal at both ends, C is a ta-
ble of squares, and D a table of roots ; consequently, opposite
to any number or division upon C is its square root upon D.
Example. — If a tower 30 feet in height is on the side of a
river which is 40 feet in width, what must be the length of a
ladder that will reach from the opposite side of the river to the
top of the tower %
Operation. — Set the slide even at both ends, and over 30 upon D is 90
on C, and at 40 on D is 160 upon C, which, when added to 90, =250;
then under 250 upon C is 50 upon D, the length of the ladder.
To Square a Number.
Set 1 upon D to 1 upon C ; then against the number upon
D will be found the square upon C.
To Cube a Number.
Set the number upon C to 1 or 10 upon D, and against the
same number upon D will be its cube upon C.
Set 6 upon C to 10 upon D, and at 6 upon D is 216 upon C.
Zand Measuring.
The Gauge points for measuring land are the number of
square chains, square perches, and square yards that are con-
tained in an acre. If the dimensions are given in chains, the
gauge point is 1, 10, 100, etc., upon A ; if in perches, it is 1G0 ;
but if it is given in yards, the gauge point is 4840, which the
length upon B must always be set to, and opposite the breadth
upon A is the result, in acres and parts, upon B.
Example. — If a field is 20 chains 50 links in length, and
4 chains 40 links broad, how many acres does it contain ?
Set 20.5 upon B to 1 upon A, and at or under 4.4 upon A is 9 upon
B=*#e result in acres.
302 carpenters' slide-rule.
Mensuration of Solids.
In measuring and weighing solid bodies, the tables of Gauge
points upon the rule are always to be made use of, and are
thus explained.
1. All gauge points are taken on the line A.
2. All lengths must be noted on the line B, and are to be
set to the gauge point on the line A.
3. All squares and diameters are found on the line D.
4. Opposite the square or diameter on the line D is the
content or result on the line C ; or, as the length upon B is
to the gauge point upon A, so is the square or diameter upon
D to the content upon C.
There are three gauge points for every element that is giv-
en in the table for square ; F.F.F. signifying that when the
length and both the squares are feet, the gauge point is to be
found under F.F.F. in the same line with that of the element
or material to be measured or weighed.
If the length is given in feet, and both the squares are
inches, then the gauge point is under F.I.I. ; but if the dimen-
sions of both length and square are in inches, then the gauge
point is under I.I.I.
There are two gauge points for every thing that is to be
measured or weighed of a cylindric form ; first, when the
length is in feet and the diameter in inches, the gauge point is
under F.I. If the length is in inches and the diameter in
inches, then the gauge point is under I.I.
There are two gauge jjoints to weigh or measure every thing
of a globular figure.
A globe having but one dimension, it must be either in feet
or inches ; if it is in feet, the gauge point is under F. ; if it is
in inches, the gauge point is under I.
Note. — In measuring or weighing square timber, stone, metals, or
any other bodies that are unequal sided, a mean j>roportion must be first
estimated, in order to obtain a true cube or square.
carpenters' slide-rule. 303
The general rule for a globe is, as the gauge point on A is
to the diameter on B, so is the content on C to the diameter
on D ; or, set the diameter upon B to the gauge point upon
A, and against the diameter upon D is the result upon C.
Example. — If a piece of timber is 16 inches broad, 6 inches
thick, and 20 feet long, how many cubic feet does it contain !
Ascertain the mean square by setting 16 upon C to 16 upon D, and
opposite to 6 upon C is 9.8 upon D, the side of a square equal to 16 by 6;
having thus ascertained the true square, look for the gauge point for
cubic feet, and under E.I.I, is 144 ; set 20, the length, upon B to 144
upon A, and against 9.8 upon D are 13.3 cubic feet upon C.
Round timber is generally measured by the girth, which
is ascertained by a line run around the middle of the tree or
log, and taking one fifth of the girth for the side of the square.
This is not precisely correct, but it is the method commonly
practiced, and is performed on this rule, after the girth is
taken, as follows.
Example. — A round log is 30 feet in length, and 40 inches
around its middle; how many cubic feet does it contain?
Here one fifth of the girth is 8 inches.
Multiply the length, 30, by 2, and set the result, 60, upon B to 144
upon A, and against 8 upon ~D is 26.6, the result, in feet, upon C.
The circumference of the above log being 40 inches, the diameter is
12.73 inches.
Set 30, the length, upon B, to 1833, the gauge point, upon A, and
against 12.73 upon D are 26.5 feet, the result, upon C. By this it will
be seen that there is but a slight difference between the customary and
true methods of measuring.
Cylinder, Globe, and Gone.
Example. — If a cylinder is 6 inches long, and 6 inches in
diameter, how many cubic inches does it contain ?
The gauge point for cubic inches is 1273.
Set 6 upon B to 1273 upon A, and against 6 upon D are 169 cubic
inches, the result, upon C.
304 carpenters' slide-rule.
Cask Gauging.
The gauging of casks is performed, after a mean diameter is
found, exactly in the same manner as in the last examples.
Casks are generally reduced to what is termed four Varieties;
and their mean diameters may be found by multiplying the dif-
erence between the head and bung diameters of the first variety
by .7, the second by .63, the third by .50, and the fourth by .52.
The respective products of these numbers, added to the head
diameter, will give the mean diameter.
Set the length upon B to the gauge point upon A, and
over the mean diameter on D is the result upon C.
Example. — In a cask of the first variety, the head diameter
is 24, the bung diameter 28, and the length 30 inches ; how
many gallons will it contain %
Set 30 upon B to 353, the gauge point for the imperial measure, upon
A, and against 26.8, the mean diameter, upon D is 61, the resuk in gal-
lons, upon C.
Ex. 2. How many gallons are contained in a cask of the
second variety, the head diameter 18, the bung diameter 23, and
length 28 inches ?
Set 28 upon B to the gauge point upon A, and against 21.15, the
mean diameter, upon D is 35.2, the result in gallons, upon C.
Ex. 3. If a cask of the third variety is 20. inches at the head,
26 at the bung, and 29 inches long, what will be its contents
in gallons ?
Set 29 upon B to the gauge point upon A, and against 23.36, the
mean diameter, upon D is 44.5, the result in gallons, upon C.
Ex. 4. A cask of the fourth variety is 34 inches long, the
head diameter 26, and the bung diameter 32 ; how many
gallons will it hold ?
Set 34 upon B to the gauge point upon A, and against 29.12, the
mean diameter, on D is 81.5, the result in gallons, upon C.
carpenters' slide-rule. 305
Miscellaneous Questions.
Under this head may be introduced a great many original
questions, as well as such as could not be introduced in regu-
lar order in the foregoing rules.
Of a Circle.
The Diameter being given, to ascertain the Area ; or the Area
being given, to ascertain the Diameter.
Set .7854, the area of unity, upon C to unity or 10 upon
D, then the lines C and D will be a table of areas and diam-
eters ; for against any diameter upon D is the area in square
inches upon C.
The Circumference being given, to ascertain the Area ; or the
Area being given, to ascertain the Circumference.
Set .0795 upon C to 1 or 10 upon D, then the lines C and
D will be a table of areas and circumferences ; for against any
circumference upon D is the area in square inches upon C.
The Circumference being given, to ascertain the Diameter; or
the Diameter being given, to ascertain the Circumference.
Set 1 upon B to 3.141 upon A, then tfce lines A and B will
be a table of diameters and circumferences ; for against any
diameter upon B is the circumference upon A.
To ascertain the Side of a Square equal in Area to any given
Circle.
Set .886 upon B to 1 upon A, then against any diameter
of a circle upon A is the side of a square that will be equal in
area upon B.
To ascertain the Side of the greatest Square that can be inscribed
in any given Circle.
Set .707 upon B to 1 upon A, and against any diameter of a
circle upon A is the side of its greatest inscribed square upon B.
306
carpenters' slide-rule.
To ascertain the Side of the greatest Equilateral Triangle that can
be inscribed in any given Circle.
Set 1 upon B to 115 upon A, and against any diameter
of a circle upon A is the length of a side of a triangle upon B.
The Weights of Metals, and various other Results, may be ob-
tained in a similar manner, for the rules of operation of which,
reference is given to the Book of Directions usually furnished
with the slide-rule.
„ 0
Illustrations.
What is the weight of a piece of cast iron 3 inches square
and 6 feet long !
The cast iron gauge point for feet long and inches square is 32. See
the rule.
Set 6 upon B to 32 upon A, and against 3 upon D is 168, the result,
in pounds, upon C.
A cylinder is 6 inches in length and 6 inches in diameter ;
what is its weight in cast iron, wrought iron, and brass?
1st. For cast iron, the gauge point is 489.
Set 6 upon B to 489 upon A, and against 6 upon D is 44 upon C.
2d. For wrought irori, the gauge point is 453.
Set 6 upon B to 453 upon A, and against 6 upon D is 47.5 upon C.
3d. For brass, the gauge point is 424.
Set 6 upon B to 424 upon A, and against 6 upon D is 511, the result,
ur on C.
TABLE OF ADDITIONAL GAUGE POINTS.
Square.
Cylinder.
Globe.
F.F.F.
F.I.I.
1. 1. 1.
p.i.
i.i.
p.
i. 1
Oak
174
15
155
591
795
8
193
252
217
243
85
115
115
278
303
2605
269
102
138
138
333
320
276
31
116
147
146
354
386
333
342
13
176
126
424
332
286
296
113
152
153
369
578
Mahogany
Box
49
512
195
263
264
637
Marble
Brick
GAUGING. 307
CASK GAUGING.
The operation of cask gauging is ordinarily performed with
the aid of five instruments, viz., a Gauging Slide-rule, a" Gauging
or Diagonal Rod, Callipers, a Bung Rod, and a Wantage Rod.
THE GAUGING SLIDE-RULE.*
The Gauging Slide-rule is a. flat rule, very similar to an or-
dinary slide-rule, except that it is not jointed, and its being
adapted for use for the purpose of measuring and gauging
casks, in addition to those of the ordinary computations effect-
ed by a slide-rule.
Upon the plain or outer face there are jive lines; the first
three are alike, being equally divided, and all *>f the same ra-
dius,! and each containing twice the length of one.
The fourth line is differently divided from the others, and
is used in the operation of gauging, in the determination of
the contents of casks when Lying, by the element of the depth
of liquor within them, which is termed the wet inches.
The fifth line is similarly divided to the fourth, and is used
in the operation of gauging, in the determination of the con-
tents of casks, when Standing, by the element of the depth of
liquor within them, which is also termed the wet inches.
Note. — The operation of gauging in this manner — that is, by the
element of wet inches — is termed Ullaging.
Upon the opposite or inner face there are four lines ; the
first js divided to represent gallons,{ the second is a line of
mean diameters, and the third and fourth lines are divided into
inches and tenths.
* As manufactured by Belcher, Brothers, & Co., New York.
t The first three lines are divided alike to the ordinary carpenters'
slide-rule, or Gunter's line, described at page 297, and the operations
of multiplication, division, etc., etc., may be performed, by inspection,
as there described. »
J 231 cubic inches, which is the U. S. standard gallon.
308 GAUGING.
The third line is divided each way from the thumb-piece,
running from 38 to G3 to the right, and from 0 to 12 to the
left.
The use of this line is to measure the diameter of the head
of a cask, and is thus operated :
Place the outside edge of the brass shoulder on the right
end'of the gauge, on the head of a cask, and close to the inside
of a stave in line with the centre of the head ; move the thumb-
piece until its perpendicular or left face is in a line with a
point at the other end, which would give the diameter of the
head on its inside ; then remove the gauge, and on the fourth
line, under the face of the thumb-piece, read off the diameter
of the head in inches and tenths.
If the diameter of the head exceeds 38 inches, then it is to
be found on the third line, at the left end of the gauge, on the
line running from 38 to 63.
On the left of the thumb-piece is a scale of 12 inches and
tenths of inches, and above it is a Scale of 1st Varieties ; that
is, varieties of the first form of casks, the use of which is here-
after explained under the head of Varieties.
The fourth line is divided into inches and tenths, running
from the right to the left, and is used for the purposes of or-
dinary measurement.
Upon one side of the instrument is a Scale of 2d Varieties,
the use of which, and of all like scales, is to obtain by inspec-
tion the mean diameter of casks of the different varieties of fig-
ure, and which are thus classed.
Varieties of Casks.*
First Variety. — Casks of the ordinary form, being that of
the Middle Frustrum of a Prolate Spheroid, as Fig.l.
* The basis of determination of a scale of varieties is that of giving
a multiplier whereby the mean diameter of arcask may be ascertained,
and the operation is effected as shown on page 310.
GAUGING.
309
Rum puncheons and whiskey barrels are fair exponents of this form,
which comprises all casks having a spherical outline of stave.
Second Variety. — Casks of the form ot the Middle Frustrum
of a Parabolic Spindle, as Fig. 2.
Fig. 2. a
Wine casks are exponents of this form, which comprises all casks in
which the curve of the staves quickens slightly at the bilge.
Third Variety. — Casks of the form of the Middle Frustrum
of a Paraboloid, as Fig. 3.
Fig. 3. a
Brandy casks and provision barrels are exponents of this form, which
comprises all casks in which the curve of the staves quickens at the
chime.
310
GAUGING.
Fourth Variety. — Casks of the form of two equal Frustrums
of Cones, as Fig. 4.
Fig. 4. a
A gin pipe is an exponent of this form, which comprises all casks in
which the curve of the staves quickens sharply at the bilge.
As the rule, however, is provided with but scales of two
varieties, it is usual to apply the first scale to all casks in
which the middle diameter (intermediate between the bung
and head), as g h, Fig. 1, approaches nearest to that of the
bung diameter.
The scale of 2d variety upon the edge of the rule is adapt-
ed to all casks in which the middle diameter approaches next
or second in order of proportion to that of the bung diameter,
as Fig. 2.
Scales of 3d and 4th varieties are not given. Such scales,
however, are wanted, and are applicable to all casks in which
the middle diameter bears a less proportion to the bung diam-
eter than in any of the other varieties.
% To ascertain the Mean Diameter of a Cash
Kule. — Subtract the head diameter from the bung diameter
in inches, and multiply the difference by the following units
for the four varieties ; add the product to the head diameter,
and the sum will give the mean diameter of the varieties re-
quired.
1st variety 7 I 3d variety .56
2d variety 63 I 4th variety 52
GAUGING.
311
Example. — The bung and head diameters of a cask of the
1st variety are 24 and 20 inches ; what is its mean diameter?
24 _ 20 = 4, and 4 X .7 = 2.8, which, added to 20, = 22.8 inches, the
mean diameter.
Ex. 2. The bung and head diameters of a cask of the 2d
variety are 23 and 20 inches ; what is its mean diameter %
Ans. 21.89 inches.
Operation by the Gauging Slide-rule.
Example 1. Subtract 20 from 24, and over 4, the difference, on the
line of inches of the scale of 1st varieties, read 2.76, which, added to 20,
=22.76, the result required.
Ex. 2. Subtract 20 from 23, and under 3, the difference, on the line
of inches of the scale of 2d varieties, read 1.88, which, added to 20, =
21.88, the result required.
THE GAUGING OK DIAGONAL ROD.
The Gauging or Diagonal Bod is a square rule having four
faces, being commonly four feet long. This instrument is used
both for gauging and measuring casks, and in gauging, the con-
tents are ascertained from one dimension only, viz., the diagonal
of the cask, or the length from the centre of the bung-hole to
the junction of the head of the cask with the stave opposite
to the bung, being the longest straight line that can be drawn
within a cask from the centre of the bung. Accordingly, on
two opposite faces of the rule are scales of inches for measur-
ing this diagonal, between which, on a third face, are placed
the contents in gallons.
To ascertain the Contents of a Cask by the Gauging or Diagonal
Bod, Fig. 5.
Fig. 5. a
I / : '
/ i
/
<4r 4x _j-
312 GAUGING.
Operation. — Introduce the pointed end of the rod into the
bung-hole of a cask (the plane face of the rod being upper-
most) until it reaches the junction of the head and bottom of
the cask at its lowest point ; then adjust the upper end of the
rod, so that the under side (divided into gallons) shall be in
the centre of the hole in a line with the under side of the
bung stave. Observe this . point, by the aid of the divisions
of inches and tenths on either of the sides of the rod ; with-
draw it, and, in a line with the division observed, read off on
the under side of the rod the contents of the cask in gallons.
Illustration. — In the preceding figure, the bung diameter, a b, is 24
inches, the head diameter, ef, is 20 inches, and the half length, /r,
is 18 inches.
Hence, by Geometry, the height, a r, of the triangle afr is 24—
24. 20
— - — =22. Consequently, the length =22, and the base = 18, the hy-
pothenuse, a f, =28.425. Then, by inspection of the rod, it will be
seen that in a line with 28.425 inches is 63, the number of gallons the
cask contains.
THE CALLIPERS.
The Callipers is a sliding rule adapted to project over the
chimes of casks to measure their inner length, and when it is
adjusted to the heads of a cask, the inner length of it may be
read off, a difference of 2 inches, being an allowance of 1 inch
for the thickness of each head, being provided for in the divis-
ions of the rule.
Note. — When the thickness of the heads is known to differ from an
inch each, the difference above or less than an inch, as the case may
be, is to be subtracted from or added to the length indicated by the cal-
lipers.
THE BUNG ROD.
The Bung Rod is alike to the diagonal rod in construction,
and is a rod for determining the inner diameter of a cask, or
the wet inches therein ; and in order to enable the divisions to
be accurately noted, there is a slide running around the rod,
with a collar upon its lower end, which is brought to bear
GAUGING.
313
upon the under side of the stave at the bung ; the rod, with
the slide retained in position, is then removed, and the divis-
ions on the rod read off.
Note. — It is customary to combine this instrument with the diagonal
rod, the Inches on the latter answering all the purposes of the measure-
ment required, and the slide is removed or adjusted as may be required.
THE WANTAGE ROD.
The Wantage Rod is a scale having four equal sides, upon
which are divisions adapted to the many varieties of vessels,
as barrels, tierces, and hogsheads.
The use of this instrument is to obtain by inspection the
number of gallons of liquor deficient in a cask, when the de-
ficiency does not reach to an extent that would class the ves-
sel as an ullage cask.
There is a metal collar upon one of its sides, which is intro-
duced into the bung-hole of a cask until its upper side is. at
the under side of the bung stave ; the instrument is then re-
moved, and the wet line indicates the number of gallons (un-
der the designation on the rod, of the cask to which it is ap-
plied) the cask is deficient, or wants of being full.
To ascertain the Contents of a Cask of the 1st Variety, Fig. 6.
Fig. 6. a
By Mensuration.
Rule. — To twice the square of the bung diameter, a b, add
the square of the head diameter, e f; multiply this sum by
the length, c d, of the cask, and the product again by .2618,
O
314 GAUGING.
and it will give the contents in cubic inches, which, being di-
vided by 231, will give the result in gallons.*
Example. — The bung and head diameters of a cask, a b and
ef, are 24 and 20 inches, and the length, c d, 36 ; what are
its contents in gallons ?
242 X 2 + 203 = 1552, which X 36 = 55872, and 55872 X .2618 =
14627.2896, which -f-231 = 63.32, the gallons required.
Ex. 2. The bung and head diameters of a cask are 36
and 30 inches, and the length 54 ; what are its contents in
gallons? Ans. 213.7 gallons.
By the Gauging Slide-rule.
Operation. — Subtract the head diameter from the bung di-
ameter, and look for the difference on the lower line (inches)
of the scale of 1st varieties, and above it, on the second line,
is the mean difference, which is to be added to the head diam-
eter'for the mean diameter.
Thus, 24— 20 = 4 = difference of diameters.
Above 4, on 1st line of scale, read 2.75, which, added to 20, =22.75,
the mean diameter required.
Then, set the left end of the slide on the inner face of the rule to the
length of the cask (36) on the first line ; look for the mean diameter
(22.75) on the second line (or top line of slide), and above it, on the first
line, read 63.6, which is the capacity of the cask in gallons.
Ex. 2. The bung and head diameters of a cask are 36 and
30 inches, and the length 54 ; what are its contents in gal-
lons?
The difference of diameters is 6, which, by the scale, =4. 16.
-. Ans. 214.6 gallons.
Ex. 3. The length of a cask is 51 inches, and its bung and
head diameters 31 and 26 inches; what are its contents in
gallons? Ans. 150.6 gallons.
* Whenever the contents are required in bushels, divide by 2150.4.
GAUGING.
315
To ascertain the Contents of a Cash of the 2d Variety, Fig. 7.
Fig. 7. a
By Mensuration.
Rule. — To the square of a head diameter add double the
square of the bung diameter, and from the sum subtract -^
of the square of the difference of the diameters ; then multiply
the remainder by the length, and the product again by .2618,
which, being divided by 231, will give the contents in gallons.
Example. — The bung and head diameters of a cask, a b and
ef, are 24 and 18 inches, and the length, c d, 36 ; what are
its contents in gallons'?
182+242x 2 = 1476, and 1476—^. of 24-18 =1461.6, which X 36
= 52617.6, and 52617.6 X. 2618 = 13775.288, which-^23 1=59.63, the
gallons required.
By the Gauging Slide-rule.
Operation. — Subtract the head diameter from the bung di-
ameter, and look for the difference on the upper line (inches)
of the scale of 2d varieties, and below it, on the second line,
is the mean difference, which is to be added to the head diam-
eter for the mean diameter.
Thus, 24 — 18 =%= difference of diameters.
Below 6, on 2d line of scale of 2d varieties, read 3.8, which, added
to 18, =21.8, the mean diameter required.
Then, set the left end of the slide on the inner face of the rule to the
length of the cask (36) on the first line; look for the .mean diameter
(21.8) on the second line (or top line of slide), and above it, on the first
line, read 58.2, which is the capacity of the cask in gallons.
316
GAUGING.
Ex. 2. The bung and head diameters of a cask are 36 and
27 inches, and the length 54 ; what are its contents in gallons?
The difference of diameters is 9, which, by the scale, =5.77.
Ans. 197.5 gallons.
Ex. 3. The length of a cask is 51 inches, its bung and head
diameters 30 and 26 inches ; what are its contents in gallons?
Ans. 141.3 gallons.
To ascertain the Contents of a Cask of the 3d Variety, Fig. 8.
Fig. 8. a
By Mensuration.
Rule. — To the square of the bung diameter add the square
of the head diameter; multiply the sum by the length, and
the product again by .3927, which, being divided by 231, will
give the contents in gallons.
Example. — The bung and head diameters of a cask, a b and
e /, are 24 and 20 inches, and the length, c d, 36; what are
its contents in gallons?
242+203x 36=35136, which X. 3927 =1379 7.907, and 13797.907-r-
231=59.73, the result required. .
By the Gauging Slide-rule.
Operation. — Subtract the head diameter from the bung di-
ameter, and multiply the difference by the unit .56, page 310,
which is to be added to the head diameter for the mean diam-
eter.
Thus, 20-24=4, which x. 56=2.24, and 20 + 2.24=22.24, the mean
diameter required.
GAUGING.
317
Then, set the left end of the slide on the inner face of the rule to the
length of the cask (36) on the first line ; look for the mean diameter
(22.24) on the second line (or top line of slide), and above it, on the first
line, read 60.16, which is the capacity of the cask in gallons.
Ex. 2. The bung and head diameters of a cask are 36 and
30 inches, and the length 54 ; what are its contents in gallons 1
The difference of diameters is 6, which, by the rule, =3.36.
Ans. 204.8 gallons.
Ex. 3. The length of a cask is 51 inches, its bung and head
diameters 31 and 26 inches ; what are its contents in gallons?
Ans. 144.9 gallons.
Ex. 4. The length of a cask is 50 inches, and its bung and
head diameters 30 and 25 inches ; what are its contents 1
Ans. 131.4 gallons.
To ascertain the Contents of a Cask of the 4th Variety, Fig. 9.
Fig. 9. a
By Mensuration.
Rule. — Add the square of the difference of the diameters
to three times the square of their sum ; then multiply the sum
by the length, and the product again by .06566, and it will
give the contents in cubic inches, which, being divided by 231,
will give the result in gallons. ■
Example. — The bung and head diameters of a cask, a b and
ef are 24 and 16 inches, and the length, c d, 36 ; what are
its contents in gallons ?
24-16 + (24 + 16)2x3=4864, which X 36 = 175104, and 175104 X
.06566 = 11497.329, whiclK-231 =49.77, the gallons required.
318 GAUGING.
By the Gauging Slide-rule.
Operation. — Subtract the head diameter from the bung di-
ameter, and multiply the difference by the unit .52, page 310,
which is to be added to the head diameter for the mean diam-
eter.
Thus, 24-16 = 8, which X. 52 =4.16, and 16+4.16=20.16, the mean
diameter required.
Then, set the left end of the slide on the inner face of the rule to the
length of the cask (36) on the first line : look for the mean diameter
(20.16) on the Second line (or top line of slide), and above it, on the first
line, read 49.8, which are the contents of the cask in gallons.
Ex. 2. The bung and head diameters of a cask are 36 and 24
inches, and the length 54 ; what are its contents in gallons?
The difference of diameters is 12, which, by the rule, =6.24.
Ans. 168.5 gallons.
Ex. 3. The length of a cask is 51 inches, its bung and head
diameters 31 and 23 inches ; what are its contents in gallons
and in bushels? Ans. 128.2 gallons,
, 128.2x231 10_, __
and ^^ An —13.77 bushels,
2150.42
To ascertain the Contents of a Cask when the Dimensions are less
than the Divisions on the Scale as numbered, viz., for a Length
of less than 25 inches, and a Mean Diameter of less than 17.3
inches.
Operation. — Determine the mean diameter; then double
both the length and the mean diameter; ascertain the con-
tents for those dimensions, and divide the result by 8.
Note. — When only one of the dimensions is doubled, divide the re-
sult by 4.
Example. *-The dimensions of a barrel of the 2d variety,
having bung and head diameters of 12 and 10 inches, is 16
inches in length ; what are its contents ?
Mean diameter = 11.26 inches.
Set the left end of the slide to 32 (16x2) on the 1st line, and over
22.52 (11.26x2), on the second line, is 55.36 on the 1st line, which
-j-8=6.92, the contents of the barrel in gallons.
GAUGING. 319
ULLAGE CASKS.
To ascertain the Contents of Ullage Casks.
When a cask is only partly filled, it is termed an ullage cash,
and is considered in two positions, viz., as lying on its side,
when it is termed a Segment Lying (S.L.), or as standing on
its end, when it is termed a Segment Standing (S.S.).
To Ullage a Lying Cask.
By Mensuration.
Divide the wet inches by the bung diameter ; find the quo-
tient in the column of versed sines in the table of circular seg-
ments, page 134, a.id take out its corresponding segment ;
then multiply this segment by the capacity of the cask in gal-
lons, and the product again by 1.25 for the ullage required.
Example. — The capacity of a cask is 90 gallons, the bung
diameter being 82 inches ; what are its contents, at 8 inches
depth?
8-^32 = .25, the tab. seg. of which is .15355, which X 90 = 13.81 £5,
which X 1.25 = 17.2744, the contents, in gallons.
By the Gauging Slide-rule.
Operation. — On the plane face,
Set the bung diameter on 3d line to 100 at the right hand
on 4th line, and on this line, under the wet inches on 3d line,
take off the number, and set it (by moving the slide) on 3d
line to 100 on 4th line, and under the capacity of the cask on
the 1st line read the contents in gallons required.
Thus, set 32 on 3d line to 100 on 4th line, and under 8 on 3d line
take off 17.75 on 4th line, and set it on 3d line at 100 on 4th line ; then
under 90 on 1st line read 16.85, the contents in gallons.
Ex. 2. The capacity of a cask being 92 gallons, and the
bung diameter 32, required the ullage of the segment when
the wet inches are 8. Ane. 17.85 gallons.
Ex. 3. The wet inches in a lying cask are 12 inches, the
bung diameter 24, and the capacity of the cask 70 gallons ;
what are the contents of the cask? Ans. 35.2 gallons.
320 GAUGING.
To Tillage a Standing Cask.
By Mensuration.
Add together the square of the diameter at the surface of
the liquor, the square of the head diameter, and the square
of double the diameter taken in the middle between the two ;
then multiply the sum by the wet inches (length between the
surface and nearest end), the product again by .1309, and
divide by 231 for the result in gallons.
Example. — The diameter at the surface of the liquor is 29
inches, the head diameter of the cask is 24, the diameter taken
in the middle of the two is 27, and the depth of the liquor, or
wet inches, is 20 ; what are the contents of the cask %
292+242+27x2 =4333, which X 10=43330, and 43330 X.1309-
231=24.554, the result, in gallons.
By the Gauging Slide-rule.
Operation. — On the plane face,
Set the length on 3d line to 100 at the right hand on 5th
line, and on this line, under the wet inches on 3d line, take off
the number, and set it (by moving the slide) on 3d line to 100
on 5th line, and under the capacity of the cask on the 1st line
read the contents, in gallons, required.
The capacity of the cask being 94.5 gallons, and the length 35 inches.
Thus, set 35 on 3d line to 100 on 5th line, and under 10 on 3d line
take off 26.8 on 5th line, and set it at 100 on 5th line ; then under
94.5 on 1st line read 25.3, the contents, in gallons.
Ex. 2. The length of a standing cask is 24 inches, the wet
inches 12, and the capacity of it 64 gallons ; what are its
contents in gallons'? Ans. 32.28 gallons.
Ex. 3. The length of a standing cask is 24 inches, the wet
inches 16, and the capacity of it 65 gallons; what are its
contents in gallons? Ans. 44.6 gallons.
GAUGING. 321
To ascertain the Contents of a Cask by four Dimensions.
Kule. — Add together the squares of the bung and head
diameters, and the square of double the diameter taken in the
middle between the bung and head ; then multiply the sum by
the length of the cask, and the product by .1309 ; then divide
this product by 231 for the result in gallons.
Example. — What are the contents of a cask, the length of
which is 40 inches, the bung diameter 32, the head diameter
24, and the middle diameter between the bung and the head
28.75 inches?
322+242 = 1600=s«ro of squares of bung and head diameters.
28.75x2 =3306.25, and 3306.25 + 1600 = 4906.25=sw7« of squares of
bung and head diameters and of double the middle diameter.
4906.25 X 40 = \§&2b0= product of above sum and the length of the ca??:
Tlien, 196250 X. 1309 =25689.125 =number of cubic inches in the casic,
which-f-231 = 11 1.2083 gallons.
Ex. 2. The bung and head diameters of a cask are 24 and
16 inches, the middle diameter 20.5, and the length of it 36
inches ; what are its contents in gallons ?
Ans. 51.26 gallons.
To ascertain the Contents of any Cash from three Dimensions only.
Rule. — Add into one sum 39 times the square of the bung
diameter, 25 times the square of the head diameter, and 26
times the product of the two diameters ; then multiply the
sum by the length, and the product by .008726 ; then divide
the quotient by 231 for the result in gallons.
Example. — The length of a cask is 35.5 inches, its bung
diameter 28.3 inches, and its head diameter 24 inches ; what
are its contents in gallons ?
28.32X 39 =31234.71 =39 times the square of the bung diameter.
242 X 25 = 14400 = 25 times the square of the head diameter.
28.3 X 24 X 26 = 17659.2=26 times the product of the two diameters.
31234.71+14400 + 17659.2 = 63293.91, which x 35.5 = 2246933.8 =
the sum of the above products X the length, which X .008726=19606.744 =
number of cubic inches, which-f-231 =84.88 —gallons required.
- 02
322 GAUGING.
Ex. 2. What are the contents of a cask, the length being
40, and the bung and head diameters 32 and 24 inches?
Ans. 112.273 gallons.
Note. — This is the most exact rule of any for three dimensions, and
agrees nearly with the result as determined by a diagonal rod.
Illustration. — If a diagonal rod was applied to a cask of the
dimensions above given (Ex. 2), the length or distance determ-
ined by it would be 34.4.
An inspection of the rod will show 110.1 gallons to be in-
dicated at this point.
Note. — The Divisor for English ale gallons is 282, and for Imperial
gallons 277.274.
THE END.
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OF 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO 50 CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.
DEC 21
fcEC'D LD
MAR Z 9 laoO
: MAY 1 0 1970 12
" ' iiilMMI* i&~
MAY 6f?Q
|OAH 0EPAJTO*a«
S^^^^i^
OCT l 5 ZGOZ
U. C. BERKELEY
LD 21-100m-7,'33
OA.
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