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PRACTICAL
MODEL CALCULATOR,
ENGINEER, MECHANIC, MACHINIST,
MANUPACTUEER OF ENGINBWOEK, NAVAL ARCHITECT,
MINER, AND MILLWRIGHT.
OLIVER BYRNE,
and Editor of Out "DUUtmar^ of Machina, MechaniiSi Sngiiuwyrk, tini Engtrteer
nrof "ISe (XmsanOm for MaAinUU, J/wJanici, und Ei\gintar>i' AvOicTimi Incinii
if a A'eio &lea«, (crnml " Tin OifcuftM iif Jbrm," a luisliiiite far (fts diffirenlM
aiid In/rgral Oolculus ; " The EWmtnts of Euclid by Coiwirs," and numemju
othiT Jfff(fiemfl«caZ and Sfwhanicol Works. SunnyorGentral of the
E«^h SetOemenls in (As Iblkhad Jjfcs, Profeimr iff
Malli£itiatici, G^gt of Cioit Enffmurs, LoR^en^
PHILADBLPniAL
PUBLISHED BY HENRY CAREY
(SUCCESSOR TO E. L. CAKEY,)
hv Google
ling to Uw ut of CopethSt Id tbe yc
HENRT CARET BAIRD,
le District Court for tho EaEtccn Bist
hv Google
THE
PRACTICAL MODEL CALCULATOR.
WEIGHTS Am) UEASUBES.
TilE UNIT OF LENGTH.
The Yaed. — If a pendulum vibrating seconds in vacuo, in Phi
ladelphia, be divided into 2509 equal parts, 2310 of Bueh equal
parts is the length of the standard yard ; the measures are taken
on brass rods at the temperature of 32° Fahrenheit. This yarJ
will not be in error the tenmiiiionth part of an inch,
2310 : 2509 as 1 to 1086142 nearly.
THE trail OP WEIGHT.
The Pound, avoirdupois, is 27*7015 cubic inches of distilled
water, weighed in air, at the temperature of maximum density,
39°82; the barometer at 30 inches,
THE LIQUID UNIT.
The Qallon, 231 cubic inches, contains 83388822 pounds avoir
dupois, equal 583721754 grains troy of distilled water, at 39°S2
Pah. ; the barometer at 30 inches.
■UNIT OP DRY CAPACITY.
The Bushel contains 215042 cubic inches, 77627412 pounds
avoirdupois, 54339189 grains of distilled water, at the temperature
of maximum density; the barometer at 30 inches.
The French unit of length or distance is the metre, and is the
tenmillionth of the quadrant of the globe, measured from the
equator to the pole.
The French ilfeire = 32808992 English /eei linear measure =
393707904 inches.
For MuUiples the following Greek
words are used :
De.ea for 10 times.
Hecto — 100 times.
Kilo — 1000 times.
Slyria— 10000 times.
Millimetre
For Divisors the following Latin
words are used :
Bed for the lOM part.
C'enti — lOOM part.
MiUi — 1000i/( part.
Thas 3. Kilometre = 1000 metres.
metro
1000
The square Deea Metre, called the Are, is the element of land
easure in France, which = 107642996 square feet English.
The Stere is a cubic m_etre = 35316582 cubic feet English.
hv Google
6 THE PRACTICAL MODEL CALCULATOR.
The Litre for liquid measure is a cubic decimetre = 1760TT
imperial pints English, at the temperature of melting ice ; a litre
of distilled water weighs 15434 grains troy.
The unit of weight is the gramme : it is the weight of a cuhic
centimetre of distilled water, or of a millilitre, and therefore equal
to 15'434 grains troy.
The kilogramme is the weight of a cuhic decimetre of distilled
water, at the temperature of maximum density, 4° centigrade.
The pound troy contains 5760 grains.
The pound avoirdupois contains 7000 grains.
The English imperial gallon contains 277"274 cuhic inches ; and
the English com bushel contains eight such gallons, or 2218*192
cuhic inches.
apothecaries' weight.
Grains marked gr.
20 Grains make 1 Scruple ■— sc. or 3
3 Scruples— 1 Dram — dr. or 5
8 Drams — 1 Ounce — oz. or ,?
12 Ounces — 1 1'ound — lb. or ft.
gr. sc.
20 = 1 dr.
60 = 3=1 OK.
480 = 24 = 8 = 1 Ih.
5700 = 288 = 96 = 12 = 1
This is the same as troy weight, only having some different
divisions. Apothecaries make use of this weight in compounding
their medicines ; but they buy and sell their drugs by avoirdupois
weight.
Drams marked dr.
16 Drams make 1 Ounce — oz.
16 Ounces — 1 Pound — lb.
28 Pounds — 1 Quarter — qr.
4 Quarters — 1 Hundred Weight... — cwt.
20 Hundred Weight... — 1 Ton — ton.
dr. oz.
16 = 1 lb.
25(5 = 16 = 1 qr.
7168 = 448 = 28 = 1 cwt.
28672 = 1792 = 112 = 4 = 1 ton.
673440 = 35840 = 2240 = 80 = 20 = 1
By this weight are weighed all things of a coarse or drossy
nature, as Corn, Bread, Butter, Cheese, Flesh, Grocery Wares,
and some Liquids ; also all Metals except Silver and Gold.
Oz. Dwt. Gr.
JWe, that 1 lb. avoirdupois = 14 11 15i troy.
1 0/. — = 18 5i —
1 dr. — = 1 Si —
hv Google
■WEIGHTS AND MEASURES.
Grains marked Gr.
24 Grains make 1 Pennyweight Divt.
20 Pennyweights 1 Onnce Oz.
12 Ounces 1 Pound Lb.
Gr. Dwt.
24 = 1 Oz.
480 = 20 = 1 Lb.
5760 = 240 = 12 = 1
By tliia weight arc weighed Gold, Silver, and Jewels.
LONa MEASURE.
Barleycorns make 1 Inch marked In.
  ■ . _ Ft.
laches.
Feet
Feet
Yards and
Furlongs
Miles
G Milca nearly..,
In.
12 =
198 =
7920 =
63360 = 5280
1 Foot...
. ITard
1 Fathom
. 1 Pole or Bod..
■ 1 Furlong
■ IMilo
■ 1 League
■ 1 Degi'ce
Yd.
Fth.
PI.
Far.
Mile.
Lea.
Deji or °
Ml
5^ = 1 Fur.
. 220  40 = 1 Mile.
 1760  320  8  1
CtOTH HEASl'EE.
2 Inches and a quarter.... make 1 Kail mariied Kl.
4 Kails ..
3 Quarters
4 Quarters
5 Quarters
4 Qrs. 1', Inoli..
Qr.
EF.
Yd.
EE.
BS.
SQUARE MEASURE.
..make 1 Sq. Foot marked Ft.
1 Quarter of a Yard..
1 Ell Flemish..
1 Yard
1 Ell English...
1 Ell Scotch ...
ISq. Yard..
1 Sq. Pole ..
lEood
1 Acre
Yd.
Pole.
. Yil.
Sq. PI.
144 Square Inches.
9 Square Feet —
30J Square Yards —
40 Square Poles —
4 Roods —
Sq. Inc. Sq. Ft.
144  1
1296 = 9 =
39204  272i .
1568160  10890 .
6272640  43560 .
When three dimensions are concerned, namely, length, breadth,
and depth or thickness, it is called cubic or solid measure, which is
used to measure Timber, Stone, &e.
The cubic or solid Foot, which is 12 inches in length, and breadth,
and thickness, contains 1728 cubic or solid inches, and 27 solid
feet make one solid yard.
. 1210 = 401
. 4840  100 = 4
Acr.
b,Google
THE PRACTICAL MODEL CALCULATOR.
, OR COEd MEASURE.
2 Pints
2 Quarts ...
2 Pottles...
2 Gallons...
8 Bushels...
5 Quarters...
2 Weya
jl Quart marked Qt.
1 Pottle..
1 Gallon
IPeck
1 Bushel
1 Quarter,
1 Weigh or Load...
1 Last
Pec.
1
Bu.
Qr.
612 = 64 = 32 = 8  1
2560 = 320 = 160 = 40 = 5 =
5120 = 640 = 320 = 80 = 10 ■■
^YcJ.
Pot.
Gal.
Pec.
Wey.
Last.
2 Pints.
2 Quarts — 1 G(
42 Gallons — 1 Tierce
63 Gallons or IJ Tier.. ■— 1 Hogshead.
i Tierces — 1 Puncheon.
.make 1 Quart marked Qt.
■ ■ Gal.
Tier.
Hhd.
Pu
2 Hogsheads...
2 Pipes
— 1 Pipe or Butt..
— ITun
Pi.
Tun.
Pts.
2 =
Qta.
1 Gal.
4 = 1 Tier.
168 = 42 = 1 Hhd.
252 = 63 = IJ = 1 Pun.
504 =
672 =
1008 = 504 = 126
2016 = 1008 = 252
 IJ = 1 Pi.
= 2 = li. = 1 Tun.
2 Pints
4 Quarts
36 Gallons —
1 Barrel and a half.... —
2 Barrels —
2 Hogsheads —
2 Butts ~
Pts. Qt.
2 = 1
8 = 4 =
288 = 144 =
ALE AKD BEER IIEASUUE.
make 1 Quart marked Qt.
1 Gallon . .
1 Barrel
1 Hogshead...
1 Puncheon..
]. Butt
ITun
Gal.
Bar.
Hhd.
Pun.
Butt.
Tun.
Gal.
Ear.
= 1 Hhd.
= 11=1 Butt.
=3 2=1
hv Google
OF TIME.
60 Seconds make 1 Minute marked M. or
eOMiEutes — 1 Hour — Hr.
24 Hours — 1 Day — Day.
7 Days — 1 Week — AVk.
4 Weeks — l.Montk — Mo.
1 3 Months,! Day, 6 Hours, 1 i t r v v
or oo6 Days, b Hours, j
See. Min.
60 = 1 Ilr.
3600 = 60 = 1 Day.
86400 = 1440 = 24 = 1 Wk.
604800 = 10080 = 168 = 7=1 Mo.
2419200 = 40320 = 672 = 28 =4 = 1
3155T600 = 525960 = 8T66 = 365^ = 1 Year.
Wk.Da.Hr. Mo. Da.Hr.
Or 52 1 6 = 13 1 6 = 1 Julian Year.
Da. Hr. M. See.
But 365 5 48 48 = 1 Solar Year.
The time of rotation of the earth on its axis is called a sidereal
day, for the following reason : If a permanent object be placed on
the surface of the earth, always retaining the same position, it may
be so located as to be posited in the same plane with the observer
and some aolected fixed star at the same instant of time; although
this coincidence may be but momentary, still this coincidence con
tinually recurs, and the interval elapsed between two consecutive
coincidences has always throughout all ages appeared the same.
It is this interval that is called a sidereal day.
The sidereal day increased in a certain ratio, and called the
mean solar day, has been adopted as the standard of time.
Thus, 366256365160 sidereal days = 866256365160  1 or
365256365160 mean solar days, whence sidereal day : mean solar
day : : 365256365160 : 366266365160 : : 0997269672 : 1 or as
1 : 1002737803, when 23 hours, 56 imnutes 40996608 sec. of
mean solar time = 1 sidereal day; and 24 hours, 3 minutes,
565461797 see. of sidereal time = 1 mean solar day.
The true solar day is the interval hetween two successive coinci
dences of the sun with a fixed object on the earth's surface, bring
ing the sun, the fixed object, and the observer in the same plane.
This interval is variable, but is susceptible of a maximum and
minimum, and oscillates ahont that mean period which is called a
mean solar day.
Apparent or true time is that which is denoted by the sundial,
from the apparent motion of the sun in its diurnal revolution, and
differs several minutes in certain parts of the ecliptic from the
mean time, or that shown by the clock. The difference is called
the equation of time, and is set down in the almanac, in order to
ascertain the true time.
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ARITHMETIC.
Arithmetic is the art or Bcience of numbering; being that
branch of Matlicmatics which treats of the nature and properties
of numbers. When it treats of whole numbers, it is called Com
mon Arithmetic ; but when of broken numbers, or parts of num
bers, it is called Fractiom.
Unity, or a Unit, is that by which every thing is called one ;
being the beginning of number ; as one man, one ball, one gun.
JVumher is either simply one, or a compound of severul units ;
as one naan, three men, ten men.
An Integer or Whole Number, is some certain precise quantity
of units ; as one, three, ten. These are so called as distinguiohed
from Fractions, which are broken numbers, or parts of numbers ;
as onehalf, twothirda, or threefourths,
NOTATIOIT AND NTTMERATION.
NOTATIOH, or Numeration, teaches to denote or express any pro
posed number, either by words or characters ; or to read and write
down any sum or number.
The numbers in Arithmetic are expressed by the following ten
digits, or Arabic numeral figures, which were introduced into
Europe by the Moors about eight or nine hundred years since :
viz. 1 one, 2 two, 3 three, 4 four, 5 five, 6 six, 7 seven, 8 eight,
9 nine, cipher or nothing. These characters or figures were
formerly all called by the general name of Ciphers; whence it
came to pass that the art of Arithmetic was then often called
Ciphering. Also, the first nine are called Significant Figures, as
distinguished from the cipher, which is quite insignificant of itself.
Besides this value of those figures, they have also another, which
depends upon the place they stand in when joined together ; as in
the following Table:
6
7 6
7 6 5
7 (5
8 7
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NOTATION AKD KUMERATIOS.
11
Here any figure in the first place, reckoning from right to left,
denotes only its own simple value ; but that in the second place
denotes tea times its simple value ; and that in the third place
a hundred times its simple value ; and so on ; the value of any
figure, in each successive place, being always ten times its former
value.
Thus, in the number 1796, the 6 in the first place denotes only
six units, or simply six ; 9 in the second place signifies nine tens,
or ninety; 7 in the third place, seven hundred; and the 1 in the
fourth place, one thousand ; so that the whole number is read thus —
one thousand seven hundred and ninetysix.
As to the cipher 0, it stands for nothing of itself, but being
joined on the righthand side to other figures, it increases their
value in the same tenfold proportion ; thus, 5 signifies only five ;
but 50 denotes 5 tens, or fifty ; and 500 is five hundred ; and
so on.
For the more easily reading of large numbers, they are divided
into periods and halfperiods, each halfperiod consisting of three
figures ; the name of the first period being units ; of the second,
millions ; of the third, millions of millions, or bimillions, contracted
to billions ; of the fourth, millions of millions of millions, or tri
millions, contracted to trillions; and so on. Also, the first part
of any period is so many units of it, and the latter part so many
thousands.
The following Table contains a summary of the whole doc
Periods. Quadrill.; TriUions; Billions; Millio
Halfper.
th. un. th. un.
Figures. 123,456; 789,098; 765,432; 101,234; 667,890.
KuMBRATiON 13 the reading of any number in words that is pro
posed or set down in figures.
Notation is the setting down in figures any number proposed in
words.
OF THK EOMAN NOTATION.
The Romans, like several other nations, expressed their numbers
by certain letters of the alphabet. The Romans only used seven
numeral letters, being the seven following capitals : viz. I for one ;
YioT five; Xiorten; Lfor fifty ; Cfor a, hundred; D for five hun
dred ; M for a thousand. The other numbers they expressed by
i repetitions and combinations of tliesc, after the following
hv Google
TUB PRACTICAL MODEL CALCULATOR.
 ir.
= 111.
= nil.
= v.
= VI.
= VIT.
= VIII.
= IX.
= x.
= h.
>c.
= DorI
3 M or C
= MM.
= Vorl
= yi.
= L or
= LX.
= C_^or
= M^or
= MM.
kc.
As often as any character is repeated,
BO many times is its value repeated,
A less character before a greater
diminishes its value.
A less character after a greater in
creases its value.
For every annexed, thia
ten times as many.
For every C and 0, placed one at
end, it becomes ten times as much.
A bar over any number incr
1000 fold.
eh
it
CCIOO.
1000.
ccciooo.
ccccioooo.
50^
100 =
500 =
1000 =
2000 =
5000 =
6000 =
10000 =
50000 =
60000 =
100000 =
1000000 =
2000000 =
&c.
EXPLANATION OF CERTAIN CHARACTERS.
There are various characters or marks used in Arithmetic and
Algebra, to denote several of the operations and propositions ; the
chief of which are as follow :
+ signifies phis, or addition.
— minus, or subtraction.
X multiplication.
j division.
Thus,
5 + 3, denotes that 8
G — 2, denotes that 2
7x3, denotes that 7
8 i 4, denotes that 8
: :: : proportion.
= equality.
V' square root.
^ cube root, kc.
to he added to 5 = S.
to be taken from 6^4.
to be multiplied by 3 = 21.
to be divided by 4 = 2.
6, shows that 2 is to 3 as 4 is to 6, and thus, 2x6=3x4.
6 + 4 = 10, shows that the sum of 6 and 4 is equal to 10.
■/Z, or 3 J denotesthesquareroot ofthenumher 3 — 1'7320508.
■^5, or 5*, denotes the cube root of the number 5 = 1709976.
7^ denotes that the number 7 is to be squared = 4f>.
8', denotes that the number 8 is to be cubed = 512.
hv Google
ETILE OF THREE. 13
RVIE OP IHE£)£.
The Role of Three teaches how to find a fourth proportional
to three numbers given. Whence it is also sometimes called the
Rule of Proportion. It is called the Rule of Three, because three
terms or numbers are given to find the fourth ; and because of its
great and extensive usefulness, it is often called the Golden Rule.
This Rule is usually considered as of two kinds, namely, Direct
and Inverse,
The Rule of Three Direct is that in which more requires more, or
less requires less. As in this : if 3 men dig 21 yards of trench in
a certain time, how much will 6 men dig in the same time ? Here
more requires more, that is, 6 men, which are more than 3 men,
will also perform more work in the same time. Or when it is thus :
if 6 men dig 42 yards, how much will 3 men dig in the same time ?
Here, then, less requires less, or 3 men will perform proportionally
less work than 6 men in the same time. In both these cases, then,
the Rule, or the Proportion, is Direct; and the stating must be
thus. As 3 : 21 : : 6 : 42,
or thus, As 6 : 42 : : 3 : 21.
But, the Rule of Three Inverse ia when more requires less, or
less requires more. As in this : if 3 men dig a certain quantity
of trench in 14 hours, in Low many hours will 6 men dig the like
quantity? Here it is evident that 6 men, being more than 3, will
perform an equal quantity of work in less time, or fewer hours.
Or thus : if 6 men perform a certain quantity of work in 7 hours,
in how many hours will 3 men perform the same ? Here less
requires more, for 3 men will take more hours than 6 to perform
the same work. In both these cases, then, the Rule, or the Pro
portion, ia Inverse ; and the stating must be
thus, As 6 : 14 : : 3 : 7,
or thus. As 3 : 7 : : 6 : 14.
And in all these statings the fourth term is found, by multiply
ing the 2d and 3d terms together, and dividing the product by the
1st term.
Of the three given numbers, two of them contain the supposi
tion, and the third a demand. And for stating and working ques
tions of these kinds observe the following general Rule :
Rule, — State the question by setting down in a straight line the
three given numbers, in the following manner, viz. so that the 2d
term be that number of supposition which is of the same kind that
the answer or 4th term ia to be ; making the other number of sup
position the 1st term, and the demanding number the 3d term,
when the question is in direct proportion ; but contrariwise, the
other number of supposition the third term, and the demanding
number the 1st term, when the question has inverse proportion.
Then, in both cases, multiply the 2d and 3d terms together, and
divide the product by the first, which will give the answer, or 4th
term sought, of the same denomination as the second term.
hv Google
14 TEE PRACTICAL MODEL CALCULATOR.
Note, If the first and third tenna consist of different denomina
tioRB, reduce them both to the same ; and if the second terra be a
compound number, it is mostly convenient to reduce it to the lowest
denomination mentioned. If, after division, there be any remainder,
reduce it to the next lower denomination, and divide by the same
divisor as before, and the quotient will be of this last denomina
tion. Proceed in the same manner with all the remainders, till
they be reduced to the lowest denomination which the second term
admits of, and the several c^uotients taken together will be the
answer required.
Note also. The reason for the foregoing Rules will appear when
we come to treat of the nature of Proportions. Sometimes also
two or more statings are necessary, which may always be known
from the nature of the question.
An engineer having raised 100 yards of a certain work in
24 days with 5 men, how many men must he employ to finish a
like quantity of work in 15 days t
da. men. da. men.
As 15 : 5 : : 24 : 8 Ans.
5
15)iM(8Answer,
120
COMPOITND PROPORTION.
Compound Proportion teaches how to resolve such questions as
require two or more statings by Simple Proportion ; and that,
whether they be Direct or Inverse.
In these questions, there is always given an odd number of terms,
either five, or seven, or nine, &c. These are distinguished into
terms of supposition and terms of demand, there being always one
term more of the former than of the latter, which is of the same
kind with the answer sought.
Rule. — Set down in the middle place that term of supposition
which is of the same kind with the answer sought. Take one of
the other terms of supposition, and one of the demanding terms
which is of the same kind with it; then place one of them for a
first term, and the other for a third, according to the directions
given in the Rule of Three. Do the same with another term of
supposition, and its corresponding demanding terra ; and so on if
there be more terms of each kind ; setting the numbers under each
other which fall all on the lefthand side of the middle term, and
the same for the others on the righthand side. Then to work.
By several Operations. — Take the two upper terms and the mid
dle term, in the same order aa they stand, for the first Rule of
Three question to be worked, whence will be found a fourth term.
Then take this fourth number, so found, for the middle term of a
second Rule of Three question, and the next two under terms in the
.1 stating, in the same order as they stand, finding a fourth
hv Google
OF COMMON FRACTIONS.
15
term from them ; and so on, as far as there are any numbers in the
genera! stating, making always the fourth number resulting from
each simple stating to he the second term of the next following one.
So shall the last resulting number be the answer to the question.
By one Operation. — Multiply together all the terms standing
under each other, on the lefthand side of the middle term ; and, in
like manner, multiply together all those on the righthand side of
it. Then multiply the middle term by the latter product, and
divide the result by the former product, so shall the quotient be
the answer sought.
How many men can complete a trench of 135 yards long in
8 days, when 16 men can dig 54 yards in 6 days ?
Creneral stating.
yds. 54 ; 16 men ; : 135 yds.
days 8 6 days
482 810
16
4860
810
432)12960(30 Ans. by one operation
1296
The same It/ two operatiom.
Ist.
2d.
As 54 ; 16 : ; 135 : 40
As 8 ; 40 ; ; 6 ; 30
16
6
810
8 ) 240 ( 30 Ans
136
24
54 ) 2160 (40
216
OF commoN fractions.
A Fraction, or broken number, is an expression of a part, or
some parts, of something considered as a whole.
It is denoted by two numberSj placed one below the other, with
a line between them:
, 3 numerator ) , . , . , ,
^''"''Tdenominator  ^^''"^ '^ "^'"'^ throe fourths.
The Denominator, or number placed below the line, shows how
many equal parts the whole quantUy is divided into ; and repre
sents the Divisor in Division. And the Numerator, or number set
above the line, shows how many of those parts are expressed by the
Fraction ; being the remainder after division. Also, both these
numbers are, in general, named the Terms of the Fractions.
hv Google
IG THE PBACIICAIi MODEL CALCULATOR.
Fractions are either Proper, Improper, Simple, Compound, or
Mixed.
A Proper Fraction is when the numerator is less than the deno
minator ; as J, or ^, or f , &c.
An Improper Fraction is when the numerator is equal to, or
exceeds, the denominator; as , or f, or I, &c.
A Simple Fraction is a single expression denoting any number
of parts of the integer ; as ^, or .
A Compovtnd Fraction is the fraction of a fraction, or several
fractions connected with the word of between them ; as J of ^, or
I of I of 3, &e.
A Mixed Number is composed of a whole number and a fraction
together; as 3J, or 12, &c.
A whole or integer number may he expressed like a fraction, by
writing 1 below it, as a denominator ; so 3 is , or 4 is *, &e.
A fraction denotes division ; and its value is equal to the quo
tient obtained by dividing the numerator by the denominator;
so '4^ is equal to 3, and ^/ is equal to 4.
Hence, then, if the numerator be less than the denominator, the
value of the fraction ia less than 1. If the numerator be the same
as the denominator, the fraction is just equal to 1. And if the
numerator be greater than the denominator, the fraction is greater
than 1.
EEDTJCTION OF FRACTIONS.
Reduction 01? Fractions is the bringing them out of one form or
denomination into another, commonly to prepare them for the opera
tions of Addition, Subtraction, &c., of which there are several cases.
The Common Measure of two or more numbers is that number
which will divide them both without a remainder ; so 3 is a com
mon measure of 18 and 24 ; the quotient of the former being 6,
and of the latter 8. And the greatest number that will do this,
is the greatest common measure : so 6 is the greatest common mea
sure of 18 and 24 ; the quotient of the former being 3, and of the
latter 4, which will not both divide farther.
Rule. — If there be two numbers only, divide the greater by
the less ; then divide the divisor by the remainder ; and so on, divid
ing always the last divisor by the last remainder, till nothing
remains ; then shall the last divisor of all he the greatest common
measure sought.
When there are more than two numbers, find the greatest com
mon measure of two of them, as before ; then do the same for that
common measure and another of the numbers ; and so on, through
all the numbers ; then will the greatest common measure last found
be the answer.
If it happen that the common measure thus found is 1, then the
numbers are said to be incommensurable, or to have no common
measure.
hv Google
REDUCTIOS OF FRACTIONS. 17
To find the greatest commoE measure of 1998, 918, and 522.
918 ) 1998 (2 So 54 is the greatest common measure
1836 of 1998 and 918.
162)918(5 Hence 54)522(9
810 486
108)162(1 3(S)54(1
108 36
54)108(2 18)36(2
108 36
So ttat 18 is the answer required.
To ahhreviate or reduce fractions to their lowest tenm.
EULE. — Divide the terms of the given fraction by any number
that will divide them without a remainder ; then divide these quo
tients again in the same manner ; and so on, till it appears that
there is no number greater than 1 which will divide them : then the
fraction will be in its lowest terms.
Or, divide both the terms of the fraction by their greatest com
mon measure, and the quotients will be the terms of the fraction
required, of the same value as at first.
That dividing both the terms of the fraction by the same num
ber, whatever it he, will give another fraction equal to the former,
is evident. And when those dirisions are performed as often as
can be done, or when the common divisor is the greatest possible,
the terms of the resulting fraction must ho the least possible.
1. Any number ending with an even number, or a cipher, is divi
sible, or can be divided by 2.
2. Any number ending with 5, or 0, ia divisible by 5.
3. If the righthand place of any number be 0, the whole is
divisible by 10; if there be 2 ciphers, it is divisible by 100; if
3 ciphers, by 1000 ; and so on, which is only cutting off those
ciphers.
4. If the two righthand figures of any number be divisible
by 4, the whole is divisible by 4. And if the tliree righthand
figures be divisible by 8, the whole is divisible by 8 ; and so on.
5. If the sum of the digits in any number be divisible by 3, or
by 9, the whole is divisible by 3, or by 9.
6. If the righthand digit be even, and the sum of all the digits
be divisible by 6, then the whole will be divisible by 6.
7. A number is divisible by 11 when the sum of the 1st, 3d,
5th, &c., or of all the odd places, is equal to the sum of the 2d,
4th, 6th, &c., or of all the even places of digits.
8. If a number cannot be divided by some quantity less than
the square of the same, that number is a prime, or cannot be
divided by any number whatever.
9. Ali prime numbers, except 2 and 5, have either 1, 3, 7, or 9,
in the place of units ; and ail other numbers are composite, or can
bo divided.
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18 THE PRACTICAL MODEL CALCULATOB.
10. When numbers, with a sign of addition or subtraction between
them, are to be divided bj any number, then each of those num
10 + 8 — 4
bers must be divided by it. Thus, ^ ' ' ' ~^ + 4 — 2 = 7.
11, But if the numbers have the sign of mnltipli cation between
, ,. . , , ^, 10 X 8 X 3
them, only one of tnem must be divided. Thus, — c v 9 — ~
10 X 4 X 3 10 X 4 X 1 10 X 2 X 1 20
6x1 ~ 2x1 ~ 1x1  I ^^■
Reduce JJ to its least terms.
IM = i^ = n= iS = A = l> the answer.
Or thus :
144 ) 240 ( 1 Therefore 48 is the greatest common measure, and
144 48 ) Jll = f the answer, the same as before.
"96)144(1
_96
"48)96(2
96
To reduce a mixed number to iU equivalent improper fraction.
Exile. — Multiply the whole number by the denominator of the
fraction, and add the numerator to the product ; then set that sum
above tlie denominator for the fraction required.
Reduce 23§ to a fraction.
Or,
23 (23 X 5) J_2 _ 117
^ 5 " "" a ■
115
2
Trf
5
To reduce an improper fraction to its equivalent whole or mixed
number.
Exile. — Divide the numerator by the denominator, and the quo
tient will be the whole or mixed number sought.
Reduce ^ to its equivalent number.
Here V or 12  3 = 4.
Reduce y to its equivalent number.
Here V or 15  7 = 2J.
Reduce ^° to its equivalent number.
Thus, 17)749(44Jf
68_
69" So that W = 44A
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EEDUOTION OP PEACTIONS. 19
To reduce a loliola number to an equivalent fraction, having a
(fiven denominator.
Rule. — Multiply tlie ivhole number by the given denominator,
then set the product over the said denominator, and it will form
the fraction reijuired.
Reduce 9 to a fraction whoae denominator shall be 7.
Here 9x7 = 63, then f is the answer.
For V = 63 5 7 = 9, the proof.
To reduce a compound fraction to an equivalent simple one.
Rule. — Multiply all the numerators together for a numerator,
and all the denominators together for the denominator, and they
wiU form the simple fraction sought.
When part of the compound fraction is a whole or mixed number,
it must first be reduced to a fraction by one of the former cases.
And, when it can be done, any two terms of the fraction may be
divided by the same number, and the quotients used instead of
them. Or, when there are terms that arc common, they may be
omitted.
Reduce I of I of I to a simple fraction.
1x2x3 6 _ 1
^^^^ 2x3x4~24~4'
1x2x3 1
' 2x3x4 ~ 4' ^^ oi^itting the twos and threes,
f of § of \l to a simple fraction.
2 X 3 X 10 _ ^ _ ^ _ £
^'^'■'^ 3 X 5 X 11 ~ 1(35 ~ 33 ~ 11"
; 3 X 10 4 ,
. t „ .I = ^pr, the same as before.
To reduce fractions of different denominators to equivalent frac
tions, having a common denominator.
Rule, — Multiply each numerator into all the denominators ex
cept its own for the new numerators ; and multiply all the denomi
nators together for a common denominator.
It is evident, that in this and several other operations, when any
of the proposed quantities are integers, or mixed numbers, or com
pound fractions, they must be reduced, by their proper rules, to
the form of simple fractious.
Reduce ^, f, and J to a common denominator.
1 X 3 X 4 = 12 the new numerator for .
2x2x4 = 16 for2
3 X 2 X 3 = 18 for.
2 X 3 X 4 = 24 the common denominator.
Therefore, the equivalent fractions are f, ^, and ^.
Or, the whole operation of multiplying may be very well per
formed mentally, and only set down the results and given fractions
thus : i, I, f = ^1, il, il = fe ^, ^i, by abbreviation.
hv Google
20 THE PRACTICAL MODEL CALCULATOK,
Wlicn the denominators of two given fractions have a coniraon
measure, let tliem lie divided by it ; then multiply the terms of
each given fraction by tho quotient arising from the other's deno
minator.
When the lesa denominator of two fractions exactly divides the
greater, multiply the terms of that which hatli the less denominator
hy the quotient.
When more than two fractions are proposed, it is sometimes con
venient first to reduce two of them to a common denominator,
then these and a third ; and so on, till they be all redaced to their
Q denominator.
To find the value of a fraction in parts of the integer.
Rule. — Multiply the integer hy tho numerator, and divide the
product by the denominator, by Compound Multiplication and
Division, if the integer be a compound quantity.
Or, if it he a single integer, multiply the numerator by the parts
in the next inferior denomination, and divide the product hy the
denominator. Then, if any thing remains, multiply it by the parts
in tho next inferior denomination, and divide by the denominator
as before ; and so on, as far as necessary ; so shall the quotients,
placed in order, be the value of the fraction required.
What is the value of  of a pound troy ? 7 oz. 4 dwta.
What ia the value of ^ of a cwt.? 1 qr. 7 lb.
What is the value of  of an acre? 2 ro. 20 po,
What is the value of j^ of a day ? 7 hrs. 12 min.
To reduce a fraction from one denomination to another.
Rule. — Consider how many of tho less denomination make
one of the greater; then multiply the numerator hy that num
ber, if the reduction be to a less name, or the denominator, if to
a greater.
Reduce f of a cwt. to the fraction of a pound.
ADDITION OF FRACTIONS,
To add fractions together that have a common denominator.
RuLT^. — Add all tho numerators together, and place the sura
over the common denominator, and that will be the sum of the
fractions required.
If the fractions proposed have not a common denominator, they
must be reduced to one. Also, compound fractions must be reduced
to simple ones, and mixed numbers to improper fractions; also,
fractions of different denominations to those of the same denomi
nation.
To add I and \ together. Here I + J = I = If
To add I and \ together. I + ^ = ii! + If = lu — la"
To add \ and 7j and ^ of  together.
i + 7i i J of f == i I ij' h 1 = i + V + ^ ^ V = 8.
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RULE OF THRBB IN FRACTI0H3. 21
SUETRACTION OF FRACTIONS.
Rule. — Prepare the fractions tfie same aa for Addition ; then sub
tract the one numerator from tte other, and set the remainder over
the common denominator, for the diEFerence of the fractions sought.
To find the difference between  and J.
Here   j =  = .
To find the difference between  and f
MULTIPLICATION OF FRACTIONS.
Multiplication of any thing by a fraction implies the taking
some part or parts of the thing ; it may therefore be truly expressed
by a compound fraction ; which is resolved by multiplying together
the numerators and the denominators.
Rule. — Reduce mixed numbers, if there be any, to equivalent
fractions ; then multiply all the numerators together for a nume
rator, and all the denominators together for a denominator, which
will give the product required.
Required the product of  and f.
Here  x J = {], ~ }.
Or, I X f = ^ X ^ = 1.
Required the continued product of , 3^, 5, and f of f .
2 13 5 3 3 _ 13 X 3 _ 39 _
Here gX ^ ^i^4^5— 4x2 ~ 8 ~ '*
DIVISION OF FEACTIONS.
Rule. — Prepare the fractions as before in Multiplication; then
divide the numerator by the numerator, and the denominator by
the denominator, if they will exactly divide ; but if not, then invert
the terms of the divisor, and multiply the dividend by it, as in
Multiplication.
Divide ^ by ^.
Here \* ^ f = i = I3, by the first method.
Divide  by ,%.
Here 5 ^ 1 5 = s X '^ = I X ^ = =5= = 4^, by the latter.
RULE OF THREE IN FRACTIONS.
Rule. — Make the necessary preparations as before directed;
then multiply continually together the second and third terms, and
the first with its terms inverted as in Division, for the answer.
This is only multiplying the second and third terms together, and
dividing the product by tlie first, as in the Rule of Three in whole
If ^ of a yard of velvet cost  of a dollar, what will /g of a
yard cost?
3 2 5 8 2 5
Hero o '■ T '■ '■ Ti^ '■ Ti y B XT5 = 'Lofa dolkr.
8 5 16 3 5 lo ^
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22 IHE PRACTICAL MODEL CALCULATOK.
DECIMAL FKACTIONS.
A Decimal Fraction ia that which has for its denominator a
unit (1) with as many ciphers annexed as the numerator has places ;
and it is usnally expressed by setting down the nnmerator only,
with a point hefore it on the left hand. Thus, f^ is '5, and ^ is
■25, and ^ is "075, and ^^^ is 00124 ; where ciphers are pre
fixed to make np as many places as are in the numerator, when
there is a deficiency of figures.
A mixed numher is made up of a whole number with some deci
mal fraction, the one being separated from the other by a point.
TiiuSj 325 IB the same as 3^, or f^.
Ciphers on the right hand of decimals make no alteration in
their value; for "5, or 50, or 500, are decimals having all the
same value, being each = ^ or J. Eut if they are placed on the
left hand, they decrease the value in a tenfold proportion. Thus,
■5 is ^ or 5 tenths, but 05 is only jl^ or 5 hundreths, and 005
is but Tiftj or 5 thousandtlis.
The first place of decimals, counted from the left hand towards
the right, is called the place of primes, or lOths ; the second is the
place of seconds, or lOOths ; the third is the place of thirds, or
lOOOths ; and so on. For, in decimals, as well as in whole num
bers, the values of the places increase towards the left hand, and
decrease towards the right, both in the same tenfold proportion ;
as in the following Scale or Table of Notation:
^a i i i i I U I I I I I 1
3333333 333333
ADDITION OP DECIMALS.
Rule. — Set the numbers under each other according to the value
of their places, like as in whole numbers ; in which state the deci
mal separating points will stand all exactly under each other.
Then, beginning at the right hand, add up all the columns of
number as in integers, and point off as many places for decimals as
are in the greatest number of decimal places in any of the lines that
are added ; or, place the point directly below all the other points.
To add together 290146, and 31465, 290146
and 2109, and 62417, and 1416. 31465
2109
■6241T
1416
529929877, the sum.
hv Google
MULTIPLICATION OF DECIMAL?, 23
The Bum of 37625 + 86125 + 6374725 + 65 + 4102 +
S58865 = 1506.2325.
The sum of 35 + 4725 + 2.0073 + 92701 + 15 = 981.2673.
The sum of 276 + 54321 + 112 + 0.65 + 125 + 0463 =
4555173.
SUBTRACTION OF DECIJIALS.
EuLE. — Place the numbers under each other according to the
value of their places, as in the last rule. Then, heginning at the
right hand, subtract as in whole numhers, and point off the deci
mals as in Addition.
To find the difference hetiveen I 9173
91.73 and 2.138. 2138
I 89592 the difference.
The difference between 19185 and 273 = 08115.
The difference hetween 21481 and 490142 = 20990858.
The difference between 2714 and 916 = 2713084.
MULTIPLICATION OF DECIMALS.
EuLE. — Place 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 fac
tors. But if there be not so many
figures in the product, then supply
the defect by prefixing ciphers.
Multiply 79347 by 2315, and ire Lave 183688305.
Multiply G3478 by 8204, and we have 520773512.
Multiply 385746 by 00464, and we have 00178986144.
CONTKACTION I.
To multipli/ decimals hy 1 with any number of ciphers, as 10, or
100, or 1000, ^e.
This is done by only removing the decimal point so many places
farther to the right hand as there are ciphers in the multiplier ;
and subjoining ciphers if need be.
The product of 513 and 1000 is 51300.
The product of 2714 and 100 is 2714.
The product of 916 and 1000 is 916.
The product of 2131 and 10000 is 213100.
COSTRACnON II.
To contract the operation, so as to retain only as many decimals in
the product as may be thought necessary, when tJie product teouJd
naturally contain several tnore places.
Set the units' place of the multiplier under that figure of the
multiplicand whose place is the same as is to be retained for th,
Multiply 321096
by 2465
1605480
1926576
1284384
642192.
■0791501640 the product.
hv Google
24 THB PRACTICAL MODEL CAICTJLATOR.
last ill the product ; and dispose of tlie rest of the figures in the
inverted or contrary order to what they are usually placed in.
Then, in multiplying, reject all the figures that are more to the
right than each multiplying figure ; and set down the products, so
that their right hand figures may fall in a column straight below
each other ; but observing to increase the first figure of every line
with what would arise from the figures omitted, in this manner,
namely, 1 from 5 to 14, 2 from 15 to 24, 3 from 25 to 34, &c. ;
and the sum of all the lines will he the product as required, com
monly to the nearest unit in the last figure.
To multiply 2714986 by 9241036, so as to retain only four
places of decimals in the product
Contracted way. Common way.
2714986 2714986
5301429 9241035
24434874 13 574930
542997 81 44958
2714 986
2715
542997 2
14 2443487 4
25089280 25089280 650510
DIVISION OF DECIMALS.
Rule. — Divide aa in whole numbers ; and point off in the quo
tient as many places for decimals, as the decimal places in the
dividend exceed those in the divisor.
When the places of the quotient are not so many as the rule re
quires, let the defect be supplied by prefixing ciphers.
When there happens to be a remainder after the division ; or
when the decimal places in the divisor are more than those in the
dividend ; then ciphers may be annexed to the dividend, and the
quotient carried on as far as required.
179) 48624097 (00271643 I 2685)2700000 (10055805
1282 15000
294 15750
1150 23250
769 17700
537 15900
000 24750
Divide 23470525 by 6425. 3653.
Divide 14 by 7854. 17825.
Divide 217568 by 100. 2175G8.
Divide 8727587 by 162. 538739.
When the divisor is an integer, with any number of ciphers an
nexed ; cut off those ciphers, and remove the decimal point in the
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REDUCTION or DECIMALS,
dividend as many places farther to the left as there are ciphers cut
off, prefixing ciphers if need he ; then proceed as before.
Divide 455 by 2100.
2100 ) 455 ( 0210, &e.
14
CONTRACTION II.
Hence, if the divisor be 1 with ciphers, as 10, or 100, or 1000,
&c. ; then the quotient 'will be found by merely moving the decimal
point in the dividend so many places farther to the left as the di
visor has ciphers ; prefixing ciphers if need bo.
So, 2173 H 100 = 2173, and 419 ^ 10 = 419.
And 516 H 100 = 0516, and 21 ^ 1000 = 00021.
CONTRACTION III.
When there are many figures in tKe divisor ; or only a certain
number of decimals are necessary to be retained in the quotient,
then 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 quo
tient, and find how many times they may he contained in the first
figures of the dividend, as usual.
Let each remainder be a new dividend ; and for every such divi
dend, leave out one figure more on the right hand side of the di
visor ; remembering to carry for the increase of the figures cat off,
as in the 2d contraction in Multiplication.
When there are not so many figures in the divisor as are required
to be in the quotient, begin the operation with all the figures, and
continue it as usaal till the number of figures in the divisor be equal
to those remaining to he found in the quotient, after which begin
the contraction.
Divide 250892806 by 9241035, so as to have only four deci
mals in the quotient, in which case the quotient will contain six
figures.
Contracted. Common way.
924103,5)2508928,06(271498
660721
13849
4608
912
92 4103,5) 2508928,06 {271408
66072106
13848610
46075750
91116100
70467850
55395T0
REDUCTION OF DECIMALS.
To reduce a common fraction to its equivalent decimal.
KoLE. — Divide the numerator by the denominator as in Division
of Decimals, annexing ciphers to the numerator as far as necessary ;
so shall the quotient be the decimal required.
hv Google
THE PRACTICAL MODEI, CALCULATOR.
1 decimal.
24 =4 X 6. Tlica4)7
G) 1750000
■291066, kc.
f reduced to a decimal, is STo.
^ reduced to a, decimal, is '04.
^ reduced to a decimal, is 015625.
iah reduced to a decimal, is OTloTT, tc.
CASE II.
To find t/ie value of a decimal in terms of the inferior denominations.
Rule. — Multiply tlie decimal by the number of parts in the next
lower denomination ; and cut off as many places for a remainder,
to the right baud, as there are places in the given decimal.
Multiply that remainder by the parts in the next lower denomi
nation again, cutting off for another remainder as before.
Proceed in the same manner through all the parts of the integer ;
then the several denominations, separated on the left hand, will
make up the value required.
What is the value of 0125 lb. troy;— S dwts.
What is the value of 4694 lb. troy :— 5 oz. 12 dwt. 15744 gi
What is the value of 625 cwt. :— 2 qr. 14 lb.
What is the value of 009943 mOes :— 17 yd. 1 ft. 598848 ii
What is the value of 6875 yd. :— 2 qr. 3 nl;
What is the value of 3375 ac. : — 1 rd. 14 poles,
What is the value of 2083 hhd. of wine :— 131229 gal.
CASE III.
To reduce integers or decimals to equivalent decimals of higher
denominations.
Rule. — Divide by the number of parts in the next higher de
nomination ; continuing the operation to as many higher denomi
nations as may be necessary, the same as in Reduction Ascending
of whole numbers.
Reduce 1 dwt. to the decimal of a pound troy.
20 1 1 dwt.
12 005 oz.
10Q04166, &c. lb.
Reduce 7 dr. to the decimal of a pound avoird.: — 02734375 lb.
Reduce 215 lb. to the decimal of a cwt. : — ■01919(3 civt.
Reduce 24 yards to the decimal of a mile: — 013636, kc. miles.
Reduce 056 poles to the decimal of an acre i — 00035 ac.
Reduce 12 pints of wine to the decimal of a Lhd. :— 00238 hhd.
Reduce 14 minutes to the decimal of a day : — 009722, kc. da.
Reduce 21 pints to the decimal of a peck: — 013125 pec.
When there are several numbers, to be reduced all to the decimal of
the highest.
Set the given numbers directly under each other, for dividends,
proceeding orderly from the lowest denomination to the highest.
hv Google
DUODECIMALS. 27
Opposite to each dividend, on the left hand, set such a number
for a divisor as will bring it to the next higher name ; drawing a
perpendicular line between all the divisors and dividends.
Begin at the uppermost, and perform all the divisions ; only ob
serving to set the quotient of each division, as decimal parts, on
the right hand of the dividend nest below it ; so shall the last quo
tient be the decimal required.
Reduce 5 oa. 12 dwts. 16 gr. to lbs. :— 46944, &c. lb.
RULE OP THREE IN DECIMALS.
Rule. — Prepare the terms by reducing the vulgar fractions to
decimals, any compound numbers either to decimals of the higher
denominations, or to integers of the lower, also the first and third
terms to the same name : then multiply and divide as in ivhole
numbers.
Any of the convenient examples in the Rule of Three or Rule of
Five in Integers, or Common I'ractions, may be taken as proper
examples to the same rules in Decimals. — The following example,
which is the first in Common Fractions, is wrought here to show the
method.
If I of a yard of velvet cost § of a dollar, what will i\ yd. cost ?
yd. S yd. $
1 = .370 375 : 4 :: 312,5 : 333, &c.
^
2 = 4 375 ) 12500 ( 833333, 33j cts.
1260
125
A = 3125.
DTJODECIMALS.
Duodecimals, or Caoss Multiplication, is a rule made use of
by workmen and artificers, in computing the contents of their works.
Dimensions aie uauiUy taken infect, inches, and quarters; any
parts smaller than these being neglected as of no consequence.
And the same in multiplying them together, or casting up the con
tents.
Rule — Set down the two dimensions, to be multiplied together,
one undei the other, so that feet stand under feet, inches under
inches, i.c
Multiply each teim in the multiplicand, beginning at the lowest,
by the feet m the multiplier, and set the result of each straight un
der its corresponding term, observing to carry 1 for every 12, from
the inches to the feet.
In like manner, multiply all the multiplicand by the inches and
parts of the multiplier, and set the result of each term one place
removed to the right hand of those in tho multiplicand ; omitting,
however, what is below parts of inches, only carrying to these the
proper number of units from the lowest denomination.
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28 THE PRACTICAL MODEL CALCULATOR.
Or, instead of multiplying by tho inches, take such parts of the
multiplicand aa these are of a foot.
Then add the two linea together, after the manner of Compound Ad
dition, carrying 1 to the feet for 12 inches, wlien these come to so many.
Multiply 4 f. 7 inc. Multiply 14 f. 9 inc.
b y 6 4 by _4__ 6
27 6 59
_1 6^ 7 4 
29 0^ 6^ 4
INTOLVTION.
Involution is the raising of Powers from any given number, aa
a root.
A Power is a quantity produced by multiplying any given num
ber, called the Root, a certain number of times continually by
itaelf. Thus, 2 = 2 is the root, or first power of 2.
3x2= 4 is the 2d power, or square of 2.
2x2x2= 8 is the 3d power, or cube of 2.
2 X 2 X 2 X 2 = 16 is the 4th power of 2, &c.
And in this manner may be calculated the following Table of the
first nine powers of the first nine numbers.
TABLE OF THE HB
ST NIKE
I'OWERS
OF KUMBERS.
1st
2d.
3d.
4th.
sa.
eth.
7t!i.
8tli.
OIL.
1
1
1
1
1
1
1
1
1
2
4
8
16
33
64
128
256
512
3
9
27
81
243
729
2187
6561
10083
4
10
64
256
1024
4096
10384
65536
202144
5
25
125
625
3126
16626
78125
890625
1953125
6
36
216
1296
7776
46656
279986
1679616
10077696
7
49
343
2401
16807
117649
823543
5764801
40353607
8
64
512
4096
32768
262144
2097152
16777216
184217728
9
SI
720
6561
59049
531441
4782969
430467:il
387i20180
The Index or Exponent of a Power is the number denoting the
height or degree of that power ; and it is 1 more than the number
of multiplications used in producing the same. So 1 is the index
or exponent of the 1st power or root, 2 of the 2d power or square,
3 of the 3d power or cube, 4 of the 4th power, and so on.
Powers, that are to be raised, are usually denoted by placing the
index above the root or first power.
So 2^ = 4, is the 2d power of 2.
2^ = S, is the 3d power of 2.
2'' = 16, is the 4th power of 2.
540^, is the 4th power of 540 = 85030560000.
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'IThen two or more powers are multiplied together, their product
will he that power whose index is the sum of the exponents of the
factors or powers multiplied. Or, the multiplication of the powers
answers to the addition of the indices. Thus, in the following
powers of 2.
1st. 2d. 3d. 4th. 5tli. 6th. 7th. 8th. 9th. 10th.
2 4 8 16 32 64 128 256 512 1024
or, 2^ 2* 2^ 2* 2' 2' 2' 2' 2« 2'"
Here, 4x4= 16, and 2 + 2 = 4 its index ;
and 8 X 16 = 128, and 3 ( 4 = 7 its index ;
also 16 X 64 = 1024, and 4 + 6 = 10 its index.
The 2d power of 45 is 2025.
The square of 416 is 173056.
The 3d power of 35 is 42875.
The 5th power of 029 is 000000020511149.
The Sijuare of § is i
The 3d power of f is ^.
The 4th power of  is •^.
EVOLTTTION.
EvoLUTios, or the reverse of Involution, is the extracting or
finding the roots of any given powers.
The root of any number, or power, is such a number as, being
multiplied into itself a certain number of times, will produce that
power. Thus, 2 is the square root or 2d root of 4, because 2^ =
2 X 2 = 4 ; and 3 is the cube root or 3d root of 27, because 3^ =
3 X 3 X 3 = 27.
Any power of a given number or root may be found exactly,
namely, by multiplying the number continually into itself. But
there are many numbers of which a proposed root can never be
exactly found. Yet, by means of decimals we may approximate
or approach towards the root to any degree of exactness.
These roots, which only approximate, are called Surd roots ; but
those which can be found quite exact, are called Rational roots.
Thus, the square root of 3 ia a surd root ; but the square root of
4 is a rational root, being equal to 2 : also, the cube root of 8 is
rational, being equal to 2 ; but the cube root of 9 is surd, or
irrational.
Roots are sometimes denoted by writing the character \/ before
the power, with the index of the root against it. Thus, the third
root of 20 is. expressed by ^20; and the square root or 2d root
of it ia v'20, the index 2 being always omitted when the square
root is designed,
AYhcn the power is expressed by several numbers, with the sign
+ or — between them, a line is drawn from the top of th e sign over
all the parts of it ; thus, the third root of 45 — 12 is ■^'45 ~ 12,
or thus, ^(45 — 12), enclosing the numbers in parentheses.
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30 THE I'KACTICAL MODEL CALCULATOR.
But all roots are now often designed like powers, with fractional
indices : thus, the square root of 8 is 8 , the cuhe root of 25 is 25 ,
and the 4th root of 45  18 is i5'^:;oII^ or, (45  IS)''.
TO EXTKACT THE SQUAKE ROOT.
Rule, — Divide the given number into periods of two figures
each, by setting a point over the place of units, another over the
place of hundreds, and so on, over every second figure, both to the
left hand in integers, and to the right in decimals.
Find the greatest square in the first period on the left hand, and
set its root on the right hand of the giv^n number, after the man
ner of a quotient figure in Division,
Subtract the square thus found from the said period, and to the
remainder annex the two figures of the next following period for a
dividend.
Double the root above mentioned for a divisor, and find how
often it is contained in the said dividend, exclusive of its righthand
figure ; and set that quotient figure both in the quotient and divisor
Multiply the whole augmented divisor by this last quotient figure,
and subtract the product from the said dividend, bringing down to
the next period of the given number, for a new dividend.
Repeat the same process over again, namely, find another new
divisor, by doubling all the figures now found in the root ; from
which, and the last dividend, find the next figure of the root as
before, and so on through all the periods to the last.
The best way of doubling the root to form the new divisor is by
adding the last figure always to the last divisor, as appears in the
following examples. Also, after the figures belonging to the given
number are all exhausted, the operation may be continued into
decimals at pleasure, by adding any number of periods of ciphers,
two in each period.
To find the square root of 29506624.
29506624 ( 5432 the root.
25
104 I 450
4 416
10862 I 21T24
2 I 21724
W7ien the root is to he extracted to many jjldcea of figures, the ivorl:
may he considerably shortened, thus :
Having proceeded in the extraction after the common method till
there be found half the required number of figures in the root, or
one figure more; then, for the rest, divide the last remainder by
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TO EXTK4CT TUB SQUABE BOOT. 31
its corresponding divisor, after the manner of the third contraction
in Division of Decimals ; thus,
To find the root of 2 to nine places of figures,
2(14142
1
24 I 100
281 1
1 1
400
281
2824
4
11900
11296
■Mam
2
: 60400
56664
28284) 3836(1356
1008
160
1'41421356 the root required.
The square root of 000729 is 027.
The square root of 3 is 1732050.
The square root of 5 is 2236068.
The square root of 6 is 2449489.
RULES rOB THE SQUARE ROOTS OP COMMON rRACTIOiSS AND JIIXED
NUMBERS.
First, prepare all common fractions hy reducing them to their
least terms, both for this and all other roots. Then,
1. Take the root of the numerator and of the denominator for
the respective terms of the root required. And this is the best
way if the denominator be a complete power ; but if it be not, then,
2. Multiply the numerator and denominator together ; take the
root of the product : this root being made the numerator to the
denominator of the given fraction, or made the denominator to the
numerator of it, will form the fractional root required.
\/a \/ab a
\/b b ^ah'
And this rule will serve whether the root be finite or infinite.
3. Or reduce the common fraction to a decimal, and extract its root,
4. Mixed numbers may he either reduced to improper fractions,
and extracted by the first or second rule ; or the common fraction
may be reduced to a decimal, then joined to the integer, and the
root of the whole extracted.
The root of t is \.
The root of ^^ is f
a'he root of /j is 0866025.
The root of ^^ is 0645497.
The root of 17 is 4168333.
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32 THE PRACTICAL MODEL CALCULATOR.
By meana of the square root, also, may readily be found the 4th
root, or the 8th root, or the 16th root, &e. ; that is, the root of any
power whose index is some power of the number 2 ; namely, by
extracting so often the square root as is denoted by that power
of 2 ; that is, two extractions for the 4th root, three for the 8th
root, and so on.
So, to find the 4th root of the number 210358, extract the
square root twice as follows :
21Qzh8m ( 145037237 ( 120431407, the 4th root.
TO EXTRACT THE CUBE ROOT.
1. Divide the page into three columns (i), (ii), (m), in order,
from left to right, so that the breadth of the columns may increase
in the same order. In column (iii) write the given number, and
divide it into periods of three figures each, by putting a point over
the place of units, and also over every third figure, from thence to
the left in whole numbers, and to the right in decimals.
2. Find the nearest less cube number to the first or lefthand
period ; set its root in column (iii), separating it from the right
of the given number by a curve line, and also in column (i) ; then
multiply the number in (i) by the root figure, thus giving the square
of the first root figure, and write the result in (ii) ; multiply the
number in (li) by the root figure, thus giving the cube of the first
root figure, and write the result below the first or lefthand period
in (ill) ; subtract it therefrom, and annex the next period to the
remainder for a dividend.
3. In (i) write the root figure below the former, and multiply
the sum of these by the root figure ; place the product in (ii), and
add the two numbers together for a trial divisor. Again, write the
root figure in (i), and add it to the former sum.
4. With the number in (ii) as a trial divisor of the dividend,
omitting the two figures to the right of it, find the next figure of
the root, and annex it to the former, and also to the number in (i).
Multiply the number now in (i) by the new figure of the root, and
write the product as it arises in (ii), but extended two places of
figures more to the right, and the sum of these two numbers will
be the corrected divisor ; then multiply the corrected divisor by the
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TO EXTRACT THE CUBE ROOT.
33
last root figure, placing the proiJuct as it arises below the cKvidcnd ;
subtract it therefrom, annex another period, and proceed precisely
as described in (3), for correcting the columns (i) and (ii). Then
with the new trial divisor in (ri), and the new dividend in (iir),
proceed as before.
When the trial divisor is not contained in the dividend, after two
figures are omitted on the right, the next root figure is 0, and there
fore one cipher must be annexed to the number in (i) ; two ciphers
to the number in (ii) ; and another period to the dividend in (lli).
When the root is interminable, we may contract the work very
considerably, after obtaining a few figures in the decimal part of
the root, if we omit to annex another period to the remainder in
(ill) ; cat off one figure from the right of (il), and two figures from
(i), which will evidently have the effect of cutting off three figures
from each column ; and then work with the numbers on the left, as
in contracted multiplication and division of decimals.
Find the cube root of 210358 to ten places of decimals.
m
(n)
■ ■ (in)
2
4
210358 (276049105o944
2
8
8
4
T»..
13035
2
469
11683
67
1669
1362800
7
518
1341576
74
2187..
11224
7
4896
9142444864
816
2 2 3 5 9 6
2081555136
6
4932
2057416281
228528. . . .
331216
2286611216
331232
228594244
7453
24139855
22860923
1278982
1143046
828112
228601697
7463
2286091511
2286092314
Required the cube roots of the following numbers : —
48228644, 46666, and 16069223. 364, 36, and 247.
64481201, «nd 28991029248. 401, and 3072.
128211191.55125, and 000076766625. 23405, and 0426.
HifJ, and 16. j, and 2519842.
91i, and7f 45, and 19
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34 THE PRACTICAL MODEL CALCULATOR.
TO EXTRACT ANT BOOT WHATEVER.
Let N be the given power or number, n the index of the power,
A the assumed power, r its root. It the required root of N.
Then, aa the sum of k + 1 times A and Ji ~ 1 times K, is to
the sum of Ji + 1 times N and n — 1 times A, so is the assumed
root r, to the required root R.
Or, as half the said sum of n + 1 times A and n — 1 times N,
ia to the difference between the given and assumed powers, so is tlie
assumed root r, to the diflerence between the true and assumed
roots ; which difference, added or subtracted, as the case rer.uivcs,
gives the true root nearly.
Thatis,(w + l).A + (Kl).N:{w + l).N + {jil)A::r:R.
Or, (« + 1) . A + (n  1) . JN : A 02 N : : » : R ai r.
And the operation may be repeated as often as we please, by
using always the last found root for the assumed root, and its Mth
power for the assumed power A.
To extract the 5th root o/ 210358.
Here it appears that the 5th root is between 7"3 and 7"4. Taking
73, its 5th power is 2073071593. Hcnco then we have,
N = 210358; r = 73; n = 5; J . (™ + 1) = 3; J.(h  1) = L'
A = 20730716
N~A = 3050S4
A 20730710 N = 210358
3 2
3 A = 62192.148 420716
2N 420716
As 1042637 : 305084 :
73
; : 73 : 0213605
915262
2186588
1042637)232711321
14184
3758
630
5
; 0213605, the Jifforeiice.
73 = r add
7321360 = E, the root, true to
the last figure.
The 6th root of 21035.8
The 6th root of 2
Tho 7th root of 210358
The 7th root of 2
The 9th root of 2
is 52.:.4037.
is 1122462.
is 4145092.
is 1104089.
is l08i:059.
OF RATIOS, PEOPORTIONS, AND PEOGEESSIOKS.
Numbers are compared to each other in two different ways : tho
one comparison considers the difference of the two numbers, nnd
is named Arithmetical Relation, and the difference sometimes
Arithmetical Ratio : the other considers their quotient, and is ca!le!.t
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ARITHMETICAL PEOPORTION AND PROGRESSION. 35
Geometrical Relation, and the quotient the Geometrical Ratio. So,
of these two numbers 6 and 3, the difference or arithmetical ratio
is 6 — 3 or 3 ; hut the geometrical ratio is f or 2.
There must be two numbers to form a comparison : the number
which is compared, being placed first, is called the Antecedent ;
and that to which it is compared the Consequent. So, in the
two numbers above, 6 is the antecedent, and 3 is the consequent.
If two or more couplets of numbers have equal ratios, or equal
differences, the equality is named Proportion, and the terms of the
ratios Proportionals. So, the two couplets, 4, 2 and 8, 6 are arith
metical proportionals, because 4 — 2 = 8 — 6 = 2; and the two cou
plets 4, 2 and 6, 3 are geometrical proportionals, because J = § = 2,
the same ratio.
To denote numbers as being geometrically proportional, a colon
is set between the terms of each couplet to denote their ratio ; and
a double colon, or else a mark of equality between the couplets or
ratios. So, the four proportionals, 4, 2, 6, 3, are set thus, 4 : 2 : : 6 : 3,
which means that 4 is to 2 as 6 is to 3 ; or thus, 4:2 = 6:3; or
thus, 1 = 1, both which mean that the ratio of 4 to 2 is equal to
the ratio of 6 to 3.
Proportion is distinguished into Continued and Discontinued.
When the difference or ratio of the consequent of one couplet and
the antecedent of the nest couplet is not the same as the common
difference or ratio of the couplets, the proportion is discontinued.
So, 4, 2, 8, 6 are in discontinued arithmetical proportion, because
4 — 2 = 86 = 2, whereas, 2  8 =  6 ; and 4, 2, 6, 3 are in
discontinued geometrical proportion, because ^ = ^ = 2, but § = ^,
which is not the same.
But when the difference or ratio of every two succeeding terms is
the same quantity, the proportion is said to he continued, and the num
bers themselves a series of continued proportionals, or a progression.
So, 2, 4, 6, 8 form an arithmetical progression, because 4 — 2=6 —
4 = 8 — 6 = 2, all the same common difference ; and 2, 4, 8, 16, a
geometrical progression, because ^ = ^ = ^ = 2, all the same ratio.
When the following terms of a Progression exceed each other,
it is called an Ascending Progression or Scries ; but if the terms
decrease, it is a Descending one.
So, 0, 1, 2, 3, 4, &c., is an ascending arithmetical progression,
but 9, 7, 5, 3, 1, &c., is a descending arithmetical progression :
Also, 1, 2, 4, 8, 16, &c., is an ascending geometrical progression,
and 16, 8, 4, 2, 1, &c., is a descending geometrical progression.
AllITHMETICAL PROPORTION AND PROGRESSION.
The first and last terms of a Progression are called the Extremes ;
and the other terms lying between them, the Moans.
The moat useful part of arithmetical proportions is contained in
the following theorems :
TiiEOBEM 1. — If four quantities he in arithmetical proportion, the
sum of the two extremes will he equal to the sum of the two means.
Thus, of the four 2, 4, 6, 8, here 2 + 8=4 + 6 = 10.
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36 THE PRACTICAL MODEL CALCULATOR.
Theorem 2. — In any continued arithmetical progression, tlie sum
of tlie two extremes is equal to the sum of any two means that are
equally distant from them, or equal to double the middle term when
there is an uneven numher of terms.
Thus, in the terms 1, 3, 5, it is 1 + 5 = 3 + 3 = 6.
And in the series 2, 4, 6, 8, 10, 12, 14, it ia 2 + 14 = 4 + 12 =
6 + 10 = 8 + 8 = 16.
Theorem 3. — The difference between the extreme terms of an
arithmetical progression, is equal to the common difference of the
series multiplied by one less than the number of the terms.
So, of the ten terms, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, the com
mon difference is 2, and one leas than the number of terms 9 ; then
the difference of the extremes is 20 — 2 = 18, and 2 X 9 = 18 also.
Consequently, the greatest term is equal to the least term added
to the product of the common difference multiplied by 1 less than
the numher of terms.
Theorem 4. — The sum of all the terms of any arithmetical pro
gression is equal to the sum of the two extremes multiplied by the
number of terms, and divided by 2 ; or the sum of the two extremes
multiplied by the number of tho terms gives double the sum of all
the terms in the series.
This is made evident by setting tho terms of the series in an
inverted order under the same series in a direct order, and adding
the corresponding terms together in that order. Thus,
in the series, 1, 3, 5, 7, 9, 11, 13, 15;
inverted, 15 , 13 , 11 , 9 , 7^ 5, _ 3, Ij
the sums are, 16 + 16 + 16 + 16 f 16 + 16 + 16 + 16,
which must be double the sum of the single series, and is equal to
the sum of the extremes repeated so often as are the number of
the terms.
From these theorems may readily be found any one of these five
parts ; the two extremes, the number of terms, the common differ
ence, and the sum of all the terms, when any three of them are
given, as in the following Problems :
PROBLESI I.
Criven the extremes and the numher of terms, to find the sum of all
the terms.
Rule. — Add the extremes together, multiply the sum by the
number of terms, and divide by 2.
The extremes being 3 and 19, and the number of terms 9 ;
required the sum of the terms ?
19
2)198
99 = the sum.
hv Google
ARITHMETICAL PEOPOIITION ASD PROGRESSION. 37
The strokes a clock strikes in one whole revolution of the index,
or in 12 hours, is 78.
PROBLEM n.
Given the extremes, and the number of terms; to find the common
difference.
EuLE. — Subtract the less extreme from the greater, and divide
the remainder hy 1 less than the numher of terms, for the common
difference.
The extremes being 3 and 19, and the number of terms 9 ; re
quired the common difference t
19
A r.. 193 _1G_
9  1 ~ 8 ~ '^■
116 0' 
If the extremes be 10 and 70, and the number of terms 21 ; what
is the common difference, and the sum of the series ?
The com. diff. is 3, and the sura is 840.
Qiven one of the extremes, the common difference, and the numher
of terms; to find the other extreme, arid the sum of the series.
Rule. — Multiply the common difference hy 1 less than the num
ber of terms, and the product will he the difference of the extremes ;
therefore add the product to the less extreme, to give the greater ;
or subtract it from the greater, to give the less.
Given the least term 3, the common difference 2, of an arith
metical scries of 9 terms ; to find the greatest term, and the sum
of the series ?
Ill the greatest term.
3 the least.
9 number of terms.
2)"198
99 the sum of the series.
If the greatest term be 70, the common difference 3, and the
number of terms 21 ; what is the least term and the sum of the
series ? The least term is 10, and the sum is 810.
PROBtEM IV.
To find an arithmetical mean proportional betteeen two c/iven terms.
Rule. — Add the two given extremes or terms together, and take
half their sum for the arithmetical mean required. Or, subtract
hv Google
38 THE PRACTICAL MODEL CALCULATOR.
the less extreme from the greater, and half the remainder will be
the common difference ; which, heing added to the less extreme, or
subtracted from the greater, will give the mean requhed.
To find an arithmetical mean hetween the two numbers 4 and 14,
Here, 14 Or, 14 Or, 14
_4 ^ 5
2)iS 2)10' 9
_9 6 the com. dif. ""
4 the less extreme.
T
So that 9 is the mean required by both methods.
PKOBLEM V.
Tojind two arithmetical means hetween two given extremes.
Rule. — Subtract the less extreme from the greater, and divide
the difference by 3, bo will the quotient be the common difference ;
which, heing continually added to the less extreme, or taken from
the greater, gives the means.
To find two arithmetical means between 2 and 8.
Hereg
2
Then 2 + 2 = 4 the one mean,
3]^ and 4 + 2 = 6 the other n
com. dif. 2
To find any number of arithmetical means hetween two given terms
Rule. — Subtract the less extreme from the greater, and divide
the difference by 1 more than the number of means required to be
found, which will give the common difference ; then this being
added continually to the least term, or subtracted from the greatest,
will give the mean terms required.
To find five arithmetical means between 2 and 14.
— Then, by adding this com. dif. continually,
6 ) 12 the means are found, 4, 6, 8, 10, 12.
com. dif. 2
GEOMETRICAL PROPORTION AND PROGRESSION.
The moat useful part of Geometrical Proportion is contained in
the following theorems ;
Theorem 1. — If four quantities be in geometrical proportion,
the product of the two extremes will be equal to the product of the
two means.
Thus, in the four 2, 4, 3, 6 it is 2 x 6 = 3 x 4 = 12.
And hence, if the product of the two means be divided by one
of the extremes, the quotient will give the other extreme. So, of
hv Google
GE0J1ETB.ICAL PROPORTION ASD PBOGKESSION", 39
the above numbers, tlie product of the means 12 § 2 = 6 the one
extreme, and 12 i 6 = 2 the other extreme; and this is the
foundation and reason of the practice in the Rule of Three.
Theorem 2. — In any continued geometrical progression, the pro
duct of the two extremes is equal to the product of any tivo means
that are equally distant from them, or equal to the square of the
middle term when there is an uneven number of terms.
Thus, in the terms 2, 4, 8, it is 2 X 8 = 4 X 4 = 16.
And in the series 2, 4, 8, 16, 32, 64, 128,
it is 2 X 128 = 4 X 64 = 8 X 32 = 16 X 16 = 256.
Theorem 3. — The quotient of the extreme terms of a geome
trical progression is equal to the common ratio of the series raised
to the power denoted by one less than the number of the terms.
So, of the ten terms 3, 4, 8, 16, 32, 64, 128, 256, 512, 1024,
the common ratio is 2, one less than the number of terms 9 ; then
1024
the quotient of the extremes is — „— = 512, and 2^ = 512 also.
Consequently, the greatest term is equal to the least term multi
plied by the said power of the ratio whose index is one less than
the number of terms.
Theorem 4. — The sum of all the terms of any geometrical pro
gression is found by adding the greatest term to the difference of
the extremes divided by one less than the ratio.
So, the sum 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, {whose ratio
1024 — 2
is 2,) is 1024 f ^_^ = 1024 + 1022 = 2046.
The foregoing, and several other properties of geometrical pro
portion, are demonstrated more at large in Byrne's Doctrine of Pro
portion. A few examples may here be added to the theorems just
delivered, with some problems concerning mean proportionals.
The least of ten terms in geometrical progression being 1, and
the ratio 2, what is the greatest term, and the sum of all the teiins ?
The greatest term is 512, and the sum 1023.
PROBLEM I.
To find one geometrical mean proportional between any two numbers.
Rule. — Multiply the two numbers together, and extract the square
root of the product, which will give the mean proportional sought.
Or, divide the greater term by the less, and extract the square
root of the quotient, which will give the common ratio of the three
terms : then multiply the less term by the ratio, or divide the
greater term by it, either of these will give the middle term required.
To find a geometrical mean between the two numbers 3 and 12.
First way. Second way.
12 ;^ ) 12 ( 4, its root, is 2, the r;itio.
o6(6 the mean. Then, 3x2 = 6 the mean.
36 Or, 12 ^ 2 = 6 also.
hv Google
■to TTIE PRACTICAL MOBEL CALCULATOR.
PaOBI.EM II.
To find two geometrical mean proportionals betiveen any two numbers.
Rtjlb. — Divide the greater number by the less, and extract the
cube root of the quotient, which will give the common ratio of the
terms. Then multiply the least given term by the ratio for the
first mean, and this mean again by the ratio for the second mean ;
or, divide the greater of the two given terms hy the ratio for the
greater mean, and divide this again by the ratio for tbo less mean.
To find two geometrical mean proportionals between 3 and 24.
Here, 8 ) 24 ( 8, its cube root, 2 is the ratio.
Then, 3x2= 6, and 6 x 2 = 12, the two means.
Or, 2i H 2 = 12, and 12 4 2 = 6, the same.
That is, the two means between 3 and 24, are 6 and 12.
To find any numher of geometrical mean proportionals between two
numbers.
Rule. — Divide the greater number by the leas, and extract such
root of the quotient whose index is one more than the number of
means required, that is, the 2d root for 1 mean, the 3d root for
2 means, the 4th root for 3 means, and so on ; and that root will
be the common ratio of all the terms. Then with the ratio multi
ply continually from the first term, or divide continually from the
last or greatest term.
To find four geometrical mean proportionals between 3 and ?G.
Here, 3 ) 96 ( 32, the 5th root of which is 2, the ratio.
Then, 8x2=6,and 6x2=12, andl2x2=24,and24x2=48.
Or, 96 J 2=48, and 48r2=24, and 242=12, and 12^2= 6.
That is, 6, 12, 24, 48 are the four means between 3 and 96.
OP MUSICAL PROPORTION.
There is also a third kind of proportion, called Musical, which,
being but of little or no common use, a very short account of it may
here suffice.
Musical proportion is when, of three numbers, the first has the
same proportion to the third, as the difference between the first and
second has to the difference between the second and third.
As in these three, 6, 8, 12 ;
where, 6 : 12 : : 8  6 : 12  8,
that is, 6 : 12 : : 2 : 4.
When four numbers are in Musical Proportion ; tlien the fiv..t
has the same proportion to the fourth, as the difference bctwcei:
the first and second has to the difference between the third ai^l
fourth.
Asia these, 6, 8, 12, 18;
whoic, C: 18:; S  G: IS  12,
that is, 6 : 18 ;: 2 ; C.
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FELLOWSHIP. 41
When numbers are in Musical Progression, their reciprocals are
in Arithmetical Progression ; and the converse, that is, when num
bers aie in Arithmetical Progression, their reciprocals are in Mu
sical Progression.
So, in these Musicals 6, 8, 12, their reciprocals I, I, ^, arc in
arithmetical progression ; for J  ^^ — ^^ = ^ ; and  + J = I — I;
that is, the sum of tl^e extremes is equal to double the mean, which
is the property of arithmeticak.
PELIOWSHIP, OR PARTNERSHIP.
Fellowship is a rule by which any sum or quantity may be
divided into any number of parts, which shall be in any given pro
portion to one another.
By this rule are adjusted the gains, or losses, or charges of part
ners in company ; or the effects of bankrupts, or legacies in case of
a deficiency of assets or effects; or the shares of prizes, or tbe
numbers of men to form certain detachments ; or the division of
waste lands among a number of proprietors.
Fellowship is either Single or Double. It is Single, when the
shares or portions are to be proportional each to one single given
number only ; as when the stocks of partners are all employed for
the same time : and Double, when each portion is to bo proportional
to two or more numbers ; as ivhen the stocks of partners are em
ployed for difi'erent times.
SINGLE FELLOWSHIP.
General Rule. — Add together the numbers that denote the
proportion of the shares. Then,
As the sum of the said proportional numbers
Is to the whole sum to be parted or divided,
So is each several proportional number
To the corresponding share or part.
Or, As the whole stock is to the whole gain or loss,
So is each man's particular stock to his particular share of
the gain or loss.
To prove the work. — Add all the shares or parts together, and
the sum will be equal to the whole number to be shared, when the
work is right.
To divide the number 240 into three such parts, as shall be in
proportion to each other as the three numbers, 1, 2, and 3.
Here 1 4 2 f 3 = 6 the sum of the proportional numbers.
Then, as 6 : 240 : : 1 : 40 the 1st part,
and, as 6 : 240 : : 2 : 80 the 2d part,
also as 6 : 240 :: 3 : 120 the 3d part.
Sum of all 240, the proof.
Three persons, A, B, C, freighted a ship with 340 tuns of wine ;
of which, A loaded 110 tuns, B 97, and C the rest : in a storm, the
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THE PEACTICAL MODEL CALCULATOR.
a were obliged to throw overboard 85 tuns ; how much must
each perBon sustain of the loss ?
Here, 110 + 97 = 20T tuns, loaded by A and E ;
tberef., 340 — 20T = 133 tuns, loaded by C.
hence, as 340 "' '""'
and, a
: 110
; 110 : 27^ tuns = A's loss;
; 97 : 24J tuns = B's loss ;
; 133 :_33i tuns = C's loss.
Sum 85 tuns, the proof.
DOUBLE FELLOWSHIP.
Double Fellowship, as has been said, is concerned in cases
in which the stocks of partners are employed or continued for dif
ferent times.
EuLE. — Multiply each person's stock by the time of its continu
ance ; then divide the quantity, as in Single Fellowship, into shares
in proportion to these products, by saying :
As the total sum of all the said products
Is to the whole gain or loss, or quantity to be parted,
So is each particular product
To the corresponding share of the gain or loss.
SIMPLE INTEEEST,
IsTEPiEST is the premium or sum allowed for the loan, or for
bearance of money.
The money lent, or forborne, is caiied the Principal.
The sum of the principal and its interest, added together, is
called the Amount.
Interest is allowed at so much per cent, per annum, which pre
mium per cent, per annum, or interest of a $100 for a year, is
called the Rate of Interest. So,
When interest is at 3 per cent, the rate is 3 ;
4 per cent 4;
5 per cent 5;
6 per cent 6.
Interest is of two sorts : Simple and Compound,
Simple Interest is that which is allowed for the principal lent or
forborne only, for the whole time of forbearance.
As the interest of any sum, for any time, is directly proportional
to the principal sum, and also to the time of continuance ; hence
arises the following general rule of calculation.
General Rule, — As $100 is to the rate of interest, so is any
given principal to its interest for one year. And again.
As one year is to any given time, so is the interest for a year just
found to the interest of the given sum for that time.
Otherwise. — Take the interest of one dollar for a year, which,
multiply by the given principal, and this product again by tlie time
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POSITION. 43
of loan or forbearance, io years and parts, for the interest of the
proposed sum for that time.
"VYhen there are certain parts or years in the time, as quarters,
or months, or days, they may be worked for either by taking the
ahquot, or like parts of tho interest of a year, or by the Rule of
Three, in the usual way. Also, to divide hy 100, is done by only
pointing off two figures for decimals.
COMPOUND INTEEEST.
Compound Interest, called also Interest upon Interest, is that
■which arises from the principal and interest, taken together, as it
becomes due at the end of each stated time of payment.
Rules. — 1. Pind the amount of the given principal, for the time
of the first payment, by Simple Interest. 'I'hen consider this
amount as a new principal for the second payment, whose amount
calculate as before ; and so on, through all the payments to the last,
always accounting the last amount as a new principal for the next
payment. The reason of which is evident from the definition of
Compound Interest. Or else,
2. Find the amount of one dollar for the time of the first pay
ment, and raise or involve it to the power whose index is denoted
by the number of payments. Then that power multiplied by the
given principal will produce the whole amount. From which the
said principal being subtracted, leaves the Compound Interest of
the same ; as is evident from the first rule.
POSITION.
Position is a method of performing certain questions which can
not be resolved by the common direct rules. It is sometimes called
False Position, or False Supposition, because it makes a supposi
tion of false numbers to work with, the same as if they were the
true ones, and by their means discovers the true numbers sought.
It is sometimes also called Trial and Error, because it proceeds
by triah of false numbers, and thence finds out the true ones by a
comparison of the errors.
Position is either Single or Double.
SINGLE POSITION.
Single Position is that by which a question is resolved by means
of one supposition only.
Questions which have their results proportional to their supposi
tions belong to Single Position ; such as those which require the
multiplication or division of the number sought by any proposed
number ; or, when it is to he increased or diminished by itself, or
any parts of itself, a certain proposed number of times.
Rule. — Take or assume any number for that required, and per
form the same operations with it as are described or performed in
the question.
Then say, as the result of the said operation is to the position
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44 THE PRACnCAL MODEL CALCULATOE.
or number assumed, so is the result in the question to the number
sought.
A person, after spending J and J of his money, has yet remain
ing $60, what had he at first ?
Suppose he had at first $120 Proof.
Now J of 120 is 40 J of 144 is 48
i of it is _30 ^ of 144 is m
their sum is 70 their sum 84
which taiien from 120 taken from 144
leaves 50 leaves tiO as per question.
Then, 50 : 120 : : 60 : 144.
What number is that, which multiplied by 1, and the product
divided by 6, the quotient may be 14 ? 12.
PERMUTATIONS AND COJIBINATIONS.
The P^rMiMMiWsof any number of quantities signify the changes
which these quantities may undergo with respect to their order.
Thus, if we take the quantities a, h, c; then, a b e, a c h, h a e,
b e a, c ah, ch a, are the permutations of thesa three quantities
taken all together; a b, a e, b a, b c, c a, e b, are the permutations
of these quantities taken two and two; a, b, c, are the permutation
of these quantities taken singly, or one and one, &e.
The number of the permutations of the eight letters, a, h, e, d,
e, f, g, h, is 40320 ; becomes,
1.2.3.4.5.6.7.8 = 40320.
DOUBLE POSITION.
Double Position is the method of resolving certain questions
by means of two suppositions of false numbers.
To the Double Rule of Position belong such questions as have
their results not proportional to their positions : such are those, in
which the numbers sought, or their parts, or their multiples, are
increased or diminished by some given absolute number, which is
no known part of the number sought.
Take or assume any two convenient numbers, and proceed with
each of them separately, according to the conditions of the ques
tion, as in Single Position ; aad find how much each result is dif
ferent from the result mentioned in the question, noting also
whether the results are too great or too little
Then multiply each of the said errors by the contrary supposi
tion, namely, the first position by the second error, and the second
position by the first error.
If the errors are alike, divide the difierence of the products by
the difference of the errors, and the quotient will be the answer.
But if the errors are unlike, divide the sum of the products by
tho sum of the errors, for the answer.
The errors are said to be alike, when they are either botli too
great, or both too little ; and unlike, ^^ hen one is too grt:i! ■■:..') rl^c
other too little.
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MBNSUEAIION OP SUPERFICIES. 45
What number ia that, ■which, being maltiplied by 6, the prodnct
increased by 18, and the sum divided by 9, the quotient shall be 20.
Suppose the two numbers, 18 and 30. Then
First position. Second position.
18 30
6 mult.
6
108
180
18 add.
18
9)126
9)198
14 results.
22
. 20 true res.
20
+ 6 errors unlike.
^72
2dpos. 30 mult.
18 1st p
E,„.{
H
Sum 8 J ai6 sum of products.
27 answer sought.
Find, by trial, two numbers, as near the true number as possible,
and operate with them as in the question ; marking tlie errors
■which arise from each of them.
Multiply the difference of tho t'wo numbers, found by trial, by
the least error, and divide the product by tlie difference of the
errors, when they are alike, but by their sum when they are unlike.
Add the quotient, last found, to the number belonging to the
least error, when that number is too little, but subtract it when too
great, and the result will give the true quantity sought.
MENSTJEATION OF SUPERFICIES.
The area of any figure is the measure of its surface, or the space
contained within the bounds of that surface, without any regard to
thickness.
A square whose side is one inch, one foot, or one yard, &c. is
called the measuring unit, and the area or content of any figure is
computed by the number of those squares contained in that figure.
To find the area of a parallelogram; whether it he a square, a
rectangle, a rhombus, or a rhowhoides. — Multiply the length by the
perpendicular height, and the product will be the area.
The perpendicular height of the parallelogram is equal to the
area divided by the base.
Required the areaof the square ABCD whose ■
side ia 5 feet 9 inches.
Here 5 ft. 9 in. = 575 : and 5'T5^ == 675 x
575 = 330625 /eei = 33/e. in. 9pa. = area
required.
hv Google
46
THE PRACTICAL MODEL CALCULATOR.
Required the area of the rectangle
AECB, whose length AB ia 1375 chains,
and hi'eadth BC 95 chains.
Here 1375 X 95 = 130625; and
'^^^ = 130625 ac. = \Zac. OmlO
fo. = area required.
Required the area of the rhombus
ABCP, whose length AB is 12 foot 6
inches, and its height DE 9 feet 3 inches.
Here 12/e. 6 in. = 125, and^fe. 3 in.
= 925.
Whence, 125 X 925 = 115625 /«. =
115 /e. 7 in. Q pa. = area required.
What is the area of the rhom
boides ABCD, whose length AE is
1062 chains, and height DE 763
chains.
Rere 1052 X 763 = 802676; /
802676 /
= 802676 acm= 8 rtc. /
10
ro. 4po. area n ^
To find the area of a triangle, — Multiply the base bj the
pendicular height, and half the product wil! be the area.
The perpendicular height of the triangle is equal to twice 1
area divided by the base.
Required the area of the triangle ABC,
whose base AB is 10 feet 9 inelies, and
height DC 7 feet 3 inches.
fferelGfe.^in. =1075, and Ife. din.
= 725.
Whence, 1075 x 725 = 779375, and
779375
^ = 3896875 feet = 38 fe. 11 in. ^ "
7^ pa. = area required.
To find the area of a triangle whose three sides only are crJra
From half the sum of the three sides subtract each side severi'
Multiply the half sum and the three remainders continually t>
ther, and the square root of the product will he the area v^jiaii
Required the area of the triangle ABC, □
whose three sides BC, CA, and AB are
24, 36, and 48 chains respectiyely.
,, 24 + 36 + 48 108
Mere s = ^ = 54 =
J sum of the sides.
Also, 54  24 = 30>s( diff. ; 5436 "*
= 18 second diff.; and 54 — 48 = 6 third diff:
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MENSURATION Off SCPERFICIE3.
4T
<18>
i = V'174£i60 = 418282 = a
Whence, \/ 54 >
required.
Any two sides of a right angled triangle being given to find
the third side. — When the two legs are given to find tlie hjpo
thennse, add the square of one of the lega to the square of the
other, and the square root of the sum will he equal to the hypo
thenuse.
When the hypothenuae and one of the legs are gtvtin to find the
other leg. — From the square of the hypothenuse take the square
of the given leg, and the square root of the remainder will be equal
to the other leg.
In the right angled triangle ABO, the n
base AB is 56, and the perpendicular EC 33,
what is the hypof henuse ?
Mere 56' + 33* = 3136 + 1089 = 4225,
and \/4225 = 65 = hypothenuse AC.
If the hypothenuse AC be 53, and the
base AB 45, what is the perpendicular BC ? '^ "
Sere 58* — 45« = 2809 — 2025 = 784, and v'784 = 28 =
perpendicular BC.
To find the area of a trapezium. — Multiply the diagonal by the
sum of the two perpendiculars falling upon it from the opposite
angles, and half the product will be the area.
Required the area of the trapezium
BAED, whose diagonal BE is 84, the
perpendicular AC 21, and DE 28.
Here 28 + 21 X 84 =49 X 84=4116,
,4116
and — q — = 2058 the area required.
To find the area of a trapezoid, or a quadrangle, two of kIiosc
opposite sides are parallel — Multiply the sum of the parallel side?
by the perpendicular distance between them, and half the product
will be the area.
Required the area of the trapezoid ABCD,
whose sides AB and DC are 321'51 and
21424, and perpendicular DE 17116.
Here 32151 + 21424 = 53575 = 3«,w
of the parallel sides AB, DC.
Whence, 53575 x 17116 (theperp. DE) =
916980700, and ^ = 45849485 the area required.
To find the area of a regular polygon. — JluUipIy half the peri
meter of the figure by the perpendicular falling from its centre
Hpon one of the sides, and the product will be the area.
The perimeter of any figure ia the sum of all its sides,
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THE PRACTICAL MODEL CALCULATOa.
Required the area of the regular p
ABCBE, whose side AB, or UC, ,
feet, and the perpendicular OP 17'2 feet.
Mere — ^ — ■ = 62'5 = half perimeter
flwd625x lT2 = Vilb square feet = arci
required.
To find the area of a regular polygon, when the side i
given. — Multiply the square of the side of the polygon by the
number standing opposite to its name in the following table, and
the product will bo the area.
^id.f
N^...
Mullipliers.
"to^
Na„^.
SlQlUplfcr^.
3
4
5
6
7
Trigon or equil, A
Tetragon or square
Pentagon
Heiagoa
Heptagon
0'4330ia
1000000
1 720177
2 ■698076
3 033912
8
10
11
12
Octagon
Nonagon
Decagon
Duodeoagon
4828427
6181824
7 ■694200
9S6.3e40
llliim52
The angle OBP, together with its tangent, for any polygon of not
more than 12 sides, is shown in tbo following table :
m,t
NamM.
Angle
Oiit>.
Tangfliils.
3
4
5
T
8
10
11
12
Trigon
Tetragon
Pentagon
Hesagon
Heptagon
Octagon
Nonogon
Decagon
Undeoagon
30°
45°
54°
60"
64°4
67''i
70'^
73° jV
75°
■57735 = J v'3
100000 = 1x1
1'37C38 = v'l+ f v^^
173205 = v/3
207652
241421 = 1 + v'^
274747
807768 = ^5 + 2 va
340568
373205 = 2 + ^3
Required the area of a pentagon whose side is 15.
The number opposite pentagon in the table is 17204V7.
Hence 1720477 x 16= = 1720477 X 225 = 387107825 =
area required.
The diameter of a circle being given to find the circumference,
or the circumference being given to find the diameter. — Multiplv
the diameter by 31416, and the product will be the circumfer
ence, or
Divide the circumference by 31416, and the quotient will be the
diameter.
As 7 is to 22, so ia the diameter to the circumference ; or as 22
is to 7, so is the circumference to the diameter.
As 113 ia to 355, so is the diameter to the circumference; or,
as 352 ia to 115, so is the circumference to the diameter.
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MENSURATION OF SUPERFICIES. 49
If the diameter of a circle be 17, what ia the circuraferenco ?
Here 31416 X 17 = 534072 = circimferenee.
If the circumference of a circle be 354, what ia the diameter ?
354000
Here ' o.j ±i a = 112681 = diameter.
To find the length of any arc of a circle. — When the chord of
the arc and the versed sine of half the arc are given :
To 15 times the square of the chord, add 33 times the square of
the versed sine, and reserve the number.
To the square of the chord, add 4 times the square of the versed sine,
and the square root of the sum will be twice the chord of half the arc.
Multiply twice the chord of half the arc by 10 times the square
of the versed sine, divide the product by the reserved number, and
add the quotient to twice the chord of half the arc : the sum will
be the length of the arc very nearly,
Wlien^e chord of the arc, and the chord of half the arc are
given. — From the square of the chord of half the arc subtract the
square of half the chord of the arc, the remainder will be the square
of the versed sine : then proceed as above.
When the diameter and the versed sine of half the arc are given :
From 60 times the diameter subtract 27 times the versed sine,
and reserve the number.
Multiply tho diameter by the versed sine, and the sqnaro root
of the product will be the chord of half the arc.
Multiply twice the chord of half the arc by 10 times the versed
sine, divide the product by the reserved number, and add the quo
tient to twice the chord of half the arc ; the sum will be the length
of the arc very nearly.
When the diameter and chord of the arc are given, the versed
sine may be found thus : From the squaro of tho diameter subtract
the square of the chord, aud extract the square root of the re
mainder. Subtract this root from the diameter, and half the re
mainder will give the versed sine of half the arc,
Tlie square of the chord of half the arc being divided by the
diameter will give tho versed sine, or being divided by the versed
sine will give the diameter.
Thelength of the arc may alsobe found by multiplying together the
number of degrees it contains, the radius and the number 01745329.
Or, as 180 is to the number of degrees in the arc, so is 3'1416
tunes the radius, to the length of the arc,
Or, as 3 is to the number of degrees in the arc, so is 05236 times
the radius to the length of the arc. ^
If the chord DE be 48, and the versed sine n / I \ ,.
CB 18, what is the length of the arc? / ^ ',
Here 48^ X 15 = 34560 \ '■ \
18^ X 33 ^ 10692 \^ !
45252 reserved number. '",. ; ,.
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50 THE PRACTICAL MODEL CALCULATOR.
48^ = 2304 = the square of the chord.
18^ X 4 = 1296 = 4 timeB the square of the versed sine.
^ 3600 = 60 = twice the ehord of half the arc.
60 X 18^ X 10 194400 „ , ,
NoiB — .nig =  .gijn = 4'2959, which added to twice
the chord of half the arc gives 642959 = the length of the arc.
50 )
2514 reserved number.
AC = v/SO X 18 = 30 = the chord of half the arc.
30"X2 X 18 X 10 10800
2514 ~ 2514
=.42959, which added to t
chord of half the arc gives 642959 = the length of the arc.
To find the area of a ciVeZe.— Multiply half the circumference by
half the diameter, and the product will be the area.
Or take ^ of the product of the whole circumference and diameter.
What is the area of a circle whose diameter is 42, and circum
ference 131946?
2 ) 131946
65973 = J circumference,
21 =  diameter.
"65973
131946
1385'433 ■= area required.
What 13 the area of a circle whose diameter is 10 1
and circumference 31 feet 6 inches ?
fe. in.
15 9 = 1575 = ^ eireumference.
5 3 = 525 = i diameter.
7875
3150
7875
826875
12
82500
82 feet 8 inches.
Multiply the square of the diameter by 7854, and the product
will be the area ; or.
Multiply the square of the circumference by 07958, and the
product will.be the area.
The following table will also show most of the useful problems
relating to the circle and its equal or inscribed square.
Diameter X 8862 = side of an equal square.
Cireumf. X '2821 = side of an equal square.
Diameter x 7071 = side of the inscribed square.
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MENSDRATION OP STIPERFICIES. 51
Circumf. X 2251 = side of the inscribed square.
Area X '6366 = sido of the inscribed square.
Side of a square x 14142 = diam. of its circums. circle.
Side of a square X 4443 = circumf, of its circums. circle.
Side of a square x 1*128 = diameter of an equal circle.
Side of a square x 3'545 = circumf. of an equal circle.
"VVhat is the area of a circle whose diameter is 6 ?
7854
25 = square of the diameter.
"39270
15708
19'6350 = tJie answer.
To find the area of a sector, or that part of a circle wh'ch w
hounded hy any two radii and their included are. — Find the length
of the arc, then multiply the radius, or half the diameter, by the
length of the arc of the sector, and half the product will be the
area.
If the diameter or radius is not given, add the square of half the
chord of the are, to the square of the versed sine of half the arc ;
this sum being divided by the versed sine, will give the diameter.
The radius AB is 40, and the chord BC
of the whole arc 50, required the area of n
the sector, ,ii;^^^:^^^^^^n "^"""^
80  k/W  50^ \ \'^ / '
2 — = 87750 = the versed \ ; /
sine of half the are. \ i /
80 X 60  87750 x 27 = 45630750 =
the reserved number.
2 X v'87750 X 80 = 529906 = twice the
chord of half the arc.
529906 X 87750 x 10 , , , ,
4XM^07"'CO ~ I'OlyO, which added to twice the chord
of half the arc gives 540096 the length of the arc.
540096 X 40
And — 2 = 10801920 = area of the sector required.
As 360 is to the degrees in the arc of a sector, so is the area of
the whole circle, whose radius is equal to that of the sector, to the
area of the sector required.
For a semicircle, a quadrant, &c. take one half, one quarter, &c,
of the whole area.
The radius of a sector of a circle is 20, and the degrees in its
arc 22 ; what is the area of the sector ?
Kere the diameter is 40.
Eenee, the area of the circle = 40= X 7854 = 1600 X 7854 ■=
125664,
Mow, 360° : 22° : : 125664 : 767947 = area of the sector.
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52 THE FBAOTICAL MODEL CALCULATOK.
To find the area of a segment of a circle. — Find the area of
the sector, having the same arc with the segment, bj the last pro
lilem.
Find the area of the triangle formed by the chord of the seg
ment, and the radii of the sector.
Then the sum, or difference, of these areas, according as the
segment is greater or less than a semicircle, will be the area re
quired.
The difference between the versed sine and radius, multiplied bj
\in\f the chord of the arc, will give the area of the triangle.
The radius OB is 10, and the chord AC 10 ;
what is the area of the segment ABC ?
^^ AG^ 100 ^ ,
CD = TTpT = "oa" = 5 = the versed
of half the arc.
20 X 60  SlTST = 1065 = the reserved
number.
rolTa xT ':
1065
to twice the chord of half the arc gives 20'9390 = the length of the arc.
209390 X 10
2 = 1046950 = area of the sector OACB.
OD = OC = CD = 5 the perpendicular height of the triangle.
AD = v/AO^  0D= = v'75 = 86603 = i the chord of the arc.
86603 X 5 = 433015 = the area of the triangle AOB.
1046950  433015 = 613935 = area of the segment required;
it being iu this case less than a semicircle.
Divide the height, or versed sine, by the diameter, and find the
quotient in the table of versed sines.
Multiply the number on the right hand of the versed sine by the
square of the diameter, and the product will be the area.
When the quotient arising from the versed sine divided by the
diameter, has a remainder or fraction after the third place of deci
mals ; having taken the area answering to 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, then the
sum will be the area for the whole quotient.
If the chord of a circular segment be 40, its versed sine 10, and
the diameter of the circle 50, what is the area ?
50 ) 10
■2 = tabular versed sine.
■111823 = tabular segment.
2500 = square of 50.
55911500
= area required.
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MEH8URATI0N OF S0PERFIOIES.
53
To find the area of a circular zone, or the epace included between
any two parallel chords and their intercepted arcs. — From the
greater chord subtract half the difference between the two, mul
tiply the remainder by the said half difference, divide the product
by the breadth of the zone, and add the quotient to the breadth.
To the square of this number add the square of the less chord, and
the square root of the sum will be the diameter of the circle.
Now, having the diameter EGr, and the two chords AB and DC,
find the areas of the segments ABBA, and DCED, the difference
of which will be the area of tho zone required.
The difference of the tabular segments multiplied by the square
of the circle's diameter will give the area of the zone.
When the larger segment AEB is greater than a semicircle, find
the areas of the segments A&B, and DCE, and subtract their sum
from the area of the whole circle ; the remainder will be the area
of the zone.
The greater chord AB is 20, the less DC 15, ,,
and their distance Dr IT^ : required the area .'"'""'''. ^
of the zone ABCD. ' "^ ^
20 — 15
^ = 25 = J = the difference ietween
the chords.
) = DF.
175
3 + i
And v'20^ + 16^ = \/Q25 = 25 = the diameter of the circle.
TJie segment AEB ieinff greater than a semicircle, we find the
versed sine o/DCE = 25, and that of AGB = 5.
Eence ^ = 100 = tabular versed sine of DEC.
And 2c = "200 = tabular versed sine of AGE.
Mw 040875 X 25' = area of seg. DEC = 25546875
And 111823 x 25^ = area of seg. AGB = 69889375
sum 9543625
•7854 X 25' = area of the whole circle, = 49087500
Difference = area of the zone ABCD = 39543875
To find the area of a circular ring, or the
space included between the circumference of
two concentric circles. — The difference between
the areas of the two circles will be the area of
the ring.
Or, multiply the sum of diameters by their
difference, and this product again by "7854,
and it will give the area required.
The diameters AB and CD are 20 and 15 : required the a
hv Google
THE PRACTICAL MODEL CALCULATOR.
ilie circular ring, or the s
rencea of those circles.
the cir cum fe
nfire AB + CD X AE CD = 85 X 5 = 1T5,«»(?176 x 7854 =
13T"4450 = area of the nng rejuind
To find the areas of lune% or tie y a es httween the tnt 'rsectinff
arcs of two eccentric circles — rm I the ireas of the two <<egmants
from which the lune is f rnip] and their difference mil he the
area required.
The following property is one of the moot eun u
If ABC be a right angled t iingle
and semicircles he described on the three
sides as diameters, then will the ii 1 tri
angle be equal to the two lunes D and F
taken together.
For the semicircles described on AC and
BO = the one described on AB, from each
take the segments cut off by AC and EC, then will the lunes AFCE
and EDC& = the triangle AOB.
The length of the chord AB is 40, the
height DC 10, and DE 4 : required the
area of the lune ACBEA.
The diameter of the circle of lohich ACB
20^ + 10^
is a part = "^'Tji = ^^^
Arid the diameter of tJie circle of which KE^ ii a part =
= 104,
Now having the diameter and versed sines, we find,
The area of seg. ACB = 111823 x 50^ = 2795575
Andareaofseg.AEB = 009955 X 104^= 1076T33
Their difference is the area of the lune \
AEBCA required, J
To find the area of an irregular polygon, or a figure of any
number of sides. — Divide the figure into triangles and trapeziums,
and find the area of each separately.
Add these areas together, and the si
of the whole polygon.
Required the area of the irre
gular figure ABCDEFGA, the fol
lowing Enes being given :
GB = 305 A»ill2, CO = 6
CD = 29 Fo =11 Cs=66
FD=248 Ejj=4
^ An + Go ^^ 112 + 6
Eere  — ^ X GB = „
= 1718842
\ will he equal to the area
X 305 + 86 X 305 = 2623
area of the trapetium ABCG,
hv Google
DECIMAL APPROXIMATIONS.
,li±9i
area of the trapezium GCDF.
., FD X Ep 248 X 4
Also, — o — — = s—
FDE.
Whence 2623 + 2552 + 406 =
figure required.
 X 29 = 88 X 20 = 2652 =
ij= 496 = area of the triangle
= arua of the whole
Lineal feet multiplied by 00019 = mifes.
— yards
—
000568
=
Squm inches
—
■007
=
square feet.
 ,.rds
—
•0002061
' =z
acres
Circular inches
—
■00546
=
square feet.
Cylindrical inches
0004546
i =
cubic feet.
— feet
■02909
=
cubic _
yards.
Cubic inclies
—
00058
=
cubic feet.
— feet
—
■03704
=
cubic yards.
— —
—
C232
=
imperial gallons.
— inches
■003607
=
Cylindrical feet
—
4895
=
—
 inches
~
■002832
—
Cubic inches
—
■263
=
tt)S. avs. of cast iron.
—
■281
=
wrought do.
—
—
283
=
—
Steel.
—
■3225
=
_
copper.
—
—
■3037
=
—
brass.
26
—
■4108
=
lead'.
—
—
■2636
=
—
tin.
—
—
■4908
=

mercury.
Cylindrical inches
■2065
cast iron,
—
—
■2168
=
—
wrought iron.
—
—
2223
=
—
steel.
—
—
■2533
=
„
copper.
■2385
brass.
•2042
=
„
zinc.
■
3223
=
lead.
—
—
■207
=
—
tin.
—
™
■3S54
=
—
mercury.
Avoirdupois lbs.
—
009
=:
cwts.
—
—
00045
=
tons.
183346 circular inches
=
1 square foot.
2200 cylindrical inches
=
1 cubic foot.
French metres X 3281
=
feet.
— kilogrammef
1 X 2
■205
=
avoirdupois lb.
— grammes X
■002205
=
avoiii
iupois lbs.
b,Google
56 lUE PKACTICAL MODEL CALCULATOR.
Diameter of a sphere X '800 = dimensions of equal cube.
Diameter of a sphere x 6667 = length of equal cjlimler.
Lineal inches X '0000158 = miles,
A Prcneh cubic foot = 20934T cuhic inches.
Imperial gallons X '7977 = New York gallons.
The average quantity of water that falls in rain and snow at
Piuladelphia is 36 inches.
At West Point the variation of the magnetic needle, Nov. 16th,
1839, was 7° 58' 27" West, and the dip 73° 26' 28".
One inch, the integer or wliole number.
■96875 J + A 625 j
•28125 1 + 1,
■9376 J + A 69375 j + ^,
•25 i
■90625 J f 3^ ■5625 i + is
•21875 S + *
■875 S j ■53126 S i + ,<,
■1875 B 5 + A
■84376 a i + A ■s a i
•15625 a S f ,i
■8125 & j + A ■46875 & f f *
125 g. s
■78125 : 1 + A 4375 ^  + A
■09376 : ,1
■75 i t ■40626 a 1 + ,',
0625 S
■71875 1 + /s 375 f
03125 J,
•6875 i f A 34376 I + ,>,
•65625 ii + A '3126 I + A
One foot, or 12 inches, the integer.
■9106 „ llinclies.'i ■4166 „ 6 inches.? ^0625 „ S of in.
■6338  10 —
•3333  4 —
0620S " i —
■75 8 9 —
■26 13
04166 i 1 —
•6666 r 8 —
■1666 ra —
03125 II 
■5833 s 7 —
•0833 • 1 —
02083 • J —
■5 = 6 — 07291 3 } —
01041 a J —
One yard, or 36 inches, the integer.
■9722 35 inches.; 6389 23 inches., ■3055 1] inelics.
■9444 34 — ! 6111 22 — I ■2778 10 —
•9167 S3 — : ■5833 21 —
26 9 
•8889 „ 32 — 1 •5566 o 20 —
2222 „ 8 —
■8611;; 81 — 1' ■5278:; 10 —
■19442 T ^
■8333 g 30 — i 5 § 18 —
1667 Be —
8066 f 29 — ! 4722? 17 — ' 1389 g* 5 —
7778 S 28  :', 4444 "16  1111 s 4 
75 "27 — / 4167 ■'15 — !i 0833 » 3 —
7222 26 — 3889 14 — t 0565 2 _
6944  25 — soil 13 — " 0278 1 —
6667 24 — 3.!::'! 12 —
b,Google
'teontamingthe areumferenaea, Sqimrel, Cube!, and Arm of
OircUiJrom 1 to 100, idmneing Ig a tenti.
„Google
J CALCULATOR.
,v Google
CIRCLES, ADVAHCING BY A 1
: iii ^
iltSS i
ill luiioLii lik
fh SJ
„Google
TUB PKACnCAL MODEL CALCULATOR.
,v Google
0IEOLE8, ADVAHCING BY A 1
s, «;.; , J
•:zs
?.!:;«
db,Google
: PRACTICAL MODEL CALCULATOR.
dhvGoogle
TABLE OF THE LENGTH OF CIRCULAR ARCS.
Di^
Ci,=^
S,^ Cb..

"
a,=c^
a,.™.
CuSs.
AM.
39M
83IiS
S74 BB
976 44
978 21
9SWM
lODOO
976191'4SS
7683.1023
Al E / 7
L
/ fC
; A " >Thu
1 bt mg un ty
r«3.
w^
Unib.
^
L.n^h
000048
■^
OO 06
2 05
00005818
K
00O0J7
00 2 9
so
396 6o4
O'OOOS J
■A
0000145
4
698 32
6 968 '
4
00011f36
00000194
ft
08 665
no
463 93
h
00014544
ft
OO00D>42
20
2 0943 5
6
0000.91
22 80
fid
000208(2
7
0000539
«
396 63
so
H
0'00.3<71
K
0000„88
w
6 96
3666 9 4
H
0'0026180
0000436
240
4 88 902
Ml
00(lfj485
8490660
i 2 8
0006S1T8
no
0000970
m
88
flOO
52 6 8 8
0008T.66
HO
00014o4
4
6 8
41 >
0116355
40
0001139
6
08 2
360
6 28 85
60
014u444
50
0002424
Reiinn ed the length of a cii culai arc of 37° 42' 58" ?
30° = O5235088
7° = 01221730
40' = 00116355
2' = 00020368
50" = 00002424
8" = 00000888
The length 06582703 required in terms of the
raiiius.
1207° Fahrenheit = 1° of Wedgewood'a pyrometer. Iron melts
at about 166° Wedgewood; 200362° Fahrenheit.
Sound passes in air at a velocity of 1142 feet a second, and in
water at a velocity of 4700 feet.
Freezing water gives out 140° of heat, and may be cooled as
low as 20°. All solids absorb heat when becoming a fluid, and the
quantity of heat that renders a substance fluid is termed its caloric
of fluidity, or latent heat. Fluids in vacuo boil with 124° leas
heat, than when under the pressure of the atmosphere.
hv Google
THE PBACTICAI. DIODEL CALCULATOE.
Aeeas of the Segments and Zones of a Circle of which the Diameter
is Unity, and supposed to be divided into 1000 equal parts.
^
Araor
j™o(
HdsM.
i^i
"iSe"'
,
A™^nt
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s^iusnl.
S^.^M.
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■052
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■051906
■102
■042080
■101288
■003
■000219
•003000
■058
•016007
062901
•103
052687
■102267
■004
•000337
■004000
■064
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■053896
■104
■043296
■103246
■005
•000470
■005000
■056
■016911
064890
■105
■043908
■104228
■006
■000018
■006000
■056
■017369
■066883
■106
■044522
■105201
■007
000779
■007000
057
■017831
■056877
■107
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■106178
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■108
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068863
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046881
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019239
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■110
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060849
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■112
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■012999
■063
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124
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122717
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133086
■026
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075707
J20
■Oo7S26
124054
■027
■005867
■026987
077
■027821
076695
127
■067991
123f21
0^8
■006194
■027986
■078
028356
077683
128
■058658
12C5S8
029
■000527
■0289S4
■079
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078670
129
059327
127j35
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■029982
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079058
130
059099
128521
■031
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080645
181
■060672
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132
061348
130451
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007918
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082618
062026
131415
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■084
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083604
134
062707
132379
■035
■008688
■034972
■085
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084589
180
068389
lo!i342
■036
■000008
■035969
■086
■032745
085G74
180
061074
134304
■037
■009388
■036967
087
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08(w9
lu7
■O64700
1uj2U6
■038
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■088
088872
087544
138
015449
136228
■039
■010148
•089
■034441
131
Oj6140
187189
•040
■010537
■039968
■090
035011
089512
140
066833
138149
■0«
■010931
■040954
■091
■035585
■09049b
141
007628
139109
■042
■011330
■041951
•092
■036162
091479
142
0682.0
1*1008
■043
■011734
■042947
•098
036741
092461
143
063924
141026
■044
■012142
043944
■094
■037323
■09C444
■144
■009625
■141984
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■012554
044940
•095
■037909
■094426
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012971
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■088496
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b,Google
AREAS OP THE SEGMENTS AND ZONES OP A CIECLB.
HriBht,
Jib or 5,.
jSjBofKoM.
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^,«.rse,.
i«„rw
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261
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■152
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■149625
207
■117460
200924
262
164019
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153
076026
■150578
■208
■118271
201835
268
164899
■360212
■154
076747
■151530
■209
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202744
264
166780
251162
■165
■077469
152481
■210
■119897
203652
65
lb6663
262011
■156
■078194
■163431
■211
120712
204559
266
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206465
267
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263704
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270
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280
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■231
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■232497
286
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173231
282
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287
186329
■270362
■178
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■174166
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188995
224269
218
187234
■271170
■179
176100
■334
139841
■225153
389
188140
■371987
■180
■096134
176033
■236
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•22608b
290
189047
■273802
■181
■096908
176966
■236
■141537
■226919
291
189966
■273616
■182
■097674
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■337
142387
■237800
392
190864
■274428
■183
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■178828
■288
148238
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398
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191
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304
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284063
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202761
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251
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285648
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307
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263
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RULES rOE, FINDIK& THE AREA OP A CIRCULAR ZONE, ETC. 67
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To find the area of a segment of a circle.
Rule. — Divide the height, or versecl sine,
hy the diameter of the circle, and finil the
quotient in the column of heights.
Then take out the corresponding area, in the column of nrcas,
and multiply it hy the square of the diameter ; this ivill give the
area of the segment.
Beqoired the area of a segment of a circle, whose height is 3^
feet, and the diameter of the circle 50 feet.
3i = 325; and 325 h 50 = 065.
■065, by the Table, = 021659; and 021659 X 50^ = 54147500,
the area required.
To find the area of a circular zone.
Rule 1. — When the zone is less than a semicircle, divide the
height by the longest chord, and seek the quotient in the column
of heights. Take out the corresponding area, in the next column
on the right hand, and multiply it bythesquareof the longest chord.
Required the area of a zone whose longest chord is 50, and height 15.
15 ^ 50 = 300 ; and 300, hy the Table, = 280876.
Hence 280876 x 50^ = 70219, 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.
Required the area of a zone, the diameter of the circle being 50,
and the height of the zone on each side of the line which passes
through the diameter of the circle 20 and 15 respectively.
20 5 50 = 400 ; 400, by the Table, = 351824 ; and 351824 x
50= = 87956.
15 ^ 50 = 300 ; 300, by the Table, =280876 ; and 280876 X
50* = 70219. Hence 87956  70219 = 158175.
Approximating rule to find the area of a segment of a circle.
Rule. — Multiply the chord of the segment by the versed sine,
divide the product by 8, and multiply the remainder by 2.
Cube the height, or versed sine, find how often twice the length
of the chord ia contained in it, and add the quotient to the former
product ; this will give the area of the segment very nearly.
Required the area of the segment of a circle, the chord being 12,
and the versed si:
12 ;
= 24;
24
= 8 ; and 8 X 2 = 16.
2^ H 24 = ;
Hence 16 i 3333163333, the ai
a of the segment very nearly.
hv Google
PROPORTIONS OF THE LESGTHS OF CIRCULAR ARCS.
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PROrORTIOi^S OP THE LENGTHS Oi' SB.MIELLirTIC AEC3.
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72 THE PRACTICAL MODEL CALCULATOK.
To find the length of an arc of a circle, or the curve of a rigid
eemi^ellipse.
EuLB. — Divide the height by the hase, and the quotient will be
the height of an arc of which the base is unity. Seek, in the
Table of Circular or of Semielliptical arcs, as the case may be,
for a number corresponding to this quotient, and take the length
of the arc from the next righthand column. Multiply the number
thus taken out by the base of the are, and the product will be the
length of the arc or curve required.
In a Bridge, suppose the profiles of the arches are the arcs of
circles ; the span of the middle arch is 240 feet and the height 24 feet ;
required the length of the arc.
24 = 240 = 100 ; and 400, by the Table, is 102645.
Hence 102645 X 24 = 24634800 feet, the length required.
The profiles of the arches of a Bridge are all equal and similar
semiellipses ; the span of each is 120 feet, and the rise 18 feet ;
required the length of the curve.
28 H 120 = 233 ; and ■233 by the Table, is 119010.
Hence 119010 x 120 = 14281200 feet, the length required.
In this example there is, in the division of 28 by 120, a remainder
of 40, or onethird part of the divisor ; consequently, the answer,
14281200, is rather less than the truth. But this difference, in
even so large an arch, is little moro than half an inch ; tlierefore,
except where extreme accuracy is required, it is not worth com
puting.
These Tables are equally useful in estimating works which may
be carried into practice, and the quantity of work to be executed
from drawings to a scale.
As the Tables do not afford the means of finding the lengths of
the curves of elliptical ares which are less than half of the entire
figure, the following geometrical method is given to supply the
defect.
Let the curve, of which the length is required to bo found, be
ABC.
Produce the height line Bd to meet the centre of the curve
in ff. Draw the right line A^, and from the centre g, with the
distance gB describe an arc EA, meeting Ag in h. Bisect Ah
in i, and from the centre g with the radius gi describe the arc ik,
meeting dB produced to k; then ik is half the arc ABC.
hv Google
TABLE OF RECIPROCALS OP SUMBERS.
T3
A Table of the Reciprocals of Mimbers ; or (fie Decimal Frac
tions eoriesponding to Vulgar Feactiojsb of which the Numera
tor is unity or 1.
[In the follow n Tabl tl e Decimal fractions are Reciprocals
of the Denominat s f tho opposite to them ; and their product
is = unity.
To find the D mal ponding to a fraction having a higher
Numerator than 1 m It j ! the Decimal opposite to the given De
nominator, by th n Numerator. Thus, the Decimal corre
sponding to ^ being 015023, the Decimal to \\ will be 015625 X.
15 = 284375.]
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hv Google
THE PEACTICAL MOBBL CALCULATOR.
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1/607
001647446
1/668
001497006
1/547
001828154
1/608
001644737
1/669
001494768
1/548
001824818
1/600
001642086
1/670
001492537
1/549
001821494
1/610
001639844
1/671
001490313
1/550
001818182
1/611
001636661
1/672
00I488O95
1/651
001814882
1/612
001638987
1/673
001485884
1/562
001811694
1/613
001681321
1/674
00148368
1/568
001808318
1/614
001628664
1/676
001481481
1/654
001805054
1/615
001626016
1/676
00147929
1/556
001801802
1/616
001623377
1/677
001477105
1/656
001798561
1/617
O0I620746
1/678
001474926
1/557
001795882
1/618
001618123
1/679
001472754
1/558
001702115
1/619
001615509
1/680
001470588
1/559
001788909
1/620
001612908
1/681
001468429
1/560
001785714
1/621
001010306
1/663
001466276
1/561
001782531
001007717
001464129
1/562
001779359
001006136
l;084
001401988
1/503
001776199
1/624
001602564
1/085
001459854
b,Google
TABLE OP BECIPHOCALS OF NUMBERS.
TrtelWniT
DetlD..! ot
ProctLnn nr
lN=lm=l M
rrsition or
Numb.
B=dr™..J.
1,686
■001457726
1/747
■001338688
1/808
001287624
1/687
001455604
1/748
■001336898
1/809
001236094
'001453488
1/749
•001335113
1,810
001234568
1/689
■001481879
1/750
■001338333
1,811
001288016
1/690
■001449275
1/751
■001331558
1/812
001231627
1/691
■001447178
1/752
■001829787
1/813
001280012
1,692
■001445087
1/753
■001328021
1/814
001228601
1,698
■001443001
1/754
00132626
1/815
001226994
l/eai
■001440923
1/756
■O0I8245OS
1/816
001225499
i^ess
■001436849
1/766
■001822751
1/817
00122899
1/696
■001486783
1/767
■001821004
1/818
001223494
1/697
■00148472
1/758
001319261
l/Bia
001231001
1/698
■001432665
1,759
■001317623
1,'820
001219512
1/699
■001480615
1/760
•001816789
1/821
O0J218O27
1/700
■001428571
I/76I
00181406
001216545
1/701
001426534
1/762
■001312386
001215067
1/703
■001424501
1/763
001810616
001218592
1/703
001422475
1/764
■001308901
001212121
1/704
001420455
1/765
■00180719
001210664
1/T05
■00141844
■001305483
1,'827
00120919
1,706
■001410481
1A67
■001803781
1/828
001207729
1/707
■001414427
1/768
■001802088
001206273
1,708
001412429
1/769
■00130039
1/830
001204819
1,709
■001410437
1/770
■001298701
1/831
001203369
IJIO
■001108451
1/771
■001297017
1/833
001201923
1/711
■00140647
1/772
■001295337
1/833
00120048
1/712
■001404404
1/778
0012S3661
1/884
001199041
1/718
■001402525
1/774
■00129199
1/885
O011S7605
1/7U
■00140056
1/776
■001290328
1/836
001196173
1/71B
■001398601
1/776
■00128866
1/887
001194743
1/716
■001896648
L777
■001287001
1/888
001193817
1/17
■0013947
1/778
001285347
1/889
001191895
1/718
001893758
1/779
■001288697
1/840
001190476
1/719
■001390821
1/780
■O0I28206I
1/841
001189061
1/720
■001888889
1^781
■00128041
1/842
001187648
1/721
■001386963
1/782
■001278773
1/843
00118624
1/732
001886043
1/783
■001277189
1^44
001184834
1/728
■001888126
1/784
■00127551
1/845
001188482
1/724
■001881215
1/785
■001278885
1/846
001182088
1/725
■00187981
1/786
001272265
1/847
001180638
1/726
■00137741
1/787
•001270648
1/848
001179245
i;727
001876516
1/788
■001269036
l,fl49
001177856
1/728
■001373626
1/789
001267427
1^
001176471
1/729
■001871742
1/790
■001265823
1/861
001175088
1/780
■001369863
1/791
■001264228
1/952
001173709
1/781
•001867989
1/792
•003262626
1/858
■001172333
1/732
00136612
1/798
001261084
1854
■00117096
1/788
■001864256
1/794
■001269446
1^5
■001169591
1/784
■001862898
1/795
•001257862
1/856
■001168224
1/735
001360644
1/798
■001266281
001166861
1/736
■001358696
1/797
■001264705
001165501
1/737
001856862
1/798
■001263138
001164144
1/738
■001355014
1/799
■001251364
001162791
1/789
00185318
1/800
■00125
1/861
00116144
1/740
■001861351
1/801
■001248489
001160093
1/741
■001349628
1/802
■001246883
1/868
Wn68749
1/742
■001847709
1/803
■00124538
1/864
001157407
1/743
■001845895
1/804
■001248781
1/866
001156069
1/744
001344086
1/805
•001242236
001154734
1/745
■001342282
1/806
■001240695
1/807
001153403
1/746
■001840488
1/807
■001339157
1/868
001152074
b,Google
THE PRACTICAL MODEL CALCULATOR.
rt«tti=n.r
DeoLiml ot
F™ n
Ff^i^or
»ecim„l or
Nnmh.
Kmin,o.»i.
1,B69
■001150748
19U
109 J
1/957
001044932
1/870
■001149425
1914
00109409"
1/958
001043841
1/871
■001148106
lIlS
O0109'>896
1/959
001042758
1/872
■001146789
1916
OOIO^I 08
1/960
001041667
1/873
■001145475
1917
001090518
1/961
001040583
1/874
■001144165
1918
001089S25
1/962
O0I089501
1/875
■001142857
1119
001088139
1/968
0010S8422
■001141553
19
00108 957
1/964
001037344
1/877
■001140251
Ifi 1
00108 6
1/965
00108C269
1/878
■001138952
19 2
001084o99
1/566
001085197
■001187656
1/9 8
0010834 3
1/967
0010S4126
1/880
■001188364
ld24
00108 ol
1/968
001033058
1^81
■001135074
1/9'>J
001081081
1/969
001031992
1/882
■001183787
1/9 6
001U79914
1/970
001080928
■001182503
1/527
0010 8 49
1/971
001029860
1/881
■001131222
J/9 8
0010 86
1/972
001028807
1/885
■001129944
19 9
0010 04''6
1/978
001027749
■001128668
1980
00107 j269
1/974
O0I020694
1/887
■001127896
1031
001074114
1/975
001025641
■001126126
193
0010 961
1/976
00102459
■O0II24859
0010 1811
1/977
■O0IO2SS41
1/890
■001123596
1/934
001070664
1/978
001032495
1/891
■001122334
1935
001069610
1/979
00102145
■001121076
19J6
001068 6
1/080
001020408
1/893
■001119821
1'937
OOlOfa "06
1/981
001019108
1/894
■001118568
001066098
1/982
00101838
1/895
■001117818
10o9
001064963
1/983
001017294
1/890
001116071
1,940
OOlOfi 83
1/984
00101626
1/897
■001114827
1/941
0010G2699
1,'985
001015228
1/898
■001118586
1/542
001061571
001014199
1/899
■001112347
1/913
001080446
1/987
001013171
1/900
■OOlllim
1/944
001059322
001012146
i;'901
■001109878
1/945
001058201
1/939
001011122
1/902
■001108647
1/946
001057082
1,990
001010101
1/903
■00110742
1/947
001055966
1/991
001009082
1/904
■001106195
1/948
0O1054S52
1/992
001008065
1/005
■001104S72
1/949
001053741
1/D93
001007049
1/906
■001108753
1/950
001052032
1/994
00100B036
1/907
■001102536
1/951
001051525
1/995
001005025
1/908
■001101322
1/962
00105042
1/996
001004010
1/909
■00110011
1/953
001049318
1/997
001008009
1/910
■001 098901
1/954
001048218
1/998
00IOO2004
1/^11
■001091695
1/955
00104712
1/909
O0!001001
1/912
■001096491
1/956
001046025
1/1000
001
Divide 80000 by 971.
By the above Tahleive find thatl divided by 9T1 g
and ■001029866 x 80000 = 8238928.
What is the sum of 5J5 and ^ ?
883
1
<I5S
+
•001132603
■001049318
2
953
X
5
=
■005602315
■002098636
6
883
■0077S1141
b,Google
MENSURATION OP SOLIDS.
D.

d™ ^
q™
4^
JO
D«
42
OS
045
43
or
44
68
4)
00
00
00
01
00
"_«
00
01
14
58
to
06
00
00
000
Tof d h s f b h
of Us i d J g — M pyh i
cube by itself, and that product again by the side,
and it will give the solidity required.
The side AB, or BO, of the cube ABODFOHE, ^
is 25'5 : what ia the solidity ?
Sere AB' = (225)'  255 X 255 X 255 
256 X 66025  16681'S76, emtent o/tlie ciiS«.
ni
bvGoogle
\
\
°
\
\
80 THE PRACTICAL MODEL CALCULATOR.
To find the eoUdity of a paralUhpipedon
— Multiply the Icngtli by the breadth, and
that product again by the depth or altitude,
and it will give the solidity required.
Required the solidity of a parallelopipedon , _
ABODFEHG, whose length AB ia 8 feet, ^
its breadth FD i^ feet, and the depth or
altitude AD 6 feet ?
Sere AB X AD X FD = 8 X 675 X 4.5 = 54 X 45 = 243 solid
feet, the contents of the parallelopipedon.
To find the solidity of a prism. — Multiply the area of the base
into the perpendicular height of the prism, and the product will be
the solidity.
What is the solidity of the triangular prism ABOF
ED, whose length AB is 10 feet, and either of the
equal aides, BC, CD, or DE, of one of its equilateral
ends BCD, 2 feet?
Rere } X 25= X ^/3 = J X 625 X x/3 = 15625
K s/d = 16625 X 1732 = 270625 = area of the
lase BCD.
0,^ 25 + 2.5 + 25 ^ r^_ g_^^ ^ ^ ^^^^ ^^
3 differences.
the sides, BC, CD, DB, of the triangle CDB.
And 375  25 = 125, .. 125, 125 and 125 =
WJience ^/375 X 125 X 125 X 125 = s/S'TS X l2o« =
v'732421875 = 27063 = area of the base as before,
And 27063 X 10 = 27063 solid feet, the content of the prism
required.
To find the convex surface of a cylinder. — Multiply the peri
phery or circumference of the base, by the height of the cylinder,
and the product will be the convex surface.
What is the convex surface of the right cylinder
ABCD, whose length BC is 20 feet, and the diame
ter of its base AB 2 feet ?
Here 31416 X 2 = 62832 = periphery of the
base AB.
And 62832 X 20 = 1256640 square feet, the ,
convexity required.
To find the solidity of a cylinder. — Multiply the area of the
base by the perpendicular height of the cylinder, and the product
will be the solidity.
What is the solidity of the cylinder ABCD, the diameter of
whose base AB is 30 inches, and the height BC 50 inches.
Here 7854 X 30^ = 7854 X 900 = 70686 = area of the lase AB.
35343
And 70686 X 50 = 35343 cubic inches; or
solid feet.
1728
= 204531
hv Google
for finding the su
MENSURATIOX OF SOLIDS.
The four following cases contain all the rules
perficiea and solidities of cylindrical ungulas.
When the section is parallel to the axis of the cylinder.
Rule. — Multiply the length of the arc line of the base
by the height of the cylinder, and the product will be
the curve surface.
Jlultiply the area of the base by the height of the ^
cylinder, and the product will be the solidity.
^Vhen the section passes obliquely through the opposite
sides of the cylinder.
Rule. — Multiply the circumference of the hase of the
cylinder by half the sum of the greatest and least lengths
of the ungula, and the product will be the curve surface.
Multiply the area of the base of the cylinder by half ^
the sum of the greatest and least lengths of the unguU, and the
product will be the solidity.
WJien tJie section passes through the hase of the cylin j r ^
der, and one of its sides. ^ ^ '^
Rule.— Multiply the sine of half the arc of the base y j,
hy the diameter of the cylinder, and from this product / ;
subtract the product of the arc and cosine. /... J
Multiply the difference thus found, by the quotient of isI^At^a
the height divided by the versed sine, and the product o
will be the curve surface.
Prom I of the cube of the right sine of half the are of the base,
subtract the prodact of the aroa of the hase and the cosine of the
said half arc.
Multiply the difference, thus found, by the quotient arising from
the height divided by the versed sine, and the product will be the
solidity. ^
When the section passes obliquely through both ends
of the cylinder.
Rttle. — Conceive the section to be continued, till it
meets the side of the cylinder produced ; then say, as
the difference of the versed sines of half the arcs of the
two ends of the ungula is to the versed sine of half the j
arc of the leas end, so is the height of the cylinder to
the part of the side produced.
Find the surface of each of the ungulas, thus formed, and their
difference will be the surface.
In lilie manner find the solidities of each of the ungulas, and
their difference will be the solidity.
To find the convex surface of a right cone. — Multiply the circum
ference of the base hj the slant height, or tho length of the side
of the cone, and half the product will be the surface required.
The diameter of the base AB is 3 feet, and the slant height
AC or EC 15 feet; required the convex surface of the cone
ACE.
hv Google
THE PRACTICAL MODEL CALCULATOR.
Here 31416 X 3  94248 .
And
94248 X 15 1413T20
,mf,r.
e of the hose AB.
; 70'686 sqiiarefeet, the convex
surface required.
To find the convex surface of the frustum of a right cone. — Mul
tiply the sam of the perimeters of the two ends, by the slant height
of the frustum, and half the product will be the surface required.
In the frustum ABDE, the circumferences of
the two ends AB and DE are 225 and 1575
respectively, and the slant height BU is 26 ; what
is the convex surface t
(225 + 1575) X 26
S^re ^ '^^^^'"J "t = 225 + 1575
X 13 = 3825 X 13 = 49725 = convex sur
face.
To find the solidity of a eone or pyramid. — Multiply the area of
the base by onethird of the perpendicular height of the cone or
pyramid, and the product will be the solidity.
Required the solidity of the cone ACB, whose
diameter AB is 20, and its perpendicular height
OS 24.
Here 7854 x 20= = 7854 x 400 = 31416
= area of the base AB.
24
And 31416 X 3 = 31416 X 8 = 251328 / ;.._.^_^ \^
= solidity required.
Required the solidity of the hexagonal pyra
mid ECBD, each of the equal sides of its base
being 40, and the perpendicular height C8 60,
Mere 2598076 {multiplier when the side is 1)
X 40'^ = 2598076 X 1600 = 41569216 = area
of the base.
60
And 41569216 x ^ = 41569216 x 20 =
■ 3
83138432 soUditi/.
To find the aoUdity of a frustum of a cone or pyramid. — For
the frustum of a cone, the diameters or circumferencea of the two
ends, and the height being given.
Add together the square of the diameter of the greater end, the
square of the diameter of the less end, and the product of the two
hv Google
MENSURATION OP SOLIBS.
S3
diameters ; multiply the sum by 7854, and the product by the
height ; ^ of tlie lust product will be the solidity. Or,
Add together tiie square of the circumference of the greater
end, the square of the circumference of the less end, and the pro
duct of the two circumferences; multiply the sum by 07958, and
the product by the height ; J of the last product will be the solidity.
For the frustum of a pyramid whose sides are regular polygons. —
Add together the square of a side of the greater end, the square
of a side of the less end, and the product of these tivo sides ; mul
tiply the sum by the proper number in the Table of Superficies, and
the product by the height ; J of the last product will be the solidity.
When the ends of the pyramids are not regular polygons. — Add
together the areas of the two ends and tho square root of their
product ; multiply the sum by the height, and J of the product
will be the solidity.
What is the solidity of the frustum of the cone
EABD, the diameter of whose greater end AB is
5 feet, that of the less end ED, 3 feet, and the
perpendicular height Ss, 9 feet ?
(5^ 4 3' + 5 X 3) X 7854 x 9 3463614 _
\ 3  3 ~ ^
115'4538 solid feet, the content of the frustum.
What is the solidity of the frustum eEDBS of a
hexagonal pyramid, the side ED of whose greater
end is 4 feet, that eh of the less end 3 feet, and
the height Ss, 9 feet ?
(4^ + 3^ + 4 X 3) X 2598076 x 9 865159308
= 28838643S solid feet, the solidity required.
The following cases contain all tho rules for finding the superficies
and solidities of conical ungulas.
Wlten the section passes through the opposite extremities of the
ends of the frustum.
Let D = AB the diameter of the greater end;
d = CD, the diameter of the less end ; /( = perpen
dicular height of the frustum, and n = '7854.
d'~ d yjyd nOh
Then — p. _ , — ■ x —5— = solidity of the greater
elliptic ungula ADB.
D ^Dd — d^ ndh , ,
— jj _ J — ■ X ^ = solidity of the less ungula ACD.
= difi"erencc of these L
J)~d ^ 3
,And jj^ v'"4"A= + {1),^) X (D"^ ^— ^/Dd = curve
surface of ADB.
hv Google
84 THE PEACTICAL MODEL CALCULATOR.
WTien the section cute off parts of the base, and makes the angle
D)B less than the angle CAB,
Let S = tabular segment, whose v ersed sine i
Bj*^D; s — tab. seg. whose versed sine is
V d, and the other letters as above.
The {SxD'sXd'x
X p _ , = solidity of the elliptic hoof EFBD.
. , 1 T, — '^ ix{D+d)Ar
And^^::^ ^/4A^ + (D  df x{Beg. FEE ^, x ^ "^ _ ^^ 
X ■/ . _ i^ X seg. of the circle AB, whose height is D X — t — )
= convex surface of EFBD,
When the section is parallel to one of the sides of the frustum.
Let A = area of the base FEE, and the other let
ters as before.
A XD — .
Then i^yzT^ ~ i"^ ^(^ ~ d) X d) X ^h = solidity
of the parabolic hoof EFBD. ^
And ^^^ ^W X (D  df X {seg. FEE  % D=d
X v'dxJ) — d) = convex surface of EFBD.
When the section cuts off part of the base, and makes the angle
DrB greater than the angle CAB.
Let the area of the hyperbolic section EDF =
and the area of the circular seg. EEF = a.
the hyperbolic ungula EFBD.
And ^ ^^ X x/U^ i (D df X (cir. seg. EEF —
d^ Br — * (D — d) Br
^ X — ^^Z^J^' V ^ = curve surface of EFBD,
D^ Br Dd BrdH
The transverse diameter of the hyp, seg, = p __ ^ _ p and the
conjugate = d V j. _ j _ t>, ) fi^oin which its area may be found
by the former rules.
To find the solidity of a euneus or wedge. — Add twice the length
of the base to the length of the edge, and reserve the number.
Multiply the height of the wedge by the breadth of the base,
and this product by the reserved number ; ^ of the last product
will be the solidity.
hv Google
MENSURATION OF SOLIDS.
How many solid feet are there in a wedge,
Tvhoso base is 5 feet 4 inches loDg, and 9 inches
broad, the length of the edge being 3 feet 6 inches,
and the perpendicular height 2 feet 4 inches?
Sere
170
(64 X 2 + 42) X 28 X 9
6
170 X 28 X 3
( 128 + 42) X 28 X 9
6
6
170 X 14 )
= 7140 solid
inches.
And 7140 ^ 1728 = 41319 solid feet, the content.
To find the solidity of aprismoid. — To the sum of the areas of
the two ends add foux times the area of a section parallel to and
equally distant from both ends, and this last sum multiplied by J
of the height will give the solidity.
The length of the middle rectangle is equal to half the sum of
the lengths of the rectangles of the two ends, and its breadth equal
to half the sum of the breadths of those rectangles.
What is the solidity of a rectangle prismoid,
the length and breadth of one end being 14 and
12 inches, and the corresponding aides of the other
6 and 4 inches, and the perpen^cular 30J feet.
Here 14 X 12 + e^Ht = 168 + 24 = 192 = c^
sum of the area of the two e
14 + 6 20
Also 
•■^ ='^^ = length of the middle rectangle.
12 + 4 16
= breadth of the middle rectangle.
:0 X 4 = 320 = 4 times the area of
Whence 10 X 8 X 4 =
the middle rectangle.
Or (320 + 192) x g^ = 512 x 61 = 31232 soUd inches.
And 31232 v 1728 = 18074 solid feet, the content.
To find the convex surface of a sphere. — Multiply the diameter
of the sphere by ita circumference, and the product will be tlie
convex superficies required.
The curve surface of any zone or segment will also be found by
multiplying its height by the whole circumference of the sphere.
What is the convex superficies of a globe
BCG whose diameter BG is 17 inches ?
Here 31416 X 17 X 17 = 534072 x 17 =
907"9224 square inches.
And 9079224 ; 144 = 6305 square feet.
hv Google
86 THE PRAOTICAI. MODEL CALCULATOR.
To find the solidity of a sphere or ghhe.— Multiply the cube of
the diameter by 5236, and the product wjll he the solidity.
What is the solidity of the sphere AEBO,
whose diameter AB is 17 inches ?
Sere IV X 5236 = 17 X 17 X 17 x 5236 =
289 X 17 X 5236 =4913 x 5236 = 25724468 '*
^»t(? 25724468 h 1728 = 148868 solid feet.
To find the solidity of the segment of a sphere. — To three times
the square of the radana of its base add the square of its height, and
this sum multiplied by the height, and the product again by '5236,
will give the solidity. Or,
From three times the diameter of the sphere subtract twice the
height of the segment, multiply by the square of the height, and
that product by 5236 ; the last product wiil he the solidity.
The radius Cw of the base of the segment *■
CAD is 7 inches, and the height An 4 inches ;
what is the solidity ?
Sere (7' X 3 + 4^ X 4 X 5236 = (49x3+4^ ^
x4x5236 = (147 + 4=)x4x5236 = (147+16)
x 4 X 5236 = 163 X 4 x 5236 = 652 x 5236
= 3413872 solid ineUs. '^Jl,''
To find the solidity of a frustum or zone of a sphere. — To the
sum of the squares of the radii of the two ends, add onethird of
the square of their distance, or of the breadth of the zone, and
this sum multiplied by the said breadth, and the product again by
15708, will give the solidity.
What is the solid content of the zone ABCD,
whose greater diameter AB is 20 inches, the
less diameter CD 15 inches, and the distance
nm of the two ends 10 inches ?
Here {W + 75= + ^) X 10 x 15708 =
(100 + 5625 + 3333) x 10 x 15708 = 18958
X 10 X 15708 = 18958 x 15708 = 297T92264 solid inches.
To find the solidity of a spheroid. — Multiply the square of the
revolving axe by the fixed axe, and this product again by '5236,
and it wiil give the solidity required.
■5236 is = i^ of S1416.
In the prolate spheroid ABCD, the
transverse, or fixed axe AC is 90, and
the conjugate or revolving axe DB is 70 ;
what is the solidity ? ^
Here DB^ x AC x 5236 = 70' x 90
X 5236 = 4900 x 90 X 5236 = 441000
X 5236 = 2309076 = solidity required.
hv Google
MENSURATION OF SOLIDS.
87
To find the content of the middle frustum of a spheroid, its
length, the middle diameter, and that of either of the ends, being
given, when the ends are circular or parallel to the revolving axis. —
To twice the square of the miildle diameter add the square of the
diameter of either of the ends, and this sum multiplied by the length
of the frustum, and the product again by 2618, will give tho solidity.
Where 2618 = ^ of 31416.
In the middle frustum of a spheroid
EFGH, the middle diameter DB is
50 inches, and that of either of the
ends EF or GH is 40 inches, and its
length nm 18 inches ; what is its soli
dity?
Mere (50' x 2 + 40') X 18 X 2618
= {2500 X 2 + 1600) X 18 X 2618 = (5000 + 1(300) x 18 x
•2618 = 6600 X IS X 2618 = 118800 x 2613 = 3110181 cuIh';
inches.
When the ends are elliptieal or perpendicular to the revolving
axis. — JIuhiply twice the transverse diameter of the middle sec
tion by its conjugate diameter, and to this product add the product
of the transverse and conjugate diameters of either of the ends.
Multiply the sum thus found by the distance of the ends or
the height of the frustum, and the product again by 2618, and it
will give the solidity required.
In the middle frustum ABCD of an ohlato «
spheroid, the diameters of the middle section
EF are 50 and 30, those of the end AD 40 _
and 24, and its height ne 18; what is the e^—
solidity ?
Here (50 X 2 x 30 f 40x24) X 18 X 2618
= (8000 + 960) X 18 X 2618 = 3960 X 18 x
■2618 = 71280 X 2618 = 18661104 = the soliditg.
To find the solidity of the segment of a spheroid, when the base
is parallel to the revolving axis. — Divide the square of the revolv
ing axis by the square of the fixed axe, and multiply the quotient
by the difference between three times the fixed axe and twice the
height of the segment.
Multiply the product thus found by the square of the height of
the segment, and this product again by 5236, and it will give the
solidity required.
In the prolate spheroid DEED, the trans "
verse axis 2 DO is 100, the conjugate AC 60,
and the height D» of the segment EDF 10 ;
what is the s
Mere (j^^, X 300  20) X 10= X
36 X 280 X lO^' X 5236 = 10080
5236 = 5277888 = tlte solidity.
■5236 
hv Google
88 THE PRACTICAL MODEL CALCULATOB.
When the base is perpendieular to the revolving axis. — Divide tlie
fixed axe by the revolving axe, and multiply the quotient by the
difference between three times the revolving axe and twice the
height of the segment.
Multiply the product thus found by the square of the height of
the segment, and this prodact again by 5236, and it will give the
solidity required.
In the prolate spheroid aEjr, the trans
verse axe EF is 100, the conjugate ab 60, !
the height an of the segment aKD 12 ; what
is the solidity ? ^
Mere 156 (= diff. of ?>ah and 2a«) X If " i
(= EF ^ a5 X 144 (= square of an) X 5236 b
= ^ "" X 144 X 5236 = 52 X 5 X 144 X 5236 = 260 x
144 X 5236 = 37440 x 5236 = 19603584 = the soUditij.
To find the solidity of aparabolio conoid. — Multiply the area of
the base by balf the altitude, and the product will be the content.
What is the solidity of the paraboloid ADB,
whose height Dra is 84, and the diameter BA
of its circular base 48 ?
Here 48^ x 7854 x 42 (= J Dm) = 2304 x
■7854 X 42 = 18095616 x 42 =760015872 ^
= the solidity.
To find the solidity of the frustum of a paraboloid, when its ends
are perpendicular to the axe of the solid, — Multiply the sum of the
squares of tbe diameters of the two ends by the height of the frus
tum, and the product again by 3927, and it will give the solidity
Required the solidity of the parabolic frus /^T'".
turn ABCd, the diameter AB of the greater end
being 58, that of the less end do 30, and the
height no 18.
Sere {58^ + 30=) X 18 X 3927 = (3364 + ^
900) X 18 X 3927 = 4264 x 18 x 3927 =
76752 x 3927 = 301405104 = the solidity.
To find the solidity of an hyperholoid. — To the square of the
radius of the base add the square of the middle diameter between
the base and the vertex, and this sum multiplied by the altitude,
and the product again by •6236 will give the solidity.
In the hyperholoid ACB, the altitude Qr
is 10, the radius Ar of the base 12, and the mid
dle diameter nm 158745 ; what is the solidity ?
< 5236 =
Here 15874 5^  12^ x 10
25199975  144 x 10 x 5236 =
lOx 5236 = 39599975 x5236 =
= the solidity.
39599975 x
2073454691 "
hv Google
MENSURATION OS SOLIDS. 89
To find the Bolidity of the frustum of an liyperholio conoid. — Add
together the squares of the greatest and least semi diameters, and
the square of the whole diameter in the middle; then this sum being
multiplied by the altitude, and the product again by '5236, will
give the solidity.
In the hyperbolic frustum ADCB, the length J?,
rs is 20, the diameter AE of the greater end 32,
that DC of the less end 24, and the middle dia
meter nm 28'1708; required the solidity.
Here {16^ + 12^ + 281708=) X 20 X 52359
= (256 + 144 + T9359S9) x 20 x 52359 = t
11935939 X 20 X 52359 = 23871878 x 62359
= 1249907660202 = s.
To find the solidity of a tetraedron. — Multiply I'j
of the cube of the linear side by the square root of
2, and the product will be the solidity.
The lineal side of a tetraedron ABCw is 4 ; what
is the solidity ?
4' ^4x4x4 ^4x4 ^16
j2Xv/2=— ^^x^2 = 3x^2=y,
16 22624
X v" 2 = g X 1414 = — g ■ = 75413 = soUditi/.
To find the solidity of an octaedron. — Multiply ^ of the cube
of the linear side by the square root of 2, and the product will be
the solidity.
"What is the solidity of the octaedron EGAD,
whose linear side is 4 V
 X ^ 2 =
21333,
v/2 =
21333 X 1414 = 3016486 = solidity.
To find the solidity of a dodecaedron. — To 21 times the square
root of 5 add 47, and divide the sum by 40 : then the square root
of the quotient being multiplied by five times the cube of the linear
side will give the solidity.
The linear side of the dodecaedron AECDE
is 3; what is the solidity ?
21^/5 + 47 ^„ , 21 X 223606+47 e/
solidity.
To find the solidity of an icosaedron. — To three times the square
root of 5 add 7, and divide the sum by 2 ; then the square root of
hv Google
90 THE PRACTICAL MODEL CALCULATOB.
this quotient being multiplied by  of the cube of the lineai side
will give the solidity.
T + 3 ^/ 5
That is ^ S^ X %/ (■ — —> } = solidity when S is = to the
linear side.
The linear side of the icosaedron ABCDEF
is 3 ; what is the solidity ?
3 n/ 5 + T 5x3^ 3 X 2'23606 + 7 <
5 X 27 670818
1370818 45
6x9
~2~
s/68540y X 225 = 261803
X 225"= 589056 = soUdity.
The superficies and solidity of any of the five regular bodies may
be found aa follows :
EULE 1. Multiply the tabular area by the square of the linear
edge, and the product will he the superficies.
2. Multiply the tabular solidity by the cube of the linear edge,
and the product will be the solidity.
Surfaces and Solidities of the Regular Bodies.
S5i'
K.....
........
........ I
4
TetraedrOQ
1.Y3205
0.H785
«
HesaGdron
6.00000
1.00000
H
Octoedron
S.46410
0.47140
12
Dodecaedron
20.64578
7.66312
20
Icosaedron
8.66025
2.18169
To find the convex superficies ofacylindrie ring. — To the thick
ness of the ring add the inner diameter, and this sum being multi
plied by the thickness, and the product again by 9.8696, will give
the superficies.
The thickness of Ac of a cylindric ring is 3
inches, and the inner diameter cd 12 inches , /
what is the convex superficies ? a I .
1213 X 3 X 98696 = 15 x 3 x 9 869i
= 45 X 98696 = 444132 = superficies
To find the solidity of a cylindria ring — To the thickness of the
ring add the inner diameter, and this sum bem^ multiplied by the
square of half the thickne&s, and the pioduct dgam ly I hbOl ,
will give the solidity.
hv Google
MENSURATION OP SOLIDS. 91
T\'hat is the solidity of an anchor ring, whoso inner diameter is
8 inclies, and thicliness in metal 3 inches ?
8"T3 X i]= X 98696 = 11 x 15^ x 98693 = 11 x 225 X
98696 = 2475 X 98696 = 2442T26
The inner diameter AB of the cylindrie ring
cdef equals 18 feet, and the sectional diameter
cA or Be equals 9 inches ; required the convex
surface and solidity of the ring.
18 feet X 12 = 216 inekes, and 216 + 9 *
X 9 X 98696 = 1998594 square inches.
216 + 9 X 9= X 24674 = 44968365 eubia
inches.
In the formation of a hoop or ring of wrought iron, it is found
in practice that in bending the iron, the side or edge which forms
the interior diameter of the hoop is upset or shortened, while at
the same time the exterior diameter is drawn or lengthened ; there
fore, the proper diameter by which to dotcrmiDe the length of the
iron in an unbent state, ia the distance from centre to centre of the
iron of which the hoop is composed : henee the rule to determine
the length of the iron. If it is the interior diameter of the hoop
that is given, add the thickness of the iron ; but if the exterior di
ameter, subtract from the given diameter the thickness of the iron,
multiply the sum or remainder by 31416, and the product is the
length of the iron, in equal terms of unity.
Supposing the interior diameter of a hoop to be 32 inches, and
the tbickneas of the iron IJ, what must be the proper length of the
iron, independent of any allowance for shutting ?
32 + 125 = 3325 x 31416 = 104458 inches.
But the same is obtained simply by inspection in the Table of Cir
cumferencea.
Thus, 3325 = 2 feet 9 in., opposite to which is 8 feet 8i inches.
Again, let it be required to form a hoop of iron ^ inch in thick
ness, and 16^ inches outside diameter.
165 — 875 = 15625, or 1 foot 3g inches;
opposite to which, in the Table of Circumferences, ia 4 feet 1 inch,
independent of any allowance for shutting.
The length for angle iron, of which to form a ring of a given di
ameter, varies according to the strength of the iron at the root ;
and the rule is, for a ring with the flange outside, add to its required
interior diameter, twice the extreme strength of the iron at the
root ; or, for a ring with the flange inside, sub cd cd
tract twice the extreme strength ; and the sum or 17^ "^
remainder is the diameter by which to determine ri ii;
the length of the angle iron. Thus, suppose two i i
angle iron rings similar to the following bo re l". —?.[
quired, the exterior diameter AB, and interior ^\ V^
diameter CD, each to be 1 foot 10 J inches, and erf cd
the extreme strength of the iron at the root cd, cd, &c, J of an inch;
hv Google
92
THE PBACnCAL MODEL CALCULATOR.
tTvice ^ = 1, and 1 ft. lOJ in. + 1 = 2 ft. ^ in., opposite to
which, in the Table of Circumferences, is 6 ft. 4^ in., the length of
the iron for CD ; and 1 ft. 10^ in.  1 = 1 ft. 8f in., opposite
to which is 5 ft. 5J in., the length of the iron for AB.
But observe, as before, that the necessary allowance for shutting
must be added to the length of the iron, in addition to the length
as expressed by the Table.
Required the capacity in gallons of a
locomotive engine tender tank, 2 feet 8
inches in depth, and its superficial di
mensions the following, with reference
to the annexed plan :
Length, or dist. between A and B = 10 ft. 2 in. or, 12275 in.
K
c
B
>— 
^ CJ
jj
Breadth
Mean breadth of coke "
C and D =
and g 
ps
lOJ
795
4675
IJ
8125
8i
3225
186
Radius of back corners vx =
Then, 12275 X 795 = 9758525 square inches, as a rectangle.
And 185^ X 7854 = 2688 " " area of circle
formed by the two ends.
Total 10027325 " " from which de
duct the area of the cokespace, and the difference of area between
the semicircle formed by the two back corners, and that of a rect
angle of equal length and breadth ;
Then 4675 x 3725 = 17314375 area of r, n, s, t, in sq. ins.
3225= X 7854 , ,
— g— = 4084 area of half the circle m.
Radius of back corners = 4 inches j
consequently 8^ X 7854 = 2513, the semicircle's area; and
8 X 4 = 32  2513 = 687 inches taken off by rounding
the c
Hence, 17314375 + 4084 + 687 = 2146T07, and
10027235  2146707 = 7880618 square inches, or
whole area in plan,
7880618 X 32 the depth = 252179776 cubic inches,
and 252179776 divided by 231 gives 10916873 the
content in gallons.
hv Google
MENSURATION OF TIMBER.
Tables hj which to facilitate the Mensuration of Timber.
1. Flat or Board Measure.
...
Ares, of*
Bre.athin A
a»of>
Bresdth in A
esofj
inohs..
aJfo^t
iniiht.. U»
Uftwt.
■0208
i
S334
8
6667
■0417
H
3542
8i
6875
■0G23
H
375
8J
7084
1
■0834
4*
8
7292
U
•10i2
6
4167
75
l
■125
H
4S75
9}
7708
l
■1450
tl
4583
9J
7917
2*
■1667
4792
9J
8125
n
■1875
6
5
!0
8334
2*
■2084
lOJ
8643
4
■2292
5416
io
875
3
■35
5625
10
8959
8
■3708
7
5888
11
9167
3
■2016
7
6042
11
9375
■3125
11
9588
7
6458
11
9792
Application, and Use of the Table.
Required the number of square feet in a board or plank 16^ feet
in length and 9f inches in breadth.
Opposite 9f is 8125 x 165 = 134 square feet.
A board 1 foot 2 inches in breadth, and 21 feet in length ; what
is its superficial content in square feet ?
Opposite 2f ia '2292, to which add the 1 foot ; then
12292 X 21 = 258 square feet.
In a board 151 inches at one end, 9 inches at the other, and
141 feet in length, how many square feet ?
'^^'^^ ^ = 12^, or 10208 ; and 10208 x 145 = 148 sq. ft.
The solidity of round or unsquared timber may be found with
much more accuracy by the succeeding Rule : — Multiply the square
of onefifth of the meao girth by twice the length, and the product
will bo the solidity, very near the truth.
A piece of timber is 30 feet long, and the mean girth is 128 in
ches, what is the solidity ?
= 27306 cubic feet.
This i
ployed.
■ the truth than if onefourth the girth he em
hv Google
THE PRACTICAL MODEL CALCULATOR.
2. Cubic or Solid Measure.
BEaanii Co
mTt^
Mf»ii"i
CuMtfMt
maaj
&^iV^
Mein^
CuUtfesl
&S. uA
^■fla.
SZS.
iii°"f«t.
^.^
iii""^6.
Si
lilK^fcot.
6
25
12
I
2 25
24
4
6
272
m
1.042
18
2S13
24
4 0S4
294
12J
1'085
2 376
24
24
lit 8
6
817
lll
1'129
18f
2 442
4''j4
7
340
1174
2 506
25
4o4
7
364
1'219
m
2 574
251
4 428
7
39
IS^
1265
2 64
2^i
2oJ
4 516
7
417
1813
i4
2 709
4 60J
8
44i
14
20'
2 777
.6
4 094
472
14^
141
201
2 898
4S5
el
601
14J
146
20J
2 917
24
48 b
531
14J
1511
20f
4 909
g
502
16
1562
21
8 '062
27
502
9
694
16^
1615
21 i
3136
27^
5 1o8
9
626
15*
15
1668
21J
3 209
27i
5.'i2
9
669
1772
3 285
211
6 818
10
694
16
1777
22
3 362
25
6 444
10
73
16
16
1833
8 438
28
5 542
10
766
189
22
3 516
28
28
6b4
10
803
16
1948
8 508
5 "4
n
81
17
2006
23
3 673
29
6 84
17i
2066
23t
^754
291
5 941
918
17i
2126
3^i
8 886
29*
6 044
11^
359
17
2187
231
8 917
29
0146
In the euliic estimation of timbei, custom has established the
rule of i, the mean girt being the side of the square considered as
the cross sectional dimensions ; hence, multiply the number of cubic
feet by lineal foot as in the Table of Cubic Measure opposite the
I girt, and the product is the solidity of the given dimensions in
cubic feet.
Suppose the mean I girt of a tree 21i inches, and its length
16 feet, what are its contents in cubic feet ?
3136 X 16 = 501T6 cubic feet.
Battens, Deals, and Planks are each similar in their various
lengths, but differing in their widths and thicknesses, and hence
their principal distinction : thus, a batten is 7 inches by 2 J, a deal
9 by 3, and a plank 11 by 3, these being what are tenned the
standard dimensions, by which they are bought and sold, the length
of each being taken at 12 feet ; therefore, in estimating for the
proper value of any quantity, nothing more is required than their
lineal dimensions, by which to ascertain the number of times 12 fL'ct,
there are in the given whole.
Suppose I wish to purchase the following :
7 of 6 feet 6 X 7 = 42 feet
5 14 14 X 5 = 70
11 Id 19 X 11 = 209
and 6 21 21 X 6 = 12fi
12 ) 447 ) 3725 standard deals.
hv Google
3IEK8URATI0N (
Table showing the number of Lineal Feet of Seantling of various
dimensions, ivhieh are equal to a Cubic Foot.
i...i,»
Ft lu
ln.h«
F. in
iml.
rt I
2
jb
4^
9
qi"
2 (
^
2& 9
Ji
?
8
7 2
m
lu"
lOi
2 ^
ih
20 7
Si
6 6
11
2 2
i
18
b
a
lU
2 1
4i
IG
^
'i
5 b
12^
2
5
14 5
^
^
1j 1
t.
7i
4 1
~1~
2 11
12
1
s'
^i
2 9
1
ej
11 1
8i
4 i
2 6
I
7
10 5
q 7
9 n
9
1?
i
S 7
1
8}
9i
2 5
2 3
2 2
8i
8 &
b
;?
3
^
10
lOS
2 1
1 11
"J
7 7
11]
11
1 10
10
10^
u
7 i
b 10
12
a
3
Y}
a
1 9
1 8
11
b G
5
6 9
1
llj
b 4.
5i
5 3
I,
s
12
b
b
£
4 10
H
—
3
1
?'
3
4 5
4 1
*
9
1
2
1 ]0
18
?
s
U 8
12
S
?
s
^ 10
3 7
1
10
1()V
S
4J
10 8
t
''I
5 5
11
6
■■J
H 7
9
8
1
aj
3 2
3
12
i b
10
a 10
^
7
7 4
6 10
j.i
2 9
2 8
J,
9
1 I
1
^*
<> 4
m
m
1 7
5
8
6
5 8
5 4
12"
2 4
!
11"
IP
1 5
1 4
I,
4
H
5
^
H
3 8
12"
1 i
10
4 10
7
lOJ
4 b
1
7J
3 2
^
10
1 J
4 4
^
S
i
lOJ
1 4
lU
4 2
h^
2 10
g
11
1 4
Ji"
4
^
2 8
o
III
Hewn and sawed timber are meaaured by the cabio foot The
unit of boaid measuio is a superficwl foot one inch thick
To measure round timbtr — Multiply the Itngth m feet by the
square of J of the mean girth in inches, and the product divided
by 144 gives the content in cubic feet.
The \ girths of a piece of timber, taken at five points, equally
distant from each other, are 24, 28, 33, 35, and 40 inches ; the
length 30 feet, what is the content ?
24 + 28 + 33 + 35 + 40
5 ^^^■
™ 32^ X 30
Then — ^3^— = 213^ cubic feet.
hv Google
96 THE PRACTICAL MODEL CALCULATOR.
Table containing the Superficies and Solid Content of Spheres, fro7n
1 to 12, and advancing By a tenth.
Diun.
BP8rat«^
SoUdlly.
DU.™.
GaperBcisB.
SoLiditj.
DL»m.
SuikiM™.
SolWilJ.
To
8'14i?
■6286
Tf
698979
548617
84
2216712
3103898
1
88013
■6969
■8
723824
579059
■5
2269606
3215558
2
45289
■9047
■9
764298
616010
6
2823527
3330389
3
68033
11603
50
785400
654500
■7
2377877
8447921
i
61676
14367
■1
817180
694560
■8
2432865
3568187
fi
70686
1^7671
■2
849488
736228
9
2488461
8691217
6
80424
21446
■3
882476
779519
90
2544696
3817044
7
90792
28724
4
916090
824481
1
2601558
3946697
8
tO1787
8 ■0636
■5
950334
871139
■2
2659130
4077210
9
113411
35913
6
986205
919528
271^7169
4211613
2
125664
41888
1020706
969670
■4
277^8917
4348937
1
188544
4^8490
1056834
1021606
■6
2835294
4489215
2
152053
55752
1093590
1075364
■6
2895298
4632477
3
166190
63706
60
1130976
1130976
■7
2966931
4777755
4
180956
72382
1 1
1168989
1188472
■8
8017192
4928081
6
196350
81812
■2
1207681
1247886
■9
3079082
5080485
«
212372
92027
1246901
1309246
100
8141600
5236000
7
229022
108060
4
1286799
1372585
■1
8204746
5394656
246300
11^4940
■6
1327826
1437936
3268520
5556485
9
264208
127700
6
1368480
1505329
8832923
5721518
3
282744
141872
■7
1410264
1574795
4
3397954
588978*
1
801907
155985
1452675
1646365
5
3463614
6061324
2
321699
171573
■9
149^6715
1720073
6
8629901
6236169
8
342120
188166
70
I53^9384
1795948
8596817
6414325
4
363168
20^5795
■1
158^3680
1874021
8
3664362
6596852
5
884846
22 '4493
■2
1628606
195^4326
9
3782534
6780771
6
407161
24^4290
8
1674158
2036898
110
3801836
6969116
7
430085
26^5219
■4
1720340
21M752
1
8870765
7160915
8
463647
287309
■5
1767160
220^8987
■2
3940823
736^6200
8
477837
310594
■6
1814588
229^8478
3
4011509
755^5008
4
502656
335104
■7
1862654
2390511
■4
4082823
7757864
1
628103
360870
.■8
1911349
248^4754
•5
4154766
7963301
2
554178
887924
9
1960672
2581562
6
4227836
8172861
3
580881
416298
80
2010624
2680832
7
4300536
8386045
4
608213
446023
1
2061203
2782626
4374363
8602915
5
636174
477130
■2
2112411
2886962
9
4448819
8828492
■6
664782
509651
■3
2164248
2993876
120
452^3904
9047808
To reduce Solid Inches into Solid Feet.
1728 Solid Inches to
one Solid Foot.
=31104
36
=60480
m
69=
=119232
53
91584
VII
120960
3 6184
W
S45G0
37
54
93312
71
122688
4 6912
VI
86288
65664
55
95040
TA
124416
152064
6 8640
n
38016
67802
56
96768
73
126144
H9
153792
6 10868
39744
40
69120
hV
98496
74
127872
91)
155520
41472
41
70848
J>K
100224
Vh
129600
91
167248
43200
4^
72576
6>l
101952
V6
131328
9 15552
26
44928
4;i
74304
60
103680
77
133056
10 17280
V7
46656
44
76032
HI
105408
7«
134784
94
162432
48384
46
62
107136
'(9
136512
95
164160
12 20786
V,H
50112
46
79488
63
108864
HO
188240
13 22404
HO
51840
47
81216
64
110593
81
189968
HV
14 24193
31
63568
48
82944
65
112320
K2
141696
HM
15 25920
56296
4H
84672
114048
K3
143424
99
171072
57024
511
86400
67
115776
84
146152
100
17 29876
34
58752
51
88128
6a
11750411
b,Google
CriTIXGS Ayo eubaxkmests.
CUTTINGS AND EMBiNKMENTS.
The angle of repose upon railways, or that incline on which a
carnage would rest in whatover situation it was placed, is said to
be at 1 ia 280, or nearly 19 feet per mile ; at any greater rise
than this, the force of gravity orercomea the horizontal traction,
and carriages will not rest, or remain quiescent upon the line, but
will of themselves run down tho line with accelerated Telocity.
The angle of practical effect ia variously stated, ranging from 1 in
To to 1 in 330.
The width of land required for a railway must vary with the
depth of the cuttings and length of embankments, together with
the slopes necessary to he given to suit the various materials of
which the cuttings are composed : thus, rock will generally stand
when tho sides are vertical ; chalk varies from ^ to 1, to 1 to 1 ;
gravel 1 to 1 ; coal 1^ to 1 ; clay 1 to 1, &c. ; but where land
can be obtained at a reasonable rate, it is always well to be on the
safe side.
The following Table ia calculated for the purpose of ascertain
ing the extent of any cutting in cubic yards, for 1 chain, 22 yards,
or 66 feet in length, the slopes or angles of the sides being those
which are most in general practice, and formation level equal SO feet.
Slopes
Itol.
Da [.in
Hiilf
o,^..
Conlenl
ConliM
T
Half
Coattnl
Content
c,„...
liDgin
>ridlK
S'Stb
f Iper
rS
vim
£S
f™u
^.hiS!'
ohiin.
bmdt).
~~r
TT
75 78
244
7.33
1467
"26"
"«"
359911
6866
19067
3S133
2
17
156 42
489
1467
2988
27
42
876200
6599
19800
39300
18
24200
2200
4400
28
48
396978
20533
41067
10
332 44
978
2B33
6867
29
44
418244
7088
21267
42588
5
20
427 78
1222
3667
7883
80
45
440000
7832
44000
b
21
628 00
1467
4400
8800
31
46
462244
7677
7
22
b3311
1711
5138
10267
47
484978
46933
8
23
74811
1956
5867
11733
83
48
508200
8067
24201
48400
24
868 00
2200
6600
18200
34
49
631911
8311
10
25
977 78
2444
7833
14667
35
50
556111
8555
51388
11
2C
1102 44
2689
8067
16133
36
61
580800
8800
26400
52800
12
27
1282 00
2933
8300
17600
37
52
605978
9044
27188 I542G7
18
28
1366 44
3178
9633
19067
38
53
681644
9239
27867156783
14
29
1505 78
8422
10267
20688
39
54
657800
9588
28800;57200
15
80
1650 00
8666
11000
22000
40
55
684444
9777
29838 58667
lb
81
1799 11
3911
11733
23467
41
56
711678
10022
30067160183
17
82
1963 11
4155
12467
24988
42
57
739200
10206
80800 61600
Ig
33
2112 00
4399
13200
26400
43
58
767311
:05ll
3153363067
19
84
2276 78
4644
18938
27867
44
69
795911
10755
8226764338
20
35
2444 44
4889
14667
29338
45
60
825000
10999
88000 66000
2618 00
5183
15400
80800
46
61
854578
11244
3S733;87467
22
87
2796 44
6377
16138
82267
47
62
884644
11488
84467,68933
.i5
2979 78
5621
16867
33733
48
63
915200
11788 1362 00 '70400
.,4
a*
dl63 00
6800
17600
35200
49
64
946244
11977 36938[71867
^
40 33bl 11
6110
18383
8S667 50
65 jD77778
12221 3666773383
b,Google
THE PRACTICAL MODEL CALCULATOR.
140800
1578 '00
1746
192500
211200
1800
371700
298883
81B700
OtSBM
102e7
11788
13200
14667
550000
679700
6101 ""
641300
073200
7058
739200
773800
808133
848700
8800
9170
954800
993300
_082688
1073500
11182'00
11546SS
05f
9777
10022
10266
10511
10756
109r
11344
11488
11788
11977
12321
227£
28467
24200
24S88
26667
36400
27138
27867
28600
29333
82267
38000
38788
18400
4B867
51333
B2800
54267
55733
57200
58667
80138
31600
067
64588
Slopes 2 to 1.
Dapth
Half
CobK«
Gontant
ConleU
Depth
Hllf
Content
Contsnt
CoDUnt
or
width
Qraleut
m.
t.S£Z
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244
733
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26
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521166
6355
19067
38183
2
19
16622
489
1467
29;33
27
69
554400
6599
19800
39600
3
21
26400
2200
4400
23
71
688622
6843
20683
41067
4
28
37156
978
2933
6867
78
7088
21267
42583
5
1232
8667
7338
30
75
660000
7332
22000
44000
6
27
61600
1467
4400
8800
31
77
697155
7577
22733
46467
7
29
76289
1711
5133
10267
32
7Q
735289
7822
23467
46933
8
81
89956
1956
5867
11738
81
774400
8067
24200
48400
83
105600
2200
6600
182 00
84
814489
8311
24933
10
35
123232
2444
7333
14067
35
855555
8665
266,67
>1333
11
87
189822
3689
8067
16138
86
87
897600
8800
36400
52800
12
168400
2938
8800
17600
87
89
94062^
9044
27133
64267
13
11
177965
8178
9533
190 67
38
91
984622
27867
55733
U
43
198489
8422
10367
20588
80
98
1029600
9638
28600
57200
16
46
220000
8686
11000
22000
40
05
1076655
9777
39383
58667
le
47
242489
8911
11788
28487
41
97
1122489
10022
80067
60183
17
49
4155
12467
24983
42
99
1170400
10266
30800
61600
18
51
290400
4399
13200
26400
43
101
1219289
10511
81533
63067
19
58
815822
4644
18033
27867
44
103
1269156
10756
32267
64533
20
55
342222
4889
14667
29833
45
105
1320000
10999
83000
66000
21
57
869600
5138
16400
30800
46
107
1371822
11244
88738
67467
22
59
397055
5377
16183
S2267
47
109
1424622
11488
34467
68988
ei
427289
6621
16867
83733
48
111
1478400
11738
35300
70400
24
63
457600
5866
17600
35200
49
118
.533165
11977
85988
71867
26
65
488889
6110
18338
36667
50
115
12221
36607
78338
b,Google
CUTTIJfGS AND EMBANKMENTS. 99
By the fourth, fifth, and sixth cohimns in each table, the niimher
of cubic yards is easily ascertained at any other ividth of formatioa
level above or below 80 feet, having the same slopes as by the
tables, thus : —
Suppose an excavation of 40 feet in depth, and 33 feet in width
at fovniatiou level, whose slopes or sides are at an angle of 2 to 1,
required the extent of excavation in cubic yards :
1075555 + 29333 = 1104888 cubic yards.
The number of cubic yards in any other excavation may be as
certained by the following simple rule :
To the width at formation level in feet, add the horizontal length
of the side of the triangle formed by the slope, multiply the sum
by the depth of the cutting, or excavation, and by the length, also
in feet ; divide the product by 27, and tho quotient is the content
in cubic yards.
Suppose a cutting of any length, and of which take 1 chain, its
depth being 14^ feet, width at the bottom 28 feet, and whose sides
have a elope of 1 to 1, required the content in cubic yards :
145 X 125 = 1M25 + 28 x 14 = 64575 X 66 =
426195
— 27 — = 15785 cubic yards.
5 { (6 + rV) ¥ + (6 + rf) 4 + 4 [S + r^^^J — g^
gives the content of any cutting. In words, this formula will be : —
To the area of each end, add four times the middle area ; the sum
multiplied by the length and divided by 6 gives the content. The
breadth at the bottom of cutting = S; the perpendicular depth of
cutting at the higher end = ^; the perpendicular depths of cutting
at the lower end = h'; I, the length of the solid ; and rh' the ratio
of the perpendicular height of the slope to the horizontal base, mul
tiplied by the height h'. rh, the ratio r, of the perpendicular
height of the slope, to the horizontal base, multiplied by the
height Ji.
Let & = 30 ; A = 50 ; /t' = 20 ; l=M feet ; and 2 to 5 or 
the ratio of the perpendicular height of the slope to the horizontal
base:
^ I (30 + § X 20) 20 + (30 + 3 X 50) 50 + 4 [30 ff ^^ t, ^^ ~\
— ^ — I = 14 I 38 X 20 + 50 X 50 + 4 X 44 X 35 I ^ 131880
131880
cubic feet. — 07 — ~ 4884'44 cubic yards.
This rule is one of the most useful in tho mensuration of solids,
it will give the content of any irregular solid very nearly, whether
it be bounded by right lines or not.
hv Google
THE PRACTICAL MODEL CALCULATOR,
Table of Sqiuires, Ouhes, Square and Ouhe Roots
of Numbers.
Kumbsr.
Sq^.tE!.
CobES.
S,n«.II.Bl..
Cql»H00lS.
ReeiptMri..
1
1
1
10000000
lOOOOOOO
■lOOOOOOOO
4
14142136
12599210
800000000
3
27
17320608
14422496
■833333333
4
16
64
20000000
1587I0I1
■260000000
5
25
125
22360680
17099769
■200000000
6
216
24494897
18171206
■166666667
49
343
26467613
19129812
■142867143
64
512
28284271
20000000
125000000
9
81
729
80O000OO
20800837
■111111111
10
100
1000
31 622777
21544847
lOOOOOOOO
11
121
1331
33166248
22289801
090909091
12
144
1728
84641016
22894280
■083338333
13
169
2197
36065518
23513847
076923077
14
196
2744
37416574
24101422
■071428571
16
225
3375
24662121
■066666667
16
256
4096
40000000
26198421
062500000
17
289
4913
41231066
26712816
■068823529
18
824
5832
42426407
26207414
065555556
19
861
6869
43588989
26684016
■062681679
20
400
8000
44721360
27144177
■050000000
21
441
9261
45825757
27589243
■047619048
22
4S4
10643
46904158
28020393
'045464646
23
629
12167
47958316
28438670
■043478261
U
676
48989795
28844991
■041666667
25
625
15625
60000000
29240177
■040000000
26
676
17676
50990196
29624960
■038461538
27
729
19683
51961524
30OO000O
■037037037
28
784
21952
62915026
80365889
■035714286
29
841
24889
58851648
30723168
■034482759
80
900
27000
54772256
81072325
■033333333
31
961
29791
56677644
31413800
082258065
32
1024
82768
66568542
81748021
031260000
1089
36937
67445626
32075343
030303030
34
1156
39304
68309519
32396118
029411765
35
1225
42875
59160798
32710663
028571429
36
1296
46656
60000000
88019272
■027777778
87
1369
60653
60827625
33322218
■027027027
1444
64872
61644140
83619754
■026816789
89
1521
59319
62449980
33912114
025641026
40
1600
64000
63245558
S4199&19
026000000
41
1681
68921
64031242
34482172
024390244
42
1764
64807407
34760266
■028809624
43
1849
79307
65574385
850S398I
023255814
44
1936
85184
66332496
35303483
■022727273
45
2026
91125
67082039
36568938
022222222
46
2116
97336
67823300
S6880479
■021789130
47
2209
103823
e 8556646
86088261
■021276600
48
2804
110592
36342411
■020833833
2401
117649
70000000
86593057
■020408163
50
2500
125000
70710678
36840814
■020000000
61
2601
132661
71414284
87084298
■019607843
62
2704
140608
72111026
87325111
■019280769
53
2809
148877
72801099
87562858
■018867925
64
2916
157464
78484692
87797631
■018618519
56
3025
166375
74161985
88029625
■0181818)8
56
8186
175616
74833148
017867143
57
S249
185193
76498344
38485011
■017648860
b,Google
TABLE OP SQUARES, CUBES, SQUARE AHD CUBE ROOTS.
S«^bzr.
S^uarss. ' CabDS.
6iumlk«».
C«l« Ro.t,.
H.=i:irr™lj.
bS
8364
195112
76167781
88708706
■017241879
69
3481
205879
7'6811457
38929965
016949153
60
3600
216000
77459667
89148676
■016066667
61
8721
226981
78102497
89304972
016393443
62
3844
288828
78740079
39578915
016129032
8969
250047
79372539
39790671
■015873016
61
40B6
262144
80000000
40000000
015625000
65
4225
274625
80622677
40207256
015384615
4856
287496
81240384
40412401
016151515
67
4489
800763
81863528
40616480
014925373
68
4624
314433
82462118
40816561
014705882
09
4761
328509
88066289
41015601
014492754
70
4900
348000
83666003
41212853
014285714
71
5041
857911
84261498
41408178
014084517
72
6184
378248
84852814
41601676
013888889
73
6329
389017
85440087
41798390
018G98680
74
6476
405224
86033253
41983864
013518514
5625
421875
86602540
42171633
■013333833
78
5776
438976
87177979
■018157895
77
5929
456583
87749644
42543210
012987018
78
6084
474552
88317609
42726586
■012820513
79
6241
493039
88881944
42908404
012658228
80
6400
512000
89442719
43088695
012500000
81
6561
531441
90000000
432S7487
■012845679
82
6724
551363
90553851
48444816
012195122
83
671787
91104336
43620707
012048193
84
7056
592704
91651514
43796191
■011904762
85
7225
614125
92195445
■011764706
7396
92736185
44140049
011627907
87
7569
658503
93273791
44310476
011494253
88
7744
6S1472
98808315
44470692
011308630
7921
704969
94389811
44647461
011236955
SO
8100
729000
94868330
44814047
011111111
91
8281
75S571
96893920
4 4979414
010989011
92
8464
778688
96916630
4 5148574
010809505
93
8649
804857
96436508
4 6300649
0107«2b88
94
830584
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170
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b,Google
THE PRACIICAL MODEL CALCDLATOR.
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257
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259
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260
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290
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294
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290
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814
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31255875
177482393
68040921
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31554496
177768888
68112847
■003164557
317
100489
31855013
178044938
68184620
■003151574
318
101124
32157432
178325545
08256242
003144654
319
101761
82461769
178605711
08327714
■003I34T96
820
102400
32768000
178885488
68399087
■003125000
821
103041
38076161
179164729
68170213
008115265
322
108084
33380248
179143584
68541240
003105590
323
104329
83698267
170722008
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104976
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68682855
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325
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326
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81645976
180564701
■003007185
327
106929
34965783
180831413
68891188
■003058101
328
107584
85287552
181107708
68964345
■0030187SO
329
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35611289
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69034369
■003039514
830
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181059021
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831
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332
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■003003003
884
111566
37259704
182750669
69382821
■002994012
835
112225
87595375
188030052
69451496
■002985075
330
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37983056
183308028
69520638
■002976190
837
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69589434
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114244
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183847763
69658198
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839
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38958219
184119526
69720826
■002949858
340
115600
89304000
184390889
69795821
■002911176
841
116281
39651821
184661858
69863681
■002932551
342
116964
40001688
184932420
69931906
■002923977
343
117649
40353607
185202592
70000000
■002915452
844
1I83S6
40707684
185472370
70067962
■00290G977
846
119025
41063025
185741756
70136791
■002898551
846
119716
41421736
186010752
70203490
■002800178
347
120409
41781328
180279800
70271058
■002881811
121104
42144192
186547581
70338497
■002878563
849
121801
42508519
186815417
704O5860
■002865830
850
122500
42876000
187082869
70472987
■002857143
351
123201
43243551
187349940
70540041
■002849008
852
128904
43614208
187016630
7O0OS967
002840909
353
124609
43980977
187882942
70073767
■0028828C1
354
12531b
44861864
188148877
70740440
■002824859
855
120023
44738875
188114437
70806988
■002816901
350
126786
45118016
188679623
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45499293
188944436
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368
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189208879
71005885
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339
128881
40268279
189472953
71071937
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860
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46666000
189736660
71137866
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361
130321
47015831
190000000
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■002770083
862
131044
47437928
190262970
71209360
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3bS
131769
47832147
190525589
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002754821
364
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48228544
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71400370
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191049782
71466695
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3bb
1339j6
49027890
191311265
7153090!
■002732240
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41430863
191572441
71595988
■002724796
b,Google
THE PHACTICAL MOLiEL CALCULATOR.
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cub...
Sq=W=ft,»tj.
Col.s 11.^.
R..,l.,oo.!,.
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71060067
002717391
369
130161
60243409
19'2093727
71726809
■002710027
870
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50653000
192353841
71790644
oo'O'os
371
187641
51061811
192613608
71856102
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372
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61478848
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400
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64000000
200000000
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20.0249811
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00 1800 4
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00 4330d0
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00 4154 J
416
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71473876
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00 4O90b9
416
173066
71991296
208960781
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00 406846
117
178889
72511713
204205779
71709991
00 3J808
418
174721
73034682
204460483
7476966*
419
175661
73560059
204694896
748 9942
■00 asorso
120
176400
74088000
204980015
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00 38010
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205182846
74948113
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178084
75151448
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75686967
205669638
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424
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76226021
206912603
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126
180626
76765625
206156281
76181 30
■0023O 141
426
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77308776
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127
182829
77851183
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7680 48
00 841 20
428
183184
78402752
206881609
75361 1
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429
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00 100
b,Google
TABLE OE SQUARES, CUBES, SQUARE AND CUBE ROOTS. 107
K.ml,.r.
S^dnr^a.
Cubes.
SqaA«Roi,^.
C01..IU,.K
Rfc,pr.«.ilB.
4iiU
184900
79507000
207364414
75478423
002325681
481
ISoTHl
80062991
207605396
75536888
002320166
432
186(i24
80621G68
207840097
75595268
002814815
43a
187489
81182787
208086520
75653648
002309409
4U
188356
81746504
208326867
75711748
002304147
435
189225
82312675
208666586
76769849
002298851
430
190096
208806180
75827865
002293678
437
190969
83453463
209046460
75886793
002288380
438
191844
64027672
209284495
75943683
002288105
439
192721
81604519
209623268
76001S85
002277904
440
193000
85184000
20'9761770
76059049
002272727
441
191481
85766121
210000000
70110626
002267674
442
195304
210237960
76174116
002262443
443
196249
86938307
210475652
76231619
002267886
444
197130
87528384
21 ■0718076
76288837
002252262
445
108025
86121125
210950281
76846007
002247191
446
198916
88716536
211187121
76403218
002242162
44
199809
89314623
211423745
76160272
002237136
448
"00 04
89915392
211660105
70517247
002282148
449
201001
90518849
211890201
76574188
002227171
4o0
^0 500
91125000
212132034
76630943
002222222
4ul
^08401
212867606
76687065
002217295
43
204304
92345408
212602916
76744303
002212389
4oJ
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212887967
76809857
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454
00116
98576064
218072768
76857328
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4oo
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94196375
213307290
76913717
002197802
4dU
20 J
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213541665
76970023
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95448903
218775583
77020246
002188181
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90071912
214009346
77082388 ,
002183406
4o9
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96702679
214242863
77188418
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40
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97336000
214476106
77194426
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214709106
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40
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214941853
77306141
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403
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99252847
215174848
77861877
002159827
404
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215406592
77417532
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210 "5
100544625
215688587
77478109
■002150538
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217156
101194696
216870831
77528006
■002145923
407
216069
101847563
216101828
77684028
■002141828
408
2190 4
102S03282
216883077
77689301
002186762
4(,9
219001
108161709
210504078
77694020
002182196
470
220900
108823000
216794834
77749801
002127000
471
2 1641
104487111
217025314
77804904
002123142
4
2 84
105154048
217255610
77859928
002118014
4 3
" 3 9
10:^828817
217485682
77914875
■002111165
4 4
224676
106406424
217715411
77969746
■002109705
4,0
2.O05
107171875
217944917
78024688
002105268
470
220o70
107850176
218174242
78079254
002100840
477
227529
108531333
218403297
78133892
002096486
4/8
228484
109215352
218682111
78188456
002092050
47'>
229441
109902239
218860686
78242942
002087668
480
230400
110592000
219089023
78207868
002083883
481
281861
111284641
219317122
78351688
002079002
482
2u2324
111980168
219544984
78405949
002074689
483
283289
112678587
219772610
78430134
002070393
484
234256
118879904
220000000
78514214
002006116
23d22o
114084125
220227166
78566281
002001656
486
236190
114791256
220464077
78622242
002067618
487
237169
11O501808
220680765
78676180
002053388
488
238144
116214272
220907220
78729944
002049180
459
239121
116930169
221133444
78783084
002044990
410
240100
117149000
221859436
78837352
002040816
411
2410S1
118*70771
221685198
78890946
002036600
b,Google
THE PRACTICAL MODEL CALCULATOR.
B«ml=™.
a^.«r=,.
ColcJ.
Sqn>'''J^t'
C=b.E„^..
I(«dpr„.ul=.
492
342064
119095488
22'1810730
78944468
002032520
493
243049
119828157
222086033
78997917
63784
222261108
79051294
002024291
2
87375
222485955
79104599
002020202
t>0
23936
222710675
79157832
002016129
00
68478
222934968
79210994
002012072
KO
05992
223159136
79264085
002008032
M
51499
223383079
79317104
002004008
00
00
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223606798
79370063
002000000
(M
61501
79422931
001990008
00
6606008
224058666
79475739
001992032
00
63527
224276615
79528477
001988072
02406*
224499443
79581144
001984127
87625
224722061
79638743
001980198
06
5
84216
224944488
79686271
001976285
23843
225166605
79788731
001972887
25
96512
225888553
79791122
001968504
225610283
79843444
001964637
51000
225831796
79895697
001960784
82881
236053091
79947883
001956947
17728
226274170
80000000
001963125
05697
226495033
80052049
001949318
96744
226715681
80104032
001945525
90876
226936114
8 ■0155946
■001941748
26
227156334
80207794
001937984
88413
227876841
80259574
■001934236
')1882
227596184
803U287
001980502
98359
227815715
80362935
001926782
400
08000
228086086
80414515
001923077
20761
228254244
80466080
001919386
.^
36648
228473193
80517479
001915709
3
56667
228691933
001912046
77824
228910468
80620180
001908397
08125
229128785
80671432
001904762
31576
80722620
001901141
68183
229564806
80773748
001897538
84
97952
229782506
80824800
00189S939
48 35889
230000000
80875794
001890359
0900
77001
230217289
80926728
001886793
21291
230434372
80977589
001883239
63
280651252
81028390
001879699
19487
230867928
81079128
001876173
78804
231084400
81129803
001872659
30375
231800670
81180414
001869159
90656
231516738
81330962
001865672
8o
54153
231732605
81281447
001862197
20872
231948270
81831870
001858736
90
90819
232168735
81382230
001855288
«
64000
232879001
8 1432529
001851852
58 40421
232594067
81482765
001848429
&42
298704
169220088
232808935
81532939
001845018
543
294849
160103007
233028604
81583051
001841621
544
295936
160989184
233238076
81688102
001888235
545
297025
161878625
233452351
81683092
001834862
546
298116
162771886
233666429
81788020
001831503
547
299209
233880311
81782888
001828154
548
300304
164566592
234093998
81832695
001824818
549
301401
165469149
234307490
81882441
001821494
550
302500
166375000
234520788
81932127
001818182
551
303601
167284151
234788892
81981753
001814883
552
804704
168196608
234946802
82031319
001811594
553
305809
169112377
235159520
82080825
001808318
b,Google
TABLE OE BCJUARES, CUBES, SQUARE AHD CUBE ROOTS.
Knmh.r.
Ecu.r...
Cuhc'.
Si...»R<,ola.
o.i„K,»t..
Bodpr^ool,,
551
806916
170031404
285372046
83180271
001805064
555
308025
170958876
236684380
82179657
001801802
55e
309186
171879618
235796522
82228985
■001798561
557
S10249
172808693
236008474
82278264
■001795882
558
811864
178741112
236220286
82827463
001792116
559
312481
174676879
236431808
82376614
001788909
560
818600
176616000
236648191
82426706
001785714
661
814731
176558481
236854386
82474740
)01782531
562
815844
177604328
237065892
82533715
668
316969
178463547
237276210
82572635
)01776199
564
318096
179406144
237486842
82631492
001778050
665
819225
180362125
237697289
82670294
001769912
566
320356
181321498
237907546
82719039
001766784
567
821489
1822842GS
288117618
82767726
001768668
568
822624
188250432
238327506
83816256
001760563
569
828761
184220009
238537209
82864938
001767469
570
324900
185193000
288746728
82918444
001754386
671
326041
18G169411
238956068
82961903
001751313
572
827184
187149248
239165215
83010304
001748252
57S
828329
188132517
239874184
83058951
001745201
574
329470
189119224
239582971
88106941
001742160
575
830625
190109876
239791576
83155175
001739130
576
331776
191102976
240000000
83203353
■001736!11
577
882927
192100088
240208243
83251475
001733102
578
334084
193100652
240416306
88299542
001730104
57B
335241
194104539
240624188
88347653
001727116
580
836400
196112000
240831891
83396609
001724138
581
887561
196122941
241039416
83443410
001731170
888724
107187868
241246762
83491256
001718213
683
3398SQ
198155287
241458929
83589047
001716266
684
841056
J99176704
241660919
83689784
001713329
585
342226
200201625
341867782
83634466
001709403
686
843896
201230056
242074369
83682095
001706485
587
344509
202262003
243280829
83729668
00170S678
588
845744
203297472
242487113
83777188
001700680
689
346921
204336469
242693222
88824658
001697793
590
848100
205879000
242899156
83872065
001694915
691
349281
206426071
243104996
83919428
001693047
592
3504fl4
207474688
248810501
83966729
001689189
593
351649
208527857
243616918
84013981
001686341
594
852836
209584584
243731153
84061180
001683602
695
354025
210644876
243926218
84108326
001680672
596
855216
211708736
244131112
84155419
001677852
597
350409
212776178
244335834
84202460
001975042
857604
218847192
244540886
84249448
■001672241
690
868801
214921799
244744765
84296883
001669449
600
860000
216000000
244948974
84343267
001696667
601
361201
217081801
246163013
84390098
001668894
362404
218167208
245856883
84436877
001661180
863609
219256227
245660583
84483606
001658875
604
864816
220348864
246764115
84530281
001655629
605
866025
221445125
246967478
84576906
001652893
606
367236
222546016
246170673
84628479
001650166
607
368449
228648548
246373700
84670001
001047446
608
369664
224755712
246576560
84716471
001644737
609
870881
225866529
246779254
84792892
001642036
610
372100
226981000
246981781
84809261
001639844
611
228099181
247184142
84855579
001936661
612
374544
247386338
84001848
001633987
618
376769
230346397
247588368
84948095
001631321
614
281475544
247790234
84094233
001628664
615
378225
232608376
247991935
85040350
001626016
b,Google
THE PRACTICAL MODEL CALCULATOR.
N>m,ta..
B,a>:...
C.b=s. B5n.r=E™«. 1 C
beltoMa.
E.=iJ.v™»l>, ,
616
379456
233744896
248198478 8
JO80417
001623377
017
880689
234885113
24'8394847 8
5132485
001620746
618
881924
248696058 8
6178403
001618128
619,
883161
287176659
24.8797106 8
5224331
■001615509
384400
■238828000
34'8997992 8
5270189
001612908
621
886641
239463061
249198716 8
5816009
001610806
622
38688*
240641848
249399278 8
5301780
001607717
623
888129
241804367
24'9599679 8
5407501
001606136
624
242970624
249799920 8
5453173
001602664
625
390025
244140626
260000000 . 8
5498797
■OO10OOOOO
891870
245134376
250199920 8
6644372
001697444
627
393129
246491888
250399681 8
6589899
001594890
628
894884
247673152
250699282 8
5636377
■001692357
629
395641
248858189
250798724 8
5680807
■001689825
830
396900
250047000
350998008 8
5726189
001587302
631
89SI61
251289591
261197134 8
5771528
001584786
633
399424
252435968
251396102 8
6816809
001582278
683
400689
268636137
251594913 8
5862247
001579779
684
401956
254840104
261798666 8
5907238
■001677287
635
403226
266047876
261992063 8
5953380
001674803
404496
267259456
262190404 8
5997476
■001573327
637
405709
258474853
25 ■2388689 8
6042525
001569859
407044
259694072
252586619 8
6087526
00] 567398
408321
200917119
262784493 8
6132480
001564945
640
409000
202144000
252982213 8
6177888
■001562500
641
410881
208874731
258179778 8
6222248
■001560002
6i2
412164
364609288
253377189 8
6267063
■001667632
643
413449
205847707
28.3674447 8
6311830
■O0156621O
644
414736
■267089984
263771551 8
6856551
001652796
645
410125
208386125
263968502 8
6401220
001550888
646
417316
269685130
26.4165302 8
6445856
001547988
647
418609
270840028
26.4861947 8
6490487
■001646595
648
419904
272097792
254558441 8
6534974
■001643210
649
421201
'273859449
254764784 8
6579405
■001540832
660
422500
274635000
254960976 8
6623911
■O01688462
651
423801
276894451
25.5147018 8
6668810
001636098
652
425104
277167808
265342907 8
6712665
■001533742
653
426409
278445077
255538647 8
6756974
■001531894
654
427716
270726204
255734287 8
6801287
001529052
655
429026
281011375
256929678 8
6845456
■001520718
656
430336
2S23004I6
260124969 8
08S963O
■001524890
657
481689
283503393
25.6330112 8
6983760
001522070
658
482964
284890312
256515107 8
6977843
■001519751
659
434281
280191179
256709953 8
7021882
00161'7461
660
435600
287496000
266904662 8
7065877
001515163
661
436921
288804781
357099208 8
7109827
■001512859
438244
290117628
267298607 8
7153784
001610674
439669
291434247
26.7487804 8
7197596
■001508296
664
. 410896
292754944
257681975 8
7241414
■001500024
442285
294079625
25.7875989 8
7285187
■001503759
666
443666
295408296
25.8069758 8
7828918
■OOI501603
667
444899
290740968
258263481 8
7372004
■001499250
440224
298077032
258456960 8
7416246
■001497006
660
447661
299418809
26.8650343 8
7459846
■001494768
070
448900
800763000
258843692 8
7503401
001492537
671
460241
302111711
259036077 8
7546913
001490313
672
451584
303464448
259229628 8
7590383
001488095
673
452929
304821217
,259422435 8
7633809
■001485884
674
454276
806182O24
259615100 8
7677193
■001483680
676
'455625
307546875
259807021 8
7720682
■001481481
676
466976
S08915776
200000000 8
7703830
001479290
677
468329
310288783
260192287 8
7807084
■001477105
b,Google
TABLE or SQUARES, CUBES, SQUABE AKD CUBE ROOTS.
r,„™i«r
Cut^.
«,u=«H„.«.
Cu>«H„«.
meim""
678
450684
811665753
260384381
87860296
00147492S
079
401041
318046889
260576284
87893466
001472754
(iSO
402400
314432000
200708096
•001470588
681
46S761
315821241
260969767
87979679
001468429
682
465124
817214568
261151297
88022721
001406275
683
460489
318611987
261342087
88065722
001464129
684
467856
320018504
261633087
88108081
001401988
685
469225
321419125
261725047
88151598
001459854
Ga3
470596
261910017
88104474
001457720
687
471969
824242703
262106848
88237807
001455604
ti88
478344
325060672
262297541
88280099
001453488
689
474721
327082769
262488005
88322850
001451879
690
476100
328509000
262678511
88865559
001449275
611
477481
829939371
262868789
88408227
O0I44T178
602
478864
331373B88
263068029
88450854
001445087
093
480249
382812557
203248082
S8493440
001443001
694
481636
884255384
203438797
88686985
001440022
695
483025
335702875
208628627
88578489
48441b
337153536
203818119
88620952
001430782
607
485800
204007676
88663876
001484720
698
487204
840068892
264106806
88705757
001432605
609
488001
341532099
264386081
88748009
001480615
700
490000
843000000
204575131
88790400
001428671
701
491401
344472101
264704046
88882661
001420534
702
492804
84504B40S
264952820
88874882
001424501
703
494209
847428927
205141473
88917063
001422475
704
495016
348913004
265820083
88959204
001420465
705
407025
350402025
265518361
80001804
001418440
703
498436
851805816
20 ■5700605
89048366
001416431
707
499849
858393248
265804716
89085387
001414427
708
501264
364894012
266O82094
89127869
001412429
700
602681
350400829
266270530
89169311
001410437
710
504100
357011000
200458252
89211314
001408451
711
505521
359425481
266646883
89253078
001406470
712
500944
860944128
266883281
89294902
001404494
718
362467097
267030598
89386687
001402625
71*
609796
863994344
207207784
8087S433
001400500
7 5
511 5
365526875
207394839
89420140
001398601
16
51 656
8b 061696
267681763
89401809
001396048
17
614089
868001818
267768567
89503488
001804700
7 8
ol5 '>i
3 0146232
267955220
80545029
001893758
19
510161
371694959
268141764
89586581
001390821
518400
873 48000
208828157
89028095
1
619841
874S05361
208514432
89660670
001366903
6 1284
6367048
268700577
89711007
001885042
23
52 a
3 7933067
89752406
001883120
7 i
5'>4176
9608424
269072481
89708706
001381215
25
5 5025
881078125
209268240
001870810
5 6
8 667176
209443872
89876373
001377410
7
5 8529
884 40583
269629375
899I7620
001375516
8
6 9984
885828352
269814751
001378626
53144
8 420489
270000000
90000000
001371742
80
58 900
889017000
270185122
90041184
001369863
731
534 61
890b 7891
270870117
90082229
001807089
7
5 8 4
89 >3168
270554985
90123288
001300120
733
53 289
398832837
270730727
90164309
001304356
34
58 6
395440904
270924344
90206298
001362398
3
540
89 065875
271108834
90246289
001800544
541 60G
398G&8256
271293199
00287149
0O1358G96
737
513169
400816553
271477149
90328021
001350853
738
544644
401 947272
271601554
90868857
001350014
730
546121
403583419
271845544
90409055
001363180
b,Google
THE PRACTICAL MODEL CALCULATOR.
N„m),«.
ScmSTM.
Cahes.
Sfloste K^ot!.
Clw Roola.
Ee^Lurotaii.
710
547600
405224000
272029140
90450119
001351361
741
549801
406869021
272218152
90491142
001319628
743
550564
408518188
90531831
001847709
713
552049
410172407
272580263
90572432
001345895
741
65S586
1118S0781
272763684
90613098
001314086
745
655025
413493625
272946881
90653677
001342282
7ie
550516
415160936
278130006
90694220
001310183
747
658009
416832723
373313007
90734726
001338688
718
55D504
418508992
273495887
90775197
001336898
749
561001
420189749
273678644
90815631
001335118
750
562500
421875000
273861279
90856030
001388333
751
564001
423564751
274043792
90896852
001331658
752
565501
425259008
274226184
90936719
001329787
753
567009
426957777
274408465
90977010
001328021
754
568616
428661064
271590604
91017265
001320260
755
570025
480368875
274772633
91057485
001821503
756
571586
432081216
274954512
91097669
001332751
757
573049
138798093
275136330
91137818
001321004
758
574561
435619612
275317998
91177931
001319261
759
576081
437245179
275499546
91218010
001317523
760
677600
438976000
275680975
91258053
■001315739
761
579121
140711081
275862281
91298061
001314060
762
680641
442150728
276043476
91838084
001312336
763
582169
114194917
276224516
91377971
001310G16
764
583696
445943744
276405499
91417874
001808901
765
58522S
447697125
276686334
9J457742
001307190
766
586756
419455096
276767050
91497576
001305483
767
588289
451217663
276947618
91687375
001303781
768
689824
152981832
277128129
91677139
O01802O8S
769
591361
154766609
377308492
91616669
001300390
770
692900
456533000
217488739
91656565
001298701
771
594441
458311011
277668868
91696225
001297017
772
595984
16U099G48
277848880
91735862
001295837
773
597529
161889917
278028775
91775115
001293661
774
599076
163684824
278208555
91815003
001291990
775
600625
465484375
278388218
91851527
001290323
773
602176
467288576
278567766
91894013
001288660
777
608729
469097483
278747197
91938474
001287001
778
606281
170910952
278926514
01972397
001285347
779
606841
472729139
279105716
92012286
001383697
780
608100
474562000
279284801
92051641
001282051
781
609961
476879541
270163772
92090962
■001280110
782
611624
478211768
279642629
92130250
001278772
783
4800486S7
279821372
92169605
001277189
781
614656
481890304
280000000
93208726
001275510
785
616225
483736626
280178515
92247914
001273886
786
617796
485587656
280356915
92287068
001272265
787
619369
487448403
28 ■0586203
92826189
001270648
788
620911
489303872
280713377
9'2365277
001269086
789
622621
491169069
280891438
92404333
001267427
790
624100
493039000
281069386
92443356
001265828
791
625681
491913671
281247222
92482311
001261223
792
627624
196793088
281424916
82521300
001262626
793
628849
498677267
281602567
92560224
001261034
794
680436
500566184
281780066
92599111
001259146
796
602159875
281957444
92637973
001257863
796
633616
5043583S6
282134720
92676798
001256281
797
685209
506261573
282311884
92716592
001354705
798
636801
508169692
282488938
92754352
001253183
799
638401
510082399
282665881
92798081
001251364
800
610000
512000000
282842712
92831777
001250000
801
641601
513922401
288019434
92870144
001248439
b,Google
TABLE OF SQUARES, CUBES, SQUARE AND CUBE ROOTS. 113
KomW.
^a™.
C,l.,,
B^»«» liout^
c=i«e™i*
K.ip.^„.
8oa
e48:;04
ei584iJ608
288196045
92903072
001246883
808
644809
617781627
283372546
92947671
001245380
804
646416
619718164
283548938
92986239
001243781
805
M8025
521660125
28S7262I9
98024776
001342236
8oe
649686
623606616
288901391
93063278
■001240695
807
651249
B25557943
284077454
93101750
■001289167 '■
808
652864
527514112
284253408
93140190
■001237624 !
809
654481
639475129
284429258
93178599
001236094
810
656100
631441000
284604989
93216975
001234668
811
657721
533411731
284780617
93256320
001233046
812
659844
635387328
284956187
93298634
001281527
813
eeo969
637367797
286131549
93331916
001230013
814
662596
639853144
285306862
93370167
001228601 1
81o
664226
541343375
285482048
001226994 i
816
b658o6
543388496
286667137
93446575
001225499 :
617
667489
545338513
285832119
93484731
■001238990 1
818
669124
647843482
93522857
■001222494 i
819
6707bl
649353259
286181761)
93560952
001231001 i
820
62400
551868000
286356421
93599016
■001219513 '
821
674041
653387661
286530976
93637049
001316037 ,
822
675684
555412248
286706424
9S675051
■001216545 !
8^3
677329
657441767
286879716
93718022
■001216067 :
824
678976
659476224
287054002
93750963
001318592 j
85
680625
561516625
287228183
93788873
001212121
826
682276
668569976
287402157
93826752
001210664 1
827
665609288
287676077
93864600
001209190 !
82S
685584
667663552
287749891
93903419
■O0I207739 
829
687241
569722789
287923601
93940206
■001206278 i
830
688000
571787000
288097206
93977964
■001204819 i
831
6d05(>l
573866191
288270706
94015691
00120S3G9
832
692224
288444102
94058887
■001201928
883
578009537
288617394
94091054
001200480
834
6J555e
580093704
288790582
94128690
■001199041
885
617225
682182875
288963666
94166297
001197606
83t.
698896
684277066
289186646
94203873
■001196172
837
700569
586376253
94241430
001194748
703244
688480472
289482297
94278986
001193317
839
703921
690589719
289654967
94316423
■001191895
810
70o600
592704000
289827535
94853800
001190476
841
707281
594828321
290000000
94391807
001189061
842
70'*964
596947688
290172363
94428704
001187648
843
710649
699077107
290344623
94466072
001186340
844
712336
601211584
290516781
9450841O
■001184834
845
714025
603351126
94540719
001183432
84(j
715716
605495786
290860791
04677999
■001182033
847
717409
607645423
291082644
94615249
001180638
848
719104
609800192
391204396
94662470
■001179245
849
720801
C11900O49
291376046
94689661
■001177856
8B0
722500
614125000
291547695
94726824
001176471
851
724201
616295061
291719043
94763967
■001175088
8a2
725904
618470208
291890390
94801061
001178709
853
727609
620660477
292061637
94888136
001172333
854
729316
622835864
292233784
94876182
■001170960
8o5
731025
626026376
292408830
94912300
001169591
86b
732736
627222016
292574777
94949188
001166224
857
734449
629422793
292746623
94986147
■001166861
868
736164
631628712
293916370
95028078
■001166601
8^9
737881
633839779
298087018
96059980
001164144
800
739600
636066000
293357566
95096864
001162791
861
741321
638277381
293428016
95133699
■001161440
862
743044
640503928
293598365
95170515
001160093
744769
642738647
293768616
95207803
001168749
b,Google
! PRACTICAL MODEL CALCULATOR.
Kuml,=r.
BqaoKS.
Cul.,a.
B,,^ R.„».
Cb. R..^..
Rooipiotal,.
864
746406
644972544
a9'8938769
95244063
001157407
805
748225
64T214625
294108823
95280704
001156069
8oe
740056
619461896
294278779
95817407
001164734
SG7
761680
651714868
294448687
D5354172
001163403
763424
053072032
29'4618397
95390818
001152074
BGS
755161
656234909
204788059
05427437
001150748
870
766900
658503000
294957624.
95464027
001149425
871
758641
660776311
295127091
96500589
001148106
872
760384
663064848
29'6296461
96537123
001146789
873
762129
666338617
295465734
05573030
001145475
763876
667627624
205634010
95610108
001144165
765625
660931875
295803989
95646659
001142857
870
767876
67222 I37G
296972972
96683782
001141563
877
769129
674526183
296141858
96719377
001 140251
878
770884
676836152
286810648
95755745
001138952
879
772641
679151439
296479342
95792085
001137656
880
774400
681472000
296647939
96828897
001136364
8S1
776181
290816442
95864683
001135074
882
777924
296084848
96900987
001183787
779680
688465387
297153159
95037169
001132603
884
781456
690807104
297321375
95973373
001131222
783226
608164125
207489496
96000548
001120944
784906
695506456
297657521
96045606
001128668
887
786769
697864103
207826462
9608181T
001127896
888
788544
T00227072
297903289
96117911
001126126
88S
790321
702595399
208161Q80
96158977
00112J850
890
792100
704960000
298828678
96100017
001128596
8'Jl
793881
707347971
298496331
96226030
001122384
802
705664
707932288
208663600
96262016
001121076
803
797449
712121957
298831056
06207978
001119821
894
799286
714516984
96383907
001118568
895
801026
716917875
299166506
96369812
001117818
89G
802816
719323136
200332591
96405690
001110071
897
804609
721784278
299499683
06441642
001114827
898
806404
724150792
299666481
96477367
001118686
839
808201
726672699
29988B287
96613166
■001112347
SOO
810000
729000000
300000000
96548938
OOIllllll
901
811801
781432701
300166621
06584684
001109878
902
813604
733870808
800333148
96620403
001108647
S03
815409
736314827
800499584
96656096
■001107420
904
817216
738763264
300665028
96691762
001106105
905
819025
741217626
300832179
96727403
001104972
906
820830
743677416
800998339
96763017
■001103753
907
822640
T461 42643
301164407
96798604
001102536
008
824464
748613312
801330383
96834166
001101822
909
826281
751089429
801496260
96809701
001100110
910
828100
753571000
801662063
96905211
001098901
9n
820921
756058031
301827765
96940694
001097695
912
881744
758550825
301993377
96076151
■001006401
913
883669
761048497
802158899
97011683
001006290
dl4
835396
763551944
302324320
97046989
001094092
915
837225
766060875
302489669
97082369
001092803
916
839056
768575296
302054910
97117723
001001708
917
840889
771096213
302820079
97153051
001090513
918
842724
773620632
302985148
97188854
001089825
919
844561
776161659
303160128
07223631
■001088130
920
846400
778688000
303316018
97258883
001086057
921
848241
78122B961
803479818
97294109
001086776
922
860084
783777448
303644629
97329309
001084699
923
851929
786330467
303809151
97364484
OOI08S423
924
853776
303078683
97399634
001082261
925
856625
791458126
304138127
97434758
001081081
b,Google
TABLE OF SQUAEES, CUBES, SQUARE AND CUBE ROOTS.
.%■„,. t=r, 1 s^u^v.,.
C.I....
S^wra IU«.ta.
Cuba Boou.
R8dp.o.al^
926
857476
794022776
301302481
97469857
O010799U
927
8598^9
796507983
80'4466747
07504030
■001078740
928
861184
790178752
804630924
97539979
001077586
929
863041
801765089
804795013
97575002
001076420
930
864000
804357000
301050014
97610001
■001075269
931
866761
800954191
305122926
97644971
■001074114
982
868624
809557508
806286750
97670922
001072901
933
870489
812160237
305150187
97711815
001071811
931
872356
811780504
805614186
07749743
■001070664
935
874225
817400375
305777697
97781016
■001060519
936
876006
820025856
805941171
07829466
■001068376
937
877969
822656958
306101557
97854288
■001067236
879844
825293672
806267857
97889087
■001066098
881721
827936019
306431069
97923861
■001064063
940
883600
880584000
306591194
97058611
■001068830
941
885481
833287621
806757283
97998886
■001062690
942
8878G4
806920185
98028030
001061571
B48
889249
838501807
307083051
08062711
■001060445
m
891136
811232384
307245830
98097362
001059322
945
893025
848908625
807408523
98131989
001058201
94G
894916
810590530
30'7571180
98166591
■001057082
947
849278123
307733651
98201169
■001055960
948
898704
851971392
307806086
98285723
001051852
949
900601
854670349
808058136
98270262
001053741
950
902500
857375000
308220700
98304757
■001062082
&51
904401
800085351
308382879
98339238
■001061526
052
906304
862801408
808644972
98378695
■001050120
953
908209
865528177
308706981
98408127
001040818
954
910116
868250664
308868001
98442536
■001048218
955
912025
870983875
809030718
98476020
■001047120
956
913036
873722816
309192477
98511280
001046025
657
915849
876467493
309354166
98545617
■001044932
958
917764
879217912
809515751
98579929
■001043841
959
919681
881974079
809677251
98614218
■001042753
960
921600
884736000
98648483
■001041667
901
923521
887603681
310000000
98682721
■001010583
962
925444
890277128
810161248
98716911
■001039501
963
893056347
810322418
98751185
001088422
964
895841844
810483491
98785305
■001037341
965
931226
898632125
310641191
98819451
■001036209
938150
901428696
310805405
98853571
■001035197
967
035089
001231063
810966236
98887673
■001084126
968
937024
907089232
311126984
98921710
■001033058
909
938961
909858209
811287648
98955801
001081992
970
940000
912673000
Sl1448230
001030928
971
942841
9161986U
811608729
99023835
■001029866
972
944731
918330018
311769115
99057817
•001028807
978
946729
921167317
811929479
99091776
■001027749
974
948676
924010424
312089781
99125712
■001026694
975
950625
926859375
312219900
99159624
■001025641
076
952576
920714176
312409087
99193513
■001021590
977
954529
932574833
812569092
99227379
■001028541
978
966484
935111352
312729915
99261222
■001022195
S79
958441
038318780
812889757
99205012
■001021460
980
960400
941192000
313049517
99328889
■001020408
981
962301
944076111
313209195
99862613
■001019168
961324
946966168
813868702
99396363
■001018330
983
066289
949862087
313528308
99430092
001017204
98*
968256
952763901
813687743
90463797
■001016260
985
970225
955671625
313847097
99497479
■001015228
972196
958585256
314006369
00531138
■001014199
987
974169
061504803
314165561
99564775
001013171
b,Google
THE PRACTICAL MODEL CALCUIATOR.
Bombst. 1
Biu»™=,
Cubs.
S,n^^BocB.
cub, ROCU.
KsriprcMla.
9 88
076144
964430272
81 ■432*678
99598389
001012146
978121
067361669
814483704
99631981
■001011122
eeo
980100
970299000
814642654
99065549
001010101
991
978242271
314801525
99099065
■001009082
993
984064
976191488
314060316
99782619
001008065
993
986049
979146657
315119025
99766120
0010070*9
994
982107784
815277655
99799599
001006086
995
990026
986074875
315436206
99833065
■001005025
992016
98S047936
816594677
99866488
■001004016
997
994009
991026973
316753068
99899900
001003009
998
996004
994011992
81'5911380
99933289
001002004
999
998001
997002999
316069618
99966656
001001001
1000
1000000
1000000000
81:6227766
lOOOOOOOO
■OOIOOOOOO
1001
1000201
1008003001
816885840
100083222
0009990010
1002
1004004
1006012008
816548866
100066622
■0009980040
1003
1006009
1009027027
816701762
100099899
0000070090
1004
1008016
1012048064
100188155
■0009960159
1005
1010025
1015075125
817017849
100166389
0009960240
1006
1010036
1018108216
317175030
100199601
0009940358
1007
1014049
1021147343
Sl7382033
100232791
0009930487
1008
1016064
1024192512
817490167
100265958
0009920636
1009
1018081
1027248729
317647603
100299104
1010
1020100
1030301000
817804972
100332228
0009900990
1011
1020121
1033364331
317062262
100865880
■0009891197
1012
1024144
1036483728
318119474
100398410
0009881423
1013
1026169
1039509197
318276609
100*31469
0009871668
1014
I028I96
1042590744
318433666
100*64506
■0009801933
1015
1030225
1045678875
318590646
100497521
0009862217
1016
1032256
1048772096
818747649
100630514
0009842520
1017
1034289
1051871913
31890437*
100563485
0009832842
1018
1036824
1054977832
819061123
100696436
■0009823188
1019
1038861
1058089869
31921770*
100629364
■0009813543
1020
1W0400
1061208000
810374388
100662271
■00O9803922
1021
1042441
1064332261
319530906
100695156
■0009794319
1022
1044484
1067462648
819687347
100728020
0000784786
1023
1046529
1070599167
319843712
100760868
■0009776171
1021
1048676
1078741824
820000000
100798684
■0000765625
1025
1060625
1076890626
820156212
100826484
0000756008
1026
1052876
1080045676
320312348
100859262
0009746589
1027
1054729
320*68407
100892019
0009787098
1028
1066T84
1086378952
82062*391
10092*756
0009727626
1029
1058841
1089647389
820780298
100957469
■O009718173
1030
1060900
1092727000
320986181
100090163
0009708738
1081
1062961
1005912791
821091887
101022835
■0009699321
1032
1066024
1099104768
321247568
101055487
■0000689922
10S3
1067089
1102802987
321*03173
101088117
■000968a542
103*
1069156
1105507304
321558704
101120726
•0000671180
1085
1071225
1108717876
821714159
101158314
■0009661836
loss
1078296
1111934656
821869539
101186882
0009652510
1037
1075369
1115157653
32202484*
101218428
0000643202
1038
1077444
1118886872
332] 8007*
101260058
0009633911
1039
1079521
1121622319
822836229
101283157
■0009624639
1040
1081600
1124864000
322490810
101315041
■0009615385
1041
1083681
1128111021
822646316
101348403
0009606148
1042
1085764
1131366088
322800248
101880845
■0009696929
1043
1087849
1134626507
322955105
101413266
0009587738
1044
1089936
1137898184
328109888
101445667
0009578544
1045
1092026
1141166125
323264598
1014780*7
0009569378
1043
1094116
1144445836
823419233
10I61O406
0009560229
1047
1096209
1147730823
32357879*
1015427*4
■0009561098
1048
1098304
1151022592
328728281
101676062
0009541985
1049
1100401
1154320640
101607359
0009532888
b,Google
TABLE OP SQUAEES, CUBES, SQUARE AND CUBE K0OT3.
NambBr.
Squire.
Cu..^
6,a.n,I.»«.
0,1,= k™* 1
RKirrorala.
1050
1102500
1157025000
824037035
101639036
0009523810
1051
1104601
1160986651
32 ■4191301
101671898
0009514746
1052
1106704
1164252608
324345496
101704129
0009505703
1053
1108809
1167676877
824499815
101736344
0009496676
1054
1110916
1170905464
824653862
101768539
0O094870U0
1055
1113125
1174241876
324807685
101800714
0009478673
1056
1116136
1177583616
82'1961636
0009469697
1057
1117249
11809821S8
826115364
101866003
0009460738
1058
1119SG4
1184287113
325260119
101897116
0O094517&6
1059
1121481
1187648379
825423802
101929209
0009442671
1060
1123600
1191016000
325576412
101961283
0009433962
1061
1125721
1194389981
825729949
101993836
0009425071
1062
1127844
1197770328
325883416
102026369
0009416196
1068
1129969
1201157047
326086807
102057382
0009407338
1064
1132096
1204650144
826190129
102089875
0009898496
1065
1134226
1207949625
826343377
102121847
0009889671
1066
1136356
1211355496
326496564
102153300
00003808G8
1067
1188489
1214767768
326649659
102185233
0009872071
1068
1140624
1218186432
826802698
102217146
0009363296
1069
1142761
1221611509
326955664
102249089
0009354537
1070
1144900
1225043000
327108544
102280912
0009845734
1071
1147041
1228480911
827261868
102312766
0009837068
1072
1149184
1231925248
327414111
102344699
0009a2&^j8
1078
1151829
1285376017
827566787
102876418
■0009319664
1074
1153478
128S83S224
827719892
102408207
0009310987
1075
1155625
1242286875
327871926
102489981
0009302326
1076
1157776
1245766979
328024398
102471785
00092936SO
1077
1159929
12492435SS
828176782
102503470
0009285061
1078
1162084
1252726562
828829108
102585186
0009276488
1079
1164241
1256216039
328481354
102666881
0009367841
1080
1166400
1259712000
328633535
102698557
0009269269
1081
1168561
1268214441
828785644
102630213
0009250694
1082
1170724
1266723868
328987684
102661860
■O009242144
1088
1172889
1270238787
829089658
103693467
•0009233610
1084
1175056
1273760704
329241553
102725065
■0009225092
1085
1177225
1277289125
829398882
102756644
0009216500
1086
1179396
1280824058
329545141
102786203
0000208103
1087
1181569
1284365503
829696830
102819743
0009190032
1088
1183744
1287913472
329848450
102851264
0009191176
1089
1186921
1291467969
880000000
102882765
0009182736
1090
1188100
1296020000
330161480
102914247
■0009174312
lOSl
1190281
1298596571
830302891
102945709
0009165903
1092
1192461
1S02170688
330454333
192977158
0009157509
1093
H9464Q
1305751357
830605605
103008577
0009149131
1094
1196833
1809838684
330756708
103039983
0009140768
109S
1109025
1812932375
830907842
108071868
0009132420
1096
1201216
1816582786
331058907
103102735
0009124008
1097
1203409
1320139673
88:20990S
108184083
0009116770
1098
1205604
1328768192
331360830
103165411
0009107468
1099
1207801
1827373299
331611689
103196721
0009099181
liOO
1210000
133I0O0O00
831662479
108228013
OOOS000909
1101
1212201
1384638301
831813200
103289284
000908:;6J2
1102
1214404
1338273208
331968868
108290537
0009074410
1103
1216609
1341919727
882114438
108821770
■00090661B3
1104
1218816
1845572864
832266955
103353985
0009057971
1105
1221025
1849282625
332415403
103384181
0009049774
1106
1223236
1352899016
332665763
103416858
0009041391
1107
1226449
1356572043
882716095
108446617
0009088424
1108
1227664
1360251712
332866339
108477657
0009035271
1109
1229881
1363938029
838016516
103508778
0009017133
1110
1282100
1367631000
833166025
103539880
00090090U9
IllI
1234321
1S7I33063I
333316666
108570964
00090009(10
b,Google
THE PRACTICAL MODEL CALCrLATOR.
1112
S,„.r».
Culas.
Sq»l,tl, nooi!.
Cal«I.o«M,
ll»cip.„.,l,.
1236544
1875036928
33'34euu40
103602029
0008992806
1118
1288769
1378749897
33361 O&W
103033076
0008984726
1114
1240996
1382469544
83 3766885
103064103
0008970661
1115
1243226
1386195875
833916157
103695118
0008968610
1116
1245456
1889928896
334065862
108726103
0008960753
1117
1247689
1393668613
334216499
103757076
■0008952551
1118
1249924
1397415082
334365070
103788030
■OD08944544
1119
1252161
1401168159
334614673
103818965
•0008936550
1120
1254400
1404928000
334664011
103849883
0008928571
1121
1256641
1408694561
384813381
1038807S1
0008960607
1122
1258884
1412467848
334962684
103911661
■0008912656
1123
1261129
1416247867
385111921
103942527
0008904720
1124
1263376
1420034624
83&261092
103978806
■0008896797
1125
1265625
1428828125
835410196
104004192
1126
1267876
1427628376
335559234
104034999
■0008880995
1127
1270129
1431435383
335708206
104065787
■0008878114
1128
1272884
1485249152
835857112
104096557
■0008865248
1129
1274641
1439069689
336005952
104127310
0008857396
1130
1270900
1442897000
336164726
104158044
0008849558
1181
1279161
1446731091
336308434
104188760
■0008841733
1132
1281424
1460571968
336452077
104219458
0008833922
1183
1283689
1454410637
386600653
104250188
■0008826125
1134
1285966
1458274104
336749165
104280800
■O008818342
1185
1288225
1462185876
SS6897610
104311443
■0008810573
1186
1290496
1466003450
837045991
104842069
■0008802817
1187
1292769
1469878353
837174806
104872677
■0008795075
1138
1295044
1473760072
337340556
10440367T
■0008787346
1139
1297321
1477648619
387490741
104433889
■0008779631
1110
1299600
1481544000
837688860
104464308
■0008771980
lUl
1801831
1485446221
837786915
104404929
■0008764242
1142
1304164
1489855288
337934905
104525448
■0008756567
1143
1808449
1493271207
338082830
104555948
OO0S7 48900
1144
J30873G
1497193984
388280691
104586431
■0008741259
1145
1811025
1501123025
838378480
104616896
■0008733624
1146
1313316
1505000136
388626218
104647848
0008726003
1147
1315609
1509008523
104077773
■0008718396
1148
1817904
1612958792
104708158
0008710801
114S
1320201
1516910049
338969026
104738579
■0008703220
1150
1322500
1620875000
389116499
104768955
■0008695652
1151
1324801
1524845951
339268909
104799314
■0008688097
1153
1327104
1528828808
839411255
104829B56
■0008680556
1153
1S29409
1582808577
339558537
104859980
0008678027
1154
1331716
1536800264
389705755
104890286
■0008665511
1155
1834025
1540798875
839852910
104920575
0908658009
1156
1544804416
840000000
104950847
■0008650619
1157
133SB49
1548816898
340147027
104981101
0008043042
1158
1340964
1552830312
340293990
105011337
0008636579
1159
1343281
1656862679
340440890
105041556
0U08U2812S
1160
1845600
1560896000
840587727
105071757
0008620690
1161
1847921
1564936281
310734501
105101042
0008613244
11B2
1350244
1568983528
340881211
105182109
0008U05852
1163
1352569
1573037749
340127858
105102259
0003598152
1164
1354896
1577098944
341174442
106192391
0008591065
1105
1357225
1581107125
341820963
106222506
0008583091
1166
1859556
1685242296
341467422
105252604
■0008676329
1167
1361889
15893:^4403
341613817
105282085
■00085689^0
1168
1364224
169841 8682
341760150
105312749
■0008501644
1109
1366561
1597509809
341906420
105342795
■0008554320
1170
1868900
100161SO0O
342052027
105872825
0003547009
1171
1371241
160 j 723211
842198773
105402837
000S539710
1172
1373584
1609810448
3423448y3
105433S32
■0008532423
UTS
1373929
1013964717
3424WU875
105462810
0008525149
b,Google
TABLE OF SQTJAILEe, CUBES, SQUARE AND CUBE ROOTS.
Kuml,«r.
.,«™.
ci..
Squ»r»K=^t.,
C»l« RooU.
IM.il>™<.l8,
1174
1878276
1018006024
34'2089884
105402771
0008517888
1175
1880026
1633234376
8427827B0
106622715
0008510688
1176
13829T6
1020379778
842928564
105552642
■0008503401
1177
1385329
183*532233
843074386
105582662
0008496177
1178
1887684
1084091762
348220046
106012445
0008488064
1179
18B0041
1638858339
843365694
105642822
0008431761
1180
1392400
1043082000
348511281
105073181
000847] 670
1181
1394761
1647213741
843G508O5
106702024
0003407401
1183
1397124
1651400508
848802268
105781849
00081602S7
1183
1899489
1655505487
843947670
105T61658
0008J530SD
1184
1401856
1060797604
344098011
105791449
■0008415040
1185
1404226
1804000826
844288289
105821225
•00084B8819
118Q
1106696
1008222856
844883507
105850988
■0008431703
1187
1408969
1072446208
344628668
105880725
■0008421000
1188
1411344
1C70076072
844078759
105910450
0008417508
1189
1418721
1680914629
844818793
105940158
000841 UJ29
1190
1416100
1085159000
344093706
105909850
■0008103301
1191
1418481
1089410871
845108678
105999525
OOOS800800 1
im
1420804
345253530
100020184
■0008389202
1193
1423249
1097930057
815398821
106058820
■0008382320
li94
1425639
1702209884
845548051
106088451
0008875200 !
1195
1428025
1706489875
345887720
106118060
■0008308201 i
1100
1430416
1710777586
846832329
100147652
0008361204
1197
1432809
1716072373
845976879
106177228
0008G5!219
llliS
1185204
1719374392
846121366
106200788
0008347345
111) J
1137001
1723688599
340205794
100280331
■0008340284
1:!00
1440000
1728000000
346410192
106205867
■00083333S8 1
l:iUl
1442401
1782323601
846654409
100295367
■0008820805 i
l:i02
1444804
1736654408
846698710
106824800
■00083 194(i8
12U3
1447209
1740992427
310842904
106854888
■0003312.152
1204
144961Q
1745837064
846987031
109383799
000830J048
laoj
1463025
1749600125
347131090
106413244
000829S755
12Uli
1454436
1754049810
847275107
106442672
■0008291874
1207
1450849
1768416748
317419055
106472085
■000828JOU4
120S
1459204
1762790912
347662944
106501480
■0008273140
1^09
1461081
1707172829
347706778
100580800
■0008271299
l:ilO
1464100
1771501000
347860643
100560233
•0008201463
1211
1406521
1776950931
847094363
100589570
0008257088
1212
14(J8944
1780360128
848187904
106018902
0W18250S25
1213
1471369
1784770597
848281495
106648217
•000824402:;
1478796
1789188844
348425028
100677616
0008237232
1215
1476225
1703613376
348508501
106706799
■0008250453
121G
1478656
1798045996
848711916
106736060
0008223684
1217
1481089
1802486318
348865271
100765317
00082] 9927
1218
1488524
1806933232
848998567
109794552
■0008210181
1219
1485961
1811389469
349141805
106828771
•0008203445
1220
1488400
1815848000
849284984
106863978
■0008190721
1221
1490841
1820316861
349428104
106882100
■0008100008
1222
1493284
1824798048
340571166
100911831
■0008183306
1223
1495729
1820276567
349714169
106940486
■00081 76 Gl 5
1224
1498176
1833764247
849857114
106969625
■0008109335
1225
1600626
1838265025
850000000
100998748
■0003103205
1220
1503276
1842771176
350142828
107027855
■0008156607
1327
1605539
1847284083
850285598
107056947
0008140D50
1228
1507984
1851804352
350428309
107080023
•0008143322
1229
1510441
1850331989
350570963
107115083
0008130090
1230
1512000
1800867000
850713558
107144127
0008130081
1231
1515361
1865400391
350850099
107173155
0006123477
1232
1517824
1809050168
850998576
107203108
0008116883
1233
1530289
I87451G837
851140907
107231105
oooaiioaoo
1234
1522756
1879080304
107260140
■0008103728
1235
1526225
1883052876
351425508
107289113
000801)7100
b,Google
THE PRACTICAL MODEL CALCULATOR.
B™b.r.
S,».reB.
Cnl«=.
SinBie Kwtfl. Cuba Bonis.
B..lpr*,^,.
iMae
1527696
1888282256
851667917
107818062
■0008090615
1287
1580169
1892819058
86'1710108
10
7346997
0008084074
1238
1632644
1897413272
351852242
10
7375916
■0008077544
1289,
1636121
19020U919
851994318
10
7404819
■0008071025
1240
1587600
1906624000
862186887
10
7438707
■0008064516
1241
1540081
1911240521
352278299
10
7462579
0008058018
1242
1642534
1915864488
852420204
10
7191436
■0008051580 
1243
1645049
1920495907
352562051
10
7520277
0008045052
1214
1547636
1925134784
352708842
10
7549108
•0008088686
1245
1660026
1929781125
852845575
10
7577918
0008082129
1248
1552621
1934484986
852987252
10
76Q670B
•0008025662
1247
1655009
1939096223
853128872
10
7635488
■0008019216,
1248
1557504
1943764992
858270435
10
7664252
0008012821
1249
1B60OO1
1948441249
353411941
10
7698001
■0008006405
1250
1662500
1968125000
868653391
10
7721785
■0008000000
1251
1565001
1957816251
358694784
10
7750458
•0007998605
1252
1667504
1962515008
353886120
10
7779156
■0007987220
1258
1570009
1967221277
85397'7400
10
7807843
■0007B80846
1254
1672516
1971935064
35 ■4118624
10
7836516
■0007974482
1265
1675025
1976656375
354269702
7965173
■0007968127
1577586
1981385216
. 864400903
IC
7898816
■0007961783
1257
1580049
1986121693
85 4641958
10
7922441
■0007955449
1258
,1582564
1990865512
854682957
10
7951063
■0007949126
1259
1585081
1995610979
854823900
10
7979049
■0007942812
1260
1587600
2000376000
854964787
10
8008230
■0007986508
1261
1600121
2005142581
, 355105618
10
8086797
■0007930214
1262
1692644
2009916728
856246393
10
8065348
■0007923930
1268
1695166
2014698447
355387113
10
8093884
•0007917656
1264
1697696
2019487744
356527777
10
8122404
■0007911392
1285
1000326
2024284625
855668885
10
8150909
■0007905138
1266
1602766
10
8179400
•0007898894
1267
1605289
2033901163
866949434
10
8207876
■0007892660
1268
1607824
2038720882
356089876
10
8286886
 0007886435
1269
1610361
2043648109
36623026a
10
8264782
■0007880221
1270
1612900.
2048883000
856370598
10
8293213
■0007874016
1271
1616441
2058225511
356510869
10
8321629
■0007867821
1272
1617984
2058075648
356651090
10
8350080
■0007861635
1278
1620529
2062983417
856791266
10
8376416
■0007855460
1274
1628076
2067798824
856931866
10
8406788
■0007819291
1275
1625625
2072671875
357071421
10
3486144
■0007848137
1276
1628176
2077552676
■ 857211422
10
8463485
■0007836991
1277
1680729
2082440938
357351867
10
8491812
■0007830854
1278
1633284
2087886952
357491258
10
8520126
•0007824726
1279
1685841
2092240689
857681095
10
8548422
■0007818608
1280
1638400
2097152000
857770876
10
8576704
■0007812500
1281
1640961
2102071841
357910603
10
8604972
■0007806401
1282
1643524
3106997768
858050276
10
8633225
■0007800312
1288
1646089
21,11982187
85 81898B4
10
8861454
■0007794282
1284
1648656
2116874304
8&8329457
10
8689687
^0007788162
1285
1651225
2121824125
358468966
10
8717897
•0007782101
1286
1658796
2126781656
858608421
10
■8746091
■0007776050
1287
1656369
2131746903
368747822
10
8774271
■0007770008
^ 1288
1658944
2136719872
868887169
10
8802436
■0007763975
1661521
2141700569
869026461
10
88805S7
0007757952
1290
1664100
2140689000
359166699
10
8858723
■0007751938
1291
1666681
2151685171
859304884
10
■00O7745S3S
1292
1669264
2156689088
859444015
10
8914962
0007789938
1208
1671849
2161700757
359588092
10
8943044
■0007733952
1294
1674436
2166720184
859722115
10
8971123
■0007727975
1295
1677026
2171747875
859861084
10
8999186
■0007722008
1296
1679616
2176782386
360000000
10
9027235
■0007716049
1297
1682209
2181826073
3601?8862
109055269
■0007710100
b,Google
TABLE OP SQUARES, CDBES, SQUAaB AHD CUBE ROOTS. 121
Numbiir.
S,«„es.
Cubea.
Sqnue Bwrts.
Cn..E„^..
E=cir.o.^=.
1298
1084804
2186876892
360277671
10'9088290
0007704160
vim
1687401
2191983899
SG'04]e426
109111296
0007698229
1300
1690000
2197000000
360556128
109139287
0007692308
1301
1692601
2202073901
86.0698776
109167266
■0007686895
1302
1696204
2207165608
860882871
109195228
■0007080492
130a
1697809
2212245127
360970913
109223177
■00076743T9
1804
1700416
2217342464
36 1109402
100251111
■0007668712 1
1305
1703026
2222417626
361247887
109279081
0007662885 \
isoe
1706636
2227560016
861886220
109806937
0007656968 i
1307
17082*9
2232681443
361624660
109384829
0007651109
1808
1710864
2237810112
861662826
109362706
0007645260 ]
1309
1713481
2242946629
361801060
109390569
0007089419
1310
1716100
2248091000
361989221
109418418
0007033588 i
1311
1718721
2253213231
862077840
109446253
0007627705 1
1812
1721844
2258403328
36 ■2216406
109475074
0007021951 :
1313
1723969
2268571297
862368419
109601880
0O0701644U ;
1814
1726696
2268747144
862491879
10<9629673
0007610350 !
1816
1720225
2273930876
3626262B7
109557461
0007004503
1316
1731856
2279122496
862767143
109585215
0007598784 ,
1317
1781489
2284822013
362901246
109612965
0007593011 ,
1818
1787124
2289529432
363042697
109640701
0007587253
1319
1739761
2294744769
868180896
109668423
0007681501
1820
1742400
22999GS00O
868818042
109696131
0007575758
1331
1745041
2306199161
863465687
109723825
•0007570023
1322
1747684
2310438248
86<859S179
109751605
■0007564297
1823
1760329
2315685267
868780670
109779171
0007558579
1824
1752070
2320910224
363868108
109800828
■0007552870
1325
1755625
2326203125
861005494
109831162
■0007547170
1326
1768276
2831478976
3G4142829
109862086
0007641478
1837
1760929
2336752783
864280113
109889696
0007536795
1823
1763584
2342089652
384117848
109917293
0007630120
1829
17G6241
23173312S9
86'4654&23
109911876
0007524454
1330
1768900
2852687000
364601660
109972445
0007618797
1831
1771561
2357917691
864828727
110000000
■0007513148
1832
1771224
2363266368
864965752
110027541
0007507608
1838
1776889
2368593087
865102T26
110066069
■0007501875
1334
1779556
2873927704
365239647
n 0082583
■0007496252
1386
1782225
2879270375
366376618
110110082
00071B003T
1836
1781896
2884621056
365518888
110187569
•00074850110
1837
1787689
2889979753
365G50106
110165041
0007479482
1338
17B0244
23953464T2
866786823
11 ■0192500
■0007173842
1389
1792921
2100T21219
365928489
110210945
0007408260
1840
1795600
2406104000
866060101
11 ■0347877
■0007462687
1341
1798281
2411494821
8G61966G8
110274796
•0007457122
1342
1800964
2416893688
860833181
11 ■0302199
■O0OT451666
1313
1803640
2422300607
866469144
11 0329590
0007116016
1844
1806836
2427715584
366606056
110856967
0007440476
1346
1809026
2433138625
860712416
110381330
0007434044
1846
181171G
2488569786
866878726
110111680
0007429421
1347
1814409
2444008923
367014986
110489017
0007423905
1348
1817104
2449466193
367161195
110466339
0007418398
1849
1819801
2454911549
867287863
I10193G49
0007112898
1350
1822500
2460876000
367423461
11 ■0620945
0007407407
1361
1825201
2165846661
867559519
110518227
0007401024
1352
1827904
2471826208
867695526
110676497
000739G450
1368
1830609
2476818977
367831483
110602752
■0007390983
1354
1883316
2482309861
367967390
110629994
€007385524
1856
1886025
2487818875
868103216
110667222
0007380074
1356
1838730
2493326016
SG8289063
110684437
0007374631
1357
1841449
2498840298
868374809
11 '0711689
0007369197
1358
1844104
2504371712
368510516
110738828
0007303770
1359
1840881
2509911279
868646172
110766003
00078&8862
b,Google
! PEACTICAL MODEL CALCnLATOIt.
Nu..b,r.
8,ar...
CaW^
s,,...™b..l..
(■„;.. ,i,.L,.
I[,o.u,.i.,.
13U0
1849600
2515456000
308781778
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■OUOTi! 3291 1
1801
18&2321
2521008881
368917330
110:^20314
■Ol*T347o3i)
1362
1855044
2526569928
86 ■9052842
110.^17449
■0007K12I41
1S63
1857769
2532339147
36 '9188299
110S74571
0O073:JO7J7
1804
1860496
2537716344
360323706
110901079
00O73ai878
136a
1868225
2548302125
369469064
110928775
0007S26OO7
1866
1865956
2548805896
36 ■9594372
110955857
00073200! 1
1367
1868689
2554497863
869729631
11 ■0082020
0007315289
1868
1871424
2560108082
360861840
111009982
■0007309912
1874161
2565726409
870000000
1110371126
■0007304002
1870
1876900
2571S58000
370135110
111064031
■0007291)270
1371
1879041
2576987811
870270172
111001070
■0007293940
1372
1882384
2582630848
870405184
111118073
■0007288030
1873
1885129
2588282117
370510146
111116064
0007288821
1374
1887876
2598041624
870676060
111172041
•0007278020
1375
18B0625
2599609375
111199004
0007272727
1898376
2605285S76
370044740
111225956
■0007207112
1377
1896129
2610969688
371079606
111262808
■00O72021C4
1878
1898884
2616662152
371214224
111279817
■0007250894
1379
1901641
2622362939
871348803
111306729
■0O072510S2
1380
1901400
2628072000
871483512
111888028
■0007240377
1S81
1907161
2683789341
371618084
U1360514
■0007241130
1B09924
26S9514908
871762606
111887386
•0007285890
1883
1912689
2615248887
371887070
111414246
■0007230058
1384
1916466
2650081104
372021505
111111008
■0007225134
1385
1918225
2656741626
872165881
11 '1407926
■O00722O217
1886
1920996
2662300156
872290209
11^1494747
■0007215007
1887
1928769
2668207603
872424480
11 ■I 521535
■0007209805
1388
1926644
2674018072
37 '2658720
11 ■1548860
■0007204011
1389
1929321
2679820869
87 ■2692903
11^157513a
■0007190424
1890
1032100
2685619000
372827087
11 ■1001003
■0007191215
1801
1934881
2601410471
872901124
11 ■1628059
■0007180073
1392
1937664
2697228288
373005162
11 ■1055408
■0007183908
isas
1940449
2703045457
878229152
1M682134
■0007178751
1394
1943236
2708870984
378868004
111708852
■0007173601
1395
1946026
2714704875
87'3496988
111735558
■0007168459
1386
1948816
2720547186
373680881
111762250
0007168324
1397
1951609
2726397773
878764682
111788930
0007158196
1398
1954404
2732256792
111815598
0007153076
1899
1967201
2738124199
874032084
111842252
0007147963
1400
1960000
2744000000
374165788
111868894
0007142857
1401
1962801
2749884201
3742B9846
111896528
■0007187769
1402
1965604
2755776808
874432904
111922139
■0007132668
1403
1068409
2761677827
374606416
111918748
■O0O71276S4
1404
1971216
2767587264
37469Q880
111975334
0007122507
1405
1974025
2773505123
874883296
112001913
0007117488
1406 ■
1976836
2779431416
374066666
112028479
■0007I12S76
1407
1970649
2785866143
875099987
112055032
0007107821
1408
1982464
2791809312
875283261
112081578
■0007102273
1409
1985281
2797260920
376366487
112108101
0007097232
1410
1988100
2808221000
875400667
112134617
0007092109
nil
1990921
2809189581
875682799
112161120
0007087172
1412
1993744
2816166528
376765885
112187611
0007082158
1413
1996569
2821151997
875808922
112214089
0007077141
1414
1999306
2827145044
87^6031913
112210054
■0007072136
1415
2002225
2833148376
87^6164867
112267007
■0007067188
1416
2005056
2889150296
376297764
112298448
■0007062147
1417
2007889
2845178718
876480604
112319876
■0007067163
1418
2010724
2851206632
376563407
112346292
■0007052186
1419
2013561
2857243059
876696164
11 '2372606
■0007047210
1420
2016400
2863288000
37^6828874
112800087
0007042254
1421
2019241
2869341461
37 6961536
112425165
0007037298
b,Google
TABLE OE SQUARES, CUBES, SQUARE AND CUBE BOOTS. 123
^■„.,l..^.
Sfloart!.
C,*.,
S,u.r.E«,«.
cube Hoct,.
itoip™,^.
U±^
202J084
2875403448
877004153
112451831
0007032349
ZU24a20
2881473907
877226722
112478185
0007027407
i4:;i
2887558024
8TTS59245
112604527
0007022472
20S0025
2898640626
877401723
112680856
0007017544
^033476
289B736776
877624152
112557173
0007012628
U27
2036320
2005841483
377756535
112688478
0007007708
14:;8
2089184
2011054762
377888873
112609770
0007002801
U2'J
2042041
2918076580
878021168
112686050
0006907901
I4S0
2044900
2924207000
878163408
112662318
0006903007
1431
20477G1
2980345991
378285606
112688578
0006988120
1432
2060624
2936408668
878417769
112711816
0006983210
1433
2053489
2042640737
878549864
112741047
0000978307
14S1
2056356
2948814504
378681924
11 '2767266
0000073501
1485
2050225
2954087875
878818988
112798472
0006068641
1436
2062096
2061169856
878045906
112810666
0006063788
1437
2064069
2067360458
379077828
112845840
0006958942
1488
2067844
2973559672
370209704
112872019
0000054103
J439
2070721
207B7075I9
879341538
112808177
0006949270
1410
2078600
2985984000
87 ■9473310
112024323
0006944414
1441
207C481
2992209121
379605058
11 ■2050457
0006939025
1442
2079364
8008442888
379736751
112976579
0006934813
1443
2082240
8004685307
87G808398
118002688
O000OSU0O7
1444
2086130
8010986884
380000000
118028786
0006925208
1445
2088025
8017196125
380131556
113054871
■0006920415
1410
2O8OBI0
3023464586
880263067
113080945
■0006915629
1417
2098800
8029741623
880894582
113107006
0006910830
1448
2096704
8030027892
380525952
118183056
0006906078
14d0
2009601
8042321849
380657826
118169004
0006901812
1450
2102500
8048625000
880788056
113185119
0006806652
Hil
aio&ioi
8064986851
880019980
118211182
0006891799
145J
2108304
3001257408
381051178
113237134
0006887062
14oa
2111200
8007586777
881182871
113268124
0006882312
1454
2114116
3073024664
88181851B
n8289102
0006877579
1465
2117026
80802713/5
88 1444622
11 3315067
0006872852
145(i
2U9936
8086626816
88 1676681
11 8341022
0006868182
1467
212284S
8092990993
38 1706693
11 3866064
0006863412
1458
2125764
8099363912
38188.662
11 8892804
0006858711
1450
2128681
310545579
38 196S685
11 8418813
0006854010
1400
2131600
3112186000
88 20004b3
11 8444719
0006849315
1461
2134521
8118685181
88 2230297
n 3470614
0006844627
1463
2137444
8124943128
38 236108O
11 8406497
0006830945
1463
2140369
8131S59847
88 2491829
11 3522368
0006885270
1464
2143206
8187785344
88 2622029
11 3548227
0006830601
14C5
2140225
3144219025
88 2763184
11 8574075
0000826989
1400
2140156
8150662606
88 2883794
11 3599011
0006821282
1107
2152080
8167114503
38 3014360
11 8625786
0006816638
1488
2155024
8163575282
38 8144881
11 3651547
0006811989
1469
2157S61
8170044709
38 3275858
11 8677347
0006807852
1470
2160900
3176523000
88 3405,00
11 3703136
0006802721
1471
2163841
8183010111
38 3536178
11 3728914
0006708097
1472
2166784
8180606048
888666522
113754670
0006793478
1473
2160720
8106010817
888796821
113780433
0006788866
1474
2172676
3202524424
888927076
113806176
0006784261
1475
2175625
3209046875
884057287
118831906
0006779661
1473
2178576
8215578176
334187454
113857625
0006775068
1477
2181529
3222118333
884317577
118883882
0006770481
1478
2184484
3228667862
884447666
113909028
0006765900
1470
2187441
8285225239
384577691
118034712
0006761825
1480
2190400
3211792000
884707681
113960381
0006756757
1481
21933G1
8248367641
113080045
■0006752104
1482
2106324
3254952168
384067530
114011605
0006747638
1488
2199:^89
3201546587
385007300
114037332
0006743U88
b,Google
THE PRACTICAL MODEL CALCULATOR.
Hnmbsr.
Squires.
Cnl,a,,
aiun.6E«,u^
Cuba Koiits.
B8cLp«™la.
1484
2202256
3268147904
385227206
114062969
■0006788844
1485
2206225
8274769125
S85866077
114088574
0006734007
1486
2208196
8281379256
88'5486705
114114177
•0006729474
1487
2211169
3288008303
385616389
114139769
■0006724930
1488
2214144
8204646272
886746080
114165349
■0006720430
1480
2217121
8301293169
885875327
114190918
0006715S17
1400
2220100"
3807949000
88'6005181
114206476
■0006711409
1491
8314613771
886184691
114242022
■0006706908
1492
3226004
3321287488
386264158
114267556
■0006702418
1493
2229049
8227070157
88689e682
114203079
■0006697924
1494
2282086
3834661784
386522962
114318591
0006698440
1406
2235025
8841362876
886652299
114344002
■00066S8963
1496
,2238016
3348071986
. 386T81593
114360581
■0006684492
1407
2241000
8364790473
386910843
114305059
■0006680027
1498
2244004
3361517992
887040050
114420525
■0006676667
1499
2247001
8368264490
387169214
U 4445980
■0006671114
1500
2280000
8375000000
S87298885
n4471424
■0006666667
1501
2253001
3381754601
887427412
114496867
•0006662225
1602
2256004
3888518008
887656447
114522278
■0006657790
1503
225^009
3S06290527
387685439
114547688
0006653360
1504
2262016
3402072064
387814380
114578087
■0008648036
1505
2265025
387948204
114508476
■0006644618
1506
3416662216
888072168
114628850
■0006640106
1507
2271048
8423470843
388300973
114649216
■0006635700
1508
2274064
3420288612
888329757
114674568
■0006631800
1609 ■
2277081
8436115229
388458491
114699911
■0006626905
1510
2280100
3442951000
388587184
114725242
■0006622517
1611
2283121
8449795831
388715884
114760662
■0006618134
1512
2286144
3466640728
888844442
11 ■4776871
0006613757
1513
2289160
8463512697
388973006
114801160
■0006609386
1514
2202196
8470884744
■ 380101520
114826455
■0006605020
1516
2295225
3477265875
389280009
114851731
■0006600660
1516
2298256
3484166096
380358447
114876095
■0006696806
1517
2301289
349! 055413
389486841
114902249
0006691958
1518
2304324
3507968832
889616194
114927401
■0006687615
1619
2307361
8504881359
380743505
114952722
■0006688278
1620
2810400
3511808000
389871774
114977942
■0006578947
1521
2313441
3518748761
390000000
116008151
■0006574622
1523
2316484
8525688648
39 ■0128184
115028848
■0006670302
1523
2S10529
3532642667
115053535
■0006565988
1524
8689605824
390384426
115078711
■0006561680
1525
8546578125
30O51248S
115103876
0006657877
1526
2828676
3553550576
300640490
116129030
■0006553080
1627
2331729
3567549562
390768473
11 ■5164178
■0000548788
1628
2334784
8660668188
' 390896406
116179306
■0006544503
1529
2337841
8674558889
801024206
115204426
■0006540222
16S0
2340900
8581677000
391152144
116229585'
■0006535948
1631
2348961
8688604291
39 1279961
115264684
■0006531679
1532
2347024
8885640768
391407716
11 6279722
■0006527416
1583
2350080
3602686437
391636439
118804799
■0006523157
1634
2358156
8600741804
891663120
116329865
■0006618905
1585
2356225
3616805375
301700760
11 ■6354920
■0000514658
1636
2369256
8628878656
391018359
115379965
■0006510417
1587
3680961163
892045915
n8404998
■0006506181
1538
2365444
8638062872
892173431 ■
116480021
■0006601951
1689
2868521
8645163819
892800905
116456033
■0006497726
1640
2371600
3652264000
392428337
115480084
■0006498506
1541
2374681
3669383421
392556728
11 ■6505025
■0006489293
1542
2377764
3666512083
892683078
115580004
O0OS485084
1543
2380849
8678650007
892810387
115654972
■0006480881
1544
2388036
3680707184
392937654
116579931
■0006476684
1545
2387025
3687068625
393084880
115604878
0006472402
b,Google
TABLE or SQUARES, CUEES, SQUARE A3D CUBE HOOTS.
Number.
S,,>=r»..
ClCE.
S,«ar. n««.
QuU R.»l.,
Ilt^lpr.c.1,.
154lj
Ii890116
3695119336
398192065
115929816
0000468305
1547
2808209
8T02294323
393319208
115664740
0006464124
1518
2396304
8709478592
393446311
116679655
■0006459948
154SI
2399401
8716872149
893573878
115704569
■0006465778
1550
2102500
3723875000
393700394
115729458
0006461613
1551
2405601
3781087161
393827873
115764336
0000447453
1552
2408704
8738308608
393^54812
115779208
0006443299
1553
2411809
8745639377
894081210
115804069
■0006439150
1551
2414916
3752779464
394208067
115828919
■0006485006
1555
2418025
3760028875
894384883
116858759
■0006430868
1556
2421186
3767287016
394461658
115878688
■0006426735
1567
2424249
3774655693
894688393
115908407
■0006422603
1658
2437304
8781888112
394716087
115928215
0006418486
155S
24S0481
8789119879
394841740
115953018
0006414868
1560
2488600
3796416000
394968353
115977799
■0006410256
1561
2436721
3808721481
896094925
116002576
■0006406150
1562
2439844
8811036828
395221457
116027342
■0006402049
1668
2442969
3818360547
896347948
116052097
■0006897953
1564
3446096
8825641444
395474399
116076841
■0006398862
1565
2449225
3888037126
896600809
116101575
0006389776
156B
2462356
3840389496
896727179
116126299
0006385696
1567
2455489
8847761268
395853608
116151012
■0006381621
loliS
2458624
3855123432
896979797
116175716
■0006377551
1569
2461761
886250S009
396106046
116200407
■O006373486
1570
2464900
3809883000
396232256
116225088
0006369427
1671
2468041
3877292411
396868424
116249759
0006365372
1572
2471184
3884701248
896484652
116274420
■0006361323
1573
2474329
8892119157
396610640
116299070
■0006357279
157*
2477476
8899647224
116823710
■0006353340
1576
2480625
3906984876
896862698
116348339
■0006349206
1676
2483776
8914430976
396988665
116372957
0006346178
1577
2486929
8921887038
397114598
116397566
■0006341154
1578
2490084
3929352552
397240481
116422164
■0006837186
1670
2498241
3936827539
897366329
116446751
■0006333122
1580
2496400
3944312000
397492138
116471329
■0006329114
1581
2499561
8951805941
897617907
116495895
0006325U1
1582
2502724
8959809368
397743636
116520452
■0006321113
1588
2505889
3966822287
897869826
116544998
■0006317119
1584
2509066
3974344704
897994976
116569584
■0006313181
16S5
2512225
3981876626
898120585
116594059
0006309148
1586
2515896
8989418056
398246155
116618574
■0006305170
1587
2518569
3996969003
898371686
116643079
■0006301197
1588
2521744
4004529472
898497177
116667574
0006297229
1589
2524921
4012099469
398622628
116692068
■0006298268
1590
2528100
4014679000
898748040
116716532
■0006289308
1591
2631281
4027268071
398873413
116740996
■0006285355
1592
2584464
4034866688
398998747
116765449
■0006281407
1503
2537649
4042474857
399124041
116789892
■0006277464
1594
2540836
4050092584
399249295
116814325
■000627852S
1595
2544025
4057719876
899374511
116838748
0006269592
1596
2547216
4065356736
899499687
116863161
0006285664
1597
2550409
4073003178
899624824
0006361741
1598
2558604
4080659192
899749922
11 691 1055
■0006257823
1599
2556801
4088324799
399874980
116936337
0006253909
1600
2560000
4096000000
400000000
116960709
■0006250000
To find the square or cube root of a number consisting of integers
and decimals.
Rule, — Multiply the difference between the root of the integer
part of the given mimher, and the root of the next higher integer
number, by the decimal part of the given number, and add the
hv Google
126 TUB PRACTICAL MODEL CALCULATOR.
product to the root of the given integer number ; the sum is the
root required.
Required the square root of 20321.
Square root of 21 = 45825
Do. 20 = 44721
1104 X 321 + 44721 = 45075384, the
square root required.
Required the cube root of 1642.
Cube root of 17 = 25712
Do. 16 = 25198
•0514 X 42 + 25198 = 2541388, the cube
root required.
To find the squares of numbers in arithmetteal progression ; or,
to extend the foregoing table of squares.
Rule. — Find, in the usual way, the squares of the first two num
bers, and subtract the less from the greater. Set down the square
of the larger number, in a separate column, and add to it the dif
ference already found, with the addition of 2, as a constant quan
tity ; the product will bo the square of the next following number.
The square of 1500 = 2250000 2250000
The square of 1499 = 2247001
Difference 2999 + 2 = 3001
The square of 1501 2253001
Difference 3001 + 2 = 3003
The square of 1502 2256004
To find the square of a greater number than is contained in the table.
Rule 1. — ^If the number required to be squared exceed by 2, 3, 4,
or any other number of times, any number contained in the table,
let the square affixed to the number in the table be multiplied by
the square of 2, 3, or 4, &c., and the product will be tho answer
sought. _
Required the square of 2595.
2595 is three times greater than 865; and tho st^uarc of 805,
by the table, is 748225.
Then, 748225 x 3^ = 6734025.
Rule 2, — If tho number required to be squared be an odd num
ber, and do not exceed twice the amount of any number contained
in the table, find the two numbers nearest to each other, which,
added together, make that sum ; then the sum of the squares of
these two numbers, by the table, multiplied by 2, will exceed the
square required by 1.
Required the square of 1865.
The two nearest numbers (932  933) = 1865.
Then, by table (932= = 868624) + (933^ = 870489) = 1739113 x
2 = 3478226  1 = 3478225.
hv Google
RULES FOR SQUARES, CUBES, SftUARB BOOTS, ETC. 127
To find the cube of a greater number than is contained in tJie table.
Rule. — Proceed, as in squares, to find hoiv many times the num
ber required to he. cubed exceeda the number contained in the table.
Multiply the cube of that number by the cube of as many times as
the number sought exceeds the number in the table, and tlie pro
duct will be the answer required.
Required the cube of 3984.
3984 is 4 times greater than 996; and the cube of 90G, by the
table, ia 988047936.
Then, 988047936 x 4^ = 63235067904.
To find the square or cube root of a highernumber than is in the table.
Role. — Refer to the table, and seek in the column of squares
or cubes the number nearest to that number whose root is sought,
and the number from which that square or cube ia derived will be
the answer required, when decimals are not of importance.
Required the square root of 542869.
In the Table of Squares, the ncareat number ia 543169 ; and
the number from which that square has been obtained is 737,
Therefore, ^"542869 = 737 nearly.
To find more nearly the cube root of a higher number than is in
the table.
Rule. — Ascertain, by the table, the nearest cube number to the
number given, and call it the assumed cube.
Multiply the aasumed cube, and the given number, respectively,
by 2 ; to the product of the assumed cube add the given number,
and to the product of the given number add the assumed cube.
Then, by proportion, as the sum of the assumed cube is to the
sum of the given number, so ia the root of the aaaumed cube to
the root of the given number.
Required the cube root of 412568555.
By the table, the nearest number is 411830784, and its cube
root is 744.
Therefore, 411830784 x 2 + 412568555 = 1236230123.
And, 412568555 X 2 + 411830784 = 1236967894.
Hence, as 1236230123 : 1236967894 : : 744 : 744369, very nearly.
To find the square or cube root of a number containing decimals.
Rule. — Subtract the square root or cube root of the integer of
the given number from the root of the next higher number, and
multiply the difference by the decimal part. The product, added to
the root of the integer of the given number will be the answer
Required tho square root of 321'62.
v^321 = 179164729, and v'322 = 179443584; the difference
(■0278855) X 62 + 179164729 = 179337619.
hv Google
128
HIE PRACTICAL MODEL CALCULATOR.
To obtain the s(iuare root or cube root of a number containing dect
mak, by inspection.
Rule. — The square or cube root of a number containing deci
mals may be found at once by inspection of the tables, by taking
the fiourcs cut off in the number, by the decimal point, in ^yairs
if for the square root, and in triads if for the cube root. The fol
lowing example will show the results obtaineiJ, by simple inspec
tion of the tables, from the figures 234, and from the numbers
formed by the addition of the decimal point or of ciphers.
■0483735465*
■13276143S
■152970585
■284J:
2340
■4837354G5
61622401
234
152970585
132761489
488785465
2860
152970585
2340
483735465
132761439
23400
152970585
2860
Tofindthe cubes of numbers in arithmetical progression, or to extend
the preceding talle of cubes.
Rule. — Find the cubes of the first two numbers, and subtract
the less from the greater. Then, multiply the least of the two
numbers cubed by 6, add the product, with the addition of 6 as a
constant quantity, to the difference ; and thus, adding 6 each time
to the sum last added, form a first aeries of differences.
To form a second series of differences, bring down, in a separate
column, tho cube of tlie highest of the above numbers, and add
the difference to it. The amount will be the cube of the next
general number.
Required the cubes of 1501, 1502, and 1503.
First series of differeTtees.
By Tat). 1500 = 3375000000
6746501 difference.
1490 X 61 6 ^ 9 000
6764501 diff. of ICOO
SOOO I 6 = 9006
6772510diff. ofl502
Then, 3375000000 Cube of 1500
Diff. for 1500 = 675450 1
3381754501 Cube of 1501
Diff. for 1502= 6772519
8895290527 Cube of 1503
• Deriyed from 002340 by mBans of 2340.
t Derived from 002340 by means of 2340.
I The nearest result by simple inspeetioa is obtained for 023 by 23. But four
places correct oan always be obtained by looking in the table of cubes for the
nearest triad or triads, in this instance for 23400; the cube beginning with the
figures 28808 ia that of 2860, whence 286013 true to the last place, and is after
warda substituted.
b,Google
TABLE OF THE FOURTH ASD FIFIH POWEKS OF NUllEERS. 129
Table of the Fourth and Fifth Powers of Numhers.
BM3058
BBS4198
iwasTB
iissiKa
20511149
282176249
Mjoasasi
M4696301
002436(43
vmnasu
19S491I6S2
20750n6B3
es5T490i
J4M5201
8UM626
SSI5334S9
wgssma
2U14(iaS6
291199921
6277S1916a
Ei9D19IHII)0D
62103ai!61
I7e231lGSS2
19261116S21
20113571876
2100MUe76
21fl3U80SS7
S28n6776«8
hv Google
THE PRACTICAL MODEL CALCULAIOK.
Table of My^erbolie Logarithm
N,
I^pLrithm,
N.
U8=rithm.
N. t
aesritlim
N.
L.,.ri,hm,
ToT
0009503
158
■4574248
215
7654678
272
I 0006318
102
■019802a
159
■4637340
216
7701082
273
10043015
103
■0295588
160
■4700036
2^17
7747271
274
10079579
104
■0892207
161
•4762341
218
7793248
2^75
1^0116008
1'05
■0487902
1^62
■4824261
219
7839015
276
10152306
106
0682689
163
■4886800
220
7884673
277
10188473
107
■0676586
164
■4946962
221
7929925
278
10224509
108
■0769610
165
■5007752
222
7975071
2^79
10260415
109
■0861777
1^66
■6068176
223
B020016
260
10396194
110
■0953102
167
■612S236
2^24
8064758
2'81
10331844
111
■1043600
168
■5187987
282
1 ■0367868
112
■1133287
1^60
■6247286
10402766
113
■1222176
170
■5806282
2^84
10488040
114
■1810283
1^71
■5364933
£■86
I 0478189
115
■1397619
172
■5428242
286
1 ■0608216
116
■1484200
173
■5481214
287
10548120
117
■1570037
174
■5638851
10577902
118
■1655144
175
■5596157
289
10612561
119
■1739533
176
•565813S
233
290
I ■0647107
120
•1828215
177
•6709795
284
8501509
291
1^0681680
121
■1906208
178
■5766133
235
8544153
2^92
10715836
132
■1988508
179
■5822156
2^86
8586616
293
10750024
\2Si
■2070141
1^80
■5877866
2^37
294
1 '0784095
124
■2151118
181
•5933268
238
8671004
295
1 ■0818051
125
■2231485
182
■5988365
239
8712983
296
10851892
123
■2811117
183
■6043159
240
8754687
297
10885019
127
■2390169
184
■6097665
241
8796267
298
10919233
128
■2468600
■6151856
2^42
8887675
2^99
10962733
129
■2546422
■6205764
243
8878912
3^00
10986123
1SO
■2028642
187
■6259884
244
8919980
801
11019400
131
■2700271
188
■6812717
2^45
302
1 ■1052568
132
■2776317
1^89
■6365768
246
9001613
303
1^1086626
183
■2851789
190
■6418588
2'47
9042181
304
MI 18575
184
■2926696
191
■6471082
2^48
9082586
306
11161416
136
■3001046
192
■6523261
2^49
3^06
11184149
■3074846
193
■6575200
250
9162907
307
1 ■1216776
137
■8148107
194
■6626879
2^61
0202827
8^08
1^1249295
■3220884
195
■6678293
2^52
9242689
309
1^1281710
■3293037
196
■6729144
263
9282198
310
11314021
t40
■3364722
1^97
■6780335
254
9821640
S^ll
1 ■1316227
141
■3435897
■6830968
255
9360933
512
1 ■1378330
142
■8506568
199
■6881846
2^56
9400072
S^18
1 ■1410830
143
■8676744
200
■6031472
257
9489058
8^14
1^1442227
144
■8646481
201
■6981347
268
9477893
815
11174024
145
■3715635
202
■7030974
259
9616578
316
1^1506720
146
■3784364
203
■7080867
260
9555114
317
1^15378I5
147
■3862624
204
■7129497
2^61
9593502
318
1 ■1568811
148
■8920420
206
■7178897
262
■9631743
319
11600209
149
■8987761
2'06
■7227069
263
9669838
320
11631508
150
■4054651
207
■7275485
264
9707789
821
1^1662709
151
■4121096
208
■7323678
265
9746596
822
11693818
152
•4187103
2^09
■7371640
266
9788261
3^23
11721821
153
■4252677
210
■7419373
2 '67
9820784
324
11755733
154
•4817824
2^11
■7466879
268
9858167
3^26
11786549
155
■4383549
212
■7514160
269
9895411
826
11817371
156
■4446858
213
7561219
270
9932517
327
11847899
157
■4510756 1 214
■7608058
2.,,
9969480
328
11878434
b,Google
TABLE OE HYPEKBOLIC LOOAPJTHMS.
X.
L^anvithi^.
H.
L»((lrtU,B.
H,
I^S^Ubm
N
Logirillim,
329
11908875
T9I
13635873
"iiT
1 5107219
6 15
16880967
11939324
392
13660916
454
1 5I^92b9
516
16409365
SBl
1 '1969481
398
13686894
466
1 6161272
5 17
16428726
832
11999647
394
13711807
456
1 6173226
618
16448060
12029722
895
13787156
457
1 6196182
519
16467886
334
12059707
396
18762440
453
1 6216990
6 20
16486586
335
12089603
397
13787661
459
1 5238800
5 21
16505798
1'211B40S
398
18812818
460
1 5260663
5 22
16524974
3'87
12149127
399
13837912
461
1 5282278
5 23
16544112
338
12178757
400
18862943
462
1 6803947
5 24
l65e32U
339
12208299
401
13887012
468
1 5326568
6 25
1 6682250
340
12237764
402
18912818
464
1 684714]
5 26
16601310
841
12267122
403
13937663
465
1 53b8672
6 27
16620308
342
12296405
404
18962446
466
1 6390154
6 28
1 6630200
343
12826605
405
18987168
467
1 6411590
5 29
16658182
B44
12364714
406
14011829
468
1 6482'>81
6 80
16677068
845
12388742
407
14036429
469
15454825
6 81
16695018
346
12412685
408
14060969
470
1 6476625
5 32
16714733
847
12141546
409
14085449
471
1 5490879
16783512
348
12470322
410
14109869
472
1 6518087
6 84
16752236
349
1 2496017
411
14134230
478
1 6689262
6 35
16770965
850
12627629
412
14I58581
474
1 5560371
6 36
16780030
351
12566160
413
14182774
475
1 5681446
6 37
16808278
352
12684609
414
14206957
476
1 6602476
6 88
16626882
12612978
416
14281083
477
16623462
539
16846453
S54
12641268
416
14255150
478
16644405
640
16863939
855
12669475
417
14279160
479
15665804
641
16882491
356
12G97605
418
14803112
480
15686159
542
1 6900058
357
12725655
419
14327007
431
15706971
643
16919391
358
12753627
420
14350846
482
16727739
544
16937700
359
12781521
421
14374626
483
16748464
545
16966163
300
12809338
422
14398351
484
15769147
546
16974487 1
361
12837077
423
14422020
485
16789787
647
16992786
862
12864740
424
14445632
15810884
648
17011051
368
12892826
425
14469189
487
15830989
649
17029282
864
12919836
426
14492691
488
15861452
650
17047481
865
12947271
427
14516138
489
15871028
551
17065646
866
12974681
428
14639680
490
15892352
552
17083778
367
18001916
429
14562867
491
15912789
553
17101878
368
■18029127
430
14586149
492
15938085
654
1.7119944
369
13056264
431
14609379
16953389
555
17187979
870
13083328
432
14632553
494
159736S8
656
17156981
13110318
433
14655675
495
16993675
567
17173950
a72
13187236
434
14678743
496
16014067
558
17191887
373
13164082
435
14701758
497
16034198
669
17209792
874
13190856
436
14724720
498
16054298
560
17227666
375
18217658
437
14747630
499
16074358
661
17245507
876
13244189
14770487
500
1 6094379
562
17268316
877
18270749
439
14793292
501
16114359
563
17281094
378
13297240
440
14816045
502
16184300
664
172988M
879
18823660
441
14888746
503
16154200
665
17310565
18360010
442
14861396
504
16174060
566
173842S8
381
13876291
443
14883995
505
16193882
667
1 ■7351601
13402504
444
14906543
506
16213664
568
17869612
383
18428648
445
14929040
507
16288408
569
17387102
884
13454723
446
14951487
508
16253112
570
17404661
385
13480781
447
14978883
609
16272778
571
17422189
886
13506671
448
14996230
510
16292405
572
17439687
387
13532544
449
15018527
611
16311994
673
17457155
18558861
450
16040774
512
16881544
574
17474591
389
13584091
451
15062971
618
16361056
675
17491998
390 ! 13609765
462
15085119
514
16370630
576
17609374
b,Google
THE PRACTICAL MODEL CALCULATOR,
N.
I.^earithm,
N.
^E="U.m.
N.
N.
Loenritbio.
577
17526720
639 1
8647342
ToT
19473376
763
20820878
578
17544086
640 1
8562979
7 02
19487682
764
20333976
67S
17561823
641 1
8578592
703
19501886
766
20347066
580
17578579
642 1
8594181
704
195] 6080
766
20860119
581
17895805
643 1
8609745
705
19580275
767
20873166
17613002
641 1
8625286
706
19544449
768
20366195
5'83
1 ■7680170
64S 1
8640801
707
19558604
769
20899207
5'84
17647308
646 1
8656298
708
19572739
770
20412203
585
17664416
647 1
8671761
709
19686853
771
20425181
o8ti
17681496
648 1
8687205
710
19600947
773
20438143
5'87
17698646
649 1
711
19615022
778
20451088
17715567
650 1
8718021
712
19629077
774
20464016
17732559
651 1
713
19648112
775
590
17749523
652 1
8748743
714
19657127
776
20489823
591
17766468
663 1
8764069
715
19671123
777
20502701
592
17783364
654 I
8779371
716
19686099
778
20515508
5'93
17800242
665 1
8794650
717
19699056
779
20523408
5'94
17817091
656 1
718
19712993
780
20541237
6'95
17833912
657 1
8825138
719
19726911
781
20654049
S'SG
17850704
668 1
8840347
720
19740810
782
20566845
697
17807469
659 1
8855583
721
19754689
20679624
5S8
17884205
660 1
8870696
722
19768549
784
20592888
6'99
17900914
661 1
8885837
723
19782890
786
20605135
GOO
17917594
662 1
8900954
724
19796212
786
20617866
(iOl
1 7934247
668 1
8916048
725
19810014
787
20030580
G02
17950872
664 1
8981119
726
19823798
20643278
17967470
665 I
8946168
727
19837662
789
20666961
604
17984040
666 1
8961194
728
19861808
790
20668627
0'03
18000582
667 1
8976198
729
19865035
791
20681277
606
18017098
668 1
8991179
730
19878743
792
20698911
G07
18038686
9006138
731
19892432
793
20706580
G08
18050047
670 1
9021075
732
19906108
794
20719132
09
18066481
671 1
9085989
733
18919754
795
20731719
e10
18082887
672 1
9050881
784
19933387
796
20744290
611
18099267
9065751
7 ■35
18947002
797
20766845
e12
18116621
674 1
9080600
19960699
798
20769384
613
18131947
675 1
9096426
737
19974177
799
20781907
614
18148247
676 1
9110228
19987786
800
20794415
615
18164520
677 1
9125011
789
20001278
801
20806907
616
18180767
678 1
9139771
740
20014800
802
30819884
617
18196988
679 1
9154509
741
20028305
808
20881845
618
18218182
9169226
742
2004179O
804
20844290
619
18229861
6'81 1
9188921
743
20065258
805
20856720
620
18245498
682 1
9198594
744
20068708
806
20869135
621
18261608
683 1
9213247
745
20082140
807
30881534
622
18277699
684 1
9227877
746
20095553
808
20803918
628
18293763
685 1
9242486
747
20108949
809
20906287
624
18309801
686 1
9267074
748
20122827
810
20918640
625
18325814
687 1
9271641
749
20135687
811
20930984
626
1 8841801
688 1
9286186
750
20149030
812
20948306
627
18357768
689 1
9300710
751
20162364
813
20966613
628
18873699
690 1
9315214
762
20175661
814
20967905
U29
18389610
691 1
9329696
753
20188950
815
20980182
630
18405496
692 1
9344167
754
20202221
816
20992444
6Sl
18421356
693 1
9358598
755
20216475
817
21004691
682
18437191
694 1
9873017
796
20228711
818
21016923
1 8453002
696 1
9387416
757
20241929
819
21029140
634
18468787
696 1
9401794
20265131
820
21041341
635
18484547
697 1
9416162
769
20268815
821
31063629
18500288
698 1
9430489
760
20281482
21065703
637
18515994
699 1
9444805
761
20294631
823
21077861
038
18531680
700 1
9469101
762
20807768
824
31089998
b,Google
TABLE OF HYPSRBOLIC LOQABITHMS.
N.
toBHlHim.
N.
Loei,.i,im.
N.
I^p^lLm.
N.
Logariam.
826
21102128
869
21621729
913
22115056
957
22586882
8'2d
2IU4243
870
21633230
014
23126603
958
22596776
827
21126348
871
21644718
916
22187688
969
22607209
828
21188428
872
21656192
916
22148461
960
22617631
8.29
21150499
873
21667653
917
23159372
961
22638042
880
21162555
874
21679101
918
22170272
962
22688442
B'31
21174596
875
21690536
919
22181160
963
22648832
8'82
21186622
876
21701959
920
32192034
964
22669211
833
2.n98684
877
21713367
921
22202898
965
22669579
884
21210633
878
21734763
922
22218750
966
22679986
885
21222616
879
21786146
923
22224590
967
22690282
21284584
880
21747617
924
22235418
963
22700618
887
Z1246539
881
21758874
925
22246386
969
22710944
838
21258479
21770218
926
22267040
970
22721258
8S9
21270405
21781550
927
22267833
971
22781562
840
21282317
884
21792868
928
22278615
972
22741856
841
21294214
885
21804174
929
22289S86
978
22752138
842
21806098
886
21815467
980
22300144
974
22762411
848
21317967
887
21836747
931
22310890
976
22772673
844
21329822
21838015
22321626
976
23782D34
845
21341664
21849270
933
22332850
977
22798165
846
21353491
890
21860512
934
22348062
978
23803395
847
21865304
891
21871742
986
22853763
fi79
328136U
848
21377104
892
21882959
22364452
980
84S
21388889
893
21894163
937
22875180
981
22834023
850
21400661
894
21905356
938
22885797
22844211
851
21412419
895
21916535
939
22896452
983
22854S89
852
21424163
896
21927702
940
22407096
984
22864556
858
21435893
897
21988856
941
22417729
22874714
8'54
21447609
898
21949998
942
22428850
986
22884861
855
21469812
899
21961128
943
22438960
987
22894D98
856
21471001
900
21972245
944
22449559
32905124
857
21482676
901
21983360
e45
22460147
22916241
8'58
21494339
902
21994443
946
22470723
9BO
22925347
8'59
21506987
908
22005523
947
22481288
991
22685448
86l>
21617622
904
22016591
948
22491843
23945329
861
21529243
905
22027647
949
23502386
993
22966604
862
21540851
906
22088691
950
32512917
994
22965670
863
21652446
907
22049722
951
22628438
fi95
22076725
8M
21664026
908
22060741
952
22583948
996
22985770
865
21575598
909
22071748
953
22544446
997
22995800
866
21587147
910
22082744
954
33554934
998
28005881
867
21598687
911
23098727
965
22665411
999
23015846
868
21610216
9.12
22104607
956
22575877
1000
28025851
Logarithms were invented by Juste Byrge, a Frenchman, and
not by Napier. See " Biographie Universelie," " The Calculus of
Form," article 822, and " The Practical, Short, and Direct Method
of Calculating the Logarithm of any given Number and the Number
corresponding to any given Logarithm," discovered by OliveirEyrno,
the author of the present work. Juste Byrge also invented the
proportional compasses, and waa a profound astronomer and ma
thematician. The common Logarithm of a number multiplied by
2 '30 2585052 994 gives the hyperbolic Logarithm of that number.
The common Logarithm of 222 ia 346353 .. 2302585 X 346355
= 7975071 the hyperbolic Logarithm. The application of Loga
rithms to the calculations of the Engineer ivill be treated of here
after.
hv Google
IS4 THE PHACTICAL MOIJEL CALCULATOR.
COMBINATIONS OF ALGEBRAIC QUANTITIES.
The following practical examples will serve to illustrate the
method of combining or representing numbers or quantities alge
braically ; the chief object of which is, to help the memory with
respect to the use of the signa and letters, or symbols.
Let a = 6, 6 = 4, c = 3, (? = 2, e = 1, and/ = 0.
Then will, (1) 2» + S  12 + 4 =. 16.
(2) «6 + 2«  li = 24 + 6  2 = 28.
(3) o  6' + « +/ 36  16 + 1+ = 21.
(4) i> X (o  6) = 16 X (6  4)  16 X 2  32.
(5) iabe  Ide  216  14  202.
(6) 2 (o  i) (So  U)  {12  8) X {16  4) = 44.
(7) ^^ X (a  o)  1^ X (6  3) = 4 X 3 = 12.
(8) V {«■  2S') + (i / ,/ (36  32) + 2  = 4.
(9) 3«S  («  6  + <i) = 72  1  Tl.
(10) 3<iS  (o  S  c  ci) = 72 + 3 = 75.
(");7^^)'<(« + ^) = ;n^4'^Tj'<P + ^) = i^
111 solving the following questiona, the letters a, h, c, &c. are
supposed to have the same values as before, namely, 6, 4, 3, tc;
hut any other values might have been assigned to them ; therefore,
do not suppose that a must necessarily be 6, nor that 6 must be 4,
for the letter a may be put for any known quantity, number, or
magnitude whatever ; thus a may represent 10 tniles, or 50 pounds,
or any number or quantity, or it may represent 1 globe, or 2 eubio
feet, &c. ; the same may he said of b, or any other letter.
{!) a + bo = 7.
(2) m~d + e = 35.
(3) 2a^+ e''d+f= 79.
{i)jx{bc + d) = 21.
(5) 5.'(i  a= + Me = 62.
In the use of algebraic symbols, 3 v
tiling as 3 (4(i — 6)'^.
4(c + df [a + i)^, or 4 X e + d' X a + b^, signifies the
same thing as 4 V c \ d • ■$' a + b.
(8)
4 („. _ J.) (c  ,)
160.
m
^■m^'.
62.
(8) »/(2(." + 2<p) + fc
/ = 20.
(9)
icfb {d'de
1  670.
^ia'
 »'= 0.
10)
s/tlOd^ — 4«(;j ^c?
», 3
^ 4a — 6 signifies ■
the same
b,Google
THE STEAM ENGINE.
The partienlar esample whieli we sfiall select is tliat of an
engine having 8 feet stroke and 64 inch cylinder.
The breadth of the web of the crank at the paddle centre is the
breadth which the web would have if it were continued to the paddle
centre. Suppose that we wished to know the breadth of the web of
crank of an engine whose stroke is 8 feet and diameter of cylinder 64
inches. The proper breadth of the web of crank at paddle centre
would in this case be about 18 inches.
To find the breadth of crank at paddle centre. — Multiply the
square of the length of the crank m inches by 1'561, and then
multiply the sqnare of the diameter of cylinder in inches by '12B5 ;
multiply the square root of the sum of these products by the square
of the diameter of the cylinder in inches ; divide the product by
45 ; finally extract the cube root of the quotient. The result is
the breadth of the web of crank at paddle centre.
Thus, to apply this rule to the particular example which wo have
selected, we have
48 = length of crank in inches.
48
2304
1*561 = constant multiplier.
35965
5058 found below.
41023
64 = diameter of cylinder.
64
4096
1235 = constant multiplier.
and s/41023 = 6405 nearly.
4096 = square of the diameter of the cylinder.
45)_
582997
and g/582997 = 18 nearly.
Suppose that we wished the proper thickness of the large eye of
crank for an engine whose stroke is 8 feet and diameter of cylinder
64 inches. The proper thickness for the large eye of crank is
5'77 inches.
hv Google
ISb THE PRACTICAL MODEL CALCULATOK.
Rule. — To find the thickness of large eye of crank. — Multiply the
square of the length of the crank in inches by 1*561, and then mul
tiply the square of the diameter of the cyliuder in inches by ISSo ;
multiply the sum of these products by the square of the diameter of
the cylinder in inches ; afterwards, divide the prodnct by 182828 ;
divide this quotient by the length of the crank in inches ; finally
extract the cube root of the quotient. The result is the proper
thickness of the large eye of crank in inches.
Thus, to apply this rule to the particular example which we have
selected, we have
48 = length of crank in inches.
48
2304
1'561 constant multiplier.
35965
5058
41023
64 = diameter of cylinder in inches,
_64 ^
4096
1235 = constant multiplier.
505^'
41023
4096 = square of diameter.
48 ) 168030208
182828 ) 35006294
191 4T
and ^19147 = 57t nearly.
The proper thickness of the web of crank at paddle shaft centre
is the thickness which the web ought to have if continued to centre
of the shaft. Suppose that it were required to find the proper
thickness of web of crank at shaft centre for an engine whose
stroke is 8 feet and diameter of cylinder 64 inches. The proper
thickness of the web at shaft centre in this case would be SOT
inches.
Rule. — To find the tJdeJcness of the weh of erank at paddle shaft
centre. — Multiply the square of the length of crank in inches by
1561, and then multiply the square of the diameter in inches by
1235 ; multiply the square root of the sum of these products hj the
square of the diameter of the cylinder in inches ; divide this quotient
by 360 ; finally extract the cube root of the quotient. The result is
the thickness of the web of crank at paddle shaft centre in inches.
Thus, to apply the rule to the particular example which wc have
selected, we have
hv Google
THE STEAM ENGINE.
48 = length of crank in inches.
2304
1*501 = constant multiplier.
41023
64 = diameter of cylinder.
64
4096
1235 = constant multiplier,
5058
Ana V 41023 = 6405 nearly.
4096 = square of diameter.
72875
And ^ 78275 = 9 nearly.
Suppose that it were required to find the proper diameter for
the paddle shaft journal of an engine whose stroke is 8 feet and
diameter of cylinder 64 inches. The proper diameter of the
paddle shaft journal in this case is 1406 inches.
Rule. — To find the diameter of the paddle shaft journal. — Mul
tiply the square of the diameter of cylinder in inches by the length
of the crank in inches ; extract the cube root of the product ;
finally multiply the result by 242. The final product is the diame
ter of the paddle shaft journal in inches.
Thus, to apply this rule to the particular example which we have
before selected, wo have
64 = diameter of cylinder in inches.
_64
4096
48 = length of crank in inches.
196608
and ^196608 = 58148
but 58148 X 242 = 1407 inches.
Suppose it were required to find the proper length of the paddlo
shaft journal for an engine whose stroke is 8 feet, and diameter of
cylinder 64 inches. The proper length of the paddle shaft journal
would be, in this case, 1759 inchea.
The following rule serves for engines of all sizes :
Rule. — To find the length of the paddle shaft journal. — Multiply
the square of the diameter of tho cylinder in inchea by the length
of the crank in inches ; extract the cube root of the quotient ;
multiply the result by 303. The product is the length of the
hv Google
ISS THE PEACIICAL MOJJEL CALCL"LATOE.
paddlo shaft journal in inches. (The length of the paddle shaft
journal is IJ times the diameter.)
To apply this rule to the example which we have eclected, we have
64 = diameter of cylinder in inches.
64
4096
48 = length of crank in inches.
196608
and ^ 196608 = 58148
• length of journal = 58148 x 303 = 1760 inches.
We shall now calculate the proper dimensions of some of those
parts which do not depend upon the length of the stroke. Suppose
it were required to find the proper dimensions of the respective parts
of a marine engine the diameter of whose cylinder is 64 inches.
Diameter of crankpin journal = 909 inches, or about 9 inches.
Length of crankpin journal = 10'18 inches, or nearly lOj
inches.
Breadth of the eye of crosshead = 264 inches, or between 2J
and 2} inches.
Depth of the eye of crosshead = 1837 inches, or very nearly
18J inches.
Diameter of the journal of crosshead = 55 inches, or 5J inches.
Length of journal of crosshead = 619 inches, or very nearly Q}
inches.
Thickness of the web of crosshead at middle = 46 inches, or
somewhat more than 4 inches.
Breadth of web of crosshead at middle = 1715 inches, or
between 17^^ and 17^ inches.
Thickness of web of crosshead at journal = 393 inches, or
veiy nearly 4 inches.
Breadth of web of crosshead at journal = 646 inches, or nearly
6^ inches.
Diameter of piston rod = 6'4 inches, or 6 inches.
Length of part of piston rod in piston = 128 inches, or 12i
inches.
Major diameter of part of piston rod in crosshead = 068 inches,
or nearly 6^ inches.
Minor diameter of part of piston rod in crosshead = 576 inches,
oP 5 inches.
Major diameter of part of piston rod in piston = 896 inches,
or nearly 9 inches.
Minor 'diameter of part of piston rod in piston = 736 inches,
or between 7i and 7J inches.
Depth of gibs and cutter through crosshead = 672 inches, or
very nearly 6 inches.
Thickness of gibs and cutter through crosshead = 135 inches,
or between IJ and 1^ inches.
hv Google
THE STEAM EKGIiSE. 139
Depth of cutter through piston = 5'45 inches, or nearly 5^ inches.
Thickness of cutter through piston = 2*24 inches, or nearly 2
inches.
Diameter of connecting roct at ends = 6'08 inches, or neaily
6i\, inches.
Major diameter of part of connecting rod in crosstail = G'27
inches, or about 6J inches.
Minor diameter of part of connecting rod in crosstail = 5'76
inches, or nearly 5 inches.
Breadth of butt = 9*98 inches, or very nearly 10 inches.
Thickness of butt — 8 inches.
Mean thickness of strap at cutter = 275 inches, or 2J inches.
Mean thickness of strap above cutter = 206 inches, or some
what more than 2 inches.
Distance of cutter from end of strap = 308 inches, or very
nearly 3^ inches.
Breadth of gibs and cutter through crosstail = 673 inches, or
very nearly 6 inches.
Breadth of gibs and cutter through hutt = 704 inches, or some
what more than 7 inches.
Thickness of gibs and cutter through hutt = 1'84 inches, or
between If and 2 inches.
These results are calculated from the following rules, which give
correct results for all sizes of engines.
Role 1. To find the diameter of crankpin journal. — Multiply
the diameter of the cylinder in inches by 142. The result is the
diameter of crankpin journal in inches.
Rule 2. To find the length of crankpin journal. — Multiply the
diameter of the cylinder in inches by 16. The product is the
length of the crankpin journal in inches.
Rule 3. To find the breadth of the eye of crosafteat?.— Multiply
the diameter of the cylinder in inches by 041. The product is
the breadth of the eye in inches.
Rule 4. To find the depth of the eye of crosshead. — Multiply
the diameter of the cylinder in inches by •286. The product is
the depth of the eye of crosshead in inches.
Rule 5. To find the diameter of the journal of crosshead. —
Multiply the diameter of the cylinder in inches by OSB. The pro
duct is the diameter of the journal in inches.
Rule 6. To find the length of the journal of crosshead. — Mul
tiply the diameter of the cylinder in inches by 097. The product
is the length of the journal in inches.
Rule 7. To find the thickness of the web of crosshead at middle.
— Multiply the diameter of the cylinder in inches by 072. The
product is the thickness of the web of crosshead at middle in
inches.
Rule 8, To find the breadth of web of crosshead at middle. —
Multiply the diameter of the cylinder in inches by 268. The
product is the breadth of the web of crosshead at middle in inches.
hv Google
140 THE PRACTICAL MODEL CALCULATOR.
EuLB 9, To find the thickness of the web of crosshead at Journal.
— Multiply the diameter of the cylinder in inches by 'Ofil, The
product is the thickness of the weh of crosshead at journal in
inches.
Rule 10, To find the breadth of weh of crosshead at journal. —
Multiply the diameter of the cylinder in inches by 'lOl. The
product is the breadth of the web of crosshead at journal in inches.
Rule 11. To find the diameter of the piston rod. — Divide the
diameter of the cylinder in inches by 10. Tho quotient is the
diameter of the piston rod in inches.
Rule 12. To find the length of the part of the piston rod in the
piston. — Divide the diameter of the cylinder in inches by 5. The
quotient is the length of the part of the piston rod in the piston in
inches.
Rule 13. To find the major diameter of the part of piston rod in
crosshead. — Multiply the diameter of the cylinder in inches by
■095. The product is the major diameter of the part of piston rod
in crosshead in inchea.
Rule 14. To find the minor diameter of the part of piston rod in
crosshead. — Multiply the diameter of the cylinder in inches by '09.
The product is tho minor diameter of the part of piston rod in
erosahead in inchea
KuLB 15. To find the major diamett r cf the pa} t ofp/sl n j od in
piston. — Multiply tho diimetei of the cylinder in inches by 14.
The product is the major diameter of the pvrt of pistDU lod in
piston in inches,
Rule 16. To find the minor diamttrr ot the pat t of piste n rod in
piston. — Multiply the diimctor of the cylinder in inches by 115.
The product ia the minor diimeter of the pait of piston lod in
piston.
Rule 17. To find the depth of gibi and cuttet though cioss
head. — Multiply the diameter of the cylinder in inches by 105.
The product is the depth of the giba and cuttei thiough cross
head.
Rule 18. To find the thukness of the qihs and cutter through
crosshead. — Multiply the dnmetci of the cylinder m inchea by
•021. The product is tho thickness of the gibs and cuttez thiough
crosshead.
Rule 1&. To find the dipth of euttt.r through pistm — Multiply
the diameter of the cylinder m inches by 08 j The product la the
depth of the cutter thiough piston m inches
Rule 20. To find the Uinkness of cutter through putrn — Mul
tiply the diameter of the cylinder m inchea by 035 The product
is the thickness of cutter through piston m inches
Rule 21. To find the diameter of connecting rod at end'i — Mul
tiply the diameter of the cylinder m inches by 0'>5 The proiuet
is the diameter of tho connecting rod at ends in inches
Rule 22. To find the majir diameter if the pait cf onnednig
rod in crosstail — Multiply the diametet of the cybndei m inches
hv Google
THE STEAM EKOINE. 141
bj 098. The product is the major diameter of the part of con
necting rod in crosstail.
Rule 23. To find the minor diameter of the part of conneeting
rod in cjosstmi.— Multiply the diameter of the cylinder in inches
by '09. The product is the minor diameter of the part of con
necting rod in crosstail in inches.
Rule 24. To find the breadth of butt. — Multiply the diameter
of the cylinder in inches by "156. The product is the breadth of
the butt in inchea.
BuLE 25. To find the tMeknese of the butt. — Divide the diameter
of the cylinder in inches by 8. The quotient is the thickness of
the butt in inches.
BuLE 26. To find the mean thickness of the strap at cutter. —
Multiply the diameter of the cylinder in inchea by 043. The pro
duct is the mean thickness of the strap at cutter.
Rule 27. To find the mean thickness of the strap above cutter. —
JIuItiply the diameter of the cylinder in inches by 032. The
product is the mean thickness of the strap above cutter.
Rule 28, To find the distance of cutter from end of strap. —
Multiply tho diameter of the cylinder in inches by 048. The
product is the distance of cutter from end of strap in inches.
Rule 29. To find the breadth of the gibs and cutter through
cjosstaiL — Multiply the diameter of the cylinder in inches by
■105. The product is the breadth of the gibs and cutter through
crosstail.
Rule 30, To find the breadth of the gibs and cutter through
butt. — Multiply the diameter of the cylinder in inches by ll.
The product is the breadth of the gibs and cutter through butt in
inches.
Rule 31. To find the thickness of the gibs and cutter through
butt. — Multiply the diameter of the cylinder in inches by 029.
TJie product is the thickness of the gibs and cutter through butt
in inches.
To find other parts of the engine which do not depend upon the
stroke. Suppose it were required to find the thickness of the small
eye of crank for an engine the diameter of whose cylinder is 64
inches. According to the rule, the proper thickness of the small
eye of crank is 4'04 inches. Again, suppose it were required to
find the length of the small eye of crank. Hence, according to
the rule, the proper length of the small eye of crank is 11'94 inches.
Again, supposing it were required to find the proper thickness of the
web of crank at pin centre ; that is to say, the thickness which it
would have if continued to the pin centre. According to the rule,
the proper thickness for the web of crank at pin centre is 7'04 inches.
Again, suppose it were required to find the breadth of the web of
crank at pin centre ; that is to say, the breadth which it would
have if it were continued to the pin centre. Hence, according to
the rule, the proper breadth of the web of crank at pin centre is
1024 inches.
hv Google
142 THE PRACTICAL MODEL CALCULATOR.
Theae results are calculated from the following rules, which give
the proper dimensions for engines of all sizes :
Kl'LE 1. To find the breadth of the small eye of crank. — Multiply
the diameter of the cylinder iu inches hy '063. The product ia
the proper breadth of the small eye of crank in inches.
Rule 2. To find the length of the small eye of crank. — Multiply
the diameter of the cylinder in inches by 'IST. The product is
the proper length of the small eye of crank in inches.
Rule 3. To find the thickness of the weh of crank at pin centre.—
Multiply the diameter of the cylinder in inches by '11. The pro
duct is the proper thickness of the web of crank at pin centre in
inches.
Rule 4. To find the breadth of the web of erank' at pin centre. —
Multiply the diameter of the cylinder in inches by •16. The pro
duct is the proper breadth of crank at pin centre in inches.
To illustrate the use of the succeeding rules, let us take the par
ticular example of an engine of 8 feet stroke and 64inch cylinder,
and let us suppose that the length of the connecting rod is 12
feet, and the side rod 10 feet. We find by a previous rule that
the diameter of the connecting rod at ends is 608, and the ratio
between the diameters at middle and ends of a connecting rod,
whose length is 12 feet, is 1'504. Hence, the proper diameter at
middle of the connecting rod = 608 X 1504 inches = 9'144
inches. And again, we find the diameter of cylinder side rods at
ends, for the particular engine which we have selected, is 410, and
the ratio between the diameters at middle and ends of cylinder
side rods, whose lengths are 10 feet, is 142. Hence, according to
the rules, the proper diameter of the cylinder side rods at middle
is equal to 4'1 X 1'42 inches = 582 inches.
To find some of those parts of the engine which do not depend
upon the stroke. Suppose we take the particular example of an
engine the diameter of whose cylinder is 64 inches. We find from
the following rules that
Diameter of cylinder side rods at ends = 41 inches, or 4,'^
inches.
Breadth of butt = 493 inches, or very nearly 5 inches.
Thickness of butt = 3'9 inches, or 3^ inches.
Mean thickness of strap at cutter = 206 inches, or a little more
than 2 inches.
Mean thickness of strap below cutter = 1'47 inches, or vcry
nearly 1^ inches.
Depths of gibs and cutter = 5'12 inches, or a little more than
5^ inches.
Thickness of gibs and cutter = 103 inches, or a little more than
1 inch.
Diameter of main centre journal = 1171 inches, or very neaily
11 J inches.
Lengdi of main centre journal = 17'6 inches, or 173 inches.
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THE STEAM ENGINE. 143
Depth of eye round end studs of lover = 475 inches, or 4^ inelies.
Thickness of eye round end studs of lever = 3'33 inches, or 3J
inches.
Diameter of end studs of lever = 4'48 inches, or very nearly 4^
inches.
Length of end studs of lever = 4'86 inches, or betireen 4f and
5 inches.
Diameter of airpump studs = 291 inches, or nearly 3 inches.
Length of airpump studs = S*16 inches, or nearly 8] inches.
These results were obtained from the folloiving rules, which will
be found to give the proper dimensions for all sizes of engines.
Rule 1. To find the diameter of cylinder side rods at ends. —
JIultiply the diameter of the cylinder in inches by '060. The
product is the diameter of the cylinder side rods at ends in inches.
Rule 2. To find the breadth of butt in inches. — Multiply the
diameter of the cylinder in inches by '077. The product is the
breadth of the butt in inches.
Rule 3. To find the thickness of the butt. — Multiply the diameter
of the cylinder in inches by "061. The product is the thickness of
the butt in inches.
Rule 4. To find the mean thiakness of strap at cutter. — Mul
tiply the diameter of the cylinder in inches by 032. The product
is the mean thickness of the strap at cutter.
Rule 5. To find the mean thickness of strap below cutter. — Mul
tiply tho diameter of the cylinder in inches by 023. The product
is the mean thickness of strap below cutter in inches.
Rule 6. To find the depth of gibs and cutter. — Multiply the
diameter of the cylinder in inches by 08. The product is the
depth of the gibs and cutter in inches.
Rule 7. To findthe thickness of gibs and cutter. — Multiply the
diameter of the cylinder in inches by •016. The product ia the
thickness of gibs and cutter in inches.
Rule 8. To find the diameter of the main centre journal. — Mul
tiply the diameter of the cylinder in inches by ISS. The product
is the diameter of the main centre journal in inches.
Rule 9, To find the length of the main centre journal. — Multiply
the diameter of the cylinder in inches by 275. The product is
the diameter of the cylinder in inches.
Rule 10. To find the depth of eye round end studs of lever, —
Multiply the diameter of the cylinder in inches by '074. The pro
duct is the depth of the eye round end studs of lever in inches.
Rule 11. To find the thickness of eye round end studs of lever.
— Multiply the diameter of the cylinder in inches by '052. The
product is the thickness of eye round end studs of lever in inches.
Rule 12, To find the diameter of the end studs of lever. — Mul
tiply the diameter of the cylinder in inches by ■07. The product
is the diameter of the end studs of lever in inches.
Rule 13. To find the length of the end studs of lever. — Multiply
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144 TUG PRACTICAL MODEL CALCrLATOR.
tlio iliamctcr of the cylinder in inches by '076. The product is the
length of the end studs of lever in inches.
UuLB 14. To find the diameter of the airpump sfuds. — Multiply
the diameter of the cylinder in inches by '045. The product is
the diameter of the airpump studs in inches.
Rule 15, To find the length of the airpump studs. — Multiply
the diameter of the cylinder in inche^by '049. The product is the
length of the airpump studs in inidjes,
The next rule gives the proper o^th in inches across the centre
of the side lever, when, as is generally the case, the side lever is
of cast iron. It will be observed that the depth is made to depend
upon the diameter of the cylinder and the fength of the lever, and
not at all upon the length of the stroke, except indeed in so far as
the length of the lever may depend upon the length of the stroke.
Suppose it were required to find the proper depth across the centre
of a side lever whose length is 20 feet, and the diameter of the
cylinder 64 inches. According to the rule, the proper depth
across the centre would be 39*26 inches.
The following rule will give the proper dimensions for any size
of engine :
Rule, — To find the depth aeross^e centre of the side lever. —
Multiply the length of the side lever In feet by 7428 ; extract the
cube root of the product, and reserve the result for a, multiplier.
Then square the diameter of the cylinder in inches ; extract the
cube root of the result. The product of the final result and the
reserved multiplier is the depth of the side lever in inches across
the centre.
Thus, to apply this rule to the particular example which ive have
selofted, we have
20 = length of side lever in feet,
•7423 = constant multigiier.
14846
and ^ 14S4(> = ^ISS nearly.
64 = diameter of wlinder in inches.
64
4096
and ,y40M ^16
Hence depth at centre = 16 x 2458 inches = 303j inches, or
between 39j and 39i^ inches.
Tlie next set of rules give the dimensions of eeffef.al of the parts
of the airpump machinery which depend upon the ^#etei of the
cylinder only. To illustrate the use of these rules, let us take the
particular example of an engine the diameter of whose cylinder is
64 inches. We find from the succeeding rules successively.
Diameter of airpump = 3S4 inches, or 38f inches.
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THE STEAM ENGINE. 145
Thickness of the eye of airpump crosshead = 1'58 inches, or
a little more than 1^ inches.
Depth of eye of airpump crosshead = 1101, or about 11 inches.
Diameter of end journals of airpump crosshead = 3'29 inches,
or somewhat more than 3 J inches.
Length of end journals of airpump crosshead = 3'7 inches, or
3^ inches.
Thictness of the web of airpump crosshead at middle = 2T6
inches, or a little more than 2J inches.
Depth of web of airpump crosshead at middle = 10'29 inches,
or somewhat more than lOJ inches.
Thickness of web of airpump crosshead at journal = 235
inches, or about 2f inches.
Depth of web of airpump crosshead at journal = 3'89 inches,
or about 3 inches.
Diameter of airpump piston rod when made of copper = 427
inches, or about 4J inches.
Depth of gibs and cutter through airpump crosshead = 4'04
inches, or a little more than 4 inches.
Thickness of gibs and cutter through airpump crosshead = 81
inches, or about  inch.
Depth of cutter through piston = 327' inches, or somewhat
more than 3J inches.
Thickness of cutter through piston = 1'34 inches, or about If
inches.
These results were obtained from the following rules, and give
the proper dimensions for all sizes of engines :
Rule 1. To jind the diameter of the airpump. — Multiply the
diameter of the cylinder in inches by "6. The product is the
diameter of airpump in inches.
Rule 2. To find the thickness of the eye of airpump crosshead.
— Multiply the diameter of the cylinder in inches by •025. The
product is the thickness of the eye of airpump crosshead in inches.
Rule 3. To find the depth of eye of airpump crosshead. — Mul
tiply the diameter of the cylinder in inches by 171. The product
is the depth of the eye of airpump crosshead in inches.
Rule 4. To find the diameter of the journals of airpump cross
head. — Multiply the diameter of the cylinder in inches by '051.
The product is the diameter of the end journals.
Rule 5. To find the length of the end Journals for airpump
crosshead. — Multiply the diameter of the cylinder in inches by
058, The product is the length of the airpump crosshead jour
nals in inches.
Rule 6. To find the thickness of the web of airpump crosshead
at middle. — Multiply the diameter of the cylinder in inches by '043.
The product is the thickness at middle of the web of airpump
cross head in inches.
Rule 7, To find the depth at middle of tlie web of airpump cross
head. — iMultiply the diameter of the cylinder in inches by IGl.
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146 THE PRACTICAL MODE!. CALCULATOR.
The product is the depth at middle of airpump crosshead Id
inches.
Rule 8, To find thp thtahiess cf the ueb of air pump cross
head atjoumaU — Multijlj the diameter of the cylinder in inches
by "037. The proiuct is the thickness of the i^eb of air pump
crosshead atjournals m mchea
Rule 9. To fitil the dfpth of the air pump u Iiead web at
journals. — Multiply the diimetei of the cylinder in inches by 061
The product ia the depth at jourmla of the web of an pump cross
head.
Rule 10. To Und the dumef^r cf tht, air pump pM n rod lihen
of copper. — Multiply the diameter of the cylindei in inches by
•067. The product is the diameter of the air pump piston lol,
when of copper, in mchea
Rule 11. To find tie d ].th ofgih and cittttr though air pump
crosshead. — Multiply the diameter of the cylinder in inches by
•063. The product is the depth of the ^ibs and cutter through air
pump crosshead lu inches
Rule 12. To find the fhiokness of the gihs and cutt t through
airpump crosshtai — Multifly the diameter of the cylm lor m
inches by 013. The product is the th cLness of the gibs an 1
cutter in inchea.
Rule 13. To in I the depth of cutter through piston — Multiply
the diameter of the cylinder in inches by 051 The product is the
depth of the cutter through piaton in inches
Rule 14. To pnd the tMrlness of eutt r though au pump
piston. — Multiply the diameter of the cylinder in inches by 021
The product is the thickness uf the cutter through air pump piston
The next seven rules give the dimensions of the remaining parts
of the engine which do not depend upon the stroke. To exemplify
their use, suppose it were required to find the corresponding dimen
sions for an engine the diameter of whose cylinder is 64 inches.
According to the rule, the proper diameter of the airpump side
rod would be 248 inches. Hence, according to the rule, the
proper breadth of butt is 2'95 inches. According to the rule, the
proper thickness of butt ia 235 inches. According to the rule,
the mean thickness of strap at cutter ought to be 1'24 inches.
Hence, according to the rule, the mean thickness of strap below
cutter is '91 inch. According to the rule, the proper depth for
the gibs and cutter is 294 inches. According to the rule, the
proper thickness of the gibs and cutter is 6.3 inchea.
The following rules give the correct dimensions for all sizes of
engines ;
Rule 1. To find the diameter of airpump side rod at ends. —
Multiply the diameter of the cylinder in inches by OSS. The
product is the diameter of the airpump side rod at ends in inches.
Rule 2, Tofindthe breadth of butt for airpump. — Multiply the
3gle
THE STEAM ENGINE. 147
diameter of the cylinder in inches by 046. The product is the
breadth of butt in inches.
BuLE 3. To find the thickness of hutt for airpump. — Multiply
the diameter of the cylinder in inches by '037. The product is
the thickness of hutt for airpump in inches,
KuLE 4. To find the mean thickness of strap at cutter. — Multiply
the diameter of the cylinder in inches by 019. The product is
the mean thickness of strap at cutter for airpump in inches.
Rule 5. To find the mean thicknegs of strap below cutter. — Mul
tiply the diameter of the cylinder in inches by 044. The product
is the mean thickness of strap below cutter in inches.
Rule 6. To find the depth of gibs and cutter for airpump. —
Multiply the diameter of the cylinder in inches hy 048. The
product is the depth of gibs and cutter for airpump in inches.
Rule 7. Tofindthe thickness of giba and cutter for airpump. —
Divide the diameter of the cylinder in inches by 100. The
quotient is the proper thickness of the gibs and cutter for airpump
in inches.
With regard to other dimensions made to depend upon the
nominal horse power of tbe engine: — Suppose that we take the
particular example of an engine whose stroke is 8 feet, and dia
meter of cylinder 64 inches. We find that the nominal horse
power of this engine is nearly 175. Hence we have successively,
Diameter of valve shaft at journal in inches = 485, or between
4f and 5 inches.
Diameter of parallel motion shaft at journal in inches = 3'91, or
very nearly 4 inches.
Diameter of valve rod in inches = 244, or about 2f inches.
Diameter of radius rod at smallest part in inches = 1'97, or
very nearly 2 inches.
Area of eccentric rod, at smallest part, in sqnare inches = 837,
or about 8f square inches.
Sectional area of eccentric hoop in square inches = 875, or SJ
sqnare inches.
Diameter of eccentric pin in inches = 2'24, or 2J inches.
Breadth of valve lever for eccentric pin at eye in inches = 5'7,
or very nearly 5f inches.
Thickness of valve lever for eccentric pin at eye in inches — S.
Breadth of parallel motion crank at eye = 4'2 inches, or very
nearly 4J inches.
Thickness of parallel motion crank at eye = 1'76 inches, or
about If inches.
To find the area in square inches of each steam port. Suppose
it were required to find the area of each steam port for an engine
whose stroke is 8 feet, and diameter of cylinder 64 inches. Accord
ing to the rule, ttie area of each steam port would be 202'26 square
inches.
"With regard to the rule, we may remark that the area of the
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148 THE PRACTICAL MODBL CALCULATOR,
Bteam port ought to depend principally upon the cnhical content of
the cylinder, which again depends entirely upon the product of the
square of the diameter of the cylinder and the length of the atroke
of the engine. It is well known, however, that the quantity of
steam admitted by a amall hole does not bear so great a proportion
to the quantity admitted by a larger one, aa the area of the one does
to the area of the other ; and a certain allowance ought to be made
for this. In the absence of correct theoretical information on this
point, we have attempted to make a proper allowance by supplying
a constant ; but of course this plan ought only to be regarded as
an approximation. Our rule is as follows :
Rule. — To find the area of each steam port. — Multiply the
square of the diameter of the cylinder in inches by tho length of
the stroke in feet ; multiply this product hy 11 ; divide the last
product by 1800 ; and, finally, to the quotient add 8. The result
is the area of each steam port in square inches.
To show the use of this rule, we shall apply it to a particular
example. We shall apply it to an engine whose stroke is 6 feet,
and diameter of cylinder 30 inches. Then, according to the rule,
we have
30 = diameter of the cylinder in inches.
_30
900 = square of diameter.
6 = length of stroke in feet.
5400
59400^1800 = 33
8 = constant to be added.
41 = area of steam port in square inches.
When the length of the opening of steam port is from any cir
cumstance found, the corresponding depth in inches may be found,
by dividing the number corresponding to the particular engine, by
the given length in inches : conversely, the length may be found,
when for some reason or other the depth is fixed, by dividing the
number corresponding to the particular engine, by the given depth
in inches : the quotient is the length in inches.
The next rule is useful for determining the diameter of the steam
pipe branching off to any particular engine. Suppose it were
required to find the diameter of the branch steam pipe for an
engine whose stroke is 8 feet, and diameter of cylinder 64 inches.
According to the rule, the proper diameter of the steam pipe
would be 13'16 inches.
The following rule will be found to give the proper diameter of
steam pipe for all sizes of engines.
Rule.— K) find the diameter of Iranah steam pipe. — Multiply
together the square of the diameter of the cylinder in inches, the
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THE STEAM ENGINE. 149
length of tlie stroke in feet, and "00498; to the prodact add 10'2,
and extract the square root of the sum. The result is the diameter
of the steam pipe in inches.
To exemplify the use of this rule we shall take an engine whose
stroke is 8 feet, and diameter of cylinder 64 inches. In this case
we have as follows : —
64 = diameter of cylinder in inches.
64
4096 = square of diameter.
8 = length of stroke in feet.
82768
■00498 = constant multiplier.
16318
10'2 = constant to he added.
17338
and v' 17338 = 1816.
To find the diameter of the pipes connected with the engine.
They are made to depend upon the nominal horse power of the
engine. Suppose it were required to apply this rule to determine
the size of the pipes for two marine engines, whose strokes are
each 8 feet, and diameters of cylinder each 64 inches. We find
the nominal horse power of each of these engines to bo 1743.
Hence, according to the rules, we have in succession.
Diameter of waste water pipe = 1587 inches, or between 15f
and 16 inches.
Area of footvalve passage = 323 square inches.
Area of injection pipe = 14'88 square inches.
If the injection pipe be cylindrical, then by referring to the
table of areas of circles, we see that its diameter would bo about
4f inches.
Diameter of feed pipe == 412 inches, or between 4 and 4J
inches.
Diameter of waste steam pipe = 1217 inches, or nearly 12J
inches.
Diameter of safety valve,
When one is used =1405 inches.
When two are used = 994 inches.
When three are used = 81^ inches.
When four are used = 704 inches
These results were obtained from the f illowmg rules, which will
give the correct dimensions fo» all sizes of engines
Rule 1. To jind the diamitet o/ nante uater pipe. — Multiply
the square root of the nominal horse powei of the engine by 12.
The product is the diameter of the waste water pipe m inches.
Bulb 2. To Jind the area of foot valve passage — Multiply the
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150 THE PRACTICAL MODEL CALCULATOR.
nominal horse power of tlic engine by 9 ; divide the product by 5 ;
add 8 to the quotient. The sum ia the area, of footvaive passage
in square inches.
BuLE 3. To find the area of injection pipe. — Multiply the nomi
nal horse power of the engine by 069 ; to the product add 281.
The sum is the area of the injection pipe in square inches.
Rule 4. To find the diameter of feed pipe. — Multiply the nomi
nal horse power of the engine by '04 ; to the product add 3 ; extract
the square root of the sum. The result is the diameter of the feed
pipe in inches.
Rule 5. To find the diameter of waste steam pipe. — Multiply
the collective nominal horse power of the engines by '375 ; to the
product add 16875; extract the square root of the sum. The
final result is the diameter of the waste steam pipe in inches.
Bulb 6, To find the diameter of the safety valve when only one
is used. — To onehalf the collective nominal horse power of the
engines add 225 ; extract the square root of the sum. The result
is the diameter of the safety valve when only one is used.
Rule 7 To find the diameter of the safett/ valve when two are
used. — Multiply the collective nominal horse power of the engines
by 25 ; to the product add 1125 ; extract the square root of the
sum. The result is the diameter of the safety valve when two
are used.
Rule 8, To find the diameter of the safety valve when three are
used: — To onesixth of the collective nominal horse power of the
engines add 75 ; extract the square root of the sum. The result
is the diameter of the safety valve where three are used.
Rule 9. To find the diameter of the safety valve when four are
waeci. ^Multiply the collective nominal horse power of the engines
by 125 ; to the product add 5625 ; extract the square root of the
snm. The result is the diameter of the safety valve when four
are used.
Another rule for safety valves, and a preferable one for low
pressures, is to allow 8 of a circular inch of area per nominal
horse power.
The next rule is for determining the depth across the web of the
main beam of a land engine. Suppose we wished to find the proper
depth at the centre of the main beam of a land engine whose main
beam is 16 feet long, and diameter of cylinder 64 inches. Accord
ing to the rule, the proper depth of the web across the centre is
46'17 inches. This rule gives correct dimensions for all sizes of
engines.
Rule. — To find the depth of the web at the centre of the main
beam of a land engine. — Multiply together the square of the di
ameter of the cylinder in inches, half the length of the main beam
in feet, and the number 3 ; extract the cube root of the product.
The result is the proper depth of the web of the main beam across
the centre in inches, when the main beam is constructed of cast
iron.
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THE STEAM ENGINE. 151
To illustrate this rule we shall take the particular example of an
engine wliose main beam is 20 feet long, and the diameter of the
cjlinder 64 inches. In this case we have
64 = diameter of cylinder in inches.
64
4096 = square of the diameter.
10 = I length of main beam in feet.
40960
3 = constant multiplier.
122880 ____
122880(49714 = ^122880
4 16 64
4 16 58880
8
4800
6231
4
1161
5112
120
5961
119
9
1242
74
129
7203
36
9
10
138
130
9
10
147 741
To find the depth of the main beam across the ends. Suppose
it were required to find the depth at ends of a castiron main beam
whose length is 20 feet, when the diameter of the cylinder is 64
inches. The proper depth will be 1989 inches.
The following rule gives the proper dimensions for all sizes of
engines.
Rule. — To find the depth of main beam at ends. — Multiply to
gether the square of the diameter of the cylinder in inches, half
the length of the main beam in feet, and the number '192 ; extract
the cube root of the product. The result is the depth in inches of
the main beam at ends, when of cast iron.
To illustrate this rule, let us apply it to the particular example
of an engine whose main beam is 20 feet long, and the diameter
of the cylinder 64 inches. In this case we have as follows :
64 = diameter of cylinder in inches.
64
4096 = square of diameter of cylinder.
10 = J length of main beam in feet.
4096?
'192 = constant multiplier.
786432
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152 THE PEACTICAL MODEL CALCOLATOR,
786432 ( 1989 = ^TSM^
1 1 1
1 1 6864
1 2 5859
"a 300 1005
X 351 _898
30 651 lOT
9 ^32
"39 1083
_9 J_
48 112
_9 J_
57 116
so that, according to the rule, the depth at ends is nearly 20 inches.
To find the dimensions of the feedpump in cubic inches. Sup
pose we take the particular example of an engine whose stroke is
8 feet, and diameter of cylinder 64 inches. The proper content of
the feedpump would te 109336 cubic inches. Suppose, now,
that the coldwater pump was suspended from the main beam at a
fourth of the distance between the centre and the end, so that its
stroke would be 2 feet, or 24 inches. In this case the area of the
pump would be equal to 109336 i 24 = 45556 square inches;
so that we conclude that the diameter is between 7J and 7f inches.
Conversely, suppose that it was wished to find the stroke of the
pump when the diameter was 5 inches. We find the area of the
pump to be 19635 square inches; so that the stroke of the feed
pump must be equal to 109336 r 19635 = 5569 inches, or very
nearly 55 inches.
This rule will be found to give correct dimensions for all sizes
Rule. — To find the content of the feedpump. — Multiply the
square of the diameter of the cylinder in inches by the length of
the stroke in feet ; divide the product by 30. The quotient is the
content of the feedpump in cubic inches.
Thus, for an engine whose stroke is 6 feet, and diameter of cylin
der 50 inches, we have,
50 = diameter of cylinder.
50
2500 = square of the diameter of the cylinder.
6 = length of stroke in feet.
Sp y 15000
500 = content of feedpump in cubic inches.
To determine the content of the coldwater pump in cubic feet.
To illustrate this, suppose we take the particular example of an en
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THE STEAM ENGINE. 153
gine whose stroke is 8 feet, and diameter of cylinder 64 inches.
Suppose, now, the stroke of the pump to be 5 feet, then the area
equal to 7'45 4 5 = I'ii) square feet = 21456 square inches;
we see that the diameter of the pump is about 16J inches. Agaiu,
suppose that the diameter of the coldwater pump was 20 inches,
and that it waa required to find the length of its stroke. The area
of the pump is 31416 square inches, or 314'16 i 144 = 218
square feet; so that the stroke of the pump is equal to *r'45 ;
218 = 342 feet.
The content is calculated from the following rule, which will he
found to give correct dimensions for all sizes of engines :
Rule, — To find the content of the coldwater pump. — Multiply
the square of the diameter of the cylinder in inches by the length
of the stroke in feet ; divide the product by 4400. The quotient
is the content of the coldwater pump in cubic feet.
To explain this rule, we shall take the particular example of an
engine whose stroke is 5^ feet, and diameter of cylinder tiO inches.
In this case we have in succession,
60 = diameter of cylinder in inches.
_60 '
S600 = square of the diameter of cylinder.
5J = length of Stroke in feet.
4400 )19800
45 = content of cold water pump in cubic feet.
To determine the proper thickness of the large eye of crank for
flywheel shaft when the crank is of cast iron. The crank is some
t'm t a tl h ft If course the thickness of the large
y h n g when the crank is only keyed on the
h f h h ho large eye at all. To illustrate the
u f 1 ul w 1 11 pply it to the particular example of an
n B wl k 8 f and diameter of cylinder 64 inches.
H n d he ul he proper thickness of the large eye
f nk 1 n f n 8'07 inches. For a marine engine
of 8 feet stroke and 64 inch cylinder, the thickness of the large
eye of crank is about 5j^ inches. The difference is thus about 2 J
inches, which is an allowance for the inferiority of cast iron to
malleable iron.
The following rule will be found to give correct dimensions for
all sizes of engines :
Rule. — To find the thieJcness of the large eye of crank for fly
wheel shaft when of cast iron. — Jlultiply the square of the length
of the crank in inches by 1561, and then multiply the square of the
diameter of the cylinder in inches by 1235 ; multiply the sum of
these products by the square of the diameter of cylinder in inches ;
divide this product by 666283 ; divide this quotient by the length
of the crank in inches ; finally extract the cube root of the quotient,
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154 THE PRACTICAL MODEL CALCULATOR.
The result is the proper thickness of the large eye of crank for
flywheel shaft in inches, when of cast iron.
As this rule is rather complicated, we shall show its application
to the particular example s,lready selected.
48 = length of crank in inches.
48
2304 = square of length of crank in inches.
1'561 = constant multiplier.
3596^
64 = diameter of cylinder in inches.
64
4096 = square of the diameter of cylinder.
•1235 = constant multiplier.
5058
35965
41023 = sum of products.
4096 = square of the diameter of cylinder.
666283 )168030208
length of crank=48 ) 25219045
525 397
and ^525397 = 807 nearly.
To find the hreadth of the web of crank at the centre of the fly
wheel shaft, that is to say, the hreadth which it would have if it
were continued to the centre of the flywheel shaft. Suppose it
were required to find the breadth of the crank at the centre of the
flywheel shaft for an engine whose stroke is 8 feet, and diameter
of cylinder 64 inches. ■ According to the rule, the proper breadth
is 2249 inches. According to a former rule, the breadth of the
web of a cast iron crank of an engine whose stroke is 8 feet, and
diameter of cylinder 64 inches, is about 18 inches. The difference
between these two is about 4^ inches ; which is not too great an
allowance for the inferiority of the cast iron.
The following rule will be found to give correct dimensions for
all sizes of engines :
TiXJiiS. — To find the breadth of the web of cranh at fiywheel shaft,
when of cast iron. — Multiply the square of the length of the crank
in inches by 1561, and then multiply the square of the diameter
of the cylinder in inches by 1235 ; multiply the square root of the
sum of these products by the square of the diameter of the cylinder
in inches ; divide the product by 2304, and finally extract the
cube root of the quotient. The final result is the breadth of the
crank at the centre of the flywheel shaft, when the crank is of
cast iron.
As this rule is rather complicated, we shall illustrate it hj show
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THE STEAM ENOINB. 155
ing its application to the particular example of an cnfjine whose
stroke is 8 feet, and diameter of cylinder 64 inches.
64 = diameter of cylinder in inches.
64
4096 = square of the diatoeter of cylinder.
■1235 = constant multiplier.
5058
48 = length of crank in inches.
48
2304 = square of the length of crank,
1561 = constant multiplier.
35965
5058
41023 = sum of products.
V 41023 = 6405 nearly.
4096 — square of the diameter of
constant divisor = 2304 )262348^ [cylinder.
1138 666 nearl y,
and 5' 1138666 = 2249.
To determine the thickness of the weh of crank at the centre of
the flywheel shaft ; that is to say, the thickness which it would
have if it were continued so far. Suppose it were required to find
the thickness of weh of crank at the centre of flywheel shaft of
an engine whose stroke is 8 feet, and diameter of cylinder 64
inches. According to the rule, tho proper thickness would be
11'26 inches. The proper thickness of web at centre of paddle
shaft for a marine engine whoso stroke is 8 feet, and diameter of
cylinder 64 inches, is nearly 9 inches. The difi'erence between the
two thicknesses is about 2^ inches, which is not too great an allow
ance for the inferiority of cast iron to malleable iron.
The following rule ■will be found to give correct dimensions for
all sizes of engines :
Rule. — To find the thickness of the web of crank at centre of
jiywheel shaft, when of east iron. — Multiply the square of the
length of the crank in inches by 1561, and then multiply the
square of the diameter of the cylinder in inches by 1235 ; multi
ply the square root of the sum of these products by the square of
the diameter of the cylinder in inches ; divide this product by
18432 ; finally extract the cube root of the quotient. The result
ia the thickness of the web of crank at the centre of the flywheel
shaft when of cast iron, in inches.
As this rule is rather complicated, we shall illustrate it by apply
ing it to the particular engine which we have already selected.
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THE PRACTICAL MODEL CALCULATOR.
48 = length of crank in inctes.
48
2304 = square of length of crank.
1561 = constant multiplier.
3596^
64 = diameter of cylinder In inches.
64
4096 = square of the diameter of cylinder.
1235 = constant multiplier.
5058
35965
41023 = sum
and v/ 41023 = 6405 nearly.
4096 = squaro of diameter.
Constant divisor = 18432) 2623485
142333
and ^ 142333 = 1124
To find the proper diameter of the flywheel shaft at its smallest
part, when, as is usually the case, it is of cast iron. Suppose it
were required to find tho diameter of the flywheel shaft for au
engine whoso stroke is 8 feet, and diameter of cylinder 64 inches.
According to the rule, the diameter would be 1759 inches. It is
obvious enough that tho flywheel shaft stands in much the same
relation to the land engine, as the paddle shaft does to the marine
engine. According to a former riJe, the diameter of the paddle
shaft journal of a marine engine whose stroke is 8 feet, and dia
meter of cylinder 64 inches, is about 14 inches. The difl'erenee
betwixt the diameter of the paddle shaft for the marine engine,
and the diameter of the flywheel shaft for the corresponding land
engine is about 3^ inches. This will be found to be a very proper
allowance for the different circumstances connected with the land
engine.
The following rule will be found to give correct dimensions for all
sizes of engines.
Rule, — To find the diameter of the flywheel shaft at smallest
part, when it is of cast iron. — Multiply the square of the diameter
of the cylinder in inches by the length of the crank in inches ;
extract the cube root of the product ; finally multiply the result
by 3025. The result is the diameter of the flywheel shaft at
smallest part in inches.
We shall illustrate this rule by applying it to the particular
engine which we have already selected.
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THE STEAM BNaiNE.
64
64
4096 
48 
diameter of cjliucler in inches.
square of tlie diameter,
length of crank in inches.
196608
5
25
196608 (6815  ^ 196608
126
5
6
10
6
150
8
168
25
50
7500
1264
8764
1328
10092
71608
70112
1496
1011
485
166 1011
8 2
174 1013
and 5815 X 3025 = 1759
Tfhich agrees ivith the number given bj a former rule.
To determine the sectional area of the flywheel rim when of
cast iron. Suppose it were required to find the sectional area of
the rim of a flywheel for an engine whose stroke is 8 feet, and
diameter of cylinder 64 inches, the diameter of the flywheel itself
being 30 feet. According to the rule, the sectional area of the
rim in square inches = 1464 x SIS = 119'03. We may remark
that this calculation has heen made on the supposition that the fly
wheel 18 BO connected with the engine, as to make exactly one revo
lution for each double stroke of the piston. If the flywheel is so
connected with the engine as to make more than one revolution for
each double stroke, then the rim does not need to be so heavy as
we make it. If, on the contrary, the flywheel does not make a
complete revolution for each double Stroke of the engine, then it
ought to be heavier than this rule makes it.
Rule. — To find the sectional area of the rim of the flywheel
when of cast iron. — Multiply together the square of the diameter
of the cylinder in inches, the square of the length of the stroke
in feet, the cube root of the length of the stroke in feet, and 6*125 ;
divide the final product by the cube of the diameter of the fiywheel
in feet. The quotient is the sectional area of the rim of flywheel
in square inches, provided it is of cast iron.
As this rule is rather complicated, we shall endeavour to illustrate
it by showing its application to a particular engine. We shall
apply the rule to determine the sectional area of the rim of fly
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158 THE PRACTICAL MODEL CALCULATOR.
wheel for an engine whose stroke is 8 feet, diameter of cylinder 50
inches; the diameter of the flywheel heing 20 feet. For this
engine we have as follows :
2500 = square of diameter of cylinder.
64 = square of the length of stroke.
160000
2 = cube root of the length of stroke.
320000
6'125 = constant multiplier.
1960000
therefore sectional area in square inches = 1960000 i 20^ =
1960000 ^ 8000 = 1960 h 8 = 245.
In the following formulas we denote the diameter of the cylinder
in inches by D, the length of the crank in inches by R, the length
of the stroke in feet, and the nominal horse power of the engine
by n.p.
MARINE ENGINES. — DIMENSIOXB OE SEVERAL OF THE PARTS OP THE
SIDE LEVER.
Depth of eye round end studs of lever = 074 x D.
Thickness of eye round end studs of lever = '052 x D.
Diameter of end studs, in inches = 07 x D,
Length of end studs, in inches = 076 x D.
Diameter of airpump studs, in inches = 045 X D,
Length of airpump studs, in inches = 049 x D.
Depth of cast iron side lever across centre, in inches = D^ x
{•7423 X length of lever in feet}
i OF AIRPUMP
Diameter of airpump, in inches = '6 X D.
Thickness of eye for airpump rod, in inches = '025 X D.
Depth of eye for airpump rod, in inches = l?! x D.
Diameter of end journals, in inches = OSl x D.
Length of end journals, in inches = '058 X D.
Thickness of web at middle, in inches = '043 x D.
Depth of web at middle, in inches = ISl x D,
Thickness of web at journal = 037 x D.
Depth of web at journal = 061 X D.
Diameter of airpump pistonrod, when of copper, in inclies =
067 X D.
Depth of gibs and cutter through crosshead, in inches =
■063 X D.
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TDK STEAM ENGINE. 15S
Thickness of gibs and cutter through crosahead, iir inches =
013 X D.
Depth of cutter through piston, in inches = '051 x D.
Thickness of cutter through piston, in inches = '021 x J).
PARTS OF THE
Diameter of airpump side rods at ends, in inches = '039 x D.
Breadth of butt, in inches = '046 X D.
Thickness of hutt, in inches = '037 x D.
Mean thickness of strap at cutter, in inches = 'Olft x D.
Mean thickness of strap below cutter, in inches = 014 x D.
Depth of gibs and cutter, in inches = '048 x D.
Thickness of gibs and cutter in inches = D ^ 100.
MARINE AND I.AND EN0IKE9, — AREA OF STEAM PORTS.
Area of each steam port, in equare inches = 11 x ^ X D^ i
1800 + 8.
MARINE AND LAND ENGINES. — DIMENSIONS OP BRANCH STEAM PIPES.
Diameter of each branch steam pipe = ^/ 00498 x I X D^ X 102.
MARINE ENGINE.
Diameter of waste .water pipe, in inches = 1'2 X x/ H.P.
Area of footvalve passage, in square inches = 1'8 X H.P.+ 8.
Area of injection pipe, in square inches = '069 X H.P. + 2'81.
Diameter of feed pipe, in inches = %/ '04 X H.P. f 3.
Diameter of waste steam pipe in inches =v''375xn.P. I 16875.
MARINE AND LAND ENGINES. — DIMENSIONS OF SAFETYVALVES.
Diam. of safetyvalve, when one only is used =n/5xH.P.22'5.
Diam. of safetyvalve, when two are used = s/'SSxH.P.MlaS,
Diam. of safetyvalve, when three are used = v'"167xH.P.+75.
Diam, of safetyvalve, when four are used =s/125 X H.P. f 5625.
LAND ENGINE. — DIMENSIONS OP MAIN REAM.
Depth of web of main beam across centre =
■^ 3 X D^ X half length of main beam in feet.
Depth of main beam at ends =
■^ 192 X D^ X half length of main beam, in feet.
LAND AND MARINE ENGINES. ^CONTENT OP PEBDPUMP.
Content of feedpump, in cubic inches = D' x Z ^ 30.
LAND ENGINES. — CONTEST OF COLD WATER PDMP.
Content of cold water pump, in cubic feet = D^ x i ^ 4400.
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.60 THE PKACTIOAL MODEL CALCULATOR.
LAND ENCilNES. — DIMENaiONS OF CRANK.
ThicknGSS of large eye of crack, in inches ^
^~D^~x~"(r56rx~:^T^1235 D') ^ (K x 666283}.
Breadth of weh of crank at flywheel shaft centre, in inches =
^D"^ X •/ (1561 X R^ + 1235 x D=) ^ 2304.
Tbieknesa of Treb of crank at flywheel shaft centre, in inches =
^nD^lw^(r56r^rRM^4235^ri)y=i^l84^.
lAND ENGINES.— DIMENSIONS OF FLYWHEEL SHAri.
Diameter of fiywheel shaft, when of cast iron = 3025 x ^BxD'.
DIMENSIONS or PAETS OF LOCOMOTIVES.
DIAMETER OF CYLINDER.
In locomotive engines, the diameter of the cylim^er varies less
than either the land or the marine engine. In few of the locomotive
engines at present in use is the diameter of the cylinder greater
than 16 inches, or less than 12 inches. The length of the stroke of
nearly all the locomotive engines at present in use is 18 inches, and
there are always two cylinders, which are generally connected to
cranks upon the axle, standing at right angles with one another.
AREA OF IKDIrCTION PORTa.
Rule, — To find the size of the steam ports for the locomotive
engine. — Multiply the square of the diameter of the cylinder hy
■068. The product ia the proper size of the steam ports in square
inches.
Required the proper size of the steam ports of a locomotive
engine whose diameter is 15 inches. Here, according to the rule,
size of steam porta = 068 x 15 X 15 = 068 X 225 = 153 square
jnehes, or between 15^ and 15^ square inches.
After having determined the area of the porta, we may easily
find the depth when the length ia given, or, conversely, the length
when the depth is given. Thus, suppose we knew that the length
was 8 inches, then we find that the depth should be 153 h 8 =
1*9125 inches, or nearly 2 inches; or suppose we knew the depth
was 2 inches, then we would find that the length was 153 r 2 =
7'65 inches, or nearly 7 inches.
AREA OP EDUCTION PORTS,
The proper area for the eduction porta may be found from the fol
lowing rule.
Rule. — To find the area of the eduction ports. — Multiply the
square of the diameter of the cylinder in inches by ■128. The
product is the area of the eduction ports in square inches.
Required the area of the eduction porta of a locomotive engine,
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THE STEAM ENGINE. 161
when the diameter of the cylinders is 13 inches. In this example
we have, according to the rale,
Area of eduction port = 128 x 13= = 128 x 169 = 21632
inches, or between 21J and 21f square inches.
BaEADTH OF BRIDGE BETWEEN PORTS,
The breadth of the hridgea between the eduction port and the
induction ports is usually between f inch and 1 inch,
DIAMETER OP BOILER.
It is obvious that the diameter of the boiler may vary very con
Bidcrably ; but it is limited chiefly by considerations of strength ;
and 3 feet are found a convenient diameter. Rules for the strength
of boilers will be given hereafter.
Rule. — To find the inside diameter of the boiler. — Multiply the
diameter of the cylinder in inches by 311. The product is the
inside diameter of the boiler in' inches.
Required the inside diameter of the boiler for a locomotive
engine, the diameter of the cylinders being 15 inches.
In this example we have, according to the rule.
Inside diameter of boiler = 15 x 311 = 4665 inches,
or about 3 feet 10 inches.
tENGTU OP BOILER.
The length of the boiler is usually in practice between 8 feet and
^ feet.
DIAMETER OF STEAM DOME, INSIDE.
It is obvious that the diameter of the steam dome may be varied
considerably, according to circumstances ; but the first indication
is to make it large enough. It is usual, however, in practice, to
proportion the diameter of the steam dome to the diameter of the
cylinder ; and there appears to be no great objection to this. The
following rule will be found to give the diameter of the dume
usually adopted in practice.
Rule. — To find the diameter of the steam dome. — Multiply the
diameter of the cylinder in inches by 143. The product is the
diameter of the dome in inches.
Required the diameter of the steam dome for a locomotive engine
whose diameter of cylinders. is 13 inches. In this example we
have, according to the rule,
Diameter of steam dome = 143 x 13 = 18'59 inches,
or about 18 J inches.
HEIGHT OE STEAM DOME.
The height of the steam dome may vary. Judging from prac
tice, it appears that a uniform height of 2J feet would answer
very well.
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162 THE PRACTICAL MODEL CALCULATOR.
rrAMETER OP SAPETY VALVE.
In practice the diameter of the safetyvalve varies considerably.
The following rule gives the diameter of the safetyvalve usually
adopted in practice.
Rule. — To jind the diameter of the safetyvalve. — Divide the
diameter of the cylinder in inches by 4. The quotient is the dia
meter of the safetyvalve in inches.
Required the diameter of the safetyvalves for the boiler of a
locomotive engine, the diameter of the cylinder being 13 inches.
Here, according to the rule, diameter of safetyvalve = 13 ^ 4 = 3 J
inches. A larger size, however, is preferable, as being less likely
to stick.
DIAMETER OF VALVE SPINDLE.
The following rule will be found to give the correct diameter of
the valve spindle. It ia entirely founded on practice.
Rule. — To findthe diameter of the valve spindle. — Multiply the
diameter of the cylinder in inches by 076. The product is the
proper diameter of the valve spindle.
Required the diameter of the valve spindle for a locomotive
engine whose cylinders' diameters are 13 inches.
In this example we have, according to the rule, diameter of valve
spindle = 13 x '076 = '988 inches, or very nearly 1 inch,
DIAMETER OF CHIMNEY.
It is usual in practice to make the diameter of the chimney equal
to the diameter of the cylinder. Thus a locomotive engine whose
cylinders' diameters are 15 inches would have the inside diameter
of the chimney also 15 inches, or thereabouts. This rule has, at
least, the merit of simplicity,
AREA or FIREr.RATil.
The following rule determines the area of the firegrate usually
given in practice. Wo may remark, that the area of the firegrate
in practice follows a more certain rule than any other part of the
cDgine appears to do ; but it is in all cases much too small, and
occasions a great loss of power by the urging of the blast it renders
necessary, and a rapid deterioration of the furnace plates from
excessive heat. Thfere is no good reason why the furnace should
not be nearly as long as the boiler : it would then resemble the
furnace of a marine boiler, and be as manageable.
Rule. — To find the areaof the firegrate, — Multiply the diameter
of the cylinder in inches by '77. The product is the area of the fire
grate in superficial feet.
Required the area of the firegrate of a locomotive engine, the
diameters of the cylinders being 15 inches.
In this example we have, according to the rule,
Area of firegrate = 77 X 15 = 11'55 square feet,
or about llj^ square feet. Though this rule, however, represents
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THE STEAM ENGINE. 163
the usual practice, the area of the firegrate shouUl not be contingent
upon the size of the cylinder, but upon tho quantity of steam to be
raised.
AREA OP UEATING SOUrACE.
In the construction of a locomotive engine, one great object is to
obtain a boiler which will produce a sufficient quantity of steam with
as little bnlk and weight aa possible. This object is admirably ac
complished in the construction of the boiler of the locomotive en
gine. This little barrel of tubes generates more steam in an hoar
than was formerly raised from a boiler and fire occupying a eon
siderable house. This favourable result is obtained simply by ex
posing the water to a greater amount of heating surface.
In the usual construction of the locomotive boiler, it is obvious
that we can only consider four of the six faces of the inside firebox
as effective heating surface ; viz. the crown of the box, and the
three perpendicular sides. The circumferences of the tubes are also
effective heating surface ; so that the whole effective heating sur
face of a locomotive boiler may bo considered to be the four faces
of the inside firebox, plus the sura of the surfaces of the tubes.
Understanding this to be the effective heating surface, the following
rule determines the average amount of heating surface usually given
in practice.
Htjle. — To find the effective heating surface. — Multiply the square
of the diameter of the cylinder in inches by 5 ; divide the product
by 2. The quotient is the area of the effective heating surface in
square feet.
Required the effective heating surface of the boiler of a locomotive
engine, the diameters of the cylinders being 15 inches.
In this example we have, according to the rule.
Effective heating surface = 15^ x 5 J 2 = 225 x 5 ^ 2 = 1125 h
2 = 5621 square feet.
According to the rule which we have given for the firegrate, the
area of the firegrate for this boiler would be about llj square feet.
We may suppose, therefore, the area of the crown of the box to be
12 square feet. The area of the three perpendicular sides of the
inside firebox ie usually three times the area of the crown ; so that
the effective heating surface of the firebox is 48 square feet. Hence
the heating surface of the tubes = 526'5 — 48 = 478'5 square feet.
The inside diameters of the tubes are generally about If inches ;
and therefore the circumference of a section of these tubes, ac
cording to the table, is 64978 inches. Hence, supposing the
tube to be 8 feet long, the surface of one = 5'4978 x 8J ^ 12 =
■45815 X 8^ = 38943 square feet ; and, therefore, the number of
tubes = 4785 h38943 = 123 nearly. The amount of heating sur
face, however, like that of grate surface, is properly a function of
the quantity of steam to be raised, and the proportions of both,
given hereafter, will be found to answer well for boilers of every
description.
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164 THE PRACTICAL MODEL CALCliLATOB.
ABEA OF TVATERLEVEL.
This, of course, varies with the different circumstaneea of the
boiler. The average area may be found from the following rule.
Rule. — To find the area of the waterlevel. — Multiply the diame
ter of the cylinder in inches by 2'08. The product is the area of
the waterlevel in square feet.
Required the area of the waterlevel for a locomotive engine,
whose cylinders' diameters are 14 inches.
In thia case we have, according to the rule,
Area of waterlevel = 14 x 208 = 2912 square feet.
CTJBICAL CONTEST OF WATER IN B01T.EE.
This, of course, varies not only in different boilers, but also in
the same boiler at different times. The following rule is supposed
to give the average quantity of water in the boiler.
Rule. — To find the cubical content of the water in the loiter. —
Mnltiply the square of the diameter of the cylinder in inches by 9 :
divide the product by 40, The quotient is the cubical content of
the water in the boiler in cubic feet.
Required the average cubical content of the water in the boiler
of a locomotive engine, the diameters of the cylinders being 14
inches. In this example wo have, according to the rule,
Cubical content of water = 9 X 14' ; 40 = 44'1 cubic feet.
CONTENT OF FEEDPUMP.
In the locomotive engine, the feedpump is generally attached to
the crosshead, and consequently it has the same stroke as the pis
ton. As we have mentioned before, the stroke of the locomotive
engine is generally in practice 18 inches. Hence, assuming the
stroke of the feedpump to be constantly 18 inches, it only remains
for us to determine the diameter of the ram. It may be found from
the following rule.
Rdle, — To find the diameter of the feedpump ™m..— Multiply
the square of the diameter of the cylinder in inches by Oil. The
product is the diameter of the ram in inches.
Required the diameter of the ram for the feedpump for a loco
motive engine whose diameter of cylinder ia 14 inches. In this
example we have, according to the rule,
Diameter of ram = Oil x 14' = Oil x 196 = 2156 inches,
or between 2 and 2J inches.
COBICAL CONTENT OE STEAil ROOM.
The quantity of steam in the. boiler varies not only for different
boilers, but even for the same boiler in different circumstances.
But when the locomotive is in motion, there is usually a certain
proportion of the boiler filled with the steam. Including the dome
and the steam pipe, the content of the steam room will be found
usually to be somewhat less than the cubical content of the water.
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THE STEAM ENGINE. 165
But as it is desirable that it should he increased, we give the fol
lowing rule.
Rule. — To find the cubical content of tfi^ steam room. — Multiply
the square of the diameter of the cylinder in inches by 9 ; divide
the product by 40. The quotient is the cubical content of the
steam room in cubic feet.
Required the cubical content of the steam room in a locomotive
boiler, the diameters of the cylinders being 12 inches.
In this example we have, according to the rule,
Cubical content of steam room = 9 X 12^ h 40 '^ 9 X 144 h 40 =
324 cubic feet.
CUBICAL CONTENT OF INSIDE PIREBOX ABOVE FIREBARS.
The following rule determines the cubical content of firebox
usually given in practice.
Rule. — To find the cubical content of inside firehox above fire
bars. — Divide the square of the diameter of the cylinder in inches
by 4, The quotient is the content of the inside firebox above fire
bars in cubic feet.
Required the content of inside firebox above firebars in a loco
motive engine, when the diameters of the cylinders are each 15
inches.
In this example wo have, according to the rule.
Content of inside firebox above firebars = 15^ ; 4 = 225 s 4 =
56^ cubic feet.
THICKNESS OP THE PLATES OP BOILER.
In general, the thickness of the plates of the locomotive boiler is
I inch. In some cases, however, the thickness is only f, inch,
INSIDE DIAMETER OF STEAM PIPE.
The diameter usually given to the steam pipe of the locomotive
engine may be found from the following rule.
Rule. — To find the diameter of the steam pipe of the loeomotivp
engine. — Multiply the square of the diameter of the cylinder in
inches by '03. The product is the diameter of the steam pipe in
inches.
Required the diameter of the steam pipe of a locomotive engine,
the diameter of the cylinder being 13 inches. Here, according to
the rule, diameter of steam pipe = 03 x 13= = 03 x 169 = 5OT
inches ; or a very little more than 5 inches. The steam pipe is
usually made too small in engines intended for high speeds.
DIAMETER OE BEANCH STEAM PIPES.
The following rule gives the usual diameter of the branch steam
pipe for locomotive engines.
Rule, — To find the diameter of the branch steam pipe for the lo
comotive engine. — Multiply the square of the diameter of the cylin
der in inches by 021. The product is the diameter of the branch
steam pipe for the locomotive engine in inches.
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IbQ TIIH PRACTICAL MODEL CALCULATOR,
Rcquiruil the diameter of the branch steam pipes for a locomo
tive engine, when the cylinder's diameter is 15 inches. Here, ac
cording to the rule, diameter of branch pipe = ■021 x 15' = 021 x
225 = 4726 inches, or about 4f inches.
DIAMETER OF TOP OF BLAST PIPE.
The diameter of the top of the blast pipe may he found from the
following rule.
Rule. — To find the diameter of the top of the blast pipe. — Mul
tiply the square of the diameter of the cylinder in inches by 0'17,
The product is the diameter of the top of the blast pipe in inches.
The diameter of a locomotive engine is 13 inches ; required the
diameter of the blast pipe at top. Here, according to the rule,
diameter of blast pipe at top = 017 X 13= = 017 X 169 =2873
inches, or between 2^ and 3 inches ; but the orifice of the blast
pipe should always be made as large as the demands of the blast
will permit.
DIAMITER OF FEED PIPES.
There appear to be no theoretical considerations which would
lead us to determine exactly the proper size of the feed pipes.
Judging from practice, however, the following rule will be found to
give the proper dimensions.
Rule. — To find the diameter of the feed pipes. — Multiply tbo
diameter of the cylinder in inches by 141. The product is the
proper diameter of the feed pipes,
Required the diameter of the feed pipes for a locomotive engine,
the diameter of the cylinder being 15 inches.
In this example we have, according to the rule,
Diameter of feedpipe = 15 X 141 = 2115 inches,
or between 2 and 2^ inches.
DIAMETER OF PISTON ROD.
The diameter of the piston rod for the locomotive engine is
usually about oneseventh the diameter of the cylinder. Making
practice onr gnide, therefore, we have the following rule.
Rule. — To find the diameter oftkepiston rod for the loeomotive
engine. — Divide the diameter of the cylinder in inches by 7. The
quotient is the diameter of the piston rod in inches.
The diameter of the cylinder of a locomotive engine is 15 inches ;
required the diameter of the piston rod. Here, according to the
rule, diameter of piston rod =15 ; 7 = 2 inches.
TUICKNEKS OP PISTON.
The thickness of the piston in locomotive engines is usually about
twosevenths of the diameter of the cylinder. Making practice our
guide, therefore, we have the following rule.
Rule. — To find the thichness of the piston in the locomotive en
jme.— Multiply the diameter of the cylinder in inches by 2 ; divide
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THE STEAM BNG:NE. IdT
the product by 7. The quotient is the thickness of the piston in
The diameter of the cylinder of a locomotive engine is 14 inches;
required the thickness of the piston. Here, according to the rule,
thickness of piston =2x14^7=4 inches.
DIAMETER OF CONNECTING ItOrS AT MinDLE.
The following rule gives the diameter of the connecting rod at
middle. The rule, we may remark, is entirely founded on practice.
Rule.— To _^m(Z the diameter of the eonneeting rod at middle of
the locomotive engine. — Multiply the diameter of the cylinder in
inches hy ■21. The product ia the diameter of the connecting rod
at middle in inches
Required the diameter of the connecting rods at middle for a
locomotive engine, the diameter of the cylinders being twelve
inches
i'or this e'^ample ne have, according to the rule.
Diameter of coDnecting rods at middle = 12 X '21 — 252 inches,
or 2 J inches.
niAMETElt or BALL ON CROSSHEAD sriNDLE.
The diameter of the ball on the crosshead spindle may he found
from the following rule.
Rule. — To find the diameter of the ball on crosshead spindle of
a locomotive engine. — Multiply the diameter of the cylinder in
inches by 23. The product is the diameter of the ball on the
crosshead spindle.
Required the diameter of the ball on the crosshead spindle of a
locomotive engine, when the diameter of the cylinder is 15 inches.
Here, according to the rule,
Diameter of ball = 23 x 15 = 345 inches, or nearly 3J inches,
niAMETER OF THE INSIDE BEARINGS OF THE CRANK AXLE.
It is obvious that the inside bearings of the crank axle of the
locomotive engine correspond to the paddleshaft journal of the
marine engine, and to the flywheel shaft journal of the landengine,
"We may conclude, therefore, that the proper diameter of these hear
ings ought to depend jointly upon the length of the stroke and the
diameter of the cylinder. In the locomotive engine the stroke is
usually 18 inches, so that we may consider that the diameter of the
bearing depends solely upon the diameter of the cylinder. The
following rule will give the diameter of the inside bearing.
Rule. — To find the diameter of the ingido hearing for the loco
motive engine. — Extract the cube root of the square of the diameter
of the cylinder in inches ; multiply the result by Oe. The product
is the proper diameter of the inside bearing of the crank axle for the
locomotive engine.
Required the diameter of the inside bearing of the crank axle
hv Google
THE PRACTICAL MODEL CALCULATOR.
for a locomotive engine whose cjlimlers are of ISiKch
In this example we have, aecordiog to the rule,
13
169 =
= diameter of cylinder in inclies.
= aquare of tlie diameter of cylinder.
5
5
5
25
25
60
169(56289  ^169
125
44000
41375
10
5
7500
J75
2625
1820
150
5
8275
800
805
726
and diameter of hearing = 5'5289 X 96 = 5'31 inches nearly; or
between 5J and 5J inches.
DIAMETER or THE OUTSIDE BEARINGS OF THE CRANK AXLE.
The crank axle, in addition to resting upon the inside bearings,
is Bometimea also made to rest partly upon outside hearings.
These outside bearings are added only for the sake of steadiness,
and they do not need to be so strong as the inside hearings. The
proper size of the diameter of these bearings may be found from
tbe following rule.
EuLE. — To find the diameter of outside hearings for the locomo
tive engine. — Multiply the square of the diameters of the cylinders
in inches by '396 ; extract the cube root of the product. The result
is the diameter of the outside bearings in inches.
Required the proper diameter of the outside bearings for a loco
motive engine, the diameter of its cylinders being 15 inches.
In this example we have, according to the rule,
15 = diameter of cylinders in inehos.
_15
225 = square of diameter of cylinder.
■396 = constant multiplier.
hv Google
THE STEAM ENGINE.
89l(4466
16
64
W
26100
32
21184
4800
3916
496
3528
■6296
388
512
358
5808
Hence diameter of outside bearing = 4466 inches, or very
nearly 4 inches.
diamt:tkr of pi.ain paet op crask axle.
It is usual to make the plain part of crank axle of the same sec
tional area as the inside bearings. Hence, to determine the sec
tional area of the plain part when it is cylindrical, we have the fol
lowing rule.
Rule. — To determine the diameter of the plain part of crank axle
for the locomotive engine. — Extract the cube root of the square of
the diameter of the cylinder in inches ; multiply the result by ■96.
The product is the proper diameter of the plain part of the crank
axle of the locomotive engine in inches.
Required the diameter of the plain part of the crank axlo for the
locomotive engine, whose cylinders' diameters are 14 inches. In
this example we have, according to the rule,
14 = diameter of cylinder in inches.
14
196 = square of the diameter of cylinder.
0 196(5808 = ^196
5 25 125
% 25 71000
5 50_ T 0112
10 7500 ^888
5 1_2_64
150 8764
8 1328
158 10092
_8
166
__8
174
hv Google
170 THE PBACTICAI, MODEL CALCULATOK.
Hence the plain part of crank axle = 5'808 X 96 — 558 nearlj,
or a little more than 5^ inchea.
DIAMETER 01' ORAKK PIN.
The following rule gives the proper diameter of the crank pin. It is
ohvious that the crank pin of the locomotive engine is not altogether
analogous to the crank pin of the marine or land engine, and, like
them, ought to depend upon the diameter of the cylinder, as it is
usually formed out of the solid axle.
Role. — To find the diameter of the crank pin for the locomotive
engine, — Multiply the diameter of the cylinder in inches by '404.
The product is the diametor of the crank pin in inches.
Required the diameter of the crank pin of a locomotive engine
whose cylinders' diameters are 15 inchea.
In this example we have, according to the rule.
Diameter of crank pin = 15 X '401 = 606 inches, or about 6
inchea.
LENGTH OF CRANK PIN.
The length of the crank pin usually given in practice may be
found from the following rule.
Rule. — To find the length of the crank pin. — Multiply the di
ameter of the cylinder in inches by •233. The product is the
length of the crank pins in inches.
Required the length of the crank pins for a locomotive engine
with a diameter of cylinder of 13 inches.
In this example wc have, according to the rule,
Length of crank pin = 13 x 233 = 3029 inches,
or about 3 inches. The part of the crank axle answering to the
crank pin is usually rounded very much at the corners, both to give
additional strength, and to prevent aide play.
These then are the chief dimensions of locomotive engines ac
cording to the practice most generally followed. The establish
ment of express trains and the general exigencies of steam locomo
tion are daily introducing innovations, the effect of which is to make
the engines of greater size and power : but it cannot be said that a
plan of locomotive engine has yet been contrived that is free from
grave objections. The most material of these defects is the neces
sity that yet exists of expending a large proportion of the power in
the production of a draft ; and this evil is traceable to the inade
quate area of the firegrate, which makes an enormous rush of air
through the fire necessary to accomplish the combustion of the fuel
requisite for the production of the steam. To gain a sufficient area
of firegrate, an entirely new arrangement of engine must be
adopted : the furnace must be greatly lengthened, and perhaps it
may be found that short upright tubes, or the very ingenious ar
rangement of Mr. Dimpfell, of Philadelphia, may be introduced
with advantage. Upright tubes have been found to be more
effectual in raising steam than horizontal tubes ; but the tube
plate in the case of upright tubes would be more liable to burn.
hv Google
THE BTEAM ENGIKE. 171
We bere give the preceding rules in formulas, in tlie belief that
those well acquainted with algebraic aymbola prefer to have a rule
expressed as a formula, as they can thus see at once the different
operations to be performed. In the following formulas we denote
the diameter of the cylinder in inches by D.
LOCOMOTIVE ENGINE. — PAETS Or THE CYLINDER.
Area, of induction ports, in square inches = '068 X B^.
Area of eduction ports, in square inches = 128 x D^.
Breadth of bridge between ports between f inch and 1 inch.
LOCOMOTIVE ESGINE. — PAETS OJ BOILER.
Diameter of boiler, in inches = 3'11 X D.
Length of boiler between 8 feet and 12 feet.
Diameter of Steam dome, inside, in inches = 1'43 X D.
Height of steam dome = 2^ feet.
Diameter of safety valve, in inches = D § 4.
Diameter of valve spindle, in inches = '076 x D.
Diameter of chimney, in inches =" D.
Area of firegrate, in square feet = 77 x D.
Area of heating surface, in square feet = 5 x D' ^ 2.
Area of water level, in square feet = 2'08 X D,
Cubical content of water in boiler, in cubic feet = 9 x D' ; 40.
Diameter of feedpump ram, in inches = 'Oil X DK
Cubical content of steam room, in cubic feet = 9 X D' ; 40.
Cubical content of inside firebox above fire bars, in cubic feet =
D=s4.
Thickness of the plates of boiler =  inch,
LOCOMOTIVE ENGINE. — DIMENSIOXS OP SEVERAL PIPES.
Inside diameter of steam pipe, in inches = '03 X IP.
Inside diameter of branch steam pipe, in inches = 021 x D^
Inside diameter of the top of blast pipe = '017 X D^
Inside diameter of the feed pipes = ■141 X D.
LOCOMOTIVE ENGINE, — DIMENSIONS OF SEVERAL MOVING PARTS.
Diameter of piston rod, in inches = D 4 7.
Thickness of piston, in inches = 2 D ^ 7.
Diameter of connecting rods at middle, in inches = '21 X D.
Diameter of tho ball on crosshead spindle, in inches = '23 X D.
Diameter of the inside bearings of the crank axle, in inches =
96 X i/ D'. _ __
Diameter of the plain part of crank axle, in inches = "SB x ^ D^
Diameter of the outside bearings of the crank axle, in inches =
^ 396 X D^
Diameter of crank pin, in inches = 404 x D.
Length of crank pin, in inches = 233 X D.
hv Google
THE PEACTICAL MODEL CALCULATOR.
Table of tie Pre
sure of Steam, in Inches of Meroury, at dif
ferent Temperatures,
t™p=
Helt
i)^i,.«.
lira.
V.,.,.
Ivotj.
Trsdgold,
^m,,..
K,i,L.,».
""'■
0°
0'08
10
012
20
017
Oll
32
026
0'20
018
017
016
000
40
0'34
025
020
024
022
010
50
0>49
036
036
036
037
033
020
60
0.65
052
053
055
048
035
70
087
073
075
073
078
06S
055
077
80
110
101
105
111
095
082
W
130
144
136
153
134
118
100
212
196
184
160
155
110
2'79
245
262
24S
279
266
225
120
36S
330
846
346
300
130
471
454
441
48S
443
395
140
6'05
678
588
621
575
615
514
ISO
753
755
742
794
746
672
160
fl79
9GO
962
1005
952
865
892
170
1231
1205
1214
12"05
1260
1214
1105
1187
180
1638
1516
1523
1567
1530
1405
1273
190
1898
1900
1896
1893
1900
1785
1900
200
2851
2360
2344
2371
2265
310
2882
2881
2881
2886
212
3000
3000
8000
3000
3000
3000
3000
2940
220
3518
8554
8519
3492
35 S
8365
230
1460
4310
4247
4363
4200
445
40
240
5345
6170
5166
5024
549
490
Table of the Temperature of Steam at different Pressures i
mospheres.
I^^^
.^•z.
r...u..
r»nne.
I..,.
TredE^H.
s™*™.
R„.i»n.
w«,.
."'■
l8t At.
2J20°
212""
212°
312°
212=
212"
212"
^2~
2cl At.
2505
2500
2408
249
250
2503
2525
3500
8d At.
2762
2750
271
274
267
2752
4th At.
3915
290
394
2984
2915
SthAt.
3088
8045
302
809
3045
euiAt.
3204
8155
S22
3155
TthAt.
3317
S255
3265
8t1> At.
3420
3360
342"
3436
3360
9th At.
3500
3450
3450
10th At.
3589
3525
11th At.
12th At.
3740
372'
13th At.
I4th At.
3369
15th At.
8928
16th At.
3985
17th At.
4038
18th At.
4089
19th At.
4139
20th At.
4185
414
405
30th At.
4572
40th At.
4666
50th At.
5106
b,Google
THE STEAM ENGIKB.
Table of the lExpannion of Air by Heat.
FBhmi
Fll,™
90 ....
.... 1182
83
1002
63
1071
91 ....
.... 1184
34
35
Sb
1004
1007
1009
63
111
65
1073
1075
37
1012
1030
95 ....
.... 1142
38
1015
07
1080
90 ....
.... 1144
S9
1018
1034
40
41
1031
1023
61
70
10M7
10B9
.... 1150
42
1025
71
1091
100 ....
.... 1152
43
1027
72
109J
110 ....
.... 1178
44
lOJO
73
lO'^t
120 ....
.... 1194
45
1032
74
1097
46
1034
75
1099
47
103G
1101
150 ....
.... 1255
48
I0o8
77
1104
100 ....
.... 1275
49
60
1040
1013
78
79
1100
1108
180 ..„
.... 1315
51
1045
80
1110
190 ....
.... 1334
52
1047
81
1112
200 ....
,... 136*
53
10'.0
82
54
1052
53
1055
84
1118
802 ....
.... 1658
S6
10j7
86
1121
892 ...
.... 1789
57
lo^y
86
1123
482 ....
.... 1919
58
1062
87
1135
572 ....
59
60
1064
1066
89
1128
1180
STKENGTH OF MATERIALS.
The chief materiaJa, of which it is necessary to record the strength
in this place, are cast and malleable iron ; and many experiments
have been made at different times upon each of these substances,
though not with any very close correspondence. The following is
a summary of them : —
: M^.i.i,,
c s
E
M
163001 omri
36000 1 ^"''*
60000 9000
800O0
69120000
91440000
5530000
6770000
1 ^'^'''^^w^:::::::::::::::.
The first column of figures, marked C, contains the mean strength
of cohesion on an inch section of the material ; the second, marked
S, the constant for transverse strains ; the third, marked E, the
constant for deflections ; and the fourth, marked M, the modulus
of elasticity. The introduction of the hot blast iron brought with
it the impression that it was less strong than that previously in use,
and the experiments which had previously been confided in as
giving results near enough the truth, for all practical purposes,
were no longer considered to be applicable to the new state of
things. New experiments Tvere therefore made. The following
Table gives, we have no doubt, results as nearly correct as can be
required or attained; —
hv Google
THE PRACTICAL MODEL CALCULATOR,
Is the following Table each bar is veducecl to exactly one incli
square ; and the transTerse strength, which may be taken aa a
criterion of the value of each Iron, is obtained froni a mean between
the experiments upon it;^fir5t on bars 4 ft. 6 in. between the
supports; and next on,t?iose of half the length, or 2 ft. 3 in. be
tween the ■ supports. All the other results are deduced from the
4 ft. 6 in. bars. In all cases the weights were laid on the middle
of the bar.
Ponker, Mo. 3. C
OMlwnj, So. S. ;
iS
WfaiUsh£r
Wiilta
Sulllfh gri
ill End, No. Z. Cold Bloat
low Jdoor, Si
BnSaty, No. 1
Brimbo, No. S
Apedale,No.:
Oiabary, Ha,
HnliUib, No. 1. tJoia Sltut*.
Adfllphi, Mo. 2. Oold KJart'
BluaiO'o. 3. Oold Bliat ■ "
Dercin, Ho. S, CoWSlaat*  .■
Gtattbenie, Noi^'Hot Blul
PnKKt, No. 4. Cold Blast
Lsne Sad, Ho. S,
CBiTon, No. a. Cold Blast*' ■
MBBslag (MMkai Bad)
Corbynska]], Ho.2
Pon^poo]» No. 2 '.■■..■... ■
WBUbrook, No. 3
MlltoiijHr " """i^
EBQt No.'
Ln^No.
o. 1. Bot Bitiat*
l.HotBlBn..
a.HotBlHt..
mloai^ Ho. 2. Cold Bis
T»rtiig,No.a.Hot;Blia
Coltham, No. 1. HOt Bl
Oanoll, Ho.3.ColdB]e
Molrktrfc No. 1. Hot B
Bierler.Ko.a
OoedT&lou, Ho. 2. Hot
CoedTalon, No. % Cold
Monklaad, No. 2. Hot 1
Wb Works, No. 1. Ho
PlasbTPaatoo, No. 2. H
The irons with asterisks a
Cold Blast Iron,
3 taken from Experiments on Hot and
hv Google
THE STEAM ENGINE. 175
Rule. — To find from the above Table tbe breakiug weight in
rectangular bars, generaDy. Calliiig b and d the breadth and
depth in inches, and I the distance between the supports, in feet,
and putting 45 for 4 ft. 6 in., we have , ~ = breaking
weight in lbs., — the value of 8 being taken from the above Table.
IFor example: — What weight would be necessary to break a bar
of Low Moor Iron, 2 inches broad, 3 inches deep, and 6 feet be
tween the supports ? According to the rule given above, we have
& = 2 inches, <? = 3 inches, ; = feet, S = 472 from the Table.
^, 45 X bd'S 45 X 2 X 3^ X 472
Then J ■ g — = 6372 lbs., the break
ing weight.
Table of the Cohesive Power of Bodies whose Gross Sectional Areas
equal one Square Inch.
M.r.,..
tohsiE Power
fewedish hir iron
t. 1 OHO
RUXSIBD do
English do
Cast steel
1 4 .iU
Blistered do
] 1,J,
Shear do
Wrought oopper
Hartl gOB metal
"(. r,(,
Cast copper
ly,072
Yellow braes, ca'it
17 %S
CaBt iron
17 (,.■'*
Tin cBBt
4 7)3
Bismuth, oaat
3 250
Lead, cast
lB4
Elastic power or direct tension of wrought iron.
medium quality
22 400
Note — A bar of iron is extended 000016, or neailv one ten
thousandth part of ita length, !oi eveiy ton of diiCLt strain per
Bquaie inch of sectionil area
CENTRE OP GRAVITY.
The centre of gravity of a body is that point within it which
continually endeavours to gain the lowest possible situation ; or it is
that point on which the body, being freely suspended, will remain
at rest in all positions. The centre of gravity of a body does not
always exist within the matter of which the body is composed,
there being bodies of such forms as to preclude the possibility of
this being the case, but it must either be surrounded by the con
stituent matter, or so placed that the particles shall be symmetri
cally situated, with respect to a vertical line in which the position
of the centre occurs. Thus, the centre of gravity of a ring is not
in the substance of the ring itself, but, if the ring be uniform, it will
be in the axis of its circumscribing cylinder ; and if the ring variea
hv Google
176 THE PRACTICAL MODEL CALCULATOR.
in form or density, it will be situated nearest to those parts where
the weight or density is greatest. Varying the position of a body
will not cause any change in the situation of the centre of gravity ;
for any change of position the body undergoes will only have the
effect of altering the directions of the sustaining forces, which will
still preserve their parallelism. When a body is suspended by any
other point than its centre of gravity, it will not rest unless that
centre be in the same vertical line with the point of suspension ;
for, in every other position, the force which is intended to insure
the equilibrium will not directly oppose the resultant of gravity
upon the particles of the body, and of course the equilibrium will
not obtain ; the directions of the forces of gravity upon the con
stituent particles are all parallel to one another and perpendicular
to the horizon. If a heavy body be sustained by two or. more
forces, their lines of direction must meet either at the centre of
gravity, or in the vertical line in which it occurs.
A body cannot descend or fall downwards, unless it be in such
a position that by its motion the centre of gravity descends. If a
body stands on a plane, and a line be drown perpendicular to the
horizon, and if this perpendicular line fall within the base of the
body, it will be supported without falling ; but if the perpendicular
falls without the base of the body, it will overset. For when the
perpendicular falls within the base, the body cannot be moved at all
without raising the centre of gravity ; but when the perpendicular
falls without the base towards any side, if the body be moved
towards that side, the centre of gravity will descend, and conse
quently the body will overset in that direction. If a perpendicular
to the horizon from the centre of gravity fall upon the extremity
of the base, the body may continue to stand, but the least force
that can be applied will cause it to overset in that direction ; and
the nearer the perpendicular is to any side the easier the body will
be made to fall on that side, but the nearer the perpendicular is to
the middle of the base the firmer the body will stand. If the
centre of gravity of a body be supported, the whole body is sup
ported, and the place of the centre of gravity must be considered
as the place of the body, and it is always in a line which ia perpen
dicular to the horizon.
In any two bodies, the common centre of gravity divides the
line that joins their individual centres into two parts that are to
one another reciprocally as the magnitudes of the bodies. The
products of the bodies multiplied by their respective distances from
the common centre of gravity are equal. If a weight be laid
upon any point of an inflexible lever which is supported at the
ends, the pressure on each, point of the support will be inversely
as the respective distances from the point whoro the weight is
applied. In a system of three bodies, if a line he drawn from the
centre of gravity of any one of them to the common centre of the
other two, then the common centre of all the three bodies divides
the line into two parts that are to each other reciprocally as the
hv Google
THE STEAM ENGISE. 177
magnitude of the body from which the line is drawn to the sum of
the magnitudes of tbe other two ; and, consequently, the single
body multiplied by its distance from the common centre of ^rai ity
ia equal to the sum of the other bodies multiplied by the distance
of their common centre from the common centre of the pjstera
If there be taken any point in the straight line or levei joinmg
the centres of gravity of two bodies, the sum of the two products
of each body multiplied by its distance from that point is erjual to
the product of the sum of the bodies multiplied by the distance of
their confmon centre of gravity from the same point. The two
bodies have, therefore, the same tendency to turn the lever about
the assumed point, as if they were both placed in their common
centre of gravity. Or, if the line with the bodies moves about the
assumed point, the sum of the momenta is equal to the momentum
of the sum of the bodies placed at their common centre of gravity.
The same property holds with respect to any number of bodies
whatever, and also when the bodies are not placed in the line, but
in perpendiculars to it passing through the bodies. If any plane
pass through the assumed point, perpendicular to the line in which
it subsists, then the distance of the common centre of gravity of
all the bodies from that plain is equal to the sum of all the
momenta divided by the sum of all the bodies. We may here
specify the positions of the centre of gravity in several figures of
very frequent occurrence.
In a straight line, or in a straight bar or rod of uniform figure
and density, the position of the centre of gravity is at the middle
of its length. In the plane of a triangle the centre of gravity is
situated in the straight line drawn from any one of the angles to
the middle of the opposite side, and at twothirds of this line dis
tant from the angle where it originates, or onethird distant from
the base. In the surface of a trapezium the centre of gravity is in
the intersections of the straight lines that join the centres of the
opposite triangles made by the two diagonals. The centre of
gravity of the surface of a parallelogram is at the intersection of
the diagonals, or at the intersection of the two lines which bisect
the figure from its opposite sides. In any regular polygon the
centre of gravity is at the same point as the centre of magnitude.
In a circular arc the position of the centre of gravity is distant
from the centre of the circle by the measure of a fourth propor
tional to the arc, radius, and chord. In a semicircular arc the
position of the centre of gravity is distant from the centre by the
measure of a third proportional to the arc of the quadrant and the
radius. In the sector of a circle the position of the centre of
gravity is distant from the centre of the circle by a fourth propor
tional to three times the arc of the sector, the chord of the arc,
and the diameter of the circle. In a circular segment, the position
of the centre of gravity is distant from the centre of the circle by
a space which is equal to the cube or third power of the chord
divided by twelve times the area of the segment. In a semicircle
12
hv Google
178 THE PRACTICAL MODEL CALCULATOR.
tlic position of the centre of gravity is distant from the centre of
the circle hy a space which is equal to four times the radius divided
by the constant number 81416 X 3 = 94248. In a parabola the
position of the centre of gravity is distant from the vertex by
threefifths of the axis. In a semiparabola the position of the
centre of gravity is at the intersection of the coordinates, one of
which is parallel to the base, and distant from it by twofifths of
the axis, and the other parallel to the axis, but distant from it by
threeeighths of the semibase.
The centres of gravity of the surface of a cylinder, a cone, and
conic frustum, are respectively at the same distances from the origin
as are the centres of gravity of the parallelogram, the triangle, and
the trapezoid, which are sections passing along the axes of the re
spective solids. The centre of gravity of the surface of a spheric seg
ment is at the middle of the versed sine or height. The centre of
gravity of the convex surface of a spherical zone is at the middle of
that portion of the axis of the sphere intercepted by its two bases.
In prisms and cylinders the position of the centre of gravity is at the
middle of the straight line that joins the centres of gravity of their
opposite ends. In pyramids and cones the centre of gravity is in
the straight line that joins the vertex with the centre of gravity
of the base, and at threefourths of its length from; the vertex, and
onefourth from the base. In a semisphere, or semispheroid, the
position of the centre of gravity is distant from the centre by three
eighths of the radius. In a parabolic conoid the position of the
centre of gravity is distant from the base by onethird of the axis,
or twothirds of the axis distant from the vertex. There are
several other bodies and figures of which the position of the centre
of gravity is known ; but as the position in those cases cannot be
defined without algebra, we omit them.
Central forces are of two kinds, centripetal and centrifugal.
Centripetal force is that force by which a body is attracted or
impelled towards a certain fixed point as a centre, and that point
towards which the body is urged ia called the centre of attraction
or the centre of force. Centrifugal force is that force by which a
body endeavours to recede from the centre of attraction, and from
which it would actually fly off in the direction of a tangent if it
were not prevented by the action of the centripetal force. These
two forces are therefore antagonistic ; the action of the one being
directly opposed to that of the other. It ia on the joint action of
these two forces that all curvilinear motion depends. Circular motion
is that afFection of curvilinear motion where the body is constrained
to move in the circumference of a circle : if it continues to move so
aa to describe the entire circle, it is denominated rotatory motion, and
the body is said to revolve in a circular orbit, the centre of which is
called the centre of motion. In all circular motions the deflection
or deviation from the rectilinear course is constantly the same at
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THE STEAM ENGINE. 179
e¥ery point of tte orbit, in which case the centripetal and centri
fugal forces are equal to one another. In circular orbits the cen
tripetal forces, by which equal bodies placed at equal distances
from the centres of force are attracted or drawn towards those
centres, are proportional to the quantities of matter in the central
bodies. This is manifest, for since all attraction takes place
towards some particular body, every particle in the attracting body
must produce its individual effect ; consequently, a body containing
twice the quantity of matter will exert twice the attractive energy,
and a body containing thrice the quantity of matter will operate
with thrice the attractive force, and so on according to the quantity
of matter in the attracting body.
Any body, whether large or small, when placed at the same dis
tance from the centre of forco, ia attracted or drawn through equal
spaces in the same time by the action of the central body. This
is obvious from the consideration that although a body two or three
times greater is urged with two or three times greater an attractive
force, yet there is two or three times the quantity of matter to be
moved ; and, as we have shown elsewhere, the velocity generated
in a given time Js directly proportional to the force by which it is
generated, and inversely as the quantity of matter in the moving
or attracted body. But the force which in the present instance is
the weight of the body is proportional to the quantity of matter
which it contains ; consequently, the velocity generated is directly
and inversely proportional to the quantity of matter in the
attracted body, and is, therefore, a given or a constant quantity.
Hence, the centripetal force, or force towards the centre of the
circular orbit, is not measured by the magnitude of the revolving
body, but only by the space which it describes or passes over in a
given time. When a body revolves in a circular orbit, and is
retained in it by means of a centripetal force directed to the
centre, the actual velocity of the revolving body at every point of
its revolution ia equal to that which it would acquire by falling
perpendicularly with the same uniform force through, one fourth of
the diameter, or onehalf the radius of its orbit ; and this velocity
is the same as would be acquired by a second body in falling
through half the radius, whilst the first body, in revolving in its
orbit, describes a portion of the circumference which is equal in
length to half the diameter of the circle. Consequently, if a body
revolves uniformly in the circumference of a circle by means of a
given centripetal force, the portion of the circumference which it
describes in any time is a mean proportional between the diameter
of the circle and the space which the body would descend perpen
dicularly in the same time, and with the same given force continued
uniformly.
The periodic time, in the doctrine of central forces, is the time
occupied by a body in performing a complete revolution round the
centre, when that body is constrained to move in the circumference
by means of a centripetal force directed to that point ; and when
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loO THE PRACTICAL MODEL CALCrLATOK.
the body revolves in a circular orbit, the periodic time, or the
time of performing a complete revolution, is expressed by the term
nt / i and the velocity or space passed over in the time ( ivill be
•/ ds; in which expressions d denotes the diameter of the circular
orbit described by the revolving body, « the space descended in any
time by a body falling perpendicularly downwards with the same
uniform force, t the time of descending through the space, s and n
the circumference of a circle whose diameter is unity. If several
bodies revolving in circles round the same or different centres be
retained in their orbits by the action of centripetal forces directed
to those points, the periodic times will be directly as the square
roots of the radii or distances of the revolving bodies, and inversely
as the square roots of the centripetal forces, or, what is the same
thing, the squares of the periodic times are directly as tlie radii,
and inversely as the centripetal forces.
CENTRE or GYRATION.
The centre of gyration is that point in which, if all the consti
tuent particles, or all the matter contained in a revolving body, or
system of bodies, were concentrated, the same angular velocity
would be generated in the same time by a given force aeting at any
place as would be generated by the same force acting similarly on
the body or system itself according to its formation.
The angular motion of a body, or system of bodies, is the motion
of a line connecting any point with the centre or axis of motion,
and is the same in all parts of the same revolving system.
In different unconnected bodies, each revolving about a centre,
the angular velocity is directly proportional to the absolute velo
city, and inversely as the distance from the centre of motion ; so
that, if the absolute velocities of the revolving bodies be propor
tional to their ridii or distances, the angular velocities will he
equil If the axis of motion passes through the centre of gravity,
then IS this centre called the principal centre of gyration.
The distiuce of the <,entre of gyration from the point of suspen
sion, or the axis of motion in any body or system of bodies, is a
geometrical mean betvieen the centres of gravity and oscillation
iiom the same point or axis , consequently, having found the dis
tances of these centres m any proposed case, the square root of
then product wdl gn e the distance of the centre of gyration. If
any part of a system be coneened to be collected in the centre of
gyration ol that particular part, the centre of gyration of the
whole system will continue the same as before ; for the same force that
moved this part of the system before along with the rest will move
it now without any change ; and consequently, if each part of the
system be collected into its own particular centre, the common
centre of the whole system will continue the same. If a circle be
described about the centre of gravity of any system, and the axis
of rotation be made to pass through any point of the circumference,
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THE STEAM ENGINE. 181
the distance of the centre of gyration from that point will always
be the same.
If the periphery of a, circle revolve about an axis passing through
the centre, and at right angles to its plane, it is the same thing as
if all the matter were collected into any one point in the peri
phery. And moreover, the plane of a circle or a disk containing
twice the quantity of matter as the said periphery, and having the
same diameter, will in an equal time acquire the same angular
velocity. If the matter of a. revolving body were actually to be
placed in the centre of gyration, it ought either to be arranged in
the circumference, or in two points of the circumference diametri
cally opposite to each other, and equally distant from the centre
of motion, for by this means the centre of motion will coincide
with the centre of gravity, and the body will revolve without any
lateral force on any aide. These are the chief properties con
nected with the centre of gyration, and the following are a few of
the cases in which its position has been ascertained.
In a right line, or a cylinder of very small diameter revolving
about one of its extremities, the distance of tho centre of gyration
from tho centre of motion is equal to the length of the revolving
line or cylinder multiplied by the square root of ^, In the plane
of a circle, or a cylinder revolving about the axis, it is equal to the
radius multiplied by the square root of . In the circumference
of a circle revolving about the diameter it is equal to the radius
multiplied by the square root of J. In the plane of a circle
revolving about the diameter it is equal to onehalf the radius. In
a thin circular ring revolving about one of its diameters as an axis
it is equal to the radius multiplied by the square root of J. In a
solid globe revolving about the diameter it is equal to the radiug
multiplied by the square root of . In the surface of a sphere
revolving about the diameter it is equal to the radius multiplied by
the square root of f . In a right cone revolving about the axis it
is equal to the radius of the base multiplied by the square root of ^.
In all these cases the distance is estimated from the centre of
the axis of motion. We shall have occasion to illustrate these prin
ciples when we come to treat of flywheels in the construction of
the different parts of steam engines.
When bodies revolving in tho circumferences of different circles
are retained in their orbits by centripetal forces directed to the
centres, the periodic times of revolution are directly proportional
to the distances or radii of the circles, and inversely as the veloci
ties of motion ; and the periodic times, under like circumstances,
are directly as the velocities of motion, and inversely as the cen
tripetal forces. If the times of revolution are equal, the velocities
and centripetal forces are directly as the distances or radii of the
circles. If the centripetal forces are equal, the squares of the
times of revolution and the squares of the velocities are as the dis
tances or radii of the circles. If the times of revolution are as
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182 THE PRACTICAL MODEL CALCULATOR.
the radii of the circles, the velocities will be equal, and the cen
tripetal forces reciprocally as the radii.
If several bodies revolve in circular orhits round the same or
difi'erent centres, the velocities are directly aa the distances or
radii, and inversely as the times of revolution. The velocities are
directly as the centripetal forces and the times of revolution. The
squares of the velocities are proportional to the centripetal forces,
and the distances or radii of the circles. When the velocities are
equal, the times of revolution are proportional to the radii of the
circles in which the bodies revolve, and the radii of the circles are
inversely as the centripetal forces. If the velocities be propor
tional to the distances or radii of the circles, the centripetal forces
■will bo in the same ratio, and the times of revolution will be equal.
If several bodies revolve in circular orbits about the same or
different centres, the centripetal forces are proportional to the dis
tances or radii of the circles directly, and inversely as the squares
of the times of revolution. The centripetal forces are directly
proportional to the velocities, and inversely as the times of revolu
tion. The centripetal forces are directly as the squares of the
velocities, and inversely as the distances or radii of the circles.
When the centripetal forces are equal, the velocities are propor
tional to the times of revolution, and the distances as the squares
of the times or as the squares of the velocities. When the central
forces are proportional to the distances or radii of the circles, the
times of revolution are equal. If several bodies revolve in circular
orbits about the same or different centres, the radii of the circles
are directly proportional to the centripetal forces, and the squares,
of the periodic times. The distances or radii of the circles are
directly as the velocities and periodic times. The distances or
radii of the circles are directly as the squares of the velocities, and
reciprocally as the centripetal forces. If the distances are equal,
the centripetal forces are directly as the squares of the velocities,
and reciprocally as the squares of the times of revolution ; the
velocities also arc reciprocally as the times of revolution. The
converse of these principles and properties are equally true ; and
all that has been here stated in regard to centripetal forces is
similarly true of centrifugal forces, they being equal and contrary
to each other.
The quantities of matter in all attracting bodies, having other
bodies revolving about them in circular orbits, are proportional to
the cubes of the distances directly, and to the squares of the times
of revolution reciprocally. The attractive force of a body is
directly proportional to the quantity of matter, and inversely as
the square of the distance. If the centripetal force of a body
revolving in a circular orbit be proportional to the distance from
the centre, a body let fall from the upper extremity of the vertical
diameter will reach the centre in the same time that the revolving
body describes onefourth part of the orbit. The velocity of the
descending body at any point of the diameter is proportional to
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THE STEAM ESGINE. 183
the ordinate of the circle at that point ; and the time of falling
through any portion of the diameter is proportional to the arc of
the circumference whose versed sine is the space fallen through.
All the times of falling from any altitudes whatever to the centre
of the orbit will he equal ; for these times are equal to onefourth
of the periodic times, and these times, under the specified condi
tions, are equal. The velocity of the descending body at the centre
of the circular orbit is equal to the velocity of the revolving body.
These are the chief principles that we need consider regarding
the motion of bodies in circular orbits ; and from them we are led
to the consideration of bodies suspended on a centre, and made to
revolve in a circle beneath the suspending point, so that when the
body describes the circumference of a circle, the string or wire by
which it is suspended describes the surface of a cone. A body thus
revolving is called a eoniaal pendulum, and this species of pendu
lum, or, as it is usually termed, the governor, is of great importance
in mechanical arrangements, being employed to regulate the move
ments of steam engines, waterwheels, and other mechanism. As
we shall have occasion to show the construction and use of this in
strument when treating of the parts and proportions of engines, we
need not do more at present than state the principles on which its
action depends. We must, however, previously say a few words
on the properties of the simple pendulum, or that which, being sus
pended from a centre, is made to vibrate from side to side in the
same vertical plane.
PEHDULUMS.
If a pendulum vibrates in a small circular arc, the time of per
forming one vibration is to the time occupied by a heavy body in
falling perpendicularly through half the length of the pendulum as
the circumference of a circle is to its diameter. All vibrations of
the same pendulum made in very small circular area, are made in
very nearly the same time. The space described by a falling body
in the time of one vibration is to half the length of the pendulum
as the square of the circumference of a circle is to the square of
the diameter. The lengths of two pendulums which by vibrating
describe similar circular ares are to each other as the squares of
the times of vibration. The times of pendulums vibrating in small
circular arcs are as the square roots of the lengths of the pendulums.
The velocity of a pendulum at the lowest point of its path is pro
portional to the chord of the arc through which it descends to ac
quire that velocity. Pendulums of the same length vibrate in the
same time, whatever the weights may be. From which we infer,
that all bodies near the earth's surface, whether they be heavy or
light, will fall through equal spaces in equal times, the resistance
of the air not being considered.
The lengths of pendulums vibrating in the same time in different
positions of the earth's surface are as the forces of gravity in those
positions. The times wherein pendulums of the same length will
vibrate by different forces of gravity are inversely as the square
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184 THE PRACTICAL MODEL CALCULATOR.
roots of the forces. The lengths of pendulums vibrating in dif
ferent places are as the forces of gravity at those places and the
squares of the times of vibration. The timea in which pendulums
of any length perform their vibrations are directly as the square
roots of their lengths, and inversely as the square roots of the gravi
tating forces. The forces of gravity at different places on the earth's
surface are directly as the lengths of the pendulums, and inversely
as the squares of the times of vibration. These are the chief proper
ties of a simple pendulum vibrating in a vertical plane, and the prin
cipal problems that arise in connection with it are the following, viz. :
To find the length of a pendulum that shall make any number
of vibrations in a given time ; and secondly, having given the length
of a pendulum, to find the number of vibrations it will make in any
time given. — These are problems of very easy solution, and the
rules for resolving them are simply as follow : — For the first, the
rule is, multiply the square of the number of seconds in the given
time by the constant number 394015, and divide the product by
the square of the number of vibrations, for the length of the
pendulum in inches. For the second, it is, multiply the square of
the number of seconds in the given time by the constant number
39'1393, divide the product by the given length of the pendulum
in inches, and extract the square root of the quotient for the num
ber of vibrations sought. The number 394015 is the length of a
pendulum in inches, that vibrates seconds, or sixty times in a minute,
in the latitude of Philadelphia,
Suppose a pendulum is found to make 35 vibrations ia a minute ;
what is the distance from the centre of suspension to the centre of
oscillation ?
Here, by the rule, the number of seconds in the given time is CO ;
hence we get 60 X 60 X 391015 = 1407654, which, being di
vided by 35 X 35 = 1225, gives 1407654 h 1225 = 1149105
inches for the length required.
The length of a pendulum between the centre of suspension and
ihe centre of oscillation is 64 inches ; what number of vibrations
will it make in 60 seconds ?
By the rule we have 60 X 60 X 394015 = 1407654, which,
being divided hy 64, gives 1407654 ^ 64 = 219946, and the
square root of this is 219946 = 46'9, number of vibrations
sought. When the given time is a minute, or 60 seconds, as in the
two examples proposed above, the product of the constant numbei'
391015 by the square of the time, or 1407654, is itself a constant
quantity, which, heingkept in mind, will in some measure facilitate the
process of calculation in all similar cases. We now return to the
consideration of the conical pendulum, or that in which the ball re
volves about a vertical axis in the circumference of a circular plane
which is parallel to the horizon.
CONICAL PENnuHTM.
If a pendulum he suspended from the upper extremity of a ver
tical axis, and be made to revolve about that axis by a conical mo
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THE STEAM ENGINE. 185
fion, ivhicli constrains the revolving body to move in the circum
ference of a circle whose plane is parallel to the horizon, then the
time in which the pendulum performs a revolution about the axis
can easily be found.
Let CD he the pendulum in question, suspended from C, the
upper extremity of the vertical axis CD,
and let the ball or body E, by revolving
about the said asis, describe the circle BE
AH, the plane of which is parallel to the
horizon ; it is proposeil to assign the time
of description, or the time in which the body
S performs a revolution about the axis CD,
at the distance BD.
Conceive the axis CD to denote the weight *(
of the revolving body, or its force in the di
rection of gravity; then, by the Compo
sition and Resolution of Forces, CB will denote the force or
tension of the string or wire that retains the revolving body in
the direction CB, and ED the force tending to the centre of the
plane of revolution at D. But, by the general laws of motion
and forces previously laid down, if the time be given, the space
described will be directly proportional to the force ; bnt, by the
laws of gravity, the space fallen perpendicularly from rest, in one
second of time, is ^ = IQ^ feet ; consequently we have CD : BD : :
1^12 = ^' , the space described towards D by the force in BD
in one second. Consequently, by the laws of centripetal forces, the
periodic time, or the time of t he bod y revolving in the circle BEAH,
is expressed by the term jt^/tL^, where « = 31416, the circum
ference of a circle whose diameter is unity ; or putting t to denote
the time, and expressing the height CD in feet, we get ( — 62832
■J . ■■ , or, by reducing the expression to its simplest form, it
becomes t = OSlflSGv/CD, where CD must be estimated in inches,
and t in seconds. Here we have obtained an expression of great
simplicity, and the practical rule for reducing it may be expressed
in words as follows :
Rule. — Multiply the square root of the height, or the distance
between the point of suspension and the centre of the plane of revo
lution, in inches, by the constant fraction 0'31986, and the product
will be the time of revolution in seconds.
In what time will a conical pendulum revolve about its vertical
axis, supposing the distance between the point of suspension and
the centre of the plane of revolution to be 391393 inches, which is
the length of a simple pendulum tiiat vibrates seconds in latitude
51° 30' ?
The square root of 391393 is 62561 ; consequently, by the rule,
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18b THE PRACTICAL MODEL CALCULATOR.
we liave, 6'2561 x 031986 = 20011 seconds for the time of revo
lution sought. It consequently revolves 30 times in a minute, as it
ought to do by the theory of the simple pendulum.
By reversing the process, the height of the cone, or the distance
between the point of suspension and the centre of the plane of revo
lution, corresponding to any given time, can easily he ascertained ;
for we have only to divide the number of seconds in the given time
by the constant decimal 031986, and the square of the quotient
will be the required height in inches, ' Thus, suppose it were re
quired to find the height of a conical pendulum that would revolve
30 times in a minute. Here tJie time of revolution is 2 seconds for
60 H 30 = 2; therefore, by division, it is 2 h 031986 = 62527,
which, being squared, gives 62527 = 390961 inches, or the )ength
of a simple pendulum that vibrates seconds very nearly. In all
coDical pendulums the times of revolution, or the periodic times, are
proportional to the square roots of the heights of the cones. This
is manifest, for in the foregoing equation of the periodic time the
numbers 62832 and 386, or 12 X 32i, are constant quantities, eon ■
sequently * varies as ^/CD.
If the heights of the cones, or the distances between the points
of suspension and the centres of the planes of revolution, he the
same, the periodic times, or the times of revolution, will be the
same, whatever may be the radii of the circles described by the re
volving bodies. This will be clearly understood by c
the subjoined diagram, where all the pendulums Ca, C5, Ce, Cd, and
Ce, having the common axis CD, will revolve in the same time ; and
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TMB STSAM ENGISE. 187
if they are all in the same vertical plane when first put in motion,
they will continue to revolve in that plane, whatever be the velocity,
so long as the common axis or height of the cone. remains the same.
This will become manifest, if we conceive an inflexible bar or rod
of iron to pass through the centres of all the balls as well as the
common axis, for then the bar and the several balls must all revolve
in the same time ; but if any one of them should be allowed to rise
higher, its velocity would be increased ; and if it descends, the ve
locity will be decreased.
Half the periodic time of a conical pendulum is equal to the
time of vibration of a simple pendulum, the length of which is
equal to the axis or height of the cone ; that is, the simple pendu
lum makes two oscillations or vibrations from side to side, or it
arrives at the same point from which it departed, in the same time
that the conical pendulum revolves about its axis. The space
descended by a falling body in the time of one revolution of the
conical pendulum is equal to 31416' maltipliod by twice the height
or axis of the cone. The periodic time, or the time of one revo
lution is equal to the product of 3'1416 */ 2 multiplied by the time
of falling through the height of the cone. The weight of a conical
pendulum, when revolving in the circumference of a circle, bears
the same proportion to the centrifugal force, or its tendency to fly
off in a straight line, as the axis or height of the cone bears to the
radius of the plane of revolution ; consequently, when the height
of the cone is equal to the radius of its base, the centripetal or
centrifugal force is equal to the power of gravity.
These are the principles on which the action of the conical pen
dulum depends ; but as we shall hereafter have occasion to con
sider it more at large, we need not say more respecting it in this
place. Before dismissing the subject, however, it may he proper to
put the reader in possession of the rules for calculating the posi
tion of the centre of oscillation in vibrating bodies, in a few cases
where it has been determined, these being the cases that are of the
moat frequent occurrence in practice.
The centre of oscillation in a vibrating body is that point in the
line of suspension, in which, if all the matter of the system were
collected, any force applied there would generate the same angular
motion in a given time as the same force applied at the centre of
gravity. The centres of oscillation for several figures of very fre
quent use, suspended from their vertices and vibrating flatwise, are
as follow : —
In a right lino, or parallelogram, or a cylinder of very small
diameter, the centre of oscillation is at twothirds of the length
from the point of suspension. In an isosceles triangle the centre
of oscillation is at threefourths of the altitude. In a circle it
is fivefourths of the radius. In the common parabola it is
fivesevenths of its altitude. In a parabola of any order it is
' S n 4 1 ^ ^ altitude, where n denotes the order of the figure.
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188 THE PRACTICAL MODEL CALCULATOR.
In bodies vibrating laterally, or in their own plane, the centres
of oscillation are situated as follows ; namely, in a circle the centre
of oscillation ia at threefourths of the diameter ; in a rectangle,
suspended at one of its angles, it is at twothirds of the diagonal ;
in a parabola, suspended by the vertex, it is fivesevenths of tho
axis, increased by onethird of the parameter ; in a parabola, sus
pended by the middle of its base, it is foursevenths of the axis,
increased by half the parameter ; in the sector of a circle it ia
three times the arc of tho sector multiplied by the radius, and
divided by four times the chord ; in a right cone it is fourfifths of
the axis or height, increased by the quotient that arises when the
square of the radius of the base is divided by five times the height ;
in a globe or sphere it is the radius of the sphere, plus the length of
the thread by which it is suspended, plus the quotient that arises
when twice the square of the radius is divided by five times the sum
of the radius and the length of the suspending thread. In all these
cases the distance is estimated from tho point of suspension, and since
the centres of oscillation and percussion are in one and the same
point, whatever has been said of the one is equally true of tho other.
THE TEMPERATURE AND ELASTIC FOHCB OP BTEAir,
In estimating the mechanical action of steam, tho intensity of its
elastic force must be referred to somo known standard measure,
such as the pressure which it exerts against a square inch of the
surface that contains it, usually reckoned by so many pounds
avoirdupois upon the square inch. The intensity of the elastic
force is also estimated by the inches in height of a vertical column
of mercury, whose weight is equal to the pressure exerted by the
steam on a surface equal to the base of the mercurial column. It
may also be estimated by the height of a vertical column of water
measured in feet ; or generally, the elastic force of any fiuid may
be compared with that of atmospheric air when in its usual state of
temperature and density ; this is equal to a column of mercury 30
inches or 2 feet in height.
When the temperature of steam is increased, respect being had
to its density, the elastic force, or the effort to separate the parts
of the containing vessel and occupy a larger space, is also increased ;
and when the temperature is diminished, a corresponding and pro
portionate diminution takes place in the intensity of the emanci
pating elFort or elastic power. It consequently follows that there
must be some law or principle connecting the temperature of steam
with its elastic force ; and an intimate acquaintance with this law,
in so far as it is known, must be of the greatest importance in all
our researches respecting the theory and the mechanical operations
of the steam engine.
To find a theorem, hy means of whieh it may he ascertained when
a general law exists, and to determine what that law is, in eases
where it is known to obtain. — Suppose, for example, that it is
required to assign the nature of the law that subsists between the
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THE STEAM ENGINE. 189
temperature of steam and its elastic force, on the supposition that
the elasticity is proportional to some power of the temperature,
and unaffected by any other constant or coofficient, except the
exponent by which the law is indicated. Let E and e be any two
values of the elasticity, and T, (, the corresponding temperatures
deducted from observation. It is proposed to ascertain the powers
of T and i, to which E and e are respectively proportional. Let n
denote the index or exponent of the required power ; then by the
conditions of the problem admitting that a law exists, wo get,
T° ; i" : : E : e ; but by the principles of proportion, it is — = — ;
and if this be expressed logarithmically, it is n X log. ^ := log. — ,
and by reducing the equation in respect of n, it finally becomes
_ log. 6 — log. E
" ^T^. (  log. r
The theorem that we have here obtained is in its form sufH
ciently simple for practical application ; it is of frequent occur
rence in physical science, but especially so in inquiries respecting
the motion of bodies moving in air and other resisting media ; and
it IB even applicable to the determination of the planetary motions
themselves. The process indicated by it in the case that we have
chosen, is pimply, To divide the difference of the logarithms of the
elasticities by the difference of the logarithms of the corresponding
temperatures, and the quotient will express that power of the tempe
rature to which the elasticity is proportional.
Take as an example the following data : — In two experiments it
was found that when the temperature of steam was 2503 and
8436 degrees of Fahrenheit's scale, the corresponding elastic
forces were 596 and 2384 inches of the mercurial column respec
tively. From these data it is required to determine the law which
connects the temperature with the elastic force on the supposition
that a law does actually exist under the specified conditions. The
process by the rule is as follows :
Greater temperature, 3436 log. 25352941
Lesser temperature, 2503 log. 23084608
Remainder = 01 368333
Greater elastic force, 2384 log. 237T3063
Lesser elastic force, 596 log. l"75g463
Remainder =0^60^0600
Let the second of these remainders be divided by the first, as
dii'ccted in the rule, and we get n = 6020600 ^ 1368333 = 43998,
the exponent sought. Consequently, by taking the nearest unit,
for the sake of simplicity, we shall have, according to this result,
the following analogy, viz. :
T":t^ ■>::£:£;
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190 THE PRACnCAL MODEL CALCrLATOR.
that is, the elasticities are proportional to the 4'4 power of the
temperatures very nearly.
Kow this law is rigorously correct, as applied to the particular
cases that furnished it ; for if the two temperatures ant! one elas
ticity he given, the other elasticity will be found as indicated by
the above analogy ; or if the two elasticities and one temperature
be given, the other temperature will be found by a similar process.
It by no means follows, however, that the principle is general, nor
could we venture to affirm that the exponent here obtained will
accurately represent the result of any other experiments than
those from which it is deduced, whether the temperature be higher
or lower than that of boiling water ; but this we learn from it, that
the index which represents the law of elasticity is of a very high
order, and that the general equation, whatever its form may be,
must involve other conditions than those which we have assumed io
the foregoing investigation. The theorem, however, is valuable to
practical men, not only aa being applicable to numerous other
branches of mechanical inquiry, but as leading directly to the
methods by which some of the best rules have been obtained for
calculating the eiasticity of steam, when in contact with the liquid
from which it is generated.
We now proceed to apply our formula to the determination of a
general law, or such as will nearly represent the cSaaa of experi
ments on which it rests ; and for this purpose we must first assign
the limits, and then inquire under what conditions the limitations
take place, for by these limitations we mnst in a great measure be
guided in determining the ultimate form of the equation which
represents the law of elasticity.
The limits of elasticity will be readily assigned from the follow
ing considerations, viz. : In the first place, it is obvious that steam
cannot exist when the cohesive attraction of the particles is of
greater intensity than the repulsive energy of the caloric or matter
of heat interposed between them ; for in this case, the change from
an elastic fluid to a solid may take place without passing through
the intermediate stage of liquidity ; hence we infer that there must
be a temperature at which the elastic force is nothing, and this
temperature, whatever may be its value, corresponds to the lower
limit of elasticity. The higher limit will be discovered by similar
considerations, for it must take place when the density of steam is
the same as that of water, which therefore depends on the modulus
of elaeticity of water. The modulus of elasticity of any substance
is the measure of its elastic force ; that of water at 60° of tempe
rature is 22,100 atmospheres. Thus, for instance, suppose a given
quantity of water to be confined in a close vessel which it exactly
fills, and let it be exposed to a high degree of temperature, then
it is obvious that in this state no steam would be produced, and the
force which is exerted to separate the parts of the vessel is simply
the expansive force of compressed water ; we therefore have tlie
following proportion. As the expanded volume of water is to the
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THE STEAM ENGINE. 191
quantity of e'^pin^mn, =«o in the tnodulus cf ela^ticitj of water to
the elaitic It ice of ^team of the sime density as watei
Having therefore assjgne I the limit'* beyond whn,h the elastic
force of =!team cmnot reach, we shall now proceed to ^pply the
principle of our formula to the determination of the general law
which connect'' the temperature with the el istic force , and for this
purpose, m addition to the notation which wo htve already laid
down, kt c denote some constant quantity that a&ects the cl isticity,
and d the temperature at which the elaaticitj vanishes , then since
this temperature must be applied suhtractively, we h^l e from the
foregoing principle c E = (T — i)', and c e = (( — *)" From
either of these e ]Uations, therefore the constant quantity e can
be determined m terms of the rest when they are known ; thus we
have c = ^ — = — i, and c = i L, and by comparing these
two independent values of c, the value of n becomes known ; for
^^ — = — '— = i— — i, and consequently
n = log  e  log. E , , ,
log. ((  i)  log. (T  6). ' ■ ■ ■ ^ ''
In this equation the value of the symbol 8 is unknown ; in order
therefore to determine it, we must have another independent
expression for the value of n ; and in order to this, iet the ciaati
cities E and e become E' and e' respectively; while the corre
sponding temperatures T and t assume the values T' and (' ; then
by a similar process to the above, we get * — •' = ^ — ^_J, and
J°SiJ
(B).
log. {f  8}  log. (T'  i]
Let the equations (A) and (B) be compared with each other, and
we shall then have an expression involving only the unknown
quantity 8, for it must be understood that the several temperatures
with their corresponding elasticities are to be deduced from experi
ment ; and in consequence, the law that we derive from them must
be strictly empirical ; thus we have
log, e — log. E. log, e'  lo g. E .„,
log. (i  B)  log. (T  8) log. {f  8}  log. (r  «) ■ ; ^ ''
We have no direct metbod of reducing expressions of this sort,
and the usual process is therefore by approximation, or by the rule
of trial and error, and it is in this way that the value of the quan
tity 8 must be found ; and for the purpose of performing the reduc
tion, we shall select experiments performed with great care, and
may consequently be considered as representing the law of elas
ticity with very great nicety.
T = 2120 Fahrenheit E = 298 inches of mercury.
t = 2503 e = 59()
T'= 2934 E'= 1192
('=3436 e'= 2.384
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192 THE PRACTICAL MODEL CALCULATOR.
Therefore, by substituting these numbers in equation (C), and
making a few trials, we find that S = — 50°, and substituting this
in either of the equations (A) or (B), we get n — 5'08 ; and
finally, by substituting these values of 8 and n in either of the
expressions for the constant quantity c, we get c = 64674730000,
the 5'08 root of which is 134'27 very nearly ; hence we have
1 134'27 / • ■ ■ ■ ^ ^
Where the symbol F denotes generally the elastic force of the
Steam in inches of mercury, and ( the corresponding temperature
in degrees of rahrcnhcit's thermometer, the logarithm of ttie
denominator of the fraction is 2X279717, which may be used as a
constant in calculating the elastic force corresponding to any given
temperature. We have thus discovered a rule of a very simple
form ; it errs in defect ; but this might have been remedied by
assuming two points near one extremity of the range of experi
ment, and two points near the other extremity ; and by substi
tuting the observed numbers in equation (C), different constants
and a more correct exponent would accordingly have been obtained.
Mr, Southern has, by pursuing a method somewhat analogous to
that which is here described, found his experiments to be very
nearly represented by
I 1357«7 /
But even here the formula errs in defect, for he has fonnd it
necessary to correct it by adding the arbitrary decimal O'l; and
thus modified, it b
{y^}"i <^'
Our own formula may also be corrected by the application of
some arbitrary constant of greater magnitude ; but as our motive
for tracing the steps of investigation in the foregoing case was to
exemplify the method of determining the law of elasticity, our end
is answered ; for we consider it a very unsatisfactory thing merely
to be put in possession of a formula purporting to be applicable to
some particular purpose, without at the same time being put in
possession of the method by which that formula was obtained, and
the principles on which it rests. Having thus exhibited the prin
ciples and the method of reduction, the reader will have greater
confidence as regards the consistency of the processes that he may
be called upon to perform. The operation implied by equation (E)
may be expressed in words as follows : —
RoLB, — To the given temperature in degrees of Fahrenheit's
thermometer add 513 degrees and divide the sum by 135 1 67 ; to
the 5'13 poiver of the quotient add the constant fraction I'fr, and
the sum will be the elastic force in inches of mercury.
hv Google
THE STEAM EHGINE. 193
The process here deserihed is that which is performed by the
rules of common arithmetic ; but since the index is affected by a
fraction, it is difficult to perform in that way : we must therefore
have recourse to logarithms as the only means of avoiding the diffi
culty. The rule adapted to these numbers is as follows :—
Rule eor Loqarithms. — To the given temperature in degrees
of Fahrenheit's thermometer add 51'3 degrees; then, from the
logarithm of the sum subtract 2'1327940 or the logarithm of
135T67, the denominator of the fraction; multiply the remainder
by the index 5'13, and to the natural number answering to the
sum add the constant fraction j\; the sum will be the elastic force
in inches of mercury.
If the temperature of steam be 2503 degrees as indicated by
Fahrenheit's thermometer, what is the corresponding clastic force
in inches of mercury ?
By the rule it is 2503 + 513 = 3016 log. 24794313
constant den. = 135767 log. 21327940 subtract
remainder = 03466373 _
SI "5 inverted
17331865
346637
103991
natural number 60013 log. 17782493
If this be increased by ^, we get 60113 inches of mercury for
the elastic force of steam at 2503 degrees of Fahrenheit.
By simply reversing the process or transposing equation (E), the
temperature corresponding to any given elastic force can easily be
found ; tlie transformed expression is as follows, viz. :
* = 135767 (F  01)^  513 .... (F).
Since, in consequence of the complicated index, the process of
calculation cannot easily be performed by common arithmetic, it is
needless to give a rule for reducing the equation in that way ; we
shall therefore at once give the rule for performing the process hj
logarithms.
Rule. — From the given elastic force in inches of mercury, sub
tract the constant fraction 01 ; divide the logarithm of the remain
der by 513, and to the quotient add the logarithm 21327940 ; find
the natural number answering to the sum of the logarithms, and
from the number thus found subtract the constant 513, and the
remainder will be the temperature sought.
Supposing the elastic force of steam or the vapour of water to
be equivalent to the weight of a vertical column of mercury, the
height of which is 2384 inches; what is the corresponding tem
perature in degrees of Fahrenheit's thermometer ¥
Here, by proceeding as directed in the rule, we have 2384 — Ol =
hv Google
194 THE PRACTICAL MODEL CALCULATOE.
238'3, and dividing the logarithm of this remainder hy the coa
staiit exponent 5'1§, we get
log. 2383 H 513 = 23771240 ^ 513 = 04633770
constant coefficient =135767   log. 21327940add
natural number =39461    log. 25961710 sum
constant temperature = 513 subtract
required temperature = 34331 degrees of Fahrenheit's ther
mometer.
The temperature by observation is 3436 degrees, giving a differ
ence of only 029 of a degree in defect. For low temperature or
low pressure steam, that is, steam not exceeding the simple pres
sure of the atmosphere, M. Pambour gives
y 004948 + (135:^) . . .(6).
In which equation the symbol p denotes the pressure in pounds
avoirdupois per square inch, and ( the temperature in degrees of
Fahrenheit's thermometer. When this expression is reduced in
reference to temperature, it is
( = 1557256 (;> 004948) ^513 .... (II).
The formula of Tredgold is well known. The equation, in its
original form, is
177/* = i + 100. . . .(I):
where / denotes the elastic force of steam in inches of mercury,
and ( the temperature in degrees of Fahrenheit's thermometer.
The same formula, as modified and corrected by M. Millet, becomes
1790773/^ = * i 103 . . . . (K).
Dr. Young of Dublin constructed a formula which was adapted
to the experiments of his countryman Dr. Dalton : it assumed a
form suf&ciently simple and elegant ; it is thus expressed—
/= (1 + 0.0029 0' . . . . (L):
where the symbol /denotes the elastic force of steam expressed in
atmospheres of 30 inches of mercury, and t the temperature in
degrees estimated above 212 of Fahrenheit. This formula is not
applicable in practice, especially in high temperatures, as it deviates
very widely and rapidly from the results of observation: it is
chiefly remarkable as being made the basis of a numerous class of
theorems somewhat varied, but of a more correct and satisfactory
character. The Commission of the Frencli Academy represented
their experiments by means of a formula constructed on the same
principles : it is thus expressed —
/=(ll 07153 ()' .... (M):
where /denotes the elastic force of the steam expressed in atmo
spheres of 076 metres or 29922 inches of mercury, and ( the tem
hv Google
THE STEAM ENGINE. 195
perature estimated above 100 degrees of the centigrade tliermo
metcr ; but when the same formula is SO transformed as to he
expressed in the usual terms adopted in practice, it is
p = (02679 + 00067585 i)' . . . . (N):
where ^ is the pressure in pounds per square inch, and t the tem
perature in degrees of Fahrenheit's scale, estimated above 212 or
simple atmospheric pressure.
The committee of the Franklin Institute adopted the exponent
6, and found it necessary to change the constant 00029 into
000333; thus modified, they represented their experiments by
the equation
p = (0460467 + 000521478 tf . . . . (0).
By combining Dr. Dalton's experiments with the mean between
those of the French Academy and the Franklin Institute, we obtain
the following equations, the one being applicable for temperatures
below 212 degrees, and the other tor temperatures above that
point as far as 50 atmospheres. Thus, for low pressure steam,
that is, for steam of less temperature than 212, it is
and for steam above the temperature of 212, it is
/=C
In consequence therefore of the high and imposing authority
from which these formulas are deduced, we shall adopt them in all
our subsequent calculations relative to the steam engine ; and in
order' to render their application easy and familiar, we shall trans
late them into rules in words at length, and illustrate them by the
resolution of appropriate numerical examples; and for the sake of
a systematic arrangement, we thmk proper to branch the subject
into a series of problems, as follows :
The temperature of steam being given in degrees of Fahrenheit's
thermometer, to find the corresponding elastic force in inches of
mercury. — The problem, as here propounded, is resolved by one or
other of the last two equations, and the process indicated by the
arrangement is thus expressed : —
Rule, — To the given temperature expressed in degrees of
Fahrenheit's thermometer, add the constant temperature 175 ; find
the logarithm answering to the sum, from which subtract the con
stant 2587711 ; multiply the remainder by the index 7'71307, and
the product will be the logarithm of the elastic force in atmospheres
of 30 inches of mercury when the given temperature is less than
212 degrees. But when the temperattu'e is greater than 212,
increase it by 121 ; then, from the logarithm of the temperature
thus increased, subtract the constant logarithm 2522444, multiply
the remainder by the exponent 642, and the product will be the
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196 THE PHACXICAI. MODEL CALCULATOR.
logaritlira of the elastic force in atmospheres of 30 inches of mer
cury ; which being multiplied by 30 will give the force in inches,
or if multiplied by 1476 the result will be expressed in pounds
avoirdupois per square inch.
When steam is generated under a temperature of 187 degrees of
Fahrenheit's thermometer, what is its corresponding elastic force in
atmospheres of 30 inches of mercury ?
In this example, the given temperature is less than 212 degrees :
it will therefore be resolved by the first clause of the preceding
rule, in which the additive constant is 175 ; hence we get
187 + 175 = 362. ..log. 2558709
Constant divisor = 387. .log. 2587711 subtract
9970998 X 771307 = 9'773393
And the corresponding natural number is 0'5934 atmospheres, or
17802 inches of mercury, the elastic force required, or if expressed
in pounds per square inch, it is 05934 x 1476 = 876 lbs. very
nearly. If the temperature be 250 degrees of rahrecheit, the pro
cess is as follows :
250 I 121 = 371. ..log. 2569374
Constant divisor = 333. ..log. 2522444 subtract
0046936 X 642 = 0301291
And the corresponding natural number is 20012 atmospheres, or
60036 inches of mercury, and in pounds per square inch it is
20012 X 1476 = 2954 lbs. very nearly.
It is sometimes convenient to express the results in inches of
mercury, without a previous determination in atmospheres, and for
this purpose the rule is simply as follows :
Rule. — Multiply the given temperature in degrees of Fahren
heit's thermometer by the constant coefficient 15542, and to the
product add the constant number 271985; then from the loga
rithm of the sum subtract the constant logarithm 2587711, and
multiply the remainder by the exponent 771307 ; the natural num
ber answering to the product, considered as a logarithm, will give
the elastic force in inches of mercury. This answers to the case
when the temperature is less than 212 degrees; but when it is
above that point proceed as follows :
Multiply the given temperature in degrees of Fahrenheit's ther
mometer by the constant coefficient 169856, and to the product add
the constant number 205526 ; then from the logarithm of the sum
subtract the constant logarithm 2522444, and multiply the re
mainder by the exponent 642; the natural number answering to
the product considered as a logarithm, will give the elastic force
in inches of mercury. Take, for example, the temperatures as
assumed above, and the process, according to the rule, is as fol
lows:
hv Google
187 X 1'5542 = 2906354
Constant = 271985 a dd
Sum = 5626204... log. 2750216
Constant = 387 log. 25 87711 subtract
0162505 X 771307 = 12
And the natural number answering to this logarithm is 17923 inches
of mercury. By the preceding calculation the result ia 17802;
the slight difference arises from the introduction "of the decimal con
stants, which in consequence of not terminating at the proper place
are taken to the nearest unit in the last figure, but the process is
equally true notwithstanding. For the higher temperature, we get
250 X 169856 = 424640
Constant = 20 5526 add
Sum = 630.166 log. 2709456
Constant = 333 log. 2522444 subtract
0277011 X 642 = 1778410
And the natural number answering to this logarithm is 60036
inches of mercury, agreeing exactly with the result obtained as
above.
It is moreover sometimes convenient to express the force of the
steam in pounds per square inch, without a previous determination
in atmospheres or inches of mercury; and when the equations are
modified for that purpose, they supply us with the following process,
viz.:
Multiply the given temperature by the constant coefiicient
141666, and to the product add the constant number 2479155;
then, from the logarithm of the sum subtract the constant logarithm
2587711, and multiply the remainder by the index 771307 ; the
natural number answering to the procluct will give the pressure in
pounds per square inch, when the temperature is less than 212 de
grees ; but for all greater temperatures the process is as follows :
Multiply the given temperature by the constant coefficient
15209, and to the product add the constant number 1840289;
then, from the logarithm of the sum subtract the constant logarithm
2522444, and multiply the remainder by the exponent 642; the
natural or common number answering to the product, will express
the force of the eteam in pounds per square inch. If any of these
results be multiplied by the decimal 07854, the product will be the
corresponding pressure in pounds per circular inch. Taking, there
fore, the temperatures previously employed, the operation is as
follows ;
187 X 141666 = 2649155
Constant = 24791 55 add
Sura = 512.8310.log. 2709974
Constant = 387 log. 2587711 subtract
0122263 X 771307 = 094265S
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198 THE PRACTICAL MODEL CALCULATOR.
And tte number answering to thia logarithm is 8'763 lbs. per square
inch, and 8763 X 07854 = 68824 lbs. per circular incb, the pro
portion in the two cases being as 1 to 07554. Again, for the
higher temperature, it is
250 X 15209 = 3802250
Constant = 1840289 add
Sum = 5642539 log. 2751475
Constant = 833 log. 2522444 subtract
0229031 X 642 = 1470279
And the number answering to this logarithm is 29568 lbs. per
sc[uare inch, or 29568 x 07854 = 232226 lbs. per circular inch.
We have now to reverse the process, and determine the tempera
ture corresponding to any given power of the steam, and for this
purpose we must so transpose the formulas (P) and (Q), as to express
the temperature in terms of the elastic force, combined with given
constant numbers; but as it is probable that many of our readers
would prefer to see the theorems from which the rules are deduced,
we here subjoin them.
Por the lower temperature, or that which does not exceed the
temperature of boiling water, we get
( = 249/^^175 .... (R).
Where t denotes the temperature in degrees of Fahrenheit's ther
mometer, and /the elastic force in inches of mercury, less than 30
inches, or one atmosphere ; hut when the elastic force is greater
than one atmosphere, the formula for the corresponding temperature
is as follows :
i = 196/*^ 121 ... . (S).
In the construction of those formulas, we have, for the sake of
simphcity, omitted the fractions that obtain in the coefficient of /;
for since they are very small, the omission will not produce an error
of any consequence ; indeed, no error will arise on this account, as
we retain the correct logarithms, a circumstance that enables the
computer to ascertain the true value of the coefficients whenever it
is necessary so to do ; but in all cases of actual practice, the results
derived from the integral coefficients will he quite sufficient. The
rule supplied hy the equations (R) and (S) ia thus expressed :
When the elastic force is less than the pressure of the atrflosphere,
that is, less than 80 inches of the mercurial column, —
Rule. — Divide the logarithm of the given elastic force in inches
of mercury, by the constant index 771307, and to the quotient add
the constant logarithm 2396204; then from the common or natural
number answering to the sum, svihtract the constant temperature
175 degrees, and the remainder will he the temperature sought in
degrees of Fahrenheit's thermometer. But when the elastic force
exceeds 30 inches, or one atmosphere, the following rule applies :
hv Google
THE STEAM ENGINE. 199
Divii^e the logarithm of the given elastic force in inches of mer
cury hy the constant index 642, and to the quotient add the con
stant logarithm 2'292363 : then, from the natural number answer
ing to the sum subtract the constant temperature 121 degrees, and
the remainder will be the temperature sought. Similar rules might
be constructed for determining the temperature, when the pressure
in pounds per square inch is given ; but since this is a less useful
case of the problem, we have thought proper to omit it. We there
fore proceed to exemplify the above rules, and for this purpose we
shall suppose the pressure in the two cases to be equivalent to the
weight of 19 and 60 inches of mercury respectively. The operations
will therefore be as follows :
Log. 19 ^ 771307 = 1278754 ; 771307 = 0165791
Constant coefficient = 249 log. 2896204 add
Natural number = 36475 log. 2561994
Constant temperature = 175 subtract
Required temperature = 18975 degrees of Fahrenheit's scale.
For the higher elastic force the operation is as follows ;
Log. 60 H 642 = 1778151 h 642 = 0276969
Constant coefficient = 196 log. 2292363 add
Natural number = 37097 log. 2569332
Constant temperature = 121 subtract
Required temperature = 24997 degrees of Fahrenheit's scale.
All the preceding results, as computed by our rules, agree as
nearly with observation as can be desired : but they have all been
obtained on the supposition that the steam is in contact i^ith the
liquid from which it is generated ; and in this case it is evident
that the steam must always attain an elastic force corresponding to
the temperature ; and in accordance to any increase of pressure,
supposing the temperature to remain the same, a quantity of it
corresponding to the degree of compression must simply be condensed
into water, and in consequence will leave the diminished space
occupied by steam of the original degree of tension ; or otherwise
to express it, if the temperature and pressure invariably correspond
with each other, it is impossible to increase the density and elas
ticity of the steam except by increasing the temperature at the same
time ; and, contrariwise, the temperature cannot be increased with
out at the same time increasing the elasticity and density. This
being admitted, it is obvious that under these circumstances the
Steam must always maintain its maximum of pressure and density :
but if it be separated from the liquid that produces it, and if its
temperature in this ease bo increased, it will be found not to possess
a higher degree of elasticity than a volume of atmospheric air simi
larly confined, and heated to the same temperature. Under this
new condition, the state of maximum density and elasticity ceases ;
for it is obvious that since no water is present, there cannot be any
hv Google
200 THE PRACTICAL MODEL CALCULATOR.
more steam generated by an increase of temperature ; and conse
quently the force of the steam is only that which confines it to its
original bulk, and is measured by the effort which it exerts to ex
pand itself. Our nest object, therefore, is to inquire what is the
law of elasticity of steam under the conditions that we have here
The specific gravity of steam, its density, and the volume which
it occupies at difi'erent temperatures, have been determined by ex
periment with very great precision ; and it has also been ascertained
that the expansion of vapour by means of heat is regulated by the
same laws as the expansion of the other gases, viz, that all gases
expand from unity to I'SYS in bulk by 180 degrees of temperature;
and again, that steam obeys the law discovered by Boyle and Mari
otte, contracting in volume in proportion to the degree of pressure
which it sustains. We have therefore to inquire what space a given
quantity of water converted into steam will occupy at a given pres
sure ; and from thence we can ascertain the specific gravity, density,
and volume at all other pressures.
When a gas or vapour is submitted to a constant pressure, the
quantity which it expands by a given rise of temperature is calcu
lated by the following theorem,
, /(' + 4.')9\ ,™j
" ="(f+45s) m
where t and t' are the temperatures, and u, v' the corresponding
volumes before and after expansion; hence this rule.
Rule. — To each of the temperatures before and after expansion,
add the constant experimental number 459 ; divide tho greater sum
by the lesser, and multiply the quotient by the volume at tho lower
temperature, and the product will give the expanded volume.
If the volume of steam at the temperature of 212 degrees of Pah
renheit be 1711 times the bulk of the water that produces it, what
will be its volume at the temperature of 250'8 degrees, supposing
the pressure to be the same in both cases ?
Here, by the rule, we, have 212 4 459 = 671, and 2503 + 459
= 709'3 ; consequently, by dividing the greater by the lesser, and
multiplying by the given volume, we get 1^^ X 1711 = 180866
671
for the volume at the temperature of 2503 degrees.
Again, if the elastic force at the lower temperature and the cor
responding volume be given, the elastic force at the higher tem
perature can readily be found ; for it is simply as the volume the
vapour occupies at the lower temperature is to the volume at the
higher temperature, or what it would become by expansion, so is the
elastic force given to that required.
If the volume which steam occupies under any given pressure
and temperature be given, the volume which it wil! occupy under
any proposed pressure can readily be found by reversing the pre
ceding process, or by referring to chemical tables containing the
hv Google
THE STEAM ENQINE. 201
specific gravity of the gases compared witli air as unitj at the same
pressure and temperature. Now, air at the mean state of the at
mosphere has a specific gravity of 1 aa compared with water at
1000 ; and the bullts are inversely as the specific gravities, accord
ing to the general laws of the properties of matter previously an
nounced ; hence it follows that air is 818 times the bulk of an
equal weight of water, for 1000 ^ If = 81818. But, by the
experiments of Dr. Dalton, it has heen found that steam of the
same pressure and temperature has a specific gravity of GSS com
pared with air as unity ; consequently, we have only to divide the
number 818'18 by '625, and the quotient will give the propor
tion of volume of the vapour to one of the liquid from which it is
generated ; thus we get 81818 5 625 = 1309 ; that ia, the volume
of steam at 60 degrees of Fahrenheit, its force being 30 inches of
mercury, is 1309 times the volume of an equal weight of water ;
hence it follows, from equation {'X), that when the temperature in
s to t', the volume t
/459 + t' \
' = 1309 X (559^^60) = 2524(459 + t');
and from this expression, the volume corresponding to any specified
elastic force /, and temperature t', may easily he found ; for it ia
inversely as the compressing force: that is,
/:30: : 2525(459 + t') : v' ;
consequently, by working out the analogy, we get
= ;[5:67(459J^). ,j,.
f ^ ''
By tliis theorem is found the volume of steam as compared with
that of the water producing it, when under a pressure correspond
ing to the temperature. The rule in words ia as follows :
Rdle. — Calculate the elastic force in inches of mercury by the
rule already given for that purpose, and reserve it for a divisor.
To the given temperature add the constant number 459, and mul
tiply the sum by 75'67 ; then divide the product by the reserved
divisor, and the quotient will give the volume sought.
When the temperature of steam is 250'3 degrees of Fahrenheit'a
thermometer, what ia the volume, compared with that of water ?
The temperature being greater than 212 degrees, the force ia cal
culated by the rule to equation (Q), and the process is as follows :
2503 + 121 = 3713 log. 25697249
Constant divisor = 333 log. 2j;5224442 subtract
00472807 x642=03035421
Atmosphere = 30 inches of mercury log. 14771213 add
Elastic force = 60348 log. 17806634 ~1
Again it is, 1 ,
459 I 2503 = 7093 log. 285083001 ,, r^'^"
Constantcoefficient = 7567 log. 18789237 ( 47297537 j
Volume = 889>39 times that of water, log. 294909'03 re
mainder.
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202 THE PEACTICAL MODEL CALCULATOR.
Thus we have given the method of calculating the ehistie force
of ateam when the temperature is given either in atmospheres or
inches of mercury, and also in pounds or the square or circular
inch : we have also reversed the process, and determined the tem
perature corresponding to any given elastic force. We have,
moreover, shown how to find the volume corresponding to different
temperatures, when the pressure is constant ; and, finally, ivo have
calculated the volume, when under a pressure due to the elastic
force. These are the chief subjects of calculation as regards the
properties of steam ; and we earnestly advise our readers to render
themselves familiar with the several operations. The calculations
as regards the motion of steam in the parts of an engine to produce
power, will be considered in another part of the present treatise.
The equation (XJ), we may add, can be exhibited in a different
form involving only the temperature and known quantities; for
since the expressions (P) and (Q) represent the elastic force in terms
of the temperature, according as it is under or above 212 degrees
of Pahrenheit, we have only to substitute those values of the elastic
force when reduced to inches of mercury, instead of the symbol/
in equation (U), and we obtain, when the temperature is less than
212 degrees,
Vol.=7567(tcm.+459)H(004016xtem.f702807)'"*" (V).
and when tlie temperature exceeds 212 degrees, the expression be
comes
Vol. =7567(tem. i459)H OOolOl x tem. f ■617195f ^ (W.)
These expressions are simple in their form, and easily reduced ;
but, in pursuance of the plan we have adopted, it becomes necessary
to express the manner of their reduction in words at length, as
follows :
Rule. — When the given temperature is under 212 degrees, mul
tiply the temperature in degrees of Fahrenheit's thermometer by
the constant fraction 004016, and to the product add the constant
increment '702807 ; multiply the logarithm of the sum by the in
dex 7'71307, and find the natural or common number answcriog to
the product, which reserve for a divisor. To the temperature add
the constant number 459, and multiply the sum by the coefncient
75'67 for a dividend ; divide the latter result by the former, and
the quotient will express the volume of steam when that of water is
unity.
Again, when the given temperature is greater than 212 degrees,
multiply it by the fraction 'OOSlOl, and to the product add the
constant increment '617195 ; multiply the logarithm of the sum
by the index 642, and reserve the natural number answering to
the product for a divisor; find the dividend as directed above,
which, being divided by the divisor, will give the volume of steam
when that of the water is unity.
How many cubic feet of steam will be supplied by one cubic foot
hv Google
THE STEAM ENGINE. 203
of water, under the respective temperatures of 187 aud 2934 de
grees of Fahrenheit's thermometer ?
Here, by the rule, we have
187x0004016=0750992
Constant incremeiit=0702807
Sum =14537^ log. 1625043 X 771307=l2534069
and the nUmher answering to this logarithm is 1792284, the di
visor. But 187 f 450 = 646, and 646 X 7507 = 4888282, the
dividend; hence, hy division, we get 488828271792284 =
27274 cubic feet of steam from one cubic foot of water.
Again, for the higher temperature, it is
2934 X 0005101 = 1496633
Constant increment = 0617195
Sum = 2113828 log. 03250696x642=20869468;
and the number answering to this logarithm is 122165, the divisor.
But 2934 4 459 = 7524, and 7524 x 7567 = 56934108, the
dividend; therefore, by division, we get 56934108 ^ 122165 =
46604 cubic feet of steam from one cubic foot of water.
The preceding is a very simple process for calculating the volume
which the steam of a cubic foot of water will occupy when under
a pressure due to a given temperature and elastic force ; and since
a knowledge of this particular is of the utmost importance in cal
culations connected with the steam engine, it is presumed that our
readers will find it to their advantage to render themselves familiar
with tlie method of obtaining it. The above example includes both
cases of the problem, a circumstance which gives to the operation,
considered as a whole, a somewhat formidable appearance : but it
would be difficult to conceive a case in actual practice where the
application of both the formulas will be required at one and the
same time ; the entire process must therefore be considered as em
bracing only one of the cases above exemplified ; and conseciucntly
it can be performed with the greatest facility by every person who
is acquainted with the use of logarithms ; and those unacquainted
with the application of logarithms ought to make themselves masters
of that very simple mode of computation.
Another thing which it is necessary sometimes to discover in
reasoning on the properties of steam as referred to its action in a
steam engine, is the weight of a cubic foot, or any other quantity
of it, expressed in grains, corresponding to a given temperature and
pressure. Now, it baa been ascertained by experiment, that when
the temperature of steam is 60 degrees of Fahrenheit, and the
pressure equal to 30 inches of mercury, the weight of a cubic foot
in grains is 3294; but the weight is directly proportional to the
elastic force, for tlie elastic force is proportional to the density :
consequently, if/ denote any other elastic force, and w the weight
in grains corresponding thereto, then we have
30:/:: 3294 : w = 1098/
hv Google
204 THE PRACTICAL MODEL CALCULATOR.
the weight of a, cubic foot of vapour at the force/, and temperature
60 degrees of Fahrenheit. Let t denote the temperature at the
<• ^ ,.. u .■ .rrx u 459 + i 459 + (
torce/; then by equation (1), we have v = T^a~sr(in ~ "^iTTq^i
the volume at the temperature t, supposing the volume at 60 de
grees to be unity ; that is, one cubic foot. Now, since the den
sities are inversely proportional to the spaces which the vapour oc
(459 + , , 5l9u> , , ,
cupies, we have — k\q — ■ : 1 : : w : w = T^jf^Ti ! "'^^ "7 ^"^
preceding analogy, the value of w ia 10'98f; therefore, by substi
tution, we get
, 669862/
•° " 459 + i ■ ■ ■ ■ (^>'
This equation expresses the weight in grains of a cubic foot of
steam at the temperature ( and force/; and if we substitute the
value of /, from equations (P) and (Q), reduced to inches of mer
cury, and modified for the two cases of temperature below and
above 212 degrees of Fahrenheit, we shall obtain, in the first case.
w' = (0012324 X temp, f 2155611)"'™^ i (temp. + 459). . . . (Y)
and for the second case, where the temperature exceeds 212, it is
w' = (001962 X temp. + 237374f ^= ~ (temp. + 459) . . . (Z)
These two equations, like those marked (V) and (AV) are suf
ficiently simple in their form, and off'er but httle difficulty in their
application. The rule for their reduction when expressed in words
at length, is as follows :
Rule. — When the temperature is less than 212 degrees, multi
ply the given temperature, in degrees of Fahrenheit's thermometer,
by the fraction 0012324, and to the product add the constant in
crement 2'155611 ; then multiply the logarithm of the sum by the
index 7*71307, and from the product subtract the logarithm of the
temperature, increased by 459 ; the natural number answering to
the remainder will be the weight of a cubic foot in grains.
Again, when the temperature exceeds 212, multiply it by the
fraction 001962, and to the product add the constant increment
237374; then multiply the logarithm of the sum by the index 642,
and from the product subtract the logarithm of the temperature in
creased by 459 ; the natural number answering to the remainder
will be the weight of a cubic foot in grains.
Supposing the temperatures to be as in the preceding example,
what will be the weight of a cubic foot in grains for the two cases ?
Here, by the rule, we have
Natural number :^ 157 '863 grains per cubic foot log. 211
b,Google
THE STEAM HKQINB. 205
For the higher temperature, it ii
2D34 X 001962 =r 575G508
Constant increment = 2873740
Sam = 8130248 log. OyiOlO58 X 642 = 58426664
293i + 459 = 7524 .... log. 2876448S , snbtraot
HatutaJ number = 92559 grains pec cubic foot . log. 29664176
Here again the operation resolves both cases of the problem ;
but in practice only one of them can be required.
THE MOTION OP ELASTIC FLUIDS.
The next subject that claims our attention is the Telocity with
which elastic fluids or vapours move in pipes or confined passages.
It is a wellknown fact in the doctrine of pneumatics, that the mo
tion of free elastic fluids depends upon the temperature and pres
sure of the atmosphere ; and, consequently, when an elastic fluid
is confined in a close vessel, it must he similarly circumstanced
with regard to temperature and pressure as it would be in an at
mosphere competent to exert the same pressure upon it. The sim
plest and most convenient way of estimating the motion of an elastic
fluid is to assign the height of a column of uniform density, capable
of producing the same pressure as that which the fluid sustains in
its state of confinement ; for under the pressure of such a column,
the velocity into a perfect vacuum will bo the same as that acquired
by a heavy body in falling through the height of the homogeneous
column, a proper allowance being made for the contraction at the
aperture or orifice through which the fluid flows.
When a passage is opened between two vessels containing fluids
of different densities, the fluid of greatest density rushes out of the
vessel that contains it, into the one containing the rarer fluid, and
the velocity of influx at the first instant of the motion is equal to
that which a heavy body acquires in falling through a certain
height, and that height is equal to the difference of two uniform
columns of the fluid of greatest density, competent to produce the
pressures under which tho fluids are originally confined ; and the
velocity of motion at any other instant is proportional to the squaie
root of the difl"erence between the heights of the uniform columns
producing the pressures at that instant. Hence we infer that the
velocity of motion continually decreases,— the density of the fluids
in the two vessels approaching nearer and nearer to an equality,
and after a certain time an equilibrium obtains, and the velocity
of motion ceases.
It is abundantly conflrmed by observation and experiment, that
oblique action produces very nearly the same efTcct in the motion
of elastic fluids through apertures as it does in the case of water ;
and it has moreover been ascertained that eddies take place under
similar circumstances, and these eddies must of course have a ten
dency to retard the motion : it therefore becomes necessary, in all
the calculations of practice, to make some allowance for the retard
ation that takes place in passing the orifice ; and this end is most
hv Google
206 THE PRACTICAL MODEL CALCUIATOK.
conveniently answered by modifying tho constant coefficient ac
cording to the nature of the aperture through which the motion
is made. Numerous experimeota have been made to ascertain the
effect of contraction in orifices of different forms and under dif
ferent conditions, and amongst those which have proved the most
successful in this respect, we may mention the experiments of Du
Buat and Eytelwein, the latter of whom has supplied us with a
series of coefficients, which, although not exclusively applicable to
the case of the steam engine, yet, on account of their extensive
utility, we take the liberty to transcribe. They are as follow : —
1. For the velocity of motion that would re
sult from the direct unretarded action of
the column of the fluid that produces it, we __^
have 3 V = s/57yA
2. For an orifice or tube in the form of the
contracted vein 10 V = </60Sih
3. For wide openings having the sill on a~l
level with the bottom of the reservoir ... 
4. For sluices with walls in a line with the V 1 V = v^5929i
orifice
5. For bridges with pointed piers J
6. For narrow openings having the sill on a^
level with the bottom of the reservoir ...
7. For small openings in a sluice with side , . .,, ■ ^^ ■ 
walls (10\ = s/iiGlh
8. For abrupt projections
9. For bridges with square piers , ^ ^
10. For openings in sluices without side walls 10 V = s/SCOlA
11. For openings or orifices in a thin plate V = \/25/i
12. For a straight tube from 2 to 3 diameters
in length projecting outwards 10 V = </4225
13. For a tube from 2 to 3 diameters in length
projecting inwards 10 V = s/291625h
It is necessary to observe, that in all these equations V is the
velocity of motion in feet per second, and h the height of the co
lumn producing it, estimated also in feet. Kos. 1, 2, 11, 12, and
13 are those which more particularly apply to the usual passages
for the steam in a steam engine ; but since all the others meet their
application in the everyday practice of the civil engineer, we have
thought it useful to supply them,
MOTIOS or STEAM IN AN ENGINE.
We have already stated that the best method of estimating tho
motion of an elastic fluid, such as steam or the vapour of water, is
to assign the height of a uniform column of that fluid capable of
producing the pressure : the determination of this column is there
fore the leading step of the inquiry ; and since the elastic force of
steam is usually reckoned in inches of mercury, 30 inches being
hv Google
THE STEAM ENGINE. 207
equal fo the pressure of the atmosphere, the subject presents but
little difSculty ; for wo have already seen that the height of a co
lumn of water of the temperature of 60 degrees, balancing a column
of 30 inches of mercury, is 34023 feet ; the corresponding column
of steam must therefore he as its relative hulk and elastic force ;
hence we have 30 : 34023 :fv:h = 1lMlfv, where / is the
elastic force of the steam in inches of mercury, v the correspond
ing volume or bulk when that of water is unity, and h the height
of a uniform column of the fluid capable of producing the pressure
due to the elastic force ; consequently, in the case of a direct un
retarded action, the velocity into a perfect vacuum, according to
No. 1 of the preceding class of formulas, is V = 8542 ^/f v ; but
for the hest form of pipes, or a conical tube in form of the con
tracted vein, the velocity into a vacuum, according to No. 2, be
comes V = 8307 '•/fv; and for pipes of the usual construction.
No. 12 gives V = 6922 v/V; No. 13 gives V = 5804 v/^
and in the ease of a simple orifice in a thin plate, we get from
No. 11 V = 5322 ^fv. The consideration of all these equa
tions may occasionally ho required, but our researches will at pre
sent be limited to that arising from No, 12, as being the best
adapted for general practice ; and for the purpose of shortening
the investigation, we shall take no further notice of the case in
which the temperature of the steam ia below 212 degrees of Fah
renheit ; for the expression which indicates the velocity into a va^
cuum being independent of the elastic force, a separate considera
tion for the two cases is here unnecessary.
It has been shown in the equation marked (TJ), that the volume
of steam which is generated from an unit of water, is i; =
7567 (temp.  459) , _, . ^ , , ,,.,,,.,.
^^ — ^ ; let this value oi v be substituted lor it in
the equation V — 6922 \/fv, and we obtain for the velocity into
a vacuum for the usual form of steam passages, as follows, viz. :
V = 602143 v/(temp. f 459).
This is a very neat and simple expression, and the object de
termined by it is a very important one : it therefore merits the
reader's utmost attention, especially if he is desirous of becoming
familiar with the calculations in reference to the motioli of steam.
The rule which the equation supplies, when expressed in words at
length, is as follows : —
Rule. — To the temperature of the steam, in degrees of Fahren
heit's thermometer, add the constant number or increment 459, and
multiply the square root of the sum by 602143 ; the product will
be the velocity with which the steam rushes into a vacuum in feet
per second.
With what velocity will steam of 2934 degrees of Fahrenheit's
thermometer rush into a vacuum when under a pressure due to the
elastic force corresponding to the given temperature.
hv Google
aUo THE PRACTICAL MODEIi CALCOLATOE.
By the rule it is 2934 1 469 = 7524 J log. 14382244
Coustant coefficient = 602143 log. 17797018 add
Velocity into a yaeunm in feet per second = 165108 log. 32179262
This is the velocity into a perfect vacuum, when the motiou is
made through a straight pipe of uniform diameter ; but when the
pipe is alternately enlarged and contracted, the velocity must ne
cessarily be reduced in proportion to the nature of the contraction ;
and it is further manifest, that every bend and angle in a pipe will
be attended with a correspondent diminution in the velocity of mo
tion : it therefore behoves ns, in the actual construction of steam
passages, to avoid these causes of loss as much as possible ; and
where they cannot be avoided altogether, such forms should be
adopted as will produce the smallest possible retarding effect. In
cases where the forms are limited by the situation and conditions
of construction, such corrections should be applied as the circum
stances of the case demand ; and the amount of these corrections
must be estimated according to the nature of the obstructions them
selves. For each rightangled bend, the diminution of velocity is
usually set down as being about onetenth of its unobstructed value ;
but whether this conclusion be correct or not, it is at least certain
that the obstruction in the case of a rightangled bend is much
greater than in that of a gradually curved one. It is a very com
mon thing, especially in steam vessels, for the main steam pipe to
send off branches at right angles to each cylinder, and it is easy to
see that a great diminution in the velocity of the steam must take
place here. In the expansion valve chest a further obstruction
must be met with, probably to the extent of reducing the velocity
of the steam twotenths of its whole amount.
These proportional corrections are not to be taken as the results
of experiments that have been performed for the purpose of deter
mining the effect of the above causes of retardation : we have no
experiments of this sort on which reliance can be placed ; and, in
consequence, such elements can only be inferred from a comparison
of the principles that regulate the motion of other fluids under simi
lar circumstances : they will, however, greatly assist the engineer
in arriving at an approximate estimate of the diminution that takes
place in the velocity in passing any number of obstructions, when
the precise nature of those obstructions can be ascertained. In the
generality of practical cases, if the constant coefficient 60'2143 be
reduced in the ratio of 650 to 430, the resulting constant 41'6868
may be employed without introducing an error of any consequence.
OP THE ABCEHI OP SMOKE AND UEATED AIR IN CIIIMNEYS.
The subject of chimney flues, with the ascent of smoke and heated
air, is another case of the motion of elastic fluids, in which, by a
change of temperature, an atmospheric column assumes a different
density from another, where no such alteration of temperature oc
curs. The proper construction of chimneys is a matter of very
great importance to the practical engineer, for in a close fireplace.
hv Google
THE STEAM ENGINE, 209
designed for the generation of steam, there must be a considerable
draught to accomplish the intended purpose, and this depends upon
the three following particulars, viz. :
1. The height of the chimney from the throat to the top.
2. The area of the transverse section.
3. The temperature at which the smoke and heated air are al
lowed to enter it.
The formula for determining the power of the chimney may be
investigated in the following manner :
Put h = the height in feet from the place where the flue enters
to the top of the chimney,
h = the number of cubic feet of air of atmospheric density
that the chimney must discharge per hour,
a = the area of the aperture in square inches through which
6 cubic feet of air must pass when expanded by a
change of temperature,
V = the velocity of ascent in feet p nl
t' = the temperature of the extern I a nd
( = the temperature of the air t b d 1 ged by the
chimney.
Now the force producing the motion n th manifestly
the difference between the weight of a lum f th tmospheric
air and another of the air discharged by th b mn y and when
the temperature of the atmospheric air at 52 1 of Fahren
heit's thermometer, this difference will b \ t 1 by the term
k L, . ^gQ ) ; the velocity of ascent will therefore be
V = JGi^ hl rrzTTKa f feet per second, and the quantity of air
discharged per second will therefore be, « J^* «^ P'+T^fl f'
supposing that there is no contraction in the stream of air ; but it
is found by experiment, that in all cases the contraction that takes
place diminishes the quantity discharged, by about threeeighths of
the whole ; consequently, the quantity discharged per hour in cu
bic feet becomes
* = 12569 .JE3.
This would be the quantity discharged, provided there were no
increase of volume in consequence of the change of temperature ;
but air expands from h to ■■■ ,  , "^tq ■■ for t' ~ t degrees of tem
perature, as has been shown elsewhere ; consequently, by compa
rison, we have
< + 469
'  12569
fl
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210 THE PRACTICAL MODEL CALCULATOR.
From this equation, therefore, any one of the quantities wticli
it involves can be found, wlien the others are given : it however
supposes that there is no other cause of diminution but th^ contrac
tion at the aperture ; but this can seldom if ever be the ease ; for
eddies, loss of heat, obstructions, and change of direction in the
chimney, wiU diminish the velocity, and consequently a larger area
will be required to suffer the heated air to pass, A sufficient al
lowance for these causes of retardation will be made, if we change
the coefficient 125'69 to 100 ; and in this case the equation for the
area of section b
a = h v'((' + 459)« ^ 100 (( + 459) ^/h (t'  t).
And if we take the mean temperature of the air of the atmo
sphere at 52 degrees of Fahrenheit, and make an allowance of 16
degrees for the difference of density between atmospheric air and
coal emoke, our equation will ultimately assume the form
a = b /((' I 459f i 51100 s/h (('( 16).
It has been found by experiment that 200 cubic feet of air of at
mospheric density aro required for the complete combustion of one
pound of coal, and the consumption of ten pounds of coal per hour
is usually reckoned equivalent to one horao power : it therefore ap
pears that 2000 cubic feet of air per hour must pass through the fire
for each horse power of the engine. This is a large allowance, bnt
it is the safest plan to calculate in excess in the first instance ; for
the chimney may afterwards be convenient, even if considerably
larger than is necessary. The rule for reducing the equation is as
follows : —
Rule.— Multiply the number of horse power of the engine by
the I power of the temperature at which the air enters the chimney,
increased by 459; then divide the product by 2555 times the
square root of the height of the chimney in feet,, multiplied by the
difference of temperature, less 16 degrees, and the quotient will be
the area of the chimney in square inches.
Suppose the height of the chimney for a 40horse engine to be
70 feet, what should be its area when the difference between the
temperature at which the air enters the flue, and that of the
atmosphere is 250 degrees ?
Here, by the rule, wo have,
250  52 = 302, the temperature at which the air enters
Constant increment = 459 [the flue.
Sum = 761 log. 28813847
2)86441541
43220770
Number of horse power = 40 log. 1'0020600
59241370
hv Google
THE STEAM ENGINE,
59241
250  16 = 234 ... . log. 23692159
height = TO feet . . log. l_8450980
2) 42143139
210T1569 1
Constant = 2555 . . log. 1.4073909 J . . . 35145478 J
Hence the area of the chimney in square inches is 25679, log.
24095892 ; and in this way may the area be calculated for any
other case ; but particular care must he taken to have the data ac
curately determined before the calculation is begun. In the above
example the particulars are merely assumed ; but even that is suffi
cient to show the process of calculation, which is more immediately
the object of the present inquiry. It is right, however, to add,
that recent experiments have greatly shaken the doctrine that it is
beneficial to make chimneys small at the top, though such is the
way in which they are, nevertheless, still constructed, and our rules
must have reference to the present practice. It appears, however,
that it would be the best way to make chimneys expand as they
ascend, after the manner of a trumpet, with its mouth turned down
wards: but these experiments require further confirmation.
The method of calculation adopted above is founded on the prin
ciple of correcting the temperature for the difi'erenee between the
specific gravity of atmospheric air and that of coalsmoke, the one
being unity and the other 105 ; there is, however, another method,
somewhat more elegant and legitimate, by employing the specific
gravity of coalsmoke itself: the investigation is rather tedious and
prolix, hut the resulting formula is by no means difficult ; and since
both methods give the same result when properly calculated, we
make no further apology for presenting our readers with another
rule for obtaining the same object. The formula is as follows :
27575 \A(t' 7755)
where a is the area of the transverse section of the chimney in
square inches, h the quantity of atmospheric air required for com
bustion of the coal in cubic feet per hour, h the height of the chim
ney in feet, and (' the temperature at which the air enters the flue
after passing through the fire. The rule for performing this pro
cess is thus expressed :
Rule. — From the temperature at which the air enters the chim
ney, subtract the constant decrement 77'55 ; multiply the remainder
by the height of the chimney in feet, divide unity by the product,
and extract the square root of the quotient. To the temperature
of the heated air, add the constant number 459 ; multiply the sum
by the number of cubic feet required for combustion per hour, and
divide the product by the number 27575 ; then multiply the quo
tient by the square root found as above, and the product will he the
number of square inches in the transverse section of the chimney.
hv Google
212 THE PEACTICAL MODEL CALODLATOB.
Suppose a mass of fuel in a state of comtiustion to require 5000
cubic feet of air per hour, what must be tlie size of the chimney
when ita height is 100 feet, the temperature at which the heated
air enters the chimney being 200 degrees of Fahrenheit's ther
mometer ?
By the rule we have 2007755=12245 . . log. 20879588
Height of the chimney=100. . . . log. 20000000
40879588
2) 59120412
79560206
200+459=659 . . . log. 28188854)
■ 5000 . . . log. 36989700 Vadd 30773399
27575 ar. co. log. 65594845 j
10333605 10798 in.
This appears to be a very small flue for the quantity of air that
passes through it per hour; but it must be observed that we have
assumed a great height for the shaft, which has the effect of cre
ating a very powerful draught, thereby drawing off the heated air
with great rapidity.
The advantage of a high flue is so very great, that the reader
may be desirous of knowing to what height a chimney of a given
base may be carried with safety, in cases where it is inconvenient
to secure it with lateral stays ; and, as an approximate rule for this
purpose is not difficult of investigation, we think proper to supply
it here.
When the chimney is equally wide throughout its whole height,
the formula is
= ; / I5tj
* '\12000JAw;
but when the side of the base is double the size of the top, the
equation becomes
8= A\
V12000042 Aw;
where 8 is the side of the base in feet, h the height, and m the
weight of one cubic foot of the material. When the chimney stalk
is not square, but longer on the one side than the other, s must be
the least dimension. The proportion of solid wall to a given base,
as sanctioned by experience, is about twothirds of its area, conse
quently w ought to bo twothirds of the weight of a cubic foot of
brickwork. Now, a cubic foot of dried brickwork is, on an average,
114 lbs. ; consequently «> = 76 lbs. ; and if this be substituted in
the foregoing equations, we get for a chimney of equal size through
out,
T. I 1^^ '
* \1200~25A;
hv Google
THE STEAM EHGINE.
and when the chimney tapers to onehalf the size at top, it is
Wn
104
'4,
V 12000  32 A;
where it may he remarked that 12000 lbs. is the cohesiye force of
one square foot of mortar ; and in the investigation of the formulas
we have assumed the greatest force of the wind on a square foot
of surface at 52 lbs. These equations are too simple in their form
to require elucidation from us; we therefore leave the reduction as
an exercise to the reader, who it is presumed will find no difficulty
in resolving the several cases that may arise in the course of his
practice.
2ffR atW
VD(2^K{L + H'
is the expression given hy M. Pikelet for the velocity of smoke in
a chimney, v, the velocity ; t, the temperature, whose maximum
value is about 300° centigrade ; g = 321 feet ; D, the diameter
of the chimney ; H, the height ; L, the length of horizontal flues,
supposing them formed into a cylinder of the same diameter
as that of the chimney. K = 0127 for brick, = 005 for sheet
iron, and — '0025 for castiron chimneys, a = '00365.
Let L=60; H=15Q; D=5 ; K=00 5; 2j?=64J; i=300^
/ 2ffHa(D
a=00365. Then v= \ jy^2a K(H+L) ^ ^^'^^^ ^^^^
A cnhic foot of water raised into steam is reckoned equivalent to
a horse power, and to generate the steam with sufficient rapidity,
an allowance of one square foot of firebars, and one square yard
of effective heating surface, are very commonly made in practice,
at least in land engines. These proportions, however, greatly vary
in different cases ; and in some of the best marine engine boilers,
where the area of firegrate is restricted by the breadth of the ves
sel, and the impossibility of firing long furnaces effectually at sea,
half a square foot of firegrate per horse power is a very common
proportion. Ten cubic feet of water in the boiler per horse power,
and ten cubic feet of steam room per horse power, have been as
signed as the average proportion of these elements ; but the fact is,
no general rule can be formed upon the subject, for the proportions
which would be suitable for a wagon boiler would be inapplicable
to a tubular boiler, whether marine or locomotive ; and good ex
amples will in such cases be found a safer guide than rules which
must often give a false result. A capacity of three cubic feet per
horse power is a common enough proportion of furnaceroom, and
it is a good plan to make the furnaces of a considerable width, as
they can then be fired more effectually, and do not produce so much
smoke as if they are made narrow. As regards the question of
draft, there is a great difference of opinion among engineers upon
the subject, some preferring a very slow draft and others a rapid
one. It is obvious that the question of draft is virtually that of
hv Google
214 THE PRACTICAL MODEL CALCULATOR.
the area of firegrate, or of the quantity of fuel consumed upon it
given area of grate surface, and the ireiglit of fuel burned on a foot
of firegrate per hour varies in diiferent cases in practice from 3J
to 80 lbs. Upon the quickness of the draft again hingea the ques
tion of the proper thickness of the stratum of incandescent fuel
upon the grate ; for if the draft be very strong, and the fire at the
same time be thin, a great deal of uncombined oxygen will escape
up through the fire, and a needless refrigeration of the contents of
the flues Tvill be thereby occasioned ; whereas, if the fire be thick,
and the draft he sluggish, much of the useful effect of the coal will
be lost by the formation of carbonic oxide. The length of the cir
cuit made by the smoke varies in almost every boiler, and the same
may be said of the area of the flue in its cross section, through
■which the smoke has to pass. As an average, about onefifth of
the area of firegrate for the area of the flue behind the bridge,
diminished to half that amount for the area of the chimney, has
been given as a good proportion, but the examples which we have
given, and the average flue area of the boilers which we shall
describe, may be taken as a safer guide than any such loose state
ments. When the flue is too long, or its sectional area is insuffi
cient, the draft becomes insufficient to furnish the requisite quantity
of steam ; whereas if the flue be too short or too large in its area,
a large quantity of the heat escapes up the chimney, and a depo
sition of soot in the flues also takes place. This last fault is one
of material consequence in the case of tubular boilers consuming
bituminous coal, though indeed the evil might be remedied by block
ing some of the tubes up. The area of waterlevel is about 5 feet
per horse power in land boilers. In many cases, however, it is
much less ; but it is always desirable to make the area of the water
level as large as possible, as, when it is contracted, not only is the
waterlevel subject to sudden and dangerous fluctuations, but water
is almost sure to be carried into the cylinder with the steam, in
consequence of the violent agitation of the water, caused by the
ascent of a large volume of steam through a small superficies. It
would be an improvement in boilers, we think, to place over each
furnace an inverted vessel immerged in the water, which might
catch the steam in its ascent, and deliver it quietly by a, pipe rising
above the waterlevel. The waterlevel would thus be preserved
from any inconvenient agitation, and the weight of water within the
boiler would be diminished at the same time that the original depth
of water over the furnaces was preserved. It would also be an
improvement to make the sides of the furnaces of marine boilers
sloping, instead of vertical, as is the common practice, for the steam
could then ascend freely at the instant of its formation, instead of
being entangled among the rivets and landings of the plates, and
superinducing an overheating of the plates by preventing a free
access of the water to the metal.
We have, in the following table, collected a few of the principal
results of experiments made on steam boilers.
hv Google
THE STEAM ENGINE.
Taele i.
— "
»
1
li
B
fl
P
{
{
li
li
!
I
1
1
3
II
i
l'li:t:",
.„
w^.
ris=
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^1r
■s
Unglb of dicuit nude t>;
US best In few ■
AreaoffiregEsteifnsquiM
2S66
..,.
725
.....
SS46
50
1425
l*45
see
Wetght of fuel lurncH on
«ieh«,uRrefcotofgr>te,
Cob. ft. of water eyaporated
byliaibsoffaS^
Cable feet of wntcr erv
poMtol per hour bom
Sqaan feet of beatel nir
liu»tbreiu:hcaUcr<»tor
water e.sporatM per
I3'31
Sqa8r.fcetofli«.teii«ui
^"bfXn2pb1n.tte"
The economical effects of expansion will be found to be very
clearly exhibited in the next table. The duties are recorded in the
fifth line from the top, and the degree of expansion in the bottom
line. It will be observed, that the order in which the different en
gines stand in respect of superiority of duty is the same as in re
spect of amount of expansion. The Holmbush engine baa a duty
of 140,484,848 lbs. raised 1 foot by 1 cwt. of coals, and the steam
acts expansively over '83 of the whole stroke; while the water
works' Cornish engine has only a duty of 105,664,118 lbs., and
expands the steam over only 687 of the whole stroke. Again,
comparing the second and last engines together, the Albion Mills
engine has a duty of 25,756,752 lbs., and no expansive action.
The waterworks' engine, again, acts expansively over onehalf of
its stroke, and has an increased duty of 46,602,333 lbs. Other
causes, of course, may influence these comparisons, especially the
last, where one engine is a doubleacting rotative engine, and the
other a singleacting pumping one ; but there can be no doubt that
the expansive action in the latter is the principal cause of its more
economical performance.
The heating surface per horse power allowed by some engineers
is about 9 square feet in wagon boilers, reckoning the total sur
face as effective surface, if the boilers be of a considerable size ;
but in the case of small boilers, the proportion is larger. The total
hv Google
THE PRACTICAL MODEL CALCULATOR,
III
Mi?.
lalJi
mi"
llslllis
P"*i.
I'S'S
hv Google
THS STEAM ENfllNE, 21T
heating surface of a two horse power wagon boiler is, according
to Fitzgerald's proportions, 30 square feet, or 15 ft. per horse
power ; whereas, in the case of a 45 horse power boiler the total
heating surface is 438 square feet, or 96 ft. per horse power.
The capaeity of steam room is 8j cubic feet per horse power, in
the two horse power boiler, and 5f cubic feet in the 20 horse power
boiler ; and in the larger class of boilers, such as those suitable for
30 and 45 horse power engines, the capacity of the steam room
does not fall below this amount, and indeed is nearer 6 than 5 cu
bic feet per horse power. The content of water is 18J cubic feet
per horse power in the two horse power boiler, and 15 cubic feet
per horse power in the 20 horse, power boiler. In marine boilers
about the same proportions obtain in most particulars. The ori
ginal boilers of one or two large steamers were proportioned
with about half a square foot of fire grate per horse power, and 10
square feet of flue and furnace surface, reckoning the total amount
as effective ; but in the boilers of other vessels a somewhat smaller
proportion of heating surface was adopted. In some cases we
have found that, in their marine flue boilers, 9 square feet of
flue and furnace surface are requisite to boil off a cubic foot of
water per hour, which is the proportion that obtains in some land
boilers ; hut inasmuch as in modern engines the nominal considera^
bly exceeds the actual power, they allow 11 square feet of heating
surface per nominal horse power in their marine boilers, and they
reckon, as effective heating surface, the tops of the flues, and the
whole of the sides of the flues, but not the bottoms. They have
been in the habit of allowing for the capacity of the steam space
in marine boilers 16 times the content of the cylinder ; hut as there
are two cylinders, this is equivalent to 8 times the content of both
cylinders, which ia the proportion commonly followed in land en
gines, and which agrees very nearly with the proportion of between
5 and 6 cubic feet of steam room per horse power. Taking, for
example, an engine with 23 inches diameter of cylinder and 4 feet
strobe, which will be 184 horse power — the area of the cylinder
will be 415*476 square inches, which, multiplied by 48, the number
of inches in the stroke, will give 19942848 for the capacity of the
cylinder in cubic inches ; 8 times this is 159542784 cubic inches,
or 923 cubic feet ; 923 divided by 184 is rather more than 5 cu
bic feet per horse power. There ia less necessity, however, that
the steam space should be large when the Sow of steam from the
boiler is very uniform, as it will be where there are two engines at
tached to the boiler at right angles with one another, or where the
engines work at a great speed, as in the case of locomotive engines.
A nigh steam' chest too, by rendering boiling over into the steam
pipes, or priming as it is called, more difficult, obviates the neces
sity for so large a steam space ; and the use of steam of a high
pressure, worked expansively, has the same operation ; so that in
modern marine boilers, of the tubular construction, where the whole
of these modifying circumstances exist, there is no necessity for so
hv Google
218
THE PRACTICAL MODEL CALCULATOR.
largo a proportion of Btcam room as 5 or 6 cubic feet per horse
power, and about half that amount more nearly represents the
general practice. Manj allow 064 of a Bquare foot per nomi
nal horse power of grate bars in their marine boilers, and a good
effect arises from this proportion ; but sometimes so large an area
of fire grate cannot be convenientlj got, and the proportion of
half a square foot per horse power seems to answer very well in
engines working with some expansion, and is now very widely
adopted. With this allowance, there will be about 22 square feet
of heating surface per square foot of fire grate ; and if the consump
tion of fuel be taken at 6 lbs. per nominal horse power per hour,
there will be 12 lbs. of coal consumed per hour on each square foot
of grate. The flues of all flue boilers diminish in their calorimeter
as they approach the chimney ; some very satisfactory boilers have
been made by allowing a proportion of 06 of a square foot of fire
grate per nominal horse power, and making the sectional area of
the flue at the largest part ^th of the area of fire grate, and the
smallest part, where it enters the chimney, ^th of the area of the
fire grate ; but in some of the boilers proportioned on this plan the
maximum sectional area is only ,?j or ^, according to the purposes
of the boiler. These proportions are retained whether the boiler is
fine or tubular, and from 14 to 16 square feet of tube surface is al
lowed per nominal horse power ; but such boilers, although they may
give abundance of steam, are generally, perhaps needlessly, bulky.
We shall therefore conclude our remarks upon the subject by
introducing a table of the comparative evaporative power of differ
ent kinds of coal, which will prove useful, by affording data for the
comparison of experiments upon different boilers when different
kinds of coal are used.
Table
of the Comparative. JSvaporative Power of different
of Goal.
kinds
K..
,..,....,0....
"Sl^
4
6
7
9
10
11
12
The best WeKh
Aatiracite \nierican
The beat small Pittsburgh
Arerage small Newcastle
PennaylTamm
Coke m Qaa works
W 1 h 9 d Newcastle, mixed ^ and 1
De Ij h Band.maUNewcaatle,iandi
A e ag large Newcastle
De hj h e
Blj he Main, NorthumberlanJ
<ti93
<H4
8 520
HU74
10 45
7 908
7 817
7 8tij
7 710
7()5&
6 772
6G0U
mgth of boilers. — The extension of the expansile method of
am to boilers of eiery denomination, and the gradual
introduction in connection therewith of a higher pieaauie than for
hv Google
THE STEAM EKGIt
219
merlj, makes the question of the strength of boilera one of great
and increasing importance. This topic was very successfully eluci
dated, a few yeara ago, by a committee of the Franklin Institute,
Philadelphia, and we shall here recapitulate a few of the more im
portant of the conclusions at which they arrived. Iron boiler plate
was found to increase in tenacity as its temperature was raised, un
til it reached a temperature of 550° above the fi'eeaing point, at
which point its tenacity began to diminish. The following table
exhibits the cohesive streogth at different temperatures.
82" t<
80° (he tenacity w
as — 58,000 lbs.
ot Iyth below its a
570=
= 6(1,500 lbs.
the madmum.
= 55.000 lbs.
the same nearly as
= 32,000 lbs.
nearly J of the mai:
124IP
— 22,000 lbs.
nearly + of the ma:
laiT"
— 0,000 lbs.
nearly l7tli of the
3000° i
■oa becomes fluid.
The difference in strength between strips of iron cut in the di
rection of the fibre, and strips cut across the grain, wiis found to
be about 6 per cent, in favour of the former. Repeated piling and
welding was found to increase the tenacity and closeness of the
iron, but welding together different kinds of iron was found to give
an unfavourable result; riveting plates was found to occasion a
diminution in their strength, to the extent of about onethird. The
accidental overheating of a boiler was found to reduce its strength
from 65,000 lbs. to 45,000 lbs. per square inch. Taking into ac
count all these contingencies, it appears expedient to limit the ten
sile force upon boilers in actua.l use to about 3000 lbs. per square
inch of iron.
Copper follows a different law, and appears to diminish in strength
by every addition of heat, reckoning from the freezing point. The
square of the diminution of strength seems to keep pace with the
cube of the temperature, as appears by tho following table : —
Table showing the Diminution of Strength of Copper Boiler
Plates by additions to the Temperature, the Cohesion at 32° being
32,800 U)s. per Square Ineh.
H»
TBmi«™w™
DimnotionoT
TenpsmniB
DLlDiDUliOBOt
Strensth.
1
90=
00175
g
660=
03425
180
00540
10
769
04398
270
00926
812
04fi44
4
01513
12
05&81
5
450
0204(i
13
081
06691
6
460
02183
14
1000
06741
7
513
0244G
1200
08801
8
629
02558
16
1300
10000
In the case of iron, the following are the results when tabulated
after a similar fashion.
b,Google
220 THE PRACTICAL MODEL CALCULATOR.
Table of ExperimenU on Iron Boiler Plate at High Tempera
ture; the Mean Maximum Tenacity leing at 550° = 65,000 ffis.
per Square Inch.
D,„„.u<,a=f
T«mp«r,(u„
Dim! notion «t
olatrveii.
ohtetvA.
550=
00000
824"
02010
570
00869
932
03324
696
00899
S47
03593
600
0096*
1030
04478
680
01047
1111
05514
562
01165
1156
06000
0'H36
1159
06011
732
01491
1187
06352
734
01535
1287
O6022
766
01589
1245
06715
770
01627
1317
07001
The application of stays to marine boilers, especially in those
parts of the water spaces which lie in the wake of the furnace bars,
has given engineers much trouble ; the f plate, of which ordinary
boilers are composed, is hardly thick enough to retain a stay with
security by merely tapping the plate, whereas, if the stay be ri
veted, the head of the rivet will in all probability be soon burnt
away. The best practice appears to bo to run the stays used for
the water spaces in this situation, in a line soniewhat beneath the
level of the bars, so that they may be shielded as much as possible
from the fire, while those which are required above the level of the
bars should be kept as nearly as possible towards the crown of the
furnace, so as to be removed from the immediate contact of the fire.
Screw bolts with a fine thread tapped into the plate, and with a
thin head upon the one side, and a thin nut made of a piece of
boiler plate on the other, appear to be the best description of stay
that has yet been contrived. The Stays between the sides of the
boiler shell, or the bottom of the boiler and the top, present little
difficulty in their application, and the chief thing that is to be at
tended to is to take care that there be plenty of them ; but we may
here remark that we think it an indispensable thing, when there is
any high pressure of steam to be employed, that the furnace crown
be stayed to the top of the boiler. This, it will be observed, is done
in the boilers of the Tagus and Infernal ; and we know of no better
E of staying than is afforded by those boilers.
AREA OF STEAM
Rule.— To the temperature of steam in the boiler add the con
stant increment 459 ; multiply the sum by 11025 ; and extract the
square root of the product. Multiply the length of stroke by the
number of strokes per minute ; divide the product by the square
root just found ; and multiply the square root of the quotient by
the diameter of the cylinder ; the product will be the diameter of
the steam passages.
hv Google
THE STEAM ENOINE. 221
Let it be required to determine the diameter of the steam pas
sages in an engine of which the diameter of the cylinder is 48
iDchea, the length of stroke 4J feet, and the number of strokes per
minute 26, supposing the temperature under which the steam is
generated to be 250 degrees of Fahrenheit's thermometer.
Here hy the rule we get s/JT025(250 + 459) = 2T9584 ; the
number of strokes ia 26, and the length of stroke 4 J feet ; hence
itia 6 = '^\ 27i)5.84 = 0"20456(i = 020456 X 48 = 9'819 inches;
so that the diameter of the steam passages is a little more than one
fifth of the diameter of the cylinder. The same rule will answer
for high and low pressure engines, and also for the passages into
the condenser.
LOSS or FOHCE BY THE DECKEASE OF TEMPERATrEE IN THE STEAM PIPES.
Rule. — From the temperature of the surface of the steam pipes
subtract the temperature of the external air ; multiply the remain
der by the length of the pipes in feet, and again by the constant
number or coefficient 1'68 ; then divide the product hy the diameter
of the pipe in inches drawn into the velocity of the steam in feet
per second, and the quotient will express the diminution of tem
perature in degrees of Fahrenheit's thermometer.
Let the length of the steam pipe be 16 feet and its diameter 5
inches, and suppose the velocity of the steam to be about 95 feet
per second, what will be the diminution of temperature, on the sup
position that the steam is at 250° and the external air at 60° of
Fahrenheit ?
Here, by the note to the above rule, the temperature of the sur
face of the steam pipe is 250 — 250 X 005 = 2375 ; hence we get
168 X 16(2375 — 60) ,r,r,AA j
/// = i—K— = 10044 degrees.
5 X 95
H we examine the manner of the composition of the above equa
tion, it will be perceived that, since the diameter of the pipe and
the velocity of motion enter as divisors, the loss of heat will be less
as these factors are greater ; but, on the other hand, the loss of
heat will be greater in proportion to the length of pipe and the
temperature of the steam. Since the steam is reduced frpm a
higher to a lower temperature during its passage through the steam
pipes, it mnst be attended with a corresponding diminution in the
elastic force; it therefore becomes necessary to ascertain to what
extent the force is reduced, in consequence of the loss of heat that
takes place in passing along the pipes. This is an inquiry of some
importance to the manufacturers of steam engines, as it serves to
guard them against a very common mistake into which they are
liable to fall, especially in reference to steamboat engines, where it
is usual to cause the pipe to pass round the cylinder, instead of
carrying it in the shortest direction from the boiler, in order to de
crease the quantity of surface exposed to the cooling effect of the
atmosphere.
hv Google
222 THE PRACTICAL MODEL CALCULATOR.
Rule. — From the temperature of the surface of the steam pipo
subtract the temperature of the external air ; multiply the remain
der by the length of the pipo in feet, aud again by the constant
fractional coefficient 0'00168 ; divide the product by the diameter
of the pipe in inches drawn into the velocity of steam in feet per
second, and subtract the quotient from unity ; then multiply the
difference thus obtained by the elastic force corresponding to the
temperature of steam in the boiler, and the product will be tlio
elastic force of the steam as reduced by cooling in passing through
the pipes.
Let the dimensions of the pipe, the temperature of the steam,
and its velocity through the passages, be the same as in the pre
ceding example, Tihat wili be the (quantity of reduction in the clastic
force occasioned by the effect of cooling in traversing the steam
pipe?
Since the elastic force of the steam in the boiler enters the equa
tion from which the above rule is deduced, it becomes necessary in
the first place to calculate its value ; and tbis is to be done by a
rule already given, ■which answers to the case in which the tempera
ture is greater than 212° ; thus we have
250 X 169856 = 424640
Constant number = 205526 add
Sum = 630166 log. 279945
Constant divisor = 3B3 log. 252^444 subtract
0277011 X 642 = 1778410,
which is the logarithm of 60036 inches of mercury.
Again, we have 250 ~ 005 X 250 = 2375 ; consequently, by
multiplying as directed in the rule, we get 2375 x 000168 x 16
= 6384, which being divided by 95 X 5 = 475, gives 001344 ; and
by taking this from unity and multiplying the remainder by the
elastic force as calculated above, the value of the reduced clastic
force becomes
/ = 60036 (1  001344) = 59229 inches of mercury.
The^loss of force is therefore 60036  59229 = 0807 inches of
mercury, which amounts to ^';th part of the entire elastic force of
the steam in the boiier as generated under the given temperature,
being a quantity of sufficient importance to claim the attention of
our engineers.
FEED "WATER.
The quantity of water required to supply the waste occasioned
by evaporation from a boiler, or, as it is technically termed, the
" feed water" required by a boiler working with any given pressure,
is easily determinable. For, since the relative volumes of water
and steam at any given pressure are known, it becomes necessary
merely to restore the quantity of water by the feed pump equiva
hv Google
THE STEAM ENOINB. Z)LA
lout to that abstracted in the form of steam, Tfhich the known rela
tion of the density to the pressure of the steam renders of easy
accomplishment. In practice, however, it is necessary that the
feed pump should be able to supply a much larger quantity of water
than what theory prescribes, as a great waste of water sometimes
occurs from' leakage or priming, and it is necessary to provide
against such contingencies. The feed pump is usually made of
such dimensions as to be capable of supplying 3^ times the water
that the boiler will evaporate, and in low pressure engines, where
the cylinder ia double acting and the feed pump single acting, this
proportion will he maintained by mating the pump a 240th of the
capacity of the cylinder. In low pressure engines the pressure in
the boiler may be taken at 5 lbs. above the pressure of the atmo
sphere, or 20 lbs. in all; and as high pressure steam is merely low
pressure steam compressed into a smaller compass, the size of the
feed pump relatively to the size of the cylinder must obviously vary
in the direct proportion of the pressnre. If, then, the feed pump
he l240th of the capacity of the cylinder when the total pressure
of the steam is 20 lbs., it must be l120th of the capacity of the
cylinder when the total pressure of the steam is 40 lbs., or 25 lbs,
above the atmosphere. This law of variation is expressed by the fol
lowing rule, which gives the capacity of feed pump proper for all
pressures : — Multiply the capacity of the cylinder in cubic inches by
the total pressure of the steam in lbs. per square inch, or the pressure
in lbs. per square inch on the safety valve, plus 15, and divide the
product by 4800; the quotient is the capacity of the feed pump in
cubic inches, when the feed pump is single acting and the engine
double acting. If the feed pump be double acting, or the engine
single acting, the capacity of the pump must be just onehalf what
is given by this rule.
CONDENSING WATER.
It was found that the most beneficial temperature of the hot
well was 100 degrees. If, therefore, the temperature of the
steam be 212°, and the latent Lcat 1000°, then 1212° may be
taken to represent the heat contained in the steam, or 1112°
if we deduct the temperature of the hot well. If the tempera
ture of the injection water be 50°, then 50 degrees of cold are
available for the abstraction of heat, and as the total quantity of
heat to be abstracted is that requisite to raise the quantity of water
in the steam 1112 degrees, or 1112 times that quantity, one degree,
it would raise onefiftieth of this, or 22'24 times the quantity of
water in the steam, 50 degrees. A cubic inch of water, therefore,
raised into steam, will require 22*24 cubic inches of water at 50
degrees for its condensation, and will form therewith 2324 cubic
inches of hot water at 100 degrees. It has been a practice to
allow about a wine pint (28'9 cubic inches) of injection water for
every cubic inch of water evaporated from the boiler. The usual
capacity for the cold water pump is ^'jth of the capacity of the
cylinder, which allows some water to run to waste. As a maximum
hv Google
224 THE PEACTICAL MODEL CALCUIATOE.
effect is obtained when the temperature of the hot well is about
100°, it will not be advisable to reduce it below that temperature
in practice. With the superior vacuum due to a temperature of 70°
or 80° the admission of so much cold water into the condenser
becomes necessary, — and which has afterwards to be pumped out
in opposition to the pressure of the atmosphere, — so that the gain in
the vacuum does not equal the loss of power occasioned by the
additional Joad upon the pump, and there is, therefore, a clear loss
by the reduction of the temperature below 100°, if such reduction
be caused by the admission of an additional quantity of water. If
the reduction of temperature, however, be caused by the use of
colder water, there is a gain produced by it, though the gain will
within certain limits be greater, if advantage be taken of the low
ness of the temperature to diminish the quantity of injection.
SAFETY VALVES.
Rule. — Add 459 to the temperature of the steam in degrees
of Fahrenheit ; divide the sum by the product of the elastic force
of the steam in inches of mercury, into its excess above the weight
of the atmosphere in inches of mercury ; multiply the square root
of the quotient by '0653 ; multiply this product by the number of
cubic feet per hour of water evaporated, and this last product is
the theoretical area of the orifice of the safety valve in square
inches.
'i'o apply this to an example — which, however, it must be remem
bered, will give a result much too small for practice.
Required the least area of a safety valve of a boiler suited for a
250 horse power engine, working with steam 6 lbs. more than the
atmosphere on the square inch.
In this case the total pressure is equal to 21 lbs. per square
inch ; and as in round numbers one pound of pressure is equal to
about two inches of mercury, it follows that / = 42 inches of
mercury.
It will be necessary to calculate t from formula (S) already given.
The operation is as follows :—
log. 42 ^ 642 = 1623249 ~ 642 = 0252842
constant coefficient = 196 2 292363
2545205
natural number = 350*92
constant temperature = 121
t = 22992
therefore J :f
J
'459 t( /459 +
/(/— 30)~"J 42 X
22992
12
. 1168;
■0653
X 1168 X N =
•075T N.
b,Google
.■043549; therefore J jf^zrin\ = ^ '042549 = 20628.
TOE STEAM EXGINE. 225
"We liave stated in a former part of this work that a cubic foot
of Tvater evaporated per hour is equivalent to one horse power;
therefore in this case N = 250 and x = 18925 sq. in.
As another example. Required the proper area of the safety
valve of a boiler suited to an engine of 500 horse power, when it
is wished that the steam should never acquire an elastic force
greater than 60 lbs. on the square inch above the atmosphere.
In this case the whole elastic force of the steam is 75 lbs. ; and
as 1 pound corresponds in round numbers to 2 inches of mercury,
it follows that / = 150. It will be necessary to calculate the
temperature corresponding to this force. The operation is as
follows : —
Log. 150 ^ 642 = 2176091 ^ 642 =
constant CO efficient = 196 log. 22
natural number = 427876 2631318
constant temperature = 121
required temperature 306876 degrees of Fahrenheit's scale
459 4 t 459 + 306876 765876 765896
therefore j^ (^f _ 30) " 150(150  30) ~ 150 x 120 ~ 18000
459 +t
Hence the required area = 0653 x 20628 X 500 =■ 01347 X
500 — 6735 square inches.
If the area of the safety valve of a boiler suited for an engine
of 500 horse power be required, when It is wished the steam should
never acquire a greater temperature than 300°, it will be necessary
to calculate the elastic force corresponding to this temperature ; and
by formula for this purpose, the required area = 0653 X 231 X
500 = 0151 X 500 — 755 square inches. It will be perceived
from these examples that the greater the elasticity and the higher
the corresponding temperature the less is the area of the safety
valve. This is just as might have been expected, for then the
steam can escape with increased velocity. We may repeat that the
results we have arrived at are much less than those used in practice.
For the sake of safety, the orifices of the safety valve are inten
tionally made much larger than what theory requires ; usually ^ of
a square inch per horse power is the ordinary proportion allowed
in the case of low pressure engines.
THE SLIDE VALVE.
The four following practical rules are applicable alike to short
slide and long D valves.
Rule I. — To find how much cover must be given on the steam
side in order to cut the steam off at any given part of the stroke.—
From the length of the stroke of the piston, subtract the length of
that part of the stroke that is to be made before the steam is cut
off. Divide the remainder by the length of the stroke of the
hv Google
226 THE PRACTICAL MODEL CALCULATOK.
piston, and extract the square root of the quotient. Multiply the
square root thus found by half the length of the stroke of the valve,
and from the product take half the !ead, and the remainder will be
the cover required.
Rule II, — To find at what part of the stroke any given aviount
of cover on the steam side will cut off the steam. — Add the cover
on the steam side to the lead ; divide the sum by half the length
of stroke of the valve. In a table of natural sines find the arc
■whose sine is equal to the quotient thus obtained. To this arc add
90° and from the sum of these two arcs subtract the arc whose
cosine is equal to the cover on the steam side divided by half the
stroke of the valve. Find the cosine of the remaining arc, add 1
to it, and multiply the sum by half the stroke of the piston, and
tho product is the length of that part of the stroke that will be
made by the piston before the steam is cut off.
Rule lU.^To find how muck before the end of the strolce, the
exhaustion of the steam in front of the piston will be cut off. — To
the cover on the steam side add the lead, and divide the sum by
half the length of the stroke of the valve. Find the arc whose sine
is equal to the quotient, and add 90° to it. Divide the cover on
the exhausting side by half the stroke of the valve, and find the arc
whose cosine is equal to the quotient. Subtract this are from the
one last obtained, and find the cosine of the remainder. Subtract
this cosine from 2, and multiply the remainder by half the stroke
of the piston. The product is the distance of the piston from tho
end of its strobe when the exhaustion is cut off.
Rule IV. — To find how far the piston is from the end of its
stroke, when the steam that is propelling it iy expansion is allowed
to escape to the condenser. — To the cover on the steam side add the
lead, divide the sum by half the stroke of the valve, and find the
arc whose sine is equal to the quotient. Find the arc whose cosine
is equal to the cover on the exhausting side, divided by half the
stroke of the valve. Add these two arcs together, and subtract
90°. Find the cosine of the residue, subtract it from 1, and mul
tiply the remainder by half the stroke of the piston. The product
is the distance of the piston from the end of its stroke, when the
steam that is propelling it is allowed to escape to the condenser.
In using these rules, all the dimensions are to be taken in inches,
and the answers will be found in inches also.
From an examination of the formulas we have given on this
subject, it will be perceived (supposing that there is no lead) that
the part of the stroke where the steam is cut off, is determined by
the proportion which the cover on the steam side bears to the
length of the stroke of the valve : so that in al! cases where the
cover bears the same proportion to the length of the stroke of the
valve, the steam will be cut off at the same part of the stroke of
the piston.
In the first line, accordingly, of Table I., will be found eight
different parts of the stroke of the piston designated ; and directly
b,Google
THE STEAM EKGINB.
227
below each, in the second line, is given the quantity of coyer requi
site to cause the steam to be cut off at that particular part of the
stroke. The different sizes of the cover are given in the second
line, in decimal parts of the length of the stroke of the valve ; so
that, to get the quantity of cover corresponding to any of the given
degrees of expansion, it is only neceaaary to take the decimal in
the second line, which stands under the fraction in the first, that
marks the degree of expansion, and multiply that decimal by the
length you intend to make the stroke of the valve. Thus, suppose
you have an engine in which you wish to have the steam cut off
when the piston is a quarter of the length of its stroke from the
end of it, look in the table, and you will find in the third column
from the left, J. Directly under that, in the second line, you have
the decimal '250. Suppose that you think 18 inches will be a con
venient length for the stroke of the valve, multiply the decimal
■250 by 18, which gives 4^. Hence we learn that with an 18 inch
stroke for the valve, 4^ inches of cover on the steam side will cause
the steam to be cut off when the piston has still a quarter of its
stroke to perform.
Half the stroke of the valve must always be at least equal to the
cover on the steam side added to the breadth of the port. By the
"breadth" of the port, we mean its dimension in the direction of
the valve's motion ; in short, its perpendicular depth when the
cylinder is upright. 'Jhe words "cover" and "lap" are synony
mous. Consequently, as the cover, in this case, must be 4^ inches,
and as half the stroke of the valve is 9 inches, the breadth of the
port cannot be moro than (9 — 4^ = 4^) 4 inches. If this
breadth of port is not enough, we must increase the stroke of the
valve ; by which means we shall get both the cover and the breadth
of the port proportionally increased. Thus, if we make the length
of valve stroke 20 inches, we shall have for the cover 250 x 20 = 5
inches, and for the breadth of the port 10 — 5 = 5 inches.
Table I.
Distance of the piston from T
tlie termination of its
stro];e, wtien tlia steam L
is cut off, in parts of tile
lengtli of its stroke. J
i
A
1
A
A
i
ft
I
A
A
■]02
Coyerontliesteamsideof
tile valTO, in deeiraal
parts of tlielongtliofits
strolie.
■289
•2T0
•250
■228
■204
■m
■144
This table, as we have already intimated, is computed on the
supposition that the valve is to have no lead ; but, if it is to have
lead, all that is necessary is to subtract half the proposed lead from
the cover found from the table, and the remainder will be the
hv Google
228
THE PRACTICAL MODEL CALCULATOR.
proper quantity of cover to give to the valve. Suppose tbat, in
the last example, the valve was to have J inch of lead, wo would
subtract J inch from the 5 inches found for the cover by the table :
that would leave 4 inches for the quantity of cover that the valve
ought to have.
Table II.
iSi.fi.'
c.
r rtqulTKi
"' °'nSd
"alS^i'fiv
I'lTZZ
 —
«— ^
..,
" Iflow"'
I
A
i
i.
I
J
i\
^h
21
694
648
6^00
5 47
4 90
425
347
245
23i
679
634
479
4^16
3^39
23
665
G21
575
5'24
469
4>07
832
234
22i
650
607
5^62
51S
459
325
229
22'
636
594
5^&0
502
■4'49
3'89
313
224
2i}
621
580
538
490
439
3 80
810
219
21
607
567
625
479
428
3^72
303
214
aoj
593
553
512
4'67
418
363
2 09
20
578
540
500
456
4 08
S^54
204
19i
6'64
526
487
445
345
282
199
Id''
549
518
47B
433
388
836
2 74
194
131
534
462
377
327
267
18*
520
450
410
8'67
319
2^60
1.83
171
5oa
472
437
399
a 57
3 10
25S
178
17''
491
400
425
347
801
2^45
173
lej
477
445
412
3'76
8'36
3'92
168
16
462
4^32
400
865
8^26
283
231
163
151
448
4'18
3 ■53
3^16
2'74
224
158
15^
405
375
342
306
265
216
153
l*i
419
3 91
862
8'31
257
209
148
1/
405
378
350
319
286
248
202
143
iSi
390
364
837
808
275
239
195
137
is'
3.76
8'5I
3^25
296
265
280
1B2
12i
361
337
812
285
£■55
221
1>80
1.27
1/
847
324
300
3 '74
245
212
173
123
111
332
B'lO
287
2^62
208
1^66
1^17
11^
318
297
275
2^51
224
195
158
1^12
101
S03
283
263
2^14
161
107
10*
270
250
2^28
304
177
1^44
I^02
9i
265
356
237
2^17
193
188
132
•96
9^
260
248
225
205
184
159
ISO
■92
Si
246
229
212
194
173
1'50
1^28
■86
8*
231
216
200
182
168
142
115
■81
'i
2'16
202
187
171
163
108
■76
7*
203
189
175
160
148
124
101
■71
6J
188
176
1^62
148
1'32
1'15
■94
66
6*
173
162
150
137
122
106
■61
H
148
137
125
1'12
■79
■SB
6
144
135
125
1^14
102
■72
51
n
1.30
121
112
1^03
■92
■80
■65
■46
4'
116
100
■91
■82
■71
58
■41
H
101
■94
87
■80
■71
62
50
■35
3
86
■81
■75
■68
■61
53
44
■80
Tablo 11. is an extension of Table I. .for the purpose of obviating,
in most cases, the necessity of even the very small degree of
trouble required in multiplying the stroke of the valve by one of
the decimals in Table I. The first line of Table II, consists, as in
Table I., of eight fractions, indicating the various parts of the stroke
hv Google
THE STEAM ENGTSE. 229
at whicli the steam may be cut off. The first column on the left
hand consists of various numbers that represent the different
lengths that may bo given to the stroke of the valve, diminishing,
by halfinches, from 24 inches to 3 inches. Suppose that you wish
the steam cut off at any of the eight parts of the stroke indicated
in the first line of the table, (say at ^ from the end of the stroke,}
you find J at the top of the sixth column from the left. Look for
the proposed length of stroke of the valve (say 17 inches) in the
first column on the left. From 17, in that column, run along the
line towards the right, and in the sixth column, and directly under
the I at the top, you will find 3'47, which is the cover required to
cause the steam to be cut off at J from the end of the stroke, if the
valve has no lead. If you wish to give it lead, (say J inch,) sub
tract the half of that, or  = ISS inch from 347, and you will have
S47 — '125 = 3345 inches, the quantity of cover that the valve
should have.
To find the greatest breadth that we can give to the port in this
case, we have, as before, half the length of stroke, 8^— 3'34.'i=5'155
inches, which is the greatest breadth wo can give to the port with
this length of stroke. It is scarcely necessary to observe that it is
not at aH essential that the port should be so broad as this ; indeed,
where great length of stroke in the valve is not inconvenient, it ia
always an advantage to make it travel farther than is just neces
sary to make the port full open ; because, when it travels farther,
both the exhausting and steam ports are more quickly opened, so
as to allow greater freedom of motion to the Steam.
The manner of using this table is so simple, that we need not
trouble the reader with more examples. We pass on, therefore, to
explain the use of Table III.
Suppose that the piston of a steam engine is making its down
ward stroke, that the steam is entering the upper part of the cylin
der by the upper steamport, and escaping from below the piston
by the lower exJiaustJngport; then, if (as is generally the case)
the slide valve has some cover on the steam side, the upper port
will be closed before the piston gets to the bottom of the stroke,
and the steam above then acts expansively, while the communica
tion between the bottom of the cylinder and the condenser still
continues open, to allow any vapour from the condensed water in
the cylinder, or any leakage past the piston, to escape into the
condenser ; but, before the piston gets to the bottom of the cylin
der, this passage to the condenser will also be cut off by the valve
closing the lower port. Soon after the lower port is thus closed,
the upper port will be opened towards the condenser, so as to allow
the Steam that has been acting expansively to escape. Thus, be
fore the piston has completed its stroke, the propelling power is
removed from behind it, and a resisting power is opposed before it,
arising from the vapour in the cylinder, which has no longer any
passage open to the condenser. It is evident, that if there is no
cover on the exhausting side of the valve, the exhausting port before
hv Google
230 THE PRACTICAL MODEL CALCULATOR.
the piston will te closed, and the one hehind it opened, at the same
time ; but, if there is any cover on the exhausting siJe, the port
before the piston will be closed before that behind it is opened ; and
the interval between the closing of the one and the opening of
the other will depend on the quantity of cover on the exhausting
side of the valve. Again, the position of the piston in the cylin
der, when these ports are closed and opened respectively, will
depend on the quantity of cover that the valve has on the steam
side. If the cover is large enough to cut the steam off when the
piston is yet a considerable distance from the end of its stroke,
these ports will be closed and opened at a pro port ion ably early part
of the stroke ; and when it is attempted to obtain great expansion
by the slidevalve alone, without an expansion valve, considerable
loss of power is incurred from this cause.
Table III. is intended to show the parts of the stroke where, un
der any given arrangement of slide valve, these ports close and open
respectively, so that thereby the engineer may be able to estimate
how much of the efficiency of the engine he loses, while he is trying
to add to the power of the steam by increasing the expansion in
this manner. In the table, there are eight double columns, and at
the heads of these columns are eight fractions, as before, represent
ing so many different parts of the stroke at which the steam may
be supposed to be cut off.
In the lefthand single column in each double one, are four deci
mals, which represent the distance of the piston (in terms of the
length of its stroke) from the end of its stroke when the exhausting
port before it is opened, corresponding with the degree of expansion
indicated by the fraction at the top of the double column and tbe
cover on the exhausting side opposite to these decimals respectively
in the lefthand column. The righthand single column in each
double one contains also each four decimals, which show in the same
way at what part of the stroke the exhausting port behind the pis
ton is opened. A few examples will, perhaps, explain this best.
Suppose we have an engine in which the slide valve is made to
cut the steam off when the piston is l3d from the end of its stroke,
and that the cover on the exhausting side of the valve is l8th of
the whole length of its stroke. Let the stroke of tho piston be 6
feet, or 72 inches. We wish to know when the exhaustingport
before the piston will be closed, and when tho ono behind it will be
opened. At tho top of the lefthand double column, the given de
gree of expansion (l3d) is marked, and in the extreme left column
we have at the top the given amount of cover (l8tb). Opposite the
l8th, in the first double column, we have ITS and '033, which
decimals, multiplied respectively by 72, the length of the stroke,
will give the required positions of the piston : thus 72X'178=12'8
inches = distance of the piston from the end of the stroke when the
exhaustingport iefore the piston is shut ; and 72 X 033 = 238
inches = distance of the piston from the end of its stroke when the
exhaustingport behind it is opened.
hv Google
THE STEAM ENGINE.
^ 'I t ^
C'uvcr on the eshoostins siiio of the Tfilvu In parts of
Uifl length oC its slrolie.
i £ i *
DiBtance of the piaton from the end of ite
Et/oTiB, when the eihauetingport before
it ia ghat (in ports of the strolte).
M
lilt
Dialanee of the piaton from the end of it!
stroke, "hen the eshanslinKport behind
it is opened (in pnrta of the atroke).
■161
■113
■101
■082
Dialnnce of the piston from the end of ita
stroke, iFhcn the exhaustingport before
it is shut (in parta of the stroke).
i i i i
Diatoncc of the piaton from the end of ita
stroke, when the exhansticgport behind
it ia opened (in parts of the stroke).
143
■100
■085
■067
Distance of the piston from the end of ita
stroke, when the exhaostingport before
it ia ahut (in parts of the stroke).
mi
'III,
* 1 i 1
Distance of the piaton from tho end of its
stroke, when the exboo stingport behind
it is opened (in pnrts of the stroke).
i i i i
Diatonee of the piaton fhjm the end of its
stroke, when the eihanating.port before
it is shut (in parts of the attcke).
'Hi
■012
■030
042
■055
DiEtance of the piston from the end of its
stroke, nhen the exboustingport behind
it is opened (in parts of the stroke).
•109
■071
053
■043
Distance of the piaton from tlie end of its
stroke, wlien the eshauetingport before
it ia sliHt (in purle of the stroke).
Hi!
i i i i
DisLnnee of the pistno fi'om the end of its
eiiiike, when tlio exhaHstitiepof'el'md
it is opened (in parti of the ptroke).
■093
■058
■043
■033
Distance of the piston fVom the end of ita
Etii)ke. nlipn tlie p:itl;aHf iinaport before
it ia ahnt (in parts of the stroke).
m
" g.
list
Distance of the piston from the end of its
stroke, when the cKhauslingport hehind
it ia ojicned (in pnrta of the stroke).
074
043
■022
Distonee of the piaton from the end of its
Etroke, when the oiiiau stingport before
it is ahnt (in parts of the stroke.)
Ills
■001
■008
■013
■022
Distonca of the piston Irom the end of its
it ia opened (in parte of the stroke).
■053
■027
■024
on
Distance of the piston from the end of its
it ia abut (in ptu'ta of the etroke).
■001
■003
■004
•on
Distance of the piston from the end of its
it is opened (in pnrts of the stroke).
b,Google
2aa THE PRACTICAL MODEL CALCULATOR.
To take another example. Let the strolte of the valve he 16
inches, the cover on the exhausting eido J inch, the cover on the
steam side 3 J inches, the length of the stroke of the piston 60 inches.
It is required to ascertain all the particulars of the working of this
valve. The cover on the exhausting side is evidently ^ of the
length of the valve stroke. Again, looking at 16 in the lefthand
column of Table II., we find in the same horizontal line 326, or very
nearly 3J under ^ at the head of the column, thus showing that the
steam will he out ofi" at J from the end of the stroke. Again, under
J at the head of the fifth double column from the left in Table III.,
and in a horizontal line with ^ in the lefthand column, we have
■053 and 033. Hence, 053 X 60 = 318 inches = distance of the
piston from the end of its stroke when the exhaustingport before
it ia shut, and '033 X 60 = 1'98 inches = distance of the piston
from the end of its stroke when the exhausting port behind it is
opened. If in this valve the cover on the exhausting side were
increased (say to 2 inches, or  of the stroke,) the effect would bo to
make the port before the valve be shut sooner in the proportion of
■109 to "053, and the port behind it later in the proportion of ■OOS
to '033 (see Table III.) Whereas, if the cover on the exhausting
side were removed entirely, the port before the piston would be
shut and that behind it opened at the same time, and {see bottom
of fifth double column. Table III.) the distance of the piston from
the end of its stroke at that time would be 043 X 60 = 258 inches.
An inspection of Table III. shows us the effect of increasing the
expansion by the slidevalve in augmenting the loss of power oeca
Bioned by the imperfect action of the eduction passages. Referring
to the bottom line of the table, we see that the eduction passage
before the piston is closed, and that behind it opened, {thus destroy
ing the whole moving power of the engine,) when the piston is ■092
from the end of its stroke, the steam being cut off at J from the
end. Whereas, if the steam is only cut off at ^^ from the end of
the stroke, the moving power is not withdrawn till only Oil of the
stroke remains uncompleted. It will also be observed that in
creasing the cover on the exhausting side has the effect of retaining
the action of the steam longer heldnd the piston, but it at the same
time causes the eductionport hefore it to be closed sooner.
A very cursory examination of the action of the slide valve is
sufficient to show that the cover on the steam side should always be
greater than on the exhausting side. If they are equal, the steam
would be admitted on one side of the piston at the same time that
it was allowed to escape from the other ; but universal experience
has shown that when this is the case, a very considerable part of
the power of the engine is destroyed by the resistance opposed to
the piston, by the exhausting steam not getting away to the con
denser with sufficient rapidity. Hence we see the necessity of
the cover on the exhausting side being always less than the cover on
the steam side ; and the difference should be the greater the higher
the velocity of the piston is intended to be, because the quicker the
hv Google
THE STEAM EHGIKB. 2i!3
piston moves tlie passage for the waste steam requires to be the
larger, so as to admit of its getting away to the condenser with as
great rapidity as possible. In locomotive or other engines, where
it 18 not wished to expand the steam in the cylinder at all, the slide
valve is sometimes made with very little cover on the steam side :
and in these circumstances, in order to get a sufficient difference
between the cover on the steam and exhausting sides of the valve,
it may be necessary not only to take away all the cover on the
exhausting side, but to take off still more, so as to make both ex
hausting passages be in some degree open, when the valve is at the
middle of its stroke. This, accordingly, is sometimes done in such
circumstances as we have described ; hut, when there is even a small
degree of cover on the steam side, this plan of taking more than all
the cover off the exhausting side ought never to be resorted to, as
it can serve no good purpose, and will materially increase an evil
we have already explained, viz. the opening of the exhaustingport
behind the piston before the stroke is nearly completed. The tables
apply equally to the common short slide threeported valves and to
the long D valves.
In fig, 1 is exhibited a common arrangement of the valves in lo
hv Google
234 THE PRACTICAL MODEL CALCULATOR.
comotive engines, and in figa. 2 aDd 3 is shown an anangement
for working valves by a shifting cam, by which the amount of ex
piinsion may be varied. This particular arrangement, however, is
anti(juated, and is now but little used.
The extent to which expansion can be carried beneficially by
means of lap upon the valve is about onethird of the stroke ; that
is, the valve may be made with so much lap, that the steam will be
cut off when onethird of the stroke has been performed, leaving
the residue to be accomplished by the agency of the expanding
steam ; but if more lap be put on than answers to this amount of
expansion, a very distorted action of the valve will be produced,
which will impair the efficiency of the engine. If a furtlier amount
of expansion than this is wanted, it may be accomplished by wire
drawing the steam, or by so contracting the steam passage, that
tiie pressure within the cylinder must decline when the speed of
the piston is accelerated, as it is about the middle of the stroke.
Thus, for example, if the valve be so made as to shut off the steam
by the time twothirds of the stroke have been performed, and the
steam be at the same time throttled in the steam pipe, the full
pressure of the steam within the cylinder cannot be maintained ex
cept near the beginning of the stroke where the piston travels
slowly ; for as the speed of the piston increases, the pressure neces
sarily subsides, until the piston approaches the other end of the
cylinder, where the pressure would rise again but that the operation
of the lap on the valve by this time has had the effect of closing
the communication between the cylinder and steam pipe, so as to
prevent more steam from entering. By throttling the steam, there
fore, in the manner here indicated, the amount of expansion due to
the lap may be doubled, so that an engine with lap enough upon
the valve to cut off the steam at twothirds of the stroke, may, by
the aid of wiredrawing, be virtually rendered capable of cutting
off the steam at onethird of the stroke. The usual manner of cut
ting off the steam, however, is by means of a sepaiate valve, termed
an expansion valve ; but such a device appears to be hardly neces
sary in many engines. In the Cornish engines, where the steam
is cut off in some cases at onetwelfth of the stroke, a separate valve
for the admission of steam, other than that which permits its es
cape, is of course indispensable ; but in common rotative engines,
which may realize expansive efiicacy by throttling, a separate ex
pansive valve does not appear to be required. In all engines thero
is a point beyond which expansion cannot be carried with advantage,
as the resistance to be surmounted by the engine will then become
equal to the impelling power ; but in engines working with a high
pressure of steam that point is not so speedily attained.
In high pressure, as contrasted with condensing engines, there is
always the loss of the vacuum, which will generally amount to 12
or 13 lbs. on the square inch, and in high pressure engines there is
a benefit arising from the use of a very high pressure over a pres
sure of a moderate account. In all high pressure engines, there is
hv Google
THE STEAM EKGINE. 235
a diminution in the power caused by the counteracting pi^ssure of
the atmosphere on the educting side of the piston ; for the force
of the piston in its descent would obviously be greater, if there was
a vacuum beneath it ; and the counteracting pressure of the atmo
splieve is relatively less when the steam used is of a very high
pressure. It is clear, that if we bring down the pressure of the
steam in a high pressure engine to the pressure of the atmosphere,
it will not exert any power at all, whatever quantity of steam may
be expended, and if the pressure be brought nearly as low as that
of the atmosphere, the engine will exert only a very small amount
of power ; whereas, if a very high pressure be employed, the pres
sure of the atmosphere will become relatively as small in counter
acting the impelling pressure, as the attenuated vapour in the con
denser of a condensing engine is in resisting the lower pressure
which is there employed. Setting aside loss from friction, and sup
posing the vacuum to be a perfect one, there would be no benefit
arising from the use of steam of a high pressure in condensing en
gines, for the same weight of steam used without expansion, or
with the same measure of expansion, would produce at every pres
sure the same amount of mechanical power. A piston with a
square foot of area, and a stroke of three feet with a pressure of
one atmosphere, would obviously lift the same weight through the
same distance, as a cylinder with half a square foot of area, a stroke
of three feet, and a pressure of two atmospheres. In the one case,
we have three cubic feet of steam of the pressure of one atmosphere,
and in the other case IJ cubic feet of the pressure of two atmo
spheres. But there is the same weight of steam, or the same quan
tity of heat and water in it, in both cases ; so that it appears a given
weight of steam would, under such circumstances, produce a definite
amount of power, without reference to the pressure. In the case
of ordinary engines, however, these conditions do not exactly apply ;
the vacuum is not a perfect one, and the pressure of the resisting
vapour becomes relatively greater as the pressure of the steam is
diminished ; the friction also becomes greater from the necessity
of employing larger cylinders, so that even in the case of condensing
engines, there is a benefit arising from the use of steam of a con
siderable pressure. Expansion cannot be carried beneficially to any
great extent, unless the initial pressure be considerable; for if steam
of a low pressure were used, the ultimate tension would be reduced
to a point SO nearly approaching that of the vapour in the con
denser, that the difference would not suffice to overcome the friction
of the piston ; and a loss of power would be occasioned by carrying
expansion to such an extent. In some of the Cornish engines, the
steam is cut off at onetwelfth of the stroke ; but there would be a
loss arising from carrying the expansion so far, instead of a gain,
unless the pressure of the steam were considerable. It is clear,
that in the case of engines which carry expansion very far, a very
perfect vacuum in the condenser is more important than it is in
other cases. Sothiug can be easier than to compute the ultimate
hv Google
236 TIIE PRACTICAL MODEL CALCULATOR.
pressure of expanded stcain, so as to see at what point expansion
ceiises to be productive of benefit ; for as tlie pressure of expanded
steam is inversely as the space occupied, the terminal pressure nhen
the expansion is twelve times is just onetwelfth of wha,t it was at
first, and so on, in all other projections. The total pressure should
be taken as the initial pressure — not the pressure on the safety
valve, but that pressure plus the pressure of the atmosphere.
In high pressure engines, working at from 70 to 90 lbs. on the
square inch, as in the case of locomotives, the efficiency of a given
quantity of water raised into steam may be considered to be about
the same as in condensing engines. If the pressure of steam in a
high pressure engine be 120 lbs., or 125 lbs. above the atmosphere,
then the resistance occasioned hy the atmosphere will cause a loss
of th of the power. If the pressure of the steam in a low pressure
engine be 16 lbs, on the square inch, or 11 lbs. above the atmo
sphere, and the tension of the vapour in the condenser be equiva
lent to 4 inches of mercury, or 2 lbs. of pressure on the square
inch, then the resistance occasioned by this rare vapour will also
cause a loss of th of the power. A high pressure engine, there
fore, with a pressure of 105 lbs. above the atmosphere, works with
only the same loss from resistance to the piston, as a low pressure
engine with a pressure of 1 lb. above the atmosphere, and with
these proportions the power produced by a given weight of steam
will be the same, whether the engine be high pressure or con
densing.
SPOEEOIDAL CONDITION OF WATER IN BOILERS.
Some of the more prominent causes of boiler explosions have
been already enumerated ; but explosions have in some cases been
attributed to the spheroidal condition of the water in the boiler,
consequent upon the flues becoming redhot from a deficiency of
water, the accumulation of scale, or otherwise. The attachment
of scale, from its imperfect conducting power, will cause the iron
to be unduly heated ; and if the scale be accidentally detached, a
partial explosion may occur in consequence. It is found, that a
sudden disengagement of steam does not immediately follow the
contact of water with the hot metal, for water thrown upon red
hot iron is not immediately converted into steam, but assumes the
spheroidal form and rolls about in globules over the surface. These
globules, however high the temperature of the metal may be on
which they are placed, never rise above the temperature of 205°,
and give off but very little steam ; but if the temperature of the
metal be lowered, the water ceases to retain the spheroidal form,
and comes into intimate contact with the metal, whereby a rapid
disengagement of steam takes place. If water be poured into a very
hot copper flask, the flask may be corked up, as there will be scarce
any steam produced so long as the high temperature is maintained;
but so soon as the temperature is sufl'ered to fall below 350" or
400°, the spheroidal condition being no longer maintainable, steam
is generated with rapidity, and the cork will be projected from the
hv Google
MB STEAM ENOIKB. 237
mouth of t!ie flask with great force. In a hoiler, no doubt, where
there is a considerable head of water, the repellant action of the
spheroidal globules will be more cifcctuallj counteracted than in
the small vessels employed in experimental researches. But it is
doubtful whether in all boilers there may not be something of the
spheroidal action perpetually in operation, and leading to efi"ects at
present mysterious or inexplicable.
One of the most singular phenomena attending the spheroidal
condition is, that the vapour arising from a spheroid is of a far
higher temperature than the spheroid itself. Thus, if a thermometer
be held in the atmosphere of vapour which surrounds a spheroid of
water, the mercury, instead of standing at 205°, as would be the
case if it had been immersed in the spheroid, will rise to a point
determinable by the temperature of the vessel in which the spheroid
exists. In the case of a spheroid, for example, existing within a
crucible raised to a temperature of 400°, the thermometer, if held
in the vapour, will rise to that point ; and if the crucible be made
redhot, the thermometer will be burst, from the boiling point of
mercury having been exceeded. A part of this effect may, indeed,
be traced to direct radiation, yet it appears indisputable, from the
experiments which have been made, that the vapour of a lii;[uid
spheroid is much hotter than the spheroid itself.
EXPANSION.
At page 131 we have given a table of hyperbolic or Byrgcan
logarithms, for the purpose of facilitating computations upon this
subject.
Let the pressure of the steam in the boiler be expressed by unity,
and let x represent the space through which the piston has moved
whilst urged by the expanding steam. The density will then be
  , and, assuming that the densities and elasticities are pro
portionate, will be the differential of the efficiency, and the
efficiency itself will be the integral of this, or, in other words, the
hyperbolic logarithm of the denominator ; wherefore the efficiency
of the whole stroke will be 1 + log. (1 + ar).
Supposing the pressure of the atmosphere to be 15 lbs., 15 + 35
= 50 lbs,, and if the steam bo cut off at i^th of the stroke, it will be
expanded into four times its original volume ; so that at the ter
mination of the stroke, its pressure will be 50^4=122 lbs., or 28
lbs. less than the atmospheric pressure.
When the steam is cut off at onefourth, it is evident that « = 3.
In such ease the efficiency is
_ 1 f log. (1 + 3), or 1 + log. 4.
The hyperbolic logarithm of 4 is 1386294, ao that the efficiency
of the steam becomes 2386294 ; that is, by cutting off the steam
at J, more than twice the effect is produced with the same consump
tion of fuel ; in other words, onehalf of the fuel is saved.
hv Google
238 THE PRACTICAL MODEL CALCULATOE.
This result may thus be expressed in words : — Divide the length
of the stroke through which the steam expands by the length of
stroke performed with the full pressure, which last portion call 1 ;
the hyperbolic logarithm of the quotient is the increase of efficiency
due to expansion. We introduce on the following page more de
tailed tables, to facilitate the computation of the power of an en
gine working expansively, or rather to supersede the necessity of
entering into a computation at all in each particular case.
The first column in each of the following tables contains the
initial pressure of the steam in pounds, and the remaining columns
contain the mean pressure of steam throughout the stroke, with the
different degrees of expansion indicated at the top of the columns,
and which express the portion of the stroke during which the steam
acts expansively. Thus, for example, if steam be admitted to the
cylinder at a pressure of 3 pounds per square inch, and be cut off
within ^th of the end of the stroke, the mean pressure during the
whole stroke will be 296 pounds per square inch. In like manner,
if steam at the pressure of 3 pounds per square inch were cut off
after the piston had gone through ^th of the stroke, leaving the
steam to expand through the remaining th, the mean pressure
during the whole stroke would be 1'164 pounds per square inch.
The friction of iron sliding upon brass, which has been oiled and
then wiped dry, so that no film of oil is interposed, is about ^ of
the pressure ; but in machines in actual operation, whore there is a
film of oil between the rubbing surfaces, the fraction is only about
onethird of this amount, or ^'jd of the weight. The tractive re
sistance of locomotives at low speeds, which is entirely made up of
friction, is in some cases jj^th of the weight ; but on the average
about ^Jijth of the load, which nearly agrees with mj former state
ment. If the total friction bo jj^th of the load, and the rolling
friction be niViith of the load, then the friction of attrition must be
^Jijth of the load ; and if the diameter of the wheels be 36 in., and the
diameter of the axles be 3 in., which are common proportions, the
friction of attrition must bo increased in the proportion of 36 to 3,
or 12 times, to represent tho friction of the rubbing surface when
moving with the velocity of the carriage, ^ths are about ^jth of
the load, which does not differ much from the proportion of ^'^d, as
previously stated. While this, however, is the average result, the
friction is a good deal less in some cases. Engineers, in some
experiments upon the friction, found tho friction to amount to
less than ^th of the weight ; and in some experiments upon the
friction of locomotive axles, it was found that by ample lubrication
the friction might be made as little as ^th of the weight, and the
traction, with the ordinary size of wheels, would in such a case be
about sJoth of the weight. The function of lubricating substances
is to prevent the rubbing surfaces from coming into contact, where
by abrasion would be produced, and unguents are effectual in this
hv Google
THE STEAM EXOINB. 239
EXPANDED STEAM. — MEAN PRESSURE AT DIrrEEEXT DENSITIES AND
KATE OE EXPAKSIOS.
Tli« column headed contains ike initial preasare in lbs., and the remaining columns
contain the mean pressure in lbs., teUh different grades of expansion.
BIBK^BS
1
i
1
§
f
1
1 1 i 1
a
296
289
275
253
222
1789
1154
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367
338
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4818
4593
4232
3708
2982
1924
5937
5782
5512
5079
4450
8579
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7
6927
6746
6431
5925
5241
4175
2694
8
7917
7710
7350
6772
69S4
4773
3079
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8906
8673
8268
7618
6675
5868
8468
9637
9187
B465
7417
5965
3848
11
10885
10601
10106
9311
8159
6561
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11875
11565
10925
10158
8901
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12865
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11943
11004
9642
7754
5003
14
13864
13492
12863
11851
10384
8631
lf>
14844
14456
18781
12697
11126
8947
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15834
15420
14700
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ia823
16383
15618
14390
12609
10140
C542
IK
17818
17347
16537
15237
13351
10787
6927
IH
18702
18811
17448
10808
14093
11833
7312
m
19792
19275
18876
16930
14835
11930
7697
m
24740
24093
22968
21162
18543
14912
9621
m
29688
28912
27562
25396
22253
11546
34G36
83731
88166
29627
25961
20877
18470
23860
15395
4S
44533
41343
38092
33378
26842
17319
5U
49481
48187
45987
42325
37067
29825
19243
BiEisam
,.,T=.
A
1%
A
A
^
A
A
A
A
a
2980
2930
2880
2710
2589
2299
1981
1608
0990
3974
8913
3780
8614
8065
2612
2087
1820
4892
4232
5901
5870
5670
5421
5079
4698
8180
1981
7
6955
6848
6615
6325
5925
5364
4624
3652
2311
7948
7560
7228
6772
6131
6284
4174
2641
8943
7618
6897
5945
10
9936
9784
S450
9036
8465
7664
6606
5218
8802
11
10929
10763
10395
9989
9311
8480
7266
5789
3632
1'',
11740
11340
10843
10158
10994
9963
8687
11851
10729
9248
15
14904
14676
14175
13554
12697
11496
9909
16
15897
15654
15120
14457
18544
12263
10569
8348
6283
17
16801
16066
16861
14051
13028
11230
8870
5013
17884
17611
15237
13795
11890
6944
1!i
18878
18589
17955
17168
16088
14661
12551
yi;
19872
19668
18900
18072
I6S80
15828
13212
10430
6600
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24840
24460
21102
19160
16515
13040
8256
ar
29808
29352
28850
27108
25395
22092
19818
15654
3)1
34776
34244
33076
31626
29627
26824
23121
18263
11557
41
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39186
87800
86144
38860
30656
26224
20872
13208
45
44912
44028
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34888
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23481
14859
50 [ 49680
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47250
45180
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33030
b,Google
240 THE PRACTICAL MODEL CALCULATOE.
respect in the proportion of their viscidity ; but if the viscidity of
the unguent he greater than what suffices to keep the surfaces
asunder, an additional resistance will be occasioned ; and the nature
of the unguent selected should always have reference, therefore, to
the size of the rubbing surfaces, or to the pressure per square inch
upon them. With oil, the friction appears to he a minimum when
the pressure on the surface of a bearing is about 90 lbs. per square
inch : the friction from too small a surface increases twice as rapidly
as the friction from too large a surface ; added to which, the bear
ing, when the surface is too small, wears rapidly away. For all
sorts of machinery, the oil of Patrick SarsSeld Devlan, of Heading,
Pa., is the best.
HORSE POWEK.
A horse power is an amount of mechanical force capable of rais
ing 33,000 lbs. one foot high in a minute. 'Jhe average force ex
erted by the strongest horses, amounting to 33,000 lbs., raised one
foot high in the minute, was adopted, and has since been retained.
The efficacy of engines of a given size, however, has been so much
increased, that the dimensions answerable to a horse power then,
will raise much more than 33,000 lbs. one foot high in the minute
now ; so that an actual horse power, and a nominal horse power
are no longer convertible terras. In some engines every nominal
horse power will raise 52,000 lbs. one foot high in the minute, in
others 60,000 lbs., and in others 66,000 lbs. ; so that an actual and
nominal horse power are no longer comparable quantities, — the one
being a unit of dimension, and the other a unit of force. The ac
tual horse power of an engine is ascertained by an instrument called
an indicator ; but the nominal power la ascertained by a reference
to the dimensions of the cylinder, and may be computed by the
following rule : — Multiply the square of the diameter of the cylin
der in inches by the velocity of the piston in feet per minute, and
divide the product by 6,000 ; the quotient is the number of nominal
horses power. In using this rule, however, it is necessary to adopt
the speed of piston which varies with the length of the stroke. The
speed of piston with a two feet stroke is, according to this system,
160 per minute ; with a 2 ft. 6 in. stroke, 170 ; 8 ft., 180 ; 3 ft., 6
in., 189 ; 4 ft., 200 ; 5 ft., 215 ; 6 ft., 228 ; 7 ft., 245 ; 8 ft., 256 ft.
By ascertaining the ratio in which the velocity of the piston
increases with the length of the stroke, the element of velocity may
be cast out altogether ; and this for most purposes is the most con
venient method of procedure. To ascertain tho nominal power by
this method, multiply the square of the diameter of the cylinder ia
inches by the cube root of the stroke in feet, and divide the pro
duet by 47 ; the quotient is the number of nominal horses power
of the engine. This rule supposes a uniform effective pressure upon
the piston of 7 lbs. per square inch ; the effective pressure upon
the piston of 4 horse power engines of some of the best makers
has been estimated at 6'8 lbs. per square inch, and the pressure
hv Google
THE STEdM EKGINE. 241
increased slightly with the power, and became 694 lbs. per square
inch in engines of 100 horse power ; but it appears to be more con
venient to take a. uniform pressure of 7 lbs. for all powers. Small
engines, indeed, are somewhat less effective in proportion than large
ones ; but the difference can be made np by slightly increasing the
pressure in the boiler ; and small boilers will bear such an increase
without inconvenience.
Nominal power, it is clear, cannot be transformed into actual
power, for the nominal horse power expresses the size of an engine,
and the actual horse power the number of times 33,000 lbs. it will
lift one foot high in a minute. To find the number of times 33,000
lbs. or 528 cubic feet of water, an engine will raise one foot high
in a minute, — or> in other words, the actual power, — we first find the
pressure in the cylinder by means of the indicator, from which we
deduct a pound and a half of pressure for friction, the loss of
power in working the air pump, &c. ; multiply the area of the
piston in square inches by this residual pressure, and by the motion
of the piston, in feet per minute, and divide by 33,000; the
quotient is the actual number of horse power. The same result is
attained by squaring the diameter of the cylinder, multiplying by
the pressure per square inch, as shown by the indicator, less a pound
and a half, and by the motion of the piston in feet, and dividing by
42,017. The quantity thus arrived at, will, in the case of nearly all
modern engines, be very different from .that obtained by multiplying
the square of the diameter of the cylinder by the cube root of the
stroke, and dividing by 47, which expresses the nominal power ; and
the actual and nominal power must hj no means be confounded, as
they are totally different things. The duty of an engine is the
work done in relation to the fuel consumed, and in ordinary mill or
marine engines it can only be ascertained by the indicator, as the
load upon such engines is variable, and cannot readily be deter
mined : but in the case of engines for pumping water, where the
load is constant, the number of strokes performed by the engine
represents the .duty ; and a mechanism to register the number of
strokes made by the engine in a given time, is a sufficient test of
the engine's performance.
In high pressure engines the actual power is readily ascertained
by the indicator, by the same process by which the actual power of
low pressure engines is ascertained. The friction of a locomotive
engine when unloaded, is found by experiment to be about 1 lb. per
square inch on the surface of the pistons, and the additional friction
caused by any additional resistance is estimated at about '14 of
that resistance ; but it will be a sufficiently near approximation to
the power consumed by friction in high pressure engines, if we
make a deduction of a pound and a half from the pressure on that
account, as in the case of low pressure engines. High pressure
engines, it is true, have no air pump to work ; but the deduction of
a pound and a half of pressure ia relatively a much smaller one
where the pressure is high than where it does not much exceed the
hv Google
242 THE PRACTICAL MODEL CALCULATOR.
pressure of the atmosphere. The rule, therefore, for the actual
horse power of a high pressure engine will stand thus :— Square
the diameter of the cylinder in inches, multiply by the pressure of
the steam in the cylinder per square inch, less 1 Ihs,, and by the
speed of the piston in feet per minute, and divide by 42,017 ; the
fjuotient is the actual horse power. The norainal horse power of a
high pressure engine has never been defined ; but it should obvi
ously hold the same relation to the actual power as that which
obtains in the case of condensing, engines, so that an engine of a
given nominal horse power may be capable of performing the same
work, whether high pressure or condensing. This relation is main
tained in the following rule, which expresses the nominal horse
power of high pressure engines : — Multiply the square of the diame
ter of the cylinder in inches hy the pressure on the piston in pounds
per square inch, and by the speed of the piston in feet per minute,
and divide the product by 120,000 ; the quotient is the power of
the engine in nominal horses power. If the pressure upon the
piston be 80 lbs. per square inch, the operation may be abbreviated
by multiplying the square of the diameter of the cylinder by the
speed of the piston, and dividing by 1,500, which will give the
same result. This rule for nominal horse power, however, is not
representative of the dimensions of the cylinder ; but a rule for the
nominal horse power of high pressure engines which shall discard
altogether the element of Yelocity, is easily constructed ; and, as
different pressures are used in different engines, the pressure must
become an element in the computation. The rule for the nominal
power will therefore stand thus : — Multiply the square of the
diameter of the cylinder in inches by the pressure on the piston in
poimds per square inch, and the cube root of the stroke in feet, and
divide the product by 940 ; the quotient is the power of the engine
in nominal horse power, the engine working at the ordinary speed
of 128 times the cube root of the stroke.
A summary of the results arrived at by these rules is given in
the following tables, which, for the convenience of reference, wc
introduce.
PAEALLEL MOTION.
Rule I. — In such a combination of two levers as is represented in
Figs. 1 and 2, ^age 245, to find the length of radius bar required
for any given length of lever (?, and proportion of parts of the link,
6rJE and F E, so as to make the point E move in a perpendicular
line. — Multiply the length of O C by the length of the segment G E,
and divide the product by the length of the segment F E. The
quotient is the length of the radius bar.
Rule II. — {Fig. 2, page 245.) The length of the radius har and
of OG being given, to find the length of the segment (FF) of the
link next the radius bar. — Multiply the length of C G by the
hv Google
THE STEAM EKGINB.
Table of Nominal Horse Power of Low Pressure Engines.
i
Lem
H6FS
IROSE » r.Bt.
1
Ij
2
H 1 3
~H~
i
H
5
_5J_
6
7
4
^
59
^
46
49
■52
^
66
~^
60
■62
■65
■02
■9e
186
8
17S
l9«
216
226
240
■47
60
2T4
304
30
}:
11?
IflS
aM
Im
S?7
Bl
*25
364
41
376
■6
■M
07
92
a06
■M
61
667
I
380
4^
6^M
tea
601
frM
8111
71
3
798
4T7
OS
860
■27
780
790
fr2S
»«6
786
961
9SO
176
es
lfr47
1138
768
879
1219
1288
1856
396
£0
1072
1227
466
1502
648
3B8
871
1046
2623
2096
2344
962
226B
22M
26sa
2848
2914
;BB9
2782
28'07
SO'40
3380
2761
3142
3308
3818
MB
4341
4469
4705
38
soia
sli?
sa71
SJ42
169
3977
4&M
4877
6fr72
62^64
4867
6011
STii
4296
4MB
4716
6691
8264
8848
8800
8181
8612
6SS6
9026
9840
T8I7
34a)
8948
W20
9819
1096
1187
f%
Tile
"la
182
141
1^1
iIm
j^ao
8179
9383
12418
12981
3603
3988
Hi
8716
»9M
118^
1267
1328
1383
m
E68
lSS8
1638
1684
1773
iaa9
338
1418
1736
1788
1882
131S
4fft
lM4
1840
1991
M
ifrs"
394
1890
lifrT
8^4
924
M^2
m^
rs
294
K
iesi
TM
18«
186a
198B
950
029
101
2285
2362
i]
1718
1934
2404
2606
^ao
1802
2448
1891
2038
2478
2138
2474
2682
2691
2880
2238
B3T6
250S
2616
281T
seo8
90
ji^
l'"^
'■a't
2339
^''
=^"2
3297
length of the link G P, and divide the product by the sum of the
lengths of the radius bar and of C G. The quotient is the length
required.
Rule III.— (K^s. 3 and 4, pages 246 and 247.) To find the length
of the radius bar (Fff), the length of CG being given. — Square the
length of C G, and divide it by the length of D G. The quotient
is the length required.
Rule IV. — (Pigi. 3 and^, pages 246 and 247.) To find the length
of the radius bar, the horizontal distance of its centre (S) from the
main centre being given. — To this given horizontal distance, add half
the versed sine (D N) of the arc described by the end of beam (D).
Square this sum. Take the same sum, and add to it the length of
hv Google
THE PRACTICAL MODEL CALCUtATOE.
Table
o/ Nominal Morse Power of High
Pressure Engines.
1
I^OTH «, S™>K« » FEE,. 1
1
_ii_
2
2J
8
JL
4
^i
5
JL
6
7
2
■as
■29
^
39
■37
38
40
■42
.44
■45
■40
■49
■39
■69
■10
95
Il
49
117
41
lis
165
216
228
196
24S
i
s
1^
2i
283
■M
U
262
293
342
36
69
330
318
3;33
405
11
«t
312
36?
3«l
42S
SBO
4oa
in
*16
m
439
667
616
'(
SM
4S3
6lfl
666
68S
621
7^20
H
fr27
fl63
89B
762
780
B13
6ie
912
B30
660
720
JSO
1017
8M
867
fl69
1053
loi
7^
11*
1M
1083
1116
1246
1820
1362
»M
1178
1380
1633
ll)63
1671
iw
"!
*S7
12^
U01
1038
!'
121l
IS'32
IMS
ir^i
a08
17e7
S43
IMC
20«
MH
^3
W4
23^91
14^
2031
2295
2S10
M39
1644
2610
leee
288G
3120
21ia
26OIi
2658
3045
3B28
260*
2»82
3141
sasa
S4W
Sa51
34^
3S'30
4068
1188
2S»3
2»^
31^
4638
3&8Y
3391
SI
4632
1988
ss01
6S3i
8063
1082
4e3B
«662
7017
W
67W
8656
683!
M03
eT4S
wra
0822
1099
M
6fr3T
fit
iK
fl*3S
9024
103
12 1
m2
131^
1260
131
1 14
IBOil
1621
16T6
18^1
i
1»B
28'B
414
i£*
31
1624
lM1
1836
jgl
lSS7
ii
2303
46
ISM
Ha
IM'6
2230
2300
mi
a464
less
ISO'S
1996
2121
22M
2BS:
242^8
260a
2ffl'2
^
im
ig^g
aoi*
2174
2186
2801
2i*8
2820
2SS3
270
2631
2720
s
313I
mro
2130
2S26
3661
3001
22S<1
2623
2^6
2705
291t
aoB6
^S
3616
sa72
60
^^■S
^p
2S95
^•»
^•^
*°^'^
1176
4396
the beam (C D). Divide the square previously found by thia last
sum, and the quotient is the length sought.
Rule Y.—{Figs. 5 and 6, pages 247, 248.) — To find the length
of the radius bar, C G and P Q heing given. — Square G G, and
multiply the square by the length of the side rod (P D) : call this
product A. Multiply Q D by the length of the aide lever (C D).
From this product subtract the product of D P into C G, and divide
A hy the remainder. The quotient is the length required.
Rule YI.—{Figs. 5 and 6, pages 24T, 248.) To find the length of
the radius bar ; P Q, and the horizontal distance of the centre Sof
the radius bar from the main centre being given. — To the given hori
zontal distance add half the versed sine (D N) of the arc described
hv Google
by tte estremity (D) of the side lever. Square this sum and mul
tiply the square by the length o? the side rod {P D). Call this pro
duct A. Take the same horizontal distance aa before added to the
same half versed sine (D N), and multiply the sum by the length of
the side rod (P D) : to the product add the product of the length of
hv Google
THE PRACTICAL MC
LiJ
tlie side lever C D into the length of Q D, and divide A by the
sum. The <iuotient wiU be the length required.
When the centre H of the radius has its position determined,
rules 4 and 6 will always give the length of the radius bar F H.
To get the length of C G, it will only he necessary to draw through
the point F a line parallel to the side rod D P, and the point where
that line cuts D C will bo the position of the pin G.
In using these formulas and rules, the dimensions must all ho
taken in the same measure ; that is, either all in feet, or all in
inches ; and when great accuracy is required, the corrections mven
in Table (A) must be added to on subtracted from the calculated
length of the radius bar, according as it is less or greater than the
length of G, the part of the beam that works it.
1. Rule 4. — Let the horizontal distance (M C) of the centre (H)
hv Google
THE STEAM EKGIS'E.
sse
a
\ "x
\
T
■il.
/ k
Jxi''\
>
of tKe radius bar from the main centre be equal to 51 inches ; the
half versed sine D N = 3 inches, and I) C = 126 inches ; then by
the rule we will have
(51 + ?if (54)^ 2916 ,,.. ,
51 + 3 + 12 6 = 180 = T80" == ^^'^ '"<^^'^^'
«hich h the required length of the radius bar (F H).
hv Google
THE PRACTICAL MOPEL CALCULATOE.
Fig. 0,
f5
\ r
\
1 1
\/
v
^'^/l ^
I
/
\
/
1
\
1 ■■■. /
,' \
, /
■••"\
JyP^,^^, '
\
!/^
k
". Rule 5. — The following dimensions are those of the Red RoV'
stefimer: C G = 32 DP = 94 QD = 74 C D = 65 P Q = 20.
Bj the rule we have, A = (32)^ X 94 = 96256 and
96256 _ 96256 _ .„ ,
74 X 65  94 X S2 ~ 1802 ~ ^^'^'
■nhich is the reqiiSred length of the radius bar.
3. Rule 6. — Tate the same data as in the last example, on
supposing that C G is not given, and that the centre H is fixed
a horizontal distance from the main centre, equal to 835 incht
Then the half versed sine of the arc D' D D" will he about
inches, and we will have by the rule
A = (835 + 2)^ X 94 = 7059635 and
A 7059635 .
855 X 94 4 65 X 74 ~ 12847 " ^* '^ '°^'"'^'
the required length of the radius bar in this case.
Table (A).
This column gives ^^ when
CG is the greater, and jrg
when F H is the greater.
Correcticn to be added to or
subtraeted from the calcu
lated length of the radins
bar, in decimal parts of its
calcatated leogth.
10
■9
8
•7
%
■0034
■01C.5
0270
■0817
b,Google
THE STEAM BNGIXE. 249
In both of the last two examples Tip = "6 nearly. Tlic correc
tion found by Table (A), therefore, would be 54 X 027 = 1458
inches, which must he subtracted from the lengths already found
for the radius bar, because it is longer than C G. The corrected
lengths will therefore be
In example 2 FH = 5194 inches.
In example 3 FH = 5334 inches.
Rule. — To find the depth of the main beam at the centre. — Divide
the length in inches from the centre of motion to the point where
the piston rod is attached, by the diameter of the cylinder in
inches ; multiply the quotient by the maximum pressure in pounds
per square inch of the steam in tho boiler ; divide the product by
202 for cast iron, and 236 for malleable iron : in either case, the
cube root of the quotient multiplied by the diameter of the cylinder
in inches gives the depth in inches of the beam at the centre of
motion. To find the breadth at the centre. — Divide the depth in
inches by 16 ; the qttotient is the breadth in inches.
An engine beam is three times the diameter of the cylinder, from
the centre to the point where the piston rod acts on it ; the force
of the steam in the boiler when about to force open the safety
valve is 10 lbs. per square inch. Required the depth and breadth
when the beam is of cast iron.
In this case m = 3, and P = 10, and therefore
The breadth = — D = 03 D.
lo
It will be observed that our rule gives the least value to the
depth. In actual practice, however, it is necessary to make allow
ance for accidents, or for faultiness in the materials. This may be
done by making the depth greater than that determined by the
rule ; or, perhaps more properly, by taking the pressure of the steam
much greater than it can ever possibly be. As for the dimensions
of tho other parts of the beam, it is obvious that they ought to
diminish towards the extremities ; for the power of a beam to resist
a cross strain varies inversely as its length. The dimensions may
be determined from the formula/?' d^ = 6 W ?.
To apply the formula to cranks, wo may assume the doptli at the
shaft to be equal to n times the diameter of the shaft ; hence, if
m X D he the diameter of the shaft, the depth of the crank will
be »i X m xD. Substituting this in the formula /6d^ = 6 W^
and it becomes /5 X n^ x m' x D' = 6 W I. Now, as before,
W = 7854 X P X D^, so that the formula becomes / x 5 X )t^ x
m^ = 47124 X P X Z. The value of n is arbitrary. In practice
it may be made equal to IJ or 15. Taking this value, then, for
hv Google
250 THE PRACTICAL MODEL CALCULATOR.
cast iron, the formula, becomes 15300 x i x  x m' = 47124 x
P X i, or t305 971= 6 = P i ; but if L denote tlie Jengtli of the crank
in feet, the formula becomes 60^ n^ 6 = PL, and .'. 6 ^ P X
L 7 609 m^ 'Ihia formula may be put into the form of a rule,
thus : —
Rule. — To find the breadth at the shaft when the depth is equal
to 1^ times the diameter of the shaft. — Divide the square of the
diameter of the shaft in inches by the square of the diameter of
the cylinder ; multiply the quotient by 609, and reserve the pro
duct for a divisor ; multiply the greatest elastic force of the steam
in lbs. per square inch by the length of the crank in feet, and
divide the product by the reserved divisor: the quotient is the
breadth of the crank at the shaft.
A crank shaft is ^ the diameter of the cylinder ; the greatest
possible force of the steam in the boiler is 20 lbs, per square inch ;
and the length of the shaft is S feet. Required the breadth of the
crank at the shaft when its depth is equal to 1^ times the diameter
of the shaft.
In this case m = 1, so that the reserved divisor — tf = 38 :
again, elastic force of steam in lbs. per square inch = 20 lbs. ;
3 X 20
hence width of crank = — hq — = 1'6 inches nearly.
Role. — To find the diameter of a revolving shaft. — Form a
reserved divisor thus : multiply the number of revolutions which
the shaft makes for each double stroke of the piston by the number
1222 for cast iron, and the number 1376 for malleable iron. Then
divide the radius of the crank, or the radius of the wheel, by the
diameter of the cylinder ; multiply the quotient by the greatest
pressure of the steam in the boiler expressed in lbs. per square
inch ; divide the product by the reserved divisor ; extract the cube
root of the quotient, and multiply the result by the diameter of
the cylinder in inches. The product is the diameter of the shaft
in inches.
STEE^fOTH Of E
Rule. — To find the diameter of a rod exposed to a tensile force
only. — Multiply the diameter of the piston in inches by the square
root of the greatest elastic force of the steam in the boiler esti
mated in lbs. per square inch ; tho product, divided by 95, is the
diameter of the rod in inches.
Required the diameter of the transverse section of a piston rod
in a single acting engine, when the diameter of the cylinder is 50
inches, and the greatest possible force of the steam in the boiler is
16 lbs. per square inch. Here, according to the formula,
50 200
<? = y^ ^/ 16 = Tj^ = 21 inches.
hv Google
THE STEAM ENGINE. 251
KuLE. — To find the strength of rods alternately extended and
compressed, suck as the piston rods of double acting engines. — Mul
tiply the diameter of the piston in inches hj the square root of the
maximum pressure of the etoam in lbs. per square inch ; divide the
product by
47 for cast iron,
50 for malleable iron.
This rule applies to the piston rods of double acting engines,
parallel motion rods, airpump and forcepump rods, and the liite.
The rule may also be applied to determine the strength of connect
ing rods, by taking, instead of P, a number P', such that P' x sine
of the greatest angle which the connecting rod makes with the
direction = P.
Supposing the greatest force of the steam in the boiler to be 16
lbs. per square inch, and the diameter of the cylinder 50 inches ;
required the diameter of the piston rod, supposing the engine to be
double acting. In this case
for cast iron t? = _ \/ P = = 5 inches nearly ;
47 47
for malleable iron d = — >/ P = 4 inches.
The pressure, however, is always taken in practice at more than 16
lbs. If the pressure be taken at 25 lbs., the diameter of a malle
able iron piston rod will be 5 inches, which is the usual proportion.
Piston rods are never made of cast iron, but airpump rods are
sometimes made of brass, and the connecting rods of land engines
are cast iron in most eases.
FORMULAS FOB TDE STEENGTH OF VAKIOCS PARTS OF MARISE ENGINES.
The following general rules give the dimensions proper for the
parts of marine engines, and we shall recapitulate, with all possible
brevity, the data upon which the denominations rest.
Let pressure of the steam in boiler = p lbs. per square inch,
Diameter of cylinder = D inches.
Length of stroke = 2 R inches..
The vacuum below the piston is never complete, so that there
always remains a vapour of Steam possessing a certain elasticity.
We may suppose this vapour to be able to balance the weight of the
piston. Hence the entire pressure on the square inch of piston in
lbs. = p + pressure of atmosphere = 15 ^ ^. We shall substi
tute P for 15  p. Hence
Entire pressure on piston in lbs. = '7854 X (15  y) X C
= 7854 X P X D^
The dimensions of the paddleshaft journal may be found from
the following formulas, which are calculated so that the strain in
ordinary working =  elastic force.
Diameter of paddleshaft journal = 08264 {R X P x D=}*
Length of ditto = IJ X diameter.
hv Google
252 THE PRACTICAL MODEL CALCULATOR.
The dimensiona of the several parts of the crank may be found
from the following formulas, whica are calculated so that the strain
in ordinary working = onehalf the elastic force ; and when one
paddle is suddenly brought up, the strain at shaft end of crank = §
elastic force, the strain at pin end of crank = elastic force.
Exterior diameter of large eye = diameter of paddleshaft +
iD[P X 1561 X E" + 00494 X D^ x P ^j^H
Length of ditto = diameter of paddle shaft.
Exterior diameter of small eye = diameter of crank pin +
02521 X v''? X D. _
Length of ditto = 0375 X ^/ P x D.
Thickness of web at paddle centre =
D' X P x ■/ U561 X R^ + 00494 x D' xT}
11000
Breadth of ditto = 2 x thickness.
Thickness of web at pin centre — 022 x v' P x D.
Breadth of ditto = f x thickness.
As these formulas are rather complicated, we may show what
hey become when ^ = 10 or P = 25.
Exterior diameter of large eye = diameter of paddle shaft f
\ D^/ (1561 X R' h 1235 x^ D^ J.
1 1512 X ^/R
Length of ditto = diameter of paddle shaft.
Exterior diameter of small eye = equal diameter of crank pin +
126 X D.
Length of ditto = '1875 X D.
Thickness of web at pin centre = 11 X D.
Breadth of ditto = f x thickness of web.
The dimensions of the crank pin journal may he found from the
following formulas, which are calculated so that strain when bear
ing at outer end = elastic force, and in ordinary working strain =
onethird of elastic force.
Diameter of crankpin journal = ■028B6 x v' P x D.
Length of ditto = I x diameter.
The dimensions of the several parts of the cross head may be found
from the following formulas, in which we have assumed, for the
purpose of calculation, the length = 14 X D. The formulas
have been calculated bo as to give the strain of we1
222i
force, and when bearing at outer end =
hv Google
THE STEAM ENGINE. 253
Exterior diameter of eye = diameter of hole + '02827 X P*^ X D.
Depth of ditto = 0979 x P* x IX_
Diameter of journal = 01716 X s/ P x D.
Length of ditto =  diameter of journal.
Thickness of ^ch at middle = 0245 x P^" x D.
Breadth of ditto = 09178 x P* x D.
Thickness of web at journal = 0122 X P^ X D.
Breadth of ditto = 0203 x P« x D.
The dimensions of the several parts of the piston rod may be
found from the following formulas, which are calculated so that the
strain of piston rod = } elastic force.
Diameter of the piston rod = — — ^^ '
Length of part in piston = 04 X D X P.
Major diameter of part in crosshead = OlS X \/P X D.
Minor diameter of ditto = '018 X v/P x D. _
Major diameter of part in piston = 028 X v'P X D.
Minor diameter of ditto = 023 X s/P X D.
Depth of gibs and cutter through crosahead = '0358 x P* x D.
Thickness of ditto = 007 X P^ X D.
Depth of cutter through piston = 017 x \/P x D.
Thickness of ditto = 007 x P^ x D.
The dimensions of the several parts of the connecting rod may
be found from the following formulas, which are calculated so that
the strain of the connecting rod and the strain of the strap are both
equal to onesixth of the elastic force. ,
Diameter of connecting rod at ends = 019 X P^ X D. .
Diameter of ditto at middle = {1 + 0035 x length in inches}
X 019 X x/P X D.
Major diameter of part in crosstail = 0196 x P^ X D.
Minor ditto = 018 X P^ x D.
Breadth of butt  0313 X P^ X D.
Thickness of ditto = 025 X P^ X D. __
Mean thickness of strap at cutter = '00854 x vT x D.
Ditto above cutter = 00634 x ^/P X D. _
Distance of cutter from end of strap = '0097 X x/P x D.
Breadth of gibs and cutter through crosstail = 0358 x P* x D.
Breadth of gibs and cutter through butt = '022 x P^ x D.
Thickness of ditto = 00564 x P^ x D,
hv Google
254 THE PRACTICAL MODEL CALCULATOR.
Tlib dimensions of the several parta of the side rods may be
found fiom the folbwmg foimulas, which aro calculated so as to
make the strain of the aide lod = oneeixth of elastic force, and
the strams ot strap ind cutter = onefifth of elastic force.
Diameter of cylinder side rods at ends = '0129 x P^ X D.
Diameter of ditto at middle = (1 + '0035 X length in inches).
X 0129 X P^ X D.
Breadth of hutt = 0154 x P^ X D.
Thickness of ditto = 0122 x P^ x D.
Diameter of journal at top end of side rod = 01716 x P^ x D.
Length of journal at top end =  diameter.
Diameter of journal at bottom end = '014 X P^ x D.
Length of ditto = 0152 x P^ X D.
Mean thickness of strap at cutter = 00643 X P^ X D.
Ditto helow cutter = 004T X P^ x D.
Breadth of gibs and cutter = 016 x P* X D.
Thickness of ditto = 0033 x P^ X D.
The dimensions of the main centre journal may he found from
the following formulas, 'which are calculated so as to make the
strain in ordinary working = one half elastic force.
Diameter of main centre journal = "0367 X P' X D.
Length of ditto = j X diameter.
The dimensions of tho several parts of the airpump may bo
found from the corresponding formulas given above, by taking for
D another number d the diameter of airpump.
TUB BEVEEAL PARTS OF FURNACES AND BOILERS.
Perhaps in none of the parts of a steam engine does the practice
of engineers vary more than in those connected with furnaces and
boilers. There are, no doubt, certain proportions for these, as well
as for the others, which produce the maximum amount of useful
effect for particular given purposes ; hut the determination of these
proportions, from theoretical considerations, has hitherto been at
tended with insuperable difficulties, arising principally from our im
perfect knowledge of the laws of combustion of fuel, and of the laws
according to which caloric is imparted to the water in the boiler.
In giving, therefore, the following proportions for the difi'erent
parts, we desire to have it understood that we do not affirm them
to be the best, absolutely considered ; we give them only as the
average practice of the best modern constructors. In most of the
cases we have given the average value per nominal horse power.
It is well known that the term horse power is a conventional unit
for measuring the size of steam engines, just as a foot or a mile is
hv Google
THE STEAM ENGIXE. 255
a unit for the measurement of extension. Tlicre is this difFcrencc,
however, in the two cases, that whereas the length of a foot is
fixed (iefiaitively, and is known to every one, the dimensions proper
to an engine horse power difi'er in the practice of every different
maker : and the same kind of confusion is thereby introduced into
engineering as if one person were to make his footrule eleven
inches long, and another thirteen inches. It signifies very little
what a horse power is defined to be ; but when onco defined, the
measurement should be kept inviolable. The question now arises,
what standard ought to be the accepted one. Por our present pur
pose, it is necessary to connect by a formula the three quantities,
nominal horses poiverj length of stroke, and diameter of cylinder.
With this intention,
Let S = length of stroke in feet,
d = diameter of cylinder in inches ;
m, , . ., , 1^ X ^S
ilien the nominal horse power = ■ ,, — nearly.
I, Area of Fire QraU. — The average practice is to give 'SS
square feet for each nominal horse power. Hence the following
rule:
Rule 1. — To find the area of the fire grate. — Multiply the num
ber of horses power by 55 ; the product is the area of the fire grate
in square feet.
Required the total area of tbe fire grate for an engine of 400
horse power. Here total area of fire grate in square feet = 400 X
•55 = 220.
A rule may also be found for expressing the area of the fire grate
in terms of the length of stroke and the diameter of the cylinder.
Por this purpose we have,
: ^g . , <?= X ^"S
— .1. leet = — i,r — leet.
This formula expressed in words gives the following rule.
Rule 2. — To find the area of fire grate. — Multiply the cube root
of the length of stroke in feet by the square of the diameter in in
ches ; divide the product )iy 86 ; the quotient is the area of Cre
grate in square feet.
Required the total area of the fire grate for an engine whose
stroke = 8 feet, and diameter of cylinder = 50 inches.
Here, according to the rule, _
50^ X ^8 2500 X 2
total area of fire grate m square feet = ■ ■ ■ „ . ■ — =  >, . — =
n^ = 5y nearly.
In order to work this example by the first rule, wo find the
nominal horse power of the engine wbose dimensions we have spe
cified is 1043 ; hence,
total area of fire grate in square feet = 1064 X 55 = 585.
hv Google
256 TUB PRACTICAL MODEL CALCULATOE,
Willi regard to these rules we may remark, not only that they
are founded on practice, and therefore erapyrical, hut they are only
applicable to large engines. When an engine is very aroall, it re
quires a niTich larger area of fire grate in proportion to its size than
a larger one. This depends upon the necessity of having a certain
amount of fire grate for the proper combustion of the coal.
II. Length ^Furnace. — The length of the furnace differs con
siderably, even in the practice of the same engineer. Indeed, all
the dimensions of the furnace depend to a certain extent npon the
peculiarity of its position. From the difBculty of firing long fur
naces efficiently, it has been found more beneficial to restrict the
length of the furnace to about six feet than to employ furnaces of
greater length,
III. Height of Furnaoe above Bars. — This dimension is variable,
but it is a common practice to mate the height about two feet.
rV. Capacity/ of Furnace Chamber above Bars. — The average
per horse power may be taken at IIT feet. Hence the following
rule:
Bole, — To find the capacity of furnace chamber above bars. —
Multiply the number of nominal horaes power by 1'17 ; the pro
duct is the capacity of furnace chambers above bars in cubic feet.
v. Areas of Flues or Tubes in smallest part. — The average value
of the area per horse power is 11'2 aq. in. Hence we have the fol
lowing rule :
Rule. — To find the total area of the flues or tubes in smallest
part. — Multiply the number of horse power by 11'2 ; the product
is the total area in square inches of flues or tubes in smallest part.
Required total area of flues or tubes for the boiler of a steam en
gine when the horse power = 400.
For this example we have, according to the rule.
Total area in square inches = 400 x 112 = 4480.
We may also find a very convenient rule expressed in terras of
the stroke and the diaraeter of cylinder. Thus, _
. , 112 X <f X ^S
Total area of tubes or flues m square inches = ■ — ■ ,^
_<P_XJ^S
 ■ — 4 ■
VI, Effective Seating Surface. — The effective heating surface of
flue boilers is the whole of furnace surface above bars, the whole
of tops of flues, half the sides of flues, and none of the bottoms ;
hence the effective flue surface is about half the total flue surface.
In tubular boilers, however, the whole of the tube surface is reckoned
effective surface.
EFFECTIVE HEATING SURFACE OF FLUB BOILERS.
Rule 1. — To find the effective heating surfaee of marine flue
boilers of large size. — Multiply the number of nominal horse
power by 5 ; the product is the area of effective heating surface in
Bi^uare feet.
hv Google
THE STEAM ENGINE. 257
Required the effective heating snrface of an engine of 400 nomi
nal horse power.
In this case, according to the rule, effective heating surface in
stjuaro feet = 400 x 5 = 2000.
The effective heating surface may he expressed in terms of the
length of stroke and the diameter of the cjlinder.
Rule 2. — To find the total effective heating surface of marine
flue boilers. — Multiply the sijuare of the diameter of cylinder in
inches hy the cube root of the length of stroke in feet ; divide the
product hy 10 : the quotient expresses the number of square feet
of effective heating surface.
Required the amount of effective heating surface for an engine
irhose stroke = 8 ft., and diameter of cylinder = 50 inches.
Here, according to Ruie 2, effective heating surface in square feet
50= X '^B 2500 X 2 5000 ^„„
= 10 = ^LO ^ ^0" = ^^'^
To solve this example according to the first rale, we have the
nominal horse power of the engine equal to 106'4. Hence, ac
cording to Rule 2, total effective heating surface in square feet =
1064 X 492 = 523^.
ErrECTIVE HEATING SURFACE OF TUBULAR BOILERS.
The effective heating surface of tubular boilers is about equal to
the total heating surface of flue boilers, or is double the effective
surface ; but then the total tube surface is reckoned effective sur
face.
It appears that the total heating surface of flue and tubular ma
rine boilers is about the same, namely, about 10 square feet per
horse power.
VII. Area of Ohimney. — Rule 1. — To find the area of chimney.
— Multiply the number of nominal horse power by 1023 ; the pro
duct is the area of chimney in square inches.
Required the area of the chimney for an engine of 400 nominii.l
horse power.
In this example we have, according to the rule,
area of chimney in square inches = 400 x 1023 = 4092.
We may also find a rule for connecting together the area of the
chimney, the length of the stroke, and the diameter of the cyJinder.
Rule 2. — To find the area of the chimney. — Multiply the square
of the diameter expressed in inches by the cube root of the stroke
expressed in feet ; divide the product by the number 5 ; the quo
tient expresses the number of square inches in the area of chimney.
Required the area of the chimney for an engine whose stroke —
8 feet, and diameter of cylinder = 50 inches.
We have in this example from the rule,
50= x ^"8 2500 X 2
area of chimney in square inches = f = r =
1000.
hv Google
258 THE PEACTICAL MODEL CALCULATOE.
To work this example according to the first rule, we find, that
the nominal horse power of this engine ia 1046 : hence,
area of chimney in square inches = 1046 x 1023 = lOTO.
The latter value is greater than the former one by 70 inches.
This difference arises from our taking too gr^at a divisor in Rule 2.
Either of the values, however, is near enough for all practical
purposes.
VIII, Water in Boiler.— Iho quantity of water in the boiler
diifers not only for different boilers, but differs even for the same
boiler at different times. It may he useful, however, to know the
average quantity of water in the boiler for an engine of a given
horse power.
Rule 1. — To determine the average quantity of tvater in the
boiler. — Multiply the number of horse power by 5 ; the product
expresses the cubic feet of water usually in the boiler.
This rule may be so modified as to make it depend upon the
stroke and diameter of the cylinder of engine.
Rule 2. — To determine the cubie feet of water usually in the
boiler. — Multiply together the cube root of the stroke in feet, the
square of the diameter of the cylinder in inches, and the number 5 ;
divide the continual product by 47 ; the quotient expresses the cu
bic feet of water usually in the boiler.
Required the usual quantity of water in the boilers of an engine
whose stroke == 8 feet, and diameter of cylinder 50 inches.
Here we have from the rule,
5 X 50= X ^8" ox 2500 x 2
cubic feet of water in boiler = — f= = jy
25000 ^_ ,
= — jii — = oo2 nearly.
The engine, with the dimensions we have specified, is of 1064
nominal horse power. Hence, according to Rule 1,
cubic feet of water in boiler = 1064 X 5 = 532.
IX. Area of Water Level. — Rule 1, — To find the area of water
level. — The area of water level contains the same number of square
feet as there are units in the number expressing the nominal horse
power of the engine.
Required the area of water level for an engine of 200 nominal
horse power. According to the rule, the answer is 200 square
feet.
We add a rule for finding the area of water level when the di
ameter of cylinder and the length of stroke is given.
Rule 2, — To find the area of water level. — Multiply the square
of the diameter in inches by the cube root of the stroke in feet ;
divide the product by 47 ; the quotient expresses the number of
square feet in the area of water level.
Required the area of the water level for an engine wlioso stroke
is 8 feet, and diameter of eyliniler 50 inelics.
hv Google
THE STEAM EKGINB. 259
lu tliiB case, according to the rule,
area of water level m sqnare leet = ■ " 47 ' "" = I'Jo.
X. iS'feflJK Room. — It is obvioua that the steam room, like the
quantity of water, is an extremely variable quantity, differing, not
only for different hollers, but even in the same boiler at different
times. It ia desirable, however, to know the content of that part
of the boiler usually filled with steam.
Rule 1. — To determine the average quantity of steam room. —
Multiply the number expressing the nominal horse power by 3 ;
the product expresses the average number of cubic feet of steam
room.
Required the average capacity of steam room for an engine of
460 nominal horse power.
According to the rule,
Average capacity of steam room = 460 X 3 cubic feet = 1380
cubic feet.
This rule may he so modified as to apply when the length of
stroke and diameter of cylinder are given.
Rl'LE 2. — Multiply the square of the diameter of the cylinder
in inches by the cube root of the stroke in feet ; divide the product
by 15 ; the quotient expresses the number of cubic feet of steam
Required the average capacity of steam room for an engine whose
stroke is 8 feet, and diameter of cylinder 5 inches.
In this case, according to the rule,
, . ^ 50' X ^8 2500 X 2 5000
Steam room m cubic feet = j  ■ = ,r"" = •■ ..r =
333f
We find that the nominal horse power of this engine is 1004 ;
hence, according to Rule 1,
average steam room in cubic feet = 106'4 X 3 = 320 nearly.
Before leaving these rules, we would again repeat that they ought
not to be considered as rules founded upon considerations for giving
the maximum effect from the combustion of a given amount of fuel ;
and consequently the engineer ought not to consider them as inva
riable, but merely to bo followed as far as circumstances will per
mit. We give them, indeed, as the medium value of the very va
riable practice of several wellknoira constructors ; consequently,
although the proportions given by tho rules may not be the best
possible for producing the most useful effect, still the engineer who
is guided by them is sure not to be very far from the common prac
tice of most of our best engineers. It has often been lamented that
the methods used by different engine makers for estimating the
nominal powers of their engines have been so various that we can
form no real estimate of the dimensions of the engine, from its re
puted nominal horse power, unless we know its maker ; but the
hv Google
260 THE PRACTICAL MODEL OALCTILATOK.
same confusion exists, also, to some extent, in the construction of
boilers. Indeed, many things may be mentioned, which have
hitherto operated aa a barrier to the practical application of any
Standard of engine power for proportioning the different parts of
the boiler and furnace. The magnitude of furnace and the extent
of heating surface necessary to produce any required rate of eva
poration in the boiler are indeed known, yet each enginemaker
has his own rule in these matters, and which he seems to think pre
ferable to all others, and there are various circumstances influ
encing the result which render facts incomparable unless those cir
cumstances are the same. Thus the circumstances that govern the
rate of evaporation, as influenced by different degrees of draught,
may be regarded as but imperfectly known. And, supposing the
difficulty of ascertaining this rate of evaporation were surmounted,
there would still remain some difficulty in ascertaining the amount
of power absorbed by the condensation of the steam on its passage
to the cylinder — the imperfect condensation of the same steam after
it has worked the piston — the friction of the various moving parts
of the machinery — and, especially, the difference of effect of these
losses of power in engines constructed on different scales of magni
tude. Practice must often vary, to a certain extent, in the con
struction of the different parts of the boiler and furnace of an en
gine ; for, independently of the difficulty of solving the general
problem in engineering, the determination of the maximum effect
with the minimum of means, practice would still require to vary
according as in any particular case tho desired minimum of means
was that of weight, or bulk, or expense of material. Again, in es
timating the proper proportions for a boiler and its appendages,
reference ought to bo made to the distinction between the " power"
or " effect" of the boiler, and its " duty." This is a distinction to
be considered also in the engine itself. The power of an engine
has reference to the time it takes to produce a certain mechanical
effect without reference to the amount of fuel consumed ; and, on
the other hand, tho duty of an engine has reference to the amount
of mechanical effect produced by a certain consumption of fuel, and
is independent of tho time it takes to produce that effect. In ex
pressmg the duty of engine', it would ha\e prevented much need
less confusion if the duty of the boiiei had been entirely separated
from that of the engine as, indeed, thoy nro two very distinct
things The duty performed by ordmaiy lind rotative steam en
gines IS —
One horse power exerted by 10 llis of fuel an hour ; or.
Quarter of a million of lbs. raised 1 foot high by 1 lb. of coal ; or,
Twenty millions of lbs. raised one foot by each bushel of coals.
Though in the best class of rotative engines the consumption is
not above half of this amount.
The constant aim of different engine makers is to increase tho
amount of the duty ; that is, to make 10 lbs. of fuel exert a greater
effect than one horse power ; or, in other words, to make 1 lb. of
hv Google
THE STEAM ENGINE. 261
coal I'aise more than a quarter of a million of lbs. one foot bigli.
To a great extent they have been successful in this. They have
caused 5 lbs. of coal to exert the force of one horse power, and even
in some cases as little as 3i^ lbs. ; but in these latter cases the
economy is due chiefly to expansive action. In some of the engines,
however, working with a consumption of 10 lbs. of coal per nominal
horse hower per hour, the power really exerted amounts to much
more than that represented by 33,000 ibs. lifted one foot high in
the minute for each horse power. Some engines lift 56,000 lbs.
one foot high in tjie minute hy each horse power, with a consump
tion of 10 lbs. of coal per horse power per hour; and even this
performance has been somewhat exceeded without a recourse to ex
pansive action. In all modern engines the actual performance
much exceeds the nominal power ; and reference must he had to
this circumstance in contrasting the duty of different engines.
MECHANICAL POWEIt OP BTEA)!.
We may here give a table of some of the properties of steam,
and of its mechanical effects at different pressures. This table may
help to solve many problems respecting the mechanical effect of
steam, usually requiring much laborious calculation.
BIE
n.i<,c.H. Effi^t .■<
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It is quite clear that although there is no theoretical limit to the
benefit derivable from expansion, there must he a limit in practice,
arising from the friction incidental to the ube of \eiy laige cylin
ders, the magnitude of the deduction due to uncondensed vapour
when the steam is of a very low pieisuie, and othei circumstances
which it is needless to relate. It is cleiJ, too, that while the effi
hv Google
262 THE PRACTICAL MODEL CALCULATOK.
cleney of the steam is increased by expansive action, the cfScieney
of the engine is diminished, unless the pressure of the steam or the
speed of the piston be increased correspondingly ; and that an en
gine of any given size will not exert the same power if made to ope
rate expansively without any other alteration that would have been
realized if the engine had been worked with the full pressure of the
steam. In the Cornish engines, which work with steam of 40 lbs.
on the inch, the steam is cut off at onetwelfth of the stroke ; but
if the steam were cut off at onetwelfth of the stroke in engines em
ploying a very low pressure, it would probably be found that tliere
would be a loss rather than a gain from carrying the expansion so
far, as the benefit might be more than neutralized by the friction
incidental to the use of so large a cylinder as would be necessary
to aceomplish this expansion ; and unless the vacuum were a very
good one, there would be but little difference between the pressure
of the steam at the end of the stroke and the pressure of the va
pour in the condenser, so that the urging force might not at that
point he sufficient to overcome the friction. In practice, therefore,
in particular cases, expansion may he carried too far, though theo
retically the amount of the benefit increases with the amount of the
expansion.
We must here introduce a simple practical rule to enable those
who may not be familiar with mathematical symbols to determine
the amount of benefit due to any particular measure of expansion.
"When expansion is performed by an expansion valve, it is an easy
thing to ascertain at what point of the stroke the valve is shut by
the cam, and where expansion is performed by the slide valve the
amount of expansion is easily determinable when the lap and stroke
of the valve are knoivn.
Rule. — To find the Increase of E^eieney arising from workiag
Steam expansively. — Divide the total length of the stroke by the
distance (which call 1) through which the piston moves before the
steam is cut off. The hyperbolic logarithm of the whole stroke ex
pressed in terms of the part of the stroke performed with the full
pressure of steam, represents the increase of efSciency due to ex
pansion.
Suppose that the pressure of the steam working an engine is 45
lbs. on the square inch above the atmosphere, and that the steam
is cut off at onefourth of the stroke ; what is the increase of effi
ciency due to this measure of expansion ?
If onefourth be reckoned as 1, then fourfourths must be taken
as 4, and the hyperbolic logarithm of 4 will be found to be 1'386,
which is the increase of efficiency. The total efficiency of the quan
tity of steam expended during a stroke, therefore, which without
expansion would have been 1, becomes 2'386 when expanded into 4
times its bulk, or, in round numbers, 24.
Let the pressure of the steam be the same as in the last example,
and let the steam be cut off at halfstroke : what, then, is the in
crease of efficiency ?
hv Google
THE STEAM EKOINE. 263
Here half the stroke is to be reckoned as 1, and tlio whole stroke
has therefore to he reckoned as 2, The hyperholic logarithm of 2
is 693, ivhich is the increase of efficieacy, and the total efficiency
of the stroke ia 1693, or 17.
We may here give a tahle to illustrate the mechanical effect of
steam under varying circumstances. The tahle shows the me
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b,Google
264 TUB PRACTICAL MODEL CALCULATOR.
chaiiical eiFect of the steam generated from a cubic inch of water.
Our foriaula gives the effect of a cuhic foot of water ; but it can he
modified to give the effect of the steam of a cubic inch by dividing
by 1728. In this manner we find, for the mechanical effect of the
steam of a cubic inch of water, about 8 (459 + t) lbs. raised one
foot high. The table shows that the mechanical effect increases
with the temperature. The increase is very rapid for temperatures
bolow 212° ; but for temperatures above this the increase is less;
and for the temperatures used in practice we may consider, with
out any material error, the mechanical effect as constant.
INDICATOR.
An instniment for ascertaining the amount of the pressure of
steam and the state of the vacuum throughout the stroke of a steam
engine. Fitzgerald and Neucuran long employed an instrument
of this kind, the nature of which was for a long time not generally
known. Boulton and Watt used an instrument acting upon the
same principle and equally accurate ; but much more portable. la
peculiarity of construction it is simply a small cylinder truly bored,
and into which a piston is inserted and loaded by a spring of suit
able elasticity to the graduated scale thereon attached.
The action of an indicator is that of describing, on a piece of
paper attached, a diagram or figure approximating more or less to
that of a rectangle, varying of course with the merits or demerits
of the engine's productive effect. The breadth or height of the
diagram is the sum of the force of the steam and extent of the va
cuum ; the length being the amount of revolution given to the paper
during the piston's performance of its stroke.
'Xo render the indicator applicable, it is commonly screwed into
the cylinder cover, and the motion to the paper obtained by means
of a sufficient length of small twine attached to one of the radius
bars ; but such application cannot always be conveniently effected,
more especially in engines on the marine principle ; hence, other
parts of such engines, and other means whereby to effect a proper
degree of motion, must unavoidably be resorted to. In those of
direct action the crosshead is the only convenient place of attach
ment ; but because the length of the engine's stroke is considerably
more than the movement required for the paper on the indicator,
it is necessary to introduce a pulley and axle, by which means the
various movements are qualified to suit each other.
When the indicator is fixed and the movement for the paper pro
perly adjusted, allow the engine to make a few revolutions previous
to opening the cock ; by which means a horizontal line will be de
scribed upon the paper hy the pencil attached, and denominated
the atmospheric line, because it distinguishes between the effect of
the steam and that of the vacuum. Open the cock, and if the en
gine he upon the descending stroke, the steam will instantly raise
the piston of the indicator, and, by the motion of the paper with the
pencil pressing thereon, the top side of the diagram will be formed.
hv Google
TUB STEAM ERQINE,
265
At the termination of the stroke and immediately previous to its
return, the piston of the indicator is pressed down by the surround
ing atmosphere, consequently the bottom side of the diagram is de
Bcribed, and by the time the engine is about to make another de
scending stroke, the piston of the indicator is where it first started
from, the diagram being completed ; hence is delineated the mean
elastic action of the steam above that of the atmospheric line, and
also the mean extent of the vacuum underneath it.
But in order to elucidate more a ^
clearly by example, take the follow
ing diagram, taken from a marine
engine, the steam being cut off after
the piston had passed through two
thirds of its stroke, the graduated
scale on the indicator, tenths of an
inch, as shown at each end of the
diagram annexed.
Previous to the cock being
opened, the atmospheric line AB
was formed, and, when opened, the
pencil was instantly raised by the
action of the steam on the piston
to C, or what is generally termed
the startiii0 comer; by the move
ment of the paper and at the ter
mination of the stroke the line CD
was formed, showing the force of
the steam and extent of expansion ;
from D to E show the moments of
eduction ; from E to F the quality of the vacuum ; and from F to
A the lead or advance of the valve ; thus every change in the en
gine is exhibited, and every deviation from a rectangle, except that
of expansion and lead of the valve show the estent of proportionate
defect. Expansion produces apparently a defective diagram, but
in reality such is not the case, because the diminished power of the
engine is more than compensated by the saving in steam. Also
the lead of the valve produces an apparent defect, but a certain
amount must be given, as being found advantageous to the working
of the engine, but the steam and eduction corners ought to be as
square as possible ; any rounding on the steam corner shows a de
fect from want of lead ; and rounding on the eduction corner that
of the passages or apertures being too small.
Rule. — To compute the power of an Engine from the Indicator
Diagram. — Divide the diagram in the direction of its length into
any convenient number of equal parts, through which draw lines
at right angles to the atmospheric line, add together the lengths of
all the spaces taken in measurements corresponding with the scale
on tlie indicator, divide the sum by the number of spaces, and the
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b,Google
266 THE PKACIICAL MODEL CALCULATOR.
quotient is the mean effective pressure on the piaton in lbs. per
square inch.
Let the result of the preceding diagram be taken as an example.
Then, the whole sum of vacuum spaces = 1220 h 10 = 122 lbs.
mean effect obtained by the vacuum ; and in a similar manner the
mean effective pressure of steam is found to be 6'23 lbs., hence the
total effective force = 1848 lbs. per square inch. And supposing
25 lbs. per square inch be absorbed by friction, What is the actual
power of the engine, the cylinder's diameter being 32 inches, and
the velocity of the piston 226 feet per minute ?
1848 — 25 = 1598 lbs. per square inch of net available force.
32^ X 7854 X 1598 x 226 , ,
Then ^flfio ~ horses power.
The line under the diagram and parallel to the atmospheric line
is 3ths distant, and represents the perfect vacuum line, the space
between showing the amount of force with which the uncondensed
steam or vapour resists the ascent or descent of the piston at every
part of the stroke.
As the mean pressure of the atmosphere is 15 lbs. per square
inch, and the mean specific gravity of mercury 13560, or 2037 cu
bic inches equal 1 lb., it will of course rise in the barometer at
tached to the condenser about 2 inches for every lb. effect of va
cuum, and as a pure vacuum would be indicated by 30 inches of
mercury, the distance between the two lines shows whether there
is or is not any amount of defect, as sometimes there ia a consider
able difference in extent of vacuum in the cylinder to that in the
condenser.
To estimate hy means of an indicator the amount of effective power
produced iy a steam engine. — Multiply the area of the piston in
square inches by the average force of the steam in lbs. and by the
velocity of the piston in feet per minute ; divide the product by
33,000, and ^ths of the quotient equal the effective power.
Suppose an engine with a cylinder of 37^ inches diameter, a
stroke of 7 feet, and making 17 revolutions per minute, or 238 feet
velocity, and the average indicated pressure of the steam 1673 lbs.
per square inch ; required the effective power.
Area ^ 11044687 in ches x 1673 lbs. X 238 feet
33000
13326 X 7
= tq ■ = 93282 horse power.
To determine the proper velocity for the piston of a steam engine. —
Multiply the logarithm of the mth part of the stroke at which the
steam is cut off by 23, and to the product of which add •7 Mul
tiply the sum by the distance in feet the piston has travelled when
the steam is cut off, and 120 times the square root of the product
equal the proper velocity for the piston in feet per minute.
hv Google
WEIGHT COKBIHED WITH MASS, VELOCITY, FOECE, AND
WOEK DONE.
CALCULATIONS ON THE rEINCIPLE OF "VIS VIVA. — MATERIALS EMPLOYED
IN THE CONBTEUCTION OF MACHINES. — STEENGTH OF JIATERIALB,
THEIR PROPERTIES. — TORSION, DEFLEXION, ELASTICITY, TENACITIES,
COMPRESSIONS, ETC. — FRICTION OS REST AHD OF MOTIOiS, COEFFJCIE.NTS
OF ALL SORTS OF MOTION. — BANDS. — ROPES.— WHEELS. — IIYDR.^fJ
Lies. — NEW TABLES FOR THE MOTION AND FRICTION OP WATER. —
WATERIVHEELS. — WINDMILLS, ETC.
1. Suppose a body resting on a perfectly smooth table, and, when
in motion, to present no impediment to the body in its couvse, but
merely to counteract the force of gravity upon it ; if this body
weighing 800 lbs. be pressed by the force of 30 lbs. acting hori
zontally and continuously, the motion under such circumstuiicea
will be uniformly accelerated : what is the acceleration ?
30
800
X 322 = 12075 feet the second.
2. IVhat force is necessary to move the abovementioned heavy
body, with a 23 feet acceleration, under the same circumstances ?
glrg X 800 = 5714285 lbs.
The second of these examples illustrates the principle that the
force which impels a body with a certain acceleration is equal to
the weight of the body multiplied by the ratio of its acceleration
to that of gravity. The first illustrates the reverse, namely, the
acceleration with which a body is moved forward with a given force,
is equal to the acceleration of gravity multiplied by the ratio of the
force to the weight,
3. A railway ear, weighing 1120 lbs., moves with a 5 feet velo
city upon horizontal rails, which, let us suppose, offer no impedi
ment to the motion, andis constantly pushed by an invariable
force of 50 lbs. during 20 seconds : with what velocity is it moving
at the end of the 20t!i second, or at the beginning of the 21st
second ?
50
5 + 322 X ^j20 >^ 20 = 3375, the velocity.
4. A carriage, circumstanced asip the Ipt question, weighs 4000
lbs. ; its initial velocity is 30 feet the secorid, and its terminal velo
city is 70 feet : with which force is the body impelled, supposing it
to be in motion 20 seconds ?
(70  30) X 4000 ,_ „
322x20 =24217 lbs.
We have before noticed that the weight (W), divided by S22, or
(?)) gives the mass; that is,
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atiO THE PHACTICAL MODEL CALCULATOR.
■ — ^— = mass,
y
And, force = masa X acceleration.
5. Suppose a railway carriage, weighing 6440 lbs., moves on a
horizontal plane offering no impediment, and is uniformly accele
rated 4 feet the secoad, ivbat continuous force is applied ¥
6440
S2'2 ~ '^^^ ^^^* ™^^^
200 X 4 = 800 Iha., the force applied.
By the four succeeding formulas, all questions may be answered
that may be proposed relative to the rectilinear motions of bodies
by a constant force.
For uniformly accelerated motions :
"F
s = rti + 161 ^^ X P.
For uniformly retarded motions :
F
F
S = at ~ lG1 X Tir ^ t^;
t = the time in seconds, W = the weight in lbs., F = the force in
lbs., a = the initial velocity, and v = the terminal velocity.
6. A sleigh, weighing 2000 lbs., going at the rate of 20 feet a
second, has to overcome by its motion a friction of 30 lbs. : what
velocity has it after 10 seconds, and what distance has it described ?
30
20  322 X ^QQ^ X 10 = lirlT feet velocity.
20 X 10  161 X ^Qjj^ X (10)^ = 17085 feet, distance de
scribed.
7. In order to find the mechanical work ivhich a draughthorse
performs in drawing a carriage, an instrument called a dynamome
ter, or measure of force, is thus used : it is put into communication
on one side of the carriage, and on the other with the traces of the
horse, and the force is observed from time to time. Let 126 Ihs.
be the initial force; after 40 feet is described, let 130 lbs. be tho
force given by the dynamometer ; after 40 feet more is described,
let 129 lbs. be the force ; after 40 feet more is passed over, let 140
lbs. be the force ; and let the next two spaces of 40 feet give forces
of 130 and 120 lbs. respectively. What is the mechanical work done ?
126 initial force.
120 terminal force.
2)246
123 mean.
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TFEIOHT COMBINED WITH MASS. VELOCITY, ETC.
1304 X 40 X 5 = 26080 units of work.
The following rule, usually given to find the areas of irregular
figures, may be applied where great accuracy is rec[uired.
Rule. — To the sum of the first and last, or extreme ordinatea,
add four times the sum of the 2d, 4th, 6th, or even ordinates, and
twice the sum of the 3d, 5th, 7th, &c., or odd ordinates, not includ
ing the extreme ones ; the result multiplied hy ^ tho ordinates'
equidistance will be the sum.
126
120
246 sum of first and last,
246 + 4 X 130 + 2 X 129 + 4 X 140 + 2 X 130 = 1844.
1844 X 40
n = 24586'66 units of work or pounds raised one foot
high. This rule of equidistant ordinates is of great use in the art of
ship building. This application we shall introduce in the proper
place.
8. How many units of work are necessary to impart to a carriage
of 3000 lbs. weif^ht, resting on a perfectly smooth railroad, a velo
city of 100 feet ?
TV1&2. ^ ^'^^'^ = 4668382 units.
A unit of work is that labour which is equal to the raising of a
poand through the space of one foot. A unit of work is done when
one pound pressure is esorted through a space of one foot, no matter
in what direction that space may lie.
Kane Fitzgerald, the first that made steam turn a crank, and
patented it, and the flywheel to regulate its motion, estimated that
a horse could perform S3000 units of work in a minute, that is,
raise 33000 lbs. one foot high in a minute. To perform 4658382
units of work in 10 minutes would require the application 14116
horse power.
9. What work is done by a force, acting upon another carriage,
under the same circumstances, weighing 5000 lbs., ivhich transforms
the velocity from 30 to 50 feet ?
(30)=
64~4 "^ 13'9907, the height due to 30 feet velocity.
(50V
g^ = 388043, the height due to 50 feet velocity.
From 388043
Take 139907
248136
5000
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270 THE PRACTICAL MODEL CALCULATOR.
.. 124068 are the units of work, and just so much workivill tlio
carriage perform if a resistance be opposed to it, and it be gradu
ally brought from a 50 feet velocity to a 30 feet velocity.
The following is without doubt a very simple formula, but the
most useful one in mechanics ; by it we have solved tlie last two
questions ;
Fs = (H  /() W.
This simple formula involves the principle tactnieally termed the
principle of vis viva, or living fobces. H is the height due to
one velocity, say v or H = q— and h, the height due to another a,
or 7( = K. The weight of the mass = \\ ; the force F, and the
space 8.
To express this principle in words, wo may say, that the working
power (Fs) which a mass either acquires when it passes from a lesser
velocity {«) to a greater velocity {v), or produces when it is com
pelled to pass from a greater velocity [v) into a less (a), is always
equal to the product of the weight of the mass and the difference
of the heights due to the velocities.
When we know the ■units of work, and the distance in which the
change of velocity goes on, the force is easily found ; and when the
force is known, the distance is readily determined. Suppose, in the
last example, that the change of velocity from 30 to 50 feet took
place in a distance of 300 feet, then
124068
 j)^^ = 41356 lbs. = F, the force constantly applied during
300 feet.
10. If a sleigh, weighing 2000 lbs., after describing a distance of
250 feet, has completely lost a velocity of 100 feet, what constant
resistance does the friction offer ?
Since the terminal velocity = 0, the height due to it = 0, hence
(100)^ 2000
G44 ^,250 '
= 12422352 1!
We have been calculating upon the principle of vis vtca; but the
product of the mass and the squaie of the velocity, without attach
ing to it any definite idea, is termed the vis viva, or living force.
11. A body weighing 2300 ibs. moves with a velocity of 20 feet
the second, required the vis viva?
2300
^^ = 7142857 lbs., mass.
7142857 X (20)= = 28571428, the amount o? vis vim.
Hence, if a mass enters from a velocity a, into another v, the
unit of work done is equal to half the difference of the vis vii:a, at
the commencement and end of the change of velocity.
For if tho mass be put = M, and W the weight,
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STRESGTH OP MATERIALS. 271
Then M = — , and tlie vis viva to velocity n = Ma^ = ■ ;
and the vis viva to velocity v = Mu' = ■
_T.,»l{f^^} = (lf;/xW(II*)W.f„,
q and q, give the heights due to the velocities v and a, respec
tively. The useful formula
Fs = {H  Ji) W,
before given, pago 270, may be applied to variable as well as to
constant forces, if, instead of the constant force F, the mean value
of the force bo applied.
STRENGTH OF MATEEIAIS.
EMPLOYED IN THE CONSTEUCTIOX OP MACHINES.
In theoretical mechanics, ire deal with imaginary quantities, which
are perfect in all their properties ; they are perfectly hard, and
perfectly elastic ; devoid of weight in statics and of friction in dy
namics. In practical mechanics, we deal with real material objects,
among which we find none which arc perfectly hard, and none, ex
cept gaseous bodies, which are perfectly elastic ; all have weight,
and experience resistance in dynamical action. Practical mechanics
is the science of automatic labour, and its objects are machines and
their applications to the transmission, modification, and regulation
of motive power. In this it takes as a basis the theoretical deduc
tions of pure mechanics, but superadds to the formulte of the ma^
thematician a multitude of facts deduced from observation, and ex
perimentally elaborates a now code of laws suited to the varied con
ditions to be fulfilled in the economy of the industrial arts.
In reference to the structure of machines, it is to be observed
that however simple or complex the machine may he, it is of im
portance that its parts combine lightness with strength, and rigidity
■with uniformity of action ; and that it communicates the power
without sliocks and sudden changes of motion, by which the passive
resistances may be increased and the effect of the engine dimi
nished.
To adjust properly the disposition and arrangement of the indi
vidual members of a machine, implies an exact knowledge and esti
tnate of the amount of strain to which they are respectively subject
in the working of the machine ; and this skill, when exercised in
conjunction with an intimate acquaintance with the nature of the
materials of which the parts are themselves composed, must con
tribute to the production of a machine possessing the highest amount
of capability attainable with the given conditions.
Materials. — The material most commonly employed in the con
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212 THE PRACTICAL MODEL CALCTJLAIOK.
atruction of macliinery is iron, in the two states of cast and wrougld
or forged iron ; and of these, there are several varieties of quality.
It becomes therefore a problem of much practical importance to
determine, at least approximately, the capabilities of the particular
material employed, to resist permanent alteration in the directions
in which they are subjected to strain ia the reception and trans
mission of the motive power.
To indicate briefly the fundamental conditions which determine
the capability of a given weight and form of material to resist a
given force, it must, in the first place, be observed, that rupture
may take place either by tension or by compression in the direc
tion of the length. To tlie former condition of strain is opposed
the tenacity of the material ; to the other ia opposed the resistance
to the cruBhing of its mhstance. Rupture, by transverse strain, is
opposed both by the tenacity of the material and its capability to
withstand compression together of its particles. Lastly, the bar
may be ruptured by torsion. Mr. Oliver Byrne, the author of the
present work, in his New Theory of the Strength of Materials has
pointed out new elements of much importance.
The capabilities of a material to resist extension and compression
are often different. Thus, the soft gray variety of cast iron offers
a greater resistance to a force of extension than the white variety
in a ratio of nearly eight to five; but the last offers the greatest
resistance to a compressing force.
The resistance of cast iron to rupture by extension varies from
6 to 9 tons upon the square inch ; and that to rupture by compres
sion, from 36 to 65 tons. The resistance to extension of the best
forged iron may be reckoned at 25 tons per inch ; but the corre
sponding resistance to compression, although not satisfactorily ascer
tained, is generally considered to be greatly less than that of cast
iron. Roudelet makes it 31J tons on the square inch. Cast iron
(and even wood) is therefore to be preferred for vertical supports.
The forces resisting rupture are as the areas of the sections of
rnpture, the material being the same ; this principle holds not only
in respect of iron, but also of wood. Many inquiries have been in
stituted to determine the commonly received principle, that the
strength of rectangular beams of the same width to resist rupturo
by transverse strain is as the squares of the depths of the beams.
In these respects the experiments, although valuable on account
of their extent and the care with which they were conducted, pos
sess little novelty ; but in directing attention to the elastic proper
ties of the materials experimented upon, it was found that the re
ceived doctrine of relation between the limit of elasticity and weight
requires modification. Tho common assumption is, that the de
struction of the elastic properties of a material, that is, the dis
placement beyond the elastic limit, does not manifest itself until
the load exceeds onethird of the breaking weight. It was found,
however, on the contrary, that its effect was produced and mani
fested in a permanent set of the material when the load did not ex
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STRENOTH OP MATEKIALE. 273
ceed onesixteenth of that necessury to produce rupture. Thus a
bar of one inch equare, supported between propa 4J feet apart, did
not break till loaded with 496 lbs. but showed a permanent deflec
tion or set when loaded with 16 lbs. In other cases, loads of 7 lbs.
and 14 lbs. were found to produce permanent sets when the break
ing weights were respectively 364 lbs. and 1120 lbs. These sets
were therefore given by j'jd and j^tb of the breai;ing weights.
Since these results were obtained, it has been found that time
and the weight of the material itself are sufficient to effect a per
manent deflection in a beam supported between props, so that there
would seem to be no such limits in respect to transverse strain as
those known by the name of elastic limits, and consequently the
principle of loading a beam within the elastic limit has no founda
tion in practice. The beam yields continually to the load, but witii
an exceedingly slow progression, until the load approximates to the
breaking weight, when rupture speedily succeeds to a rapid deflection.
As respects the effect of tension and compression by transverse
strain, it was ascertained by a very ingenious experiment that equal
loads produced equal deflections in both cases.
Another moat important principle developed by experiments, fa
that respecting the compression of supporting columna of different
heights. When the height of the column exceeded a certain limit,
it was found that the crushing force became constant, and did not
increase as the height of the column increased, until it reached
another limit at which it began to yield, not strictly by crushing,
but by the bending of the material. The first limit was found to be
a height of little leas than three times the radius of the column ;
and the second double that height, or about six times the radius of
the column. In columns of different heights between these limits,
having equal diameters, the force producing rupture by compression
was nearly constant. When the column was less than the lower
limit, the crushing force became greater, and when it was greater
than the higher limit, the crushing force became less. It was fur
ther found that in all cases, where the height of the column was
exactly above the limits of three times the radius, the section of
rupture was a plane inclined at nearly the same constant angle of
55 degrees to the axis of the column. These facts mutually ex
plain each other ; for in every height of column above the limit,
the section of rupture being a plane at the same angle to the axis
of the colutan, must of necessity bo a plane of the same size, and
therefore in each case the cohesion of the same number of particles
must bo overcome in producing rupture. And further, the same
number of particles being to be overcome under the same circum
stances for every different height, the same force will be required
to overcome that amount of cohesion, until at double the height
(three diameters) the column begins to bead under ita load. This
height being surpassed, it follows that a pressure which becomes
continually lesa as the length of the column is increased, will be
sufficient to break it.
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2V4 TEE PRACTICAL MODEL CALCULATOR.
This property, moreover, is not confined to cast iron ; the ex
periments of M.'Rond.elet show that with coluaiiis of wrought iron,
wood, and stone, similar results are obtained.
Ejom theae facts then, it appears that if supporting columns be
taken of different diameters, and of heights so great aa not to allow
of their bending, yet suf&ciently high to allow of a complete sepa
ration of the planes of fracture, that is, of heights intermediate to
three times and sis times their radius, then will their strengths be
as the number of particles in their planes of fracture ; and the
planes of fracture being inclined at eijnal angles to the axes of tbe
columns, their areas will be as the transverse sections of the .co
lumns, and consequently the strengths of the columns will be as
tbeir transverse sections respectively. Taking the mean of three
experiments upon a column \ inch diameter, the crushing force was
6426 lbs. ; whilst the mean of foiu' experiments, Conducted in ex
actly the same manner, upon ^ column of f of an inch diameter,
gave 14542 lbs. The diameters of the columns being 2 to 3, the
areas of transverse section were therefore 4 to 9, which is very
nearly the ratio of the crushing weights.
When the length of the column ia so great that jta fracture ia
produced wholly by bending of its material, the limit has been fixed
for columns of cast iron, at 30 times the diameter when the ends
are flat, and 15 times the diameter when the ends are. rounded. ,In
shorter columns, fracture takes place partly by crushing and partly
by bending of the material. When the column is enlarged in the
middle of its length from one and a half to two times the diameter
of the ends, the strength was found by the same experimenter to be
greater by oneseventh than in solid columns containing the same
quantity of iron, in the same length, with their extremities rounded ;
and stronger by an eighth or a ninth when the extremities were flat
and rendered immovable by disks.
The following formulas give the absolute strength of cylindrical
columns to sustain pressure in the direction of their length. In
these formulas
D = the external diameter of the column in inches.
d = the internal diameter of hollow columns in inches.
L = the length of the column in feet,
W = the breaking weight in tons.
CliMaolMoEtoBSoIniim.
MdinglS
Length ofths oolumi oM
=dii>B30
Solid cylindiioal col
lumn of oast iron, /
HolloiT ojlindricftl co "I
Inmnof oast iron, . /
Solid cylindrical ■ 001
lumn of wtoaght iron, j
W = ^'8l^
rf»
Both ends tst.
T 11 "1^"'
d^"
W..SS..J
For shorter columns, if W' represent the wfeight in tons which
would break the column by bending alone, as given by tbe preced
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STEENGTII OF MATERIALS. 275
ing formulas, and W" tlio weight in tons which would crush the co
lumn without bending it, as determined from the subjoined table,
then the absolute breaking weight of the column W, is represeuted
in tons by the formula,
" ~w + w
These rules require the use of logarithm? in their application^.
When a beam is deflected bj transverse strain, the material on
that side of it on which it sustains the strain is compressed, and the
material on the opposite side is exteyided. Tbe imaginary surface
at which the compression terminates and the extension begins — at
which there is supposed to be neither extension nor compression —
is termed the neutral axis of the beam. What constitutes the
strength of a beam is its resistance to compression on the one side
and to extension on the other side of that axis — the forces acting
about the line of axis like antagonist force at the two extremities
of a lever, so that if either of them yield, the beam will he broken.
It becomes, however, a question of importance to determine the re
lation of these forces ; in other words, to determine whether the
beam of given form and material will yield first to compression or
to extension. This point is settled by reference to the columns of
the subsequent table, page 280, in which it will be observed that the
metals require a much greater force to crush them than to tear them
asander, and that the woods require a much smaller force.
There is also another consideration which must not be overlooked.
Bearing in mind the condition of antagonism of the forces, it is ob
vious, that the further these forces are placed from the neutral
axis, that is, from the fulcrum of their leverage, the greater must
be their effect. In other words, all the material resisting compres
sion will produce its greatest effect when collected the farthest possi
ble from the neutral axis at the top of the beam ; and, in like man
ner, all the material resisting extension will produce its greatest
effect when similarly disposed at the bottom of the beam. We are
thus directed to the first general principle of the distribution of the
material into two flanges — one forming the top and the other the bot
tom of the beam — joined by a comparatively slender rib. Associat
ing with this principle the relation of the forces of extension and
compression of the material employed, we arrive at a form of beam
in which the material is so distributed, that at the instant it is about
to break by extension on the one Side, it i« ibout to bicik by n
pression on the other, and consequently is of the
strongest form. Thus, supposing that it is le
quired to determine that foim m a girder of
cast iron: the ratio of the crushing foice of
that metal to the force of extension raxj be •
taken generally as 6^ to 1, which ii theietore al«o the ratio of the
lower to the upper flange, as in the annexed sectional dngrim
A series of nine castings weie made, giadually mcieasmg the
lower flange at the expense of the uppei one, and m the fiist aight
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276 THE PRACTICAL MODEL CALCULATOE.
experiments the beam broke by tbe tearing asunder of the longer
flange ; and in the last experiment the beam yielded by the crush
ing of the upper flange. In the eight experiments the upper flange
was therefore the weakest, and in the ninth the strongest, so that
the form of maximum strength was intermediate, and very closely
allied to that form of beam employed in the last experiment, nhich
was greatly the strongest. The circumstances of these experiments
are contained in the following table.
No »t siperi
Rilio of surfaces of oom
1
1 tol
282
2368
2
Ito 2
287
2567
3
lto4
302
2737
4
lto4i
337
3183
5
Ito 4
450
3214
fl
4 to 51
600
3346
7
Ito 31
4628
3246
8
4 to 43
58()
3317
»
1 to 61
64
4075
To determine the weight necessary to hreak beams cast according
to the form described :
Multiply the area of the section of tbe lower flange by the depth
of the beam, and divide the product by the distance between the
two points on which the beam is supported : this quotient multi
plied by 536 when the beams are east erect, and by 514 when they
are cast horizontally, will give the breaking weight in cwts.
From this it is not to be inferred that the beam ought to have
the same transverse section throughout its length. On the con
trary, it is clear that the section ought to have a definite relation
to the leverage at which the load acts. From a mathematical con
sideration of the conditions,
it indeed appears that the
effect of a given load to
break the beam varies when
it is placed over different yr —
points of it, as the products ^—^
of the distances of these points from the points of support of \\\:i
beam. Thus the eff'ect of a weight pVced at the point W^ is to the
effect of the same weight acting upon the point W^, as the product
AWj X W^ B is to the product AW, X W^ E ; the points of sup
port being atA andB. Since then the effect of a weight increnses
as it approaches the middle of the length of the beam, at which it
is a maximum, it is plain that the beam does not require to have
the same transverse section near to its extremities as in the middle ;
and, guided by the principle stated, it is easy to perceive that its
strength at different points should in strictness vary as the products
of the distances of these points from the points of support. By
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STREKKTH OP MATERIALS. 2TY
taking this law as a fundamental condition in the distribution of
the strength of a beam, whose load we may conceive to be accumu
lated at the middle of ita length, we arrive at the strongest form
which can be attained under given circumstances, with a given
amount of material; we arrive at that form which renders the beam
equally liable to rupture at every point. Kow this form of masi
mum strength may be attained in two ways ; either by varying the
depth of the beam according to the law stated, or by preserving
the depth everywhere the same, and varying the dimensions of the
upper and lower flanges according to the same law. The conditions
are manifestly identical. We may therefore assume generally the
condition that the section is rectangular, and that the thickness of
the flanges is constant; then the outline determined by the law in
question, in the one case of the elevation of the beam and in the
other of the plan of the flanges, is the geometrical curve called a
parabola — rather, two parabolas joined base to base at the middle
between the points of support. The annexed diagram represents
the plan of a castiron girder according to this form, the depth
being uniform throughout. Both flanges are of the same form,
but the dimensions of the upper one are such as to give it only a
sixth of the strength of the other.
This, it will be observed, is also the form, considered as an ele
vation, of the beam of a steam engine, which good taste and regard
to economy of material have rendered common.
It must, however, be borne in mind, that in the actual practice
of construction, materials cannot with safety be subjected to forces
approaching to those which produce rupture. In machinery espe
cially, they are liable to various and accidental pressures, besides
those of a permanent kind, for which allowance must be made.
The engineer must therefore in his practice depend much on expe
rience and consideration of the species of work which the engine is
designed to perform. If the engine be intended for spinning,
pumping, blowing, or other regular work, the material may be sub
jected to pressures approaching twothirds of that which would ac
tually produce rupture ; but in engines employed to drive bone
mills, stampers, breakingdown rollers, and the like, double that
strength will often be found insufficient. In cases of that nature,
experience is a better guide than theory.
It is also to be remarked that we are often obliged to depart
from the form of strength which the calculation gives, on account
of the partial strains which would be put upon some of the parts
of a casting, in consequence of unequal cooling of the metal when
the thicknesses are unequal. An expert founder can often reduce
the irregular contractions which thus result ; but, even under the
best management, fractm'e is not unfrequently produced by irregu
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278 THE PRACTICAL MODEL CALCULATOR.
larity of cooling, and it is at all times better to avoid tlie danger
entirely, than to endeavour to obviate it by artifice. For this rea^
son, the parts of a casting ought to be as nearly as possible of such
thickness as to cool and contract regularly, and by that means all
partial strain of the parts will be avoided.
With respect to design, it is also to be remarked, that mere theo
retical properties of parts will not, under all the varieties of circum
stances which arise in the working of a machine, insure that exact
adjustment of material and propriety of form so much desired ia
constructive mechanics. Every design ought to take for its basis
the mathematical conditions involved, and it would, perhaps, be im
possible to arrive at the best forms and proportions by any more
direct mode of calculation ; but it is necessary to superacid to the
mathematical demons tratiftn, the exercise of a wellmatured judg
ment, to secure that degree of adjustment and arrangement of parts
in which the merits of a good design mainly consist. A purely
theoretical engine would look strangely deficient to the practised
eye of the engineer ; and the merely theoretical contriver would
speedily find himself lost, should he venture beyond his construction
on paper. His nice calculations of the " work to be performed," of
the vis viva of the mechanical organs of his machine, and of the modu
lus of elasticity of his material, would, in practice, alike deceiie him.
The first consideration in the design of a machine is the quantity
of work which each part has to perform — in other words, the forces,
active and inactive, which it has to resist ; the direction of the
forces in relation to the crosssection and points of support ; the
velocity, and the changes of velocity to which the moving parts are
subject. The calculations necessary to obtain these must not be
confined to theory alone ; neither should they be entirely deduced
by " rule of thumb ;" by the first mode the strength would, in all
probability, be deficient from deficiency of material, and by the
second rule the material would be injudiciously disposed ; weight
would bo added often where least needed, merely from the deter
mination to avoid fracture, and in consequence of a want of know
ledge respecting the true forms best adapted to give strength.
To the following general principles, in practice, there are but
few real exceptions :
I. Direct Strain.—To thib a utiaight line n
if the part be of consideidble
length, vibration ought to be coun
teracted by intersection of planes,
(technically feathers,) as repre
sented in the annexed diagiams,
or some such form, consistent with the purpo=ie for i\hich the pirt
is intended.
II. Transverse Strain. — To this a parabolic form of section must
be opposed, or some simple figure including the parabolic form.
For economy of material, the vertex of the curve ought to bo at
the point where the force is applied; and when the strain passes
hv Google
STREKGTH OP MATERIALS. 279
alternately from one side of the part to the other, the curve ought
to be on both sides, as in the beam of a steam engine.
When a loaded piece is supported at one end only, if the breadth
bo everywhere the same, the form of equal strength is a triangle ;
but, if the section be a circle, then the solid will be that generated
by the rcYolution of a semiparabola about its longer axis. In prac
tice, it Tvill, however, be sufficient to employ the frustum of a cone,
of which, in the case of cast iron, the diameter at the unsupported
end is onethird of the diameter at the fixed end.
III. Torsion. — The section most commonly opposed to torsion
ia a circle ; and, if the strain be applied to a cylinder, it is obvious
the rupture must first take place at the surface, where the torsion is
greatest, and that the further the material is placed from the neutral
axis, the greater must be its power of resistance ; and hence, the
amount of materials being the same, a shaft is stronger when made
hollow than if it were made solid.
It ought not, however, to be supposed that the circle ia the only
figure which gives an axis the property of off'ering, in every direc
tion, the same resistance to flexure. On the contrary, a square sec
tion gives the same resistance in the direction of its sides, and of
its diagonals ; and, indeed, in every direction the resistance is equal.
This is, moreover, the case with a great number of other figures,
which may be formed by combining the circle and the square in a
symmetrical manner ; and hence, if the axis, strengthened by salient
sides, as in feathered shafts, do not answer as well as cylindrical
ones, it must arise from their not being so well disposed to resist
torsion, and not from any irregularities of flexure about the axis
inherent in the particular form of section.
This subject has been investigated with much care, and, accord
ing to M. Cauchy, the modulus of rupture by torsion, T, is con
nected with the modulus of rupture by transverse strain S, by the
simple analogy T = J S.
'I'he forms of all the parts of a machine, in whatever situation
and under every variety of circumstances, may be deduced from
these simple figures ; and, if the calculations of their dimensions
be correctly determined, the parts will not only possess the requi
site degree of strength, but they will also accord with the general
principles of good taste.
In arranging the details of a machine, two circumstances ought
to be taken into consideration. The first is, that the parts subject
to wear and influenced by strain, should be capable of adjustment ;
the second is, that every part should, in relation to the work it has
to perform, be equally strong, and present to the eye a figure that
is consistent with its degree of action. Theory, practice, and taste
must all combine to produce such a combination. No formal law
can be expressed, either by words or figures, by which a certain
contour should be preferred to another ; both may be equally strong
and equally correct in reference to theory ; custom, then, must be
appealed to as the guide.
hv Google
THE PRACTICAL MODEL CALCULATOR.
hv Google
STaENGTII OP MATERIALS.
THE COHESIVE STRENOTH t
The following Table contains the result of experiments on the
cohesive Btrength of various hodies in avoirdupoie pounds ; also,
onethird of the ultimate strength of each body, this being consi
dered sw^cient, in most cases, for a permanent load:
Kim^ofada
=^aar8Bsr
Oiiath ni
Bound KiT
Uu. tlinl.
1.00D3.
v..
»
a.
a
2onoo
66b7
15, OS
5230
Ah
17000
5667
13357
4452
Teak
15000
5000
11791
3927
Fir
12000
4000
J424
3141
Beach
UjOO
3&ff
90o2
soil
Oal:
IIUOO
361)7
8tio9
2b80
Ca«tiron
1&656
6219
I46o2
4884
English wrought iron
65872
ia24
43881
14027
Swedish do <]o
720b4
24021
56599
iBaeo
Blistered steel
183152
44381
104d7T
34859
Shear do
124400
418bb
97703
32jG8
Cnst do
lS42o6
44T52
10a4o4
85151
10072
6357
14979
4993
Wrought do
3^7J2
112b4
20540
8837
Telbw brass
ITOfS
1989
14112
404
Cait tm
4 (>
15 9
sit
1 39
Cast leai
IS 4
14 2
47
IROLLE'U I
Rule — To find thi. ultimate cohem e stt enjth of squ ire, round,
and rectangular hats, of any of the vat lous bodies, as specified in
the table — Multiply the strength of an incli bai, (as in the table,)
of the body required, by the ero?a sectional aiea of square ana
rectangulai bar=!, or by the square of the diameter of round bars ;
and the product it ill be the ultim do coht'^ne stiengtb
A bar of ca^t iion being 1^ inches square, lequiied its cohesiyo
power.
15 X 15 X 18656 = 41976 lbs.
Required the cohesive force of a bar of English wrought iron,
2 inches broad, and f of an inch in thickness.
2 X 375 X 55872 = 41904 lbs.
Required the ultimate cohesive strength of a round bar of
wrought copper I of an inch in diameter.
■75^ X 26540 = 1492875 lbs.
PROBLEM ir.
Rule, — Tlie weight of a body being given, to find the eross sec
tional dimensions of a bar or rod capable of sustaining that weight. —
Por square and round bars, divide the weight given by ouethird
of the cohesive strength of an inch bar, (as specified in tlie table,)
and the square root of the quotient will he the side of the square,
or diameter of the bar in inches.
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282
THE PRACTICAL MODEL OALCL'LATOE,
And if rectangular, divide the quotient by tlie breadth, and the
result will be the thickness.
What must be the side of a square bar of Swedish iron to bus
tain a permanent weight of 18000 lbs ?
■ynTfWV ^ '^^' ^^ nearly  of an inch square.
Required the diameter of a round rod of cast copper to carry
a weight of 6800 lbs.
6800
"^4993 "
A bar of English wrought iron is to be applied to carry a weight
of 2760 lbs. ; required the thickness, the breadth being two inclies.
2760
j = 1.16 inches diameter.
: = 142 
■071 of an inch in thickness.
A Table skoiving the circumference of a rope equal to a chain
made of iron of a given diameter, and the tveight in tonn that
each is proved to carry ; also, the weight of a foot of chain made
from iron of tlmt d '
^
Provofl b tjrrj
nvisht of . lintal
■in tons.
3
JandJ,
1
108
i
i
2
16
4f
land J,
3
2
6i
i
4
27
6
} md ;,
5
33
6J
^
G
4
7
fi and X
8
4'6
n
f
n
55
8
(and J,
111
61
9
i
13
7.2
9i
{and ft
15
84
101
1 inch.
18
94
ON THE TEANSVEnSE STRENGTH OF BODIES.
The tranverse strength of a body is that power which it exerts
in opposing any force acting in a perpendicular direction to its
length, as in the case of beams, levers, &c., for the fundamental
principles of which observe the following : —
That the transverse strength of beams, &c. is inversely as their
lengths, and directly as their breadths, and square of their depths,
and, if cylindrical, as the cubes of their diameters ; that is, if a
beam 6 feet long, 2 inches broad, and 4 inches deep, can carry
2000 lbs., another beam of the same material, 12 feet long, 2 inches
broad, and 4 inches deep, will only cairy 1000, being inversely as
their lengths. Again, if a beam 6 feet long, 2 inches broad, and
4 inches deep, can support a weight of 2000 lbs., another beam of
hv Google
ETEENQTH OP MATEEIALS, 283
tht ame niateiiil, C ttetlong, 4 inclica Lioad, ml 4 iiidies deep,
ynll suppoit double that weight, being dnectlj 15 then bieadths,
— tut a heim of that mitenal, G ftet long, 2 inches bro"d, and
8 inches deep, will austain a weight of &0U0 Iba , being as the
squaie of their depths
Fiom a mean ot experiments made, to asceinm the ti ms^trse
stength of various bodie", it appear'* that the ultimite stieu^th
of an inch scjuaie, and an mcii round bar of eath, 1 font long,
loaded m the middle, and lying looie at both end., is neailj a'^
follows, m lb=! a\o]idupoi«
Nsmes ru 1 „
,a.reEar
Ca lied
Hnond Usr
Ons 11 rd
800
217
6 3
201
Ash.
11 "
8=15
519
1 <l
417
!49
Pitch pine
S16
SOj
719
239
Deal
4J4
14S
Cast iron
2j80
8fO
202r
f75
4013
lo3S
31)^
10^0
PROBLEM I
Rl/LC — To find th ultiniate hansietie Birength of any iectan
gula) beam, auppoited at bcfk ends, and loaded in the mtddh, or
supported tn the middle, and loaded at both ends; also, when the
weight is between the middle and the end; likewise when fixed at
me end and loaded at the other. — Multiply the strength of an inch
square bar, 1 foot long, {as in the table,) by the breadth, and square
of the depth in inches, and divide the product by the length in
feet ; the quotient will be the weight in lbs. avoirdupois.
What weight will break a beam of oak 4 inches broad, 8 inches
deep, and 20 feet between the supports ?
800 X 4 X 8^
20 — ^ = ^^^*^ ^^^■
When a beam is supported m the middle, and loaded at each
end, it will bear the same i\eight as when supported at both ends
and loaded in the middle , that is, each end will bear half the
weight.
When the weight is not situated m the middle of the beam, but
placed somewhere between the middle and the end, multiply twice
the length of the long end by twice the length of the short end, aud
divide the product by the whole length of the beam ; the quotient
will be the effectual length,
Required the ultimate transverse strength of a pitch pine plank
24 feet long, 3 inches broad, 7 inches deep, and the ^veight placed
8 feet from one end.
2 X 16
24 '^
 213 eifective length.
hv Google
284 IIIB PRACTICAL MODEL CALCULATOR.
Again, wben a beam is fixed at one end and loaded at tlic otlier,
it will only bear J of the weight aa when supported at both eads
and loacled in the middle.
What is the weight requisite to break a deal beam 6 inches broad,
9 inches deep, and projecting 12 feet from the wall ?
506^j<jP _ 22923 + 4 = 57307 lb..
The same rules apply aa well to beams of a cylindrical form,
with this exception, that the strength of a round bar (as in the
table) is multiplied by the cube of the diameter, in place of the
breadth, and square of the depth
Required the ultimate transverse strength of a solid cylinder of
cast iron 12 feet long and 5 inches diameter.
2026 X 5^
 j^ = 21104 lbs.
IVhat is the ultimate transverse strength of a hollow shaft of
cast iron 12 feet long, 8 inches diameter outside, and containing
the same cross sectional area as a solid cylinder 5 inches diameter 't
v/8^  5' = 624, and 8=  624^ = 269.
2028 X 269
Then, j^ = '^^^^^ ''^s
When a beam is fixed at both ends, and loaded in the middle, it
will bear onehalf more than it will when loose at both ends.
And if a beam is loose at both ends, and the weight laid uni
formly along its length, it will bear double ; but if fixed at both
ends, and the weight laid uniformly along its length, it will bear
triple the weight.
raoBLEJi II,
Rule. — To find the breadth or depth of beams intended to suj)
port a permanent wie^j'Ai.— Multiply the length between the sup
ports, in feet, by the weight to be supported in lbs., and divide tho
product by onethird of the ultimate strength of an inch bar, (as
in the table,) multiplied by the square of the depth ; the quotient
wilt be the breadth, or, multiplied by the breadth, the quotient will
be the square of the depth, both in inches.
Required the breadth of a cast iron beam IG feet long, 7 inches
deep, and to support a weight of 4 tons in the middle.
, 8960 X 16
4 tons = 8960 lbs. and ^^ ^ .^ ■ = 34 inches.
What must be the depth of a cast iron beam 34 inches broad,
16 feet long, and to bear a permanent weight of four tons in the
middle ?
8960 "x 'Tir
hv Google
STREKGTH OP MATERIALS. 285
When a beam is fixed at both ends, the divisor must be multi
plied by 15, on account of it being capable of bearing onehalf
When a beam is loaded uniformly throughout, and loose at both
ends, the divisor must be multiplied by 2, because it nill bear
double the weight.
If a beam is fast at both ends, and loaded uniformly throughout,
the divisor must be muitipled by 3, on account that it will bear
triple the weight.
Required the breadth of an oak beam 20 feet long, 12 inches
deep, made fast at both ends, and to be capable of supporting a
weight of 12 tons in the middle.
26880 X 20
12 tons = 26880 lbs., and 266 x !'>' x 15 ^ ^'"^ inches.
Again, when a beam is fixed at one end, and loaded at the other,
the divisor must be multiplied by '25 ; because it will only bear
onefourth of the weight.
Required the depth of a beam of ash 6 inches broad, 9 feet
projecting from the wall, and to carry a weight of 47 cwt.
5264 X 9
47 cwt. = 5264 lbs., and v^ ora v fi x ■S'i ~ ^'^^ inches deep.
And when the weight is not placed in the middle of a beam, the
effective length must be found as in Problem I.
Required the depth of a deal beam 20 feet long, and to support
a weight of 63 cwt. 6 feet from one end.
28 X 12
"" 2C} — ~ ^^'^ eifective length of beam, and
63 cwt. = 7056 lbs. ; hence
%/ iQQ X B ~ I0'24 inches deep.
Beams or shafts exposed to lateral pressure are subject to all the
foregoing rules, but in the case of waterwheel shafts, &c., some al
lowances must be made for wear ; then the divisor may be changed
from 6T5 to 600 for cast iron.
Required the diameter of bearings for a waterwheel shaft 12
feet long, to carry a weight of 10 tons in the middle.
10 tons = 22400 lbs., and
92400 ^
f,f.r. = ^448 = 7'65 inches diameter.
And when the weight is equally distributed along its length, the
cube root of half the quotient will be the diameter, thus :
448
g = ^224 = 607 inches diameter.
Required the diameter of a solid cylinder of cast iron, for the
shaft of a crane, to he capable of sustaining a weight of 10 tons ;
hv Google
286 THE PBACTIOAL MOBEL CALCULATOR.
one cnfl of the shaft to he made fast in the ground, the other to
projfcct GJ feet; and the effective leverage of the jih as IJ to 1.
10 tons = 22400 lbs., and
22400 X 65 X 175
^^75^^25 = ^^0^
And ^150y = 1147 inches diameter.
The strength of cast iron to wrought iron, ia this direction, is as
9 is to 14 nearly ; hence, if wrought iron is taken in place of cast
iron in the last example, what must be its diameter ?
ISUil X y ^„„. ,
V TT — ■ = y'o9 inches diameter.
ON TORSION OK TWISTING.
The strength of bodies to resist torsion, or wrenching asunder,
is directly as the cubes of their diameters ; or, if square, as the
cube of one side ; and inversely as the force applied multiplied into
the length of the lever.
Hence the rule. — 1. Multiply the strength of an inch bar, by
experiment, {as in the following table,) by the cube of the diameter,
or of one side in inches ; and divide by the radius of the wheel, or
length of the lever also in inches ; and the quotient will be the ul
timate strength of the shaft or bar, in lbs, avoirdupois.
2. — Multiply the force applied in pounds by the length of the
lever in inches, and divide the product by onethird of tlie ultimate
strength of an inch bar, (as in the table,) and the cube root of the
quotient will be the diameter, or side of a square bar in inches ;
that is, capable of resisting that force permanently.
The following Table contains the result of experiments on ineh hars,
of various metals, in
lbs. avoirdupois.
N=m»„0^i»
Rl.miB:.t.
0.»UlM.
S,=»r. Bar.
0«.>,I.M.
11943
120G3
11400
20025
20508
211 U
6549
4825
1G98
1206
SflSl
4021
3800
6876
7037
1850
1C08
563
402
15206
15360
14502
25i97
26112
T065
0144
2150
1536
5069
5120
4864
8409
8701
8960
2355
2048
717
English wrouglit iron
SweiUsl. <lo. do.
Cast do
Tin
What weight, applied on the end of a 5 feet lever, will wrench
asunder a 3 inch round bar of cast iron ?
.Tfj = 5374 lbs. avoirdupois.
Required the side of a square bar of wrought iron, capable of re
sisting the twist of 600 lbs. on the end of a lover 8 feet long.
()60~x 116
^"""51:^0"" =^i inches.
hv Google
STRESGTn OP MATERIALS. 287
In the case of revolving shafts for machinery, &c., the strength
is directly as the cuhes of their diameters, and revolutions, and in
versely C5 the resistance they have to overcome ; hence,
From praetiee, we find that a 40 horse power steam engine,
making 25 revolutioaa per minute, requires a shaft (if made of
•wrouffhtiron) to be 8 inches diameter : now, the cube of 8, multi
plied by 25, and divided by 40 = 320 ; which serves as a constant
multiplier for all others in the same proportion.
What must be the diameter of a wrought iron shaft for an engine
of G5 horse power, making 23 revolutions per minute ?
65 X 320 „„„ .
■&" ng = 9'67 inches diameter.
James Glenie, the mathematician, gives 400 as a constant mul
tiplier for cast iron shafts tliat are intended for first movers in ma
chinery ;
200 for second movers ; and
100 for shafts connecting smaller machinery, &c.
The velocity of a 30 horse power steam engine is intended to be
19 revolutions per minute. Kerjuired the diameter of bearings for
the flywheel shaft.
400 X 30
= 8579 inches diameter.
Required the diameter of the hearings of shafts, as second movers
from a 30 horse engine ; their velocity being 36 revolutions pet
minute.
200 X 30
■^ OQ — = 55 inches diameter.
"When shafting is intended to be of ivrougbt iron, use 100 as the
multiplier for second movers ; and 80 for shafts connecting smaller
machinery.
Table of the Proportionate Length of Searings, c
Shafts of V, ■ " ■
Journals for
ni^.i.7a=b=».
I.en,inl»d.w
Bii.fBl»eha
La=.i„I»cW
1
li
61
8
11
2i
1
91}
3
71
10
2i
H
8
lOJ
21
31
81
111
S
4L
9
12
31
4
91
121
4
5
10
131
41
6
lOJ
14
6>
11
14}
61
71
11}
15.
6
8i
12
10
b,Google
THE PEACTICAL MODEL CALCULATOE.
uiti't, Ris fiances to Compression, and other Properties of tlic
•.oimnon Materials o/ Construction.
Al«tof
coDir
isd^ltbC
=t .c™.
''*"" ^'f^'"
'^ni::t
to Ita per Ki.
Ita Hrsnilh
tiii'/t'
I««iff„e«i»
A«li
uua
_
023
26
0089
Betth
12225
8i,48
015
21
0073
17308
10304
0435
09
049
Bnok
275
562
Cast iron
13434
80^97
I'OOO
10
1000
Copper (wTOnglit)
3u(J0O
Elm
9720
1038
021
29
0073
Fir, or Pine, white
1J346
2028
023
24
01 
— — red
llbOO
6^75
03
24
01
  yellDw
llS3o
5445
025
29
0087
Granite, Aberdeen
10910
Griin metal (copper 8
andtml)'
GSSOS
065
125
053B
MaUeable iron
56000
113
086
Laieli
12240
5jb8
0136
23
0058
I*ad
1824
0096
25
00385
Mahogany, Honduras
1145
8000
024
29
0487
Majble
5e>1
faOuu

Oak
118M)
1j04
025
28
OO08
Rope (1 in m oircum )
200

Steel
128000

Stone Baih
478
_
— Craigleith
772
6410
_
— Dundee
2661
6630
_
— Portland
857
3720
Tin (cBsil
47^1.
0182
075
025
Zino (slice c)
0120
—
0S6i
05
076
Comparative Strength and Weight of Ropes and Chains.
1
II
A
..„,
1
1
■1
3 a
T"
'4
:2«;:
n
■4
a
ii
(i
»
u
J3
43
3!
2f
ft
5J
1 6S
10
23
i
10
l',
4>,
8
1 16J
XOj
28
if
40
11 11
h
S
;,
IIH
2 10
lU
30!
lin.
hli
13 8
S!
7
s
M
3 5i
121
36
lA
63
14 18
61
S)i
A
IK
4 3.,
13
30
n
71
16 14
7
in
■ffl
6 2
13}
46
1*
70
18 11
H
IS
«
•27
6 4J
141
4K1
1i
Kl
20 8
«i
i»
1
32
7 7
l.H
66
ift
06
22 13
»i
21 la
37
8 13S
16
60
■If
106
24 18
It must be understood and also borne in minii, that in estimating
the amount of tensile strain to which a body is subjected, the weight
of the body itself must also be taken into account ; for according
to its position so may it approsimate to its whole weight, in tcnd
hv Google
STRENGTH OP MATBMALS. 289
ing to produce tension within itself; as in the almost constant
application of ropes and chains to great depths, considerable
heights, ko.
Alloys that are of greater Tenacity than the sum of their Constitu
ents, as determined hy the Experiments of Musehenbroeh
Swedish oopper 6 parts, Malacca tin 1 — teoacitj per aquire jnch 64,000 lbs,
CbiU copper 8 parts, Malacca tin 1 60,000
Japan copper 5 parts, Banca tin 1 57,000
Anglesea oopper 6 parts, Cornish tin 1 41,000
Common Mock tin 4 lead 1 lino ! 18,000
Malacca tin 4, regnlus of antimony 1 12,000
Block tin o, lead 1 10.200
Block tin S, imo 1 10,000
Lead 1, aino 1 4,500
tttify and Stiengtk of v
trious Species of Timber
Species otTimliw.
TllUSOfE.
V»1.8»fS. SpwUsotTiiDber.
V^lueofE.
Vllneofd.
I74T
122 'Se
105
1555
. 862
705
119
98
2463
2221
1672
1766
1457
1383
2026
1556
5064
8868
133
1585
90
76
10547
1013
1632
1341
1102
1100
1200
800
1474
English oak.
Canadian do
Dantzic do
Adriatic do
New England fir
Mar Forest do.
Norway spruce,. .
Rule. — To find the dimensions of a beam capable of sustaining
a given weight, with a given degree of deflection, when supported
at both ends. — Multiply the weight to be supported in Iha. by the
cube of the length in feet ; divide the product by 32 times the
tabular value of E, multiplied into the given deflection in inches,
and the quotient is the breadth multiplied by the cube of the depth
When the beam is intended to bo square, then the fourth root
of the quotient is the breadth and depth required.
If the beam is to be cylindrical, multiply the quotient by 1T,
and the fourth root of the product la the diameter.
The distance hotiveen the supports of a beam of Riga fir is
16 feet, and the weight it must be capable of sustaining in the
middle of its length is 8000 ibs., with a deflection of not more
than I of an inch ; what must be the depth of the beam, suppos
ing the breadth 8 inches ?
16 X 8000 
90 X 32 X 7 5 ^ ^^^^^ "^ ^ = "^^^^"^ == 1^'^^ '"• t''^ ''epth.
Rule. — To determine the absolute strength of a rectangular beam
of timber when supported at both ends, and loaded in the middle
of its length, as beams in general ought to be calculated to, so that
they may be rendered capable of withstanding all accidental cases
of emergency. — Multiply the tabular value of S by four times the
depth of the beam in inches, and by the area of the cross section
in inches ; divide the product by the distance between the supports
hv Google
290 THE PRACTICAL MODEL CALCULATOR.
in inches, anJ the quotient iviU be the ahsoliite strength of the
beam in lbs.
If the beam be not laid horizontally, the distance between the
supports, for calculation, must be the horizontal distance.
Onefourth of the weight obtained by the rule is the greatest
weight that ought to be applied in practice aa permanent load.
If the load is to be applied at any other point than the middle,
then the strength will he, as the product of the two distances is to
the square of half the length of the beam between the supports ;
or, twice the distance from one end, multiplied by twice from the
other, and divided by the whole length, equal the effective length
of the beam.
In a building 18 feet in width, an engine boiler of 5^ tons is to
be fixed, the centre of which to be 7 feet from the wall ; and having
two pieces of red pine 10 inches by 6, which I can lay across the
two walls for the purpose of slinging it at each end, — may I with
sufficient confidence apply them, so as to effect this object?
2240 X 55
2 ■ — 6160 lbs. to carry at each end.
And 18 feet 
— 17 feet, or 204 inches, effective length of beam.
1341 X 4 X 10 X 60 ,_^^
Tabular value of 8, red pme = nnj ' = l'Ji 1 6
Iba., the absolute strength of each piece of timber at that point.
Rule. — To determine the dimensions of a rectangular beam capa
ble of supporting a required weight, with a given degree of deflection,
when fixed at one end. — Divide the weight to be supported, in lbs.,
by the tabular value of E, multiplied by the breadth and deflection,
both in inches ; and the cube root of the quotient, multiplied by
the length in feet, equal the depth required in inches.
A beam of ash ia intended to bear a load of 700 lbs. at its ex
tremity ; its length being 5 feet, its breadth 4 inches, and the de
flection not to exceed ^ an inch.
Tabular value of E = 119 x 4 x 5 = 238, the divisor ; then
700 H 238 = ^2^ X 5 = 725 inches, depth of the beam.
Rule. — To find the absolute strength of a rectangular beam, when
fixed at one end, and loaded at the other. — Multiply the value of S
by the depth of the beam, and by the area of its section, both in
inches ; divide the product by the leverage in inches, and the quo
tient equal the absolute strength of the'beam in lbs.
A beam of Riga fir, 12 inches by 4^, and projecting G} feet from
the wail; what ia the greatest weight it will support at the ex
tremity of its length ?
Tabular value of S = 1100
12 X 4'5 = 54 sectional area,
1100 X 12 X 54 „ „„ .
Then, ■ yg = 91384 lbs.
hv Google
STRESGTH OF MATERIALS. 291
"When fracture of a beam is producoil by vertical pressure, the
fitires of the lower section of fracture are separated by exteasion,
whilst at the same time those of the upper portion are destroyed
by compression ; hence exists a point in section where neither the
one nor the other takes place, and which ia distinguished as the
point of neutral axis. Therefore, by the law of fracture thus esta
blished, and proper data of tenacity and compression given, as in
the Table (p. 281), we are enabled to form metal beams of strongest
section with the least possible material : thus, in cast iron the re
sistance to compression is nearly as 6^ to 1 of tenacity ; conse
quently a beam of cast iron, to be of strongest section, must be
of the form TB, and a parabola in the direction of its ^^
length, the quantity of material in the bottom flange jf
being about 6} times that of the upper ; but such is not , [
the case with beams of timber ; for although the tenacity
of timber he on an average twice that of its resistance to compres
sion, its flexibility is so great, that any considerable length of beam,
where columns cannot be situated to its support, requires to he
strengthened or trussed by iron rods, as in the following manner :
An 1 these applications of pnnciple not only tend to diminish de
flection lut the required purpose is also more effectively attained,
and thit by lighter pieces of timber.
Ktjlb — To ascertain tie ah olute strength of a oast iron beam of
tkt pj uedinff form, oj that of strongest section. — Multiply the sec
tional area of the bottom flange in inches by the depth of the beam
in inches, and divide the product by the distance between the sup
ports also in inches ; and 514 times the quotient equal the absolute
strength of the beam in cwts,
The strongest form in which any given quantity of. matter can
be disposed is that of a hollow cylinder ; and it has been demon
strated that the maximum of strength is obtained in cast iron, when
the thickness of the annulus or ring amounts to Jth of the cylinder's
external diameter ; the relative strength of a solid to that of a
hollow cylinder being as the diameters of their sections.
The following table shows the greatest weight that ever ought
to be laid upon a beam for permanent load, and if there bo any
liability to jerks, &c., ample allowance must be made; also, the
weight of the beam itself must be included.
KuLE. — To find the weight of a cast iron beam of given dimen
sions. — Multiply the sectional area in inches by the length in feet,
and by 32, the product equal the weight in lbs.
Required the weight of a uniform rectangular beam of cast iron,
16 feet in length, 11 inches in breadth, and 1^ inch in thickness.
H X 15 X 16 X 32 = 8448 lbs.
hv Google
Sya THE PRACTICAL MODEL CALCDLATOK.
A Table showing the Weight or Pressure, a Beam of Cast Iron,
1 inch in breadth, will sustain without destroying its elastic force,
when it is supported at each end, and loaded in the middle of its
length, and also the deflection in the middle which that weight
will produce.
E^asH.
et«Dt
7 tea.
SM.
9t^el.
10 f«.
io'.
3
1278
'24
1089
■33
954
■426
855
■64
765
■66
H
1739
■205
1482
■28
12«8
■365
1164
■46
1041
■57
4
2272
■18
■245
1700
■32
1520
■405
1360
■5
H
2875
■16
2450
■217
2146
■284
1924
■86
1721
■443
5^
3560
■144
3050
■196
2650
■256
2375
2125
■4
6
5112
■12
4856
■163
3816
■213
8420
■27
8060
T
6958
■103
5929
■14
5194
■183
4655
23
4165
8
9088
■09
7744
■123
6784
■16
6080
■203
6440
■25
9801
■103
8586
■142
7696
■18
■22
10
12100
10600
■128
9500
■162
8500
■2
li
12826
■117
11495
■15
10285
■18
12
15264
■107
18680
186
12240
■17
13
16100
■125
14400
■154
U
6


~
—
—

18600
■115
16700
■148
lift
liu
T.
ISftst.
18te=t.
Wtett. 1
2548
■48
2184
■65
1912
■85
1699
1'08
1530
1'34
7
3471
■41
2975
■58
2603
■73
2314
■93
2082
114
8
4532
■36
3884
49
8396
■64
3020
81
2720
I'OO
a
5738
■33
4914
44
4302
■57
8826
72
8488
■89
10
7083
6071
■39
5312
■51
4722
■64
4250
11
8570
■26
7846
6428
■47
5714
■59
5142
■73
12
10192
■24
8736
7648
■48
6796
■54
6120
■67
IS
11971
■22
■81
8978
39
7980
49
7182
CI
14
1SS83
■21
11900
■28
10412
9253
■46
8330
■57
15
15987
■19
18660
■26
11952
■31
10624
43
9502
■53
16
18128
■18
15586
■24
18584
■82
12080
■10
10880
■5
17
20500
■17
17500
■28
15353
■8
13647
12282
■47
IS
22982
■18
19656
■21
17208
■28
15700
■36
13752
■44
Resistance of Bodies to Flexure hy Vertical Pressure. — When a
piece of tinfber is employed as a column or support, its tendency
to yielding by compression is different according to the proportion
between its length and area of its cross section ; and supposing the
form that of a cylinder whose length is less than seven or eight timea
its diameter, it is impossible to bend it by any force applied longi
tudinally, as it will be destroyed by splitting before that bending
can take place ; but when the length exceeds this, the column will
bend under a certain load, and be ultimately destroyed by a similar
kind of action to that which has place in the transverse strain.
Columns of cast iron and of other bodies are also similarly cir
cumstanced.
Wlien the length of a cast iron column with flat ends c([ual3
about thirty times its diameter, fracture will be produced wholly by
bending of the material ; — when of less length, fracture takes place
partly by crushing and partly h^ bending : but, when the column
hv Google
STREHGTH OP MATERIALS. 293
is enlarged in the middle of its length from one and a half to twice
its diameter at the ends, hy heing cast hollow, the strength is
greater by ^th than in a solid column coataining the same quantity
of material.
Rule. — To determine the dimensions of a support or column to
hear without sensible curvature a given pressure in the direction of
its axis. — Multiply the pressure to be supported in lbs. by the
square of the eolumu's length in feet, and divide the product by
twenty times the tabular value of E ; and the quotient will be equal
to the breadth multiplied by the cube of the least thickness, both
being expressed in inches.
When the pillar or support is a square, its side will be the fourth
root of the quotient.
If the pillar or column be a cylinder, multiply the tabular
value of E by 12, and the fourth root of the quotient equal the
diameter.
What should be the least dimensions of an oak support, to bear
a weight of 2240 lbs. without sensible flexure, its breadth being 3
inches, and its length 5 feet ?
2240 X 5^
Tabular value of E = 105, and i,Q— jng^— o = ^8'8tt8 =
205 inches.
Required' the side of a square piece of Riga fir, 9 feet in length,
to bear a permanent weight of 6000 lbs.
6000 X 9= —
Tabular value of E = 96, and gn v ' qii " ~ '^^^53 = 4 inches
nearly.
IHmensions of Oylindrioal Columns of Oast Iron to sustain a t/iven
load or pressure with safety.
s
li
L«w
orheiBhtlnftM
;
4
6
8
10
12
14 1 le
18
20
1 22
1 24 1
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fl
T"
60
49
40
82
2fi
22
18
15
13
11
n
m
10!1
91
77
65
55
47
40
34
29
25
17«
loa
145
iy«
111
97
84
7H
64
56
49
M
■Ml
232
2X4
IHJ
172
156
106
94
83
810
28S
■;m
242
220
19H
17fi
160
144
130
400
«;.+
327
KOI
275
251
229
208
189
fi
4.1'/
427
365
337
810
285
262
«
m
599
673
550
.125
497
469
440
413
3K6
360
7
Mm
1013
989
fift9
924
887
H4K
808
7Hf.
»
M\m
1315
V/hH
im4
1185
1142
1097
1052
1003
959
1H4II
1603
IfiKI
1515
1467
1416
1364
1311
m
?UM
2100
2077
VO+5
2007
1964
1916
1885
1S11
17.15
1697
11
7r>7(i
2550
2520
V4WI
2450
2410
23.18
2m
H24H
Ml 89
2127
12
SUaO
3040
■im
2U70
2930
■i'm
2830
2780
2730
2670
2600
Practical utility of thepreceding Table. — Wanting to support the
front of a building with cast iron columns 18 feet in length, 8 inches
in diameter, and the metal 1 inch in thickness ; what weight may
hv Google
294 TDE PRACTICAL MODEL CALCULATOR.
I confidently expect each column capable of supporting without
tendency to deflection ?
Opposite 8 inches diameter and under 18 feet = 1097
Also opposite 6 in. diameter and under 18 feet = 440
= 657 civts.
The strength of cast iron as a column being = 10000
~ steel — = 2518
— wrought iron — = 1745
— oak (Dantzic) — = 1088
— red deal — = 0785
£lastieity of torsion, or resistance of bodies to twisting. — The
angle of flexure by torsion is as the length and extensibilitj of the
body directly, and inversely as the diameter ; hence, the length
of a bar or shaft being given, the power, and the leverage the
power acts with, being known, and also the number of degrees of
torsion that will not affect the action of the machine, to determine
the diameter in cast iron with a given angle of flexure.
Rule. — Multiply the power in lbs. by the length of the shaft in
feet, and by the leverage in feet ; divide the product by fiftyfivo
times the number of degrees in the angle of torsion, and the fourth
root of the quotient equal the shaft's diameter in inches.
Required the diameters for a series of shafts 35 feet in length,
and to transmit a power equal to 1245 lbs., acting at the circum
ference of a wheel 2^ feet radius, so that the twist of the shafts
on the application of the power may not exceed one degree.
1245 X 35 X 25 .^ . .« . ,
cr ^ = '>/Vd!i\ = 667 inches in diameter.
Relative strength of metals to resist torsion.
Cast iron = 1 Swedish bar iron ...= 105
Copper = 48 English do = 112
Yellow brass = '511 Shear steel — 196
Gunmetal = 55 Cast do = 21
Deflexion of Rectasgui^^r Beams.
Rule. — To ascertain the ainount of deflexion of a uniform beam
of cast iron, supported at both ends, and loaded in the middle to the
extent of its elastic force. — Multiply the square of the length in feet
by '02, and the product divided by the depth in inches equal tho
Required the deflection of a cast iron beam 18 feet long between
the supports, 128 inches deep, 256 inches in breadth, and bear
ing a weight of 20,000 lbs. in the middle of its length.
18= X 02
— Yg .Q •• = '506 inches from a straight line in the middle.
For beams of a similar description, loaded uniformly, the rule is
the same, only multiply by 025 in place of 02.
Rule. — To find the deflection of a beam when fixed at one end
hv Google
ETKENGTH OP MATEKIAIS. 295
and loaded at the other. — Divide the length in feet of the fixed part
of the heam by the bngth in feet of the part which yields to the
force, and add 1 to the quotieat ; then multiply the square of the
length in feet by the quotient so increased, and also by 13 ; divide
this product by the mitldle depth in inches, and the quotient will
be the deflection, in inches also.
Multiply the deflection so obtained for cast iron by SG, the pro
duct equal the deflection for wrought iron ; for oak, multiply by
28; and for fir, 24.
A Table of the Depths of Square Beams or Bars of Cast Iron,
calculated to mj^^iort from 1 Cwt. to 14 Tons in the Middle, the
Deflection not to exceed ^^th of an Inch for each Foot in Length.
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THE PRACIICAL MODEL CALCULATOR.
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^ fc3 illustrative of the Table. — 1. To find the depth of a
rectangular bar of east iron to support a weight of 10 tons in the
middle of its length, the deflection not to exceed ^ of an inch per
foot in length, and its length 20 feet, also let the depth be 6 times
the breadth.
Opposite 6 times the weight and under 20 feet in length is 15'3
inches, the depth, and J of 15"3 = 2'6 inches, the breadth.
2. To find the diameter for a cast iron shaft or soUd cylinder
that will bear a given pressure, the flexure in the middle not to ex
inch for each foot of its length, the distance of the
20 feet, and the pressure on the middle equals 10
ceed ^'jth of
bearings be.
tons.
Constant
nltiplier 17 for round shafts, then 10 X 17 = IT.
And opposite 17 tons and under 20 feet is 11'2 inches for the di
ameter.
But half that flexure ia quite enough for revolving shafts : hence
17 X 2 = 34 tons, and opposite 34 tons is 133 inches for the di
ameter.
3. A body 256 lbs. weight, presses against its horizontal sup
port, so that it requires the force of 52 lbs. to overcome its friction ;
if the body be increased to 8750 lbs., what force will cause it to
pass from a state of rest to one of motion ?
52_
2,50 '
■ •203125 = , in this case, the eoeffieient offnciion;
.: 8750 X 203125 = 177734375 lbs., the force required.
This calculation is based upon the law, that friction is propor
tional to the normal pressure between the rubbing surfaces. Twice
the pressure gives twice the friction ; three times the pressure gives
three times the friction ; and so on. With light pressures, this law
may not hold, but then it is to he attributed to the proportionately
greater effect of adhesion.
4. If a sleigh, weighing 250 lbs., requires a force of 28 lbs. to
draw it along ; when 1120 lbs. are placed in it, required the units
of work expended to move the whole 350 feet '!
hv Google
STREKGTIl OF MATERIALS. 297
28 „
^TQ = ■112, the coefficient of friction.
Then (1120 + 250) x 112 = 15344 lbs., the force required to
move the whole.
.. 15344 X 350 = 53704, the uuits of work required.
A CNIT OP WORK IB the labour which ia equal to that of raising
one pound a foot high. It is supposed that a horse can perform
33000 units of work in a minute.
It may also be remarked that friction is independent of the ex
tent of the surfaces in contact, except with trifling pressures and
large surfaces, which is on account of the effect of adhesion. The
friction of motion is independent of velocity, and is generally leas
than that of quiescence. a
5. Required the co
efficient of friction, for
a sliditig motion, of
cast iron upon wrought,
lubricated with Dev
lin's oil, and under
the following circum
stances : the load A,
and sledge nm, weighs
8420 lbs., and requires
a weight W, of 1200 lbs. to cause it to pa*?? from a state of rest
into one of motion : the sledge and load pass over 22 feet on the
horizontal way rs, in 8 seconds.
In this ease the coefficient of sliding motion will bo
1200 1200 I 8420 2 x 2 2
8420 ~~ 8420 ^ </ x 8""
in which ^ = 32'2 feet; the acceleration of the free descent of
bodies brought about by gravity. The above expression becomes
44
142515  1142515 X goM^ = ■118121.
Hence the coefficient of the friction of motion is •118121, and the
coefficient of the friction of quiescence ia "142515.
OF IRICTION,
In the years 1831, 1832, and 1833, a very extensive set of ex
periments were made at Metz, by M. Morin, under the sanction
of the French government, to determine as nearly as possible the
laws of friction ; and by which the following were fully est.abli3hed :
1. When no unguent is interposed, the friction of any two sur
faces (whether of quiescence or of motion) is directly proportional
to the force with which they are pressed perpendicularly together ;
so that for any two given surfaces of contact there is a constant
ratio of the friction to the perpendicular pressure of the one surface
upon the other. AVhilst this ratio is thus the same for the same
hv Google
ays THE PEACTICAL MODEL CALCULATOR.
surfaces of contact, it is different for JifFercnt surfaces of contact.
The particular value of it iti respect to a,ny two given surfaces of
contact is called the coefficient of friction in respect to those sur
faces.
2. When no unguent is interposed, the amount of the friction is,
in every case, wholly independent of the extent of the surfaces of
contact ; so that, the force with which two surfaces are pressed to
gether being the same, their friction is the same, whatever may be
the extent of their surfaces of contact.
3. That the friction of motion is wholly independent of the velo
city of the motion.
4. That where unguents arc interposed, the coefficient of friction
depends upon the nature of the unguent, and upon the greater or
less abundance of the supply. In respect to the supply of the un
guent, there are two extreme cases, that in which the surfaces of
contact are but slightly rubbed with the unctuous matter, as, for
instance, with an oiled or greasy cloth, and that in which a con
tinuous stratum of unguent remains continually interposed between
the moving surfaces ; and in this state the amount of friction is
found to be dependent rather upon the nature of the unguent than
upon that of the surfaces of contact. M. Morin found that with
unguents (hog's lard and olive oil) interposed in a continuous stra
tum between surfaces of wood on metal, wood on wood, metal on
wood, and metal on metal, when in motion, have all of them very
near the same coefficient of friction, being in all cases included be
tween 07 and 08.
The coefficient for the unguent tallow is the same, except in that
of metals upon metals. This unguent appears to be less suited for
metallic substances than the others, and gives for the mean value
of its coefficient, under the same circumstances, 10. Hence, it is
evident, that where the extent of the surface sustaining a given
pressure is so great as to make the pressure less than that which
corresponds to a state of perfect separation, this greater extent of
surface tends to increase the friction by reason of that adhesiveness
of the unguent, dependent upon its greater or less viscosity, whose
effect is proportional to the extent of the surfaces between which
It was found, from a mean of experiments with different unguents
on axles, in motion and under different pressures, that, with the
unguent tallow, under a pressure of from 1 to 5 cwt., the friction
did not exceed ^th of the whole pressure ; when soft soap was ap
plied, it became ^th ; and with the softer unguents applied, such
as oil, hog's lard, &c., the ratio of the friction to the pressure in
creased ; but with the harder unguents, as soft soap, tallow, and
antiattrition composition, the friction considerably diminished ;
consequently, to render an unguent of proper efficiency, the nature
of the unguent must be measured by the pressure or weigtit tend
ing to force the surfaces together.
hv Google
STRERGTH OF MATERIALS.
Table of tlm Hesiilts of Experiments on the Friction, of Unctuous
Surfaces. By M. Mows.
Oak upon oak, the fibres being paralle! to the motion
Ditto, the libreB of the moving body being perpendicu
lar to the motion
Oak apon elm, fibres parallel
Elm upon oafc, do
Eeeeh upon oak, do
Elm upon elm, do
Wrought iron upon cJm, do
Ditto npon wrought iron, do
Ditto upon oast iron, do
Cast iron upon ncoaght iron, do
Wrought iron npon brosa, do
Brass upon wrougiit iron, do
Cast iron upon onk, do
Ditto upon, elm, do., tha unguent being tallow.
Ditto, do., lie unguent being hog's lard and blaok
Elm upon oast iron
Ditto upon bcasa ,
Copper npon oak
Yellow copper upon cast iron
Leather (oiMde), well tanned, upon
Ditto upon brass, wetted
OIGO
O'liiG
O'lCT
O'lSo
0144
0'132
0391)
0314
0'420
In these esperiments, the surfaces, after having been smeared
ivith an unguent, were wiped, so that no interposing layer of the
unguent prevented intimate contact.
Taele of the Results of Experiments on, FHction, ivith Unguents
interposed. By M. MoRiK,
C06ffl=lsnU
of FrlcMon.
0'164
0440
0075
0164
0067
0083
0254
0072
0250
0136
0073
0178
0066
0080
0098
0055
0187
041t
0170
0142
0OfiO
0]30
0217
0066
02o6
0649
0214
...
Oiik upon oak, libres parallel..
upon wrought iron
Beeoh npon oak, fibres parallel.
Elm upon oak, do
Do. do
Do. do
Elm upon elm, do
Do. upon cast iron
Wrought iron upon oak, fibres ■)
parallel J
Dry soap.
Tallow.
Hog's lard.
Tallow.
Dry soap.
TalloiT.
Hog's lard.
Dry soap.
Tallow.
t Greased and s:
\ rated wilb wat
Dry Boa,p.
b,Google
THE PIUOTICAL MODEL CALCULATOR.
C»fflcl^„l>rfFri«i.n. 
Wrought iron upon oak, fibres 1
parallel i
085
0108
Tallow.
Do. upon elm, do
0078
Tallow.
Ho. do
007S
Hos's !a.rd.
Do. do
056
Olive oil.
Do. upon i.a=t iron do
Oi08
Tallow.
Do. do
073
Hoe's Inrd.
Do. do
00(16
0100
Olive oil.
Do. ui on wiought iron, do
082
Tullow.
Do, do
081
Ilog's lard.
Do. do
070
011s
Olive oil.
Wrought irm upon brass, do
0103
Tallow.
Do. do
073
Hog's lard.
Do. do
078
Oliva oil.
Cast iron upon oak, do
0189
Dry soap.
Do. do
218
01b
f Greased and satu
\rated with water.
Do. do
078
0100
Tallow.
Do. do
075
Hog's lard.
Do. do
075
100
Olive oil.
Do. upon elm, do
077
Tdlow.
Do. do
OObl
Olive oil.
Do. do
091
; Hog's lard and
\ plumbago.
j Do. iipouwrougbt lion
0100
Tallow.
Do. upon CJ.t lion
314
Water.
Do. do
197
Soap.
Do, do
0100
100
Tallow.
Do. do
070
100
Hog's lard.
Do, do
OOUJ
Olive oil.
Do. do
0^5
f Hog'a lard and
\ plumbago.
Do, upon bi a"
103
Tallow.
Do. do
07o
Hog's lard.
Do. do
078
Olive oil.
Copper upon oak, fibres paralle
001
0100
Tallow.
Yellow copper upon cast iron.
072
0103
Tallow.
Do. do
0l>8
Hog's lard.
Do do
O'OfjO
Olive oil
ErasB upon cast lion
086
010b
Tallow
Do do
077
Olive cid
Do upon wrought iron
081
1illow
Do do
089
( lird lud plum
Ihftgo
Do do
072
Olive oil
Bras? upon brass
OOjS
Oliie oil
Steel upon cast iron
0105
0108
Tallow
Do do
0081
Hog Elm d.
Do do
0070
01i>e oil
Du upon wruuglit iron
093
Tallow
Do do
076
HogslirJ.
Do upon brass
056
Tillow
Do do
OOjS
Obi e ml
Ho do
0(,7
f Lard and plum
1 li i„o
Tanned oshide upon east iron..
oati5
...
Jf.i™elam(satu
1 Tilted with water.
The esteut of the surfaces in these experiments bore such aiolati"u to II
sure as to cause them to be separated from one another throughout by m
posed stratum of the unguent.
b,Google
STKENOTH OF MATERIALS.
Table ofthi. Bi&uHs of Expinments on the Fiution ot Gudgeons
or Axleends, in motion vpon their bearings By M ftlORlH,
SurWilnCuBlat
saUoniifSurf^es
LosfficeaL fFritU™.
Conted with oil of olneo )
withhogsUrd tallow, \
OT to 008
Cast iion a\!ef jn
and soft gome J
roitfd with asphaltum
0U54
Creasy
Greasy and wetted.
Coated with oil of olives, 1
witlihogalard,tallow \
07 to 008
Cast jjon ailp'j in
cast iron be itiDj^
Greasy
Greasy and damped
OIG
16
Startely greaev
Coated with oil of oiiYss,
tallow lioga Inrd oc
07 to 008
ings
soft gome
Toated with oil of olivea
07 to 008
n09
019
Wrought iron a^ka
ii!bia3 be^rmga
hogs lard or tillow, /
Coaled with hard gome
tieasy ani wetted
Iron axles in lignum
Coated with oil or hog s \
Oil
■vitiE beoimgs
Greasy
019
Coated with oil
bearrogs
Hithhngc lard
Table of Coefficients of Friction under Pressures increased continu
ally up to limits of Abrasion.
C™ffi<:i=»U
.fFrL,.U»n.
3261bs.
■140
■174
■166
■157
I'BlJowts.
■250
■275
■300
■225
200
■271
■292
■333
219
233
■E21
■340
■329
■344
■211
8O0
■312
■333
■347
■215
333
■300
■351
3S1
■206
S66
■876
■353
353
■205
400
■395
■365
■354
■208
4'33
■403
■366
■856
■221
iM
■409
■366
■357
5'00
■367
■858
■238
■367
■351)
566
■367
367
■235
600
■376
■403
■233
■434
666
■235
700
733
;;;;;;
■232
b,Google
302 THE PRACTICAL MODEL CALCULATOR.
Comparative friction of steam engines of different modificntion?,
if the beam engine be taken aa the stanctard of comparison : —
The vibrating engine has a gainof ll percent.
The directaction engine, with slides — loss of 18 —
Ditto, with rollers — gain of 08 —
Ditto, with a parallel motion — gain of 1'3 —
Excessive allowance for friction lias liitherto been made in cal
culating the effective power of engines in general ; as it is found
practically, by experiments, that, where the pressure upon the pis
ton is about 12 lbs. per square inch, the friction does not amount
to more than 1 lbs. ; and also that, by esperiments with an indi
cator on an engine of 50 horse power, the whole amount of friction
did not exceed 5 horse power, or onetenth of the whole power of
the engine.
RECENT EXPEUlMtlNTS MADE BY 31. MORIS ON TUB STIFFNESS OF EOrEfi,
OH. TilE RESISTANCE OP BOrES TO EENDINO UPON A CIECIILAR ARC.
The experiments upon which the rules and table following are
founded were made by Coulomb, with an apparatus the invention
of Amonton, and Coidomb himself deduced from them the follow
ing results : —
1. That the resistance to bending could be represented by an
p nut f tw t rms, the one constant for each rope
nd a h 11 1 h w hall designate by the letter A, and
wl 1 tb jl 1 f h n m d the natural stiffness, because it de
p ! n 1 m d f f b ion of tho rope, and the degree of
t f t y n nd t ands ; the other, proportional to the
t n n T ft! nd f th ope which is being bent, and which
p d >y th p 1 t BT, in which B is also a number
ntntf 1 padah roller,
2 Tb t h ta t 1 nding varied inversely as the diame
t ftl 11
Thus the complete resistance is represented by the expression
A + BT
D " '
where D represents the diameter of the roller.
Coulomb supposed that for tarred ropes the stiffness was pro
portional to the number of yarns, and M. Navier inferred, from
examination of Coulomb's experiments, that the coeflicionta A and
E were proportional to a certain power of the diameter, which de
pended on the extent to which the cords were worn. JI. Morin,
however, deems this hypothesis inadmissible, and the following is
an extract from his new work, "Lei^ons de &Kcanique Pratique,"
December, 1846 : —
" To extend the results of the experiments of Coulomb to ropes
of different diameters from those which had been experimented
upon, M. Navier has allowed, very explicitly, what Coulomb had
but surmised : that tho coefficients. A, were proportional to a cer
hv Google
STRENGTH OP MATEKIALS. 303
tain power of the diameter, which depended on the state of wear
of the ropes ; but this supposition appears to us neither borne out,
nor even admissible, for it would lead to this consequence, that a
worn rope of a metre diameter would have the same stiifness as a
new rope, which is evidently wrong ; and, besides, the comparison
alone of the values of A and B shows that the power to which the
diameter should be raised would not be the same for the two terms
of the resistance."
Since, then, the form proposed by M. Navier for the expression
of the resistance of ropes to bending cannot be admitted, it is ne
cessary to search for another, and it appears natural to try if the
factors A and B cannot be expressed for white ropes, simply accord
ing to the number of yarns in the ropes, as Coulomb has inferred
for tarred ropes.
Now, dividing the values of A, obtained for each rope by M.
Navier, by the number of yarns, we find for
n = dO d = 0"'200 A = 02224GO  = 0'0074153.
lu d = 0^444 A =
^ 6d = 0"0088 A = 0010604 
= 00042343.
= 00017673.
It is seen from this that the number A is not simply propor
tional to the number of yarns.
Comparing, then, the values of the ratio — corresponding to
the three ropes, wo find the following results : —
' "nroa.
Dileroiitea of tl.c numVst. of
rv'S";/
80
15
00074153
00042343
OO0T7G73
From SO to 15, 15 jams
— 15 to 6. 9 —
0W31810
00024770
000ofi400
0000212
000372
0000352
Mean difference per jara, 0000245
It follows, from the above, that the values of A, given by the
experiments, will be represented with sufficient exactness for all
practical purposes by the formula
A = M [00017673 I 0000245 (« — 6)].
= n [00002973 f 0000245 k].
An expression relating only to dry white ropes, such as were used
by Coulomb in his experiments.
With regard to the number E, it appears to be proportional to
the number of yarns, for we find for
hv Google
TOE PRACTICAL MODEL CALCULATOK.
= SO d = 0"0200 B 
n = e d = O^OOSS B =
Moan 00003630
Whence
B = 0000363 n.
Consequently, the results of the experiments of Coulomb on dry
white ropes will he represented ivith sufEcient exactoess for prac
tical purposes hy the formula
K = M [000029r + 0000245 n + 0>0003G3 T] kil.
which will give the resistance to bending upon a drum of a metre
in diameter, or by the formula
R = ^ [0000297 + 0000245 n + 0000363 T] kil.
for a drum of diameter D metres.
These formulas, transformed into the American scale of weights
and measures, become
U = n [00021508 + 000n724 n + 000119090 T] lbs.
for a drum of a foot in diameter, and
E = g [00021508 + 00017724 «, + 0.00119096 T] lbs.
for a clrum of diameter D feet.
With respect to worn ropes, the rule given by M. Navier cannot
be admitted, as we have shown above, because it would give for tlio
stiffness of a rope of a diameter equal to unity the same stiifness
as for a new rope.
The experiments of Coulomb on worn ropes not being sufficiently
complete, and not furnishing any precise data, it is not possible,
without new researches, to give a rule for calculating the stiffness
of these ropes.
TARRKD aOPES.
In reducing the results of the experiments of Coulomb on tarred
ropes, as we have done for white ropes, we find the following
values : —
n = 30 yarns A = 034982 E = 00125605
„ = 15 _ A = 0106003 B = 0006037
„ = 6 — A = 00212012 E = 00025997
which diifcr very slightly from those which JI, Navier has given.
But, if we look for the resistance corresponding to each yarn, we
find
hv Google
:kgth of materials.
30 yarns
~ =
. 001166l)3
 = 0000418683
15 —
A
 00070662
?  0000402466
6 —
A _
= 00035335
2 = 0000433283
Mean
0000418144
We see by this that the value of B is for tarred ropes, as for
white ropes, sensibly proportional to the number of yarns, hut it
13 not BO for that of A, as M. Kavier has supposed.
Comparing, as we have done for ivhite ropes, the values of —
corresponding to the three ropea of 30, 15, and 6 yarns, wo obtain
the following results : —
JMM.
"'?•'
„,._
;;,::
sp
Ipir'
80
16
6
0'0116603
00070662
00035335
From 30 1.
 ISt
— eot
IS.
15 yarns
25 —
00045941
00035327
0'0081268
0000306
0000392
0O00339
Mea
..0000346
It follows from this that the value of A can be represented by
the formula
A = K [00035335 + 0000346 (n ~ 6)]
= n [00014575 + 0000346 m]
and the whole resistance on a roller of diameter D metres, by
R
= j5 [00014575 + 0000346 n + 0000418144 T] kil.
Transforming this expression to the American scale of weights and
measures, we have
R = ^ [001054412 + 000250309 n + 0001371889 T] lbs.
for the resistance on a roller of diameter D feet.
This expression is exactly of the same form as that which relates
to white ropes, and shows that the stiffness of tarred ropes is a little
greater than that of new white ropes.
In the following table, the diameters corresponding to the differ
ent numbers of yarns are calculated from the data of Coulomb, by
the formulas,
d cent. = v'01338 n for dry white ropes, and
d cent, = ^/0.18^ n for tarred ropes,
which, reduced to the American scale, become
d inches = v^ 00207"B9 n for dry white ropes, and
d inches = ^/6■02883 for tarred ropes.
hv Google
6m THE PRACTICAL MODEL CALCULATOR.
Note,— The diameter of the rope is to be included in D ; thus,
ivith an inch rope passing round a pulley, 8 inches in diameter in
the groove, the diameter of the roller is to be considered as 9
1
1
67
—
^
«B,
^
"'r.^.T'
«:=
yn
3^
eo
300
776
072135T
«'37323T
f OOlOMJIJii
l+OOOaiKOflnl
flOSOB W*
0KH3T1889B
Apjilioatiurh of the 'preceding Tables or Formulas.
To find the stiffness of a rope of a given diameter or number of
yarns, we must first obtain from the table, or by the formulas, tho
values of the quantities A and B corresponding to these given
quantities, and knovring the tension, T, of the end to he wound
up, we shall have its resistance to bending on a drum of a foot in
diameter, by the formula
R = A + ET.
Then, dividing tbis quantity by the diameter of the iolbr or
pulley round which the rope is actually to be bent, we shall have
the resistance to bending on this roller.
What is the stiffness of a dry white rope, in good condition, of
60 yarns, or 0928 diameter, which passes over a pulley of 6 inches
diameter in the groove, under a tension of 1000 lbs. ? The table
gives for a dry white rope of 60 yarns, in good condition, bent
upon a drum of a foot in diameter,
A = 05097T B = 0OT14576
and WO have D = 0'5 + 0'0928 ; and consequently.
05928
The whole resistance to be overcome, not including the f:ictioi
on the axis, is then
Q + R = 1000 + 128 = 1128 lbs.
The stiffness in this case augments the resistance by moi'O thai
oneeighth of its value.
hv Google
STRRKOTH OF MATERIALS. 307
FtllLTHER EUCEKT EXPEIdllENTS MADE BY JI. JIORIN, ON TOE TRAC
TION" OP CAKUIACiES, AND THE DESTUCCTIVE EEEECTS WHICH THEY
PRODUCE DPON THE ROADS.
The study of the effects which ave produced ivhen a. carriage 13
set in motion can be divided into two distinct parts : the traction
of carriages, properly so called, and their action upon the roads.
The researches relative to the traction of carriages have for their
object to determine the magnitude of the effort that the motive
power ought to exercise according to the weight of the load, to the
diameter and breadth of the wheels, to the velocity of the carriage,
and to the state of repair and nature of the roads.
The first experiments on the resistance that cylindrical bodies
offer to being rolled on a level eurfaco are due to Coulomb, who
determined the resistance offered by rollers of lignum vitie and
elm, on plane oak surfaces placed horizontally.
ilis experiments showed that the resistance was directly propor
tional to the pressure, and inversely proportional to the diameter
of the rollers.
If, then, P represent the pressure, and r the radius of the roller,
the resistance to rolling, R, could, according to the laws of Cou
lomb, be expressed by the formula
in which A would be a number, constant for each kind of ground,
but varying with different kinds, and with the state of their
The results of experiments made at Vineennes show that the
law of Coulomb is approximately correct, but that the resistance
increases as the width of the parts in contact diminishes.
Other experiments of the same nature have confirmed these con
clusions; and we may allow, at least, as a lavf sufficiently exact
for practical purposes, that for woods, plasters, leather, and gene
rally for hard bodies, the resistance to rolling is nearly —
1st. Proportional to the pressure.
2d. Inversely proportional to the diameter of the wheels.
Sd. Greater as the breadth of the zone in contact is smaller.
S UrON CAEBIAGES TEAVEIJ.ISa ON ORDINAUY ROADS.
These experiments were not considered sufBcient to authorise
the extension of the foregoing conclusions to the motion of car
riages on ordinary roads. It was necessary to operate directly on
the carriages themselves, and in the usual circumstances in which
they are placed. Experiments on this subject were therefore un
dertaken, first at Metz, in 1837 and 1838, and afterwards at Coor
bevoie, in 1839 and 1841, with carriages of every species ; and
attention was directed separately to the influence upon the magni
tude of the traction, of the pressure, of the diameter of the wheels,
of their breadth, of the speed, and of the state of the ground.
In heavily laden carriages, which it is most important to take
hv Google
SOO THE PEACTICAL MODEL CALCULATOR.
into consideration, tbe weight of the wheels may be neglected in
cornpiirison with the total load ; and the relation between the load
and the traction, upon a level road, ia approximately given by the
equation —
F, 2 (Ax/?,)
p= — ^ ,, for carnages with four wheels,
F,' A X /r, ^
and p= — ■ for carriages with two wheels,
in which F, represents the horizontal component of the traction ;
P, the total pressure on the ground ;
t' and /' the radii of the fore and hind wheels ;
r, the mean radius of the boxes ;
/ the coefficient of friction ;
and A the constant multiplier in Coulomb's formula for the
resistance to rolling.
These expressions will serve us hereafter to determine, by aid of
experiment, the ratio of the traction to the load for the most usual
cases.
Influence of the Pressure.
To observe the influence of the pressure upon the resistance to
rolling, the same carriages were made to pass with different loads
over the same road in the same state.
The results of some of these experiments, made at a walking pace,
are given in the following table: —
a.,.,.„ „.,,.
„..„_
PrMSQia.
.™.,.
Chariotporte corps
d'artillerie.
Road from Conrbe
Toie to Colomber,
dry, in good re
pair, dusty.
Eoad from Courbe
Toie to Bezoua,
solid. *harJ gra
vel, very dry.
6140
4580
iso'ti
1599
1137
1/886
1/392
1/402
ChariotderonlagE,
wjtlioutapriaga.
7126
5458
4450
3430
1389
1155
932
684
1/513
1/489
1/477
1/502
Cbariotderoalage,
Vfith apringa.
Eoad from Colomber
to Courbevoie,
pitohtd, inordina
ry repair, f mud dy
Boad from Courbe
voie to Colomber,
deep ruts, with
muddy detritus.
1600
3292
4300
393
892
I860
1/408
1,369
1/368
Cunisgwi with ail
equnl wheels.
Tw<icafriag«3nith
sii equal wheela,
hooked on, one
behind the other.
3000
4692
6000
eooo
1389
2240
2858
2867
I/2I6
1/210
1,210
1/210
From the examination of this table, it appears that on Jsolid
gravel and on pitched roads the resistance of carriages to traction
is sensibly proportional to the pressure.
• En gcavier dur. f Pav^ en 6tat ordinaire. J En
b,Google
BTRENSTH OF MATEIilALS. 6\}y
We remark that tte experiments made upon one and upon two
sixwheeled carriages have given the same traction for a load of
6000 kilogrammes, including the vehicle, whether it waa borne
upon one carriage or upon two. It follows thence that the trac
tion is, ca;teris paribus and between certain limits, independent of
the number of wheels.
Influence of the Diameter of the Wheeh.
To observe the influence of the diameter of the wheels on the
traction, carriages loaded with the same weights, having wheels
with tires of the same width, and of which the diameters only were
varied between very extended limits, were made to traverse the
same parts of roads in the same state. Some of the results obtained
are given in the following table.
These examples show that on solid roads it may be admitted aa
a practical law tl h n n erselj proportional to the
diameters of the w
C.rri«.>™.r"j..L
■


I
S~
1
i
~:
H
il
Ciffilon.
Camion.
Road ft.
bevoie
701, dn
60 SW
St
86
"2
/A
OB'6
I88(
1/60'
1/455
HiSri
i/as8
17
67
4241
l>013[
1
UIH494
aosoje
l029»
Influence of the Width of the Felloes.
Experiments made upon wheels of different breadths, having the
same diameter, show, 1st, tfeat on soft ground the resistance to
rolling increases aa the width of the felloe ; 2dly, on solid gravel
and pitched roads, the resistance is very nearly independent of the
width of the felloe.
Influence of the Velocity.
To investigate the influence of the velocity on the traction of
carriages, the same carriages were made to traverse different roads
in various conditions ; and in each series of experiments the velo
cities, while all other circumstances remained the same, underwent
successive changes from a walk to a canter.
Some of the results of these experiments are given in the follow
ing table : —
* Empierrement eolido.
t Pav6 en grfes.
hv Google
THE PKACTICAL MODEL CALCULATOR.
c...,.„.,..,.,.
— ■
....
....
w
t™
"3"
Apparatus npon a
Ground of the po
104'>
Wall!
3 13
165'0
1/632
brass shaft.
]ygoii at Mctz,
Trot
tiati
lti80
1/82
\Mh
Walk
SfifiO
2150
1/621
Trot
7560
1070
1/6*8
Asisteenpounder
Road from Metz
Walk
carriage and
to Montigny,
'Brisk walk
34110
92
1/408
piece.
solid graTel,
very eyen and
Tery dry.
Pitoiied road of
1,^1
fCimter
8450
121
Cbaiiot des Mes
R2Rfi
Walk
2770
144
1/228
Fontainebleau,
x^rvA
»Brisk walk
mi
153
1/219
eprings.
1/183
tBrisk trot.
805
1835
We see, by these examples, tliat the traction undergoes no sen
Bible augmentation with the increase of velocity on soft grounds ;
but that on solid and uneven roads it increases with an increase of
velocity, and in a greater degree as the ground is more uneven, and
the carriage has less spring.
To find the relation between the resistance to rolling and the ve
locity, the velocities were set ofi' as abscissas, and the values of A
furnished by the experiments, as ordinates ; and the points thus
determined were, for each series of experiments, situated very
nearly upon a straight line. The value of A, then, can be repre
sented by the expression,
A = a + d(V2)
in which a is a number constant for each particular state of each
kind of ground, and which expresses the value of the number A for
the velocity, V = 2 miles, {per houi',) which is that of a very slow
walk.
d, a factor constant for each kind of ground and each sort of
carriage.
The results of experiments made with a carriage of a siege train,
■with its piece, gave, on the Montigny road, §very good solid gravel, —
A = 003215 X 0'00295 (V  2).
On the llpitched road Of Metz, A = 001936 X 0'08200 (V — 2).
These examples are sufficient to show —
Ist. That, at a walk, the resistance on a good pitched road is
less than that on very good solid gravel, very dry.
2d. That, at high speeds, the resistance on the pitched road in
creases very rapidly with the velocity.
On rough roads the resistance increases with the velocity much
more slowly, however, for carriages with springs.
t Grand trot.
hv Google
STEENOTH OP MATERIALS. 311
TIius, for a clianotdcs Messagenes G^n(;rales, on a pitched roaiJ,
the experiments gave A = 00117 X 000361 (V — 2) ; while, with
the springs wedged so as to prevent their action, the experiments
gave, for the same carriage, on a similar road, A = 002723 X
001312 (V — 2). At a speed of nine miles per hour, the springs
diminish the resistance by onehalf.
The experiments further showed that, while the pitched road was
inferior to a *3olid gravel road when dry and in good repair, the
latter lost its superiority when muddy or out of repair.
INPLUINCE OF THE INCLINATION OF TIIE TRACES.
The inclination of the traces, to produce the maximum effect, is
given by the expression —
AxO9 6/r'
in which /; = the height of the fore extremity of the trace above
tte point where it is attached to the carriage ; 6 = the borizontat
distance between these two points, r' is the radius of interior of
the boxes, and r the radius of the wheel.
The inclination given by this expression for ordinary carriages
is very small ; and for trucks with wheels of Bmall diameter it is
much less than the construction generally permits.
It follows, from the preceding remarks, that it is a
to employ, for all carriages, wheels of as large a diameter
be used, without interfering with the other essentials to the pur
poses to which they are to be adapted. Carts have, in this respect,
the advantage over wagons ; but, on the other hand, on rough roads,
the thill horse, jerked about by the shafts, is soon fatigued. Kow,
by bringing the hind wheels as far forward as possible, and placing
the load nearly over them, the wagon is, in effect, transformed into
a cart ; only care must be taken to place the centre of gravity of
the load so far in front of the hind wheels that the wagon may not
turn over in going up hill.
ON THE PESTEDCnvil El'FECTS PRODUCED BY CAUttIACE3 ON TUB IIOAIIS.
If we take stones of mean diameter from 1\ to 3^ inches, and,
on a road slightly moist and soft, place them first under the small
wheels of a diligence, and then under the large wheels, we find that,
in the former case, the stones, pushed forward by the small wheels,
penetrate the surface, ploughing and tearing it up ; while in the
latter, being merely pressed and leant upon by the large wheels,
th^ undergo no displacement.
From this simple esperiment we are enabled to conclude that
the wear of the roads by the wheels of carriages is greater the
smaller the diameter of the wheels.
Experiments having proved that on hard grounds the traction
was bat slightly increased when the breadths of the wheels was
* En erapierrement.
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312 THE PRACTICAL MODEL CALCULATOK.
diminisliecl, we might also conclude that the wear of thi Hir.d wouhl
be but slightly increasecl by dinouishing the width nf the felloes.
Lastly, the resistance to rolling increasing with the velocity, it
was natural to think that carriages going at a trot would do liiore
injury to the roads than those going at a walk. But s[>nng=(, by
diminishing the intensity of the impacts, are able to compensate,
in certain proportions, for the effects of the velocity.
Experiments, made upon a grand scale, and having for their
object to observe directly the destructive effects of carnages upon
the roads, have confirmed these conclusions.
These experiments ahoived that with equal loads, on a solid gra
vel road, wheels of two inches breadtli produced considerably moi o
wear than those of 4^ inches, but that beyond the latter width there
was scarcely any advantage, so far as the preservation of the road
was concerned, in increasing the size of the tire of the wheel.
Experiments made with wheels of the same breadth, and of dia
meters of 286 ft., 477 ft., and 669 ft., showed that after the
carriage of 100182 tons, over tracks 21872 yards long, the track
passed over by the carriage with the smallest wheels was by far
the most worn ; while, on that passed over by the carriage with
the wheels of 669 ft. diameter, the wear was scarcely perceptible.
Experiments made upon two wagons exactly similar in all other
respects, but one with and one without springs, showed that the
wear of the roads, as well as the increase of traction, after the
passage of 4577'36 tons over the same track, was sensibly the same
for the carriage without springs, going at a walk of from 2'237 to
2684 miles per hour, and for that wjtii springs, going at a trot of
from 7'158 to 8053 miles per hour.
HYDRAULICS,
THE DISCHARGE OF WATER BY SIMPLE ORIFICES AND TUBES.
The formulas for finding the quantities of water discharged in a
given time are of an extensive and complicated nature. The more
important and practical results are given in the following Deduc
tions.
When an aperture is made in the bottom or side of a vessel con
taining water or other homogeneous fluid, the whole of the particles
of fluid in the vessel will descend in lines nearly vertical, until they
arrive within three or four inches of the place of discharge, when
they will acquire a direction more or less oblique, and flow directly
towards the orifice.
The particles, however, that are immediately over the orifice, de
scend vertically through the whole distance, while those nearer to
the sides of the vessel, diverted into a direction more or less oblique
as they approach the oriflce, move with a less velocity than the
former ; and thus it is that there is produced a contraction in the
size of the stream immediately beyond the opening, designated the
vena contracta, and bearing a proportion to that of the orifice of
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HYDEAtfLICS. 313
about 5 to 8, if it pass through a thin plate, or of 6 to 8, if through
a short cylindrical tube. But if the tube be conical to a length
equal to half its larger diameter, having the issuing diameter less
than the entering diameter in the proportion of 26 to 33, the stream
does not become contracted.
If the vessel be kept constantly full, there will flow from the
aperture twice the quantity that the vessel is capable of contain
ing, in the same time in which it would have emptied itself if not
kept supplied.
1. How many horsepower (H. P.) is required to raise 6000
cubic feet of water the hour from a depth of 300 foct ?
A cubic foot of water weighs 625 lbs. avoirdupois.
(3000 X 625
• on " = 6250, the weight of water raised a minute.
6250 X 300 = 1875000, the units of work each minute.
Then ~<^ oqqq = 56818 = the horsepower required.
2. What quantity of water may be discharged through a cylin
drical moutlipiece 2 inches in diameter, under a head of 25 feet ?
2 1
^2 = g of a foot; ,•. the area of the cross section of the
mouthpiece, in feet, is ^ x ^ x 7854 = 021816.
Theory gives 021816 \/2 ^ X 25 the cubic feet discharged each
second ; but experiments show that the effective discharge is 97 per
cent, of this theoretical quantity: g = 322.
Hence, 97 X 021816 v'644 x 25 = 84912, the cubic feet
discharged each second.
84912 X 625 = 530688 lbs. of water discharged each second.
Effluent water produces, by its vis viva, about 6 per cent. less me
chanical effect than does its weight by falling from the height of
the head.
3. What quantity of water flows through a circular orifice in a
thin horizontal plate, 3 inches in diameter, under a head of 49 feet ?
Taking the contraction of the fluid vein into account, the velo
city of the discharge is about 97 per cent, of that given hy theory.
The theoretic velocity is v/2^~x49 = 7 ^"644 = 5621.
■97 X 5621 = 54523 = the velocity of the discharge.
The area of the transverse section of the contracted vein is 64
of the transverse section of the orifice.
3 1
jg = ^ = 25, and (25)2 X 7854 = 0490875 = arcaof orifice.
.. 64 X 0490875 = 031416, the area of the tr.ansversc section
of the contracted vein.
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314 THE paACTICAL MODEL CALCULATOH.
Hence, 54>523 X 031416 = 17129, the cubic feet of water
discharged each second. The later experiments of Poncebt,
Bidone, and Leshros give SSS for the coefficient of contraction,
"Water issuing through lesser orifices give greater coefficients of
contraction, and become greater for elongated rectangles, than for
those which approach the form of a square.
Observations show that the result above obtained is too great ;
5^5 of this result aro found to be very near the truth.
jgOfl'ri29 = 10541.
4. AYhat quantity of water flows through a rectangular aperture
7'87 inches broad, and 3'94 inches deep, the surface of the water
being 5 feet above the upper edge ; the plate through which the
water flows being 125 of an inch thick.
787
jn" = ■C5583, decimal of afoot.
394
j^ = 32833, decimal of a foot.
5 and 532833 are the heads of water above the uppermost and
lowest horizontal surfaces.
The theoretical discharge will bo
g X G5583 >/2y((5328)^  (5)') = 39268 cubic feet.
Table I. gives the coefficient of eflux in this case, 615, which
is found opposite 5 feet and under 4 inches ; for 394 is nearly
equal 4.
39268 X 615 = 2415 cubic feet, the effective discharge.
5. What water is discharged through a rectangular orifice in a
thin plate 6 inches broad, 3 inches deep, under a head of 9 feet
measured directly over the orifice ?
:jg = 5, decimal of a
foot.
12 "
, decimal of a foot.
The theoretical discharge will be
I X 6 v/2^ I (925/  (9f I = 3033 cubic feet.
Table II. gives the coefficient of efflux between 604 and 606 ;
we shall take it at 605, then
3033 X 605 = 1833 cubic feet, the efi'ective discharge.
6. A weir 82 feet broad, and 492 feet head of water, hovr many
cubic feet are discharged each second ?
The quantity will be
c X 82 ^/2i?(4■92)^; g = 322;
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HYDRAULICS.
Table I,— The Ooeficients for the JUjKux through rcetangvlar ori
fices in a thin vertical plate. The heads are measured where the
water may he considered still.
aa •»»
ARM h
Height o
Ohifke.
8
i
1
4
1
79
J d
619
634
656
686
«0l
620
638
b 4
G8I
60J
6 1
640
653
C b
4
606
6
639
6
11
607
6 7
6oO
b66
594
eo9
11
4
6 o
63j
634
649
648
bb"
G18
6 3
b3
(47
656
^M^
615
6 7
64i>
10
593
610
6 8
6
644
650
bOO
f 1
bl
m
6 8
6
b 8
6''6
641
638
647
644
GO
616
Ci
601
bl5
621
621
630
635
eo
6 3
613
618
618
6 5
630
fO
611
615
bl
1
6 J
bO
601
609
612
613
617
619
fto
C
bO
e 9
4
CIS
10
4
9
1
Ml
Table II— The Co ^i t ntsfo tlelJffl z t? oujJ r da juh
fices in at! n verti il plat tie lea la o/ ater hei j
8
4
2
1
8
4
■1
593
613
(37
059
6&>
biZ
612
f3b
bu6
6»0
K
u«
bl3
635
653
06
694
I
594
bl4
6^4
650
087
5
595
614
633
647
681
597
615
632
644
064
b7o
/
615
633
641
660
661
599
616
630
638
6j5
•A
601
616
f29
650
6^7
1)
617
629
632
644
651
?
II
604
617
626
628
040
646
"
b05
616
622
f27
6G6
'.
II
604
en
618
624
632
636
6
n
604
613
bl6
b21
631
II
603
612
613
618
b24
6.0
7
(1
GOi
blO
611
hi 6
020
C.l
f
(1
608
009
614
C!6
617
!
1)
(01
m
607
C12
bl3
(13
100
FOl
bOS
606
610
610
t,09
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iSTo TilE PRACTICAL MODEL CALCULATOH.
c is termed the cociEcIent of efflux, aad on an average may be taken
at 4, It 13 found to varj from 385 to lW,
Then 4 X 82 v'(ti44) (492p = 2670033, the cubic feet dis
charged each second.
7. What breadth must be given to a notch, in a thin plate, with
a head of water of 9 inches, to allow 10 cubic feet to flow each
second *r
The breadth will b
10
10
.  • ,, = 479C3 feet.
c ^2g X (75)^ 4 X v/644 x (75)^
Changes in the coefiicients of efflux through convergent sides
often present themselves in practice : they occur in dams which are
inclined to the horizon.
Poneelet found the coefficient '8, when the board was inclined
45°, and the coefficient '74 for an inclination of 63° 34', that is
for a slope of 1 for a base, and 2 for a perpendicular.
8. If a sluice board, inclined at an angle of 50°, which goes
across a channel 2'25 feet broad, is drawn out ■§ feet, what quan
tity of water will be discharged, the surface of the water standing
4 feet above the surface of the channel, and the coefficient of efflux
taken at '78 ?
The height of the aperture = 5 sin. 50° =. 3850222 ; 4 and
4 3830222 = 36169778, aie the heads of water.
.. gx 225 X 78 X ^2^Uif  {3G17)H = 10'5257 cu
bic feet, the quantity discharged.
The calculations just made appertain to those cases where the
water flows from all sides towards the aperture, and forms a con
tracted vein on every side. We shall next calculate in cases where
the water flows from one or more sides to the aperture, and hence
produces a stream only a
partially contracted, ni,
n, 0, p, are four orifices in
the bottom AECD of a
vessel ; the contraction by
efflux through the orifice
o, in the middle of the bot
tom, isgeneral, as the water
can flow to it from all
sides ; the contraction c
from the efflux through m, n, p, is partial, as the water can only
flow to them from one, two, or three sides. Partial contraction
gives an oblique direction to the stream, and increases the quantity
discharged.
9. What quantity of water is delivered through a flow 4 feet
broad, and 1 foot deep, vertical aperture, at a pressure of 2 feet
above the upper edge, supposing the lower edge to coincide with
hv Google
UYDRAULICS.
the lower side of the channel, so tliat ther
bottom ?
The theoretical discharge will be
4 .^ f,„J .,,l\
817
contraction at the
( J X ^Ig I (3f  (2)^ I = 50668 cubic feet.
1 the table page 315, may
The coefficient of contraction g
be taken at 603.
I. — Comparison of the Theoretical withthe Real Dhohargesfror.
Orijice.
"'Hf2r
°SS'
'""Slices," '
4381
2722
1 to 062133
619(i
3846
1 to 062078
7589
4710
1 to 062004
i
8763
6430
1 to 06203*
5
9797
6075
I to O'6201O
10732
6654
1 to 062000
11502
7183
1 to 061965
12392
1 to 061911
1 to 061892
10
13855
8574
11
14530
8990
1 to 061873
12
15180
9381
1 to 061819
13
15797
9764
1 to 061S10
10130
1 to 061795
15
1G968
10472
1 to 061716
II. — Comparison of the Tlieoretkal with the Real Discharges fro
a Tube.
Cr>iiEtant height
*.rS6 throuih 1
£'Sfe
thsoiiUcS 10 Iha raJ
Piris FD^t.
Cubit IiwhoB
1
4381
3539
1 to 081781
2
6196
5002
1 to 080729
7589
6126
I to 080724
8763
7070
1 to 080681
6
0797
7900
1 io 080638
10732
8654
1 to 080638
7
11592
9340
1 to 080577
8
12392
9976
1 to 080496
9
13144
10579
1 to 080485
10
13855
11161
1 to 080488
11
l'!530
1169S
J to 080477
12
15180
12205
1 to 080403
13
15797
12699
1 to 080390
14
16393
13177
1 to 080382
16968
13620
1 to 080270
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318 THE PRACTICAL MODEL CALCCLATOR.
THE DISOIIAEGB BY DIFFERENT APERTURES AiSD lUEES, UKDER DIF
The velocity of water flowing out of a hoHzontal aperture, is as
the square root of the height of the head of the water. — That is, the
pressure, and consequently the height, is as the square of the ve
locity ; for, the quantity flowing out iu any short time is aa the
velocity ; and the force required to produce a velocity in a certain
quantity of matter in a given time is also as that velocity ; there
fore, the force must be as the square of the velocity.
Or, supposing a very small cylindrical plate of water, imme
diately over the orifice, to be put in motion at each instant, by the
pressure of the whole cylinder upon it, employed only in generat
ing its velocity ; this plate would he urged by a force as much
greater than its own weight as the column is higher than itself,
through a space shorter in the same proportion than that height.
But where the forces are inversely as the spaces described, the
final velocities are equal. Therefore, the velocity of the water
flowing out must be equal to that of a heavy body falling from the
height of the head of water ; which is found, very nearly, by mul
tiplying the square root of that height in feet by 8, for the number
of feet described in a second. Thus, a head of 1 foot gives 8 ; a
head of 9 feet, 24. This is the theoretical velocity ; but, in con
sequence of the contraction of the stream, we must, in order to ob
tain the actual velocity, multiply the square root of the height, in
feet, by 5 instead of 8.
The velocity of a fluid issuing from an aperture is not affected
by Its density being greater or less. Mercury and water issue
with equal velocities at equal altitudes.
The proportion of the theoretical to the actual velocity of a fluid
issuing through an opening in a thin substance, according to M,
Eytelwein, is as 1 to '619 ; but more recent experiments make it
aa 1 to '621 up to ■645.
APPLICATION OP THE TABLES IN THE PRECEDINO PAGE.
Table I. — To find the quantities of water discharged hy orifices
of different sizes under different altitudes of the fluid in the reser
voir.
To find the quantity of fluid discharged by a circular aperture
3 inches in diameter, the constant altitude being 30 feet.
As the real discharges are in the compound ratio of the area of
the apertures and the square roots of the altitudes of the water,
and aa the theoretical quantity of water discharged by an oriSce
one inch in diameter from a height of 15 feet is, by the second co
lumn of the table, 16968 cubic inches in a minute, we have this
proportion ; 1 ^/lb : 9 v/30 : : 16968 : 215961 cubic inches ; the
theoretical quantity required. This quantity being diminished in
the ratio of 1 to 62, being the ratio of the theoretical to the ac
tual discharge, according to the fourth column of the table, gives
133896 cubic inches for the actual quantity of water discharged by
hv Google
HYDRAULICS. S19
tbe given aperture. Hence, the quantity sliould be rather greater,
because large orifices discharge more in proportion than small ones ;
while it should be rather less, because the altitude of the fluid
being greater than that in the table with which it is compared, the
flowing vein of water becomes rather more contracted. The quan
tity thus found, therefore, is nearly accurate as an average.
When the orifice and altitude are less than those in the table, a
few cubic inches should be deducted from the result thus derived.
The altitude of the fluid being multiplied by the coefficient 8'016
will give its theoretical velocity; and as the velocities are as the
qnantities discharged, the real velocity may be deducted from the
theoretical by means of the foregoing results.
Table II, — To find the quantities of water disaharged ly tubes
of different diameter, and under different heights of water.
To find the quantity of water discharged by a cylindrical tube,
4 inches in diameter, and 8 inches long, the constant altitude of
the water in the reservoir being 25 feet.
Find, in the same manner as by the example to Table I., the
theoretical quantity discharged, which is furnished by this analogy.
1 v^l5 : 16 ^/25 : : 16968 : 350490 cubic inches, the theoretical
discharge. This, diminished in the ratio of 1 to 81 by the 4th
column, will give 28473 cubic inches for the actual quantity dis
charged. If the tube be shorter than twice its diameter, the
quantity discharged will be diminished, and approximate to that
from a simple orifice, as shown by the production of the vena eoTi
tracta already described.
According to Eytelwein, the proportion of the theoretical to tlie
real discharge through tubes, is as follows :
Through the shortest tube that will cause the stream to adhere
everywhere to its sides, as 1 to 08125.
Through short tubes, having their lengths from two to four
times their diameters, as 1 to 082.
Through a tube projecting within the reservoir, as 1 to 050.
It should, however, be stated, that in the contraction of the
stream the ratio is not constant. It undergoes perceptible varia
tions by altering the form and position of the orifice, the thickness
of the plate, the form of the vessel, and the velocity of the issu
ing fluid.
Dedtietions Jrom experiments made iy Bossut, MiclieUoti.
1. That the quantities of fluid discharged in equal times from
differentsized apertures, the altitude of the fluid in the reser
voir being the same, are to each other nearly as the area of the aper
tures.
2, That the quantities of water discharged in equal times by
the same orifice under different heads of water, are i?early as the
square roots of the corresponding heights of water in the reservoir
above the centre of the apertures.
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320 TUE PRACTICAL MODEL CALCOLATOR.
3. That, in general, the quantities of water discharged, in the
same time, bj different apertEres under different heights of water
in the reservoir, are to one another in the compound ratio of the
areas of the apertures, and the square roots of the altitudes of the
water in the reservoirs.
4. That on account of the friction, the smallest orifice discharges
proportionally less water than those which are larger and of a
similar figure, under the same heads of water.
5. That, from the same cause, of several orifices whose areas
are equal, that which has the smallest perimeter will discharge
more water than the other, under the same altitudes of water in
the reservoir. Hence, circular apertures are most advantageous, as
they have less rubbing surface under the same area.
6. 'Xhat, in consequence of a slight augmentation which the
contraction of the fluid vein undergoes, in proportion as the height
of the fluid in the reservoir increases, the expenditure ought to be
a little diminished.
7. That the discharge of a fluid through a cylindrical horiaontal
tube, the diameter and length of which are equal to one another,
is the same as through a simple orifice.
8. That if the cylindrical horizontal tube be of greater length
than the extent of the diameter, the discharge of water is much
increased.
i). That the length of the cylindrical horizontal tube may be
increased with advantage to four times the diameter of the onfice.
10. That the diameters of the apertures and altitudes of water
in the reservoir being the same, the theoretic discharge through a
thin aperture, which is supposed to have no contraction in the vein,
the discharge through an additional cylindrical tube of greater
length than the extent of its diameter, and the actual discharge
through an aperture pierced in a thin substance, are to each other
as the numbers 16, 13, 10.
11. That the discharges by diff'erent additional cylindrical tubes,
under the same head of water, are nearly proportional to the areas
of the orifices, or to the squares of the diameters of the orifices.
12. That the discharges by additional cylindrical tubes of the
same diameter, under different heads of water, are nearly propor
tional to the square roots of the head of water,
13. That from the two preceding corollaries it folloivs, in gene
ral, that the dischai'ge during the same time, by ditlerent addi
tional tubes, and under different heads of water in the reservoir,
are to one another nearly in the compound ratio of the squares
of the diameters of the tubes, and the square roots of the heads
of water.
The discharge of fluids by additional tubes of a conical figure,
when the inner to tlie outer diameter of tlie orifice is as '6'd to
2(), is augmented very nearly oneseventeenth and seveutcnths
wore tlwu by cylindrical tubes, if the enlargement be not curried
too far.
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HYDRAULICS. 321
DISCHARGE BY COMrOUND TUBES.
Deductions from the experiments of M. Venturi.
In the discharge by compound tubes, if the part of the addi
tional tube nearest the reservoir have the form of tho contracted
vein, the expenditure will be the same as if the fluid were not con
tracted at all ; and if to the smallest diameter of this cone a cylin
drical pipe be attached, of the same diameter as the least section
of the contracted vein, the discharge of the fluid will, in a horizon
tal direction, be lessened by the friction of the water against the
side of the pipe ; but if the same tube be applied in a vertical
direction, the expenditure will be augmented, on the principle of
the gravitation of falling bodies; consequently, the greater the
length of pipe, the more abundant is the discharge of fluid.
If the additional compound tube Lave a cone applied to the op
posite extremity of the pipe, the expenditure will, under the same
head of water, bo increased, in comparison with that through a
simple orifice, in the ratio of 24 to 10.
In order to produce this singular effect, the cone nearest to the
reservoir must be of the form of the contracted vein, which will
increase the expenditure in the ratio of 121 to 10. At the other
extremity of the pipe, a truncated conical tube must be applied,
of which the length must be nearly nine times the smaller diameter,
and its outward diameter must be 1"8 times the smaller one. This
additional cone will increase the discharge in tho proportion of
24 to 10. But if a great length of pipe intervene, this additional
tube has little or no efl"oct on the quantity discharged.
According to M. Venturi's experiments on the discharge of
water by bent tubes, it appears that while, with a height of water
in the reservoir of 32'5 inches, 4 Paris cubic feet were discharged
through a cylindrical horizontal tube in the space of 45 seconds,
the discharge of the same quantity through a tube of tho same
diameter, with a curved end, occupied 50 seconds, and through a
like tube bent at right angles, 70 seconds. Therefore, in making
cocks or pipes for the discharge or conveyance of water, great
attention should be paid to the nature and angle of the bondings ;
right angles should be studiously avoided.
The interruption of the discharge by various enlargements of
the diameter of the tubes having been investigated by Si. Venturi,
by means of a tube with a diameter of 9 lines, enlarged in several
parts to a diameter of 24 lines, the retardation was found to in
crease nearly in proportion to the number of enlargements ; the
motion of the fluid, in passing into the enlarged parts, being
diverted from its direct course into eddies against the sides of the
enlargements. From which it may be deduced, that if the inter
nal roughness of a pipe diminish the expenditure, the friction of
the water against these asperities does not form any considerable
part of the cause. A rightlined tube may have its internal sur
face highly polished throughout its whole length, and it may every
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,122 THE PRACTICAL MOI>EL CALCULATOE.
whore possess a diameter greater than the orifice to wliich it is
applied; but, nevertheless, the expenditure will be greatly retarded
if the pipe should have enlarged parts or swellings. It is not
enough that elbows and contractiona be avoided ; for it may hap
pen, by an intermediate enlargement, that the whole of the other
advantage may be lost. This will be obvious from the results in
the following table, deduced from experiments with tubes having
various enlargements of diameter.
Head of wnter
in inches.
Number of en
iarged parts.
Seconds in vliieli
4 cubic feet were
disoiiarged.
325
325
325
325
1
3
5
109
147
192
240
DISCHARGE BY CONDUIT PIPES.
On account of the friction against the sides, the less the dia
meter of the pipe, the less proportionally is the discharge of fluid.
And, from the same cause, the greater the length of conduit pipe,
the greater the diminution of the discharge. Hence, the dis
charges made in equal times by horizontal pipes of different lengths,
hut of the same diameter, and under the same altitude of water,
are to one another in the inverse ratio of the square roots of the
lengths. In order to have a perceptible and continuous discharge
of fluid, the altitude of the water in the reservoir, above the axis
of the conduit pipe, must not he less than If inch for every ISO
feet of the length of the pipe.
The ratio of the difl'erence of discharge in pipes, 16 and 24 lines
diameter respectively, may be known by comparing the ratios of
Table I. with the ratios of Table II., in the following page.
The greater the angle of inclination of a conduit pipe, the
greater will be the discharge in a given time ; but when the angle
of the conduit pipo is 6° 31', or the depression of the lower extre
mity of the pipe is oneeighth or oneninth of its length, the rela
tive gravity of the fluid will ho counterbalanced by the resistance
or friction against the sides ; and the discharge is then the same
as by an additional horizontal tube of the same diameter.
A curvilinear pipe, the altitude of the water in the reservoir being
the same, discharges less water when the flexures lie horizontally,
than a rectilinear pipe of the same diameter and length.
The discharge by a curvilinear pipe of the same diameter and
length, and under the same head of water, is still further dimi
nished when the flexures lie in a vertical instead of a horizontal plane.
When there is a number of contrary flexures in a large pipe, the
air sometimes lodges in the highest parts of the flexures, and greatly
retards the motion of the water, unless prevented by airholes, or
stopcocks.
hv Google
HYDEAULICS.
Table I, — Oomparison of the di»charge by conduit pipes of different
lengths, 16 lines in diameter, with the discharge ly additional
tubes inserted in the same reservoir. — By M. Eosstjt.
2778
JOO to 43'39
1957
100 to 3091
1687
100
2507
1351
100
2134
1178
100
1861
1052
100
1662
4068
100
o 4518
2888
100
3231
2353
100
2631
2011
100
2250
1762
100 to 1971
1583
100
1770
Table II. — Comparison of the di
ferent lengths, 24 lines in diameter,
tional tubes inserted in the i
conduit pipes of dif
ith the discharge by addi
■By M. BossuT.
IJamititj.ofW
aim diKbirgti
wSiwa'i'h,
I.«iELhr>r
B«Wol»li.aaothe
'^m^i^Tl^'
IoJ»,24liiJ'An
^■lilS''"
F8«
F6!l
CoWc IntSas.
Cabiolnths.,
30
14243
100 to 6892
60
14243
5564
100 to 8906
90
14243
4534
100 to 3183
120
14243
3944
100 to 2769
14243
3486
100 to 2448
14243
3119
100 to 2190
80
20113
11219
100 to 5578
60
20112
8190
100 to 4072
2
SO
20112
6812
lOO to 8387
120
20112
5885
100 to 2923
2
20113
6232
100 to 2601
2
180
20112
4710
100 to 2341
BISCHAKOE BY WEIKS AND EECTAKGULAS ^
Rectangular orifices in the side of a reservoir, extending to the surface.
Tlie velocity varying nearly as tho square root of the height,
may here be represented hj tho ordinates of a parabola, and the
quantity of water discharged by the area of the parabola, or
twothirda of that of the circumscribing rectangle. So that the
quantity discharged may be found by taking twothirds of the velo
city due to the mean height, and allowing for tlie contraction of
the stream, according to the form of the opening.
In a lake, for example, in the side of which a rectangular open
ing is made without any oblique lateral walls, three feet wide, and
hv Google
324 THE PRACTICAL MODEL CALCULATOR.
extending two feet below the surfaco of the water, the coefficient
of the velocity, corrected for contraction, is 5'1, and the corrected
mean velocity  \/2 X 5'1 = 4'8 ; therefore the area being 6, the
discharge of water in a second is 28'8 cubic feet, or nearly four
The same coefficient serves for determining the discharge over
a weir of considerable breadth ; and, hence, to deduce the depth
or breadth requisite for the discharge of a given quantity of water.
For example, a lake baa a weir three feet in breadth, and the sur
face of the water stands at the height of five feet above it : it is
required how much the weir must be ividened, in order that the
water may be a foot lower. Here the velocity ia f v/5 X 5'1, and the
quantity of water § \/5 X 51 x 3 X 5 ; but the velocity must be re
2 /^ X 5 '1 X 3 X 5
ducedtof /i X 51, andthen the section will bo ■'^ = — —
§ ■/4 X 51
= — — = = 7'5 X \/5 ; and the height being 4, the breadth
must be 7 \/5 = 419 feet.
The diacbargc from reservoira, with lateral orifices of consider
able magnitude, and a constant head of water, may be found by
determining the difi'erence in the discharge by two open orifices of
different heights ; or, in most cases, with nearly equal accuracy,
by considering the velocity due to the distance, below the surface,
of the centre of gravity of the orifice.
Under the same height of water in the reservoir, the same quan
tity always flows in a canal, of whatever length and declivity ; but
in a tube, a difference in length and declivity has a great effect on
the quantity of water discharged.
The velocity of water flowing ia a river or stream varies at dif
ferent parts of the same transverse section. It ia found to be
greatest where the water is deepest, at somewhat less than one
naif the depth from the surface ; diminishing towards the sides
and shallow parts.
Jtemtance to bodies moving infinids. — The deductions from the
experiments of C. Colles, (who first planned the Croton Aqueduct,
Now York,) and others, on this intricate subject, are, as stated, thus :
1. The confirmation of the theory, that the resistance of fluids
to passing bodies is as the squares of the velocities.
2. That, contrary to the received opinion, a cone will move
through the water with much less resistance with its apex foremost,
than with its base forward,
3. That the increasing the length of a solid, of almost any form,
by the addition of a cylinder in the middle, diminbhes the resist
ance with which it moves, provided the weight in the water remains
the same.
hv Google
HYDRAULICS, 325
4, That the greatest breadth of the moving hody should be
placed at tho distance of twofifths of the whole length from the
how, when applied to the ordinary forms in naval architecture.
5. That the bottom of a floating solid should be made triangu
lar ; as in that case it will meet with the least resistance when
moving in the direction of ita longest axis, and with the greatest
resistance when moving with its broadside foremost.
F]iction. of fluids. — Some experiments have been made on this
subject, with reference to the motion of bodies in water, upon a
cylindrical model, 30 inches in length, 26 inches in diameter, and
weighing 255 lbs. avoirdupois. The cylinder was placed in a cis
tern of salt water, and made to vibrate on knifeedges passing
through its axis, and was deflected over to various angles by means
of a weight attached to the arm of a lever. The esperiments were
then repeated without the water, and the following are the angles
of deflection and vibration in the two cases.
In the
■'—'"■ ,
luthd
22° 30'
22° 24'
22° 30'
20° 0'
22 10
22 6
21 36
21 3
21 54
21 48
20 48
2016
21 3S
21 30
Sc.
4c.
&c.
fa.
Showing that the amplitude of vibration when oscillating in water
is considerably less than when oscillating without water. In the
experiments there is a falling off in the angle of 24', or nearly
half a degree. The amount of force acting on the surface of the
cylinder necessary to cause the above difference was calculated;
and tho author thinks that it is not equally distributed on the
surface of the cylinder, but that the amount on any particular
part might vary as the depth. On this supposition, a constant
pressure at a unit of depth is assumed, and this, multiplied by the
depth of any other point of the cylinder immersed in the water,
will give the pressure at that point. These forces or momenta
being summed by integration and equated with the sum of the
moments given by the experiments, we have the value of the con
stant pressure at a unit of depth = ■0000469. This constant, in
another experiment, the weight of the model being 197 lbs. avoir
dupois, and consequently the part immersed in the water being dif
ferent from that in the other experiment, was '0000452, which
differs very little from the former, — indicating the probability of
the correctness of the assumption.
The drainage of water through pipes. — The experiments made
under the direction of the Metropolitan Commissioners of Sewers,
on the capadties of pipes for the drainage of towns, have presented
some useful results for the guidance of those who have to make
hv Google
326 THE PRACTICAL MODEL CALCULATOR.
calculations for a similar purpose. The pipes, of various dia
meters, from 3 to 12 inches, were laid on a platform of 100 feet
in length, the declivity of which could be varied from a horizontal
level to a, fall of 1 in 10, The water was admitted at the head
of the pipe, and at five junctions, or tributary pipes on each side,
so regulated as to keep the main pipe full.
The results were as follow : —
It was found — to mention only one result — that a line of 6inch
pipes, 100 feet long, at an inclination of 1 in 60, discharged 75 Cubic
feet per minute. The same experiment, repeated with the line of
?ipes reduced to 50 feet in length, gave very nearly the same result.
Vithout the addition of junctions, the transverse sectional area of
the stream of water near the discharging end was reduced to one
fifth of the corresponding area of the pipe, and it required a sim
ple head of water of about 212 inches to give the same result as
that accruing under the circumstances of the junctions. AVith
regard to varying sizes and inclinations, it appears, sufSciently for
practical purposes, that the squares of the discharges are as the
fifth powers of the diameters ; and again, that In steeper declivi
ties than 1 in 70, the discharges are as the square roots of the
inclinations ; but at less declivities than 1 in 70, the ratios of the
discharges diminish very rapidly, and are governed by no constant
law. At a certain small declivity, the relative discharge is as the.
fifth root of the inclination ; at a smaller declivity, it is found as
the seventh root of the inclination ; and so on, as it approaches the
horizontal plane. This may be exemplified by the following results
found by actual experiment :
Discharges of a 6mchpipe at several inclinations.
D'itLuaf s in 1*1
lin 60
75
lin 320
49
lin 80
68
lin 400
486
1 in 100
63
lin 480
48
1 in 120
69
lin 640
475
1 in 160
64
lin 800
472
1 in 200
62
1 in 1200
467
1 in 240
60
Level
46
The conclusion arrived at is, that the requisite sizes of drains
and sewers can be dctcrmLned (near enough for practical purposes,
as an important circumstance has to he considered in providing for
the deposition of solid matter, which disadvantageously alters the
form of the aqueduct, and contracts the waterway) by taking the
result of the 6inch pipe, under the circumstances before mentioned
as a datum, and assuming that the squares of the discharges are
as tho fifth powers of the diameters.
That at greater declivities than 1 in 70, tho discharges are as
the square roots of the inclinations.
hv Google
WATER WHEELS. 02 i
That at less declivities ttan 1 in 70, titc usual law will not
obtain ; but near approximations to the trutli may be obtained by
observing the relative discltarges of a pipe laid at various small
inciinatioas.
That increasing the number of junctions, at intervals, accele
rates the velocity of the main stream in a ratio which increases as
the square root of the inclination, and which is greater than the
ratio of resistance due to a proportionable increase in the length
of the aqueduct. The velocity at which the lateral streams enter
the main line, is a most important circumstance governing the flovt'
of water. In practice, these velocities are constantly variable,
considered individually, and always different considered collectively,
so that their united effect it is difficult to estimate. Again, the
same sewer at different periods may be quite filled, but discharges
in a given time very different quantities of water. It should be
mentioned that in the case of the 6inch pipe, which discharged
75 euhic feet per minute, tho lateral streams had a velocity of
a few feet per second, and the junctions were placed at an angle
of about 35° with the main line. It is needless to say that all
junctions should be made as nearly parallel with the main lino as
possible, otherwise the forces of the lateral currents may impedo
rather than maintain or accelerate the main streams.
WATER WHEELS.
TUE IJXDtaSIIOT WHEEL.
The ratio between the power and effect of an undershot T;heel
is as 10 to 3'18 ; consequently 31'43 lbs. of water must be expended
per second to produce a mechanical effect equal to that of the esti
mated labour of an active man.
The velocity of the periphery of the undershot wheel should bo
equal to half the velocity of the stream ; the floatboards should be
ao constructed as to rise perpendicularly from the water ; not more
than onehalf should ever be below the surface ; and from 3 to 5
should be immersed at once, according to the magnitude of the
wheel.
The following maxims have been deduced from experiments : —
1. The virtual or effective head of water being the same, the
effect will he nearly as the quantity expended ; that is, if a mill,
driven by a fall of water, whose virtual head is 10 feet, and which
discharges 30 cubic feet of water in a second, grind four bolls of
corn in an hour ; another mill having the same virtual head, hut
which discharges 60 cubic feet of water, will grind eight hulls of
corn in an hour.
2. The expense of water being the same, the effect will be nearly
as the height of the virtual or effective head.
3. The quantity of water expended being the same, the effect is
nearly as the square of its velocity ; that is, if a mill, driven by a
hv Google
328 THE PRACTICAL MODEL CALCULATOR.
certain quantity of water, moving with the velocity of four feet per
second, grind three bolls of corn in an hour; anotlici mill, driven
by tho same quantity of water, moving with the velocity of five
feet per second, will grind nearly i^/^ bolls in the hour, because
a : 4^ : : 4* : 5^ nearly.
4. The aperture being the same, the effect will be nearly as the
cube of the velocity of the water ; that is, if a mill driven by water,
moving through a certain aperture, with the velocity of four feet
per second, grind three holla of corn in an hour; another mill,
tliiven by water, moving through the same aperture with the velo
city of five feet per second, will grind 5jg bolls nearly in an hour ;
for as 3 : 5^g : : 4' : 5' nearly.
The height of the virtual head of water may be easily deter
mined from the velocity of the water, for the heights are as the
squares of the velocities, and, consequently, the velocities are as
the square roots of the height.
To calculate the proportions of undershot wheels. — Find the per
pendicular height of the fall of water above the bottom of the mill
course, and having diminished this number by onehalf the depth of
the water where it meets the wheel, call that the height of the fall.
Multiply the height of the fall, so found, by 64348, and take the
square root of the product, which will be the veloeity of the water.
Take onehalf of the velocity of the water, and it will be the
velocity to be given to the fioatboards, or the number of feet they
must move through in a second, to produce a maximum effect.
Divide the circumference of the wheel by the velocity of its fioat
boards per second, and the quotient will be the number of seconds
in which the wheel revolves. Divide 60 by the quotient thus found,
and the new quotient will be the number of revolutions made by
the wheel in a minute.
Divide 90, the number of revolutions which a millstone, 5 feet
in diameter, should make in a minute, by the number of revolutions
made by the wheel in a minute, the quotient will be the number of
turns the millstone ought to make for one turn of the wheel.
Then, as the number of revolutions of the wheel in a minute is to
the number of revolutions of the millstouo in a minute, so must
the number of staves in the trundle bo to the number of teeth in
the wheel, (the nearest in whole numbers.) Multiply the number
of revolutions made by the wheel in a minute, by the number of
revolutions made by the millstone for one turn of the wheel, and
the product will be the number of revolutions made by the millstone
in a minute.
The efi'ect of tlie water wheel is a maximum, when its circum
ference moves with onehalf, or, more accurately, with three
sevenths of the velocity of the stream.
THE BKEAST WUEEL.
The efi'ect of a breast wheel is equal to the efi'ect of an under
shot wheel, whose head of water is equal to the difference of level
hv Google
WATER WnBELS.
329
between the surface of water in tlio reservoir, and the part whero
it strikes tlie wheel, added to that of an overshot, wlioso Iicight is
equal to the difference of level between tlie part where it strikes
the wheel and the level of the tail water.
When the fall of water is between 4 and 10 feet, a breast
wheel should be erected, provided there be enough of water ; an
undershot should be used when the fall is below 1 feet, and an
overshot wheel when the fall exceeds 10 feet. Also, i\hen the fall
exceeds 10 feet, it should be divided into two, and two breast wheels
be erected upon it.
Table for breast wheels.
3i
ti
Ills
1.
IL
111
%
hi
34
Is
il
111
Ti
Mt
V.i
nt

ss
k
a
.f
{i
^ss
FeM.
F8«.
rsel.
FtBl.
E=o.
IM, svr.
CaWs n.
1
017
198 ■§
075
218
192
480
1536
7430
034
351
150
309
680
1084
3715
051
127
8 78
333
882
886
2477
069
62
301
486
384
060
762
1857
5
080
3S7
376
488
428
1070
680
1486
103
225
451
58S
470
1176
626
1238
r
120
153
526
577
1270
581
1061
8
137
110
602
617
513
1858
543
929
l5i
081
677
655
576
U40
512
826
10
171
077
752
C90
607
1518
436
743
It is evident, from the preceding table, that when the height of
the fall is less than 3 feet, the depth of tho floatboarda is so great,
and their breadth so small, that the breast wheel cannot well be
d ; and, on the contrary, when the height of the full ap
! to 10 feet, the depth of the floatboards is too small in
proportion to their breadth ; these two extremes, therefore, must
be avoided in practice. Tho ninth column contains the quantity
of water necessary for impelling tho wheel ; but the total expense
of water should always exceed this by the quantity, at least, which
escapes between the millcourse and the sides and extremities of
the floatboards.
THE OVERSHOT WHEEL.
The ratio between the power and effect of an overshot wheel, is
aa 10 to 6'6, when the water is delivered above the npex of the
wheel, and is computed from the whole height of the fall ; and as
10 to 8 when computed from the height of the wheel only ; con
sequently, the quantity of water expended per second, to produce
a mechanical eff'ect equal to that of the aforesaid estimated labour
of an active man, is, in the first instance, 15'15 lbs., and in the
second instance, 125 lbs.
Hence, the effect of the overshot wheel, under the same circum
hv Google
o30 THE PKACTICAL MODEL CALCULATOR.
Stances of quantity and fall, is, at a medium, double that of the
undershot.
The velocity of the periphery of an overshot wheel should be
from 6^ to 8^ feet per second.
The higher the wheel is, in proportion to the whole descent, the
greater will be the effect.
And from the equality of the ratio between the power and effect,
subsisting where the constructions are similar, we must infer that
the effects, as well as the powers, are as the quantities of water and
perpendicular heights multiplied together respectively.
Worhmg machinery hyhydraulio pressure. — The vertical pressure
of water, acting on a piston, for raising weights and driving machi
nery, is coming into use in many places where it can be advantage
ously applied. At Liverpool, Newcastle, Glasgow, and other places,
it is applied to the working of cranes, drawing coalwagons, and other
purposes requiring continuous power. The presence of a natma! fall,
like that of Golway, Ireland, which can be conducted to the engine
through pipes, is, of course, the most economical situatioit for the
application of such power ; in other situations, artificial power must
be used to raise the water, which, even under this disadvantage, may,
from its readiness and simplicity of action, be often serviceably em
ployed. WTierever the contiguity of a steam engine would be dan
gerous, or otherwise objectionable, a water engine would afford the
means of receiving and applying the power from any required dis
tance, precautions being taken against the action of frost on the fluid.
Required the horse power of a centre discharging Turbine water
wheel, the head of water being 25 feet, and the area of the open
ing 400 inches.
The following table shows the woikmg horse power of both the
inward and outward discharging Turbine w ater wheels ; they are
calculated to the square inch of opening
Disch^S^ne
Oalwarinatliiirj
DlariUis
He»a.
UnnsPoKEr.
IlOHPo^sr.
H<.d
II Tis rm«T.
Horse PowBr.
g
■00821
■012611
22
19d23
■339972
4
■01483
■025145
23
20T87
•864182
6
■02137
■038124
24
22315
■384615
6
02685
■045618
25
23367
■112018
7
03414
■058314
2G
25125
■487519
8
■04198
■074113
27
26482
■455698
■Oo206
28
28135
4844:>7
10
058S3
■106215
20
2Jot.3
■610838
11
06921
■118127
30
80817
■537721
12
■07H51
■135610
31
32316
■5GI425
13
■08882
■160638
32
83617
■587148
H
10054
■173158
o3
34823
■611018
15
■11002
■192234
34
3a»4
■638174
■12093
■211592
35
37123
■665164
17
■13196
■231161
31874
■692150
18
■14275
■2S7U5
37
10118
■726148
19
■15613
■273325
38
41762
■764115
20
■le927
■296618
33
42]oa
■804479
21
■18109
■3171S7
' 40
43718
■849814
b,Google
WATER WHEELS. 331
Opposite 25 in the column marked " Head," tlio working horse
power to the square inch is found to lie '25667, which, multiplied
by 400, gives fll60S, the horse power required.
What is the working horse power of an outward discharging
Tuibine, under the effective head of 20 feet ; the area of all the
openings being 325 square inches. In the table, opposite 20, wo
find 2^6618, then 296618 X 325 = 964, the required horse power.
What is the number of revolutions a minute of an outward
discharging Turbine wheel, the head being 19 feet and tlie dia
meter of the wheel 60 inches ?
In the table for the outward discharging wheel, opposite 19, and
under 60 inches, we find 97, the number of revolutions required.
What is the number of revolutions a minute of an inward dis
charging Turbine, under a head of 21 feet, the diameter being
72 inches ?
In the table for the inward discharging wheel, opposite 21 feet,
and under 72 inches, we find 95, the number of revolutions a
minute.
These Turbine tables were calculated by the author's brother,
the late John O'Bjrne, C. E., wlio died in Kew York, on the tith
of AprU, 1851.
TiS
—
Out'ward discharging
Turh
me.
—
—
—
— 1
1"
■li
~w
"IT
~^
48
54
sr
^
11
7884
Z
•Jb
__
100
so
70
60
52
42
37
o5
82
30
28
27
11
4
111
73
57
49
44
41
37
«4
32
30
28
5
123
100
82
71
62
55
51
4t.
42
37
81
6
135
109
91
78
62
66
60
io
il
Stj
36
7
146
118
96
84
73
65
59
53
49
47
42
40
JH
8
156
125
105
90
79
71
63
57
5
49
43
4J
39
S
IGG
133
111
95
75
67
bl
67
50
49
41
10
1T5
UO
117
100
87
70
64
59
50
ol
47
40
H
183
147
122
106
92
81
74
b7
62
57
o4
49
4b
Vi
191
156
127
110
96
79
70
64
o9
5.
•)i
51
IB
200
169
115
100
89
81
7^
57
5j
6.,
li
206
166
138
n8
104
92
76
64
69
5j
16
213
171
142
122
107
95
86
78
72
60
bl
58
Ob
le
222
177
148
126
111
98
89
8J
74
69
64
59
67
n
227
182
152
131
115
101
91
ss
77
71
66
02
j9
18
234
187
156
134
117
105
94
85
78
73
t3
61
19
193
161
138
120
107
97
88
bl
74
o4
bJ
20
247
197
164
141
124
110
99
90
84
00
€4
ai
252
202
168
145
126
114
101
"iZ
86
73
(&
Oj '
'Z2
259
208
172
149
129
116
105
94
87
80
74
m
07 1
23
2B3
212
no
151
183
119
106
9b
8>
84
70
21
270
216
180
156
135
120
109
98
92
8^
7b
74
72
25
277
222
184
158
188
123
111
101
93
&b
80
76
74
20
282
226
189
161
141
125
113
108
9j
87
81
78
27
280
229
191
165
148
129
116
105
97
S3
79
77
28
291
233
195
167
146
ISO
118
107
99
91
85
78
2W7
'2Z7
199
170
149
lis
109
100
81
80
ao
303
241
202
174
162
135
111
102
94
»a
i^
_!L
b,Google
HE PRACTICAL MODEL CALCULATOR.
Inward discharging Turbine.
1l
■n
30
30
42
48
54
60
66
72
78
TT
00
"ss
~^
111
86
74
62
54
48
47
40
30
~32~
31
30
"27"
i
125
96
83
70
62
55
51
45
41
37
34
31
5
141
112
94
78
61
55
50
46
43
40
36
152
122
101
80
76
67
62
55
5i
47
48
42
88
7
166
181
108
98
82
72
65
60
54
51
47
44
42
S
175
189
U6
87
76
71
67
54
49
47
45
9
188
149
123
106
93
81
74
68
63
57
53
51
47
10
135
136
129
111
99
ee
78
71
61
56
52
49
11
208
167
130
117
102
91
82
74
63
5S
62
12
217
169
142
122
107
97
85
78
71
66
61
57
54
13
221
178
148
127
112
99
82
74
69
64
01
56
14
231
184
153
133
110
104
92
76
71
62
58
15
191
159
130
119
107
05
87
SO
73
68
64
01
16
245
198
105
144
111
09
00
76
71
66
68
17
253
203
108
148
127
114
102
92
85
78
73
64
18
260
209
173
ISO
132
116
104
95
87
82
75
69
60
267
215
176
153
134
120
108
89
77
72
67
20
276
222
183
157
138
122
111
101
93
85
79
74
69
ai
288
226
180
162
141
125
lis
103
95
86
SO
75
71
22
200
102
164
145
129
116
107
96
89
77
23
209
236
190
1G7
140
13S
118
100
97
84
79
74
2i
303
210
201
171
151
136
122
HI
101
86
80
75
■25
810
247
206
176
155
138
123
112
104
88
82
76
20
314
248
210
180
157
189
126
115
100
97
00
84
79
27
SIO
254
218
183
102
142
128
117
108
09
85
80
28
827
261
218
186
164
140
129
119
109
102
87
83
20
833
265
221
189
16«
148
111
103
95
80
30
;«6
271
'^'2*
1"3
168
151
13(5
124
lit
105
97
00
85
wnrDMins.
1. The velocity of windmill sails, whether unloaileil or loaded,
so as to produce a maximum effect, is nearly as the velocity of the
wind, their shape and position being the same.
2. The load at the maximum is neiudy, hut somewhat less than,
as the square of the velocity of the wind, the shape and position
of the sails being the same.
3. The effects of the same sails, at a maximum, are nearly, but
somewhat less than, as the cubes of the velocity of the wind.
4. The load of the same sails, at the maximnm, is nearly as the
squares, and their effect as the cubes of their number of turns in a
given time.
5. When sails are loaded so as to produce a maximum at a given
velocity, and the velocity of the wind increases, the load continu
ing the same, — 1st, the increase of effect, when the increase of the
velocity of the wind is small, will be nearly as the squares of those
velocities ; 2dly, when the velocity of the wind is double, the ef
fects will be nearly as 10 to 27^ ; but, 3dly, when the velocities
compared are more than double of that when the given load pro
duces a maximum, the effects increase nearly in the simple ratio
of the velocity of the wind.
hv Google
■WINDMILL?.
333
6. In sails where the figure and position are similar, and the ve
locity of the wind the same, the number of turns, in a given time,
will be reciprocally as the radius or length of the sail.
7. The load, at a maximum, which sails of a similar figure and
position will overcome, at a given distance from the centre of mo
tion, will be as the cube of the radius.
8. The effects of sails of similar figure and position are as the
square of the radius.
9. The velocity of the extremities of Dutch sails, as well as of the
enlarged sails, in all their usual positions when unloaded, or even
loaded to a maximum, is considerably greater than that of the wind.
The results in Table 1 are for Dutch sails, in their common posi
tion, when the radius was 30 feet. Table 2 contains the most
efficient angles.
Kumler of
R.HO bsntfen
POM! .f the
■iS'
sll
"£""
ijsl'jri*
Angle ot™tiEr.
3
2 miles
ome
1
2
72°
71
IS
IS
6
4 miles
osoo
3
4
72
74
18 middle
16
6
5 miles
0833
5
6
77*
83'
121
7
Supposing the radius of the sail to be 30 feet, then the sail* will
commence at i, or 5 feet from the axis, where the angle of inclina
tion will be 72 degrees ; at , or 10 feet from the axis, the angle
will be 71 degrees, and so on.
Results of Experiments on the effect of Windmill Sails in grind
ing corn. — By M. Coulomb.
A windmill, with four sails, measuring 72 feet from the ex
tremity of one sail to that of the opposite one, and 6 feet 7 inches
wide, or a little more, was found capable of raising 1100 lbs. avoir
dupois 238 feet in a minute, and of working, on an average, eight
hours in a day. This is equivalent to the work of 34 men, 30 square
feet of canvas performing about the daily work of a man.
When a vertical windmill is employed to grind corn, the mill
stone makes 5 revolutions in the same time that the sails and the
arbor make 1.
The mill does not begin to turn till the velocity of the wind is
about 13 feet per second.
When the velocity of the wind is 19 feet per second, the sails
make from 11 to 12 turns in a minute, and the mil! will grind from
880 to 990 lbs. avoirdupois in an hour, or about 22,000 lbs. in 24
hours.
hv Google
THE APPLICATION OF LOGAKITHMS.
The practice of performing calculations by Logarithms is an ex
ercise so useful to computers, that it requires a more particular ex
planation than could have been properly given in that part of the
work allotted to Arithmetic.
A few of the various applications of logarithms, best suited to
the calculations of the engineer and mechanic, have therefore been
collected, and are, with other matter, given, in hopes that they will
come into general use, as the certainty and accuracy of their re
sults can be more safely relied upon and more easily obtained
than with common arithmetic.
By a slight examination, the student will perceive, in some de
gree, the nature and effect of these calculations; and, by frequent
exercise, will obtain a dexterity of operation in every case admitting
of their use. He will also more readily penetrate the plans of the
different devices employed in instrumental calculations, which are
rendered obscure and perplexing to most practical men by their ig
norance of the proper application of logarithms.
Logarithms are artificial numbers which staad for natural num
bers, and are so contrived, that if the logarithm of one number be
added to the logarithm of another, the sum will be the logarithm
of the product of these numbers ; and if the logarithm of one num
ber be taken from the logarithm of another, the remainder is the
logarithm of the latter divided by the former ; and also, if the loga
rithm of a number be multiplied by 2, 3, 4, or 5, &c., we shall have
the logarithm of the square, cube, &o., of that number ; and, on the
other hand, if divided by 2, 8, 4, or 5, &c., we have the logarithm
of the square root, cube root, fourth root, &c., of the proposed num
ber ; so that with the aid of logarithms, multiplication and division
are performed by addition and subtraction ; and the raising of
powers and extracting of roots are effected by multiplying or di
viding by the indices of the powers and roots.
In the table at the end of this work, are given the logarithms of
the natural numbers, from !■ to 1000000 by the help of differences ;
in large tables, only the decimal part of the logarithm is given, as
the index is readily determined ; for the index of the logarithm of
any number greater than unity, is equal to one less than the num
ber of figures on the left hand of the decimal point ; thus,
The index of 12345 is i;
12345 _ Z;
12g4o  2;
12345  1,
12345  0
hv Google
THE APPLICATION OP LOGAEITIIMS. 335
The index of any decimal fraction is a negative number equal to
one and the number of zeros immediately following the decimal
point ; thus,
The index of 00012345 is 4 or 4
0012345 is 3 or ^
012345 ii 2 or 2
12345 is 1 or 1
Eecanse the decimal part of the logarithm is always positive, it
is better to place the negative sign of the index above, instead of
before it; thus, 3" instead of 3. For the log. of 00012345 is
better expressed by 40914911, than by — 4'0914911, because only
the index is negative — i. e., 4 is negative and 0914911 is positive,
and may stand thus, —4 + ■0914911.
Sometimes, instead of employing negative indices, their comple
ments to 10 are used :
for J0914911 is substituted 60914911
— 30914911 70914911
— 20914911 80914911
&c. ko.
"When this is done, it is necessary to allow, at some subsequent
stage, for the tens by which the indices have thus been increased.
It is so easy to take logarithms and their corresponding numbers
out of tables of logarithms, that we need not dwell on the method
of doing so, but proceed to their application.
MULTIPLICATION BY LOGAEITHMS.
Take the logarithms of the factors from the tabic, and add them
together ; then the natural number answering to the sum is the
product required : observing, in the addition, that what is to be
carried from the decimal parts of the logarithms is always positive,
and must therefore be added to the positive indices ; the differencehe
tween this sum and the sum of the negative indices is the index of the
logarithm of the product, to which prefix the sign of the greater,
This method will be found more convenient to those who have
only a slight knowledge of logarithms, than that of using tJic aritli
metical complements of the negative indices.
1. Multiply 37153 by 4086, by logarithms.
37153 156&9
4086 .0611298 4
Prod. 1518071 21812923
2. Multiply 112246 by 13958, by logarithms.
112246 20501709
13958 11448232
Prod. 1566729 31949941
hv Google
33S THE PRACTICAL MODEL CALCULATOR,
3. Multiply 467512 by 3275, by logarithms.
Kbs. Logs.
467512 16697928
■3275 r51.52113
Prod. 1531102, 11850041
Here the +1 that is to be carried from the decimals, eanccla
the —1, and consequently there remains 1 in the upper line to be
set down.
4. Multiply 37816 by 04782, by Jogarithraa.
Nos. Logs.
■37816 .T57T6756
■04782 2^6796096
Prod. 0^018083G .22572852
Here the +1 that is to be carried from the decimals, destroys
the —1 in the upper line, as before, and there remains the —2
to be set down.
5. Multiply 3768, 2053, and 007G93, together.
3768 0576lioD
2053 03123889
■007693 38S60957
Prod. 0595108 .27745955
Here the +1 that is to be carried from the decimals, ivhon ad
ded to —3, makes —2 to be set down.
6. Multiply 3586, 21046, 8372, and 0294, together.
3586 05546103
21046 03231696
8372 .19228292
0294 24683473
Prod. 1857618 12689564
Here the +2 that is to be carried, cancels the —2, and there
remains the —1 to be set down.
DIVISION EY LOGAKITHMS.
From the logarithm of the dividend, subtract the logarithm of
the divisor ; the natural number answering to the remainder will be
the quotient required.
Observing, that if the index of the logarithm to be subtracted is
positive, it is to be counted as negative, and if negative, to bo con
sidered as positive ; and if one has to be carried from the decimals,
it is always negative : so that the index of the logarithm of the
quotient is equal to the sum of the index of the dividend, the index
hv Google
THE APPLICATION OP LOGARITHMS. 661
of the divisor witli its sign changed, and —1 when 1 is to be
carried from the decimal part of the logarithms.
1. Divide 47682 hj 36954, by logarithms.
Nbs. Logs.
47682 36783545
36954 15676615
Qaot. 129032 .21106930
2. Divide 21754 by 24678, hy logarithms.
Nos. Logs.
21754 13375391
24078 . 03923100
Quot. 881514 09452291
3. Divide 46257 by 17608, by logarithms.
J\^. Logs.
46257 06651775
■17608 12457100
Quot. 2627045 14194675
Here the —1 in the lower index, is changed into +1, which ia
then taken for the index of the result.
4. Divide 27684 by 51576, by logarithms.
Nos. _ Logs.
■27684 14422288
51576 07124477
Quot. 0536761 27297811
Here the 1 that is to be carried frtfm the decimals, is taken as
—1, and then added to —1 in the'' upper index, which gives —'I
for the index of the result.
5. Divide 69875 by 075789, by logarithms.
Nos. Logs.
69875 .08443218
075789 . 28796062
Quot. 921967 19647156
Here the 1 that is to he carried from the decimals, is added to
—2, which maltes — 1, and this put down, with its sign ehangeii,
is +1.
6. Divide 19876 hy 0012345, bj logarithms.
Nos. Logs.
•19876 r2983290
0012345 30914911
Quot. 1610043 22068379
Here — 3 in the lower index, is changed into +3, and this ad
ded to 1, the other index, gives + 3 — 1, or 2.
b,Google
338 THE PRACTICAL MODEL CALCULATOE.
PRO PORTION; OR, THE RULE OF THREE, BY LOGARITHMS.
From the sum of the logarithms of the numbers to be multiplied
together, take the sum of the logarithms of the divisors : the re
mainder is the logarithm of the term sought.
Or the same may be performed more conveniently, for any
single proportion, thus — Find the complement ot the logarithm
of the hrst term, or what it ^ants of HI, by hegmnmg at the left
hand and taking each of the figures fiom 9, except the last figure
on the right, which must he taken from 10 , then add this result
and the logaiithma of the other two figures togethei : the sum,
abating 10 m the mdex, will be the kgaiithm of the fourth term.
1. Fmd % fourth proportional to >1 12:,, 14 768, and 135279,
by logarithms
Log of 37 125 1 5b966C5
Complement 8 4303335
Log of 14 7b8 1 16t3217
Log of 135 279 2 1312304
An« 53 812h 1 73U'<856
2. Pind a fourth propoitional to 0j7H 7186, and 34721, by
logarithms
Log. of 05764 27607240
Complement 112392760
Log. of 7186 18564872
Log. of 34721 15405922
Ana. 432868 06363554
3. Find a third proportional to 12796 and 324718, by logarithms,
Log. of 12796 11070742
Complement 88929258
Log. of 32471S 05115064
Log. of 324718 05115064
Ans. 8240216 .T9159386
INVOLUTION; OB, THE RAISING OF POWERS, BY LOGARITHMS.
Multiply the logarithm of the given number by the index of
the proposed power ; then the natural number answering to the
result will be the power required. Observing, if the index be nega
tive, the index of the product will be negative ; but as what is to
be carried from the decimal part will be affirmative, therefore the
difference is the index of the result.
1. Find the square of 2'7568, by logarithms.
Log. of 27568 04404053
2
Square 7599947
hv Google
THE APPLICATION OF LOGARITHMS. 339
2. Find the cube of 70851, by logarithms.
Log. of 70851 08503460
Cube 3556625 25510380
Therefore 3556625 is the answer.
3. Find the fifth power of 87451, by logaritbms.
Log. of 87451 r9417648
5
Fifth power 5114695 1^7088240
Where 5 times the negative index 1, being —5, and +4 to
carry, the index of the power is 1.
4. Find the 365th power of 10045, by logarithms.
Log. of 10045 00019499
365
97495
116994
58497
Power 5148888 Log. 07117135
EVOLUTION; OR, THE EXTRACTION OP ROOTS, BY LOGARITHMS.
Divide the logarithm of the given number by 2 for the square
root, 3 for the cube root, &c., and the natural number answering
to the result will be the root required.
But if it be a compound root, or one that consists both of a root
and a power, multiply the logarithm of the given number by the
numerator of the index, and divide the product by the denomina
tor, for the logarithm of the root sought.
Observing, in either case, when the index of the logarithm is
negative, and cannot be divided without a remainder, to increase
it by such a number as will render it exactly divisible ; and then
carry the units borrowed, as so many tens, to the first figure of the
decimal part, and divide the whole accordingly.
1. Find the square root of 27465, by logarithms.
Log. of 27465 2 } 14387796
Root 52407 7193898
2. Find the cube root of 356415, by logarithms.
Log. of 356415 S ) 15519560
Root 329093 5173186
3. Find the fifth root of 70825, by logarithms.
Log. of 70825 5 ) 08501866
Root 1479235 1700373
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340 THE PRACTICAL MODEL CALCULATOR.
4. Eiad the 365th root of 1045, by logarithma.
Log. of 1045 365) 00191163
Koot 1000121 00000524
6. Find the value of (001234)* by logarithms.
Log. of 001234 30913152
3) 61826304
Ans. 00115047 .20608768
Here the divisor 3 being coataiaed exactly twice in the negative
index —6, the index of the quotient, to be put down, will he —2.
6. Find the value of (024554)^, by logarithms.
Log of 024554 .13901223
2) 61703669
Ans. 00384754 .'35851834
Here, 2 not being contained exactly in —5, 1 is added to it,
which gives —3 for the quotient ; and the 1 that is borrowed being
carried to the next figure makes 11, which, divided by 2, gives
5851834 for the decimal part of the logarithm.
METHOD or CALCULATING THE LOGARITHM OF ANY GIVEN NUMBER,
AND THE NUMBER CORRESPONDING TO ANY GIVEN WGARITHM. DIS
COVERED BY OUVER BYRNE, THE AUTHOR 01 TUE PRESENT WORK.
The succeeding numbers possess a particular property, which is
worth being remembered.
log. 1371288574238642  01371288574238542
log. 1000000000000000  1000000000000000
log. 2375812087593221 = 2375812087593221
log. 3550260181586691  3550260181686591
log. 4669246832877758 = 4669246832877758
loo. 5760456934135527 = 5760456934135627
log. 6834720776754367 = 6834720776764357
log. 7897489031398144  7897489031398144
log. 8961915998267852  8951915998267839
In these numbers, if the decimal points be changed, it is evident
the logarithms corresponding can also be set down without any cal
culation whatever.
Thus, the log. of 1S71288574238542  21371288574238542;
the log. of 3550260181586591 = 1;550260181586591;
log. 002375812087693221 = 3376812087593221 ;
log. 0008951915998267852  4951916998267852;
b,Google
THE APPUCATION OP L0GAE1THM3. 341
and St) on in similar cases, since the change of the decimal
point in a number can only affect the whole number of its loga
rithm.
These numbers whose logarithms are made up of the same digits
will bo found extremely useful hereafter. We shall next give a
simple method of multiplying any number by any power of 11 , 101,
1001, 10001, 100001, kc.
This multiplication is performed by the aid of coefficients of a
binomial raised to the proposed power.
ix + yV ^x + ^, the coefficients are 1, 1.
X h ^f =x^ + 2xy + y^, the coefficients are 1, 2, 1.
xiryY = 7? \ Z3?y + Zxy^ + f, the coefficients are 1, 3, 3 1.
The coefficients of fa; + ?/)''are 1, 4, 6, 4, 1,
— ~ {x\yf~ 1, 5, 10, 10, 5, 1.
— — {x + yf~ 1, 6, 15, 20, 15, 6, 1.
— — h\yY— 1, 7, 21, 35, 35, 21, 7, 1.
— — (a; + 3')=— 1,8, 28, 56, 70, 56, 28, 8,1.
— — {x+ yf— 1, 9, 36, 84, 126, 126, 84, 36, 9, 1.
Let it be required to multiply 54247 by (101)'.
The number must be divided into periods of two figures when the
multiplier is 101 ; into periods of three figures when the multiplier
is 1001 ; into periods of four figures when the multiplier is 10001 ;
and so on.
e ct
b
"1
54
24
70
0000
1
3
25
48
2000
6
8
13
70150
I 15
10
84'94
c 20
1 814
d 15
1 3
e 6
true to 10 pi
(542«) X (101)'
= m
58
42
83 61
This operation is readily understood, since the multipliers for the
6th power are 1, 6, 15, 20, 15, 6, 1 ; we begin at a, a period in ad
vance, and multiply by 6 ; then we commence at b, two periods in
advance, and multiply by 15 ; at c, three periods in advance, and
multiply by 20 ; at d, four periods in advance {counting from the
right to the left), and multiply by 15 ; the period, e, should be
multiplied by 6, but, as it is blank, we only set down the 3 carried
from multiplying d, or its first figure by 6.
As it is extremely easy to operate with 1, 5, 10, 10, 5, 1, the
multipliers for the 5th power, it may be more convenient first to
multiply the given number by (101)', and then by (101)' ; because,
to multiply any number by 5, we have only to affix a cipher (or
suppose it affixed) and to take the half of the result.
The above example, if worked in the manner just described, will
stand as follows :
hv Google
THE PRACTICAL MODEL CALCULATOR.
4247t
2 712:
(54247) X (101)'
. 570141142
570141
..5..0
)...10..6
! ...1D..C
..5..d
67 58 42 83 61  (54247)' X (lOl)'.
The truth of tliis is readily shown by common multiplication, but
the process is cumbersome. However, for the sake of comparison,
we shall in this instance multiply 54247 by (101) raised to the 6th
power.
101
101
(101).
101
1010
10201 =
101
10201
102010
1030301  (101)'.
101
10S0301
10303010
104060301 .
101
104060401
1040604010
10510100501 .
101
10510100501
105101005010
1061520150601 .
54247
7430641054207
4246080602404
2123040301202
4246080602404
5307600753005
575842836019652447 the required product,
(101)<.
(101)'.
(101).
b,Google
I'HB APPLICATION OF LOGAKITlIJtS.
343
which shows that the former process gives the result true to 10
places of figures, of whiuh we shall add another example.
Multiply 34567812 hy (1001)', so that the result may he true to
12 places of figures.
3456
I 2
78120000
76542496
96790
19
....8,.a
..28..S
3459 5475 9305 the required product.
The remaining multipliers, 70, 56, 28, 8, 1, are not necessary in
obtaining the first 12 figures of the product of 34567812 by 10001
in the 8th power.
As 28 and 56 are large multipliers, the work may stand thus
3456
...b.. 8/
...C..501
Result, = 345954759305 the same as before.
Perhaps this product might be obtained with greater ease by first
multiplying 34567812 by {lOOOlJ^ and the prodact by (10001)^
the operation will stand thus :
345850093631
103755298.
10376.
= 34567812 x (10001)'.
345954759305 = twelve places of the product of
34567812 by (10001)* X (10001)^ = ^34567812) X (10001)^
Although these methods are extremely simple, yet cases will oc
cur, wbea one of them will have the preference.
Oar next object is to determine the logarithms 1*1 ; 101 ; 1001 ;
10001; 100001; &c.
It is we!i known that
log. (1 + ») = M (n  ^n? + in' ~ ^n' + in' — Jn* + &c.)
M being the modulus, = 432944819032618276511289, &c.
It is evident that when n is ^i j^j, jiga, rMay ^cj the calcula
tion becomes very simple.
hv Google
844 THE PEACTICAL MODEL CALCULATOR.
M = 4342944819032518
J M  2171472409516269
1 M  1447648273010889
J M = 1085736204758130
i M  0868588968806504
J M = 0723824136505420
{ M = 0720420788433217
i M  0542868102379065
i M = 0482549424336946
i,M  0434294481903252
&c. &e., are constants employed to determine
tte logarithms of 11, 101, 1001, 100001, ke.
To compute the log. of 1001. In this ease n = jJp.
M
+ jjjg  0004342944819033 positive
 TlOMp " 0000002171472410 negative
■0004340773346623
+ njjjjji  0000000001447648 positive
0004340774794271
iM
 rgjfjffa = 0000000000001086 negative
0004340774793185
jM
■ (1000)'
OQOOOOQOOOOOOOOl positive
0004840774793186  the log. of 1001 ;
true to sixteen places.
It is almost unnecessary to remark, that, instead of adding and
suhtracting alternately, as above, the positive and negative terms
may be summed separately, which will render the operation more
concise.
Posilii^ Terms.
0004342944819038
1447648
+ 0004342945266682
— 000000217473496
Negative Terms.
0000002171472410
1086
0000002171473496
■0004340774793186 = log. 1001.
In a similar manner the succeeding logarithms may be obtained
to almost any degree of accuracy.
b,Google
THE APPLICATIOH OF LOGARITHMS.
345
Log. 11
101
1001
10001
100001
1000001
10000001
100000001
1000000001
10000000001
100000000001
1000000000001
 041392685168225 &c. which ti
 004321373782643
= 000434077479319
 000043427276863
= 000004342923104
 000000434294265
=. 000000043429447
= 000000004342945
: 000000000434296
: 000000000043430
: 000000000004343
. 000000000000434
10000000000001  000000000000043
100000000000001  000000000000004
&c. &c.
Without further formality or paraphernalia, for it is presumed
that such is not necessary, we shall commence operating, as the
method can be acquired with ease, and put in a clearer point of
view by proper examples.
" ' 1 the logarithm of 542470, to seven places of decimals.
e call A
— B
— — C
D
E
F
G
H
I
J
K
L
ke.
5 412 4
32 5
i 8
70
48
13
10
20
71
85
8
 6B
= 3D
5 7 5 8)4 2 8 4
i72 7 5
3
= 02692824
Tale 57601562
Prom 5 7 6 4 5 6 9
= 00013028.
576) ■ ■ • • 310 7
28 8
= 6E
= 00002171
112 7
ll5
_2i'
 00000087
12
12
2G
= 00000009
02608119 Tske
676045693 From
Hence we have log. 542470 = 573437574, which is correct
to seven decimal places.
6B is written to represent 6 times the log. of 101.
The nearest nuKuber to 542470, whose log. is composed of the
same digits as itself, being 576045'6934, &c,, onr object was to
raise 542470 to 57604569 by multiplying 542470 by some power
or powers of 11, 101, 1001, 10001, &c.
hv Google
346 THE PEACXICAL MODEL CALCULiTOE.
It 13 here necessary to remark, that A is not employed, because
the given number multiplied by I'l, would exceed 576045'69 ; for
a like reason C is omitted.
Again, ■when half the figures coincide, the process may be per
formed (as above) by common division ; the part which coincides
becoming the divisor ; thus, in finding 5 E, 576 is divided into 3007,
it goes 5 times, the E showing that there are five figui'es in each
period at this step. For A, there is but one figure in each period ;
for B, there are two figures ; for 0, there are three figures in each
period, and so on.
Let it be required to calculate the logarithm of 2785'9, true to
seven places of decimals.
It will be found more convenient, in this instance, to bring the
given number to 355026018, the log. of which is 355026908.
2[785I90IOjO
5571800
278590
3 317 0!9 3,9 = 2 A = 08278537
16854170
33 7 09
3:3 7
3 5 4I2 8 90 8 
It 85 8
3 5
Take 354 919 801 = 20 = 00086815
Prom 3 5 5 2 6 2
■ ■ • aisoi
2]4 85
TE
. 00003040
3116
2;8 4
= 8P =
= 0000034r
%
 9G
Tako
From
7859 
. 00000039
■10529465
355026018
log. 2'
344496653
At the Observatory at Paris, g = 9*80896 metres, the second
being the unit of time, what is the logarithm of 980896 1
In this example, we shall bring 980896 to 999999, &c.
hv Google
THE APPLICATION OF LOGARITHMS.
98OS[9 60000
9 8l0 89 6i0
990:70419600  IE  0043213;
89163446
856654
832
9996:5705
29989
99995:69804
3 9 9 9 8 3
= 3 D = 0001302818
Tike 9999969793 4E = 0000173717
From 10000000000
30207
From which wo hwe 3F = 0000013029
2 H  0000000087
7 J  0000000003
Take 0083770365
Log. 9
From 10000000000
•9016229635
As before observed, 9 C might have been obtained in the follow
ing manner :
8 9 017 49 6 0;0 = 1 B, as above.
4l9 5 3l5 24!8
99 70
9P)
5 times 9 9 5:6 6 8'4 0117
89826736
597 39
^ 40
4times9996570632  9 C.
A French metre is equal to 32808992 English feet, required
the log. of 32808992.
e\ d
192 00. ..once
i2944... 7 times from ffl
198 88.. .21 — b
.4831... 35 — e
1148. ..35 — d
L7...21 — e
B7.
3517 56 8018 ■
b,Google
348 THE PEACirCAL MODEL CALCULATOR.
The manner in which B 7 ia obtained is worthy of remarli : the
multipliers being 1, 7, 21, 35, 35, 21, 7, 1, when 7 times the first
line (commencing with the period marked a) is obtained, 21 times
the same line (commencing with the period marked 6) is determined
by multiplying the 2d line by 3. If the 2d lino be again multiplied
by 5, we have the 4th line of the multiplier 35; but to multiply
by 5, we have only to take tho half the product produced by mul
tiplying by 7, advancing the result one figure to the right. Hence,
to find the result for 35 is almost as easy as to find the result
for 5.
But the object in this case being to bring the proposed number
to 35502601815, the process must be continued.
c \ b\ a\
1 351 ToOiSOl 8 = B 7, as above.
9 3l6581l!2
36 12 663 2
84 29'6
354 935 3058 = C9
The 2d (or 9) line is produced by beginning at a, but the multi
plication may be performed by subtracting 3517568 from 35175680 ;
the 36 line is produced by beginning at f>, observing to carry from
the preceding figure, making the usual allowance when the number
is followed by 5, 6, 7, 8, or 9. The 36 line may be produced by
multiplying tho 9 line by 4, beginning one period more to the left.
To multiply by 84 is not apparently so convenient, for 84 x 352 =
29568 ; and as only one figure of the period 568 is required, when
the proper allowance is made, the result becomes 296.
But, since 84 is equal to 36 x 2^, we have only to multiply tho
36 line by 2, aud add J of it ; with such management, the work
will stand thus : —
351]756r801l8 = E 7, as before
3165Sll2 = 9 times
1260.3j2 = 36 times
24 3 = 72 times \
42 = 12 times/
84 times
354 035 305 8 = C 9
This amounts to very little more than adding tho above numbers
together.
Many other contractions will suggest themselves, when the mul
pliers are large: thus, to multiply any number 57837 by 9, as
alluded to above, is easily effected, by the following wellknown
process : — Subtract the first figure to the right from 10, the second
from the first, the third from the second, and so on.
C 578370. ..ten times
Thus, 57837 x 9 =! 57837 . ..once
520533. ..nine times
hv Google
THE APPLICdTION OS LOOjIEITHMS. 349
Such simple observations are to be found in eyerj bool; on men
tal arithmetic, and therefore require but little attention here.
The whole work of the previous esampie will stand thus : —
8 018 9:920
944
22 96 62 9
618 89 8
1]4 8
11
31
+ T
351175618 018 = 0302496165 
Sll 6 68 112
1 26 6 8 2
296
■354913530
17098
5 8 = 0039066973 
71
36
= 365006 2964 = 0000868546 = 2 D
1:7 7 5 3
TaheE5 = 3550240471 = 0000217146  5E
From 3550260182
3550) 119 7 1 1
F 6 ll7 7 6  0000021715  5 F
119 6 1
a 5 ll77_5  0000002172  5 G
118 6
H 6 ll7 8  0000000217  5 H
12 l7  0000000009  I 2
II
J 3 ll  0000000001 = J 3
Take 0342672944
From 35502601816
Log. 32808992 = 36159928972
.. log. 32808992 = 05159928972.
The constant sidereal year consists of 36525636516 days ; what
is the log. of this number ?
In this case it is better to bring the constant 35502601816 to
36525636516, instead of bringing the given number to the con
stant, as in the former examples.
b,Google
THE PRACTICAL MODEL CALCULATOE.
35
50;2 6 018 16
710 0520 36
3660260
■0086427476 =
■0034726298 =
32 = 36 2116204112
28 9 7 2 9 6 3 3
1014054
2028
C 8 = 3 6 5 016 9 4 9;8 2 7 =
lis 2 5 3476
36 61
.2U
= 80
Take D6 = 36626206963
Ftom 36625636516
•0002171364 =
■0000043429
•0000004343 .
= 5D
S66262
El
429563
365262 =
IE
Fl 
16 4 3 1 1
18 6 52 6
IP
G7 
27786
25568 
•0000003040 =
»7e
H6 =
2218
2;i 9 1 
•0000000261 
. 6H
10
J7 
27
25 =
■0000000003 =
= 7J
•0123376214
Add 35502601816
Hence, log. 36525636516 = 35625978030
.. log. 36525636516 = 256259T803.
M. Regnault determined with the greatest care the density of
mercury to be 1359593 at the temperature 0°, centigrade. It ia
required to calculate the log, of 1359593, to eight places of decimals.
In thia case it is better to bring the given number to the constant
1371288574. 13 5'969300
1,0 8 7 6 7 4
38 7
C8 13706078'8 .
l6 8 5 2'5
14
■003472030 80
Subtract D6137119328 ■000217136 = 5 D
From 137128867
9,5 2 9  000026058 = S 6
E6 82 2 7
r302
F 9  1 213 4  ■000003909 = F 9
H5.
68
69 .
■ 000000022 H5
■003719765
b,Google
APPLICATION OB LOGARITHMS. 351
Take 003719755
from 137128857
log. 1359593 = 133409102
.. log. 1359593 = 1133409102.
TO DETBBMIHB THE NUMBER CORRESPOKDISG TO A GIVEN LOGARITHM.
This problem has been very mueli neglected — so much so, that
none of our elementary books ever allude to a method of comput
ing the number answering to a given logarithm. When an opera
tion is performed by the use of logarithms, it ia very seldom that
the resulting logarithm can be found in .the table ; we have, there
fore, to find the nearest less logarithm, and the next greater, and
correct them by proportion, so that there may be found an inter
mediate number that will agree with the given logarithm, or nearly
80. But although the proportional parts of the difference abridge
thia process, we can only find a number appertaining to any loga
rithm to seven places of figures when using our best modern tables.
As, however, the tabular logarithms extend only to a degree of
approximation, fixed generally at seven decimal places, all of which,
except those answering to the number 10 and its powers, err, either
in excess or defect, the maximum limit of which is ^ in the last
decimal, and since both errors may conspire, the 7th figure cannot
be depended on as strictly true, unless the proposed logarithm falls
between the Hmits of log. 10000 and log. 22200.
Indubitably we are now speaking of extreme cases, but since it
is not an unfrequcnt occurrence that some calculations require the
most rigid accuracy, and many resulting logarithms may be ex
tended beyond the limits of the table, this subject ought to have
a place in a work like the present. It is not part of the present
design to enter into a strict or formal demonstration of the follow
ing mode of finding the number corresponding to a given logarithm,
as the operation will be fully explained by suitable examples.
What number corresponds to the logarithm 34449555 '(
The next less constant log. to the one proposed is 237581209,
r rather, 337581209, when the characteristic or index is ii
a unit.
First from 344486555
talis 337681209
213 7 5 8 1 2 9 constant
2 3758121=.A1
06915346
•04139269 lA
■02776077
•02592824 6E
2 6.1 3 3 9 3 3
l6 6 8 0'3 6
3 9 20
6!2 2
3
2 7 714 1 69 6
ijl 96 6
lie 6
9
7
9
5 = B0
8
4
1
8 C4
. .183253
17S631  4 C
.... 9622
8686 = 2 D
937
2 7 8 62 8 2 'J
b,Google
THE PRACTICAL MODEL CALCULATOR.
. . . 937
22  5 a
S7H
2786282 918 C4
65706
Isi
27 8 6 8400l7'D2
5!5 7 2 = E 2
2l79 El
l'3 9 05
lis = 117
278690016
.. 278690016 is the number sought.
What number correapouds to the logarithm 673437674 i
When the index of this log. is reduced by a unit, the
next less constant is 466924683.
From 473437674
Take 466924683
•6512891
4139269 1 A
2160687 6B
••212035
173631 4
. . . 39804
39085 9D
219 There is neither the Cfjuaj of
217 5 F this number, nor a
772 Cr less, obtainable from
2 4 H ^^ •'■ E9, or B, is
omitted.
Then, 4166924683
46692468 A]
6113 6117
25680
613
5
1611
868
617
136
26
5 3 918 1617 8 8 B5
2l 5 92 6 7
32 3 9
2
64197 9 2 9I6 4
48 7 7 81
19!5
64246721712 D9
12712 F6
I2I2 H4
542470006
. 542470006 is the number whoso logarithm ia 673437574.
b,Google
THE APPLICATION OF lOQARITHMS.
353
Had the given logarithm represented a decimal with a positive
index, the required number would bo 0'000054247, &c. ; or if
written with a negative index, as 573437574, the result would be
the same, for the characteristic 5, shows how many places the first
significant figure is below unity.
Required the number corresponding to log. 23727451.
The constant 100000000 is the one to be employed in this case.
13727451 the given log. minus 1 in the index.
10000000
■3727461
3726342..
. . . 2109
1737 . 4D
....372
347 8K
.26
22..
..5F
1:010
90
7
Constant.
86 AS
I9 4B2
B 81911
189
ll
1
D4
E8
E5
6 G7
23690949
.. 23590949 is the required number, and the seconds in the di
urnal apparent motion of the stars.
23590949" = 3' 5590949".
Let it be required to find the hy^jerhoUc logarithm of any
number, as 31415926536. The common log. of this number is
49714987269 (33), and the common log. of this log. isT6964873.
The modulus of the common system of logarithms is 4342944819,
kc.
.. 1 : 4342944819 : : hyperbolic log. N : common log. N.
b,Google
354
THE PRACTICAL MODEL CALCULATOIl.
To distingmsh the hyperbolic logarithm of the number N from
its common logarithm, it is necessary to tvrite the hyp. log. Log. Jy,
and the common logarithm log. N.
Hence, 4342941819 X Log. N = log. N ;
or log. (4342944819) + log. (log. N)  log. (log. N).
.. log. (Log. N) = log. (log. N)  1637784.3 ; for 16377843 
log. 4342944819.
Now, to work the above example, from 16964873
take F6377843
"0587030, the number
corresponding to this com. log. will be the hyp. log. of 31415927.
0587030 must be reduced to 0000000 which is known to be the
log. of 1.
■0587030
0413927 lA
. 173103
172855 4B
248
217 5E
1 A  1 1!0 OiO 0:0 0:0
440 0,0
6 6'0
14 40
.31
30 71?
..1 2G
. 114472988 i
1144(1,64411 = B4
!6,7 2'3 = E 5
18 01 =F7
!2i3  G2
114472988
the hyperbolic log. of 31415927, true to the
last figure ; for the hyp. log. 31415926535898 = 11447298858494.
The reason of this operation is very clear, because
1 X 11 X (101)' X (100001)' X (1000001)' X (10000001)' 
114472988.
This example answers the purpose of illustration, but the hyp.
log. of S1415927 can be more readily found by dividing its com.
log. 49714987269 by the constant 4342944819, which is termed
the modulus of the common system of logarithms.
Suppose it is known that 13426139 is the log. of the decimal
which a French litre is of an English gallon. Reiiuiied the decimal.
The index, 1, may be changed to any other characteristic, so as
to suit any of thi ecmstants, as the alteration is easily allowed for
when the work is completed. In this instance, it is best to put
■f 1 instead of 1.
From 1
Take 10000000
3426139
3311415  8 A
•0114724
. .86427  2 B
28297
2604560
2262
1'0'0'0 000;0^0 Constant
l8'0 0lo0 0'0
280 0:0 0'0'0
66
700
0,0:0
000
00
800
8:0
1
21:4315 8:8 8,1 = A8
b,Google
THE APPLICATION OF LOGAKITHMS.
22o2
2171 
^81
5D
IE
8P
7(J
2 1)4 35 S
4 2l8 7
21
881 = A8
178
436
43 
38
35 =
"3
3 =
218,667495B2
l'3120 5
' 3280
4
2198,82784 = 06
10 9 9 91
22
2 2 92;7
7
97D 5
01 =E1
61 F8
54 = 67
220096913
.. The French litre = 2200969 English gaUons.
In measuring heights hy the barometer, it is necessary to know
the ratio of the density of the mercury to that of the air.
At Paris, a litre of air at 0° centigrade, under a pressure of 760
millimetres, weighs 1293187 grammes. At the level of the sea,
in latitude 45°, it weighs 1292697 grammes. A litre of water,
at its maximum density, weighs 1000 grammes, and a litre of mer
cury, at the temperature of 0° cent., weighs 1359593 grammes;
1359593
■"■ 1292697 ~ ratio at 46"
Now, log. 1369593  4133409102 (29)
and log. 1292697  0111496744 (30)
4021912358 = the log. of the ratio at 45°.
To find thenumber corresponding to thislog., it is necessary to reject
the index for the present, and reduce the decimal part to zero. By
this means the necessity of using any of the constants is superseded.
•021912358
•021606869  5 B
, 305489
303991 = 7D
. . . 1498
1303 = 3 F
lOIOOjOO
'sOiOO
10
OOjOO
00
00
000
10511010
7 8 57
.195
174 =
 91
1359593
1292697
veritied hy common division.
10517 415 98 D7
3 1 6 = F 3
42 =G4
4 = 114
119
. hy logarithms.
105174 9 61
= 1051749, &c., which is easily
b,Google
366 THE PRACTICAL MODEL CALCULATOK.
M. Regnault found that, at Paris, the litre of atmospheric air
weighs 1'293187 grammes ; the litre of nitrogen. 1'256167 grammes;
& litre of oxygen, 1'429802 grammes; of hydrogen, 0089578
grammes ; and of carhonic acid, 1'977414 grammes. But, strictly
considered, these numbers are only correct for the locality in which
the experiments were made ; that is for the latitude of 48° 50' 14"
and a height about 60 metres above the level of the sea ; M. Eeg
nault finds the weight of the litre of air under tho parallel of 45°
latitude, and at the same distance from the centre of the earth as
that which the experiments were tried, to be 12'926697.
Assuming this as the standard, ho deduces for any other latitude,
any other distance from the centre of the earth, the formula,
1292697 (100001885) (1  0002837) cos. 2^
Hero, te is the weight of the litre of air, R the mean radius of
the earth = 6366198 metres, A the height of the place of observa
tion above the mean radius, and x tho latitude of the place.
At Philadelphia, lat. 39° 56' 515", suppose the radius of the
earth to be 6367653 metres, the weight of the litre of air will he
1'2914892 grammes. The ratio of the density of mercury to that
of air at the level of the sea at Philadelphia is 10527735 to 1;
required the number of degrees in an arc whose length is equal to
that of the radius.
360
As 31415926535898 : 1 : ; g : the required degrees.
Log. 360 = 2556302500767
log. 314159265859 = 0497149872694
2059452623073
log. 2 = 0301029995664
1758122632409 = the log. of the
number required.
When the index of this log. is changed into 4, the nearest next
less constant is 4669246832878.
= Constant
From 4758122e32409
Take 4669246832878
■088875799531
46;6 92I4 6 813
9!3:38493 6
46 69 24!6
2878
6676
8329
7;7 83
6;6 7;8
2 A  ■82786370816
. . 6090429216
6 614 9,7 88 6,6
6e497 8!8
IB 4321373783
..1769055432
4 0 1736309917
....32745615
7 E = 30400462
6701628,655
2128 2514
3J4 2 3
2
1461
6218
7719
2825
6
2345053
5729114 696
12 2 9
b,Google
.298
261
THE APPLICATION OP LOGARITHMS, 357
...2345053 5729i;459(il;229  04
2171471 401040J217
...! 173582 ISjQSl
130288 5729547013477 E7
....^43294 218647735
"" 39087 51
'^^ 5729576!66il2l69F5
3509 l718873e3
5l 5 6 6 2 = H 9
S'l 566 = 19
3 438 J 6*
37 45S = K8
8K 35 ^L6
T2 5729577951295 tie num
5 L = 2 ber required.
But the original index is 1; .. 5729577951295» are tlie num
ber of degrees in an arc the length of which is equal to that of the
radius.
The above result may be easily verified by common division, a
method, no doubt, which would be preferred by many, for loga
rithms are seldom used when the ordinary rules of arithmetic can
be applied with any reasonable facility. However, this example,
like many others, is introduced to show with what ease and correct
ness the number corresponding to a given log. can be obtained.
The extent, also, hj far exceeds that obtainable by any tables
extant.
Other computations give,
i"  672957795130°  57° 17' 44" 80624
the degrees in an arc = radius.
/ =. 34377467707849' = 3437' 44" 80624
the mmutes in an are = radius.
/' = 2062648062470963
the number of seconds in an arc = radius.
The relative mean motion of the moon from the sua in a Julian or
fictitious year, of 365J days, is 12 eir. 4 signs, 12°40' 15977315' =
16029615977315".
.. 16029615977315" : 1 circumference { 129600")
: : 36525 days
; 295305889216 days = the mean synodic month.
This proportion may, for the sake of example, be found by loga
rithms.
Log. 36525 256259022460634
log. 1296000 6112605001534 57
867619522614091
log. 16029615977315 = 720492311805406
147027210808685
b,Google
358 THG PRACTICAL MODEL CALCULATOR.
If the index of this log. be made 2 instead of 1, the neaiest nest
less constant will be 2376812087693221.
Prom 247027210808685
Take 237681208769S22
2A =
09446002049363
08278537031645
2B =
.1167466017718
864274758529
6C =
. . 303190261189
260446487591
9D
...42743778598
39084549177
8E =
....3669224421
3474338483
4P =
184886938
173717706
2G =
11168232
8686889
5H =
2482348
2171478
71 =
310870
304006
IJ =
4343
5K =
2520
2172
348
8L =
2N
317
1
213 7,61811:2 0187:5 9 3i22 Const.
4 7 6 1 6:24 1 716:1 8 64
2375(8il2o;8i75;93
.2A
2 817417 3 2 6(2 6 9 817 79
J67494 6 6 25ll9 7 6
2 874i7 3;26!2 6;0
293251475]177
015
1759508861
{)6 2
43981772
128
5i8 6 5
2!)
4
399
296015388669
26 55
1016205
3506
80S
640
296281008749763
23622480700
826787
17
2953041632057
ll81218
1 7 7 2
2963 58J1327756J7 =F4
2953058712138:8,73
114 7 6'6:2 9
20 61711 4
29163
ll47 6
236
302
H6
1 = 17
1Jl
6K6
2L8
6N2
295305889217832
.. 296305889218 is the number required.
To perform by logarithms the ordinary operations of multipli
cation, divisim proportion oi even the e\tiiction of the square
root, except in the ^ ay of illustiatio is not the design of these
pages ; for such an appl cat on ot logarithms m a particular man
ner only dimin »h the labour of the operator It is not necessary,
however to examine mi utely hcie the instauces in which common
arithmet c is piefeiable to a till al numbe s besides, much will
depend on tl e sL li and fac 1 ty oi the p atoi.
b,Google
TRIQONOMETEY.
SPHERICAL TRIGONOJIETEY.
ANGULAR MAGNITUDES.
Plane trigonometry treats of the relations and calculations of
the sidesand angles of plane triangles.
The circumference of every circle is supposed to be divided into
360 equal parts, called degrees ; also each degree into 60 minutes,
each minute into 60 seconds, and so on.
Hence a semicircle contains 180 degrees, and a quadrant 90 de
grees.
The measure of any angle is an arc of any circle contained be
tween the two lines which form that angle, the angular point being
the centre ; and it is estimated by the number of degrees contained
in that arc.
Hence, a right angle being measured by a quadrant, or quarter
of the circle, is an angle of 90 degrees; and the sum of the three
angles of every triangle, or two right angles, is equal to 180 de
grees. Therefore, in a rightangled triangle, taking one of the
acute angles from 90 degrees, leaves the other acute angle ; and
the sum of two angles, in any triangle, taken from 180 degrees,
leaves the third angle ; or one angle being taken from 180 degrees,
leaves the sum of the other two angles.
Degrees are marked at the top of the figure with a small ", mi
nutes with ', seconds with ", and so on. Thus, 57° 30' 12" de
note 57 degrees 30 minutes and 12 seconds.
The complement of an arc, is what it wants of
a quadrant or 90°, Thus, if AD he a quadrant,
then BD is the complement of the arc AB ; and,
reciprocally, AB is the complement of BD. So ^
that, if AB he an arc of 50°, then its complement
BD will be 40°.
The supplement of an arc, is what it wants o
a semicircle, or 180°. Thus, if ADE be a semicircle, then BDE
is the supplement of the arc AB ; and, reciprocally, AB is the sup
plement of the arc BDE. So that, if AB he an arc of 50°, then
its supplement BDE will be 130°.
The sine, or right sine, of an arc, is the line drawn from one
extremity of the arc, perpendicular to the diameter passing tiirough
the other extremity. Thus, BF is the sine of the arc AB, or of
the arc BDE.
Hence the sine (BF) is half the chord (BG) of the double arc
(BAG).
The versed sine of an arc, is the part of the diameter intercepted
between the arc and its sine. So, AF is the versed sine of the arc
AB, and EF the versed sine of the arc EDB.
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3{!0
THE PRACTICAL MODEL CALCULATOE.
The tangent of an are is a line touching the circle in one ex
tremity of that arc, continued from thence to meet a Ime Jidivu
from the centre through the other extremity: fthich last line is
called the secant of the same arc. Thus, AH is the tangent, anJ
CII the secant, of the arc AB. Also, EI is the tangent, ami CI
the secant, of the euppleiaental are BDE. And this latter tangent
and secant are equal to the former, but are accounted negative, as
being drawn in an opposite or contrary direction to the former.
The cosine, cotangent, and cosecant, of an arc, are the sine,
tangent, and secant of the complement of that arc, the co being
only a contraction of the word complement. Thus, the arcs AB,
ED being the complements of each other, the sine, tangent or se
cant of the one of these, is the cosine, cotangent or cosecant of the
other. So, BF, the sine of AE, is the cosine of ED ; and BK,
the sine of ED, is the cosine of AB : in iike manner, AH, the
tangent of AB, is the cotangent of ED ; and DL, the tangent of
DB, is the cotangent of AB : also, CH, the secant of AB, is the
cosecant of BD ; and CL, the secant of ED, is the cosecant of AB.
Hence several remarkable properties easily follow from these
definitions ; as.
That an arc and its supplement have the same sine, tangent, and
secant ; but the two latter, the tangent and secant, are accounted
negative when the are is greater than a quadrant or 90 degrees.
When the arc is 0, or nothing, the sine and tangent are nothing,
but tho secant is then the radius CA. But when the arc is a
quadrant AD, then the sine is the greatest it can be, being the ra
dius CD of the circle ; and both the tangent and secant are infinite.
Of any arc AE, the versed sine AF,
and cosine BK, or CF, together make
up the radius CA of the circle. The
radius CA, tangent AH, and secant
CH, form a rightangled triangle CAH.
So also do the radius, sine, and cosine,
form another rightangled triangle
CBF or CEK. As also the radius,
cotangent, and cosecant, another right
angled triangle CDL. And all these
rightangled triangles are similar to
each other.
The sine, tangent, or secant of an
angle, is the sine, tangent, or secant
of the arc by which the angle is mea .
sured, or of the degrees, &c. in the same ^
arc or angle.
The method of constructing the scales
of chords, sines, tangents, and secants,
usually engraven on instruments, for
practice, is exhibited in the annexed
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TEIOONOMETRT. 301
A trigonometrical canon, is a table exhibiting the length of the
sine, tangent, and secant, to every degree and minute of the quad
rant, with respect to the radius, which is expressed by unity, or 1,
and conceived to he divided into 10000000 or more decimal parts.
And further, the logarithma of these sines, tangents, and secants
are also ranged in the tables ; which are most commonly used, as
they perform the calculations by only addition and subtraction,
instead of tho multiplication and division by the natural sines, kc,
according to the nature of logarithms.
Upon this table depends the numeral solution of the several
cases in trigonometry. It will therefore be proper to begin with
tho mode of constructing it, which may he done in the following
To find the sine and cosine of a given are.
This problem is resolved after various ways. One of these is as
follows, viz. by means of the ratio between the diameter and cir
cumference of a circle, together with the known series for the sine
and cosine, hereafter demonstrated. Thus, the semicircumference
of the circle, whose radius is 1, being 31415926535897S3, kc,
the proportion will therefore be,
As the number of degrees or minutes in the semicircle,
Is to the degrees or minutes in the proposed arc.
So is 314159265, &c., to the length of the said arc.
This length of the arc being denoted by the letter a; also its
sine and cosine by g ami c; then will these two be expressed by the
two following series, viz. :—
^ = '^ 2l'^2:SAl ~ 2.3.4.5.6.7 + ^^•
= "^ ~ "6 + 120 ~ 5040 + ^'^•
■ 2.3.4 2.3.4.5.6 '
+ ^TM + ^
24 720
If it be required to find the sine and cosine of one minute.
Then, the number of minutes in 180° being 10800, it will be first,
as 10800 : 1 : : 314159265, &c. : 000290888208665 = the length
of an arc of one minute. Therefore, in this case,
a = 0002908882
■000000000004, &c.
■0002908882 the sine of 1 minute.
and ^a^ =
the dilFerence is s =
Also, from
take Jd^ =
leaves c =
■0000000423079, &c.
9999999577 the cosim
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3G2 THE PRACTICAL MODEL CALCL^LATOR.
For the sine and cosine of 5 degrees.
Here, as 180° : 5° : : 314159265, &c., : 08720
length, of 5 degrees.
Hence, a = 08726646
_ ifflS =  00011076
+ Jjffl* = 00000004
these collected give s = 08715074 the si
And, for the cosine, 1=1
 la^ = _ 00380771
+ la' = 00000241
these collected, give c = 99619470 the consine of 5°.
After the same manner, the sine and cosine of any other arc
may he computed. But the greater the arc is, the slower the series
will converge, in ivhich case a greater number of terms must be
taken to bring out the conclusion to the same degree of exactness.
Or, having found the sine, the cosine will be found from it, by
the property of the rightangled triangle CEF, viz. the cosine
CF = ^/CB'  EF^ or e = v'T^^=.
There are also other methods of constructing the canon of sines
and cosines, which, for brevity's sake, are here omitted.
To conijiute the tangents and secants.
The sines and cosines being known, or found, by the foregoing
problem ; the tangents and secants will be easily found, from the
principle of similar triangles, in the following manner : —
In the first figure, where, of the are AB, BF is the sine, CF or
BK the cosine, AH the tangent, CH the secant, DL the cotangent,
and CL the cosecant, the radius being CA, or CB, or Cl>; the
three similar triangles CFB, CAH, CJDL, give the following pro
portions :
1. CF : FB ; : CA : AH ; Trhence the tangent is known, being
a, fourth proportional to the cosine, sine, and radius.
2. CF ; CB : : CA : CH; whence the secant is known, being a
third proportional to the cosine and radius.
3. BF : FO ; : CD : DL ; whence the cotangent is known, being
a fourth proportional to the sine, cosine, and radius.
4. BF : EC : : CD : CL ; whence the cosecant is known, being
a third proportional to the sine and radius.
Having given an idea of the calculations of sines, tangents, and
secants, we may now proceed to resolve the several cases of trigo
nometry; previous to which, however, it may be proper to add a
few preparatory notes and observations, as below.
There are usually three methods of resolving triangles, or the
cases of trigonometry — namely, geometrical construction, arith
metical computation, and instrumental operation.
In tlie first method. — The triangle is constructed by making the
parts of the given magnitudes, namely, the sides from a scale of
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laiGOKOMETET. 6b6
equal parts, and the angles from a scale of cliords, or by some
other instrument. Then, measuring the unknown parts by the
same scales or instruments, the solution will be obtained near
the truth.
In the second method. — Having stated the terms of the propor
tion according to the proper rule or theorem, resolve it like any
other proportion, in which a fourth term is to he found from three
given terms, by muItiplyiDg the second and third together, and
dividing the product by the first, in working with the natural num
bers; or, in working with the logarithms, add the logs, of the
second and third terms together, and from the sum take the log.
of the first term ; then the natural number answering to the re
mainder is the fourth term sought.
In the third method. — Or ins t rumen tally, aa suppose by the log.
lines on one side of the common twofoot scales ; extend the com
passes from the first terra to the second or third, which happens to
be of the same kind with it ; then that extent will reach from the
other term to the fourth term, as required, taking both extents
towards the same end of the scale.
In every triangle, or case in trigonometry, there must be given
three parts, to find the other three. And, of the three parts that
are given, one of them at least must be a side ; because the same
angles are common to an infinite number of triangles.
All the cases in trigonometry may be comprised in three vari
eties only ; viz.
1. When a side and its opposite angle are given.
2. When two sides and the contained angle are given.
3. When the three aides are given.
For there cannot possibly be more than these three varieties of
cases ; for each of which it will therefore be proper to give a sepa
rate theorem, as follows :
When a side and its opposite angle are two of the given parts.
Then the sides of the triangle have the same proportion to each
other, aa the sines of their opposite angles have.
That is.
As any one side,
Is to the sine of its opposite angle ;
So is any other aide,
To the sine of its opposite angle. ^
¥or, let ABO he the proposed triangle, hai
AB the greatest aide, and BO the least. Take
AD = EC, considering it as a radius ; and let ^
fall the perpendiculars DE, CF, which will evi a ^ • □
dently be the sines of the angles A and B, to the radius AD or
BO. But the triangles ADE, ACF, are equiangular, and there
fore AO : CF ; : AD or BC : DE ; that is, AC is to the sine of its
opposite angle B, as BC to the sine of its opposite angle A.
In practice, to find an angle, begin the proportion with a side
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364 THE PILAGirCAL MODEL CALCULATOR.
opposite a given angle. And to find a side, begin ivitli an angle
opposite a given side.
An angle found by this rule is ambiguous, or uncertain whether
it be acute or obtuse, unless it be a right angle, or unless its mag
nitude be such as to prevent the ambiguity ; because the sine an
swers to two angles, which are supplements to each other ; and
accordingly the geometrical construction forms two triangles with
the same parts that are given, as in the example below ; and when
there is no restriction or limitation included in the question, either
of them may be taken. The degrees in the table, answering to the
sine, are the acute angle ; but if the angle be obtuse, subtract those
degrees from 180°, and the remainder will be the obtuse angle.
When a given angle is obtuse, or a right one, there can be no am
biguity ; for then neither of the other angles can be obtuse, and
the geometrical construction will form only one triangle.
In the plane triangle ABC,
rAB 345 yards
Given, ^BC 282 yards
I angle A 37° 20'
Required the other parts.
Geometrically . — Draw an indefinite line, upon which set off AB
= 345, from some convenient scale of equal parts. Make the
angle A = 37^°. With a radius of 232, taken from the same
scale of equal parts, and centre B, cross AC in the two points C, C.
Lastly, join BC, BC, and the figure is constructed, which gives
two triangles, showing that the case is ambiguous.
Then, the sides AC measured by the scale of equal parts, and
the angles B and C measured by the line of chords, or other in
strument, will be found to be nearly as below; viz.
AC 174 angle B 27° angle C 115°
or 374^ or 78i or 64i
Arithmetically. — First, to find the angles at :
As side BO 232 log. 23654880
To sin. opp. angle A 37° 20' 97827958
So side AB345 25378191
To sin. opp. angle C 115° 36' or 64° 24 99551269
Add angle A 37 20 37 20
The sum 152 56 or 101 44
Taken from 180 _00 _1?0_00
Leaves angle B 2T 04 or 78 16
Then, to find the side AC :
As sine angle A 37° 20' log. 97827958
To opposite side BC 232 2.365488
^ . , r, /27°04' 96580371
So sine angle E  .^g ^^ 99908291
To opposite side AC 17407 22407293
or, 37456 25735213
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TRIGONOMETRY.
In the plane triangle ABO,
r AB 365 poles
Given, < angle A 57° 12'
(. angle B 24 45
Required the other parts.
In the plane triangle ABC,
( AC 120 foot
Given, ^ BC 112 feet
Ungle A 67° 27'
Required the other parts.
Ans. r angle C 98° 3'
{ AG 15433
I BC 30986
igle B 64=" 34' 21"
or, 115 25 39
iglc C 57 58 39
or, 7 7 21
AB 11265 feet
or, 1647 feet
When two aides
Then it will be,
As the sum of those two sides.
their contained angle are given.
Is to the difference of the s
So ia the tang, of half the sum of their opposite angles,
To the tang, of half the difference of the same angles.
Hence, because it is known that the half sum of any two quan
tities increased by their half difference, gives the greater, and di
minished by it gives the less, if the half difference of the angles,
so found, be added to their half sum, it will give the greater angle,
and subtracting it will leave the less angle.
Then, all the angles being now known, the unknown side will be
found bj the former theorem.
Let ABG be the proposed triangle, having r
the two given sides AG, BC, including the given ,
angle C. With the centre 0, and radius CA, 
the less of these two sides, describe a semicircle,
meeting the other side BC produced in D and E,
Join AE, AU, and draw DF parallel to AE.
Then, BE is the sum, and BD the difference c
sides CB, GA, Also, the sum of the two angl . . _
equal to the sum of tho two CAD, CBA, these suras being each
the supplement of the vertical angle C to two right angles ; but
the two latter CAD, CDA, are equal to each other, being opposite to
the two equal sides CA, CI): hence, either of them, as CD A, is equal
to half the sum of the two unknown angles CAB, CBA. Again,
the exterior angle CDA is equal to the two interior angles B and
DAB ; therefore, the angle DAB is equal to the difference between
CDA and B, or between CAD and B; consequently, the same
angle DAB is equal to half the difference of the unknown angles
B and CAB ; of which it has been shown that CDA is the half sum.
Now the angle DAE, in a semicircle, is a right angle, or AE is
perpendicular to AD ; and DF, parallel to AU, is also perpendicular
2r2
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Sbb THE PRACTICAL MODEL CALCULATOR.
to AD : consequently, AE is the tangent of CDA tte half aum
and DF the tangent of DAB the half difference of the angles, to
the same radius AD, by the definition of a tangent. But, the tan
gents AE, DF, being parallel, it will he as BE : BD : : AE : DF ;
that ia, as the sum of the sides is to the difference of the sides, so
is the tangent of half the sum of the opposito angles, to the tan
gent of half their difference.
The sum of the unknown angles is found, by taking tho given
angle from 180°.
In the plane triangle ABO,
f AE 345 yards 5.
Given, J AC 174OT yards y^ \,
(angle A 37° 20' ^ ' i,
Required the other parts.
Qeometrically. — Draw AE = 345 from a scale of equal parts.
Make the angle A = 37° 20'. Set off AC = 174 by the scale of
equal parts. Join BC, and it is done.
Then the other parts being measured, they are found to he nearly
as follows, viz. the side EC 232 yaada, the angle B 27°, and the
angle C 115^°.
Arithmetically.
As sum of sides AE, AC 51907 log. 27152259
To difference of sides AB, AC 17093 22328183
So tangent half sum angles C and E 71° 20' 104712970
To tangent half difference angles C and E 44 16 99888903
Their sum gives angle C 115 36
Their diff. gives angle B 27 4
Then, by the former theorem.
As sine angle 115° 36', or 64° 24' log. 90551259
To its opposite side AB 345 25378191
So sine angle A 37° 20' 97827958
To its opposite side EC 232 23654890
In the plane triangle ABC,
r AB 365 poles
Given, ^ AC 15433
[angle A 57° 12' ( BC 30986
Required the other parts. < angle E 24° 45'
(angle C 98° 3'
In the plane triangle ABC,
f AC 120 yards
Given,^ EC 112 yards
(angle C57°58'39" ( AE 11265
Required tho other parts. < angle A 57° 27' 0"
igle B 64 34 21
f ^
\ angle
(angle
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TRIGONOMETRY. oOT
When the three sides of the triangle are given.
Then, having let fall a perpendicular from the greatest angle
upon the opposite side, or base, dividing it into two segments, and
the whole triangle into two rightangled triangles ; it will be,
As the base, or sum of the segments,
la to the sura of the other two sides ;
So is the difference of those sides,
To the difference of the segments of the base.
Then, half the difference of the segments being added to the
half sum, or the half base, gives the greater segment ; and the
same subtracted gives the less segment.
Hence, in each of the two rightangled triangles, there will be
known two sides, and the angle opposite to one of them ; conse
quently, the other angles will be found by the first problem.
The rectangle under the sum and difference of the two sides, is
equal to the rectangle under the sum and difference of the two seg
ments. Therefore, by forming the sides of these recta.ngles into
a proportion, it will appear that the sums and differences are pro
portional, as in this theorem.
In the plane triangle AEO,
fAB 345 yards
Given, the sides ^ AC 232
(_BG 1T407
To find the angles.
Geometrieally. — Draw the base AB = 345 by .a scale of equal
parts. With radius 232, and centre A, describe an arc ; and with
radius 174, and centre B, describe another arc, cutting the former
in C. Join AC, BC, and it is done.
Then, by measuring the angles, they will be found to be nearly
as follows, via. angle A 27°, angle B 37J°, aad angle C 115J°.
Arithmetically. — ^Ilaving let fall the perpendicular OP, it will be.
As the base AB : AG + BC : : AC  BC : AP  BP
that is, as 345 : 40607 : : 5793 : 6818 = AP  BP
its half is 3409
the half base is 17250
the sum of these is 20659 = AP
and their difference 13841 = BP
Then, in the triangle APC, rightangled at P,
As the side AC 232 log. 23054880
To sine opposite angle 90° 100000000
So is side AP 20659 23151093
To sine opposite angle AGP 62° 56' 99496213
Which taken from 90 00
Leaves the angle A 27 04
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368 THE TEACTICAL MODEL CALCULATOR.
Again, in the triangle BPC, riglitarc'led at P,
As the side of BC 17407 log. 2240T239
To sine opposite angle P.. . 90° 100000000
So is side BP 13841 21411675
To sin. opposite angle BCP 52° 40' 99004436
Which taken from 90 00
Leaves the angle B... 37 20
Also, the angle ACP... 62° 56
Added to angle BCP... 52 40
Gives the whole angle ACB...115 36
So that all the three angles are as follow, viz.
the angle A 27° 4'; the angle E 3T° 20'; the angle C 115° 36'.
In the plane triangle ABC,
fAB 365 poles
GiveD the sides, ^ AC 15433
(EC 30986 (angle A 57° 12'
To find the angles. ■{ angle B 24 45
U»gleC98 3
In the plane triangle ABC,
(AB 120
Given the sides J AC 11265
(BC 112 (angle A 57° 27' 00"
To find the angles. { angle B 57 58 39
(_ angle C 64 34 21
The three foregoing theorems include all the cases of plane tri
angles, both rightangled and obiiqae ; besides which, there are
other theorems suited to some particular forms of triangles, ivhich
are sometimes more expeditious in their use than the general ones ;
one of which, as the case for which it serves so frequently occurs,
may be here taken, as follows : —
When, in a rightangled triangle, there are given one leg and the
angles ; to find the other leg or the hy^othenuse; it will be.
As radius, i. e. sine of 90° or tangent of 45°
Is to the given leg.
So is the tangent of its adjacent angle
To the other leg ;
And so is the secant of the same angle
To the hypothenuse.
AB being the given leg, in the rightangled tri
angle ABC ; with the centre A, and any assumed ra
dius, AD, describe an arc DE, and draw DF perpen
dicular to AB, or parallel to BC. Now it is evident,
from the definitions, that DF is the tangent, and AF
the secant, of the arc DE, or of the angle A which j
is measured by that arc, to the radius AD. Then, because of the
parallels BC, DF, it will be as AD : AB :: DF : BC : : AF : AC,
which is the same as the theorem is in words.
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OE HEIGHTS AND DISTANCES.
In the right ai
Given <
iglcd triangle ABC,
Geometrically. — Make AB =162 equal parts, and the angle A =
53° 7' 48" ; then raise the perpendicular BO, meeting AC in C
So sliall AC measure 270, and BC 216.
Arithmeticalh/ .
..log. lOOOOOOOO
22096150
101249371
23344521
102218477
24313627
As radius tang. 45°
TolegAB 162
So tang, angle A 53° 7' 48"
TolegEC 216
So secant angle A 53° 7' 48"
Tohjp. AC 270
In the rightangled triangle ABC,
c™/ the leg AB 180
''"™ t the angle A 62° 40' , ^„ jg^OMT
To find tiie other two sides. < Tjri qioOdi^A
There 13 sometimes given another method for rightangled tri
angles, which is this : o
ABC being such a triangle, make one leg AB ra
dius, that is, with centre A, and distance AB, de
scribe an arc BF. Then it is evident that the other
leg BC represents the tangent, and the hypother
AC the secant, of the arc EF, or of the angle A.
In tike manner, if the leg BC be made radius ;
then the other leg AB will represent the tangent, and the hypo
thenuse AC the secant, of the arc EG or angle C.
But if the hypothonuse he made radius ; then each leg will re
present the sine of its opposite angle ; namely, the log AB the sine
of the arc AE or angle C, and the leg BC the sine of the arc CD
or angle A.
And then the general rule for all these cases is this, namely,
that the sides of the triangle hear to each other the same propor
tion as the parts which they represent.
And this is called. Making every side radius.
OP HEIGHTS AND DISTANCES.
Bt the mensuration and protraction of lines and angles, are de
termined the lengths, heights, depths, and distances of bodies or
objects.
Accessible lines are measured by applying to them some certain
measure a number of times, as an inch, or foot, or yard. But in
accessible lines must be measured by taking angles, or by some
such method, drawn from the principles of geometry.
When instruments are used for taking the magnitude of the
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370 THE PRACTICAL MODEL CALCULATOR.
angles in degrees, the lines are then calculated bj trigonometry :
in the other methods, the linos are calculated from the princi
ple of similar triangles, without regard to the measure of the
angles.
Angles of elevation, or of depression, are usually taten either
with a theodolite, or with a quadrant, divided into degrees and mi
nutes, and furnished with a plummet suspended from the centre,
and two sides fixed on one of the radii, or else with telescopic
sights.
To take an angle of altitude and depression with the quadrant.
Let A be any object, as the sun, a
moon, or a star, or the top of a tower, ,'
or hill, or other eminence; and let it ^.'■''^
be required to find the measure of the y''
angle ABC, which a line drawn from ^,''
the object makes with the horizontal
line BC.
Fix the centre of the quadrant in
the angular point, and move it round
there aa a centre, tilt with ono eye at
D, the other being shut, jou perceive the object A through the
sights : then will the arc Gil of tho quadrant, cut off by the plumb
line BH, be the measure of the angle ABC, as required.
"<F
The angle ABC of depression of any ob
ject A, is taken in the same manner ; except
that here the eye is applied to the centre, and
tho measure of the angle is the arc GH, on \
the other side of the plumb line. \
\.
The following examples arc to be constructed and calculated by
the foregoing methods, treated of in trigonometry.
Having measured a distance of 200 feet, in a direct horizontal
line, from the bottom of a steeple, the angle of elevation of its top,
taken at that distance, was found to be 47° 30': from hence it is
required to find the height of the steeple.
Construction. — Draw an indefinite line, upon which set off AC =
200 equal parts, for the measured distance. Erect the indefinite
perpendicular AB ; and draw CB so as to make the angle C ™
47° 30', the angle of elevation; and it is done. Then AB, mea
sured on the scale of equal parts, is nearly 218j. n
Calculation. /R
As radius 100000000 / 1 1 ,
To AC 200 23010300 / iAWy
So tang, angle C 47° 30' 100379475 / P.mm
To AB 21826 required 23389775 ^ ' " ' T
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OF HEIGHTS AND DISTANCES. 371
"What was the perpendicular height of a cloud, or of a balloon,
when its angles of elevation were 35° and 64°, as taken by two
observers, at the same tirae, both on the same side of it, and in
the same vertical plane ; tlieir distance, as under, being half a mile,
or 880 yards. And what was its distance from the said two ob
Oonstruction. — Draw an indefinite ground line, upon which set
off the given distance AB = 880; then A and E are the places
of the observers. Make the angle A = 35°, and the angle E =
64° ; and the intersection of the lines at C will be the place of the
balloon ; from whence the perpendicular CD, being let fall, will be
its perpendicular height. Then, by measurement, are found the
distances and height nearly, as follows, viz. AC 1631, BC 1041,
DC 936.
C'aleulation. .''/^
First, from angle B 64° ,''
Take angle A 35 ^•"'' ,•''
Leaves angle ACB 29 ."'
Then, ia the triangle ABC, abb
Assine angleACB 29° 96855712
To opposite side AB 880 29444827
Sosine angleA 35° 97585913
To opposite side BC 1041125 30175028
Assine angleACB 29° 96'855712
To opposite side AB 880 29444827
Sosine angleB n0°or64° 99536602
TooppositesideAC1631442 32125717
And, in the triangle BCD,
Assine angle D 90° 100000000
To opposite side BC 1041125 30175028
Sosine angleB 64° 99536602
To opposite side CD 935757 29711630
Having to find the height of an obelisk standing on the top of a
declivity, I first measured from its bottom, a distance of 40 feet,
and there found the angle, formed by the oblique plane and a line
imagined to go to top of the obelisk 41° ; but, after measuring on
in the same direction 60 feet farther, the like angle was only 23° 45'.
What then was the height of the obelisk ?
Construction. — Draw an indefinite line for the sloping plane or
declivity, in which assume any point A for the bottom of the
obelisk, from whence set off the distance AC = 40, and again
CD ^ 60 equal parts. Then make the angle C = 41°, and the
angle D = 23° 45'; and the point B, where the two lines meet,
will be the top of the obelisk. Therefore AB, joined, will be its
height.
hv Google
! THE PRACTICAL MODEL CALCUIATOR.
Calculation.
From the angle C 41° 00'
Take the angle D 23 45
Leaves the angle DEC 17 15
Then, in the triacgle DBC,
As sine angle DBG 17° 15'
To opposite side DC 60
So sine angle D 24 45
To opposite side CB Sl^S'^
9 4720856
1 7781513
y eojosao
1 ')110977
1 6179225
10 4272623
9 9606516
9 6582842
1 6020000
9 8169429
1 7(;0Y187
And, in the triangle ABO,
As sum of sides CB, C 4 121 488
To difference of sidca CB, CA 41 488
So tang, half sum angles A, B 69° 30'
To tang, half diff. angles A, B 42_24J
The diff. of these is angle CBA 27 5J
Lastly, as sine angle CBA 27° 5y
To opposite side CA 40
So sine angle G 41° 0'
To opposite side AB 57 633
Wanting to know the distance hetween two inaccessible trees, or
other objects, from the top of a tower, 120 feet high, which lay in
the same right line with the two objects, I took tho angles formed
hy the perpendicular wall and lines conceived to be drawn from the
top of the tower to the bottom of each tree, and found them to be
33° and 64^°. What then may be the distaiice between the two
objects? *
Oonstruetion. — Draw the indefinite
ground line BD, and perpendicular to
it BA = 120 equal parts. Then draw
the two lines AC, AD, making the two
angles EAC, BAD, equal to the given
angles 33° and 64^°. So shall C and D bo the places of the two
objects,
Qaleulation. — First, In the rightangled triangle ABC,
As radius 100000000
ToAB 120 20791812
So tang, angle BAG 33° 98125174
ToBG 77929 18916986
And, in the rightangled triangle ABD,
As radius 100000000
ToAB 120 20791812
So tang, angle BAD.... 64i° 103215039
ToBD 251585 24006851
From which take BO 77929
Leaves the dist. CD 173656 as required.
hv Google
SPHERICAL TBXGOSOMETKY.
373
Being on the side of a river, and wanting to know the distance
to a house which was seen on the other side, I measured 200 yards
in a straight line by the side of the river ; and then at each end
of this line of distance, took the horizontal angle formed between
the house and the other end of the line ; which angles were, the
one of them 68° 2', and the other 73° 15'. What then were the
distances from each end to the house ?
Construction. — Draw the line AB = 200 equal parts. Then
draw AC so aa to make the angle A = 68° 2', and BC to make
the angle B = 73° 15'. So shall the point C be the place of the
house reqiuired,
Calculation.
To the given angle A (58° 2'
Add the given angle B 73 16
Then their sura 141 17
Being taken from 180
Leaves the third angle 38 43
Hence, As sin. ang!e C 38° 43'. 97962062
To op. side AB 200
So sin. angle A 68° 2'
To op, "
To op.
So sin.
To op.
3010300
■9672679
BC296'54 24720917
!e C 38° 43' 97962062
AB 200 23010300
le B73°15' 99811711
AC 30619 24859949
SPHEEICAL TRIGONOMETEY.
TkU Arttde is taken Jrom a short Practical Treatise on Spheiical Trigonomeirii,
III/ Oliver Byrne, the author of the present v>ak. PiihUshed by J. A. Valpy.
London, 1835.
As the sides and angles of spherical triangles are measured by
circular arcs, and as these arcs are often greater than 90°, it may
be necessary to mention one or two particulars respecting them.
The arc CB, which when added to
AB makes up a quadrant or 90°, is
called the complement of the arc AB ;
every arc will have a complement,
even those which are themselves
greater than 90°, provided wo con
sider the ares measured in the direc
tion ABCD, &e., as positive, and
consequently those measured in the
opposite direction as negative. The
complement BC of the arc AE com
mences at B, where AB terminates,
and may be considered aa generated by the motion of B, the ex
2G
hv Google
374 TUE PRACTICAL MODEL CALCULATOR.
tremity of the radius OB, iu the direction EC. But the eomplo
meat of the arc AD or DC, commencing in like manner at the ex
tremity D, must be generated by the motion of D in the opposite
direction, and the angular magnitude AOD will hero be diminished
by the motion of OD, in generating the complement ; therefore
the complement of AOD or of AD may with propriety be consi
dered negative.
Galling the arc AE or AD, e, the complement will he 90° — e ;
the complement of 36° 44' 83" is 53° 15' 27" ; and the complement
of 136° 2T' 39" is negative 46° 27' 39".
The arc BE, which must be added to AB to make up a semi
circle or 180°, is called the supplement of the arc AB. IF the arc
is greater than 180°, as the arc ADP its supplement, FE mea
sured in the reverse direction is negative. The expression for the
supplement of any arc o is therefore 180° ~ o; thus the supple
ment of 112° 29' 35" is 67° 30' 25", and the supplement of 205°
42' is negative 25° 42'.
In the same manner as the complementary and supplementary
arcs are considered as positive or negative, according to the di
rection in which they are measured, so are the arcs themselves
positive or negative ; thus, still taking A for the commencement,
or origin, of the arcs, as AB is positive, All will be negative. In
the doctrine of triangles, we consider only positive angles or arcs,
and the magnitudes of these are comprised between o = and o =
180° ; hut in the general theory of angular quantity, we consider
both positive and negative angles, according as they are situated
above or below the fixed line AO, from which they are measured,
that is, according as the ares hy which they are estimated are posi
tive or negative. Thus the angle BOA is positive, and the angle
AOH negative. Moreover, in this move extended theory of angular
magnitude, an angle may consist of any number of degrees what
ever ; thus, if the revolving line OB set out from the fixed line OA,
and make n revolutions and a part, the angular magnitude gene
rated is measured by n times 360°, plus the degrees in the ad
ditional part.
In a rightangled spherical triangle we are to recognise but five
parts, namely, the three sides a, 5, c, and the two angles A, ,
BO that the right angle C is omitted.
hv Google
SPHERICAL Tr.IGOKOMETKT.
Let A', c', B,' be the comple
ments of A, c, B, respectively,
and suppose b, a, W, c', A', to be
placed on the hand, as in the
annexed figure, and that the
fingers stand in a circular order,
the parts represented by the
fingers thus placed are called *
circular parts.
If we take any one of these as
a middle part, the two which lie
next to it, one on each side, will
be adjaaent parts. The two parts
immediately beyond the adjacent
parts, one on each side, are called
the opposite parts.
Thus, taking A' for a middle part, h and c' will be adjacent parts,
and a and B' are opposite parts.
If we take c' as a middle part. A' and B' are adjacent parts, and
6, a, opposite parts.
When B' is a middle part, c', a, become adjacent parts, and A',
h, opposite parts.
Again, if we take « as a middle part, then B', h, will be adjacent
parts, and c', A', opposite parts.
Lastly, taking 6 as a middle part. A', a, are adjacent parts, and
c', B', opposite parts.
This being understood, Napier's two rules may be expressed as
follows : —
I. Ead. X sin. middle part = product of tan. adjacent parts.
II. Rad. X sin. middle part = product of cos. opposite parts.
Both these rules may be comprehended in a single expression, thus,
Rad. sin. mid. = prod, tan, adja. = prod. cos. opp. ;
and to retain this in the memory we have only to remember, that
the vowels in the contractions sin., tan., cos., are the same as those
in the contractions mid., adja., opp., to which they are joined.
These rules comprehend all the succeeding equations, reading
from the centre, R = radius.
In the solution of rightangled spherical triangles, two parts are
given to find a third, therefore it is necessary, in the application of
this formula, to choose for the middle part that which causes the
other two to become either adjacent pa,rts or opposite parts.
In a rightangled spherical triangle, the hypothenuse
e = 61" 4' 56" ; and the angle
A =■ 61° 50' 29". Required the adjacent leg?
90°
61
0'
P0°
0'
A = 01 50
hv Google
THE PEACTICAI. MODEL CALCULATOR.
, «
' ?v.
.# 'i
■ • ^ SOO 'DSOJ 
Tan.i'tait.':
In this example, A' is selected for the middle part, because then
6 and v' become adjacent parts, as in the annexed figure.
Kad. X sin. A' = tan. h X tan. c'.
rad. X sin. A'
By Logarithms.
Kad.  lOOOOOOOO
Sin. A^28°9'21"  96738628
19'6738628
Tan. c'28°55'4"  974228 08
Tan.5'40°30'16"99315820
'Ihe side adjacent to the given
angle is acute or obtuse, accord
ing as the hypothenuse is of the
same, or of different species with the given angle,
.. the leg h = 40° 30' 16", acute.
Supposing the hypothenuse c = 113° 65', and the angle A = 31° 51',
then tiie adjacent leg 6 would he 117° 34', obtuse.
hv Google
SPUEKICAL TltlGONOMEIKT.
In the rightangled spherical triangle ABC, are given tlie hypo
thenuse c = 113° 55', and the angle A = 104° 08'; to find the
opposite leg a.
14 08= A'.
In this example, a is taken for the middle part, then A' and c'
are opposite parts. (See the subjoined £
From the general formula, we
have,
Rad. X sin. a = cos. A' X cos. c'.
COS. A' X COS. c'
.. sm, a = f,— 5 .
Had.
Sy Logarithms.
COS. A'  14° 08' 99860509
cos.c' 2fJ 55 99610108
199476617 / L 
Radius 100000000 ^ \\
sin «/l^^°^*'l 9.947661T '^
The obtuse side 117° 34' is the leg required, for the side oppo
site to the given angle is always of the same species with the
given angle.
If in a rightangled spherical
triangle, the hypothenuse were
78° 20', and the angle A =
37° 25', then the opposite leg
a = 36° 31', and not 143° 29',
because the given angle is acute.
In arightangled spherical tri
angle, are given c = 78° 20', and
A = 37° 25', to find the angle B.
90° 0'
c = 78_20
11 40 = c'.
90° 0'
A = 37 25
i>2 35= A'
hv Google
378 THE PHAGTICAL MODEL CALCULATOR.
Here the complement of the hypothenuse {«') is the middle part;
and the complement of the ^^ '
angle opposite the perpen
dicular (A'), and the com
plement of the angle oppo
site the base (B'J are the
adjacent parts. This wUl
readily be perceived by
reference to the usual
figure in the margin.
Rad. X sin. c' = tan. A'
X tao. B' ;
_ Rad, X sin. c'
.■.tan.B' =
tan. A' '
B^ Logarithms.
100000000
 11° 40'. 93058189
Bad.
8in. c
193058189
tan. A'  52° 35' 101163279
..tan. B' 8° 48' 91894910
But 90 — B = B'
hence 90  E' = B.
90° 0'
B = 81° 12'.
When the hypothenuse and an angle are given, the other angle is
acute or obtuse, according as the given parts are of the same or of
different species.
In the above example, both the given parts are acute, therefore
the required angle is acute; but if one be acute and the other ob
tuse, then the angle found would be obtuse : — Thus, if the hypo
thenuse be 113° 55', and the angle A = 31° 51' ; tlien will B' =
14° 08', and the angle B = 104° 08'.
Given the hypothenuse e = 61° 04' 56", and the aide or leg,
a = 40° 30' 20", to find tlie angle adjacent to a. c' •&■
90° 0' 0"
e = 61 04 56
28 55 04" = c".
The three parts are here
connected ; therefore the com
plement of .B is the viiddle
part, a and the complement of
are the adjacent parts.
Hence we have.
Bad. X sin. B' = tan. a X tan.
Rad.
hv Google
SPHERICAL TRIGONOMETEY,
By Logarithms.
tan. a  40° 30' 20" = 99315841
tan. c' ~ 28 55 04 = 9 •7422801
196738642
Bad 1000000 00
sin. B'....28° 09' 31" 96738642
90° 0' 0"
E' = 28 09 31
61 50 29 = B.
The angle adjacent to the given side is acute or ohtuse accord
ing as the hypothenuse is of the same or of different species with
the given side.
Before ivorking the above example, it was easy to foresee that
the angle B would he acute ; but suppose the hypothcnuse = 70°
20', and the side a = 117° 34', then the angle E would be obtuse,
because a and c are of different species.
Rule V. — In a spherical triangle, rightangled at c, are given
c = 78° 20' and b == 117° 34', to find the angle E ; opposite the
given leg, (see the next diagram.)
In this example, b becomes the middle part, and e' and B' oppo
site parts ; and therefore, by the rule,
Rad. X sin. b = cos. E' X cos. c' ; that is,
Ead. X sin. b
cos. B' = ■ i ■.
cos. c'
90°  78° 20' = 11° 40' = c'.
Sence, hy Logarithms. i ai i i .
Ead 100000000 / '^l l\ ^
sin. 6 = sin. lir 34' I 9.947^555
or Bin. 62 2o J
199476655 / ' ^^J
COS. «• 11° 40' 99909338 ' ^T
COS. B'25°09' 9956731T
b,Google
380 THE PRACTICAL MODEL CALCULATOR.
But since the angle
opposite tlie given
Bide is of the same
species with the given
side, yO° must be
added to W, to pro
duce E :— via. 90° +
25° 09' = 115° 09'.
Given c= 61° 04'
56", and 6 = 40° 30'
20", to find tte other
side a.
Here c' is the mid
dle part, a and b the
opposite parts ; hence
by position i, a = 50° 30' 30".
Given the side 5 = 48° 24' 10", and the adjacent angle A =
66° 20' 40", to find the side a.
In this instance, b is the middle part, the complement of A and
a are adjacent parts. Consequently, a = 59° 38' 27".
In the rightangled spherical triangle ABC,
r,. ( The side a = 59° 38' 27" 1 , e ^ ^i i *
Given i Ti. 3 I 1 It cfio ■>!)/ ccvi ?to find the angle A.
[^ Its adjacent angle B = 52° 32' 55" J °
Answer, 66° 20' 40".
The required angle is of the same species as the ^iven side, and
Given the side l = 49° 17', and its adjacent angle A = 23° 28',
to find the hypothenuse.
Making A' the middle part, the others will be adjacent parts,
and, therefore, by the first rule we have c = 51° 42' 37".
In a spherical triangle, rightangled at C, are given b = 29° 12'
50", and B = 37° 26' 21", to find the side a.
Taking a for the middle part, the other two will be adjacent parts ;
hence by the rule,
Rad. X sin. a = tan. b x tan. B'
that is, rad. X sin. a — tan. b X cot. B
tan. b X cot. B
.■. sin. a — — — — ^
rad.
In this case, there arc two solutions, i. e. a and the supple
ment of a, because both of them have the same sine. As sin. a
is necessarily positive, b and B must necessarily be always of
the same species, so that, as observed before, the sides including
the right angle are always of the same species as the opposite
angles.
hv Google
SPHERICAL TKIGONOMETKT.
381
In working this example,
we find the log. sin. a —
98635411, which corre
sponds to 46° 55' 02",
or, 133° 04' 58".
It appears, therefore,
that a is ambiguous, for J
there exist two rightangled I
triangles, having an ohlique I
angle, and the opposite side 1
in the one equal to an
oblique angle and an oppo
site side in the other, but
the remaining oblique angle
in the one the supplement
of the remaining oblique
angle in the other. These triangles are situated with respect
to each other, on the sphere, as the triangles AEO, ADC,
in the annexed diagram, in which, with tho exception of the
common side AC, and the equal angles B, D, the parts of the
one triangle are supplements of the corresponding parts of the
other.
In a rightangled spherical triangle are
f, /the side a = 42° 12', \ to find the adjacent
\ its opposite angle A = 48° J angle B.
The complement of the given angle is the middle part; and
neither a nor E' being joined to A', they are consequently opposite
parts ; hence, the angle E = 64° 35', or 115° 25' ; this case, like
the last, being ambiguous, or doubtful.
Given a = 11° 30', and A = 23° 30', to find the hypothenusc c.
e = 30°, or 150°, being ambiguous.
In a rightangled triangle, there are given the two perpendicu
lar sides, viz. a = 48° 24' 16", b = 50° 38' 27", to find tho
angle A.
A = 66° 20' 40".
Given a =
142° 31', b = 54° 22', to find c
c = 117° 33'.
Given ■! d _ oi ig > Required the sidaa.
n fA = 66° 20'40"K c 4 ^1, 1, .1.
Given ■{ Tj ~ i^2 '!'' SS ( hypothenuso c.
hv Google
THE PRACTICAL MODEL CALCULATOR.
MEASUREMENT OY ANGLES,
" CibU Engineer and Archileet's Jotirnai," Jbr Oct. and Kov. 1847.
A NEW METHOD OF MEASUEING THE DEGKEE8, MINUTES, ETC., IN ANY
EECTILINEAR ANGLE BY COMPASSES ONLY, WtTHOOT USING SCALE OR
PROTRACTOR .
Apply AB = z, from B to 1 ; from 1 to 2 ; from 2 to 8 ; from
3 to 4 ; from 4>to 5. Then take B 5, in the compasses, and apply
it from B to 6 ; from 6 to 7 ; from 7 to 8 ; from 8 to 9 ; and from
9 to 10, near the middle of the arc AE, With the same opening,
B 5 or A 4, or ^, which we shall terra it, lay off 4,11, 11,12, and
12,13. Then the arc betiveen 13 and 10 is found to be contained
23 times in the arc AB.
hv Google
MEASUREMENT OF ANGLES.
Hence, we have,
5»y .
23 z =
"23
  73° 33'82.
a — 22 »
Ev substituting tilis value in the iirst equation, we obtain,
22 a:
6.25^^ = 360.
1013 a; „„„ , 360 X 207
Wr " ^™' '"* " lOlT
Apply AB = X, from B to 1 ; from 1 to 2 ; from 2 to 3 ; from
3 to 4. Then take E 4, in the compasses, and apply it on the arc,
from B to 4 ; from 4 to 5 ; from 5 to 6 ; from 6 to 7 ; and from
7 to 8, near the middle of the are AE. With the same opening,
B4 = ^, lay 6ff A9, 9,10, 10,11, 11,12, 12,13, and 13,14. The
arc between 14 and 8 is found to be contained nearly 24 times in
the are AB. Therefore, we have,
ix + y = Zm;
11^2 =^; X
24 e = x; or, 2 = gl
X mx
■■■ll!'24 = ^' ■■•!' = 264
Substituting this value of y in the first equation,
2b X
4aT4 9fiT =360;
860 )
264 "
1071
 = 88° 44'333.
How to lay off an angle of any number of degrees, minutes, §■€..
with compasses only, without the use of scale or protractor.
Let it bo required
to lay off an angle of
36° 40' = 3. Take any
small opening of the
compasses less tban
onetenth of the ra
dius, and lay off any
number of equal email ,
ares, from A to '
from 1 to 2 ; from 2 to
3, &c., until we have
laid off an arc, AB,
greater than the one
required. Draw B b
through the centre o,
then will the are a 5 =
arc AB, which we shall
hv Google
384 THE PRACTICAL MODEL CALCULATOR.
put = 20 * in this example, and proceed to measure ah aa in tKe
(irs^ example. Lay off a 6 from 6 to c ; from c? to d ; from t? to e ;
from c to/; from /to ^. Patting g a = A„ t'len,
108
6 X 20* + A. = 360° = TjP; because,
360° _ 21600 _ 108
36° 40' ~ 2200 ~ 11 ■
Lay off, as before directed, g a, = Ai, from a to li, from /( to s,
and btot; then calling a (, A„ we have
3 A, + A, = 20 f ;
and we find that s ( is contained 28 times in the a
108
iah;
.. 120 ^> + A. = jr ^ ; 3 Ai + As = 20 ?> ; and 28 A^ = 20 1.
Eliminating A, and As) we find
29205 ^„ .
3 = ^29gg"^ = 129 times ^ nearly ;
,■, 36° 40' = /, A N is laid off with a3 much ease and certainty
as by a protractor.
As a second example, let it be required to lay off an angle of
132° 2T'. From 180° 0' take 132° 27' = 47° 33', which put = s.
360° 2400 , V ^
7, = "oiy' ivhen put = ;, then « 3 = oo'J = "■
We have laid off 29 small arcs from A to B ; 29 = j. AB =
ib = be = cd = de = cf. And a g = hh = af = l\^; hg = Ab
..5 X 29* + Ai = 360° = ^u = me?± A, (1)
2 A,  A, = 29<., or m A. ± A, = ^* (2)
13 Aa = 29 f, or ^ A. = '^ (3)
hv Google
MBAStTREMBNT OF ANGLES. 385
Eliminating Ai smd A™t we have
_ {mnq±{q=pl)}U _ {5213 + (13 + 1)}29317
^ ~ ^nq *~ 2400213 *■ ""
1323729
■ irifAoo ? = 21 J times t very nearly. Hence the line o N deter
mines the angle aoN = 132° 27'.
In the expression
^ ■ ,„g ?■ (^)
substituting the numerals of the first example, then
{6328 + (28  1)12011 29205
^=' 108328 ^ = ^268" ^ = ^^'^ ^''^'^^ ^ "'^''y
the result before obtained.
The ambiguous signs of (R) cannot be mistaken or lead to error,
if the manner in which it is deduced from (1), (2), (3), be attended
to. From (3)
Ao = ~~ i substituting this value of Aai '■i (2),
M A.= f* q^ A, =
in (1), gives
ji = mi^±{^i^=P —) ; from which (K) is found.
This method of measuring angles is more exact than it may ap
pear ; for if, in the first example, we take
5x — y = 360; 9 J/ + z = a^ ; and 20 z = 3:,
64800 „„„ „„, ,.
then X = gg. = 73° 33' 8a.
The first equations gave 73° 33' 82 when 23 s = r, so it does
not matter much whether 20, 21, 22, 23, 24, or 25 times z make
X. This fact is partieiilarly worth attention.
Given the three angles to find the three sides,
The following formulas give any side a of any spherical triangle.
— cos. J S cos. {^ S — A)
on. i a = y sGTluErc > "''
cos  » = ,55!iiiA^B)_coMi_sq
■ ^ sin. B sin, C.
Given tile tliree sides to iind tlie three angles.
sin. (1 S  6) sin. (} S  c)
sm. jA^ sin. Ssin. «."
. sin. } S .in. (i S  <■)
COS. J A = </ sin b sin
b,Google
6RATITYWEIGHTMASS.
BPECIFIC GRAVITY, CENTRE OJ GRAVITY, AND OTTITB CKNTRES Or BODIES.
— WEIGHTS or ENGINEERING AND MECHANICAL MATERIALS. — ItKASS,
COPl'EE, STEEL, IRON, "WATER, BTONE, LEAB, TIN, ROUND, SQUARE, PLAT,
ANGDLAR, ETC.
1. In a second, the acceleration of a body falling freely in vacuo
ia 32'2 feet ; what velocity has it acquired at the end of 5 seconds ?
322 X 5 = 161 feet, the velocity.
2. A cylinder rolling down an inclined plane with an initial velo
city of 24 feet a second, and suppose it to acquire each second 5 ad
ditional feet velocity ; what is its velocity at the end of 37 seconds ?
24 + 3T X 5 = 425 feet.
3. Suppose a locomotive, moving at the rate of 30 feet a second,
(as it is usually termed, with a 30 feet velocity,) and suppose it to lose
5 feet velocity every second ; what is its velocity at the end of 3"33
seconds ?
The acceleration is — 333, negative,
.. 30  5 X 333 = 1335 feet.
4. If a body has acquired a velocity of 36 feet in 11 seconds,
by uniformly accelerated motion ; what is the space described?
36 X 11
2^— = 198 feet.
5. A carriage at rest moves with an accelerated motion over a
space of 200 feet in 45 seconds; at what velocity does it proceed
at the beginning of the 46th second ?
200 X 2
— 7P — = 88889 feet, the velocity at the end of the 45th second.
The four fundamental formulas of uniformly accelerated motion are
v=pt; 8=2; « = T' '^^'
V the velocity, p the acceleration, * the time, and s the space.
6. What space will a body describe that moves with an accele
ration of 115 feet for 10 seconds.
ii:^">* = 5T5fes..
7 A body commences to move with an acceleration of 55 feet,
and moves on until it is moving at the rate of 100 feet a second ;
what space has it described ?
^ir? = 90909 feet.
2 X 05
hv Google
GEAVITY — WEIGHT — MASS. 387
8. A body is propelled witli an initial velocity of 3 feet, and with
an acceleration of 8 feet a second ; what space is described in
13 seconds?
8 X 13 + ^^ = 715 fee..
9. What distance will a body perform in 35 seconds, commenc
ing with a velocity of 10 feet, and being accelerated to move with
a velocity of 40 feet at the beginning of the 36th second ?
2 X 35 = 8Y5 feet, the distance.
The formulas for a uniformly accelerated motion, commencing
with a velocity c, are as follow : —
pe c + V if — e
The succeeding formulas are applicable for a uniformly retarded
motion with an initial velocity c.
pe c + V e ~v^
10. A body rolls up an inclined plane, with an initial velocity
of 50 feet, and suffers a retardation of 10 feet the second ; to what
height will it ascend ?
60
10
= 5 seconds, the time.
q — yy = 125 fcet, thc height required.
The free vertical descent of bodies in vacuo offers an important
example of uniformly accelerated motion. The acceleration in the
previous examples was designated by p, but in the particular mo
tion, brought about by the force of gravity, the acceleration is
designated by the letter g, and has the mean value of 322 feet.
If this value of g be substituted for p, in the preceding formula,
v = 'd22xt; I! = 8024964 xv^; 8 = 161 X (^; s= 015528 Xv^;
t = 031056 X v; a.iiit= 2492224 X ^/3.
11. What velocity will a body acquire at the end of 5 seconds,
in its free descent?
322 X 5 = 161 feet.
12. What velocity will a body acquire, after a free descent
through a space of 400 feet ?
8024064 X >/400 = 16049928 feet.
13. What space will a body pass over in its free descent during
10 seconds?
161 X (10)= = 1610 feet.
hv Google
388 THE PEAOTICAL MODEL CALCULATOR.
14. A body falling freely in vacuo, has in its free descent
acquired a velocity of 112 feet ; what space is passed over ?
■015528 X (112)' = 194783232 feet.
15. In what time will a body falling freely acquire the velocity
of 30 feet ?
■031056 X 30 = 93168 seconds.
16. In what time will a body pass oyer a space of 16 feet, fall
ing freely in vacuo ?
■2492224 x ^/TQ = 9968896 seconds.
If the free descent of bodies go on, with an initial velocity,
which we may call c, the formulas are,
ft if — ,^
s = ct + g^ = ct + lQlxe; « = ~^ = 015528 {^^'^c^).
If a body be projected vertically to height, with a velocity which
wc shall term c, then the formulas become,
V = e — 322 X ( ; « = v^e' — 644 xs; s = ct — g ^ =
ct  161 X i=; s = ^ ~'" . = 015528 (c=  v%
17. What space is described by a body passing from 18 feet velo
city to 30 feet velocity during its free descent in vacuo.
From the annexed table, we find that the height due to 30 feet
velocity = 1397516
The height due to 18 = 508106
Space descrihcd 894410
Since this problem and table are often required in practical me
chanics, we shall enter into more particulars respecting it.
^^~ 2g 2g~2g'
if we put h = height due to the initial velocity c; that is,
A == 2^ ; and ft, — the height due to the terminal velocity v ; that is,
ft, = o~ ; then,
s = ftj — ft, for falling bodies, as in tlie last example ; and
g = A — ftj, for ascending bodies.
Although these formulas are only strictly true for a free descent
in vacuo, they may be used in air, when the velocity is not great.
The table will be found useful in hydraulics, and for other heiglits
and veiocities besides those set down, for by inspection it is seen
that the height 201242 answers to the velocity 36 ; and the height
2012423 to 36 ; and the height 2012423 to 360 ; and so on.
hv Google
WEIGHT— GRAVIIY—
Table of the Heights corresponding to different Velocities, i
the second.
',feet
fi
CossEsrasDisD llnoiT in t^m. 
I
2
3
^
5
6
8 1 9 1
?ii
■0X8789
■169006
S9aS94
'Os:iii3
IflBOOB
mm
'0349378
■097050
■C6D060
■B9B89ft
■044571
■113189
■21257]
■Moeli
■060311 06 5066:
■121730 130590:
■K4224 ■238180;
•6^360 66053
14SI304; lS21S0i
The following extension is obtained from the foregoing table,
by mere inspection, and moving the decimal point as before di
rected.
is,
CorrMnondiTig
HUslll>»F«t.
1^
CMTfaponatnr
Jfi
S^sj'^^t' :
pa
71)
1552796
li)
500559
23
12^17392
87
21 ■25777 i
11
1878882
?0
621118
23
18^O50O1
88
12
2065218
21
684783
SO
13 ■97616
2301802 1
13
2824224
22
751553
SI
1492337
40
2484472 i
14
3 ■013478
23
821429
82
1590062
15
8^4e379
24
804410
16 ■60994
1H
8^975I6
25
970497
S4
1878883
43
2857143 i
17
418753
'^11
1O49G00
85
]9^02174
41
3006212 1
13
50310G
27
1131988
2012423
45
311441 !
18. AVhat mass does a body weighing 30268 lbs. contain 1
80268 302680 , „
"32T = ~3^^ = ^**' ^^^■
For the mass is equal to the weight divided by^r. And g U
taken equal to 32'2; but the acceleration of gravity is somewhat
variable ; it becomes greater the nearer we approach the poles of
the earth. It is greatest at the poles and least at the equator,
and also diminishes the more a body is above or below the level of
the sea. The mass, so long as nothing is added to or taken fron.'
it,' is invariable, whether at the centre of the earth or at any dis
tance from it. If M be the mass and W the weight of a body,
Then M = 
W
W
■03105
i9 W.
200 lbs ?
" 322 "
19. What is the mass of a body whose weight ii
■031055 X 200 = 621118 lbs.
The weight of a body whose mass is 200 lbs. is 322 X 200 =
64400 lbs. It may be remarked, that one and the same steel
spring is differently bent by one and the same weight at different
places.
The force which accelerates the motion of a heavy body on aii.
inclined plane, is to the force of gravity as the sine of the inclina
hv Google
390 THE PRACTICAL MODEL CALCULATOR.
tion of the plane to the radius, or as the height of the plane to its
lenjrth.
The velocity acquired by a body in falling from rest through a
given height, is the same, whether it fail freely, or descend on a
plane at whatever inclination.
The space through which a body will descend on an inclined
plane, is to the space through which it would fall freely in the same
time, as the sine of the inclination of the plane to the radius
The velocities which bodies acquire by descending along chords
of the same circle, are aa the lengths of those chords.
If the body descend in a curve, it suffers no loss of velocity.
The centre of gravity of a hody is a point about which all its
parts are in equilihrio.
Hence, if a body be suspended or supported by this point, the
body will rest in any position into which it is put. We may, there
fore, consider the whole weight of a body as centred in this point.
The common centre of gravity of two or more bodies, is the point
about which they would equiponderate or rest in any position. If
the centres of gravity of two bodies be connected by a right line,
the distances from the common centre of gravity are reciprocally
aa the weights of the bodies.
If a line be drawn from the centre of gravity of a body, perpen
dicular to the horizon, it is called the line of direction, being the
line that the centre of gravity would describe if the body fell freely.
The centre of gyration is that part of a body revolving about an
axis, into which if the whole quantity of matter were collected, the
same moving force would generate the same angular velocity.
To find the centre of Gyration. — Multiply the weight of the
several particles by the squares of their distances from the centre
of motion, and divide the sum of the products by the weight of the
whole mass; the square root of the quotient will be the distance
of the centre of gyration from the centre of motion.
The distances of the centre of gyration from the centre of mo
tion, in different revolving bodies, are as follow : —
In a straight rod revolving about one end, the length. X STTS.
In a circular plate, revolving on its centre, the radius X TOTl.
In a circular plate, revolving about one diameter, the radius X 'O.
In a thin circular ring, revolving about one diameter, radius X
■7071.
In a solid sphere, revolving about one diameter, the radius x
■6325.
In a thin hollow sphere, revolving about one diameter, radius X
•8164.
In a cone, revolving about its axis, the radius of the base x
■547T.
In a rightangled cone, revolving about its vertex, the height X
■86ti.
hv Google
8PECIPI0 GRAVITY. 391
In a paraboloid, revolving about its axis, the radius of the base
X 5773.
The centre of •percussion is that point in a hody revolving about
a jkced axi», into which the whole of the force or motion is collected.
It is, therefore, that point of a revolving body which would strike
any obstacle with the greatest effect ; and, from this property, it
has received the name of the centre of percussion.
The centres of oscillation and percussion are in the same point.
If a heavy straight bar, of uniform density, he suspended at one
extremity, the distance of its centre of percussion is twothirds of
its length.
In a long slender rod of a cylindrical or prismatic shape, the
centre of percussion is nearly twothirds of the length from the
axis of suspension.
In an isosceles triangle, suspended by its apex, the distance of
the centre of percussion is threefourths of its altitude. In a line
or rod whose density varies as the distance from the point of sus
pension, also in a flywheel, and in wheels in general, the centre
of percussion is distant from the centre of suspension threefourths
of the length.
In a very slender cone or pyramid, vibrating about its apex, the
distance of its centre of percussion is nearly fourfifths of its length.
Pendulums of the same length vibrate slower, the nearer they
are brought to the equator. A pendulum, therefore, to vibrate
seconds at the equator, must be somewhat shorter than at the poles.
When we consider a simple pendulum as a ball, which is sus
pended by a rod or line, supposed to be inflexible, and without
weight, we suppose the whole weight to be collected in the centre
of gravity of the ball. But when a pendulum consists of a ball,
or any other figure, suspended by a metallic or wooden rod, the
length of the pendulum is the distance from the point of suspension
to a point in the pendulum, called the centre of oscillation, which
does not exactly coincide with the centre of gravity of the ball.
If a rod of iron were suspended, and made to vibrate, that point
iu which all its force would be collected is called its centre of oscil
lation, and is situated at twothirds the length of the rod from the
point of suspension.
SPECIFIC GRAVITY.
The comparative density of various substances, expressed by the
term specific gravity, affords the means of readily determining the
bulk from the known weight, or the weight from the known bulk ;
and this will be found more especially useful, in cases where the
substance is too large to admit of being weighed, or too irregular
in shape to allow of correct measurement. The standard with
which ail solids and liquids are thus compared, is that of distilled
water, one cubic foot of which weighs 1000 ounces avoirdupois ;
hv Google
392 THE PRACTICAL MODEL CALCULATOR.
and the specific gravity of ii solid body is determined by the dif
ference between its weight in the air, and in water. Thus,
If the body be heavier than water, it will displace a quantity of
fluid equal to it in bulk, and wil! lose as much weight on immersion
as that of an equal bulk of the fluid. Let it be weighed first,
therefore, in the air, and then in water, and its weight in the air
be divided by the difference between the two weights, and the quo
tient will be its specific gravity, that of water being unity.
A piece of copper ore weighs 56J ounces in the air, and 43
ounces in water ; required its specific gravity.
6625 — 4375 = 125 and 5625 ^ 125 = 45, the specific gravity.
If the body be lighter than water, it will float, and displace a
quantityoffloidequal toit in we^^Ai, the bulk of which will be equal
to that only of the part immersed. A heavier substance must,
therefore, be attached to it, so that the two may sink in tlie fluid.
Then, the weight of the lighter substance in the air, must be added
to that of the heavier substance in water, and the weight of both
united, in water, be subtracted from the sum ; the weight of the
lighter body in the air must then be divided by the difference, and the
quotient will be the specific gravity of the lighter substance required.
A piece of fir weighs 40 ounces in the air, and, being immersed
in water attached to a piece of iron weighing 30 ounces, the two
together are found to weigh 3'3 ounces in water, and the iron alone,
258 ounces in the water ; required the specific gravity of the wood.
40 + 258 = 658  33 = 625 ; and 40 j 625 = 064, the
specific gravity of the fir.
The specific gravity of s, fluid may be determined by taking a,
solid body, heavy enough to sink in the fluid, and of known spe
cific gravity, and weighing it both in the air and in the fluid. The
difference between the two weights must be multiplied by the spe
cific gravity of the solid body, and the product divided by the
weight of the solid in the air : the quotient will be the specific
gravity of the fluid, that of water being unity.
Required the specific gravity of a given mixture of muriatic acid
and water ; a piece of glass, the specific gravity of which is 3,
weighing 3J ounces when immersed in it, and 6 ounces in the air.
6  375 = 225 X 3 = 675 r 6 = 1125, the specific gravity.
Since the weight of a cubic foot of distilled water, at the tem
perature of 60 degrees, (Fahrenheit,) has been ascertained to be
1000 avoirdupois ounces, it follows that the specific gravities of all
bodies compared with it, may be made to express the weight, in
ounces, of a cubic foot of each, by multiplying these specific gra
vities (compared with that of water as unity) by 1000. Thus, that
of water being 1, and that of silver, as compared with it, being
10474, tho multiplication of each by 1000 will give 1000 ounces
for the cubic foot of water, and 10474 ounces for the cubic foot
of silver.
hv Google
SPECIFIC GRAVITY, 393
In the following tables of specific gravities, the niimbers in the
first column, if taken as whole numbers, represent the weight of a,
cubic foot in ounces ; but if the last three figures arc taken aa deci
mals, they indicate the specific gravity of the body, water being
considered as unity, or 1.
To ascertain the number of cubic feet in a substance, from its
weight, the whole weight in pounds avoirdupois must be divided by
the figures against the name, in the second column of the table,
taken as whole numbers and decimals, and the ijuoticnt will be the
contents in cubic feet.
, Required the cubic content of a mass of castiron, weighing 7 cwt.
1 qr. = 812 lbs.
812 lbs. ^ 4505 (the tabular weight) = 1803 cubic feet.
To find the weight from the measurement or cubic content of a
substance, this operation must be reversed, and the number of cubic
feet, found by the rules given under "Mensuration of Solids,"
multiplied by the figures in the second column, to obtain the weight
in pounda avoirdupois.
Required the weight of a log of oak, 3 feet by 2 feet 6 inches,
and 9 feet long.
9x3x25 = C75 cubic feet.
And 675 x 582 (the tabular weight) = 39285 lbs., or 35 cwt.
qr. 8 lbs.
The velocity g, which is the measure of the force of gravity,
varies with the latitude of the place, and with its altitude above
the level of the sea.
The force of gravity at the latitude of 45° = 321803 feet ; at
any other latitude L, g = 321803 feet — 00821 cos. 2 L. If
ff' represents the force of gravity at the height h above the sea,
and r the radius of the earth, the force of gravity at the level of the
sea will be ^=^'(1 + j^).
In the latitude of London, at the level of the sea, </ = 32191 feet.
Do. Washington, do. do., y = 32155 feet.
The length of a pendulum vibrating seconds is in a constant
ratio to the force of gravity.
I = 98696044.
Length of a pendulum vibrating seconds at the level of the sea, in
various latitudes.
At the Equator ,...390152 inches.
Washington, lat.38° 53' 23" 390958 —
New York, lat. 40° 42' 40" 391017 —
London, iat.51°31' 391393 —
lat.45° 391270 —
kt.i 391270 in.— 009982 cos. 2 X.
hv Google
394 THE PRACTICAL MODEL CALCULATOR.
Specific Gravity of various Substances.
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SPECIFIC GRAVITY. 395
Taelb of the Weight of a Foot in length of Flat and Rolled Iron.
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m THE PKACTICAL MODEL CALCL'LATOE.
Table of the Weight of one Mot Length of Malleable Iron.
EUC.«E ,BOK.
Eov»D IBM. 1
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Weislit
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lothsa.
Founds
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026
047
037
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134
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H
082
1
189
148
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2&7
202
2
134
33*i
263
2
166
H
425
833
2
201
412
237
ll
635
408
279
593
3
324
6%
309
H
lO'ag
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3709
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1921
8400
4221
s
12096
H
5341
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2643
7
16464
6593
12
B!'99
The following tables are rendered of great utility hj moan's of
this table :—
Suppose it he required to ascertain the weight of a east iron
pipe 26J inches outside and 23 inside, the length being 6J feet.
Opposite 26J in the table is
2348576 X 72 x 65 = 10991135.
And opposite 23 in the t0,b!e is
192285G X 72 x 65 = 8998966 subtract
1992169 lbs. avr.
The succeeding table contains the surface and solidity of spheres,
together ivith the edge or dimensions of equal cubes, the length
of equal cylinders, and the weight of water in avoirdupois pounds : —
hv Google
SPECIFIC QEiVITT.
Surface and Solidity of Spheres
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■THE PRACTICAL MODEL CALCULATOR.
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SPECIFIC GRAVITY.
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THE PRACTICAL MODEL CALCULATOR.
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9047808
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327259
4714363
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5309304
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548014
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116870
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1
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1680265
118885
98332
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15 in.
7068600
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120900
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640178
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7547694
1949821
124930
103332
705250
1
7793131
2045697
126940
105000
739929
16 in.
8042496
2144665
128960
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775725
Table containing the Weight of Flat Bar Iron, 1 foot in length,
of various breadths and thicknesses.
jl
TBIC^KISS IH rlRTS or AI. IT^CH. I
^~
A
1
A
i
A
1
LtE.
J
1 inch.
Lb,.
LbB.
LW.
Lbs.
Lta,
IM.
LUs.
Lts.
lin.
083
104
125
145
166
187
208
250
291
333
U '
09S
M7
140
164
187
200
234
281
828
104
180
156
182
208
284
260
312
374
416
114
]43
171
2 00
329
257
343
401
458
125
166
187
218
250
281
312
375
487
500
135
203
804
338
406
541
145
218
255
291
364
437
5iO
156
195
234
273
812
851
390
468
646
625
2 m.
166
2 08
250
291
838
375
416
500
583
666
H
177
221
265
309
354
442
581
619
708
2^
187
234
281
328
875
421
468
562
G66
750
2
197
247
296
846
395
445
494
593
e92
791
208
260
864
416
468
520
625
729
833
2
218
328
382
437
492
546
666
765
875
2
229
286
S43
401
458
515
572
G87
802
016
2
239
299
359
419
479
639
588
718
958
am.
250
8 12
375
437
500
562
625
750
876
1000
3
270
838
4.06
178
641
609
677
812
947
1083
S
291
364
437
510
583
656
729
876
1020
1166
3
4in.
812
390
468
546
625
708
781
937
1098
1250
S3S
419
600
666
750
833
1000
H66
]333
4
354
442
531
619
708
796
1062
1239
1416
4
376
468
662
666
750
843
937
1125
1312
1500
4
395
494
593
692
791
890
1187
1365
1588
6 n.
417
520
625
729
883
937
1041
1250
1458
1666
5
437
546
656
765
875
984
1093
1312
1581
1750
6
458
572
802
916
1031
1145
1375
1604
1833
479
698
718
838
958
1078
1197
1437
1677
1916
6 n.
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626
750
875
1000
1125
1250
1500
1750
2000
b,Google
SPECIFIC QEAvnr.
(fc Smdik Gramtm and other Projxrtie, of
'ider the ttxndard of comparinn, or 1000.
tli illi
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:» Ungth, artel of VaHota DiomeUn, in Us. am
each one fool
s "■■ * '■;■■ J "■■ If ,
„Google
402 THE PRACTICAL MODEL CALCULATOR.
Table containing the Weight of Square Bar Iron, from 1 to 10 feet
in length, and from J of an inch to 6 inches square.
7T
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3 hit.
iitti.
4 feel.
5twt.
6tt*t.
7 (eft.
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10 (Ml.
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Ma.
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Lbs.
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i 02
05
04
06
08
11
13
15
17
19
21
10
14
19
24
29
33
38
43
48
0>8
17
26
84
42
51
59
68
76
85
18
26
40
63
66
79
82
106
110
133
1'9
57
76
95
114
133
152
171
190
26
62
78
104
129
166
181
207
233
259
34
68
101
185
169
203
237
270
804
1
43
86
128
171
214
257
299
842
385
428
1
53
106
158
211
264
317
870
422
475
1
64
128
192
820
883
447
511
575
639
1
76
152
228
804
380
466
632
608
760
1
89
179
367
446
626
714
803
1
104
207
31 1
414
518
621
725
932
1035
l
119
238
356
475
594
713
832
951
1069
1188
a'in.
136
270
406
641
676
81 J
946
1082
1217
1352
2*
21
153
805
458
611
768
91G
1068
1221
1374
1526
174
342
513
G84
866
1027
1198
1369
1640
1711
iif
191
572
763
953
1144
1335
1625
1716
lSO7
U
211
428
634
845
1056
1267
1478
169
1901
2112
2
233
466
699
932
1165
1398
1630
18G3
2096
2329
2J
a66
611
767
1022
1278
1684
1789
2045
2300
2656
270
55'9
1897
1676
1957
2515
2794
304
608
912
1217
1521
2129
2488
2737
3042
3i '
33
660
990
1820
1651
1981
2311
2641
2971
8301
867
714
1071
1428
1786
2142
2499
2856
8213
8570
3
386
77
1156
1540
1926
2310
8080
8465
8850
8
■3
414
1242
1656
2070
2484
331 S
3727
4141
444
888
188 S
1777
2221
2665
310e
3653
3096
4442
3
3
47 '5
951
1426
1901
2377
2852
3327
8803
4278
4763
508
1015
1623
2030
2538
3045
8563
4060
4568
5076
/in.
641
1082
1623
2168
2704
8246
8786
4327
4868
5408
^i
675
1150
172a
2801
2876
3461
4026
4601
5177
5752
4i
4
611
1221
1832
3442
8053
8863
4274
4884
5495
6106
647
1294
1941
2588
3285
3882
4529
5176
6823
6470
4
684
186S
2053
2788
3422
4107
4791
5476
6160
6845
723
1446
2169
2892
8615
4338
6061
5784
6507
7231
4
76'3
1525
2288
8061
3818
4576
5338
6101
7626
803
1607
2410
3213
4017
4830
5623
6427
7280
8033
5 m.
845
1690
2534
8379
4224
5069
6914
6758
7603
8448
6
1863
2796
8727
4658
6590
6522
7468
6
102'2
2045
8067
4090
6112
6184
7157
8179
9202
10224
5
6 Q.
ni8
2235
3358
4470
5588
6706
7823
8940
10058
11176
1217
2433
3650
4867
6083
7800
8416
9733
10095
12166
;Tablb of the Weight of a Square Foot of Sheet Iron in lbs. avoirdu
pois, the thickness being the number on the wiregauge. No. 1
is 1*5 of an inch; No. 4, \; No. 11, \, ^e.
No. on wiregauge  1 j 2
3
4 5 1 6 7
8
10
11
Pounds nvoir [125 12
11
10 1 9 8 75
7
6
568
5
No. on wiregauge  12 13
14
16 16[ 17 18
19
20
21
32
Pounds avoir. l46243l
*
8a& 3 25218
193
15
137
b,Google
SPECIFIC GBAVITT.
Table of the Wdylit of a Square Foot of Boiler Plate Iro
^ to 1 inch thick, in lbs. avoirdupois.
ilMH fV M A
i A i\\i flu i\ U\iin.
575]lCI125 15
7 '5
20 226 •J6276 30 1 325 35875 40
Table containing the Weight of Bound Bar Iron, from 1 to 10 f
in length, and from J of an inek to 6 inches diameter.
1
1
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ifi,.
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3f<,^
4fMt.
Efrt.
6 fast.
7f«l.
Btet
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"ib^
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Lbs.
Lbs.
Lb=.
Lh..
2 O'a
05
07
10
12
13
^I!
17
1 07
Il
15
19
22
26
80
34
 13
20
27
40
46
68
60
66
1
21
31
42
62
63
73
83
94
104
1
5 30
45
60
75
00
106
119
134
J49
2
41
a1
81
102
122
142
163
183
203
in.
2
7 5a
80
106
138
159
186
212
239
265
4 67
101
134
lfl8
302
285
269
^2'?
386
4
2 83
12u
167
209
250
292
834
417
a
100
ISl
201
251
301
351
402
452
603
110
179
299
418
478
537
597
14
210
280
351
421
491
661
701
1 163
244
825
406
488
569
660
732
813
B
3 187
280
373
4G7
560
653
747
840
933
2ui.
10
212
818
425
581
637
743
849
956
1063
2J
la
24'0
860
480
599
71 9
889
959
1079
1199
^
IS
4 269
403
638
672
806
941
1076
1310
1344
15
300
449
600
749
899
1018
1198
1818
1498
16
7 334
501
608
835
1001
1168
1836
1502
1669
2
18
3 . 866
549
732
915
1098
1281
1463
1646
1829
2
20
1 402
002
803
1004
1205
1405
1606
1807
2008
2
31
9 439
6o8
878
1097
1817
1586
175Q
1875
2194
Sin.
28
9 478
717
956
1194
1488
1672
1911
2150
2389
3
2S
9 61 9
778
103T
1296' 1550
1815
2074
2593
3
28
661
841
1122
1102 , J682
1968
2243
2584
2804
8
30
2 eo5
907
1210
1612 1 1814
2117
2410
3722
8024
8
82
5 650
975
1800
1626
1951
2270
2601
2926
8251
3
34
1047
1395
1744
2413
2791
3140
3489
3
37
3 747 [ 1120
1498
1867
2240
2618
2987
3360
8738
3
39
9 797 1 1196
1595
1993
2892
2790
8189
8588
3086
4 in.
42
5 849 ' 1274
1699
2128
2548
2972
3397
3823
4210
4
45
2 903 ; 1355
1807
3259
2710
3162
8614
4066
4517
i
48
959 1 1439
1918
2398
2877
3857
3836
4316
4795
4
CO
8 lOlG, 1624
2033
2541
S049 1 8557
4005
4573
5082
4
53
8 107B
lGl8
2160
3226
S768
4301
4838
5376
41
66
8 1130
1704
2272
2B8Q
8407
3975
4643
5111
5679
i
6(
1198
1797
2306
2995
8594
4198
4792
5SB1
6990
4r
1 1262
1893
2524
3165
3780
4417
6048
5678
6309
bm.
61
8 1835
2003
2070
4005
4678
5340
6008
GC75
H
73
2 1463
21S6
2927
3659
4390
5122
5854
6585
7317
n
6ln.
80
3 1606
2109
3212
4016
4818
5621
6434
7227
8030
87
8 1756
2633
3511
4389
6144
702 '2
7000
8778
95
6 1911
2867
3822
4778
5733
6689
7644
8600
9556
Table of the Weight of Cast Iron Plateg, per Superficial Foot, from
oneeighth of an inch to one inch thick.
>im.i,. [ rC''l'
Jitoch.
!^in,.h.
KiMh.
MK^^t. TiiDch.
1 tn»h.
4 i3f 1 a'Voj
14 8
19 sj
24 2S
29 33 ISf
lbs. OS.
88 10=
b,Google
404 THE PRACnCAL MOflEL CALCULATOK.
Table containing the Weight of Cast Iron Pipes, 1 foot in length.
ut
TB.CKNESS m I^CHE^. 1
1
i
i
t
i
1 inch.
li
n 1
Lbl.
Ll«.
Lta.
Lbs.
Lb!,
Lbs.
Ll«
Lbs.
J'
69
99
123
le'i
203
106
147
192
239
124
172
222
276
33'3
393
ai
142
196
26'3
81 '8
87'6
442
511
168
284
35'0
419
491
566
644
180
245
314
46'2
540
70C
198
270
345
42'8
50'5
58'9
^
216
295
87 '6
48^0
54'8
638
235
319
40'7
497
691
68'7
787
6*
253
844
437
584
634
734
842
951
272
468
568
677
785
1012
u
29'0
S9'l
499
607
720
835
1074
H
BO8
417
52 '9
644
762
884
1008
1135
29
44 4
56"
68 3
935
1065
1199
84 5
4b b
591
718
848
1258
HI
49 1
6 1
55
1031
1174
131 9
88
615
00 2
98 4
1080
1228
1881
54
68
8 8
97 7
1129
1284
1442
56 4
86 5
10
1178
1339
1503
5S9
4
901
106 8
1564
bl3
4
93 6
110 b
1276
ie26
13
82
1012
118
1374
1541
1735
14
108 2
1 65
1462
1658
1852
15
95
115 7
135 3
1562
1762
1981
123 3
145 1
1661
187 5
2113
17
130
15 5
1785
1982
2284
IH
19
1
lbl2
169 2
1853
1957
2091
2223
2356
2471
20
1 81
2053
2590
21
2141
2435
2782
■ 22
2230
2548
2854
1 aa
2834
265 5
2983
1 24
2452
2775
8106
Table contit7n7tg tie Weight of Solid Cyl nders of Oa»t Iron, one
foot in length, and from f of an inch to 14 inches diameter.
DliniBWrin
^S""
"'iTh':^'"
™'^e'"
KmrtM in
w.^u.
■"irbtr
w.^..
1
139
^
■ 2048
^
5872
n
14887
188
Bin.
2235
Bin.
6196
15863
247
2420
6
6466
^i
16815
813
2618
5
6831
8i
17908
387
H
2823
5
7100
8J
18900
468
8036
5
7498
9 in.
20077
557
s
8257
5
6
7865
3
21112
654
3485
8195
9
9
22370
769
5
8581
28631
4 m.
3966
6 u.
10 in.
24787
2 in.
991
4
4180
6i
9682
lOi
27327
2
1119
4
4477
6*
10472
29992
2
1254
i
4700
6^
11293
32781
2
■1398
4
5019
12146
85698
1549
4
5271
1\
13
41890
17 08
4
5592
13942
48583
2%
1874
b,Google
SPECIFIC GRAVITY.
Table containing the Weight of a Square Foot of Copper and
Lead, in lbs. avoirdupois, from ^ta ^ an inch in thickness, ad
vancing hy ^.
11.ick„,..
CoiT".
L,id.
A
145
185
290
370
«
435
554
580
739
I + A
726
924
i + A
87X
1108
i + A
1016
1293
J
1161
1477
* + A
1307
1662
i + A
1452
1847
1 +ft
1597
2031
1
1741
2216
i + A
1887
2400
i + A
2032
2585
1 + A
2177
2770
i
2322
2955
Table for finding theWeight of Malleable Iron, Copper, andLead
Pipes, 12 inches long, of various thicknesses, and any diameter
required.
Thiatnca. tl^
ei,M«:r™.
Coiper.
LMd.
. 5^ of an inch.
104
121
1539
A
208
2419
3078
A
3108
3628
4616
414
4838
■6155
! +ft
618
6047
7694
! +1,
621
7258
9232
i +A
725
8466
10771
i
828
9678
1231
Rdle. — Multiply the circumference of the pipe in inches hj the
numbers opposite the thickness rei^uired, and by the length in feet ;
the product will be the weight in avoirdupois lbs. nearly.
Required the weight of a copper pipe 12 feet long, 15 inches in
circumference, J + ^ of an inch in thickness.
•7258 X 15 = 10'817 x 12 = 130644 Iba. nearly.
Table of the Weight of a Square Foot ofMillboardin lbs. avoirdupois.
Ttiokness in incbes
J 1 A J A 1 i
688 j 1032 1 1376 172 j 2064
b,Google
406 THE PRACTICAL MODEL CALCULATOR.
Table containing the Weight of Wrought Iron Bars 12 inches long
in lbs. avoirdupois.
I„th.
,^,„„d. 1 S,a.r,,
Inth,
Kounii.
?4U«a.
■163
■208
2
1632
2080
■467
2
1800
2289
■653
1976
2512
103
130
2
2159
2746
147
1'87
2352
2992
200
255
3
2760
8512
261
a 32
3200
4080
H
331
421
3
8672
4672
i
i08
520
4
4176
5812
494
628
4
4725
6000
588
748
4
5298
6724
680
878
4;
5892
7495
S'OO
1020
5
6528
8320
918
1168
7200
9156
1044
1328
4
7904
10048
2J
11 80
1&00
01
8636
10982
4
1323
lS81
9408
11968
£
1478
1874
7
12800
16320
LEom^
Prouortio
lal Dimen
torts of 6
■sided Na
S for Bolts
\ to 2J inches diameter.
Diameter of bolts
s
1
^
t
1
i
1
11
li
H
a
1
1ft
1!
1*
1
ISi
2f
Breadth over the angles
»
M
1,
1
Ift
IK
2
2S
2A
I'hickness
A
ft
A
f
!
1
11
IJ
1ft
Diameter of bolts
15
li
1
IJ
1*
2
2i
2!
2*
A
2Hi2i
3A
3!
3i
4
Breadth over the angles
2ii
2t
H SA
34
3f
4A
n

Thickness
lA
1«
1« 2
21
2i
2J
2}
Table of the Speeifie Qravity ofWater at different temperatures,
that at 62° being taken as unit^.
70° F,
99913
52= F.
100076
68
50
100087
99958
48
100095
46
100102
G2 1
44
100107
30035
i%
100111
30O50
40
100118
54 1
00064
38
1 00116
The difference of temperatures between G2° and S9°2, where
water attains its greatest density, will vary the bulk of a gallon
rather loss than the third of a cubic inch.
hv Google
SPECIFIC GILAVITY.
from 1 to 12 incheB diameter, advancing hy an eighth.
luthSB.
IM.
Inchaa.
Ll«.
Inchc..
i.b..
■14
ii
1476
8*
8456