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INDUSTRIAL SERIES
SHOP MATHEMATICS
PART I
SHOP ARITHMETIC
PREPARED IN THE
EXTENSION DIVISION OF
THE UNIVERSITY OF WISCONSIN
BY
EARLE B. NORRIS, M. E.
ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING, IN CHARGE
OF MECHANICAL ENGINEERING COURSES IX THE
UNIVERSITY EXTENSION DIVISION
AND
KENNETH G. SMITH, A. B., B. S.
ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING, DISTRICT
REPRESENTATIVE IN CHARGE OF FIRST DISTRICT, THE
UNIVERSITY EXTENSION DIVISION, MILWAUKEE
FIRST EDITION
SECOND IMPRESSION
McGRAW-HILL BOOK COMPANY
239 WEST 39TH STREET, NEW YORK
6 BOUVERIE STREET, LONDON, E. C.
1912
$883
COPYRIGHT, 1912, BY THE
BOOK COMPANY
7113
THE. MAPLE. PRESS- YORK. PA
N n
\r
PREFACE
The aim of this book is to teach the fundamental principles of
mathematics to shop men, using familiar terms and processes,
and giving such applications to shop problems as will maintain
the interest of the student and develop in him an ability to apply
the mathematical and scientific principles to his every day
problems of the shop. The problems and applications relate
largely to the metal working trades. It has, however, been the
aim in preparing this volume not to apply the work to these
particular trades so closely but that it shall be of interest and
value to men in other lines of industry.
This volume presents the first half of the instruction papers in
Shop Mathematics as developed and used by the Extension
Division of the University of Wisconsin. As here offered, it
embodies the point of view obtained through apprenticeship
and shop experience as well as the experience gained through its
use during the past four years as a text for both correspondence
and class room instruction. It is believed that the book will
be found suitable for home study and for use as a text in trade,
industrial, and continuation schools.
The instruction in arithmetic ends with Chapter XII. The
remaining chapters are introduced to give further practice in
calculation and to develop an ability to handle simple formulas,
as well as to impart a knowledge of the principles of machines.
The second volume will take up more fully the use of formulas
and will teach the principles of geometry and trigonometry as
applied to shop work.
The authors are indebted to Mr. F. D. Crawshaw, Professor of
Manual Arts in The University of Wisconsin, for a careful
reading of the proof and for valuable criticisms and suggestions.
E. B. N.
THE UNIVERSITY OF WISCONSIN,
MADISON, Wis.
June 1, 1912.
CONTENTS
CHAPTER I
COMMON FRACTIONS
ART. PAGE
1. Why we use fractions 1
2. Definition of a fraction 2
3. The denominator 2
4. The numerator 2
5. Writing and reading fractions 2
6. Proper fractions 3
7. Improper fractions 3
8. Mixed numbers 3
9. Reduction of fractions 3
10. Reduction to higher terms 4
11. Reduction to lower terms 4
12. Reduction of improper fractions 5
13. Reduction of mixed numbers 6
CHAPTER II
ADDITION AND SUBTRACTION OF FRACTIONS
14. Common denominators 9
15. To find the L. C. D 10
16. To reduce to the L. C. D 10
17. Addition of fractions 11
18. Subtraction of fractions 12
CHAPTER III
MULTIPLICATION AND DIVISION OF FRACTIONS
19. A whole number times a fraction 17
20. "Of" means times 17
21. A fraction times a fraction 18
22. Multiplying mixed numbers 18
23. Cancellation 19
24. Division the reverse of multiplication 21
25. Compound fractions 21
26. How to analyze practical problems 22
CHAPTER IV
MONEY AND WAGES
27. U. S. money 24
28. Addition 25
vii
viii CONTENTS
ART. PAOE
29. Subtraction '. 26
30. Multiplication , 26
31. Division . 26
32. Reducing dollars to cents '27
33. Reducing cents to dollars 28
34. The mill 28
35. Wage calculations 29
CHAPTER V
DECIMAL FRACTIONS
36. What are decimals? 33
37. Addition and subtraction 35
38. Multiplication 36
39. Short cuts 37
40. Division 37
41. Reducing common fractions to decimals 39
42. Complex decimals 40
43. The micrometer 40
CHAPTER VI
PEKCENTAGE
44. Explanation 44
45. The uses of percentage 46
46. Efficiencies 47
47. Discount . .' 47
48. Classes of problems 48
CHAPTER VII
CIRCUMFERENCES OF CIRCLES: CUTTING AND GRINDING SPEEDS
49. Shop uses 51
50. Circles 51
51. Formulas 52
52. Circumferential speeds 54
53. Grindstones and emery wheels 55
54. Cutting speeds 57
55. Pulleys and belts 58
CHAPTER VIII
RATIO AND PROPORTION
56. Ratios 59
57. Proportion '.'... 60
58. Speeds and diameters of pulleys 63
59. Gear ratios 64 '
CONTENTS ix
CHAPTER IX
PULLEY AND GEAR TRAINS; CHANGE GEARS
Am. J'ACib;
60. Direct and inverse proportions 65
61. Gear trains 60
62. Compound gear and pulley trains 68
63. Screw cutting 72
CHAPTER X
AREAS AND VOLUMES OF SIMPLE FIGURES
64. Squares 75
65. Square root 75
66. Cubes and higher powers 75
67. Square measure 76
68. Area of a circle 77
69. The rectangle 79
70. The cube 80
71. Volumes of straight bars 80
72. Weights of metals 82
73. Short rule for plates 83
74. Weight of casting from pattern 83
CHAPTER XI
SQUARE ROOT
75. The meaning of square root 85
76. Extracting the square root 86
77. Square roots of mixed numbers 87
78. Square roots of decimals 87
79. Rules for square root 88
80. The law of right triangles 89
81. Dimensions of squares and circles 91
82. Dimensions of rectangles 91
83. Cube root 92
CHAPTER XII
MATHEMATICAL TABLES (CIRCLES, POWERS, AND ROOTS)
84. The value of tables 94
85. Explanation of the tables 95
86. Interpolation 96
87. Roots of numbers greater than 1000 5)7
88. Cube roots of decimals ! 98
89. Square root by the table 99
x CONTENTS
CHAPTER XIII
LEVERS
ART. I'A<;K
90. Types of machines 115
91. The lever 115
92. Three classes of levers 117
93. Compound levers 118
94. Mechanical Advantage 119
95. The wheel and axle 120
CHAPTER XIV
TACKLE BLOCKS
96. Types of blocks . . . , 123
97. Differential pulleys 126
CHAPTER XV
THE INCLINED PLANE AND SCREW
98. The use of inclined planes 130
99. Theory of the inclined plane 130
100. The wedge 132
101. The jack screw 133
102. Efficiencies 134
CHAPTER XVI
WORK POWER AND ENERGY; HORSE-POWER OF BELTING
103. Work 137
104. Unit of work . . . 137
105. Power 138
106. Horse-power of belting 139
107. Widths of belts 140
108. Rules for belting 141
CHAPTER XVII
HORSE-POWER or ENGINES
109. Steam engines 145
110. Gas engines 147
111. Air compressors 149
112. Brake horse-power 11!)
113. Frictional horsc-pbwer 151
114. Mechanical efficiency 151
CONTENTS xi
CHAPTER XVIII
MECHANICS OF FLUIDS
ART. PAGE
115. Fluids 153
116. Specific gravity 153
117. Transmission of pressure through fluids 154
118. The hydraulic jack 155
119. Hydraulic machinery 158
120. Hydraulic heads 158
121. Steam and air 159
CHAPTER XIX
HEAT
122. Nature of heat 163
123. Temperatures 164
124. Expansion and contraction 168
125. Allowances for shrink fits 171
HAPTER XX
STRENGTH OF MATERIALS
126. Stresses 173
127. Ultimate stresses 174
128. Safe working stresses 174
129. Strengths of bolts 175
130. Strengths of hemp ropes 177
131. Wire ropes and cables 177
132. Strengths of chains 177
133. Columns . .178
SHOP ARITHMETIC
CHAPTER I
COMMOX FRACTIONS
1. Why We Use Fractions. When we find it necessary to deal
with things that are less than one unit, we must use fractions.
A machinist cannot do all his work in full inches because it is
generally impossible to have all measurements in exact inches.
Consequently, for measurements less than 1 in., he uses fractions
of an inch; he also makes use of fractions for measurements
between one whole number of inches and the next whole number.
If a bolt is wanted longer than 4 in. but shorter than 5 in., it
would be 4 in. and a fraction of an inch. This fraction of an
inch might be nearly a whole inch or it might be a very small part
of an inch. The system used to designate parts of a unit is
FIG. 1.
easily seen by looking at a machinist's scale or at a foot-rule of
any sort. Each inch on the scale is divided into a number of
equal parts. A wooden foot-rule usually has eight or sixteen
parts to each inch, while a machinist's steel scale has much, finer
divisions. Now, if we want to measure a piece of steel which is
not an inch long, we hold a scale against it, as dn Fig. 1, and find
out how many of these divisions of an inch it takes to equal the
length of the piece. The scale in Fig. 1 is 3 in. long and each
inch is divided into eight parts. We see that this piece 'is as
1
2 SHOP ARITHMETIC
long as five of these eight parts of an inch, or we say that it is
"five-eighths " of an inch long.
2. Definition of a Fraction. A Fraction is one or more of the
equal parts into which anything may be divided. Every fraction
must contain two numbers, a numerator and a denominator.
These are called the terms of a fraction.
3. The Denominator. The Denominator tells into how many
equal parts the unit is divided. In the case shown in Fig. 1,
1 in. was the unit and it was divided into eight equal parts.
The denominator in this case was eight.
4. The Numerator. The Numerator shows how many of these
parts are taken. In giving the length of the piece of steel in
Fig. 1, we divided the inch into eight parts and took five of
them for the length. Five is the numerator and eight is the
denominator.
5. Writing and Reading Fractions. In writing fractions, the
numerator is placed over the denominator and either a slanting
line, as in 5/8, or a horizontal line, as in f , drawn between them.
The horizontal line is the better form to use, as mistakes are
easily made when a whole number and a fraction with a slanting
line are written close together.
-j is read one-fourth or one-quarter.
-= is read one-half.
2 is read three-fourths or three-quarters.
-= is read five-eighths.
o
3
= is read three-sevenths.
We can have fractions of all sorts of things besides inches.
An hour of time is divided into sixty equal parts called minutes.
A minute is merely -fa of an hour. Likewise, 20 minutes is f $
of an hour. In the same way, 1 second is -^ of a minute.
In the early days, before we had the unit called the inch, the
foot was the common unit for measuring lengths. When it was
necessary to measure lengths less than 1 ft., fractions of a foot
COMMON FRACTIONS 3
were used. This got to be too troublesome, so one-twelfth of a
foot was given the name of inch to avoid using so many
fractions. For instance, where formerly one said f\ of a foot,
we can now say 5 in. This shows how the use of a smaller unit
reduces the use of fractions. In Europe, a unit called the
millimeter is used in nearly all shop work. This is so small, being
only about T T of an inch, that it is seldom necessary in shop
work to use fractions of a millimeter.
6. Proper Fractions. If the numerator and denominator of a
fraction are equal, the value of the fraction is 1, because there
are just as many parts taken as there are parts in one unit.
4_! 8 10
4~ 8~ 10"
In each of these cases, the numerator shows that we nave taken
the full number of parts into which the unit has been divided.
Consequently, each of the fractions equals a full unit, or 1.
A Proper Fraction is one whose numerator is less than the
denominator. The value of a proper fraction, therefore, is
always less than 1.
3 5 7 27
A' T' Q' QO are a ^ P r P er fractions.
4 ID o ou
7. Improper Fractions. An Improper Fraction is one whose
numerator is equal to or larger than the denominator. There-
fore, an improper fraction is equal to, or more than 1.
24 14 17 64
To' IP T' 7 are a ^ im proper fractions.
1 _. o It) C)4
8. Mixed Numbers. A Mixed Number is a whole number and
a fraction written together: for example, 4 is a mixed number.
4 is read four and one-half and means four whole units and
one-half a unit more.
9. Reduction of Fractions. Quite often we find it desirable to
change the form of a fraction in order to make certain calcula-
tions; but, of course, the real value of the fraction must not be
changed. The operation of changing a fraction from one form
to another without changing its value is called Reduction.
By referring to the scale in Fig. 2 it will be seen that, if we
take the first inch and divide it into 8 parts, each \ in. will con-
SHOP ARITHMETIC
tain 4 of these parts. Hence, ^ in. = in. In this case, we
make the denominator of the fraction 4 times as large, by making
4 times as many parts in the whole. It then takes a numerator
4 times as large to represent the same fractional part of an inch.
This relation holds whether we are dealing with inches or with
any other thing as a unit.
FIG. 2.
10. Reduction to Higher Terms. When we raise a fraction to
higher terms, we increase the number of parts in the whole, as
just shown, and this likewise increases the number of parts
taken. Therefore, the numerator and denominator both become
larger numbers.
1 . 4.
- m. = in.
A
16
10.
= 32 m '
A fraction is raised to higher terms by multiplying both numer-
ator and denominator by the same number.
Examples :
1X4
Similarly,
-!= 5X2 = 10
16 16X2 32
2
Suppose we want to change y^ of an inch to 64ths. To get 64 for the
denominator, we must multiply 16 by 4 and, therefore, must multiply 3 by
the same number.
3 = 3X4 ^12
lo = 16X4 64
11. Reduction to Lower Terms. When we reduce a fraction
to lower terms, we reduce the number of parts into which the
whole unit is divided. This likewise reduces the number of
parts which are taken.
4 . 1 . 2 . 1 .
A fraction is reduced to lower terms by dividing both numer-
ator and denominator by the same number. When there is no
COMMON FRACTIONS 5
number -which will exactly divide both numerator and denomina-
tor, the fraction is already in its lowest terms.
Example :
Reduce ._ to its lowest terms.
36 36 4-2^18-4-2^ 9
128 128-5-2 64-J-2 32
There is no number that will exactly divide both 9 and 32 and, therefore,
the fraction is reduced to its lowest terms.
12. Reduction of Improper Fractions. When the numerator
of a fraction is just equal to the denominator, we know that the
value of the fraction is 1 (see Art. 6) :
8 = 1 64 = 1 !? = i
8 64 10
In each of these cases we have taken the full number of parts
into which we have divided the unit. Consequently, each of
these fractions is one whole unit, or 1.
When the numerator is greater than the denominator, the
value of the fraction is one or more units, plus a proper fraction,
or a whole plus some part of a whole.
Examples :
12 8.4 .4 ,1
-8- = 8 + 8 = 1 8 rl 2
47 = 36 11 11
12 12 + 12 12
From these examples we may see that to reduce an improper
fraction to a whole or mixed number the simplest way is as
follows:
Divide the numerator by the denominator. The quotient will
be the number of whole units. If there is anything left over, or
a remainder, write this remainder over the denominator since
it represents the number of parts left in addition to the whole
units. We now have a mixed number, or an exact whole number,
in place of the improper fraction.
Examples :
27 _ 27 . 7 _Q 6 7)27(3
y =27 ^ 7=3 7 21
6
45 3 1 6)45(7
-6 =45 ^ 6 = 7 6 2 f 2
3
6 SHOP ARITHMETIC
These show that a fraction represents unperformed division.
In fact, division is often indicated in the form of a fraction. The
numerator is the dividend and the denominator is the divisor.
24
24 -f- 3 can be written
o
2
2-7-8 can be written =
13. Reduction of Mixed Numbers. It is often necessary or
desirable to change mixed numbers to improper fractions. The
method of doing this may be seen from the following examples.
Examples :
Reduce 5- to an improper fraction.
4-+J
In one unit there are two halves. Therefore,
, 5X2 10
1 = 10 1 = 11
2 22 2
If 7 ^ were to be reduced to an improper fraction we would say: "Since
there are 4 fourths in 1, in 7 there are 4X7, or 28 fourths. 28 fourths
plus 1 fourth equals 29 fourths."
1 = 28 1 = 29
4~ 4 4~ 4
The rule which this gives us is very simple: Multiply the whole
number by the denominator of the fraction and write the prod-
uct over the denominator. This reduces the whole number to
a fraction. Add to this the fractional part of the mixed number.
The sum is the desired improper fraction.
In working problems like the above, the work should be
arranged as in the following example.
Example :
Reduce 5 5 in. to eighths of an inch,
o
_ 1 40 1 41
5 H = ~Q~ + fi = ^~' Answer -
o o o o
COMMON FRACTIONS 7
USEFUL TABLES
Measures of Length
12 inches (in.) = 1 foot (ft.)
3 ft. or 36 in. = 1 yard (yd.)
5J yd. or 164 ft. = 1 rod (rd.)
320 rd. or 5280 ft. = 1 mile (mi.)
Measures of Time
60 seconds (sec.) =1 minute (min.)
60 minutes =1 hour (hr.)
24 hours =1 day (da.)
7 days =1 week (wk.)
365J days =; 1 average year (yr.)
100 years = 1 century
Note. Thirty days are generally considered as one month,
though the number of days differs for different months.
Miscellaneous Units
12 things =1 -dozen (doz.)
12 dozen or 144 things = 1 gross (gr.)
12 gross = 1 great gross
20 things = 1 score
QUESTIONS AND PROBLEMS
1. What is a fraction?
2. Name some fractions of an inch commonly used.
3. Write the following as fractions or mixed numbers.
Five-sixteenths
Nine thirty-seconds
Twenty and one-eighth
Twenty-one eighths
Three and three-fourths
4. Write out in words the following:
ol 5 7 8 5 1 21
3 6' 8' 16' 8' 24' 2 ~4' 4
6. Indicate the proper fractions, the improper fractions, and the mixed
numbers among the following:
3 _1 16 21 5 7 9
4' 6 ~&' 16' 16' ^e' 8' 16
6. Change yg of an inch to eighths of an inch.
4
Change of an inch to fourths of an inch.
8 SHOP ARITHMETIC
g
7. How many sixteenths of an inch in j in.?
3
How many thirty-seconds of an inch in -r in.?
13 7 33
8. Which is greater, -r^ in. or n in.? 77j or To ?
9. Reduce the following mixed numbers to improper fractions:
,1 ^1 o3 ! ,1
V V V V '16
10. Reduce the following improper fractions to whole or mixed numbers:
21 8 24 7 121
16' 8* 3' 2' 12
11. I want to mix up a pound of solder to be made of 5 parts zinc, 2
parts tin, and 1 part lead. What fraction of a pound of each metal zinc,
tin and lead must I have?
12. If a train is running at the rate of a mile a minute, how many feet
does it go in 1 second?
7
13. An apprentice who is drilling and tapping a cylinder for ^ in. studs,
3 1
tries a -7 in. drill but the tap binds, so he decides to use a drill TTT in. larger.
W 7 hat size drill does he ask for?
14. The tubes in a certain boiler are 15 ft. 11 in. long. How many inches
long are they?
15. How many seconds in an hour? 40 seconds is what fraction of an
hour?
16. An 8-ft. bar of steel is cut up into 16 in. lengths. AVhat fraction of
the whole bar is one of the pieces?
17. When a man runs 100 yd. in 10 seconds, how many feet does he go
in 1 second?
18. Wood screws are generally put up in boxes containing one gross.
If 36 screws are taken from a full box for use on a certain job, what fraction
of the gross is used on this job and what fraction is left in the box? Reduce
both fractions to their lowest terms.
19. A steel plate 2 ft. 6 in. wide is to be sheared into four strips of equal
width. How wide will each strip be in inches?
20. In one plant all drawings are dimensioned in inches, while in another
all dimensions above 2 ft. are given in feet and inches. If a dimension is
given as 89 in. in the first plant, how would the same dimension be stated
in the other plant?
CHAPTER II
ADDITION AND SUBTRACTION OF FRACTIONS
14. Common Denominators. Fractions cannot be added
unless they contain the same kind of parts, or, in other words,
have the same denominator. When fractions having different
denominators are to be added, they must first be reduced to frac-
tions having a common denominator. A number of fractions are
said to have a common denominator when they all have the
same number for their denominators, f and cannot be added
as they stand, any more than can 3 bolts and 5 washers. Both
the fractions must be of the same kind, that is, must have the
same denominator, may be changed to f . By making this
change, the fractions are given a common denominator and can
now be added. 6 eighths plus 5 eighths equals 1 1 eighths, in just
the same manner as 6 inches plus 5 inches equals 11 inches.
The work of this example would be written as follows:
4X2~8
6 5 11 ,3
+ = ~~ =1 > Answer -
8 is called the Least Common Denominator (L. C. D.) of f
and f , because it is the smallest number that can be used as a
common denominator for these two fractions. In this case, the
least common denominator is apparent at a glance; in many
other cases it is more difficult to find, especially if there are several
fractions to be added. In the case just given, the denominator
of one fraction can be used for the common denominator. When
we have two denominators like 5 and 8, neither of them is an
exact multiple of the other number, and so neither can be the
common denominator. In such a case, the product of the two
numbers can be used as a common denominator.
9
10 SHOP ARITHMETIC
Example :
5X8 = 40, theL. C. D.
3^3X8^24
5 5X8 40
5^5X5^25
8~8X5~40
24 25 49 9
We can always be sure that the product of the denominators
will be a common denominator, to which all the fractions can be
reduced, but it will not always be the least common denominator.
For example, if we wish to add T 5 ^- and -j^-, we can use 12X16 =
192 for the common denominator, but we readily see that 48
will serve just as well and not make the fractions so cumbersome.
In this case 48 is the least common denominator.
15. To Find the L. C. D. If the L. C. D. cannot be easily seen
by examining the denominators, it may be found as follows:
Suppose we are to find the L. C. D. of \, f , f , and T \- First
place the denominators in a row, separating them by commas.
2)4, 3, 9, 16
2)2, 3, 9, 8
3)1,3,9, 4
1, 1, 3, 4
L. C. D. =2X2X3X3X4 = 144
Select the smallest number (other than 1) that will exactly
divide two or more of the denominators. In this case, 2 will
exactly divide 4 and 16. Divide it into all the numbers that are
exactly divisible by it, that is, may be divided by it without
leaving a remainder. When writing the quotients below, also
bring down any numbers which are not divisible by the divisor
and write them with the quotients. Now proceed as before,
again using the smallest number that will divide two or more
of the numbers just obtained. Continue this process until
no number (except 1) will exactly divide more than one of the
remaining numbers. The product of all the divisors and all the
numbers (except 1's) left in the last line of quotients is the Least
Common Denominator.
16. To Reduce to the L. C. D. Having found the least common
denominator of two or more fractions, the next step is to reduce
ADDITION AND SUBTRACTION OF FRACTIONS 11
the given fractions to fractions having this least common
denominator. Let us take $, f, , and We first find the
L. C. D., which turns out to be 120. We next proceed to reduce
the fractions to fractions having the L. C. D. Divide the com-
mon denominator by the denominator of the first fraction.
Multiply both numerator and denominator of the fraction by
the quotient thus obtained. Do this for each fraction, as illus-
trated here.
2)3, 5, 4, 8
2)3, 5, 2, 4
3, 5, 1, 2
L.C.D. =2X2X3X5X2 = 120
1 1X40 40
3 3X40 120
120-i-3 =
120.5 = 24 rsx24 - 120
120.4 = 30 i= 1X3 30
120.8 =
4 4X30 120
3 = 3X15^ 45^
8~~8X15~ 120
17. Addition of Fractions. Addition of fractions is very
simple after the fractions have been reduced to fractions with
a common denominator. Having done this it is only necessary
to add the numerators and place this sum over the common
denominator. The sum should always be reduced to lowest
terms and if it turns out to be an improper fraction it should
be reduced to a mixed number.
Example :
Find the sum of _5^ 3 9 7
16 4 32 32
Common denominator =32
.
16~16X2~32
3 3X8 = 24
4 4X8 32
10 24 _9_ 7 = 50
:u 32 32 32 32
50 25 .9
If there are mixed numbers and whole numbers, add the whole
numbers and fractions separately. If the sum of the fractions
12 SHOP ARITHMETIC
is an improper fraction, reduce it to a mixed number and add
this to the sum of the whole numbers.
Example :
How long a steel bar is needed from which to shear one piece
each of the following lengths:
1731
7 S in., 5^ in., 4 7 in., 65 in. ?
2 o 4 o
7 1 4
? 2 8
5 7 7
Explanation: The sum of the whole numbers is 22.
,3 6 18
4 7 o The sum of the fractions is -^-t which reduces to 2-r-
o 4
A* 1 Adding this to the sum of the whole numbers (22),
8 8 1
gives 24 -r as the sum of the mixed numbers. Hence
22 18 ,-.1 4
== 2 ~~ 1
2\ 84 we mus t have a bar 24- inches long.
4
24-r' Answer.
4
18. Subtraction of Fractions. Just as in addition, the fractions
must first be reduced to a common denominator. Then we can
subtract the numerators and write the result over the common
denominator.
Example :
Subtract - from -=-
o 16
Common denominator = 16
5 = 10
8 16
15 10 5
-T^=TS' Answer.
16 16 16
In subtracting mixed numbers, subtract the fractions first
and then the whole numbers.
Example :
in. long?
1
How much must be cut from a 15 in. bolt to make it
L. C. D. =16
7^' Answer.
Ib
ADDITION AND SUBTRACTION OF FRACTIONS 13
Sometimes, in subtracting mixed numbers, we find that the
fraction in the subtrahend (the number to be taken away) is
larger than the fraction in the minuend (the number from which
the subtrahend is to be taken) . In this case, we borrow 1 from
the whole number of the minuend and add it to the fraction of
the minuend. This makes an improper fraction of the fraction
in the minuend and we can now subtract the other fraction
from it.
Example :
3 i
Take 9 7 from 12 5
4 8
19 3
12 5 = 11 5 Explanation: -7 cannot be substracted from
o o 4
1 / 8\
Q 3 = 6 -, so we borrow 1 (or^) from 12 and write the
4 _8 9
3 minuend 11~-
2 5 > Answer.
o
( or^j
If the minuend happens to be a whole number, borrow 1 from
it and write it as a fractional part of the minuend. Then sub-
tract as before.
Example :
10-4-T
-r^> Answer.
ID
PROBLEMS
53 9 10 24 9
21. Reduce to the L. C. D. jg j> and ^H- Answer, ^ oo Q-X-
22. Reduce to the L. C. D. | |. ^> and ~-
4 o 1U lo
, 10 11 , 15 109 _13
23. Add 25, y^ and ^- Answer, -^ = 2^-
24> + ++" 7
14
SHOP ARITHMETIC
25. Add 2g> 5^ and 7^-
15 3
26. Find the sum of 8, 3o 4g> and JTT-
4 7
27. Subtract from -
3 1
29. Find the difference between 13j and 202"
30. 15-ll|=?
o
31. The weights of a number of castings are: 412^ lb., 270^ lb., 1020 lb.,
75^ lb., 68^ lb. What is their total weight?
32. Four studs are required: 2- in., lg in., 2-r^ in., and loo in. long; how
3
long a piece of steel will be required from which to cut them allowing -7 in.
altogether for cutting off and finishing their ends?
33. Monday morning an engineer bought 48n gallons of cylinder oil;
2
on Monday, Tuesday, and Wednesday he usedj gallon per day; on Thurs-
7 1
day he used g gallon; and on Friday H gallon. How much oil had he left
on Saturday?
94--
_ 2 i-l"_
"8
Fia. 3.
34. Find the total length of the roll shown in the sketch in Fig. 3.
35. A piece of work on a lathe is 1 ft. in diameter; it is turned down in
3
five cuts; in the first step the tool takes off 09 in- from the diameter; then
jg in. ; then oo m - > then ^o m -> an< ^ the fifth time gj in. What is the diam-
eter of the finished piece?
36. How long must a machine shop be to accommodate the following
machines installed in a single line: lathe, 8?> ft. long; planer, 14^ ft. long;
17 1
milling machine, 4g ft. long; engine, 7o ft. long; tool room, 12g ft. long?
Allow 3j ft. between a wall and a machine, and 3^ ft. between two machines.
The tool room is to be placed at the end of the shop.
37. In doing a certain piece of work one man puts in lq hours, a second
man ?> hour, a third works 2~- hours, and a fourth man works lj hours.
How long would it take one man to do the work?
38. By mistake, the draftsman omitted the thickness of the flange on the
drawing of a gas engine cylinder in Fig. 4. From the other dimensions
given, calculate the thickness of the flange.
-J-
fr*
1C
Fia. 4.
39. A millwright has to rig up temporarily a 6 in. belt to be 367g in. long.
In looking over the stock of old belting he finds the following pieces of the
17 3
right width; one piece 126^ in. long, one 142g in. long, and one 133g in. long.
16 SHOP ARITHMETIC
How many inches must be cut from one of the pieces so that these pieces can
be laced together to give the right length?
40. The time cards for a certain piece of work show 2 hours and 15
minutes lathe work, 3 hours and 10 minutes milling, 1 hour and 10 minutes
planing, and 1 hour and 15 minutes bench work; what is the total number of
hours to be charged to the job?
CHAPTER III
MULTIPLICATION AND DIVISION OF FRACTIONS
19. A Whole Number Times a Fraction. In the study of
multiplication, we learn that multiplying is only a short way of
adding. 4x7 is the same as four 7's added together. Either
4X7, or 7+7+7+7 will give 28. If we apply this same principle
to the multiplying of fractions, we see that 4 X | is the same as
four of these fractions added together.
4X 7 = V + V = 28
A 8 8^8^8^8 8
This shows that multiplying a fraction by a whole number is
performed by multiplying the numerator by the whole number
and placing the product over the denominator of the fraction.
In other words, the size of the parts is not changed, but the
number of parts is increased by the multiplication. After
multiplying, the product should be reduced to lowest terms and,
if an improper fraction, should be reduced to a whole or mixed
number.
Example :
What would be the total weight of 12 brass castings each
weighing f of a pound?
f\ *3f\
12X~=-r=9 lb.. Answer.
4 4
20. "Of" Means "Times." The word "of" is often seen in
problems in fractions, as for instance, "What is of 5 in.?"
In such a case, we work the problem by multiplying, so we say
that " of" means " times." You can see that this is so by taking
a piece of wood 5 in. long and cutting it into four equal parts
and then taking three of these parts. These three parts will be
| of 5 in., and by actual measurement will be 3J in long, so we
know that of 5 = 3f . Now see what times 5 is'
3 vr 15 3
TXo = - r =3j
4 44
which is the same value. Therefore, we see that the word
"of" in such a case signifies multiplication.
2 17
18 SHOP ARITHMETIC
21. A Fraction Times a Fraction. To multiply two or more
fractions together, multiply the numerators together for the
numerator of the product and multiply the denominators together
for the denominator of the product.
Example :
Multiply gxf-
o o
7 2_14_ 7 Explanation: The numerator of the product is
8 3~24~T2 obtained from multiplying the numerators to-
gether: 7X2 = 14. The denominator of the
product, in the same manner, is 8 X 3 = 24. This
14 7
gives the product,, - which can be reduced to j-_-
24 12
Let us see what multiplication of fractions really means,
and why the work is done as just shown. Suppose we are to
find | of | in. This means that of an inch is to be divided into
4 equal parts and 3 of these parts are wanted. If we divide
| in. into 4 equal parts, each part will be one-fourth as large as
| in. and, therefore, can be considered as being made up of
7 parts, each one-fourth as large as | in. Then \ of f =-jV-
Three of these parts will naturally contain three times as many
thirty-seconds, or - 8 ^- = f^. Therefore:
? 7 3 7 3X7 21
4 8~4 X 8~4X8~32'
22. Multiplying Mixed Numbers. This is one of the most
difficult operations in the study of fractions, unless one adopts
a fixed rule and follows it in all cases. The student will have no
trouble if he will first reduce the mixed numbers to improper
fractions, and then multiply these like any other fractions.
Example :
Find the product of 3^ and 2^-
1 13 1 5
113 513X565 1
~~ ~*' *
To multiply a mixed number by a whole number, we can
reduce the mixed number to an improper fraction and then
multiply it; or we can multiply the fractional part and the
whole number part separately by the number and then add
the products.
MULTIPLICATION AND DIVISION OF FRACTIONS 19
Example :
3
What would be the cost of ten J in. by 6 in. machine bolts at 1 Q
o
cents a piece?
First method:
10x1^ = 10 X-Q- =- 8 -= 13x = 13 r cents.
Second Method:
10 3
Explanation: First multiply 10 by 5- This gives
8 30 6 3
. or 3 6 . or 3.- Set this down. Then multiply
6 o o 4
i 3
4 lObyl. This gives 10, and we add this to the 3
l.'V* cents, Answer, giving a total of 13^>
23. Cancellation. Very often the work of multiplying frac-
tions may be lessened by cancellation, as it avoids the necessity
of reducing the product to lowest terms. To get an idea of
cancellation we must first understand what a "factor" is. A
Factor of a number is a number which will exactly divide it.
Thus, 2 is a factor of 8, 3 is a factor of 27, 5 is a factor of 35, etc.
When the same number will exactly divide two or more numbers
it is called a common factor of those numbers. Thus, 2 is a
common factor of 8 and 12, because it will divide both 8 and 12
without leaving a remainder. 4 is also a common factor of 8
and 12. Similarly, 7 is a common factor of 14 and 21.
This idea of common factors we have already used in reducing
fractions to lowest terms. Thus, when we have -fa we divide
both 8 and 12 by 4 and get
8 = 8-h4 = 2
12 12+4 3
Cancellation is a process of shortening the work of reduction by
removing or cancelling the equal factors from the numerator
and denominator.
Example :
Suppose we have several fractions to multiply together, as
?x 2 x 3 x 21 -
4*3 X 14 X 32
,, . . . 3X2X 3 X21 378
The,r product is --
20 SHOP ARITHMETIC
This is not in its lowest terms so we divide both numerator and denominator
by 2, 3, and 7, and get
378-^-2 189-i-3 63-r-7 9
5376^2 2688*3 ~ 896 -h 7 ~ 128
Now, if we had struck out the common factors from the numerator and
denominator before multiplying the fractions, we would have shortened the
work and our answer would have been in its lowest terms without reducing.
Thus:
1 1 3
9
2X1X2X32 128
2 1 2
Explanation: First the 3 in the numerator is cancelled with the 3 in
the denominator. This merely divides the numerator and denominator
by 3 at the outset instead of waiting until the terms are all multiplied
together; and, as 3 -f-3 = 1, we cancel a 3 from both numerator and denomin-
ator and place 1's in their stead. Next we divide both terms by 2. The
gives 1 in place of the 2 in the numerator and 2 in place of the 4 in this
denominator. Next we see that 7 is a common factor of the numerator and
denominator, so we divide the 21 and 14 each by 7 and place 3 and 2 in
their places. There are no more common factors; so we multiply together
9
the numbers we now have and get -=-==
iZo
Another Example :
1
2465
= 120
"
Explanation: First we cancel 250 out of 500 and 250; and then 9 out of 36
and 63; then 7 out of 42 and 7; then 10 out of 50 and 20; and finally 2 out
of 2 and 2. This removes all the common factors and we get 120 for the
answer.
PROBLEMS IN MULTIPLICATION
41. 7X^ = ?
42. 8 X g = ?
23 5
43. Multiply ^ by Tg'
3 5
44. Find the product of j and
45. What is % of 2^?
o o
46. What is | of | of ^ of |?
MULTIPLICATION AND DIVISION OF FRACTIONS 21
24. Division The Reverse of Multiplication. Division is just
the opposite of multiplication and this fact gives us the cue to a
very simple method of dividing fractions.
To divide one fraction by another, invert the divisor and then
multiply. To invert means to turn upside down. Invert f
and we get f ; invert and we get .
Example :
TV -J 27 U 3
Divide 09 by--
9
27^3 = ?J * = 9 J
8
3 4
Explanation: The divisor is -: Inverting this gives - In multiplying,
9 1
we make use of cancellation to simplify the work, and we get - or 1 for
o o
the result.
Suppose we have a fraction to divide by a whole number; as
-h2 = ?
Therefore,
14 2 1
4- 2 = ? 2 is the same as n If we invert this we get ?
lu 1 2i
7
14 2 Ml 7
-
25. Compound Fractions. Sometimes we see a fraction which
has a fraction for the numerator and another fraction for the
denominator. This is called a Compound Fraction. If we
remember that a fraction indicates the division of the numerator
by the denominator, we will see that a compound fraction can be
simplified by performing this division.
Example :
27
W 1 i;it i s iii ?
T
4
27 3
This means the same as --*- and, therefore, would be solved as follows:
' 4
27 9 1
27 *
x
8 1
32 27 3 27 * 9 ,1
--- x - 1 Answer -
22 SHOP ARITHMETIC
PROBLEMS IN DIVISION
47. Divide 175 by |-
15 7
48. Divide yg by ITT-
49. Divide s| by 5g-
60 27-A_ ?
&0< 32^10"'
7
61. Find the quotient of 21 -J-K-
62. - = ?
26. How to Analyze Practical Problems. The chief trouble
that students have in working practical problems is in analyzing
the problems to find out just what operations they should use to
work them. Problems in multiplication or division of fractions
will fall in one of the three following cases:
1. Given a whole; to find a part (multiply).
2. Given a part; to find the whole (divide).
3. To find what part one number is of another (divide).
Example of Case I :
The total weight of a shipment of steel bars is 3425 Ib. ^ of this consists
&O
3 1
of j in. round bars and the balance is ?> in. round. What weight is there
of each size?
137 Explanation : In this example we
7 3>f2S _., .3. , have the whole (3425 Ib.) ; to find a
^ X ~ =959 Ib. of j in. bars. / 7 \
* part \-~-A If the whole is 3425, then
3425 -959 = 2466 Ib. of ~ in. bars. 7 V '
~o{ 3425 = 959 Ib. The balance, which
or ^o -
137 1
1 8 3423 1 consists of in. bars, will be 3425 959,
^ X^^ =2466 Ib. of 5 in. bars.
rv 1 * ., M1 , 1 7 ^o 7 lo ,,,
or it will be l-25=25-25 = 25 ofthe
18
whole, g of 3425 = 2466 Ib.
Example of Case 2 :
3
The base of a dynamo weighs 270 Ib.; the base is yr of the total weight;
find the total weight.
MULTIPLICATION AND DIVISION OF FRACTIONS 23
4- of the whole = 270 lb. Explanation: Here we have a
part given to find the whole, -jy
whole = 270 H- jj- of the whole is 270 j b Thig means
90 that if the whole machine were
3 270 11 divided into eleven equal parts,
270 -5-jj = -y- X-^ =990 lb., Answer. three of these parts together would
weigh 270 lb. Then one part would
weigh 270 H- 3 =90 lb. Since there
are 11 of these equal parts, the
whole machine weighs
HX90 = 9901b.
This is the same as dividing 270 by
g
the fraction vy
Example of Case 3 :
A molder who is on piece work sets up 91 flasks, but the castings from
7 of them are defective. What fractional part of his work does he get
paid for?
91 7 = 84 sound castings. ' Explanation: The problem is: What part
84 84-*-? 12 of 91 is 84? There are 91 parts in his whole
= ~ = TT Answer. g4
work and he gets paid for 84 parts, or Q-T of
12
the whole. This can be reduced to-pr; which
is the answer.
PROBLEMS
2
53. A gallon is about JF of a cubic foot. If a cubic foot of water weighs
62^ lb., how much does a gallon of water weigh?
2 4
64. Aluminum is 2^ times as heavy as water; and copper is 8v times as
heavy as water. Copper is how many times as heavy as aluminum?
66. If a certain sized steel bar weighs 2-= lb. to the foot, how long must a
piece be to weigh 8j lb.?
66. What is the cost of a casting weighing 387 lb. at 4j cents a pound?
67. How many steel pins to finish lg in. long can be cut from an 8 ft.
3
rod if we allow -jg in. to each pin for cutting off and finishing?
68. A certain piece for a machine can be made of steel or of cast iron.
3 1
If drop forged from steel it would weigh 7j lb. and would cost 6g cents
per pound. If made of cast iron, it would have to be made much hetvier
24 SHOP ARITHMETIC
and would weigh 14 Ib. and cost 2g cents per pound. Which would he the
cheaper and how much?
69. I want to measure out 2-r gallons of water, but I have no measure at
hand. However, there are some scales handy and I proceed to weigh out
7
the proper amount in a pail that weighs ITT Ib. What should be the total
weight of the pail and the water, if one gallon of water weighs 8^ Ib. ?
60. I want to cut 300 pieces of steel, each 112 in. long for wagon tires.
I have in stock a sufficient number of bars of the same size, but they are
120 in. long; and I also have a sufficient number 235 in. long. Which length
should I use in order to waste the least material? Calculate the total num-
ber of inches of stock that would be wasted in each case.
CHAPTER IV
MONEY AND WAGES
27. U. S. Money. Nearly every country has a money system
of its own. The unit of money in the United States is the
dollar. To represent parts of a dollar, we use the cent, which is
yfo of a dollar. Fifty cents is $ of a dollar; it is also one-half
dollar (y 5 ^ = J) . Likewise, twenty-five cents is T 2 ^ dollar,
or one-quarter dollar.
In writing United States money, the dollar sign ($) is written
before the number; a period called the decimal point, is placed
after the number of dollars; following this decimal point is
placed the number of cents.
Two dollars and seventy cents is written $ 2 . 70
Fifteen dollars and seven cents is written $ 15.07
One Hundred twenty-five dollars is written $125.00
One dollar and twenty-five cents is written $ 1 . 25
Thirty-five cents is written $ . 35
Eight cents is written $ . 08
Since one cent is T ^ dollar, it follows that the figures to the
right of the decimal point represent a fraction of a dollar. These
figures are the numerator, and the denominator is 100.
$ 2 . 70 is the same as $ 2- i y -
$15.07 is the same as $15 r J- -
$ . 08 is the same as $
MONEY AND WAGES 25
The first figure following the decimal point can be said to
indicate the number of dimes, because 1 dime = 10 cents, and
this figure indicates the number of tens of cents. Also this
number represents tenths of a dollar, because 1 dime = 10 cents
= i l <nr dollar = 1 1 U - dollar. The second figure after the decimal
point indicates cents, or hundredths of a dollar.
This decimal system of writing amounts of money has great
advantages in performing the operations of addition, subtraction,
multiplication, and division, because we can perform these opera-
tions just as if we were dealing with whole numbers, which makes
the work much simpler than if we had fractions to deal with.
28. Addition. We can add numbers made up of dollars and
cents and carry forward just as in simple addition. The number
of tens of cents will represent dimes (10 cents = 1 dime; 30 cents
= 3 dimes, etc.) and thus can be carried forward and added into
the dime column. Likewise, the number of tens of dimes will
represent dollars (10 dimes = 1 dollar) and, therefore, this number
can be added into the dollar column.
Example :
Add $5.20, $2.65, $3.25, and $.35.
Explanation: Adding the cents column, we get
5 + 5 + 5 + = 15 cents. Put down the 5 and crary
the 1 into the next column (since 15 cents = 1 dime
and 5 cents.) Adding the dimes, we get 1+3+.
2 + 6 + 2 = 14 dimes. Put down the 4 and carry
the 1 into the dollar column (since 14 dimes = $1.4).
1+3 + 2 + 5 = 11 dollars, which we put down com-
$11.45, Answer, plete. The decimal point we now place in the
sum exactly as it was in the numbers added, so
that it properly separates dollars from cents.
The only precaution to be observed is to see that the dollars,
dimes, and cents are properly lined up vertically before adding.
To do this it is only necessary to see that the decimal points are
kept in a straight vertical line.
Example :
What is the total cost of three articles priced as follows: $2.25,
SI, and $1.75?
Explanation: Here the $1 does not have any
$2.25 decimal point or cents after it, and care should
be taken to see that it is put down in the dollar
l.~-'> column and not in the cents column. $1 can be
$5.00, Answer. written $1.00, if desired, to avoid any danger of a
mistake.
26 SHOP ARITHMETIC
29. Subtraction. The same rules should be followed in sub-
tracting. If any figure in the subtrahend is larger than the
corresponding figure in the minuend, we can borrow 1 from the
figure next to the left, just as in ordinary subtraction.
Example :
A man draws $24.75 on pay day and immediately pays bills
amounting to $8.86. How much does he have left to put in the bank?
Explanation: Having set down the numbers
properly, we subtract j ust as if we were subtract-
$24.75 ing whole numbers. The only difference is that,
8.86 after subtracting, we put the decimal point in
$15 gg Answer. tn remainder directly below the other decimal
points, to separate the dollars and cents in the
remainder.
30. Multiplication. In multiplying an amount of money by
any number, the process is the same as in simple multiplication,
remembering, however, to keep the decimal point to separate the
dollars from the cents. The reasoning for this is just the same as
in addition. Multiplying cents, gives cents; multiplying dimes
gives dimes, and multiplying dollars gives dollars. The figures
left over are carried forward just as in plain multiplication,
because 10 cents = 1 dime, and 10 dimes = 1 dollar.
Examples :
1. What should a machinist receive for finishing 48 gas engines pistons
at 25 cents each?
$25 Explanation: The 25 cents is put down as $.25.
The multiplication of 25 by 48 is performed as
200 usual. Then the decimal point is placed in the
100 product to separate the dollars and cents by leaving
<12 00 Answer. the * wo places, counting from the right, to repre-
sent cents.
2. If you owe 2$ weeks board at $3.25 a week, how much money will
it take to settle the bill?
Explanation: First, we put down the numbers
$3.25 and then multiply as if we were simply multi-
_2i plying 325 X2$. The product is 812$, but since
162$ we are dealing with dollars and cents, we must
650 put the decimal point in the product to show
<jg 124 Answer what part of it represents dollars and what part
cents.
31. Division. In dividing an amount of money by any number,
the division is carried out as in ordinary division. The decimal
point is then placed in the quotient in the same position (from
the right) that it had in the dividend.
MONEY AND WAGES 27
Example :
The weekly pay roll of a company employing 405 men is $4880.25.
What is the average amount paid to each man?
405)$4880.25($12.05, Answer. Explanation: The division is carried
405 out just as if we were dividing 488025 by
~830 405. The quotient obtained is 1205.
g JO Then, since two of the figures in the
dividend were cents, we place the deci-
ma l point in the quotient so as to point
off the cents, making the quotient read
$12.05.
Another way of locating the decimal point is to place it in
the quotient as soon as the number of dollars in the dividend has
been divided. Taking the same example; we first divide 405
into 488 and get 1, with a remainder of 83. Annexing the next
figure and dividing again, we get 2 for the quotient. We have
now divided the number of whole dollars (4880) and have $12
for the quotient, with a remainder of 20. The 12 is, therefore,
the number of whole dollars in the quotient. We now bring
down the next figure (2) from the dividend and find that 405
will not go into 202, so we have dimes. Then, bringing down
the five cents, we get 2025 cents, which, divided by 405, gives
just 5 cents. The men, therefore, get an average of $12.05
each, per week.
32. Reducing Dollars to Cents. Sometimes we find it desirable
to change a number of dollars and cents all into cents. To do
this, merely remove the decimal point from between the dollars
and cents and you will have the number of cents. Every one
knows that:
$1.00 is 100 cents
$1.25 is 125 cents
$ . 25 is 25 cents
Likewise :
$ 12. 75 is 1275 cents
$ 247 . 86 is 24786 cents
$1000.00 is 100000 cents
What we have really done in making these changes is to
multiply the dollars by 100 to get the equivalent cents. We
have taken a mixed number and multiplied it by 100 because
there are 100 cents in a dollar. This operation is performed
by moving the decimal point two figures to the right, or placing
28 SHOP ARITHMETIC
it after the cents, where it is, of course, useless and is seldom
written.
In many problems it is quite desirable to change the dollars
to cents and carry the work through as cents. The following
example shows clearly such a case.
Example :
During one month a foundry turned out 312,000 Ib. of iron
castings. The total cost of the iron used, including the cost of melting
and pouring was $3900. What was the cost, in cents, of 1 Ib. of iron,
melted and poured?
Explanation: Since the cost
$3900 = $3900.00 = 390000 cents. of iron, melted and poured, is
,. ,. .78 ,1 but 1 or 2 cents, we might as
390000-^312000 = 1^2 = 14 cents, Answer. well change the ' t?tal c * st to
cents before we divide by the
number of pounds. Then we
will get the cost directly in
cents per pound, as we want it.
33. Reducing Cents to Dollars. The reduction of cents to
dollars is really performed by dividing the number of cents by
100, since there are 100 cents in 1 dollar.
217 17
217cents= ^dollars = $2^ = $2. 17
Hence, 2 1 7 cents = $2 . 1 7
This shows us that the following simple rule can be adopted
for this reduction:
To reduce cents to dollars, place a decimal point in the number
so as to have two figures to the right of the decimal point.
34. The Mill. There is another division of U. S. money called
the mill. A Mill is one-tenth of a cent or one one-thousandth
of a dollar.
10 mills = 1 cent
100 mills = 10 cents = 1 dime
1000 mills = 100 cents = 10 dimes = 1 dollar
There is no coin smaller than the cent and, therefore, the mill
is merely a name applied in calculations where it is desirable
to have some unit smaller than the cent. For example, tax
rates are usually given in mills per dollar. A tax rate of 15
mills on the dollar would mean that a person would have to pay
15 mills (or 1 cents) on each dollar of assessed valuation.
Cost accountants generally figure costs down as fine as mills
and even, in some cases, to tenths of mills or finer.
MONEY AND WAGES 29
In sums of money containing mills, there are three figures
following the decimal point. The first and second figures after
the decimal point indicate dimes and cents, as before. The
third figure indicates mills.
$ . 014 is 1 cent and 4 mills
It is also 14 mills (since 1 cent = 10 mills).
In multiplying or dividing numbers containing mills, we must
place the decimal point in the answer in the same position,
that is, three places from the right of the number.
Example :
If your house and lot were assessed at $2000, and the tax rate
was 15 mills on the dollar, what would be the amount of your taxes?
15 mills = $.015.
Explanation: 15 mills is 1 cent and 5 mills, or
$.015 $.015. This is the amount you must pay on each
2000 dollar of assessed value. For an assessment of
I 30 000 ' Answer $2000, you would pay 2000 times $.015, which
is $30.
35. Wage Calculations. The chief use that shop men have,
within the shop, for calculations concerning money is in connec-
tion with their time and wages. With some wage systems the
calculations of one's earnings is comparatively simple; with
others it seems rather complicated until the systems are thor-
oughly understood. The simplest systems are, of course, the
well-known day-rate and piece-rate systems. By the old day-
rate system, the men are paid according to the time put in,
without any reference to the work accomplished. The rate
may be so much per hour, per day, or per week, but the method
of calculating is the same, and the time keeper will use exactly
the same process in each case. The pay roll calculations consist
merely in multiplying the number of units of time which each
man has to his credit by his rate per unit of time; hours by rate
per hour, or days by rate per day.
Examples :
1. A machinist puts in 106 hours at 32 J cents per hour. How
much money is due him from the company?
106
. 32J Explanation: The amount due is the product of
53 106X32J. Since the amount will run into dollars,
212 we write the 32$ cents as dollars, $.32 J. The prod-
318 uct is $34.45, which is the amount due.
$34.45, Answer.
30 SHOP ARITHMETIC
2. The tool-room foreman gets $6.00 a day and has worked 12 days.
What is the amount due him?
12 Explanation: The same process is carried out
6.00 here except that we have the rate per day times
$727067 Answer. the number of days.
The piece-work system, by setting prices for certain pieces of
work and paying according to the work done, rewards the man
in exact proportion to the work that he does. In this system,
an account is kept of the amount of a man's work and his pay is
calculated by multiplying the numbers of pieces by the piece-
rates.
Example :
The price for assembling a certain sized commutator is 55 cents
each. If a man assembles an order of 14 of them, how much does he
receive? What will he average per day if he does the job in three days?
550=$ .55
$ . 55 X 14 = $7 . 70 Amount he receives for the j ob.
$7.70-j- 3= 2.56$ Amount he averages per day.
There are many other wage systems, too numerous to be all
explained here. They all are planned to take into account
both the amount of work a man does and the time that he puts in
to do it. One of these systems, the Premium System, is so well
known and so successful as to warrant brief mention here. In
its most common form this system is as follows:
The men are all placed on a time rate, usually a rate per hour.
A record is kept of every man's time and also of the work done
by him.
Every job has a standard time (sometimes called " the limit")
allowed for its completion.
On each job, the man's actual time is recorded and also the
limit for the job.
The man is paid straight wages for the time put in on the job
and, in addition, is paid a premium of usually one-half of the time
that he saves below the limit.
If it takes a man a longer time than the limit, he is paid full
wages for the time he puts in.
This system is planned to satisfy both parties by giving the
efficient man an increased earning, and by giving the firm a
share of the time saved, thus giving them a reduced cost when-
ever they pay higher wages.
Let us take an example to see how one's pay would be figured
on this plan.
MONEY AND WAGES 31
Example :
In one day, a man whose rate is 27 cents an hour, does the fol-
lowing premium jobs: the first has a limit of 8 hours and is done in 5 hours;
the second has a limit of 5 hours and is done in 3 hours; the third has a
limit of 3 hours and is finished in 2 hours. What is his pay for the day?
5 + 3 + 2 = 10 hours, actually put in.
85= 3 hours saved on the first job.
53= 2 hours saved on the second job.
32= 1 hour saved on the third job.
3 +2 + 1 = 6 hours saved on the days work.
He will get paid for the 10 hours and, in addition, for half of the 6 hours
that he saved. Altogether he will be paid for 10 + ^ of 6 =13 hours.
2i
13X$.272 = $3.5?2, Answer.
He gets $3.58 for his 10 hours work and, therefore, makes a premium of
83 cents. Meanwhile, the company gets the work done for $3.58, instead
of paying $4.40, which it would have cost if the workman had taken the
full limit for the work.
PROBLEMS
61. Write in figures the following sums of money:
One dollar and twelve cents
Two dollars and twenty-five cents
Eight cents
Fifteen dollars and thirty-seven and one-half cents
Twenty-fi ve^mills
62. Read the following and write them out in words:
$ 2.75 $ .008
$ .03J $1.08
$16.25
63. A young man makes the following purchases: Suit of clothes $25,
shoes $3.75, hat $2.25, necktie 50 cents. What is the total cost of his
purchases?
64. A certain job calls for four f in. by 3 in. machine bolts, two f in. by 1 J
in. set screws, and two i in. by 2 in. cap screws. What would be the total
cost of the bolts and screws, if the machine bolts are worth 2J cents each,
the set screws 1^ cents each, and the cap screws If cents each?
65. If you bought a house for $3000 and it was assessed at two-thirds
of what it cost you, what taxes would you have to pay if the tax rate was
14 mills on the dollar?
66. How much would a man who is paid $4.25 a day earn in a month of
26 working days?
67. If the piece price for a certain job is 4 cents, how many pieces must
a man do in one day to make $5.00?
68. An apprentice and a machinist are working together on a piece-
work job and they earn $30. They are prorated on the job, which means
that the money is divided according to their day-work rates. If the
apprentice's rate is 10 cents an hour and the machinist's is 30 cents, what
fraction of the money does each get, and how much money would each get?
32
SHOP ARITHMETIC
69. A man rated at 25 cents an hour is working under the premium
system. In one day of 9J hours he performs 3 operations the limits of
which are 4J hours each. If he gets paid a premium of half the time saved,
how much will he make for the day?
70. The following figure represents a page from a time book of a shop
working entirely on day work. Calculate the total time of each man,
the amount of money due him, and the total pay roll for the shop.
SHOP- TWO WEEKS
. I9I./....
NAME
AL
RS
AMOUNT
FOR THr
Z WEEKS
7-Lfo
(O /O fcr
/O
/o
to
10 /O
.0
to
/o
to
fO
/O
/o
fo
/o
fO
fO 10
to
'0
10
to
iff. /&.
/o
/o
10 fO
/o
/o
/O fO
/o
fO
fO
fo
fo
/o
10
to
10
fo
fO
/o
to
/o
/o
10
10
/o
/o
/o
/c
/o
fO
.35-
fO
(0
to
/o
/o
/o
/o
/o
/o
/o
/O fO
fo
.20
10
/o
FIQ. 5.
/o
/o
/o
/o
to
TOTAL.
CHAPTER V
DECIMAL FRACTIONS
36. What are Decimals? In the old days, when no machinist
pretended to work much closer than ^ in. and the micrometer
was unknown, the mechanic had little use for decimals except in
figuring his pay. Now, however, we find that micrometer
measurements are used so generally that a knowledge of decimal
fractions is essential.
A Decimal Fraction is merely a fraction having a denominator
of 10, 100, 1000, or some similar multiple of 10. The denomi-
nator is never written, however, but a system similar to that used
in writing U. S. money is used. A decimal fraction is written
by first putting down a period or "decimal point" and then
writing the numerator of the fraction after the decimal point in
such a manner that the denominator can be understood. Every-
thing that comes after the decimal point (to the right of it) is a
fraction, or part of a unit.
In writing sums of money, the first figure after the decimal
point indicates dimes or tenths of a dollar; the second figure
indicates cents, or hundredths of a dollar; the third figure, if
any, indicates mills or thousandths of a dollar. This system
has proved so handy that it has been extended to representing
fractions of any sort of a unit (not necessarily dollars).
.08 in. means y^ in.
25 .
.25 in. means T ^. in.
1UU
256
1 . 256 in. means Ivrx in.
Let us take a decimal, say .253, and find out its meaning.
We said that the first figure was tenths; the second, hundredths;
the third, thousandths, and so on. Then .253 would be & -f
rib" + ToW This is not a very handy system unless there is
3 33
34 SHOP ARITHMETIC
some easier way to read it. If we reduce these to a common
denominator and add them, we get:
, ____ = -.-|_J!L. 3 253
10 100 1000 1000 1000 1000 1000
253
Then . 253 is
1000
This shows that one place to the right of the decimal point indi-
cates a denominator of 10, two places a denominator of 100,
three places a denominator of 1000, and so on. The number
at the right of the decimal point can, therefore, be taken as
the numerator, and the denominator obtained as follows: Put
down a 1 below the decimal point and a cipher (0) after it for
each figure in the numerator. This will give the denominator.
In the case just given, we would have i%%l showing that the
denominator is 1000.
In the same manner:
2
. 2 is ^> or two-tenths.
37
.37 is y^:> or thirty-seven one-hundredths.
526
. 526 is > or five hundred twenty-six one-thousandths.
2749
. 2749 is > or two thousand seven hundred
forty -nine ten-thousandths.
42
.042 is > or forty-two one-thousandths.
1000
The last case (.042) presents an interesting problem. Here
we have a numerator so small in respect to the denominator
that it is necessary to have a cipher, or zero (0) between it and the
decimal point, in order that the denominator can be indicated
correctly. Let us see how we would go about writing such a
common fraction into a decimal. Take -nnnr- If we merely
wrote .5 that would be -^ and would, therefore, not be right.
From the rule for finding denominators of decimals we see that
DECIMAL FRACTIONS 35
there must be as many figures after the decimal point as there
are ciphers in the denominator. In this case the denominator
(1000) has 3 ciphers, so we must have three figures in our decimal.
We, therefore, put two ciphers to the left of the 5 and then put
down the decimal point. We now have .005, which can be
easily seen to be -nmr-
One thing that must be carefully borne in mind is that adding
ciphers after a decimal does not change the value of the fraction.
.5 is the same in value as .50 or .500 because T 5 T is the same in
value as T 5 ^ or -nnnr- O n tne other hand, ciphers immediately
following the decimal point do affect the value of the fraction,
as has just been shown.
Mixed numbers are especially easy to handle by decimals,
because the whole number and the fraction can be written out
in a horizontal line with the decimal point between them. We
read mixed decimals just as we would any mixed number; first
the whole number, then the numerator, and lastly, the
denominator.
Example :
42137.24697
In this example, 42137 is the whole number, and .24697 is the fraction.
The number reads "forty-two thousand one hundred thirty-seven and
twenty-four thousand six hundred ninety-seven hundred-thousandths.
The names and places to the right and left of the decimal
point are as follows:
5 i
^
^ <5
7654321. 123456
37. Addition and Subtraction. Knowing that all figures to
the right of the decimal point are decimal parts of 1 thing and
that all figures to the left are whole numbers and represent whole
things, it will be seen readily that in addition and subtraction the
figures must be so placed that the decimal points come under
36 SHOP ARITHMETIC
each other. As was shown under U. S. Money, the operations
can then be carried out just as if we were dealing with whole
numbers.
Examples :
Addition Subtraction
783.5 22.7180
21.473 1.7042
804.973 21.0138
Pay no attention to the number of figures in the decimal.
Place the decimal points in line vertically. You can, if you desire,
add ciphers to make the number of decimal places equal in the
two numbers. Remember, however, that the ciphers must
be added to the right of the figures in the decimal. Proceed as
in ordinary addition and subtraction, carrying the tens forward
in addition and borrowing, where necessary, in subtraction just
as with whole numbers.
38. Multiplication. In multiplication forget all about the
decimal point until the work is finished ; multiply as usual with
whole numbers. Then point off in the product as many decimal
places, counting from the right, as there are decimal places in the
multiplier and multiplicand together.
Example :
6 . 685 Multiplicand (3 places)
5.2 Multiplier (1 place)
13370
33425
34.7620 Product (4 places)
Since there are three decimal places in one number, and one in the other,
we count off in the product four (3 + 1) places from the right and place the
point between the 7 and the 4. The last can be dropped after pointing
off the product, giving the result 34.762 (or 34-,^). The reason for this
can be seen from the following: The whole numbers are 6 and 5. The result
must be a little more than 6X5 = 30, and less than 7X6 = 42, since the
numbers are more than 6 and 5, and less than 7 and 6. The actual result
is 34.762.
The position of the decimal point can be reasoned out in this
way for any example, but the quickest way is to point off from
the right a number of decimal places equal to the sum of the
numbers of decimal places in the multiplier and multiplicand.
DECIMAL FRACTIONS 37
Examples :
(a) .0045X2.7 (b) .000402x4.26
. 0045 (4 places) . 000402 (6 places)
2.7(1 place) 4 . 26 (2 places)
:U5 2112
90 804
.01215 (5 places) 1608
.00171252 (8 places)
In the above examples it was necessary to put ciphers before
the product in order to get the required number of decimal
places. To see the reason for this take a simple example such as
.2X-3 The product is .06 or -j-f-g-, as can be readily seen if
they are multiplied as common fractions (^X^\ = ^^i).
This checks with the rule of adding the number of decimal places
in the two numbers to get the number in their product. The
product of two proper fractions is always less than either of the
fractions, because it is part of a part.
39. Short Cuts. If we want to multiply or divide a decimal by
10, 100, 1000, or any similar number, the process is very simple.
Suppose we had a decimal .145 and then moved the decimal
point one place to the right and made it 1.45. The number
would then be 1-j 4 ^ or {%% instead of iV 4 Tsir5 so we see that mov-
ing the decimal point one place to the right has multiplied the
original number by 10. Therefore, we see that:
To multiply by 10 move the decimal point one place to the
right.
To multiply by 100 move the decimal point two places to the
rigJiT.
For other' similar multipliers move the decimal point one place
to' the right for each cipher in the multiplier. This process is
reversed in division, the rules being:
To divide by 10 move the decimal point one place to the left.
To divide by 100 move the decimal point two places to the
left, etc.
Example :
Reduce 10275 cents to dollars.
10275-^-100 = $102. 75 (Decimal point moved two places to left).
40. Division. The division of decimals is just as easy as the
multiplication of them after one learns to forget the decimal
point entirely until the operation of dividing is finished. Divide
38 SHOP ARITHMETIC
as in simple numbers. Then point off from the right as many
decimal places in the quotient as the number of decimal places
in the dividend exceeds that in the divisor. In other words, we
subtract the number of decimal places in the divisor from the
number in the dividend and point this number off from the right
in the quotient.
Example : Explanation: The number of
Divide 105.587 bv .93 places in the dividend is 3, and
ooMnc KC7/-i"iQ K_L A c in the divisor 2. Hence, when
.93) 105. 587(113.5 + , Answer. thfi division ^ complet ' ed) we
_ point off 3-2 = 1 place in the
quotient. The + sign after the
Dividend, 3 places quotient means that the division
328 Divisor, 2 places ^ not come out even> but that
279 Quotient, 1 place there was a remainder and that
497 the quotient given is not com-
465 plete. If desired, ciphers could
~32 Remainder. , e PJ. ac ^ after the dividend and
the division carried farther, giv-
ing more decimal places in the
quotient.
It makes no difference if the divisor is larger than the dividend,
as in the following example. In such a case the quotient will
be entirely a decimal.
Example :
22.762-5-84.25 = ?
84.25)22.762000(.2701+ or .2702, Answer.
16 850
5 9120
5 8975
14500
8425
6075
Explanation: The divisor being larger than the dividend, the quotient
turns out to be an entire decimal. In this case we will presume that we
wanted the answer to four decimal places. We have, therefore, added
ciphers to the dividend until we have six decimal places. When these
have all been used in the division, we have 6 2 = 4 places in the quotient.
The remainder is more than half of the divisor, showing that if we had
carried the division to another place, the next figure would have been
more than 5. We, therefore, raise the last figure (1) of the quotient to 2,
because this is nearer the exact quantity.
In stopping any division this way, if the next figure of the
quotient would be less than 5, let the quotient stand as it is, but,
if the next figure would be 5 or more, as in the example just
worked, raise the last figure of the quotient to the next higher
figure.
DECIMAL FRACTIONS 39
Sometimes the decimal places are equal in dividend and divisor,
as for instance, if we divide .28 by .07.
.07). 28
4
As the numbers of decimal places in the dividend and divisor are
the same, the difference between them is zero, and there are no
decimal places in the quotient. The answer is simply 4. The
decimal point would come after the 4 where it would, of course,
be useless.
If there are more decimal places in the divisor than in the
dividend, add ciphers at the right of the decimal part of the
dividend as far as necessary. In counting the decimal places,
be sure to count only the ciphers actually used.
Examples :
l.H-.025 = ? 4.2 + 38.25 = ?
. 025) 1 . 000(40, Answer. 38 . 25)4 . 200000( . 1098 + , Answer.
1 00 3 825
37500
34425
30750
30600
150
41. Reducing Common Fractions to Decimals. Common
fractions are easily reduced to decimals by dividing the numer-
ator by the denominator. In the case of , we divide 1.0 by 2
and get .5 All that is necessary is to take the numerator and
place a decimal point after it, adding as many ciphers to the
right of the decimal point as are likely to be needed, four being a
common number to add, as four decimal places (ten thousandths)
are accurate enough for almost any calculations.
If -g^- is to be reduced to a decimal, the work is simply an
example in long division, the placing of the point being the main
thing to consider. Simply divide 1.00000 by 32. This gives
.03125 or 3125 one hundred-thousandths.
32)1.00000(.03125
96
40
32_
80
64_
16,0
160
40
SHOP ARITHMETIC
42. Complex Decimals. A complex decimal is a decimal with
a common fraction after it, such as .12J, .0312^, etc. The frac-
tion is not counted in determining the number of places in the
decimal. .12 is read "twelve and one-half hundredths."
.0312 is read "three hundred twelve and one-half ten-thou-
sandths." To change a complex decimal to a straight decimal,
reduce the common fraction to a decimal and write it directly
after the other decimal, leaving out any decimal point between
them.
Examples :
.06^=. 065 .8|=.875 .03g= .03125
43. The Micrometer. The micrometer is a device to measure
to the thousandth of an inch and is best known to shop men in
the form of the micrometer caliper shown in Fig. 6. The whole
principle of the micrometer, as generally made, can be said to
depend on the fact that ^V of TV~TtW- The micrometer, as
shown in Fig. 6, is made up of the frame or yoke 6, the anvil c,
the screw or spindle a, the barrel d, and the thimble e. The
1 2 3 4 50 1
-
20
e
15
a
inliniiiiHu
,.
niiliiii
)
r-
1
FIQ. 6.
spindle a is threaded inside of d. The thimble e is attached to
the end of the spindle a. The piece to be measured is inserted
between c and a, and the caliper closed on it by screwing a
against it. The screw on a has 40 threads to the inch, so if it is
open one turn, it is open ^ in., or T^, or .025. Along the
barrel d are marks to indicate the number of turns or the number
of fortieths inch that the caliper is open. Four of these divisions
(A) wn l represent one-tenth of an inch, so the tenths of an inch
are marked by marking every fourth division on the barrel.
41
Around the thimble e are 25 equal divisions to indicate parts of a
turn. One of these divisions on e will, therefore, indicate -^ of a
turn, and the distance represented will be -^ of ^ = T^ns m -
To read a micrometer, first set down the number of tenths inch
as shown by the last number exposed on the barrel. Count the
number of small divisions on the barrel which are exposed be-
tween this point and the edge of the barrel. Multiply this
number by .025 and add to the number of tenths. Then observe
how far the thimble has been turned from the zero point on its
edge. Write this number as thousandths of an inch and add
to the reading already obtained. The result is the reading in
thousandths of an inch.
Example :
Let us read the micrometer shown in Fig. 6.
. 7 Explanation: First we find the figure 7 exposed on
.025 the barrel, indicating that we have over ^ in. This
.018 we put down as a decimal. In addition, there is one
7743 in. Answer. f * ne smaller divisions uncovered. This is .025 in
more. And on the thimble, we find that it is 3
divisions beyond the 15 mark toward the 20 mark.
This would be 18, and indicates .018 in. more. Adding the three,
.7 + .025 + .018 = .743 in., Answer. This can perhaps be better under-
stood as being 7 thousandths less than ^ in. Lots of men locate a
decimal in their minds by its being just so far from some common fraction.
Most micrometers have stamped in the frame the decimal
equivalents of the common fractions of an inch by sixty-fourths
from -fa in. to 1 in. A table of these decimal equivalents is given
in this chapter, and will be found very useful. Everyone
should know by heart the decimal equivalents of the eighths,
quarters, and one-half, or, at least, that one-eighth is .125. Then
f = 5 X. 125 = .625; and $ = 7 X. 125 = .875, etc. Also, if possible,
learn that -^- = .062^, or .0625. To get the decimal equivalent
of a number of sixteenths, add .062^ to the decimal equivalent
of the eighths next below the desired sixteenths.
Example :
13
What is the decimal equivalent of -^ in.?
|| = |+^=. 750+. 062^=. 812 Jin., or .8125 in.
To set a micrometer to a certain decimal, first unscrew the
thimble until the number is uncovered on the barrel correspond-
ing to the number of tenths in the decimal. Divide the remainder
by .025. The quotient will be the additional number of the
42
SHOP ARITHMETIC
divisions to be uncovered on the barrel and the remainder will
give the number of divisions that the thimble should be turned
from zero.
Example :
Calculate the setting for ft in. (= .4375 or .437$).
First unscrew the micrometer until the 4. is uncovered on the barrel.
Then divide the remainder .0375 by .025. This gives 1 and leaves a re-
mainder of .0125. The thimble should, therefore, be unscrewed one full
turn or 1 division beyond 4 on the barrel, plus 12.5 divisions on the thimble.
TABLE OF DECIMAL EQUIVALENTS FROM & TO 1.
Fraction
Decimal
Equivalent
Fraction
Decimal
Equivalent
&....
i
.015625
.03125
||....
4J. .
.515625
.53125
,
.046875
.0625
f|....
9
.546875
.5625
16 5
751-
A.
.078125
.09375
10 II...-
1|. .
.578125
. 59375
A----
.109375
.125
5
.609375
.625
5
A.
. 140625
. 15625
8 iiv
SI
.640625
. 65625
||....
3
.171875
.1875
11
.671875
.6875
10 i,
04 '
A.
.203125
.21875
10 ....
.703125
.71875
H,...
1
.234375
.25
if...;
3
.734375
.75
4 J7
84 '
A.
.265625
.28125
4 I!-.
64
. 765625
78125
19
b4 ' ' ' '
5
.296875
.3125
-...
13
. 796875
.8125
16 S1
6 4 '
u. .
.328125
.34375
16 If..-.
.828125
84375
I!....
3
.359375
.375
If---.
7
.859375
875
8 -...
n
.390625
.40625
8 ....
n. .
.890625
90625
7 ;;
.421875
.4375
69
??.
15
.921875
9375
16 J9
ft . . . .
.453125
.46875
16 ::::
H
.953125
96875
,
.484375
.5
&....
1
.984375
1 00000
2
DECIMAL FRACTIONS 43
PROBLEMS
71. Write the following as decimals:
One and twenty-five one-hundredths.
Three hundred seventy-five one-thousandths.
Three hundred and seventy-five one-thousandths.
Sixty-two and one-half one-thousandths.
Seven hundred sixty-five and five one-thousandths.
72. Read the following decimals and write them out in words:
.075
.137
100.037
.121
1.09375
73. Find the sum of .2143, 783.5, 138.72, and 10.0041.
74. From 241. 70 take 215.875.
76. a. Find the product of 78.8763 X .462.
b. Multiply 21.3 by .071.
76. a. Divide 187.2421 by 123.42.
b. Divide 25 by .0025.
77. Reduce in. to a decimal and compare with the table.
78. Reduce or, in. to a decimal and compare with the table.
79. Calculate the decimal equivalent of HT in.
80. Write .8125 as a common fraction and reduce it to the lowest terms.
81. If an alloy is .67 copper and .33 zinc, how many pounds of each metal
x would there be in a casting weighing 75 lb.?
v 82. A steam pump delivers 2.35 gallons of water per stroke and runs 48
strokes per minute; how many gallons will it deliver in one hour?
83. The diameter of No. 8 B. W. G. wire is .165 in. and of No. 12 wire is
.109 in. What is the difference in diameter of the two wires? What do
the letters B. W. G. stand for?
I 84. A machinist whose rate is 27.5 cents per hour puts in a full day of
10 hours and also 3 hours overtime. If he is paid "time and a half" for
overtime, how much should he be paid altogether?
3
85. The depth of a thread on a v in- bolt with U. S. Standard threads is
.065 in. What is the diameter at the bottom of the threads?
86. I want 5000 ft. of $ in. Q (square) steel bars. I find from a table
that this size weighs 1.914 lb. per foot of length. How many pounds must
I order and what will it cost at $1.85 per 100 lb.?
2 i
87. Explain how you would set a micrometer for JQQQ in. over g in.
\
44
SHOP ARITHMETIC
88. A 28-tooth 7-pitch gear has an outside diameter of 4.286 in. The
diameter at the bottom of the teeth is 3.67 in. How deep are the teeth cut?
89. A 2 in. pipe has an actual inside diameter of 2.067 in. The metal of
the pipe is .154 in. thick. What is the outside diameter of the pipe?
90. Read the micrometer shown below in Fig. 7.
Fio. 7.
CHAPTER VI
PERCENTAGE
44. Explanation. Percentage is merely another kind of
fractions or, rather, a particular kind of decimal fractions, of
which the denominator is always 100. Instead of writing the
denominator, we use the term "per cent" to indicate that the
denominator is 100. When we speak of "6 per cent" we mean
T | 7 or .06. These all mean the same thing; namely, six parts out
of one hundred. Instead of writing out the words "per cent"
we more often use the sign % after the number, as, for instance,
6%, which means "6 per cent." Since per cent means hun-
dredths of a thing, then the whole of anything is 100% of itself,
meaning ]---, or the whole. If a man is getting 40 cents an hour
and gets an increase of 10%, this increase will be 10% (or -j^V or
.10) of 40 cents and this is easily seen to be 4 cents, so his new
rate is 44 cents. Another way of working this would be to say
that his old rate is 100% of itself and his increase is 10% of the
old rate, so that altogether he is to get 110% of the old rate.
Now 110% is the same as 1.10 and 1.10x40 = 44 cents, the new
rate.
PERCENTAGE 45
Any decimal fraction may be easily changed to per cent.
07 K
875-^-87.5%.
Here we first change the decimal to a common fraction having
100 for a denominator. Then we drop this denominator and
use, instead, the per cent sign (%) written after the numerator.
This sign indicates, in this case, 87.5 parts out of 100, or
87.5
100
The change from a decimal to percentage can be made without
changing to a common fraction as was just done. Having a
decimal, move the decimal point two places to the right and
write per cent after the new number.
.625=62.5% .06=6% 1.10=110%
If it is desired to use a certain number of per cent in calcula-
tions, it is usually expressed as a decimal first and then the
calculations are made. For example, when figuring the interest
on $1250 at the rate of 6%, we would first change 6% to .06 and
multiply $1250 by .06 which gives $75.00.
$1250
.06
$75.00
A common fraction is reduced to per cent by first reducing it
to a decimal and then changing the decimal to per cent.
Example :
The force in a shop is cut down from 85 men to 62. What per cent
of the original number of men are retained?
no
62 is | of 85.
oo
' AO
^=.729 = 72.9%
62 is 72. 9% of 85.
Therefore, the number of men retained is 72.9% or nearly 73% of the
original number of men.
If we want to reduce the fraction | to per cent, we first get
= .125 and then, changing this decimal to per cent, we have
.125 = 12.5%. Then | of anything is the same as 12% of it,
because
46
SHOP ARITHMETIC
The following table gives a number of different per cents with
the corresponding decimals and common fractions:
Per cent
Decimal
Fraction
Per cent
Decimal
Fraction
1%
.01
1
100
25%
.25
25 1
100 4
2%
.02
2 1
100 " 50
33J%
.88|
33^ 1
100 3
2*%
.025
2i .. 1
100 40
37}%
.375
37J 3
100 ~ 8
5%
.05
5 1
100 20
50%
.50
50 1
100 2
61%
.0625
6i _ 1
100 16
75%
.75
75 3
100 ~ 4
10%
.10
10 1
100 10
90%
.90
90 9
100 10
12J%
.125
12^ 1
100 ~ 8
100%
1.00
100
100 ~
16f%
-16|
163 l
100 6
200%
2.00
200
100~ *
45. The Uses of Percentage. In shop work, the chief use of
percentage is to express loss or gain in certain quantities or to
state portions or quantities that are used or unused, good or bad,
finished or unfinished, etc. Very often we hear expressions like:
"two out of five of those castings are bad;" or "nine out of ten
of those cutters should be replaced." If, in the first illustration,
we wanted to talk on the basis of a hundred castings instead of
five, we would say "40 per cent of those castings are bad,"
because "two out of five" is the same as f, = T \\> =40%. And
in the second case: "90 per cent of those cutters should be
replaced." Here, "nine out of ten" =^,=^,=90%. If a
piece of work is said to be 60% completed, it means that, if we
divide the whole work on the job into 100 qual parts, we have
already done 60 of these parts or -$ of the whole.
If a shop is running with 50% of its full force, it means that
T'W or \ f the full force is working. If the full force of men is
1300, then the present force is 50% of 1300 = .50X1300 = 650.
If the full force were 700 men, then the 50% would be 350.
Another very common use of percentage is in stating the por-
tions or quantities of the ingredients going to make up a whole.
We often see formulas for brasses, bronzes, and other alloys in
PERCENTAGE 47
which the proportions of the different metals used are indicated
by per cents. For example, brass usually contains about
65% copper and 35% zinc. Then, in 100 Ib. of brass, .there
would be 65 Ib. of copper and 35 Ib. of zinc. Suppose, however,
that instead of 100 Ib. we wanted to mix a smaller amount, say
8 Ib. The amount of copper needed would be 65 % or .65 of 8 Ib.
.65x8 = 5.20 Ib., or 5 T \ Ib., the copper needed.
.35X8 = 2.80 Ib., or 2 T 8 T Ib., the zinc needed.
Sometimes, in dealing with very small per cents, we see a
decimal per cent such as found in the specifications for boiler
steel, where it is stated that the sulphur in the steel shall not
exceed .04%. Now this is not 4%; neither is it .04; but it is
.04%, meaning four one-hundredths per cent, or four one-
hundredths of one one-hundredth. This is -j^-g- of T ^- = y o o o o > so
if we write this .04% as a decimal, it will be .0004. It is a very
common mistake to misunderstand these decimal per cents, and
the student should be very careful in reading them. Likewise,
be careful in changing a decimal into per cent that the decimal
point is shifted two places to the right.
46. Efficiencies. Another common use of percentage is in
stating the efficiencies of engines or machinery. The efficiency of
a machine is that part of the power supplied to it, that the machine
delivers up. This is generally stated in per cent, meaning so
many out of each hundred units. If it requires 100 horse-power
to drive a dynamo and the dynamo only generates 92 horse-power
of electricity, then the efficiency of the dynamo is i\\ or 92%.
If the engine driving a machine shop delivers 250 horse-power
to the lineshaft, but the lineshaft only delivers 200 horse-power
to the machines, then the efficiency of the lineshaft is ff = .80
= 80%. The other 50 horse-power, or 20%, is lost in the
friction of the shaft in its bearings and in the slipping of the
belts. The efficiencies of all machinery should be kept as high
as possible because the difference between 100% and the effi-
ciency means money lost. The large amount of power that is
often lost in line shafting can be readily appreciated when we
try to turn a shaft by hand and try to imagine the power that
would be required to turn it two or three hundred times a minute.
47. Discount. In selling bolts, screws, rivets, and a great
many other similar articles, the manufacturers have a standard
list of prices for the different sizes and lengths and they give their
48 SHOP ARITHMETIC
customers discounts from these list prices. These discounts or
reductions in price are always given in per cent. Sometimes
they -are very complicated, containing several per cents to be
deducted one after another. Each discount, in such a case, is
figured on the basis of what is left after the preceding per cents
have been deducted.
Example :
The list price of i in. by 1} i Q - stove bolts is $1 per hundred.
If a firm gets a quotation of 75, 10 and 10% discount from list price, what
would they pay for the bolts per hundred?
100 X . 75 = 75 cents Explanation: 75, 10 and 10% discount means
100 75 = 25 cents 75% deducted from the list price, then 10% de-
1 ducted from that remainder, then 10% taken from
= 2 ce the second remainder.
1 1 Starting with 100 cents, the list price, we de-
25-2^ = 22- cents duct the first discount of 75%. This leaves 25
1 , cents. The next discount of 10% means 10% off
22- X . 10 = 2- cents from this balance. Deducting this leaves 22* cents.
Next, we take 10% from this, leaving 20\ cents
22! _ 2! = 20- cents per hundred as the actual cost of these stove bolts.
24 4
48. Classes of Problems. Nearly all problems in percentage
can be divided into three classes on the same basis as explained
in Article 26. There are three items in almost any percentage
problem: namely, the whole, the part, and the per cent. For
example, suppose we have a question like this: "If 35% of the
belts in a shop are worn out and need replacing, and there are
220 belts altogether, how many belts are worn out?" In this
case, the whole is the number of belts in the shop, 220. The part
is the number of belts to be replaced, which is the number to be
calculated. The per cent is given as 35%.
Any two of these items may be given and we can calculate the
missing one. We thus have the three cases:
1. Given the whole and the per cent, to find the part.
2. Given the part and the per cent that it is of the whole, to
find the whole.
3. Given the whole and the part, to find what per cent the,
part is of the whole.
The principles taught under common fractions will apply
equally well in working problems under these cases, the only
difference being that here a per cent is used instead of a common
fraction. In working problems, the per cent should always be
changed to a decimal.
One difficulty in working percentage problems is in deciding
PERCENTAGE 49
just what number is the whole (or the base, as it is often called).
The following illustration shows the importance of this.
If I offer a man $2000 for his house, but he holds out for $3000,
then his price is 50% greater than my offer, while my offer is
33^% less than his price. The difference is $1000 either way
but, if we take my offer as the base, it would be necessary for me
to raise it i, or 50%, to meet his price. On the other hand, for
him to meet my bid, he would only have to cut his price J, or
Examples of the three types of problems, before mentioned,
may help somewhat in getting an understanding of the processes
to be used.
Example of Case i :
How many pounds of nickel are there in 1 ton of nickel-steel containing
2.85% nickel? Explanation: Here we have the whole
1 ton = 2000 Ib. ( 200 lb -) and th e per cent (2.85%) to
2. 85% =.0285 find the part. After changing the 2.85%
.0285X2000 = 57 lb., Answer. to a decimal fraction, the problem be-
comes a simple problem in multiplica-
tion of decimals.
Example of Case 2 :
The machines in a small pattern shop require altogether 12 horse-power
and are to be driven from a lineshaft by a single electric motor. If we
assume that 20% of the power of the motor wfll be lost in the line shaft
and belting, what size motor must we install?
100% 20% = 80% Explanation: Here the per cent given (20%)
80%= .80, or .8 is based on the horse-power of the motor, which
12 -T-. 8 = 15, Answer. is, as yet, unknown. The horse-power of the
motor is 100% of itself and, if 20% is lost,
then the machines will receive 80%, or .8 of
the power of the motor. This is 12 horse-
power. The whole will be 12 -f-. 8 = 15 horse-
power. This is the size of motor to install.
Example of Case 3 :
If the force in a shop is increased from 160 to 200 men, what per cent
is the capacity of the shop increased?
200-5-160 = 1.25 Explanation: The present force is 1} or
1.25 = 125% 1.25 of the old force, because 200 is :';; of
125% -100% = 25%, Answer. 160, and fgg = l}. This is the same as 125%.
The increase is, therefore, 25% of the former
force.
Note. To reduce the present force back to the old number would require
a reduction of only 20%, because now the base is different on which to figure
the per cent. 40 + 200= .20, or 20%.
PROBLEMS
91. Write 4% as a decimal and as a common fraction.
92. Write 25% as a common fraction and reduce the fraction to its lowest
terms.
4
50 SHOP ARITHMETIC
93. What per cent of an inch is ^ in.?
94. If there are 240 men working in a shop and 30% of them are laid off,
how many men will be laid off and how many will remain at work?
96. Out of one lot of 342 brass castings, 21 were spoiled and out of an-
other lot of 547, 32 were spoiled. Which lot had the larger per cent of
spoiled castings?
96. 500 Ib. of bronze bearings are to be made; the mixture is 77% copper,
8% tin, and 15% lead. How many pounds of copper, tin, and lead are
required?
Note. This is a standard bearing mixture used by the Pa. R. R.
and by some steam turbine manufacturers.
97. The boss pattern maker is given a raise of 25% on Christmas, after
which he finds that he is receiving $130 a month. How much did he get
per month before Christmas?
98. In testing a shop drive it was found that the machines driven by one
motor required horse-power as follows:
60 in. mill 3.31 horse-power
20 in. lathe 75 horse-power
48 in. lathe 2.42 horse-power
42 in. by 42 in. by 12 ft. planer 4.82 horse-power
16 in. shaper 33 horse-power
The total power delivered by the motor was 13.65 horse-power. What
per cent of the total power was used in belting and lineshaft? What per
cent by the machines?
99. A man who has been drawing $2.50 a day gets his pay cut 10% on
May 1, and the following September he is given an increase of 10% of his
rate at this time. How much will he get per day after September?
100. The following weights of metals are melted to make up a solder:
18 Ib. of tin, 75 Ib. of bismuth, 37.5 Ib. of lead, and 19.5 Ib. of cadmium.
What per cent of the total weight is there of each metal?
CHAPTER VII
CIRCUMFERENCES OF CIRCLES; CUTTING AND GRINDING
SPEEDS
49. Shop Uses. In the running of almost any machine, judg-
ment must be used in order to determine the speed which will
give the best results. Lathes, milling machines, boring mills,
etc., are provided with means for changing the speed, according
to the -judgment of the operator. Emery wheels and grind-
stones, however, are often set up and run at any speed which
the pulleys happen to give, regardless of the diameter.
If an emery wheel of large size is put on a spindle that has been
belted to drive a smaller wheel, the speed may be too great for
the larger wheel and, if the difference is considerable, the large
wheel may fly to pieces. Every mechanic should know how to
calculate the proper sizes of pulleys to use for emery wheels or
grindstones, the correct speed at which to run the work in his
lathe, or the most economical speeds to use for belts and pulleys.
A little data on this subject may be useful and will afford
applications for arithmetical principles.
50. Circles. To understand what has just been mentioned, it
is necessary to get a knowledge of circles and their properties.
The distance across a circle, measured straight through the center,
is called the Diameter. Circles are generally designated by their
diameters. Thus a 6 in. circle means a circle 6 in. in diameter.
Sometimes the radius is used. The Radius is the distance from
the center to the edge or circumference and is, therefore, just
half the diameter. If a circle is designated by the radius, we
should be careful to say so. Thus, there would be no misunder-
standing if we said " a circle of 5-in. radius" ; but unless the word
radius is used, we always understand that the measurement
given is the diameter. The Circumference is the name given to
the distance around the circle, as indicated in Fig. 8. The
circumference of any circle is always 3.1416 times the diameter.
In other words, if we measure the diameter with a string and lay
this off around the circle, it will go a little over three times.
5 51
52 SHOP ARITHMETIC
This number 3.1416 is, without doubt, the most used in practical
work of any figure in mathematics. In writing formulas, it is
quite common to represent this decimal by the Greek letter
TT (pronounced "pi")> instead of writing out the whole number.
For this reason, the number 3.1416 is given the name "pi."
FIG. 8.
Where it is more convenient and extreme accuracy is not
quired, the fractio
exact value 3.1416.
22
required, the fraction -- may be used for n instead of the more
= 3 = 3.1429
It, therefore, gives values of the circumference slightly too
large, but in many cases it is sufficiently accurate and saves time.
Examples :
1. What length of steel sheet would be needed to roll into a drum 32 in.
in diameter?
When rolled up, the length of the sheet will become the circumference of a
32-in. circle. The circumference must be x times 32.
3.1416X32 = 100.5+ in.
The length of the sheet must, therefore, be 100 in. and, if it is to be lapped
and riveted, we would have to add a suitable allowance of 1 in. or so for
making the joint.
2. A circular steel tank measures 37 ft. 8$ in. in circumference. What is
its diameter?
If the circumference of a circle is 3.1416 times the diameter, then the
diameter can be obtained by dividing the circumference by 3.1416.
37 ft. 8^ in. = 37^ ft. = 37.7+ ft.
2i 1
37.7 -4- 3.1416 = 12 ft., Answer.
61. Formulas. A formula, in mathematics, is a rule in which
mathematical signs and letters have been used to take the place
of words. We say that "the circumference of a circle equals
CIRCUMFERENCES OF CIRCLES 53
3.1416 times the diameter." This is a rule. But suppose we
merely write
This is the same rule expressed as a formula. We have used C
instead of the words "the circumference of a circle;" the sign =
replaces the word "equals;" the symbol it is used instead of the
number 3.1416; X stands for "times;" and D stands for "the
diameter."
We found in the second example under Article 50 that, when
the circumference is given, we can obtain the diameter by divid-
ing the circumference by n. As a formula this would be written
D = - or D = C + K
7T
This arrangement is useful when we want to get \he diameters
of trees, chimneys, tanks, and other large objects. We can
easily measure their circumferences and, by dividing by 3.1416,
we get the diameters.
Formulas do not save much, if any space, because it is necessary
usually to explain what the letters stand for. They have,
however, the great advantage that intricate mathematical opera-
tions can be shown much more clearly than if they were written
out in a long sentence or statement. One can usually see in one
glance at a formula just what is to be done, with the numbers
that are given in the problem, to find the quantity that is
unknown.
Example :
What is the circumference in feet of a 16-in. emery wheel?
C = 3.1416 X 16 = 50.2656 in.
50.2656 in. -T- 12 = 4.1888 ft., Answer.
Explanation: We have the diameter given and want to get the circum-
ference. We, therefore, use the formula which says that C = nXD. x is
always 3.1416 and D in this case is 16 in. Then C comes out 50.2656 in.
But the problem calls for the circumference in feet. This is T ' ? of the number
of inches, or it is the number of inches divided by 12.
In the work of this chapter, the circumferences of circles
are always used in feet, and, consequently, should always be
calculated in feet. If we use D in feet, we will get C in feet, while,
if D is in inches, C will come out in inches. If the diameter can
be reduced to exact feet, it is easiest to use the diameter in feet
when multiplying by TT, rather than to reduce to feet after
multiplying.
54 SHOP ARITHMETIC
Example :
What is the circumference of a 48-in. fly wheel?
48 in. -5- 12 = 4 ft., the diameter.
C = nXD
(7 = 3.1416X4 = 12.5664 ft., Answer.
This is much shorter than it would be to multiply 3.1416 by 48 and then
divide the product by 12.
52. Circumferential Speeds. When a fly wheel or emery
wheel or any circular object makes one complete revolution,
each point on the circumference travels once around the circum-
ference and returns to its starting-point. When the wheel turns
ten times, the point will have travelled a distance of ten times
the circumference. In one minute, it will travel a distance equal
to the product of the circumference times the number of revolu-
tions per minute. The distance, in feet per minute, travelled
by a point on the circumference of a wheel is called its Circum-
ferential Speed, Rim Speed, or Surface Speed. It is also some-
times called Peripheral Speed, because the circumference is
sometimes given the name of periphery. It is the surface speed
by which we determine how to run our fly wheels, belts, emery
wheels, and grindstones, and what speeds to use in cutting
materials in a machine.
Written as a formula:
S = CXN
where :
S is the surface speed
C is the circumference
N is the number of revolutions per minute (R.P. M.).
Expressed in words this formula states that the surface speed of
any wheel is equal to the circumference of the wheel multiplied
by the number of revolutions per minute.
Example :
What would be the rim speed of a 7 ft. fly wheel when running at
210 revolutions per minute?
C = xXD Explanation: First we find the circum-
^_22 ference of the wheel, by multiplying the
~^f diameter by x. Here is a case where it is
S = CXN much easier to use *f- for n than to use the
5=^2x210 = 4620 decimal 3.1416, and the result is sufficiently
4620 ft. per min., Answer. accurate for our purposes. We get 22 ft. for
the circumference. We can now get the rim
speed, which is equal to the product of the
circumference times the number of revolutions per minute; orS = CxN.
C being 22 ft. and N being 210 revolutions per minute, we find that S is
4620 ft. per min. Hence, the rim of this fly wheel travels at a speed of 4620
ft. per minute.
CIRCUMFERENCES OF CIRCLES 55
If we have given a certain speed which is wanted and have the
circumference of the wheel, then the R. P. M. (revolutions per
minute) will be obtained by dividing the desired speed by the
circumference. In the example just worked, if we want to give
the fly wheel a rim speed of 5280 ft. per minute, it requires
no argument to show that the wheel will have to run at
5280-^-22 = 240 revolutions per minute. In such a case, we
would use our formula in the form
A7 S
N= c
This formula expresses the same relation as S = NxC, but now
it is rearranged to enable us to find the R. P. M. when the rim
speed and the circumference are given.
Sometimes, especially with emery wheels, we know the proper
surface speed and we have an arbor belted to run a certain
number of R. P. M. The problem then is to find the proper size
of stone to order.
The desired speed divided by the number of R. P. M. will give
the circumference, and from this we can figure the diameter of
the stone.
c- s
C ~N
Here again we have merely rearranged the formula S = CXN so
as to be in more suitable form for finding the circumference when
the surface speed and the R. P. M. are given.
53. Grindstones and Emery Wheels. Makers of emery wheels
and grindstones usually give the proper speed for the stones in
feet per minute. This refers to the distance that a point on the
circumference of the stone should travel in 1 minute and is
called the "surface speed" or the "grinding speed."
The proper speed at which to run grindstones depends on the
kind of grinding to be done and the strength of the stones.
For heavy grinding they can be run quite fast. For. grinding
edge tools they must be run much slower to get smooth sur-
faces and to prevent heating the fine edges of the tools. The
following surface speeds may be taken as representing good
practice:
Grindstones:
For machinists' tools, 800 to 1000 ft. per minute.
For carpenters' tools, 550 to 600 ft. per minute.
56 SHOP ARITHMETIC
Grindstones for very rapid grinding:
Coarse Ohio stones, 2500 ft. per minute.
Fine Huron stones, 3000 to 3400 ft. per minute.
Sometimes the rule is given for grindstones as follows: "Run
at such a speed that the water just begins to fly." This
is a speed of about 800 ft. per minute and would be a good
average speed for sharpening all kinds of tools.
Examples :
1. A 36-in. grindstone, used for sharpening carpenters' and pattern-
makers' tools, is run at 60 R. P. M. Is this speed correct?
We must first find the circumference and then the surface speed to see if
it falls between the allowable limits.
36 in. -i- 12 = 3 ft., the diameter Explanation: First we find the
C = 7tXD circumference, which comes out
C = 3. 1416X3 = 9.4248 ft. 9.4248ft. Using this and the R. P.
S = CxN M., we find S to be 565 F. P. M.
5 = 9.4248X60 = 565.488 (feet per minute). As this lies be-
S = 565.488 ft. per minute. tween the allowed limits (550 to 600
F. P. M.) the speed of the stone is
correct.
2. At what R. P. M. should a 50-in. Huron stone be run if it is to be used
for rough grinding?
C nXD Explanation: First we find the
C = 3. 1416X50 = 157.08 in. circumference of the stone in feet,
C = 157.08 in. = 13.09 ft. which turns out to be a little over 13
~~To ft. The proper speed is given as
g 3000 to 3400 F. P. M. Trying 3200
N=Y? .we find that N comes out 246 R.
TVI Q Qonn v P M P. M. The stone should, therefore,
Take S = 320C F. P. M. be be]ted to fun about 24Q Qr 25Q
AT = 3200 = 246.+ R.P.M. R.P.M.
lo
Emery wheels are usually run at a speed of about 5500 ft.
per minute. A good, ready rule, easy to remember, is a speed of
a mile a minute. Most emery wheel arbors are fitted with two
pulleys of different diameters. When the wheel is new, the larger
pulley on the arbor should be used and, when the wheel becomes
worn down sufficiently, the belt should be shifted to the smaller
pulley. Never shift the belt on an emery wheel, however, with-
out first calculating the effect on the surface speed of the wheel.
Many serious accidents have been caused by emery wheels
bursting as a result of being driven at too great a speed. Before
cutting a new wheel on an arbor the resultant surface speed
CIRCUMFERENCES OF CIRCLES 57
should be calculated, to see if the R. P. M. is suitable for the size
of the wheel.
Example :
What size wheel should be ordered to go on a spindle running
1700 R. P.M.?
-I
3 - 106
D = 12 in. wheel, Answer.
Note. A wheel of exactly 12 in. diameter would, at 1700 R. P. M., have*
a surface speed of 5340 F. P. M. (1700 XT: = 5340).
54. Cutting Speeds. Cutting speeds on lathe and boring mill
work may be calculated in the same way that grinding speeds are
calculated. The life of a lathe tool depends on the rate at which
it cuts the metal. This cutting speed is the speed with which
the work revolves past the tool and is, therefore, obtained by
multiplying the circumference of the work by the revolutions
per minute. The same formulas are used as in the calculations
for emery wheels and grindstones but, of course, the allowable
speeds are much different. Tables of proper cutting speeds are
given in many handbooks in feet per minute. To find the
necessary R. P. M., divide the cutting speed by the circumference
of the work.
The cutting speeds used in shops have increased considerably
with the advent of the high speed steels. No exact figures can
be given for the best speeds at which to cut different metals.
The proper speed depends on the nature of the cut, whether
finishing or roughing, on the size of the work and its ability to
stand heavy cuts, the rigidity and power of the lathe, the nature
of the metal being cut, and the kind of tool used. If the work
is not very rigid it is, of course, best to take a light cut and run
at rather high speed. On the other hand, it is generally agreed
that more metal can be removed in the same time if a moderate
speed is used and a heavy cut taken.
As nearly as any general rules can be given, the following
table gives about the average cutting speeds.
58 SHOP ARITHMETIC
CUTTING SPEEDS IN FEET PER MINUTE
Kind
of tool
Material
Carbon steel
High speed steel
Cast iron
30 to 40
60 to 80
Steel or wrought iron
25 to 30
50 to 60
Tool steel. . .
20 to 25
40 to 50
Brass. . . . . .
80 to 100
160 to 200
Cutting speed per minute (in feet)
revolutions per minute.
^Circumference of work (in feet)
Example :
A casting is 30 in. in diameter. Find the number of R. P. M.
necessary for a cutting speed of 40 ft. per minute.
94.248 _ _
^,^ = 7.804 ft., circumference.
\.Zi
N = ~ =^ ^7 = 5.09 R. P. M., Answer.
/ . oo4
The same principles apply to milling and drilling, except that
in these cases the tool is turning instead of the work. Conse-
quently, the cutting speeds are obtained from the product of the
circumference of the tool times its R. P. M.
In calculating the cutting speed of a drill, take the speed of the
outer end of the lip or, in other words, the speed of the drill
circumference.
Example :
A ^-in. drill is making 300 revolutions per minute; what is the
cutting speed?
3.1416X^ = 1.5708, circumference in inches
a
=.131 ft. (nearly)
.131X300 = 39.3 ft. per minute, cutting speed.
55. Pulleys and Belts. If the rim of a pulley is run at too
great a speed, the pulley may burst. The rim speeds of pulleys
are calculated in the same manner as are grinding and cutting
speeds. A general rule for cast iron pulleys is that they should
not have a rim speed of over a mile a minute (5280 ft. per minute) .
RATIO AND PROPORTION 59
This speed may be exceeded somewhat if care is taken that the
pulley is well balanced and is sound and of good design.
The proper speeds for belts is taken up fully in a later chapter
under the general subject of belting. It is well, however, to
point out now that the speed at which any belt is travelling
through the air is practically the same as that of the rim of
either of the pulleys over which the belt runs; and, if we neglect
the small amount of slipping which usually occurs between a
belt and its pulleys, we can say that the speed of a belt is the
same as the rim speed of the pulleys. It will be seen from this
that if two pulleys are connected by a belt, their rim speeds are
practically the same.
PROBLEMS
101. A stack is measured with a tape line and its circumference found to
be 88 in. What is the diameter of the stack?
102. An emery wheel 16 in. in diameter runs 1300 R. P. M. Find the sur-
face speed.
103. The Bridgeport Safety Emery Wheel Co., Bridgeport, Conn., build
an emery wheel 36 in. in diameter and recommend a speed of 425 450
revolutions. Calculate the surface speeds at 425 and at 450 revolutions.
104. An emery wheel runs 1000 R. P. M. What should be its diameter
to give a surface speed of 5500 ft.?
105. A grindstone 3 \ ft. in diameter is to be used for grinding carpenters'
tools; how many R. P. M. should it run?
106. Calculate the belt speed on a high-speed automatic engine carrying
a 48 in. pulley and running at 250 R. P. M.
107. How many revolutions will a locomotive driving wheel, 72 in. in
diameter, make in going 1 mile?
108. What would be the rim speed in feet per minute of a fly wheel 14
ft. in diameter running 80 R. P. M.?
109. At how many R. P. M. should an 8 in. shaft be driven in a lathe to
give a cutting speed of 60 ft. per minute?
110. At what R. P. M. should a 1^ in. high speed drill be run to give a
cutting speed of 80 ft. per minute? If the drill is fed .01 in. per revolution,
how long will it take to drill through 2 in. of metal?
CHAPTER VIII
RATIO AND PROPORTION
66. Ratios. In comparing the relative sizes of two quantities,
we refer to one as being a multiple or a fraction of the other. If
one casting weighs 600 lb., and another weighs 200 lb., we say
that the first one is three times as heavy as the second, or that
GO SHOP ARITHMETIC
the second is one-third as heavy as the first. This relation
between two quantities of the same kind is called a Ratio.
In comparing the speeds of two pulleys, one of which funs 40
revolutions per minute and the other one 160 revolutions per
minute, we say that their speeds are " as 40 is to 160," or " as 1 is
to 4." In this sentence, "40 is to 160" is a ratio, and so also is
"1 is to 4" a ratio.
Ratios may be written in three ways. For example, the ratio
of (or relation between) the diameters of two pulleys which are
12 in. and 16 in. in diameter can be written as a fraction, -J-f ; or,
since a fraction means division, it can be written 12-7-16; or,
again, the line in the division sign is sometimes left out and it
becomes 12:16. The last method, 12:16, is the one most used
and will be followed here. It is read "twelve is to sixteen."
A ratio may be reduced to lower terms the same as a fraction,
without changing the value of the ratio. If one bin in the stock
room contains 1000 washers, while another bin contains 3000,
then the ratio of the contents of the first bin to the contents of
the second is " as 1000 is to 3000." The relation of 1000 to 3000
can be reduced by dividing both by 1000. This leaves the ratio
1 to 3.
1000 -s- 1000 1
Hence, the ratio between the contents of the bins is also as 1 is
to 3.
Likewise, the ratio 24:60 can be reduced to 2:5 by dividing
both terms by 12. If we write it as a fraction we can easily see
that
24 = 24--12_2
60~60-f-12~5
Therefore, 24:60 = 2:5.
The ratio of the 1000 washers to the 3000 washers is 1000:3000
or 1:3.
The ratio of 8 in. to 12 in. is 8:12 or 2:3.
The ratio of $1 to $1.50 is 1:1 J or 2:3.
The ratio of 30 castings to 24 castings is 30:24 or 5:4.
57. Proportion. When two ratios are equal, the four terms
are said to be in proportion. The two ratios 2:4 and 8:16 are
clearly equal, because we can reduce 8:16 to 2:4 and we can
RATIO AND PROPORTION 61
therefore write 2:4 = 8:16. When written thus, these four
numbers form a Proportion.
Likewise, we can say that the numbers 6, 8, 15, and 20 form
a proportion because the ratio 6:8 is equal to the ratio of 15:20.
6:8 = 15:20
Now, it will be noticed that, if the first and fourth terms of
this proportion be multiplied together, their product will be
equal to the product of the second and third terms:
/First\ /SecondN /Third\ /Fourth\
Vterm/ V term ) \ term / \ term /
6:8 15 , : 20
6X20 = 120
8X15 = 120
Therefore, 6X 20 = 8X15
This is true of any proportion and forms the basis for an easy
way of working practical examples, where we do not know one
term of the proportion, but know the other three. The first and
fourth terms are called the Extremes, and the second and third
are called the Means. Then we have the rule: " The product of
the means is equal to the product of the extremes."
This relation can be very nicely and simply expressed as a
formula.
Let a, &, c, and d represent the four terms of any proportion
so that
a : b=c : d
Then, according to our rule, we have
aXd=bXc
Let us now see of what practical use this is. We will take this
example:
If it requires 137 Ib. of metal to make 19 castings, how many
pounds will it take to make 13 castings from the same pattern?
Now very clearly the ratio between the number of castings
19:13 is the same as the ratio of the weights, but one of the
weights we do not know. Writing the proportion out and put-
ting the word " answer" for the number which we are to find, we
have
19:13 = 137 :Answer
02 SHOP ARITHMETIC
From our rule which says the product of the means equals the
product of the extremes:
13 X 137 1781, product of " means."
This must equal the product of the extremes which would be
19 X Answer.
Then: 19 X Answer = 1781
Answer = 1781-*- 19
Answer = 93. 7+ Ib.
In using proportion keep the following things in mind:
(1) Make the number which is the same kind of thing as the
required answer the third term. Make the answer the fourth
term.
(2) See whether the answer will be greater or less than the
third term; if less, place the less of the other two numbers for the
second term; if greater, place the greater of the other two numbers
for the second term.
(3) Solve by knowing that the product of the means equals the
product of the extremes, or by this rule: Multiply the means
together and divide by the given extreme; the result will be the
other extreme or answer.
Let us see how these rules would be applied to a practical
example.
Example :
A countershaft for a grinder is to be driven at 450 R. P. M. by a
lineshaft that runs 200 R. P. M. If the pulley on the countershaft is 8 in.
in diameter, what size pulley should be put on the lineshaft? A proportion
can be formed of the pulley diameters and their revolutions per minute.
Applying the rules of proportion, we get the following analysis and solution
to the problem.
(1) The diameter of the lineshaft pulley is the unknown answer. The
other number of the same kind is the diameter of the countershaft pulley
(8 in.). So we have the ratio.
8 :Answer
(2) If the countershaft pulley is to run faster, its diameter must be
smaller than the other one. Therefore, the answer is greater than 8.
Hence, the greater revolutions (450) will be placed as the second term and
the other R. P. M. (200) will be the first term. Therefore, we have the
completed proportion:
200:450 = 8 : Answer
(3) Solving this we get:
450X8 = 3600, product of means.
3600-^-200 = 18, Answer.
Hence, an 18-in. pulley should be put on the lineshaft to give the desired
speed to the countershaft.
RATIO AND PROPORTION 03
Sometimes the letter X is used to represent the unknown
number whose value is sought. The following is an example of
such a case.
6:40 = 5:X. Find what number X stands for.
40 X 5 = 200, product of the means.
Hence, 6XX = 200
200
X=~ =33.3+, Answer.
58. Speeds and Diameters of Pulleys. As shown in an example
previously worked, if two pulleys are belted together, their
diameters and revolutions per minute can be written in a pro-
portion having diameters in one ratio and R. P. M. in the other
ratio of the proportion. It will be noticed from the example
which was worked, that the numbers which form the means apply
to the same pulley, while the extremes both refer to the other pul-
ley. Then, since the product of the means equals the product
of the extremes, we obtain the following simple relation for
pulleys belted together: The product of the diameter and revolu-
tions of one pulley equals the product of the diameter and revolu-
tions of the other. This gives us the following simple rule for
working pulley problems.
Rule for Finding the Speeds or Diameters of Pulleys. Take the
pulley of which we know both the diameter and the R. P. M.,
and multiply these two numbers together. Then divide this
product by the number that is known of the other pulley. The
result is the desired number.
Examples :
1. A 36-in. pulley running 240 R. P. M. is belted to a 15-in. pulley. Find
the R. P. M. of the 15-in. pulley.
36 X 240 = 8640, the product of the known diameter and revolutions.
8640 -=-15 = 576, the R. P. M. of the 15-in. pulley, Answer.
2. A 36-in. grindstone is to be driven at a speed of 800 R. P. M. from a
6-in. pulley on the lineshaft which is running 225 R. P. M. \Vhat size pulley
must be put on the grindstone arbor?
S Explanation: First we must find
C the R. P. M. for the grindstone as
800 explained in Chapter VII. To get
= Q . .-- ~ =85 R. P. M., nearly, the required surface speed we find
v99* ,o- n 85 R. P. M. necessary.
, Now we have the R. P. M. and
the size of the lineshaft pulley.
Use a 16-m. pulley on the arbor. The product of these two nu J jnber8
is 1350. Dividing this by the R.
P. M. of the grindstone arbor gives 16 in. as the nearest even size of pulley,
so we will use that size.
64
SHOP ARITHMETIC
69. Gear Ratios. The same principles as are applied to pulleys
can be applied to gears. If we have two gears running together
as shown in Fig. 9, the product of the diameter 'and R. P. M. of
one gear will be equal to the product of the diameter and R. P. M.
of the other. In studying gearing, we do not deal with the diam-
eters so much as we do with the numbers of teeth. We find that
gears are generally designated by the numbers of teeth. For
example, we talk of 16 tooth gears and 24 tooth gears, etc., but
we seldom talk about gears of certain diameters.
In making these calculations for gears, we can use the numbers
of teeth instead of the diameters. When a gear is revolving, the
number of teeth that pass a certain point in one minute will be
the product of the number of teeth times the R. P. M. of the
gear.
FIG. 9.
If this gear is driving another one, as in Fig. 9, each tooth on
the one gear will shove along one tooth on the other one. Conse-
quently, the product of the number of teeth times R. P. M. of
the second gear will be the same as for the first gear. This gives
us our rule for the relation of the speeds and numbers of teeth of
gears.
Rule for Finding the Speeds or Numbers of Teeth of Gears.
Take the gear of which we know both the R. P. M. and the num-
ber of teeth and multiply these two numbers together. Divide
their product by the number that is known about the other gear.
The quotient will be the unknown number.
PULLEY AND GEAR TRAINS 65
Example :
A 38 tooth gear running 360 R. P. M. is to drive another gear at
190 R. P. M. What must be the number of teeth on the other gear?
38 X 360 = 13,680, the product of the number of teeth and revolutions
of one gear.
1 90 X Answer = 13, 680
13,680
Answer -
190
Answer = 72 teeth.
PROBLEMS
111. (a) If you draw $33.00 on pay day and another man draws $22.00,
what is the ratio of your pay to hist
(b) What is the ratio of his pay to yours?
112. The speeds of two pulleys are in the ratio of 1:4. If the faster one
goes 260 R. P. M., how fast does the slower one go?
113. Two castings are weighed and the ratio of their weights is 5:2.
If the lighter one weighs 80 lb., what does the heavier one weigh?
114. Find the unknown number in each of the following proportions:
(a) 2:10 = 5:Answer
(b) 6:42 = 5:Answer
(c) 7:35 = 10:X
(d) 6:72 = 8 iX"
115. If it takes 72 lb. of metal to make 14 castings, how many pounds are
required to make 9 castings?
116. A 14 tooth gear is driving a 26 tooth gear. If the 14 tooth gear
runs 225 revolutions per minute, what is the speed of the 26 tooth gear?
117. A 12 in. lineshaft pulley runs 280 revolutions and is belted to a
machine running 70 revolutions. What must be the size of the pulley on
the machine?
118. A lineshaft runs 250 R. P. M. A grinder with a 6 in. pulley is to
run 1550 R. P. M. Determine size of pulley to put on the lineshaft to run
the grinder at the desired speed.
119. An apprentice was given 100 bolts to thread. He completed three-
fifths of this number in 45 minutes and then the order was increased so
that it took him 2 hours for the entire lot. How many bolts did he thread?
120. A 42 in. planer has a cutting speed of 30 ft. per minute and the ratio
of cutting speed to return speed of the table is 1 : 2.8. What is the return
speed in feet per minute?
CHAPTER IX
PULLEY AND GEAR TRAINS CHANGE GEARS
60. Direct and Inverse Proportions. A proportion formed of
numbers of castings and the weights of metal required to make
them is a direct proportion, because the amount of metal required
increases directly as the number of castings increases.
66 SHOP ARITHMETIC
When two pulleys (or gears) are running together, one driving
the other, the larger of the two is the one that runs the slower.
The proportion formed from their diameters and revolutions is,
therefore, called an Inverse Proportion, because the larger pulley
runs at the slower speed. The number of revolutions of one
pulley is said to vary inversely as its diameter, since the greater
the diameter, the less the number of revolutions it will make.
In every pair of gears one of them is driving the other, so the
one can be called the driving gear, or the driver, and the other the
driven gear, or the follower. These names are in quite general
use to designate the gears and to assist in keeping the propor-
tions in the right order. Accordingly, we have the proportion:
R. P. M. of
driven
/R. P. M. of\ = /No. of teethN /No. of teethN
' y driver / \ on driver / ' \ on driven /
This is an inverse proportion because the driver and the driven
are in the reverse order in the second ratio from what they are
in the first ratio. Perhaps this can be seen better if the ratios are
written as fractions.
R. P. M. of driven _ No. of teeth on driver
R. P. M. of driver No. of teeth on driven
Fio. 10.
Here the reason for the name "inverse proportion" is easily
seen. The second fraction has the driver and the driven inverted
from what they are in the first fraction. This method of writing
proportions as fractions is much used in solving problems in
gears or pulleys.
61. Gear Trains. A gear train consists of any number of
gears used to transmit motion from one point to another. Fig.
PULLEY AND GEAR TRAINS 67
10 shows the simplest form of gear train, having but two gears.
Fig. 1 1 shows the same gears A and J3, as in Fig. 10, but with a
third gear, usually called an intermediate gear, between them.
The intermediate gear C can be used for either of two reasons:
1. To connect A and B and thus permit of a greater distance
between the centers of A and B without increasing the size of
the gears; or
2. To reverse the direction of rotation of either A or B. If
A turns in a clockwise direction, as shown in both Figs. 10 and
11, B in Fig. 10 will turn in the opposite, or counter-clockwise
direction, but in Fig. 11, B will turn in the same direction as A.
FIG. 11.
The introduction of the intermediate gear C has no effect on
the speed ratio of A to B. If A has 48 teeth and B 96 teeth, the
speed ratio of A to B will be 2 to 1 in either Fig. 10 or Fig. 11.
In Fig. 10 suppose A to be the driver.
R. P. M. of driver _ No. of teeth on driven
R. P. M. of driven" No. of teeth on driver
R. P. M. of driver _ 96 _2
R. P. M. of driven" 48~I
Hence, the speed ratio of A to B is 2 to 1.
In the case shown in Fig. 11 when A moves a distance of one
tooth, the same amount of motion will be given to C, and C must
at the same time move B one tooth. To move B 96 teeth, or
one revolution, will require a motion of 96 teeth on A, or two
revolutions of A. Hence, A will turn twice to each one turn of
6
68 SHOP ARITHMETIC
B, or the speed ratio of A to B is 2 to 1, just as in the case of
Fig. 10.
62. Compound Gear and Pulley Trains. Quite often it is
desired to make such a great change in speed that it is practically
necessary to use two or more pairs of gears or pulleys to accom-
plish it! If a great increase or reduction of speed is made by a
single pair of gears or pulleys, it means that the difference in the
diameters will have to be very great. The belt drive of a lathe is
an example of a compound train of pulleys, though here the
train is used chiefly for other reasons. In the first step, the
pulley on the lineshaft drives a pulley on the countershaft; then
another pulley on the countershaft drives the lathe. The back
gearing on a lathe is an example of compound gearing, two
pairs of gears being used to make the speed reduction from the
cone pulley to the spindle and face plate.
FIG. 12.
Fig. 12 shows a common arrangement of compound gearing.
Here A drives B and causes a certain reduction of speed. B
and C are fastened together and therefore travel at the same
speed. A further reduction in speed is made by the two gears
C and D. A and C are the driving gears of the two pairs and
B and D are the driven gears.
In making calculations dealing with compound gear or pulley
trains, we might make the calculations for each pair as explained
in Chapter VIII and then proceed to the next pair, etc., but this
can be shortened to form a much simpler process.
The speed ratio for a pulley or gear train is equal to the product
of the ratios of all the separate pairs of pulleys or gears making up
the train.
PULLEY AND GEAR TRAINS 69
In using this principle for calculations, the ratios are written
as fractions and Ave have the following formula:
R. P. M. of last driven gear Product of Nos. of teeth of all drivers
K. P. M. of first driver 1 roduct of Nos. of teeth of all driven gears
Or, if we want the ratio stated the other way around
R. P. M. of first driver Product of Nos. of teeth of all driven gears
R. P. M. of last driven gear" Product of Xos. of teeth of all drivers
Example :
Let us calculate the speed ratio for the train of gears in Fig. 12.
This would be the ratio of the speed of A to the speed of D.
A is the first driver and D the last driven gear, and the ratio of their speeds
is the ratio for the whole train.
Speed of A ^ Teeth on Bx Teeth on D
Speed of~Z> "" Teeth on A X Teeth on C
2
5 I
10
Hence,
Speed of A: Speed of D = 10:l
In other words, A revolves 10 times as fast as D.
Problems in getting the speed ratios of pulley trains are solved
in the same way except that diameters are used instead of num-
bers of teeth.
Speed of last driven pulley _ Product of diameters of all driving pulleys
Speed of first driving pulley Product of diameters of all driven pulleys
Example :
Let us take the pulley train of Fig. 13 and calculate the ratio of
the speeds of pulleys A and D. A and C are the drivers and B and D are the
driven pulleys.
Speed of A _ Diameter of B X Diameter of D
Speed of D Diameter of A X Diameter of C
5
_ 5 __ i_
~36 7.2
3 12
Hence,
Speed of A : Speed of D = 1 : 7.2
Trains are frequently used having combinations of pulleys and
gears. In nearly all machine tools, we will find both pulleys and
gears between the lineshaft and the work. In wood-working
70 SHOP ARITHMETIC
machinery, on the other hand, we usually find only pulleys and
belts, on account of the high speeds at which the machines are
run. In calculating the speed ratios of these combined trains,
we can use the diameters of the pulleys and the numbers of
teeth of the gears in the same formula.
/R. P. M. of\ /Product of diameters\ x /Product of teeth oA
\first driver/ _ \of all driven pulleys/ \ all driven gears /
/R. P. M. of\ = /Product of diametersX /Product of teeth of\
yast driven j \of all driving pulleys j x \^ all driving gears )
Fio. 13.
If a problem calls for the calculation of the size of one pulley
or gear in a train, all the others and the speed ratio being known,
start at one or both ends of the train and work toward the gear or
pulley in question until you get a proportion which will give the
desired quantity.
Example:
The punch shown m Fig. 14 is to be set up so that it will make 20
strokes per minute. (The 80 tooth gear must, therefore, run 20 R. P. M.)
The punch is to be driven from a countershaft and we want to calculate the
size of the pulley to put on the countershaft, to drive the punch at the desired
speed. We find that the main lineshaft runs 240 R. P. M. and carries a 16-in.
pulley which drives a 24-in. pulley on the countershaft.
Working from the lineshaft:
24 in.xR. P. M. of countershaft = 16X240.
R. P. M. of countershaft=^p = 160 R. P. M.
24
PULLEY AND GEAR TRAINS
Working from the punch:
20XR. P. M. of 20 T gear = 80X20.
R. P. M. of 20 T gear = -^-- = 80
71
This is also the R. P. M. of the 24-in. pulley and this pulley is driven
by the unknown pulley on the countershaft, which we have found
runs 160 R. P. M.
IGOxDiam. of pulley = 80X24
1920
Diam. of pulley = -^r- = 12 in., Answer.
Fio. 1 1.
Another way to solve this would be to write put an equation for the entire
train, using X to represent the pulley whose size we want to find.
20 = 16 XXX 20
240 "24X24X80
? 1
==
12 6X24 6 24
XI IX 1
Now, if ^ X is to equal ^-.y then must be which would be when A"
12
24
is 12.
Hence, X 12 in., Answer.
72 SHOP ARITHMETIC
63. Screw Cutting. Most lathes are equipped with a small
plate giving the necessary gears to use for cutting different
threads, but every good machinist should know how to calculate
the proper gear setting for such work. This is a simple problem
in gear trains and should cause no difficulty for the man who
understands the principles of gear trains.
The lathe carriage and tool are moved by a "lead screw"
having usually 2, 4, 6, or 8 threads per inch. If a lathe has a
lead screw having 6 threads per inch, each revolution of the lead
screw will move the carriage in. ; a 4 pitch screw would move
the carriage \ in. for each revolution of the screw. Then, if the
spindle of the lathe and the lead screw turn at the same speed, the
lathe will cut a thread of the same pitch as that on the lead screw.
If a finer thread is wanted than that on the lead screw, the
spindle should make more turns than does the lead screw.
Suppose we want to cut 24 threads per inch and have a 6 thread
per inch lead screw. It will require 6 turns of the lead screw to
move the carriage 1 inch. Meanwhile, the work should revolve
24 times. Then the ratio of spindle speed to lead screw speed
should be 4:1
Speed of spindle Threads per inch to be cut
Speed of lead screw Threads per inch on lead screw
The first driving gear is that on the spindle, while the last driven
gear is that on the end of the lead screw. Hence,
Threads per inch to be cut _ Product of Nos. of teeth on driven gears
Threads per inch on lead screw Product of Nos. of teeth on driving gears
PROBLEMS
121. In Fig. 10, if we removed the 48 tooth gear and put a 64 tooth gear
in its place, what would be the speed ratio of A to B?
122. In Fig. 11, if B makes 6 revolutions, how many turns will C make
and how many will A make?
123. What would be the speed ratio of the train of Fig. 12 if we put a
32T gear on at C and a 48T gear at D?
124. The lineshaft in Fig. 15 runs 250 R. P. M. Determine the size of
lineshaft pulley to run the grinder at 1550 R. P. M. using the countershaft
as shown in the figure.
126. A machinist wishes to thread a pipe on a lathe having 2 threads per
inch on the lead screw. There are to be 11^ threads per inch on the pipe.
What is the ratio of the speeds of the spindle and the lead screw?
126. Two gears are to have a speed ratio of 4.6 to 1. If the smaller gear
has 15 teeth, what must be the number of teeth on the larger gear?
PULLEY AND GEAR TRAINS
73
COUNT CJ?SHAFT
GRINDLR
FIQ. 15.
FIG. IS.
74 SHOP ARITHMETIC
127. It has been decided to equip the punch in Fig. 14 with a motor drive
by replacing the fly wheel with a large gear to be driven by a small pinion
on the motor. If the motor runs 800 R. P. M., and has a 16 tooth pinion,
what must be the number of teeth on the other gear? Speed of the punch
to be 20 strokes per minute.
128. A street car is driven through a single pair of gears, a large gear on
the axle being driven by a smaller one on the motor shaft. If a car has
33-in. wheels and a gear ratio of 1 :4, how fast would the car go when the
motor is running 1200 R. P. M.?
129. Fig. 16 shows the head stock for a lathe. The cone pulley carries
with it the cone pinion A, which drives the back gear B. B is connected
solidly with the back pinion C which drives the face gear D. If the gears
have the following numbers of teeth, determine the back gear ratio (speed
of A : speed of D) :
Teeth on cone pinion, A 28
Teeth on back gear, B 82
Teeth on back pinion, C 25
Teeth on face gear, D 74
130. If you were to cut a 20 pitch thread on a lathe having a 4 pitch lead
screw, what would be the ratio of the speeds of the spindle and the lead
screw?
CHAPTER X
AREAS AND VOLUMES OF SIMPLE FIGURES
64. Squares. In taking up the calculation of areas of surfaces
and the volumes and weights of objects, the expressions "square"
and "square root" will be met and must be understood. To
one unfamiliar with these names and the corresponding operations
the signs and operations themselves seem difficult. They are in
reality very simple. The square of a number is simply the prod-
uct of the number multiplied by itself; the square of 2 is 2 X2 = 4;
the square of 5 is 5X5 = 25; the square of 12.5 is 12.5X12.5 =
156.25. Instead of writing 2x2 or 5X5, it is customary to
write 2 2 and 5 2 . These are read "2 squared" and "5 squared."
12.5 2 =12.5 squared, and so on. The little 2 at the upper
right hand corner is called the Exponent.
65. Square Root. The square root of a given number is
simply another number which, when multiplied by itself (or
squared), produces the given number. Thus, the square root
of 4 is 2, since 2 multiplied by itself (2X2) gives 4. The square
root of 9 is 3, since 3X3=3 2 = 9. Square root is the reverse of
square, so if the square of 5 is 25 the square root of 25 is 5. The
mathematical sign of square root, called the radical sign, is \/7
Then V9 = 3; V 25 = 5. These expressions are read "the square
root of 9=3"; "the square root of 25 = 5". Square roots of
larger numbers can usually be found in handbooks and the
actual process of calculating them, which is somewhat compli-
cated, will be taken up later on.
66. Cubes and Higher Powers. In the same way that 2 2
(2 squared) =2X2 =4, 2 3 (2 cubed) =2x2x2 = 8. The exponent
simply indicates how many times the number is used as a factor,
or how many times it is multiplied together. 4 s = 4x4x4 = 64.
3 3 =3X3X3 = 27.
Just as square root is the reverse of square, so cube root is
the reverse of cube. The sign for cube root is V7 So if 3 8 =
3X3X3=27, then ^27 = 3. Sometimes a factor is repeated
more than 3 times, in which case, the exponent indicates the
7 75
76
SHOP ARITHMETIC
number of times. 2* means 2x2x2x2 and is read "2 to the
fourth power." 2 s = 2X2X2X2X2 and is read "2 to the fifth
power," and so on.'" The roots are indicated in the same way.
^16 = fourth root of 16 = 2; 5 4 = 5X5X5X5 = 625, etc.
67. Square Measure. Before going further, it will be well
to get clearly in mind just what the term "Square" means in
terms of the things we see. Areas of figures are measured in
terms of the "square" unit. For instance, if the dimensions of
the base of a milling machine are 3 ft. by 5 ft., the floor space
covered by this base is 15 square feet. In this case the area is
5'-
FIQ. 17.
measured by the unit known as the square foot. A Square
Foot is a surface bounded by a square having each side 1 ft.
in length. In case of the milling machine base represented in
Fig. 17, there are by actual count 15 sq. ft. in this surface and
this is readily seen to be the product of the length and the breadth
of the base, since 3X5 = 15.
The Square Inch is another common unit of area. This is
much smaller than the square foot, being only one-twelfth as
great each way. If a square foot is divided into square inches
it will be seen to contain 12X12 or 144 sq. in. (see Fig. 18). It
will be readily seen that the area of any square is equal to the
product of the side of the square by itself. In other words, the
area of a square equals the side "squared" (referring to the
process explained in Article 64). Looking at it the other way
around, the square of any number can be represented by the
area of a square figure, one side of which represents the number
itself. The actual things which the number represents makes no
difference whatever. If the side of a square is 5 in., the area is
AREAS AND VOLUMES OF SIMPLE FIGURES 77
25 sq. in.; if the side is 5 ft., the area is 25 sq. ft. If we simply
have the number 5, its square is 25, no matter what kind of
things the 5 may refer to.
As mentioned before, 1 sq. ft. is the area of a square 1 ft. on
each side and, if divided into square inches, will be found to
contain 12 2 or 144 sq. in. Likewise, a square yard is 3 ft. on
each side and, therefore, contains 3* = 9 sq. ft. The following
table gives the relation between the units ordinarily used in
measuring areas:
FIG. 18.
MEASURES OF AREA (SQUARE MEASURE)
144 square inches (sq. in.) = 1 square foot (sq. ft.)
9 square feet=l square yard (sq. yd.)
30 J square yards = 1 square rod (sq. rd.)
160 square rods=l acre (A)
640 acres = 1 square mile (sq. mi.)
68. Area of a Circle. If a circle is drawn in a square as shown
in Fig. 19, it is easily seen that it has a smaller area than the
square because the corners are cut off. The area of the circle
is always a definite part of the area of the square drawn on its
diameter, the area of the circle being always .7854 times the
area of the square. This number .7854 happens to be just one-
fourth of the number 3.1416 given in Chapter VII. Just why
this is so will be shown later on. If the diameter of the circle
= 10 in., as in Fig. 19, the area of the square is 100 sq. in. and the
area of the circle is .7854x100 = 78.54 sq. in. You can prove
this to your own satisfaction in the following manner. Cut a
78
SHOP ARITHMETIC
square of carHboard of any size, and from the center describe a
circle as shown just touching on all four sides. Weigh the square,
and then cut out the circle and weigh it. The circle will weigh
. 7854 X weight of the square. A pair of balances such as are
found in a drug store are the best for this experiment.
10"-
FIQ. 19.
Rule for Area of Circle. The area of any circle is obtained by
squaring the diameter and then multiplying this result .by .7854.
If written as a formula this rule would read
where A = area of a circle
of which D = the diameter.
Example :
Find the area of a circle 3 in. in diameter.
A= .7854 XD 2
A= .7854 X3 2
= .7854X9 _
= 7.0686 sq. in., Answer.
If you think a little you will see that, if the diameter is doubled,
the area is increased four times. This can also be seen from Fig.
20. The diameter of the large circle is twice that of one of the
small circles, but its area is four times that of one of the small
circles. This is a very important and useful law and may be
stated as follows: "The areas of similar figures are to each other
as the squares of their like dimensions." A 2 in. circle contains
2 2 X. 7854 = 3. 1416 sq. in., while a 6 in. circle contains 6 2 X
.7854 = 28.2744 sq. in., or nine times as much. This we can find
AREAS AND VOLUMES OF SIMPLE FIGURES 79
by saying the 6 in. circle has three times the diameter of the
2 in. circle and, therefore, the area is 3 2 , or nine times as great.
A piece of steel plate 6 in. in diameter weighs nine times as much
as a piece 2 in. in diameter of the same thickness. Likewise a
10 in. square has four times the area of a 5 in. square. If we
Fio. 20.
let A represent the area of the larger circle, a the area of the
smaller circle, D the diameter of the larger circle, and d the diam-
eter of the smaller circle, then we have the direct proportion:
69. The Rectangle. When a four-sided figure has square
corners it is a Rectangle. Each side of a brick is a rectangle.
3"
Fio. 21.
A Square is a special kind of rectangle having all the sides equal.
The area of a rectangle is obtained by multiplying the length by
the breadth. In Fig. 21 the area is 2X3 = 6 sq. in., as can be
seen by counting the 1-in. squares, which each contain 1 sq. in.
80
SHOP ARITHMETIC
70. The Cube. Just, as the square of a number is represented
by the area of a square, one side of which represents the number,
so the cube of a number is represented
by the volume of a cubical block, each
edge of which represents the number.
The volume of a cube which is 10 in.
on each edge is 10 X 10 X 10 - 1000 cubic
inches, and since. this is obtained by
"cubing" 10 (10 3 = 10X10X10 = 1000),
we can see that the cube of a number
FIQ 22 can be represented by the volume of a
cube, the edge of which represents the
number. If the edge of the cube is one-half as long, that is 5 in.,
the volume is 5X5X5 = 1 25 cubic inches, or only the volume of
the 10 in. cube.
^ s> ^>
I
, "p\
1
1
\
^
Fia. 23.
MEASURES OF VOLUME (CUBICAL MEASURE)
1728 cubic inches (cu. in.) = 1 cubic foot (cu. ft.)
27 cubic feet = l cubic yard (cu. yd.)
(Larger units than cubic yards are seldom, if ever, used.)
71. Volumes of Straight Bars. A piece 1 in. long cut from a bar
will naturally contain just as many cu. in. as there are sq. in.
on the end of the bar. In the billet shown in Fig. 24, there are
3x4 = 12 sq. in. on the end of the bar, and a piece 1 in. long
contains 12 cu. in. The entire billet, contains 10 slices just like
this one, so there are 12X10 = 120 cu. in. in the entire billet.
AREAS AND VOLUMES OF SIMPLE FIGURES 81
Therefore, we see that to find the number of cu. in. in any
straight bar we proceed as follows:
Calculate the area of one end of the bar in square inches;
then multiply this result by the length of the bar in inches;
the result will be the number of cubic inches in the bar. For
bars of square or rectangular section, the volume is the product
of the three dimensions, length, breadth, and thickness. If L,
B, and T represent the length, breadth, and thickness, and V
stands for the volume, then
V=LXBXT
L
10"-
Fio. 24.
Example :
How many cubic inches of steel in a bar 2 in. square and 4 ft. long?
4 ft. =48 in.
V=LXBXT
= 48X2X2 = 192 cu. in., Answer.
For round bars, the area on the end is .7854 times the square of
the diameter, and this, multiplied by the length, gives the volume.
Then, if D represents the diameter of the bar and'L its length,
the volume V will be
7=.7854X 2 XL
This will apply equally well to thin circular plates or to long
bars or shafting. With thin plates, we would naturally speak
of thickness (T) instead of length (L). Fig. 25 shows that the
two objects have the same shape except that their proportions
are different.
Examples :
1. How many cubic inches of steel in a shaft 2 in. in diameter and 12 ft.
long?
12 ft. = 12X12 = 144 in., length of bar.
F = .7854XD 2 XL
V = . 7854 X2 2 X 144
V = .7854 X 4 X 144 =452.39 cu. in., Answer.
82
2. How many cubic inches in a blank for a boiler head 60 in. in diameter
and TTT in. thick?
7 = .7854X60 2 X ^
F = . 7854 X 3600 XTF = 1237 cu. in., Answer.
Fio. 25.
72. Weights of Metals. The chief uses in the shop for calcu-
lations of volume are in finding the amount of material needed
to make some object; in finding the weight of some object that
cannot be conveniently weighed; or in finding the capacity of
some bin or other receptacle. Having obtained the volume of
an object, it is only necessary to multiply the volume by the
known weight of a unit volume of the material to get the weight
of the object. In the case of the shaft of which we just got the
volume, 1 cu. in. will weigh about .283 lb., so the total weight
of the shaft will be
452 . 39 X . 283 = 128 . + pounds.
The weight of the boiler head will be
1237 X . 283 = 350 + pounds.
The following table gives the weights per cubic inch and per
cubic foot for the most common metals and also for water:
Material
. 1 cu. in.
1 cu. ft.
Cast iron
.260 Ib.
450 Ib.
Wrought iron
.278 Ib.
480 Ib.
Steel
.283 Ib.
489 Ib.
Brass
. 301 Ib.
520 Ib.
Copper. .
.3181b.
550 Ib.
Lead
.411 Ib.
711 Ib.
Aluminum
. 094 Ib.
162 Ib.
Water
.036 Ib.
62.4 Ib.
73. Short Rule for Plates. A flat wrought iron plate in.
thick and 1 ft. square will weigh 5 Ib., since 12X12x^ = 18 cu.
in., and 18 X. 278 = 5 Ib. The rule obtained from this is very
easy to remember and is very useful for plates that have their
dimensions in exact feet.
Rule. Weight of flat iron plates = area in square feetX number
of eighths of an inch in thickness X 5. This rule can also be used
for steel plates by adding 2 per cent, to the result calculated from
the above rule.
Example :
3
Find the weight of a steel plate 30 in. X96 in. X s in.
o
1 3
30 in. = 2^ ft., 96 in. =- 8 ft., g in. = 3 eighths.
2^X8X3X5 = 300 (weight if it were of wrought iron).
2% of 300 = 6 Ib.
300 + 6 = 306 Ib.; weight of steel plate, Answer.
If this weight is calculated by first getting the cubic inches of steel, we get :
Q
30 X 96 X 77 = 1080 cu. in.
o
1080 X. 283 = 305.64 Ib. weight of steel plate, Answer.
We see that the results check as closely as could be expected
and, in fact, different plates of supposedly the same size would
differ as much as this because of differences in rolling.
74. Weight of Casting from Pattern. In foundry work, it is
often desired to get the approximate weight of a casting in order
to calculate the amount of metal needed to make it. The prob-
84
SHOP ARITHMETIC
able weight of the casting can be obtained closely enough by
.weighing the pattern and multiplying this weight by the proper
number from the following- table. In case the pattern contains
core prints, the weight of these prints should be calculated and
subtracted from the pattern weight before multiplying; or else the
total pattern weight can be multiplied first and then the weight
of metal which would occupy the same volume as the core print
be subtracted from it.
PROPORTIONATE WEIGHT OF CASTINGS TO WEIGHT OF
WOOD PATTERNS
For each 1 Ib. weight of
pattern when made of
(less weight of core prints)
Casting will weigh if made of
Cast iron
Copper or
bronze
Brass
Aluminum
White pine
16.
12.
10.2
10.6
0.84
2.6
0.95
19.6
14.7
12.5
13.
1.
3.2
1.3
18.5
14.
11.7
12.3
0.95
3.1
1.2
5.9
4.5
3.8
3.9
0.32
0.95
0.38
Mahogany
Pear wood
Birch
Brass
Aluminum
Cast iron
PROBLEMS
131. Find the weight of a piece of steel shafting 2 in. in diameter and
20 ft. long.
132 . What is the weight of a billet of wrought iron 4 in. square and 2 ft.
8 in. long?
133. What would a steel boiler plate 36 in. by 108 in. by \ in. weigh?
134. A cast steel cylinder is 42 in. inside diameter, 4 ft. 6 in. long and
1J in. thick. Find its weight.
135. A steam engine cylinder 4 in. inside diameter has the cylinder
head held on by four studs. When the pressure in the cylinder is 125 Ib.
per square inch, what is the total pressure on the cylinder head and what
is the pull in each stud?
136. 50 studs 1\ in. long and 1 in. in diameter are to be cut from cold
rolled steel. Find the length and weight of bar necessary, allowing \ in.
per stud for cutting off.
137. What would be the weight of a \ in. by 3 in. wagon tire for a 40 in.
wheel? (Length of stock = circumference of & 39 J in. circle.)
SQUARE ROOT
85
138. A copper billet 2 in. by 8 in. by 24 in. is rolled out into a plate of
No. 10 B. & S. gage. The thickness of this gage is .1019 in. What would
be the probable area of this plate in square feet?
139. The steel link shown in Fig. 26 is made of f in. round steel (round
steel f in. in diameter). Find the length of bar necessary to make it and
then find the weight of the link.
140. A steel piece is to be finished as shown in the sketch below (Fig. 27).
The only stock available from which to make it is 4 in. in diameter. Com-
pute the length of the 4 in. stock which must be upset to make the piece and
have -5*5 extra stock all over for finishing.
T
Fio. 26.
Fio. 27.
CHAPTER XI
SQUARE ROOT
76. The Meaning of Square Root. The previous chapter
showed the usefulness of squares in finding areas and of cubes in
finding volumes. Problems often arise in which it is necessary
to find one edge of a square or cube of which only the area or
, volume is given. For instance, what must be the side of a square
so that its area will be 9 sq. in.? The length of the side must be
such that when multiplied by itself it will give 9 sq. in. A
moment's thought shows that 3X3 = 9, or 3 2 = 9. Therefore,
3 is the necessary side of the square. Finding such a value is
called Extracting the Square Root, and is represented by the sign
V called the square root sign or radical sign. Thus v9 = 3 ;
\/16 = 4. To make clear the idea of extracting square roots,
the student should consider it as the reverse or "the undoing"
86 SHOP ARITHMETIC
of squaring, just as division is the reverse of multiplication or as
subtraction is the reverse of addition.
5 2 = 25, and its reverse is: \/25 = 5.
The square roots of some numbers, like 4, 9, 16, 25, 36, 49,
64, 81, etc.', are easily seen, but we must have some method that
will apply to any number. There are several methods of finding
square root, of which two are open to the student of shop arith-
metic: (1) by actual calculation; (2) by the use of a table of
squares or square roots. A third method which uses logarithms
will be explained in the chapters on logarithms. In many
handbooks will be found tables giving the square roots of
numbers, but we must learn some method that can be used
when a table is not available and the method that will now
be explained should be used throughout the work in this
chapter.
76. Extracting the Square Root. The first step in finding the
square root of any number is to find how many figures there are
in the root. This is done by pointing off the number into
periods or groups of two figures each, beginning at the decimal
point and working each way.
12 = 1 10 2 =1'00 100 2 =1'00'00
From these it is evident that the number of periods indicates
the number of figures in the root. Thus the square root of
103684 contains 3 figures because this number (10'36'84) con-
tains three periods. Also the square root of 6'50'25 contains
three figures since there are three periods. (The extreme left
hand period may have 1 or 2 figures in it.) We must not forget
that, for any number not containing a decimal, a decimal point
may be placed at the extreme right of the number. Thus the
decimal point for 62025 would be placed at the right of the number
(as 62025.)
The method of finding the square root of a number can best
be explained by working some examples and explaining the work
as we go along. The student should take a pencil and a piece
of paper and go through the work, one step at a time, as he reads
the explanation.
Example :
Find the square root of 186624.
Point off into periods of two figures each (18'66'24) and it will be seen
that there are 3 figures in the root. The work is arranged very similarly to
division.
SQUARE ROOT 87
18'66'24(432 Explanation: First find the largest
number whose square is equal to or
2X40 = 80
JJ
83
2X430 = 860
2
862
9~fiS~ l ess ^an I**, *ke ^ rst P ef iod. This is
4, since 5 2 is more than 18. Write the
4 to the right for the first figure of the
root just as the quotient is put down
17 24 in long division. The first figure of
the root is 4. Square the 4 and write
17 24 its square (16) under the first period
(18) and subtract, leaving 2.
Bring down the next period (66) and annex it to the remainder, giving
266 for what is called the dividend. Annex a cipher to the part of the root
already found (4) giving 40; then multiply this by 2, making 80, which is
called the trial divisor. Set this off to the left. Divide the dividend (266)
by the trial divisor (80). We obtain 3, which is probably the next figure of
the root. Write this 3 in the root as the second figure and also add it to the
trial divisor, giving 83, which is the final divisor. Multiply this by the figure
of the root just found (3) giving 249. Subtract this from the dividend (266)
leaving 17.
Bring down the next period (24) and annex to the 17, giving a new
dividend 1724. Repeat the preceding process as follows: Annex a cipher
to the part of the root already found (43) giving 430; and multiply by 2,
giving 860, the. trial divisor. Divide the dividend by this divisor and ob-
tain 2 as the next figure of the root. Put this down as the third figure of
the root and also add it to the trial divisor, giving 862 as the final divisor.
Multiply this by the 2 and obtain 1724, which leaves no remainder when
subtracted from the dividend. As there are no more periods in the original
number, the root is complete.
77. Square Roots of Mixed Numbers. If it is required to find
the square root of a number composed of a whole number and
a decimal, begin at the decimal point and point off periods to
right and left. Then find the root as before.
Example :
Find the square root of 257.8623
2' 5 7.86' q 23'00(16.058 + , Answer.
20
_6
26
_1 to or less than 2, the first period. Proceeding as
I 57 before, we get 6 for the second figure. After
subtracting the second time (at a) we find that
, eg the trial divisor 320 is larger than the dividend
186. In this case, we place a cipher in the root,
annex another cipher to 320 making 3200, annex
the next period, 23, to the dividend and then
1 60 25 proceed as before. If the root proves, as in this
25 98 00 case, to be an interminable decimal (one that
does not end) continue for two or three decimal
5
320.5
32100
8
32108
25 68 64
places and put a + sign after the root as in divi-
sion. In this example the decimal point comes
29 36 after 16, because there must be two figures in the
whole number part of the root since there are
two periods in the whole number part of our
original number.
78. Square Roots of Decimals. Sometimes, in the case of a
decimal, one or more periods are composed entirely of ciphers.
The root will then contain one cipher following the decimal point
for each full period of ciphers in the number.
88 SHOP ARITHMETIC
Example :
Take .0007856 as an example.
Beginning at the decimal point and pointing off into periods of two
figures each, we have .00'07'85'60. Hence, the first figure of the root must
be a cipher. To obtain the rest of the root we proceed as before.
.00'07'85'60(.0280 + , Answer.
4
40
_8
48
385
384
560 | 1 60
It will be noticed that the square root of a decimal will always
be a decimal. If we square a fraction, we will get a smaller
fraction for its square, () 2 = ^; or as a decimal, .25 2 = .0625.
Therefore, the opposite is true; that, if we take the square root
of a number entirely a decimal, will get a decimal, but it will be
larger than the one of which it is the square root. Notice the
example just given: .0007856 is less than its square root .028.
79. Rules for Square Root. From the preceding examples
the following rules may be deduced:
1. Beginning at the decimal point separate the number into
periods of two figures each. If there is no decimal point begin
with the figure farthest to the right.
2. Find the greatest whole number whose square is contained
in the first or left-hand period. Write this number as the first
figure in the root; subtract the square of this number from the
first period, and annex the second period to the remainder.
3. Annex a cipher to the part of the root already found and
multiply by 2; this gives the trial divisor. Divide the dividend
by the trial divisor for the second figure of the root and add this
figure to the trial divisor for the complete divisor. Multiply
the complete divisor by the second figure in the root and sub-
tract this result from the dividend. (If this result is larger than
the dividend, a smaller number must be tried for the second
figure of the root.)
Bring down the third period and annex it to the last remainder
for the new dividend.
4. Repeat rule 3 until the last period is used, after which,
if any additional decimal places are required, annex cipher
periods and continue as before. If the last period in the decimal
should contain but one figure, annex a cipher to make a full
period.
SQUARE ROOT
89
5. If at any time the trial divisor is not contained in the
dividend, place a cipher in the root, annex a cipher to the trial
divisor and bring down another period.
6. To locate the decimal point, remember that there will be
as many figures in the root to the left of the decimal point as
there were periods to the left of the decimal point in our original
number.
80. The Law of Right Triangles. One of the most useful laws
of geometry is that relating to the sides of a right angled triangle.
Fig. 28 shows a right angled triangle, or "right triangle," so
FIQ. 29.
called because one of its angles (the one at C) is a right angle, or
90. The longest side (c) is called the hypotenuse. "In any
right triangle the square of the hypotenuse is equal to the sum of
the squares of the other two sides." Written as a formula this
would read
This can be illustrated by drawing squares on each side, as in
Fig. 29, and noting that the area of the square on the hypotenuse
is equal to the sum of the areas of the other two.
In using this rule, however, we do not care anything about these
90
SHOP ARITHMETIC
areas and seldom think of them except as being the squares of
numbers. It is used to find one side of such a triangle when the
other two are known.
Examples :
1. If the trolley pole in Fig. 30 is 24 ft. high, and the guy wire is anchored
7 ft. from the base of the pole, what is the length of the guy wire?
The guy wire is the hypotenuse of a right triangle whose sides are 24 ft.
and 7 ft.
c 2 = b' + a 2
c 2 = 24-< + 7 2
= 576 + 49 = 625
c = v X 625 = 25 ft., Answer.
2. If the triangle of Fig. 31 is a right triangle having the hypotenuse
c = 13 in. and the side a = 5 in., what is the length of the side 6?
Hence,
2 = 132-52
= 169-25 = 144
b =\/144 = 12 in., Answer.
This property of right triangles is also useful in laying out
right angles on a large scale more accurately than it can be done
with a square. This is done by using three strings, wires, or
chains of such lengths that when stretched they form a right
triangle. A useful set of numbers that will give this are 3, 4, and
5, since 3 2 + 4 2 = 5 2 (9 + 16-25).
Any three other numbers having the same ratios as 3, 4 ?
and 5 can be used if desired. 6, 8, and 10; 9, 12, and 15; 12, 16 ;
and 20; 15, 20, and 25; any of these sets of numbers can be used.
SQUARE ROOT 91
A surveyor will often use lengths of 15 ft., 20 ft., and 25 ft. on
his chain to lay out a square corner; this method can also be used
in aligning engines, shafting, etc.
81. Dimensions of Squares and Circles. Square Root must be
used in getting the dimensions of a square or a circle to have
a given area. If the area of a square is given, the length of one
side can be obtained by extracting the square root of the area.
If we wish to know the diameter of a circle which shall have a
certain area, we can find it by the following process:
The area is . 7854 X the square of the diameter or, briefly,
If we divide the given area by .7854, we will get the area of
the square constructed around the circle (see Fig. 19).
One side of this square is the same as the diameter of the circle
and is equal to the square root of the area of the square.
Then, to find the diameter of a circle to have a given area:
Divide the given area by . 7854 and extract the square root of the
quotient.
n- P 1 "
= \77854~
82. Dimensions of Rectangles. Occasionally one encounters
a problem in which he wants a rectangle of a certain area and
knows only that the two dimensions must be in some ratio.
It may be that a factory building is to cover, say 40,000 sq. ft.
of ground and is to be four times as long as it is wide, or some
problem of a similar nature. Suppose we take the case of this
factory and see how we would proceed to find the dimensions
of the building.
Example :
Wanted a factory building to cover 40,000 sq. ft. of ground.
Ratio of length to breadth, 4:1. Find the dimensions.
I
1
1
1
1
1
40000 -s-4
b
Fio. 32.
= 10000
= 10000
= 100
= 4X100 = 400.
92 SHOP ARITHMETIC
Explanation: If we divide the total area by 4, we get 10,000 as the area of
a square having the breadth 6 on each side. From this we find the breadth
or width 6 to be the square root of 10,000 or 100 ft. If the length is four
times as great it will be 400 ft. and the dimensions of the building will be
400 by 100.
83. Cube Root. The Cube Root of a given number is another
number which, when cubed, produces the given number. In other
words, the cube root is one of the three equal factors of a number.
The cube root of 8 is 2, because 2 3 = 2x2x2 = 8; also the cube
root of 27 is 3 (since 3 3 = 27) and the cube root of 64 is 4 (since
4 3 = 64).
The sign of cube root is V placed over the number of which
we want the root. Thus we would write
4 ^1000= 10
If we consider the number of which we want the cube root
as representing the volume of a cubical block, then the cube
root of the number will represent the length of one edge of the
cube. The cube root of 1728 is 12 and a cube containing 1728
cu. in. will measure 12 in. on each edge.
There are four ways of getting cube roots: (1) by actual
calculation, (2) by reference to a table of cubes or cube roots,
(3) by the use of logarithms, and (4) by the use of some calcu-
lating device like the slide rule.
The use of a table is the simplest way of finding cube roots,
but its value and accuracy is limited by the size of the table.
Tables of cubes or cube roots are to be found in many handbooks
and catalogues and should be used whenever they give the desired
root with sufficient accuracy.
Logarithms give us an easy way of getting cube roots, but
here also a table is necessary and the accuracy is limited by the
size of the table of logarithms. 'The use of logarithms will be
explained in a later chapter. The ordinary pocket slide rule
will give the first three figures of a cube root and for many
calculations this is sufficiently accurate. The method of actually
calculating cube roots is very complicated and is used so seldom
that one can never remember it when he needs it. Consequently,
if it is necessary to hunt up a book to find how to extract the
cube root, one might just as well look up a table of cube roots
or a logarithm table, either of which will give the root much
quicker. The next chapter contains tables of cube roots and a
chapter further on explains the use of logarithms.
SQUARE ROOT
PROBLEMS
93
141. Extract the square roots of
(a) 64516 Answer 254
(6)198.1369 Answer 14.076 +
(c) .571428 Answer .7559 +
Note. These answers are given so that the student can see if he
understands the operations of square root before proceeding
further.
142. The two sides of a right triangle are 36 and 48 ft. ; what is the length
of the hypotenuse?
143. A square nut for a 2 in. bolt is 3 in. on each side. What is the
length of the diagonal, or distance across the corners?
144. A steel stack 75 ft. high is to be supported by 4 guy wires fastened
to a ring two-thirds of the way up the stack and having the other ends
anchored at a distance of 50 ft. from the base and on a level with the base.
How many feet of wire are necessary, allowing 20 ft. extra for fastening the
ends?
145. What would be the diameter of a circular brass plate having an
area of 100 sq. in.?
FIG. 33.
146. A lineshaft and the motor which drives it are located in separate
rooms as shown in Fig. 33. Calculate the exact distance between the
centers of the two shafts.
147. I want to cut a rectangular sheet of drawing paper to have an area
of 235 sq. in. and to be one and one-half times as long as it is wide. What
would be the dimensions of the sheet?
148. A 6 in. pipe and an 8 in. pipe both discharge into a single header.
Find the diameter of the header so that it will have an area equal to that of
both the pipes.
149. What would be the diameter of a 1 Ib. circular cast iron weight
J in. thick?
94
SHOP ARITHMETIC
150. How long must be the boom in Fig. 34 to land the load on the 12 ft.
pedestal, allowing 4 ft. clearance at the end for ropes, pulleys, etc.?
FIG. 34.
CHAPTER XII
MATHEMATICAL TABLES (CIRCLES, POWERS, AND ROOTS)
84. The Value of Tables. There are certain calculations that
are made thousands of times a day by different people in different
parts of the world. For example, the circumferences of circles
of different diameters are being calculated every day by hundreds
and thousands of men. To save much of the time that is thus
wasted in useless repetition, many of the common operations
and their results have been "tabulated," that is, arranged in
tables in the same way as are our multiplication tables in arith-
metics. These tables are not learned, however, as were the
multiplication tables, but are consulted each time that we have
need for their assistance.
Just what tables one needs most, depends on his occupation.
The machinist has use for tables of the decimal equivalents of
common fractions, tables of cutting speeds, tables of change
gears to use for screw cutting, etc. The draftsman would use
tables of strengths and weights of different materials, safe loads
for bolts, beams, etc., tables of proportions of standard machine
parts of different sizes, etc. The engineer uses tables of the prop-
erties of steam, and of the horse-power of engines, boilers, etc,
MATHEMATICAL TABLES 95
There are certain mathematical tables that are of value to
nearly everyone. Among these are the tables given in this
chapter: Tables of Circumferences and Areas of Circles; Tables
of Squares, Cubes, Square Roots, and Cube Roots of Numbers.
85. Explanation of the Tables. The first table is to save the
necessity of always multiplying the diameter by 3.1416 when we
want the circumference of a circle, or of squaring the diameter
and multiplying by .7854 when the area of a circle is wanted.
To find the circumference of a circle: Find, in the diameter
column, the number which is the given diameter; directly
across, in the next column to the right, will be found the corre-
sponding circumference.
Examples :
Diameter 1J Circumference 3.9270
Diameter 27, Circumference 84.823
Diameter 90J, Circumference 284.314
To find the area of a circle: Find, in the diameter column,
the number which is the given diameter; directly across, in the
second column to the right (the column headed "Area") will be
found the area.
Examples :
Diameter 66, Area 3421.2
Diameter 17, Area 226.98
Diameter f , Area 0.3067
If the area or circumference is known and we want to get the
diameter, we find the given number in the area or circumference
column and read the diameter in the corresponding diameter
column to the left.
Examples :
Area 78.54, Diameter 10
Area 706.86, Diameter 30
Circumference 281. Diameter 89 J
The second table, that of squares, cubes, square roots, and
cube roots, is especially valuable in avoiding the tedious process
of extracting square or cube roots. The table is read the same as
the other one. Find the given number in the first column; on
a level with it, in the other columns, will be found the corre-
sponding powers and roots, as indicated in the headings at the
tops of the columns.
96 SHOP ARITHMETIC
Examples :
^25 = 2.924
\/260 - 16.1245
295 3 = 25,672,375
865 2 = 748,225
86. Interpolation. This is a name given to the process of
finding values between those given in the tables. For example,
suppose we want the circumference of a 30 \ in. circle. The table
gives 30 and 30^ and, since 30J is half way between these, its
circumference will be half way between that of a 30 in. and a
30J in. circle.
Circumference of 30 \ in. circle = 95. 8 19
Circumference of 30 in. circle = 94 . 248
The Difference = 1.571
Then the circumference of the 30 \ in. circle is just half this
difference more than that of the 30-in. circle,
94.248 + ^ of 1.571 = 95.034
A
This method enables us to increase greatly the value of tables.
For most purposes the interpolation can be done quickly, and
while it requires some calculating, is much shorter than the
complete calculation would be. This is especially true in finding
square or cube roots.
Example :
Find from the table the cube root of 736.4
3/737 = 9.0328
= 9.0287
Difference =41
.4 X the difference = .4X41 = 16.4
9.0287
16
^736>4= 9.0303, Answer.
Explanation: The root of 736.4 will be between that of 736 and that of
737, and will be .4 of the difference greater than that of 736. In making
this correction for the .4, we forget, for the minute, that the difference is a
decimal and write it as 41 merely to save time. We then multiply it by .4,
and, dropping the decimal part, add the 16 to the 90287. This gives 9.0303
as the cube root of 736.4
Hence,
MATHEMATICAL TABLES 97
87. Roots of Numbers Greater than 1000. For getting the
cube roots of numbers greater than 1000, the easiest and most
accurate way is to look in the third column headed "cubes" for
a number as near as possible to our given number. Now, we
know that the numbers in the first column are the cube roots of
these numbers in the third column. If we can find our number
in the third column, there is nothing further to do because its
cube root will be directly opposite it in the first column.
Examples :
3/1728 = 12
3/6967871 = 191
3/166375 =55
Likewise, the numbers in the first column are the square roots of
the numbers in the second column. But suppose we want the
cube root of a number which is not found in the third column,
but lies somewhere between two consecutive numbers in that
column. In this case we pursue the method shown in the
following example:
Example :
Find 3/621723
In the column headed " Cube" find two consecutive numbers, one larger and
one smaller than 621723. These numbers are
636056 whose cube root is 86
and 614125 whose cube root is 85
Hence, the cube root of 621723 is more than 85 and less than 86; that is, it
is 85 and a decimal, or 85 + .
The decimal part is found as follows: Subtract the lesser of the two num-
bers found in the table from the greater and call the result the First
Difference.
636056-614125 = 21931, First difference.
Then subtract the smaller of the two numbers in the table from the given
number and call the result the Second Difference.
621723-614125 = 7598, Second Difference
Now the first difference, 21931, is the amount that the number increases
when its cube root changes from 85 to 86. Our given number is only 7598
more than the cube of 85, so its cube root will be approximately 85 zWVr-
We do not want a fraction like this, so we reduce it to a decimal as follows:
Divide the second difference by the first difference and annex the quotient
to 85. This will give us the cube root of our number, approximately.
Second Difference _ 7598 _ ., .
First Difference " 2l93l "
This is the decimal part of the root sought and the whole root is 85.346 + .
Hence 3/621723 = 85.346 + .
This method is not exact and the third decimal place will usually be
slightly off, so it is best to drop the third decimal if less than 5, or raise it to
10, if more than 5. In this case we will call the root 85.35.
98 SHOP ARITHMETIC
88. Cube Roots of Decimals. In getting the cube root of
either a number entirely decimal, or a mixed decimal number,
it is best to move the decimal point a number of periods, that is,
3, 6, 9, or 12 decimal places, sufficient to make a whole number
out of the decimal. After finding the cube root, shift the decimal
point in the root back to the left as many places as the number of
periods that we moved the decimal point in our original number.
For example, suppose that we had .621723 of which to find the
cube root. Moving the decimal point two periods (of three
places each) to the right gives us 621723, of which we just found
the cube root to be 85.35. We moved the decimal point of our
original number two periods to the right, so we must move the
decimal point back two places to the left in the root; we then
have ^.621723 = .8535. The following illustrations will show
the principle:
4/621723 =85.35
V62 1.723 = 8.535
V. 621723 .8535
</ 00062 1723 = .08535
Notice that there is no such similarity between the cube roots of
numbers if we move the decimal point any other number of
places than a multiple of three.
VQ= 1.817+ V60= 3.915 ^600= 8.434
But ^6000 = 18.17 ^60,000 = 39. 15 ^600,000 = 84.34
Care should be taken, therefore, that, if necessary to move the
decimal point in finding a cube root, it should be moved an exact
multiple of 3 places. If we have a decimal such as .07462, it is
necessary to attach a cipher at the right, making the decimal
.074620, so we can shift the decimal point 2 periods or six places.
We can now find v 74620 as follows:
43 3 = 79507 74620
42 3 = 74088 74088
5419 First Difference. 532 Second Difference.
532
Hence ^74620 = 42.01
and V. 074620= .4201
MATHEMATICAL TABLES 99
89. Square Root by the Table. The same methods as have
been explained for cube root can be applied to finding square
roots by the use of the table. The only difference is in the case
of decimals, in which case the decimal point is shifted by mul-
tiples of two places in the number; then, after we have the root, we
shift the decimal point back one place for each period of two
places that we moved the decimal point in our original number.
PROBLEMS
The tables are to be used wherever possible in working these problems.
151. What would be the length of a steel sheet from which to make a
28 in. circular drum, allowing 1^ in. extra for lapping and riveting the ends?
The circumference of the drum is measured in the direction of the length of
the sheet.
152. The small sprocket of a bicycle contains 8 teeth, the large sprocket
24 teeth. The rear wheel is 30 in. in diameter. Find the distance travelled
over by the bicycle for one revolution of the pedals.
153. The diameter of a If in. bolt at the bottom of the threads is 1.16 in.
What is the sectional area at the bottom of the threads?
154. A No. 000 copper trolley wire has a diameter of 0.425 in. What
would be the cost of 1 mile of the wire at 33 cents a pound?
155. A circular piece of boiler plate (Fig. 35) in. thick and 60 in. diam-
eter has an elliptical man hole in it 14 in. by 10 in. Find weight of plate.
Note. Area of an ellipse =.7854 X X&, where a and 6 are the long and
short diameters of the ellipse.
156. The paint shop wants a cubical dip tank built to hold, when full,
400 gallons of varnish. What will be the dimensions of the tank? (There
are 231 cu. in. in a gallon.)
157. How many square feet of galvanized iron will be needed to line the
tank of problem 156 on four sides and the bottom, allowing 10% extra for
the joints?
100
SHOP ARITHMETIC
158. A high carbon steel contains the following items in the percentages
given: Carbon, .60%; silicon, .10%; manganese, .40%; phosphorus, .035%;
sulphur, .025%. The rest is pure iron. Calculate the weights of carbon,
silicon, manganese, phosphorus, sulphur, and iron in one ton (2000 Ib.) of
the steel.
159. I want to get a cast iron block 18 in. long and of square cross-section
so that it will weigh 200 Ib. How many cubic inches of metal must be in
the block and what will be its dimensions?
160. Fig. 36 shows a steam hammer having an 8000 Ib. ram. If we
assume that this ram is a rectangular block of steel, four times as high and
twice as wide as it is thick, what will be the dimensions of the ram?
. 36.
MATHEMATICAL TABLES
CIRCUMFERENCES AND AREAS OF CIRCLES
101
Diameter
Circumference
Area
Diameter
Circumference
Area
i
.3927
0.0123
191
61.261
298.65
I
.7854
0.0491
20
62.832
314.16
*
1.1781
0.1104
201 64.403
330.06
1
1.5708
0.1963
21
65.973
346.36
i
1.9635
0.3067
211
67.544
363.05
i
2.3562
0.4417
22
69.115
380.13
i
2.7489
0.6013
221
70.686
397.61
i
3.1416
0.7854
23
72.257
415.48
11
3.5343
0.9940
231
73.827
433.74
ii
3.9270
1.227
24
75.398
452.39
U
4.3197
1.484
241
76.969
471.44
H
4.7124
1.767
25
78.540
490.87
li
5.1051
2.073
251
80.111
510.71
U
5.4978
2.405
26
81.681
530.93
ii
5.8905
2.761
261
83.252
551.55
2
6.2832
3.141
27
84.823
572.56
21
7.0686
3.976
271
86.394
593.96
21
7.8540
4.908
28
87.965
615.75
2|
8.6394
5.939
281
89.535
637.04
3
9.4248
7.068
29
91.106
660.52
3i
10.210
8.295
291
92.677
683.49
31
10.996
9.621
30
94.248
706.86
3!
11.781
11.044
301
95.819
730.62
4
12.566
12.566
31
97.389
754.77
41
14.137
15.904
311
08.960
779.31
5
15.708
19.635
32
100.531
804.25
51
17.279
23.758
321
102.102
829.58
6
18.850
28.274
33
103.673
855.30
6i
20.420
33.183
331
105.243
881.41
7
21.991
38.485
34
106.814
907.92
71
23.562
44.179
341
108.385
934.82
8
25.133
50.265
35
109.956
962.11
81
26.704
56.745
351
111.527
989.80
9
28.274
63.617
36
113.097
1017.9
91
29.845
70.882
361
114.668
1046.3
10
31.416
78.540
37
116.239
1075.2
101
32.987
86.590
371
117.810
1104.5
11
34.558
95.033
38
119.381
1134.1
H*
36.128
103.87
381
120.951
1164.2
12
37.699
113.10
39
122.522
1194.6
121
39.270
122.72
391
124.093
1225.4
13
40.841
132.73
40
125.664
1256.6
131
42.414
143.14
401
127.235
1288.2
14
43.982
153.94
41
128.805
1320.3
141
45.553
165.13
411
130.376
1352.7
15
47.124
176.71
42
131.947
1385.4
151
48.695
188.60
421
133.518
1418.6
16
50.265
201.06
43
135.088
1452.2
161
51.836
213.82
431
136.659
. 1486.2
17
53.407
226.98
44
138.230
1520.5
171
54.978
240.53
441
139.801
1555.3
18
56.549
254.47
45
141.372
1590.4
181
58.119
268.80
451
142.942
1626.0
19
59.690
283.53
46
144.513
1661.9
102 SHOP ARITHMETIC
CIRCUMFERENCES AND AREAS OF CIRCLES. Continue*
Diameter
Circumference
Area
Diameter
Circumference
Area
46J
146.084
1698.2 731
230.907
4242.9
47
147.655
1734.9
74
232.478
4300.8
47J
149.226
1772.1
74}
234.049
4359.2
48
150.796
1809.6
75
235.619
4417.9
481
152.367
1847.5
75}
237.190
4477.0
49
153.938
1885.7
76
238.761
4536.5
491
155.509
1924.4
76}
240.332
4596.3
50
157.080
1963.5
77
241.903
4656.6
501
158.650
2003.0
77}
243.473
4717.3
51
160.221
2042.8
78
245.044
4778.4
511
161.792
2083 . 1
78}
246.615
4839.8
52
163.363
2123.7
79
248.186
4901.7
521
164.934 2164.8
79}
249.757
4963.9
53
166.504 2206.2
80
251.327
5026.5
531
168.075
2248.0
80}
252.898
5089.6
54
169.646
2290.2
81
254.469
5153.0
541
171.217
2332.8
81}
256.040
5216.8
55
172.788
2375.8
82
257.611
5281.0
551
174.358
2419.2
82} 259.181
5345.6
56
175.929
2463.0 83 260.752
5410.6
561
177.500
2507.2
83}
262.323
5476.0
57
179.071
2551.8
84
263.894
5541.8
571
180.642
2596.7
84}
265.465
5607.9
58
182.212
2642 . 1
85
267.035
5674.5
581
183.783
2687.8
85}
268.606
5741.5
59
185.354
2734.0
86
270.177
5808.8
591
186.925
2780.5
86}
271.748
5876.5
60
188.496
2827.4
87
273.319
5944.7
601
190.066
2874.8
87}
274.889
6013.2
61
191.637
2922.5
88
276.460
6082.1
611
193.208
2970.6
88}
278.031
6151.4
62
194.779
3019.1
89
279.602
6221.1
621
196.350
3068.0
89}
281.173
6291 .2
63
197.920
3117.2
90
282.743
6361.7
631
199.491
3166.9
90}
284.314
6432.6
64
201.062
3217.0
91
285.885
6503.9
641
202.633
3267.5
91}
287.456
6575.5
65
204.204
3318.3
92
289.027
6647.6
651
205.774
3369.6
92}
290.597
6720.1
66
207.345
3421.2
93
292.168
6792.9
661
208.916
3473.2
93}
293.739
6866.1
67
210.487
3525.7
94
295.310
6939.8
671
212.058
3578.5
94}
296.881
7013.8
68
213.628
3631.7
95
298.451
7088.2
681
215.199
3685.3 951
300.022
7163.0
69
216.770
3739.3 96
301.593
7238.2
691
218.311
3793.7 961
303.164
7313.8
70
219.911
3848.5 97
304.734
7389.8
701
221. 4S2
3903.6 971
306.305
7466.2
71
223.053
3959.2 98
307.876
7543.0
711
224.624
4015.2 981
309.447
7620.1
72
226.195
4071.5 99
311.018
7697.7
721
227.765
4128.2 991
312.588
7775.6
73
229.336
4185.4
100
' 314.159
7854.0
MATHEMATICAL TABLES
103
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
1
1
1
1.
1.
46
2116
97336
6.7823
3.5830
2
4
8
1.4142
1.2599
47
2209
103823
6.8557
3.6088
3
9
27
1.7321
1.4422
48
2304
110592
6.9282
3.6342
4
16
64
2.
1.5874
49
2401
117649
7.
3.6593
5
25
125
2.2361
1.7100
50
2500
125000
7.0711
3.6840
6
36
216
2.4495
1.8171
51
2601
132651
7.1414
3.7084
7
49
343
2.6458
1.9129
52
2704
140608
7.2111
3.7325
8
64
512
2.8284
2.
63
2809
148877
7.2801
3.7563
9
81
729
3.
2.0801
54
2916
157464
7.3485
3.7798
10
100
1000
3.1623
2.1544
55
3025
166375
7.4162
3.8030
11
121
1331
3.3166
2.2240
56
3136
175616
7.4833
3.8259
12
114
1728
3.4641
2.2894
57
3249
185193
7.5498 3.8485
13
169
2197
3.6056
2.3513
58
3364
195112
7.6158 3.8709
14
196
2744
3.7417
2.4101
59
3481
205379
7.6811 3.8930
15
225
3375
3.8730
2.4662
60
3600
216000
7.7460
3.9149
16
256
4096
4.
2.5198
61
3721
226981
7.8102
3.9365
17
289
4913
4.1231
2.5713
62
3844
238328
7.8740 3.9579
18
324
5832
4.2426
2.6207
63
3969
250047
7.9373
3.9791
I'.i
361
6859
4.3589
2.6684
64
4096
262144
8.
4.
20
400
8000
4^4721
2.7144
65
4225
274625
8.0623
4.0207
21
441
9261
4.5826
2.7589
66
4356
287496
8.1240
4.0412
22
484
10648
4.6904
2.8020
67
4489
300763
8.1854
4.0615
23
529
12167
4.7958
2.8439
68
4624
314432
8.2462
4.0817
21
576
13824
4.8990
2.8845
69
4761
328509
8.3066
4.1016
25
625
15625
5.
2.9240
70
4900
343000
8.8666
4.1213
26
676
17576
5.0990
2.9625
71
5041
357911
8.4261
4.1408
27 729 19683 5.1962
Q
72
5184
373248
8.4853
4.1602
28 781 21952 5.2915
3.0366
73
5329
389017
8.5440
4.1793
29 84 li 24389 5.3852
3.0723
74
5476
405224
8.6023
4.1983
30 900
27000
5.4772
3.1072
75
5625
421875
8.6603
4.2172
31 961
29791
5.5678
3.1414
76
6776
438976
8.7178
4.2358
32 1024
32768
5.6569
3.1748
77
5929
456533
8.7750
4.2543
33 1089
35937
5.7446
3.2075
78
6084
474552
8.8318
4.2727
34 1156
39304
5.8310
3.2396
79
6241
493039
8.8882
4.2908
35 1225
42875
5.9161
3.2711
80
6400
512000
8.9443
4.3089
36 1296
46656
6.
3.3019
81
6561
631441
9.
4.3267
37
1369
50653
6.0828
3.3322
82
6724
551368
9.0554
4.3445
38
1444
54872
6.1614
3.3620
83
6889
571787
9.1104
4.3621
39
1521
59319
6.24.50
3.3912
84
7056
592704
9.1652
4.3795
40
1600
64000
6.3246
3.4200
85
7225
614125
9.2195
4.3968
41
1681
68921
6.4031
3.4482
86
7396
636056
9.2736
4.4140
42
1761 74088
6.4807 3.4760
87
7569 658503
9.3274
4.4310
43
1849 79507 6.5574 3.5034J 88
7744
681472
9.3808
4.4480
44
1936 85184! 6.6332 3.5303]
89
7921
704969
9.4340
4.4647
45
2025
91125 6.7082
3.55691 90
8100
729000
9.4868
4.4814
104
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
91
8281
753571
9.5394
4.4979
136
18496
2515456
11.6619
5.1426
92
8464
778688
9.5917
4.5144
137
18769
2571353
11.7047
5.1551
93
8649
804357
9.6437
4.5307
138
19044
2628072
11.7473
5.1676
91
8836
830584
9.6954
4.5468
139
19321
2685619
11.7898
5.1801
95
9025
857375
9.7468
4.5629
140
19600
2744000
11.8322
5.1925
96
9216
884736
9.7980
4.5789
141
19881
2803221
11.8743
5.2048
97
9409
912673
9.8489
4.5947
142
20164
2863288
11.9164
5.2171
98
9604
941192
9.8995
4.6104
143
20449
2924207
11.9583
5.2293
99
9801
970299
9.9499
4.6261
144
20736
2985984
12.
5.2415
100
10000
1000000
10.
4.6416
145
21025
3048625
12.0416
5.2536
101
10201
1030301
10.0499
4.6570
146
21316
3112136
12.0830
5.2656
102 10404
1061208
10.0995
4.6723
147
21609
3176523
12.1244
5.2776
103 10609
1092727
10.1489
4.6875
148
21904
3241792
12.1655
5.2896
10J 10816
1124864
10.1980
4.7027
149
22201
3307949
12.2066
5.3015
105
11025
1157625
10.2470
4.7177
150
22500
3375000
12.2474
5.3133
106
11236
1191016
10.2956
4.7326
151
22801
3442951
12.2882
5.3251
107 11449
1225043
10.3441
4.7475
152
23104
3511808
12.3288
5.3368
108 11664
1259712
10.3923
4.7622
153
23409
3581577
12.3693
5.3485
109
11881
1295029
10.4403
4.7769
154
23716
3652264
12.4097
5.3601
110
12100
1331000
10.4881
4.7914
155
24025
3723875
12.4499
5.3717
111
12.321
1367631
10.5357
4.8059
156
24336
3796416
12.4900
5.3832
112
12544
1404928
10.5830
4.8203
157
24649
3869893
12.5300
5.3947
113
12769
1442897
10.6301
4.8346
158
24964
3944312
12 . 5698
5.4061
114
12996
1481544
10.6771
4.8488
159
25281
4019679
12.6095
5.4175
115
13225
1520875
10.7238
4.8629
160
25600
4096000
12.6491
5.4288
116
13456
1560896
10.7703
4.8770
161
25921
4173281
12.6886
5.4401
117
13689
1601613
10.8167
4.8910
162
26244
4251528
12.7279
5.4514
118 13924
1643032
10.8628
4.9049
163
26569
4330747
12.7671
5.4626
119 14161
1685159
10.9087
4.9187
164
26896
4410944
12.8062
5.4737
120
14400
1728000
10.9545
4.9324
165
27225
4492125
12.8452
5.4848
121
14641
1771561
11.
4.9461
166
27556
4574296
12.8841
5.4959
122
14884
1815848
11.0454
4.9597
167
27889
4657463
12.9228
5.5069
123
15129
1860867
11.0905
4.9732
168
28224
4741632
12.9615
5.5178
124
15376
1906624
11.1355
4.9866
169
28561
4826809
13.
5.5288
125
15625
1953125
11.1803
5.
170
28900
4913000
13.0384
5.5397
126
15876
2000376
11.2250
5.0133
171
29241
5000211
13.0767
5.5505
127
16129
2048383
11.2694
5.0265
172
29584
5088448
13.1149
5.5613
128] 16384
2097152
11.3137
5.0397
173
29929
5177717
13.1529
5.5721
129
16641
2146689
11.3578
5.0528
174
30276
5268024
13.1909
5.5828
130
16900
2197000
11.4018
5.0658
175
30625
5359375
13.2288
5.5934
131
17161
2248091
11.4455
5.0788
176
30976
5451776
13.2665
5.6041
132
17424
2299968
11.4891
5.0916
177
31329
5545233
13.3041
5.6147
133
17689
2352637
11.5326
5.1045
178
31684
5639752
13.3417
5.6252
134
17956
2406104
11.5758
5.1172
179
32041
5735339
13.3791
5.6357
135
18225
2460375
11.6190
5.1299
180
32400
5S32000
13.4164
5.6462
MATHEMATICAL TABLES
105
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
181
32761
5929741
13.4536
5.6567
226
51076
11543176
15.0333
6.0912
182
33124
6028568
13.4907
5.6671
227
51529
11697083
15.0665
6.1002
183
33489
6128487
13.5277
5.6774
228
51984
11852352 15.0997
6.1091
184
33856
6229504
13.5647
5.6877
229
52441
12008989 15.1327
6.1180
185
34225
6331625
13.6015
5.6980
230
52900
12167000
15.1658
6.1269
186
34596
6434856
13.6382
5.7083
231
53361
12326391
15.1987
6.1358
187
34969
6539203
13.6748
5.7185
232
53824
12487168 15.2315
6.1446
188
35344
6644672
13.7113
5.7287
233
54289
12649337 15.2643
6.1534
189
35721
6751269
13.7477
5.7388
234
54756
12812904
15.2971
6.1622
190
36100
6859000
13.7840
5.7489
235
55225
12977875
15.3297
6.1710
191
36481
6967871
13.8203
5.7590
236
55696
13144256
15.3623
6.1797
192
36864
7077888
13.8564
5.7690
237
56169
13312053
15.3948
6.1885
193
37249
7189057
13.8924
5.7790
238
56644
13481272") 15.4272
6.1972
194
37636
7301384
13.8284
5.7890
239
57121
13651919 15.4596
6.2058
195
38025
7414875
13.9642
5.7989
240
57600
13824000
15.4919
6.2145
196
38416
7529536
14.
5.8088
241
58081
13997521
15.5242
6.2231
197
38809
7645373
14.0357
5.8186
242
58564
14172488
15.5563
6.2317
198
39204 7762392
14.0712
5.8285
243
59049
14348907
15.5885
6.2403
l-.r.i
39601 7880599 14.1067
5.8383
244
59536
14526784
15.6205
6.2488
200
40000
8000000
14.1421
5.8480
245
60025
14706125
15.6525
6.2573
201
40401
8120601
14.1774
5.8578
246
60516
14886936
15.6844
6.2658
202
40804
8242408
14.2127
5.8675
247
61009
15069223
15.7162
6.2743
203
41209
8365427
14.2478
5.8771
248
61504
15252992
15.7480
6.2828
204
41616
8489664
14.2829
5.8868
249
62001
15438249
15.7797
6.2912
205
42025
8615125
14.3178
5.8964
250
62500
15625000
15.8114
6.2996
206
42436
8741816
14.3527
5.9059
251
63001
15813251
15.8430
6.3080
207
42849
8869743
14.3875
5.9155
252
63504
16003008
15.8745
6.3164
208
43264
8998912
14.4222
5.9250
253
64009
16194277
15.9060
6.3247
209
43681
9129329
14.4568
5.9345
254
64516
16387064
15.9374
6.3330
210
44100
9261000
14.4914
5.9439
255
65025
16581375
15.9687
6.3413
211
44521
9393931
14.5258
5.9533
256
65536
16777216
16.
6.3496
212
44944
9528128
14.5602
5.9627
257
66049
16974593
16.0312
6.3579
213
45369
9663597
14.5945
5.9721
258
66564
17173512
16.0624 6.3661
214
45796
9800344
14.6287
5.9814
259
67081
17373979
16.0935
6.3743
215
46225
9938375
14.6629
5.9907
260
67600
17576000
16.1245
6.3825
216
46656
10077696
14.6969
6.
261
68121
17779581
16.1555
6.3907
L'17
47089
10218313
14.7309
6.0092
262
68644
17984728
16.1864
6.3988
218
47524
10360232
14.7648
6.0185
263
69169
18191447
16.2173
6.4070
219
47961
10503459
14.7986
6.0277
264
69696
18399744
16.2481
8.4151
220
48400
10648000
14.8324
6.0368
265
70225
18609625
16.2788
6.4232
221
48841
10793861
14.8661
6.0459
266
70756
18821096
16.3095
6.4312
222
49284
10941048 14.8997
6.0550
267
71289
19034163
16.3401
6.4393
223
49729 11089567 14.9332
6.0641
268
71824
19248832
16.3707
6.4473
224
50176 11239424 14.9666
6.0732
269
72361
19465109
16.4012
6.4553
225
50625 11390625 15.
6.0822
270
72900 19683000
16.4317
6.4633
106
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
271
73441
19902511
16.4621
6.4713
316
99856
31554496
17.7764
6.8113
272
73984
20123648
16.4924
6.4792
317
100489
31855013
17.8045
6.8185
273
74529
20346417
16.5227
6.4872
318
101124
32157432 17.8326
6.8256
274
75076
20570824
16.5529
6.4951
319
101761
32461 759 j 17.8606
6.8328
275
75625
20796875
16.5831
6.5030
320
102400
32768000' 17.8885
6.8399
276 76176
21024576
16.6132
6.5108
321
103041
33076161
17.9165
6.8470
277 76729
21253933
16.6433
6.5187
322
103681
33386248 17.9444
6.8541
278 77284
21484952
16.6733
6.5265
323
104329
33698267
17.9722
6.8612
279 77841
21717639
16.7033
6.5343
324
104976
34012224
18.
6.8683
280 78400
21952000
16.7332
6.5421
325
105625
34328125
18.0278
6.8753
281
78961
22188041
16.7631
6.5499
326
106276
34645976
18.0555
6.8824
282 79524
22425768 16.7929
6.5577
327
106929
34965783 18.0831
6.8894
283 80089
22665187 16.8226
6.5654
328
107584
35287552 18.1108
6.8964
284 80656
22906304 16.8523
6.5731
329
108241
35611289
18.1384
6.9034
285 81225
23149125 1 16.8819
6.5808
330
108900
35937000
18.1659
6.9104
286
81796
23393656
16.9115
6.5885
331
109561
36264691
18.1934
6.9174
287 82369
23639903 16.9411
6.5962
332
110224 36594368
18.2209
6.9244
288 82944
23887872 16.9706
6.6039
333
110889 36926037
18.2483
6.9313
289 83521
24137569 17.
6.6115
334
111556
37259704
18.2757
6.9382
290 84100
24389000
17.0294
6.6191
335
112225
37595375
18.3030
6.9451
291 84681
24642171J 17.0587
6.6267
336
112896
37933056
18.3303
6.9521
292 85264
24897088 17.0880
6.6343
337
113569
38272753
18.3576
6.9589
293: 85849
25153757 17.1172
6.6419
338
114244
38614472
18.3848
6.9658
294 86436
25412184 17.1464
6.6494
339
114921
38958219
18.4120
6.9727
295
87025
25672375
17.1756
6.6569
340
115600
39304000
18.4391
6.9795
296
87616
25934336
17.2047
6.6644
341
116281
39651821
18.4662
6.9864
297 88209
26198073
17.2337
6.6719
342
116964
40001688 18.4932
6.9932
298 88804
26463592
17.2627
6.6794
343
117649
40353607 18.5203
7.
299 89401
26730899
17.2916
6.6869
344
118336
40707584
18.5472
7.0068
300 90000
27000000
17.3205
6.6943
345
119025
41063625
18.5742
7.0136
301
90601
27270901
17.3494
6.7018
346
119716
41421736
18.6011
7.0203
302
91204 27543608
17.3781
6.7092
347
120409
41781923 18.6279
7.0271
303; 91809 27818127
17.4069
6.7166
348
121104
42144192! 18.6548
7.0338
304 92416 28094464
17.4356
6.7240
349
121801
42508549
18.6815
7.0406
305 93025
28372625
17.4642
6.7313
350
122500
42875000
18.7083
7.0473
306
93636
28652616
17.4929
6.7387
351
123201
43243551
18.7350
7.0540
307 94249 28934443
17.5214
6.7460
352
123904
43614208
18.7617
7.0607
308 94864 29218112
17.5499
6.7533
353
124609
43986977
18.7883
7.0674
309 95481
29503629
17.5784
6.7606
354
125316
44361864
18.8149
7.0740
310
96100
29791000
17.6068
6.7679
355
126025
44738875
18.8414
7.0807
311 96721
30080231
17.6352
6.7752
356
126736
45118016
18.8680
7.0873
312
97344
30371328
17.6635
6.7824
357
127449
45499293
18.8944
7.0940
313
97969 30664297
17.6918
6.7897 358
128164
45882712; 18.9209
7.1006
314
98596 30959144
17.7200
6.7969 3.59
128881
46268279 ' 18.9473
7.1072
315
99225
31255875
17.7482
6.8041
360
129600
46656000 18.9737 7.1138
MATHEMATICAL TABLES
107
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
361
130321
47045881
19.
7.1204
406
164836
66923416
20.1494
7.4047
362
131044
47437928 19.0263
7.1269;
407 165649 67419143 20.1742
7.4108
363
131769 47832147 19.0526 7.1335|
408! 166464 67917312 20.1990
7.4169
364
132196 48228544 19.0788 7.1400
409
167281] 68417929 20.2237
7.4229
365
133225
48627125
19.1050
7.1466
410
168100
68921000 20.2485
7.4290
366
133956
49027896' 19.1311
7.1531
411
168921
69426531
20.2731
7.4350
367
134689
49430863 19.1572
7.1596
412 169744 69934528 20.2978
7.4410
368
135424
49836032 19.1833
7.1661
413 170569
70444997 20.3224
7.4470
369
136161
50243409 19.2094
7.1726
414
171396
70957944 ' 20.3470
7.4530
370
136900
50653000 19.2354
7.1791
415
172225
71473375 20.3715
7.4590
371
137641
51064811
19.2614
7.1855
416
173056
71991296 20.3961
7.4650
372
138384
51478848 19.2873
7.1920
417
173889
72511713, 20.4206
7.4710
373
139129
51895117 19.3132
7.1984 418
174724
73034632; 20.4450
7.4770
374
139876
52313624 19.3391
7.2048
419 175561
73560059: 20.4695
7.4829
375
140625
52734375
19.3649
7.2112
420
176400
74088000
20.4939
7.4889
376
141376
53157376
19.3907
7.2177
421
177241
74618461
20.5183
7.4948
377
142129
53582633 19.4165 7.2240
422 178084 75151448
20.5426
7.5007
378
142884
51010152 19.4422 7.2304
423 178929
75686967
20.5670
7.5067
379
143641
54439939 19.4679
7.2368
424
179776
76225024
20.5913
7.5126
380
144400
54872000
19.4936
7.2432
425
180625
76765625
20.6155
7.5185
381
145161
55306341 19.5192
7.2495
426
181476
77308776
20.6398
7.5244
382
145924
55742968 19.5448 7.2558
427
182329
77854483 20.6640
7.5302
383
146689
56181887 19.5704
7.2622
428
183184
78402752
20.6882
7.5361
384
147456
56623104 19.5959
7.2685
429
184041
78953589
20.7123
7.5420
385
148225
57066625
19.6214
7.2748
430
184900
79507000
20.7364
7.5478
386
148996
57512456
19.6469
7.2811
431
185761
80062991
20.7605
7.5537
387
149769
57960603 19.6723
7.2874 432
186624
80621568; 20.7846
7.5595
388
150544
58411072 19.6977
7.2936 433
187489
81182737 20.8087
7.5654
389
151321
58863869 19.7231
7. 2999 ' 434
188356 81746504 20.8327
7.5712
390
152100
59319000 19.7484
7.3061
435
189225
82312875
20.8567
7.5770
391
152881
59776471
19.7737
7.3124
436
190096
82881856
20.8806
7 . 5828
392
153664 60236288 19.7990
7.3186
437
190969
83453453 20.9045
7.5886
393
154449 60698457 19.8242
7.3248
438
191884
84027672 20.9284
7.5944
391
155236 61162984 19.8494
7.3310
439 192721
84604519
20.9523
7.6001
395
156025
61629875
19.8746
7.3372
440
193600
85184000
20.9762
7.6059
396
156816 62099136 19.8997
7.3434
441
194481
85766121
21.
7.6117
397
157609 62570773 19.9249
7.3496
442
195364
86350888 21.0238
7.6174
398
158404 63044792
19.9499 7.3558
443
196249
86938307 1 21.0476
7.6232
399
159201 63521199 19.9750 7.3619
444
197136
87528384 21.0713
7.6289
400
160000 64000000
20.
7.3681
445
198025
88121125
21.0950
7.6346
401
160801 64481201
20.0250
7.3742
446
198916
88716536 21.1187
7.6403
402
161604 64964808 20.0499 7.3803
447
199809 89314623 21.1424
7.6460
403
162409
65450827 20.0749 7.3864
448
200704
89915392 21.1660
7.6517
404
163216
65939264! 20.0998
7.3925
449
201601
90518S49 21.1896
7.6574
405
164025
66430125
20.1246
7.3986
450
202500
91125000
21.2132
7.6631
108
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
451
203401
91733851
21.2368
7.6688
496
246016
122023936
22.2711
7.9158
452
204304
92345408
21.2603
7.6744
497
247009
122763473
22.2935
7.9211
453
205209
92959677
21.2838
7.6801
498
248004
123505992 22.3159
7.9264
454
206116
93576664
21.3073
7.6857
499
249001
124251499
2-2.3383
7.9317
455
207025
94196375
21.3307
7.6914
500
250000
125000000
22.3607
7.9370
456
207936
94818816
21.3542
7.6970
501
251001
125751501
22.3830
7.9423
457 208849
95443993 21.3776
7.7026
502
252004
126506008 22.4054
7.9476
458 209764
96071912 21.4009
7.7082
503
253009
127263527 22.4277
7.9528
459
210681
96702579 21.4243
7.7138
504
254016
128024064 22.4499
7.9581
460
211600
97336000
21.4476
7.7194
505
255025
128787625
22.4722
7.9634
461
212521
97972181
21.4709
7.72.50
506
256036
129554216
22.4944
7.9686
462
213444
98611128
21.4942
7.7306
507 257049
130323843 22.5167
7.9739
463
214369
99252847
21.5174
7.7362
508
258064
131096512 22.5389
7.9791
464
215296
99897344
21.5407
7.7418
509
259081
131872229
22.5610 7.9843
465
216225
100544625
21.5639
7.7473
510
260100
132651000
22.5832
7.9896
466
217156
101194696
21.5870
7.7529
511
261121
133432831
22.6053
7.9948
467
218089
101847563
21.6102
7.7584
512
262144
134217728
22.6274 8.
468
219024
102503232
21.6333
7.7639
513
263169
135005697
22.6495 j 8.0052
469
219961
103161709
21.6564
7.7695
514
264196
135796744
22.6716 8.0104
470
220900
103823000
21.6795
7.7750
515
265225
136590875
22.6936 8.0156
471
221841
104487111
21.7025
7.7805
516
266256
137388096
22.7156 8.0208
472 222784
105154048
21.7256
7.7860,
517
2672S9
138188413 22.7376 8.0260
473
223729
105823817
21.7486
7.7915
518
268324
138991832 22.7596 8.0311
474
224676
106496424 21.7715
7.7970
519
269361
139798359
22.7816 8.0363
475
225625
107171875
21.7945
7.8025
520
270400
140608000
22.8035
8.0415
476
226576
107850176
21.8174
7.8079
521
271441
141420761
22.8254
8.0466
477
227529
108531333 21.8403
7.8134
522
272484
142236648 22.8473 8.0517
478
228484
109215352 21.8632
7.8188
523
273529
143055667 22.8692 8.0569
479
229441
109902239 21.8861
7.8243
524
274576
143877824 22.8910 8.0620
480
230400
110592000
21.9089
7.8297
525
275625
144703125
22.9129
8.0671
481
231361
111284641
21.9317
7.8352
526
276676
145531576
22.9347
8.0723
482
232324
111980168 21.9545 7.8406
527
277729
146363183 22.9565 8.0774
483
233289
112678587 21.9773
7.8460
528
278784
147197952 22.9783
8.0825
484
234256
113379904
22.
7.8514
529
279841
148035889 23.
8.0876
485
235225
114084*25
22.0227
7.8568
530
280900
148877000
23.0217
8.0927
486
236196
114791256
22.0454
7.8622
531
281961
149721291
23.0434
8.0978
487
237169
115501303 22.0681
7.8676
532
283024
150568768 23.0651 8.1028
488
238144
116214272 22.0970
7.8730
533
284089
151419437 23.0868 8.1079
489
239121
116930169 22.1133
7.8784
534
285156
152273304 23.1084
8.1130
490
240100
117649000
22.1359
7.8837
535
286225
153130375
23.1301
8.1180
491
241081
118370771
22.1585
7.8891
536
287296
153990656
23.1517
8.1231
492
242064
119095488
22.1811
7.8944
537
288369
154854153 23.1733
8.1281
493
243049
119823157
22.2036
7.8998
538
289444
155720872 23.1948
8.1332
494
244036
120553784
22.2261
7.9051
539
290521
156590819 23.2164
8.1382
495
245025
121287375
22.2486
7.9105
540
291600
157464000 23.2379
8.1433
MATHEMATICAL TABLES
109
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
N UM BE RS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
541
292681
158340421
23.2594
8.1483
586
343396
201230056
24.2074
8.3682
642
293764
159220088
23.2809
8.1533
587
344569 202262003
24.2281
8.3730
543
294849
160103007 23.3024
8.1583
588
345744 203297472
24.2487
8.3777
544
295936
160989184 23.3238
8.1633
589
346921 204336469
24.2693
8.3825
545
297025
161878625
23.3452
8.1683
590
348100
205379000
24.2899
8.3872
r> 10
298116
162771336
23.3666
8.1733
591
349281
206425071
24.3105
8.3919
547 299209
163667323 23.3880
8.1783
592
350464
207474688
24.3311
8.3967
548 300304
164566592 23.4094
8.1833
593
351649 208527857
24.3516
8.4014
549 301401
165469149 23.4307
8.1882
594
352836
209584584
24.3721
8.4061
550
302500
166375000
23.4521
8.1932
595
354025
210644875
24.3926
8.4108
551
303601
167284151
23.4734
8.1982
596
355216
211708736
24.4131
8.4155
552 304704 168196608
23.4947
8.2031
597
356409
212776173
24.4336
8.4202
553 305809 169112377
23.5160
8.2081
598
357604
213847192
24.4540
8.4249
554 306916 170031464
23.5372
8.2130
599
358801
214921799
24.4745
8.4296
555
308025
170953875
23.5584
8.2180
600
360000
216000000
24.4949
8.4343
556
309136
171879616
23.5797
8.2229
601
361201
217081801
24.5153
8.4390
557
310249
172808693 23.6008
8.2278
602
362404
218167208
24.5357
8.4437
558 311364 173741112 23.6220
8.2327
603
363609
219256227
24.5561
8.4484
559 312481 174676879
23.6432
8.2377
604
364816
220348864
24.5764
8.4530
560 313600
175616000
23.6643
8.2426
605
366025
221445125
24.5967
8.4577
561 314721
176558481
23.6854
8.2475
606
367236
222545016
24.6171
8.4623
562 315844
177504328
23.7065
8.2524
607
368449
223648543
24.6374
8.4670
563 316969
178453547
23.7276
8.2573
608
369664
224755712
24.6577
8.4716
564 318096
179406144
23.7487
8.2621
609
370881
225866529
24.6779
8.4763
56.')
319225
180362125
23.7697
8.2670
610
372100
226981000
24.6982
8.4809
566
320356
181321496
23.7908
8.2719
611
373321
228099131
24.7184
8.4856
567
321489 182284263
23.8118
8.2768
612
374544
229220928
24.7386
8.4902
568
322624 183250432
23.8328
8.2816
613
375769
230346397
24.7588
8.4948
569
323761 ] 184220009
23.8537
8.2865
614
376996
231475544
24.7790
8.4994
570
324900
185193000
23.8747
8.2913
615
378225
232608375
24.7992
8.5040
571
326041
186169411
23.8956
8.2962
616
379456
233744896
24.8193
8.5086
672
327184
187149248
23.9165
8.3010
617
380689
234885113
24.8395
8.5132
573
328329 188132517
23.9374
8.3059
618
381924
236029032
24.8596
8.5178
571
329476, 189119224
23.9583
8.3107
619
383161
237176659
24.8797
8.5224
575
330625
190109375
23.9792
8.3155
620
384400
238328000
24.8998
8.5270
576
331776
191102976
24.
8.3203
621
385641
239483061
24.9199
8.5316
577
332929
192100033
24.0208
8.3251
622
386884
240641848
24.9399
8.5362
578
334084
193100552
24.0416
8.3300
623
388129
241804367
24.9600
8.5408
579
335241
194104539
24.0624
8.3348
624
389376
242970624
24.9800
8.5453
580
336400
195112000
24.0832
8.3396
625
390625
244140625
25.
8.5499
581
337561
196122941
24.1039
8.3443
626
391876
245314376
25.0200
8.5544
582
338724 197137368
24.1247
8.3491
627
393129
246491883
25.0400
8.5590
583
339889; 198155287
24.1454
8.3539
628
394384
247673152
25.0599
8.5635
584
341056
199176704
24.1661
8.3587
629
395641
248858189
25.0799
8.5681
585
342225
200201625
24.1868
8.3634
630
396900
250047000
25.0998
8.5726
110
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
631
398161
251239591
25.1197
8.5772
676
456976
308915776
26.
8.7764
632
399424
252435968
25.1396
8.5817
677
458329
310288733
26.0192
8.7807
633
400689
253636137
25.1595
8.5862
678
459684
311665752
26.0384
8.7850
634
401956
254840104
25.1794
8.5907
679
461041
313046839
26.0576
8.7893
635
403225
256047875
25.1992
8.5952
680
462400
314432000
26.0768
8.7937
636
404496
257259456
25.2190
8.5997
681
463761
315821241
26.0960
8.7980
637
405769
258474853
25.2389
8.6043
682
465124
317214568
26.1151
8.8023
638
407044
259694072
25.2587
8.6088
683
466489
318611987
26.1343
8.8066
639
408321
260917119
25.2784
8.6132
684
467856
320013504
26.1534
8.8109
640
409600
262144000
25.2982
8.6177
685
469225
321419125
26.1725
8.8152
641
410881
263374721
25.3180
8.6222
686
470596
322828856
26.1916
8.8194
642
412164
264609288
25.3377
8.6267
687
471969
324242703
26.2107
8.8237
643
413449
265847707
25.3574
8.6312
688
473344
325660672
26.2298
8.8280
644
414736
267089984
25.3772
8.6357
689
474721
327082769
26.2488
8.8323
645
416025
268336125
25.3969
8.6401
690
476100
328509000
26.2679
8.8366
646
417316
269586136
25.4165
8.6446
691
477481
329929371
26.2869
8.8408
647
418609
270840023
25.4362
8.6490
692
478864
331373888
26.3059
8.8451
648
419904
272097792
25.4558
8.6535
693
480249
332812557
26.3249
8.8493
649
421201
273359449
25.4755
8.6579
694
481636
334255384
26.3439
8.8536
650
422500
274625000
25.4951
8.6624
695
483025
335702375
26.3629
8.8578
651
423801
275894451
25.5147
8.6668
696
484416
337153536
26.3818
8.8621
652
425104
277167808
25.5343
8.6713
697
485809
338608873
26.4008
8.8663
653
426409
278445077
25.5539
8.6757
698
487204
340068392
26.4197
8.8706
654
427716
279726264
25.5734
8.6801
699
488601
341532099
26.4386
8.8748
655
429025
2S1011375
25.5930
8.6845
700
490000
343000000
26.4575
8.8790
656
430336
282300416
25.6125
8.6890
701
491401
344472101
26.4764
8.8833
657
431649
283593393
25.6320
8.6934
702
492804
345948408
'26.4953
8.8875
658
432964
284890312
25.6515
8.6978
703
494209
347428927
26.5141
8.8917
659
434281
286191179
25.6710
8.7022
704
495616
348913664
26.5330
8.8959
660
435600
287496000
25.6905
8.7066
705
497025
350402625
26.5518
8.9001
661
436921
288804781
25.7099
8.7110
706
498436
351895816
26.5707
8.9043
662
438244
290117528
25.7294
8.7154
707
499849
353393243
26.5895
8.9085
663
439569
291434247
25.7488
8.7198
708
501264
354894912
26.6083
8.9127
664
440896
292754944
25.7682
8.7241
709
502681
356400829
26.6271
8.9169
665
442225
294079625
25.7876
8.7285
710
504100
357911000
26.6458
8.9211
666
443556
295408296
25.8070
8.7329
711
505521
359425431
26.6646
8.9253
667
444889
296740963
25.8263
8.7373
712
506944
360944128
26.6833
8.9295
668
446224
298077632
25.8457
8.7416
713
508369
362467097
26.7021
8.9337
669
447561
299418309
25.8650
8.7460
714
509796
363994344
26.7208
8.9378
670
448900
300763000
25.8844
8.7503
715
511225
365525875
26.7395
8.9420
671
450241
302111711
25.9037
8.7547
716
512656
367061696
26.7582
8.9462
672
451584
303464448
25.9230
8.7590
717
514089
368601813
26 . 7769
8.9503
673
452929
304821217
25.9422
8.7634
718
515524
370146232
26.7955
8.9545
674
454276
306182024
25.9615
8.7677
719
516961
371694959
26.8142
8.9587
675
455625
307546875
25.9808
8.7721
720
518400
373248000
26.8328
8.9628
MATHEMATICAL TABLES
111
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
721
519841
374805361 26.8514
8.9670
766
586756
449455096
27.6767
9.1498
722
521284 376367048 26.8701
8.9711
767
588289 451217663 27.6948
9.1537
723
522729 377933067 26.8887
8.9752
768
589824 452984832 27.7128
9.1577
724
524176
379503424 26.9072
8.9794
769
591361
454756609 27.7308
9.1617
725
525625
381078125
26.9258
8.9835
770
592900
456533000
27.7489
9.1657
726
527076
382657176
26.9444
8.9876
771
594441
458314011
27.7669
9.1696
727
528529 384240583 26.9629
8.9918
772
595984
460099648 27.7849
9.1736
728
529984 385828352
26.9815
8.9959
773
597529
461889917 27.8029
9.1775
729
531441 387420489 27.
9.
774
599076
463684824
27.8209
9.1815
730
532900
389017000
27.0185
9.0041
775
600625
465484375
27.8388
9.1855
731
534361
390617891
27.0370
9.0082
776
602176
467288576
27.8568
9.1894
732
535824
392223168J 27.0555
9.0123
777
603729
469097433
27.8747
9.1933
733
537289
393832837 27.0740
9.0164
778
605284
470910952
27.8927
9.1973
734
538756
395446904
27.0924
9.0205
779
606841
472729139
27.9106
9.2012
735
540225
397065375
27.1109
9.0246
780
608400
474552000
27.9285
9.2052
736
541696
398688256
27.1293
9.0287
781
609961
476379541
27.9464
9.2091
737
543169
400315553
27.1477
9.0328
782
611524
478211768 27.9643
9.2130
738
544644
401947272 27.1662
9.0369
783
613089
480048687; 27.9821
9.2170
739
546121
403583419 27.1846
9.0410
784
614656
481890304
28.
9.2209
740
547600
405224000
27.2029
9.0450
785
616225
483736625
28.0179
9.2248
741
549081
406869021 27.2213
9.0491
786
617796
485587656
28.0357
9.2287
742
5505641 408518488
27.2397
9.0532
787 619369 487443403
28.0535
9.2326
743
552049
410172407
27.2580
9.0572
788 620944 489303872
28.0713
9.2365
744
553536
411830784
27.2764
9.0613
789
622521; 491169069
28.0891
9.2404
745
555025
413493625
27.2.947
9.0654
790
624100
493039000
28.1069
9.2443
746
556516
415160936
27.3130
9.0694
791
625681
494913671
28.1247
9.2482
747
558009
416832723
27.3313
9.0735
792
627264
496793088 28.1425
9.2521
748
559504
418508992
27.3496
9.0775
793
628849
498677257 28.1603
9.2560
749
561001
420189749
27.3679
9.0816
794
630436
500566184 28.1780
9.2599
750
562500
421875000
27.3861
9.0856
795
632025
502459875
28.1957
9.2638
751
564001
423564751
27.4044
9.0896
796
633616
504358336
28.2135
9.2677
752
565504 425259008
27.4226
9.0937
797 635209 506261573 28.2312
9.2716
753
567009
426957777
27.4408
9.0977
798 636804 508169592 28.2489
9.2754
754
568516
428661064
27.4591
9.1017
799 638401 510082399 28.2666
9.2793
755
570025
430368875
27.4773
9.1057
800
640000
512000000 28.2843
9.2832
756
571536
432081216
27.4955
9.1098
801
641601
513922401
28.3019
9.2870
757
573049 433798093
27.5136
9.1138
802 643204
5158496081 28.3196
9.2909
758
574564 435519512
27.5318 9.1178
803: 644809
517781627 28.3373
9.2948
759
576081 437245479
27.5500 9.1218
804 646416
519718464 28.3549
9.2986
760
577600
438976000
27.5681
9.1258
805
648025
521660125
28.3725
9.3025
761
579121
440711081
27.5862
9.1298
806 649636
523606616 28.3901
9.3063
762
580644 442450728 27.6043
9.1338
807 651249; 525557943 28.4077
9.3102
763
-582169 444194947 27.6225
9.1378
808 652864 527514112 28.4253 9.3140
764
583696; 445943744 27.6405
9.1418
809 654481
529475129 28.4429 9.3179
765
585225 447697125 27.6586 9.1458
810 656100
531441000 28.4605 9.3217
112
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
N UM BERS . Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
811
657721
533411731
28.4781
9.3255
856
732736
627222016
29.2575
9.4949
812
659344
535387328
28.4956
9.3294
857
734449
629422793 29.2746
9.4986
813
660969
537367797
28.5132
9.3332
858 736164 631628712 29.2916
9.5023
814
662596
539353144
28.5307
9.3370 859 737881' 633839779 29.3087
9.5060
815
664225
541343375
28.5482
9.3408
860
739600
636056000
29.3258
9.5097
816
665856
543338496
28.5657
9.3447
861
741321
638277381
29.3428
9.5134
817
667489
545338513
28.5832
9.3485
862 743044
640503928
29.3598
9.5171
818
669124
547343432
28.6007
9.3523
863 744769
642735647
29.3769
9.5207
819
670761
549353259
28.6182
9.3561
864 746496 644972544
29.3939
9.5244
820
672400
551368000
28.6356
9.3599
865
748225
647214625
29.4109
9.5281
821
674041
553387661
28.6531
9.3637
866
749956
649461896
29.4279
9.5317
822! 675684
555412248
28.6705
9.3675
867 751689 T 651714363
29.4449
9.5354
823
677329
557441767
28.6880
9.3713
868 753424
653972032
29.4618
9.5391
824
678976
559476224
28.7054
9.3751
869 755161
656234909
29.4788
9.5427
825
680625
561515625
28.7228
9.3789
870
756900
658503000
29.4958
9.5464
826
682276
563559976
28.7402
9.3827
871
758641
660776311
29.5127
9.5501
827
683929
565609283
28.7576
9.3865
872 760384
663054848
29.5296
9.5537
828
685584
567663552 28.7750
9.3902 873
762129
665338617
29.5466
9.5574
829
687241
569722789
28.7924
9.3940
874
763876
667627624
29.5635
9.5610
830
688900
571787000
28.8097
9.3978
875
765625
669921875
29.5804
9.5647
831
690561
573856191
28.8271
9.4016
876
767376
672221376
29.5973
9.5683
832
692224
575930368
28.8444
9.4053
877
769129
674526133
29.6142
9.5719
833
693889 578009537
28.8617
9.4091
878
770884
676836152
29.6311
9.5756
834
695556
580093704
28.8791
9.4129
879
772641
679151439
29.6479
9.5792
835
697225
582182875
28.8964
9.4166
880
774400
681472000
29.6648
9.5828
836
698896
584277056
28.9137
9.4204
881
776161
683797841
29.6816
9.5865
837
700569
586376253
28.9310
9.4241
882
777924
686128968
29.6985
9.5901
838
702244
588480472
28.9482
9.4279
883
779689
688465387
29.7153
9.5937
839
703921
590589719
28.9655
9.4316
884
781456
690807104
29.7321
9.5973
840
705600
592704000
28.9828
9.4354
885
783225
693154125
29.7489
9.6010
841
707281
594823321
29.
9.4391
886
784996
695506456
29.7658
9.6046
842 708964
596947688
29.0172
9.4429
887
786769
697864103
29 . 7825
9.6082
843 710649
599077107
29.0345
9.4466
888
788544
700227072
29.7993
9.6118
844
712336
601211584
29.0517
9.4503
889
790321
702595369
29.8161
9.6154
845
714025
603351125
29.0689
9.4541
890
792100
704969000
29.8329
9.6190
846
715716
605495736
29.0861
9.4578
891
793881
707347971
29.8496
9.6226
847
717409
607645423
29.1033
9.4615
892
795664
709732288
29.8664
9.6262
848
719104
609800192
29.1204
9.4652
893
797449
712121957 29.8831
9.6298
849
720801
611960049
29 . 1376
9.4690
894
799236
714516984 29.8998
9.6334
850
722500
614125000
29.1548
9.4727
895
801025
716917375 29.9166
9.6370
851
724201
616295051
29.1719
9.4764
896
802816
719323136 29.9333
9.6406
852
725904
618470208 29.1890
9.4801
897 804609 721734273 29.9500 9.6142
853
727609
6206.50477 29.2062
9.4838
898
806 10 i 724150792 29.9666 9.6477
854
729316
622835864
29.2233
9.4875
899
808201 726572699 29.9833
9.6513
855
731025
625026375
29.2404
9.4912
900
810000- 729000000 30.
9.6549
MATHEMATICAL TABLES
113
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
NUMBERS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
901
811801
731432701
30.0167
9.6585
946
894916
846590536
30.7571
9.8167
902 813604
733870808
30.0333 9.6620 947
896809
849278123 30.7734
9.8201
903 815409
736314327
30.0500 9.6656 948 898704
851971392 30.7896
9.8236
901 817216
738763264
30.0666 9.6692
949 900601
854670349 30.8058
9.8270
903 819025
741217625
30.0832
9.6727
950 902500
857375000 30.8221
9.8305
906 820836
743677416
30.0998
9.6763
951
904401
860085351
30.8383
9.8339
907 822619, 746142643
30.1164 9.6799 952 906304
862801408 30.8545
9.8374
908 824461
748613312
30.1330 9.6834 953 908209 865523177 30.8707
9.8408
909 826281
751089429
30.1496 9.6870 954 910116 86S250664 30.8869
9.8443
910 828100
753571000
30.1662
9.6905
955
912025
870983875 30.9031
9.8477
911 829921
756058031
30.1828
9.6941
956
913936
873722816 30.9192
9.8511
912 831744
758550528
30.1993
9.6976 957
915849
876467493 30.9354
9.8546
913 833569
761048497
30.2159
9.7012 958
917764| 879217912 30.9516
9.8580
914 835396
763551944
30.2324
9.7047
959
919681 881974079 30.9677
9.8614
<Jlo
837225
766060875
30.2490
9.7082
960
921600
884736000 30.9839
9.8648
916
839056
768575296
30.2655
9.7118
961
923521
887503681
31.
9.8683
917 840889
771095213
30.2820
9.7153
962
925444
890277128 31.0161
9.8717
918 842724
773620632
30.2985
9.7188 903
927369 893056347 31.0322
9.8751
919 844561
776151559
30.3150
9.7224 964 929296 895841344 31.0483
9.8785
920 846400
778688000
30.3315
9.7259
965
931225
898632125
31.0644
9.8819
921 848241
781229961
30.3480
9.7294
966
933156 901428696 31.0805
9.8854
922 850084
783777448
30.3645
9.7329
967 935089 904231063 31.0966
9.8888
923 851929
786330467 30.3809
9.7364
968 937024 907039232 31.1127
9.8922
921 853776
788889024
30.3974
9.7400
969 938961 909853209
31.1288
9.8956
925
855625
791453125
30.4138
9.7435
970 940900 912673000
31.1448
9.8990
926
857476
794022776
30.4302
9.7470
971 942841
915498611
31.1609
9.9024
927 859329
796597983
30.4467
9.7505
972 944784 918330048 31.1769
9.9058
928 861184
799178752
30.4631
9.7540
973 946729! 921167317 31.1929
9.9092
929 863041
801765089
30.4795
9.7575
974' 948676 924010424 31.2090
9.9126
930
864900
804357000
30.4959
9.7610
975 950625. 926859375
31.2250
9.9160
931
866761
806954491
30.5123
9.7645
976
952576
929714176
31.2410
9.9194
932
868624
809557568
30.5287 9.7680
977 954529
932574833 31.2570
9.9227
933 870489
812166237
30.5450
9.7715
978 956484
935441352 31.2730
9.9261
0:u 872356
814780504
30.5614
9.7750
979 958441 938313739
31.2890
9.9295
935
874225
817400375
30.5778
9.7785
980
960400
941192000
31.3050
9.9329
936 876096
820025856
30.5941
9.7819
981
962361 944076141
31.3209
9.9363
937 877969
822656953
30.6105
9.7854
982
964324
946966168
31.3369 9.9396
938 879844
825293672
30.6268
9.7889
983
966289
949862087 31.3528 9.9430
939 881721
827936019
30.6431
9.7924
984
968256
952763904 31.3688 9.9464
940
883600
830584000
30.6594
9.7959
085
970225
955671625
31.3847
9.9497
911
885481
833237621
30.6757
9.7993
986
972196
958585256 31.4006
9.9531
942 887364
835896888
30.6920
9.8028 987 974169 961504803 31.4166 9.9565
943
889249
838561807
30.7083
9.8063 988 976111 96-4430272 31.4325 9.9598
944
891138
841232384
30 . 7246
9.8097 989 978121 967361669 31.4484 9.9632
945
893025
843908625
30.7409
9.8132. 990
980100 970299000 31.4643 9.9666
114
SHOP ARITHMETIC
SQUARES, CUBES, SQUARE ROOTS, AND CUBE ROOTS OF
N U M BE RS. Continued
No.
Square
Cube
Square
root
Cube
root
No.
Square
Cube
Square
root
Cube
root
991
9S20S1
973242271
31.4802
9.9699
996
992016
988047936
31.5595
9.9866
992 984064
976191488
31.4960
9.9733
997
994009
991026973
31.5753
9.9900
993 986049 979146657 31.5119
9.9766
998
996004
994011992
31.5911
9.9933
994
988036 982107784 31.5278
9.9800
999
998001
997002999
31.6070
9.9967
995
990025
985074875
31.5436
9.9833
1000
1000000
1000000000
31.6228
10.
CHAPTER XIII
LEVERS
90. Types of Machines. All machines consist of one or more
of the three fundamental types of machines the Lever, the
Cord, and the Inclined Plane or Wedge. Any piece of mechanism
can be proved to be of one or more of these types. Pulleys,
gears, and cranks will be shown to be forms' of Levers; belts and
chains come under the type called the Cord; while screws, worms,
and cams are forms of Inclined Planes. They are all used to
transmit power from one place to another and to modify it, as
desired.
91. The Lever. The lever is probably the most used and the
simplest type of machine. We are all familiar with it in its
simplest forms, such as crow bars, shears, pliers, tongs, and the
numerous simple levers found on machine tools.
A lever is a rigid rod or bar so arranged as to be capable of
turning about a fixed point. This fixed point about which the
lever turns is called the Fulcrum. In Fig. 37 the fulcrum is
Fia. 37.
represented by the small triangular block F. The position of
this fulcrum determines the effect which the force P applied at
one end has toward lifting the weight W at the other end. If
F is close to W, a comparatively small force P may be able to
raise the weight W, but if F is moved away from W and placed
close to P, then a greater force will be required at P. If F is in
the middle, P and W will be just equal.
In every lever there are two opposing tendencies: first, that
of the load or weight W tending to descend; and second, that of
the force P tending to raise W. The ability of W to descend or
to resist being lifted depends on two things its weight and its
distance from the fulcrum F. The product of these two is the
measure of the tendency of W to descend. This product is
10 115
116 SHOP ARITHMETIC
called, in books on mechanics, the Moment. Likewise, the force
P has a moment, which is the product of the force P and the
distance from P to the fulcrum F. If the force and the weight
just balance each other, their moments are equal.
The length from P to F is called the force arm and the length
from W to F, the weight arm. Then, for balance, we have the
equation:
Force X force arm = Weight X weight arm
If we let P stand for the force
a stand for the force arm
W stand for the weight
and b stand for the weight arm
as shown in Figs. 39, 40, and 41, we will have the formula
Although the force and weight are really balanced when this
formula is fulfilled, still we use the formula for calculating the
forces necessary to lift weights. The very slightest increase in
the force above that necessary for balance will cause W to rise
and, therefore, we can say practically that P will lift W if
p= WXb
a
If it is the length a that is wanted, we can see that P would
lift W when
Wxb
a = ...
14.'-
FIG. 38.
Example :
We have a lever 14 ft. long, with the fulcrum placed 2 ft. from the
end, as shown in Fig. 38; how much force must we exert to lift 1800 lb.? In
this problem a is 12 ft., b is 2 ft., and W is 1800 lb.
= 1800X2 =
Then, if Pxl2(or 12 XP) is 3600
P will be 3600 -H! 2
P = 3600 -nl2 = 300 lb., Answer.
LEVERS
117
It will be seen that the relation between force, weight, force
arm, and weight arm, can be written as an inverse proportion.
Force : weight = weight arm : force arm
or
P : W = 6 : a
This form of expressing the relation is not generally as useful as
the other form, PXa = WXb. It is very useful, however, in
cases where neither the force arm nor weight arm are known.
Example :
If a man wanted to lift a 750 Ib. weight by means of a 12 ft.
timber used as a lever, where would he place the fulcrum so that his whole
weight of 150 Ib. would just raise it?
Explanation: The total length of the timber
(12 ft.) is the sum of a and 6 (see Fig. 39). We
can find the ratio of b to a which is the same
as P:W and reduces to 1:5. If the ratio is
1 : 5, then the whole length is 6 parts of which
o is 5 parts and 6, 1 part. Hence a = 10 ft. and
6'= 2 ft., and the fulcrum must be placed 2 ft.
from the weight.
P
150
150
b
W =b
750 = 6
750 = 1
a =1
= of 12 = 10 ft.
=iof 12 = 2 ft.
92. Three Classes of Levers. Levers are divided into there
kinds or classes according to the relative positions of the force,
fulcrum, and weight.
Those shown so far are of the first class, Fig. 39, in which the
fulcrum is between the force and the weight. The weight is
lifted by pushing down at P.
FIQ. 39.
W
Fia. 40.
In the second class, Fig. 40, the weight is between the fulcrum
and the force, and the weight is lifted by pulling up at P.
In the third class, Fig. 41, the force P is between the weight
and the fulcrum and, therefore, P must be greater than the load
that it lifts. The weight is lifted by an upward force at P.
118
SHOP ARITHMETIC
In all these types the same rule holds that :
Force X force arm = weight X weight arm
or
p x a = W X b
V
/
F
57
h
1 J
FIQ. 41.
Particular attention should be given to the fact that the force
arm and weight arm are always measured from the fulcruhi.
In levers of class 2, the force arm is the entire length of the lever.
In class 3, the force arm is shorter than the weight arm. This
type may be seen on the safety valves of many boilers and is
used so that a small weight can balance a considerable pressure
at P.
Quite often there appear to be two weights, or two forces, on a
lever, and it is difficult to decide which to designate as the force
and which as the weight. It really makes no difference which
we call the force and which the weight; the relations between
them would be the same in any case.
\z-
f
n
1 W 1
L_+_J
M
a-
u
-- in 1 *.
Fia. 42.
93. Compound Levers. We frequently meet with compound
levers; but problems concerning them are easily reduced to
repeated cases of single levers, the force of one lever correspond-
ing to the weight of the next, etc. To illustrate this we will solve
the following example:
LEVERS 119
Example :
We wish to lift 8000 Ib. with a compound lever as shown in Fig. 42,
the first one being 10 ft. long with the weight 2 ft. from the end; the second
16 ft. long with the fulcrum 4 ft. from the end; what will be the necessary
force, P 2 ?
PXa=Wxb Explanation: Taking the first, or lower
p X10 = 8000X2 lever, we find it to be an example of the
16000 second class. W has a weight arm 6 of 2 ft.
PI = TQ = 1600 Ib. The force has an arm equal to the whole
w p _ iron lh length of the lever, or 10 ft. The necessary
P v- w vfc force on the end o f this lever we find to be
^
IK
v19 ID.
.
6400 1 The second lever must pull upward through
P 2 = - = 533= Ib., Answer, the connection with a force of 1600 Ib. In
other words, the 8000 Ib. weight on the first
lever is equivalent to a 1600 Ib. weight on the short end of the second
lever. The first lever pulls downward the same amount that the second
pulls up, or P l = W 2 . Having this 1600 as the weight, we find that a force
of 533J Ib. is needed on the end of the second lever.
94. Mechanical Advantage. The ratio of the weight to the
force is often called the Mechanical Advantage of the lever; this
ratio is equal to the force arm -r- the weight arm. In the com-
pound lever of Fig. 42 the M. A. (mechanical advantage) of the
first lever equals 10 -f- 2 or 5; of the second, 12-^4 or 3; the M. A.
of a compound lever is equal to the product of the M. A. of the
separate single levers; hence, of the given compound lever the
M. A. is 5 X3 or 15. This means that a 1 Ib. force will lift 15 Ib.;
10 Ib. will lift 150; or 100 Ib. will lift 1500. The force multiplied
by the M. A. gives the weight that can be lifted, or the weight
divided by the M. A. gives the necessary force. In the case
shown in Fig. 42, the mechanical advantage is 15 and conse-
quently the necessary force is 8000 -7-15 = 533 Ib.
If the mechanical advantage of a lever is 10, then 1 Ib. will
lift 10 Ib., or 800 Ib. will lift 8000 Ib., etc.; but it must be remem-
bered that the 1 Ib. or the 800 Ib. must travel 10 times as far as
the 10 Ib. or the 8000 Ib.
If a lever has a mechanical advantage of 10, the force must
travel 10 times as far as it lifts the weight, and consequently a
lever effects no saving in work. Work is the product of force,
or weight, times the distance moved, and is the same for either
end of the lever. It is similar to carrying a lot of castings to the
top floor of a building. If I carry half of them at a time, I must
make two trips; if I carry one-fourth of them at a time, I must
make four trips. The lighter the load, the more trips I must
make. The work done is the same whatever way I carry them
120
SHOP ARITHMETIC
and is equal to the product of the total weight times the height
to which the load must be carried.
95. The Wheel and Axle. This is a name given in mechanics
to the modification of levers that enables them to be rotated
continuously. Fig. 43 shows the principle of this: By wrapping
a belt or rope around each of the two circular bodies, we find
that the pulls in the cords are in inverse proportion to the radii
of the circles. A little consideration shows that the wheel and
axle may easily be studied as a force, P, with lever arm R t equal
to the radius of the wheel; and a weight, W, with a lever arm r,
equal to the radius of the axle. Two pulleys on a countershaft
FIG. 43.
might be likened to a lever in the same way. The belt which
drives the countershaft furnishes the force P. The radius of this
pulley is the force arm. The radius of the other countershaft
pulley, which transmits the power to the machine, is the weight
arm and the pull in this belt is the weight. Gears also are levers
that can be rotated continuously. The simplest example of
the use of the axle is probably the windlass, which we see used for
hoisting, house-moving, etc. See Figs. 48 and 50.
A geared windlass, such as shown in Fig. 50, is a case of com-
pound levers. The crank and pinion form the first lever and the
load on the gear teeth is transmitted to the teeth of the larger
gear and becomes the force of the other lever, which consists of
the large gear and the drum.v
LEVERS 121
Example :
A geared windlass, such as shown in Fig. 50, has a crank 20 in.
long; the small gear is 6 in. in diameter, the large gear is 30 in. in diameter,
and the diameter of the drum is 6 in.
What load could be raised by a man exerting a force of 25 Ib. on the crank?
6-r-2 = 3 in., radius of pinion
20n-3 = 6, M. A. of crank and pinion
30-5-2 = 15 in., radius of gear
6n-2 = 3 in., radius of drum
15-:- 3 = 5, M. A. of gear and drum
6jf X5 = 33J in., total mechanical advantage
25X33^=833 + Ib., Answer.
The solution of this problem might be shortened by writing all the work in a
single equation:
30
25X20X-2" 25X20X15
-e-y- 3^3" =833+ lb " Answer -
2 X 2
If we do not know the sizes of the gears in inches, but know the
numbers of teeth, we can figure that the mechanical advantage
of the pair of gears is the ratio of the numbers of teeth. In such
a case with a hoist as shown in Fig. 50, we would first find the
mechanical advantage of a simple windlass with the crank at-
tached directly to the drum; then find the M. A. of the pair of
gears, and by multiplying these two quantities together we
would get the mechanical advantage of the entire hoist.
PROBLEMS
ZOO
teoo
. *
-i _.. . 12 .
FIG. 44.
161. The lever shown in Fig. 44 is 12 ft. long. Where should the fulcrum
be placed so that a weight of 200 lb. will lift a weight of 1800 lb.?
,00
1800
Fia. 45.
162. In Fig. 45 where should the weight of 1800 lb. be placed so that it
can be lifted by a force of 200 lb.?
122
SHOP ARITHMETIC
163. Fig. 46 shows a safety valve V loaded with a 50 Ib. weight at W.
Find the total steam pressure on the bottom of V necessary to lift the valve.
F
-f
FIG. 46.
164. From the result of problem 163, find the steam pressure per square
inch if V is 1 J in. in diameter on the bottom where exposed to the steam.
#
FIG. 47.
FIG. 48.
165. Fig. 47 shows the clutch pedal for an automobile. What must be
the length of the power arm a in order that a foot pressure of 15 Ib. can open
the clutch against a spring pressure of 60 Ib. having an arm of 3 in.?
5lb 3 .
1000 I bs.
Fio. 49.
TACKLE BLOCKS
123
166. Fig. 48 shows an old fashioned windlass for raising water. If the
crank is 15 in. long, and the drum is 5 in. in diameter, what pressure would
be needed on the crank to raise a pail of water weighing 30 lb.?
167. Fig. 49 represents in an elementary way the levers of a pair of plat-
form scales. How far from the fulcrum must the 5 lb. weight be placed to
balance the 1000 lb. weight located as shown?
FIG. 50.
168. The hoist of Fig. 50 has an 18 in. crank; the drum is 10 in. in diam-
eter; the diameter of the large gear is 30 in., and of the small gear 6 in.
What weight can be raised by a force of 25 lb. on the crank?
CHAPTER XIV
TACKLE BLOCKS
96. Types of Blocks. When a heavy weight is to be raised or
moved through any considerable distance, either a windlass,
such as described in Chapter XIII, or tackle blocks can be used.
Referring to the figures in this chapter, the revolving part is
called the Pulley or Sheave; the framework surrounding the
pulleys is called the Block and, as generally used, includes both
the frame and the sheaves contained in it.
In Fig. 51 we have a single pulley which serves merely to give
a change of direction. There is no mechanical advantage in a
single fixed pulley such as this. The pull on the rope at P is
transmitted around the pulley and supports W on the other side.
We can look at the pulley in this case as a lever with equal arms.
P on one end must equal W on the other end. Such a block
would be used solely for the convenience it affords, since it is
usually easier to pull down than up.
124
SHOP ARITHMETIC
In Fig. 52 the pull P is in the same direction that W is to be
moved and is only one-half of W. As explained in Art. 94, the
mechanical advantage of a machine can be obtained by compar-
ing the distances moved by the force and the weight. If a force
must move five times as far as it lifts the weight, then the mechan-
ical advantage is 5, and the force is one-fifth of the weight. In
Fig. 52 the mechanical advantage is 2. This can be seen by
raising W a certain distance. The rope on each side will be
Blacked this same distances and, therefore, P must be drawn up
twice this distance in order to remove the slack. Since P moves
twice as far as W, the force P will be one-half of W, and the
mechanical advantage will be 2.
I
3
J
FIG. 51.
FIG. 52.
In Fig. 53 we have merely added, to the device of Fig. 52, a
fixed block above to change the direction of P. It makes no
change in the relation of P and W, except as to direction.
In Fig. 54 we have two pulleys in the fixed block and two in
the movable block. Other cases might have even more pulleys,
but the principle is the same, and a general rule for calculating
their mechanical advantages will be worked out for all cases.
Proceeding as before, let us imagine that W and the movable
block of Fig. 54 are lifted 1 ft. The four ropes supporting W
will each be slacked 1 ft., and it will be necessary to move P 4 ft.
to remove this slack. Hence, the mechanical advantage of this
system is 4, and P is \ of W , or W is 4 times P.
TACKLE BLOCKS
125
In general, we can say that the mechanical advantage is equal
to the number of ropes supporting the movable block and the
load. The best way to find the mechanical advantage is to draw
a sketch of the blocks and to count the number of ropes that are
pulling on the movable block. This number represents the
mechanical advantage.
Whenever convenient, it is best to use as the movable block
the one from which the free end of the rope runs. This means
that P will pull in the same direction that W is to be moved. The
I
J
Fio. 53.
FIG. 54.
mechanical advantage is greater by 1 if P is pulling in the direc-
tion of motion. Notice in Fig. 54 that, if we turned these blocks
around and pulled the other way, fastening W to what is in the
figure the fixed block, the mechanical advantage would be 5 in-
stead of 4.
In all problems where there is any doubt, draw a rough sketch
and count the number of ropes pulling on the movable block
(see Fig. 55).
Example :
How great a weight can be lifted by a pull of 150 Ib. with a pair of
pulley blocks, one being a three sheave and the other a two sheave block?
Calculate, first, using the three sheave as the movable block and, second,
using the two sheave block as the movable one.
126
SHOP ARITHMETIC
Explanation: To avoid confusion, the sheaves are drawn one above the
other, instead of parallel. The free end of the rope must run from the three
sheave block. Starting from P, we wind the rope in and find that the inner
end must be fastened to the two sheave block. We count the ropes pulling
on each block and find that, with the three sheave block as the movable one,
the mechanical advantage is 6, and the weight lifted would be
150X6 = 900 lb., First Answer.
With the two sheave block as the movable one, the mechanical advantage
is 5 and the weight would be
150X5 = 750 lb., Second Answer.
In practice, about 60% of these theoretical weights would be
raised, the rest being lost in overcoming friction. Likewise, to
lift a certain load, the actual pull required will be about V^r or 1 i
of the theoretical pull.
Fia. 55.
Fia. 56.
97. Differential Pulleys. In lifting heavy weights by hand, a
very satisfactory apparatus to use is a Differential Hoist. This
is a very simple and cheap apparatus but it is not very efficient
and is, therefore, not to be recommended for continuous use.
The pulleys are arranged as shown in Fig. 56. In the fixed
block are two pulleys, A and B, A being somewhat larger than B.
TACKLE BLOCKS 127
These pulleys, A and B, are fastened solidly together and rotate
as one about a fixed axis; the pulley C is in the movable block.
An endless chain passes over the pulleys as shown, the rims of
the pulleys being grooved and fitted with lugs to prevent the
chain from slipping. The loop np hangs free and is the pulling
loop.
From the figure it is easily seen that, if we pull down on p
until pulley A is turned once around, the branch m will be
shortened a length equal to the circumference of A. Since B
is attached to A, it also will turn once around and the branch o
will be lengthened a distance equal to the circumference of B.
Hence, the loop mo will be shortened by an amount equal to the
difference of the circumferences of A and B; and the pulley C
will rise one-half this amount.
We can express the difference in the distances moved by m
and o as
where D and d represent the diameters of large and small pulleys,
A and B. Hence C will move up one-half of this or of TT X (D d) .
To cause this motion of C upward, the chain p was moved a dis-
tance of 7TX-D.
The mechanical advantage of the hoist is obtained by dividing
the motion of P by the motion of W.
Mech. Adv. =
2'
This can be simplified by cancelling n out of both numerator
and denominator of the fraction, leaving
Mech. Adv. =^
This formula might be written as a rule in the following words:
" The mechanical advantage of a differential hoist is obtained by
dividing the diameter of the larger pulley in the upper block by
half the difference between the diameters of the larger and
smaller pulleys."
A differential hoist can actually lift about 30% of the theoret-
ical load with a given pull; that is, the efficiency is about 30%.
Likewise, to lift a given weight will require about -^ or 3J
128
SHOP ARITHMETIC
times the theoretical force. In other words, the actual force
must be such that the 30% that is really effective will equal the
theoretical force.
Example :
Calculate the actual pull required to lift 600 Ib. with a differential
hoist having 10 and 8 in. pulleys and an efficiency of 30%.
D = 10in.
rf = 8 in.
D-rf = 2 in.
of CD-d) = l in.
Mech. Adv. = :
7)
10
of (D-d)
600-7-10 = 60 Ib. theoretical pull
60 -=-.30 = 60X^ = 200 Ib. actual pull.
oO
Explanation: In this case the letters D and d of the formula are 10 in.
and 8 in., and we find the M. A. to be 10. It should, therefore, only require
a force of 60 Ib. to raise the 600 Ib. weight. But we find that this type of
hoist has only an efficiency of 30%, that is, it only does 30% of what we
might expect it to do from pur theories. Then to lift 600 Ib. will require a
force such that 30% of it. will be 60 Ib. This necessary force is 200 Ib.
Of Sheovas
Fia. 57
TACKLE BLOCKS
129
PROBLEMS
169. A weight of 2000 Ib. is to be lifted with a four sheave and three
sheave pair of blocks, as shown in Fig. 57. The four sheave is used as the
movable block. Neglecting friction and assuming each man to be capable
of pulling 125 Ib., how many men are necessary?
Fio. 58.
i\ 3 Sheaves
Z Sheaves
Fio. 59.
170. A windlass and tackle blocks, as shown in Fig. 58, are used for mov-
ing a house. If the team can exert a steady pull of 200 Ib. at the end of the
sweep, find the theoretical pull on the house. Also find the actual pull on
the house if the efficiency of the whole mechanism is 65%.
171. Draw a sketch of a pair of blocks, each having 3 pulleys, and indicate
which should be the movable block in order to secure the greatest mechanical
advantage. What would the mechanical advantage be?
130 SHOP ARITHMETIC
172. When the geared windlass of the dimensions shown in Fig. 59 is
used with the pair of pulley blocks, find the weight at W that can be lifted
by a force of 25 Ib. on the crank.
173. A differential hoist has pulleys 1\ in. and 6 in. in diameter. We
attach a weight of 200 Ib. to the hoist and find that a pull of 58 Ib. is re-
quired to raise the weight.
(a) Find the theoretical force required to raise 200 Ib. with
this hoist.
(b) From this and the force actually required, calculate the
efficiency of the hoist.
174. Three men pull 70 Ib. apiece on a pair of pulley blocks, two sheaves
above and one below. The single block is movable. Find the weight that
can be lifted:
(a) Neglecting friction;
(b) Assuming that 40% of the work is lost in friction.
175. A load of 2 tons is to be lifted with a differential hoist. The pulleys
are 12 in. and 10J in. in diameter.
(a) What is the theoretical pull required to lift the load?
(b) What is the actual pull required, if the efficiency of the
hoist is 30%?
CHAPTER XV
THE INCLINED PLANE AND SCREW
98. The Use of Inclined Planes. An Inclined Plane is a surface
which slopes or is inclined from the horizontal. Any one who
has had experience in raising heavy bodies from one level to
another knows that inclined planes are very useful for such work.
The Wedge is a form of an inclined plane, the powerful effect of
which in splitting wood, quarrying stone, aligning machinery,
and performing many other heavy duties is well known. The
inclined plane, like the lever and the tackle block, enables us to
lift a heavy weight with a smaller force.
99. Theory of the Inclined Plane. The work done in moving a
body up an inclined plane is merely the work of raising the body
vertically. If we skid an engine base from the shop floor onto a
flat car, the work accomplished is the raising of the base from
the floor level to the car level, and is the same as if it was raised
straight up by a crane, or by tackle blocks. The effect of the
long incline is similar to that of a long force arm on a lever. It
enables the force doing the work to use a greater distance, and
hence the force will be smaller than the weight raised.
Neglecting friction or, in other words, supposing bodies to be
perfectly smooth and hard, no work is done in moving the bodies
THE INCLINED PLANE AND SCREW
131
in a horizontal direction; hence the work done upon a body when
it is moved equals the weight of the body times the vertical
height to which it is raised. In studying the theory of the
inclined plane, we find that the force generally acts in one of two
directions in raising the body: either parallel to the incline or
parallel to the horizontal base.
In Case I (Fig. 60) the force P is exerted along the incline, and,
in raising the weight to the top, will act through a distance I.
Meamvhile, it will raise W a distance h. Consequently, the
I
mechanical advantage will be -
h
If we remember that the work put in equals the work got out
of a machine (neglecting what is lost in friction) we see that we
have the formula
To sum up, when the force is exerted parallel to the surface of
the inclined plane, the force times the length of the inclined
plane equals the weight times the vertical height through which
I F~OftO f t-MTAt.t.E.i. TO
Fio. 60.
TOTHX: BASE..
FIG. 61.
the weight is raised by the plane; or the mechanical advantage
equals the length of the inclined surface divided by the height.
If the weight to be raised is great as compared with the force
available, a comparatively long incline must be used to give the
necessary mechanical advantage.
In Case II (Fig. 61) the force acts parallel to the base of the
inclined plane; that is, along the horizontal. This case is not
often found in this elementary form, but is seen in jack screws, in
worm gearing, in wedges, and in cams, all of which are modifica-
tions of inclined planes. When the force acts parallel to the
base, the work expended by it is the product of the force times
the length of the base; the work accomplished is, as before, the
product of the weight times the height.
11
Mechanical advantage = -r
132 SHOP ARITHMETIC
A comparison of these formulas with those for Case I shows
that the mechanical advantage is greatest where the force P is
exerted along the incline, as in Case I, because I, the length of the
incline or the hypotenuse of a right triangle, is greater than 6,
the length of the base.
There may be other cases where the force acts in some other
direction, but they are seldom seen in practice.
100. The Wedge. The Wedge consists of two inclined planes
placed base to base, the force acting parallel to the base, as
shown in Fig. 62, where the horizontal center line of the wedge is
FIG. 62.
the common base of the two inclined planes. Usually the wedge
is moved instead of the object to be raised, but the effect is the
same and the force relations are the same as if the object itself
were being moved up a stationary incline. In Fig. 62 it will be
seen that the weight W will be raised a distance h when the
wedge is driven a distance I. The work expended in driving the
wedge is PXl; the work accomplished in raising the weight is
WXh; and, neglecting friction, these are equal, or
From this we see that the mechanical advantage of the wedge is :
Mechanical advantage = j-
The relation of P and W might, if desired, be written as a pro-
portion, as follows:
P:W = h:l
Example :
Fig. 63 shows an adjustable pillow block for a Corliss engine, the
bearing being raised or lowered by means of the wedge underneath. If the
weight of the shaft and the fly-wheel upon this bearing is 6000 lb., and the
wedge has a'taper of 1 in. per foot of length, what pressure must be exerted
on the wedge by the screw S in raising the shaft?
Mech. Adv. = 12
6000 -4-12 =500 lb. Answer.
THE INCLINED PLANE AND SCREW
133
Explanation: If the taper of the wedge is 1 in. in 12 in., then a motion
of 1 ft. would raise the bearing 1 in.; or a motion of 1 in. in the wedge would
raise the bearing fa in. Hence, the Mech. Adv. of the wedge is 12 and
W
p= ^ = 500 Ib.
\2t
Note. There would also be required, in addition to this 500 lb., a force
sufficient to overcome the friction on the top and bottom of the wedge,
which is neglected in this solution.
FIG. 63.
101. The Jack Screw. A screw is nothing but an inclined
plane which, instead of being straight, is wrapped around or cut
into a round rod or bar. Turning the screw gives the same
effect as giving a straight push on an inclined plane or wedge.
When we raise an object with a jack screw, such as shown in
Fig. 64, the weight presses down on the screw and, consequently,
I////////////J '///////////,
P <- ->
Fia. 84.
is borne by the threads (which are the inclined planes). The
threads are advanced and the weight is raised by a pull (which
we will call P) on the end of the rod whose length is marked R.
The distance the weight W is moved for one revolution of the
screw equals the lead of the screw expressed as a fraction of an
Inch. The lead of a screw is the distance it advances lengthwise
in one turn or revolution. The force P moves through a distance
134 SHOP ARITHMETIC
equal to the circumference of a circle whose radius is the length
of the handle; if we represent the length of the handle or lever by
R, then the distance traversed by P in one revolution is TT X 2 X R.
If we let the letter L represent the lead of the screw then we will
have the work accomplished in one revolution of the screw =
WxL. Meanwhile, the work expended in doing it
= PXKX2XR.
Assuming that there is no friction in the screw, we have
Distance P moves 7TX2X.R
or Mech. Adv. =^ - ^r~- -- = r= -
Distance W is raised L
If stated in words, these formulas would read: "The force
multiplied by the circumference of the circle through which it
moves equals the weight multiplied by the lead of the screw."
"The mechanical advantage of a jack screw equals the cir-
cumference of the circle through which the force moves divided
by the lead of the screw" (the amount the screw advances in one
turn) .
Example :
With a 1$ in. jack screw having 3 threads per inch and a pull of
50 Ib. at a radius of 18 in., calculate:
(a) The theoretical load that can be lifted by the screw;
(6) The actual load if the efficiency of the screw is 18%.
(a) Mech. Adv. =
.R = 18 in. and
L
1 .
= 3 in.
Hence,
3.1416X2X18
,. , . ,
Mech. Adv. =
=
3
_
OOJ7
_!
3
W = 339X50 = 16050 Ib., Answer.
(b) 18% of 16950 = 3051 Ib., Answer.
Explanation: If there are 3 threads per inch, the lead is $ in., and if the
radius is 18 in., we have the Mech. Adv. =339.3. In theory then we should
be able to lift 50 Ib. X 339 = 16950 Ib. with this screw. But a screw has con-
siderable friction and for this reason only 18% of the energy expended in
this case is effective, the remaining 82% being all lost in friction. The
actual weight lifted is, therefore, only 18% of 16950 Ib. or 3051 Ib.
102. Efficiencies. In explaining the machines of this chapter
and of Chapters XIII and XIV, it was assumed that no work is
lost in friction within the machines. In a properly mounted lever
there is little energy lost. In a tackle block the loss depends on
THE INCLINED PLANE AND SCREW 135
the size of the pulleys as compared with the size of the rope and
on the nature of the pulley bearings. The efficiency may vary
from 60 to 95 %. The more pulleys there are, the lower will be
the efficiency, because each bend in the rope and each pulley
means a loss in friction.
With inclined planes, the efficiency may vary all the way
from to nearly 100%. It will be lowest if the weight is merely
slid on the plane and will be much higher if wheels or rollers are*
used.
In any machine, if the weight will start back of its own accord
when the force is removed, the friction is less than 50% and the
efficiency is greater than 50%. If the weight will not start back,
the efficiency is less than 50%. This can be shown as follows:
Of the force applied to a machine, part of it is absorbed in over-
coming the friction within the machine. The balance goes
through the machine and is effective in accomplishing the work
to be done. Of the whole force applied, the per cent which this
effective force represents is called the Efficiency. If the efficiency
is less than 50%, it shows that the friction absorbs more than
half of the total force and, therefore, that the friction is greater
than the effective force. Now, suppose we had a simple machine
such as a jack-screw, being used to raise a weight. If the applied
force is removed, the friction will remain the same, but will now
act to hold the weight from 'going back. If the friction is suffi-
cient to hold the weight, it must at least equal the effective or
theoretical force required to raise the weight. Therefore, if a
machine does not run backward when the force is removed, the
friction must be more than one-half of the total force required to
raise the weight, and the effective force must be less than one-
half of this total force. Hence, the efficiency in such a case is
less than 50%. A jack-screw will not go down of its own accord
when the force is removed and therefore its efficiency is less than
50%. In reality, for the usual dimensions of screws, it has been
found to be only from 15 to 20%. Mr. Wilfred Lewis has
derived, from experiment, a simple formula which gives the
average efficiency for a jack-screw under ordinary conditions.
L
in which E is the efficiency, as a decimal,
where L is the lead of the screw,
and D is the diameter of the screw.
12
136
SHOP ARITHMETIC
Example :
Find the probable efficiency of the screw given in the example
under Article 101.
T l A n i 1 -
jL/=77 in., and D = IH in.
o ~
.33
.33 + 1.5
.# = 18%, Answer.
One can get an approximate idea of the efficiency of any
machine by observing, as before explained, whether or not it will
run backward of its own accord when the force is removed.
This will tell whether the efficiency is above or below 50%. If
it is above 50% and a considerable force is required to keep the
weight from going back, then the efficiency is high. If, however,
a very slight pull will hold it from going back, then the efficiency
is not very much above 50%. If we find the efficiency to be
under 50% but find that only a very small pull will start the
weight down, then the efficiency is not far under 50%. On the
other hand, if it seems as if almost as great a force is required to
lower the weight as to raise it, this signifies that the efficiency of
the machine is extremely low.
PROBLEMS
176. An engine weighing 5 tons is to be loaded onto a car, the floor of
which is 6 ft. from the ground. If 16 ft. timbers are used for the runway,
find the pull necessary to draw the engine up the slope, neglecting friction.
177. How many pounds must a locomotive exert to pull a train of 50
cars, each weighing 50 tons, up a grade of 3 in. in 100 ft.?
178. A building is to be raised by means of 4 jack-screws; the screws are
2 in. in diameter, with 4 threads to the inch. The lever is 20 in. long and a
30 Ib. force is required on each handle. Calculate th6 theoretical weight
which the four screws should lift under these conditions.
179. Calculate the probable efficiency of these jack-screws from Lewis'
formula and estimate the probable weight of the building.
FIG. 65.
180. A windlass with an axle 8 in. in diameter and crank 18 in. long is
used in connection with an inclined plane 20 ft. long and 5 ft. high, as shown
in Fig. 65. Neglecting friction, what weight can be pulled up the slope
with a force of 150 Ib. on the crank?
CHAPTER XVI
WORK, POWER, AND ENERGY; HORSE-POWER OF BELTING
103. Work. Whenever a force causes a body to move, work
is done. Unless the body is moved, no work is accomplished.
A man may push against a heavy casting for hours and, unless
he moves it, he does no work, no matter how tired he may feel at
the end of the time. It is evident that there are two factors to
be considered in measuring work force and distance. In
the study of levers, tackle blocks, and inclined planes we dealt
with the problem of work. In any of these machines the work
accomplished in lifting a weight is measured by the product of
the weight and the distance it is moved. The work expended
or put into the machine to accomplish this is the product of the
force exerted times the distance through which this force must
act. We found that, if we neglect the work lost in friction, the
work put into a machine is equal to the work accomplished by it.
The actual difference between the work put in and the work
accomplished is the amount that is lost in friction. The follow-
ing expressions may make these relations clearer:
Work lost in Friction = Work put in Work got out.
_~ . Work got out
Efficiency = ^ r-&
Work put in
104. Unit of Work. The unit by which work is measured is
called the Foot-pound. This is the work done in overcoming a
resistance of one pound through a distance of 1 ft.; that is, if a
weight of 1 Ib. is lifted 1 ft., the work done is equal to 1 foot-
pound. All work is measured by this standard. The work in
foot-pounds is the product of the force in pounds and the distance
in feet through which it acts. In lifting a weight vertically, the
resistance, and hence, the force that must be exerted, is equal to
the weight itself in pounds. The work done is the product of
the weight times the vertical distance that it is raised. If a
weight of 80 Ib. is lifted a distance of 4 ft., the work done is 80X4
or 320 foot-pounds. It would require this same amount of work
to lift 40 Ib. 8 ft., or to lift 20 Ib. 16 ft.
13 137
138 SHOP ARITHMETIC
When a body s moved horizontally, the only resistance to be
overcome is the friction. When a team of horses pulls a loaded
wagon, the only resistances which it must overcome are the
friction between the wheels and the axles, and the resistance on
the tires caused by the unevenness of the road.
The work necessary to pump a certain amount of water is the
weight of the water times the height through which it is lifted or
pumped (plus, of course, the work lost in friction in the pipes).
The work necessary to hoist a casting is the weight of the casting
times the height to which it is lifted. The work done by a belt
is the effective pull of the belt times the distance in feet which
the belt travels. The work done in hoisting an elevator is the
weight of the cage and of the load it carries times the height of
the lift. Numerous other illustrations of work will suggest
themselves to the student.
105. Power. Power is the rate of doing work; that is, in cal-
culating power the time required to do a certain number of foot-
pounds of work is considered. If 10,000 Ib. are lifted 7 ft. the
work done is 70,000 foot-pounds, regardless of how long it takes.
But, if one of two machines can do this in one-half the time that
the other machine requires, then the first machine has twice the
power of the second.
The engineer's standard of power is the Horse-power, which may
be defined as the ability to do 33, 000 foot-pounds of work per minute.
The horse-power required to perform a certain amount of work
is found by dividing the foot-pounds done per minute by 33,000.
If an engine can do 1,980,000 foot-pounds in a minute, its horse-
power would be 1,980,000^-33,000 = 60. An engine that can
raise 66,000 Ib. to a height of 10 ft. in 1 minute will do 66,000 Ib.
XlO ft. =660,000 foot-pounds per minute, and this will equal
a ff$=20 horse-power. If another engine takes 4 minutes to
do this same amount of work, it is only one-fourth as powerful;
the work done per minute will be &&S>AQ. = 165,000 foot-pounds
per minute; and its horse-power is - L H7nnj- = 5 horse-power.
Example :
An electric crane lifts a casting weighing 3 tons to a height of 20 ft.
from the floor in 30 seconds; what is the horse-power used?
3 tons = 6000 Ibs.
6000 Ib. X20 ft. = 120,000 foot-pounds done.
120,000 foot-pounds done in 30 seconds =
240,000 foot-pounds per 1 minute.
= 7.27 horse-power used.
WORK, POWER, AND ENERGY 139
106. Horse -power of Belting. A belt is an apparatus for the
transmission of power from one shaft to another. The driving
pulley exerts a certain pull in the belt and this pull is transmitted
by the belt and exerted on the rim of the driven pulley.
The power transmitted by any belt depends on two things
the effective pull of the belt tending to turn the wheel, and the
speed with which the belt travels. From the preceding pages, it
is easily seen that these include the three items necessary to
measure power. The pull of the belt is the force. The speed,
given in feet per minute, includes both distance and time. Force,
distance and time are the three items necessary for the measure-
ment of power.
The total pull that a belt will stand depends on its width and
thickness. It should be wide enough and heavy enough to
stand for a reasonable time the greatest tension put upon it.
This is, of course, the tension on the driving side. This tension,
however, does not represent the force tending to turn the pulley.
The force tending to turn the pulley (or the Effective Pull, as it
is called) is. the difference in tension between the tight and the
slack sides of the belt.
The effective pull that can be allowed in a belt depends prima-
rily on the width, thickness, and strength of the leather, or what-
ever material the belt is made of. Besides, we must consider
that every time a belt causes trouble from breaking or becoming
loose, it means a considerable loss in time of the machine, of
the men who are using it, and of the men required to make the
repairs and, therefore, it should not be loaded as heavily as
might otherwise be allowed. Leather belts are called "single/'
"double," "triple," or "quadruple," according to whether they
are made of one, two, three, or four thicknesses of leather.
Good practice allows an effective pull of 35 Ib. in a single leather
belt per inch of width. In a double belt a pull of 70 Ib. per inch
of width may be allowed. The pull times the width gives the
total effective pull or the force transmitted by the belt.
The force times the velocity, or speed, of the belt in feet per
minute will give the foot-pounds transmitted by it in 1 minute.
One horse-power is a rate of 33,000 foot-pounds per minute;
hence, the horse-power of a belt is obtained by dividing the
foot-pounds transmitted by it per minute by 33,000. The
velocity of the belt is calculated from the diameter and revolu-
tions per minute of either one of the pulleys over which the belt
140 SHOP ARITHMETIC
travels, as explained in Chapter VII. From these considerations,
the formula for the horse-power that a belt will transmit may
be written
PXWXV
H =
33000
where H = horse-power
P = effective pull allowed per inch of width
W = width in inches
V = velocity in feet per minute
Stated in words, this formula would read as follows: "The horse-
power that may be transmitted by a belt is found by multiplying
together the allowable pull per inch of width of the belt, the
width of the belt in inches, and the velocity of the belt in feet per
minute and then dividing this product by 33,000.
Example :
Find the horse-power that should be carried by a 12-in. double
leather belt, if one of the pulleys is 14 in. in diameter and runs 1100 R. P. M.
Explanation: To get the horse-
P = 70 Ib. power, we must first find the values
W = 12 in. O f Pt W> and F. We will take P
v ,N/^ vunn as 70 Ib. since this is a double belt.
r 71 A Vr * * AUU Tr , . . , . T , .-, | .,
12 W is given, 12 in. V, the velocity,
= 4032 ft. per min. is obtained by multiplying the cir-
TT_PXWX V cumference of the pulley by the R.
33000 P- M., which gives us 4032. Multi-
70X12X4032 plying these three together gives
= - oonnn 3,386,880 foot-pounds per minute,
-102+ horse-Dower Answer and divi ding by 33,000 we have
>e-power, Answer. 102 + as the horse-power that this
belt might be required to carry.
107. Widths of Belts. It is possible, also, to develop a
formula with which to calculate the width of belt required to
transmit a certain horse-power at a given velocity.
One horse-power is 33,000 foot-pounds per minute. Then the
given number of horse-power multiplied by 33,000 gives the
number of foot-pounds to be transmitted per minute.
Foot-pounds per minute = 33000 X H
If we know the velocity in feet per minute, we can divide the
foot-pounds per minute by the velocity; the quotient will be the
force or the effective pull in the belt.
Force ^
X \J V/V> * y-
WORK, POWER, AND ENERGY 141
Now the force can be divided by the allowable pull per inch of
width of belt. The result will be the necessary width.
, 33000 XH
Stated in words, this formula would read: "To obtain the width
of belt necessary for a certain horse-power; multiply the horse-
power by 33,000 and divide by the product of the allowable pull
per inch of width of belt times the velocity of the belt in feet per
minute."
Example :
Find the width of a single belt to transmit 10 horse-power at a
speed of 2000 ft. per minute.
II = 10 Explanation: We have given the horse-
V = 2000 power and the velocity, and we know that
P = 35 for a single belt a pull of 35 Ib. per inch is al-
33000 XH lowable. This data is all that is needed to
PXV calculate the width, which comes out 4 in.
The next larger standard width is 5 in., so
2 that is the size that would be used.
33000 X^ 33 .
~ ??X#W> = 7
7
Use 5 in. belt, Answer.
108. Rules for Belting. 1. Belt Thickness. It is generally
advisable to use single belting in all cases where one or both
pulleys are under 12 in. in diameter, and double belting on pulleys
12 in. or larger. Triple and quadruple belts are used only for
main drives where considerable power is to be transmitted and
where a single or double belt would have an excessive width. A
triple belt should not be run on a pulley less than 20 in. in diam-
eter, nor a quadruple belt on a pulley less than 30 in. in diameter.
2. Tension per Inch of Width. An effective pull of 35 Ib. per
inch of width of belt is allowable for single belts. For double
belts an effective pull of 70 Ib. per inch is allowable unless the
belt is used over a pulley less than 12 in. in diameter, in which
case only 50 Ib. per inch should be allowed. A prominent manu-
facturer of rubber belting recommends 33 Ib. per inch of width of
belt for 4-ply belts and 43 Ib. for 6-ply rubber belts.
3. Belt Speeds. The most efficient speed for belts to run is
from 4000 to 4500 ft. per minute. Belts will not hug the pulley
and therefore will slip badly if run at a speed of over one mile
per minute. These figures are seldom reached in machine shops.
142
SHOP ARITHMETIC
Belts for machine tool drives run from 1000 to 2000 ft. per min-
ute, while main driving belts for line shafts are more often run
about 3000 ft. per minute. On wood-working tools we find
higher speeds, usually 4000 ft. per minute or over.
4. Distance between Centers. The best distance to have be-
tween the centers of shafts to be connected by belting is 20 to 25
ft. For narrow belts and small pulleys this distance should be
reduced.
5. Arrangement of Pulleys. It is desirable that the angle of
the belt with the floor should not exceed 45 degrees; that is, the
belt should be nearer horizontal than vertical. Fig. 66 shows
Driver
Fio. 67.
the effect of having a belt nearly vertical. Any sag in the belt
causes it to drop away from the lower pulley and lose its grip on
it. Fig. 67 shows the best arrangement. Have the belt some-
where near horizontal and have the tight side of the belt under-
neath, if possible. This will increase the wrap of the belt around
the pulleys. If the lower side is the loose side, the wrap will be
decreased by the sag.
It is also desirable, whenever possible, to arrange the shafting
and machinery so that the belts will run in opposite directions
from the shaft, as shown in Fig. 68. This arrangement balances
somewhat the belt pulls, and reduces the friction and wear in
the bearings.
WORK, POWER, AND ENERGY
143
For belts which are to be shifted, the pulley faces should be
flat; all other pulleys should have the faces crowned (high in the
center) about T 8 in. per foot of width.
I. !
FIG. 68.
6. Grain and Flesh Sides. The grain side of the leather is the
side from which the hair is removed. It is the smoothest but
weakest side of the leather, and should run next to the pulley
surface. It will wrap closer to the pulley surface and thus get a
better grip on the pulley. Furthermore, the flesh side, being
stronger, is better able to stand the stretching which must occur in
the outside of the belt in bending around a pulley.
oursioc.
Fio. 69.
7. Belt Joints. Whenever possible, the ends of belts should
be fastened together by splicing and cementing. Never run a
wide cemented belt onto the pulleys as one side is liable to be
stretched out of true. Rather lift one shaft out of the bearings,
144 SHOP ARITHMETIC
place the belt on the pulleys, and force the shaft back into place.
Of other methods of fastening belts, the leather lacing is un-
doubtedly the best when properly done. In lacing a belt, begin
at the center and lace both ways with equal tension. Fig. 69
shows an excellent method of lacing belts. The lacing should
be crossed on the outside of the belt. On the inside, the lacing
should lie in line with the belt. Holes should be about 1 in.
apart and their edges should be at least $ in. from the ends of
the belt. The holes should be punched, preferably with an
oval punch, the long dimension of the oval running lengthwise of
the belt so as not to weaken the belt too much.
PROBLEMS
181. A casting weighs 300 Ib. How much work is required to place it on
a planer bed 3 ft. 5 in. above the floor?
182. How much work is required to pump 5000 gallons of water into a
tank 150 ft. above the pump?
183. Find the horse-power that may be transmitted per inch of width by
a single belt running at 2500 ft. per minute. How does this compare with
a double belt running at the same speed?
184. A 6 in. double belt is carried by a 48 in. pulley running 250 R. P. M.
Find the horse-power that may be transmitted.
185. A shop requires 50 horse-power to run it. The main shaft runs
250 R. P. M. Select a main driving pulley and determine width of double
belt to run the shop.
186. A foundry fan runs 3145 R. P. M., and requires 24 horse-power to
run it. There are two single belts on the blower running over pulleys 7 in.
in diameter. Determine the necessary width of belt.
Note. (Each belt should be wide enough to drive the fan so that in case
one breaks, the other will carry the load.)
187. A belt is carried by a 36 in. pulley running at 150 R. P. M. The
effective pull in the belt is 240 Ib. Find the horse-power.
188. A pumping engine lifts 92,500 gallons of water every hour to a
height of 150 ft. What is the horse-power of the engine?
189. If a freight elevator and its load weigh 5000 Ib., what horse-power
must be exerted to raise the elevator at a rate of 2 ft. per second?
190. A touring car is travelling on a level road at a rate of 45 miles an
hour. If it is shown by actual test that a force of 200 Ib. is required to
maintain this rate of speed, what horse-power must the engine deliver at
the wheels?
CHAPTER XVII
HORSE-POWER OF ENGINES
109. Steam Engines. In the last chapter, the meaning of the
term horse-power was explained and its application to belting
was discussed. We will now take up the calculations of the
horse-powers of steam and gas engines.
One horse-power was given as the ability to do 33,000 foot-
pounds of work in 1 minute. From this we see that the best way
to get the horse-power of any engine is to find out how many
foot-pounds of work it does in 1 minute and then to divide the
number of foot-pounds delivered in a minute by 33,000.
Let us study the action of the steam in the cylinder of the
ordinary double-acting steam engine. In Fig. 70 is shown a
FIQ. 70.
section of a very simple boiler and engine. We find that steam
enters one end of the cylinder behind the piston and pushes the
piston toward the other end of the cylinder. Meanwhile, the
valve is moved to the other end of the valve chest. The opera-
tion is then reversed and the piston is pushed back to the starting-
point. It has thus made two strokes, or one revolution. The
steam pressure on the piston is not the same at all points in the
stroke, but varies according to the action of the valve in cutting
off the admission of steam into the cylinder. However, it
145
146 SHOP ARITHMETIC
is possible to obtain the average or "mean effective pres-
sure" per square inch during a stroke, and, if we multiply
this by the piston area in square inches, we will have the average
total pressure or force exerted during one stroke. Now reduce
the length of the stroke to feet and multiply this by the total
pressure just found, and we have the number of foot-pounds of
work done during one stroke. This result, when multiplied by
the number of working strokes per minute, gives the foot-pounds
per minute and this divided by 33,000 gives the horse-power.
The following are the symbols generally used :
H. P. Horse-power.
P = Mean pressure in pounds per square inch.
A = Area of piston in square inches.
L = Length of stroke in feet .
N = Number of working strokes per minute.
PX A = Total pressure on piston.
PX A xL = it. Ib. of work done per stroke.
PXA XLX N it. Ib. of work done per minute, and hence
PXAXLXN
= 33000 " or ' as usuall y wntten >
PXLXAXN
- . In the latter form, the letters in the numerator
33000
spell the word Plan and the formula is thus easily remembered.
In the common steam engine, there are two working strokes
for every revolution of the engine, that is, the engine is \vhat is
called double acting, and N is twice the revolutions per minute.
A few steam engines, like the vertical Westinghouse engine, are
single acting and, hence, have only one working stroke of each
piston per revolution. Unless otherwise stated, it will be as-
sumed in working problems that a steam engine is double acting.
Example :
Find the horse-power of a 32 in. by 54-in. steam engine running at
94 R. P. M. with an M. E. P. (Mean Effective Pressure) of 60 Ib.
Note. In giving the dimensions of an engine cylinder, the first number
represents the diameter and the second number the stroke.
P = 60 Ib.
L = 54 in. =4^ ft.
A =Area of 32 in. piston = 804. 25 sq. in.
N = Number of strokes per minute = 94 X 2 = 188
H p _-PxLxAxJV
33000
60X4.5X804.25X188
1237+ Jiorse-power, Answer.
HORSE-POWER OF ENGINES 147
Notice particularly that the area of the piston is expressed in
square inches, because the pressure is given in pounds per square
inch; but that the stroke is reduced to feet because we measure
work in foot-pounds and, consequently, must express in feet the
distance which the piston moves.
If an engine has more than one cylinder, the horse-power of
each can be calculated and the results added; or, if the cylinders
are arranged to do equal amounts of work, we can find the horse-
power of one cylinder and multiply this by the number of
cylinders.
The mean effective pressure can be obtained for any engine
by the use of a device called an "indicator," which draws a
diagram showing just what the pressure is in the cylinder at each
point in the stroke. From this diagram, we can calculate the
average or mean effective pressure for the stroke. This pressure
must not be confused with the boiler pressure or the pressure in
the steam pipe. For instance, when the steam comes from the
boiler to the engine at 100 Ib. pressure, the mean pressure in the
cylinder will not be 100 Ib., as it would be very wasteful to use
steam from the boiler for the full stroke. Instead, the M. E. P.
(Mean Effective Pressure) will be from 20% to 85% of the boiler
pressure depending on the type of the engine and the load it is
carrying. Horse-power calculated as explained here is called
Indicated Horse-power because an indicator is used to determine
it. The indicated horse-power represents the power delivered
to the piston by the steam.
110. Gas Engines. The most common type of gas or gasoline
engine works on what is called the four stroke cycle. Such
an engine is called a four-cycle engine. Fig. 71 shows in four
views the operation of such an engine. Four strokes, or two
revolutions, are required for each explosion that occurs in the
cylinder. Consequently, in calculating the horse-power of a
single cylinder gas engine, the number of working strokes (or N
in the horse-power formula) is one-half of the R. P. M. There
is another type of gasoline engine called the two-cycle engine.
A single cylinder two-cycle engine has one working stroke for
each revolution of the crank shaft and N is therefore the same
as the number of R. P. M.
The mean effective pressure of a gas engine is from 40 to 100
Ib. per square inch, depending chiefly on the fuel used. For
148
SHOP ARITHMETIC
gasoline or natural gas or illuminating gas it is usually between
80 and 90 Ib. per square inch.
2. Compression
4-. Exttausr
Fio. 71
Example :
What horse-power could be delivered by a single cylinder 5 in.
by 8 in. four-cycle gasoline engine running 450 R. P. M.?
Note. Use a value of P = 80 Ib. per square inch.
P = 80
A = . 7854 X5 2 = 19.6 sq. in.
Then, H. P.
33000
80x|xl9.6x225
O
" 33000
- = 7.13 horse-power, Answer.
HORSE-POWER OF ENGINES 149
111. Air Compressors. An air compressor is like a double
acting steam engine in appearance; but, instead of delivering up
power, it requires power from some other source to run it. This
power is stored in the air and later is recovered when the air is
used. An air compressor takes air into the cylinder, raises its
pressure by compressing it, and then forces it into the air line or
the storage tank. In calculating the horse-power of a com-
pressor, the same formula can be used as for a steam engine.
The value of P to use is not the pressure to which the air is raised,
but is the average or mean pressure during the stroke. It is
usually somewhat less than half the final air pressure; for ex-
ample, when an air compressor is delivering air at 80 Ib. pressure,
the mean pressure on the piston is about 33 Ib.
Most air compressors are double acting, though there are many
small single acting ones.
Example :
A double acting 12 in. by 14 in. air compressor is running 150
R. P. M. It is supplying air at 100 Ib. and the mean pressure in the cylinder
is 37 Ib. per square inch. Calculate the horse-power necessary to run it.
P = 37 Ib.
A = Area of 12 in. piston = 113.1 sq. in.
N = Strokes = 150 X 2 = 300 per minute.
ThPn ff P
Then, H.P. -
6 110
In this case 12 appears in the denominator in order to reduce the 14 inches
to feet.
112. Brake Horse-power. The Brake Horse-power of an
engine is the power actually available for outside use. It,
therefore, is equal to the indicated horse-power minus the power
lost in friction in the engine. Brake horse-power can be readily
determined by putting a brake on the rim of the fly-wheel and
thus absorbing and measuring the power actually delivered.
Fig. 72 shows such a brake arranged for use. This form is
known as the "Prony Brake." It consists of a steel or leather
band carrying a number of wooden blocks. By tightening the
bolt at A, the friction between the blocks and the rim of the wheel
can be varied at will. The corresponding pull which this friction
gives at a distance R ft. from the shaft is weighed by a
150
SHOP ARITHMETIC
platform scale or spring balance. From the scale reading must
be deducted the weight due to the unbalanced weight of the
brake arms, which can be determined by reading the scales when
the brake is loose and the engine is not running. If an engine
is capable of maintaining a certain net pressure W on the scale,
and meanwhile maintains a speed of N revolutions per minute,
we can readily see that this is equivalent to an effective belt pull
of W pounds on a pulley of radius R running at N revolutions;
or it can be considered as being equivalent to raising a weight
FIG. 72.
equal to W by means of a rope wound around a pulley of radius
R turning at N revolutions per minute. This weight would be
lifted at the rate of
3.1416 X2XRXN ft. per minute
and the brake horse-power will be
' <P ' = 33,000
The brake and wheel rim will naturally get hot during a test, as
all of the work done by the engine is transformed back into heat
at the rubbing surfaces of the pulley rim and the brake. It is
necessary to keep a stream of water playing on the rim to remove
this heat and it is best to have special brake wheels for testing.
These have thin rims and inwardly extending flanges on the
rims so that a film of water can be maintained on the inner sur-
face of the rim.
HORSE-POWER OF ENGINES 151
Example :
Suppose that, at the time of testing the 5x8 gas engine in article
110, we also determined the brake horse-power by means of a Prony brake
having a radius of 3 ft. and that a net pressure of 22 Ib. was exerted on the
scales (the speed of the engine was 450 II. P. M.). Let us calculate the brake
horse-power.
Explanation: Our data is equi-
} l41fiv9VW4 { in-84R2 valent to that of hoisting a weight
22X8482-186 604?t Ib f 22 lb " ^ a r P e windin ^ u P on
K k J c a pulley of 3 ft. radius turning at
186,604 + 33,000 = 5.65, Answer. ^ ^p M The 22 Jb we f ght
would rise 8482 ft. per minute
and the work done per minute would be 22 Ib.X 8482 ft. = 186604 foot-
pounds per minute. Hence, the brake-horse power of the engine is 5.65.
113. Frictional Horse-power. If this engine gave 7.13 indi-
cated H. P. (I. H. P.), but the power available at the fly-
wheel was only 5.65, it stands to reason that the difference, or
1.48 H. P., was lost between the cylinder and fly-wheel.
The explanation is that this power is expended in simply over-
coming the friction of the engine; and this horse-power is, there-
fore, called the Frictional Horse-power. At zero brake horse-
power, the entire I. H. P. is used in overcoming friction.
114. Mechanical Efficiency. The ratio of the Brake Horse-
power to the Indicated Horse-power gives the mechanical
efficiency, meaning the efficiency of the mechanism in trans-
mitting the power through it from piston to fly-wheel. This is
usually expressed in per cent. In the case of the engine of
which we figured the I. H. P. and B. H. P., the mechanical
efficiency was
K ^
= = .792 = 79.2%
The mechanical efficiency of a gas engine is lower than that of a
steam engine on account of the idle strokes which use up work
in friction while no power is being generated, but at full load a
well built gas engine should show over 80 per cent, mechanical
efficiency. The mechanical efficiency of a steam engine should
be above 90% at full load.
PROBLEMS
191. The cage in a mine weighs 2200 Ib. and the load hoisted is 3 tons,
The hoisting speed is 20 ft. per second. Calculate horse-power necessary.
allowing 25% additional for friction and rope losses.
192. A 10 in. by 12 in. air compressor runs 150 R. P. M. The M. E. P.
is 30 Ib. Calculate the horse-power required to run it.
193. A pump lifts 2000 gallons of water per minute into a tank 150 ft.
above it. Find the horse-power of the pump.
152 SHOP ARITHMETIC
194. Find the horse-power of a 10 in. by 12 in. steam engine running
250 R. P. M. with a M. E. P. of 60 Ib.
195. What will be the horse-power of a single cylinder, four cycle gas
engine with the following data:
Size of cylinder, 12 in. by 16 in.
Rev. per minute, 225
Mean effective pressure, 78 Ib. per square inch?
Numhrer of working strokes = J of the number of revolutions.
196. A body can do as much work in descending as is required to raise it.
Knowing this fact, calculate the horse-power that could be developed by a
water-power which discharges 800 cu. ft. of water per second from a height
of 13.6 ft., assuming that 25% of the theoretical power is lost in the wheel
and in friction.
197. What would be the brake horse-power of a steam engine which
exerted a net pressure of 100 Ib. on the scales, at a radius of 4 ft., when
running at 250 R. P. M.?
198. How many foot-pounds of work per hour would be obtained from
a 60 H. P. engine?
199. A centrifugal pump is designed to pump 3000 gallons of water per
minute to a height of 70 ft. If the efficiency of the pump is 60%, what
horse-power will be required to drive it?
200. The pump of problem 199 is to run 1500 R. P. M. and is to be belt-
driven from a 48 in. pulley on a high speed automatic engine, running 275
R. P. M. What should be the diameter and width of fa n e of the pulley on
the pump, if the pulley is to be 1 in. wider than the belt?
CHAPTER XVIII
MECHANICS OF FLUIDS
116. Fluids. Nearly every shop of any size contains some
devices which are operated by water or air pressure, so every up-
to-date mechanic should have a knowledge of how these machines
work.
A Fluid is any substance which has no particular form, but
always shapes itself to the vessel which contains it. Water, oil,
air, steam, gas all are fluids. In some ways water, oil, and
similar substances are different from the lighter substances air,
steam, etc. To separate these, we give the name of Liquids to
such substances as water and oil; while air, steam, etc., are given
the general name of Gases. In some respects liquids and gases
are alike and in others they are different. The chief difference is
that liquids have definite volumes; they cannot be compressed
or expanded any visible amount, while gases can be readily
compressed or expanded to almost any extent. For all practical
purposes we can say that liquids cannot be compressed. The
third form of matter Solids needs no explanation. The
differences in these three forms can be stated as follows:
A Solid has a definite shape and volume.
A Liquid has no definite shape, but has a definite volume.
A Gas has neither a definite shape nor volume.
There are some substances that exist in states in between the
solid and the liquid form. Among these are tar, glue, putty,
gelatine, etc.
116. Specific Gravity. By Specific Gravity of a substance we
mean its relative weight as compared with the same volume of
water. Thus we say that the specific gravity of cast iron is 7.21,
meaning that cast iron is 7.21 times as heavy as water. A cubic
foot of water weighs 62.4 Ib. and a cubic foot of cast iron weighs
about 450 Ib. The quotient * 2 =7.21 is the specific gravity
of the iron.
In many hand books we find tables of specific gravities and,
when we wish to get the actual weight per cubic inch or per cubic
H 153
154.
SHOP ARITHMETIC
foot of some substance, we must multiply the weight of the unit
of water by the specific gravity of the substance.
Example :
The specific gravity of alcohol is .8. How many pounds would
there be to a gallon of alcohol?
One gallon of water =8J Ib.
One gallon of alcohol = .8 X8J = 6f Ib., Answer.
If a substance has a specific gravity less than 1, it will float in
water, because it is lighter than the same volume of water. If
the specific gravity is greater than 1, the substance is heavier
than water and will sink. A substance that will float in water
may sink in some other liquid if it has a greater specific gravity
than the liquid in question. For example, a piece of apple-wood
will float in water but will sink when placed in gasoline, the
specific gravity of the wood being about .76 and that of gasoline
about .71. A piece of iron will sink in water but will float when
placed in mercury (quick silver) , the specific gravity of mercury
being 13.6 and that of iron about 7.21.
117. Transmission of Pressure Through Fluids. One of the
most useful properties of all fluids is the ability to transmit
FIG. 73.
pressure in all directions. If we have a vessel filled with water,
as shown in Fig. 73, and apply a pressure to the water by means
of a piston, as shown, this pressure will be transmitted through
the water in all directions. If the sides of the vessel are flat,
they will bulge out, showing that there is a pressure on the sides;
and if the piston is loose, the water will escape upward around
it, showing that there is a pressure in.this direction also.
MECHANICS OF FLUIDS
155
If the total force on the piston is W Ib. and the area on
the bottom of the piston is A sq. in., then there will be a pressure
W
of -j- Ib. exerted on each square inch of the water beneath the
piston. This pressure will be transmitted equally in all direc-
tions and the pressure on each square inch of the top, sides, and
W
bottom of the vessel will be -j- Ib.
Example :
If the piston of Fig. 73 is 6 in. in diameter, and has a total weight
of 1000 Ib., what would be the water pressure per square inch?
W
1000
.7854 ><6*
p
Explanation: As the piston
rests on the water, the pressure
f the water on the bottom of
1000 the piston must be sufficient to
-5^-== 35.4 Ib. per sq. in., Answer, support the weight. The area
of the bottom is 28.27 sq. in.,
1000
and the pressure on each sq. in. will be ^nvr or 35.4 Ib. persquare inch.
Zo.Z I
This pressure is transmitted throughout the water and is exerted by it with
equal force in all directions.
118. The Hydraulic Jack. This property of water of trans-
mitting pressure in any direction is made use of in many ways.
The same property is, of course, common to other fluids such as
FIG. 74.
oil, air, etc. Wherever we find a powerful, slow-moving force
required in a shop, we usually find some hydraulic machine.
(The word " hydraulic " refers to the use of water but it is often
applied to machines using any liquid water, oil, or alcohol.)
Fig. 74 shows the principle of all these hydraulic machines.
156 SHOP ARITHMETIC
A small force P is exerted on a small piston and this produces a
certain pressure in the water. This pressure is transmitted to
the larger cylinder where the same pressure per square inch is
exerted on the under side of the large piston. If the large piston
has 100 times the area of the small piston, the weight supported
(W) will be 100 times P. If the water pressure produced by P
is 100 Ib. per square inch and the large piston has an area of 100
sq. in., then the weight W that can be raised will be 10,000 Ib.
Like the lever, the jackscrew, and the pulley, this increase in
force is obtained only by a decrease in the distance the weight is
moved. The work done on the small piston is theoretically the
same as the work obtained from the large piston. For example,
suppose that the large piston has 100 times the area of the small
one and we shove the small piston down 1 in. ; the water that is
thus pushed out of the small cylinder will have to spread out
over the entire area of the large piston; the large piston will,
therefore, be raised only one one-hundredth of the distance that
the small piston was moved. Thus, we have, here also, an applica-
tion of the law that the work put into a machine is equal,
neglecting friction, to the work done by it.
Force X distance moved = weight X distance raised.
The Mechanical Advantage of such a machine will be seen to
be the ratio of the areas of the pistons. In the case just men-
tioned, the ratio of the areas of the pistons was 100:1; hence,
the mechanical advantage was 100.
In Fig. 74, the motion that can be given to W is very limited,
but by using a pump with valves, instead of the simple plunger
P, we can continue to force water into the large cylinder and
thus secure a considerable motion to W.
Fig. 75 shows a common form of hydraulic jack which operates
on this principle. The top part contains a reservoir for the
liquid, and also has a small pump operated by a hand lever on
the outside of the jack. By working the lever, the liquid is
pumped into the lower part of the jack between the plunger and
the casing, thus raising the load. The load may be lowered by
slacking the lowering screw Y. This opens a passage to the
reservoir, and the load on the jack forces the liquid to flow back
through this passage to the reservoir.
In calculating the mechanical advantage of a hydraulic jack,
we must consider the mechanical advantage of the lever which
MECHANICS OF FLUIDS
157
operates the pump as well as the advantage due to the relative
sizes of the pump and the ram.
Fia. 75.
Example :
If the ram of Fig. 75 is 3 in. in diameter and the pump is 1 in. in
diameter, while the lever is 15 in. long and is connected to the pump at a
distance of 1 J in. from the fulcrum, what is the mechanical advantage of the
entire jack?
Explanation: The areas of the ram and pumps
are as 9:1, hence their mechanical advantage is 9.
The lever has a mechanical advantage of 10.
Hence, that of the whole jack is 9X10 = 90, and
a force applied at the end of the lever would be
multiplied 90 times. This force would, however,
move through a distance 90 times as great as the
distance the load would be raised.
7854X3 2 _9
.7854 Xl 1 ""!"
9X10 = 90, Answer.
158
SHOP ARITHMETIC
The hydraulic jack has usually an efficiency of over 70% and
is, therefore, a much more efficient lifting device than the jack
screw. A mixture containing one-third alcohol and two-thirds
water should be used in jacks. The alcohol is added to prevent
freezing.
119. Hydraulic Machinery. In the shop, we often find water
pressure used to operate presses, punches, shears, riveters,
hoists, and sometimes elevators. These machines are seldom
operated by hand power but have water supplied under pressure
from a central pumping plant. The admission of the water and
the consequent motion of the machine is controlled by hand
operated valves. Most of these hydraulic machines are used
where tremendous forces are required. Therefore, very high
water pressures are used, occasionally as high as 3000 Ib. per
square inch. 1500 Ib. per square inch is a common working
pressure for hydraulic machines.
FIG. 76.
FIG. 77.
Fig. 76 shows a press operated by hydraulic pressure. It will
be noticed that the movable head is connected to two pistons
a large one for doing the work on the down stroke, and a smaller
one above, used only for the idle or return stroke of the press.
120. Hydraulic Heads. Quite often we use a high tower or
MECHANICS OF FLUIDS 159
tank to secure a water pressure, or we make use of some natural
source of water which is at some elevation. This is most often
seen in the water supplies for towns and cities. Water tanks
are put upon the roofs of some factories for the same purpose.
The higher the tank, the greater will be the pressure which it will
maintain in the system. Let Fig. 77 represent such a system.
The water at the bottom has the weight of a column of water
h ft. high to support and, consequently, will be under a pressure
equal to the weight of this column of water. A volume of water
1 in. square and 1 ft. high weighs .434 lb., so the pressure per
square inch at the base of the column in Fig. 77 will be .434 X h.
Notice particularly that the shape and size of the tank has no
influence on the pressure, it being used merely for storage and to
keep the pressure from falling too fast if the water is drawn off.
The water in the tank on either side of the outlet is supported by
the bottom of the tank and has no effect on the pressure in the
pipe. The pressure at the bottom of the pipe would be the same
if the pipe alone extended up to the height h without the tank.
Also the size of the pipe has no effect on the pressure per square
inch. The water in a large pipe will weigh more than in a small
pipe, but the pressure will be spread over a larger area and if,
the heights are the same, the pressure per square inch will be the
same.
In pumping water to an elevated tank or reservoir, the pressure
required per square inch is also determined in the same manner
and is .434 times the height to which the water is raised, plus an
allowance for friction in the pipes. Thus, to pump water to a
height of 100 ft. requires a pressure somewhat greater than .434 X
100 = 43.4 lb. per square inch.
121. Steam and Air. Steam and air are likewise used to pro-
duce pressures in shop machinery. Being more elastic than
water, they are preferred where the machines are to be operated
quickly. Devices using air are called "pneumatic appliances,"
among the most common of which are pneumatic drills, hammers,
and hoists. The air for operating these is supplied by air com-
pressors located in the power house. These take the air from out
of doors and compress it into a smaller volume. The resistance
of the air to this compression causes it, in its effort to escape, to
exert a pressure on the walls of the tank or pipe containing it.
The more the air is compressed, the greater is the pressure exerted
by it. The air pressure used in shop work is usually about
160
SHOP ARITHMETIC
80 Ib. per square inch. The air is conducted through pipes and
hose to the point where it is to be used and there allowed to
exert its pressure on the piston of the appliance which is to be
driven.
PROBLEMS
201. The specific gravity of Lignum Vitse (a hard wood) is 1.328.
this wood float or will it sink in water?
Will
TTT
fit
FIG. 78.
FIG. 79.
202. What weight on the small piston of Fig. 74 would support a weight
of 30,000 Ib. on the large piston if the small piston is 1 in. in diameter and
the large one 12 in. in diameter?
203. If a hydraulic press works with a water pressure of 1500 Ib. per
square inch, what must be the diameter of the ram if a total pressure of
75,000 Ib. is to be produced?
204. If the air hoist of Fig. 78 has a cylinder 10 in. in diameter inside,
and the piston rod is 1J in. in diameter, and an air pressure of 80 Ib. per
square inch is exerted on the bottom of the piston, what weight can be
lifted by the hoist?
205. If a city wishes to maintain a water pressure of 80 Ib. per square
inch at their hydrants, how high above the streets must be the water level
in the stand pipe?
MECHANICS OF FLUIDS
161
206. Fig. 79 shows the principle of one form of hydraulic elevator, the
car being fastened directly to a long ram which is raised by water pressure.
The weight of the ram and car are partially balanced by a counterweight.
If this elevator is operated with water from the city mains at 80 Ib. pressure
per square inch and the ram has a diameter of 10 in., what load can be
lifted allowing 30% for losses in friction, etc.?
FIQ. 80.
Fia. SI.
207. If a steam pump, such as shown in Fig. 80, has a 12 in. steam piston
and an 8 in. water piston, what water pressure can be produced with a steam
pressure of 90 Ib. per square inch? How many gallons would be pumped
per minute when the pump is running at 100 strokes per minute, the stroke
of the pump being 12 in.? (1 gallon = 231 cu. in.)
208. What water pressure must a pump be capable of producing in order
to force the water to a reservoir at an elevation of 300 ft. above the pump?
15
162 SHOP ARITHMETIC
209. A gravity oiling system has an oil tank placed 10 ft. above the
bearings to be lubricated. The tank is connected by small tubes to the
various bearings. If the specific gravity of the oil is .88, what pressure will
the oil have at the bearings?
210. In Fig. 81 we have a hoist operated by a hydraulic ram in the top
of the crane post. The motion of the ram is multiplied by the system of
pulleys shown in the figure. What size must the ram be that a load of
10,000 Ib. can be lifted with a water pressure of 72 Ib. per square inch, the
efficiency of the whole apparatus being 70%?
CHAPTER XIX
HEAT
122. Nature of Heat. Some of the effects of heat are very-
useful in shop work and every mechanic should know something
of the nature of heat and of the laws which govern its applica-
tions to shop work.
Heat is a form of energy; that is, it is capable of doing work.
This we see amply illustrated in the steam engine and the gas
engine, where heat is used in producing work. The steam engine
uses heat which has been imparted to the steam in the boiler.
Part of the heat of the steam is changed to work in the engine and
the rest is rejected in the exhaust. Heat is not a substance as
was formerly supposed it cannot be weighed and cannot exist
by itself.
It is always found in some substances. We generally get heat
by burning some fuel such as coal, wood, gas, or oil. In burning,
the fuel unites with oxygen, one of the constituents of air, and
this process, called combustion, generates the heat. We cannot
get heat by this process, therefore, without air. No fuel will
burn without a supply of air, and as soon as we shut off the air
from a fire, combustion stops and no more heat is generated.
A fire may continue to give off heat for some time after the air is
cut off, but this heat comes from the cooling of the hot fuel in
the fire. Of the heat generated during combustion, some of it
goes through the furnace walls to the surrounding air; some goes
to heat up the bed of coals and any object that may be placed in
the fire to be heated; but the greater part of the heat goes off in
the gases that are formed by the union of the fuel with the air.
It is to save this heat that we sometimes see steam boilers set up
in connection with the furnaces of large forge shops.
There are other ways of generating heat besides that of com-
bustion. One method, that is coming into considerable use and
which is especially interesting to shop men because of the ease
with which it can be controlled, is by the use of electricity. We
now have electric annealing and hardening furnaces for use in
16 163
164 SHOP ARITHMETIC
tool rooms, where a close regulation of the heat is very desirable.
Then there are the electric furnaces by which aluminum and
carborundum are produced. We also have electric welding as
an example of the production of heat from electricity.
Another method of heat generation that is frequently encoun-
tered in shops, often where it is not desired, is the production of
heat from work. We have seen how heat is turned into work,
but here we have work returned into heat. One common case
of this is in bearings, where heat is produced from the work that
is spent in overcoming the friction. Another example is seen in
the heating of a lathe tool when it is taking a heavy cut, or in
the heating of the tool when it is being ground. In either event,
the work spent in removing the metal goes into heat.
123. Temperatures. Temperature is the indication of the
height or intensity of the heat in a body. Lowering the tem-
perature means a removal of heat from a body, and raising the
temperature means the addition of more heat. The common
method of measuring temperature is by means of an instrument
known as a thermometer, which usually consists of a glass tube
which is partly filled with mercury and which has the air ex-
hausted from the other part of it. The mercury expands and
contracts as the temperature rises or falls and, therefore, the
height of the column of mercury is a measure of the temperature.
Alcohol is often used instead of mercury for outdoor thermome-
ters where the mercury might freeze.
There are two kinds of thermometer scales in common use
the Centigrade (abbreviated C.) and the Fahrenheit (abbreviated
Fahr. or F.) . On the Centigrade thermometer the space between
the freezing-point of water and the boiling-point at atmospheric
pressure is divided into 100 equal parts called Degrees (represented
by ) the freezing-point being marked zero (0) and the boiling-
point 100. The balance of the scale is then divided into spaces
of equal length below zero and above 100 in order that tem-
peratures higher than 100 and lower than zero may be read.
On the Fahrenheit scale the freezing-point of water is marked
32 and the boiling-point 212, so the space between is divided
into 180 (212 -32 = 180). This scale is also marked with
divisions below 32 and above 212 in order to make the thermom-
eter read through a wider range. The Fahrenheit thermometer
is used more commonly in the United States than the Centigrade,
which is used extensively in Europe. The Centigrade scale is,
HEAT
165
however, used in this country for most scientific work and is
becoming so common that it is desirable to understand the rela-
tions between the two scales. Fig. 82 shows the relation of the
two scales up to 212 F. or 100 C.
Since the same interval of temperature is divided into 100
parts in the Centigrade scale, and 180 parts in the Fahrenheit
scale, each Centigrade degree is }$, or
f Fahrenheit degrees. Similarly, one
Fahrenheit degree is f of a Centigrade
degree. A change of 30 in tempera-
ture on the Centigrade scale would
equal f of 30, or 54 change on a
Fahrenheit thermometer. Likewise,
when the mercury moves 27 on a
Fahrenheit thermometer, it would move
only J- of 27 = 15 on a Centigrade scale.
In changing a reading on one thermom-
eter scale to the corresponding reading
on the other, it is necessary to remem-
ber that the zero points are not the
same. The Centigrade zero is at 32 F.
In other words, the two zeros are 32
Fahrenheit degrees apart.
To change a reading on the Centigrade
scale to the corresponding Fahrenheit
reading: First multiply the degrees C.
by f . This gives an equivalent number
of degrees on the F. scale. To this add
32, in order to have the reading from
the F. zero.
To change a reading on the Fahrenheit
scale to the corresponding Centigrade
reading: First subtract 32. This gives the number of F. de-
grees above freezing (which is the C. zero) . Multiply the result
by f , thus obtaining the desired C. reading.
Examples :
1. Change 30 C. to the corresponding Fahrenheit reading.
30 X r = 54, the equivalent number of F. degrees.
54+32-86 F., the reading on a F. thermometer.
2. What would a Centigrade thermometer read when a Fahrenheit ther-
mometer stood at 72?
210
200
190
180
170
160
150
140
130
120
110
100
so-
so
70
60
BO
40
F ^0=
20
10
-10
-20
-30
-40
1
212F = 100C
BOILING i
100
- 90
- 80
- 70
60
50
1
40
|
30
-
- 20
;
10
32 F. 0C 1
FREEZING
o C
\
10
\
20
1
30
-
1G6 SHOP ARITHMETIC
72 32 = 40, the number of F. degrees above freezing.
5 2
40 X = 22 C., Answer.
These rules or relations are often expressed by the following
formulas, in which C stands for a reading on the Centigrade
thermometer and F for a reading on the Fahrenheit thermometer.
The parenthesis ( ) when used as above means that the work
indicated inside of it is to be done first and then the result
multiplied by f . In the second formula, the C is first to be
multiplied by f and then the 32 is added to the product. It is
always to be understood that multiplications and divisions are
to be performed before additions and subtractions unless the
reverse is indicated, as was done in the first of these formulas, by
the use of the parenthesis ( ) .
When it comes to measuring the temperatures in furnaces, as
is often desirable in fine tool work, a thermometer is clearly out
of the question. As the mercury thermometer is ordinarily
made, it should not be used for temperatures above 500, but,
by filling the glass tube above the mercury with nitrogen gas
under pressure, a thermometer can be made that may be read
up to 900. For higher temperatures, devices called Pyrometers
are used. There are numerous kinds of pyrometers, but the one
most used in the shops for furnace temperatures is what is called
the Le Chatelier pyrometer. In this pyrometer, one end of a
porcelain tube about f in. in diameter and from 12 in. to 40 in.
long is thrust into the furnace and held there or, if frequent
readings are to be taken, it may be placed there permanently.
Inside this tube are some wires of special composition that
generate an electric current when they get hot. From the other
end of the tube a couple of wires run to a small box containing
a "galvanometer," that is, a device for indicating the strength of
the electric current generated. This has a needle swinging over
a dial and the dial is usually laid off in degrees so the temperature
is read direct. Most of these pyrometers have centigrade gradu-
ations, but one should be sure which scale a pyrometer has
before he uses it.
HEAT 167
For example, suppose we wanted to get 1000 F. and by mis-
take had 1000 C. instead.
? X 1000 + 32= 1832 F.
o
1000 C. = 1832 F.
This shows that it would be pretty serious to use the w r rong scale.
It has for years been the practice of the older shop men to tell
the temperature of steel or iron by its color. This method has
its disadvantages, however, as so much depends on the sensitive-
ness of the man's eye and on whether the work is being done in
bright sunlight or in a dark corner of the shop. A bar will show
red in the dark when it would still be black in the sunlight.
For the lower range of temperatures (those used in tempering
tools) we can judge the temperature by the color which will
appear on a polished steel surface when heated in the air. These
tempering colors and their uses for carbon tool steels are about
as follows:
430 F. Very pale yellow Scrapers
Hammer faces
Lathe, shaper, and planer tools
460 Straw yellow Milling cutters
Taps and dies
480 Dark straw color Punches and dies
Knives
Reamers
f>00 Brownish-yellow Stone cutting tools
Twist drills
.">20 Yellow tinged with purple Drift pins
530 Light purple Augers
Cold chisels for steel
oo() Dark purple Hatchets
Cold chisels for iron
Screw drivers
Springs
570 Dark blue Saws for wood
610 Pale blue
630 Blue tinged with green
More uniform results can be obtained if the steel is heated in
a bath of sand or of oil, the bath being maintained at the
desired temperature and a pyrometer being used to observe the
temperature. For higher temperatures, molten lead or mineral
1G8 SHOP ARITHMETIC
salts such as common salt, barium chloride, potassium chloride,
and potassium cyanide are used.
When steel and -iron are heated to higher temperatures, they
successively become red, orange, and white. These colors and
the corresponding temperatures are about as follows:
957 F. First signs of red
1290 Dull red
1470 Dark cherry
1655 Cherry red
1830 Bright cherry
2010 Dull orange
2190 Bright orange
2370 White heat
2550 Bright white welding heat
2730 ^
2910 t0 } Dazzling white.
124. Expansion and Contraction. Nearly all substances
expand when heat is applied to them and contract when heat is
removed. This phenomenon is greatest in gases and least in
solids, but even in solids it is of enough moment to be extremely
useful at times or to cause considerable trouble when allowance
is not made for it.
There are a few metal alloys which, within certain limits, do
not change their volumes with changes of temperature, and there
are also some which between certain temperatures will even
expand when cooled and contract when heated. A nickel steel
containing 36% nickel has practically no expansion or contrac-
tion with changes in temperature and is, therefore, used in some
cases for accurate measurements where expansion of the measur-
ing instruments would introduce serious errors.
When a solid body is heated it expands in all directions, if
free to do so, but as a rule we are concerned only with the change
of one dimension and not with the change in volume. Thus, in
the case of a steam pipe we do not care about the change in
thickness or in diameter, but we are concerned with the change in
length. On the other hand, when a bearing gets hot and seizes,
it is the change in diameter that causes the trouble. There are
few machinists who have not had the experience, in boring a
sleeve to fit a certain shaft, of having a free fit when tested just
after taking a cut through the sleeve, and then later of finding
that the sleeve fitted so tightly that it had to be driven off the
shaft. Of course, the explanation is that the sleeve becomes
HEAT 169
warm when being bored in the lathe, while the shaft is much
cooler. When the sleeve cools to the temperature of the .shaft, it
contracts and seizes or "freezes" to the shaft. In accurate tool
work the effect of differences in temperature between the measur-
ing instruments and the work may become serious. For this
reason many gages are provided with rubber or wooden handles
which do not conduct heat readily. They thus prevent the heat
of the hand from getting into the gages and expanding them.
But this is enough to give some idea of the troubles caused by
this property of materials; let us now see of what benefit it is.
We have already seen the use that is made of the expansion of
mercury in thermometers. There are numerous heat regulating
devices" (called thermostats) which depend on the expansion
or contraction of a bar to perform the desired operations. We
find these used for regulating house heating boilers and furnaces,
incubators, anti. other devices where uniform temperatures are
required. Probably the greatest shop use of expansion and
contraction is in making shrink fits. When we want to fasten
securely and permanently one piece of metal around another, we
generally shrink the first onto the second. This process is used
for attaching all sorts of bands and collars to shafts, cylinders,
and the like, for putting tires on locomotive wheels, and for
similar work. The erecting engineer uses it to put in the links in
a sectional fly-wheel rim or to draw up bolts in the hub or in any
other place where he wants to make a rigid permanent joint.
The amount of linear expansion which a body undergoes
depends upon the kind of material of which the body is made,
upon the amount of the temperature change and, of course,
upon the original length.
The coefficient of linear expansion of a substance is that part
of its original length which a body will expand for each degree
change in temperature. Coefficients for different metals have
been determined for our use by careful experiments, and can
be found in hand books or tables under the head of "Coefficients
of Expansion." The values given in different books do not al-
ways agree. In fact, the exact compositions of the metals used
in the tests were undoubtedly different for the different tests that
are on record. Hence, different tables give slightly different
rates of expansion. The following values are taken from the
most reliable authorities and are sufficiently accurate for most
purposes.
170
SHOP ARITHMETIC
COEFFICIENTS OF EXPANSION
Metal
Coefficient
Aluminum
. 00001234
Brass
. 00001
Cast iron
.0000055 to
Wrought iron and machine steel .
36% nickel steel
.000006
.0000065
. 0000003
The above values are based on a temperature rise of 1 F.
For one Centigrade degree change in temperature the coefficients
would be f of those just given. The student is not expected to
memorize these values. Remember that if the length is given
in feet the expansion calculated will be in feet, and if the length is
in inches the expansion calculated will be in inches. To get
the actual expansion per degree for any certain length, multiply
the coefficient of expansion by the length. If the temperature
change is 100, the expansion will be 100 times that for 1.
Example :
The head of a gas engine piston in operation has a temperature of
about 400 higher than the cylinder in which it is running. What allowance
must be made for this expansion in a 12 in. piston? (The piston is made of
cast iron.)
.000006X400 = .0024 in. expansion per inch
.0024X12 =.0288 in. expansion in 12 in., Answer.
The head of the piston must, therefore, be turned at least .0288 in. small to
allow for the expansion to take place without the piston seizing in the
cylinder.
The law of expansion and contraction may be expressed by a
formula as follows:
where
E = TxCxL
E is the change in length
T is the change in temperature
C is the coefficient of linear expansion
L is the original length of the body.
HEAT 171
Example :
What will be the expansion in a steam pipe 200 ft. long when
subjected to a temperature of 300, if erected when the temperature was 60?
7 7 = 300-60 = 240; C = .0000065; L = 200
E=TXCXL
= 240 X. 0000065X200 = .312 ft., Answer.
Notice particularly that here we use L in feet and, consequently, the expan-
sion E comes out in feet. This can be reduced to inches if desired, giving
12 X. 312 = 3. 744 in. or 3| in. nearly.
125. Allowances for Shrink Fits. In making a shrink fit, the
collar or band, or whatever is to be shrunk on, is bored slightly
smaller than the outside diameter of the part on which it is to
be shrunk. It is then heated and thus expanded until it can be
slipped into place. When it cools, it cannot return to its original
size but is in a stretched condition. It, therefore, exerts a power-
ful grip on the article over which is has been shrunk.
Practice differs considerably in the allowances that are made
for shrink fits. A rule which has been widely and successfully
used is to allow oVu" m< f r each inch of diameter. According
to this rule, if we were shrinking a crank on a 6 in. shaft, the
crank should be bored . 006 in. small or else the shaft turned . 006
in. oversize and the crank bored exactly 6 in. For a 10-in. shaft
we would allow .010 in, and so on for other sizes.
This could be expressed by the following formulas:
A = .OOlXl>, or, since -001=77)7)7), this could be written
D
1000
where A stands for " allowance "
and D for the diameter
Assuming that an allowance of .001 XD is made, let us see
what temperature is necessary in order to give the necessary
expansion so that a steel tire can be put over a locomotive driving
wheel.
For each degree that the tire is heated, it will expand .0000065
in. per inch of diameter. We must have an expansion of at least
.001 in. The number of de'grees necessary to get this will be
.001
. 0000065
17
= 154
172 SHOP ARITHMETIC
It would look as if a difference of 154 would be sufficient.
However, a greater difference is necessary in practice. There
must be sufficient clearance so that the tire can be slipped quickly
into place before it has time to cool off or to warm the wheel.
Once in place, the tire will grip the wheel when a temperaUire
difference of 154 exists.
PROBLEMS
211. In testing direct current generators, it is customary to specify that
under full load the temperature of the armature shall not rise more than 40
Centigrade above a room temperature of 25 C. ; that is, the temperature
of the armature under these conditions should not exceed 65 C.
In making a test a Fahrenheit thermometer was used. The room was at
a temperature of 77 F. and the temperature of the armature at the end of
the run was 180 F. Did the generator meet the specifications? What
was the temperature change, Centigrade?
212. In erecting a long steam line that will have a variation in tempera-
ture of 320, how far apart should the expansion joints be placed if each
joint can take care of a motion of 3 inches?
213. If a brass bushing measures 2 in. just after boring, when its tem-
perature is 95 F., what will it caliper when it has cooled to 65 F.?
214. A steel link 2 ft. long is made ^ in. too short for the slot in the fly-
wheel rim into which it is to be shrunk. How hot must the link be before
it will go in?
215. If a hub bolt is heated until it just begins to show red and is im-
mediately screwed up snug and allowed to cool, what shrinkage allowance
per inch of length would we be allowing by such a plan?
216. If we wished to maintain a tempering bath at a temperature of
500 F., what should be the reading on a Centigrade pyrometer?
217. If the brass bearings for a 2 in. steel crank shaft are given a running
clearance of .002 in. at a temperature of 60 F., what would be the clearance
when running at a temperature of 100 F.?
218. A horizontal steam turbine and dynamo are to be direct-connected,
their shaft centers being 3 ft. above the bed plate. If the bearings are
lined up at a temperature of 70, how much will they be out of alignment
under running conditions when the temperature of the dynamo frame is
80 F. and that of the turbine is 215 F., both frames being of cast iron?
CHAPTER XX
STRENGTH OF MATERIALS
126. Stresses. When a load is put upon any piece of material,
it tends to change the shape of the piece. The material naturally
resists this and, therefore, exerts a force opposite to the load. If
the load is not too heavy, the material may be able to exert a
sufficient force to hold it, but often the strength of the material
is exceeded and the piece breaks.
The resistance which is set up when a piece of material is loaded
is called the Stress. For instance, if a casting weighing 3 tons or
6000 Ib. is suspended by a single rope, the stress in the rope will
be 6000 Ib.
x\\\\\\ \v\\\\
W
tension
\\\\\\\\\\\VC
compression
FIG. 83.
shear
There are three different kinds of stresses that can be produced,
depending on the way the load is applied
(1) Tensile stress (pulling stress).
(2) Compressive stress (crushing or pushing stress) .
(3) Shearing stress (cutting stress).
Fig. 83 shows how these different stresses are produced.
We sometimes recognize two other kinds of stresses, but these
are really special cases of the three just given. These two others
are:
173
174
SHOP ARITHMETIC
(a) Bending Stress (really a combination of tension on one
side and compression on the other).
(b) Torsional or Twisting Stress (a form of shearing stress) .
127. Ultimate Strengths. By taking specimens of the different
materials and loading them until they break, it has been possible
to find out just what each kind of material will stand. The load
to which each square inch of cross-section must be subjected in
order to break it, is called the Ultimate strength of the material.
The strength of most materials differs for the different methods
of loading shown in Fig. 83.
The Tensile Strength of a material is the resistance offered by
its fibers to being pulled apart.
The Compressive Strength of a material is the resistance
offered by its fibers to being crushed.
The Shearing Strength of a material is the resistance offered
by its fibers to being cut off.
The following table gives the average values for the most used
materials.
ULTIMATE STRENGTHS POUNDS PER SQUARE INCH
Material
Tension
Compression
Shear
Timber
10000
8000
3000 (across grain)
Cast iron
20000
90000
20000
Wrought iron
Machine steel
50000
65000
50000
65000
40000
50000
128. Safe Working Stresses. Having found how great a stress
is required to break one square inch of material, we naturally
would not allow anywhere near this stress to come on a piece
of material in actual service. The Ultimate Strength is usually
divided by some number, known as the Factor of Safety, and the
quotient is used as the Safe Working Stress.
For example, if 60,000 Ib. per square inch will break a piece of
soft steel and we use a factor of safety of 5, this would give:
(\C\OC\O
Safe working stress = = = 12000 Ib. per square inch.
o
The following table gives the Safe Working Stresses of the most
used materials.
STRENGTH OF MATERIALS 175
SAFE WORKING STRESSES POUNDS PER SQUARE INCH
Material
8 t
Tension
S c
Compression
s,
Shear
Timber
700
700
500
Cast iron ....
3- 4000
15-18000
3- 4000
Wrought iron ....
Machine steel. . . .
8-10000
10-16000
8-10000
10-16000
7- 9000
8--12000
Instead of writing " safe loads in pounds per square inch " for
tension, compression, or shear, the symbols S t , S c , and S 8 are used.
So if A = area in square inches, then the load W which can be
carried safely = Area X safe load per square inch or
AxS t = W (tension)
A X S c = W (compression)
AxS a = W (shear)
Or, in general, for all stresses
AXS = W
Perhaps more often we would want to find the area necessary in
order to support a certain weight or load. In this case, we would
want a formula which would give A.
If we divide the total load by the safe stress, we will get the
necessary area; or
W
A =
S
This simply says, area of metal necessary = total weight to be
carried divided by safe load in pounds per square inch. From
the area of a bolt or rod, its diameter can be easily found.
129. Strengths of Bolts. There is a well-known saying that
" a chain is only as strong as its weakest link." This means, in
general, that any mechanism must be so designed that its weakest
part will be strong enough to stand the greatest load that may
come on it. In figuring the size of a bolt to hold a certain load,
we would not calculate the full diameter of the bolt and make
the area there just sufficient, but we must see to it that the bolt
has a cross-sectional area at the root of the threads large enough to
support the load. Then the body of the bolt will have a surplus
of strength.
176
SHOP ARITHMETIC
Example :
' What size of steel eyebolt will support a weight of 5000 lb.?
Take 12,000 lb. as the safe load in tension.
W 5000
Then ' A= T = l2ooo
5
A = 2 sq. in. = .416 sq. in.
.416 sq. in. is then the necessary area to support the weight. Of course,
the example could be completed by saying .7854 D 2 = .416 sq. in., where
D = diameter at the root of the thread. By then solving for D we would get
the diameter at the root of the threads. But the Bolt Tables afford an
easier method than this. In the following table, .4193 is given as the area
of a | in. bolt at the root of the thread. Therefore, a | in. eyebolt would
probably be used.
In figuring the allowable loads for steel bolts, it is best not
to allow over 12,000 lb. stress per square inch and 10,000 lb. is
perhaps even more usual on account of the sharp root of the
threads, which makes a bolt liable to develop cracks at this point.
BOLT TABLE. U. S. S. THREADS
Diam.
Threads
to inch
Diam. at
bottom of
thread
Area of bolt
Area at
bottom of
thread
tin.
20
.1850
.0491
.0269
fs in.
18
.2403
.0767
.0454
1 in.
16
.2938
.1104
.0678
T 7 T in.
14
.3447
.1503
.0933
i in.
13
.4001
.1963
.1257
-, 9 ,T in.
12
.4542
.2485
.1621
fin.
11
.5069
.3068
.2018
J in.
10
.6201
.4418
.3020
I in-
9
.7307
.6013
.4193
1 in.
8
.8376
.7854
.5510
IJin.
7
.9394
.9940
.6931
Hin.
7
1 . 0644
1 . 2272
.8899
If in.
6
1.1585
1.4849
1.0541
liin.
6
1.2835 1.7671 1.2938
If in.
5i
1.3888 2.0739 1,5149
If in.
5
1.4902 2.4053 1.7441
2 in.
4i
1.7113 3.1416 2.3001
2\ in.
4i
1.9613 3.9761 3.0213
2iin.
4
2.1752 4.9087 3.7163
2| in.
4
2.4252 5.9396 4.6196
3 in.
34
2.6288 7.0686 5.4277
STRENGTH OF MATERIALS 177
130. Strength of Hemp Ropes. It is quite common in calcu-
lating the strength of ropes and cables to assume that the section
of the rope is a solid circle. Of course, the strands of the rope
do not completely fill the circle but, if we find by test the allow-
able safe strength per square inch on this basis, it will be perfectly
safe to make calculations for other sizes of ropes on the same
basis. The safe working stress based on the full area of the
circle is 1420 Ib. per square inch. The Nominal Area (as the
area of the full circle by which the rope is designated is called) is
A = .7854X.D 2 . The safe stress is 1420 Ib. per square inch and,
consequently, the weight that can be supported by a rope of
diameter D is
W = SxA
= 1420 X. 7854 X# 2
Here we have two constant numbers (1420 and .7854) that
would be used every time we were to calculate the safe strength
of a rope. If this were to be done often we would not want to
multiply these together every time, so we can combine them now,
once and for all.
1420 X. 7854 = 1120, approximately
Hence
* TF = 1120XZ) 2
Example :
Find the safe load on a hemp rope of in. diameter.
= 1120X4 = 280 Ib., Answer.
131. Wire Ropes and Cables. For wire ropes made of crucible
steel, a safe working load of 15,000 Ib. per square inch of nominal
area is allowable. For cables of Swedish iron but half this value
should be used.
132. Strength of Chains. It has been demonstrated by re-
peated tests that a welded joint cannot be safely loaded as heavily
as a solid piece of material. Of course, there are often welds
that are practically as strong as the stock, but it is not safe to
depend on them. For this reason, the safe working load per
178 SHOP ARITHMETIC
square inch for chain links is often given as 9000 lb., which is
just f of 12,000 lb.
If D = the diameter of the rod of which the links are made
A=2X.7854X> 2
W = S t XA
F = 9000X2X.7854X> 2
Combining the constant numbers, this can be simplified into
W = 14,000 X> 2
This is used in the same way as the formula for a rope.
133. Columns. The previous examples were cases of tension.
The size of a rod or timber subjected to compression is computed
in the same way unless it is long in comparison with its thickness.
When a bar under compression has a length greater than ten
times its least thickness, it is called a Column and must be con-
sidered by the use of complicated formulas which take account
of its length. It can be seen by taking a yardstick, or similar
piece, that it is much easier to break than a piece of shorter
length but otherwise of the same dimensions. A long piece,
when compressed, will buckle in the center and break under a
light thrust or compression. An example of this can be found
in the piston rod on a steam engine, where, on account of the
length of the rod, it is necessary to use much lower stresses than
those given in the tables. The compressive stress allowed in
piston rods varies with the judgment of different designers but
is generally about 5000 lb. per square inch, using a pressure on
the piston of 125 lb. per square inch.
Example :
Find the size of rod for a 30 in. by 52 in. Corliss engine with 125 lb .
steam pressure.
30 in. is the diameter of the cylinder and 52 in. is the stroke, which is not
considered in the problem except in that it has reduced the allowable stress
in the rod.
.7854 X30 2 = 706.86 sq. in., area of piston.
706.86X125 = 88357.5 lb., total pressure on piston.
Using 5000 lb. per square inch, allowable stress in the rod.
88358-7-5000 = 17.67 sq. in. sectional area of rod,
From the table of areas of circles, it is seen that this is the area of a circle
nearly 4f in. in diameter, so we would use a 4 in. rod.
PROBLEMS
Note. In all examples involving screw threads, to get areas at root of
thread, use the table given in this chapter. Give sizes of bolts always as
diameters.
STRENGTH OF MATERIALS
179
219. If the generator frame shown in Fig. 84 weighs 3000 lb., what size
steel eyebolt should be used for lifting it, allowing a stress of 10,000 lb. at
the root of the thread?
220. What would be the safe load for a J in. chain?
221. What size hemp rope would be necessary to lift a load of 4000 lb.?
Fio. 84.
222. What force would be necessary to shear off a bar of machinery
steel 2 in. in diameter?
223. A certain manufacturer of jack screws states that a 2J in. screw is
capable of raising 28 tons. If the diameter of the screw at the base of the
threads is 1.82 in., what is the stress per square inch at the bottom of the
threads when carrying 28 tons?
224. A soft steel test bar having a diameter of .8 in. is pulled in two by a
load of 31,500 lb. What was the breaking tensile stress per square inch?
Fio. 85.
225. The cylinder head of a small steam engine (Fig. 85) having a cylinder
diameter of 7 in. is held on by 6 studs of $ in. diameter. When there is a
steam pressure of 125 lb. per square inch in the cylinder, what will be the
pull on each stud? And what will be the stress per square inch in each stud,
due to the steam pressure?
226. With a cylinder diameter of 10 in. and an air pressure of 100 lb.
per square inch, find the greatest weight that can be lifted by the air hoist,
180
SHOP ARITHMETIC
shown in Fig. 86. Also find the size of piston rod necessary, assuming that
it is screwed into the piston. Notice that this rod is subject only to tension
and, therefore, a greater stress is allowable than in steam engine piston rods.
227. Work out a formula for the strength of crucible steel cables on the
same plan as that given for hemp rope.
228. What is the greatest load that should be lifted with a pair of tackle
blocks having 3 pulley in the movable block and 2 in the fixed block, and
having a f in. rope.
m
m
FIG. 86.
INDEX
Addition of decimals, 35
of fractions, 1 1
Air compressors, 149
steam and, 159
Allowances for shrink fits, 171
Analyzing practical problems in fractions, 22
Area of a circle, 77
Areas of circles, table of, 101
Arrangement of pulleys, 142
Belting, horse power of, 139
rules for, 141
Belt joints, 143
Belts, grain and flesh sides, 143
speeds of, 141
tension per inch of width, 141
thickness of, 141
width of, 140
Blocks, types of tackle, 123
Bolts, strength of, 175
Bolt table, U. S. S. threads, 170
Cables, strength of wire rope and, 177
Cancellation, 19
Casting, weight of, 83
Centigrade thermometers, 164
Chains, strength of, 177
Circle, area of, 77
circumference of, 51
diameter of, 51
radius of, 51
Circumference of a circle, 51
Circumferences of circles, table of, 101
Circumferential speeds, 54
Classes of levers, 117
Coefficient of linear expansion, 169
Columns, 178
Common denominator, 9
fractions reduced to decimals, 39
Complex decimals, 40
Compound fractions, 21
gear and pulley trains, 68
Contraction, expansion and, 168
19 181
182 INDEX
Cube, the, 80
root, 75, 92
roots of decimals, 98
of numbers greater than 1000, 97
table, 103
Cubes, 75
and higher powers, 75
of numbers, table of, 103
Cubical measure, units of, 80
Cutting speeds, 57
Decimal equivalents, table of, 42
fractions, 33
Decimals, addition of, 35
complex, 40
cube roots of, 98
division of, 37
multiplication of, 36
short cuts, 37
subtraction of, 35
Denominator, common, 9
least common, 10
of a fraction, 2
Diameter of a circle, 51
from area, calculation of, 91
Differential pulleys, 126
pulley, mechanical advantage of, 127
Dimensions of circles, 91
rectangles, 91
squares, 91
Direct and inverse proportions, 65
Division of decimals, 37
fractions, 21
Efficiencies, 134
Efficiency of engines, mechanical, 151
of hydraulic jack, 158
of jack screw, 135
Emery wheels, 55
Expansion and contraction, 168
Extracting square root, 86
cube root, 96
Fahrenheit thermometers, 164
Fluids, 153
transmission of pressure through, 154
Foot pound, 137
Formulas, 52
INDEX 183
Fractions, addition of, 11
common fractions, reducing to decimals, 39
compound, 21
decimal, 33
definition of, 2
denominators of, 2
division of, 21
improper, 3
multiplication of, 17
numerators of, 2
proper, 3
reduction of, 3
subtraction of, 12
whole number times a fraction, 17
writing and reading of, 2
Frictional horse power, 151
Gas engines, 147
horse power of, 147
Gear ratios, 64
Gears, relation of sizes and speeds of, 64
Gear trains, 66
compound, 68
Grindstones and emery wheels, 55
Heat, nature of, 163
Horse power, 138
brake, 149
frictional, 151
of belting, 139
of gas engines, 147
of steam engines, 145
Hydraulic heads, 158
jack, 155
mechanical advantage of, 156
efficiency of, 158
Hypotenuse, 89
Improper fractions, 3
reduction of, 5
Inclined planes, mechanical advantage of, 131
theory of', 130
use of, 130
Interpolation, 96
Inverse proportion, 65
Iron and steel, colors at different temperatures, 167
Jack screws, 133
efficiency of, 135
mechanical advantage of, 134
184 INDEX
Law of right triangles, 89
Least common denominator, 10
Levers, 115
classes of, 117
compound, 118
mechanical advantage of, 119
Linear expansion, coefficient of, 169
Machines, types of, 115
Mean effective pressure, 146
Measure, cubical, 80
square, 76
Measures of length, 7
of time, 7
of volume, 80
Mechanical advantage of differential pulley, 127
hydraulic jack, 156
inclined plane, 131
jack screw, 134
lever, 119
tackle blocks, 125
wedge, 132
efficiency of an engine, 151
Metals, weights of, 82
Micrometer, the, 40
Mill, the, 28
Mixed numbers, 3
multiplication of, 18
reduction of, 6
Money, U. S., 24
addition, 25
division, 26
multiplication, 26
reducing cents to dollars, 28
reducing dollars to cents, 27
subtraction, 26
table of, 28
the "mill," 28
Multiplication of decimals, 36
fractions, 17
mixed numbers, 18
whole numbers and fractions, 17
Nature of heat, 163
Numbers, mixed, 3
multiplication of, 18
Numerator of a fraction, 2
Percentage, 44
INDEX 185
Percentage, classes of problems under, 48
uses of, 46
Peripheral speed, 54
Periphery, 54
Planes, inclined, 130
Plates, short rule for weights of, 83
Power, 138
Powers, cubes and higher, 75
Pressure through fluids, transmission of, 154
Proper fractions, 3
Proportion, 59
direct and inverse, 65
Pulleys and belts, 58
Pulleys, arrangement of, 142
diameters of, 63
distance between centers of, 142
speeds of, 63
Pulley trains, compound, 68
Pyrometers, 166
Radius of a circle, 51
Ratio and proportion, 59
Rectangle, the, 79
Rectangles, dimensions of, 91
Reduction of fractions, 3
of improper fractions, 5
of mixed numbers, 6
Right triangles, 89
Rim speed, 54
Roots of numbers, by table, square, 99
Roots of numbers, cube, 75, 92
square, 75
table of, 103
Ropes, strengths of, 177
Rules for area of a circle, 78
belting, 141
gears, 64
pulleys, 63
square root, 88
weights of plates, 83
Safe working stresses, 174
Screw cutting, 72
Shrink fits, allowances for, 171
Specific gravity, 153
Speeds of pulleys, 63
circumferential, 54
cutting, 57
peripheral, 54
186 INDEX
Speeds, rim, 54
surface, 54
Square measure, 76
table of, 77
root, 75
by table, 99
extracting, 86
meaning of, 85
rules for, 88
roots of numbers, table of, 103
Squares of numbers, table of, 103
Steam and air, 159
engines, 145
horse power of, 145
Strengths of bolts, 175
of cables and wire ropes, 177
of chains, 177
of hemp ropes, 177
of wire ropes and cables, 177
Stresses, definition of kinds of, 173
safe working, 174
Subtraction of decimals, 35
of fractions, 12
Surface speed, 54
Tables, areas of circles, 101
bolt table^-U. S. standard thread, 176
circumferences and areas of circles, 10 1
coefficients of expansion, 170
cube roots of numbers, 103
cubes of numbers, 103
cutting speeds, 58
decimal equivalents, 42
explanation of, 95
measures of length, 7
of time, 7
miscellaneous units, 7
square measure, 77
roots of numbers, 103
squares of numbers, 103
U. S. money, 28
weights of castings from patterns, 84
of materials, 83
Tackle blocks, mechanical advantage of, 125
types of, 123
Temperatures, 164
of iron and steel by color, 167
Thermometers, Centigrade, 164
Fahrenheit, 164
INDEX 187
Thermometers, relation of Centigrade to Fahrenheit, 165
for temperatures above 500, 166
Thermostats, 169
Threads, cutting of, 72
Triangles, right, 89
Types of machines, 115
of tackle blocks, 123
Ultimate strengths, 174
Unit of work, 137
U. S. money, 24
Volume, measures of, 80
Volumes of straight bars, 80
Wage calculations, 29
Wedge, 132
mechanical advantage of, 132
Weights of castings from patterns, 83
of materials, 83
of metals, 82
Wheel and axle, 120
Widths of belts, 140
Work, unit of, 137
2883