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IN MEMORIAM
FLORIAN CAJORI
Digitized by tine Internet Archive
in 2007 with funding from
IVIicrosoft Corporation
http://www.archive.org/details/dooleymathOOdoolrich
VOCATIONAL
MATHEMATICS
BY
WILLIAM H. DOOLEY
AUTHOR OP " VOCATIONAL MATHEMATICS FOB
GIRLS," "textiles," ETC.
REVISED
D. C. HEATH & CO., PUBLISHERS
BOSTON NEW YORK CHICAGO
Copyright, 191 5,
By D. C. Heath & Co.
2 B O
CAJORI
PREFACE
The author has had, during the last ten years, considerable
experience in organizing and conducting intermediate and sec-
ondary technical schools. During this time he has noticed the
inability of the regular teachers in mathematics to give the
pupils the training in commercial and rule of thumb methods
of solving mathematical problems that are so necessary in
everyday life. A pupil graduates from the course in mathe-
matics without being able to " commercialize " or apply his
mathematical knowledge in such a way as to meet the needs
of trade and industry.
It is to overcome this difficulty that the author has prepared
this book on vocational mathematics. He does not believe in
doing away with the regular course of mathematics but in
supplementing it with a practical course. This course may take
the place of the first year algebra and the first year geometry
in vocational classes in which it is not desirable to give the
traditional course in algebra and geometry.
This book may be used by the regular teacher in mathe-
matics and by the shop teacher. It can be used in the shop
in teaching mathematics and in providing drill problems upon
the shop work. A course based upon the contents of the book
should be provided before pupils finish their training, so that
they may become skillful in applying the principles of mathe-
matics to the daily needs of manufacturing life.
In revising the manuscript the author has had the assistance
of his teachers in the Lawrence Industrial School, the Lowell
Industrial School, the Fall River Technical High School, and
9i6287
IV PREFACE
of many other teachers, practical men, and manufacturing firms.
Valuable material has also been obtained from standard hand-
books, such as Kent's.
To mention the names of all persons to whom the author is
indebted is impossible. Acknowledgment should be made to
the following persons and firms who have kindly consented
to furnish cuts and information, and have offered valuable
suggestions and problems : Mr. George W. Evans, Principal
Charlestown (Mass.) High School; Dr. David Snedden;
Mr. Charles E. Allen ; Mr. Fred. W. Turner ; Mr. Peter Gart-
land. Principal South Boston High School ; Mr. John Casey,
Worcester (Mass.) Trade School ; Mr. John L. Sullivan, Princi-
pal Chicopee (Mass.) Industrial School; Mr. William Hunter,
Fitchburg Industrial Course; Mr. J. Gould Spofford, Princi-
pal Quincy Industrial School ; Mr. Edward K. Markham, Cam-
bridge Technical High School; Principal Joseph J. Eaton,
Saunders Trades School, Yonkers, N, Y. ; Miss Bessie King-
man, Brockton (Mass.) High School ; Mr. E. W. Boshart,
Director of Industrial Arts, Mt. Vernon, N. Y. ; Mr. G. A.
Boate, Technical High School, Newton, Mass. ; Mr. G. R.
Smith, Bradford, England ; Mr, H. R. Carter, Belfast, Ireland ;
New York Central Lines, Apprenticeship Department ; Mr. H.
E. Thomas, Tuskegee Institute, Ala. ; Brown & Sharp Co. ;
Simond's Guide for Carpenters ; R. M. Starbuck & Sons ;
Garvin Machine Co. ; The Crane Co. ; The William Powell Co. ;
Crosby Steam Gage and Valve Co. ; Mr. Peter Lobben ; Mr. H.
P. Faxon ; Bardons and Oliver ; Stanley Rule and Level Co. ;
Pittsburgh Valve Foundry and Construction Co. ; Becker
Milling Machine; Braeburn Steel Co.; Fore River Ship Build-
ing Co. ; Engineering Workshop Machines and Processes ;
American Injector Co. ; B. F. Sturtevaut Co. ; E. W. Bliss Co. ;
Dillon Boiler Works; Whitcomb-Blaisdell Co.; Hoefer Drill
Co. ; Bradford Lathe Co. ; and Detroit Screw Works.
The author will be very thankful for any suggestions relating
to the work,
CONTENTS
PART I — REVIEW OF ARITHMETIC
OUXPTXR PAUS
I. Essentials of Arithmetic 1
Fundamental Processes. Fractions. Decimals. Com-
pound Numbers. Percentage. Ratio and Proportion.
Involution. Evolution.
II. Mensuration 62
Circle. Triangles. Quadrilaterals. Polygons. El-
lipse. Pyramid. Cone. Sphere. Similar Figures.
III. Interpretation of Results 78
Reading of Blue Print. Drawing to Scale. Methods
of Solving Examples.
PART II — CARPENTERING AND BUILDING
IV. Woodworking 83
Board Measure. Short Methods. Tables.
V. Construction 89
Frame and Roof. Lathing.
VI. Building Materials 94
Mortar. Plaster. Bricks. Stone Work. Cement.
Shingles. Slate. Clapboards. Flooring. Stairs.
PART III — SHEET METAL WORK
Vn. Die Cutting 107
Blanking Dies. Combination Dies. Standard Gauge.
Tables.
V
VI CONTENTS
PART IV — BOLTS, SCREWS, AND RIVETS
CHAPTER PAGE
VIII. Bolts, Screws, and Rivets 126
Kinds of Bolts. Rivets. Nails. Tacks. Kinds of
Screws. Screw Threads. V-Shaped. U. S. Stand-
ard. Acme Standard. Square. Micrometer.
PART V — SHAFTS, PULLEYS, AND GEARS
IX. Shafts and Pulleys . 146
Belting. Arc of Contact. Speed. Countershafts.
X. Gearing 159
Pitch. Trains of Gears.
PART VI — PLUMBING AND HYDRAULICS
XL Plumbing and Hydraulics . . . . . 172
Tanks. Drainage Pipes. Weight of Lead Pipe.
Capacity of Pipe. Atmospheric Pressure. Water
Pressure. Water Traps. Velocity of Water. Water
Power Density of Water. Specific Gravity.
PART VII — STEAM ENGINEERING
XII. Heat 195
Heat Units. Temperature. Value of Coal. Intrinsic
Heat. Latent Heat. Value of Water and Steam.
Steam Heating.
XIIL Boilers 203
Kinds of Boilers. Tensile Strength. Safe Working
Pressure. Boiler Tubes. Safety Valves. Super-
heated Steam. Boiler Pumps.
XIV. Engines 222
Kinds of Engines. Fly Wheel. Horse Power. Di
ameter of Cylinder. Diameter of Supply Pipe. Steam
Indicator.
CONTENTS Vll
PART VITI. ELECTRICAL WORK
caAPTBB PA«B
XV. CoMMKKriAi. Elkctricity 231
Amperes. Volts. Ohms. Ammeter. Voltmeter.
Series and Parallel Circuits. Power Measurement.
Resistance. Size of Wire.
PART IX — MATHEMATICS FOR MACHINISTS
XVI. Matkkials 251
Cast Iron. Wrought Iron. Steel. Copper. Brass.
Weight of Steel.
XVn. Lathes 255
Engine Lathe, (iear and Pitch. Adjusting Gears.
Compound Lathe. To cut Double or Multiple
Threads. Machine Speeds. Net Power for Cutting
Iron or Steel. Surface Speeds.'
XVIII. Planers, Shapers, and Drilling Machines . 268
Planer. Planer and Shaper. Drilling Machine.
PART X — TEXTILE CALCULATIONS
XIX. Yarns 276
Materials. Kinds. Weight.
APPENDIX
Metric System 291
Graphs 296
Formulas 298
Logarithms 315
Trigonometry 320
Tables of Formulas 330
Index 337
NOTE TO TEACHERS
The author has found that it adds to the interest of the
subject if the teacher will provide models, instruments, etc.,
in the class and explain the subject matter and problems in
terms of the actual subject.
If the pupils have not had a course in Algebra, it is better
to take up the subject of Formulas in the Appendix before
beginning the subject of Mensuration.
viU
VOCATIOIN^AL MATHEMATICS
PART I — REVIEW OF ARITHMETIC
CHAPTER I
Notation and Numeration
A unit is one thing ; as, one book, one pencil, one inch.
A number is made up of units and teUs how many units are
taken.
An integer is a whole number.
A single figure expresses a certain number of units and is said to be in
the units column. For example, 5 or 8 is a single figure in the units
column ; 63 is a number of two figures and has the figure 3 in the units
column and the figure 5 in the tens column, for the second figure
represents a certain number of tens. Each column has its own name,
as shown below.
1 I S 1
3 8, 6 9 6, 4 O 7, 1 2 5
Reading Numbers. — For convenience in reading and writing
numbers they are separated into groups of three figures each
by commas, beginning at the right :
138,695,407,125.
The first group is 125 units.
The second group is 407 thousands.
The third group is 695 millions.
The fourth group is 138 billions.
1
VOfSATjLONAL MATHEMATICS
' ^^e; ^rec^ding fiui»be,r is read one hundred thirty-eight
billion, six hundred ninety-five million, four hundred seven
thousand, one hundred twenty-five ; or 138 billion, 695 million,
407 thousand, 125.
Standard Mathematics Sheet. — To avoid errors in solving problems
the work should be done in such a way as to show each step, and should
make it easy to check the answer when found. A sheet of paper of
standard size, 8| in. by 11 in., should be used. Rule this sheet as in the
following diagram, set down each example with its proper number in the
margin, and clearly show the different steps required for the solution.
To show that the answer obtained is correct, the proof should follow the
example itself.
Standard Mathematics Sheet
8h in.
1.
t
It
John Smith — 100
Vocational Arithmetic
10-2-m No. 10
1,203 20
2,672 2^
31,118 23
480 10
39
19,883
55,395 Ans.
2.
Proof:
3.
The pupil should write or print his name and class, the date when the
problem is finished, and the number of the problem on the Standard
REVIEW OP ARITHMETIC 3
Mathematics Sheet. If the question contains several divisions or prob-
lems, they should be tabulated — (a), (6), etc. — at the left of the prob-
lems inside the margin line. Tiiere should be a line drawn between
problems to separate them.
Addition
Addition is the process of finding the sum of two or more
numbers. The result obtained by this process is called, the
sum or amount.
The sign of addition is an upright cross, +, called plus. The
sign is placed between the two numbers to be added.
Thus, 9 inches + 7 inches (read nine inches plus seven inches).
The sign of equality is two short horizontal parallel lines, =,
and means equals or equal to.
Thus, the statement that 8 feet + 6 feet = 14 feet, means that six feet
added to eight feet (or 8 feet plus 6 feet) equals fourteen feet.
To find the sum or amount of two or more numbers.
Example. — What is the total weight of a machine made up
of parts weighing 1203, 2672, 31,118, 480, 39, and 19,883 lb.,
respectively ?
[The abbreviation for pounds is lb.]
The sum of the units column is 3 + 9 -f
-1-8 + 2 + 3 = 26 units or 20 and 5 more ;
20 is tens, so leave the 6 under the units
column and add the 2 tens in the tens column.
The sura of the tens column is 2 + 8 + 3 + 8
+ 1 + 7 + = 29 tens. 29 tens equal 2 hun-
dreds and 9 tens. Place the 9 tens under
the tens column and add the 2 hundreds to
the hundreds column, 2 + 8 + 4 + 1+6
+ 2 = 23 hundreds ; 23 hundreds are equal to 2 thousands and 3 hundreds.
Place the 3 hundreds under the hundreds column and add the 2 thousands
to the next column. 2 + 9 + 1+2 + 1 = 15 thousands or 1 ten-thousand
and 6 thousands. Add the 1 ten-thousand to the ten-thousands column
1,203
2p
2,672
29
31,118
2;^
480
Ip
39
19,883
55,395 lb.
4 VOCATIONAL MATHEMATICS
and the sum isl+l-)-3 = 5. Write the 5 in the ten-thousands column.
Hence, the sum or weight is 55,395 lb.
Proof. — Repeat the process, beginning at the top of the right-hand
column.
Exactness is very important in arithmetic. There is only-
one correct answer. Therefore it is necessary to be accurate
in performing the numerical calculations. A check of some
kind should be made on the work. The simplest check is to
estimate the answer before solving the problem. A comparison
can be made between them. If there is a great discrepancy
then the work is probably wrong. It is also necessary to be
exact in reading the problem.
EXAMPLES
1. Write the following numbers as figures and add them :
Seventy-five thousand three hundred eight ; seven million two
hundred five thousand eight hundred forty -nine.
2. In a certain year the total output of copper from the
mines was worth $58,638,277.86. Express this amount in
words.
3. Solve the following:
386 -h 5289 -f 53666 + 3001 -f 291 + 38 = ?
4. On a shelf there were three kegs of bolts. The first keg
weighed 203 lb., the second 171 lb., and the third 93 lb. How
many pounds of bolts were there on the shelf ?
5. On the platform in an electrical shop there were a motor-
generator and two motors. The motor-generator weighed 275
lb., one of the motors 385 lb., and the other motor 492 lb.
What weight did the platform support ?
6. Solve the following :
6027 + 836 -f 4901 + 3,800,031 -f 28,639 + 389,661 = ?
7. Four coal sheds in a shop contained respectively 1498
lb., 4628 lb., 6125 lb., and 12,133 lb. What was the whole
amount of coal in the shop ?
REVIEW OF ARITHMETIC 5
8. An engineer ordered coal and found that the first load
weighed 5685 lb., the second 5916 lb., the third 6495 lb., and
the fourth 5280 lb. What was the total weight ?
9. Wire for electric lights was run around four sides of
three rooms. If the first room was 13 ft. long and 9 ft. wide ;
the second 18 ft. long and 18 ft. wide ; and the third 12 ft. long
and 7 ft. wide, what was the total length of wire required?
Remember that electric lights require two wires.
10. Find the sum :
46 lb + 135 lb. + 72 lb. + 39 lb. + 427 lb. + 64 lb.
-h 139 lb.
Subtraction
Subtraction is the process of finding the difference between
two numbers, or of finding what number must be added to a
given number to equal a given sum. The minuend is the num-
ber from which we subtract ; the subtrahend is the number sub-
tracted; and the difference or remainder is the result of the
subtraction.
The sign of subtraction is a short horizontal line, — , called
minuSf and is placed before the number to be subtracted.
Thus, 12 — 8 = 4 is read twelve minus (or less) eight equals four.
To find the difference of two numbers.
Example. — A reel of wire contained 8074'. If 4869' were
used in wiring a house, how many feet remained on the reel ?
[Feet and mches are represented by ' and " respectively.]
Minuend 8074 ft. Write the smaller number under the
Subtrahend 4869 greater, with units of the same order in
Pfmn 'ndpr S^H)5 ft ^^^ same vertical line. 9 cannot be taken
from 4, so change 1 ten to units. The 1 ten
that was changed from the 7 tens makes 10 units, which added to the 4
units makes 14 units. Take 9 from the 14 units and 6 units remain.
Write the 5 under the unit column. Since 1 ten was changed from 7 tens,
there are 6 tens left, and 6 from leaves 0. Write under the tens col-
unm. Next, 8 hundred cannot be taken from hundred, so 1 thousand
6 VOCATIONAL MATHEMATICS
(ten hundred) is changed from the thousands column. 8 hundred from
10 hundred leave 2 hundred. Write the 2 under the hundreds column.
Since 1 thousand has been taken from the 8 thousand, there are left 7
thousand to subtract the 4 thousand from, which leaves 3 thousand.
Write 3 under the thousand column. The whole remainder is 3205 ft.
Proof. — If the sum of the subtrahend and the remainder equals the
minuend, the answer is correct,
EXAMPLES
1. Subtract 1001 from 79,999.
2. A box contained one gross (144) of wood screws. If 48
screws were used on a job, how many screws were left in
the box ?
3. What number must be added to 3001 to produce a sum
of 98,322?
4. A machinist had castings of different kinds on hand in
the morning, weighing 6018 lb. He used during the day 911
lb. of castings. How many pounds were left ?
5. The value of electrical equipment produced by one firm
in a certain year was $ 102,000,000, and in the previous year it
was $93,000,000. What was the difference between the two
years ?
6. In the coal shed of a factory there were 52,621 lb. of
coal at the beginning of the week. Monday 6122 lb. were
used ; Tuesday 5928 lb ; Wednesday 2448 lb. were received
and 5831 lb. used, (a) How much coal was used during those
days ? (6) How many pounds of coal were there in the shed
on Thursday morning ?
7. Monday morning an engineer had 72 gallons of cylinder
oil. Monday he used 8 gallons, Tuesday 12 gallons, and
Wednesday 9 gallons, (a) How many gallons did he use ?
(b) How many gallons of oil did he have on hand Thursday
morning ?
a 69,221-3008 = ?
REVIEW OF ARITHMETIC 7
9. A reel contained 13,211' of wire; 1112' were used in
wiring a house and 341' in wiring another house, (a) How
many feet of wire were used for the houses ? (6) How many
feet of wire were left on the reel ?
10. Several castings were placed on a platform. Their total
weight was 1625 lb. One casting weighing 215 lb. and an-
other weighing 75 lb. were taken away, (a) What was tlie
weight of the castings taken away ? (b) What was the weight
of the remaining castings ?
Multiplication
Multiplication is the process of finding the product of two
numbers. The numbers multiplied together are called factors.
The vudtipUcand is the number multiplied ; the multiplier is tlie
number multiplied by; and the result is called the product.
The sign of multiplication is an oblique cross, x , which means
midtiplied by or times.
Thus, 8x3 may be read 8 multiplied by 3, or 8 times 3.
To find the product of two numbers.
Example. — An iron beam weighed 24 lb. per foot of length.
What was the weight of a beam 17 feet long ?
Midtiplicand 24 lb. Write the multiplier under the multipli-
jif jf J- -I- cand, units under units, tens under tens,
^ — — etc. 7 times 4 units equal 28 units, which
^^^ are 2 tens and 8 units. Place the 8 under
^4 the units column. The 2 tens are to be
Product 408 lb. added to the tens product. 7 times 2 tens
are 14 tens -j- the 2 tens are 16 tens, or 1
hundred and 6 tens. Place the 6 tens in the tens column and the 1 hun-
dred in the hundreds column. 168 is a partial product. To multiply by
the 1, proceed as before, but as 1 is a ten, write the first number, which
is 4 of this partial product, under the tens column, and the next number
under the hundreds column, and so on. Add the partial products and
their sum is the whole product, or 408 lb.
8 VOCATIONAL MATHEMATICS
EXAMPLES
1. An electrical job required the following labor : 5 men, 42
hours each ; 6 men, 4 hours each ; 12 men, 14 hours each ; 7
men, 12 hours each ; and 3 men, 6 hours each. Find the total
number of hours time on the job.
2. Multiply 839 by 291.
3. A machinist sent in the following order for bolts : 12
bolts, 6 lb. each; 9 bolts, 7 lb. each; 11 bolts, 3 lb. each; 6
bolts, 2 lb. each; and 20 bolts, 3 lb. each. What was the
total weight of the order ?
4. Find the product of 1683 and 809.
' To multiply by 10, 100, 1000, etc.y annex as many ciphers to
the multiplicand as there are ciphers iii the multiplier.
Example. — 864 x 100 = 86,400.
EXAMPLES
Multiply and read the answers to the following :
1. 869 X 10 a 100 X 500
2. 1011 X 100 9. 1000 X 900
3. 10,389 X 1000 10. 10,000 x 500
4. 11,298 X 30,000 11. 10,000 x 6000
5. 58,999 X 400 12. 1,000,000 x 6000
6. 681,719 X 10 13. 1,891,717 x 400
7. 801,369 x 100 14. 10,000,059 x 78,911
Division
Division is the process of finding how many times one number
is contained in another. The dividend is the number to be
divided ; the divisor is the number by which the dividend is
divided; the quotient is the result of the division. When a
number is not contained an equal number of times in another
number what is left over is called a remainder.
REVIEW OF ARITHMETIC 9
The sign of division is -j-, and wlieii placed between two
numbers signifies tliat the first is to be divided by the second.
Thus, 66 -!- 8 is read 56 divided by 8.
Division is also indicated by writing the dividend above the
divisor with a line between.
Thus, *^ ; this is read 56 divided by 8.
In division we are given a product and one of the factors to
find the other factor.
To find how many times one number is contained in another.
Example. — A machinist had 8035 rivets which he wished to
arrange in groups of three. How many such groups did he
have ? How many rivets did he have left over ?
Quotient Place the numbers in the manner in-
2678 dicated at the left. 8 thousand is in the
Divisor 3)8035 Dividend thousands column. The nearest 8 thou-
g sand can be divided into groups of 3 is 2
OA (thousand) times, which gives (> thousand.
^^ Write 2 as the first figure in tlie quotient
over 8 in the dividend. Place the 6 (thou-
•^"^ sand) under the 8 thousand and subtract ;
21 the remainder is 2 thousand or 20 hundred.
25 8 is contained in 20 himdred 6 hundred
24 times or 18 hundred and 2 hundred re-
R m ind ~T mainder. Write 6 as the next figure in
the quotient. Add the 3 tens in the divi-
dend to the 2 hundred or 20 tens, and 23 tens is the next dividend to be
divided. 3 is contained in 23 tens 7 times or 21 tens with a remainder
of 2 tens. Write 7 as the next figure in the quotient. 2 tens or 20 units
plus the 6 units from the quotient make 25 units. 3 is contained in 26,
8 times. Write 8 as the next figure in the quotient. 24 units subtracted
from 25 units leave a remainder of 1 unit. Then the answer is 2678
groups of rivets and 1 rivet left over.
Proof. — Find the product of the divisor and quotient, add the re-
mainder, if any, and if the sum equal the dividend, the answer is correct.
10 VOCATIONAL MATHEMATICS
EXAMPLES
1. A strip of plate measures 85" in length. How many
pieces 6" long can be cut from it? AVould there be a re-
mainder ?
2. How many pieces 2" long can be cut from a brass plate
62' long, if no allowance is made for waste in cutting ?
3. If the cost of constructing 362 miles of railway was
$4,561,200, what was the cost per mile ?
4. If a job which took 379 hours was divided equally
among 25 men, how many even hours would each man w^ork,
and how much overtime would one of the number haye to put
in to complete the job ?
5. The " over-all " dimension on a drawing was 18' 9".
The distance was to be spaced off into 14-inch lengths, beginning
at one end. How many such lengths could be spaced ? How
many inches would be left at the other end ?
6. If a locomotive consumed 18 gallons of fuel oil per mile
of freight service, how far could it run with 2036 gallons of oil ?
7. If 48 screws w^eigh one pound, how many cases each
containing 36 screws could be filled from a stock of 29 lb. of
screws ?
8. A plate measures 68" in length. How many pieces 15"
long can be cut from it ? How much will be left over?
9. A machinist desired to know the number of lots of 42
lb. each, contained in 3276 lb. of screws. Solve.
10. If a freight train made a distance of 112 miles in 8
hours, what was the average speed per hour ?
11. If the circumference of the driving wheel of a loco-
motive was 22 feet, how many turns did it make in going 88
miles ?
12. Divide 38,910 by 3896.
REVIEW OF ARITHMETIC 11
REVIEW EXAMPLES
1. A load of castings came to a machine shop. It was not
desirable to weigh the castings on the wagon, so they were
weighed in 6 lots as follows: 196 lb., 389 lb., 876 lb., 899 lb.,
212 lb., and 847 lb. What was the total weight?
2. Five steel bars are placed end to end. If each bar is
29 ft. long, what is the total length ?
3. A steam fitter found that the weight per foot of 3{"
wrought iron pipe is 9 lb. ; what is the weight of a piece of
3k" wrought iron pipe 12 feet long ?
4. An accident happened in a mill. A number of men were
sent out to make the repairs. The following number of hours
was reported :
8 men 10 hours each 3 men 5 hours each
4 men 65 hours each 4 men 21 hours each
7 men 14 hours each 6 men 11 hours each
What was the total number of hours worked?
5. The consumption of water for a city during the month of
December was 116,891,213 gallons and for January 115,819,729
gallons. How much was the decrease in consumption ?
6. An order to a machine shop called for 598 machines each
weighing 1219 pounds. What was the total weight ?
7. If an I beam weighs 24 lb. per foot of length, find the
weight of one measuring 16' 9" long.
8. Multiply 641 and 225.
9. Divide 24,566 by 319.
10. An order was sent for ties for a railroad 847 miles long.
If each mile required 3017 ties, how many ties would be
needed ?
11. How many gallons j)er minute are discharged by two
pipes if one discharges 25 gallons per minute and the other
6 gallons less ?
12 VOCATIONAL MATHEMATICS
Factors '
The factors of a number are the integers which when multi-
plied together produce that number.
Thus, 21 is the product of 3 and 7 ; hence, 3 and 7 are the factors of 21.
Separating a number into its factors is (tailed factoring.
A number that has no factors but itself and 1 is a prime
number.
The prime numbers up to 25 are 2, 3, 5, 7, 11, 13, 17, 19 and 23.
A prime number used as a factor is 2i prime factor.
Thus, 3 and 5 are prime factors of 15.
Every prime number except 2 and 5 ends with 1, 3, 7, or 9.
To find the prime factors of a number.
Example. — Find the prime factors of 84.
2 )84 The prime number 2 divides 84 evenly, leaving the quotient
2)42 ^'^' which 2 divides evenly. The„next quotient is 21 which 3
divides, giving a quotient 7. 7 divided by 7 gives the last
quotient 1 which is indivisible. The several divisors are the
prime factors. So 2, 2, 3, and 7 are the prime factors
of 84.
3)21
7)7
1
Phoof. — The product of the prime factors gives the number.
EXAMPLES
Find the prime factors :
1. 63 4. 636 7. 1155
2. 60 5. 1572 8. 7007
3. 250 6. 2800 9. 13104
Cancellation
To reject a factor from a number divides the number by that
factor ; to reject the same factors from both dividend and divisor
does not affect the quotient. This process is called cancellation.
This method can be used to advantage in many everyday cal-
culations.
Example, — Divide 12 x 18 x 30 by 6 x 9 x 4.
REVIEW OF ARITHMETIC 13
1
2 ;2 16 By ^^^ method it is not
Dividend ;S2 X ;? X 3P OA ^ .. . neccBsary to multiply be-
Tk' • a w Q w 7 = ^^ Quotient. fore dividing. Indicate
\^ ^ ' the division by writing
r the divisor under the divi-
1 dend with a line between.
Since 6 is a factor of 6
and 12, and 9 of 9 and 18, reapectively, they may be cancelled from both
divisor and dividend. Since 2 in the dividend is a factor of 4 in the
divisor it may be cancelled from both, leaving 2 in the divisor. Then the
2 being a factor of 30 in the dividend, is cancelled from both, leaving 15.
The product of the uncancelled factors is 30. Therefore, the quotient
is 30.
Proof. — If the product of the divisor and the quotient equal the
dividend, the answer is correct.
EXAMPLES
Indicate and find quotients by cancellation :
1. Divide 36 X 27 x 49 x 38 x 50 by 70 x 18 x 15.
2. What is the quotient of 36 x 48 X 16 divided by 27 x 24
X8?
3. How many pounds of washers at 50 cents a pound must
be given in exchange for 15 pounds of bolts at 40 cents a
pound ?
4. There are 16 ounces in a pound ; 30 pounds of steel will
produce how many horseshoes, if each weighs 6 ounces ?
5. How many pieces of steel rod, each weighing 10 pounds,
at 20 cents a pound, must be given in exchange for 10 bars of
jY' iron rod, each weighing 5 pounds, at 4 cents per pound ?
6. Divide the product of 10, 75, 9, and 96 by the product of
5, 12, 15, and 9.
7. If 24 men, working 9 hours a day, can do a piece of work
in 12 days, how many days will it take 18 men, working 8
hours a day, to do the work ?
14 VOCATIONAL MATHEMATICS
Greatest Common Divisor
The greatest common divisor of two or more numbers is the
greatest number that will exactly divide each of the numbers.
To find the greatest common divisor of two or more numbers.
Example. — Find the greatest common divisor of 90 and
150.
90 = 2x3x5x3 2 )90 150 First Method
150 = 2 X3x5x5 5 )45 75 The prime factors com-
Ans. 30 = 2 X 3 X 5 3)9 15 ^^^^ ^o both 90 and 150
"o K are 2, 3, and 5. Since
2 X 3 X 5 = 30 Ans. ^^^ greatest common di-
visor of two or more num-
90)150(1 hers is the product of
gQ their common factors, 30
TTT^NQp..-. is the greatest common
' ^^^ divisor of 90 and 150.
bO
Greatest Common Divisor 30)60(2 '^^^^^^ ^^^^^^
(jQ To find the greatest
— common divisor when
the numbers cannot jbe readily factored, divide the larger by the smaller,
then the last divisor by the last remainder until there is no remainder.
The last divisor will be the greatest common divisor. If the greatest com-
mon divisor is to be found of more than two numbers, find the greatest
common divisor of two of them, then of this divisor and the third num-
ber, and. so on. The last divisor will be the greatest common divisor of
all of them.
EXAMPLES
Find the greatest common divisor :
1. 270,810. 3. 504,560. 5. 72,153,315,2187.
2. 264, 312. 4. 288, 432, 1152.
Least Common Multiple
The product of two or more numbers is called a multiple of
each of them; 4, 6, 8, 12 are multiples of 2. The multiple
RKVIKW OF AHITHMKTIC 15
of two or more numbers is called the common multiple of the
numbers ; 00 is a common multiple of 4, 5, 6.
The least common multiple of two or more numbers is the
smallest common multiple of the number; 30 is the least
common multiple of 8, 5, 6.
To find the least common multiple of two or more numbers.
ExAMPLK. — Find the least common multiple of 21, 28,
and 30. _.. , ,. ,. ^
, First Method
-1=3x7 Take all the factors of the firet number, all of
28 = 2 X 2 X 7 ^h^ second not already represented in the first, etc.
30 = 2 X i} X 5 'f has,
3 x7x2x2x5 = 420i.C.3f.
Second Method
2 )21 28 30
3)21 14 15
7 )7 14 5
2 X 3 x 7 X 1 x 2 X 5 = 420 X. (7. M.
Divide any two or more numbers by a prime factor contained in them,
like 2 in 28 and 30. Write 21 which is not divided by the 2 for the next
quotient together with tl>e 14 and 16. 3 is a prime factor of 21 and 15
which gives a quotient of 7 and 5 with 14 written in the quotient undi-
vided. 7 is a prime factor of 7 and 14 which gives a remainder of 1, 2 ;
and 6 undivided is written down as before. The product 420 of all these
divisors and the last quotients i^ the least common multiple of 21, 28,
and 30.
EXAMPLES
Find the least common multiple :
1. 18, 27, 30. 2. 15, 60, 140, 210. 3. 24, 42, 54, 360.
4. 25, 20, 35, 40. 5. 24, 48, 96, 192.
6. What is the shortest length of rope that can be cut into
pieces 32', 36', and 44' long?
16 VOCATIONAL MATHEMATICS
Fractions
A fraction is one or more equal parts of a unit. If an apple
be divided into two equal parts, each part is one-half of the
apple, and is expressed by placing the number 1 above the
number 2 with a short line between : i A fraction always
indicates division. In i, 1 is the dividend and 2 the divisor ;
1 is called the nuinerator and 2 is called the denominator.
A common fraction is one which is expressed by a numerator
written above a line and a denominator below. The nu-
merator and denominator are called the terms of the fraction.
A proper fraction is a fraction whose value is less than 1 ; its
numerator is less than its denominator, as -|, -f, f, ^. An
improper fraction is a fraction whose value is 1 or more than 1;
its numerator is equal to or greater than its denominator, as f,
■}-f. A number made up of an integer and a fraction is a
mixed number. Read with the word and between the whole
number and the fraction : 4:^\, S^, etc.
The value of a fraction is the quotient of the numerator
divided by the denominator.
EXERCISE
Read the following :
1. I 3. 121- 5. ^ 7. 9^ 9. I
2. \i 4. ^ 6. ^ 8. 12^
Reduction of Fractions
Reduction of fractions is the process of changing their form
without changing their value.
To reduce a fraction to higher terms.
Multiplying the denominator and the numerator of the given
fraction by the same number does not change the value of the
fraction.
REVIEW OF ARITHMETIC 17
Example. — Reduce | to thirty -seconds.
The denominator must be multiplied by 4 to
5 ^1 _ ^ jins, obtain 32 ; so the numerator must be multiplied
8 4 32 by the same number so that the value of the
fraction may not be changed.
EXAMPLES
Change the following :
1. f to 27ths. 6. ^^ to 75th8.
2. \^ to 60ths. 7. Jl to 144ths.
3. f to 40ths. 8. ^ to 168ths. -
4. Jto56ths. 9. |Jto522ds.
5. 3% to 50ths. 10. ^ to 9375ths.
A fraction is said to be in its lowest terms when the numersr
tor and the denominator are prime to each other.
To reduce a fraction to its lowest terms.
Dividing the numerator and the denominator of a fraction
by the same number does not change the value of the fraction.
The process of dividing the numerator and denominator of a
fraction by a number common to both may be continued until
the terms are prime to each other.'
Example. — Reduce -f| to fourths.
The denominator must be divided by 4 to give
12 _ 3 J, the new denominator 4 ; then the numerator must be
16 4 divided by the same number so as not to change the
value of the fraction.
If the terms of a fraction are large numbers, find their
greatest common divisor and divide both terms by that.
Example. — Reduce |Jff to fourths.
(1) 2166)2888(1 (2) 2166^3 .
2166 2888 4 ^ '
O. a D. 722)2166(3
2166
18 VOCATIONAL MATHEMATICS
EXAMPLES
Reduce to lowest terms :
1- A 3- Hil s- H '• IS* 9- Hi
2- m *■ U 6. -rVj 8. iff 10. T-VA
To reduce an integer to an improper fraction.
Example. — Reduce 25 to fifths.
25times4 = l*i Ans. , '" ' "'f""'[!,^- ^ 25 there must be
5 5 25 tunes 5, or i|^.
To reduce a mixed number to an improper fraction.
Example. — Reduce 16^ to an improper fraction.
i_ sevenths Since in 1 there are }, in 16 there must
112 be 16 times ^, or ^2^
4 sevenths lp + f=^K
116 sevenths, = J-^.
EXAMPLES
Reduce to improper fractions :
1. 3J 3. 171 5. 13J 7. S59j%
2. 16^ 4. 12^ 6. 27t«^ 8. 482i|
9. 25-^ - 10. Reduce 250 to 16ths.
11. Change 156 to a fraction whose denominator shall be 12.
12. In $ 730 how many fourths of a dollar ?
13. Change 12 1 to 16ths. 14. Change 24f to ISths.
To reduce an improj^er fraction to an integer or mixed number
divide the numerator by the denominator.
Example. — Reduce -\%^- to an integer or mixed number.
24
16)385
^^ Since || equal 1, ^^ will equal as many
times 1 as 16 is contained in 385, or 24^15
65 24-jL Ans. ^^^^^^^
64
1
REVIEW OF ARITHMETIC 19
EXAMPLES
Reduce to intecjers oi* mixed numbers :
1. H *• 'iV 7. VV 10. i^i
. Wlieii fractions have the same denominator their denomi-
nator is called a common denominator.
Thus, |§, ^, ^j have a common denominator.
The smallest common denominator of two or more fractions is
their least common denominator.
Thus, \^, ^«2, ^ become ^, |, J when changed to their least common
denominator.
The common denominator of two or more fractions is a
common multiple of their denominators.
The least common denominator of two or more fractions is
the least common multiple of their denominators.
Example. — Reduce } and f to fractipi^s having a common
denominator.
a w 6 _- 18 The common denominator must be a
, 4 _ 20 common multiple of the denominators 4
^ T ~ tT and 0, and since 24 i.s the product of the
? ~ Ti ^"^ t — TX denominators, it is a common multiple of
them. Therefore. 24 is a common denominator of J and f .
To reduce fractions to fractions having the least common denominator.
Example. — Reduce |, |, and j^ to fractions having the
least common denominator.
2^^ S 6 12 "^'^^ least common de-
^— — — nominator must be the
1 — z. least common multiple of
1 1 -^ the denominators 3, 6, 12,
2x3x2 = 12 L. r'. M. which is 12.
f = -(8^ ; ^ = 1^ ; Vj = f^. Ans. l^'vide the least common
multiple 12 by the denom-
inator of each fraction, and multiply both terms by the quotient. If the
20 VOCATIONAL MATHEMATICS
denominators should be prime to each other, their product would be theii
least common denominator.
EXAMPLES
Reduce to fractions having a common denominator :
1- h i 5- f> I, I
4- f , tV, i 8- i> A, I, i
Reduce to fractions having least common denominator :
1- I, J, A 5- 1,1, A, 4
2- I, h A 6. I, f , I, I
^- A> TO i 7. Which fraction is larger,
4- iA.A.A fori?
Addition of Fractions
Only fractions with a common denominator can be added.
If the fractions have not the same denominator, reduce them
to a common denominator, add their numerators, and place
their sum over the common denominator. The result should be
reduced to its lowest terms. If the result is an improper
fraction, it should be reduced to an integer or mixed number.
The least
common multi-
ExAMPLE. — Add I, |, and y\.
1. 2 )4 6 16
2}2__3__8 pie of the de-
13 4 48 Z/. C. M. nominators is
o 3i5i9 36i40i27_i03 J*n, 48. Dividing
this by the de-
nominator of each fraction and multiplying both terms by the quotient
give If, f§, f|. The fractions are now like fractions, and are added by
adding their numerators and placing the sum over the common denomi-
nator. Hence, the sum is ^5^/-, or 2j7j.
REVIEW OF ARITHMETIC 21
Example. — Add 5J, 7y^, and 6^j^-
^i = ^a First find the sum of the fractions,
7^= TfJ- which is fj, or 1|^. Add this to the
QJL = 6lr4t sum of the integers, 18. 18 -|- 1 J^ =
i8|f=19H. Ana. ^m-
EXAMPLES
1. Find the "over-all" dimension of a drawing if the
separate parts measure -j^", f", ^", and -^"j respectively.
2. Find the sum of ^, J, J, j-f , and fj.
a Find the sura of 3|, 4}, and 2^.
4. Four castings weigh respectively 8J lb., 5^ lb., llf lb.,
and 7| lb. What is their total weight ?
5. The diameter of two holes is 3|" and the distance be-
tween the sides of the holes is J". What is the distance from
the outside of one hole to the outside of the other ?
6. Two brass rods measure 8-j^" and 5^" What is their
combined length ?
7. A board was cut into two pieces, one 8f" and the other
6^" long. If ^" was allowed for waste in cutting, what was
the length of the board ?
8. Three pieces of rod contain 38^, 12^, and 53| feet re-
spectively. What is their total length in feet?
9. Add: lOi, 7|, 11, i|.
Subtraction of Fractions
Only fractions with a common denominator can be sub-
tracted. If the fractions have not the same denominator,
reduce them to a common denominator, and write the differ-
ence of their numerators over the common denominator. The
result should be reduced to its lowest terms.
22 VOCATIONAL MATHEMATICS
Example. — Subtract J from |.
5_2_i5._i2_ 3 The least common denominator of
^ _/ = i' ^]^. ^ I and I is 6. t = f, and f = ^.
^^ ^' ' Their difference is \.
Example. — From 11^ subtract 5|.
11 1 ^ 10 8 When the fractions are changed to
^ 5^ __ ^5 their least common denominator, they
^ — ^ _ fii A ' ^^^ ^^^ ~ ^t- I cannot be subtracted
*> ~ 2* • • from I, hence 1 is taken from 11 units,
changed to sixths, and added to the f which makes f . 10| — 4| = 6|=6|.
EXAMPLES
1. The distance between two holes is 5f " measured from the
centers. If the holes are -^^" in diameter, what is the length
of metal plate between them ?
2. From a steel bar 26|" long were cut the following pieces :
one 7\", one 6^", one 3|" long. If after cutting these pieces,
the length of the bar was 8f", what was the amount of waste
in cutting?
3. A piece of steel on a lathe is 1" in diameter. In the first
cut -^" were taken off, in the second cut g-\", in the third cut
-jig-", in the fourth -^-q". What was the diameter of the finished
piece ?
4. A bolt is 15^" long. How much must be cut from it to
make it llf " long ?
5. In wiring a house five men work 14 hours ; one man works
1 hour and 20 minutes ; a second man works 2 hours and 15
minutes ; a third man works 5 hours ; and a fourth man, 4J
hours. How many hours did the fifth man work ?
6. It is S\" between the centers of two holes of the same
size. The distance between the sides of the holes is 1^"
What k the diameter gf each hole ?
REVIEW OF ARITHMETIC 23
7. There were 48^ gallons in the tank. First 4^ gallons
were used, then 5^ gallons, and last 2'j gallons. How many
gallons were left in the tank ?
8. \Vhat is the difference between ^ and JJ ?
9. What is the difference between 32J and .SJJ ?
10. An electrician had a reel of .SOO feet of copper wire. He
used at various times 50^', 32 J', 1091', and 2737^'. How much
wire was left?
Multiplication of Fractions
To multiply fnictiona, mnltiply the numerators together for the
neiv numerator and multiply the denominators together /or the
new denominator.
Cancel when possible. The word of between two fractions
is equivalent to the sign of multiplication.
To multiply a mixed number by an integer, nmltiply the whole
number and the fraction separately by the integer then add the
prrxlucts.
To multiply two mixed numbers, change each to an improper
fraction and multiply.
Example. — Multiply | by J.
I multiplied by J is the same as 5 0/ 1. 3 and 5 are prime to each other
so that answer is |. This method of solution is the same as multiplying
the numerators together for a new numerator and the denominators for
a new denominator. Cancellation shortens the process.
Example. — Find the product of 124} and 5.
124}
- If the fi-action and integer are mul-
— ^ «; 4 — L5 — i ^^^ divisor becomes 11 halves and the dividend
^ ' 7364 halves. Multiplying both dividend and
""^TT Ans. divisor by the same number does not change
the quotient. Dividing, the quotient is 6603^.
A fraction having a fraction for one or both of its terras is
called a complex fraction.
To reduce a complex fraction to a simple fraction.
42
Example. — Reduce ^ to a simple fraction.
|| = ^ = ¥^V- = ¥x^=if Ans.
Change 4f and 7^ to improper fractions, ^* and V, respectively. Per-
form the division indicated with the aid of cancellation and the result will
bejf
EXAMPLES
1. Divide ^ by |.
7. 296-10i = ?
2. Divide ^^ by f .
8. 28,769 -^7f=?
3. Divide |f by \.
•■?r'
4. Divide J by \.
5. Divide } by J.
6. 384J^5 = ?
- -m-'
26
VOCATIONAL MATHEMATICS
Drill in the Use of Fractions
Addition
1.
i + 4 =?
19.
1
8
+ i =? .
37.
A
+ i =?
2.
i + \ =^
20.
i
+ 1 =?
38.
A
+ i =?
3.
i + i =''
21.
i
+ i =?
39.
iV
+ i =?
4.
i + TV = ?
22.
i
+ ,v = ?
40.
A
+ tV = ?
5.
i+^=?
23.
I
+ .v=?
41.
3V
+ A=?
6.
i+^v=?
24.
1
+ 6^4=?
42.
A
+ 6V = ?
7.
i+i =?
25.
tV
+ i =?
43.
A
+4 =?
8.
i + i =^
26
tV
+ i =?
44.
eV
+i =?
9.
i + i =?
27.
tV
+ i =?
4S.
A
+i =?
10.
i + iV = ?
28.
tV
+ tV = '''
46.
A
+ tV = ?
11.
i + A = ^
29.
-A
+ A=?
47.
A
+A=?
12.
i+6V = '^
30.
t's
+ «V=?
48.
A
+A=?
13.
l + i =?
31.
i
+ 4 =?
49.
*
+1 =?
14.
l + i =?
32.
3
+ i =?
SO.
i
+ i =?
15.
l + i =?
33.
i
+ i =?
51.
i
+ i =?
16.
I + tV = ?
34.
i
+ tV = ?
52.
1
+ tV = ?
17.
I + ^V = ?
35.
8
4
+ A = ?
53.
i
+A=?
18.
l + -6^=?
36.
f
+ eV = ?
54.
i
+A=?
Subtraction
1.
i-i =?
8.
i-
-i =?
15.
5
8"
-i =?
2.
i-i =?
9.
i-
-i =?
16.
i
-tV = ?
3.
i-i =?
10.
\
-tV = ?
17.
1-
-A=?
4.
i-TV=?
11.
\-
-sV = ?
18.
f-
-Vt = ?
5.
i-^V = ?
12.
i-
-^v=?
19.
i-
-f =?
6.
i-^V = ?
13.
-4 =-?
20.
i-
-A = ?
7.
i-^=?
14.
f
-i =?
21.
1 '
-i =?
REVIEW OF ARITHMETIC
27
22- i -.', = ?
23. i -3'l = ''
24. i -ff'4 = -"
25-4 -T%=-'
26. ^ -tV=-
27. A\--tV = '^
28. -jV — tV = ^'
29- tV - /2 = ■••
30. T*!
31. f
32. f
1
ITT
1 — .»
-i =
1. 4
2. ^
3- 4
4. *
6. i
a i
9. i
10. J
11. J
12. i
13. I
14. f
15. I
16. J
17. »
18. I
X J — .
xi =?
X^ =?
XA = ?
V 1 *>
xi =?
xi =?
xi =?
Xt', = '^
x,V = ^^
X^ = ?
Xi =?
X J =?
xj .= ?
xA = ?
V 1. — 9
33. J -J =
34. 3 -l'« = -'
35. J -5>j = V
36. I -j^f =
37. U-jV =
38. i -A =
39. H-A =
40. H-A =
42. i?-s^ =
43. 4 -« =
Multiplication
19. I X .V = ?
__ 9
20. J xi =?
21. i xi
22. J X j5
23.
i X^j = ?
1 s^ \ •>
24. i Xj't
44. 1 -A =
45. \ -i,=
46. t*-."!"
47. s't-j^ =
48. A-,'i =
49. i -i =
50. J -V =
51- i -i =
52. J -t'j =
53. J -A =
54. i -A = ?
37. ,V
38. ,>j
39. ^j
40. ,V
41. ^j
42. ^2
25. ^Vxi =? 43. ,V
26. T»jXi =? 44. ,>j
27. iVxi =?
28. AXt>j = ?
29. T'ffX^j = ?
30. T'irX5'i = ?
31. I X 4 = ?
32. J X } = ?
33. I Xj =?
34. J X ,V = ?
35. f X jV = '.'
36. J Xji = ?
45. sV
46. ,V
47. i^T
48. sV
49. I
50. i
51. J
52. J
53. i
54. J
X4 =?
X i =?
Xi =?
Xt'5 = ?
Xi^l = V
X ^I = V
xi =?
X} =?
xi =?
xA = "''
x^=?
X «V = ■''
xi =?
xi =?
xi =?
Xt',= V
X j'l = '!
XVj=?
28
VOCATIONAL MATHEMATICS
Division
37.
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
48.
49.
50.
51.
52.
53.
54.
Decimal Fractions
A power is the product of equal factors, as 10 x 10 = 100.
10 X 10 X 10 = 1000. 100 is the second power of 10. 1000 is
the third power of 10.
A decimal fraction or decimal is a fraction wliose denominator
is 10 or a power of 10. A common fraction may have any
number for its denominator, but a decimal fraction must always
have for its denominator 10, or a power of 10. A decimal is
written at the right of a period (.), called the decimal point.
A figure at the right of a decimal point is called a decimal
figure.
1.
i^i =?
19.
\
-^i =?
2.
i^i =?
20.
i
-i =?
3.
i^i =?
21.
i
^i =?
4.
i-^TV = ?
22.
i
^t'^ = ?
5.
i-^A = ?
23.
i
-A = ?
6.
i^^V=?
24.
i
^7V = ?
7.
i^i =?
25.
iV
^i =?
8.
i^i =?
26.
tV
^i =?
9.
i^i =?
27.
tV
^i =?
10.
i^A = ?
28.
tV
-tV=?
11.
i^A = ?
29.
tV
^A = ?
12.
i^A = ?
30.
1^6
^F't = ?
13.
I^i =?
31.
8
^i =?
14.
I^i =?
32.
1
-i =?
15.
I^i =?
33.
f
^i =?
16.
I^tV = ?
34.
*
^tV = ?
17.
I^A = ?
35.
f
^A=?
18.
I^^ = ?
36.
1
^TrV = ?
A-
-i =?
^v-
-i =?
A-
-i =?
Ti-
-tV = ?
W-
-A = ?
7V-
-A = ?
^v-
-i =?
.v^
-i =?
A-
-i =?
A-
-tV = ?
A-
-^\ = -f
e'l-
-A = ?
J -
-i =?
i -
-i =?
i -
-i =?
i -
-tV = ?
i -
-A = ?
i -
-A = ?
REVIEW OF ARITHMETIC
29
A mixed decimal is an integer and a decimal ; as, 16.04.
To read a decimal, read the decimal as an integer, and give
it the denomination of the right-hand figure. To tvrite a deci-
mal, write the numerator, prefixing ciphers when necessary to
express the denominator, and place the point at the left.
There must be as many decimal places in the decimal as there
are ciphers in the denominator.
EXAMPLES
Read the following numbers:
1.
.7
7.
.4375
13.
.0000054
19.
9.999999
2.
.07
8.
.03125
14.
35.18006
20.
.10016
3.
.007
9.
.21875
15.
.0005
21.
.000155
4.
.700
10.
.90625
16.
100.000104
22.
.26
5.
.125
11.
.203125
17.
y.1032002
23.
.1
6.
.0625
12.
.234375.
la
30.3303303
24.
.80062
Express decimally :
1. Four tenths.
2. Three hundred twenty-five thousandths.
3. Seventeen thousand two hundred eleven hundred-thou-
sandths.
4. Seventeen hundredths. 6. Five hundredths.
5. Fifteen thousandths. 7. Six ten-thousandths.
8. Eighteen and two hundred sixteen hundred-thousandths.
9. One hundred twelve hundred-thousandths.
10. 10 milliouths. 11. 824 ten-thousandths.
12. Twenty-nine hundredths.
13. 324 and one hundred twenty-six millionths.
14. 7846 hundred-millionths.
30 VOCATIONAL MATHEMATICS
1 c 5_6 3 1 2 123 3 2 8 6 5 4
17. One and one tenth.
18. One and one hundred-thousandth.
19. One thousand four and twenty-nine hundredths.
Reduction of Decimals
Ciphers annexed to a decimal do not change the value of
the decimal; these ciphers are called decimal ciphers. For
each cipher prefixed to a decimal, the value is diminished ten-
fold. The denominator of a decimal — when expressed — is
always 1 with as many ciphers as there are decimal places in
the decimal.
To reduce a decimal to a common fraction.
Write the numerator of the decimal omitting the point for the
numerator of the fraction. For the denominator write 1 with as
many ciphers annexed as there are decimal places in the decimal.
Then reduce to lowest terms.
Example. — Reduce .25 and .125 to common fractions.
1 Write 25 for the numerator and
nK 25 'l^ 1 A 1 for the denominator with two O's,
"" 100 ~~ ^00 ~~ 4 * which makes j^^-^ ; -^-^^ reduced to
4 lowest terms is \.
1
^ oe- 125 J2f> 1 A '^25 is reduced to a common frac-
.125 = = ^ - = - Ans. ^. . ^^
1000 ^p0j^ ^ ^^^^ ^" ^^^ ^^"^^ ^^y-
8
Example. — Reduce .37^ to a common fraction.
37i has for its denominator 1
3 371
37i^i^l5^_jL = ? Ans "^^^^^^'"^"^^^^^^^«I^-
100 100 2 ;0p 8 ' This is a complex fraction
4 which reduced to lowest terms
isf.
REVIEW OF ARITHMETIC 31
EXAMPLES
Reduce to cominon trartions :
1. .01)375
6. 2.25
11.
.16J
16.
m
2. .15625
7. 16.144
12.
..S3J
17.
.66J
3. .015625
a 25.0000100
13.
.06J
18.
.36i
4. .609.375
9. 1084.0025
14.
.140625
19.
.83i
5. .578125
10. .12.V
15.
.984375
20.
.62^
To reduce a cominon fraction to a decimal.
Annex decimal ciphers to the numerator and divide by the de-
nominator. Point off from the right of the quotient as many
places as there are ciphers annexed. If there are not figures
enough in the quotient, prefix cijthers.
The division will not always be exact, i.e. | = .142f or .142+.
Example. — Reduce | to a decimal.
.75
4)3.00
28
20
I = .76
EXAMPLES
Reduce to decimals :
1.
^V
6.
i
u.
A
16.
H
21.
-^iji
2.
TiTT
7.
w
12
^icT
17.
16i
22.
25.12i
3.
TSU
8.
M
13.
T2oJi
18.
66|
23.
33'
4.
i
9.
A
14.
m
19.
U
24.
^i
5.
i
10
jh
15.
i\
20.
a
IT
25.
Ti^
Addition of Decimals
To add decimals^ write them ,so that their decimal points are in
a column. Add as in integers, and place the point in the sum
directly under the points above it.
32 VOCATIONAL MATHEMATICS
Example.— Find the sum of 3.87,2.0983, 5.00831, .029,
.831.
3.87
2 0983 Place these numbers, one under the other, with
n fW'iS'?! decimal points in a column, and add as in addition
of integers. The sum of these numbers should
'^^^ have the decimal point in the same column as the
•831 numbers that were added.
11.83661 Ans.
EXAMPLES
Find the sum :
1. 5.83, 7.016, 15.0081, and 18.3184.
2. 12.031, 0.0894, 12.0084, and 13.984.
3. .0765, .002478, .004967, .0007862, .17896.
4. 24.36, 1.358, .004, and 1632.1.
5. .175, 1.75, 17.5, 175., 1750.
6. 1., .1, .01, .001, 100, 10., 10.1, 100.001.
7. Add 5 tenths; 8063 millionths; 25 hundred-thousandths ;
48 thousandths; 17 millionths; 95 ten-millionths ; 5, and 5
hundred-thousandths ; 17 ten-thousandths.
8. Add 24|, 17}, .0058, 7i, 9^^.
9. 32.58, 28963.1, 287.531, 76398.9341.
10. 145., 14.5, 1.45, .145, .0145.
Subtraction of Decimals
To subtract decimals, ivrite the smaller number under the
larger icith the decimal point of the subtrahend directly under the
decimal point of the minuend. Subtract as in integers, and place
the point directly under the points above.
Example. — Subtract 2.17857 from 4.3257.
Write the lesser number under the greater,
4.32570 Minuend y^^^y^ ^^^ decimal points under each other
2.17857 Subtrahend Add a to the minuend, 4.3257, to give it the
2.14713 Remainder same denominator as the subtrahend. Then
subtract as in subtraction of integers. Write
the remainder with decimal point under the other two points,
REVIEW OF ARITHMETIC 33
EXAMPLES
Subtract :
1. 69.0364-30.8691 = ? 3. .0625 - .03125 = ?
2. 48.7209 - 12.0039 = ? 4. .0001 1 - .000011 = ?
5. 10 - .1 -H .0001 = ?
6. From one thousand take five thousandths.
7. Take 17 hundred-thousandths from 1.2.
8. From 17.37^ take 14.16f
9. Prove that ^ and .500 are equal.
10. Find the difference between yYA ^°^ rlfH*
Multiplication of Decimals
To multiply decimals jnoceed as in integers, and give to the
product as many decimal figures as there are in both multiplier
and multiplicand. When there are not figures enough in the
product, prefix ciphers.
Example. — Find the product of 6.8 and .63.
6.8 Multiplicand
.63 Multiplier ^-^ ^^ ^^® multiplicand and .63 the multiplier.
"^777 Their product is 4,284 with three decimal figures,
^ the number of decimal figures in the multiplier
and multiplicand.
4.284 Product
Example. — Find the product of .05 and .3.
.05 Multiplicand The product of .05 and .3 is .016 with a cipher
.3 Multiplier prefixed to make the three decimal figures re-
.015 Product quired in the product.
EXAMPLES
Find the products :
1. 46.25 X. 125 3. .015 x .05
2. 8.0625 X. 1875 4. 25.863 x 4i
34 VOCATIONAL MATHEMATICS
5. 11.11 X 100 8. .325 X 12i
6. .5625 X 6.28125 9. .001542 x .0052
7. .326 x 2.78 10. 1.001 x 1.01
To multiply btj 10, 100, 1000, etc., remove the point one place
to the right for each cipher m the multiplier.
This can be performed without writing the multiplier.
Example.— Multiply 1.625 by 100.
1.625 x 100 = 162.5
To multiply by 200, remove the point to the right and multiply
by 2.
Example. — Multiply 86.44 by 200.
86.44.
2
17,288
EXAMPLES
Find the product of :
1. 1 thousand by one thousandth.
2. 1 million by one millionth.
3. 700 thousands by 7 hundred-thousandths.
4. 3.894 x 3000 5. 1.892 x 2000.
Division of Decimals
To divide decimals proceed as in integers, and give to the quo-
tient as many decimal figures as the number in the dividend ex-
ceeds those in the divisor.
Example. — Divide 12.685 by .5.
The number of decimal figures in
Divisor . 5)12.685 Dividend the quotient, 12.685, exceeds the num-
25.37 Quotient ber of decimal figures in tlie divisor, .5,
by two. So there tnust be two deQi-
pial figures in the quotient.
REVIEW OF ARITHMETIC
35
Example. — Divide 899.552 by 192.
When the divisor is an integer*
the point in
tlie dividend, and tli<» diviMfoii |H»r-
formeoint
in both divisor and dividend as
many place.s U) the right as
tlicre are decimal placcB in tlie
divisor, which is equivalent to
multiplyin<{ l)oth divisor and
dividend by the same number
and does not change the (}Uo.
tient. Then place the point in
the quotient as if the divisor
were an integer. In this ex-
ample, the multiplier of both
dividend and divisor is 100.
EXAMPLES
Eind the quotients :
1. .0625 -.125 5. 1000 -I- .001
2. 315.432 -.132 6. 2.490 + . 136
3. .75 -.0125 7. 28000 + 16.8
4. 125 -5- 12^
a 1.225 + 4.9
9. 3.1416 + 27
10. 8.33 + 6
To divide by 10, 100, 1000, etc.^ remove tin- inimt nue place to
the left for each cipher in the didaor.
7V> divide by 200, remove the point ttco places to the le/t^ and
divide by 2.
36 VOCATIONAL MATHEMATICS
EXAMPLES
Find the quotients :
1. 38.64-10 6. 865.45-5000
2. 398.42-1000 7. 38.28-400
3. 1684.32-1000 8. 2.5-500
4. 1.155-100 9. .5-10
5. 386.54-2000 10. .001-1000
Parts of 100 or 1000
1. What part of 100 is 12^ ? 25 ? 33^ ?
2. What part of 1000 is 125 ? 250 ? 333^ ?
3. How much is i of 100 ? Of 1000?
4. How much is J of 100 ? Of 1000 ?
5. What is 1 of 100 ? Of 1000?
Example. — How much is 25 times 24 ?
100 times 24 = 2400.
25 times 24 = |^ as much as 100 times 24 = 600. Ans.
Short Method in Multiplication
To multiply by
25, multiply by 100 and divide by 4 ;
33|, multiply by 100 and divide by 3 ;
16|, multiply by 100 and divide by 6 ;
121 multiply by 100 and divide by 8 ;
9, multiply by 10 and subtract the multiplicand ;
11, if more than two figures, multiply by 10 and add the
multiplicand;
11, if two figures, place the figure that is their sum between
them.
63 X 11 = 693 74 X 11 = 814
Note that when the sum of the two figures exceeds nine, the one in the
tens place is carried to the figure at the left.
REVIEW OF ARITHMETIC 37
EXAMPLES
Multiply by the short process :
1. 81 by 11 =? 10. 68 by 16J = ?
2. 76 by 33i = ? 11. 112 by 11 = ?
3. 128 by 12J = ? 12. 37 by 11 = ?
4. 87 by 11 = ? 13. 4183 by 11 = ?
5. 19 by 9 = ? 14. 364 by 33^ = ?
6. 846 by 11 = ? 15. 8712 by 12J = ?
7. 88 by 11 = ? 16. 984 by 16J = ?
8. 19 by 11 = ? 17. 36 by 25 = ?
9. 846byl6J = ? la 30by333J=?
Aliquot Parts of $1.00
The aliquot parts of a nuinber are the numbers that are
exactly contained in it. The aliquot parts of 100 are 5, 20,
12J, 16f, 334, etc.
The monetary unit of the United States is the dollar, con-
taining one hundred cents which are written decimally.
25 cents = $ ^ = quarter dollar
33^ cents = $ ^
50 cents = $ ^ = half dollar
n
cents = $^^
«i
cents = $ 3^5
12^
cents = $ i
16J
cents = $ i
10 mills
5 cents
10 cents
10 dimes
= 1 cent, ct. = $ .01 or $ 0.01
= 1 " nickel " = $ .05
= 1 dime, d. = $ .10
= 1 dollar, $ = $1.00
10 dollars = 1 eagle, E. = $ 10.00
Example. — What will 69 drills cost at 16J cents each ?
69 drills will cost 69 x 16i cts., or 69 x $i=V = f Hf = -^11.60.
38 VOCATIONAL MATHEMATICS
Example. — At 25^ a box, ho^v many boxes of nails can be
bought lor $ 8.00 ?
8^i = 8xf = 32 boxes. Ans.
Review of Decimals
1. For work on a job one man receives $ 13.75, a second
man $ 12.45, a third man $ 14.21, and a fourth man $ 21.85.
What is the total amount paid for the work ?
2. A pipe has an inside diameter of 3.067 inches and an
outside diameter of 3.428 inches. What is the thickness of
the metal of the pipe ?
3. At 21 cts. a pound, what will be the cost of 108 castings
each weighing 29 lb. ?
4. A man receives $ 121.50 for doing a piece of work. He
gives $12.25 to one of his helpers, and S 10.50 to another.
He also pays $75.75 for material. How much does he make
on the job ?
5. An automobile runs at the rate of 9^ miles an hour.
How long will it take it to go from Lowell to Boston, a dis-
tance of 26.51 miles ?
6. A f square steel bar weighs 1.914 lb. per foot. What
will be the cost of 5000 feet of |" steel bars, if it costs $ 1.75
per 100 lb. ?
7. Which is cheaper, and how much, to have a 13i cents
an hour man take 13J hours on a piece of work, or hire a 17^
cents an hour man who can do it in 9^ hours ?
8. On Monday 1725.25 lb. of coal are used, on Tuesday
2134.43 lb., on Wednesday 1651.21 lb., on Thursday 1821.42
lb., on Friday 1958.82 lb., and on Saturday 658.32 lb. How
many pounds of coal were used during the week ?
9. If in the example above there were 10,433.91 lb. of
coal on hand at the beginning of the week, how much was left
at the end of the week ?
RKVIEW OF ARITHMETIC 30
10. A foot length of }" round steel bar weighs 1.50^5 lb.,
10-foot length of J" square steel bar weiglis 19.140 lb., 1-foot
length of y^" round steel bar weighs 3.017 lb. What is the
total weight of the three pieces ?
U. An alloy is made of copper and zinc. If .(iO is copper
and .34 is zinc, how many pounds of zinc and how many
pounds of co})per will there be in a casting of the alloy
weighing 98 lb. ?
12.. A train leaves New York at 2.10 p.m., and arrives in
Philadelphia at 4.15 p.m. The distance is 90 miles. What is
the average rate per hour of the train ?
13. The weight of a foot of ^^" steel bar is 1.08 lb. Find
the weight of a 21-foot bar.
14. A steam pump pumps 3.38 gallons of water to each
stroke and the pump makes 51.1 strokes per minute. How
many gallons of water will it pump in an hour?
15. At 12^ cents per hour, what will be the pay for 23^ days
if the days are 10 hours each ?
Compound Numbers
A number which expresses only one kind of concrete units
is a simple number; as, 5 ])k., 4 knives, 6.
A number composed of different kinds of concrete units that
aie related to each other is a compound number ; as, 3 bu. 2 pk.
1 qt.
A denomination is a name given to a unit of measure or of
weight.
A number having one or more denominations is also called a
denominate number.
Reduction is the process of changing a number from one
denomination to another without changing its value.
Changing to a lower denomination is called reduction descend-
ing; as, 2 bu. 3 pk. = 88 qt.
40 VOCATIONAL MATHEMATICS
Changing to a higher denomination is called reduction
ascending ; as, 88 qt. = 2 bu. 3 pk.
Linear Measure is used in measuring lines or distance.
Table
12 in. (in.) = 1 foot, ft.
3 feet = 1 yard, yd.
5^ yards or 16^ feet = 1 rod, rd.
40 rods = 1 furlong, fur.
8 furlongs = 1 mile, mi.
320 rods, or 5280 feet = 1 mile.
1 mi. = 320 rd. = 1760 yd. = 5280 ft. = 63,360 in.
Surveyors' Measure is used in measuring land.
Table
7.92 inches = 1 link, li.
100 links = 1 chain, ch.
80 chains = 1 mile, mi.
A chain, or steel measuring tape, 100 feet long, is generally used by
engineers. The feet are usually divided into tenths instead of into
inches.
Square Measure is used in measuring surfaces.
Table
144 square inches = 1 square foot, sq. ft.
9 square feet = 1 square yard, sq. yd.
30i square yards 1 , , ,
160 square rods = 1 acre, A.
640 acres = 1 square mile, sq. mi.
1 sq. mi. = 640 A. = 102,400 sq. rd. = 3,097,600 sq. yd.
Cubic Measure is used in measuring volumes or solids.
Table
1728 cubic inches = 1 cubic'foot, cu. ft.
27 cubic feet = 1 cubic yard, cu. yd.
16 cubic feet = 1 cord foot, cd. ft.
8 cord feet, or 128 cu. ft. = 1 cord, cd.
1 cu. yd. = 27 cu. ft. = 46,666 cu. in.
REVIEW OF ARITHMETIC 41
Liquid Measure is used in measuring liquids.
Table
4 gills (gi.)= 1 P'"^ pt"
2 pints = 1 quart, qt.
4 quart* = 1 gallon, gal.
1 gal. = 4 qt. = 8 pt. = 32 gi.
A gallon contains 231 cubic inches.
The standard barrel is 31 J gal., and the hogshead 63 gal.
Dry Measure is used in measuring roots, grain, vegetables,
etc.
Table
2 pints = 1 quart, qt.
8 quarts = 1 peck, pk.
4 pecks = 1 bushel, bu.
1 bu. = 4 pk. = 32 qt. = 64 pints.
The bushel contains 2150.42 cubic inches; 1 dry quart contains
67.2 cu. in. A cubic foot is |J of a bushel.
Avoirdupois Weight is used in weighing all common articles ;
as, coal, groceries, hay, etc.
Table
16 ounces (oz.) = 1 pound, lb.
100 pounds = 1 hundredweight, cwt. ;
or cental, ctl.
20 c\n., or 2000 lb. = 1 ton, T.
1 T. = 20 cwt. = 2000 lb. = 32,000 oz.
The long ton of 2240 pounds is used at the United States Custom
House and in weighing coal at the mines.
Measure of Time.
Table
60 seconds (sec.) = 1 minute, min.
60 minutes = 1 hour, hr.
24 hours
= 1 day, da.
7 days
= 1 week, wk.
365 days
= 1 year, yr.
366 days
= 1 leap year.
100 years
= 1 century.
42 VOCATIONAL MATHEMATICS
Counting.
Table
12 things = 1 dozen, doz.
12 dozen = 1 gross, gr.
12 gross = 1 great gross, G. gr.
Paper Measure.
Table
24 sheets = 1 quire 2 reams = 1 bundle
20 quires = 1 ream 5 bundles = 1 bale
Reduction Descending
Example. — Reduce 17 yd. 2 ft. 9 in. to inches.
1 yd. = 3 ft.
17 yd. = 17 X 3 = 51 ft.
51 + 2 = 53 ft.
1 ft. = 12 in.
53 ft. = 53 X 12 = 6.36 in.
636 + 9 = 645 in. Ans.
EXAMPLES
Reduce to lower denominations :
1. 46 rd. 4 yd. 2 ft. to feet.
2. 4 A. 15 sq. rd. 4 sq. ft. to square inches.
3. 16 cu. yd. 25 cu. ft. 900 cu. in. to cubic inches.
4. 15 gal. 3 qt. 1 pt. to pints.
5. 27 da. 18 hr. 49 min. to seconds.
Reduction Ascending
Example. — Reduce 1306 gills to higher denominations.
4 )1.306 gi. Since in 1 pt. there are 4 gi., in 1306 gi.
2 )326 pt. + 2 gi. there are as many pints as 4 gi. are contained
4 )163 qt.. times in 1306 gi., or .326 pt. and 2 gi. remainder.
40 gal. + 3 qt. In the same way the quarts and gallons are
40 gal. 3 qt. 2 gi. Ans. found. So there are in 1306 gi., 40 gal. 3 qt.
2gi.
REVIEW OF ARITHMETIC 43
EXAMPLES
Reduce to higher denominations :
1. Reduce 225,932 in. to miles, etc.
2. Change 1384 dry pints to higher denominations.
3. In 139,843 sq. in. how many square miles, rods, etc. ?
4. How many cords of wood in 3692 cu. ft. ?
5. How many bales in 24,000 sheets of paper ?
A denominate fraction is a fraction of a unit of weight or
measure.
To reduce denominate fractions to integers of lower denominations.
Change the fraction to the next lower denomination. Treat
the fractional part of the product in the same ivai/, and so pro-
ceed to the required denomination.
Example. — Reduce 4 of a mile to rods, yards, feet, etc.
^ of 320 rd. ^ J-V^ rd. = 228| rd.
^of Vyd. = nyd=3Jyd.
^ f of 3 ft. = Of ft.
f of 12 in. = 5^ in. =5f in.
f of a mile = 228 rd. 3 yd. ft. 6f in.
The same process applies to denominate decimals.
To reduce denominate decimals to denominate numbers.
Example. — Reduce .87 bu. to pecks, quarts, etc.
.87 bu. .84 qt.
Change the decimal fraction to
the next lower denomination. Treat
the decimal part of the product in the
same way, and so proceed to the re-
quired denomination.
3.48
pk. 1.68
.48 pk.
8
3.84 qt.
pt.
3pk.
, 3 qt. 1.68 pt.
Ans.
44 VOCATIONAL MATHEMATICS
EXAMPLES
Reduce to integers of lower denominations :
1. J of an acre. 3. ^ of a ton.
2. .3125 of a gallon. 4. .51625 of a mile.
5. Change f of a year to months and days.
6. .2364 of a ton.
7. What is the value of ^ of 1| of a mile ?
8. Reduce |^ bu. to integers of lower denominations.
9. .375 of a month.
10. y^y acre are equal to how many square rods, etc. ?
Addition of Compound Numbers
Example. — Find the sum of 7 hr. 30 min. 45 sec, 12 hr.
25 min. 30 sec, 20 hr. 15 min. 33 sec, 10 hr. 27 min. 46 sec.
The sum of the seconds = 154 sec. =
2 min. 34 sec. Write the 34 sec. under
the sec. column and add the 2 rain, to
the min. column. Add the other columns
50 39 34 in the same way.
60 hr. 39 min. 34 sec. Ans.
Subtraction of Compound Numbers
Example. — From 39 gal. 2 qt. 2 pt. 1 gi. take 16 gal. 2 qt.
3 pt. 3 gi.
As 3 gi. cannot be taken from 1 gi., 4 gi.
or 1 pt. are borrowed from the pt. column
and added to the 1 gi. Subtract 3 gi. from
the 5 gi. and the remainder is 2 gi. Continue
in the same way until all are subtracted.
Then the remainder is 22 gal. 3 qt. pt. 2 gi.
hr.
min.
sec.
7
30
45
12
25
30
20
15
33
10
27
46
gal.
qt.
pt.
gi.
39
2
2
1
16
2
3
3
22
3
2
22 gal.
3qt
.2
gi. Ans.
yd.
ft.
In.
4
2
8
8
39
4
39 yd
.4i
in. Ans.
REVIEW OF ARITHMETIC 45
Multiplication of Compound Numbers
Example. — Multiply 4 yd. 2 ft. 8 in. by 8.
8 times 8 In. = 64 in. = 5 ft. 4 in. Place the
4 in. under the in. column, and add the 5 ft. to
the product of 2 ft. by 8, which equals 21 ft. = 7 yd.
Add 7 yd. to the product of 4 yd. by 8 = 39 yd.
Division of Compound Numbers
Example. — Fiud ^ of 42 rd. 4 yd. 2 ft. 8 in.
-^ of 42 rd. = 1 rd. ; re-
mainder, 7 rd. = 38^ yd. ;
add 4 yd. = 42} yd. ^5 of
42^ yd. = 1 yd. ; remainder,
7i yd., = 22i ft. = 24J ft.
5^ of 24 J ft. = ft. 24 i ft.
=294 in. ; add 8 in. =302 in.
^ of 302 in. = 8Jf in.
Ans.
Difference between Dates
Example. — Find the time from Jan. 25, 1842, to July 4,
1896.
1896 7 4 It is customary to consider 30 days
1842 1 25 to a month. July 4, 1896, i.s the 1896th
54 yr. 5 mo. 9 da. Ans. yr., 7th mo., 4th da., and Jan. 25, 1842,
is the 1842d yr., Ist. mo., 25th da.
Subtract, taking 30 da. for a month.
Id. yd.
ft. in.
36)42 4
2 8(1 rd.
36
i
^
H
35)24^(0 ft.
35
12
38i
294
+ 4
+ 8
35)42J yd. (1 yd. 36)iM(8|^ in.
35
280
7i
22
3
22i ft.
1 rd. 1 yd. Sji in.
12
46 VOCATIONAL MATHEMATICS
Example. — What is the exact number of days between
Dec. 16, 1895, and March 12, 1896 ?
Dec. 15 Do not count the first day mentioned. Tliere
Jan. 31 are 15 days in December, after the 16th. Jan-
Feb. 29 uary has 31 days, February 29 (leap year),
Mar. 12 and 12 days in March ; making 87 days.
87 days. Ans.
EXAMPLES
1. How much time elapsed from the landing of the Pil-
grims, Dec. 11, 1620, to the Declaration of Independence,
July 4, 1776?
2. Washington was born Feb. 22, 1732, and died Dec. 14,
1799. How long did he live?
3. Mr. Smith gave a note dated Feb. 25, 1896, and paid it
July 12, 1896. Find the exact number of days between its date
and the time of payment.
4. A carpenter earning $2.50 per day commenced Wednes-
day morning, April 1, 1896, and continued working every week
day until June 6. How much did he earn ?
5. Find the exact number of days between Jan. 10, 1896,
and May 5, 1896.
6. John goes to bed at 9.15 p.m. and gets up at 7.10 a.m.
How many minutes does he spend in bed ?
To multiply or divide a compound number by a fraction.
To multiply by a fraction, multiply by the numerator, and
divide the product by the d^nomiriator.
To divide by a fraction, multiply by the denominator , and divide
the product by the numerator.
When the multiplier or divisor is a mixed number, reduce to
an improper fraction, and proceed as above.
REVIEW OF ARITHMETIC 47
EXAMPLES
1. How much is 4 of 16 hr. 17 min. 14 sec. ?
2. A field contains 10 acres 12 sq. rd. of lan«l, which is J of
the whole farm. Find the size of the farm.
3. If a train runs 60 mi. 35 rd. 16 ft. in one hour, how far
will it run in 12f hr. at the same rate of speed ?
4. Divide 14 bu. 3 pk. 6 qt. 1 pt. by J.
5. Divide 5 yr. 1 mo. 1 wk. 1 da. 1 hr. 1 min. 1 sec. by 3J.
EXAMPLES
1. A time card on a piece of work states that 2 hours and
15 minutes were spent in lathe work, 1 hour and 12 minutes in
milling, 2 hours and 45 minutes in planing, and 1 hour and 30
minutes on bench work. What was the number of hours spent
on the job ?
2. How many castings, each weighing 14 oz., can be
obtained from 860 lb. of metal if nothing is allowed for
waste?
3. How many feet hmg must a machine shop be to hold a
lathe 8' 6", a planer 14' 4", a milling machine 4' 2", and a
lathe 7' 5", placed side by side? 3' 3" were allowed between
the machines and between the walls and the machines.
4. How many gross in a lot of 968 screws ?
5. Find the sum of 7 hr. 30 min. 45 sec, 12 hr. 26 min.
30 sec, 20 hr. 15 min. 33 sec, 10 hr. 27 min. 46 sec.
6. If a train is run for 8 hours at the average rate of
50 mi. 30 rd. 10 ft. per hour, how great is the distance covered?
7. A telephone pole is 31 feet long. If 4 ft. 7 in. are
under ground, how high (in inches) is the top of the pole above
the street?
8. If 100 bars of iron, each 2}' long, weigh 70 lb., what is
the total weight of 2300 ?
48 VOCATIONAL MATHEMATICS
9. If 43 in. are cut from a wire 3 yd. 2 ft. 6 in. long, what
is the length of the remaining piece ?
10. If a rod of iron 18' 8" long is cut into pieces 6J" long
and Jj" is allowed for waste in each cut, how many pieces
can be cut ? How much remains ?
11. I have 84 lb. 14 oz. of salt which I wish to put into
packages of 2 lb. 6 oz. each. How many packages will there be ?
12. If one bottle holds 1 pt. 3 gi., how many dozen bottles
will be required to hold 65 gal. 2 qt. 1 pt. ?
13. How many pieces 5J" long can be cut from a rod 16' 8"
long, if 5" is allowed for waste ?
14. What is the entire length of a railway consisting of
five different lines measuring respectively 160 mi. 185 rd. 2 yd.,
97 mi. 63 rd. 4 yd., 126 mi. 272 rd. 3 yd., 67 mi. 199 rd. 5 yd.,
and 48 mi. 266 rd. 5 yd. ?
Percentage
Percentage is a process of solving questions of relation by
means of hundredths or per cent (%).
Every question in percentage involves three elements : the
rate per cent, the base, and the percentage.
The rate per cent is the number of hundredths taken.
The base is the number of which the hundredths are
taken.
The percentage is the result obtained by taking a certain per
cent of a number.
Since the percentage is the result obtained by taking a cer-
tain per cent of a number it follows that, the percentage is the
product of the base and the rate. The rate and base are always
factors, the percentage is the product.
Example. — How much is 8 % of $200?
8% of $ 200 = 200 X .08 = $ 16. (1)
REVIEW OF ARITHMETIC 49
In (1) we have the three elements: 8% is the rate, 8200 is the base,
and $ 16 is the percentage.
Since $200 x .08 = $ 16, the percentage ;
$ 16 -i- .08 = $ 200, the base ;
and $ 16 -i- $ 200 = .08, the rate.
If any two of these elements are given, the other may be
found :
Base X Rate = Percentage
Percentage -5- Rate = Base
Percentage -5- Base = Rate
Per cent is commonly used in the decimal form, but many
operations may be much shortened by using the common frac-
tion form.
1 % = .01 = ^^ i%= .00^ or .005
10%= .10 = ^V 33J% = .33J = i
100 % = 1.00 = 1 8^ % = .08^ = .0825
12^ % = .12^ or .125 = ^ | % = .00^ = .00125
There are certain per cents that are used so frequently that
we should memorize their equivalent fractions.
10%=^ 37-^%=! 75% = I
12i%=i 40% = J 80%= J
16i%=i 50% =i 83i%=J
20% =i 60%= I 87i%=J
EXAMPLES
1. Find 75 % of $ 368.
2. Find 15% of S412.
3. 840 is 33 J % of what number ?
4. 615 is 15% of what number?
5. What per cent of 12 is 8 ?
50 VOCATIONAL MATHEMATICS
6. What per cent of 245 is 5 ?
7. What per cent of 195 is 39 ?
8. What per cent of 640 is 80 ?
9. What per cent of 750 is 25 ?
10. What per cent of 819 is 45 ?
Trade Discount
Merchants and jobbers have a price list. From this list
they give special discounts according to the credit of the
customer and the amount of supplies purchased, etc. If they
give more than one discount it is understood that the first
means the discount from the list price, the second denotes the
discount from the remainder.
EXAMPLES
1. What is the price of 200 No. 1 cleats at $ 36.68 per M.
at 40% off?
2. Supplies from a hardware store amounted to $ 58,75. If
121 % were allowed for discount, what was the amount paid ?
3. A dealer received a bill amounting to $ 212.75. Succes-
sive discounts of 75%, 15%, 10%, and 5% were allowed.
What was amount to be paid ?
4. 2 % is usually discounted on bills paid within 30 days.
If the following are paid within 30 days, what will be the
amounts due?
a. $2816.49 d. $1369.99 g. $4916.01
b. 399.16 e. 2717.02 h. 30.19
c. 489.01 /. 918.69
5. What is the price of 20 fuse plugs at $ .07 each, 30 %
off?
6. Hardware supplies amounted to $ 127.79 with a discount
of 40 and 15 %. What was the net price?
REVIEW OF ARITHMETIC 51
7. Which is better, for a merchant to receive a straight dis-
count of 95 % or a successive discount of 75, 15, 5 % ?
8. Twenty per cent is added to the number of workmen in
a machine shop of 575 men. What is the number employed
after the increase ?
9. A steam pressure of 180 lb. per square inch is raised
to 225 lb. per square inch. What is the per cent of in-
crease ?
10. If 31 out of 595 wheels are rejected because of defects,
what per cent is rejected ?
11. A clerk's salary was increased OJ %. If he now receives
$ 850, what was his original salary ?
12. A company lost 12|^ % of its men and had 560 left.
How many men were there before ?
Simple Interest
Money that is paid for the use of money is called interest.
The money for the use of which interest is paid is called the
principal^ and the sum of the principal and interest is called the
amojint.
Interest at 6 % means 6% of the principal for 1 year; 12
months of 30 days each are usually regarded as a year in com-
puting interest.
Example. — What is the interest on $ 100 for 3 years at 6 % ?
8100
.06
$ 6.00 interest for one year. Or, yg^ x Jf *i x f = $ 18. Ans.
3
% 18.00 interest for 3 years. Ana.
8 100 -I- § 18 = $ 118, amount.
Pnncfpal X Kate x Time = Literest.
52 VOCATIONAL MATHEMATICS
Example. — What is the interest on $ 297.62 for 5 yr. 3 mo
at6%?
1297.62
ni:sM o,, 1_ ^ $ 297.62 21 ^ 118750.06 ^ ,3^,
., 100 1 ^ 200
44643
892860 Note. — Final results should not include mills.
$93.7503 Mills are disregarded if less than 5, and called another
$ 93.75. Ans. cent if 5 or more.
EXAMPLES
1. What is the interest on $ 586.24 for 3 months at 6 % ?
2. What is the interest on $ 816.01 for 9 months at 5 % ?
3. What is the interest on $ 314.72 for 1 year at 4 % ?
4. What is the interest on $ 876.79 for 2 yr. 3 mo. at 41 fo ?
The Six Per Cent Method
By the 6 % method it is convenient to find first the interest
of $ 1, then multiply it by the principal.
Example. — What is the interest of $ 50.24 at 6 % for 2 yr.
8 mo. 18 da. ?
Interest on $; 1 for 2 yr. =2 x $ .06 = .$ .12
Interest on § 1 for 8 mo. =8 x $ .00^ = .04
Interest on $ 1 for 18 da. = 18 x $ .000^ = .003
Interest on $ 1 for 2 yr. 8 mo. 18 da. .$ .163
Interest on ^50.24 is 50.24 times $ .163 = .$8.19. Ans.
Find the interest on $ 1 for the given time, and multiply it by the prin-
cipal, considered as an abstract number.
EXAMPLES
Find the interest and amount of the following :
1. $ 2350 for 1 yr. 3 mo. 6 da. at 5 %.
2. $ 125.75 for 2 mo. 18 da. at 7 %.
3. $ 950.63 for 3 yr. 17 da. at 41 %.
4. $ 625.57 for 1 yr. 2 mo. 15 da. at 6 %.
REVIEW OF ARITHMETIC 53
Exact Interest
When the time includes clays, interest computed by the 6%
method is not strictly exact, by reason of using only .'^0 days
for a month, which makes the year only 360 days. The day is
therefore reckoned as -^ of a year, whereas it is ^^-y of a year.
To compute exact interest^ find the exact time in daya^ and coiir
aider 1 day^s interest as ^^ of 1 yea^s interest.
Example. — Find the exact interest of $ 358 for 74 days at
7 %.
$:368 X .07 = $26.06, 1 year's interest.
74 days* interest is j\^ of 1 year's interest.
3Vir of $26.06 = $6.08. Ans.
' 1 100 366
EXAMPLES
Find the exact interest of :
1. $324 for 15 da. at 5 %.
2. $253 for 98 da. at 4%.
3. $624 for 117 da. at 7%.
4. $ 620 from Aug. 15 to Nov. 12 at 6 %.
5. $ 153.26 for 256 da. at 5^ %.
6. S 540.25 from June 12 to Sept. 14 at 8 %.
Rules for Computing Interest
The following will be found to be excellent rules for finding the inter*
est on any principal for any number of days.
Divide the principal by 100 and proceed as follows:
2 % — Multiply by number of days to run, and divide by 180.
2^ % — Multiply by number of days, and divide by 144.
3 % — Multiply by number of days, and divide by 120.
3^ % — Multiply by number of days, and divide by 102.86.
54
VOCATIONAL MATHEMATICS
4 % — Multiply by number of days, and divide by 90.
5 % — Multiply by number of days, and divide by 72.
6 % — Multiply by number of days, and divide by 60.
7 % — Multiply by number of days, and divide by 51.43.
8 % — Multiply by number of days, and divide by 45.
Savings Bank Compound Interest Table
Showing the amount of $ 1, from 1 year to 15 years, with compound
interest added semiannually, at different rates.
Pkb Cent
3
4
5
6
7
8
9
h year
101
102
102
103
103
1 04
104
1 year
103
104
105
106
107
108
109
1^ yeare
104
106
107
109
110
1 12
1 14
2 yeai*s
106
108
1 10
1 12
1 14
116
119
2^ years
107
1 10
113
1 16
1 18
1 21
124
3 years
109
1 12
1 15
1 19
1 22
1 26
130
3^ years
110
1 14
118
122
127
131
136
4 years
1 12
1 17
121
126
131
1 36
142
4| years
1 14
1 19
124
130
136
142
148
5 years
1 16
121
128
134
141
148
165
5^ years
1 17
124
131
138
145
163
162
6 years
1 19
126
134
142
1 61
160
169
6^ years
121
129
137
146
156
166
1 77
7 years
123
131
141
1 51
161
1 73
185
7^ years
124
1 34
144
166
167
180
193
8 years
126
137
148
1 60
173
187
2 02
8| years
128
139
152
166
179
194
2 11
9 years
1 30
142
1 66
170
185
2 02
2 20
9^ years
132
146
1 59
176
192
2 10
2 30
10 years
1 34
148
163
180
1 98
2 19
2 41
11 years
138
1 54
1 72
191
2 13
2 36
2 63
12 years
142
160
180
2 03
2 28
2 56
2 87
13 years
147
167
1 90
2 16
2 44
2 77
3 14
14 years
1 61
1 73
1 99
2 28
2 62
2 99
3 42
16 years
1 66
1 80
2 09
2 42
2 80
3 24
3 74
REVIEW OF ARITHMETIC
EXAMPLES
Solve the following problems according to the tables given
above :
1. What is the compound interest of $ 1 at the end of
8.1 years?
2. What is the compound interest of $1 at the end of 11
years ?
3. How long will it take $400 to double itself at 4%,
compound interest ?
4. How long will it take $580 to double itself at 4^ %,
compound interest?
5. How long will it take $615 to double itself at 6%,
simple interest ?
6. How long will it take $784 to double itself at 5| %,
simple interest ?
7. Find the interest of S684 for 94 days at 3 %.
a Find the interest of $ 1217 for 37 days at 4 %.
9. Find the interest of $681.14 for 74 days at 4^ %.
10. Find the interest of $414.50 for 65 days at 5 %.
11. Find the interest of $384.79 for 115 days at 6 %.
Ratio and Proportion
Ratio is the relation between two numbers. It is found
by dividing one by the other. The ratio of 4 to 8 is 4 -§- 8 = i.
The terms of the ratio are the two numbers compared. The
first term of a ratio is the antecedent^ and the second the con-
sequent. The sign of the ratio is (:). (It is the division sign
with the line omitted.) Ratio may also be expressed fraction-
ally, as J/ or 16:4; or yV or 3 : 17.
A ratio formed by dividing the consequent by the antece-
dent is an inverse raJtio : 12 : 6 is the inverse ratio of 6 : 12.
56 VOCATIONAL MATHEMATICS
The two terms of the ratio taken together form a couplet.
Two or more couplets taken together form a compound ratio.
Thus, * 3 : 6 = 23 : 46
A compound ratio may be changed to a simple ratio by tak-
ing the product of the antecedents for a new antecedent, and
the product of the consequents for a new consequent.
Antecedent -i- Consequetit = Ratio
Antecedent -;- Ratio = Consequent
Ratio X Consequent = Antecedent
To multiply or divide both terms of a ratio by the same
number does not change the ratio.
Thus 12 : 6 = 2
3x12:3x6 = 2
EXAMPLES
Find the ratio of
1. 20 : 300 Fractions with a common de-
2. 3 bu. : 3 pk. nominator have the same
3 21-16 ratio as their numerators.
4. 12 : 4- 7 8_.16 28.7 15.30
• -^*^ • 4 '• IT • IT? TT- 7 5J 11 • TT
5- i = l a |:|,B:|,.2.5
6. 16:(?) = |
Proportion
An equality of ratios is a proportion.
A proportion is usually expressed thus : 4 : 2 : : 12 : 6, and is
read 4 is to 2 as 12 is to 6.
A proportion has four terms, of which the first and third are
antecedents and the second and fourth are consequents. The
first and fourth terms are called extremes, and the second and
third terms are called means.
The product of the extremes equals the product of the
means.
REVIEW OP ARITHMETIC 57
Tojind an extreme, divide the product of the means Ini the given
extreme.
Tojind a mean, divide the proilact of the extremes by the given
mean.
EXAMPLES
Supply the missing term :
1. 1:836::25:() 4. 10 yd. : 50 yd. : : $ 20 : ($ )
2. 6:24::( ) : 40 5. $}:S3|::( ):5
3. ( ):15::60:6
Simple Proportion
An equality of two simple ratios is a simple proportion.
Example. — If 12 bushels of charcoal cost $4, what will 60
bushels cost ?
12 • 60 • • S4 • f.«5 ^ There is the same relation between the cost
„ ■ , " '^'' '' of 12 bu. and the cost of 60 bu. as there is be-
— -|— =$20. Am. tween the 12 bu. and the 60 bu. $4 is the
third term. The answer is the fourth term.
It must form a ratio of 12 and 60 that shall equal the ratio of § 4 to the
answer. Since the third term is lass than the required answer, the first
must be less than the second, and 12 : 60 is the first ratio. The product
of the means divided by the given extreme gives the other extreme, or $ 20.
EXAMPLES
Solve by proportion :
1. If 150 fuses cost $ 6, how much will 1200 cost ?
2. If 250 pounds of lead pipe cost S 15, how much will 1200
pounds cost ?
3. If 5 men can dig a ditch in 3 days, how long will it take
2 men ?
4. If .4' men can shingle a shed in 2 days, how long will it
take 3 men ?
5. The ratio of Simon's, pay to Matthew^s is |. Simon
earns $ 18 per week. What does Matthew earn ?
58 VOCATIONAL MATHEMATICS
6. What will llf yards of cambric cost if 50 yards cost
$ 6.75 ?
7. A spur gear making 210 revolutions per minute is en-
meshed with a pinion. The gear has 126 teeth and the pinion
has 42 teeth. How many revolutions does the pinion make ?
8. In a velocity diagram a line 3|" long represents 45 ft.
What would be the length of a line representing 30 ft. velocity ?
9. How many pounds of lead and tin would it take to make
4100 pounds of solder if there are 27 pounds of tin in each 100
pounds of solder ?
10. It is necessary to obtain a speed reduction of 7 to 3 by
use of gears. If the pinion has 21 teeth, how many teeth
must the gear have ?
11. A bar of iron 3^ ft. long and |" diameter weighs 6.64
pounds. W^hat would a bar 4|^ ft. long of the same diameter
weigh ?
12. In a certain time 15 workmen made 525 pulleys. How
many pulleys will 32 men make in the same length of time ?
13. When a post 11.5 ft. high casts a shadow on level
ground 20.6 ft. long, a telephone pole near by casts a shadow
59.2 ft. long. How high is the pole ?
14. The diameter of a driving pulley is 18". This pulley
makes 320 revolutions per minute. What must be the diam-
eter of a driving pulley in order to make 420 revolutions per
minute ?
15. A ditch is dug in 14 days of 8 hours each. How many
days of 10 hours each would it have taken ?
16. If in a drawing a tree 38 ft. high is represented by IJ",
what on the same scale will represent the height of a house 47
ft. high?
17. What will be the cost of 21 motors if 15 motors cost
$ 887.509 ?
REVIEW OF ARITHMETIC 59
18. The main drive pulley of a machine is 6 inches in diam-
eter and makes 75G revolutions per minute. A pulley on the
line shaft is belted to a machine. What is the diameter of the
line shaft pulley if the line shaft makes 252 revolutions per
minute ?
19. If a pole 8 ft. high casts a shadow 4 J ft. long, how high
is a tree which casts a shadow 48 ft. long ?
Involution
The product of equal factors is a power.
The process of finding powers is involution.
The product of two equal factors is the second power, or
square, of the equal factor.
The product of three equal factors is the third power, or cube,
of the factor.
4^ = 4 X 4 is 4 to the second power, or the square of 4.
2* = 2 X 2 X 2 is 2 to the third power, or the cube of 2.
3^ = 3x3x3x3 is 3 to the fourth power, or the fourth power of 4.
EXAMPLES
Find the powers :
1. 5» 3. 1^ 5. (2^)2
7. 9^
2. 1.1' 4. 2o' 6. 2*
a .15«
Evolution
One of the eqiuil factors of a power is a root.
One of two equal factors of a number is the square root.
One of three equal factors of a number is the cube root of it.
The square root of 16 = 4. The cube root of 27 = 3,
The radical sign (^) placed before a number indicates that
its root is to be found. The radical sign alone before a number
indicates the square root.
Thus, Vd = 3 is read, the square root of 9 = 3.
60 VOCATIONAL MATHEMATICS
A small figure placed in the opening of the radical sign is
called the index of the root, and shows what root is to be
taken.
Thus, -v^ = 2 is read, the cube root of 8 is 2.
Square Root
The square of a number composed of tens and units is equal
to the square of the tens, plus twice the product of the tens by
the units, plus the square of the units.
ten^ + 2 X te7is X units + U7iits^
Example. — What is the square root of 1225?
12^25( 30 + 5^35 Separating
Tens% 302 = 900 into periods of
325 two figures
325 each, by a
2 X tens = 2 x 30 =60
2 X tens + units = 2 x 30 + 6 = 65
checkmark ('),
beginning at units, we have 12 '25. Since there are two periods in tlie
power, there must be two figures in the root, tens and units.
The greatest square of even tens contained in 1225 is 900, and its
square root is 30 (3 tens). Subtracting the square of the tens, 900, the
remainder consists of 2 x (tens x units) + units.
325, therefore, is composed of two factors, units being one of them,
and 2 x tens — units being the other. But the greater part of this factor
is 2 X tens (2 x 30 = 60). By trial we divide 325 by 60 to find the other
factor (units), which is 5, if correct. Completing the factor, we have
2 X tens + units =; 66, which, multiplied by the other factor, 5, gives 325.
Therefore the square root is 30 + 5 = 35.
The area of every square surface is the product of two equal
factors, length, and width.
Finding the square root of a number, therefore, is equivalent
to finding the length of one side of a square surface, its area
being given.
1. Length X Width —Area
2. Area -r- Length = Width
3. Area -^• Width = Jjength ,
REVIEW OP ARITHMETIC 61
Short Mkthod
Example. — Find the square root of 1300.0990.
13'0«.(W9<5 (8rtJ4 BeginninK at the decimal point, separate the
9 number into i)eri X i C
In this formula D equals the diameter and C the circum-
ference,
4 4
Example. — What is the area of a circle whose radius is
3ft.?
4
^=irx9 ^=^^ = ir9 = 28.278q. ft. Ans.
i
Example. — What is the area of a circle whose circumfer-
ence is 10 ft. ?
2) = -^ A=Idx-C
3.1416 2 2
- X -^~ X 1 X 10 = -^^ = 7.1 sq. ft. Ans.
2 3.141« 2 3.1410
Area of a Ring. — On examining a flat iron ring it is clear that
the area of one side of the ring may be found by subtracting
the area of the inside circle from the area of the outside circle.
Let D = outside diameter
d = inside diameter
A — area of outside circle
a = area of inside circle
(1) ^ = :^=.7854Z>*
4
64
VOCATIONAL MATHEMATICS
(2)
a = ^ = .7854(^2
(3) A-a = ^~^
Let B = area of circular ring = A — a
^ = — - ^ = - (Z>2 - d2w .7854 (Z>2 _ (^2)
4 4 4 ^
Example. — If the outside diameter of a flat ring is 9" and
the inside diameter 7", what is the area of one side of the
ring?
^=.7854 (D2 _(f2)
B = .7854 (81 - 49) = .7854 x 32 = 25.1328 sq. in. Ans.
Angles
Mechanics make two uses of angles : (1) to measure a cir-
cular movement, and (2) to measure a difference in direction.
A circle contains 360°, and the angles at the center of the
circle contain as many degrees as their corresponding arcs on
the circumference.
Angle FOE has as many degrees as arc PE.
A right angle is measured by a quarter
of the circumference of the circle, which ^
is 90°.
The angle AOG is a right angle.
The angle AC, made with half the cir-
cumference of the circle, is a straight angle, and the two right
angles, AOG and GOC, which it contains, are supplementary
to each other. When the sum of two angles is equal to 90°,
they are said to be complementary angles, and one is the com-
plement of the other. When the sum of two angles equals 180°,
they are supplementary angles, and one is said to be the supple-
ment of the other.
MENSURATION
65
The number of degrees in an angle may be measured by a
protractor. The distance around a semicircular protractor is
Protractor — Semicircular, having 180°.
divided into 180 parts, each division measuring a degree. It
is used by placing the center of the protractor on the vertex
and the base of the protractor on the side of the angle to be
3G0° Protractor.
A a is a circle divided into degrees.
measured. Where the other side of the angle cuts the circular
piece, the size of the angle may be read.
EXAMPLES
1. What is the area of a circular sheet of iron 8" in
diameter ?
2. What is the distance around the edge of a pulley 6" in
diameter ?
66 VOCATIONAL MATHEMATICS
3. What is the area of one side of a flat iron ring 14" inside
diameter and 18" outside diameter ?
4. A driving wheel of a locomotive has a wheel center of
56" in diameter ; if the tires are 3" thick, what is the circum-
ference of the wheel when finished ?
5. Find the area of a section of an iron pipe which has an
inside diameter of 17" and an outside diameter of 17|".
6. Name the complements of angles of 30°, 45°, 65°, 70°,
85°.
7. Name the supplements of angles of 55°, 140°, 69°, 98° 44',
81° 19'.
8. What is the diameter of a wheel that is 12' 6" in circum-
ference ?
Triangles
A triangle is a plane figure bounded by three straight lines.
Triangles are classified according to the relative lengths of
their sides and the size of their angles.
A triangle having equal sides is called equilateral. One
having two sides equal is isosceles. A triangle having no
sides equal is called scalene.
If the angles of a triangle are equal, the triangle is equi-
angular.
If one of the angles of a triangle is a right angle, the tri-
angle is a right triangle. In a right triangle the side opposite
the right angle is called the hypotenuse and is the longest side.
The other two sides of the right triangle are the legs, and are
at right angles to each other.
Equilateral Isosceles Scalene Right
MENSURATION
67
Kinds of Triangles
Right Triangles
In a right triangle the
square of the hypotenuse
equals the sum of the
squares of the other two
sides or legs.
If the length of the hy-
potenuse and one leg of a
right triangle is known,
the other side may be
found by squaring the
hypotenuse and squaring
the leg, and extracting the
square root of their dif-
ference.
Example. — If the hypotenuse of a right angle triangle is
30" and the base is 18", what is the altitude?
J
302 = 30 X 30 = 900
182 = 18 X 18 = 324
900 _ 324 = 576
\/670=24". Ana.
Areas of Triangles
The area of a triangle may be found when the length of the
thi'ee sides is given by adding the three sides together, divid-
ing by 2, and subtracting from this sum each side separately.
Multiply the four results together and find the square root of
their product.
68 VOCATIONAL MATHEMATICS
Example. — What is the area of a triangle whose sides
measure 15, 16, and 17 inches, respectively ?
16
16
17 V24 X 9 X 8x7 = \/l2096
2^ V12096 = 109.98 sq. in. Ans.
24 - 15 = 9
24 - 16 = 8
24-17 = 7
Area of a Triangle = ^ Base x Altitude
Example. — What is the area of a triangle whose base is
17" and altitude 10"?
1 ^
^ = - X 17 X ;^ = 85 sq. in. Ans.
EXAMPLES
1. A ladder 17 ft. long standing on level ground reached to
a window 12 ft. from the ground. If it is assumed that the
wall is perpendicular, how far is the foot of the ladder from
the base of the wall ?
2. Find the area of a triangular sheet of metal having the
base 81" and the height measured from the opposite angle 56".
3. Find the length of the hypotenuse of a right triangle
with equal legs and having an area of 280 sq. in.
4. Find the length of a side of a right triangle with equal
legs and an area of 72 sq. in.
5. Find the hypotenuse of a right triangle with a base of
8" and the altitude of 7".
6. What is the area of a triangle whose sides measure 12,
19, and 21 inches ?
7. What is the altitude of an isosceles triangle having sides
8 ft. long and a base 6 ft. long ?
MENSURATION 69
Quadrilaterals
Four-sided plane figures are called quadrilaterals. Among
them are the trapezoid, trajyezium, rectangle, rhotubus, atid rhom-
boid.
SquA&B Rkctamolb Rhomboid Rhombus
Trafbzium Trapezoid Parallelugbam
Kinds of Quadrilaterals
A rectangle is a quadrilateral which has its opposite sides
parallel and its angles right angles. Its area equals the prod-
uct of its base and altitude.
A= ha
A trapezoid is a quadrilateral having only two sides parallel.
Its area is equal to the product of the altitude by one half the
sum of the bases.
A=(b-\-c)xl^a ^
In this formula c = length of longest side J ^ V
b = length of shortest side / i \
a = altitude *
A trapezium is a four-sided figure with no two sides parallel.
The area of a trapezium is found by dividing the trapezium
into triangles by means of a diagonal. Then the area may* be
found if the diagonal and perpendicular heights of the triangles
are known.
70
VOCATIONAL MATHEMATICS
Example. — In the trapezium ABCD if the diagonal is 43'
and the perpendiculars 11' and 17', respectively, what is the
area of the trapezium ?
43 X V-
43 X
H^
¥- = ^f
: 236i- sq. ft., area of ABC
: 36oj sq. ft.,areaof ^Z>(7
602 sq. ft., total area
Ans.
To find the areas of irregular figures,
draw the longest diagonal and upon this
diagonal drop perpendiculars from the ver-
tices of the figure. These perpendiculars will form trapezoids
and right triangles whose areas may be determined by the pre-
ceding rules. The sum of the areas of the separate figures will
give the area of the whole irregular figure.
Polygons
A plane figure bounded by straight lines is a polygon. A
polygon which has equal sides and equal angles is a regular
polygon.
The apothem of a regular polygon is the line drawn from the
center of the polygon perpen-
dicular to one of the sides.
A five-sided polygon is a
pentagon.
A six-sided polygon is a
hexagon.
An eight-sided polygon is an octagon.
The shortest distance between the flats of a regular hexagon
is the perpendicular distance between two opposite sides, and is
equal to the diameter of the inscribed circle. The diameter of
the -circumscribed circle is the long diameter of a regular hexa-
gon.
The perimeter of a polygon is the sum of its sides.
Pentagon
Hexagon
MENSURATION 71
The area of a regular polygon equals one half the product of
the apothem and the perimeter.
Formula .1 = ^(1/'
In this formula P = perimeter
a = apothem
Ellipse
Only the approximate circumference of an ellipse can be ob-
tained.
The circumference of an ellipse equals one half the product of
the sum of two diameters and tt.
If c/i = major diameter
c/j = minor diameter
C = circumference
then c = 4±^7r
The area of an ellipse is equal to one fourth the product of
the major and minor diameters by tt.
If A = area
dj = major diameter
dj = minor diameter
then A = TT^
4
EXAMPLES
1. Find the area of a trapezium if the diagonal is 93' and
the perpendiculars are 19' and 33'.
2. What is the area of a trapezoid whose parallel sides are
18 ft. and 12 ft., and the altitude 8 ft. ?
3. What is the distance around an ellipse whose major
diameter is 14" and minor diameter 8" ?
72 VOCATIONAL MATHEMATICS
4. In the map of a country a district is found to have two
of its boundaries approximately parallel and equal to 276 and
216 miles. If the breadth is 100 miles, what is its area ?
5. If the greater and lesser diameters of an elliptical man-
hole door are 2' 9" and 2' 6", what is its area ?
6. Find the area of a trapezium if the diagonal is 78" and
the perpendiculars 18" and 27".
7. The greater diameter of an elliptical funnel is 4 ft. 6 in.,
and the lesser diameter is 4 ft. (a) What is its area?
(b) How many square feet of iron will it contain if its height
is 16 ft., allowing 4" for the seams ?
8. What is the area of a pentagon, whose apothem is 4i"
and whose side is 5" ?
Volumes
The volume of a rectangular-shaped bar is found by multi-
plying the area of the base by the length. If the area is in
square inches, the length must be in inches.
The volume of a cube is equal to the cube of an edge.
The contents or volume of a cyliJidrical solid is equal to the
product of the area of the base by the height.
If S = contents or capacity of cylinder
R = radius of base
H= height of cylinder
TT = 3.1416+ or ^/ (approx.)
S = ttB^H
Example. — Find the contents of a cylindrical tank whose
inside diameter is 14" and height 6'.
H=6' = 72"
/S' = V X 7 X 7 X 72 = 11,088 cu. in.
MENSURATION
73
The Pyramid
The volume of a puramid eciuals one
third of the product of the area of the base
and the altitude.
V=\ba
The volume of a frustum of a pyramid
equals the product of one third the alti-
tude and the sum of the two bases and the
square root of the product of the bases.
The surface of a regular pyramid is equal to the product of
the perimeter of the bases and one half the slant height.
S=Px^sh
The Cone
A cone is a solid generated by a right triangle revolving on
one of its legs as an axis.
The altitude of the cone is the perpendicular distance from
the base to the apex.
The volume of a cone equals the product of the area of the
base and one third of the altitude.
or F= .2618 D'H
Example. — What is the volume of a cone 1^" ^
in diameter and 4" high ?
Area of base = .7864 x |
7.0686
= 1.7671 sq. in.
F= ^6181)2^
= .2618 X I X 4 = 2.3662 cu. In. Ans,
The lateral surface of a cone equals one half the product of
the perimeter of the base by the slant height.
74
VOCATIONAL MATHEMATICS
Example. — What is the surface of a cone having a slant
height of 36 in., and a diameter of 14 in. ?
3^
14 X V = 44"
^i-^^ = 702 sq. in. Ans.
Frustum of a Cone
The frustum of a cone is the part of a cone included between
the base and a plane or upper base which is parallel to the
lower base.
The volume of a frustum of a cone equals the product of one
third of the altitude and the sum of the two bases and the
square root of their product.
When H = altitude
B^ = upper base
B = lower base
V=iH{B-hB'-\- VBB')
The lateral surface of a frustum of a cone equals one half the
product of the slant height and the sum of the perimeters
of the bases.
The Sphere
The volume of a sphere is equal to
3
where B is the radius.
The surface of a sphere is equal to
The Barrel
To find the cubical contents of a barrel, (1) multiply the
square of the largest diameter by 2, (2) add to this product
MENSURATION 75
the square of the head diameter, and (3) multiply this sum by
the length of the barrel and that product by .2618.
Example. — Find the cubical contents of a barrel whose
largest diameter is 21" and head diameter 18", and whose
length is 33".
V= 1{D'^ X 2) + d'^] X L X .2618
31)798
.2618
21* =
= 441 X
2 = 882
182 =
= 324
.324
1206
33
3618
3618
10419.11 cu. in
10419.11
39798 281
= 46.10 gal. Ans.
Similar Figures
Similar figures are figures that have exactly the same shape.
The areas of similar figures have the same ratio as the
squares of their corresponding dimensions.
Example. — If two boilers are 15' and 20' in length, what is
the ratio of their surfaces ?
J^ = I, ratio of lengths
— = — , ratio of surfaces
42 16
One boiler is ^5 as large as the other. Ans.
The volumes of similar figures are to each other as the cubes
of their corresponding dimensions.
Example. — If two iron balls have 8" and 12" diameters,
respectively, what is the ratio of their volumes ?
^ = f , ratio of diameters
= /y, ratio of their volumes. Am.
One ball weighs ^ as much as the other.
76 VOCATIONAL MATHEMATICS
EXAMPLES
1. Find the volume of a rectangular iron bar 8'' by 10"
and 4' long.
2. Find the weight of a rectangular steel bar 31" x 49" x 3"
thick, if the metal weighs .28 lb. per cubic inch.
3. The radius of the small end of a bucket is 4 in. Water
stands in the bucket to a depth of 9 in., and the radius of the
surface of the water is 6 in. (a) Find the volume of the
water in cubic inches. (6) Find the volume of the water in
gallons if a cubic foot contains 7.48 gal.
4. What is the volume of a steel cone 2i" in diameter and
6" high ?
5. Find the contents of a barrel whose largest diameter is
22", head diameter 18", and height 35".
6. What is the volume of a sphere 8" in diameter ?
7. What is the volume of a pyramid with a square base,
4" on a side and 11" high ?
8. What is the surface of a steel cone with a 6" diameter
and 14" slant height ?
9. Find the surface of a pyramid with a perimeter of 18"
and a slant height of 11".
10. Find the volume of a cask whose height is 3^' and the
greatest radius 16" and the least radius 12", respectively.
11. What is the weight of a cast-iron cylinder 2.75" in
diameter and 12j" long, if cast iron weighs 450 lb. per
cu. ft. ?
12. How many gallons of water will a round tank hold
which is 4 ft. in diameter at the top, 5 ft. in diameter at the
bottom, and 8 ft. deep ? (231 cu. in. = 1 gal.)
13. What is the volume of a cylindrical ring having an
outside diameter of 6 J", an inside diameter of 5^", and a
height of 3|" ? What is its outside area ?
MENSURATION 77
14. A sphere has a circumference of 8.2467". (a) What is
its area? (6) What is its volume ?
15. If it is desired to make a conical oil can with a base
3.5" in diameter to contain \ pint, what must the approximate
height be ?
16. What is the area of one side of a flat ring if the inside
diameter is 2^" and the outside diameter 4J" ?
17. There are two balls of the same material with diameters
4" and 1", respectively. If the smaller one weighs 3 lb., how
much does the larger one weigh ?
18. If the inside diameter of a ring must be 5 in., what
must the outside diameter be if the area of the ring is 6.9
sq. in. ?
19. How much less paint will it take to paint a wooden
ball 4" in diameter than one 10" in diameter ?
20. What is the weight of a brass ball 3J" in diameter if
brass weighs .303 lb. per cubic inch ?
21. A cube is 19" on its edge, (a) Find its total area.
(6) Its volume.
22. If the area of a J" pipe is .049 sq. in., what will be the
diameter of a pipe having 8 times the area?
23. What is the weight of a cast-iron cylinder 2.75' in
diameter and 12}' long, if the cast iron weighs 450 lb. per
cubic foot ?
24. A conical funnel has an inside diameter of 19.25" at the
base and is 43" high inside, (a) Find its total area. (6) Find
its cubical contents.
25. If a bar 3" in diameter weighs 24.03 lb. per foot of
length, what must be the weight per foot of a bar 3" square of
the same material?
CHAPTER III
Reading a Blue Print
Every skilled worker in wood or metal must know how to
read a " blue print," which is the name given to working plans
Screw Table Frame
Simple Blue Prints or Working Drawings
and drawings with white lines upon a blue background. The
blue print is the language which the architect uses to the
78
WORKING DRAWINGS 79
builder, the machinist to the pattern maker, the engineer to the
foreman of construction, and the designer to the workman,
Through following the directions of the blue print the carpenter,
metal worker, and mechanic are able to produce the object
wanted by the employer and his designer or draftsman.
Two views are usually necessary in every working drawing,
one the plan or top view obtained by looking down upon the
object, and the other the elevation or front view. When an object
is very complicated, a third view, called an end or profile view,
is shown.
All the information, such as dimensions, etc., necessary to construct
whatever is represented by the blue print, must be supplied on the draw-
ing. If the blue print represents a machine it is necessary to show all the
parts of the machine put together in their proper places. This is called
an assembly drawing. Then there must be a drawing for each part of
the machine, giving information as to the size, shape, and number of
the pieces. Then if there are interior sections, these must be represented
in section drawings.
Drawing to Scale
As it is impossible to draw most objects full size on paper, it
is necessary to make the drawings proportionately smaller.
This is done by making all the dimensions of the drawing a
certain fraction of the true dimensions of the object. A draw-
ing made in this way is called drawn to scale.
Triangular Scale
The dimensions on the drawing are designated the actual
size of the object — not of the drawing. If a drawing were
made of an iron bolt 25 inches long, it would be inconvenient
to represent the actual size of the bolt, and the drawing might
be made half or ipiarter the size of the bolt, but the length
would read on the drawing 25 inches.
80 VOCATIONAL MATHEMATICS
In making a drawing "to scale," it becomes very tedious to
be obliged to calculate all the small dimensions. In order to
obviate this work a triangular scale is used. It is a ruler with
the different scales marked on it. By practice the student will
be able to use the scale with as much ease as the ordinary
ruler.
QUESTIONS AND EXAMPLES
1. Tell what is the scale and the length of the drawing of
each of the following :
a. An object 14" long drawn half size.
h. An object 26" long drawn quarter size.
c. An object 34" long drawn one third size.
d. An object 41" long drawn one twelfth size.
2. If a drawing made to the scale of |" = 1 ft. is reduced \ in
size, what will the new scale be ?
3. The scale of a drawing is made \ size. If it is doubled,
how many inches to the foot will the new scale be ?
4. On the yy scale, how many feet are there in ^d>
inches ?
5. On the y scale, how many feet are there in 26 inches ?
6. On the ^" scale, how many feet are there in 21 inches ?
7. If the drawing of a bolt is made -J- size and the length of
the drawing is 8^', what will it measure if made to scale 3"
= lft.?
8. What will be the dimensions of the drawing of a machine
shop 582' by 195' if it is made to a scale of yV = 1 ft. ?
Arithmetic and Blue Prints. — Mechanics are obliged to read from blue
prints. In order to verify the necessary dimensions in the detail of the
work, it becomes necessary to do more or less addition and subtraction of
the given dimensions. This involves ability to add, and subtract mixed
numbers and fractions.
METHODS OF SOLVING EXAMPLES 81
Methods of Solving Examples
Every mechanical problem or operation has two distinct
sides : the collecting of data and the solving of the problem.
The first part, the collecting of data, demands a knowledge
of the materials and conditions under which the problem is
given, and calls for considerable judgment as to the necessary
accurateness of the work.
There are three ways by which a problem may be solved :
1. Exact method.
2. Rule of thumb method, by the use of a two-foot rule or
a slide scale.
3. By means of tables.
The exact method of solving a problem in arithmetic is the
one usually taught in school and is the method obtained by
analysis. Every one should be able to solve a problem by the
exact method.
The rule of thumb method. — Many of the problems that arise
in industrial life have been met before and very careful judg-
ment has been exercised in solving them. As the result of
this experience and the tendency to abbreviate and devise
shorter methods that give sufficiently accurate results, we find
many rule of thumb methods used by the mechanics in daily
life. The exact method would involve considerable time and
the use of pencil and paper, whereas in cases that are not too
complicateid the two-foot rule or the slide scale will give a
quick and accurate result.
In solving problems involving the addition and subtraction of fractions,
by the rule of thumb, use the two-foot rule or steel scale to carry on the
computation. To illustrate : if we desire to add \ and ^, place the
thumb on \ division, then slide (move) the thumb along a division cor-
responding to ^, and then read the number of divisions passed over by
the thumb. In this case the result is ^. For fractions involving ^, ^^,
^, and j^ use the steel scale. The majority of machinists, carpenters,
etc., use this method of sliding the thumb over the rule, in adding and
subtracting inches and fractions of inches.
82 VOCATIONAL MATHEMATICS
The use of tables. — In the commercial and industrial world
the tendency is to do a thing in the quickest and the most
economical way. To illustrate : hand labor is more costly
than machine work, so wherever possible, machine work is
substituted for hand labor. The same condition applies to
the calculations that are used in the shop. The methods of
performing calculations are the most economical — that is,
the quickest and most accurate — that the ordinary mechanic
is able to perform. Since a great many of the problems in
calculation that arise in the daily experience of the mechanic
are about standardized pieces of metal and repeat themselves
often, it is not necessary to wot-k them each time if results are
kept on file when they are once solved. This tiling is done
by means of tables that are made from these problems.
See pages 105, 117, and 121, for tables used in this book.
PART II — MATHEMATICS FOR CARPENTERING
AND BUILDING
CHAPTER IV
MEASURING LUMBER
The carpenter or builder is often required to give an estimate
of the cost of the work to be done for his prospe(;tive custom-
ers. People who contemplate building have several estimates
submitted to them by different builders and generally give the
work to the lowest bidder. An architect usually draws plans
of the building and from these plans the contractor or builder
makes his estii^ate of the cost. In doing repairing or cabinet-
making the carpenter makes his own plans and estimates. In
order to make a proper estimate of the cost, one must know the
market price of materials, the cost of labor, the amount of
material needed, and the length of time required to do the work.
Preparation of Wood for Building Purposes
In winter the forest trees are cut and in the spring the logs floated
down the rivers to sawmills, where they are sawed into boards of different
thickness. To square the log, four slabs are first sawed off. After these
slabs are off, the remainder is sawed into boards.
As soon as the boards or planks are sawed from the logs, they are piled
on prepared foundations in the open air to season. Each layer is separated
from the one above by a crosspiece, called a strap, in order to allow free cir-
culation of air about each board to dry it quickly and evenly. If lumber
were to be piled up without the strips, one board upon another, the ends
of the pile would dry and the center would rot. This seasoning or drying
out of the sap usually lasts several months. *' Air dried " lumber is used
for most building purposes except in buildings or places where there is
a warm, dry atmosphere.
88
84 VOCATIONAL MATHEMATICS
Wood that is to be subject to a warm atmosphere has to be artificially
dried. This artificially dried or kiln-dried lumber has to be dried to a point
in excess of that of the atmosphere in which it is to be placed after being
removed from the kiln. This process of drying must be done gradually
and evenly or the boards may warp and then be unmarketable.
Definitions
Board Measure. — A board one inch or less in thickness is said
to have as many board feet as there are square feet in its surface.
If it is more than one inch thick, the number of board feet is
found by multiplying the number of square feet in its surface
by its thickness measured in inches and fractions of an inch.
The number of board feet = length {in feet) x width (in feet) x thick-
ness (in inches).
Board measure is used for plank measure. A plank 2" thick, 10" wide,
and 15' long, contains twice as many square feet (board measure) as
a board 1" thick of the same width and length.
To measure a board that tapers, the width is taken at the
middle, where it is one half the sum of the widths of the
ends.
Boards are sold at a certain price per hundred (C) or per
thousand (M) board feet.
The term lumber is applied to pieces not more than four
inches thick ; timber to pieces more than four inches thick ; but
a large amount taken together often goes by the general name
of lumber. A piece of lumber less than an inch and a half
thick is called a board and a piece from one inch and a half to
four inches thick a plank.
Rough Stock is lumber the surface of which has not been
dressed or planed.
The standard lengths of pieces of lumber are 10, 12, 14, 16,
18 feet, etc.
In measuring and marking large lots of lumber in which
there are a number of pieces containing a fraction of a foot, in case
of one half foot or more, 1 is added, and in case of less than one
half foot, it is disregarded. This is especially true with boards.
CARPENTERING AND BUILDING 85
A board 1" x 4" x 16' contains 6 J feet (board measure). Two
boards would be marked 6 ft., and every third board would be marked
(i ft. So two boards may be exactly the same length and one marked 6 ft.
and the other ft. In the purchasing of a single board there might be a
small undercharge or overcliarge, but in large lots the average would be
struck.
EXAMPLES
1. How many board feet in a board 1 in. thick, 15 in. wide,
and 15 ft. long ?
2. How many board feet of 2-inch planking will it take to
make a walk 3 feet wide and 4 feet long ?
3. A plank 19' long, 3" thick, 10" wide at one end and 12"
wide at the other, contains how many board feet ?
4. Find the cost of 7 2-inch planks 12 ft. long, 16 in. wide
at one end, and 12 in. at the other, at S 0.08 a board foot.
5. At $ 12 per M, what will be the cost of 2-inch plank for
a 3 ft. 6 in. sidewalk on the street sides of a rectangular corner
lot 56 ft. by 106 ft. 6 in. ?
Quick Method for Measuring Boards
To measure boards 1" thick, multiply length in feet by
width in inches and divide by 12, and the result will be the
board measure in feet.
For boards IJ" thick, add one quarter of the quotient to the
result as above.
For boards 1|" thick, add one half of the quotient to the
r(!sult as found above.
For plank 2" thick, divide by 6 instead of 12.
For plank 3" thick, divide by 4 instead of 12.
For plank 4" thick, divide by 3 instead of 12.
For timber 6" thick, divide by 2 instead of 12.
EXAMPLES
1. Find the number of board feet in 10 planks, 3" thick,
12" wide, 14' long.
86
VOCATIONAL MATHEMATICS
2. Find the number of board feet in 4 timbers, 8" thick,
10" wide, 17' long.
3. Find the number of board feet in 18 joists, 2" thick,
4" wide, 14' long.
4. Find the number of board feet in 16 beams, 10" thick,
12" wide, 11' long.
5. Find the number of board feet in 112 boards, J" thick,
.8" wide, 14' long.
Note. — Ordinarily fractions of a foot less than one half are omitted,
but when the fraction is one half or larger it is reckoned as a foot. This
is sufficiently accurate for all practical purposes.
Board measure of one lineal foot of timber may be found
from the following table :
Contents (Board Measure) of One Lineal Foot of Timber
5r.
T
HICKNE88 IN InOIIES
2
3
4
5
6
7
8
9
10
11
12
13
14
18
3.
4.5
6.
7.5
9.
10.5
12.
13.5
15.
16.5
18
19.5
21.
17
2.83
4.25
5.66
7.08
8.5
9.92
11.33
12.75
14.17
15.58
17
18.42
19.83
16
2.67
4.
5.33
6.67
8.
9.33
10.67
12.
13.33
14.67
16
17.3
18.66
15
2.5
3.75
5.
6.25
7.5
8.75
10.
11.25
12.5
13.75
15
16.25
17.6
14
2.33
3.5
4.67
5.83
1 .
8.17
9.33
10.5
11.67
12.83
14
15.17
16. ()6
13
2.17
3.25
4.33
5.42
6.5
7.58
8.67
9.75
10.83
11.92
13
14.08
12
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12
11
1.83
2.75
3.67
4.58
5.5
6.42
7.33
8.25
9.17
10.08
10
1.67
2.5
3.33
4.17
5.
5.88
6.67
7.5
8.33
9
1.5
2.25
3.
3.75
4.5
5.25
6.
6.75
8
1.33
2.
2.67
3.33
4.
4.67
5.33
7
1.17
1.75
2.33
2.92
3.5
4.08
6
1.
1.5
2.
2.5
3.
5
.83
1.25
1.67
2.08
4
.67
1.
1.33
3
.5
.75
2
.33
CARPENTERING AND BUILDING
87
To ascertain the contents of a piece of timber, find in the
table the contents of one foot and multiply by the length in
feet of the piece.
EXAMPLES
By means of the above table find the board measure of the
following :
1. One timber 8" x 9", 14' long.
2. One timber 8" X 11", 13' long.
3. One timber 9" x 10", 11' long.
4. One timber 8" x 10", 16' long.
5. One timber 7" x 9", 11' long.
6. Two planks 2" x 3", 10' long.
7. Two planks 4" x 4", 13' long.
8. Two timbers 10" x 11", 15' long.
9. Two timbers 10" x 18", 14' long.
10. Two timbers 8" X 16", 13' long.
The weight per cubic foot of different woods can be readily
seen from the following table :
Weight of One Cubic Foot of Timber
Wood
Weight
I'ER CiT. Ft.
Wood
Wbiuut
PER Cv. Ft.
White Pine . .
Georgia Pine
Hemlock . . .
Cypress . . .
Spruce . . .
White Oak . .
Red Oak . . .
Maple. . . .
28 1b.
38 1b.
24 1b.
331b.
28 1b.
481b.
46 1b.
42 lb.
Whitewood ....
Ash
Hickory . • . . .
Chestnut
Cedar
Birch .......
Ebony
Boxwood
30 lb.
45 1b.
48 1b.
36 1b.
39 1b.
411b.
76 1b.
70 lb.
Lumber is bought and sold in the log by cubic measure.
Lumber used for framing buildings and for building bridges,
docks, and ships is also sold by the cubic foot.
88 VOCATIONAL MATHEMATICS
The rule that is most extensively used for computing the
contents in board feet of a log is as follows :
Rule. — SvMract 4 inches from the diameter of the log at the
small end, square one quarter of the remainder, and multiply the
result by the length of the log in feet.
EXAMPLES
Find the board feet of lumber in the following logs :
Diameter in inches ^- 2. 3. 4. 5. 6. 7. 8. 9.
at small end: 12 13 14 15 16 17 18 20 24
Length in feet : 10 12 14 16 18 20 16 20 24
CHAPTER Y
CONSTRUCTION
Excavations. — After the })lans for a building are drawn by
the architect and the work given to the contractor on a bid,
usually the lowest, the work of excavating begins. In esti-
mating excavations the cubic yard, or 27 cubic feet, is used.
EXAMPLES
1. What will it cost to excavate a cellar that is 32' x 28'
and 5' deep at 34 cents per cubic yai-d ?
2. What will the cost be of excavating a lot 112' by 58' and
averaging 12' deep at $1.65 per yard ?
3. In excavating a tunnel 374,166 cubic feet of earth were
removed. If the length of the tunnel was 492 ft. and th«
width 39 ft., what was the height of the tunnel ?
4. How many cubic yards of earth must be removed to
build a cellar for a house when the measurements inside the
wall are 28' long and 16' wide, the wall being 1'.8" thick
and 8' deep, with 2' of the wall above the ground level?
5. In making a bid on some excavating a contractor notes
that the excavation is in the shape of a rectangle 8' deep, 11'
wide at the top and bottom, and 483' long. What will it cost
him to excavate it at 29 cents per cubic yard ? What must
his bid be to make 10 % profit ?
Frame and Roof
After the excavation is finished and the foundation laid, the
construction of the building itself is begun. On the top of the
foundation a large timber called a sill is placed. The timbers
90
VOCATIONAL MATHEMATICS
running at right angles to the front sill are called side sills ;
The sills are joined at the corners by a half-lap joint and held
together by spikes.
a. Outside studding
b. Rafters
c. Plates
il. Ceiling joists
de. Second floor joists
def. First floor joists
g. Girder or cross sill
h. Sills
i. Sheathing
j. Partition studs
k. Partition heads
l. Piers 1
Then the building has its walls framed by placing corner
posts of 4" by 6" on the four corners. Between these corner
posts there are placed smaller timbers called studding, 2" by
4", 16" apart. Later the laths, 4' long, are nailed to this stud-
ding. The upright timbers are mortised into the sills at the
bottom. When these uprights are all in position a timber
called a plate is placed on the top of them and they are spiked
together.
On the top of the plate is placed the roof. The principal
timbers of the roof are the rafters. Different roofs have a dif-
1 If made of brick or stone, '' shones " or "supports " if made of wood.
CARPENTERING AND BUILDING
91
fereiit pitcli or slope — that is, form
different angles with the plate. To
get the desired pitch the carpenter
uses the steel square.
When we speak of the pitch of a roof we
mean the slope or slant of the roof. A
roof with oi>€ half pitch means that the
height of the ridkn|wt i iHi ' iiviPjer but a few more should be added for
waste, etc. Verify answer by use of table on page 98.
Another method is : Since it takes approximately 3 bunches
(250 shingles each) laid 5" to the weather to cover a square,
multiplying the number of squares in the roof by 3 will give
the number of bunches required.
Example. — If a roof contains 50 squares, how many
shingles will the roof need to cover it ?
50 X 3 = 150 bunches = 37,500 shingles, Ans.
EXAMPLES
1. How much will it cost for shingles to shingle a roof
50 ft. by 40 ft., if 1000 shingles are allowed for 125 square
feet and the shingles cost $ 1. 18 per bundle ?
2. Find the cost of shingling a roof 38 ft. by 74 ft, 4" to the
weather, if the shingles cost $ 1.47 a bundle, and a pound and a
half of cuti nails at $.0G a i)ound are used with each bundle.
100 VOCATIONAL MATHEMATICS
3. How many shingles would be needed for a roof having
four sides, two in the shape of a trapezoid with bases 30 ft. by
10 ft., and altitude 15 ft., and two (front and back) in the
shape of a triangle with base 20 ft. and altitude 15 ft.?
(1000 shingles will cover 120 sq. ft.)
Slate Roofing
Slates make a good-looking and durable roof. They are put
on with nails similarly to shingles. Estimates for slate roof-
ing are made on 100 sq. ft. of the roof.
The following are typical data for building a slate roof :
A square of No. 10 x 20 Monson slate costs about 38.35,
Two pounds of galvanized nails cost $.16 per pound.
Labor, $S per square.
Tar paper, at 2| cents per pound, 1^ lb. per square yard.
EXAMPLES
Using the above data, give the cost of making slate roofs
for the following :
1. What is the cost of laying a square of slate ?
2. What is the cost of laying slate on a roof 112' by 44' ?
3. What is the cost of laying slate on a roof 156' by 64' ?
4. What is the cost of laying slate on a roof 118' by 52' ?
5. What is the cost of laying slate on a roof 284' by 78' ?
Weight of Roof Coverings
Every builder should know the weight of different roof
coverings in order to make a roof strong enough to support its
covering. Besides, allowance must be made for ice and snow.
The weight in pounds of roof covering is usually expressed in
pounds per 100 sq. ft., or square of roof.
CARPENTERING AND BUILDING 101
Approximate WEUiiiT ov Roof Covkrinos
Na.mk Wkioiit pkr 1()0 bq. pt.
Sheathing, Pine 1 inch thick, yellow northern . . . 300
Sheathing, Pine 1 inch thick, yellow southern . 400
Spruce, 1 inch thick 200
Sheathing, Chestnut or Maple, I inch thick .... 400
Sheathing, Ash, Hickory, or Oak, 1 inch thick . . . 500
Sheet iron, ^\ inch thick 800
Sheet iron, ^5 inch thick, and laths 500
Shingles, Pine 200
Slates, \ inch thick 1)00
Skylights (Glas-s, f\ to I inch thick) 250-700
Sheet Lead 500-800
Cast Iron Plates, | inch thick 1500
Copper 80-126
Felt and Asphalt 100
Felt and Gravel 800-1000
Iron, Corrugated 100-2375
Iron, Galvanized Flat . . • 100-350
Lath and Plaster 900-1000
Thatch 650
Tin 70-125
Tiles, Flat 1500-2000
Tiles (Grooves and Fillets) 700-1000
Tiles, Pan 1000
Tiles, with Mortar 2000-3000
Zinc 100-200
EXAMPLES
What is the approximate weight of:
1. 800 sq. ft. of \ inch slate ?
2. 1645 sq. ft. of flat tiles ?
3. 23.32 sq. ft. of tiles with mortar?
4. 3184 sq. ft. of lath and plaster ?
5. 2789 sq. ft. of sheathing pine 1" thick ?
6. 1841 sq. ft. of pine shingles ?
102 VOCATIONAL MATHEMATICS
7. 1794 sq. ft. of thatch?
8. 3279 sq. ft. of felt and gravel ?
9. 1973 sq. ft. of asphalt ?
10. 1589 sq. "ft. of skylight glass i in. thick ?
Clapboards
Clapboards are used to cover the outside walls of frame
buildings. Most clapboards are 4 ft. long and 6 in. wide. They
are sold in bundles of twenty-five. Three bundles will cover
100 square feet if they are laid 4" to the weather.
To find the number of clapboards required to cover a given
area, find the area in square feet and divide by IJ. One
quarter the area should be deducted to allow for openings.
EXAMPLES
1. How many clapboards will be required to cover an area
of 40 ft. by 30 ft. ?
2. How many clapboards will be necessary to cover an area
of 38' by 42' if 56 sq. ft. are allowed for doors and windows ?
3. How many clapboards will a barn 60 ft. by 50 ft. require
if 10% is allowed for openings and the distance from founda-
tion to the plate is 17 ft. and the gable 10 ft. high ?
•; * ; \ ; : y r \ ' Flooring
Mcrst floors in hou'seft, are made of oak, maple, birch, or pine,
i^his flooring is grooved so that the boards fit closely together
without cracks between them.
The accompanying figure shows the ends of ^ ^ j— 1
pieces of matched flooring. Matched boards are ^ ^— '— '
also used for ceilings and walls. In estimating for matched flooring
enough stock must be added to make up for what is cut away from the
width in matching. This amount varies from |" to |" on each board ac-
cording to its size. Some is also wasted in squaring ends, cutting up, and
CARPENTERING AND BUILDING 103
fittinjj to exact lengths. A common floor is made of unmatched boards
and is usually used ;us an under floor. Not more than \ is allowed for
wastr.
ExAMi'LK. — A room is 12 ft. square and is to have a floor
laid of matched boards IV' wide; one third is to be added for
waste. What is the number of scjuare feet in the floor ? What
is the number of board feet required for laying the floor?
12 X 12 = 144 sq. ft. = area. 144 x J = 48
144 Ans. 144
192 board measure for
matched floor.
192 Ans.
EXAMPLES
1. How much J in. matched flooring 3" wide will be re-
quired to lay a floor 16 ft. by 18 ft. ? One fourth more is al-
lowed for matching and 3 % for squaring ends.
2. How much hard pine matched flooring J" thick and 1^"
wide will be required for a floor 13' 0" X 14' 10" ? Allow J for
matching and add 4 % for waste.
3. An office floor is 10' 6" wide at one end and 9' 6" wide at
the other (trapezoid) and 11' 7" long. What will the material
cost for a maple floor |" thick and 1^" wide at $60 per M,
if 4 sq. ft. are allowed for waste ?
4. How many square feet of sheathing are required for the
outside, including the top, of a freight car 34' long, 8' wide,
and 7^' high, if 12^% is allowed for waste and overhang?
5. In a room 50' long and 20' wide flooring is to be laid;
how many feet (board measure) will be required if the stock
is J" X 3" and \ allowance for waste is made?
Stairs
The perpendicular distance between two floors of a building
is called the rise of a flight of stairs. The width of all the
steps is called the run. The perpendicular distance between
steps is called the width of riser. Nosing is the slight projec-
104
VOCATIONAL MATHEMATICS
tion on the front of each
step. The board on
each step is the tread.
To find the number of
stairs necessary to reach
from one floor to an-
other: Measure the rise
first. Divide this by 8
inches/ which is the most
comfortable riser for
stairs. The run should
be 8i inches or more
to allow for a tread of
9| inches with a nosing of li inches.
Example. — How many steps will be required, and what
will be the riser, if the distance between floors is 118 inches ?
118 -- 8 = 14f or 15 steps.
118 -=- 15 = 7^1 inches each riser. Ans.
Stairs
EXAMPLES
1. How many steps will be required, and what will be the
riser, (a) if the distance between floors is 8' ? (b) If the dis-
tance is 9 feet ? 104
2. How many steps will be required, and what will be the
riser, (a) if the distance between floors is 12' ? (b) If the dis-
tance is 8' 8" ?
Carpenters' Table of Wages
To find the amount due at any rate from 30 cents to 55 cents
per hour, look at the column containing the rate per hour and
the column opposite that containing the number of hours, and
the amount will be shown. Time and a half is counted for
overtime on regular working days, and double time for Sundays
and holidays.
1 Other distances may also be used if required.
CARPENTERING AND BUILDING
105
X
X
i
s
04
H
91
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$0 30
♦0 15
♦0 224
$0 80
♦0 824
♦Olfii
♦0 24g
♦0 324
♦0 4:)
♦0 224
♦0 83}
♦0 45
1...
SO
80
45
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824
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45
46
674
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60
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65
974
1 80
46
90
1 85
1 1-0
s...
80
90
1 85
1 80
824
974
1464
1 95
45
1 36
2 024
2 70
4...
80
1 20
1 80
2 40
824
1 80
1 95
2 60
46
1 80
2 70
3 ro
6...
80
1 50
2 25
8 00
824
1 624
2 48}
3 26
46
2 26
8 874
4 60
6...
80
1 SO
2 70
8 60
824
1 95
2 924
3 90
46
2 70
4 06
6 40
7...
80
2 10
8 15
4 20
824
2 274
8 41J
466
46
8 16
4 724
6 30
8...
80
2 40
8 60
4 80
324
2 60
3 90
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3 60
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7 20
9...
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2 70
4 05
540
324
2 924
4 38}
6 86
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4 05
6 074
8 10
10 ..
80
8 00
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6 00
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46
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$0 47i
$0 28}
$0 85|
♦0 474
10 50
♦0 26
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♦0 274
♦0 41}
♦0 56
1...
474
474
714
95
50
60
76
1 00
55
56
824
1 10
2...
47*
95
1 424
1 90
50
1 00
1 50
2 00
55
1 10
1 65
2 20
8...
474
1424
2 13}
2 85
50
1 50
2 26
8 00
56
1 65
2 474
8 80
4...
474
1 90
2 85
8 80
50
2 00
8 00
4 00
55
2 20
8 80
4 40
6...
474
2 874
8 564
4 75
50
2 50
8 75
500
Ki
2 76
4 124
5 50
6...
474
2 85
4 274
5 70
50
8 00
4 60
6 00
56
8 80
4 96
6 60
7...
474
8 824
4 98}
6 65
50
8 50
5 25
7 W
55
3 85
6 774
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8...
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880
6 70
7 60
50
4 00
6 00
8 00
56
440
6 60
8 80
9...
474
4 274
6 41}
8 55
50
4 50
6 75
9 00
65
4 96
7 424
9 90
10...
474
4 75
7 124
9 50
50
5 00
7 50
10 00
Ki
5 50
8 26
11 00
EXAMPLEvS
1. Find the amount due a carpenter who has worked 8 hours
regular time and 2 hours overtime at 55 cents per hour.
2. A carpenter worked on Sunday from 8 to 11 o'clock. If his
regular wages are 45 cents per hour, how much will he receive ?
3. A carpenter received 55 cents an hour. How much money
is due him for working July 4th from 8-12 a.m. and 1-4.30 p.m. ?
4. A carpenter works six days in the week ; every morning
from 7.30 to 12 m. ; three afternoons from 1 to 4.30 p.m. ; two
afternoons from 1 to 5.30 ; and one from 1 until 6 p.m. What
will he receive for his week's wage at 50 cents per hour ?
106 VOCATIONAL MATHEMATICS
Painting
Paint, which is composed of dry coloring matter or pigment mixed
with oil, drier, etc., is applied to the surface of wood by means of a
brush to preserve the wood. The paint must be composed of materials
which will render it impervious to water, or rain would wash it from the
exterior of houses. It should thoroughly conceal the surface of whatever
it is applied to. The unit of painting is one square yard. In painting
wooden houses two coats are usually applied.
It is often estimated that one pound of paint will cover 4 sq. yd. for
the first coat and 6 sq. yd. for the second coat. Some allowance is made
for openings ; usually about one half of the area of openings is deducted,
for considerable paint is used in painting around them.
Table
1 gallon of paint will cover on concrete . . 300 to 375 superficial feet
1 gallon of paint will cover on stone or brick
work 190 to 225 superficial feet
1 gallon of paint will cover on wood . . . . 375 to 525 superficial feet
1 gallon of paint will cover on well-painted sur-
face or iron 600 superficial feet
1 gallon of tar will cover on first coat ... 90 superficial feet
1 gallon of tar will cover on second coat . . 160 superficial feet
EXAMPLES
1. How many gallons of paint will it take to paint a fence
6' high and 50' long, if one gallon of paint is required for
every 350 sq. ft. ?
2. What will the cost be of varnishing a floor 22' long and
16' wide, if it takes a pint of varnish for every four square
yards of flooring and the varnish costs $ 2.65 per gal. ?
3. What will it cost to paint a ceiling 36' by 29' at 21 cents
per square yard ?
4. What will be the cost of painting a house which is 52'
long, 31' wide, 21' high, if it takes one gallon of paint to cover
300 sq. ft. and the paint costs $1.65 per gallon ?
PART III — SHEET AND ROD METAL WORK
CHAPTER VII
Blanking or Cutting Dies
Many kinds of receptacles are pressed or cut from difTerent
kinds of sheet metal, such as copper, tin, and aluminum. Cans,
pots, parts of metal boxes, and all sorts of metal novelties are
punched out of sheet metal most economically by the punch
and die operated by the hydraulic press. So skillfully can die
makers produce dies and punches to cut out articles that thou-
sands of everyday necessities in the household are made by
this method. Parts of watches, parts of automobiles, and parts
of machinery are punched out. Some presses that operate the
Blanking" or "Cutting" Diks
Cutting dies consist of an upper " male" die or "punch," and the lower
or "female" die. Circumstances determine whether any or how much
"shear" shall be given to the cutting eecial nnu-hinery. In some rases it is prefer-
able to fasten the steel dies in cast-iron chucks or die-beds by means of keys
107
108 VOCATIONAL MATHEMATICS
or screws. This applies more particularly
to small dies. Cutting dies may be made
to tit any size and style of press. For
cutting thick iron, steel, brass, and other
heavy metals, both the die and punch
should be hard and provided with strip-
pers.
Punch and Die without Stripper Punch and Die with Strippkr
punch and die are run by foot power, but those most generally
used are run by electricity.
A blanking or cutting die is a metal plate or disk having
an opening in the center used in a punching machine or press
which is supplied with ample power and which also supports
the metal from which pieces are punched. Dies are made in
almost any size and shape for cutting flat blanks in tin, iron,
steel, aluminum, brass, copper, zinc, silver, paper, leather, etc.
Holes are punched in thick sheets of metal or in heavy
plates by means of great pressure exerted by a hydraulic press.
This pressure, in pounds, is usually about 60,000 times the
area (expressed in square inches) of the surface cut out, or in
other words about 60,000 lb. per square inch.
EXAMPLES
1. How much pressure will be necessary for a hydraulic
press to exert on a sheet of boiler plate ^y thick, if it is de-
sired to punch holes ^" in diameter ?
2. How much pressure will it be necessary to use in order
to punch holes f" in diameter out of \" boiler plate ?
3. How much pressure will it be necessary to use in order to
punch jV holes out of j^" boiler plate?
SHEET AND HOD METAL WORK
109
Combination Dies
Double dies for blanking; and perforating are extensively
used in the manufacture of washers, key blanks, electrical
instruments, hardware, etc. The blanking and perforating
punches act simultaneously. At a single stroke of the ])ress
Double" Dies for Blankino and Perforating
Tlie perforating and blanking punches act simultaneously in such a manner
that the lioles are punched first, whereupon the strip or slieet being fed for-
ward, the blank is cut out around them and the
holes for the next blank perforate*! at the same /"
stroke. In this manner a blank is < ' " '
forated at every stroke of tlie press.
The same principle may be extended so as to
punch a number of perforated blanks at a time.
ate— ^ \ ^^^
is completely per- f } O ) O ( O )
Perfobatino Dibs with Stripper Plates
110 VOCATIONAL MATHEMATICS
one punch perforates the holes and the other punch cuts out
the metal between the holes punched at the previous stroke.
These dies are usually so arranged that the finished article is
automatically pushed out from the dies by the action of the
springs. An expert operator can punch many thousands of
pieces in a day.
Example. — How much metal will be required for 2000
blacking box covers, 6" in diameter, and f deep ?
6" + f " + f" = 7i", diameter of one cover.
.7854 X 7.5 X 7.5 = 44.1787 sq. in., area.
44.1787 X 2000 = 88,357.4 sq. in. = 613.6 sq. ft. Ans.
EXAMPLES
1. How large must the blank be cut for a pail cover that is
to be (a) 5" in diameter and |" deep ? (6) 7" in diameter and
ly deep ? (c) 6" in diameter and V' deep ? (d) 8" in diameter
and li" deep ?
2. How large must the blank be cut for the pail bottom in
each of the above examples (a, b, c, and d) ?
The blank must be i" larger in diameter than the diameter of the part
in order to allow |" all around for forming into the sides.
3. How large must the piece be to form the sides of the
pail in each of the above examples ? Pail (a) to be 6" high,
(6) 9" high, (c) 8" high, and (d) 10'' high.
Allow ^" in height, and for the lock seam allow f" in circumference.
4. How much metal will be necessary for 850 complete pails
in example (a) above ?
5. How much metal will be necessary for 1500 complete
pails in example (6) above ?
6. How much metal will be necessary for 840 complete pails
in example (c) above ?
7. How much metal will be necessary for 1000 complete
pails in example (d) above ?
SHEET AND ROD METAL WORK
111
Mah I >i.\«; Dies
8. How many square feet of sheet copper will be required
to make a rectangular tank 7' long, 3' wide, IV deep, allowing
10 S for WiKK (iAHOK
Dimensions ofSiKos in Decinml Parts of an Inch
BlRMINO-
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22
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.0258
.024
.153
.028125
23
24
.0201
.022
.0230
.022
.151
.025
24 >
26
.0179
.02
.0204
.020
.148
.021875
25
26
.01594
.018
.0181
.018
146
.01875
26
27
.014195
.016
.0173
.0164
.143
.0171875
27
28
.012641
.014
.0162
.0149
.139
.015625
28
29
.011257
.013
.0150
.01.36
.134
.0140625
29
30
.010025
.012
.0140
.0124
.127
.0125
30
31
.008928
.01
.0132
.0116
.120
.0109375
31
32
.00795
.009
.0128
.0108
.115
.01015625
32
33
.00708
.008
.0118
.0100
.112
.009375
33
34
.006304
.007
.0104
.0092
.110
.00859375
34
35
.005614
.005
.0095
.0084
.108
.0078125
34
36
.005
.004
.0090
.0076
.106
.00703125
36
37
.004453
....
....
.0068
.103
.006640625
37
38
.003965
.0060
.101
.00625
38
39
.003531
....
.0052
.099
39
40
.003144
.0048
.097
40
The American or B. «k 8. frauge l« the standard for sheet brass, copper, or German silver,
and for wire of the same material.
The Binnin^'ham (range is use85
6.875
.
9
5-32
.15(525
6.350
6.25
^ 8% per cent.
10
•M)4
.14062
5.715
5.(525
1 10 per cent.
11
1-8
.125
5.080
5.00
12
7-(54
.10937
4.445
4.375
13
3-32
.09374
3.810
3.75
14
5-64
.07812
3.175
3.125
15
9-128
.07031
2.857
2.812
1(5
1-16
.0625
2.540
2..'>0
17
9-1(50
.05625
2.28(5
2.25
18
1-20
.a5
2.032
2.
\\)
7-160
.04375
1.778
1.75
20
;i-«o
.0.375
1.524
l.-W
21
11-.'V20
.03437
1.397
1.375
22
1-32
.03125
1.270
1.2.5
23
i>-320
.02812
1.143
1.125
24
1-40
.025
1016
25
7-320
.02187
1.389
!875
26
3-160
.01875
.762
.75
27
\um
.01718
.698
.(587
28
1-64
.01562
.635
.623
29
9-(i40
.01406
.571
.5(52
30
1-80
.0125
.508
.5
31
l-iy^O
.010t)3
.694
.437
.S2
13-1280
.01015
.413
.40(5
33
3-320
.00937
.381
.375
. 34
11-1280
.00859
..349
..343
35
5-640
.00781
.317
.312
.36
9-1280
.00703
.28.'>
.281
37
17-2560
.00f5(ht
.271
.2(55
38
1-160
.00625
.254
.25
122
VOCATIONAL MATHEMATICS
Weights and Arbas of Round, Square, and Hexagon Steel
Weight of one cubic inch = .2836 lb. Weight of one cubic foot = 490 lb.
c«
Abea
= DlAM.« X
.7854 1
Akea = Side2 X 1
Area = DiAM.2x. 866
c -
Hound
Sqtiare
Ilea'agon
Weight
Per Inch
Area
Square
Inches
Circum-
ference
Inches
Weight
Per
Inch
Area
Square
Inches
Weight
Per
Inch
Area
Square
Inches
1-32
1-16
S-S2
1-8
.0002
.0009
.0020
.0035
.0008
.0031
.0069
.0123
.0981
.1963
.2995
.3927
.0003
.0011
.0025
.0044
.0010
.0039
.0088
.0156
.0002
.0010
.0022
.0038
.0008
.0034
.0076
.0135
5-32
8-16
7-32
1-4
.0054
.0078
.0107
.0139
.0192
.0276
.0376
.0491
.4908
.5890
.6872
.7854
.0069
.0101
.0136
.0177
.0244
.0352
.0479
.0625
.0060
.0086
.0118
.0154
.0211
.0304
.0414
.0540
9-32
5-16
11-32
3-8
.0176
.0218
.0263
.0313
.0621
.0767
.0928
.1104
.8835
.9817
1.0799
1.1781
.0224
.0277
.0335
.0405
.0791
.0977
.1182
.1406
.0194
.0240
.0290
.0345
.0686
.0846
.1023
.1218
13-82
7-16
15-82
1-2
.0368
.0426
.0489
.0557
.1296
.1503
.1726
.1963
1.2762
1.3744
1.4726
1.5708
.0466
.0.543
.0623
.0709
.1651
.1914
.2197
.2500
.0405
.0470
.0540
.0614
.1428
.1658
.1903
.2161
17-82
9-16
19-32
5-8
.0629
.0705
.0785
.0870
.2217
.2485
.2769
.3068
1.6689
1.7671
1.8653
1.9635
.0800
.0897
.1036
.1108
.2822
.3164
.3526
.3906
.0693
.0777
.0866
.0959
.2444
.2743
.3053
.3383
21-32
11-16
23-32
3-4
.0959
.1053
.1151
.1253
.3382
.3712
.4057
.4418
2.0616
2.1598
2.2580
2.3562
.1221
.1340
.1465
.1622
.4307
.4727
.5166
.5625
.1058
.1161
.1270
.1382
.3730
.4093
.4474
.4871
25-32
13-16
27-32
7-8
.1359
.1470
.1586
.1705
.4794
.5185
.5591
.6013
2.4543
2.5525
2.6507
2.7489
.1732
.1872
.2019
.2171
.6103
.6602
.7119
.7656
.1499
! .1620
.1749
.1880
.5286
.5712
.6165
6631
29-32
15-16
31-32
1
.1829
.1958
.2090
.2227
.6450
.6903
.7371
.7854
2.8470
2.9452
3.0434
3.1416
.2329
,2492
.2661
.2836
.8213
.8789
.9384
1.0000
.2015
.2159
.2305
.2456
.7112
.7612
.8127
.8643
1 1-16
1 1-8
1 3-16
1 1-4
.2515
.2819
.3141
.3480
.8866
.9940
1.1075
1.2272
3.3379
3.5343
3.7306
3.9270
.3201
.3589
.4142
.4431
1.1289
1.2656
1.4102
1.5625
.2773
.3109
.3464
.3838
.9776
1.0973
1.2212
1.3531
1 5-16
1 3-8
1 7-16
1 1-2
.3837
.4211
.4603
.5012
1.3530
1.4849
1.6230
1.7671
4.1233
4.3197
4.5160
4.7124
.4885
.5362
.5860
.6487
1.7227
1.8906
2.0664
2.2500
.4231
.4643
.5076
.5526
1.4919
1.6373
1.7898
1.9485
1 9-16
1 5-8
1 11-16
1 3-4
.5438
.5882
.6343
.6821
1.9175
2.0739
2.2365
2.4053
4.9087
5.1051
5.3014
5.4978
.6930
.7489
.8076
.8685
2.4414
2.6406
2.8477
3.0625
.5996
.6480
.6994
.7521
2.1143
2.2847
2.4662
2.6522
1 13-16
1 7-8
1 15-16
2
.7317
.7831
.8361
.8910
2.5802
2.7612
2.9483
3.1416
5.6941
5.8905
6.0868
6.2832
.9316
.9970
1.0646
1.1342
3.2852
3.5156
3.7539
4.0000
.8069
.8635
.9220
.9825
2.8450
3.0446
3.2509
3.4573
SHEET AND ROD METAL WORK
123
Wkiohts and Areas of Round, Square, and Hexagon Stebl. — Con-
tinued
Weight of one cubic inch » .2S8C lb. Weight of one cubic foot - 490 lb.
8«
Area
= DiAM.* X
.7854
Area = Side* x 1
Arka»Diaii.*x .866
IE
Hound 1
Square
, Ilevagon
55
Weight
Per Inch
Are*
Square
Indies
CHrcum-
ference
Inches
Weight
Per
Inch
Area
Square
I iiuhes
Weight
Per
Inch
Area
Square
Inches
S 1-1«
S 1-8
S S-16
11-4
.9475
1.0058
1.0658
1.1276
3.3410
3.5466
3.7583
3.9761
6.4795
6.6759
6.8722
7.0686
1.2064
1.2806
1.3570
1.4357
4.2539
4.5156
4.7852
5.0625
1.0448
1.1091
1.1753
1.2434
3.6840
3.9106
4.1440
4.3892
S 6-16
t S-8
S 7-16
% 1-S
1.1911
1.2564
1.3234
1.3921
4.2000
4.4301
4.6664
4.9087
7.2649
7.4613
7.6575
7.8540
1.5165
1.6569
1.6849
1.7724
5.3477
5.6406
5.9414
6.2500
1.3135
1.3854
1.4593
1.5351
4.6312
4.8849
5.1454
5.4126
S6-8
S S-4
S 7-8
S
1.5348
1.6845
1.8411
2.0046
5.4119
5.9396
6.4918
7.0636
8.2467
8.6394
9.0321
9.4248
1.9541
2.1446
2.3441
2.5548
6.8906
7.5625
8.2656
9.0000
1.6924
1.8574
2.0304
2.2105
5.9674
6.5493
7.1590
7.7941
S 1-8
S 1-4
S S-8
Sl-1
2.1752
2.3527
2.5371
2.7286
7.6699
8.2958
8.9462
9.6211
9.8175
10.2102
10.6029
10.9956
2.7719
2.9954
3.2303
3.4740
9.7656
10.5625
11.3906
12.2500
2.3986
2.5918
2.7977
3.0083
8.4573
9.1387
9.8646
10.6089
S 8-8
SS-4
S 7-8
4
2.9269
3.1323
3.3446
3.5638
10.3206
11.0447
11.7932
12.5664
11.3883
11.7810
12.1737
12.5664
3.7265
3.9880
4.2582
4.5374
13.1407
14.0625
15.0156
16.0000
3.2275
3.4539
3.6880
3.9298
11.3798
12.1785
13.0035
13.8292
4 1-8
4 1-4
4 S-8
4 1-S
3.7900
4.0232
4.2634
4.5105
13.3640
14.1863
15.0332
15.9043
12.9591
13.3518
13.7445
14.1372
4.8254
5.1223
5.4280
5.7426
17.0156
18.0625
19.1406
20.2500
4.1792
4.4364
4.7011
4.9736
14.7359
15.6424
16.5761
17.5569
4 6-8
4 S-4
4 7-8
8
4.7645
5.0255
5.2935
5.5685
16.8002
17.7205
18.6655
19.6350
14.5299
14.9226
15.3153
15.7080
6.0662
6.6276
6.7397
7.0897
21.3906
22.5625
23.7656
25.0000
5.2538
5.5416
5.8371
6.1403
18.5249
19.5397
20.5816
21.6503
8 1-8
6 1-4
6 S-8
6 1-S
5.8504
6.1392
6.4351
6.7379
20.6290
21.6475
22.6905
23.7583
16.1007
16.4934
16.8861
17.2788
7.4496
7.8164
8.1930
8.5786
26.2656
27.5624
28.8906
30.2500
6.4511
6.7697
7.0959
7.4298
22.7456
23.8696
25.0198
26.1971
6 8-8
6 S-4
6 7-8
6
7.0476
7.3643
7.6880
8.0186
24.8505
25.9672
27.1085
28.2743 1
17.6715
18.0642
18.4569
18.8496
8.9729
9.3762
9.7883
10.2192
31.6406
.33.0625
34.5156
36.0000
7.7713
8.1214
8.4774
8.8420
27.4013
28.6361
29.8913
31.1765
6 1-4
6 1-S
6S-4
7
8.7007
9.4107
10.1485
10.9142
30.6796
33.1831
35.7847
38.4845
19.6350
20.4204
21.2058
21.9912
11.0877
11.9817
12.9211
13.8960
39.0625
42.2500
45.5625
49.0000
9.5943
10.3673
11.1908
12.0351
33.8291
36.5547
39.4584
42.4354
Tl-1
8
12.5291
14.2553
44.1786
50.2655
23.5620
25.1328
15.9520
18.1497
56.2500
64.0000
13.8158
16.7192
48.7142
66.3169
Pupils should practice the use of tables in order to obtain accurate results quickly.
Additional tables like these may be obulned from such standanl handbooks as Kents.
Multiply above weights by .998 for wrought iron, .918 for ca.st iron. ^^Y.V^\ for cast brass,
.1209 for copper, and 1.1748 for phos. bronze.
124 VOCATIONAL MATHEMATICS
EXAMPLES
By means of the table of weights and areas of round, square,
and hexagonal steel, solve the following problems :
1. Find the circumference of a steel bar (a) -^^" in thickness ;
(6) ^" in thickness ; (c) \l" in thickness ; {d) 111" in thick-
ness; (e) 3|" in thickness.
2. Find the area of a steel bar (a) |^" in diameter ; (6) 1^^"
in diameter; (c) 3|" in diameter; (d) 4^' in diameter; (e) 5 J"
in diameter.
3. Find the weight per inch of a steel bar (a) ||'' in
diameter; (h) ly^' in diameter ; (c) 2^^" in diameter.
4. Find the area of a square bar (a) f" per side ; (h) If"
per side ; (c) 5^ per side ; (d) 3|'' per side.
5. Find the weight per inch of a square bar (a) j^' in
thickness ; (6) f|" in thickness ; (c) Iff" in thickness.
6. Find the area of a hexagonal bar (a) i|" in thickness ;
(6) 1\^" in thickness ; (c) 3J" in thickness.
7. Find the weight per in. of a hexagonal bar (a) |f" in
thickness ; (6) |^'' in thickness ; (c) 2^^ in thickness.
8. Find the weight of a round bar of steel (a) 7' long -f^"_
in diameter ; (&) 11' long lyV in diameter ; (c) 16' long 2^%"
in diameter ; {d) IV long 2J" in diameter ; (e) 13' long 4|" in
diameter.
9. Find the weight of a round bar of steel (a) 8' long 2|"
in diameter; (h) 14' long 1|" in diameter; (c) 11' long lif" in
diameter.
10. Find the weight of a hexagonal bar of steel (a) 8' long
3f" in diameter; {h) T long 2J" in diameter; (c) 9' long Ifi"
in diameter.
11. What is the weight of a square bar of wrought iron 16'
long 4^" in thickness ?
SHEET AND ROD METAL WORK 125
12. What is the weight of a hexagonal bar of wrought iron
18' long 2yV' i" diameter ?
13. What is the weight of a round bar of cast iron 14' long
Si" in diameter?
14. What is the weight of a square bar of cast iron 13' long
1|" in thickness ?
15. What is the weight of a square bar of cast brass 15' long
l|f " in thickness ?
16. What is the weight of a hexagonal bar of cast brass 8'
long If" in diameter ?
17. What is the weight of a hexagonal bar of copper 7' long
2yV' in diameter ?
la What is the weight of a round bar of copper 6' long l^y
in diameter?
PART IV— BOLTS, SCREWS, AND RIVETS
CHAPTER VIII
BOLTS
The most common forms of fastenings are bolts, screws,
rivets, pins, and nails. These are turned out in large numbers,
usually by feeding long lengths of iron or steel rod into auto-
matic machines. In order to make these fasteners of the right
size one has to be familiar with the problems that are con-
nected with them.
The common bolt is made in many different sizes and is
usually held in place by a nut screwed on the end. There are
many different kinds of bolts for the different uses to which
they are put.
Rough Bolts
The small diameter of a rough bolt head, that is, the dis-.
tance across the flats, is 1^ times the diameter of the bolt, plus
1 inch ; or, it may be stated as follows : The diameter of a
rough bolt head = li^Z) plus ^ inch, D being diameter of the
bolt.
Sometimes bolts or nuts are made from round stock and cut
either square or hexagon. In such cases it is necessary to hud
the proper diameter to which the stock must
be turned in order that it may be milled to
size. Let A represent the distance across the
flats on the head of a flat bolt, and B the
diameter of the round stock required to make
a square-head bolt. Square Head
126
BOLTS, SCREWS, AND RIVETS
127
Wlieii the dimension A is given and it is necessary to turn a
piece that will mill down to this size and leave full corners,
multiply A by 1.414. The product is the desired size B,
B = y/WTA^ Why ?
B = y/¥A^ = AVi = 1.414 A
When ^1, the distance across the flats, is given, and it is
necessary to turn a piece that will mill down to this size and
leave full corners, multiply A by 1.155. The product is the
desired size {B) of the hexagon head.
Let
X
2
(ir =(fr-(fy -^^
4 ■■ 16 4
4 52 = //-' -f 4 X2
3 J?^ = 4 X2
4
B^
3
X2
B=2X^2_XV3=1.165X
V3 3
Hexagonal Head
Example, — To what diameter should a piece of stock be
turned in order that it may have full corners when milled
down six-sided to \\" across the flats?
li" X 1.155 = 1.7325", diameter of the blank. Ans.
EXAMPLES
1. What is the size of stock required to make the following
bolts having hexagonal heads ? (a) Body .}" diam. (6) Body
J" diam. (c) Body I" diam. (d) Body V diam.
2. What is the size of stock required to make the following
bolts with square heads? (a) Body \" diam. (6) Body \"
diam. (c) Body |" diam. {d) Body \" diam.
128
VOCATIONAL MATHEMATICS
EXAMPLES
1. To what diameter should a piece of stock be turned so
that it ma}^ have full corners when milled down square to 1|"
across the fiats ?
2. To what diameter should a piece of stock be turned in
order that it may have full corners when milled down six-sided
to 1|" across the flats ?
Sizes of Standard Hexagon Head Bolts
Size of
DiAM. OK
Bolt
Thickness
OF Heads
Hexagon or
UlSTANCE
Across
Corners
Threads
Tap Drill
In.
In.
Across the
Flats
In.
In.
Per Inch
In.
i
i
1%
20
t\
t\
H
u
n
18
C
M
H
¥i
16
N
tV
M
11
If
14
S
tV
i
1
13
if
T?
n
M
V.
12
If
H
ii*.
1/^
11
II
f
n
i-A
10
f
II
IrV
m
9
H
if
n
ii
8
II
M
m
2/2
7
fi
1
2
h%
7
1/t
h\
2A
n
Ui
ifV
2|
2|
6 .
m
1/3.
2/.
2-it
5^
111
If
2|
•3A
5
H
HI
m
3M
5
If
2
1^
H
3t
H
111
^
H
n
4t^.
41
1.962 = If ^
^
iH
H
4i
4
2.176 = 2/^
2f
2i
H
4|f
4
2.426 = 2j.V
3
2A
^
H
3^
2.629 = 2f ^
Notice that size of hexagon is equal to diameter of bolt + \ diameter of
bolt + I of an inch, and also that thickness of head is \ of hexagon in
every case. The thickness of nut is equal to the diameter of bolt.
BOLTS. SCREWS, AKD UIVKTS
129
3. To what diameter should a piece of stock be turned so
that it may have full corners when milled down sc^uare to 2J"
across the flats ?
4. To what diameter should a piece of stock be turned so
that it may have full corners when milled down six-sided to
2 J" across the flats ?
Size across Corners of Squares
SiXK OF SqUARK, In.
DiAUONAL
Size of Sqitabr, In.
Diagonal
}
.177
1.4141
^.
.205
1.590 .
\
.354
1.768
^ .
.442
1.945
i
.530
2.121
A
.619
2.298
i
.707
2.476
A
.706
2.652
f
.884
2
2.828
n
.972
^
3.005
i
1.061
2\
3.182
H
1.149
^
3.535
1
1.237
2f
3.889
H
1.326
3
4.243
EXAMPLES
Use the table to obtain the size of bolts.
1. Find the distance across the corners of a hexagon head
bolt (a) with a diameter of 1 J" ; (6) with a diameter of 2J" ;
(c) with a diameter of 1\".
2. Find the diagonal distance across a square bolt with size
ofsquare(a)-fV'; (/>) A"-
3. (a) If a bolt^heading machine has the following daily
output, 23:^0, 2060, 1950, 2420, 2310, 2030, what is the average
130 VOCATIONAL MATHEMATICS
daily output ? (6) What would be the daily wage at 13 cents
per 100 bolts ? (c) The weekly wage ?
4. A blacksmith requires six pieces of steel of the following
lengths: If", 2|", 2j\'\ ^\\", 1\^", 2". How long a piece of
steel will be necessary to make them, if ^' is allowed for each
in finishing ?
5. A machinist has to make five bolts from the same size
bar. One bolt is to be 1-^' over all, another 2^", another 2-^%",
another 3//', and the last 2f".. How much stock will he need
if he allows ^" for each cut-off ?
Rivets
One of the simplest and most efficient metal fastenings which
has been extensively used is the rivet. It resembles the bolt,
but it can be removed only by chipping off the head, while the
bolt can be taken off by removing the nut. Rivets, like bolts
and nails, are quickly turned out by the thousand with the
aid of automatic machines.
EXAMPLES
1. What must be the length of a bolt under the head, to go
through 9^y thickness of plank and allow IJ" outside for tak-
ing a nut ?
2. Whgit must be the length of a bolt under the head, to go
through 7^\" thickness of plank and allow If" outside for tak-
ing a nut ?
3. (a) How many rivets can be made in a bolt machine,
from a round iron rod 6' long, if each rivet requires 2 J" of bar ?
(6) If the bar weighs | lb. per ft., how many rivets weighing
\ lb. apiece can be made from it ? (c) How much waste will
there be in each case ?
4. What is the total thickness of three plates riveted to-
gether, each plate i^" thick ? What would be the total thick-
ness of five such plates ?
BOLTS, SCREWS, AND RIVETS 131
5. What is the thickness of a steel plate that is only | the
thickness of a -j^" plate ?
6. A blacksmith and his helper made 192 bolt dogs in 18J^
hours. They received 6.} cents apiece for them, (a) How much
did they both receive per hour? (b) If the blacksmith re-
ceived G65 % of the money, how much did each receive per
hour?
7. How many feet of round iron weighing 2.67 lb. per foot
will be required to make 87 rivets weighing 2^ lb. apiece, not
counting waste ? Give answer in feet and decimals of foot.
8. How many rivets weighing 7.J ounces each can be made
from 15' of round iron weighing 1.5 lb. per foot ? How much
waste will there be ? ,
9. If the drawing of an armor bolt is made J size and the
length of the drawing measures 7J", what will it measure if
made to the scale of 3'' = 1' ? What is the actual length of the
bolt?
10. If the drawing of an armor bolt is made \ size and the
length of the drawing measures 9|", what will it measure if
made to the scale of 4" = 1'? What is the actual length of
the bolt?
11. The over-all length of a threaded bolt is 7f", the thick-
ness of the head is I", and the other end of the bolt is threaded
for a distance of 2^" ; what is the length of the shank between
the under side of the head and the threaded part of the bolt ?
Nails
W^ooden objects are often held together by means of nails.
There are two kinds of nails: cut and wire. The wire nails
are more commonly used, as they penetrate the wood without
splitting it as the cut nails do. They have different kinds of
heads, according to the use for which they are intended.
132
VOCATIONAL MATHEMATICS
The origin of the common terms " sixpenny," " tenpenny," etc., as
applied to nails, though not generally known, is involved in no mystery.
Nails have been made a certain number of pounds to the thousand for
many years and are still reckoned in that way in England, a tenpenny
behig a thousand nails to ten pounds, a sixpenny a thousand nails to six
pounds, a twentypenny weighing twenty pounds to the thousand ; and in
ordering buyers call for the three-pound, six-pound, or ten-pound variety,
etc., until, by the Englishmen's abbreviation of "pun" for "pound,"
the abbreviation has been made to stand for penny, instead of pound, as
originally intended.
Length and Number of Cut Nails to the Pound
SiZK
in
d
1
K
■<
S
■<
s
^
M
<
H
f
|in.
800
f
lin.
500
2d
1 in.
800
1100
1000
376
3d
li in.
480
720
760
224
4d
11 in.
288
623
368
180
398
5d
If in.
200
410
130
6d
2 in.
1H8
95
84
268
224
126
96
7d
2^ in.
124
74
64
188
98
82
8d
21 in.
88
62
48
146
128
75
68
9d
2| in.
70
53
36
130
110
65
lOd
3 in.
58
46
30
102
91
55
28
12d
3Jin.
44
42
24
76
71
40
16d
31 in.
35
38
20
62
54
27
22
20d
4 in.
23
33
16
54
40
14^
30d
4^ in.
18
20
33
m
40d
5 in.
14
27
^
60d
6^ in.
10
8
60d
6 in.
6^ in.
7 in.
8 in.
8
6
BOLTS, SCREWS, AND RIVETS
133
EXAMPLES
1. How many Jd nails are there in 3 pounds ?
2. How many Jd nails are there in 4 pounds ?
3. How many 2d nails are there in 2 pounds ?
4. How many 7d nails are there in 8 pounds ?
5. How many 16d nails are there in 1(> pounds ?
6. How long is (a) an 8d nail? (b) a 40d? (o) a 16d?
(d) a 9d ?
7. How long is (a) a 7d nail? (6) a 5d ? (c) a 12d ?
Tacks
Tacks are used to fasten thin pieces of material to wood.
They vary in form and size. The size is represented by a
number, as 16 oz. tacks, 24 oz. tacks ; a No. 1 tack is called a
one-ounce tack.
Ndmbbr of Tacks in a Pound
Title
Length
No. PER Lb.
Title
Lexotii
N(». PER Lll.
1 ounce
A inch
10,000
10 ounce
H inch
1,600
IJ ounce
y'j inch
10,606
12 ounce
1 inch
1,332
2 ounce
i inch
8,000
14 ounce
\i inch
1,143
2\ ounce
A i"^^
6,400
10 ounce
1 inch
1,000
3 ounce
f inch
5,3.32
18 ounce
U inch
888
4 ounce
^ inch
4,000
20 ounce
1 inch
800
6 ounce
A inch
2,060
22 ounce
ly^g inch
727
8 ounce
f inch
2,000
24 ounce
1\ inch
666
EXAMPLES
1. How many tacks are there in 8 lb. of 1 oz. or No. 1 tacks ?
2. How many tacks are there in 13 lb. of 2^ oz. or No. 2^
tacks?
3. How many tacks are there in 7 lb. of 8 oz. or No. 8 tacks ?
134 VOCATIONAL MATHEMATICS
4. How many tacks are there in 6 lb. of 12 oz. or No. 12
tacks ?
5. How many tacks are there in 17 lb. of 20 oz. or No. 20
tacks ?
Screws
Pieces of wood and of metal are often fastened together by
means of screws instead of by nails, especially if it is desirable
to separate the parts at any time. Screws are made of iron,
steel, or brass and have either a flat, fillister, or round head.
When it is desired to have the head of the screw flush with the
surface, the flat head type is used. Screws are made on auto-
matic screw machines into which wire of various sizes may be
fed. The machines are so constructed that they turn out
large numbers of screv/s all complete in a very short time.
i Mmmm i Q^ UiUMlliillig !^ iuiiimiiiiiirf i
^^^^^^MHH| ^Wfffff'^^B^^ ^^^^^^^^^^jg
flat head round head fillister head
Iron Machine Screws
Screw threads are divided into two classes : first, those used
for fastening ; and second, those used in large machines for
communicating motion. The screw threads used for communi-
cating motion with which the mechanic has to deal are pro-
duced by a cutting process in which the thread is formed from
the solid piece of stock by means of a single pointed cutting
tool in a lathe. Screws used for fastening are made by means
of taps and dies. The tap is a tool used to produce internal
threads, and the die is a tool used to cut the external threads.
The screw thread is applied in many ways, but the most com-
mon use is that of fastening together the various parts of
machines, etc.
We find the mechanic using many different forms of bolts
and screws to meet the needs of industries. In order to
BOLTS. SCREWS. AND RIVETS
135
specify a particular f?rade of bolt or screw it is necessary to
mention (a) shape or form of head, (b) pitch or number of
threads to the inch, (c) shape of thread, (rf) outline of body,
barrel or stem, (e) size of diameter, (/) direction of thread, as
right or left hand, (g) length, (/i) material, as brass, iron, etc.
There are four different-shaped threads in common use in
the United States : 1. The V thread ; 2. The U. S. standard ;
3. The Acme standard or worm ; 4. The square thread.
Sharp V Thread
*PITCH-
U. S. Standard Thread
AcMR Standard or Worm Thread
TTTTZy
P/TCH
Squabb Thread
1 Sometimes called modified square thread.
136
VOCATIONAL MATHEMATICS
The different screws are formed by cutting a spiral groove
around a cylinder. The projecting stock between the grooves
is called the land or thread. There may be any number of
threads ; to every groove there is an accompanying thread.
RiGHT-ILvxD, Single V-Thread, 8 Threads to the Inch
Nut
A screw that has one thread is called single-threaded ; one
having two threads, double-threaded; three threads,
triple-threaded; etc. The cutting of the spiral
groove or grooves is called cutting or threading a
screw. A nut is a piece of iron or steel with a
threaded hole which goes over a screw, and will
turn off and on the screw.
Lead of a Screw. — The distance that a thread advances in
one turn is called the lead of the screw. In a single-threaded
screw the lead is equal to the distance occupied by one thread ;
and when the nut has made one complete turn, it has advanced
one thread upon the screw. In a double-threaded screw the
nut advances two threads with each complete turn ; when the
lead is three threads, the nut advances three threads in one
turn. In general, the lead can be divided by any number of
threads, the advance of any one of these threads in one turn
being always equal to the lead.
One complete revolution of a single-threaded screw or the
lead of a screw, if the screw had twelve threads to the inch,
would be one twelfth of an inch.
BOLTS, SCREWS, AND RIVETS 137
Threads to an Inch. — By placing a scale upon a screw as in
the figures on page 130, the number of thread-windings or coils
in an inch can be counted. These windings or coils are called
threads, and the number of coils to an inch is called the number
of threads to an indi. The thread commences at the root or
bottom of the screw. To measure screws for the number of
threads per inch, the measurement must begin at the point
of the thread. Place the end of the scale in line with this
portion and then count the number of threads within the one-
inch line.
Pitch. — T?ie distance from the center of one thread to the cen-
ter of the next thread, measured in a line parallel to the axis, is
tJie pitch of the thread, or the thread-pitch. Divide 1" by the
number of threads to 1" and the quotient is the thread-pitch.
The threads to 1" and the thread-pitch are reciprocals of each
other.
In a single-threaded screw the pitch is equal to the lead. In
a double-threaded screw the pitch is half the lead; in a triple-
threaded screw the pitch is \ the lead; and so on. When the
thread inclines so as to be nearer the right hand at the under
side, or clockwise, it is a right-hand thread. When the under
side is toward the left, or counter clockwise, the thread is left-
handed. Again, when a right-hand screw turns in a direction
to move its upper side away from the eye, the thread appears
to move toward the right; while a left-hand thread moves
toward the left.
The term turns to an inch means the number of times a
screw must be turned around to advance one inch. If a screw
makes four turns in advancing one ioch, its lead is \, and it
has 4 turns to the inch. Divide one inch by the lead, and the
quotient is the number of turns that the screw makes in ad-
vancing one inch. If a screw does not advance exactly an inch
in a whole number of turns, or if it does not advance some
whole number of inches in one turn, it is said to have a frac-
tional thread. In any screw divide any number of turns by
138 VOCATIONAL MATHEMATICS
the number of inches occupied by these turns and the quotient
will be the turns to an inch. Thus, a screw that turns 96
times in 12.005 inches, turns 96 -^ 12.005, or 7.9967 turns in
one inch.
EXAMPLES
1. How many turns to the inch has (a) a screw that ad-
vances 25 inches in 200 turns ? (b) a single-threaded screw
of 9 threads to the inch ? (c) a double-threaded screw with
8 threads to the inch ? (d) a 'double-threaded screw with 6
threads to the inch ? (e) a single-threaded screw with 24
threads to the inch ?
2. AVhat is the lead of a single-threaded screw if it has
(a) five threads to the inch ? (6) eight threads to the inch ?
(c) fifteen threads to the inch ? (d) twenty-eight threads to
the inch ?
3. What is the lead of a double-threaded screw if it has
eight threads to the inch ?
4. What is the lead of a single-threaded screw if it has
twenty-two threads to the inch ?
5. What is the lead of a single-threaded screw if it has
twenty-four threads to the inch ?
6. A jack screw has three threads to the inch ; how far does
it move in ^ of a revolution ?
7. A jackscrew has three threads to the inch ; how far does
it move in i of a revolution ?
8. What is the thread pitch of a single-threaded screw that
has 18 threads to the inch ?
9. What is the thread pitch of a double-threaded screw
that has 8 threads to the inch ?
10. Let each pupil have five different kinds of screws, and
tell the number of threads to the inch of each screw.
BOLTS, SCREWS, AND RIVETS 139
The Micrometer
Accurate mathematical work in measuring diameter is done
with the micrometer caliper. With this instrument thousandths
of an inch may easily be found. The micrometer is easily
adjusted, finely graduated, and has stamped on its yoke the
fractions and decimal equivalents which may be needed in
close measuring. The parts of the micrometer best known are
the screw J the linhj the thimble.
MiCKOMKTER CALIPER
The screw of the micrometer is covered by the thimble to
protect it from dust and wear. By turning the thimble we
move the screw back and forward, increasing or decreasing
the distance between the measuring points of the micrometer
and so opening or closing the instrument for larger or smaller
diameters. One complete revolution of the thimble changes
the opening of the caliper .025, and as the pitch of the screw
in the caliper is 40 per inch and the circumference of the
thimble graduated into 25ths, the turn from one of these to
the next makes the caliper opening .001.
The heel is graduated in a straight line parallel with the
screw length and conforms to the pitch of the screw, each
division being .025 inch, and the fourth division, which is
.100, is made on the frame with the figure 1, the eighth, with
2, etc. When the thimble is turned one complete revolution,
140 VOCATIONAL MATHEMATICS
the screw advances one fortieth of an inch and one twenty-fifth
of one fortieth is .001. In using the micrometer care must be
taken to get the proper touch with the instrument or it may
be crowded over with an error of one half thousandth more
or less than the actual size of the work required. The microm-
eter is a very delicate instrument and must be kept away from
excessive heat or cold as expansion or contraction of the metal
will cause it to become inaccurate. In close work the heat of
the hand, when it is held too long, will change a micrometer
reading.
EXAMPLES
1. If the screw of a micrometer has 40 threads to the inch,
how far will it move in (a) one complete revolution ; (b) ^ rev-
olution ; (c) \ revolution ; (cZ) f revolution ; (e) J^ of a revo-
lution ?
2. What is. the lead of the screw in the above micrometer ?
3. If the screw of a micrometer has 60 threads to the inch,
what is the lead ?
4. In example 3, how far will it advance in (a) ^ revolu-
tion ; (b) \ revolution ; (c) f revolution ?
V-Shaped Thread
The common V-shaped thread is a thread having its sides at
an angle of 60 degrees to each other, and perfectly sharp, top
and bottom. The objections to using this thread are that the^
top is so sharp that it is injured by the slightest accident and
in using the taps and dies in making it the fine, sharp edge is
quickly lost, causing constant variation in fitting.
Formula:
P= Pitch =
No. of threads per inch
JD = Depth = P X .8660
BOLTS, SCREWS, AND RIVETS 141
The diameter of the root (effective diameter) of the thread
is found by multiplying the product above by 2, and then sub-
tracting this double depth from the diameter of the screw.
A formula is used to tind the size of a tap drill to use in
connection with a tap of a given size which has a given number
of threads per inch, as :
Let 7= diameter of the tap, or size of the thread the nut
is to tit;
N= number of threads per inch ;
S = size at root of the thread, or size of the tap drill.
N
Example. — What must the size be of a tap drill for a 1-inch
V-thread tap or 1-inch bolt having 8 threads per inch ?
According to the formula :
^ = 1 _ hl^ OT S = l- .216 or S = .784 inch, Ans.
8
By referring to the Table of Decimal Equivalents ^ the drill
nearest in size to .784 is ^ inch, which will cut a trifle larger
and therefore will be right.
United States Standard Thread
The United States Standard Thread has its sides also at an
angle of 60 degrees to each other, but the top is cut off to the
extent of one eighth of its pitch and the same quantity filled
in at the bottom. The advantages claimed for this thread are
that it is not so easily injured, that the taps and dies retain
their size longer, and that the bolts and screws made with
this thread are stronger and have a better appearance. This
system has been recommended by the Franklin Institute of
Philadelphia and it is often called the Franklin Institute
Standard. Although the V-shaped thread is the strongest
^ See Appendix for Table of Decimal Equivalents.
142 VOCATIONAL MATHEMATICS
form of screw thread, yet as the thrust between the screw
and the nut is parallel to the axis of the screw, there is a
tendency to burst the nut. So this form of thread is unsuitable
for transmitting power.
As \ of the height of the U. S. standard thread is taken from
the top and ^ from the bottom, the thread is only J as deep as
the V-form, so in the formula for finding the diameter at the
root of the thread we use a numerator which is but f of the
numerator used for the V-thread : f of 1.733 is 1.3. So
the formula is
Example. — What should the size of a tap drill be for a one-
inch U. S. S. tap?
As the U. S. standard is not only a thread of a certain form
but also of a given pitch for each diameter of screw, by re-
ferring to the table of United States Standard Screw Threads
we find that one-inch screws have eight threads to the inch.
S = l-— ov S = l- .1625 ovS= .8375
8
In the Table of Decimal Equivalents .8375 has no common
fraction equivalent among the sizes given, but as f f is only
.006 larger we would select a drill of that size.
Formula : P = Pitch = ^ — -
No. of threads per inch
Z>=Depth = Px .6495
i^=Flat = -
EXAMPLES
Solve the following examples by the use of the table.
1. How many threads are there per inch of the U. S. S. screw
with (a) I" diameter ? (b) f " diameter ?
BOLTS, SCREWS, AND RIVETS
143
2. What is the diameter of a screw with the U. S. S. thread
with (a) 5 threads per inch ? (6) 4^ threads per inch ? (c) 13
threads per inch ?
Table of United States Standard Screw Threads
DiAM. or SCRKW
TiiRKAne I'KB Incu
DiAM. OK SCRBW
Threads i'kb Ikcii
i in.
20
l}in.
6
A in.
18
n in.
6
i in.
16
If in.
6i
Ai°-
14
If in.
5
J in.
13
l|in.
6
A in.
12
2 in.
^
f in.
11
2\ in-.
n
i in.
10
2Jin.
4
i in.
9
2f in.
4
l.in.
8
3 m.
3i
IJin.
7
l^in.
7
Acme Standard Thread
The Acme standard thread is an adaptation of the most
commonly used worm thread and is intended to take the place
of the square thread. It is more shallow than the worm
thread but was the same depth as the square thread and is
much stronger than the latter.
The worm thread has sides with an angle of 29° ; the top is
flat, .335 P, and the bottom is .31 P. The depth is .6866 P,
the double depth being 1.3732 P; d=D- ^"^''^'^ , that is, the
N
diameter at the bottom of a worm thread is equal to the
diameter of the worm minus 1.3732" divided by the number
of threads to one inch.
144 VOCATIONAL MATHEMATICS
Square Thread
A square thread has parallel sides; the thickness of the
thread and its depth are each one half the pitch : d = D — —,
that is, the diameter at the bottom of a square thread is equal
to the diameter of the screw minus one inch divided by the
number of threads to one inch. The thrust on the square-
threaded screw is parallel to the axis of the screw; couse-
quently the frictional losses are not so great in this form as
in the V form, but it is not so strong in the base as the V thread.
EXAMPLES
1. What is the diameter of the root of a |" V-threaded
bolt with 20 threads to the inch?
2. What is the diameter of the root of a |" U. S. S.
threaded bolt with 16 threads to the inch?
3. What is the thread pitch of a V-threaded screw with
20 threads to an inch ?
4. What is the thread pitch of an Acme worm screw with
20 threads to an inch ?
5. What is the thread pitch of a square-threaded screw
with 14 threads to an inch ?
6. What is the diameter of the root of a y\" V-threaded
bolt with 18 threads to the inch ?
7. What is the diameter of the root of a |" V-threaded
bolt with 14 threads to the inch ?
8. A U. S. S. screw has 20 threads. What is its thread
pitch ? What is its depth ?
9. What is the diameter of the root of a f " U. S. S. screw
with 20 threads to the inch ?
10. What is the thread pitch of a square-threaded screw
with 8 threads to the inch?
BOLTS, SCREWS, AND RIVETS 145
11. A 2 J" diameter bolt has a diameter of 1.962" at the root
of the thread. What is the depth of the thread ? If the bolt
has 4.V threads per inch, what is the pitch of the threads ?
12. What is the depth of a U. S. S. threaded screw of 7
threads ?
13. What is the diameter of the root of a V-threaded J"
screw with 11 threads to the inch ?
14. What is the depth of a V-threaded screw with 11
threads to the inch?
15. What is the diameter of the root of a U. S. S. threaded
bolt of 11 threads ?
16. What is the diameter of the root of a U. S. S. J"
threaded bolt of 8 threads?
17. What is the thread pitch of a U. S. S. threaded screw
of 9 threads ?
18. What is the depth of a worm-threaded screw with 16
threads to the inch ?
19. What is the depth of a square-threaded screw with 8
threads to the inch ?
20. What is the depth of a U. S. S. threaded screw with 8
threads to the inch ?
21. What is the depth of a U. S. S. 1 J" threaded bolt of
9 threads ?
22. What is the diameter of the root of a U. S. S. threaded
screw of 9 threads ?
23. What is the depth of a V-threaded screw with 14
threads to the inch?
24. What is the thread pitch of a V-threaded screw with
16 threads to the inch ?
25. A U. S. S. threaded bolt has 9 threads to the inch.
What is its thread pitch ?
PART V — SHAFTS, PULLEYS, AND GEARING
CHAPTER IX
SHAFTS AND PULLEYS
In a machine shop one notices at once the revolution of the
shafting. There is one long cylindrical bar called the main
line attached to the ceiling. The power that drives the ma-
chinery is taken from this main line by means of pulleys and
belts to smaller shafts called countershafts. The machinery
is driven directly from the countershafts while the main line
is driven from a motor or engine and flywheel.
Shafts of different sizes are used according to the horse
power required. ,To determine the horse power (H. P.) of a
shaft, multiply the speed (revolutions per minute) by the
cube of the diameter of the shaft, and divide the product by
84 for a steel shaft or by 160 for an iron shaft, and the
quotient is the H. P. (For further discussion of horse power
see pages 188 and 225.)
EXAMPLES
1. What is the H.P. of a 2Jg" steel shaft having 280 revo-
lutions per minute ?
2. What is the H.P. of a 2J" iron shaft having 245 revolu-
tions per minute ?
3. What is the H.P. of a 2|" steel shaft having 290 revolu-
tions per minute ?
4. What is the H.P. of a 2f^'' iron shaft having 350 revolu-
tions per minute ?
146
SHAFTS AND PULLEYS 147
Belting
Belts for transmitting ix)wer are divided into two general
classes : leather belts and canvas belts. Both are sold by
the foot. The material of the belt, the thickness, and width
determine its value. Coils of belting need not be stretched
out to measure their length (A). To do this first count the
number of coils (N), measure the diameter of the hole in the
center of the coil (d), and the outside diameter of the roll (/>).
Then L = 0.1309 N(D + d)
In this formula Z=: length in feet, and D and d diameter
in inches.
This rule is used in estimating.
The formula is obtained as follows :
— ^^— = average diameter of coil
The length L = circumference of average diameter
C =
2
^-12
— = 0.1309
24
Substituting this in the formula ^^(-P + <^) = 0.1309 iV^(Z? -h d)
EXAMPLES
1. How many feet of belting are there in a coil that has a
diameter of 18", if the hole is 3" in diameter and there are 36
coils?
2. How many feet of belting are in a coil that has a 16" di-
ameter, if the hole is 2^" in diameter and there are 38 coils?
148 VOCATIONAL MATHEMATICS
Length of Belting on Pulleys
To find the length of belting on pulleys, add together the
diameters of the pulleys in inches and divide the sum by 2.
Multiply this quotient by 3.25. Add this product to twice
the distance in inches between the centers of the pulleys, and
divide by 12. The final quotient is the length in feet of the
belting on the pulleys. This is an approximation and applies
to open belts only.
Example. — The diameters of two pulleys are 24" and 12"
respectively, and the distance between their centers is 108
inches. Find the length of the belting.
24" + 12" = 36" 18" X 3.25 = 58.50" ^J^!! ^ 22.8 feet
36" ^ 2 = 18" 58.5" +216"- = 274.5" 22.8' = 22' 11". Ans.
EXAMPLES
1. What is the length of the belting connecting two pulleys
having diameters of 18'' and 12" respectively, if the distance
between their centers is 92" ?
2. What is the length of the belting on two pulleys having
diameters of 16" and 22" respectively, if the distance between
their centers is 86" ?
3. Find how many feet of belting are needed to make a belt
to run over two pulleys each 30" in diameter if the distance
between their centers is 13'.
Arc of Contact
In setting up machinery where there are pulleys, it is some-
times desirable to find the arc of contact on the smaller pulley.
The Boston Belting Company gives this rule, which holds when
the pulleys are nearly of the same diameter : Divide the dif-
ference between the diameters of the two pulleys by the dis-
tance between the centers of the shafts, both being in the same
SHAFTS AND PULLEYS
149
denomination ; multiply the quotient by 57, and subtract this
product from 180 ; the result will be the number of degrees
in the arc of contact.
Multiply the entire circumference of the smaller pulley in feet
by the degrees of the arc of contact as above, divide by 360,
and the result "will be the number in feet of the arc of contact
of the belt on the smaller pulley.
EXAMPLES
1. Two pulleys, one 18" and the other 24" in diameter, are
connected by a belt. If the distance between the centers of
the shafts is 8' 6", what is the number of feet in the arc of
contact of the belt on the smaller pulley ?
2. Find the number of lineal inches that a belt touches a
pulley when the arc of contact is 240°, if the diameter of the
pulley is 4 feet.
3. If the distance between the centers of two shafts is 9' 8"
and the pulleys are 22" and 16", what is the number of feet of
arc of contact of the belt on the smaller pulley ?
D
A common way for one shaft to drive another is by means of
a belt running upon two pulleys, one on the driving shaft and
the other on the driven, as in the above figure. In solving
problems the pulley on the driving shaft is called the driver
and the one on the driven shaft the driven.
In order to install machinery and have it run at the proper
speed, the relations between the driving and the driven pulleys
150 VOCATIONAL MATHEMATICS
and the different methods of transmitting power from one
shaft to another must be thoroughly understood. To make
the shafts and pulleys run at the proper speed the correct
diameter and circumference of the driving and the driven
pulleys must be known.
At every revolution of the driver the belt is pulled through
a distance equal to the circumference of the driver; in moving
a distance equal to the circumference of the driven pulley, the
belt turns the driven pulley one revolution. When two pulleys
are connected by a belt, their rim speeds are equal. If we
divide the distance the belt has moved by the circumference of
the pulley, the quotient gives the number of revolutions of
the pulley. The smaller pulley revolves at the higher speed,
a fact that is usually stated mathematically by saying that the
revolutions of the pulleys are inversely proportionate to their
circumferences.
Example. — A pulley 12 inches in diameter makes 300 revo-
lutions per minute. How fast is the rim traveling in feet per
minute ?
The circumference equals the diameter multiplied by 3.1416, or approx-
imately 3|. ■*■
12 X 3.1416 = 37.6992 inches circumference
l^JiAlil^^ 3.1416' feet
12
Since it is running 300 revolutions per minute,
300 X 3.1416 = 942.48 feet per minute
It is possible to express the operations of the above in a for-
mula by letters :
Let D = diameter of the pulley in inches
C = circumference of the pulley in inches
R = revolutions per minute (abbreviated R. P. M.)
^ = 3.1416
F= feet per minute that the rim travels (circumference
speed)
SHAFTS AND PULLEYS 151
Feet traveled per minute is equal to the circumference in
inches, multiplied by revolutions per minute, and divided by 12.
CR
That is, F— — — , or substituting for C its equal ttZ),
1 w
If we know the value of any two of the quantities F, D, or
Bf we can find the others:
(I) P = ^ (2) /> = ^ (3) 7J = 1-2,^
EXAMPLES
1. If a ten-inch pulley is making 300 revolutions per
minute, how fast is a point on the rim traveling in feet per
minute ?
2. The rim of a 14" pulley is running 1048 feet per minute.
How many revolutions are made per minute ?
3. A pulley 18" in diameter makes 375 revolutions per
minute. How fast is the rim traveling in feet per minute ?
4. What is the diameter of a pulley making 286 revolutions
per minute if a point on the rim is traveling 984 feet per
minute ?
5. A 9" pulley is making 198 revolutions por minute. How
fast is a point on the rim traveling?
6. A pulley 32" in diameter is making 198 revolutions per
minute. How fast is the rim traveling ?
Speed
The speed of any machine from the driving shaft or motor
may be traced as follows :
The circumference of a pulley equals its diameter multiplied
by 3.1416.
152 VOCATIONAL MATHEMATICS
This may be abbreviated when C stands for the circumfer-
ence and D for the diameter :
C = 3.1416 D or 7rZ>
Let C = circumference of driver pulley
C" = circumference of driven pulley
D = diameter of driver pulley
Z>' = diameter of driven pulley
N = revolutions of driver pulley per minute
jV' = revolutions of driven pulley per minute
(C is read C prime.)
C X N =z distance belt moves per minute on driver wheel
C X A^ = distance belt moves per minute on driven wheel
The distance represented by C x N h called rim speed of
driver wheel.
The distance represented by C" X N^ is called rim speed of
driven wheel.
Since the surface speeds are equal,
(1) CxN^C^xN', (2) g = ^,
Since C ^-jtB ovD^N[ ^ ^ D^N'
C'=^7rD' UN N
If we know any three of the above four quantities, the
fourth can be found :
If i) = ^^' N==^^
N D
The proportion may be easily remembered by noting that
the primes come together as middle terms:
D:D'::N':N
The above formula may be expressed in the form of rules:
Whenever one pulley drives another we have four quantities
to consider — the diameters of the two pulleys and the revo-
SHAFTS AND PULLEYS 153
lutions per minute of the two pulleys. The above equation
shows us that there is a definite relation between them. If
we know any three of the quantities, we may find the fourth
by transferring the factors of the equation.
NoTK. — II is often difficult to remember these niles. In that case
draw a sketch and place the given infonnation about each pulley. Let x
represent the unknown quantity. Then multiply the two numbers known
about one pulley and divide by the number given in the other pulley.
The quotient will represent x.
Example. — Find the size of a pulley on a countershaft that
runs 120 revohitions per minute, if the diameter of the pulley
on line shaft is 18" and runs 180 revolutions per minute.
18 : Z>' : : 120 : 180
120 D' = 18 X 180
2 Z>' = 54
D' = 27
EXAMPLES
1. The diameter of a pulley on the line shaft is 30" and it
runs 158 revolutions per minute; the countershaft runs 400
revolutions per minute. What is the size of the pulley on the
countershaft ?
2. What is the size of a pulley on a countershaft of an
engine lathe if the diameter of the pulley on the line shaft is
15" and it runs 150 revolutions per minute while the counter-
shaft runs 140 revolutions per minute ?
3. What is the size of a pulley on a countershaft of a planer
if the diameter of the pulley on line shaft is 26" and it runs
305 revolutions per minute while the countershaft runs 793
revolutions per minute ?
4. What size pulley should be placed on the countershaft of
a band saw if the diameter of the pulley on the main line shaft
is 14", if it runs 250 revolutions per minute, and the counter-
shaft runs 225 revolutions per minute ?
154 VOCATIONAL MATHEMATICS
5. A pulley 30" in diameter on a main shaft running 180
revolutions per minute is required to drive a countershaft 450
revolutions per minute.. What will be the diameter of the
pulley on the countershaft ?
Example. — Find the number of revolutions tha£ a counter-
shaft is running if the line shaft runs 140 revolutions per minute,
and the diameter of the pulley on it is 30" and the diameter
of the pulley on the countershaft 12".
Z> : D' : : .V : .V
30 : 12 : : N' : 140
2N' = 700
JV'=360. Ans.
EXAMPLES
1. Find the number of revolutions per minute of a counter-
shaft of an engine lathe if driven by an 18" pulley on the line
shaft which runs 168 revolutions per minute, if the diameter
of the countershaft pulley of lathe is 12''.
2. What is the speed of a countershaft of a speed lathe, if
the pulley on the main shaft is 10" with 305 revolutions per
minute and the diameter of the pulley on the countershaft is
4"?
3. Find the number of revolutions of a countershaft of a
band saw if the pulley on the main shaft is 19" with 215 revo-
lutions per minute and the diameter of the pulley on the
countershaft is 16".
4. What is the speed of the countershaft of a cutting-off
saw if the pulley on the main shaft is 15" with 230 revolutions
per minute and the diameter of the pulley on the countershaft
is 12"?
5. A pulley 30" in diameter making 180 revolutions per
minute drives a countershaft with a 12" pulley. What is the
speed of the countershaft ?
SHAFTS AND PULLEYS 155
6. The main driving pulley of an engine is 12' G" in diam-
eter and makes 96 revolutions per minute; it is belted to a
48" pulley on the main shaft. Find the speed of the latter.
Example. — The line shaft of a machine shop runs 120 rev-
olutions per minute, the diameter of the pulley on the counter-
shaft is 15", and the countershaft runs 240 revolutions per
minute. Find the size of the pulley on the main line.
D:D'::N':.y |=^
D : 1 V : : 240 : 120
120 Z> = lo'' x240
D = 30". Ans. N'D' = ND
EXAMPLES
1. The main line runs 160 revolutions per minute; the
countershaft pulley is 9" in diameter and runs 320 revolutions
per minute. What is the diameter of the pulley on the main
line ?
2. A pulley 24" in diameter running 144 revolutions per
minute is to drive a shaft 192 revolutions per minute. What
must the diameter of the pulley be on the driven shaft ?
3. A driving shaft runs 140 revolutions per minute; the
driven pulley is 10" in diameter and is to run 350 revolutions
per minute. What must the diameter of the driving pulley be ?
4. A countershaft with a 12" pulley runs 450 revolutions
per minute ; the revolutions of the main shaft are 180. What
size pulley must be used on the main shaft ?
5. A main line runs 189 revolutions per minute and the
countershaft pulley is 10" in diameter and runs 385 revolutions
per minute. What is the size of the pulley on the main line ?
6. A pulley 36" in diameter running 168 revolutions per
minute is to drive a shaft 212 revolutions per minute. What
must be the diameter of the pulley on the driven shaft ?
156 VOCATIONAL MATHEMATICS
7. A driving shaft runs 184 revolutions per minute; the
driven pulley is 12" in diameter and is to run 350 revolutions
per minute. What must be the diameter of the driving pulley ?
8. What is the speed of the main shaft if the pulley is 12"
and the revolutions per minute on the other shaft are 228 with
a pulley 8" ?
9. What is the speed of a countershaft if the pulley is 11"
and the main shaft runs 196 revolutions per minute with a
14" pulley ? •
10. A 14' flywheel running 98 revolutions per minute drives
a 9" pulley. What is the speed of the pulley ?
Countershafts or Jackshafts
The first shaft belted off from a flywheel is often called a
jackshaft. In the figure below the jackshaft carries the
pulley d ; on the main line is the large pulley D and the small
pulley F on the jackshaft. On another main line is the pulley
/. Pulley B is the first driver and by means of pulley F on
the jackshaft drives pulley d. Pulley d is the second driver.
Pulley F is the first driven and pulley /is the second driven
pulley.
D
Main Line Jackshaft
Example. — The main shaft runs 160 R. P. M. (revolations
per minute) ; D is 60" in diameter and drives F 140 E,. P. M.
What is the diameter of 2^?
The second main / is to run 186 R. P. M. What should be
the diameter of/ if the 72" driver d is running 140 R. P, M, ?
SHAFTS AND PULLEYS
157
An examination of the data in connection with D and / in
the figure above will show that the relative speed,of /to D is
XxDxd=^N'xFxf
where N= number of revolutions of driver
and N' = number of revolutions of driven
That is, the continued product of the speed of the first driver
and the diameters of all the drivers is equal to the continued
product of the speed of the last driven by the diameters of all
the driven pulleys. In this combination of driving f hy D
there are six quantities, any of which can be found when we
know the other five, by figuring from one shaft to the next
step by step.
EXAMPLES
1. The R.P. M. of a 26" driving pulley is 270. What are
the revolutions per minute of an 18" driven pulley ?
2. If there is a shaft with
a speed of 270 R. P. M. upon
which there is a 26" pulley
driving a 16" pulley, and on the
shaft with the 16" pulley is a
30" pulley driving an 18" pulley,
what is the speed of the shaft
which carries the 18" pulley ?
3. A flywheel which is 30 feet in diameter drives a coun-
tershaft by means of a pulley
6 feet in diameter ; the flywheel
makes 50 R. P. M. What size
of pulley must be used on the
countershaft to give 300 R. P. M.
to a pulley 2 feet in diameter?
4. If 3 flywheels, respectively 13', lOf , and 9' di-
ameter have the same circumferential speed of 2750'
per minute, how many revolutions per minute does each make ?
270
158 VOCATIONAL MATHEMATICS
5. A flywheel which is 40 feet in diameter drives a counter-
shaft by means of a pulley 6^ feet in diameter ; the flywheel
makes 54 R. P. M. What size of pulley must be used on the
countershaft to give 341 R. P. M. to a pulley 3 feet in
diameter?
6. The size of a main driving pulley is 20 feet in diameter
on a shaft with a speed of 70 K. P. M., driving a pulley 4 feet
in diameter and two other pulleys 5.6 and G.36 feet in diameter,
respectively. What is the speed of each shaft ?
CHAPTER X
GEARING
In a machine shop power is transmitted from one part of a
machine to another part by means of tooth-shaped interlocking
wheels. The train of toothed wheels for transmitting motion
is called gearing.
Gears are nothing but pulleys with teeth, and are made to
drive one another by the teeth coming in contact with each
other. If a small gear drives a
larger gear, the larger gear will go
more slowly than the smaller ;
that is, the larger gear will make
fewer turns in a minute. Just
the reverse is true if a larger gear
drives a smaller one. The num-
ber of revolu-
tions which
a gear makes
is always pro-
portional to
the number
of its teeth.
As in pul-
leys, there
are the driver and the driven gears. The driver may be dis-
tinguished from the driven by examining the gears and notic-
ing that it is the gear that is bright or worn on the front of the
tooth — that is, the side of the wheel moving. The driven wheel
is worn on the side away from the direction of the motion.
Bkvel Gear
Spur 6bab
160 VOCATIONAL MATHEMATICS
In order to express the relation between the driver and
driven gears it is necessary to use abbreviations.
Let Z) = number of teeth in driver
D' = number of teeth in driven
N= number of revohitions of the driver
N" = number of revolutions of the driven
D:D' :: N': ^Vor DN= D'N'
That is, the product of the teeth in the driver by its revolutions
equals the tooth transits of the driver, which in turn equal the
tooth transits of the driven or follower. If any three of these
quantities are known, the fourth can be found.
Example. — How many revolutions does a 12-tooth follower
make to five revolutions of a 24-tooth driver ?
12iV^=24x6
iV=10 Ans.
EXAMPLES
1. A driver has 98 teeth and its follower 42. How many
revolutions will the follower make to one revolution of the
driver ?
2. In Example 1 how many revolutions of the driver will
drive the follower one revolution ?
3. How many teeth must a follower have in order to make
three revolutions while a 96-tooth driver makes one ?
4. How many teeth must a gear have to revolve 16 times
while a 60-tooth mate revolves 12 times ?
5. A 96-tooth gear drives a 48-tooth gear. What is the ratio
of their speeds ?
6. A 48-tooth gear drives a 120-tooth gear. What is the ratio
of their speeds ?
7. Two shafts are connected by gears, one of which turns 55
times a minute and the other 11 times a minute. If the small
gear has 32 teeth, how many teeth has the larger gear ?
GEARING
161
Pitch
In order to solve problems connected with the use of gears
it is necessary to know the different terms used in connection
with gearing.
The pitch circle is shown in the illustration and is the circle
which runs around the teeth. It is the same size as the fric-
Gear Teethe
L Pitch
r^ lAat
Diameter
tion rollers or cylinders would be if no teeth were there.
When two spur gears roll together their pitch circles are con-
sidered to be always in contact. The jy^tch diameter is the
diameter of the pitch circle. The word diameter when applied
to gears always means the pitch diameter.
The circular pitch is the distance measured on the pitch
circle from the center line of one tooth to the center of the
next. This is illustrated in the diagram. In solving gearing
problems the circular pitch is not nearly so important as the
diametral pitch.
If the distance from the center of a tooth to the center of tlie next
tooth is i", the gear is I" circular pitch.
The diametral pitch is the number of teeth for each inch of
pitch diameter.
For example, if a gear has thirty teeth and the pitch diameter is three
inches, then the diametral pitch is .SO -=- .3 = 10, or a 10 diametral pitch
gear. Using P for the diametral pitch, D for the pitch diameter, and ^V
for the number of teeth, we liave a formula P = N ^ D.
162 VOCATIONAL MATHEMATICS
To find the thickness of tooth at the pitch line when the
diametral pitch is given, divide 1.57 by the diametral pitch.
For example, if the diametral pitch is 3, divide 1.57 by 3, and the quo-
tient, which is .523 inches, is the thickness of the tooth.
To find the circular pitch when the diametral pitch is given
divide 3,1416 by the diametral pitch.
For example, if the diametral pitch is 4, divide 3.1416 by 4 and the
quotient, which is .7854, is the circular pitch.
Diametral pitch is found when the circular pitch is given by
dividing 3.1416 by the circular pitch. Since the circumference
of any circle is equal to 3.1416 times its diameter, every inch of
diameter of any circle is equal to 3.1416 inches of circum-
ference, or in this instance, for every inch of pitch diameter
we have 3.1416 inches of circumference measured on the
pitch circle. But the diametral pitch is by definition the num-
ber of teeth for each inch of pitch diameter, and by. the above
reasoning we can also say that the diametral pitch is the num-
ber of teeth for each 3.1416 inches of circumference of pitch
circle.
Since the circular pitch is the distance from the center of
one tooth to the center of the next, it will also be equal to
3.1416 of circumference of pitch circle divided by the number
of teeth in that 3.1416. But tlie number of teeth in 3.1416 is
equal to the diametral pitch, and therefore the distance from
the center of one tooth to the center of the next, or the circular
pitch, is equal to 3.1416 divided by the diametral pitch.
Therefore, using P for circular pitch and P for diametral
pitch, P = 3.1416 - F or P' = 5:111^.
EXAMPLES
1. What is the diametral pitch of a gear liaviug (a) 56 teeth
and a pitch diameter of 8" ? [b) 60 teeth and a pitch diameter
of 12" ?
GEARING
163
2. What is the thickness of the tooth of a gear having
diametral pitch of (a) 8? (6) 4? (c) 14?
3. What is the eirciiUir pitch if the diametral pitch of
geAris(a) 8? (6) 10? (c) 5?
4. What is the diametral pitch if the circular pitch of
gear is (a) 1.5708? (b) .5230? (c) .2618*^ ((J) .7854?
To find the pitch di-
ameter when the number
of teeth and the diame-
tral pitch aregiven, divide
the number of teeth by
the diametral pitch. For
example, if the number
of teeth is 40 and the
diametral pitch is 4,
divide 40 by 4, and
the quotient, which is
10", is the pitch di-
ameter.
To make a gear the
outside diameter must
first be known. If the
diametral pitch and the
number of teeth are
given, the outside di-
ameter may be found
by the following rule:
D =
N-\-2
where D = outside di-
ameter, N= number of
teeth, and P= diametral
pitch.
164 VOCATIONAL MATHEMATICS
For example, if a gear has 24 teeth and its diametral pitch is 2, the
outside diameter will be 13"; O = ^£+2 ^ 26 ^ ^g,, ^^^ ^ :^±2.
2 2'
To find the diame^raZj[)iYc7i when the number of teeth and the
diameter of the blank are given, add 2 to the number of teeth
and divide by the diameter of the blank.
For example, if the number of teeth is 40 and the diameter of the blank
is 10^", add 2 to the number of teeth, making 42, and divide by 10^, and
the quotient, which is 4, is the diametral pitch.
To find the thickness of tooth at the pitch line when the cir-
cular pitch is given, divide the circular pitch by 2.
For example, if the circular pitch is 1.047", divide this by 2, and the
quotient, which is .523, is the thickness of tooth.
To find the number. of teeth when the pitch diameter and the
diametral pitch ai-e given, multiply the pitch diameter by the
diametral pitch.
For example, if the pitch diameter is 10" and the diametral pitch is 4,
multiply 10 by 4, and the product 40 will be the number of teeth in the
gear.
To find the whole depth of tooth divide 2.157 by the diametral
pitch.
For example, if the diametral pitch is 6, divide 2.157 by 6, and the
quotient, .3595", is the whole depth of the tooth.
Formulas for Determining the Dimensions of Gears bt
Diametral Pitch
P = the diametral pitchy or the number of teeth to one inch of
diameter of pitch circle.
D' = the diameter of pitch circle.
D = the whole (outside) diameter or the diameter of the blank.
JV = the number of teeth.
V= the velocity.
t = the thickness of tooth or cutter on pitch circle.
D" = the working depth of tooth.
GEAKING 165
/ = the amount added to depth of tooth for rounding the corners
and for clearance.
D" 4-/ = the whole depth of tooth.
r denotes the constant 3.1416.
P' denotes the circular pitch or the distance from the center of one tooth
to the center of the next on the pitch circle.
FormuUm KjrnmpleH '
^±1 = 71+^, or T?+_2_ 12.
6.166 ' 6fV
= 1:^^ = 12.
6
6.166 X 72 ^g
72 + 2
= ^ = 6.
12
= 12 X 6 = 72.
= 12 X 6.166 - 2, or 12 X 6j-\ - 2 = 72.
72 + 2
D
P--
N
= —
D'
DxX
N-^2
D'
N
P
X.
= PD^
X.
= PD-2
n.
_.V+2
P 12
rr
= 6 166, or (jj\.
D=D'-h- = 6 + :p^,or 6 + .166=6.166.
^ = 1:51 = M[ = .130. •
P 12
9 O O
f = — = :^ = .013.
10 10
/>"+/ = .166 + .013 = .179.
rw TT 3.1416 nf,n
^ = P = -12- = •2^2-
3.1416
P = ^ = ^•"" :^ = 12.
F .262
1 The examples placed opposite the formulas above are for a single gear of
12 pitch, 6.166" or Gft" diameter.
166 VOCATIONAL MATHEMATICS
EXAMPLES
Solve by using the above fonnulas and the rules on pages
161-165 :
1. If the number of teeth of a gear is 46 and the diametral
pitch is 8, what will the pitch diameter be ?
2. If a gear is to have 54 teeth and the diametral pitch is 6,
what will be the pitch diameter ?
3. If a gear has 38 teeth and the diametral pitch is 6, what
is the pitch diameter ?
4. If the pitch diameter of a gear is 6" and the diametral
pitch is 8, what will be the number of teeth?
5. How many teeth has a gear if the pitch diameter is 7f "
and the diametral pitch is 6 ?
6. How many teeth has a gear if the pitch diameter is 8J"
and the diametral pitch is 4?
7. If the pitch diameter of a gear is 12" and the diametral
pitch 7, how many teeth has the gear ?
8. If the diametral pitch is 16 and the pitch diameter of a
gear is 6J", how many teeth has the gear ?
9. What is the whole depth of the tooth if the diametral
pitch of a gear is (a) 8? . (6) 4 ? (c) 7 ? (d) 10?
10. If a gear has 48 teeth and the diameter of the blank
is 6y, find the diametral pitch.
11. If a gear has 61 teeth and the diameter of the blank
is 101", find the diametral pitch.
12. If a gear has 35 teeth and the diameter of the blank
is Syy, find the diametral pitch.
13. If a gear has 78 teeth and the diameter of the blank
is 5", find the diametral pitch.
14. If a gear has 54 teeth and the diameter of the blank
is 6|", find the diametral pitch.
15. What is the thickness of the tooth if the circular pitch
of a gear is (a) .1848" ? (b) .1428" ? (c) 6.2832" ? (d) 2" ?
GEARING 167
If one examines closely the movements of two gears on a
machine, one will notice that all the teeth pass a given point at
every revolution of the wheel. That
is, the teeth marked Tty on the gear f^J lyO
shown in the illustration pass an out- ^ ^)
side point like P in making one com- 2'— CT 7^^
plete revolution. So when a tooth and (\ /O— *
the space adjacent to it have passed a (j^ r\j
given point like P, one transit of a
tooth has occurred. There are as many tooth transits at every
complete revolution of a gear as there are teeth in the gear.
If we multiply the number of teeth by the number of
revolutions, the product will be the number of transits. If
this product (number of transits) be divided by the number
of teeth, the quotient will be the revolutions of the gear. For
example, if a 12-tooth gear makes CO transits, then it has made
(^J = 5) five revolutions.
Two shafts, D and Fy are to be connected by gears so that
shaft D will make one revolution while shaft F makes two.
To do this a gear must be put on shaft D having twice the
number of teeth that the gear on shaft F has. If a gear hav-
ing 24 teeth is put on shaft D, the gear on shaft F will have
half as many. Each time the gear on D turns around once the
168 VOCATIONAL MATHEMATICS
gear on F will turn twice ; that is, the 24 teeth on gear D will
have to turn gear F twice in order to mesh with the 24 teeth
on F.
The relation or ratio of the speed of F to the speed of D is
2 to 1. This is called the ratio of the gearing. The ratios
between the speeds and number of teeth can be written in the
form of a proportion :
24:12:: 2:1
A^nd the number of teeth on gear D is to the number of teeth
on gear F as the speed of F is to the speed of Z>.
Train of Gears
When two gears mesh, as in the illustration on page 167, one
revolves in the opposite direction from the other. Three or
more gears running together, as in the illustration above, are
often called a train of gears. In a train of spur gears like
these, one gear, J, which is called an intermediate gear, meshes
with the other two gears, D and F, and causes the revolutions
of both Z) and F to be made in one direction, while the inter-
mediate revolves in the opposite direction. The intermediate
does not change the relative speeds of D and F, so that they
can be figured as explained on page 167. An intermediate
gear is also called an idler.
GEARING 169
We may calculate the nuinbsr of revolutions of any follower
for any number of revolutions of a driver, in a train of gears,
step by step. To find the number of revolutions of the last
follower when the number of revolutions of the first driver
and the teeth in all the gears are known, take the continued
product of the revolutions of the first driver and all the driv-
ing gears, and divide it by the continued product of all the
followers ; the quotient is the number of revolutions of the last
follower. In other words, the product of the revolutions of
the first driver and the teeth of all the driving gears is equal
to the continued product of the revolutions of the last follower
and the teeth of all the driven gears.
EXAMPLES
1. If a 60-tooth wheel is to mesh with one having 46 teeth
and the 60-tooth gear makes 25 revolutions per minute, how
many will the 46-tooth gear make ?
2. A 168-tooth gear drives a 28-tooth gear. What is the
ratio of the gearing ?
3. What would be the ratio of the gearing in the example
above if the 28-tooth gear were the driver ?
4. If the 28-tooth gear is making 48 revolutions per min-
ute, how many revolutions per minute is the 168-tooth gear
making ?
5. How could the gearing be changed to make the 28-tooth
gear turn in the same direction as the 168-tooth gear ?
6. Two gears running together have a speed ratio of 7 to 1.
If the smaller one turns 14 times, how many times will the
larger one turn ?
7. If a 144-tooth gear makes one complete turn, how many
turns will a 32-tooth gear make working with it ?
8. In the above example if the 32-tooth gear turned once,
how many turns will the 144-tooth gear make ?
170 VOCATIONAL MATHEMATICS
9. A train of 3 gears has 69, 30, and 74 teeth. If the 69-
tooth gear makes 100 revolutions per minute, how many
revolutions per minute will the 74-tooth gear make ? Make
a sketch showing the direction in which each gear turns.
10. A train of gears is made up of 6 gears having teeth as
follows: 46, 60, 32, 72, 56, and 48; while the first gear in the
train makes 10 turns, how many turns will the last gear make ?
11. What two gears will give a ratio of speeds so that the
driver will make if as many turns as the follower ; that is,
while the driver makes 13 turns the follower will make 14 ?
12. If 24 gears work in a train, in what direction will the
last one turn if the first turns right-handed ?
REVIEW EXAMPLES
1. Two gears working together have pitch circles 14^' and
26f " in diameter. What is the distance between their centers ?
The distance between centers may be obtained from the following
formulas :
D' -\-d' h
«= — ^r — '•<''' = 7r^
2 2P
where a — distance between centers
d' = diam. of pitch circle of smaller gear
h = number of teeth on both wheels
2. Two gears in mesh have pitch circles 17.082" and
31.3124" in diameter. What is their center distance?
3. Two gears working together are four pitch, the larger
has 36 teeth and the smaller has 14 teeth. What is their
center distance ?
4. The circular pitch of a planer table gear is 1". What
should be the total depth to cut the tooth space ?
5. The circular pitch of a milling machine gear is |^" and it
has 96 teeth. What is its outside diameter ?
6. A flue rattler gear is to be renewed. Its outside diam-
eter is 34" and it has 100 teeth. What size shall we draw its
pitch circle ?
GEARING 171
7. Two gears mesh together, one has 28 teeth and an
outside diameter of 2|" and the other 68 teeth and an outside
diameter of 5|". Find the center distance.
8. A gear has a circular pitcli of |f ". What is the thick-
ness of the tooth at the pitch line ?
9. A pitch circle is 18" in circumference. If the teeth are
I" thick, what is the circular pitch ?
10. A gear is 10 pitch. What is the total depth of its
teeth ?
11. The diametral pitch of a gear is 4. What is the cir-
cular pitch ?
12. The circular pitch of a gear is .157. What is the di-
ametral pitch ?
13. A gear is 12 pitch. What is its circular pitch ?
14. A gear is 20 pitch. If it has 105 teeth, what is its out-
side diameter ?
15. A gear of 10 pitch has 44 teeth. Find the pitch
diameter.
16. A gear has 28 teeth and a pitch diameter of 8. What
is the pitch ?
17. A gear is set in a milling machine ready to cut the
teeth, and if the pitch is to be one, what is the depth of the
cut?
la A gear is 14 pitch and has an outside diameter of 4|".
Find the depth of cut for the teeth.
19. A gear has 75 teeth and is 18 pitch. What is the work-
ing depth of the tooth ?
20. A pinion has 44 teeth and is 10 pitch. What is the out-
side diameter ?
21. A lathe back gear has 108 teeth and is 5 pitch. What
is its outside diameter ?
PART VI — PLUMBING AND HYDRAULICS
CHAPTER XI
RECTANGULAR AND CYLINDRICAL TANKS
Water Supply. — The question of the water supply of a city or a town
is very important. Water is usually obtained from lakes and rivers
which drain the surrounding country. If a lake is located in a section
of the surrounding country higher than the city (which is often located
in a valley), the water may be obtained from the lake, and the pressure
of the water in the lake may be sufficient to force it through the pipes
'm&WW'
Water Supply
into the houses. But in most cases a reservoir is built at an elevation as
high as the highest portion of the town or city, and the water is pumped
into it. Since the reservoir is as high as the highest point of the town,
the water will flow from it to any part of the town. If houses are
built on the same hill with the reservoir, a stand-pipe, which is a steel
tank, is erected on this hill and the water is pumped into it.
Water is conveyed from the reservoir to the house by means of iron
pipes of various sizes. It is distributed to the different parts of the house
by small lead, iron, or brass pipes. Since water exerts considerable pres-
sure, it is necessary to know how to calculate the exact pressure in order
to have pipes of proper size and strength.
172
PLUMBING AND HYDRAULICS 173
EXAMPLES
1. Water is measured by means of a meter. If a water
meter measures for tive hours 7()0 cubic feet, how many gal-
lons does it indicate ?
Note. — 231 cubic inches = 1 gallon.
2. If a water meter registered 1845 cubic feet for 3 days,
how many gallon? were used ?
3. A tank holds exactly 12,852 gallons ; what is the capacity
of the tank in cubic feet ?
4. A tank holds 3841 gallons and measures 4 feet square on
the bottom ; how high is the tank ?
Rectangular Tanks. — To find the contents in gallons of a square or
rectangular tank, multiply together the length, breadth, and height in
feet ; multiply the result by 7.48.
I = length of tank in feet
b = breadth of tank in feet
h = height of tank in feet
Contents = Ibh cubic feet x 7.48 = 7.48 Ibh
gallons
(Note. — 1 cu. ft. = 7.48 gallons.)
If the dimensions of the tank are in inches, multiply the length,
breadth, and height together, and the re.sult by .004829.
5. Find the contents in gallons of a rectangular tank having in-
side dimensions (a) 12' x 8' x 8'; (b) 15" x 11" X 6" ; (c) 3' 4"
X2'8"x8"; (d) 5'8"x4'3"x3'5"; (e) 3' 8" x 3' 9" x 2' 5".
Cylindrical Tank. — To find the contents of a cylin-
drical tank, square the diameter in inches, multiply
this by the height in inches, and the result by .0034.
d = diameter of cylinder
h = height of cylinder
Contents = d^h cubic inches x .0034 = d^h .0034 gallons
6. Find the capacity in gallons of a cylindrical tank (a) 14"
in diameter and 8' high; (b) (>" in diameter and 5' high;
174 VOCATIONAL MATHEMATICS
(c) 15" in diameter and 4' high; (d) 1' 8" in diameter and
5' 4" high; (e) 2' 2" in diameter and 6' V high.
Inside Area of Tanks. — To find the area, for lining purposes, of a
square or rectangular tank, add together the widths of the four sides of
the tank, and multiply the result by the height. Then add to the above
the area of the bottom. Since the top is usually open, we do not line
it. In the following problems find the area of the sides and bottom.
7. Find the amount of zinc necessary to 'line a tank whose
inside dimensions are 2' 4"x 10" x 10".
8. Find the amount of copper necessary to line a tank
whose inside dimensions are 1'9" x 11" x 10", no allowance
made for overlapping.
9. Find the amount of copper necessary to line a tank
whose inside dimensions are 3' 4" x 1' 2" x 11", no allowance
for overlapping.
10. Find the amount of zinc necessary to line a tank
2' 11" X 1' 4" X 10".
Drainage Pipes
In order to have pipes of a proper size to carry away the
water from the roof of a building, it is necessary to know the
number of gallons of water that will vbe drained. To find the
number of gallons that will drain from a roof in a month,
multiply the number of square feet of the roof by the average
number of inches of rainfall per month, and multiply this prod-
uct by .623. When the roof is not flat, or very nearly so, its
area should be considered as the area which it actually covers.
^ EXAMPLES
11. A roof is 48' by 62'. How many gallons of water will
drain from it in a month if the rainfall is 6" ?
12. A roof is 29' by 74'. How many gallons of water will
drain from it if the rainfall is 4^^" ?
PLUMBING AND HYDRAULICS 175
Note. — The rainfall per month in Massachusettjs varies from J" to
12'^ In calculating the following problems consider 10" rainfall as the
average amount.
13. Find the number of gallons that will drain from a flat
roof (a) 112' by 64'; (6) 88' by 49'; (c) 120' by 80'.
14. Find the number of gallons that will drain from a slant-
ing roof each half of which is (a) 52' by 34' ; {b) 49' by 28' ;
(c) 112' by 54'; (d) 57' by 33'; (e) 54' by 31'.
Weight of Lead Pipe
Pipes, particularly those of lead, are sold by weight, aiid this
depends upon the diameter and thickness of the metal in the
pipe.
To find the weight of a length of pipe, subtract the square
of the inner diameter in inches from the square of the outer
diameter in inches, and multiply the remainder by the weight
of 12 cylindrical inches. This product multiplied by the
length in feet gives the required weight.
Example. — What is the weight of 1450 feet of lead pipe,
the outer diameter being J" and the inner diameter being ^"?
D = outer diameter
d = inner diameter
I = leiijith of pipe
3.8698 lb. = weight of 12 cylindrical inches of lead pipe 1' long
and 1" in diameter. Weight in lb. = (Z>- — d-) x 3.8698 x I
Then ^^ = LJLL - J^ (sciuare of outer diameter)
(8)2 8 X 8 64 ^ * - ^
VJ « = ^ ^ ^ = — ^ (square of inner diameter)
(16)2 16 X 16 256 ^ ^ ^
15 _ i5. = iL = .5742 (difference)
64 266 256 ^ ^
.6742 X 3.8698 x 1450 = 3221.94 lb.
EXAMPLES
1. What is the weight of 364' of lead pipe, outer diameter
}", inner diameter ^y ?
176 VOCATIONAL MATHEMATICS
2. What is the weight of 1189' of lead pipe, outer diameter
If", inner diameter ly\" ?
3. What is the weight of 2189' of lead pipe, outer diameter
2f' ', inner diameter 2^'' ?
4. What is the weight of 112' of lead pipe, outer diameter
3", inner diameter 2^%" ?
5. What is the weight of 212' of lead pipe, outer diameter
3i", inner diameter 3^^" ?
Capacity of Pipe
Rules for Finding the Capacity in Gallons of a Foot of Pipe
of any Diameter :
1. Find the cubical contents in inches, and divide by 231,
C=D2x.7854x 12-- 231
2. Multiply the square of the inside diameter by .0408.
C=D^X .0408
Rules for Findiyig the Capacity of a Pipe of any Length and
any Diameter :
1. Multiply the number of gallons in a foot of the pipe by
the number of feet in the length of the pipe.
0=Z)2 X .0408 X L
2. Find the cubical contents in inches, and divide by 231.
,(7=7>2 X .7854 X i.^231
Note. — In the above formulas, let C = the capacity of the pipe in
gallons, D = the diameter in inches, and L = the length in feet.
EXAMPLES
1. What is the capacity of a pipe having an inside diameter
of (a) J inch, 16 feet in length ; (b) 1} inches, 20 feet in length ;
(c) 1.^ inches, 1 foot in length; (d) 3 J inches, 1 foot in length;
(e) 14 inches, 10 feet in length ; (/) 5 inches^ 22 feet in length ?
PLUMBING AND HYDRAULICS
177
Number of U. S. Gallons in Round Tanks
for onb foot in defth
DiA. or
Tanks
Ft. 1«.
No.
U. S.
Gals.
CiTBic Ft.
AND Area
in 8q. Ft
DlA. OF
Tanks
Ft. In.
No.
U.S.
Gals.
Cub in Ft.
ANo Area
IN 9«i. Ft.
DlA. Of
Tanks
Fr. In.
No.
U.S.
Gals.
Cdhio Ft.
ANi» Arka
iH 8m. Ft.
1
5.87
.7S5
3
52.88
7.0
,
O
r
0°
—
""
— 17.8°
c
1
II
Thermometers
90
32°
Example. — Convert 212 degrees F. to C. reading.
5f212°-32°) ^ 5(180°) ^ 900^ ^ j^^o q
9 9 9'
Example. — Convert 100 degrees C. to F. reading.
^ ^ ^^^° + 32° = ^^ + 32° = 180° + 32° = 212° F.
6 6
HEAT 197
If the teinperatiire is below the freeziuj^ point, it is usually
written with a minus sign before it : thus, 15 degrees below the
freezing point is written — 15**. In changing — 15° ( -. into F.
we must bear in mind the minus sign.
Thus, F =— + 32^. F = ~^'^° ^ ^ + 32" = - 27-^ + 32° = 6*^'
o 5
Example. — Change - 22° F. to C.
C. = § (F. - 32°)
C. = I (- 22° - 82°) = |( - 64°) = - 30°
EXAMPLES
1. Change 36° F. to C. 6. Change 225° C. to F.
2. Change 89° F. to C. 7. Change 380° C. to F.
3. Change 289° F. to C. 8. Change 415° C. to F.
4. Change 350° F. to C. 9. Change 580° C. to F.
5. Change 119° C. to F.
Value of Coal in Producing Heat ^
There "are different kinds of coal on the market. Some
grades of the same coal give off more heat than others in
burning. The heating value of a coal may be found in
three ways : (1) By chemical analysis to determine the amount
of carbon, (2) by burning a definite amount in a calorimeter
and noting the rise of temperature of the water, (3) by actual
trial in a steam boiler. The first two methods give a theoreti-
cal value, the third gives the real result under the actual con-
ditions of draft, heating surface, combustion, etc.
EXAMPLES
1. If 125 lb. of ashes are produced from one ton of coal,
what is the percentage of ashes in that coal ?
198 VOCATIONAL MATHEMATICS
2. Twelve tons of coal are burned per day, and twenty -two
baskets of ashes, each weighing 65 lb., are removed; what per-
centage of ashes does the coal contain ?
3. If twelve tons of coal are burned per day, and 1450 lb. of
ashes are produced, what percentage of ashes does the coal
contain ?
4. A quantity of coal is built into a rectangular stack 50 ft.
long, 25 ft. broad, and 6 ft. high ; what is the weight of the
coal, allowing 45 cubic feet per ton ?
5. If 92,400 lb. of coal are consumed in 60 hours and the
engines regularly develop 480 1. H. P. (Indicated Horse Power),
how much coal is consumed per H. P. per hour ?
6. With the price of coal at $3.25 per ton of 2000 lb., and
the power produced earning a profit of 25 % on the cost of
production, what would be the amount of profit in Ex. 5 when
running full power ?
Latent Heat
By latent heat of water is meant that heat which water ab-
sorbs or discharges in passing from the liquid to the gaseous,
or liquid to solid state, without affecting its own temperature.
Thus, the temperature of boiling water at atmospheric pressure
never rises above 212 degrees F., because the steam absorbs the
excess of heat which is necessary for its gaseous state. Latent
heat of steam is the quantity of heat necessary to convert a
pound of water into steam of the same temperature as the
steam in question.
To find the latent heat of steam, subtract ten times the
square root of the gauge pressure from 977.
Example. — Find the latent heat of steam at 169 lb. gauge
pressure.
Vim = 13
13 X 10 = 130
977 - 180 = 847 B. T. U.
HEAT 199
EXAMPLES
Find the latent heat of steam at the following gauge
pressures :
1. 132 lb. gauge pressure. 6. 39 lb. gauge pressure.
2. 116 lb. gauge pressure. 7. 41 lb. gauge pressure.
3. 208 lb. gauge pressure. 8. 160 lb. gauge pressure.
4. 196 lb. gauge pressure. 9. 159 lb. gauge pressure.
5. 84 lb. gauge pressure. 10. 180J^ lb. gauge pressure.
Heat Units Required to Produce a Given Pressure
To lind the number of units of heat required to raise the
temperature corresponding to one gauge pressure to that of
another, find the square roots of the gauge pressures, subtract
these values, and multiply by 14J.
Example. — Find the number of heat units required to raise
the temperature of 64 pounds gauge pressure to 169 pounds
gauge pressure.
\/(l4 = 8 13-8 = 6 14^x6 = 711
Vl«9 = 13 Approximately 72 B. T. U. Am.
EXAMPLES
Find the number of heat units required to raise the tempera-
ture between the following gauges:
1. From 64 to 128 lb. gauge. 5. From 42 to 121 lb. gauge.
2. From 26 to 131 lb. gauge. 6. From 28 to 132 lb. gauge.
3. From 39 to 149 lb. gauge. 7. From 33 to 144 lb. gauge.
4. From 49 to 165 lb. gauge. 8. From 55 to 164 lb. gauge.
Volume of Water and Steam »
According to steam tables one cubic foot of steam at 100
pounds' pressure weighs 0.2307 lb., one cubic foot of water
weighs 62^ lb., and one gallon of water may be taken as 8^ lb.
200 VOCATIONAL MATHEMATICS
At atmospheric pressure one cubic foot of steam has nearly
the weight of one cubic inch of water, and the weight increases
very nearly as the pressure. Hence, to find the number of cubic
inches of water required to make a certain amount of steam,
multiply the number of cubic feet of steam by the absolute
pressure in the atmosphere ; the product is the number of
cubic inches of water required. In all such calculations for
practical purposes, a liberal allowance must be made for loss
and leakage.
Absolute pressure is the total pressure, or the gauge pressure
plus the atmospheric pressure (which at sea level is 14.7 lb.
per sq. in.).
EXAMPLES
1. How much water will it take to make 800 cu. ft. of steam
at 10 lb. pressure ?
2. How much water will it take to make 3020 cu. ft. of
steam at 65 lb. pressure ?
3. How much water will it take to make 4812 cu. ft. of
steam at 8 lb. pressure ?
4. How much water will it take to make 512 cu. ft. of
steam at 75 lb. pressure ?
5. How much water will it take to make 1213 cu. ft. of
steam at 80 lb. pressure ?
Solve the following problems according to the table on the
next page :
6. What is the total steam pressure if the steam gauge reads
55 lb. ?
7. How many cubic feet of steam from 2 lb. of water at a
steam gauge pressure of 65 lb. ?
8. ,What is the latent heat of 1 lb. of water at a total
pressure of 75 lb. at 307.5° F. ?
9. What is the total heat required to generate 1 lb. of steam
from water at 32° F. under total pressure of 90 lb. ?
HEAT
201
Properties of Saturated Steam
Pbbmdeb
VOLUMB
Total Heat
Tempera-
ture In
Fahrenbeit
Dejfrees
Latent Heat
in B. T. U.
required to
XT
ToUl
Compared
with Water
(^Hblc Ft. of
Steam fi*»ttn
1 Lb. of
peneralc 1 Lb.
of Steum from
Water at «tio
under Conhtant
Water
I're.smire
Heat unltH
16
212.0
nm
26.36
966.2
1146.1
5
20
228.0
1220
10.72
952.8
1150.9
10
25
240.1
9JK5
15.99
945.3
1154.6
16
30
260.4
838
18.40
937.9
1157.8
20
36
269.3
726
11.65
931.6
1160.5
26
40
267.3
640
10.27
926.0
1162.9
30
46
274.4
672
9.18
920.9
1165.1
86
60
281.0
618
8.31
910.3
1167.1
40
66
287.1
474
7.61
912.0
1169.0
46
60
292.7
437
7.01
908.0
1170.7
60
65
298.0
405
6.49
904.2
1172.3
66
70
302.9
378
6.07
900.8
1173.8
60
76
307.5
353
5.68
897.5
1175.2
66
80
312.0
333
5.35
894.3
1176.6
70
86
316.1
314
5.05
891.4
1177.9
76
90
320.2
298
4.79
888.5
1179.1
80
95
324.1
283
4.55
885.8
1180.3
86
100
327.9
270
4.33
888.1
1181.4
00
106
331.3
267
4.14
880.7
1182.4
96
110
334.6
247
3.97
878.3
1183.6
100
116
338.0
237
3.80
875.9
1184.5
110
125
344.2
219
3.50
871.5
1186.4
120
135
350.1
203
3.27
867.4
1188.2
130
146
355.6
liK)
3.06
863.5
1189.9
140
165
361.0
179
2.87
859.7
1191.6
160
166
366.0
169
2.71
866.2
1192.9
Steam Heating
A steam heating system with steam having a pressure of less
than 15 lb. by the gauge is called a low pressure system. If
the steam pressure is higher than 15 lb. it is called a high
pressure system. When the water of condensation flows back
to the boiler by gravity alone, the apparatus is known as a
gravity circulating system. When the boiler is run at a high
pressure and the heating system at a low pressure, the con-
densed steam must be returned to the boiler by a pump,. steam
return trap, or injector.
202 VOCATIONAL MATHEMATICS
The quantity of heat given off by the radiators of steam pipes, in the
ordinary methods of heating buildings by direct radiation, varies from If
to 3 heat units per hour per square foot of radiating surface for each degree
of difference in temperature ; an average of from 2 to 2i is a fair estimate.
One pound of steam at about atmosplieric pressure contains 1146 heat
units, and if the temperature in the room is to be kept at 70"^ F., while the
temperature of the pipes is 212 degrees, the difference in temperature is
142 degrees. Multiplying this by 2J, the emission of heat will be 319^ heat
units per hour per square foot of radiating surface. A rule often given is
to allow one square foot of heating surface in the boiler for every eight to
ten square feet of radiating surface.
In steam heating the following rule is used: To find the
amount of direct radiating surface required to heat a room,
basing the calculation upon its cubic contents, allow one square
foot of direct radiating surface for the cubic feet shown in
the following table.
Proportion of Radiating Surface to Volume of Rooms
Bathrooms or living rooms, with 2 or 3 exposures . . . 40 cu. ft.
Living rooms, with 1 or 2 exposures 50 cu. ft.
Sleeping rooms 65-70 cu. ft.
Halls 50-70 cu. ft.
Schoolrooms 60-80 cu. ft.
Large churches and auditoriums 65-100 cu. ft.
Lofts, workshops, and factories 75-150 cu. ft.
The above ratios will give reasonably good results on ordi-
nary work if proper judgment is exercised.
EXAMPLES
1. How much radiating surface is necessary to heat a bath-
room containing 485 cu. ft. ?
2. How much radiating surface is necessary to heat a bath-
room 10^' X 5i' X 9' ?
3. How much radiating surface is necessary to heat a living
room with three windows, and containing 2798 cu. ft. ?
4. How much radiating surface is necessary to heat a living
room 18' X 16|-' X 10', with three windows?
CHAPTKK XIII
BOILERS
There are two classes of boilers — water tube and fire tube.
The difference between the two is that water flows through
water tube boilers and the fire to heat the water is outside,
while in the fire tube boiler the conditions are reversed.
Return Tubular Boilers. — The boiler most widely used in
America is the return tubular, which is a type of fire tube
boiler. It is a closed tube, simple in construction, inexpensive
to make, and easy to clean and repair. The first horizontal
tubular boiler was an ordinary storage tank made of iron.
Now horizontal tubular boilers, 16 to 20 feet long, and 4 to 8
feet in diameter, and even larger, are used and can withstand
a pressure of 150 lb. per sq. in.
Boilers up to fourteen feet in length are constructed of two plates, each
forming the entire circumference of the boiler. Above fourteen feet long
the shell is constructed of three plates which make the required length of
the boiler shell. These plates are J, f, ^, or ^ in. thick, having from
45,000 to 86,000 lb. tensile strength.
Tensile Strength. — The tensile strength is the pull applied
in the direction of its length required to break a bar of
boiler plate one square inch in area. Different pieces are
taken from the various parts of the boiler plate, reduced to \
inch square, and subjected to pressure on a testing machina
The average strength of the samples is thus obtained and multi-
plied by 16 to determine the strength of one square inch. The
tensile strength is usually stamped on the boiler steel. If it is
not stami)ed on it, the tensile strength is 48,000 lb.
203
§
as 45
03 g
^ a
mi
en
ts as
pd c3
O
N
o
W
BOILERS
205
If four samples give tests of 3998 lb., 4001 lb., 4001 lb., and 4000 lb.,
then the avera«;e is 4000. Therefore, for one square inch the tensile
strength is 4000 x 10 (the numlu'r of quartt-r sq. in. in one aq. in.) =
64,000 lb.
O
^######JJ i
SlNOLK RiVETKD LAP JoiVT ; ^ ErFirTEXCY ABOUT 56%
o
#--### #H
Double Riveted Lap Joint; Efficiency about 70%
Triple Riveted Butt-strap Joint; Efficiency about 85%
Safe Working Pressure. — In order to know the safe working
pressure of a single riveted boiler, it is necessary to multiply
one sixth of the tensile strength by the thickness of the shell,
and divide this product by the inside radius of the shell. If
1 The efficiency of a riveted joint id' the ratio of the strength of the joint to
that of the solid plate.
206 VOCATIONAL MATHEMATICS
a boiler is double riveted, add 20 per cent to the safe pressure
of a single riveted boiler of the same dimensions.
Example. — What is the safe working pressure of a single
riveted boiler having 72" diameter, i" shell, if the boiler plate
has a tensile strength of 66,000 lb. ?
I X 66,000= 11,000
11,000 X .5" = 5500
5500 H- 36" (radius) = 152| lb. approx. Ans.
EXAMPLES
1. (a) What is the safe working pressure of a single riveted
boiler of 60" diameter, j%'' shell, the T. S. being 60,000 lb.
per sq. in. ? (b) What is the safe working pressure of a double
riveted boiler of the same dimensions?
2. (a) What is the safe working pressure of a single riveted
boiler of 54'' diameter, f " shell, the T. S. being 60,000 lb. ?
(6) What is the safe working pressure of a double riveted
boiler of the same dimensions ?
3. (a) What is the safe working pressure of a single riveted
boiler of 48" diameter, f " shell, the T. S. being 60,000 lb. ?
(b) What is the safe working pressure of a double riveted
boiler of the same dimensions?
4. (a) What is the safe working pressure of a single riveted
boiler of 42" diameter, j\" shell, the T. S. being 60,000 lb. ?
(6) Of a double riveted boiler of the above dimensions?
The Plate. — The diameter of the rivet used for boiler plate
is generally double the thickness of the plate. The distance
between the rivet holes of a boiler is found by dividing the
area of the rivet hole by the thickness of the plate. The pitch
or distance between the rivet hole centers in a boiler plate
is found by dividing the area of the rivet hole by the thick-
ness of the plate and adding the diameter of one hole. The
pitch in a rivet hole is found, when the shearing strength is
208 VOCATIONAL MATHEMATICS
known, by multiplying the area of the rivet hole by the shear-
ing strength, then multiplying the thickness of the plate by
the tensile strength, dividing the first product by this product,
and adding one rivet hole diameter to the quotient.
Example. — What thickness of plate should be used on a
40-inch boiler to carry 125 lb. pressure, if the tensile strength
of the plate is 60,000 lb. ?
125 = steam pressure
6 = factor of safety
40 = diameter of boiler
20 = i diameter of boiler
60,000 = tensile strength of plate
126 X 6 = 750
750 X 20 = 15,000
— 5 — = .25 or i" thickness of plate. Ans.
60,000 * ^
The Boiler Inspection Department of Massachusetts recom-
mends the following fornuila for determining the thickness of
boiler plate :
rp^Px Rx F.S.
T.S, X %
T = thickness of plate
P = pressure
E = radius (| diameter of boiler)
F. S. - safety factor
T. S. = tensile strength
% = strength of joint
Example. — What thickness of plate should be used on a
40-inch boiler to carry 125 lb. pressure, if the strength of the
plate is 60,000 lb., using 50 % as the strength of the joint?
r = ^f><^Qxe = |// sheet. Ans.
60,000 X .50 ^
Find the safe working of the same boiler with the above
figures :
Safe working pressure = 60,000 x .5 x .50 ^ ^35 lb. Ans.
20 X 6
BOILERS
209
Small Vertical Boilkr
Size. — The engineer often has to calculate the size of a
boiler to carry a definite steam pressure.
The size of a single riveted l)oiler may be found by multi-
210 VOCATIONAL MATHEMATICS
plying J of the tensile strength by the thickness of the shell,
and dividing this product by the steam pressure. The quo-
tient is the radius of the boiler. Multiply this radius by two
to obtain the diameter (twice this radius equals the theoretical
diameter) ; add one fifth of the diameter just found as a safety
factor, and this sum gives the working diameter of a boiler
that will safely carry the required pressure.
Example. — What is the diameter of a ^Ijoiler that will with-
stand 150 lb. pressure, if made of f" steel, 60,0U0 lb. T. S.?
I X 60,000 = 10,000
10,000- X f = 3750
-^■^-^Jt = 2.j" tlieoretical radius
50' = theoretical diameter
50" 4- 10" = 60" working diameter. A7is.
The Boiler Inspection Department of Massachusetts recom-
mends the following formula for finding the diameter of a boiler
when the pressure, thickness, tensile strength, and per cent are
known.
rf^ Tx T.S.x % 2
P.F.
D = diameter of boiler
T = thickness of plate
T. S. = tensile strength
0/, = strength of joint
P = pressure
F = safety factor
Example. — What is the diameter of a boiler having |" shell,
allowing 50 per cent for the strength of the joint, with a tensile
strength of 60,000 lb., when the factor of safety is 6 and the
pressure of steam is 125 lb. ?
^ .l>.4-
7854
226 VOCATIONAL MATHEMATICS
Diameter of Supply Pipe. — The diameter of the steam supply-
pipe for a given engine may be calculated from the H. P. of
the engine by dividing the H. P. by 6, and extracting the
square root of the quotient.
hTp.
0=4
6
Example. — What is the diameter of a steam supply pipe
of a 216 H. P. engine.?
216 - 6 = 36
\/36 = 6", diameter of supply pipe. Ans.
EXAMPLES
1. Find the diameter of the steam supply pipe of a 180
H. P. engine.
2. What is the H. P. of an engine whose area of piston is
114 sq. in., mean effective pressure 80, and velocity of piston
112 ft. per minute ?
3. What is the approximate diameter of a cylinder of an
engine of 50 H. P. ?
4. What is the approximate H. P. of an engine the cylinder
diameter of which is 28" ?
5. What is the effective area, for power calculation, of the
piston of a steam engine, the bore of the cylinder being 28",
and the diameter of the piston rod which passes through both
ends of the cylinder 2" ?
6. What is the H. P. of an engine that can raise 3 tons of
coal (1 ton = 2240 lb.) from a mine 289 ft. deep in 9 minutes?
7. What is the approximate H. P. of an engine the cylinder
diameter of which is 16" ?
8. Find the diameter of the steam supply pipe of a 24 H. P.
ENGINES 227
9. What is the approximate diameter of a cylinder of an
engine of 80 H. P. ?
10. What is the approximate H. P. of an engine the cylinder
diameter of which is 20" ?
11. What is the H. P. of an engine whose diameter of piston
is 15", mean effective pressure 110, velocity of piston per
minute 189' ?
12. How many pounds of water per half minute can an 8 H. P.
fire pump raise to a height of 86 ft. ?
13. What is the effective area in square inches of the piston
of a steam engine if the diameter of the cylinder is 20" and
the diameter of the piston rod is 3" ?
14. What is the horse power of an engine that is required
to pump out a basement 51' x 22' x 10' deep, full of water, in
20 minutes ?
15. (a) Find the diameter of the steam supply pipe of a
98 H. P. engine.
(6) Find .the approximate diameter of a cylinder of an eugine
of 48 H. P.
(c) What is the approximate H.P. of an engine the cylinder
diameter of which is 28" ?
Steam Indicator
In order to know the condition of the steam within the
cylinder of an engine an indicator is used. This consists of
a small cylinder containing a piston, the rod of which is en-
closed in a spiral spring which opposes the motion of the piston.
The piston rod, after i)assing through the top of the cylinder
cover, is connected with a long light lever, on the end of which
is a pencil. This pencil moves in a vertical straight line when-
ever the piston moves.
Another cylinder with an axis parallel to the first carries a
paper drum, and this drum is connected to the crosshead of
the engine by means of a cord and a reducing motion, so that
228
VOCATIONAL MATHEMATICS
the movement of the drum is proportional to that of the cross-
head. When the pipe between the indicator and the engine
is closed by means of a cock, the pencil, when held against the
drum, makes a horizontal line called an atmospheric line. If
Cross Section of
Steam Indicator
the cock is opened, admitting to the small cylinder of the in-
dicator the pressure that exists in the engine cylinder, the pen-
cil will trace a figure, every point of which is at a height from
the atmospheric line proportional to the number of pounds'
pressure in the engine cylinder at every point in the stroke.
ENGINES
229
Operating Power. — To find the power required to drive a
certain machine when driven direct from the shaft or engine :
Indicate the engine with the machinery running and calcu-
late from the card. Then indicate the engine (from formula
on p. 225) without the machinery running and from this obtain
the H. P. The difference will give the power for operating.
To indicate an engine means to place a steam engine indicator on the
steam engine and record on the diagrams the pressure in the steam cyl-
inder at every part of the stroke. The diagrams are measured by means
of a planimeter or by means of ordinutes. The latter way may be done
by dividing the diagram into 10 or 20 equal spaces by vertical lines.
On these verticals measure the length between the back pressure line and
mark each length on a long strip of paper. Dividing the sum of all these
lengths by 10 or 20 will give the average length of ordinate which multiplied
by the scale of the spring will give the mean effective pressure (M.E.P.).
Diagram from Hartford Engine
Cylinder, IH X 24 inches. B<»iler pressure, 87 pounds. Vacuum by gauge, 23i
inches. 130 revolutions per minute. Scale, 50. The vertii-al lines from A
are called ordinates.
Example. — If the total length is 9" and the spring used
is 40 lb., how is the mean effective pressure (M. E. P.) found ?
40 X .9 = 36 lb. M. E. P. Ans.
230 VOCATIONAL MATHEMATICS
EXAMPLES 1
1. If the total length of the ordinates is 11" and the spring
used is 56 lb., what is the M. E. P. ?
2. A 9 X 10 engine has a 48 lb. average pressure on the
piston, the length of rod is 35", and the crank is 5". Find the
maximum thrust at right angles.
3. Find the weight of a flywheel of an engine 12'' x 24",
diameter of wheel 7', with 168 K. P. M.
4. What is the necessary steam lap of an engine with a
45" stroke, valve travel 5", and cut-off at half stroke, and valve
i" lead ?
5. If the total length of ordinates is 10" and the spring
used is 49 lb., what is the M. E. P. ?
6. A 12'' X 24" engine has a 50 lb. average pressure on the
piston, the length of the rod is 59", and the crank is 12" ; what
is the maximum thrust at right angles ?
7. What is the necessary steam lap of an engine with a
64" stroke, valve travel 7, cut-off at half stroke, and valve ^"
lead ?
8. If the total length of ordinates is 8" and the spring used
is 38 lb., what is the M. E. P. ?
9. Find the weight of a flywheel of an engine 24" x 60",
the diameter of wheel being 24' and having 75 R. P. M.
10. If the total length of ordinates is 12" and the spring
used is 61 lb., what is the M. E. P. ?
11. Find the weight of a flywheel of an engine 30" x 60",
the diameter of the wheel 30', with 63 K P. M.
12. An engine 10" x 12" has a 35 lb. average pressure on the
piston, the length of the rod is 42", and the crank is 6". Find
the maximum thrust at right angles.
1 Use 10 ordinates in solving problems.
PART VIII — MATHEMATICS FOR ELECTRICAL
WORK
CHAPTER XV
COMMERCIAL ELECTRICITY
Amperes. — What electricity is no one knows. Its action,
however, is so like that of flowing water that the comparison
is helpful. A current of water in a pipe is measured by the
amount which flows through the pipe in a second of time, as
one gallon per second. So a current of electricity is measured
Water Analogy of Fall of Potential
by the amount which flows along a wire in a second, as one
coulomb per second, — a coulomb being a unit of measurement
of electricity, just as a gallon is a unit of measurement of
water. The rate of flow of one coulomb per second is called one
ampere. The rate of flow of five coulombs per second is five
amperes.
Volts. — The quantity of water which flows through a pipe
depends to a large extent upon the pressure under which it
flows. The number of amperes of electricity which flow along
231
232 VOCATIONAL MATHEMATICS
a wire depends in the same way upon the pressure behind it.
The electrical unit of pressure is the volt. In a stream of
water there is a difference in pressure between a point on the
surface of the stream and a point near the bottom. This is
called the difference or drop in level between the two points.
It is also spoken of as the pressure head, " head " meaning the
difference in intensity of pressure between two points in a body
of water, as well as the intensity of pressure at any point.
Similarly the pressure (or voltage) between two points in an
electric circuit is called the difference or drop in pressure or
the poteyitial. The amperes represent the amount of electricity
flowing through a circuit, and the volts the pressure causing
the flow.
Ohms. — Besides the pressure the resistance of the wire
helps to determine the amount of the current: — the greater
the resistance, the less the current flowing under the same
pressure. To the electrical unit of resistance the name ohm
is given. A wire has a resistance of one ohm when a pressure
of one volt can force no more than a current of one ampere
through it.
Ohm's Law. — The relation between current (amperes),
pressure (volts), and resistance (ohms) is expressed by a law
known as Ohm^s Law. This is the fundamental law of the
study of electricity and may be stated as follows :
An electric current flowing along a conductor is equal to
the pressure divided by the resistance.
Current (amperes) = - — — ^— — ^
Resistance (ohms)
Letting /= amperes, -EJ = volts, i? = ohms,
I=E-r- Rov I = ~
R
E=IR
R = ^
I
ELECTRICAL WORK 233
Example. — If a pressure of 110 volts is applied to a re-
sistance of 220 ohms, what current will flow ?
/ = ^ = li2 = 1 = .6 ami^re. Ans.
R 220 2
Example. — A current of 2 amperes flows in a circuit the
resistance of which is 300 ohms. ^Vhat is the voltage of the
circuit ?
IIi = E
2 X 300 = 600 volte. Ans.
Example. — If a current of 12 amperes flows in a circuit
and the voltage applied to the circuit is 240 volts, find the
ijesistance of the circuit.
^=Ji ?i2 = 20 ohms. Ans.
Ammeter and Voltmeter. — Ohm's Law may be applied to a
circuit as a whole or to any part of it. It is often desirable to
/
Ammbtkb Voltmetke
know how much current is flowing in a circuit without calcu-
lating it by Ohm's Law. An instrument called an ammeter is
used to measure the current. This instrument has a low
resistance so that it will not cause a drop in pressure. A
voltmeter is used to measure the voltage. This instrument has
high resistance so that a very small current will flow through
234 VOCATIONAL MATHEMATICS
it, and is always placed in shunt, or parallel (see p. 235) with
that part of the circuit the voltage of which is to be found.
Example. — What is the resistance of wires that are carry-
ing 100 amperes from a generator to a motor, if the drop or
loss of potential equals 12 volts ?
Drop in voltage = IB
I = 100 amperes
Drop in volts = 12
B = ? ohms
-f
B-
= — = 0.12 ohm. J
100
Ans.
Example. — A circuit made up of incandescent lamps and
conducting wires is supplied under a pressure of 115 volts.
The lamps require a pressure of 110 volts at their termiiials
+ lead
Wiring of Incandescent Lamp Circuit
and take a current of 10 amperes. What should be the resist-
ance of the conducting wires in order that the necessary cur-
rent may flow ?
Drop in conducting wireS = 115 — 110 = 5 volts
Current through wires = 10 amperes
E 5
B = — zz — = 0.5 ohm resistance. Ans.
I 10
EXAMPLES
1. How much current will flow through an electromagnet
of 140 ohms' resistance when placed across a 100-volt circuit ?
2. How many amperes will flow through a 110-volt lamp
which has a resistance of 120 ohms ?
3. What will be the resistance of an arc lamp burning
upon a 110-volt circuit, if the current is 5 amperes ?
ELECTRICAL WORK
235
4. If the lamp in Example 3 were to be put upon a 150-
volt circuit, how much additional resistance would have to be
put into it in order that it might not take more than 5
amperes ?
Motor
Electric Road System
5. In a series motor used to drive a street car the resistance
of the field equals 1.06 ohms; the current going througli equals
30 amperes. What would a voltmeter indicate if placed
across the field terminals ?
6. If the load upon the motor in Example 5 were increased
so that 45 amperes were flowing through the field coils, what
would the voltmeter then indicate ?
Series and Parallel Circuits
Pieces of electrical apparatus may be connected in two ways.
When the pieces are connected so that the current passes
through them in a single path, they are said to be in series.
Wmf\ -
B
Cells Connected in Parallel
^}\m\
Cells Connected in Series
WTien the pieces are connected so that the current is divided
between them, they are said to be in jxtrallel with one another..
The total resistance of a series circuit is equal to the sum of
the resistances of the separate parts of the circuit. The total
236 VOCATIONAL MATHEMATICS
voltage of a series circuit is equal to the sum of the voltages
across the separate resistances.
A /?i /?2 B
Total resistance A to B = Ei -\- R.^-
Example. — If there is a circuitof 240 volts and the lamps
are of 240 ohms' resistance but made to carry only ^ an ampere,
two lamps would have to be put in series in order to use them
on the 240-volt circuit.
The resistance of the two lamps in series would then be 480 ohms,
the voltage of the circuit 240 volts, and the current by Ohm's Law
I= — = — = - ampere.
H 480 2
A i?i R-2 . B
In any closed circuit, the algebraic sum of the products
found by multiplying the resistance of each part by the current
passing through it, is equal to the voltage of the circuit.
Example. — If three lamps of 110 ohms' resistance are con-
nected in series and take i ampere, the voltage of the circuit
IiBi 4- IoIi2 + Is^i =
(^x 110) + (i X 110) + (^x 110) =
55 + 55 -f 55 =: 165 volts. Ans.
Example. — A current of 50 amperes flowed through a
circuit when the voltage was 550. What resistance should be
added in series with the circuit to reduce the current to 11
amperes ? _ .
Resistance = ^■^^- = 11 ohms
Resistance = ^f^- = 50 ohms
Additional resistance = 50 — 11 = 39 ohms. Ans.
Example. — The voltage required by 15' arc lamps connected
in series is 900 and the current is 6 amperes. If the resistance
of the connecting wires is 5 ohms, how much additional volt-
age will be necessary so that the lamp voltage may not drop
below 900 ?
Drop in voltage in connecting wires = E = IE
6 X 5 = 30 volts = additional voltage necessary. Ans.
ELECTRICAL WORK 237
Example. — The field coil of a motor having 4 poles is
measured for voltage across the terminals and the following
readings are taken :
Voltage across line = 220
Voltage across coil No. 1 = 78.33
Voltage across coU No. 2 = 00.00
Voltage across coil No. 3 = 73.33
Voltage across coil No. 4 = 73.33
Current flowing =1.5 amperes.
What is the total resistance and what is the trouble at coil
No. 2 ?
Total resistance = 7? = ^ = ^=^ = 140.6 ohms. Am.
I 1.5
As thei-e is no drop across the terminal of coil No. 2, there is prac-
tically no resistance and the current is not going around the coil but
throusjli a path of extremely low resistance.
EXAMPLES
1. If three electromagnets are connected in series and the
resistances are 3, 5, and 17 ohms, respectively, what is the
total resistance of this set ?
2. The field coils of a series motor have a resistance of 10
ohms and the armature has a resistance of 7 ohms ; what is the
total resistance of the motor ?
3. («) What would be the total resistance of two 110-volt
incandescent lamps placed in series across a 110-volt line if
each lamp has a resistance of 220 ohms? {h) Would these
lamps light on this voltage in this position ? (c) Why ?
4. Three coils are connected in series and have a resistance
of 3, 5, and 8 ohms, respectively. What current will flow if
the voltage of the circuit is 64 ?
5. Five arc lamps, each having a resistance of 4 ohms, are
connected in series. The resistance of the connecting wires
and the other apparatus is o ohms. What must be the voltage
of the circuit so that a current of 10 amperes may flow?
238 VOCATIONAL MATHEMATICS
6. A current of 10 amperes was passing througli a circuit
under a pressure of 550 volts. The circuit was made up of
three sections connected in series, and the resistance of two
sections was 8 and 12 ohms, respectively. What was the
resistance of the third section ?
Example. — If 5 electromagnets are arranged in series and
marked A, B, C, D, and E, the resistance of the circuit is 45
ohms and the resistance of each coil is : ^, 5 ohms ; B, 10 ohms ;
C, 7 ohms; D, 8 ohms; E, 15 ohms. How much E. M. F.
would be required to cause 10 amperes to flow through the
coils, and what would be the E. M. F. across the terminals of
each coil ?
Voltage across ^ = 50
Voltage across jB = 100
Voltage across C = 70
Voltage across i> = 80
Voltage across E = 150
450
10 X 45 = 450, total voltage. Ans.
Example. — What E. M. F. is necessary to send a current
through 10 field coils connected in series, if each has a resist-
ance of 10 ohms and 3 amperes are required to produce the
necessary magnetization ?
10 ohms = resistance of each coil
10 coils are in series
10 X 10 = 100 ohms, total resistance
B = 100 ohms
7=3 amperes
E = IB =3x100= 300 volts. Ans.
In any closed circuit the algebraic sum of the products found
by multiplying the resistance of each part by the current
passing through it, is equal to the voltage of the circuit. This
is practically an inverse statement of the law of series circuits.
Example. — If we have five lamps of 110 ohms' resistance,
connected in series and taking ^ ampere, the voltage of the
circuit is :
ELECTRICAL WORK 239
E = IxRi + hRi + 7,^3 + URi + /»/?» = i X 110 + i X 110 -f i X 110
+ \ X 110 + i X 110 = 66 + 66 + 66 -h 65 -f- 66 = 276 volts. Ans.
In a parallel circuit the voltage across each branch is the
same as the voltage across the combination. The current is
equal to the sum of the currents in the separate parts.
The resistance is equal to the reciprocal of the sum of the
conductances of the separate parts.
Conductance is the reciprocal of resistance and is equal to
— . The unit of conductance is mho (ohm spelled backwards).
mho= — — or ohm =
ohm mho
Series and parallel circuits may be combined and may exist
in the same circuit. In parallel circuits the reciprocal of the
total resistance is equal to the sum of the reciprocals of the
paralleled resistance.
If Rq = resistance of circuit and
rj, r2, and r^ = parallel resistances,
ri _
A r^ B
rg
resistance between A and B = Rq.
R n r^ 'i\
Example. — Suppose that ?•,, Vz, and r^ are lamps of 300
ohms' resistance each, then
Bo 300 300 300 iWO 100
Bo = 100 ohms. Ans.
Or, ri = 100 ohms J__J_ ,_L,_L_ 11L__L
rj= 60 ohms Bo 100 60 300~300~30
rs = 300 ohms Bo = 30 ohms. Ans.
When it is necessary to get tlie resistance of parallel
circuits, it is often more convenient to use the sum of the
conductances as the total conductance of the circuit.
240 VOCATIONAL MATHEMATICS
Example. — Three resistances in parallel :
7*1 = 2 ohms
^2 = 5 ohms
rg = 1 olim
A
I'n
Conductance
» H-
.5 mho
<^>H'
: .2 mho
<" rr
: 1 mho
1.7 mho'conductanc
Resistance of circuit AB = — - = .58^
Ans.
EXAMPLES
1. Ten arc lamps of 200 ohms' resistance are connected in
series and the voltage of the circuit is 300. How much current
does each lamp take ?
2. What will be the E. M. F. necessary to supply 60 Thomp-
son-Houston arc lamps arranged in series, the resistance of each
lamp being 5 ohms when burning, making a total resistance of
300 ohms in the circuit, if the current required is 10 amperes ?
3. Four parallel circuits of 2, 4, 5, and 10 ohms' resistance,
respectively, have 40 volts impressed upon their terminals,
(a) What is the total current flowing ? (b) How much current
flows in each branch ?
4. Three incandescent lamps of different sizes are placed in
parallel on a circuit. These have respective resistances of 100,
150, and 300 ohms. What is the total current flowing through
these lamps when the pressure applied is 100 volts?
ELECTRICAL WORK 241
5. What is the total resistance of four incandescent lamps
placed in parallel, if each lamp has a resistance of 220 ohms ?
6. What is the resistance of a shunt-wound generator, if the
field and the armature are respectively 25 and 10 ohms ?
7. Two coils on a large electromagnet for lifting iron ore are
connected in parallel and the resistance of each coil is 40 ohms.
What is the whole resistance ?
8. The resistance of a line frou) a power house to a mill is
6 ohms; there are 50 lamps in the mill, each lamp having a re-
sistance of 220 ohms. What is the total resistance from the
power house ?
9. In an electric street car 4 heaters are all connected in
series and each has a resistance of 20 ohms with a voltage of
600 across the circuit, (a) What is the total resistance of
these ? (b) How many amperes will go through them ?
Power Measurement
The flow of an electric current has been compared to the flow
of water through a pipe. The water current is measured by
the number of gallons or pounds flowing per minute. A cur-
rent of electricity is measured by the number of amperes or
coulombs per second. When a gallon of water is raised a foot
by means of a pump a certain amount of work is done. So
when a coulomb of electricity is passed through a wire under
the pressure of one volt, a certain amount of work is done. In
the case of water the work done is measured in foot pounds.
A foot pound is the work done in raising a weight of one pound
through a distance of one foot.
Work = Force x Distance
When one coulomb of electricity is passed through a wire
under a pressure of one volt, the amount of work done is called
one joule.
The ix)wer required to keep a current of water flowing is the
product of the current in pounds per minute by distance in feet.
242
VOCATIONAL MATHEMATICS
This gives the power in foot pounds per minute. Mechanical
power is usually expressed in horse power (H. P.). The power
required to keep a current of electricity flowing is the product
of the current in amperes by the pressure in volts and is ex-
pressed in watts.
1 H. P. = 746 watts
1000 watts = 1 kilowatt
1 kilowatt (K. W.) = 1.34 or li- H. P.
Volts X Amperes ^j^.j^^^^^^
lOUO
Let
P = power in watts
/ = current in amperes
E = pressure in volts
then (1)
F= IE (equation for power)
but
E=TB (Ohm's Law)
therefore
P= I (IB) or PB (by substitution)
(2)
P=I^B
but
7 = ^ (Ohm's Law)
B
therefore
P = ^ (E) or ^ (by substitution)
B B
(3)
■ -f
thus
P= IE= DB =
E^
B
To measure power a wattmeter is used, which is a combina-
tion of a voltmeter and an ammeter.
In order to find the amount of work done by a certain engine,
it is necessary to know the time it has been running and the
power it has been supplying, i.e. its rate of doing work. If
the power is measured in horse power and the time in hours,
the work is done in horse power hours. Similarly, if the
power is measured in kilowatts and the time in hours, the work
done is measured in kilowatt hours (K. W. H.).
1 H. P. H. = 0.746 K.W. H.
1 K. W. H. = 1.84 H. P. H.
ELECTRICAL WORK 243
These units are too large to be used conveniently in all problems,
so a smaller electrical unit called the watt secondf or joulCf is
used.
EXAMPLES
1. If the resistance of a circuit is 1 ohm and the current 30
amperes, what energy is expended in one half hour ?
2. With a potential difference of 95 volts and a current of
15 amperes, what energy is expended in 20 minutes?
3. If a current of 100 amperes flows for 2 minutes under a
pressure of 500 volts, what is the work done in joules ?
4. If 12 incandescent lamps burn for 10 hours under a
pressure of 110 volts, each lamp consuming J an ampere, how
many kilowatt hours are used ?
5. Fifty horse power expended continuously for one hour
will produce how many kilowatt hours ?
6. If 4000 watts are expended in a circuit, how much horse
power is being developed ?
7. If 20 horse power of mechanical energy were converted
into electrical energy, how many watts would be developed ?
8. If a current of 50 amperes flows through a circuit under
a pressure of 220 volts, what is the power ?
9. If 200 watts are expended in a circuit by a current of 4
amperes, what is the voltage required to drive the current
through the wire ?
10. If an incandescent lamp requires \ an ampere of current
and the resistance of its filament is 220 ohms, how many watts
are required for it ?
Measurement of Resistance
The amount of current in a circuit depends upon the voltage
and upon the resistance. To control the current it is necessary
to change one of these two factors. The resistance to the flow
of water through a pipe depends upon the shape of the pipe and
244 VOCATIONAL MATHEMATICS
its length. The electrical resistance of a conductor depends
upon the nature of the metal from which the conductor is made,
its size, its length, and the temperature. The greater the size
of the conductor, the greater is its power for conducting elec-
tricity, and therefore the less its resistance. The longer the
wire, the less its conducting power, and therefore the greater
the resistance. As the resistance of a large pipe is less than
the resistance of a small one, so the resistance of a large wire
is less than the resistance of a small one.
Copper is the material generally used for wires, and its con-
ductivity, or capacity for conducting current, is taken as the
standard. The conductivity of pure copper is expressed as
100 % . Commercial copper usually has from 98 to 99 % con-
ductivity. Other materials used in electric wires are iron,
aluminum, brass, etc. Iron has a conductivity of about 16 %
and brass of about 25 %. It would require an iron wire with
over 6 times the cross section of a copper wire to give the same
conductivity, and brass wire would have to be about 4 times as
large in its cross section for the same conductivity. Generally
it is assumed that all electric wires are copper.
In measuring the length of wires the unit used is feet, while
the cross section area is measured in circular mils; y^Vir ^^
an inch is called a m^7, and a round wire one mil in diameter
is said to have a cross-section area of one circular mil. A
wire 1 foot long, with a cross-section area of 1 circular mil,
is called a mil-foot wire. The area of any wire in circular mils
may be found by squaring the number of thousandths of an inch
in the diameter.
If B = resistance of wire in ohms
L = resistance of wire
D = diameter of wire in mils
cP = area of wire in circular mils
K = resistance of 1 mil foot in ohms called
resistivity or specific resistance of material
then B = ^
ELECTRICAL WORK 245
Note. — /T varies with the niAterial and the temperature. The resist-
ance of 1 mil foot of soft copper wire at 60° F. is 10.4 ohms.
Example. — What is the area of a wire 0.1 inch in diameter ?
0.1 inch = 100 thouBandths of an inch
100 X 100 = 10,000, number of circular ihils. Ans.
Example. — What is the resistance at ordinary temi)erature
of a copper wire 2500 ft. long with a cross-section area of
10,000 circular mils ? (K= 10.)
^^^^10^^^500^2.6 ohms. Aus.
d^ 10000
EXAMPLES
1. What must be the diameter of a wire in mils in order
that it may have a cross-section area of 200 circular mils ?
2. How many circular mils are there in a wire 50 mils in
diameter ?
3. How many circular mils are there in a wire 150 mils in
diameter ?
4. What is the diameter in inches of a copper wire which
has a cross-section area of 20,000 circular mils ?
5. If it is desired to have a copper wire of J ohm resistance
and 2000 ft. long, what must its cross section be ?
6. What pressure is required to force a current of 50 am-
peres over a copper wire IGOO ft. long which has a cross-
section area of 20,000 circular mils ?
7. A current is forced through a copper wire 2000 ft. long
under a pressure of 50 volts. If the wire has an area of 5000
circular mils, what is the value of the current flowing?
8. If a wire 5000 ft. long carries a current of 5 amperes
under a pressure of 100 volts, what is the cross-section area of
the wire ?
246 VOCATIONAL MATHEMATICS
9. A wire 100 ft. long is in series with another wire 500 ft,
long and the first wire is ^" in diameter. If the second wire
has a cross-section area of 20,000 circular mils, what is the
resistance of the circuit ?
10. How many amperes will a wire ^" in diameter carry if
a wire 1000 mils in diameter will carry 650 amperes ?
Size of Wire
If an electrician wishes to know the size of a wire to carry
a certain current a certain distance with a certain drop of
voltage, he may ascertain it by substituting values in the foL
lowing formula, which is called a two-wire formula,
e
CM = size of wire in circular mils
D = distance from distribution
7= amperage
e = drop or volts lost
Example. — What size of wire will be required for a motor
situated 85 ft. from the center of distribution, if the motor is
5 H. P., operating at difference of potential of 110 volts, allow-
ing 3 % drop ?
85 ft.
Ans.
P=ET
1 H. P. = 746 watts
1=^ j^3730 g:= no X. 03=3.30 D=
P = 746 X 6 = 3730
E=no
21.6x85x^^3^
110 _ 21.6 X 85 X 3730
3.30 3.30 X 110
= 18865.7 wire.
REVIEW EXAMPLES
1. What size of wire will be required for a 15 H. P. motor
operating at 550 volts and situated 35 ft. from the center of
distribution, allowing a 2 % drop ?
ELECTRICAL WORK 247
2. J low many coulombs are delivered in a minute when
the current is 17^ amperes ?
3. What is the current when 480 coulombs are delivered
per minute ?
4. In what time will 72,000 coulombs be delivered when
the current is 80 amperes?
5. What size of wire will be required for a 7^ H. P. motor
operating at 220 volts and situated 65 ft. from the center of
distribution and allowing a 5 % drop ?
6. If a current of 20 amperes flows through a circuit for
21.2 hours, what quantity of electricity is delivered?
7. How many ampere hours pass in a circuit in 2^ hours
when the current is 1 6 amperes ?
8. What size of wire will be required for a 10 H. P. motor
operating at 110 volts and situated 105 ft. from the center of
distribution, allowing a drop of 3 volts?
9. If 200 coulombs of electricity are passed through an
electrolytic vat each second under a pressure of 10 volts, how
many joules of work are expended in an hour ?
10. What quantity of electricity must flow under a pressure
of 5 volts to do 125 joules of work ?
11. If 10 coulombs do 10 joules of work flowing through
a wire, what is the pressure ?
12. What must be the diameter of a wire in mils in order
that it may have a cross section of 200 circular mils?
13. A wii-e 5000 ft. long carries a current of 6 amperes
under a pressure of 100 volts. What is the cross-section area
of the wire ?
14. A copper wire ^ inch in diameter and 100 ft. long is in
series with another copper wire 500 ft. long with a cross
section of 20,000 circular mils. What is the resistance of the
circuit ?
248 VOCATIONAL MATHEMATICS
15. What is the resistance at ordinary temperature of a
copper wire 2500 ft. long having a cross-section area of 10,000
circular mils ?
16. If it is desired to have a copper wire of i ohm resistance
and 2000 ft. long, what must be its cross-section area?
17. A voltmeter which measures the pressure on a circuit
registers 500 volts and the ammeter on the same circuit shows
25 amperes. What is the resistance of the circuit ?
18. An electromagnet has a resistance of 25 ohms, and there
must be 41 amperes passing through it in order that the
magnetism may be strong enough. What must be the voltage ?
19. What current is needed to light a 16 C. P. lamp, if the
hot resistance of the lamp is 220 ohms and the voltage is 110?
20. A storage battery gives 2.3 volts and it is connected
with a coil having a resistance of 25 ohms. What current
will flow through the circuit if the internal resistance of the
cell is zero?
21. What is the power necessary to drive a current of 500
amperes through a resistance of 5 ohms ?
22. How many watts of power are going to an electric motor
if the voltage of the line is 500 and there are 7 amperes enter-
ing the motor ?
23. How many H. P. are needed to run a dynamo that is
lighting 258 lamps in parallel, if each lamp takes | an ampere
at 110 volts ?
24. How many 40-watt electric glow lamps can be run on a
110- volt circuit with an expenditure of 48 amperes of current ?
Brown and Sharpe Wire Table
The unit chosen for this table is a copper wire .1 inch in
diameter. This is called No. 10 wire and has the following
characteristics: No. 10, B & S wire, diameter, .1 inch; area in
circular mils, 10,000 cm.; resistance, per 1000 ft., 1 ohm;
ELECTRICAL WORK 249
weight per 1000 ft., 31. o lb. Since this table was made, the
standard has changed slightly, so that at present No. 10 wire
is .1019 inch in diameter and the other vahies are changed
l)roportionately, but for all ooiumercial work the values as
originally given are sufficiently accurate.
The table is so arranged that the area of the wire is doubled
every three gauges down and halved every three numbers up.
Example. — Find the area of No. 7 wire.
No. 7 is three numbers below No. 10, whose area is 10,000 cm., so that
its area is 2 x 10,000 = 20,000 cm.
Area of No. 4 wire = 2 x 20,000 = 40,000 cm.
Area of No. 13 wire = I x 10,000 = 5000 cm.
The resistance is reduced to one half every three numbers
down and doubled every three numbers up. The weight
doubles every three numbers down and halves every three
numbers up.
Example. —
R per 1000 ft. of No. 10 H & S wire equals 1 ohm
R per 1000 ft. of No. 7 H & S wire equals .5 ohm
R per 1000 ft. of No. 4 H & S wire equals .25 ohm
R per 1000 ft. of No. 13 B & S wire equals 2 ohms
R per 1000 ft. of No. 16 B & S wire equals 4 ohms
Example. —
TTper 1000 ft. of No. 10 B & S wire equals 31.5 lb.
W per 1000 ft. of No. 7 B & S wire equals (k^ lb.
IF per 1000 ft. of No. 4 B & S wire equals 126 lb.
TKper 1000 ft. of No. 13 B & S wire equals 15.8 lb.
IK per 1000 ft. of No. 16 B & S wire equals 7.9 lb.
If the gauge number is not three or a multiple of three be-
low or above No. 10, get the area, resistance, or weight desired
which is less than the value for the wire required, and if it is
one below the required number, multiply by 1.26, and if two
below, by 1.59.
250 VOCATIONAL MATHEMATICS
In computing the resistance and weight of cables the follow-
ing formula is used :
Resistance = R = in ohms per 1000 ft.
cm.
Weight = .00305 x cm. in lb. per 1000 ft.
EXAMPLES
Find the resistance, area, and weight of the following wires :
1. No. 16. 6. No. 14.
2. No. 13. 7. No. 15.
3. No. 7. 8. No. 10.
4. No. 3. > 9. No. 2.
5. No. 1. 10. No. 11.
PART IX — MATHEMATICS FOR MACHINISTS
CHAPTER XVI
MATERIALS
Every machinist should be familiar with the strength and
other properties of the materials that he uses — such metals as
cast iron, wrought iron, steel, copper, bronze, and brass.
Cast Iron. — Since much of the machinist's work is on cast iron, he
should know something of its nature and manufacture. Iron ore as
found in the earth generally contains many impurities, such as silicon,
sulphur, phosphorus, manganese, combined carbon, and graphitic carbon.
To free the iron from the grosser impurities, the ore is crushed and mixed
with coke and limestone and intense heat applied in a blast furnace. The
melted iron, being heavier than the other materials, falls to the bottom of
the furnace. When a sufficient quantity has accumulated, it is allowed
to flow out of a tap hole into molds of sand. After it has cooled it is
broken into lengths suitable to be remelted in foundries and made into
iron castings. Such iron is called pig iron.
During the process of smelting in the blast furnace, the liquid iron
combines with a considerable quantity of carbon, sulphur, silicon, phos-
phorus, and manganese from the impurities in the ore and coke. Some
of the carbon combines with the iron chemically and forms iron carbide,
while the remainder exists in the iron as a mixture of carbon and is
known as graphite. The amount of carbon may weaken the iron by
making it soft, and it may also make the iron too brittle to work. So
the man in charge of the foundry must use his judgment in mixing
different grades and quantities of pig iron to obtain a casting of the de-
sired strength, hardness, toughness, and clearness of grain.
Castings. — Machines are made of iron castings, forgings, steel parts,
etc. Castings and forgings can be distinguished from each other by the
appearance of the fractures in them. After the machines are designed
and the wooden patterns made in the pattern shop, the patterns are sent to
251
252 VOCATIONAL MATHEMATICS
the foundry, where aa impression of the machine is made in sand. Dur-
ing this operation of molding, the sand is confined in an iron or wooden
device of two or more parts called a flask. The lower or bottom part of
the flask is the drag or nowel, while the top or upper part is the cope;
and other parts are the checks.
Sometimes pig iron and old scrap iron are melted together in a furnace
called a cupola. The liquid iron is taken from the furnace in ladles and
poured into different molds. As the hot iron flows into the mold and
cools, it becomes solid and takes the shape of the mold.
When the castings are removed from the mold they present a rough
surface and have to be cleaned and smoothed or "machined" before
they can be put to the use intended for them. They are cleaned in
various ways — by means of emery wheels and revolving wire brushes,
by being rotated in " tumblers," by chipping with pneumatic chisels, or
by means of a sand blast. The scales on the castings are removed by
wetting them with diluted sulphuric acid. This process is called pickling.
After this the casting is attached to the plate or table of the machine
tool that is to perform the necessary work upon it. Special devices are
made to hold castings when they are being machined.
Wrought Iron. — One of the valuable qualities of wrought iron is the
comparative ease with which it can be united with another piece by weld-
ing. When two pieces of wrought iron are heated to a white heat, they
assume a viscous condition, and when hammered together become united.
Wrought iron differs from cast iron in that it is capable of assuming any
shape under the hammer. It is readily made from cast iron by heating
in & puddling furnace. In this furnace the cast iron is subjected to great
heat and constant stirring, which allows the carbon to pass off as a gas
and the other impurities to rise to the surface, where they can be removed.
When the impurities 'are removed the iron is hammered to remove
particles of slag and then rolled in order to make it more compact. After
this it is heated again and rolled into bars for different purposes.
Wrought iron is sometimes case-hardened when it is used in machine
parts that need to be harder than the common iron. After the piece has
been finished and properly sized it is heated a bright red and the surface
rubbed with prussiate of potash. When it has cooled to a dull red, it is
immersed in water. Three parts of prussiate of potash and one of sal
ammoniac is a good case-hardening mixture. The temperature (Fahren-
heit) for cherry red is 1832°. If there are holes in such iron work,
the hardening by this process reduces them slightly.
Steel. — Steel is a form of iron which contains, as a rule, more car-
bon and other elements than wrought iron and less than cast iron.
MACHINISTS' WORK 253
There are many grades of steel, and each one is made by a 8i)ecial process,
steel may be recognized by the appearance of a dark spot when nitric
acid is placed on its surface. The darker the spot, the harder the steel.
Iron, on the contrary, shows no sign when touched with nitric acid,
(iootl steel will not stand a high heat, but will crumble under the hammer
blow at a bright red heat, while at a moderate heat — a full dull red or
cherry red — it may be drawn out to a fine edge tool. Steel that has
once been overheated or burned cannot be restored.
Steel for cutting-tools on lathes and i)laners should be drawn to a straw
color, or 430^, while for wood tools, taps, and dies, dark straw color, or
41(f ; for chisels for chipping, brown yellow or SW ; for springs, dark
purple, or 650°.
Steel may be softened or annealed if heated to a low red and placed in
a box of slaked lime and well covered, or in a box of fine bonedust, care
being taken in either case to cover the piece all around and on top to a
depth of not less than one and one half inches.
Copper. — Copper is used to a great extent because it may be easily
forged when cold. It may also be pressed into different shapes by means
of molds. Its strength is greatly increased by hammering and rolling.
It is used principally in wires and plates.
Brass. — Brass is an alloy or mixture of copper and zinc. Its tensile
strength is nearly equal to that of copper.
Weight of Bars of Steel. — The weight of bars of steel, as
they are usually made, is found by multiplying the area of
the ci'oss section or end in inches by the length in inches, and
multiplying the resulting nural^er of cubic inches in the bar
by 0.3. This will give practically an accurate result, since all
bars, unless otherwise ordered, will be rolled or hammered
slightly "full" to the dimensions given, or a bar that is
" full " to size may be trimmed to exact dimensions if neces-
sary. Whereas, if the bar is slightly under size, it cannot
easily be made larger.
To find the weight of a triangular bar of steel, multiply the
area of the base in square inches by the height in inches and
then by 0.3. The base of the triangular bar is found by multi-
plying the length of one of the bars or sections by one half the
l)erpendicular height; that is, by the distance to the opposite
vertex of the cross section.
254 VOCATIONAL MATHEMATICS
EXAMPLES
1. What is the weight of a triangular bar of steel when the
base contains 16 sq. in. and the height is 6 ft. ?
2. What is the weight of a triangular bar of steel when the
base contains 13 sq. in. and the height is 4 ft. ?
3. What is the weight of a triangular bar of steel when the
base contains 21 sq. in. and the height is lo feet ?
4. What is the weight of a triangular bar of steel when the
base contains 23 sq. inches and the height is 17 feet ?
5. What is the area of the base or section of a triangular
bar one side of which is 8 inches and the altitude 6 inches ?
6. What is the area of the base or section of a triangular
bar one side of which is 11 in. and the altitude 9 in. ?
7. What is the weight of a triangular bar of steel one side
of which and altitude of whose base or section are respectively
13 in. and 11 in., and the length of the bar 23 ft. ?
8. What is the weight of a triangular bar whose section is
24 sq. in. and whose length is 22 ft. ?
9. The weight of a cast-iron wheel is approximately sixteen
times as heavy as the white pine pattern from which it is cast.
What is the probable weight of a casting if the pattern for it
weighs 2J pounds ?
10. A white pine pattern weighs 12.5 pounds. What will be
the weight of an iron casting from it ? (Use data in Ex. 9.)
11.' If the weight of a brass casting is approximately fifteen
and a half times that of its white pine pattern, what will be
the weight of a casting if the pattern weighs 15 oz. ?
12. A white pine pattern weighs 1.75 pounds. What will be
the weight of 50 brass castings made from it ? (Use data in
Ex. 11.)
13. Since the shrinkage of brass castings is about ^ inch in
10 inches, what length would you make the pattern for a brass
collar which is required to be 6 inches long ?
CHAPTER XVII
LATHES
The Engine Lathe. — The lathe is one of the most important
machines in the machine shop. There are different kinds of
lathes, each adapted to certain kinds of work, the engine
lathe being the most important. The motion of the tool is
controlled by power speed; that is, the tool is moved automati-
cally parallel with and at right angles to the center line of the
lathe spindle. Most lathes are furnished with a series of
clutch gears and lead screws by means of which threads of
different pitch may be cut. All lathes have a series of stepped
cones in order to obtain a variety of speeds which are neces-
sary in order to work on hard and soft metals and to obtain
constant surface speeds for different diameters. The slower
speed makes the deeper cut.
Size. — The size of the lathe is expressed by stating the
length of the bed and the largest diameter it will swing on
centers. The swing is found by measuring from the point of
the headstock center to the ways on the bed and then multi-
plying by 2. The English measurement is from the center to
the way. The next important measurement is the length of
the bed, which is the entire amount of distance the tailstock
will move backward. If, however, accuracy is desired in this
measurement, the figure given should be the distance between
the two centers when the tailstock is in its extreme backward
position, as the lathe will turn no longer piece than will go
between centers.
Gear and Pitch. — Lathes that will cut a thread the same
pitch as the lead screw, with gears having the same number
of teeth on the stud and screw, are called geared even. If a
255
256
VOCATIONAL MATHEMATICS
lathe will not do this, find what thread will be cut with even
gears on both stud and lead screw and consider that as the
pitch of the lead screw.
Adjusting Gears. — A simple or single-geared lathe is one
having a straight train of gearing from its spindle to its feed
Engine Lathe
screw, excepting intermediate gears, which only serve as idlers
to take up the distance between the driver and driven gears
or spindle and screw gears. Index plates are usually found
on lathes giving the change gear used for different threads,
LATHES 257
but when threads are called for that are not indexed, or when
those ending in fractions are to be cut, the machinist must
make his own figures.
Refer to the screw cutting table and see what number of
threads to an inch are cut with equal gears. This number is
the number of turns to an inch that we a^ume the lead screw
has, no matter what its real number of turns to an inch is.
Write above the line the number of turns to an inch of the
lead screw and below the line the number of turns to an inch
of the screw to be threaded, thus expressing the ratio in the
form of a fraction, the lead screw being the numerator and the
screw to be threaded the denominator. Now find an ecjual
fraction in terms that represent numbers of teeth in available
gears. The numerator of this new fraction will be the spindle
or stud gear and the denominator the lead screw gear. The
new fraction is usually found by multiplying the numerator
and denominator of the first fraction by the same number.
PlxAMPLE. — It is required to cut a screw having 11^ threads
l)er inch.
The index gives 48 to 48 cuts 4 threads per inch.
_L X -* = —
Hi 6 69
Put the 24-tooth gear on the stud and the tt9-tooth gear on the lead
screw to cut 11^ threads per inch.
EXAMPLES
1. What gears should be used to cut a screw having 16
threads per inch, if a 40-gear on the stud and an 80-gear on
the screw will cut 8 threads to the inch ?
2. What geai's should be used to cut a screw having 3
threads per inch, if a 48-gear on the stud and a 56-gear on the
screw will cut 14 threads to the inch ?
* If multiplying by 6 will not give the gears available, use any other
number.
258 VOCATIONAL MATHEMATICS
3. What gears should be used to cut a screw 32 threads per
inch, if the pitch of the lead screw is 12 ?
4. What gears should be used to cut the following threads
per inch, if the pitch of the lead screw is 12 ?
a. 36 thread* d. 64 threads
b. 42 threads e. 3\ threads
c. 56 threads / 3^ threads
g. 12 threads
Compound Lathes. — The term compound applied to a lathe
means that in its train of gearing from its spindle to lead
screws there is a stud or spindle having two different sized
gears, both connected in such a way as to change the link of
revolution between the spindle and the lead screw to a different
number of revolutions from that which would take place if
the straight line of gears were used.
First Method. — Write the number of turns to an inch of the
lead screw as the numerator of a fraction and the turns of the
screw to be threaded as the denominator. Factor this fraction
into an equal compound fraction. Change the terms of this
compound fraction either by multiplying or dividing into
another equal compound fraction whose terms represent num-
bers of teeth in available gears. Then the two terms in the
numerator represent the number of teeth in the gears to be
used as drivers and those in the denominator the gears to be
used as driven gears.
Example. • — It is required to cut a screw having 3^ inches
lead or -^^ turns to an inch. The lead screw is 1^ inches lead
or I turns to an inch.
3 ' 13 3x4
Multiply numerator and denominator by 5,
2 X (13 X 5) _ 2 X 65
1 X (12 X 6) 1 X 60
>r ,.. 1 * a' A • . V. o. (24 X 2) X 65 48 X 65
Multiply numerator and denominator by 24, (24 x 1) x 60 " 24x60
LATHES 259
The 48-tootli and the 05-tooth gears will be tlie drivers and the 24-
tooth and GO-tooth gears the driven.
NoTK. — Any multiplier may be used to obtain the gear that is avail-
able.
Second Method. — Another satisfactory method of working
out the change gears is by proportion. If it is desirable to
cut a screw having 10 threads per inch and the lead screw
has 6 threads per inch, the first two terms of the proportion
would be 10: 6. As a rule, the smallest gear in the gear box
is used on the spindle, if it will serve the purpose, and as the
number of teeth on this gear is generally a multiple of the
number of threads per inch on the lead screw, in the present
case it would probably be 24. As the number of teeth on the
lead screw is to be the third term of the proportion, and as
this is unknown, x is used to represent it, and then the pro-
portion is 10: 6: : x: 24. By multiplying the first and fourth
terms together and the second and third terms together, the
result is 6x = 240. Then ic = 40, the number of teeth on the
screw gear.
If the lathe is compound geared, it is necessary to find the
proportional speed of spindle and stud. If the stud makes
three quarters of a revolution while the spindle makes a com-
plete revolution, it is necessary to use a gear on the screw
with but three quarters the number of teeth represented by x
in the proportion.
Example. — What gear should be used on the screw of a
compound geared lathe with the stud turning only three
quarters as fast as the spindle, in order to cut a screw having
13 threads per inch, if the lead screw has 6 threads per inch
and the stud gear 48 teeth ?
13 : 6 : : a: : 48
8
Cancelling, 13 : ^ : : x : ^ = 104
Three quarters of 104 = 78
With a 78-tooth gear on the screw, a 48-tooth gear on the stud of the
compound geared lathe will cut a screw having 13 threads per inch.
260 VOCATIONAL MATHEMATICS
Note. — If the stud turned but one half as fast as the spindle, then a
gear should be used on the screw with one half as many teeth as shown
under the method for simple geared lathes. ,
QUESTIONS AND EXAMPLES
1. Is the lathe in the classroom simple or compound
geared ?
2. What gears should be used to cut a screw having 18
threads per inch, if a 40-gear on the stud and an 80-gear on
the screw will cut 8 threads to the inch ?
3. How many threads per inch has the lead screw ? Is it
a square, V, or acme thread ? Is it right or left hand ?
4. What gears should be used to cut a screw having 6
threads per inch, if a 48-gear on the stud and a 56-gear on the
screw will cut 14 threads to the inch ?
5. How many gears are there between the spindle gear and
the lead screw gear ?
6. If the pitch of the lead screw is 16, what gears should
be used to cut a screw with 38 threads per inch ?
7. Has this lathe reversing gears? If not, state briefly
how the reversing is accomplished.
8. If the pitch of the lead screw is 18, what gears should
be used to cut a screw with 44 threads per inch ?
9. Put even gears on the first change gear stud and lead
screw and put a smooth round piece of wood on the lathe
centers. Clamp a pencil on the tool post so that it will mark
on the wood, then turn the lathe spindle until the carriage has
moved 1 inch. How many threads did the pencil draw on the
wood ?
10. What gears should be used to cut the following threads
per inch, if the pitch of the lead screw is 18 ?
a. 42 threads c. 16 threads
h. 8 threads d. 5^ threads
LATHES 261
11. What is the true pitch of the lead screw ? Is it the
same as the actual pitch ?
12. A 1" bolt is to have 8 threads per inch cut in it. If a
56-tooth gear is on the lead screw, what gear must be put on
the stud ?
13. A lathe has a feed rod turning at the same rate as the
lead screw, while the carriage travels one quarter as fast as
it would when screw cutting. If geared for 12 threads per inch
and the feed shaft is used, what will be the feed in fractions of
an inch per revolution ?
14. At 64 revolutions per minute (R. P. M.) how long will it
take to make a roughing cut with y\" feed and a finishing cut
with y\" feed, if both cuts are 21" long and 1 minute is
allowed for changing tools ?
To Cut Double or Multiple Threads
In modern machine construction there are many studs,
screws, and feed rods having threads for rapid travel, and
instead of a single spiral thread, there are two and sometimes
three spirals. If a machinist is called upon to replace or
duplicate such a thread, the method would not be to cut the
multiple threads at one time but to cut one thread at a time.
If the pitch of the double thread is measured, the pitch of
every second thread will be measured and the lathe set for 4
threads per inch. The cut of 4 threads is chased out to size
and the lathe left geared and the tool unchanged, and by turn-
ing the gears on the lathe spindle one half revolution, the tool
position for beginning the second thread is gained.
Before fixing the gear wheel position, the carriage should be
reversed in the position of the cut, thus taking up all lost
motion in screws, lathe nuts, and gears. Then make the exact
position of the mesh on the teeth of the spindle and on the
lead screw and count the number of teeth on the spindle gear,
which equals one half the entire number, and make this
262 VOCATIONAL MATHEMATICS
tooth. Now take off the spindle and turn the lathe one half a
revolution, bringing the second marked tooth to the position
of the first, and the lathe is then ready for cutting the second
thread. It is necessary in cutting multiple threads to select a
driving gear wheel having a number of teeth exactly divisible
by the number of threads cut.
Machine Speeds
Eveiy casting is cut into a definite shape by removing a certain portion
of it, called a cut or turning, by one of the machine shop tools. The
casting must be cut in the most economical way and in the shortest pos-
sible time. The tool that does the cutting must attain the highest pos-
sible speed, but there is a limit to the speed of a tool on account of the
heat generated as it moves against the casting. If too great speed is
used, the heat generated takes the temper out of the tool, renders it use-
less, and causes the casting to expand. The effect of the expansion on
the casting is to make it no longer true. The machine that is doing the
cutting should be accurate to .005 of an inch.
The cutting capacity of a machine depends on (1) the speed
of the cut, (2) the distance traversed by the tool in passing
from one cutting portion to the next, (3) the depth of cut, i.e.,
the thickness of the strip removed from the casting.
The volume of metal removed from a casting may be calcu-
lated as follows:
The volume of metal removed from good steel in one minute
equals : Cutting speed (length) times the feed (width) times
the depth of cut (thickness).
Example. — If the speed of a cut is 18 ft. per minute, the
feed .06 in., and the depth of cut is \ in., what is the volume
of metal removed in one hour ?
18' X 12" X .06 X .25 = 3.24 cu. in.
weight of 1 cu in. of good steel = .277 lb.
3.24 X .277 = 0.897 lb.
.897 X 60 = 53.8 lb. in 1 hr. Ans.
LATHES 263
Speeds for Different Metals. — Various materials, such as iron, copper,
ami wrought iron, possess different standards of hardness. A hanler
metal will wear away the tool and strain the machine more than a softer
metal. Therefore, there are different sj^eeds for each metal. With car-
bon steel cutting tools, the surface speed varies from 30 to 40 feet per
minute for cast iron, wrouglit iron, and soft steel ; 15 to 26 feet for well
annealed tool steel, and from 60 to 80 feet per minute for brass, while
the speed for dies and taps varies from 12 to 18 feet per minute on steel,
and from 30 to 50 feet per minute on brass, depending upon the quality
of the metal and the shape of the piece. With cutting tools of high-speed
steel these speeds, except for dies and taps, can be nearly doubled.
Cast iron should be worked at a speed which is j^j to j'j of that for
copper, or J to 1*5 of that for wrought iron.
Net Power for Cutting Iron or Steel
To find the net power for cutting cast iron and steel, mul-
tiply the section of cut or chip in square inches by 230,000
pounds for steel, or 168,000 pounds for cast iron, to get the
pressure on the tool ; and multiply this product by the cutting
speed in feet per minute, and divide the result by 33,000 to
obtain the horse power required.
Example. — Steel is being cut with \" cut, 1-64" feed at
a speed of 20' per minute. What is the H. P. ?
\" X 1-64" = .0039 sq. in.
.0039 X 230,000 lb. = 897 lb., pressure on tool
897 X 20 =-- 17,940
17,940 -T- 33,000 = .64 horse power. Ans.
Example. — If the cutting speed at the rim of iron stock
should be 40 feet per minute, at what speed should the lathe
(spindle) be driven for a piece of stock 3" in diameter ?
3" diameter
8 X 8.1416 = 9.4248", nearly 9.6", or f '
40' H- 1'
40 X J = ifft = 53 R. P. M. Ans.
264 VOCATIONAL MATHEMATICS
EXAMPLES
1. What is the amount of metal removed in one hour from
a casting with a cutting speed of 69' per minute and a feed
of j\" and a depth of -f ?
2. As a tool will stand a cutting speed of 35 feet per
minute when turning cast iron, how many revolutions per
minute should the lathe spindle make when a piece of cast
iron 8'' in diameter is being turned ?
3. Write the formula for the feed of a lathe tool, making
JV"= number of revolutions, D = distance the tool moves, and
i^=the feed.
4. A lathe tool moves 2.1-" along the work in one minute,
and the speed of the lathe is 400 R. P. M. What is the feed ?
(Use the formula.)
5. In turning up in the lathe a gun metal valve, 4 inches
in diameter, it is desirable that the surface speed shall not
exceed 45 ft. per minute. How many revolutions per minute
may the w^heel make ?
6. Find the time required and the speed of the lathe in
turning one 20-foot length of 3-inch wrought iron shafting,
one cut, traverse 28 per inch, cutting speed 20 ft. per minute,
no allowance being made for grinding or breaking of tools or
for setting stays.
7. How many revolutions per minute may be made in
turning np a steel shaft 6 inches in diameter, if the surface
speed must not exceed 12 ft. per minute ?
8. A piece of tool steel 1|'' in diameter is turned in a
lathe at 74 K. P. M. What is the cutting speed ?
9. What is the amount of metal removed in one hour from
a casting with a cutting speed of 64 ft. per minute and a feed
of jY and a depth of i" ?
10. If a tool will stand a cutting speed of 237 F. P. M. when
LATHES 265
turning cast iron, how many R. 1*. M. should the hithe (spindle)
make when a piece of cast iron iV in diameter is being turned ?
11. A valve yoke, stem 2" in diameter is being turned in a
lathe. If the lathe spindle makes 50 H.P. M., what is tlie
cutting speed in F. P. M. ?
12. What number of revolutions must a lathe spindle make
to cut 15 ft. per minute, in turning up an iron shaft 7J inches
in diameter ?
13. (a) With a tool steel that can stand a cutting speed of
30' per minute, how many revolutions per minute may a
lathe be run in turning one cut off a piece of shafting 15" in
diameter ?
(b) To hold the same cutting speed, how many revolutions
per minute would be required if the shaft were but 7^" in
diameter ?
14. In turning a cast iron piston head 15 inches in diameter
in the lathe, it is desired that the surface speed shall not ex-
ceed 15 ft. per minute. How many revolutions per minute
may the work make ?
15. How many revolutions per minute should the lathe
spindle make in turning up a cast iron pulley 33J inches in
diameter, at a cutting speed of 15 feet per minute ?
Table of Surface Speeds
The table on page 266 has been computed to facilitate the
figuring of speeds for machines and is used as follows :
Having first determined the proper surface speed, refer in
the table to the column of revolutions per minute corresponding
to this surface speed, and in this column opposite the diameter
corresponding to that of the work under consideration will be
found the required revolutions per minute of spindle or work.
Other surface speeds than those for which the table is com-
puted can be readily obtained from the table by multiplying
or dividing, as the case may require.
266
VOCATIONAL MATHEMATICS
Table of Surface Speeds
32^
3TJ
47^
DiAM.
Bevolntions per
Mimitc
■i^
3667.8
3973.5
4279.1
4584.8
4890.4
5196.1
6501.8
6807.4
6113.0
l\
1838.9
1986.7
2139.5
2292.4
2445.2
2598.0
2750.9
2903.7
3056.5
^
1222.6
1324.5
1426.3
1528.2
1630.1
1732.0
1833.9
1936.8
2037.7
i
916.9
993.3
1069.8
1146.2
1222.6
1299.0
1375.4
1451.8
1628.3
A
733.6
794.7
855.8
916.9
978.0
1039.2
1100.3
1161.4
1226.6
t\
611.3
662.2
713.2
764.1
815.7
866.0
916.9
967.9
1018.8
j\
523.9
567.6
611.3
654.9
698.6
742.3
785.9
829.0
873.3
k
458.4
496.6
534.9
673.1
611.3
649.5
687.7
725.9
764.1
^
407.5
441.5
475.4
509.4
543.3
677.3
611.3
645.2
679.2
t\
366.7
397.3
427.9
458.4
489.0
519.6
560.1
680.7
611.3
M
333.4
361.2
389.0
416.9
444.5
472.3
500.0
627.9
555.7
f
305.6
331.1
356.6
382.0
407.6
433.0
458.6
483.9
509.4
il
282.1
305.6
329.1
352.3
376.1
399.7
423.2
446.7
470.2
tV
261.9
283.8
305.6
327.4
349.3
371.1
392.9
414.8
436.6
M
244.5
264.9
285.1
305.6
326.0
346.4
366.7
387.8
407.5
i
229.2
248.3
267.4
286.5
305.6
324.7
343.8
362.9
382.0
^
203.7
220.7
237.7
254.7
271.6
288.6
306.6
322.6
340.0
f
183.4
198.6
213.9
229.2
244.5
259.8
276.0
290.3
306.2
ii
166.7
180.6
194.5
208.4
222.3
236.1
250.0
263.9
277.8
1
152.8
165.5
178.3
191.0
203.7
216.6
229.2
241.9
254.2
H
141.0
152.8
164.5
176.2
188.0
199.8
211.6
223.3
234.8
1
130.9
141.9
152.8
163.7
174.6
186.6
196.4
207.4
218.2
if
122.2
132.4
142.6
152.8
163.0
173.2
183.4
193.9
203.8
1
114.6
124.2
133.7
143.2
152.8
162.3
171.9
181.4
191.0
IrV
108.2
117.2
126.2
135.2
143.2
163.2
162.2
170.8
180.4
H
101.9
110.3
118.8
127.3
136.8
144.3
152.8
161.3
170.0
ii\
96.8
104.8
113.0
121.0
129.2
137.0
145.0
152.8
161.2
H
91.7
99.3
106.9
114.6
122.2
129.9
137.6
145.1
153.1
lA
87.5
94.7
102.1
109.4
116.8
124.0
131.2
138.2
146.0
LATHES 267
EXAMPLES
1. If the cutting speed of soft steel is 25, find the speed
(spindle) for ly stock.
2. If the cutting speed of soft steel is 25, find the speed
(spindle) for f J^" stock.
3. If the cutting speed of tool steel is 75, find the speed
(spindle) for ^4'' stock.
4. According to the table find the speed of (a) soft steel
1^" in diameter; (b) soft steel ^" in diameter; (c) tool steel
2 J" stock ; (d) tool steel 2V' stock ; (e) soft steel ^" in
diameter ; (/) soft steel if" in diameter.
CHAPTER XVIII
PLANERS, SHAPERS, AND DRILLING MACHINES
Planers. — The planer is usually one of the heaviest machines in the
machine shop. Planing is rough and heavy v^ork and stiffness is needed
in order to make the heavy cuts the planer usually has on castings and
heavy forgings. In many shops much of this rough surface cutting is
saved, however, by the use of a pickling solution of one part of sulphuric
acid to eight parts of water. This is painted over the casting and, after
four or five hours, washed off with clear water, removing the sand and
hard grit from the surface.
Machine Shop Planer
Care must be taken, during the operation of a planer, that chips and
dirt from the platen are not swept or allowed to blow into the V-ways, as
this will cause injury to the machine. Great care must be taken, also,
that oil is freely supplied to the machinery under the planer platen, as
these parts are hidden and likely to be neglected.
A planer 24 inches by 24 inches by 6 feet will consume an average of
0.085 horse power for every pound of cast iron removed per hour, and a
consumption of 0.066 horse power for every pound of n)achinery steel
removed per hour under good operation with sharp tools. The operator
268
PLANERS AND DRILLING MACHINES 269
of a planer may, by poor grinding and not setting his tools, waste much
power in driving the machine.
The cutting speed for planers is about the same as for lathe tools, the
advantage of the planer being ihat heavier cuts can be produced on
account of the rigid support of the platen.
The principal objection to planing machines is that they perform
useful work in only one half of the motion. When the work is drawn
back, no cutting takes place. Then again, siiice the forces are com-
pletely changed when the motion is changed, the slides upon which the
table moves, suffer. In this way accurate work is difficult on account
of the backlash. The return on capital invested in such machines is
smaller than in the case of machines with continuous motion.
Planer and Shaper. — The planer and shaper, and such modified ma-
chines as the slotting machine and key seater, are in a distinct cla.ss. Their
use is to machine and plane irregular surfaces that can be machined by a
straight line cut. The cutting tools of the planer and shaper are practi-
cally the same as those on the lathe. In the planer the work moves, and
vertical and lateral feeds are given to the tools. In the sliaper the tools
move over the work, lateral to the work and vertical to the tool. The
shaping machine is designed for small pieces and short travels. It is
nicely adapted for cutting grooves, slots, and dovetails.
Shaper tools should be kept in the best condition as to shear, clearance,
and cutting edge, as they are called upon to do accurate shaping of metal
parts of machinery not possible on other machines. Shear is a certain
amount of angle given the face of a tool, which throws its cutting edge
forward into the metal to be cut. Tools without clearance drag and pull
heavily through the metal.
For work on the shaper, the student should have a good knowledge of
the small try-square and the surface block. These tools are constantly
used, the square showing when finished pieces are square with the shap-
ing machine vise or when one cut is square with another. A universal
bevel or a bevel protractor should also be used, enabling angles to be laid
out for planing; A scriber and a four-inch outside caliper will enable the
beginner to grasp the first operations of shaping and planing. The ram
should be adjusted to proper length of stroke to save time. If a piece of
work which measures two and three fourths inches is to be placed on
the machine, one fourth inch is sufficient for the tool to enter and one
eighth inch is even more than enough to allow the head to overreach.
These additions then give us a total stroke of three and one eighth inches,
and any amount over this loses time and causes unnece8.sary wear on the
machine.
270 VOCATIONAL MATHEMATICS
EXAMPLES
1. If a planer has a cutting speed of 30 ft. per minute and
a return speed of 147 ft. per minute, what is the ratio of the
cutting to the return speed ?
2. On a 36" planer the ratio of the cutting speed to the re-
turn speed of the table is 1 to 2.94, with a cutting speed of 68
ft. per minute. What is the return speed ?
3. It is necessary to plane a bench block on the top and
bottom surfaces. An equal amount of stock is to be removed
from each side, and the thickness of the casting should be re-
duced from ly% in. to If in. What thickness of stock should
be cut from the top surface ?
4. A piece of work on the planer is 10 in. thick ; it is re-
duced to a certain size in 5 cuts ; at the first cut the tool takes
off ^ in. from the thickness ; then i in. ; g\ in., ^i^- in., and the
fifth cut is ^ly in. What is the thickness of the finished
piece ?
5. On a cast iron block 6 in. square^ a groove y\" wide and
I" deep must be planed. The planer makes 12 strokes per
minute and the down feed is set at -^" for every stroke. How
long will it take to cut the groove ?
Drilling Machines. — Machines for drilling holes in the different pieces
made in a machine shop are divided into two general classes — vertical
and horizontal. The vertical drilling machines include those with a
number of drill spindles called multiple spindles. Besides, there are
special machines of both classes, as portable drills, hand drills, etc.
The most common form of drill is the vertical drilling machine or
drill-press. The machine consists of a frame supporting the drill spindle
and the drilling table, and an arrangement for feeding the tool into the
work by hand or power. On this machine the work to be drilled is
1 The information 6" square is not necessary for the solution of the
problem, but is given to add interest to it. The same thing applies to some
of the problems in drilling on the following pages, where the size of
hole is given but not required.
PLANERS AND DRILLING MACHINES 271
Slwimq Head Drillino Machine
272 VOCATIONAL MATHEMATICS
placed on the drilling table, and is held stationary by means of a clamp
or vise, while the revolving drill is fed through the work by hand or by
power. The feeding mechanism is similar to that of the lathe. It is
more convenient, usually, to drill small work in a speed lathe and to use
the drill-press on heavy work.
There are two classes of drills — straight and twist ; the twist drill being
the latest and most approved in the leading shops. Whenever a manu-
facturer is making standardized pai-ts, that is, uniform parts of the same
machine, day after day, he designs and builds special drilling machines,
also special milling and planing machines, gauges called jigs^ and gauges
to produce the standard or duplicate parts.
Power is transmitted from the pulley on the shaft to the countershaft
at the bottom of the machine. Here are tight and loose pulleys so that
the machine may be stopped by shifting the belt from the tight to the
loose pulley. Different materials have qualities which make it necessary
to use different cutting speeds in order to remove the metal quickly and
efficiently. In order to provide for this a cone pulley giving a number of
different speeds is used. The cone pulley on the countershaft is the same
as the cone pulley on the headstock. The power is transmitted from the
headstock to the drill spindle by bevel gears. The power is transmitted
from the headstock shaft to the feeding spindle — which in turn lets the
drill spindle down .
Reamers. — Drills cannot be relied upon to make holes as round,
straight, smooth, or uniform in diameter as are required in the construc-
tion of accurate machinery. To make these holes accurately a tool
called a reamer is used. It has two or more teeth, either parallel or at
an angle with each other. The periphery of a drill or reamer is the distance
around the outside, and the peripheral speed is the distance a point
in the circumference travels in a minute. It is equal to the number of
turns that the tool makes in a minute multiplied by the circumference.
The proper speed at which to run a drill depends upon the kind of drill,
its size, and the material to be drilled. A large drill must run more
slowly than a smaller one, the turns per minute becoming less the larger
the size of the drill. For instance, a drill \" in diameter should make
twice as many revolutions per minute as a I" drill and four times as
many as one 2" in diameter, used on the same material. The peripheral
speeds usually recommended for carbon steel drills are as follows :
Wrought iron or steel, 30ft. per min. High speed drills, 60 to 70 ft. per min.
Cast iron, 35 ft. per min. High speed drills, 60 to 80 ft. per min.
Brass, 60 ft. per min. High speed drills, 100 to 140 ft. per min.
PLANERS AND DRILLING MACHINES 273
The feed of a drill is the amount the drill enters the hole for each turn
or revolution, and in good practice the feed may be set to give 1" depth
of the hole for every 96 to 125 revolutions, according to the size of the
drill, the material of which the drill is made and the kind of metal drilled.
Example. — How long will it take a one-inch drill, making
134 R. P. M., with a feed of .012" per revolution, to drill a
hole 1^" deep in cast iron ?
— ^— = 126 rev. 1 rev. = — of a minute
.012 134
125 rev. = 126 X -^ = — = .933 of a minute. Ans.
134 134
Example. — A piece of wrought iron 2.69" thick is to have
two If" holes drilled through it. If the drill makes 112
R. P. M., what must the feed be to drill each hole in two
minutes ?
112 X 2 = 224 revs, in 2 min.
^i, of 2.69" = .012" feed. Ans.
Example. — A 2" drill is used in drilling an electrical gen-
erator bed plate. If the drill is making 67 R. P. M., and is
being fed to the work at the rate of .015" per revolution, how
deep will the hole be when the drill has worked 4| minutes ?
67 X .016" = 1.006" depth per min.
1.005x4.5=4.5225". Ans.
Example. — What will be the R. P. M. of a drill If" in
diameter used for drilling out a lathe spindle 30.24" long, the
feed being .015" per revolution, and the time given the mar
chinist to do the job being 42 minutes ?
80.24 OA1A 2016 Aan r> njr a
:5T5- = 2016 .^ = 48R.P.M. Ans.
Example. — Conditions being the same as in the above ex-
ample, what length of a spindle could be drilled in 50 minutes ?
48 X .016 = .720" per min.
.720" X 50 = 36" in 60 min. Ans.
274 VOCATIONAL MATHEMATICS
EXAMPLES
1. How long will it take a carbon steel drill to drill a J"
hole through a piece of wrought iron l|f" thick, if the drill
makes 105 E. P. M., with the feed at .011'' per revolution ?
2. With the feed at .015'' per revolution and the speed at
115 E. P. M., how long will it take to drill a hole with a 2"
carbon steel drill through a brass bushing 4|" long?
3. A machinist using a high speed drill If" in diameter, is
to drill 4 bolt holes in the base of an electrical generator, the
holes to be 2\" long, the drill to make 164 R. P. M., with the
feed at .018" per revolution. How long will it take him to do
the job if 3 minutes are allowed for the setting for each hole ?
4. A I" drill working on brass is running at 306 E. P. M.
and advancing at the rate of .015" per revolution. How long
will it take to drill through a piece lyV' thick ?
5. The E. P. M. of a carbon steel drill ly%" in diameter
is 113, and the feed is .013" per revolution. How much time
will a machinist require to drill 18 holes in a cylinder head
1^" thick, allowing one minute for the setting of each hole ?
6. A high speed drill If" in diameter is advancing at the
rate of .018" per revolution, making 229 E. P. M. while drill-
ing an angle iron If" thick. How long will it take the drill to
go through it ?
7. In Example 6, what is the periphery speed in feet per
minute of the drill ?
8. A cast iron casing for a steam turbine is to have a series
of holes li" in diameter drilled in it 2^" deep with a high
speed drill making 250 E. P. M. What must the feed be per
revolution to drill each hole in f of a minute ?
9. The holes in a face plate 4" thick are to be drilled
with a 1-in. carbon steel drill which makes 134 E. P. M.
(a) What feed will be required to drill through the plate in
PLANERS AND DRILLING MACHINES 275
2J minutes? (6) What will be the periphery speed in feet
per minute of the drill ?
10. What must be the R. T. M. of a l.J" drill, feeding' at
the rate of .013" per revolution, to drill a hole 3.'/' deep in an
engine bed in 3 minutes ?
11. A 2" drill is used in drilling an electrical generator bed
platen. If the drill is making 87 R. P. M., and is being fed
to the work at the rate of .015" per revolution, how deep will
the hole be when the drill has worked 4 J minutes ?
12. What will be the R. P. M. of a 1^" carbon drill used in
drilling out an engine lathe spindle 24.05" in 24.GG minutes ?
What will be the total number of revolutions? (Same rate
of feed as in Ex. 11.)
PART X — TEXTILE CALCULATIONS
CHAPTER XIX
YARNS
Worsted Yarns. — All kinds of yarns used in the manufacture
of cloth are divided into sizes based on the relation between
weight and length. To illustrate : Worsted yarns are made
from combed wools, and the size, technically called the counts, is
Roving or Yarn Scales
These scales will weigh one pound by tenths of grains or one seventy-
thousandth part of one pound avoirdupois, rendering them well adapted for
use in connection with yarn reels, for the numbering of yarn from the weight
of hank, giving the weight in tenths of grains to compare with tables.
based upon the number of lengths (called hanks) of 560 yards
required to weigh one pound. Thus, if one hank weighs one
pound, the yarn will be number one counts, while if 20 hanks
276
TEXTILE CALCULATIONS
277
are required for one pound, the yarn is the 20's, etc. The
greater the number of hanks necessary to weigh one pound,
the higher the counts and the finer the yarn. The hank, or
500 yai'ds, is the unit of measurement for all worsted yarns.
Lknotu for Worsted Yarns
No.
Yards
PRR Lb.
N.».
Yabdh
PKR Lb.
No.
Yariw
PER Lb.
No.
Yardk
PKR Lh.
1
2
3
4
560
1120
1680
2240
6
6
7
8
2800
3360
3920
4480
10
11
12
6040
6600
6160
6720
13
14
16
16
7280
7840
8400
8960
Woolen Yams. — In worsted yams the fibers lie parallel to
each other, while in woolen yarns the fibers are entangled.
This difference is due entirely to the different methods used
Yarn Reel
For reeling and measuring lengths of cotton, woolen, and worsted yams.
in working up the raw stock. In woolen yarns there is a great
diversity of systems of grading, varying according to the dis-
tricts in which the grading is done. Among the many systems
278
VOCATIONAL MATHEMATICS
are the English skein, which differs in various parts of Eng-
land ; the Scotch spyndle ; the American run ; the Philadelphia
cut; and others. In these lessons the run system will be used
unless otherwise stated. This is the system used in New
England. The run is based upon 100 yards per ounce, or
1600 yards to the pound. Thus, if 100 yards of woolen yarn
weigh one ounce, or if 1600 yards weigh one pound, it is
technically termed a No. 1 run ; and if 300 yards weigh one
ounce, or 4800 yards weigh one pound, the size will be No. 3
run. The finer the yarn, or the greater the number of yards
necessary to weigh one pound, the higher the run.
Length for Woolen Yarns (Run System)
No.
Taros
PER Lb.
No.
Yards
PER Lb.
No.
Yards
PER Lb.
No.
Yards
PER Lb.
f
200
400
800
1200
1
If
1600
2000
2400
2800
2
^
2f
3200
3600
4000
4400
3
H
4800
5200
. 5600
Raw Silk Yarns. — For raw silk yarns the table of weights
is:
16 drams = 1 ounce
16 ounces = 1 pound
256 drams = 1 pound
The unit of measure for raw silk is 256,000 yards per pound.
Thus, if 1000 yards — one skein — of raw silk weigh one
dram, or if 256,000 yards weigh one pound, it is known as
1-dram silk, and if 1000 yards weigh two drams the yarn is
called 2-dram silk, hence the following table is made :
1-dram silk = 1000 yards per dram, or 256,000 yards per lb.
2-dram silk = 1000 yards per 2 drams, or 128,000 yards per lb.
4-dram silk = 1000 yards per 4 drams, or 64,000 yards per lb.
TEXTILE CALCULATIONS
279
■
Dramr i'br 1000 Yards
Yards prr Pound
Yardh prr Ounck
1
2r)12 yards of 60's cotton.
39. Find the weight in grains of 118 yards of 44's linen.
40. Find the weight in pounds of 315 yards of 32's linen.
41. Find the weight in grains of 84 yards of 25's worsted.
42. Find the weight in grains of 112 yards of 20's woolen.
43. Find the weight in grains of 197 yards of 16's woolen.
44. Find the weight in grains of 183 yards of 18's cotton.
45. Find the weight in grains of 134 yards of 28's worsted
46. Find the weight in grains of 225 yards of 34's linen.
47. Find the weight in pounds of 369 yards of 16's spun silk.
48. Find the weight in pounds of 484 yards of 18's spun silk.
To find the Size or the Counts of Cotton Yarn of Known
Weight and Length
Example. — Find the size or counts of 84 yards of cotton
yarn weighing 40 grains.
Since the counts is the number of hanks to the pound,
^ X 84 = 14,700 yd. in 1 lb.
14,700 H- 840 = 17.5 counts. Ans.
Rule. — Divide 840 by the given number of yards ; divide
7000 by the quotient obtained ; then divide this result by the
weight in grains of the given number of yards, and the quotient
will be the counts.
840 - 84 = 10
7000 -f- 10 = 700
700 -^ 40 = 17.5 counts. Ans.
To find the Ran of a Woolen TJiread of Knoivn Length and
Weight
Example. — If 50 yards of woolen yarn weigh 77.77 grains,
what is the run '/
1«00 ^ 50 = 32
7000 -f- 32 = 218.75
218.75 - 77.77 = 2.812-run yarn. Ans.
TEXTILE CALCULATIONS 285
Rule. — Divide 1600 (the number of yards per pound of 1-
run woolen yarn) by the given number of yards ; then divide
7000 (the grains per pound) by the quotient; divide this
(juotient by the given weight in grains and the result will be
the run.
To find the Weight in Ounces for a Oiven Number of Yards of
Worsted Yarn of a Known Count
Example. — What is the weight in ounces of 12,650 yards
of 30's worsted yarn ?
12,650 X 16 = 202,400
202,400 ^ 16,800 = 12.047 oz. Ans,
Rule. — Multiply the given number of yards by 16, and
divide the result by the yards per pound of the given count,
and the quotient will be the weight in ounces.
To find the Weight in Pounds for a Oiven Number of Yards of
Worsted Yai'n of a Known Count
Example. — Find the weight in pounds of 1 ,500,800 yards
of 40's worsted yam.
1,500,800 -4- 22,400 = 67 lb. Ans.
Rule. — Divide the given number of yards by the number
of yards per pound of the known count, and the quotient will
be the desired weight.
EXAMPLES
1. If 108 inches of cotton yarn weigh 1.5 grains, find the
counts.
2. Find the size of a woolen thread 72 inches long which
weighs 2.5 grains.
3. Find the weight in ounces of 12,650 yards of 2/30's
worsted yarn.
4. Find the weight in ounces of 12,650 yards of 40's
worste = 1.1283 xV]4
11. The number of miles in a given length, expressed, in
feet, may be obtained from the formula
3f=.00019xF
12. The number of cubic feet in a given volume expressed
in gallons may be obtained from the formula
0=. 13367 X Q
13. Contractors express excavations in cubic yards; the
number of bushels in a given excavation expressed in yards
may be obtained from the formula
C=.0495x Y
14. The circumference of a circle may be obtained from the
area by the formula
C= 3.5446 X V2
15. The area of the surface of a cylinder may be expressed
by the formula A = (^C X L) -\-2a
When C = circumference
L = length
a = area of one end
16. The surface of a sphere may be expressed by the formula
S=D'x 3.1416
17. The solidity of a sphere may be obtained from the
formula
S = D'x .5236
18. The side of an inscribed cube of a sphere may be ob-
tained from the formula
S = R X 1.1547, where S = length of side,
R = radius of sphere.
304 VOCATIONAL MATHEMATICS
19. The solidity or contents of a pyramid may be expressed
by the formula
F
S = Ax^, where A = area of base,
F = height of pyramid.
20. The length of an arc of a circle may be obtained from
the formula
L = Nx .017453 E, where L = length of arc,
N= number of degrees,
a = radius of circle.
21. The horse power of a single leather belt may be deter-
mined by the formula
DRW
HP = ^ , where D = diameter of pulle}^ in inches,
W— width of belt in inches,
M = revolutions per minute,
HP = horse power transmitted.
22. The formula for finding the weight of an iron ball may
be calculated by the following :
Tr= 2)3x0.1377
23. The formula for finding the diameter of an iron ball
when the weight is given is
D = 1.936 Vl^ where D = diameter of the ball in inches,
W= weight of ball in pounds.
24. The volume of a sphere when the circumference of a
great circle is known may be determined by the formula
25. The diameter of a circle the circumference of which is
known may be found by the formula
FORMULAS 305
26. The area of a circle the circumference of which is known
may be found Jby the formula
4 tr
Coefficients and Similar Terms
When a quantity may be separated into two factors, one of
these is called the coefficient of the other ; but by the coefficient
of a term is generally meant its numerical factor.
Thus, 4 6 is a quantity composed of two factors 4 and 6 ; 4 is a coef-
ficient of h.
Similar terms are those that have as factors the same letters
with the same exponents.
Thus, in the expression, 6 a, 4 6, 2 a, 6 a6, 5 a, 2 6. 6 a, 2 a, 5 a are
similar terms ; 4 6, 2 & are similar terms ; 5 ab and 6 a are not similar
terms because they do not have the same letters as factors. 8 a&, 6 ah,
1 aft, 8 ah are similar terms. They may be united or added by simply
adding the letters to the numerical sum, 17 aft.
In the following, 8 ft, 6 ft, 3 aft, 4 a, aft, and 2 a, 8ft and 5ft are similar
terms ; 3 aft and aft are similar terms ; 4 a and 2 a are similar terms ; 8 ft,
3 aft, and 4 a are dissimilar terms.
In addition the numerical coefficients are algebraically added ;
in subtraction the numerical coefficients are algebraically sub-
tracted ; in multiplication the numerical coefficients are alge-
braically multiplied ; in division the numerial coefficients are
algebraically divided.
EXAMPLES
State the similar terms in the following expressions :
1. 5 a;, 8 OLc, 3 a;, 2 ax. 6. 15 ahc, 2 aJbc, 4 dbc^ 2 aft,
2. 8aftc^ 7c, 2aft, 3c, ^ah, 3aZ>.
9«*c. 7. Saj, 6a;, 13xy, 6x, 7y.
3. 2pq, 5p,Sq, 2p, 3g, ^pq. a 7y, 2y, 2xy, 3y, 2xy,
4. 4.V, 5y2, 2y,ir)2,52, 2.V2. ir . „
« 1ft r K A o ^ 2ir, 5xr*, -, irr», 2xr.
5. lo mn, m, 5 7i, 4 mny 2 m. J
306 VOCATIONAL MATHEMATICS
Equations
A statement that two quantities are equal may be expressed
mathematically by placing one quantity on the left and the
other on the right of the equality sign (=). The statement
in this form is called an equation.
The quantity on the left hand of the equation is called the
left-hand member and the quantity on the right hand of the
equation is called the right-hand member.
An equation may be considered as a balance. If a balance
is in equilibrium, we may add or subtract or multiply or divide
the weight on each side of the balance by the same weight and
the equilibrium will still exist. So in an equation we may
perform the following operations on each member without
changing the value of the equation :
We may add an equal quantity or equal quantities to ea/ih memr
her of the equation.
We may subtract an equal quantity or equal quantities from
each member of the equation.
We may multiply each member of the equation by the same or
equal quantities.
We may divide each member of the equation by the same or
equal quantities.
We may extract the square root of each member of the equation.
We may raise each member of the equation to the same poiver.
The expression, A = ttR^ is an equation. Why ?
If we desire to obtain the value of R instead of A we may do
so by the process of transformation according to the above
rules. To obtain the value of R means that a series of opera-
tions must be performed on the equation so that R will be left
on one side of the equation.
(1) A = 7rJ?2
A
(2) — = R^ (Dividing equation (1) by the coefficient of B^.)
IT
(^) -\/~ = -^ (Extracting the square root of each side of the equation.)
FORMULAS 307
Methods of Representing Operations
Multiplication
The multiplication sign ( X ) is used in most eases. It should
not be used in operations where the letter (x) is also to be em-
ployed.
Another method is as follows :
2.3 a. 6 2 a. 36 4a;. 5a
This method is very convenient, especially where a number
of small terms are employed. Keep the dot above the line,
otherwise it is a decimal point.
Where parentheses, etc., are used, multiplication signs may
be omitted. For instance, (a + b)x{a — h) and (a -\- b'){a — b)
are identical ; also, 2'(x — y) and 2{x —y).
The multiplication sign is very often omitted in order to
simplify work. To illustrate, 2 a means 2 times a ; 5 xyz means
5 • X • y • z ; x{a — b) means x times (a — b), etc.
A number written to the right of, and above, another (a:*) is
a sign indicating the special kind of multiplication known as
involution.
In multiplication we add exponents of similar terms.
Thus, «* . a^ = a:»+' = ar*
(ibc ' ab • aVj = a*bh
The multiplication of dissimilar terms may be indicated.
Thus, a-b ' C' x-y 'Z = abcxyz.
Division
The division sign (-*-) is used in most cases. In many
cases, however, it is best to employ a horizontal line to indicate
division. To illustrate, means the same as (a -\' b) -i-
x-y
(x — y) in simpler form. The division sign is never omitted.
308 VOCATIONAL MATHEMATICS
A root or radical sign (V^> y/^) is a sign indicating the special
form of division known as evolution.
In division, we subtract exponents of similar terms.
Thus, ar»H-aj' = - = a^-2 = a;
a^lfif^ ^ a'^bc^ = ^^^^^ = a^bc.
a^bc^
The division of dissimilar terms may be indicated.
Thus, (abc) -i- xyz =
xyz
Substituting and Transposing
A formula is usually written in the form of an equation.
The left-hand member contains only one quantity, which is
the quantity that we desire to find. The right-hand member
contains the letters representing the quantity and numbers
whose values we are given either directly or indirectly.
To find the value of the formula we must (1) substitute for
every letter in the right-hand member its exact numerical
value, (2) carry out the various operations indicated, remem-
bering to perform all the operations of multiplication and
division before those of addition and subtraction, (3) if there
are any parentheses, these should be removed, one pair at a
time, inner parentheses first. A minus sign before a parenthesis
means that when the parenthesis is removed, all the signs of
the terms included in the parenthesis must be changed.
Find the value of the expression
3d-\-b(2a-b-\- 18), where a = 5,b = 3.
Substitute the value of each letter. Then perform all addition or
subtraction in the parentheses.
3x6-1- 3(10 - 3 + 18)
16-^.3(28-3)
15 + 3(26)
15 4-75 = 90
FORMULAS 309
EXAMPLES
Find the value of the following expressions :
1. 2 ^ X (2 + 3 ^) X 8, when J = 10.
2. 8 o X (6 — 2 a) X 7, when a = 7.
3. 8 6 + 3 c + 2 rt (a -f /> + c) - 8, when a = 9; 6 = 11 ; c = 13.
4. 8(a; + y), when a? = 9; y = 11.
5. 13 {x- y), when a; = 27 ; y = 9.
a 24y-|-8z(2 + y)-3y, when jy = 8; z = ll.
7. Q{(^M-\-^N)-^2 0, when Jlf=4, xV=5, Q=6, = 8.
a Find the value of X in the formula X = ^^^^^"^^^
when J»f = 11, iV^= 9, P = 28.
9. .c = ^.^L±^, when n = 5, m = 6, P = 8, Q = 7.
10. Find the value of T in the equation
(x + y)(x-y)
11. 3a+4(6-2a + 3c)-c, when a = 4, 6 = 6, c = 2.
12. op — Sq(p + r — S) — qy when /) = 5, g = 7, r = 9, i5 = 11.
13. S^^t*-p'i-3(S-\-t + p)j whenp = 5, S=Sy t = 9.
14. a* - &>+ c», when a = 9, 6 = 6, c = 4.
15. (a + 6) (a + 6 - c), when a = 2, 6 = 3, c = 4.
16. (a* - b*) (a* + 6*), when a = 8, 6 = 4.
17. (c-» + cP) (c» - rf»), c = 9, (/ = 5.
18. Va* + 2 a6 + 6«, when a = 7, 6 = 8.
19. ^c*-61, when c = 5.
310 VOCATIONAL MATHEMATICS
PROBLEMS
Solve the following problems by first writing the formula
from the rule on page 300, and then substituting for the answer.
1. How many electrical horse power in 4389 watts ?
2. How many kilowatts in 2389 watts ?
3. (a) Give the number of watts in a circuit of 110 volts
and 25 amperes.
(6) How many electrical horse power ?
4. What is the voltage of a circuit if the horse power is
2740 watts and the quantity of electricity delivered is 25
amperes ?
5. What is the resistance of a circuit if the voltage is 110
and the quantity of electricity is 25 amperes ?
6. What is the pressure per square inch of water 87 feet
high?
7. What is the capacity of a cylinder with a base of 16
square inches and 6 inches high ? (Capacity in gallons is
equal to cubical contents obtained by multiplying base by the
height and dividing by 231 cubic inches.)
8. What is the length of a 30° arc of a circle with 16"
diameter?
9. What is the area of a sector with an arc of 40° and a
diameter of 18"?
10. What is the weight of the rim of a flywheel of a 25 H. P.
engine ?
11. What is the area of a cylinder of a 50 H. P. engine with
the piston making 120 ft. per minute ?
12. What is the H. P. of a 2ff'' shaft making 180 R. P. M. ?
13. What is the capacity of a pail 14'' (diameter of top),
11" (diameter of bottom), and 16" in height ?
14. What is the area of an ellipse with the greatest length
16" and the greatest breadth 10" ?
FORMULAS
311
Interpretation of Negative Quantities
The quantity or number — 12 lias no meaning to us according
to our knowledge of simple arithmetic, but in a great many
problems in practical work the minus sign before a number
assists us in understanding the different solutions.
To illustrate :
FaBRSNHKIT TllF.BlIOllBTKR
Cbntiorape Tiibrmombtkr
Boilinf?
point of
Freezinjf
point of ■
water
212*
Roilinsr
point uf
water
82*
2. a.
Freezing
point uf '
water
100»
On the Centigrade scale the freezing point of water is marked
0°. Below the freezing point of water on the Centigrade scale
all readings are expressed as minus readings.
— 30° C means thirty degrees below the freezing point. In
other words, all readings, in the direction below zero are
expressed as — , and all readings above zero are called -|-.
Terms are quantities connected by a plus or minus sign.
Those preceded by a plus sign (when no sign precedes a quan-
tity plus is understood) are called positive quantities, while
those connected by a minus sign are called negative quantities.
312 VOCATIONAL MATHEMATICS
Let us try some problems involving negative quantities.
Find the corresponding reading on the Fahrenheit scale cor-
responding to — 18° C.
F = I C + 32°
F= |(- 18°)+32°
Notice that a minus quantity is placed in parenthesis when it is to be
multiplied by another quantity.
F =- ip° + 32" = - 32f° + 32° ; F = - |°.
The value — §° is explained by saying it is § of a degree below zero
point on Fahrenheit scale.
Let us consider another problem. Find the reading on the Centi-
grade scale corresponding to — 40° F.
Substituting in the formula, we have
C = s (_ 40° _ 32C) ^ |(_ 72) = _ 40^
Since subtracting a negative number is equivalent to adding
a positive number of the same value, and subtracting a posi-
tive number is equivalent to adding a negative number of the
same value, the rule for subtracting may be expressed as fol-
lows: Change the sign of the subtrahend and proceed as in
addition.
For example, 40 minus — 28 equals 40 plus 28, or 68.
40 minus + 28 equals 40 plus — 28, or 12.
— 40 minus 4- 32 equals — 40 plus — 32 = - 72.
(Notice that a positive quantity multiplied by a negative quantity or
a negative quantity multiplied by a positive quantity always gives a
negative product. Two positive quantities multiplied together will give
a positive product, and two negative quantities multiplied together will
give a positive product.) To illustrate :
5 times 5 = 5 x 5 = 25
5 times — 5=r5x(— 5) = — 25
(-5) times (-5) = + 25
In adding positive and negative quantities, first add all the
positive quantities and then add all the negative quantities
FORMULAS 313
together. Subtract the smaller from the larger and prefix the
same sign before the remainder as is before the larger number.
For example, add :
2a, 5a, - Oa, 8a, -2a
2a + 6a -I- 8a = 16a; -6a-2a=-8a
T5a-8a = 7a
EXAMPLES
Add the following terms :
1. 3Xf —X, 7 Xf AXj —2x.
2. 6y, 2y, 9y, -7y.
3. 9 aby 2 a6, 6 aft, — 4 aft, 7 a6, — 5 ab.
Multiplication of Algebraic Expressions
Each term of an algebraic expression is composed of one or
more factors, as, for example, 2 ab contains the factors 2, a, and
b. The factors of a term have, either expressed or understood,
a small letter or number in the upper right-hand corner, which
states how many times the quantity is to be used as a factor.
For instance, ab\ The factor a has the exponent 1 understood
and the factor b has the exponent 2 expressed, meaning that a
is to be used once and b twice as a factor, ab'^ means, then,
a X 6 X 6. The rule of algebraic multiplication by terms is as
follows: Add the exponents of all like letters in the terms
multiplied and use the result as exponent of that letter in the
product. Multiplication of unlike letters may be expressed
by placing the letters side by side in the product.
For example : 2 a6 x .3 6^ = a6«
4ax86 = 12a6
Algebraic or literal expressions of more than one term are
multiplied in the following way : begin with the first term to
the left in the multiplier and multiply every term in the multi-
plicand, placing the partial products underneath the line. Then
314 VOCATIONAL MATHEMATICS
repeat the same operation, using the second term in the multi-
plier. Place similar products of the same factors and degree
(same exponents) in same column. Add the partial products.
Thus, a -\- b multiphed by a — 5.
a + h
a — h
a^+ ab- b-^
-ah
a^ - b-^
Notice the product of the sum and difference of the quantities is equal
to the difference of their squares.
EXAMPLES
1. Multiply a-\-hhy a-\-h.
State what the square of the sum of the quantities equals.
2. Multiply X — y hy X — y.
State what the square of the difference of the quantities equals.
3. Multiply (j) + q)(p — q). 7. Multiply (x — y)(x — y).
4. Multiply (p -f q)(p + q). 8. (x -\-yy=?
5. Multiply (r + s){r - s). 9. {x - yf = ?
6. Multiply (a ± 6)(a ± 6). 10. (x + y){x-y) = ?
LOGARITHMS
Thk logarithm of a iuiiuImt to the base 10 is defined as the povrer
to Nvhicli 10 must be raiseil in order to equal the number.
Thus the logarithm (or log, as it is more generally written) of 10
Ls 1, for the first power of 10 is 10. The log of 100 is 2, for the second
power of 10 is 100. Hence the log of a number between 10 and
100 is a number between 1 and 2, and is, therefore, 1 plus a decimal.
The whole number 1 is called the characteristic, and the decimal part
is called the mers 7.21, 72.1, and 721.
721. = 10 X 72.1 = 10* X 7.21. Therefore, log 721 = log 10 + log 72.1
= log 10* + log 7.21. (In order to multiply the numbers, we may
add the logarithms of these numbers.) Therefore, log 721 = 1 +
log 72.1 = 2 + log 7.21, (log 10 = 1 ; log 10* = 2). Therefore, we see
that the logarithms of numbers made up of the same sequence of
315
316 VOCATIONAL MATHEMATICS
digits, but differing in the position of the decimal point, differ only
in the characteristic, the mantissce remaining the same.
For this reason, in making tables of logaritlims, only the mantissse
are given, the characteristics being added according to the rules
above. The tables we are to use are made for three significant digits
in the number and four digits in the mantissa of the log.
To Find the Logarithm of a Number
A. When the number has three significant figures.
The first two figures are found in the column headed No.
(pages 318-319), and the third figure in the top row. The mantissa
is stated in the column and row so determined.
To illustrate, find the log of 62.8. In the 62 row and in the col-
umn under 8 is the mantissa, 7980. By Rule I we find the charac-
teristic to be 1 ; therefore, the log of 62.8 is 1.7980.
Find the log of .00709. The significant figures are 709; in the
70 row and in the column under 9, we find the mantissa 8506. By
Rule II the characteristic is — 3. Therefore the log of .00709 is
— 3.8506. To prevent confusion in the operations of arithmetic — 3 is
usually written 3, so the log is stated to be 3.8506, or 7.8506 — 10.
EXAMPLES
Find log of 117 ; of 1280 ; of 16.5 ; of 2.09 ; of .721 ; of .0121.
B. When the number has four or more significant figures.
Ex. 1. — Find the log of 7987. The tables give only the mantissa of
numbers of three figures, so the mantissa of 7987 cannot be found in
the table ; 7987, however, lies between 7980 and 7990, hence its log is
between the log of 7980 and the log of 7990. We find the mantissa
7980 to be 9020, and the mantissa of 7990 to be 9025. Now the
difference between these mantissas is 5, while the difference between
the two numbers, 7980 and 7990, is 10. But the difference between
7980 and 7987 is ^jj of the difference between 7980 and 7990, and the
mantissa of 7987 will be /^ of the difference between the mantissa of
7980 and the mantissa of 7990. Therefore it will equal 9020 (the
mantissa of 7980) plus yV of 5 (the difference between the mantissse
of 7980 and 7990) : 5 x .7 = 3.5, hence the mantissa of 7987 is 9024.
In such reckoning all decimal parts less than .5 are counted as 0, and
all decimal parts greater than .5 are counted as 1 ; .5 is counted as
either 1 or 0. Log 7987 is .3^024.
LOGARITHMS 317
Ex. 2. — Find the log of 12.564.
First find the mantissa. In looking for the mantissa the decimal
{H)int need not be considered.
Solution. — 12554 lies between 12500 and 12600.
Mantissa of 12600 is 1004.
Mantissa of 12.500 is 0969.
35 is the difference between the mantissa;.
12600 - 12500 = 100 = difference between numbers.
12554 — 12500 = 54 = difference between original numbers.
The multiplier is f*^ = .54.
35 X .54 = 18.90 or 19.
0969 -f 19 = 0988.
log 12.564 is 1.0988. Ans.
EXAMPLES
Find log of 17.89 ; of 2172 ; of 652.12 ; of 4213 ; of 3342000.
C To find the number corresponding to a given logarithm.
Ex. 1. — What number has the log 1.6085?
The mantissa 6085 is found in the 40 row and in the column
under 6. Tiie number corresponding to mantissa 6085 is therefore
406. The characteristic 1 states that there are two digits to the left
of the decimal point. The number is therefore 40.6.
Ex. 2.— What number has the log 5.8716?
The mantissa 8716 is in the 74 row and in the column under 4.
The characteristic 5 states that there are six digits to left of decimal
point. The number is therefore 744000.
Ex. 3. — What number has the log ,2.6538?
The mantissa 6538 is not found in the tables, but lies between
6532 and 6542, hence the number (not considering the decimal point)
lies between 4.50 and 451. The difference between the mantissa; of
4.50 and 451 is 10; the difference between the mantissa of 450 and of
the number to be found is 6. The difference between 450 and 451 is 1.
1 X .6 = .6. The number corresponding to mantissa 65.38 is 4506,
hence the numl)er corre.sponding to log 2.6538 is 4.50.6.
EXAMPLES
Find number corresponding to log 1.5481; to log 0.6681; to log
1.9559; to log 2.9324.
31S
LOGARITHMS OF NUMBERS
No.
1
2
3
4
7
8
9
10
OOCX)
0043
0086
0128
0170
0212
0253
0294
0334
0374
IX
0414
0453
0492
0531
0569
0607
0645
0682
0719
0755
12
0792
0828
0864
0899
0934
0969
1004
1038
1072
1 106
13
1 139
1173
1206
1239
1271
1303
1335
■^3^1
1399
1430
14
1 461
1492
1523
1553
1584
1614
1644
1673
1703
1732
15
1 761
1790
1818
1847
1875
1903
1931
1959
1987
2014
16
2041
2068
2095
2122
2148
2175
2201
2227
2253
2279
17
2304
2330
2355
2380
2405
2430
2455
2480
2504
2529
18
2553
2577
2601
2625
2648
2672
2695
2718
2742
2765
19
2788
2810
2833
2856
2878
2900
2923
2945
2967
2989
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
3222
3243
3263
3284
3304
3324
3345
3365
3385
3404
22
3424
3444
3464
3483
3502
3522
3541
3560
3579
3598
23
3617
3636
3655
3674
3692
3711
3729
3747
3766
3784
24
3802
3820
3838
3856
3874
3892
3909
3927
3945
3962
25
3979
3997
4014
4031
4048
4065
4082
4099
4116
4133
26
4150
4166
4183
4200
4216
4232
4249
4265
4281
4298
27
43H
4330
4346
4362
4378
4393
4409
4425
4440
4456
28
4472
4487
4502
4518
4533
4548
4564
4579
4594
4609
29
4624
4639
4654
4669
4683
4698
4713
4728
4742
4757
30
4771
4786
48CX)
4814
4829
4843
4857
4871
4886
4900
31
4914
4928
4942
4955
4969
4983
4997
501 1
5024
5038
32
5051
5065
5079
5092
5105
5119
5132
5145
5159
5172
33
5185
5198
5211
5224
5237
5250
5263
5276
5289
5302
34
5315
5328
5340
5353
5366
5378
5391
5403
5416
5428
35
5441
5453
5465
5478
5490
5502
5514
5527
5539
5551
36
55^3
5575
5587
5599
561 1
5623
5635
5647
5658
5670
37
5682
5694
5705
5717
5729
5740
5752
5763
5786
38
5798
5809
5821
5832
5843
5855
^Hl
5888
5899
39
59"
5922
5933
5944
5955
5966
5977
5988
5999
6010
40
6021
6031
6042
6053
6064
6075
6085
6096
6107
6117
41
6128
6138
6149
6160
6170
6180
6191
6201
6212
6222
42
6232
6243
6253
6263
6274
6284
6294
6304
6314
6325
43
6335
6345
6355
6365
6375
6385
6395
6405
6415
6425
44
6435
6444
6454
6464
6474
6484
6493
6503
6513
6522
45
6532
6542
6551
6561
6571
6580
6590
6599
6609
6618
46
6628
6637
6646
6656
6665
6675
6684
6693
6702
6712
47
6721
6730
6739
6749
6758
6767
6776
6785
6794
6803
48
6812
6821
6830
6839
6848
6857
6866
6875
6884
6893
49
6902
691 1
6920
6928
6937
6946
6955
6964
6972
6981
50
6990
6998
7007
7016
7024
7033
7042
7050
7059
7067
51
7076
7084
7093
7IOI
7110
7118
7126
7135
7143
7152
52
7160
7168
7177
7185
7193
7202
7210
7218
7226
7235
53
7243
7251
7259
7267
7275
7284
7292
7300
7308
7316
54
7324
7332
7340
7348
3
7356
7364
7372
7380
7388
7396
No.
1
2
4
5
6
7
8
9
LOGARITHMS OF NUMBERS
310
Ko.
1
2
3
4
5
6
7
8
9
55
7404
7412
7419
7427
7435
7443
745J
7459
7466
7474
56
74S2
7490
7497
7505
75U
7520
7528
7536
7543
755»
57
7559
7566
7574
7582
7589
7597
7604
7612
7619
7627
58
7634
7642
7649
7657
7664
7672
7679
7686
7694
7701
59
7709
7716
7723
7731
7738
7745
7752
7760
7767
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
6x
7853
7S60
7808
7875
78S2
7889
7896
7903
7910
7917
6a
63
7924
7993
?^
7938
8007
7945
8014
7952
8021
^51
7966
8035
7973
8041
c
79S7
8055
64
8062
8069
8075
8082
8089
8096
8102
8109
8II6
8122
65
8129
8136
8142
8149
8156
8162
8169
8176
8182
8189
66
8195
8202
8209
821S
8222
8228
8235
8241
8248
8254
67
8261
8267
8274
8280
8::S7
8293
8299
8306
8312
8319
C8
8325
8331
8338
8344
8351
8357
8363
8370
8376
8382
69
8388
8395
8401
8407
8414
8420
8426
8432
8439
8445
70
8451
8457
8463
8470
8476
8482
8488
8494
8500
8506
71
8513
8519
8525
8531
8537
8543
8549
8555
8561
8567
7a
8573
8579
8585
8591
8597
8603
8609
8615
8621
8627
73
8633
8639
8645
8651
8657
8663
8669
8675
8681
8686
74
8692
8698
8704
8710
8716
8722
8727
8733
8739
8745
75
8751
8756
8762
8768
8774
8779
8785
8791
8797
8802
76^
8808
8814
8820
8825
8831
8837
8842
8848
8854
8859
77
8865
8871
8876
8882
8887
8893
8899
8904
8910
8915
78
8921
8927
8932
8938
8943
8949
8954
8960
8965
8971
79
8976
8982
8987
8993
8998
9004
9009
9015
9020
9025
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
8x
9085
9138
9090
9096
9101
9106
9112
9117
9122
9128
9»33
8a
9143
9149
9154
9159
9165
9170
9175
9180
9186
83
9191
9196
9201
9206
9212
9217
9222
9227
9232
9238
84
9243
9248
9253
9258
9263
9269
9274
9279
9284
9289
85
9294
9299
9304
9309
9315
9320
9325
9330
9335
9340
86
9345
9350
9355
9360
9365
9370
9375
9380
9385
9390
87
9395
9400
9405
9410
9415
9420
9425
9430
9435
9440
88
9445
9450
9455
9460
9465
9469
9474
9479
9484
9489
89
9494
9499
9504
9509
9513
9518
9523
9528
9533
9538
90
9542
9547
9552
9557
9562
9566
9571
9576
9581
9586
91
9590
9595
9600
9605
9609
9614
9619
9624
9628
9633
ga
9638
9643
9647
9652
9657
9661
9666
9671
9675
9680
93
968s
9689
9694
9699
9703
9708
9713
9717
9722
9727
94
9731
9736
9741
9745
9750
9754
9759
9763
9768
9773
95
9777
9782
9786
9791
9795
9800
9805
9809
9814
9818
96
9823
9827
9832
9836
9841
9845
9850
9854
9859
9863
S7
9868
9872
9877
9881
9886
9890
9894
9899
9903
9908
98
9912
9917
9921
9926
9930
9934
9939
9943
9948
9952
99
9956
9961
9965
9969
9974
9978
9983
9987
9991
9996
9
Ho.
1 1 2 ! 3
4
5
6
7
8
TRIGONOMETRY
There are many problems in the shop that involve a knowledge of
angles and the relation between the parts of triangles — angles and
sides. Such problems include laying out angles, depth of screw
threads, diagonal distance across bolts, etc.
Trigonometry is the subject which deals with the properties and
measurement of angles and sides of triangles, and is spoken of in the
machine shop as simply " trig."
Trigonometric Functions (Ratios). — Since the sum of all the angles
of a triangle equals 180^, and since a right triangle is composed of a
right angle and two acute angles, it follows that, if we know one
acute angle, we may obtain the other by subtracting it from 90°.
The angle found by subtracting a definite angle from 90° is called
the complement of the given angle.
The complement of 45° is 45°, for 90° - 45° = 45°.
The complement of 30° is 60°, for 90° - 30° = 60°.
The supplement of an angle is the difference between it and 180°.
The supplement of 60° is 120°, for 180° - 60° = 120°.
Examine a right angled triangle very carefully. Notice the position
of the two acute angles and of the right angle. The longest side of
the triangle, called the hypotenuse, is op-
posite the right angle. The sides of the
triangle called the legs, are the two
smallest sides. These sides are perpen-
dicular to each other, and the longer side
is opposite the greater angle.
Draw a right angle with sides re- .^
spectively 3 and 4 inches long. _ . _
„, , .1 , , , Right Angled Triangle
Then draw the hypotenuse and meas-
ure the length of it, which will be 5 inches.
Divide the length of the side opposite Z A hy the hypotenuse,
3 ^ 5 = .6.
Divide the length of the side adjacent to Z A by the hypotenuse,
4 -f- 5 = .8.
Divide the length of the side opposite Z A hy the adjacent side,
3 -i- 4 = .75.
320
TRIGONOMETRY 321
These values are called ratios. The ratio of the length of the side
opposite Z A to the length of the hypotenuse is called the sine oi ^ A.
The ratio of the adjacent Z ^ to the length of hypotenuse is
called the cosine of Z A.
The ratio of the length of side opposite Z A to the length of
the adjacent leg is called the tangent oi Z A.
The ratio of the length of the adjacent side of Z ^4 to the opposite
side is called the cotangent.
The ratio of the length of the hypotenuse to the adjacent side of
Z A IB called the secant.
The ratio of the length of the hypotenuse to the opposite side of
Z A is called the cosecant.
These six ratios or constants represent a definite relation between
the sides and angles of right triangles.
Since all triangles may be divided into right angled triangles, by
dropping a perpendicular from one of the vertices of the triangle, we
might say these relations apply to all triangles. The same relations
apply to all figures, since any figure may be divided into triangles,
and then into right angled triangles.
These definite relations between the sides and angles are called
the functions of the angles, and are i*eally constants representing the
fixed proportions between the sides and angles of a triangle. The
exact values of these functions, or the proportion of the various parts
of the triangles, have been figured out for every degree that will be
used in daily practice. They are given in the tables for sine, cosine,
and tangent at the end of this chapter.
By means of a protractor construct an angle of 45°, Z CAB. At
a point on ^ C 1" from A^ erect a perpendicular,
CB. Connecting parts A and B by the line AB,
we have a rt. A with BC = AC, because Z A —
45° and Z B = 45°.
Measure accurately the length of all the sides
of the triangle.
Divide the length of side opposite Z A by the
hypotenuse, — '- — .7.
^^ AB
AC
Divide the length of side adjacent to ^A by the hypotenuse, — = .7.
AB
322 VOCATIONAL MATHEMATICS
Divide the length of side opposite Z A by the side adjacent to ZA,
AC
Divide the length of the adjacent side by the side opposite ZA,
BC
Divide the length of the hypotenuse by the side adjacent to ZA,
i^ = 1.4.
AC
Divide the length of the hypotenuse by the side opposite ZA,
^^1.4.
BC
If the work is carried out accurately, the above values should be
obtained.
Find the value of sine, cosine, and tangent when angles of 25^, 30°,
60°, 75°, are constructed.
The reciprocal of the tangent of an angle {Z A) is called the co-
tangent.
The reciprocal of the sine of an angle (^Z A) is called the cosecant.
The reciprocal of the cosine of an angle {ZA) is called the secant.
If the angles of the triangles are always marked with capital letters,
and the length of the sides opposite the angles are marked with small
letters as shown below, it will be possible to abbreviate functions of the
angle in terms of the sides.
Trigonometrical Formulas, etc.
Geometrical Solution of Right
Angled Triangles.
c = Va^ + 62
h = Vc^ - a2
a = Vc2 - 62
. a opposite side „ j
sin A=- = -i-s- cos A
c hypotenuse c hypotenuse
A _a_ opposite side + j _ ^ _ adjacent side
b adjacent side a opposite side
. c hypotenuse ^^c«« a c hypotenuse
sec ^ = - = --4-t^ ^^ cosec A = -= — •^^^-^~ — ^^
b adjacent side a opposite side
THIGONOMETHY 323
Practical Value of a Trigonometric Ratio
Since each function of an angle may be expressed as the ratio of two
sides, it follows that if we know the size of an angle and the length of
one side of a right triangle, we may determine the other sides by sub-
stituting in the equation.
Ex. If one angle of a right angled triangle is 30^, and the adjacent
side is 25 inches, what is the length of the other leg ?
D
Since we desire the other leg we should ust
(Formula involving two legs)
Tangent ^A=^
25
Ix>oking up tables for tangent .1, we find it to be .5773.") ; substitut-
ing, we have
.577 = — a = 14.4
25
EXAMPLES
1. The simplest application of trigonometry in the shop is the
problem for finding the depth of a V screw thread. Since the angle at
the depth of the screw thread is 60° and the sides are equal, an equi-
lateral triangle is formed. Forming right angled triangles by drop-
ping a perpendicular from the angle of the thread, we divide the angle
of 00° into two angles of 30'^ each. If we consider the pitch 1 inch,
then the sides are each an inch and the perpendicular divides the base
into two parts of \ inch each.
The perpendicular, or depth of the thread, is the cosine of the
angle of 30°. Looking in the table for 30°, then across until we
come to the column headed cosine, we find .8. This gives us the
depth directly.
2. Find the side of the thread if the depth is 1 inch.
3. What is the depth of a V thread with 8 threads to the inch ?
4. What is the distance across the corners of a square bar 3^" on
each side ?
324 VOCATIONAL MATHEMATICS
6. What size circular stock will be required to mill a square nut
lj\" on a side?
6. What is the distance across the corners of a bar 2^" by |"?
What angle does the diagonal make with the base?
Natural Trigonometric Functions
The tables on pages 326-329 give the numerical values of the
trigonometric functions of angles between 0° and 90° with inter-
vals of 10 minutes. For angles from 0° to and including 44°, read
from the top of the table downwards; for angles from 45° to and in-
cluding 89°, read from bottom of table upwards. The degrees and
minutes are found in the column marked A or Angle, and the value of
the function in the column marked by the name of the function. For
instance, the value of the sine of 15° 40' is found, in the column
marked sine and in the row 40 under 15, to be .2700. The value of the
cotangent of 63° 10' is found (reading from the bottom of the table
upwards, since the angle is between 45° and 89°), in the column
marked cotangent (at bottom of table) and in the row marked 10
above 63, to be .5059. The values of the sine, tangent, and secant in-
crease as the angle increases, while the values of the cosine, cotangent,
and cosecant decrease as the angle increases, so that for the former
functions the correction must be subtracted from the value given in
the table.
I. To find the value of a function of an angle which is not given in
the table.
Ex. 1. Find the value of sine 35° 21'.
Angle 35 is given in table, but 21' is not. 21', however, lies between
20' and 30', hence the value of sine of 35° 21' will lie between the value
of sine of 35^ 20' and the value of sine of 35° 30'. The value of sine of
35° 20' is given in table as .5783, and the value of sine of 35° 30' is
.5807 ; the difference between these two values is .0024. The tabular
difference in the angles (that is, the angular difference between 35° 20'
and 35° 30') is 10', \\4iile the angular difference between the smaller
angle given in the table (35° 20') and the angle of which we are trying
to find the value (35° 21') is 1'. The correction which we will have to
add to .5783, the value of sine of 35° 20', is j\ x .0024 = .00024. Hence
the value of sine of 35° 21' is .5783 -1- .00024 = .57854.
TRIGONOMETRY 325
Ex. 2. Find the value of cotangent 82° 11'.
82*' 41' lies between 82° 40' and 82° 50
cotan82°40' = .12869
cotan 82° oO' = .12574
Difference = .00295
The tabular difference l>etween angles 82° 40' and 82° 50' is 10'. The
difference between the smaller angle and given angle (82° 41' — 82° 40')
is 1'. The correction is therefore y^^ x .00295 or .000295, but since the
last figure is 5 we called it .00300. This correction must be subtracted
from the value of cotan 82° 40' since, as above, the value of the cotan
decreases as the angle increases.
cotan 82° 41' = .12869 - .00.3 = .12569
EXAMPLES
Find value of tan 45° 19' ; cosine 32° 8' ; cotan 78° 51' ; sine 62° 37'.
II. To find the value of an angle, when the value of a function is
given.
Ex. 1. Find the angle whose cotan is .6873.
In the column marked cotan look for the numbers .6873. Since it
is found in the column marked cotan at the bottom of the table, the
degrees and minutes will be found at the right-hand side of the table,
reading from the bottom upwards. It is in the row marked 30 above
5."). Hence, angle 55° 30' has the cotan .6873.
Ex. 2. Find the angle whose cosine is .9387.
This number is found in the column marked cosine at the top of the
table, therefore the degrees and minutes will be found on the left of
the table reading from top downwards. It is in the row marked 10
under 20. Hence, the angle 20"^ 10' has the cosine .9387.
EXAMPLES
Find the angle that has the sine .7642; cotan 1..5607; cosine .4746;
tangent 1.5108.
326
NATURAL SINES AND COSINES
A.
Sin.
Cos.
A.
Sin.
Cos.
|A.
Sin.
Cos.
0°
lO'
20'
30'
40'
50'
1°
10'
20'-
50'
2°
10'
20'
30'
40;
50'
3°
10'
20^
30;
40
50'
40
10'
20'
30;
40'
50'
5°
icy
20'
30;
40'
50'
6°
10'
20'
30;
40'
50'
70
10'
20'
30'
.000000
1. 0000
90°
50;
40'
30'
20'
10'
89°
50;
40'
30;
20'
10'
88°
50;
40'
30;
20'
10'
87°
50;
40'
30;
20'
10'
86°
50;
40'
30'
20'
10'
85°
50;
40'
30;
20'
10'
84°
50;
40'
30;
20'
10'
83°
50;
40'
30'
30;
40
50'
8°
10'
20'
30;
40'
50'
9°
10'
20'
30'
40'
50'
10°
10'
20'
30'
40'
50^
11°
10'
20'
30;
40
50'
12°
10'
20^
30;
50^
13°
10'
20'
^^'
40^
50'
14°
10'
20'
30^,
40'
50'
15°
•1305
.1334
.1363
.9914
•99 1 1
.9907
30;
20'
10'
82°
50'
40'
30;
20'
10'
81°
50;
40'
30'
20'
10'
80°
50;
40'
30;
20'
10'
79°
50;
30^
20'
10'
78°
50;
40'
30'
20'
10'
77°
50;
40
30
20'
10'
76°
50;
40'
30;
20'
10'
75°
15°
10'
20^
30;
40'
SO'
16°
10'
20'
30;
40
50'
17°
10'
20'
40^
50'
18°
10'
20'
30;
40'
50'
19°
10
20'
30'
40'
50'
20°
10'
20'
30'
40;
50'
21°
10'
20'
30/
40
50'
22°
lo'
20'
30'
.2588
•9659
75°
50;
40'
30;
20'
10'
74°
50;
40'
30;
20'
10'
73°
50;
40'
30'
20'
10'
72°
50'
40'
30'
20'
10'
710
40^
30'
20'
10'
70°
40^
30'
20'
10'
69°
5°,'
40'
30;
20'
10'
68°
50;
40'
30^
.002909
.005818
.008727
.011635
.014544
1. 0000
I.UUOU
1. 0000
.9999
•9999
.2616
.2644
.2672
.2700
.2728
.9652
.9644
.9636
.9628
.9621
.1392
■9903
.1421
.1449
.1478
•1507
•1536
.9899
•9894
.9890
.9886
.9881
.017452
.9998
.2756
.9613
.02036
.02327
.02618
.02908
.03199
.9998
•9997
•9997
•9996
•9995
.2784
.2812
.2840
.2868
.2896
.2924
•9605
•9596
.9588
•9580
•9572
•9563
.1564
■1593
.1622
.1650
.1679
.1708
•9877
.9872
.9868
.9863
•9858
•9853
.03490
•9994
.03781
.04071
.04362
.04653
•04943
•9993
.9992
.9990
.9989
.9988
.2952
.2979
.3007
•3035
.3062
•9555
•9546
•9537
•9528
•9520
•1736
.9848
•1765
.1794
.1822
•9843
.9838
•9833
.9827
.9822
.05234
.9986
.3090
•95"
.05524
.05814
.06105
.06685
•9985
•9983
.9981
.9980
•9978
.3118
•3145
•3173
.3201
.3228
•9502
.9492
•9483
•9474
•9465
.1908
.9816
•1937
.1965
.1994
.2022
.2051
.9811
•9805
•9799
.9793
.9787
.06976
•9976
•3256
•9455
.07266
•07556
.07846
.08136
.08426
•9974
.9971
•9969
.9967
•9964
•3283
•331 1
•3338
•3365
•3393
•9446
•9436
.9426
•9417
•9407
.2079
.9781
.2108
.2136
.2164
•2193
.2221
•9775
•9769
•9763
•9757
.9750
.08716
.9962
.3420
•9397
.09005
.09295
.09585
.09874
.10164
•9959
•9957
•9954
•9951
.9948
•3448
•3475
.3502
•3529
•3557
•9387
•9377
•9367
•9356
•9346
.2250
•9744
.2278
.2306
•2334
•2363
.2391
•9737
•9730
.9724
.9717
.9710
•10453
•9945
•3584
•9336
.10742
.11031
.11320
.11609
.11898
.9942
•9939
•9936
•9932
.9929
.3611
•3638
•3665
.3692
•3719
•9325
•9315
•9304
•9293
9283
.2419
•9703
.2447
.2476
.2504
•2532
.2560
.9696
.9689
.9681
.9674
.9667
.12187
.9925
•3746
.9272
.12476
.12764
.13053
.9922
.9918
.9914
.3800
•3827
.9261
•9250
•9239
.2588
•9659
Cos.
Sin.
A.
Cos.
Sin.
A.
Cos.
Sin.
A.
NATURAL SINES AND COSINES
327
A.
8in.
Con.
A.
Sin.
Cos.
A.
Sin.
Coi..
i
50'
lOf
20*
40'
50'
240
10'
2&
^i
40'
50'
25°
10'
20'
30'
40'
50'
26^
10'
20'
30'
40;
50'
27°
10'
2cy
30'
40'
so'
28°
i2 _
-cF)
See page 64.
Area of a triangle
See page 68.
^ = i Base X Altitude
Area of a rectangle
See page 69.
A = ha
Area of a trapezoid
See page 69.
^=(& + c)xi
a
Area of a polygon
See page 71.
A = \aP
Area of an ellipse
4
Circumference of an ellipse
See page 71.
c- 2 .
Contents of a cylinder
See page 72.
S = ttB^H
Volume of a pyramid
F=i6a.
Volume of a frustum of a pyramid
330
TABLE OF FORMULAS 331
Surface of a regular pyramid See page 73.
Volume of a cone See page 73
Lateral surface of a coue See page 73.
Volume of a frustum of a cone See page 74.
Volume of a sphere
3
Surface of a sphere See page 74.
Lateral surface of a frustum of a cone See page 74.
S = \shx(P-\- P^)
Volume of a barrel See page 75.
V=(D^x 2)-{-d^x Xx.2618
Diameter of blank for square bolt See page 127.
5 = 1.414^
Diameter of blank for hexagonal bolt See page 127.
B = 1.155 X
Pitch of a screw with V-shaped thread See page 140.
No. of threads per inch
Depth for V-shaped thread See page 140.
D=Px .8660
Size of tap drill for V-shaped thread See page 141.
c» rp l.i3*t
332 VOCATIONAL MATHEMATICS
Size of tap drill for U. S. Standard Thread See page 142.
Depth of thread of U. S. Standard See page 142.
D = Px .6495
Flat
-r
Acme Standard Thread See page 143.
, n 1-3732
^ = ^— ^
Square Thread See page 144.
Belting See page 147.
L = 0.1309 N{D-{-d)
Pulleys See page 161.
ttDR
F =
12
irli irD
Surface speed of pulleys See page 152.
N " D
Thickness of tooth of gearing See page 162.
y^ 1.57
Diametral Pitch
Circular pitch See page 162.
^p^ 3.1416
Diametral Pitch
Diametral pitch See page 162.
3.1416
DP
Circular Pitch
TABLE OP FORMULAS 333
Dimensions of gears by diametrical pitch See page 105.
p_ iV-h2 p_N .1.57
D ly p
N+2 P P
N=Piy N=PD-2 f=Tr.
jy ^f= whole depth of tooth
P= — P' =z —
p, p
Distance between centers of two gears See page 168.
Volume of rectangular tank See page 173.
LxBxHx 7.48 = V
Volume of cylindrical tank See page 174.
C = d^hx .0034 or .0034 dVi
Weight in lb. of lead pipe See page 175.
W= (D'-iP) X 3.8697 x I
Cubical contents of a foot of pipe See page 176.
C=I)^ X.7854 X 12 ^231
C^W X .0408
Capacity of a pipe of any length and any diameter See page 176.
C^LC X.0408 xL
C=Z>»x.7854x L-!-231
Pressure of water per sq. in. See page 188.
P=h X 0.434 lb. per sq. in.
334 VOCATIONAL MATHEMATICS
Head of water in feet See page 183.
0.434 0.434 ^ ^
Thickness of pipe
h X .s
T =
750
Velocity of water See page 186.
F=V^
X 2500
; 13.9
I X
d
Head to produce a given velocity See page 187.
V xLx^
h=- ^
2500
Twaddell scale into specific gravity See page 194.
(5 X N) + 1000 ^^
1000 ^^^
To change specific gravity into Twaddell scale See page 195.
(sa X 1000) - 1000 ^ ij^g^^^^ .,.^^^^,1
5
To convert Centigrade to Fahrenheit See page 196.
jP=— +32°
5
To convert Fahrenheit to Centigrade See page 208.
(7=^(i^-32°)
Thickness of boiler plate See page 208.
T. S. X %
Diameter of boiler See page 210.
/; = 7^ X T. .S. X % ^
TABLE OF FORMULAS 335
Size of safety valve See page 21C.
n=^K^^n^:^^
p
D = J ^-^ ^
\(P+ 8.62) X. 7854
Horse power of an engine See page 225.
jj J, _ AxPx V
33,000
//.P.(api)rox.)=|^
Diameter of cylinder See page 225.
^^ / 55()0 X H. P.
V .7854 X V
Diameter of supply pipe
See page 226.
z,-vv
Electric current
See page 232.
I=E^Ii
or / =
' R
Power in watts
See page 242.
"-f
-
Resistance of wire in ohms
See page 244.
7? ^^
Size of wire
See page 246.
e
DI
Resistance of cables
See page 250.
R = ^N}^ in ohms per 1000 ft.
cm.
Weight of cables See page 250.
ir= .00305 X c. m. in lb. per 1000 ft.
336
VOCATIONAL MATHEMATICS
Table of Decimal Equivalents of the Fraction of an Inch
By 8ths, 16ths, 32ds, and 64ths
8ths
82d8
64th s
64ths Continued
i = .126
xfV = 03125
^\ = .015626
If = .515625
\ = .250
xf\ = .09376
/j = .046875
II = .546876
1 = .375
,\ = .16625
^^ = .078125
II = .578125
| = .500
xjV = .21875
^\ = .109375
If = .609376
1 = .625
^^ = .28126
g\ = . 140625
II = .640625
f = .750
11 = .34376
li = .171875
If = .671875
i = .876
If = .40625
II = .46875
II = .63126
If = .203125
^1 = .234375
II = .265625
II = .703126
leths
II = .734375
tV = .0625
If = .765625
^5 = .1876
II = ..59375
If = .296875
II = .796875
j', = .3125
fl = .65625
II = .328125
If = .828125
^ = .4375
If = .71875
If = .359375
11 = .859375
^■, = .6625
If = .78125
II = .390625
II = .890625
jl = .6875
II = .84375
II = .421875
If = .921875
H = .8125
II = .90626
If = .453125
II = .953125
if = .9375
II = .96875
II = .484375
If = .984375
By 64ths ; from ^j to 1 inch
^ = .Q15626
II = .265625
If = .516625
If = .765626
-g\ = .031250
j% = .281250
II = .631250
If = .781260
^^ = .046875
11 = .296875
II = .546875
II = .796875
^5 = .062600
t\ = .312500
j% = .562500
II = .812500
^\ = .078125
II = .328125
II = .678126
If = .828125
^\ = .093750
^1 = .343750
II = .693750
II = .843750
^\ = .109375
If = .369375
II = .609375
II = .869375
1 = .125000
1 = .375000
1 = .625000
1 = .875000
^% = .140625
II = .390625
II = .640625
II = .890625
^\ = .156250
II = .406260
II = .656250
If = .906250
II =.171876
II = .421875
II = .671875
II = .921875
j\ = .187600
j\ = .437500
\l = .687500
II = .937500
If = .203125
If = .453125
II = .703126
II = .953125
j\ = .218750
It = .468750
If = .718750
II = .968760
if = .234376
II = .484375
II = .734376
If =.984375
1 = .250000
I = .500000
f = .760000
1 = 1.000000
INDEX
Addition, 3
Compound Nunabere, 44
Decimals, 31
Fractions, 20
Adjusting Gears, 256
Aliquot Parts. 37
Alligation, 287
Alternate, 288
Ammeter, 233
Amperes, 231
Angles, 64
Complementary, 64
Right. 64
Straight, 64
Supplementary, 64
Antecedent, 55
Apothem, 70
Arc, 62
Arc of Contact, 148
Area. 70
Circles, 62
Irregular Figures, 70
Rectangles, 69
Rings. 63
Triangles. 70
Atmospheric Pressure, 181
Avoirdupois Weight, 41
Barometer, 181
Base, 48
Belting, 147
Bevel Wheels, 163
Mitre Wheels. 163
Blanking Dies. 107
Blue Prints, 78
Board Measure, 84.
Boiler Plates. 200
Pumps, 218
Tubes, 212
Boilers, 203
Bolts, 126
Bracket, 298
Brass, 253
Brickwork, 95
Brown and Sharpe Wire Table,
248
Building. 83
Building Materials. 94
Buying Cotton, 286
Rags, 286
Wool, 286
Yarn, 286
Cancellation, 12
Capacity of Pipe, 176, 179
Capacity of Pump, 219
Carpentering, 83
Carpenter's Table of Wages, 104
Castings, 251
Cast Iron, 251
Cement, 97, 190
Centigrade Thermometer, 196
Circles, 62, 112
Circuits, 235
Parallel, 235
Series, 235
Circular Pitch, 161
Circumference, 62
Clapboards, 102
CoeflBcient. 304
Common Fraction. 16
Common Multiple, 15
Complex Fraction, 25
Compound Interest Table, 54
Compound Lathes, 258
Compound Numbers, 39
Cone, 73
Consequent, 56
387
338
INDEX
Construction, 89
Copper, 253
Cosecant, 321
Cosine, 321
Cotangent, 321
Cotton Yarns, 279
Countershaft, 156
Cube Root, 59
Cubic Measure, 40
Cutting Dies, 107
Decimal Fraction, 28
Addition, 31
Division, 34
Multiplication, 33
Reduction, 30
Subtraction, 32
Denominate Numbers, 39
Denomination, 39
Denominator, 16
Density of Water, 191
Diameter, 62
Diameter of Cylinder, 225
Diameter of Supply Pipe, 225
Diametral Pitch, 161, 162
Die, 134
Division, 8
Division of Compound Numbers, 45
Decimals, 34
Fractions, 24
Drainage Pipes, 174
Drilling Machines, 270
Driven Pulley, 149
Driver, 149
Dry Measure, 41
Efficiency of Water Power, 189
Ellipse, 71
Engine Lathes, 255
Engines, 222
Equations, 305
Equilateral Triangle, 66
Evolution, 59
Exact Method of Solving Example,
81
Exactness, 4
Excavations, 89
Factoring, 11
Factors, 11
Fahrenheit Thermometer, 196
Flooring, 102
Formulas, 297
Fractions, 16
Addition, 20
Common, 16
Complex, 25
Decimal, 28
Division, 24
Improper, 16
Multiplication, 23
Proper, 16
Reduction, 16
Simple, 25
Subtraction, 24
Framework, 89
Friction in Water Power, 189
Frustum of a Cone, 74
Fusible Plug, 213
Gear and Pitch, 255
Gearing, 159
Girders, 90
Graphs, 295
Greatest Common Divisor, 14
Hand Hole and Blow-Off, 213
Heat Units, 195
Hexagon, 70
Horse Power, 224
Hydraulics, 172
Hypotenuse, 66
Idler, 168
Inside Area of Tanks, 174
Interest, 51
Interpretation of Negative Quanti-
ties, 311
Involution, 59
Isosceles Triangles, 66
Jack Shaft, 156
Joule, 241
Kilowatt, 242
Latent Heat, 198
Lateral Pressure, 183
Lathes, 255
Lathing, 93
Laths, 90
Law of Squares, 179, 180
INDEX
339
Least Common Multiple. 14
Lever Safety Valve, 214
Linear Measure, 40
Linen Yarns, 279
Liquid Measure, 40
Logarithms, 315
Lumber, 84
Machine Speeds, 262
Manhole. 213
Mantissa, 315
Maximum Pressure, 223
Measurement of Resistance, 243
Measure of Time, 41
Mensuration. 62
Methods of Representing Operations,
306
By Means of Tables. 81
Exact Method, 81
Of SoUnng Examples. 81
Rule of Thumb Method, 82
Metric Equivalent Measures, 292
Measure of Capacity, 292
Length, 292
Surface, 292
Volume, 292
Weight, 293
Metric System, 291
Micrometer, 139
Mixed Decimal, 29
Mixed Number, 16
Mortar, 95
Multiplication, 7
Multiplication of Algebraic Expres-
sions, 313
Multiplication of Compound Num-
bers, 45
Decimals. 33
Fractions, 23
Multiple, 14
Multiplier, 7
NaUs. 131
Net Power for Cutting Iron or Steel,
263
Numerator, 16
Octagon, 70
Ohms, 232
Ohm's Law. 232
Operating Power, 228
Painting, 106
Paper Measure, 42
Parallelogram, 69
Parenthesis, 298
Pentagon, 70
Percentage, 48
Perimeter, 70
Pitch, 161
Pitch Circle, 161
Pitch Diameter, 161
Planers, 268
Planks, 84
Plaster, 95
Plates, 90
Plumbing. 172
Polygons, 70
Pop Safety Valve, 215
Power, 28, 59, 188
Power Measurement, 241
Prime Factors, 12
Proportion, 57
Protractor, 65
Pulleys, 146
Driven, 149
Driver, 149.
Pyramid, 73
Quadrilaterals, 69
Quotient, 8
Radii, 62
Radius. 62
Rafters. 90
Raising Water, 188
Rate Per Cent. 48
Ratio, 55, 62
Raw Silk Yarns, 278
Reamers, 272
Rectangle, 69
Rectangular Tanks, 173
Reduction Ascending, 40
Reduction Descending, 39
Reduction of Decimals, 30
Reduction of Fractions, 16
Regular Polygon, 70
Remainder. 5. 8
Return Tubular Boilers, 203
Rhomboid. 69
Rhombus. 69
Right Angle. 64
Right Triangle. 66, 67
340
INDEX
Rivets, 130
Roofing, 89
Root, 59
Rough Stock, 84
Rule of Thumb Method,
81
Safety Valves, 214
Safe Working Pressure, 205
Scale Drawings, 79
Scalene Triangle, 66
Screws, 134
Lead, 136
Pitch, 137
Threads, 137
Turns, 137
Secant, 321
Sector, 62
Shafts, 146
Shaper, 269
Sheet and Rod Metal Work, 107
Shingles, 98
Similar Figures, 75
Similar Terms, 304
Simple Fractions, 25
Simple Interest, 51
Simple Number, 39
Simple Proportion, 57
Sine, 321
Size of a BoUer, 209
Size of Lathes, 255
Size of Pump, 219
Slate Roofing,. 100
Soldering, 190
Specific Gravity, 192
Speed, 151
Speeds for Different Metals, 263
Sphere, 74
Spun Silk, 279
Spur Wheels, 163
Square Measure, 40
Square Root, 59, 60
Stairs, 103
Standard Gauge for Sheet Metal and
Wire, 116
Steam Engineering, 195
Steam Heating, 200
Steam Indicator, 227
Steam Lap, 224
Steel, 252
Stonework, 96
Studding, 90
Substituting, 308
Subtraction, 5
Compound Numbers, 44
Decimals, 32
Fractions, 21
Subtrahend, 5
Superheated Steam, 217
Supplement, 64
Surveyors' Measure, 40
Tables containing Number of U. S.
Gallons in Round Tanks for
One Foot in Depth, 177
Tables for Sheet Metal Workers, 117,
118
Tables of Metric Conversion, 293
Tacks, 133
Tangent, 321
Taps and Dies, 134
Temperature, 196
Tensile Strength, 203
Textile Calculations, 276
Thickness of Pipe, 184
Threads, 140
Acme Standard, 143
Square, 144
U. S. Standard, 141
V-shaped, 140
Trade Discount, 50
Train of Gears, 168
Translating Formulas into Rules,
301
Rules into Formulas, 298
Transposing, 308
Trapezium, 69
Trapezoid, 69
Triangle, 66
Equilateral, 66
Isosceles, 66
Right, 66
Scalene, 66
Triangular Scale, 79
Trigonometry, 320
Uses of Tables, 81
Value of Coal to produce Heat, 197
Velocity through Pipes, 186
Velocity of Water, 186
Voltmeter, 233
Volts, 231
INDEX
341
Volume, 72
Cone. 73
Cube. 72
Cylindrical Solid. 72
Rectangular Bar. 72
Water and Steam. 200
Water Gauge. 213
Power, 189
Pressure. 182
Supply, 172
Traps. 185
Wattmeter. 242
Watts. 242
Weight of Bars of Steel. 253
Flywheel. 223
Lead Pipe, 175
Roof Coverings, 100
Weights and Areas, 120,
123
Woodworking. 8.3
Woolen Yarns. 277
Worsted Yarns. 276
Wrought Iron. 252
Yarns. 276
Cotton. 279
Linen, 279
Raw Silk. 278
Two or More Ply, 279
W^oolen, 277
Worsted. 276
121, 122.
THIS BOOK IS DUE ON THE LAST DATE
STAMPED BELOW
AN INITIAL FINE OP 25 CENTS
WILL BE ASSESSED FOR FAILURE TO RETURN
THIS BOOK ON THE DATE DUE. THE PENALTY
WILL INCREASE TO 50 CENTS ON THE FOURTH
DAY AND TO $1.00 ON THE SEVENTH DAY
OVERDUE.
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LOAN DEPT.
JUN27 ^942
SEP 15 -342
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JAW 3 1968 3
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THE UNIVERSITY OF CALIFORNIA UBRARY
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