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ARITHMETIC

FOR

CARPENTERS AND BUILDERS

R. BURDETTE DALE, M.E.

Professor, Rensselaer Polytechnic Institute, Troy, New York
Formerly in Charge of Vocational Courses in Engineering and Cor-
respondence Instruction, Department of Engineering
Extension, Iowa State College

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PREFACE

This book was written for the purpose of presenting the
subject of arithmetic, as used in the daily work of the car-
penter and builder, in a simple form. It is intended for
^ the practical man as well as for the beginner and the student.
^ The material is such that the work can be followed success-

•>.\Afully by those who have had an eighth-grade education.
Upon the completion of this study the student should be
prepared to take up the problem of estimating the cost of
buildings,

^1 The author has not attempted to treat the subject
\ exhaustively. Though the first chapters may seem element-
ary to some, they will furnish a much-needed review to
others. Geometry is touched upon merely to serve as a
foundation for the work in mensuration. Practical appli-
cations of geometric truths are emphasized, while rigid
proofs and developments are omitted. Two chapters on
the steel square, that most useful tool of the carpenter,
are included.

Many of the problems do not admit of exact answers
for the reason that the judgment of the student plays an
important part in the solution. Different results will be
obtained, depending upon the lengths and widths of the
boards chosen and other similar details. Careful planning
for the economical use of material is required, for this is
an essential training for the work of the practical carpenter.

-i

IV PREFACE

Students will do well to take advantage of this and every
similar opportunity to develop their judgment.

It is believed that the reader who follows this text care-
fully as a guide may learn much that may not strictly be
classed as arithmetic. If the manner in which the subject
is treated and the many applications made inspire the prac-
tical worker and the beginner to seek a better understanding
of the fundamentals of carpentry, the author's most san-
guine hopes will be justified.

As a text in vocational mathematics it is believed that
this little volume should find a place in vocational and in-
dustrial schools, trade schools, manual training schools
and night schools. It is also well adapted for use in corre-
spondence instruction. The material has been used by the
author in the Vocational Courses in Engineering at the
Iowa State College and also in the extension classes of the
Department of Engineering Extension in various cities in
the state of Iowa.

No single book is the entire and original product of one
man's mind; it consists rather of accumulated knowledge
interpreted and adapted for a particular purpose. The
author gratefully acknowledges assistance rendered know-
ingly and unknowingly by many others. Material borrowed
from other sources is acknowledged in the text. The
author's appreciative thanks are due his colleague, Mr. H. E.
Freund, for reading the manuscript, and to Prof. K. G.
Smith for many suggestions and kindly criticisms. In-
debtedness is also acknowledged to Sargent & Company
for permission to use illustrations of their steel square.

R. BuRDETTE Dale.

\me8, Iowa,
April, 1915.

TABLE OF CONTENTS

CHAPTER I

UNITS OP LENGTH. ADDITION, SUBTRACTION, MULTIPLICATION AND
DIVISION OF UNITS OF LENGTH.

Units of Length — ^Abbreviations — Adding Feet and Inches — Subtract-
ing Feet and Inches — Multiplying Feet and Inches — Division of
Feet and Inches 1

CHAPTER II

FRACTIONS. MIXED NUMBERS. LEAST COMMON DENOMINATOR.
ADDITION AND SUBTRACTION OF FRACTIONS.

Fractions — Definitions — Reading and Writing Fractions — Proper and
Improper Fractions — Mixed Numbers — Fractions of an Inch —
Reduction of Fractions to Lower Terms — Reduction of Fractions
to Higher Terms — Changing Whole Numbers to Improper Fractions
— Changing Mixed Numbers — Changing Improper Fractions —
Least Common Denominator — To Find the Least Common Denomi-
nator— To Reduce to the Least Common Denominator — Adding
Fractions — Subtracting Fractions 10

CHAPTER III

MIXED NUMBERS. ADDITION, SUBTRACTION AND MULTIPLICATION OF
MIXED NUMBERS. CANCELLATION.

Working with Mixed Numbers — Adding Mixed Numbers — Sub-
tracting Mixed Numbers — Multiphcation of Fractions and Mixed
Numbers — The Product of a Whole Number and a Fraction — The
Product of Two Fractions — " Of " means " Times " — Multiplying
a Group of Whole Numbers and Fractions — Multiplying Mixed
Numbers — Factors — Cancellation 26

VI TABLE OF CONTENTS

CHAPTER IV

DIVISION OF FRACTIONS AND MIXED NUMBERS. COMPOUND FRACTIONS.

Division of Fractions — Compound Fractions — Dividing a Whole
Number by a Fraction — Dividing Mixed Numbers — Solution of
Problems — Classes of Problems 39

CHAPTER V

MONEY. ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION OF
MONEY QUANTITIES. CHECKS, STATEMENTS, RECEIPTS, ESTIMATES.

Money — Writing and Reading Money Sums — The Decimal Point —
Addition — Subtraction — Multiplication — Division — Bank Accounts
— The Bank Check — The Statement — The Receipt — Assets and
Liabilities — Wage Accounts — Building Estimates 47

CHAPTER VI

DECIMALS. ADDITION, SUBTRACTION, MULTIPLICATION AND DIVISION
OF DECIMALS. CHANGING FROM COMMON FRACTIONS TO DECIMAL
FRACTIONS.

Decimal Fractions — Writing Decimals — Reading Decimals — Signifi-
cance of Position — Addition of Decimals — Subtraction of Lecimals
— Multiplication of Decimals — Division of Decimals — Accuracy of
Results — Changing a Decimal Fraction to a Common Fraction —
Changing a Common Fraction to a Decimal Fraction — Decimal
Equivalents of the Fractions of an Inch 60

CHAPTER VII

THE USE OF RULES. PULLEY SIZES. WIDTH OF BELTS. FORMXTLAS.
SHORT METHODS OF MULTIPLICATION AND DIVISION.

Rules — Rules for Pulley Sizes — Rule for the Width of Belts — Using
Letters in Rules — How to Use Formulas — Short Methods — Addition
— Subtraction — MultipUcation — Division 73

TABLE OF CONTENTS Vli

CHAPTER VIII

PERCENTAGE. DISCOUNTS. PROFIT. INTEREST.

Percentage — Changing a Fraction to a Per Cent — Meaning of Terms-
Analysis of Problems — Solution of Problems — List Prices and Dis-
counts— Profits — Interest 82

CHAPTER IX

RATIO. PROPORTION. CEMENT AND CONCRETE MIXTURES. SLOPES.

LEVERS.

Ratio — A Ratio is a Fraction — Cement and Concrete Mixtures —
Proportion — Inverse Proportion — Solving a Proportion — The
Statement of a Proportion — Slopes — Levers — Arrangement of
Levers 91

CHAPTER X

LINEAR MEASURE. AREA. VOLUME. LIQUID MEASURE. DRY MEASURE.
WEIGHT. THE CIRCLE. WEIGHT OF BUILDING MATERIALS. BEAR-
ING POWER OF SOILS. FOOTINGS. CONCRETE MIXTURES.

Weights and Measures — Measures of Length — Measures of Surface —
Measures of Volume — Measures of Capacity — Measures of Weight
— Square or Surface Measure — Area of the Square and the Rectangle
— The Circle — Volume or Cubic Measure — Weight of Lumber —
Weight of Building Material — Bearing Power of Soils — Quantities
of Material Required per Cubic Yard of Concrete 105

CHAPTER XI

POWERS. ROOTS. RIGHT TRIANGLES. AREA OF CIRCLES. CAPACITY
OF TANKS AND CISTERNS.

Powers — Exponents — Powers of Common Fractions — Roots — Square
Root — Directions for Extracting Square Root — Proof of Square
Root — Cube Root — Practical AppUcations — The Side of a Square
— The Right Triangle — The Circle — Capacities of Circular Tanks
and Cisterns 122

viii TABLE OF CONTENTS

CHAPTER XII

LUMBER TERMS AND DIMENSIONS. FLOORING. SHINGLES.

Lumber Trade Customs — Lumber Terms — Allowance for Dressing —
Width of Lumber — Standard Lengths — Sizes of Common Boards
— Sizes of Dimension Lumber — Sizes of Dressed Finishing Lumber
— Sizes of Framing Lumber — Measurement of Lumber — Counting
Lumber — Flooring — Shingles — Material Lists 142

CHAPTER XIII
builders' geometry, geometric constructions, angles and

ANGULAR measure. ELLIPSE. POLYGONS.

Use of Geometry — Geometric Points and Lines — Geometric Surfaces
and Solids — Geometric Angles — Circular or Angular Measure —
Sectors and Segments — To Bisect a Line and to Erect a Perpen-
dicular— To Bisect an Angle — -To Erect a Perpendicular at any
Point on a Line — To Construct a Right Angle — To Construct
Various Angles — To Find the Center of an Arc or Circle — The
Ellipse — Polygons — To Construct a Hexagon — To Construct an
Octagon 160

CHAPTER XIV

AREAS OF SURFACES AND VOLUMES OF SOLIDS. AREA OF TRIANGLE,
PYRAMID, CYLINDER AND CONE. VOLUME OF CYLINDER, PRISM,
PYRAMID AND CONE. MEASURING SURFACES AND VOLUMES.

Area of a Triangle — ^Area of Pyramid — ^Area of a Cylindrical Surface
— Area of the Surface of a Cone — Volume — Volume of a Cylinder
— Volume of a Prism — Volume of a Pyramid — Volume of a Cone —
Volume of the Frustum of a Cone or Pyramid — Measuring Surfaces
and Volumes 174

TABLE OF CONTENTS IX

CHAPTER XV

USE OF THE carpenter's SQUARE. BRACE MEASURE. OCTAGONAt
SCALE. ESSEX BOARD MEASURE. DIAGONAL SCALE. POLYGON
SCALE. GEOMETRIC APPLICATIONS.

Carpenter's Square — The Brace Measure — The Octagonal Scale —
The Essex Board Measure — The Diagonal Scale — The Polygon
Scale — Geometric Applications — Hypotenuse of a Right-angled
Triangle — To Divide a Line — To Bisect an Angle 195

CHAPTER XVI

USE OF THE carpenter's SQUARE IN FRAMING. FLOOR BRIDGING.
STAIR BEAMS. RAFTER FRAMING. ROOF PITCHES. RAFTER TABLE.
ROOF FRAMING. HIP OR VALLEY RAFTERS. JACK RAFTERS.

To Find the Cuts for Floor Bridging — To Find the Cuts on a Stair
Beam — Rafter Framing — Pitch — To Find the Cuts for a Common
Rafter — The Rafter Table — Roof Framing — Length of Hip or
Valley Rafters — Cuts for Hip or Valley Rafters — Hip Rafter Tables
— Jack Rafters — Side Cut for Hip Rafters — Length of Hip or
Valley Rafters — Cuts for Cripple Rafters — Actual Lengths .... 206

ARITHMETIC

FOR

CARPENTERS AND BUILDERS

CHAPTER I

UNITS OF LENGTH. ADDITION, SUBTRACTION, MUL-
TIPLICATION AND DIVISION OF UNITS OF LENGTH

1. Units of Length. We are all familiar with the com-
mon foot rule, by which we measure the length, width or
thickness of a board. The /oof is the unit of length measure-
ment used by English-speaking people, just the same as the
dollar is the unit of money in the United States. The foot
unit is, however, too long to make small measurements
conveniently, therefore it is divided into twelve equal parts
called inches. With these units of length, the foot and the
inch, we can make a great many rough measurements with
all the accuracy necessary.

The foot is the unit of length measurement.

The inch is one- twelfth of a foot.

There are three other units of length based on the foot
which are in common use. They are the yard, the rod and
the mile.

The yard equals three feet or thirty-six inches.

2 ARITHMETIC FOR CARPENTERS AND BUILDERS

The rod equals sixteen and one-half feet or five and
one-half yards.

The mile equals five thousand two hundred eighty feet.

2. Abbreviations. The word foot is often abbreviated
to Jt., and the word inch to in. The mark (') also means

Fig. 1. — Section of Timber Framing for Garage.

feet and (") means inches,
expressed thus:

These abbreviations may be

20 feet 4 inches = 20 ft. 4 in. = 20' 4".

The abbreviation for yard is yd., and for mile is mi.

3. Adding Feet and Inches. You will recollect that
you cannot add two unlike quantities and get a result which
has any meaning. If you try to add two nails to five screws
you have seven things, but they are still two nails and five

UNITS OF LENGTH

screws. If you have two nails and five screws in one box
and three nails and four screws in another box and you put
them all in the first box, you will have five nails and nine
screws in that box. On paper it would look like this:

2 nails 5 screws in first box.

3 nails 4 screws in second box.

5 nails 9 screws in one box.

It- 4'8'-^4<28^3'6M^285iU-3'6^4^^4' 8^

FT I i'f

Fig. 2. — Ground Floor Plan of Garage.

When we add feet and inches we must do it in a similar
manner.

Example. 7 feet 6 inches

10 feet 3 inches

17 feet 9 inches. Ans.

The inch is a part of the foot. If we have more than
twelve inches, we can change some of the inches to feet and
the part left over will be the odd inches. We could not

4 ARITHMETIC FOR CARPENTERS AND BUILDERS

do this with the nails and screws because a screw is not a
part of a nail. The following example will make this clear.
Example. Add 9' 6" and 10' 10".

10'

10"

19'

+
1

16"
~12

20' 4" Ans.

Explanation. Sixteen inches are more than one foot,
therefore if we subtract 12" from 16" we must also add
1' to the 19'. The result will be 20' 4" as shown.

4. Subtracting Feet and Inches. In subtracting feet
and inches, we use the same method of combining the units
as we did in adding.

Example. From 12' 10" take 4' 6".

12' 10" minuend
4' 6" subtrahend

8' 4" difference. Ans.

Notice that in the above example in subtraction the
larger of the two numbers is called the minuend; the num-
ber to be subtracted is called the subtrahend and the result
is called the difference.

Example. From 9' 2" take 6' 9".

9' 2" = 8' 14"
6' 9"

2' 5" Ans.

Explanation. Before we can subtract, we must borrow
r from 9' and change it into inches.

UNITS OF LENGTH 5

5. Multiplying Feet and Inches. Multiplication is a
short method of addition. The multiplicand is the number
to be multiplied and the multiplier indicates how many
times the multiplicand is to be taken. The result is called
the product.

To multiply feet and inches, we first multiply the inches
and then the feet by the multiplier. If there are 12" or
more in the resulting product, we change the inches to feet
so that the result will be expressed in the simplest form.

Example. What is the total length of board required
to give 5 pieces each 2' 6" long?

2' 6" multiplicand
5 multiplier

product

10'

30"

—

24"

12' 6" simplified product. Ans.

Explanation. 5X6" = 30" and 5X2' = 10'. In order to
express this result in its simplest form, we must subtract
24" from 30" and add 2' to 10'. This gives the result 12' 6".

6. Division of Feet and Inches. Division is the opposite
of multiplication. The number to be divided is called the
dividend. The number by which the dividend is divided
is called the divisor. The result is called the quotient.
To divide feet and inches we first divide the feet and then
the inches, expressing the result in its simplest form.

Example. A board 15' 8" long is to be cut into four
equal lengths. How long will each piece be?

4 divisor) 15' 8" dividend
3|' 2" quotient
^' = 9"
Therefore, 31' 2" = 3' 11" Ans.

6 ARITHMETIC FOR CARPENTERS AND BUILDERS

Explanation. Four goes into 15' three times and 3 over,
or 3|'. One-quarter of a foot is 3" and three-quarters of
a foot is 3 X 3" or 9". Four goes into 8" two times with-
out a remainder. Now add the 9" which represents the f
to 2" and the simpUfied quotient is 3' 11".

Another very good way to divide feet and inches is
first to reduce the feet to inches and then divide.

Example. A board 5' 10" long is to be cut in the middle;
what is the length of each piece?

5' 10" = 70"

+
10

70"

2 )70"

35" = 2' 11" Ans.
~24
U"

Explanation. To reduce 5' to inches multiply by 12.
Then add the 10" to get 70". Divide 70" by 2, the number
of equal pieces into which the board is to be cut, to get 35".
This is more than 2', so we can subtract 24" from 35" and
we get 11" as the difference. The simplified form is, there-
fore, 2' 11".

UNITS OF LENGTH

Summary of Chapter I

1. To change feet to inches, multiply by 12.

2. To change inches to feet, divide by 12.

3. To add measurements, separate the feet and inches
and add them separately, changing the result to its simplest
form. (Sec. 3.)

4. To subtract feet and inches, subtract each separately
and reduce the result to its simplest form. (Sec. 4.)

^BuEz saw table

-A

Fig. 3.

5. To multiply feet and inches, multiply first the inches
and then the feet by the multipher, reducing the result to
its simplest form; or reduce the feet to inches and mul-
tiply, changing the product again to feet and inches for
the answer, (Sec. 5.)

6. To divide feet and inches, divide first the feet and
then the inches, reducing the result to its simplest form; or
reduce the feet to inches and divide, changing the quotient
again to feet and inches for the answer. (Sec. 6.)

PROBLEMS

1. A board is cut up in the factory into three pieces. One
piece is 3' 4" long; another is 5' 2" long, and another is 7' 5" long.
What was the original length of the board if 2' 1" were wasted?

8 ARITHMETIC FOR CARPENTERS AND BUILDERS

K*;*,

2. A farmer has a field 3G rods long, and 15 rods wide; how long
and how wide is this in feet and inches? What is the total dis-
tance around the field in feet and inches?

3. Fig. 1 shows the section of the timber framing for a two
story garage. What length of studding must be ordered for the
job? To what length must the studding be cut? What is the
total distance from the grade Une to the ridge of the roof; in
other words, what is the distance marked A in the figure?

4. A carpenter wishes to cut four shelves each 2' 9" long. He
has three boards which he may use. One of these is 10' long,
another 12' and another 14'. Which board will cut with the least
waste? How much waste will there be?

5. Fig. 2 shows the ground floor plan of the
garage mentioned in Prob. 3. What is the total

>*■><' length and width of the building?

6. In Fig. 2 how far apart must the windows
in the end of the building be placed so that the
distances from the edge of each window to the
end of the wall and the distance between the
windows will be equal?

7. A flag pole is 96' 3" long. If it is set
9' 8" in the ground, how much of it projects
above the ground?

8. A workman in going from his home to the
mill walks ten blocks and a half and crosses
.eleven streets. The blocks are each 300' long

and the streets are 60' Avide. If he makes the trip morning and
evening for 300 days in the year, how many miles will he have
walked at the end of that time?

9. If it takes the man mentioned in Prob. 8 an average of 17
minutes to make one trip to the mill, how much time will he spend
in walking in the course of a year if he makes two trips each day?

10. If telegraph poles are spaced 176' apart, how many of
them are there to the mile?

11. What must be the width of the shop to properly accommo-
date the buzz saw shown in Fig. 3 if the clearance for the board
at the right is the same as that at the left?

12. A shipping box is 77" long, 43" wide and 58" high. Give
the dimensions of this box in feet and inches.

Fig. 4.
Door Frame.

UNITS OF LENGTH 9

13. The steps in a cellar stairs are 2' 9" wide and there are
15 of them. What is the total number of Uneal feet of stepping
required? How many boards 14' long will be necessary to cut
the steps?

14. If 16 sticks each 16' long and 6 sticks each 10' long are
ordered to cut 16 door frames like the one shown in Fig. 4, how
many lineal feet of lumber will be wasted?

Fig. 5. — Cement Walks for Residence and Lawn.

15. The windows in the side of a house 21' 10" long are 28"
wide. There are four of them and they are spaced so that the
distance between each of them and the distance between the end
windows and the ends of the wall are all equal. What is the
distance between them?

16. How many lineal feet of two by four edge guides will be
required for the cement walks shown in Fig. 5, if all of the work
is done in one day so that all the guides are in use at the same
time? Allow 50' for waste and cutting.

CHAPTER II

FRACTIONS. MIXED NUMBERS. LEAST COMMON DE-
NOMINATOR. ADDITION AND SUBTRACTION OF
FRACTIONS

7. Fractions. If anything is divided into equal parts
we call the parts fractions. The fraction may be either one,
or several of these parts. Fig. 6 shows a circle made up
of segments by the method often used by woodworkers.
If you count the segments you will see that there are eight
of them. The circle has been divided into eight equal parts
and each one of these parts is one-eighth of the whole
circle. If we select three of these segments they will
yepresent three-eighths of the whole.

The expression 8-f-4 = 2 means that 8 divided by 4 equals
2. The sign -j- means divided by. Now if we replace the
dots above and below the line in this sign by the figures,
we have f = 2, which is a simpler way of expressing the same
thing. But this f is a fraction. Thus we see that a fraction
is another way of expressing division, and it means that the
number above the line is to be divided by the number below
the line.

8. Definitions. A fraction is one or more of the equal
parts into which anything may be divided. For example
three-eighths means that the whole is divided into eight
parts and three of these are taken. A fraction also indicates
division.

A fraction has two parts called terms. One is the

FRACTIONS

denominator which is written below the line and this tells
into how many parts the whole has been divided. The
numerator or the number above the line tells how many
of these parts are taken.

9. Reading and Writing Fractions. To pronounce a
fraction, name the numerator first and then the denominator
followed by the ending th or ths. When the numbers 2, 3

Fig. 6. — Circle made up of Segments.

and 4 are used for the denominator we say one-half, one-third
or one-quarter. Also when the number of the denominator
ends in 1, 2 or 3, we say one twenty-first, one thirty-second
or five fifty-thirds, as the case may be.

To write a fraction, place the numerator over the denomi-
nator with a line between. For example, three-eighths is
written |; seven-sixteenths is written ye-

10. Proper and Improper Fractions. A proper fraction
is one the numerator of which is less than the denominator.
The value of a proper fraction is less than unity or one.

12 ARITHMETIC FOR CARPENTERS AND BUILDERS

^, §, f and ^ are all proper fractions because the numer-
ators are less than the denominators.

An improper fraction is one the numerator of which
is equal to or greater than the denominator. If the numer-
ator is equal to the denominator, the value of the fraction
is unity or one, for the reason that all of the parts into which
the whole has been divided are taken. For example, |, f , |
and jl are improper fractions the value of each of which is
1. If the numerator of the fraction is greater than the
denominator, the value of the fraction is greater than one.

Fig. 7. — Portion of an Ordinary Rule.

For example, |, | and -y- are all improper fractions, the value
of each of which is greater than 1.

11. Mixed Numbers. A mixed number is a number
composed of a whole number and a fraction written together.
For example, 5^ is a mixed number and denotes five and a
half units. If the units are feet, this would mean five and
one-half feet.

12. Fractions of an Inch. If you will examine the ordi-
nary rule carefully, a representation of which is shown in
Fig. 7, you will notice that the inch is divided first into two
parts or fractions, each being one-half of an inch. The
halves are again equally divided into quarters of an inch
and these again into eighths and sixteenths of an inch. For
very accurate measurements some rules are divided into
thirty-seconds and sixty-fourths of an inch.

FRACTIONS 13

13. Reduction of Fractions to Lower Terms. Everyone
knows that ^"+5" = !"= 1". We can also say,

4 +4 — i —2 ana 8 -f-j — | — j .

This shows that the same value can be expressed by frac-
tions in different ways. In spite of this fact we do not
usually say two-eighths of an inch when we mean one-
quarter of an inch, although they both have the same value.
, We are obUged to say three-sixteenths of an inch, but we
usually say one-eighth rather than two-sixteenths. Unless
it is necessary to do otherwise we like to reduce the frac-
tion to its lowest terms.

Example, Reduce ^" to its lowest terms.

i4-"-^4 — ^" Ant

Explanation. Divide both the numerator and the denom-
inator of the fraction by the same number and use the largest
number that will go into both an even number of times. In
this case 4 is the largest number which will go into both 12
and 32.

Thus we see that if we divide both the numerator and
the denominator of the fraction by the same number, we
change the form of the fraction but we do not change its
value.

14. Reductions of Fractions to Higher Terms. If we
multiply both the numerator and denominator of a fraction
by the same number, we will change its form but we will not
change its value. We can, therefore, reduce a fraction to
higher terms. In this way we can increase the denominator
of any fraction.

Example. Change f " to sixteenths of an inch.

8 X"j= iV) Ans.

14 ARITHMETIC FOR CARPENTERS AND BUILDERS

Explanation. To change eighths to sixteenths we must
multiply the denominator by 2, but we cannot do this with-
out changing the value of the fraction unless we also multi-
ply the numerator by 2. Multiplying both the numerator
and denominator by 2 gives the required result, -^".

Example. Change |" to thirty-seconds.

f"Xl = f|" Ans.

Explanation. We must multiply the denominator 8 by 4
to get 32. If we multiply the denominator by 4 we must
also multiply the numerator by 4. The result is |^".

15. Changing Whole Numbers to Improper Fractions.
A whole number can be changed into the form of a fraction
by using the number 1 for a denominator. This does not
change the value of the number. For example, the whole
number 6 is equal to -f . We can multiply both the numera-
tor and denominator of this fraction by the same number
without changing its value.

Example. Change the whole number 6 to eighths.

XI = ^ Ans.

Explanation. We first changed the whole number into
the form of a fraction by using the denominator 1. To
reduce | to 8ths we must multiply both numerator and de-
nominator by 8. The result is an improper fraction, ^.

16. Changing Mixed Nmnbers. A mixed number can
be changed to an improper fraction by multiplying the
whole number by the denominator of the fraction, adding
to this the numerator of the fraction and placing the sum
over the denominator. The result will be an improper
fraction which is equivalent to the mixed number,

FKACTIONS 15

Example. Change 5§ to an improper fraction having
2 for a denominator.

Explanation. In one unit there are 2 halves; therefore,
in 5 units there are 5 times 2 halves or 10 halves. Adding
the 5 to the ^- we get y- for the result. If you will examine
the rule given above you will see that it has been followed
in the solution of this example.

17. Changing Improper Fractions. Changing an im-
proper fraction to a mixed number is just the reverse of
changing a mixed number to an improper fraction. The
process is as follows: Divide the numerator by the denomi-
nator to find the whole number and annex to this the frac-
tion obtained by using the remainder for a numerator and
the denominator of the improper fraction for a denominator.
The result is a mixed number.

Example. Change -^J to a mixed number.

8 )39( 4| Ans.

Explanation. Dividing the numerator by the denomi-
nator, 8 goes into 39 four times with a remainder of 7.
Placing this remainder over the denominator and annexing
this fraction to the whole number in the quotient, we have
4| for the result.

18. Least Common Denominator. Suppose you are
required to find the combined thickness of two boards one
of which is I of an inch thick and the other is f of an inch
thick. If you measured these two boards together with
your rule you would see at once that the combined width
is I of an inch. How would you add these two fractions

16 ARITHMETIC FOR CARPENTERS AND BUILDERS

on paper? You will see that since the answer is |, you must
change both of the fractions to eighths before you can add
them, because you cannot add two things that are not of
the same kind. You must, therefore, reduce the fractions
to a common denominator.

Example. What is the sum of j" and |"?

r'x^^r

Ans.

Explanation. Change j" to eighths by multiplying both
numerator and denominator by 2. This gives f". The

Fig. 8. — Panel for a Door

other fraction is already expressed in eighths so that we
can now add the numerators and place their sum over the
common denominator.

The Least Common Denominator is the smallest denomi-
nator to which all of several fractions may be changed.
The abbreviation for the least common denominator is
L.C.D.

If you are working with ordinary fractions of an inch,
it is not difficult to find the L.C.D. provided all of the frac-
tions are first reduced to their lowest terms. The ordinary
fractions of an inch commonly used are halves, quarters,
eighths, sixteenths, thirty-seconds and sixty-fourths. The
L.C.D. of a number of fractions having these denomina-

FRACTIONS 17

tors will be the largest denominator which occurs in the
group.

Example. Find the L.C.D. of the following fractions:

5 _7. 3 _9_ 1 1 „ J _5_
8) 16) 4> 32> 4> 2 '*^l'-l 16'

5 y 4 -_ 2_fl.. -I-V2 — 14- IvS — 24. _g_.

4Af — ^^, 2A1B— F?) i6/^T~TT- /ins.

Explanation. Since all of the fractions are expressed
in their lowest terms, we can determine the L.C.D. by in-

-^^~

1 1

■ T"

Fig. 9. — Board Grooved for Splines.

spection. It will be the largest number among the denomi-
nators which, in this case, is 32. Now taking each fraction
in turn we divide its denominator mentally into 32 to get
the number 'by which it must be multiplied. We then
multiply both the numerator and denominator by this num-
ber.

19. To Find the Least Common Denominator. Now
and then it is necessary to find the L.C.D. of several frac-
tions when it cannot be determined by inspection. In that
case, we first place the denominators in a row, separating
them by dashes. Then select the smallest number, other
than 1, which will be exactly contained into two or more

18 ARITHMETIC FOR CARPENTERS AND BUILDERS

of the denominators. Draw a line under this row and per-
form the division, writing the quotients below the line. Also
write below this line any numbers which are not divisible
by the divisor. Then repeat the operation, selecting a new
divisor that will be exactly contained into two or more of
these new numbers. Continue this process until "no number
will exactly divide more than one of the numbers below the
line. The product of all of the divisors and all of the num-
bers left in the last line of quotients is the L.C.D.
Example. Find the L.C.D. of j, f , |, and ■^.

2 )4— 3— 9— 16
2)2—3—9—8
3)1—3—9— 4

1—1—3— 4
L.C.D. = 2X2X3X3X4 = 144 Ans.

Explanation. First place the denominators of the frac-
tions in a row, separating them by dashes. Then select
the number 2 for a divisor which will be contained into both
4 and 16. Perform the division and write down the quotients,
bringing down the 3 and 9 also. Select 2 again and divide
it into 2 and 8, writing down the quotients and the 3 and
the 9 as before. The number 2 will no longer be contained
into more than one of the numbers remaining; hence, we
will try 3. This will go into 3 and 9, and after writing the
quotients of this division we find that no number will be
contained into more than one of the numbers remaining.
To find the L.C.D. then we multiply all of the divisors and
all of the remaining numbers together. This gives the
result 144, which is the smallest number into which all
of the denominators will be exactly contained without a
remainder.

FRACTIONS

20. To Reduce to the Least Common Denominator.

After we have found the L.C.D. we next reduce each fraction
in turn to that denominator. To do this we multiply both
the numerator and the denominator by such a number as
will raise the fraction to the required L.C.D. To find the
number by which to multiply, divide the denominator of
the fraction into the L.C,D. The quotient will be the
required number.

|*^->i B^D i1k"H

Fig. 10. — Section of Partition Material.

Example. Reduce the fractions in the previous example
to their L.C.D.

144-

- 4 = 36

iX|| = T¥r

144-

- 3 = 48

2 s/ 4 8 _ 9 6

3 A^g — ITT

144-

- 9 = 16

ixu=m

144-

-16= 9

TeX |=TTT

Ans.

Explanation. In the previous example we found that
the L.C.D. of l,i, I, and ys was 144. We must then divide
each denominator into this number in turn and multiply
both numerator and denominator of the fraction by this
quotient. Taking the first fraction, 4 will go into 144
thirty-six times. Multiplying both the numerator and the
denominator of the fraction j by 36 we get -fj^ for the
result. We proceed in exactly the same way for each of the
other fractions. v.

20 ARITHMETIC FOR CARPENTERS AND BUILDERS

21. Adding Fractions. To add fractions change them
to their L.C.D., then add the numerators and place their
sum over the L.C.D. Reduce the resulting fraction to its
lowest terms if necessary.

Example. Add the following fractions of an inch: f",
A", ¥', f " and ^".

3" 8 24

4 ^8 ~

32"

5" 2 10
16^2

32"

1" 16 16
2 ^16

32"

3" 4 12
8 ^4 ~

32"

9-1 _ 9
32^1

32"

"^)71(2^

Ans.

Explanation. By inspection we see that 32 is the L.C.D.
Changing each fraction to this L.C.D. and adding the
numerators, we have |^". Reducing this improper fraction
to a mixed number gives 2^" as the result.

22. Subtracting Fractions. To subtract fractions we
must first change them to their L.C.D. the same as we did
in adding fractions. Then subtract the numerator of the

FRACTIONS 21

subtrahend from the numerator of the minuend and write
the result over the common denominator. If necessary,
reduce the result to its lowest terms.
Example. Subtract a^" from |",

7 28 ■

8 32

25" Ans.

Explanation. Changing |" to thirty-seconds, we multiply
both numerator and denominator by 4. This gives ff".
We can now subtract •^" and obtain ff " for the result.

22 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter II

7. The numerator and the denominator of a fraction
may both be divided by the same number without changing the
value of the fraction. (Sec. 13.)

8. The numerator and the denominator of a fraction
may both be multiplied by the same number without changing
the value of the fraction. (Sec. 14.)

Fia. 11. — Mortise and Tenon Joint for Table Leg.

9. To reduce a fraction to its lowest terms, divide
both the numerator and the denominator by the largest
number which will exactly go into each of them. (Sec. 13.)

10. To reduce a fraction to higher terms, multiply both
the numerator and denominator by the number which will
give the required denominator. (Sec. 14.)

11. To change a whole nimiber to an improper fraction,
having a given denominator, supply the denominator 1 to

FRACTIONS

the whole number and multiply both numerator and denomi-
nator of this resulting fraction by the number which will
give the required denominator. (Sec. 15.)

12. To change a mixed number to an improper fraction,
multiply the whole number by the denominator of the frac-
tion, add to this the numerator of the fraction and place
the sum over the denominator. (Sec. 16.)

13. To change an improper fraction to a mixed number,
divide the numerator by the denominator to find the whole

Fig. 12. — Dovetail Drawer Joint.

number and annex to this the fraction obtained by using
the remainder for a numerator and the denominator of the
improper fraction for a denominator. (Sec. 17.)

14. To find the least common denominator of several
fractions, find the least number into which each of the de-
nominators will, be exactly contained without a remainder.
(Sec. 18 and 19.)

15. To reduce several fractions to their least common
denominator, multiply both the numerator and the denomi-
nator of each fraction by such a number as will give the
required L.C.D. (Sec. 20.)

16. To add fractions, first reduce them to their L.C.D.,

24 ARITHMETIC FOR CARPENTERS AND BUILDERS

then add the numerators and place their sum over the com-
mon denominator. SimpHfy this result, if possible. (Sec.
21.)

17. To subtract fractions, first reduce them to their
L.C.D., then subtract their numerators and place the result
over the common denominator. Simplify, if possible. (Sec.
22.)

Fig. 13. — Dovetail Mortise Joint.

PROBLEMS

17. (a) Change f|" to a mixed number.

(6) Change 9|" to an improper fraction.

18. Which is greater, H" or |"?

19. Reduce the following fractions to their lowest terms: ^,

« 18 8

ie» 32) T2'

20. Change the following fractions to their L.C.D. : A", fj",

i", A".

21. Add the following fractions: A", {", I" and ^".

22. From f take A".

FRACTIONS 25

23. Fig. 8 shows a panel for a door. What is the thickness of
the panel?

24. The door stile shown in Fig. 8 is to be 1 1" thick. If the
veneers are each ^" thick, how thick must the core be?

25. A drawing board is built of three plies; one is f" thick,
one is H" thick and one is ^" thick. What is the total thick-
ness of the board?

26. A dressed board is required f|" thick. How much must
be planed off from a rough board 1" thick to give the required
thickness?

27. A floor is laid by first putting down the rough sheathing
which is I" thick. On this are placed furring strips |" thick and
then the top or finish floor is laid which consists of boards \i"
thick. What is the total thickness of this floor?

28. Fig. 9 shows a board grooved for splines. If a A"X|"
sphne is to be used, what will be the width of the projection A on
each side of the groove?

29. Fig. 10 shows 1X4 partition material in section. Give
the missing dimensions indicated by the letters A, B, C and D.

30. Flooring I5" nominal size is worked to Hi" in the mill.
How much is it scant?

31. Fig. 11 shows a mortise and tenon joint for a table leg.
How far back from the face of the leg must the outside edge of
the mortise be? In other words, what is the distance A7

32. What is the missing dimension on the dovetail drawer joint
indicated by the letter A in Fig. 12?

33. What must be the size of the square key used for the dove-
tail mortise joint shown in Fig. 13?

34. A lumber dealer has in stock 2000 bundles of shingles.
Of these 400 bundles are graded Extra Clear and 1600 are Extra
*A*. Each grade is what fractional part of the whole stock?
Give the fractions in their lowest terms.

CHAPTER III

MIXED NUMBERS. ADDITION, SUBTRACTION AND
MULTIPLICATION OF MIXED NUMBERS. CAN-
CELLATION

23. Working with Mixed Numbers. In performing opera-
tions with mixed numbers, there are two possible methods
of procedure; either the whole numbers and the fractions
may each be handled separately or the mixed numbers may
be reduced to improper fractions and the operations per-
formed as though working directly with fractions. In
general, the first method is used when adding or subtracting
mixed numbers and the second method is used when multi-
plying or dividing them.

24. Adding Mixed Numbers. It is often necessary to
add mixed numbers. Although this operation is somewhat
long it is not at all difficult. To add mixed numbers we first
add the whole numbers and then the fractions, reducing the
final result to its lowest terms.

Example. What is the sum of the following dimensions:

61", 3A", 9ii", Qh" and 4^"?

28
6|" (Reduce fractions to L.C.D.) = — ^

^^" " 32

^i« 32

MIXED NUMBERS 27

4^"

3
32

+
23T

30H" Ans.

Reducing = 32)75(2i|"

Explanation. The first step is to write the numbers in
a vertical column for adding. It is then necessary to reduce
the fractions to a common denominator before we can add
them. The common denominator in this case is easily seen
to be 32. We must, then, multiply both the numerator and
denominator of each fraction by such a number as will make
the denominator of the new fraction 32. After this has been
done we add the numerators and place their sum over the
common denominator, finally reducing the resulting improper
fraction to a mixed number, which is 2f^". We next add the
whole numbers together to get 28 and to this we add the
number obtained by adding the fractions. This gives 30|^"
as the final result.

25. Subtracting Mixed Numbers. To subtract mixed
numbers we treat the whole numbers and the fractions
separately, just as we did in adding them. We subtract
the fractions and the whole numbers separately and reduce
the resulting expression to its lowest terms.

Example. How much more is Qfl" than 7f "?

911" (Reduce fractions to L.C.D.) =9if"

75'/ _710"

2^' Ans.

28 ARITHMETIC FOR CARPENTERS AND BUILDERS

i___

Fig. 14.— Turned
Porch Post.

Explanation. The first step, just as in
adding fractions, is to write the figures in
a vertical column. We must then reduce
the fractions to a common denominator
before we can subtract them. The L.C.D.
in this case is 16. We do not have to do
anything with the xt" but we do have to
change |" to sixteenths. Multiplying both
the numerator and the denominator of this
fraction by 2 gives ] g . We can now sub-
tract both the fractions and the whole
numbers to get the result, 2^", which tells
how many inches longer is a stick which
measures 9x1" than a stick which measures

•8 •

Sometimes in subtracting mixed num-
bers we find that the fraction in the sub-
trahend (the number to be taken away)
is larger than the fraction in the minuend
(the number from which we subtract). In
this case we may borrow one whole unit
or 1 from the whole number of the minu-
end and add it to the fraction of the
minuend after it has been reduced to the
same denominator. This makes an im-
proper fraction out of it and we can then
subtract the other fraction from it.

Example. How much wider is a board
which measures 8j" than a board which
measures 5|" ?

21'

Ans.

MIXED NUMBERS 29

Explanation. Here we cannot take |" from I" until we
change |" to eighths and add the 1 or |^ to it which we
have borrowed from the whole number 8. This makes the
minuend read 7-^-" and we can now proceed to subtract.
The result is 2f ", which tells how much wider the one board
is than the other.

If the minuend happens to be a whole number without
a fraction we can borrow 1 and change it into a fraction
as we did in the previous example.

Example. What is the difference between 5|" and 8"?

8" = 7f"

2r Ans.

26. Multiplication of Fractions and Mixed Numbers.

We have already learned that multiplication is a short
method of addition. In multiplying fractions and whole
numbers, fractions and fractions, fractions and mixed num-
bers, and mixed numbers together, certain precautions must
be taken which lead to speed and accuracy.

27. The Product of a Whole Number and a Fraction.
To multiply a whole number by a fraction we multiply the
number by the numerator and divide this product by the
denominator, or if it is possible without a remainder, simply
divide the denominator by the number.

Multiplying a whole number by a fraction is the same
thing as multiplying the fraction by the wholo number;
thus, 5 X| is the same thing as I X 5.

Example: What is 5 times f?

5Xf = -V=3f Ans.

Explanation. Multiplying the whole number 5 by the
numerator of the fraction 5 gives the product 25, and

30 ARITHMETIC FOR CARPENTERS AND BUILDERS

dividing this by the denominator 7 to reduce the improper
fraction to a mixed number gives 3f. Let us test this
problem by addition: -fH-f+f+4+4 = -¥- = 34, which is
the same result as found before.

Example. What will be the combined thickness of 4
boards each |" thick?

^x| = ^ = 3i Arts.

Explanation. This is a case in which the whole number
will be contained exactly into the denominator, without
a remainder. Performing this division, 4 goes into 8 two
times. Cross out the 4 and the 8 and write the quotient
of this division near the 8. Since 7 is left above the line
and 2 is left below the line, the result of the multiplication
is I, which, when simplified, equals 3^. It is suggested that
the student test this result by multiplying the whole number
by the numerator and dividing this product by the denomi-
nator. Which method is easier?

28. The Product of Two Fractions. To multiply one
fraction by another, we multiply the numerators together
for the new numerator and multiply the denominators
together for the new denominator.

Example. What is the product of f and f ?

jXg^^Ta A.ns.

Explanation. Multiplying the numerators, 3X3 equals
9, which is the numerator of the product. Multiplying the
denominators, 4X8 equals 32, which is the denominator of
the product. The result is, therefore, ^.

29. " Of " means " Times." The word of is often seen
in problems in fractions. For example, the statement some-

MIXED NUMBERS

times appears like this: What is | of 5"? In such a case
we work the problem by multiplying, so we say that of
means times. You can see that this is so by taking a piece
of wood 5" long and cutting it into four equal parts and then
taking three of these parts. These three parts will be f of 5",
and by actual measurement, they will be 3|" long. Thus
we know that f of 5" = 3f ". Let us now see what f times 5"
3t", which is the same value that we found

X5" = J^

Fig. 15. — Four-light Window Sash.

before. Therefore, we see that the word of when used thu3
signifies multiplication.

30. Mtiltipljring a Group of Whole Numbers and Frac-
tions. If we wish to multiply any number of whole numbers
and fractions together, we multiply the whole numbers and
the numerators together for the numerator of the product
and multiply the denominators together for the denominator
of the product. Reduce the result to its simplest form.

Example. What is the product of the following:

2X|X5Xt?

„ 1 , 3 2X1X5X3 30 15

Ans.

32 ARITHMETIC FOR CARPENTERS AND BUILDERS

Explanation: The product of the whole numbers an'i
the numerators is 30 and the product of the denominators
is 32. This gives fl, which, when reduced to its lowest
terms, equals yf.

31. Multiplying Mixed Numbers. To multiply a mixed
number by a fraction or to multiply two mixed numbers
together, reduce the mixed numbers to improper fractions
and proceed as in multiplying fractions, reducing the prod-
uct to its lowest terms. Never try to multiply mixed
numbers without first reducing them to improper fractions.

Example. Multiply 5| by 7^.

51=^

71= —

* 2 ^

15
2

43 15^645
S'^ 2 16

645
Reducing, -r^ = 16 ) 645(40 1^ Ans.

^^ 64

Explanation. First reducing the mixed numbers to im-
proper fractions, 5f = \^- and 7^ = J/. We then multiply
these just as we did when working with fractions. This
gives W^-, which may be reduced to the mixed number, 40]^.

32. Factors. The factors of a quantity are the numbers
which multiplied together will make that quantity. The
factors of 4 are 2 and 2 because 2X2 = 4. If we multiply
2X2X3 we get 12 and, therefore, the factors of 12 are 2
and 2 and 3. Notice that every factor of a quantity will

MIXED NUMBERS

-3'0^

^Ji-

exactly divide the quantity; that is, will divide it without
a remainder.

33. Cancellation. Now that we know what a factor is
we can work many prob-
lems in the multiplication
of fractions and whole num-
bers by the short method of
cancellation. You remem-
ber from a previous lesson
that we can divide both
the numerator and de-
nominator of a fraction by
the same number without
changing its value. Can-
cellation consists in separat-
ing the numerator and the
denominator of the fraction
into their factors and di-
viding by (canceling out)
their common factors. If
a number of fractions and
whole numbers or mixed
numbers are to be multi-
plied together, the method
of canceling can be used
to great advantage. To
show the advantage of the
method of cancellation we
will work the same example out both ways.

Example. Multiply the following fractions together:

35 34 12
48^21^17*

Fig. 16.— Door.

34 ARITHMETIC FOR CARPENTERS AND BUILDERS

35X34X12 14,280

48X21X17 17,136

7 =

2040

2448

6 =

340
408

4 =

85
102

17 = ^
6

Ans.

Explanation. Multiplying all of the numerators together

to get the new numerator and all of the denominators to get

the new denominator, as we have been previously directed,

14 280
gives us _' . This fraction must be reduced to its

lowest terms and we do this by dividing both the numerator
and the denominator by the same number. The fraction
is so large that we cannot see at first the greatest number
that will be contained exactly in both the numerator and
the denominator; we will try 7. We write 7 at the right
of the vertical line so that we can remember what the
divisor is and proceed. The result of the first division is
2040
2448'

We next try 6 as a divisor. The result of this divi-

340
sion is jT^. Then we divide by 4 and finally by 17 to get
40o

the result, |.
by trial.

We must determine each of these divisors

Example. Try the same example given above by can-
cellation.

1
5^1

^^^Z^;j 6

Ans.

Explanation. Divide 7 into 35 in the numerator and 21
in the denominator. 7 is a factor which is common to both

MIXED NUMBERS

35 and 21. Cancel the 35 and write 5 above it. Cancel the
21 and write 3 below it. 12 will divide into 12 in the numer-
ator and 48 in the denominator. It goes into 12 once and
into 48 four times. 17 will go into 34 in the numerator and
17 in the denominator. It goes into 34 twice and into 17
once. Each time a number is divided we cross it out and
write the quotient obtained near it. This is called canceling.

Fig. 17. — Forms for a Concrete Wall.

The numbers now left in the numerator are 5, 2 and 1 and
in the denominator 4, 3 and 1. 2 will divide into 2 and 4,
leaving 5, 1 and 1 in the numerator and 2, 3 and 1 in the
denominator. Now multiply these remaining factors to-
gether for the final result, which is |. It has taken quite
a while to tell about this, but in actual work it is much
easier and quicker than the method used in the previous
example.

36 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter III

18. To add mixed numbers, first add the whole numbers
and then the fractions, reducing the result to its simplest
form. (Sec. 24.)

19. To subtract mixed numbers, subtract the fractions
and the whole numbers separately and reduce the result
to its simplest form. (Sec. 25.)

20. To multiply a whole number by a fraction mul-
tiply the number by the numerator and divide the product
by the denominator, or if it is possible without a remainder,
divide the denominator by the number and reduce the result
to its simplest form. (Sec. 27.)

21. To multiply one fraction by another, multiply the
numerators together for the numerator of the product and
the denominators together for the denominator of the prod-
uct. Reduce the result to its simplest form. (Sec. 28.)

22. To multiply a group of whole numbers and fractions
together, multiply the numbers and the numerators together
for the numerator of the product and multiply the denomi-
nators together for the denominator of the product. Sim-
plify the result. (Sec. 30.)

23. To multiply mixed numbers together, first reduce
them to improper fractions and then proceed as in multi-
plying fractions. (Sec. 31.)

24. To multiply by cancellation, write the numerators
as a series of factors and also the denominators. Then
cancel out the factors which are common to both the numera-
tors and denominators. Multiply the remaining factors
in the denominator for the denominator of the product.
Simplify the result, if possible. (Sec. 33.)

MIXED NUMBERS 37

PROBLEMS

35. Add the following mixed numbers:

6r, 2A", 111", 5H".

36. Multiply:

(a) 2iX8XA.

(fc) fxfxuxi

37. If a carpenter receives 65 cents per hour how much
does he receive per day of 8 hours? How much does he get per
week of 5^ days?

38. If a bricklayer receives 72| cents per hour, how much
will he earn per year of 230 days of 8 hours each?

39. What is the cost of 27f sq.ft. of plate glass at 66f cents
per square foot?

40. If 20d spikes are worth 4^ cents per pound, how much
will 23 pounds cost?

41. A turned porch post is shown in Fig. 14. What is the total
length of the post? If a 10' post is desired, how long must the
distance marked A be if we do not change any other dimensions?

42. A special 4-light window sash is shown in Fig. 15. What
is the total height and width of the sash?

43. A 3'X7' door is shown in Fig. 16. What are the widths of
the rails and stiles indicated by letters?

44. Each rise in a certain stair is 7^" high and there are 15
risers. What is the total rise of the stair?

45. The sum of the rise and the tread should about equal 17
in. What should be the width of the tread in Prob. 44 and what
is the total run of the stair?

46. Fig. 17 shows an arrangement for -the forms for a concrete
wall. A contractor finds that he has 32 pieces of form material,
each board being 9|" wide actual measurement and 12' long.
How high can he build the forms for a 12' section of the wall?

47. A gang of men can mix and place 8^ cu. yds. of concrete
per hour. How much can they place per daj' of 8 hours?

48. The size of the Western brick is 8|"X4|"X2|". If the

125658

38 ARITHMETIC FOR CARPENTERS AND BUILDERS

mortar joint is A" thick and the brick are laid flat, how high will
a wall be which has 45 courses?

49. A case of shelves is shown in Fig. 18. What is the total
height and width of the case?

Fig. 18.— A Case of Shelves.

50. The actual face of a l"X3" flooring board is 2^'. A
builder has on hand 80 pieces l"X3" flooring 8 ft. long. Will
these cover a porch floor 8 ft. wide by 14 ft. long ? How many
pieces will be left over ?

CHAPTER IV

DIVISION OF FRACTIONS AND OF MIXED NUMBERS.
COMPOUND FRACTIONS

34. Division of Fractions. If we were required to di-
vide anything by two we would separate it into two equal
parts. One of these parts would be one-half of the whole.
Suppose that we wish to divide | by 2, we would write it
thus: (a) |-j-2 = f or, in other words, one-half of three-
quarters is three-eighths. But of means the same thing as
times as explained before. We could write this same opera-
tion thus: (6) f X| = |. Thus we see that the first expres-
sion and the second are equivalent. You will notice that
the whole number 2 may be written |, as previously ex-
plained, since 2 divided by 1 equals 2, Now let us write
these two expressions directly under one another and see
what conclusion we may draw.

(a) f-l = f
(&) fXHI

Do you see that the first looks like the second except
that the division sign in (a) is changed to the multiplication
sign in (6) and the second fraction in (6) is the second fraction
in (a) only it is turned upside down ?

We can, therefore, write the following rule for the division
of fractions: To divide one fraction by another, invert
(turn upside down) the divisor (the second fraction) and
proceed as in multiplication.

40 ARITHMETIC FOR CARPENTERS AND BUILDERS

Example. A man left f of his land to his children,
and ^ of this to his eldest son. What fraction of the land
did his eldest son get?

5 =

_ 3

X i — tV

Ans.

Explanation. The estate is first divided into three equal
parts. Each of these is then divided into five equal parts

Fig. 19.

to determine the son's share. This division of the land is
shown in Fig. 19. If we divide f by 5 this would be equiva-
lent to multiplying! by ^and the result of this multiplication
is ^, which is the fraction of the original property which
the son got.

Example. Divide H by ^.

64 • 32 04^ 3 6'

Ans.

DIVISION OF FRACTIONS 41

Explanation. In this example we follow the rule exactly.
Notice also that we can apply the method of cancellation.
Do not fail to take advantage of any method by which you
can shorten your work, ei^ sfe is the same as HX V-

35. Compound Fractions. Division is indicated by a

fraction. If we divide one fraction by another, we may

express the operation as a compound fraction. A compound

fraction has a fraction for a numerator and a fraction for

a denominator.

Example. Simplify ^.

il_llx?-2-ll-l-^ Ans

Explan£ition. This is the same example as the one just

preceding but the division is first expressed as a compound

11
^Taction. The expression ~ means that ^ is to be divided

by ^.

36. Dividing a Whole Number by a Fraction. To divide
a whole number by a fraction we write the whole number
over the denominator 1, which does not change its value,
and proceed as in dividing fractions.

Example. Divide 3 by 4.

3 = f.

r"^7"~l^^~2~ =* ^'

37. Dividing Mixed Nmnbers. To divide mixed num-
bers, we first reduce them to improper fractions and then
proceed as in dividing fractions.

42 ARITHMETIC FOR CARPENTERS AND BUILDERS
Example. Divide 2^ by 61.

5^25_^ 4_2 .

2 • 4~^^52^~5 '^^'

38. Solution of Problems. Problems involving addition,
subtraction, multiplication and division or combinations of
these operations may be easily solved provided the con-
ditions are properly analyzed. A problem consists of three
parts: the statement, the operation and the conclusion or
answer. The statement of the problem contains the con-
ditions and these must be carefully thought out before an
attempt is made to perform the operations. Be sure that
you understand fully what is required by the problems;
then you can proceed intelligently with the necessary
mathematical operations. The conclusion or answer should
be expressed in its simplest form.

Example. If 6 men can do the framing for a certain
house in 4 days, how long will it take 4 men to do it?

6X4 = 24 days for 1 man to do the work.
lof24= 6 days for 4 men Ans.

Explanation. If 6 men do the work in 4 days, it would
take 1 man 6X4 = 24 days at the same rate of speed. If
4 men are working together, they can do the work in I of
this time, or j of 24 equals 6 days.

39. Classes of Problems. Most problems involving
fractions fall into one of three classes.

DIVISION OF FRACTIONS 43

Case 1. Given the whole to find a part: the whole
multiplied by the fraction equals the part.

Case 2. Given a part to find the whole : the part divided
by the fraction equals the whole.

Case 3. To find what part one number is of another:
the part divided by the whole equals the fraction.

Example of Case 1. A contractor purchases 15,000
lineal feet of 2X4 scantling. Three-fifths of this order are
pieces 18 ft. long and the remainder are 12 ft. long. How
many pieces of each length did he receive?

|(the fraction) X 15,000 (the whole) =9000 ft. (the part).
9000 ^ 18 = 500 pieces 18 ft. long Ans.

15,000-9000 = 6000 ft.

6000-1-12 = 500 pieces 12 ft. long Ans.

Explanation. If the whole amount is 15,000 lineal feet
and the part is | of the whole, we must find | of 15,000 or
multiply 15,000 by | to get 9000 ft., the required part. If
this 9000 ft. is divided into pieces each 18 ft. long, we must
divide by 18 to find the number of pieces. To find the part
which is in 12 ft. lengths, we may subtract the part already
found from the whole. 9000 from 15,000 gives 6000 ft.
in 12 ft. lengths. Divide 6000 by 12 to get 500 pieces.

Example of Case 2. A contractor orders a certain num-
ber of brick, two-sevenths of which are to be face brick.
He receives 16,000 face brick. How many brick does he
receive in the entire order? How many common brick
does he receive?

16,000 (the part)^f (the fraction) = 16,000X1 = 56,000

(the whole).
56,000 brick- 16,000 face brick = 40,000 common brick.
56,000 total; 40,000 common brick; 16,000 face brick. Ans.

44 ARITHMETIC FOR CARPENTERS AND BUILDERS

Explanation. If 16,000 is ^ of the whole, \ will be 5 of
16,000 or 8000, and ^ will be 7 times 8000 or 56,000, which
is the total number of bricks ordered. This is equivalent
to dividing 16,000 by f . The difference between the whole
number of bricks and the number of face bricks will give the
number of common bricks.

Example of Case 3. A lumber inspector finds that a
board 18 ft. long shows sap for 18 in. of its length and wane
for 24 in. What fractional part shows sap and what part
wane?

18 ft.Xl2 ins. = 216 ins., the length of the board in inches.

[ Ans.

^,„ ,., — —. — r^ =-7r (the fraction) shows wane. J
216 (the whole) 9

Explanation. In order that we may not have to deal
with mixed numbers it is convenient to reduce the length
of the board in feet to inches. Then the part divided by
the whole will equal the fraction. 18 divided by 216
(expressed in the form of a fraction and reduced to its lowest
terms) gives ^\, the fractional part which shows sap. 24
divided by 216 gives ^, the fractional part which shows wane.

DIVISION OF FRACTIONS 45

Summary of Chapter IV

25. To divide one fraction by another, invert the divisor
and proceed as in multiplication. (Sec. 34 and 35.)

26. To divide a whole number by a fraction, supply the
denominator 1 to the whole number and divide as with
fractions. (Sec. 36.)

27. To divide mixed nimibers, first reduce them to
improper fractions and proceed as in dividing fractions.
(Sec.' 37.)

28. To find the part when the whole is given, multiply
the whole by the fraction. (Sec. 39.)

29. To find the whole when a part is given, divide the
part by the fraction. (Sec. 39.)

30. To find what part one number is of another, divide
the part by the whole to find the fraction. (Sec. 39.)

PROBLEMS

51. Divide U by 3f.

2K10.

52. A mortise |" wide is to go in the exact center of a piece
of If" lumber. How much wood will be left on each side of it?

53. The distance from the first floor of a dwelling house to
the second floor is 9' 10". If 16 risers are required in the stair,
what will be the rise of each step?

54. A carpenter has a roof surface 22 ft. long and 15 ft. wide
to cover with sheathing which measures 5^" wide. The boards
are spaced 2" apart. How many lineal (running) feet of sheath-
ing will he need?

55. How many times may a board 12" wide be ripped if the
strips are to be 1" wide and the saw cut |" wide? How wide a
strip will be left?

56. If shingles are laid 4|" to the weather, how many courses
will be required to cover a roof slope 15' 9" long?

46 ARITHMETIC FOR CARPENTERS AND BUILDERS

57. A certain job requires 36 hours to finish it. If a man works
at it 1 ^ hours a day, how many days will it take to complete it?

58. If I of a thousand feet of lumber costs $3.00, how much
will 1000 feet cost?

59. A pile of lumber contains 12,000 feet. A contractor pur-
chased § of it. How many feet did he buy?

60. A man builds three houses which cost him a total of $12,000.
The first cost 5 of the whole, the second cost § of the whole and
the third cost the remainder. What is the cost of each house?

61. A builder buys 240 boards, J of which are graded as No. 1
Clear and the remainder are No. 2. How many No. 1 boards
and how many No. 2 boards did he receive?

62. A man bought a pile of lumber estimating it to contain
2000 ft. On measuring he found that it contained 1960 ft. What
fractional part did he lose?

63. Three-eighths of a thousand feet of lumber costs $ii4.
What is the price of the lumber per thousand?

64. A board weighs 42 pounds green: in drying the weight
decreases \. What does it weigh when dried?

65. In cutting material to lay a certain floor the waste amounts
to 50 ft. in every thousand. What fractional part is wasted?

66. A man puts in an eight-hour day. One-third of the day
he is employed in laying floor, one-eighth in setting bridge braces
and the remainder in finishing. How many hours and minutes
does he spend at each kind of work?

67. A fence post is 9 ft. long and ys of its length is below
ground. How many feet are in sight?

68. A man does one-third of a piece of work in 51 hours. How
long will it take him to complete the entire job?

CHAPTER V

MONEY. ADDITION, SUBTRACTION, MULTIPLICATION
AND DIVISION OF MONEY QUANTITIES. CHECKS.
STATEMENTS. RECEIPTS. ESTIMATES

40. Money. The money unit in the United States is
the dollar. The dollar is divided into cents each of which
has a value of one hundredth of the dollar. Fifty cents are
-jW or one-half of one dollar. Twenty-five cents are ^W
or a quarter of a dollar. The dime is one-tenth of a dollar
and the five-cent piece or nickel is one-twentieth of a dollar.
To facilitate business transactions, currency is issued in
gold, silver, nickel, bronze and paper money of various
denominations.

The builder should learn to be very accurate in his com-
putations with money and he should be able to keep his
accounts in a neat and orderly manner. If he does this,
he will save himself much trouble and worry. It is not
necessary to know how to keep a set of business books in
order to manage your accounts. Much of the business of
to-day is done without the use of complicated systems of
bookkeeping. A satisfactory record may be made by filing
visible evidences of money transactions such as checks,
receipts, statements, inventories, etc.

41. Writing and Reading Money Sums. When express-
ing money in numerals, the dollar sign ($) is placed before
the number of dollars, a period is placed after this and the
number of cents follows: thus, eight dollars and seventy-

48 ARITHMETIC FOR CARPENTERS AND BUILDERS

five cents is written $8.75. If cents only are to be written,
no figures are written to the left of the period: thus, the
sum of eight cents is written $.08. It is also sometimes
written 8^.

42. The Decimal Point. This system gives us our first
introduction to decimal fractions. The period mentioned
above is the decimal point. The first figure to the right of
the decimal point indicates the number of dimes or tenths
of a dollar in the sum. The second figure to the right of the
decimal point indicates the number of cents or hundredths
of a dollar in the sum. This method of writing money
quantities greatly simplifies the mathematical operations
because we may handle parts or fractions of a dollar in the
same manner that we handle whole numbers.

43. Addition. Money quantities are added in the same
manner that we add whole numbers. In writing down money
quantities for adding, it is necessary to see that all of the
periods are directly under one another in a vertical colunm.
If this is done, the cents, dimes and the dollars will be directly
under one another. The number of tens of cents may be
carried forward and added to the dime colunm. The number
of hundreds of cents will represent dollars and may be added
to the dollar column.

Example. Add $3.25, $5.56, $1.00 and $.85.

$3.25

5.56

1.00

.85

$10.66 Ans.

Explanation. Add the column to the right or the cents
column first, 5+0+6+5 = 16 cents. This is 6 cents and 1

MONEY 49

dime. Put down the 6 and carry the dime to the next
column. 1+8+0+5+2 = 16 dimes. Put down the 6 and
carry the 1 to the dollar column. 1 + 1+5+3 = 10 dollars,
which we put down to complete the sum.

44. Subtraction. Subtraction of money is performed
as in the subtraction of whole numbers, the only difference
being the use of the decimal point separating dollars and
cents.

Example. Subtract $6.28 from $9.10.

$9.10
6.28

$2.82 Ans.

Explanation. We set the numbers down with the decimal
points under each other just as in addition. We then sub-
tract just as if we were working with whole numbers. In
the remainder we place the point directly under the other
points to separate dollars and cents.

45. Multiplication. Multiplication of quantities repre-
senting money is performed in the same manner as for whole
numbers except for the location of the decimal point separat-
ing dollars and cents.

Example. If one shipping crate is worth $2.50, how
much are 12 crates worth?

$2.50
12

500
2500

$30.00 Ans.

Explanation. The multiplication is performed exactly
as though we were multiplying whole numbers and the

60 ARITHMETIC FOR CARPENTERS AND BUILDERS

decimal point is placed in the product by counting two places
from the right to mark off the cents.

46. Division. The operation of division of money quan-
tities is the same as for whole numbers. The decimal point
is placed in the quotient in the proper position to separate
dollars and cents.

Example. A contracting firm bought 5 wheelbarrows
for $19.25. What was the price of one?

5) $19.25 ($3.85 Ans.
15

~42
40

25
25

Explanation. We divide exactly as if we were working
with whole numbers and place the decimal point before the
last two figures to the right because they represent cents.
Another way to locate the point is to place it in the quotient
as soon as the number of dollars in the dividend has been
divided. For example, in dividing the $19.25 above, we
divide 5 into $19 first and get $3. We have then divided
all the figures up to the decimal point and we therefore put
a point in the quotient after the 3 before we divide again.

47. Bank Accounts. When a man opens a checking
account at a bank, he makes a deposit sufficient to fully cover
all of his expected disbursements for a reasonable period.
He is also expected to satisfy the bank that he is conducting
a legitimate business. When his application is accepted
and his first deposit made, he is given a pass book in which
his account with the bank is kept. This pass book should

MONEY

be presented at frequent intervals for a balance. A page
from a contractor's pass book is shown in Fig. 20. On the
left-hand side of the page the abbreviation Dr. indicates
that the bank is debtor to the depositor on account of cash
deposits. On the right-hand side the abbreviation Cr.

/Dr. MECHANICS BANK

with d.lS.ef<yttlaii)

CrN

\<\\4-

\qi4- (1

OjWVUL

feoWa^

144

5>9

)

^Uk

2Gl

48^

(Wb

150

BoJi/utC^

^07

4^1

(\uw

feoftiuwue.

Zoy

iQWfea-

iqi

6<]

0 1

'Bofiojuai^

1^2-

^07

l^oJ^ouae^

\(o

^2.

S^juJu—

l4o

^oAj

/4o

^ ()

»»

1^0

]

Fig. 20. — ^Page from a Contractor's Pass Book.

means that the bank credits its account with the depositor
because of checks drawn against his account. The balance
indicates the difference between the amount of cash de-
posited and the sum of the checks drawn against the account.
48. The Bank Check. Fig. 21 shows a bank check
with the stub attached, properly drawn and signed. This
check is an order on the bank to pay the owner of the check

52 ARITHMETIC FOR CARPENTERS AND BUILDERS

the sum specified and to charge the transaction to the
signer of the check. The payee must endorse or sign the
check on the back when he presents it to the bank for pay-
ment. The check may also be transferred to another party
by endorsement. It is the duty of the payee to present
the check at the bank for payment as promptly as possible.
The stub indicates the number of the check for purposes
of identification. It also indicates to whom the check was
drawn, the amount and for what purpose it was paid. The

No. ZqG

To .

For ly^iUa^

i^:

BAL.

3Z6j

DEP.

TOTAL

CHECK

BAL.

j3a,o

.1.0

PAY TO THE
ORDER OF

DES MOINES, IOWA^_^Jtt._5_191^

MECHANICS BANK

^h^(xnd>^^

No. /goG

foo-

DOLLARS

^. \£2. ^9Tdffni)

Fig. 21.— Bank Check with Attached Stub.

person who draws the check should also determine the
remaining balance to his credit after signing each check
so that there will be no danger that he will unknowingly
overdraw his account at the bank.

49. The Statement. When a service is performed under
verbal or written contract it is customary for the contractor
to render an itemized statement on the first of the month
immediately following the date on which the work was com-
pleted. Such an itemized statement is shown in Fig. 22.
This statement serves as a reminder to the person for whom
the work was done that he owes the contractor the sum
mentioned. Unless there is specific agreement to the con-

MONEY

DES MOINES, IOWA,

Mr.

Terms

AiU^. k30 1Q1 ^

^cMj maJ <^o d/SMO

2o&u&al4id/y^3

15-

Z-2^S-/W

S'S

3-Zx/o-Zo

5(?

l-lh, /hoAj^

0£_

f)^. -^aJlhrr Q^

d'o

Fig. 22. — Itemized Statement.

54 ARITHMETIC FOR CARPENTERS AND BUILDERS

trary, the contractor has a right to expect that the account
will be settled promptly.

50. The Receipt. When payment has been made for a
service performed it is proper to offer a receipt. This should
be done even though a receipt is not requested. Fig. 23
shows a common form of receipt for money paid. If the
payment is made by check, the endorsement on the back
of the check is considered to be sufficient evidence that the
money has been paid. It is also often convenient to write

No sy

From S^ G.

'h(yzU4-.

For

cSc/rf^!ige^

No. tSy DES MOINES, IOWA,

RECEIVED OF <S^ A ^oJdi^

.DOLLARS

FOR 'PouJtiiia M£/ri^>r^ QjjAy

hCUUHtuf ^J^jJajj/t- iuj )M^^

%4oM. \1MA

Fig. 23. — Form for a Receipt with Attached Stub.

the receipt in the form of a notation on the statement or
bill. Fig. 22 shows an example of such a notation.

51. Assets and Liabilities. There is nothing more im-
portant in the successful management of the business of
the contractor and builder than a frequent and careful
estimate of his assets and liabilities. A man's assets are his
credits and his liabilities are his debts. By this means the
contractor is able to tell whether he is making money or
not. In making a statement of the condition of his business,
the contractor should be careful not to include in his assets
any items which are really not credits. Unless this point

MONEY

is carefully watched, a misleading statement may easily
be made.

52. Wage Accounts. Every mechanic should know how
to keep time and make out a pay roll. Wage accounts
should be made out with care and should be properly

TIME Week Ending

5^. /3 19/3

DATE

NAMES

,„„L-L

Wed

rhu

Fri.

Sat.

Total
Time

Hate
per

nat

Amount |

null

cts.

^/uUi . ^. -Tu,

,Y°

rio'^vjA-o^c . i&Znt~LJL^

flso

&/,J>J-'^^./0

S'.eo-i^i!^. ^-^ju-y

"/^

,550

"^fM^d^. ^JlM>uJ<-y

"^k

fyS^

QjuO-^^f

'7fi^^v-a'<y>-y duuAjhdy

SZi

T^JuUuAC. y<^

QZi

£4Er^.H7-^'

,325

/5&J-0-T-E/J Ko<aX4—

5i5

'M^-ncL.. 4ifnJU

^X^

•57t/Cia//cA<o'. <^- S",

Z'^

''/tiAjiy^t^^vu , J^CUuJL^>-^

^d;L^, /^.

s-k.

Zso

S^L^nA^kX^^ . V^<v»^

/so

jfd^Mii^ %.^.

hUoFro-

Fig. 24. — Page of a Time Book.

arranged and preserved. The best way to do this is to use
a time book. Such a book may be purchased in a form
similar to that shown in Fig. 24. In making up pay rolls
it is first necessary to find the total amount of time that a
workman puts in during a given period. This should be
multiplied by his rate to find the amount due him. The

56 ARITHMETIC FOR CARPENTERS AND BUILDERS

WvK'.'wSllU VsbuAMjJtSL^ 11

^G ^-K;^<JjZ^ cWe^^ C

JUuj:^i?AAxl>uuu;^

-^g/nJomyr^ c^^lS^^

OoJy^ qwAj

~^feii-^4^

1 (Luyg.1^ c£ediL/t- ^1.^4qQQ

Fig. 25. — Contractor's Estimate Sheet.

MONEY 67

rate is usually given either by the hour or by the day.
Adding the amounts due each workman will give the total
amount of the pay roll.

53. Building Estimates. The contractor is often called
upon to prepare estimates and submit proposals for work
about to be done. The utmost care should be taken to make
these estimates in a systematic manner and to preserve them.
This material is not only valuable for the present purpose,
but it has a 'value for future reference. P'ig. 25 shows a
contractor's typical estimate sheet.

58 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter V

31. To add or subtract money quantities, treat them as
whole numbers, placing a period in the result to separate
dollars and cents. (Sec. 43 and 44.)

32. To multiply or divide money quantities, proceed as
with whole numbers, separating the dollars and cents in
the result with a period. (Sec. 45 and 46.)

PROBLEMS

Note. When working with dollars and cents always find the answer
to the nearest cent; that is, if it is more than half a cent over call it
a whole cent and if it is less than half a cent, do not count it.

69. Add the following sums of money: $1.75, $2.38, $9.75, $22,
$2.08, $25.00, $11.66.

70. A carpenter agrees to build a wardrobe for $8.50. The
lumber costs $4.00 and the hardware $1.30. If he spends 16 houra
making it, how much does he earn per hour?

71. A contractor agrees to build a barn for $250. He uses
6000 ft. of lumber at $20 per thousand, 12 squares of roofing at
$1.75 per square, and hardware at a cost of $15. Four men put
in 60 hours each at 30^ per hour. What is the profit of the con-
tractor?

72. A man sets and cases 4 door frames in a day. If his rate
is $3.25 per day, how much will it cost for him to do this work in
a house wliich has 31 doors?

73. At 3|fi per pound, how many nails can be bought for $6?

74. If lOd common wire nails are $2.50 per hundred pounds,
how much will 35 pounds cost?

75. W^hat will be the price of 28 lineal feet of 14 in. valley tin
at $2.75 per 50 feet?

76. The price of one 26X24 check-rail window, glazed double-
strength glass, is quoted at $1.84. How^much will 19 such windows
cost?

77. A set of front porch steps 6 ft. long with 4 risers and treads
will cost $7.50. How much will steps 7 ft. long cost figuring at
the same rate?

MONEY 59

78. A porch 7 ft. wide by 16 ft. long is estimated to cost $95.
What is the cost per foot of length? What would be the cost of
a porch 18 ft. long at the same rate?

79. Determine the amount of money due each man and the
total amount of the pay roll shown in Fig. 24.

80. The pass book shown in Fig. 20 gives an open account.
Indicate a balance and close the account.

81. The following material and labor was used in making a
medicine case 16"X23" inside measurement: 6 ft. 1X4 jambs,
10^; 2' 4" of stool 1|X6, 10^; 1 sheet of tin for back, 2o/; 4 ft.
casing, 12^; 2 ft. apron, 5^; 2' 2" head casing, 7^; 3 ft. head
mold, 6^; 1 door, 80^; 1 beveled plate glass mirror, $1.10; hinges
and catch, 25^; labor, 4 hours at 40j^ per hour. What was the cost
of the case?

82. Frank Allan bought the following hardware from the J. C.
Kelly Hardware Co., on April 15, 1914: 1 Sliding door set for double
doors, $2.25; 5 Mortise latch sets at 55^ each; 2 Push plates at
30)^ each; 1 Cyhnder front door set, $7.25; 8 Inside lock sets at
$1.65 each; 28 Flush sash lifts at 3^; 16 Ball-tipped loose-pin
butts 4X4 at 22^ each; 1 set Parlor door hangers for double doors,
$4.00; 1 Double-acting floor hinge, $1.10. Make out a statement
for the above material and find the total amount of the bill.

83. What is the total of the estimate shown in Fig. 25?

84. If one concrete finisher at $5.00 per day and five laborers
at $2.50 per day are required to lay 700 sq.ft. of cement sidewalk,
what will be the labor cost for 100 sq.ft. of walk?

85. To lay 100 sq.ft. of cement sidewalk, the following material
is required: 2| barrels cement at $1.50 per barrel; Ij cu.yds. of
stone at $1.50 per cu.yd.; f cu.yd. of sand at $1.25 per cubic yard;
2| cu.yds. of cinders at 50^ per cubic yard. What is the total cost
of the material? What is the cost of 100 sq.ft. of sidewalk including
the labor cost as found in the previous problem? What is the cost
per running yard for a walk 4 ft. wide?

CHAPTER VI

DECIMALS. ADDITION, SUBTRACTION, MULTIPLICA-
TION AND DIVISION OF DECIMALS. CHANGING
FROM COMMON FRACTIONS TO DECIMAL FRAC-
TIONS

64. Decimal Fractions. Decimal fractions are different
from common fractions in that they have one of the numbers
10, 100, 1000, etc., for a denominator. Thus j\, j\\, ^^|^,
etc., are really decimal fractions expressed as common frac-
tions. The advantage of using decimal fractions is that
computations are often much simplified.

55. Writing Decimals. Decimals are written in the
same way that we wrote money quantities. In fact, money
quantities are only a special kind of decimals in which we
have only to deal with tenths and hundredths of the dollar:
thus, $1.75 means one and seventy-five hundredths dollars.
In the same manner we may write 1.75 lbs., which means
one and seventy-five hundredths pounds.

In writing decimals we omit the denominator, but we
indicate what the denominator is by means of the decimal
point. The figure which is written after the decimal point
is put down is the numerator of the fraction. For each
figure written to the right of the decimal point we under-
stand that the denominator has 1 and a 0 (cipher): thus,
.5 means that the numerator is 5 and the denominator is 1
and one 0 (cipher) or 10. This decimal expressed as a com-
mon fraction is, therefore, ■^%. Taking another example,
.19 means that the numerator is 19 and the denominator

DECIMALS 61

is 1 and two ciphers or 100. This fraction, therefore equals
^y^. It is worth while to state this as a rule: The denomi-
nator of a decimal fraction has the unit 1 and as many-
ciphers following as there are figures to the right of the
decimal point.

56. Reading Decimals. The whole number 5280 is read
" five thousand two hundred eighty." The word " and "
is not used in reading a whole number. A decimal is read
like a whole number except that the name of the right-
hand number is added: for example, .528 is read " five
hundred twenty-eight thousandths." You will notice that
the denominator of this decimal is 1000, In reading a
whole number and a decimal the word " and " is placed
between the two: thus, 45.6235 is read "forty-five and
six thousand two hundred thirty-five ten thousandths."
When one person reads numbers containing decimals for
another to write, it is customary to say " point " to indicate
where the decimal begins; for example, 468.59 would be read
" four sixty-eight, point, fifty-nine."

57. Significance of Position. If we multiply the unit
1 by 10, we increase its value. If we multiply it by 100,
we increase its value still more. If we divide 1 by 10, we
decrease its value and if we divide by 100 we decrease its
value still more. We express these changes in value by the
position which the figure occupies in the expression: thus,
10 has two places of figures; 100 has three places of figures;
.1 has one place to the right of the decimal point. The
values and places of numbers ordinarily met with are shown
in the column below.

1,000,000 Millions

100,000 Hundred thousands

10,000 Ten thousands

62 ARITHMETIC FOR CARPENTERS AND BUILDERS

,000

One thousands

100

Hundreds

Tens

Units

Tenths

.01

Hundredths

.001

Thousandths

.0001

Ten thousandths

.00001

Hundred thousandths

.000001

MilUonths

For every place that the unit is moved to the left its
value is increased ten times. For every place the unit is
moved to the right its value is decreased ten times. It is
to be noticed that the location of the decimal point deter-
mines the value of the expression. For example, 345 has
a value ten times as great as 34.5 because the number of
places to the left of the decimal point has been decreased
by one. Thus we see that the location of the decimal
point is a very important matter.

58. Addition of Decimals. To add numbers containing
decimal fractions, it is first necessary to write them so that
the decimal points all come directly under one another in
a straight vertical line. We then proceed to add as with
whole numbers and we place the decimal point in the answer
directly under the other decimal points.

Example. Add 692.25, 427.201, .026, 1.5, .3415.

692.25
427.201

.026
1.5

.3415

1121.3185 Ans.

DECIMALS 63

Explanation. We must first write the numbers in a
vertical column in such a manner that all the decimal
points are in the same vertical line. Where the column
is not filled out to the right, we are to add as though ciphers
were in these places. We place the decimal point in the
answer directly under the other decimal points.

59. Subtraction of Decimals. To subtract decimals we
first write them so that the decimal points will be directly
under each other and then proceed as in subtracting whole
numbers. We can " borrow," if necessary, exactly as
though working with whole numbers.

Example, Subtract 4.3975 from 10.2.

10.2000 minuend
4 . 3975 subtrahend

5 . 8025 difference Ans.

Explanation. There are four decimal places in the
subtrahend. We must, therefore, also have four places
in the minuend and we write three ciphers after the 2.
We can then borrow and subtract as when dealing with
whole numbers. We must not forget to place the deci-
mal point in the result directly under the other decimal
points. To check the result we have only to add the
subtrahend and difference together and if this sum exactly
equals the minuend, we know the result to be correct.

60. Multiplication of Decimals. To multiply decimals,
write the numbers so that the figures at the extreme right
in both the multiplicand and multiplier are directly under
each other. We then proceed as in multiplying whole num-
bers, forgetting about the decimal point for the moment.
When the product has been obtained we point off as many

64 ARITHMETIC FOR CARPENTERS AND BUILDERS

places from the right as there are decimal places in both the
multiplicand and multiplier.

Example. Multiply 62.53 by .124.

62 . 53 multiplicand
. 124 multiplier

25012
1 2506
6 253

7.75372 Ans.

Explanation. We write the two numbers so that the two
figures at the extreme right of both the multiplicand and
multiplier are in the vertical Une. Multiplying as with whole
numbers we get the result as shown. To point off the
decimal places in the product, we add the number of places
in the multiplicand, 2, to the number of places in the multi-
plier, 3, to get 5. In the answer, therefore, we will point
off 5 places beginning at the right and place the decimal
point in front of the fifth figure.

61. Division of Decimals. To divide decimal numbers
we proceed as with whole numbers, forgetting about the
decimal point for the moment. If there are not sufficient
figures in the dividend so that the division may be made,
we simply add ciphers to the right of the decimal point.
This does not change the value of the number. We may
carry the operation as far as we please, adding ciphers if
necessary. To locate the decimal point in the quotient we
point off as many places from the right as there are more
decimal places in the dividend than in the divisor.

Example. Divide 528.01 by 9.5.

DECIMALS 65

Divisor Dividend Quotient
9.5) 528.01 (55.5 Ans.

475

530
475

551
475

Explanation. This example will give no difficulty until
we come to locate the decimal point. There are 2 places in
the dividend and 1 place in the divisor. Subtracting the
number of places in the divisor from the number in the
dividend, 2 minus 1, gives 1 place to point off from the right
in the quotient. In the last division there is a remainder,
but since this is less than half of the divisor, we neglect it.

Example. Divide 2 by 1.43.

1.43) 2.0000 (1.39 Ans.

1 43

570
429

1410

1287

Explanation. We must add ciphers to the dividend before
we can divide. We place a decimal point after the 2 and
add the ciphers to the right. We add as many ciphers as
are necessary to get the required number of places in the
quotient. If we desired to carry the division farther, we
would add more ciphers. To locate the decimal point in
the quotient, we subtract the number of places in the divisor
from the number in the dividend, 4 — 2 = 2, and point off
two places in the answer.

66 ARITHMETIC FOR CARPENTERS AND BUILDERS

62. Accuracy of Results. When we were working with
common fractions, we could always express a remainder in
division by means of a fraction. You will notice that in
the above examples in division we might continue to add
ciphers to the dividend and carry the division out as far as
we please. There will always be a remainder, but the
further we carry the operation the more accurate will
be the result. For all practical problems we must put a
limit to this or the labor involved will be out of all pro-
portion to the accuracy obtained. For all practical prob-
lems four places to the right of the decimal point are all
that are necessary, and for most cases two or three places
are enough. When more than four, or three, or two, as the
case may be, occur, we may neglect all of the figures to the
right of the fourth, or third, or second, providing the next
figure is less than five. If the next figure is more than five,
we add one to the last figure that we count and neglect all
the others. If the first figure that we neglect happens to be
five, we either do or do not add one as our judgment dic-
tates. For example, the number 52.345278 may be written to
the fourth decimal place as 52.3453— . The — sign following
the number means that the real value is something less than
the value given. Taking another example, the decimal
0.02534 may be written to the fourth decimal place as
0.0253 +. The + sign means that the real value is some-
thing more than that given. It is not always necessary to
write the + and — signs as has been done here, but it is
frequently done if it makes the work clearer.

63. Changing a Decimal Fraction to a Common Fraction.
To change a decimal fraction to a common fraction it is only
necessary to supply the denominator and to reduce the
resulting fraction to its lowest terms.

Example. Reduce .625" to a common fraction.

•^^^ 1000

DECIMALS 67

K^125
200

^~40

5 = 1" Ans.

Explanation. We first supply the denominator, which
will be 1, with as many ciphers following as there are decimal
places to the right of the decimal point, or 1000. Reducing
this fraction to its lowest terms by dividing by 5 three times,
gives the result f .

64. Changing a Common Fraction to a Decimal Fraction.
To change a common fraction to a decimal fraction, place
a decimal point after the numerator and add as many ciphers
after it as are needed, usually four or less, depending upon
the number of decimal places desired in the quotient. Then
divide the numerator by the denominator and point off as
many places in the quotient as there are ciphers added in
the dividend.

It is not possible to reduce every common fraction to
an exactly equivalent decimal fraction. Fractions having
only the prime numbers 2 and 5 in the denominator may be
reduced to an exactly equivalent decimal. However, we
may reduce any common fraction with sufficient exactness
for all practical purposes.

Example. Change y&" to a decimal fraction.

10) 7.0000 ( .4375 Ans.

60
48

120
112

80
80

68 ARITHMETIC FOR CARPENTERS AND BUILDERS
Table of Decimals of an Inch fob Each i/^th

Ads.

Atha.

Decimal.

Fraction.

Ad8.

Aths.

Decimal.

Fraction.

.015625

.515625

.03125

.53125

.046875

.546875

.0625

.5625

.078125

.578125

.09375

. 59375

. 109375

.609375

.125

.625

. 140625

.640625

.15625

.65625

. 171875

.671875

.1875

.6875

.203125

.703125

.21875

.71875

.234375

.734375

.25

.75

.265625

.765625

.28125

.78125

.296875

.796875

.3125

.328125

.828125

.34375

.84375

.359375

.859375

.390625

.890625

.40625

.90625

.421875

.921875

.4375

.453125

.953125

.46875

.96875

.484375

..984375

DECIMALS 69

Explanation. Since the denominator contains only the
prime factor 2 it may be reduced to an exactly equivalent
decimal. It is found necessary to add four ciphers to the
dividend and, therefore, there will be four decimal places
in the answer.

65. Decimal Equivalents of the Fractions of an Inch.
It is often necessary to find the decimal equivalent of a given
fraction of an inch. It is convenient to have a table giving
these values. Such a table is shown on page 68, but all of
the values are not filled in, for the reason that the student
is expected to fill in the missing ones as part of the problem
assignment of this lesson.

70 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter VI

33. To add decimals, write the numbers so that the
decimal points are in a vertical line, add as with whole
numbers and place the decimal point in the result directly
under the other decimal points. (Sec. 58.)

34. To subtract decimals, write the minuend and sub-
trahend so that the decimal points are in a vertical line,
subtract as with whole numbers and place the decimal point
in the result directly under the other decimal points. (Sec.
59.)

35. To multiply decimals, proceed as though multiply-
ing whole numbers and point off as many places in the
product as there are places in both the multiplicand and
multiplier. (Sec. 60.)

36. To divide decimals, proceed as though dividing
whole numbers, adding ciphers to the right of the decimal
point in the dividend if necessary. Point off as many places
in the quotient as the difference of the number of places in
the dividend less the number of places in the divisor. (Sec.
61.)

37. To change a decimal fraction to a common fraction,
supply the denominator 1, followed by as many ciphers as
there are figures to the right of the decimal point, and reduce
the resulting common fraction to its lowest terms. (Sec. 63.)

38. To change a common fraction to a decimal fraction,
place a decimal point after the numerator and add as many
ciphers after it as are needed. Then divide the numerator
by the denominator and point off as many places in the
quotient as there are ciphers added in the dividend. (Sec.
64.)

DECIMALS 1\

PROBLEMS

86. Write the following decimals in figures: Sixty-five thou-
sandths, two hundred twenty-five thousandths, fifty-two and two
hundredths, five hundred ninety-six and two-tenths, one half of
one-thousandth, seven hundred sixty-five and five one-thou-
sandths.

87. Read the following decimals and write them out in words:
622.215, .075, .2865, 1.024, 100.005.

88. Add up the following decimals an^ check the result by
adding down:

(a) 532.1

(b)

5.22

52.75

.496

.025

42.1325

987.

27.25

.43

39.5

.025

llowing :

5.02

89.

Subtract the fo

(a)

.021 from 1

(6)

2.2

from 10.75

(c)

92.25 from 201.2

90.

Multiply :

(a)

2.2

by 31.75

(&)

5.25

by 38.2

91. Divide 4.2 by 3.25 and express the result to three decimal
places.

92. Divide 9.035 by 13 and express the result to three decimal
places.

93. Reduce the following common fractions of an inch to
decimal equivalents: |", W', i" and M". These are the values
which are missing in the table on page 68 and the student should
supply these results in that table.

94. Change the following decimals to common fractions and

72 ARITHMETIC FOR CARPENTERS AND BUILDERS

reduce the result to lowest terms: (a) .0221, (6) .0638, (c) .6862,
id) .8255.

95. How many times is the length .0725' contained into 16.862'?

96. A carpenter agrees to build a fence 120 ft. long for $45.50.
The cost of the lumber is 21.8ji per lineal foot; the labor cost 7.2ji
per lineal foot and the hardware cost f^ per lineal foot. What was
the actual cost of the fence and what was the profit?

97. California redwood weighs 26.23 pounds per cubic foot.
How much will 15 cu.ft. weigh?

98. A circle is 3.1416 times as far around as across it. Find
the number of feet around a circle which is 3.1667' across.

Fig. 26.— Section of a Steel Pipe.

99. One cubic foot of water weighs 62.5 pounds. Find the
part of a cubic foot occupied by one pound of water and express
the result in a decimal.

100. If it costs Sfl25 per day to run a gang and a concrete mixer
laying 23 cu.yds. of concrete paving material, what is the cost per
cubic yard?

101. A screen used for sifting sand is made out of wire which
is .0925" in diameter. What is the size of the opening between
the wires if there are 4 wires to the inch?

102. The external diameter of a steel pipe used for a column
is 5.563". The thickness of the metal is .259". What is the inside
diameter of this pipe? (See Fig. 26.)

CHAPTER VII

THE USE OF RULES. PULLEY SIZES. WIDTH OF BELTS.
FORMULAS. SHORT METHODS OF MULTIPLICA-
TION AND DIVISION

66. Rules. The practical carpenter is often obliged to
use a rule in working out some of his problems. Such rules
are of two kinds; those which have a rational mathematical
basis and those which are the result of experience. Both
of these kinds are valuable so long as they are true. Some
of the rules are exact and others merely give an approximate
result. You must be sure that you know whether a rule
is reliable or not before you use it. It pays to be on the
lookout for good, reliable rules which will give the desired
result with but little work.

67. Rule for Pulley Sizes. The mill-man frequently
has to find the speed of a saw, grindstone or belt-pulley.
The following rule is a good one to use in a case of that kind.

To find the number of revolutions of a driven pulley in a
given time, multiply the diameter of the driving pulley by its
number of revolutions in the given time and divide by the
diameter of the driven pulley.

Example. A pulley on the main shaft 48" in diameter
drives a pulley 26" in diameter on the counter shaft. The
main shaft makes 65 revolutions per minute (r.p.m.). How
many r.p.m. does the counter shaft make?

H ARITHMETIC FOR CARPENTERS AND BUILDERS

48"
66

240

288

26)3120( 120r.p.m. Ans.
26

52
52

Explanation. Multiplying the diameter of the driving
pulley, 48", by the r.p.m. of the driving pulley, 65, gives
3120. Dividing this by 26, the diameter of the driven
pulley, gives 120 r.p.m. for the speed of this pulley. Notice
that we multiply the two things that belonged to the same
pulley together. If you keep this fact in mind, it will help
you to remember the rules.

68. Rule for the Width of Belts. The question some-
times comes up in the mill : What width of belt must be used
to drive a certain machine? If you know how many horse-
power are required you can use the following rule:

To find the width of belt necessary to transmit a given horse-
power, multiply the horse-power by 33,000 and divide this
product by the product of the speed of the belt in feet per rriinute
times 40, for a single leather belt.

Example. A steam engine used to furnish power for
a mill is rated at 75 horse-power. The driving pulley is
54" in diameter and the engine runs at 180 r.p.m. How
wide a belt of single thickness of leather will be required?

THE USE OF RULES 75

54'
3.1416
4.5

'=4.5 ft.

ft. around the

ft. around the
r.p.m.

ft. per min. sp

61875.0 ( 24"
50894

1 57080
12 5664

14.13720

14.1372
180

pulley,
pulley.

1130 976
1413 72

825

2544.6960

61875
) 2544.7)

leed of belt,

25U.7XM

width of b(

109810
101788

Ans.

Explanation. With the information given, it is first
necessary to find the speed of the belt in feet per minute.
Every time the pulley travels once around, t-he belt will
travel the same distance. To find the distance around the
pulley we multiply the distance across it in feet by 3.1416.
This gives 14.1372 ft, around the pulley. The pulley re-
volves 180 times in a minute and, therefore, the belt will
travel 14.1372 times 180 or 2544.7 ft. per minute. From
now on we can follow the rule directly. Multiply the horse-
power 75 by 33,000 and divide this by the product of 2544.7

76 ARITHMETIC FOR CARPENTERS AND BUILDERS

times 40. The result when reduced to its lowest terms
gives a 24" belt.

69. Using Letters in Rules. It is possible to shorten
the statement of a rule by using letters instead of words.
Usually the' first letter of the word is used and the mathe-
matical operations are indicated by using the mathematical
signs. It is sometimes hard for a mechanic to understand
this arrangement and he immediately gets confused and
troubled when he sees these formulas, as they are called.
To show just how this is done and what it means, let us take
the following example:

To find the number of board feet in a piece of lumber multi-
ply the width in feet by the length in feet and this again by its
thickness in inches, counting a thickness less than one inch
as a full inch.

Ft.B.M. = WXLXt

The above rule is stated in two ways: first, in words
and second with letters. The abbreviation, " Ft. B.M."
means " feet board measure." The sign = is used to indi-
cate that the quantity of Ft. B.M. is equal to the value of
the expression to the right of the sign. W stands for the
width of the board in feet and L stands for its length in feet.
WXLXt indicates that the value of W is to be multiplied-
by the value of L and this again by the value of t. i stands
for the thickness of the board in inches. Notice that W
and L are capital letters and that t is a small letter. This
is a short way, used in this case, to show the difference
between values given in feet and those given in inches.

70. How to Use Formulas. To show how formulas may
be used we will make use of this one which we have just
found: Ft. BM. = WXLxt. Usually we will know the
width, length and thickness of the board and we will want

THE USE OF RULES

to find how many feet B.M. the board contains. In this
event, we will write the left side of the formula just as it
stands, but in writing the right side we will substitute the
values for the width, length and thickness of the board
which we know. We will then perform the operations
indicated on the right side of the formula and reduce the
expression to its lowest terms. The result will be a value

Fig. 27.

for the feet B.M. contained in the board. To illustrate this
more fully, let us take a specific example.

Example. How many feet board measure are contained
in the board shown in Fig. 27, if W = 9",L = U', and t = i"?

Then ¥t.BM. = WXLXt

or, substituting the values of W, L and t as given above:
Ft.B.M. = 1X14X1

78 ARITHMETIC FOR CARPENTERS AND BUILDERS

21
Then Ft.B.M. = ^Xl=4-Xl=y=10^

2
Therefore Ft.B.M. = 10| Ans.

Explanation. The rule states that W and L must be in
feet, so we must change the 9" to feet. L is already in feet.
The rule also says to count the thickness less than one inch
as a full inch; therefore |" must be counted the same as 1".
Now let us substitute these values in the formula. For W
we will write f ', for L we will write 14 and for t we will write
1. Then we solve the right-hand side by multiplying f X14
and this again by 1. This gives us the result IO5. The
board, therefore, contains 10^ ft. B.M.

71. Short Methods. When making computations in con-
nection with this work, the mechanic wishes, first, to get
an accurate result and, second, to get the result quickly.
All of this requires practice. If you do a little figuring once
a month or so, it will take quite a while to make a small
calculation and get it accurate. If, on the other hand,
you figure a little every day, you will soon be able to do it
more accurately and rapidly.

72. Addition. There is no short cut in addition. You
can, however, by following a few hints add very rapidly and
accurately. When adding, try to pick out combinations
that make 10, such as 4 and 6, 7 and 3, etc. Divide a long
column of figures into several parts and add each separately.
When adding long columns, put down the total of each
column separately and add these together for the final result.
To check addition, add again beginning at the other end of
the column.

73. Subtraction. To check subtraction add the differ-
ence and the subtrahend together to get the minuend.

THE USE OF RULES 79

74. Multiplication. To multiply a number by 10, add
one cipher to the right or move the decimal point one place
to the right. To multiply by 100 add two ciphers or move
the decimal point two places. To multiply by 5 (Y-)divide
by 2 and multiply by 10. To multiply by 25 (i^) divide
by 4 and multiply by 100. To multiply by 6^, 8^ 12^,
16f , 33i divide by 16, 12, 8, 6 or 3 and multiply by 100.
To multiply by 9 (10—1) first multiply by 10 and then sub-
tract the multipUcand from this product. To multiply by
11 (10+1) multiply by 10 and add the multipUcand to
the product.

To check multiplication, interchange the multiplicand
and multiplier and multiply again.

75. Division. Since division is the reverse of multipli-
cation all hints given for multiplication will work for division
if reversed.

To check division, multiply the divisor by the quotient
to get the dividend.

80 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter VII

39. To find the speed of the driven pulley, multiply the
diameter of the driving pulley by its number of revolutions
and divide by the diameter of the driven pulley. (Sec. 67.)

40. To find the width of single leather belt necessary
to transmit a given horse-power, multiply the horse-power
by 33,000 and divide this by the product of the speed of the
belt in feet per minute times 40. (Sec. 68.)

41. To find the number of board feet in a given piece
of lumber, multiply the width in feet by the length in feet
and multiply this product by the thickness in inches. (Sec.
69.)

PROBLEMS

103. The Roman designation M is used for 1000 when speaking
of board feet: For example, 5M. B.M. means 5000 ft. board
measure. Write a rule similar to Rule 41 to find M.B.M.

104. Write a rule to find the diameter of pulley to use when
the diameter and speed of the driving pulley and the speed of the
driven pulley are known.

105. A main shaft pulley 18" in diameter makes 100 r.p.m.
If the counter shaft must make 220 r.p.m., what must be the
diameter of the pullej' on the counter shaft?

106. When using a double leather belt we use the same rule
as for the single belt, but we use 60 instead of 40 in the rule. What
must be the width of belt for a 75 horse-power engine, having a
belt wheel 52" in diameter making 320 r.p.m.?

107. Add the following table of figures both across the page
and down the page. What is the grand total? How long did
it take you to get it correctly?

4562

4379

8256

4972

3459

7824

4652

3784

9286

4785

5672

8439

5648

7231

2657

4962

1435

4798

6245

3941

1126

2297

4456

6522

4733

THE USE OF RULES 81

108. Multiply the following without using a pencil:

(a) 76X50 (6) 79.2X5000

109. Write out a set of hints for division similar to those foi
multiplication given in the lesson.

110. Divide the following without using a pencil:

(a) 9^16§ (6) 450 -5- 6i

111. Multiply 624728 by 645 and check.

112. Divide 68,492 by 356 and check.

CHAPTER VIII
PERCENTAGE. DISCOUNTS. PROFIT. INTEREST

76. Percentage. To one who has thoroughly mastered
fractions and decimals, the subject of percentage should not
present any serious difficulties. The words " per centum"
are Latin and mean " by the hundred," The words "per
cent" are an abbreviation of " per centum " but they are
so common that they are now scarcely recognized as an
abbreviation. The sign % is used in place of the words:
thus, 10% means 10 per cent.

77. Changing a Fraction to a Per Cent. A per cent may
be expressed as a decimal or •as a common fraction. In
fact, a per cent is a fraction. Thus, 10% is the same thing
as .10 or J/^ or ^^^ If this is true we may change any frac-
tion into an equivalent form in per cent. This may be done
by reducing the fraction to a decimal, setting the decimal
point two places to the right and affixing the per cent sign.

Example. One-fifth of the brick taken from the kiln
after a certain burning were soft or salmon brick. What
was the per cent of soft brick?

I = W?j = -20 = 20 per cent soft brick. Ans.

Explanation. The soft brick were ^ of the whole number.
If we change ^ to a per cent we will get the per cent of soft
brick. I reduced to a decimal equals .20 and this is the
same as 20%.

PERCENTAGE 83

Notice that the sign % indicates two decimal places.
The expression 4.2% means four and two-tenths per cent
and this expressed as a decimal fraction would be written
.042.

Example. Express the fraction | as a per cent.

1 = 8) 7.000 ( .875 = 87.5% Ans.

60
56

40
40

Explanation. First change | to a decimal fraction by
dividing the numerator by the denominator. This gives
.875. To change the decimal to a per cent, move the
decimal point two places to the right and add the per cent
sign: thus, .875 = 87.5%. The result is read " eighty-seven
and one-half per cent " or " eighty-seven and five-tenths
per cent."

78. Meaning of Tenns. Before we can work problems
in percentage we must get a clear idea of the meaning of the
terms used. The base is the number of which the per cent
is taken. The rate is the per cent taken, often called rate
per cent. The percentage is the part of the base deter^
mined by the rate. The amount is the sum of the base and
the percentage. The difference is the base minus the per-
centage.

79. Analysis of Problems. Problems involving the use
of percentage usually fall into one of the three following
classes :

84 ARITHMETIC FOR CARPENTERS AND BUILDERS

Case 1. What is 10% of 250? The base and rate are
here given to find the percentage. Multiply the base by
the rate to get the percentage.

Percentage = base X rate.

Case 2. 25 is what per cent of 250? The base and per-
centage are here given to find the rate. Divide the percent-
age by the base to get the rate.

Rate = percentage -^ base.

Case 3. 25 is 10% of what number? The percentage and
rate are here given to find the base. Divide the percentage
by the rate to get the base.

Base = percentage -r- rate.

Example of Case 1. In ordering 6| M.B.M. of siding,
the builder agrees with the dealer that not more than 8%
of the entire order may be in pieces less than 8' long. How
many feet B.M. may be less than 8' long?

8% (rate) X 6500 (base) = .08X6500 = 520' (percentage)
B.M. Ans.

Explanation. It is easier to change the per cent to a
decimal before we multiply. We could multiply first and
then divide by 100 to get the result. In this case the word
of in the statement of the problem means the same thing
as times just as it did in common fractions. .08 times 65
thousand or 6500 equals 520 ft. B.M.

Example of Case 2. In taking off quantities an esti-
mator finds that of 16,000 sq.ft. of flooring, 5600 sq.ft. are
to be quarter-sawed oak. What per cent of the total is the
oak?

PERCENTAGE 85

5600 (percentage)

16,000 (base)

= .35 = 35% (rate) quartered oak. Ans.

Explanation. 5600 is the percentage and 16,000 is the
base. Divide 5600 by 16,000 to get the rate. This gives
.35 or 35% of the order to be quartered oak.

Example of Case 3. A furniture maker builds desks in
mahogany and oak. The number of mahogany desks turned
out of the factory in one year was 22, which was 4% of the
total output of the desks. What was the total output of
the desks?

22 (percentage) h-4% (rate) =22-^ .04 = 550 (base) desks.

Ans.

Explanation. The 22 mahogany desks represent the
percentage and 4% is the rate. It is easier to reduce the
4% to a decimal before dividing. 22 divided by .04 gives
550, which is the total output of desks, including both oak
and mahogany.

80. Solution of Problems. The solution of these prob-
lems does not appear very difficult when the statement of
the problem is simple. It is very important and not always
easy to determine which quantity is the base, which the
percentage and so on. For example, in the following state-
ment it makes a great deal of difference which quantity is
used as the base: One contractor bids S4200 on a certain
set of plans. A second contractor bids S4800 on the same
plans. The first man's price is 14.3% lower than the second
man's, using the lower price as a base, but the second man i^
only 12.5% higher than the first, using the higher price as
a base.

It is also important to determine exactly which quan-
tity is the rate. For example, consider the following state-

86 ARITHMETIC FOR CARPENTERS AND BUILDERS

ment: When solid rock is crushed, it occupies more space
than it did before. The per cent of small open spaces or
voids is from 30% to 55%. How much space will one
cubic yard of solid rock occupy after it is crushed? In
this case the volume of the solid rock is 30% less than or
70% of the total space occupied by the crushed rock.
70% is the rate and 1 cu.yd. is the percentage.

81. List Prices and Discounts. In business transactions
many quotations are given as list or catalog prices and are
subject to discounts. Discounts are usually given thus:
" 60% and 20% and 10% off for cash in thirty days." This
does not mean that the total discount is 90%, but it means
that a discount of 60% is taken from the list price and then
a second discount of 20% is taken from what remains.
Finally 10% is taken off this last price for cash in thirty days.

Example. The list price on a certain store door lock
is $18 with 40% and 10% discounts and 5% for cash in
thirty days. What is the cost of the lock if it is paid for
within thirty days?

$18.00 list
40%

00 list
7 . 20 discount

80 1st price
1 . 08 discount

$7.2000 discount. $10.80 1st price. $9.72

2d price

10%

$1.0800 discount

$9.72 2d price $9.72

2d price

5% off for cash .48

discount

. 4860 discount for cash.

24 final price
Ans.

Explanation. To find the final cost of the lock we must
find the discounts and subtract them from the price in

PERCENTAGE 87

consecutive order. The first discount allows 40% off
from the list price. 40% of $18.00 is $7.20. To get the
first discount price we subtract $7.20 from $18.00 to get
$10.80. The next discount is 10% off from this price. 10%
of $10.80 is $1.08 and we subtract this from $10.80 to get
the second discount price. This is $9.72. Now 5% from
this price is allowed for cash in thirty days. 5% of $9.72
is 48 cents and $9.72 less 48 cents gives $9.24 as the final
price.

82. Profits. Profit is often expressed as a per cent of
the actual cost. If a contractor estimates the actual cost
of construction to be a certain sum, he adds a certain per
cent of the cost for his profit.

Example. The estimated actual cost of a building is
is $4280. If the contractor adds 10% for profit, what price
does he bid on the job?

$4280X10% = $428

Adding the profit: $4280
428

$4708 price. Ans.

Explanation. 10% of $4280 is $428. Adding this to the
actual cost gives $4708 for the price which the contractor
bid.

83. Interest. When one man borrows money from
another, he pays interest for the use of the money. This
interest is usually reckoned at a certain rate per cent per
year. The base in this case is called the principal. The
only difference between figuring percentage and figuring
interest is that in figuring interest time must be taken into
account. The interest on a certain principal for one year
is a certain sum and for two years it is twice that sum.

88 ARITHMETIC FOR CARPENTERS AND BUILDERS

Interest is compounded by adding the interest to the prin-
cipal at the end of the year and figuring the interest on this
amount for the next year.

Example. A man deposits $500 in the bank at 3%
interest. What is the mount of the principal and interest
at the end of the year?

$500 principal $500 principal

.03 rate 15 interest

$15.00 interest $515 amount. Ans.

Ej'planation. For convenience we express the per cent
as a decimal late. Multiplying the principal, $500, by the
rate, .03, gives $15 for the interest for one year. Adding
this to the principal gives the amount, or $515.

Example: A contractor borrows $1200 for 6 months
at the rate oil\% per annum. What interest does he pay?

$1200 principal
.075

6 000
84 00

2 ) $90.000 interest for one year.

$45 . 00 interest for six months. Ans.

Explanation. Reducing the rate 7\% gives .075 as the
rate. Multiplying the principal $1200 by this rate gives
$90.00 interest for one year. The money was loaned only
for 6 months and the interest for that period will be one-half
of the interest for a whole year. One-half of $90 is $45,
the interest on $1200 for 6 months at 7|% per annum.

PERCENTAGE 89

Summary of Chapter VIII

42. The words " per cent " are an abbreviation of " per
centum," meaning " by the hundred." The sign % means
per cent. (Sec. 76.)

43. To change a fraction to a per cent, find its decimal
equivalent. Then move the decimal point two places to the
right and affix the per cent sign. (Sec. 77.)

44. To find the percentage when the base and rate are
given, multiply the base by the rate. (Sec. 79.)

45. To find the rate when the base and the percentage
are given, divide the percentage by the base. (Sec. 79.)

46. To find the base when the percentage and rate are
given, divide the percentage by the rate. (Sec. 79.)

47. To find the cost price when a series of discounts
are given, find the amount of the discounts consecutively,
subtracting each in its turn. The final result will be the
cost price. (Sec. 81.)

48. To figure profit, multiply the estimated cost by the
desired rate and add this allowance for profit to the cost.
(Sec. 82.)

49. To figure interest, multiply the principal by the
rate per annum. If the period stipulated is other than for
one year do not fail to take this into account. (Sec. 83.)

PROBLEMS

113. What is 25% of 16? of 48? of 90? of 240?

114. 8 is what per cent of 16? of 40? of 80?

115. 30% off of a number leaves 350. What is the number?

116. A pattern maker allows I" per foot for shrinkage. What
per cent does he allow?

117. A contractor figures that 1890 ft. B.M. will be required
for the studding in a house. If he add 25% for waste, what
B.M. must he order?

90 ARITHMETIC FOR CARPENTERS AND BUILDERS

118. In receiving an order for siding, a contractor gets 650 ft.
B.M. in 6' and 8' lengths out of a total of 5000 ft. B.M. What
per cent is this?

119. The actual face of a 4" floor board is only 3i". What
p)er cent must be added for matching?

120. Maple flooring " Clear " grade will allow 7% of lengths
2' to 35' according to the grading rules of the Maple Flooring Manu-
facturers' Association. How many feet B.M. may be of this
length in an order for 7^ M.B.M. Clear maple flooring?

121. In laying the floor for a house the contractor allowed 320
ft. B.M. for waste, which was 12% of the total. How many feet
B.M. were in the total order?

122. What is the net price of an enameled iron wash bowl if
the list price is $18.00 with 30% and 10% off?

123. A contractor figures the actual cost of a job to be $5820.
He was awarded the contract at $6550. What per cent of the
actual cost was profit?

124. In figuring on a certain job a contractor estimates the
actual cost to be $10,490. If he adds 8% for contingencies and
10% of this price for profit, what price will he bid?

125. What is the interest on $500 for 3 years 2 months at 65%
per annum?

126. What will be the total amount of $1200 in 5 years at 5%
per year compounded annuallj^?

127. A contractor owns a concrete mixer which cost him $1400.
It costs him $15 per day to run the mixer. Figuring money worth
5% and depreciation on the machine at 20%, what will be the
total cost of operating the machine per year of 200 days? What
will be the cost per day?

CHAPTER IX

RATIO. PROPORTION. CEMENT AND CONCRETE
MIXTURES. SLOPES. LEVERS.

84. Ratio. We are constantly comparing things with
each other. Among many other things we may compare
weight, distance or size. In fact, this is exactly what we
do whenever we measure the length of a board. We com-
pare the length of a foot measure with the length of the
board and we say that the board is 9 times as long as the foot
measure or 9 ft. long. If one plank is 9 ft. long and another
is 3 ft. long, we say the first is three times as long as the
second or the second is one-third as long as the first. This
comparison may also be stated by what we call a ratio.
That is, we say the length cf the first plank is to the length of
the second as 9 is to 3. This is written mathematically,
9 : 3 or f .

85. A Ratio is a Fraction. When the ratio is written
in the form of a fraction, it may be reduced to its lowest
terms: thus, |=i =3, which means that the first board is
three times as long as the second. Reversing our com-
parison we have 1=^, which means that the second board
is one-third as long as the first. This ratio is said to be
the inverse of the one just preceding. Thus we see that a
ratio may be written like a fraction and, like a fraction, it
may be reduced to its lowest terms.

Please notice that the things to be compared must be of
91

92 ARITHMETIC FOR CARPENTERS AND BUILDERS

the same kind. For example, it would be absurd to compare
feet with bushels.

A ratio is a mathematical comparison of two things of,
the same kind. The two numbers of the ratio are called
its terms.

Example. A room is 12' wide and 18' long. What is
the ratio of the width to the length?

-i| = ?=2:3 Ans.

Explanation. The comparison of the width to the length
will be as 12 is to 18. Reduced and expressed as a ratio
this is 2 : 3.

86. Cement and Concrete Mixtures. We frequently see
the statement that cement mortar is to be mixed in the
ratio 1: 3. This means that 1 part by volume of cement
is to be mixed with 3 parts by volume of sand. A sack of
cement weighing 94 pounds contains 1 cu.ft. of cement, very
nearly. Thus for every sack of cement we must have 3
cu.ft of sand. Likewise in mixing concrete the directions
sometimes state that it shall be mixed 1:2^:5 or 1:3:6.
Taking the first ratio as an example, this means that the
mixture shall be composed of 1 part by volume of cement
to 2^ parts by volume of sand to 5 parts by volume of
gravel or broken stone. In order to insure that the mixture
shall be uniform and shall completely fulfill the require-
ments, great care should be taken in measuring and mixing.
In Fig. 28 is shown a bottomless measuring box which con-
tains 1 cu.ft. of material for every 3" of height. A box
like this, which is 12" high, will contain 4 cu.ft. Such a
box is more accurate and frequently more convenient than
the box of an ordinary wheelbarrow for measuring the
material.

RATIO

Example: In making a concrete mixture in the ratio
1 : 2§ : 5 a contractor expects to use 25 sacks of cement.
How many cubic feet of sand and gravel does he require?

25 sacks cement

12i
50

Q2h cu.ft. sand.

25 sacks cement
5

125 cu.ft, gravel'. Ans.

Ans.

Fig. 28. — Measuring Box.

Explanation. For every sack of cement he will need
2^ cu.ft. of sand. For 25 sacks he will need 2^ times 25
or 62| cu.ft. of sand. Also for every sack of cement he will
need 5 cu.ft. of gravel. For 25 sacks he will need 25 times
5 or 125 cu.ft. of gravel.

94 ARITHMETIC FOR CARPENTERS AND BUILDERS

87. Proportion. If two ratios are equal, the statement
of their equaUty is a proportion. The ratio 2 : 3 is equal to
the ratio 12 : 18 because the last one mentioned can be
reduced to the first: thus, 12 : 18= l| = f =2 : 3. We can,
therefore, write these two ratios equal to each other: thus,
2 : 3 = 12 : 18. Another way of writing this same expres-
sion would be 2 : 3:: 12 : 18.

To illustrate this further, let us consider the statement:
For each 100 sq.ft. of roof surface, 833 shingles are required
to lay 4|" to the weather. Then for 500 sq.ft. of roof
surface, 4165 shingles will be required. We can write this
statement in the form of a proportion. 100 sq.ft. are to
500 sq.ft. as 833 shingles are to 4165 shingles, or, more
briefly, 100 sq.ft. : 500 sq.ft.:: 833 shingles : 4165 shingles.
Such a statement as this can, as we have just seen, be
expressed mathematically in the form of a proportion.

88. Inverse Proportion. If the four numbers of a pro-
portion are so related that an increase in one of the four
causes a corresponding decrease in another, the proportion
is an inverse proportion. For example, the following is
the statement of an inverse proportion: If it takes 4 men
12 days to do a certain job, 6 men working at the same
rate can do it in 8 days. Written as a proportion this would
be, 4 men : 6 men : : 8 days : 12 days. Increasing the num-
ber of men from 4 to 6 decreases the number of days from
12 to 8.

The first and last terms of a proportion are called the
extremes. The second and third terms are called the
means.

89. Solving a Proportion. If we multiply the two means
of a proportion together and also the two extremes, the
results will be equal. To prove this let us try it with the
following proportion :

RATIO 05

(Extreme) : (Mean) : : (Mean) : (Extreme)
2 : 3 :: 12 : 18

Product of the means, 3X12 = 36.
Product of the extremes, 2 X 18 = 36.
The two products are equal.

Let us now multiply the two means together and divide
the product by one extreme.

(Mean) X (Mean) = (Product)
3 X 12 = 36

(Product) -7- (Extreme) = (Extreme)
36 -T- 2 =18

This gives the other extreme.

We will now multiply the two extremes together and
divide by one of the means.

(Extreme) X (Extreme) = (Product)
2 X 18 = 36

(Product) -7- (Mean) = (Mean)
36 -^ 12 = 3

This gives the other mean.

We may write the following statements:

(1) The product of the two means of any proportion is
equal to the product of the two extremes.

(2) The product of the two means divided by either extreme
gives the other extreme.

(3) The product of the two extremes divided by either mean
gives the other mean.

Thus we see if one term of a proportion is unknown it
can readily be found by means of these relations.

96 ARITHMETIC FOR CARPENTERS AND BUILDERS

Example.

missing term?

In tlic proportion, 25 : a:::5 : 11, find the

1 1 extreme
25 extreme

55
22

Mean 5)275
55

product
mean. Ans.

Explanation. The missing term is a mean. Therefore,
if we multiply the two extremes together and divide the
product by the known mean, the result will be the unknown
mean, which is indicated in this proportion by the letter x.
The product of the extremes, 11 and 25, is 275. Dividing
this by the known mean, 5, gives 55 for the unknown term
of the proportion.

90. The Statement of a Proportion. Care must be used
in making up a proportion from the statement. A good
rule to follow is: Put the two things of the same kind in
their proper order as the first and second term ; then write the
known thing of the other kind as the third term and indi-
cate the unknown thing as the last term of the proportion.

By the use of the principle of proportion many practical
problems may easily be solved.

Example. Joists 2"X12", 16" on center, are to be set
for a floor surface having 960 sq.ft. of area. How many feet
B.M. will be required for the joists?

16" = li'

1^ : 2:: 960: a:

240

960X2 r60X2 900X2X3 . ..n f+ ti a/t • • * a
— :rT — = — T — = -^--^ = 1440 ft. B.M. joists. Ans.

RATIO

Explanation. For every foot of floor length there will be
a corresponding foot of joist length. For every 16" or I5'
of floor width there will be 2X1 = 2 or 2 board feet of joist,
A board foot is one square foot of board 1" thick. We can,
therefore, write this in the form of a proportion. We solve
the proportion by multiplying the two means together and

Fig. 29.— Slope of a Rafter.

dividing by the extreme. Expressing this in the form of

960 X 2 X 3
a fraction and simplifying gives which equals 1440

ft. B.M. of 2X12 joists. The use of the proportion in this
problem shortens the work considerably.

91. Slopes. Proportion can frequently be used to ad-
advantage in connection with embankments, roof slopes and
other similar slopes and grades. The total horizontal run
is to the unit horizontal distance as the total rise is to the rise
for the unit run.

Example. Fig. 29 shows a rafter which rises 6' 4" in
16' of horizontal measurement. How many inches does it
rise per foot of horizontal measurement?

98 ARITHMETIC FOR CARPENTERS AND BUILDERS

6' 4" = 76"

16 : l::76:x

76X1 76 .3,, .

^^ = -=41 Ans.

Explanation. First change the 6' 4" to inches because
the answer will be in inches. Then writing the proportion,
we have: 16 ft. are to 1 ft, as 76 ins, are to the answer.
Solving this proportion gives 4f" rise for every horizontal
foot.

Example. An earth embankment rises 1| ft, on every
foot of level ground. How much will the embankment rise
for 18 ft. of level ground?

1 : 18::U '• ^

18XU

= 27 ft. rise in 18' run, Ans.

Explanation. In this case the proportion can be stated
thus: The unit run is to the total run as the rise for the unit
run is to the total rise. Writing this in the form of a pro-
portion and solving we get 27 ft, rise in 18 ft, run.

92. Levers. Every mechanic is familiar with the ordi-
nary crow-bar when used as a lever. There are many other
familiar examples of the lever. The wheelbarrow is a good
example of a lever. The scale beam on which a small
weight balances a large weight on the scale platform is
another example.

The lever is a stiff bar or rod supported or pivoted at
a point on which it can move freely. The pivot is called
the fulcrum. The distance from the center of the weight
to the fulcrum is called the weight ann. The distance from
the point where the force is applied to the fulcrum is called
the force arm.

RATIO

93. Arrangement of Levers. There are three arrange-
ments of levers. The first type is shown in Fig. 30(a). The
fulcrum is between the weight and the acting force. Fig.
30(6) shows a lever of a second type in which the weight is

Weight

Weight ^j^
Arm [^

Force

Arm

Fia.30(a)
Weight

ZS"

■

Weight

Arm

Force

Arm
FiG.30(6)

Forco

f
H

Force

Weight

FoTce

Arm

Weight

Arm
Fig. 30(0)

between the fulcrum and the force. A lever of a third type
is shown in Fig. 30(c). Here the force is applied between the
weight and the fulcrum.

In each of these levers the following proportion holds
good :

Force : Weight : : Weight Arm : Force Arm

100 ARITHMETIC FOR CARPENTERS AND BUILDERS

Example. A man pushes down on the end of a crow-bar
with a force of 140 pounds. The distance from his hand to
the fulcrum is 5 ft. How large a weight can he lift at a
distance of 1 ft. from the fulcrum?

l:5::140:a;

^^^ = 700 lbs. Ans.

Explanation. In writing this proportion we put down

the value of the weight arm first. This arrangement brings

the answer last. It is not always necessary to do this, but

it is convenient. Multiplying the means together and

140X5
dividing by the extreme gives — ij which, when solved,

gives 700 lbs. that the man can lift at the end of the bar.

RATIO

101

Summary of Chapter IX

50. To change a ratio to a fraction, write the first term
over the second and reduce this fraction to its lowest terms.
(Sec. 85.)

51. To find the number of cubic feet of sand and gravel
required for any given mixture of concrete, multiply the
number of sacks of cement to be used by the second term
of the ratio to get the cubic feet of sand required and by

Fig. 31.— Roof Rafter and Strut.

the third term to get the cubic feet of gravel required.
(Sec. 86.)

52. To solve a proportion, multiply the two means
together and divide the product by the known extreme or
multiply the two extremes and divide the product by the
known mean. The result will be the unknown term or
answer of the proportion. (Sec. 89.)

53. To write the statement of a proportion, write the
two things of the same kind that are known in their proper
order as the first and second terms, write the known thing
as the third term and indicate the answer as the fourth
term. This rule is not rigid but it is convenient. (Sec. 90.)

102 ARITHMETIC FOR CARPENTERS AND BUILDERS

54. To solve slopes, write the proportion: The total
horizontal run is to the unit horizontal run as the total rise
is to the rise for the unit run. Then substitute the given
values and solve the proportion. (Sec. 91.)

55. To solve a problem in levers, write the proportion:
Force in pounds is to the weight in pounds as the weight
arm in feet or inches is to the force arm in feet or inches.
Then substitute the given values and solve the proportion.
(Sec. 93.)

Fig. 32. — Wheelbarrow as a Jjever.

PROBLEMS

128. A building is three times as long as it is high. What is
the ratio of the height to the length?

129. Reduce the ratio 1^ : 12 to its lowest terms.

130. Concrete for a certain job is to be mixed 1:3:5. If
250 sacks of cement are used, how many cubic feet of sand and
gravel are required?

131. A drawing is shown to a scale of 1" = 1'. What is the
ratio of the actual size of the piece to the size of the drawing?

132. If 5 men complete a barn in 18 days, how long will it take
7 men to complete it?

133. If 3 men can make 85 ft. of cement sidewalk 5 ft. wide
in a day, how much sidewalk of the same width can 5 men make
in one day? In two days?

134. A contractor knows from past experience that he uses 30
ft. B.M. of 4X4's and 45 ft. B.M. of 2X4's for every 150 ft. B.M.
of 1" stuff for concrete forms. On a job requiring 22,000 ft. B.M.

RATIO

103

of I" stuff, how many feet B.M. 4X4's and 2X4's will be re-
quired?

135. In estimating a certain job a builder finds that there are
2500 sq.ft. of partitions requiring 2X4 studding set 16" on centers.
How many board feet of studding will be required?

136. For a hanging ceiling of 4200 sq.ft. 1 X6's set 20" on center
are used for joists. How many feet B.M. of 1 X 6's will be required?

Fig. 33. — The Hammer as a Lever.

137. In making a rough estimate a builder remembers that a
certain residence containing 32,500 cu.ft cost $5600. How much
will a residence which contains 47,000 cu.ft. cost at the same rate?

138. The rise of a truck approach at the end of a platform is
3" for every horizontal foot. What is the height of the platform
if the total horizontal run is 12 ft.?

139. If a roof rafter rises 9' 6" in 18' horizontal run, what
must be the length of a vertical strut 11' from the lower end as
shown in diagram in Fig. 31?

140. A wheelbarrow may be considered as a lever of the second

104 ARITHMETIC FOR CARPENTERS AND BUILDERS

class. This is shown diagramatically in Fig. 32. With what
force must a man Uft in order to move the wheelbarrow when it
is loaded with 350-pounds of material?

141. A claw-hammer is a bent lever as shown in Fig. 33. In
drawing a nail a carpenter pulls with a force of 20 lbs. What is
the resistance of the nail in pounds?

142. Two men on the end of a bar can push down with a force
of 120 lbs. each. If the distance from their hands to the fulcrum
is 6 ft. and the distance from the fulcrum to the weight is 18",
how large a weight can they lift?

CHAPTER X

LINEAR MEASURE. AREA. VOLUME. LIQUID MEAS-
URE. DRY MEASURE. WEIGHT. THE CIRCLE.
WEIGHT OF BUILDING MATERIALS. BEARING
POWER OF SOILS. FOOTINGS. CONCRETE MIX-
TURES

94. Weights and Measures. There are two systems of
weights and measures in common use. The one most gener-
ally used among English-speaking nations is the British sys-
tem. The other system in use is the metric or decimal system.
The metric system is much more convenient. Although it
is used to a considerable extent in some of the metal-working
trades, it is doubtful if it will ever be used extensively in the
building trades. We will, therefore, confine ourselves to
the British system. Tables of values and equivalents of
the British system most commonly met with in the building
trades are given herewith.

95. Measures of Length. Linear measure is used for
comparing lines and distances.

TABLE I

COMMON LINEAR MEASURE

12 inches (in. ") = 1 foot (ft. ')•
3 feet =1 yard (yd.) =36 ins.

51 yards =1 rod (rd.) = 16nt. =198 ins.

320 rods = 1 mile (mi.) = 1760 yds. =5280 ft.

105

106 ARITHMETIC fOR CARPENTERS AND BUILDERS

96. Measures of Surface. Surface has two dimensions,
length and breadth.

TABLE II

97. Measures of Volume. Volume has three dimen-
sions, length, breadth and thickness.

TABLE III

COMMON CUBIC MEASURE

1728 cubic inches (cu.ins.) =1 cubic foot (cu.ft.)
27 cubic feet = l cubic yard (cu.yd.) =46,656 cu.ins.

TABLE IV

STONE MEASURE

24f cubic feet = 1 perch.

A perch of stone in masonry is I65 ft. long, 1^ ft.
wide and 1 ft. high.

98. Measures of Capacity. Capacity is expressed in
units of liquid and dry measure.

TABLE V
COMMON LIQUID MEASURE

4 gills (gi.) = l pint (pt.)
2 i)ints =1 quart (qt.)

4 quarts =1 gallon (gal.) =8 pints = 231 cu.ins.
31i gallons =1 barrel (bbl.) = 126 qts. =252 pts.
2 barrels = 1 hogshead (hhd.) = 63 gals. = 252 qts. = 504 pts.

LINEAR MEASURE 107

Dry measure is used for measuring grains, seeds, fruits,
lime, etc.

TABLE VI

DRY MEASURE

2 pints = 1 quart.

8 quarts = 1 peck (pk.) =16 pts.

4 pecks = 1 bushel (bu.) =32 qts. =64 pts.

The following table gives the value of some of these
units in cubic inches.

TABLE VII

COMPARATIVE VALUES

U. S. liquid measure, 1 gal. =231 cu.ins.
U. S. liquid measure, 1 qt. =57 J cu.ins.
U. S. dry measure, 5 pk=268T cu.ins.
U. S. dry measure, 1 qt. = 67i cu.ins.

From the above you will see that a quart liquid measure
is smaller than a quart dry measure. The values used in
Great Britain are different from the values given here.

99. Measures of Weight. The system used for meas-
uring the weight of all ordinary articles is called avoirdupois.
Druggists use the apothecary system and jewelers use troy
weight. We will use the avoirdupois system.

TABLE VIII

AVOIRDUPOIS WEIGHT

16 ounces foz.) =1 pound (lb.)
100 pounds = 1 hundredweight (cwt.)

20 hundredweight = 1 ton (T.) = 2000 pounds.
2240 pounds =1 long ton.

With these units we are able to make all the measure-
ments commonly used by the carpenter and builder. We

108 ARITHMETIC FOR CARPENTERS AND BUILDERS

are already familiar with the measures of length; let us
now find out how to use measures of surface.

100. Square or Surface Measure. If we cut out a piece
of paper in the form of a square one inch on a side, we say
that the sheet contains one square inch of surface. If we
cut out a square, one foot on a side, we say that this contains
one square foot of surface. The square inch and the square
foot are the units of surface measurement. Surface is also
often called area.

101. Area of the Square and the Rectangle. A square
surface is one the four sides of which are of equal length
and the four angles of which are right angles. To find the
area of a square we multiply the length of one side either in
feet or inches by the length of the other side in feet or inches.
The product gives the number of square feet or of square
inches in the square. A rectangle is a figure similar to the
square except that two of its opposite sides are longer than
the other two. To find. the area of a rectangle we multi-ply its
length by its width. It is important to notice that the length
and width must be in the same units before you can multiply.
For example, feet times feet will give square feet; inches
times inches will give square inches. You can change square
inches to square feet by dividing by 144 and you can change
square feet to square inches by multiplying by 144.

Example. How many square feet does a floor 16'X19'
contain?

16 ft.

19 ft.

144

304 sq.ft. Ana.

LINEAR MEASURE

109

Explanation. Since the length and width are in feet
the answer will be in square feet. Multiply the length by
the width to get the area of the rectangle. 16' X 19' = 304
sq.ft. in the floor.

Example. What is the area of a table top 64|" long
by 28|" wide?

641x281-64.5"
28.75"

225

516 0

1290

1854.

375 sq.

ins,

Ans.

Explanation. For convenience first change the common
fractions to decimals. Then multiply the length in inches
64.5" by the width 28.75" to get the number of square inches
in the area.

102. The Circle. A circle is a plane figure bounded by
a curved line called a circumference. A line drawn through
the center of the circle and ending on
each side at the circumference is called
a diameter. A line drawn from the
center to the circumference is called a
radius. The radius is equal to one-
half the diameter. The meaning of
these terms is further illustrated in
Fig. 34.

The diameter of a circle bears a
definite relation to the circumference.
This relation is such that if we divide the circumference of
any circle by its diameter we will get Z\ for an answer. If

^^tcu^ilfere^

Fig. 34.

no ARITHMETIC FOR CARPENTERS AND BUILDERS

we divide one quantity by another we have a ratio. This
value is, therefore, the ratio of the circumference to the
diameter. The decimal equivalent of this number is 3.1416.
We use this figure so often that we have assigned a special
character to name it. This character is the Greek letter pi
(pronounced pie) and it is written

7r = 3.1416=V=3f

It is to be noticed that these values given are not exact,
but they are close approximations. We could carry the
decimal fraction out to as many places as we pleased,
but we would still get a remainder. Four decimal places
are all that are necessary for practical purposes. For most
purposes the value of 3^ or \^- is sufficiently accurate. Since
7r is the ratio of the circumference to the diameter, we can
find the circumference when the diameter is given if we
multiply the diameter by tt.

Example. The diameter of a circle is 56 '. What is the
length of the circumference?

Solution No. 1 ^^Xy=176" Ans.

Solution No. 2 56

3.1416

336

2 24

5 6

168

175.9296" Ans.

LINEAR MEASURE 111

Explanation. Here is an example in which the value y-
is very convenient. Using this value and canceling we get
the result 176" in solution No. 1. Compare this to solution
No. 2 in which we have multiplied by 3.1416. The result is
175.9296" or very nearly 176". This shows that there is
no great error in using -^^j?- instead of 3.1416 for the value of r.

Example. The circumference of a circle is 77". What
is its radius?

22 ^ 7 4Q
77^y = JJX^=^ = 24i" diam.

2
24^^-2 = 121" rad. ' Ans.

Explanation. Divide 77 by 3^ or \2 to get the diameter.
The radius is one-half of the diameter, so we divide by 2 to
get the radius.

Example. A wooden stave silo is to be built 16' in
diameter. How many lineal feet of 1^" rod will be necessary
for each band, allowing about 9" for the joint?

3.1416
16

18 8496
31 416

50.2656' = 50i' about.

9"= ^'

51' Ans.

Explanation. Multiply the diameter 16' by % to get
the circumference of the silo. This gives 50.26/ which is
a little more than 50j', Adding 9" or f ' for the joint gives
51' required in each band.

112 ARITHMETIC FOR CARPENTERS AND BUILDERS

103. Volume or Cubic Measure. When we wish to know
volume or contents of a cube or rectangular block we find
out how many cubic inches or cubic feet it contains. A
cubic inch is the volume of a cube all of where edges are one
inch in length. A cubic foot is a cube all of whose edges
are one foot in length. To find the volume of a cube we multi-
ply its length by its breadth and this again by its thickness.
Since all the edges of a cube are equal, this is equivalent
to multiplying the length of one side of the cube by itself
three times.

Example. How many cubic inches does a cube each of
whose edges are 5" long contain?

5"X5" = 25sq.ins.

25 sq.ins. X5" = 125 cu.ins. Ans.

Explanation. If we multiply 5" times 5" we get the
number of square inches in one face of the cube or 25 sq.ins.
Multiplying this again by 5" gives 125 cu.ins. in the cube.

A rectangular block is similar to a cube except that the
length of all the sides is not the same. To find the volume
of a rectangular block, we multiply the length by the width and
this again by the height.

Example. What is the cubic contents of a room 12' X
16'X9' high?

12'
16'

72
12

192 sq.ft.
9'

1728 cu.ft. Ans,

LINEAR MEASURE 113

Explanation. To find the contents of the room in cubic
feet, we multiply the width 12' by the length 16' by the
height 9'. This gives 1278 cu.ft. as the contents of the room.

104. Weight of Lumber. If we know the weight of a
cubic foot of lumber we may estimate the weight of a quan-
tity by multiplying the volume in cubic feet by the weight
per cubic foot. The weights of some of the common kinds
of lumber are given in the table on page 114. The weights
given are for perfectly dry lumber. Green timbers weigh
from one-fifth to one-half more than the figures given here.
Ordinary building timbers, fairly well seasoned, weigh about
one-sixth more than these values.

Example. What will be the weight of a pile of closely
stacked Northern pine lumber, air seasoned, which is 6 ft.
wide, 18 ft. long and 5 ft. high?

18'

108

540

cu.ft.

34.3

lbs. per cu.ft.

162 0

2160

1620

6)18522.0

lbs. dry

3087

lbs, to add for moisture.

18522

21609

lbs, weight,

Ans.

114 ARITHMETIC FOR CARPENTERS AND BUILDERS
TABLE IX

WEIGHT OF LUMBER PER CUBIC FOOT

Cypress, American, dry 34.3 lbs. per cu.ft.

Hemlock, perfectly dry 25 "

Maple, dry 49 "

Oak, white, dry 48 "

Oak, red, dry 32-45 "

Pine, white, dry 25 "

Pine, yellow. Northern 34. 3 "

Pine, yellow. Southern 45 "

Poplar, dry 29

Spruce, perfectly dry 25 "

(Kidder's Pocket Book.)

Explanation. First find the cubic contents of a pile of
lumber by multiplying the length by the width by the height.
This gives 540 cu.ft. for the volume of the pile. Multiply-
ing this by the weight per cubic foot of Northern pine,
which is 34.3 lbs., we get 18,522 lbs. for the weight of the
pile if the lumber were perfectly dry. It is only air seasoned
and so we must add one-sixth. Dividing by 6 we get 3087
lbs. to add on account of the moisture in the lumber. This
gives 21,609 lbs. as the estimated weight of the pile.

105. Weight of Building Material. It is often con-
venient to know the weight of definite quantities of building
materials of various kinds. If we know the weight of a
cubic foot of these materials we can find the weight of any
quantity just as we did in the case of lumber.

TABLE X

WEIGHT OF VARIOUS BUILDING MATERIALS

PER CUBIC FOOT Average weight.

pouud.s per cu.ft_

Brick, best pressed 150

Brick, common hard 125

Brick, soft inferior 100

Brickwork, pressed brick 140

Brickwork, ordinary , 112

LINEAR MEASURE 115

Average weight,
pounds per cu.ft.

Concrete 150

Granite 170

Limestones and marbles 165

Masonry of granite or limestone, well dressed 165

Masonry of granite rubble 154

Masonry of sandstone, well dressed 144

Sandstone, fit for building 151

(Kidder's Pocket Book.)

Example: How much will an ordinary brick wall 9"
thick by 14' high by 32' long weigh?

9" = r 14'

32'

28
42

4)448 sq.ft.

112
3

336 cu.ft. in wall.

112 weight of brickwork.

672
336
336

37632 lbs. weight of wall. Ans.

Explanation. We first find the cubic contents of the wall.
There are 336 cu.ft. in the wall. Ordinary brickwork,
according to the table, weighs 112 lbs. per cubic foot. Hence
multiplying the number of cubic feet in the wall by the
weight per cubic foot gives, 336X112 = 37,632 lbs. as the
weight of the wall.

106. Bearing Power of Soils. The bearing power of
various soils depends upon the form of the structure and the

116 ARITHMETIC FOR CARPENTERS AND BUILDERS

character of the soil. The following table of allowable soil
pressures is given by Prof. Ira O. Baker. To be on the safe
side the minimum values are recommended. To deter-
mine the size of footings we first find the weight of one lineal
foot of the wall in tons. Divide this by the allowable pres-
sure per square foot and the result will give the number of
square feet per lineal foot required for the footings.

TABLE XI

BEARING POWER OF SOILS

Kind of Material.

Bearing power in tons

per sq.ft.

Min.

Max.

200

0.5

Rock, the hardest — in thick layers, native bed .

Rock, equal to the best ashlar masonry

Rock, equal to best brick masonry

Rock, equal to poor brick masonry

Clay on thick beds, always dry

Clay on thick beds, moderately dry

Clay, soft

Gravel and coarse sand, well cemented

Sand, compact and well cemented

Sand, clean, dry

Quicksand, alluvial soils, etc

Example. What size footings will be required for a 12'
concrete wall 46' high to be placed on clean dry sand?

46 cu.ft. per lineal foot of wall.
150 wt. per cu.ft. concrete.

2300

00 lbs.) 6900

lbs.

2 tons) 3. 45

tons

, 725 sq.ft. of footings or,

say, 2 sq.ft. of footings.

Ans.

LINEAR MEASURE

117

Explanation. The ^vall will contain 46 cu.ft. per foot of
length. The weight of concrete is given as 150 lbs, per cubic
foot. 46X150 = 6900 lbs. that the wall weighs per foot of
length. Dividing by 2000 gives 3.45 tons. Allowing 2
tons per square foot of bearing surface gives 1.725 sq.ft.
of bearing surface required or, to be on the safe side, we will
say 2 sq.ft. of bearing surface per lineal foot of wall. The
footing should, therefore, be 2 ft. wide.

107. Quantity of Materials Required per Cubic Yard of
Concrete. Concrete for building purposes is usually meas-
ured in cubic yards. If the number of cubic yards of con-
crete required is known, the quantities of cement, sand and
gravel necessary may be obtained by using the following
table. The table is based on one sack being equal to one
cubic foot of cement and four sacks to the barrel.

TABLE XII

QUANTITIES OF MATERIALS PER CUBIC YARD OF CONCRETE *

Mixture.

Bbls. Cement.

Cu.yds. Sand.

Cu.yds. Gravel.

1 : U :3

1.91

.42

.85

1:2:3

1.74

.52

.77

1:2:4

1.51

.45

.89

1 : 2i : 4|

1.31

.48

.87

1 : 2| : 5

1.24

.46

.92

1:3:5

1.16

.52

.86

1 : 1

4.88

.72

l:U

3.87

.86

1 :2

3.21

.95

* Adapted from a similar table in 'Concrete, Plain and Reinforced,''
by Taylor and Thompson.

Example. A foundation wall is to be made of concrete
1 : 2i : 5 mixture. The wall is 9" thick, 8' high and 96'
long. How much material will be required?

118 ARITHMETIC FOR CARPENTERS AND BUILDERS

9"=r

96'
8'

4)768

192

27)576

cu.ft. (21| cu.yds. in walL

1
^3

21.3

21.3 21.3

1.24

.46 .92

852 1 278 426

4 26 8 52 1917

21 3

9.798 19.596

26.412

26^ bbls. cement. 10 cu.yds. sand. 20 cu.yds. gravel.
Ans. Ans. Ans.

Explanation. First find the contents of the wall in
cubic feet and divide this by 27 to get the contents in cubic
yards. We see from the table that a 1 : 2^ : 5 mixture
requires 1.24 bbls. of cement for every cubic yard of con-
crete. Multiply 21.3 cu.yds. by 1.24 to get 26.4112 or,
say, 265 bbls. of cement. In a similar manner it is found
that 10 cu.yds. of sand and 20 cu.yds. of gravel are neces-
sary.

LINEAR MEASURE 119

Summary of Chapter X

56. To find the area of a square or rectangle, multiply
the length by the width. (Sec. 101.)

57. To find the length of the circumference of a circle
when the diameter is given, multiply the diameter by 3.1416.
(Sec. 102.)

58. To find the diameter of a circle when the circum-
ference is given, divide the circumference by 3.1416. (Sec.
102.)

59. To find the volume of a cube or rectangular block,
multiply the length by the width and this again by the
thickness. (Sec. 103.)

60. To find the weight of lumber, multiply the number
of cubic feet in the closely stacked pile by the weight per
cubic foot. (Sec. 104.)

61. To find the weight of any quantity of building
material, multiply the quantity in cubic feet by the weight
par cubic foot. (Sec. 105.)

62. To determine the size of footings, divide the weight
of the wall in tons per lineal foot by the bearing power of
the soil in tons per square foot. The result will be the
required area of the footing in square feet per lineal foot
of wall. (Sec. 106.)

63. To find the quantities of materials for concrete work,
multiply the number of cubic yards of concrete by the value
for the required mixture given in the table. (Sec. 107.)

PROBLEMS

143. How many square feet in a cement sidewalk 6 ft. wide
and 180 ft. long?

144. A house is approximately 22' X 30' X 18' high. Allowing
30 sq.yds for gables and dormers, how many square yards of paint-
ing surface are there on the house?

120 ARITHMETIC FOR CARPENTERS AND BUILDERS

145. If a gallon of paint will cover 100 sq.yds. two coats, how
many gallons of paint will be required for the above house?

146. If a silo is 18' across, how many feet are there around it?

147. A farmer can measure the distance around his silo with
a tape, but he cannot measure across it. He finds that the distance
around is 50J'. Tell him the distance across it.

148. How many cubic feet does a bin 6 ft. square and 8 ft. high
contain? How many bushels?

149. How many barrels of water will a cistern that is 4 ft. square
and 8 ft. high hold?

Fig. 35.— Testing Tank.

150. A testing tank shown in Fig. .35 is built of concrete. It is
6 ft. deep, 14 ft. wide and 25 ft. long. How many barrels of water
will it hold?

151. If water weighs 62j pounds per cubic foot, how much will
the water in the above-mentioned tank weigh?

152. How much will a stick of green white oak 12"X12"X10'
weigh? Add one-fourth to the dry weight to allow for moisture.

153. How much will 500 pieces 2"X4"X16' seasoned Northern
pine scantlings weigh?

154. If the freight in carload lots from St. Louis to Des Moines
is 22jii per hundred pounds, what will the charges on a load of
red oak which measures 8' 4" wide, 32 ft. long and 6 ft. high when
closely piled? Figure 38 pounds per cubic foot.

LINEAR MEASURE

121

155. A carload of common hard brick measures 8' 4" wide,
5 ft. high and 30 ft. long. What is the weight of the load?

158. A certain concrete chimney contains about If cu.j'ds. cf
'■•oncrete for every foot of height. How much will a stack 150 ft,
high weigh in tons?

157. What must be the size of
the footing for the above-men-
tioned chimney if it is to be
placed on gravel and coarse sand
well cemented?

158. A chimney for a residence
is 35' liigh and is made of com-
mon hard brick of the dimensions
shown in Fig. 36. What must
be the size of the footings allow-
ing J ton to the square foot?

159. A concrete retaining wall
is 18" wide, 8' high and 60' long.
A 1 : 3 : 5 mixture is used. How
many barrels of cement and
cubic yards of sand and gravel
;vill be required?

160. The walls of the testing
tank shown in Fig. 35 are 9"

thick and the floor is 12" thick. What will be the weight of the
tank when it is full of water? What will be the pressure per square
foot of bearing surface?

161. A cement sidewalk is 6' wide, 180' long and 4" in total
thickness. The base is a 1 : 3 : 5 mixture and the top coat, which
is 1" thick, is a 1 : H mixture. How much cement, sand and
gravel will be required?

Fig. 36.— Chimney.

CHAPTER XI

POWERS. ROOTS. RIGHT TRIANGLES. AREA OF
CIRCLES. CAPACITY OF TANKS AND CISTERNS

108. Powers. It is sometimes convenient and neces-
sary to multiply a number by itself one or more times. This
is performed just like any ordinary multiplication, but we
have a special name for the product. When a number
is multiplied by itself one or more times we call the result-
ing product the power of the number. The number which is
used as a factor is call the base.

Thus in the expression, 2X2 = 4, we have raised 2 to
second power. The base is 2 and 4 is the power. Again
in the expression 2X2X2 = 8, we have raised 2 to the third
power. Instead of writing 2X2X2 = 8, we may shorten
it by writing 2^ = 8. The small figure above and to the right
of 2 indicates how many times 2 is to be used as a factor.

109. Exponents. The exponent of a power is a small
figure placed above and to the right of the base to indicate
how many times it is to be used as a factor to get the power.
The second power of a number is called the square of the
number. The third power of a number is called the cube
of that number. The powers above the third have no
special names. They are simply called the fourth power,
the eighth power, etc. The process of finding the powers
of numbers is called involution.

122

POWERS 123

Example. Find the square of 25.

25
25

125

625
252 = 625 Ans.

Explanation. To find the square of a number we must
use it twice as a factor. 25 times 25 is 625.
Example. Raise 9 to the fifth power.

9
9

81
9

729
9

6561
9

59049
95 = 59049 Ans.

Explanation. To raise 9 to the fifth power means that
we must use 9 as a factor five times.

To raise a number to a given power we must use it as a
factor as many times as indicated by the exponent.

110. Powers of Common Fractions. To raise a common
fraction to a given power we raise the numerator to the

124 ARITHMETIC FOR CARPENTERS AND BUILDERS

required power for the new numerator and raise the de-
nominator to the required power for the new denominator.
Example. Find the value of (f)^.

tX| = -5'V Am.

Explanation. To raise a fraction to the second power
we multiply it by itself. This is in effect multiplying the
two numerators together 3X3, to get 9 for the numerator
of the power and multiplying the denominators together,
5X5, to get 25 for the denominator of the power.

111. Roots. Finding the root of a number is just the
reverse of finding its power. If we separate 9 into its two
factors 3X3, the factor 3 is said to be the square root of 9.
In a similar manner, if we separate the number 125 into
its three factors 5X5X5, the factor 5 is said to be the cube
root of 125.

The square root of a number is one of the two equal
factors into which the number may be separated. The
cube root is one of the three equal factors into which a
number may be separated. The fourth root is one of the
four equal factors, and so on for the higher roots. To indi-
cate that a root is to be taken of a number we use a sign
y/~ called the radical sign. A small figure placed in the
opening of the sign indicates what root is to be extracted.
This figure is called the index of the root. For example
'V^27 indicates the cube root of 27. The small figure 3 is
the index of the root. When the square root is intended
the index figure is omitted, since it is understood. The
operation of finding the root of a given quantity is called
evolution.

Roots are of two kinds. Those which may be extracted
without a remainder are called perfect roots. Those which
have a remainder are imperfect roots. It is impossible to

POWERS 125

find the exact value of an imperfect root, but it may be found
correctly to any required number of decimal places. Three
decimal places are usually sufficient.

112. Square Root. If we square the numbers 1, 2, 3,
4, 5, 6, 7, 8, 9, the squares of these numbers will be 1, 4, 9, 16,
25, 36, 49, 64 and 81. Thus 1 is the square root of 1, 2 is the
square root of 4, 3 is the square root of 9, 4 of 16 and so on.
It is worth while to remember these figures and to be able
to recognize them when you see them. The square root of
numbers above one hundred are not so easy to determine.
To do this we must go through a somewhat complicated
process.

Example. Find Vl 19025.

Vll'90'25 ( 345 Ans.

9
64 1290
|2_56
3425
3425

685

Explanation. First point off the number into periods
of two figures each beginning at the right and placing a
mark between them.

The next step is to find the largest perfect square which ■\<^,^c
is equal to or less than the first period at the right. This ^ ' '^
may readily be determined by inspection. In this case
the perfect square is 9 and the root of the square is 3. Put
the 9 under the first period, 11, and put its square root, 3,
to the right as the first figure in the root or answer. Now
subtract 9 from 11 to get 2. Bring down the next period,
90, and write it with this remainder to get 290.

Multiply the first figure of the root, which is 3, by 2 and
write the product, 6, at the left of 290 as a trial divisor.

126 ARITHMETIC FOR CARPENTERS AND BUILDERS

Divide the first two figures of the number 290, which are 29,
by this 6 and write the result as the next figure of the root.
Six will go into 29 four times and so we write 4 as the second
figure of the root to the right of the figure 6 to get 64 as the
first divisor.

Multiply the trial divisor, 64, by 4, the second figure of
the root, and subtract the product, 256, from 290. The
remainder is 34. Bring down the next period to get 3425.

Multiply the first two figures of the root as found, 34,
by 2 and write the product, 68, to the left of the number
3425. Using the first three figures, 342, find how many times
68, the second trial divisor, is contained into 342. It is con-
tained 5 times. Write 5 as the third figure in the root. Also
write 5 to the right of the 68 to get 685 as the second divisor.
685 is contained into 3425 five times without a remainder.
The square root of the number 119,025, is, therefore, 345.

The process of finding the square root of a number
which is not a perfect square is similar to that just described.
The root of such a number must include a decimal, A
decimal point and ciphers must, therefore, be written to
the right of the number. We add as many pairs of ciphers
as there are to be decimal places in the root.

Example. Find V596.

V5'96.00'00'00 ( 24.413 Ans.

484

4881

48823

196
176

2000
1936

6400

4881

151900
146469

POWERS 127

Explanation. This is not a perfect square; that is to
say there is no number which, multipHed by itself, will give
596.

To extract the root we proceed to point off the number
into periods as before explained. Since the root is to con-
tain a decimal we must place a decimal point at the right
of the number and add as many pairs of ciphers as there
are to be decimal places in the root. We will carry the
result to three decimal places and, therefore, we will add
three pairs of ciphers to the right of the decimal point.
In all other respects the solution is similar to the example
given above.

113. Directions for Extracting Square Root. Begin at
the decimal point and separate the number into periods of
two figures each, going both to the right and to the left
of the decimal point. If there is no decimal point begin
with the figure farthest to the right. Place a mark between
these periods.

Find the largest whole number whose square will be con-
tained in the first left-hand period. Write this number as
the first figure of the root. Subtract the square of this
number from the first period and annex the second period
to the remainder thus found.

Multiply the part of the root already found by 2, and
write the result at the left of the first remainder. This
gives the first trial divisor. Divide that portion of the
dividend except the last figure by the trial divisor, and write
the result as the second figure of the root. Annex this
figure to the right of the trial divisor to get the first divisor.
Perform the division and obtain the remainder, which
together with the next period forms the next dividend.

Repeat the process just described above until the last
period of the number is used and to the last figure of the

128 ARITHMETIC FOR CARPENTERS AND BUILDERS

root obtained. If more decimal places are required in the
root add as many pairs of ciphers as there are decimal places
desired. If the last period of a decimal should contain but
one figure, add one cipher to complete the period.

If at any time the trial divisor will not be contained in
the dividend, write a cipher as the next figure of the root,
bringing down another period and proceed.

There should be as many decimal places in the root as
there are periods to the right of the decimal point in the
number of which the root is to be extracted.

114. Proof of Square Root. To prove a root multiply
it by itself to get the square. If the number is not a perfect
square, the proof will not give the exact square, but it will
be very close.

Example. Prove the root in the last example given
above.

24.413

73239

24413

7652

652

488

595.994569

Ans»

Explanation. Multiplying the root by itself to get the
square we get very nearly 596. Since this is not a perfect
square the proof will never be exact.

116. Cube Root. To find the cube root of a number
means to find a factor which when used three times will
give the number. Thus 2 is the cube root of 8 because

POWERS 129

2X2X2 = 8. The cube root of 27 is 3, of 64 is 4, of 125 is
5, etc.

To find the cube root of larger numbers we may perform
an operation similar to the one for square root. This is
somewhat more complicated than the method of finding the
square root. Instead of working it out we usually make
use of a table of cubes or cube roots or solve the problem
by means of logarithms. The carpenter has httle occasion
to find the cube root and for that reason the process is
omitted from these lessons.

116. Practical Applications. The practical applications
of the processes of involution and evolution which occur
in the daily work of the carpenter have to do mostly with
the properties of regular geometric figures.

117. The Side of a Square. If the area of a square
is given we can find the length of the side by extracting
the square root of the area. This is self-evident, because
we find the area by squaring the length of the side and to
find the side we would naturally expect to extract the square
root of the area.

Example. A builder has sufficient material to lay 1250
sq.ft. of floor. What is the length of the side of the largest
square that can be covered?

Vl2' 50 ( 35.3 ft. Ans.

65 350

325

703

2500
2109

Explanation. Extracting the square root of 1250 gives
35.3 approximately. This is the length in feet of the side
of the square which can be covered by the material.

130 ARITHMETIC FOR CARPENTERS AND BUILDERS

118. The Right Triangle. A triangle is a figure which

has three straight sides and three angles. A right triangle

is a triangle one of whose angles is a right

angle. Fig. 37 represents a right triangle.

The side marked a is called the altitude;

the side marked h is called the base and

the side marked c, which is the longest side,

is called the hj^otenuse. To find the

length of the hypotenuse when the length

of the altitude and base are known, we

may square the value of the length of the

base and the length of the altitude, add

A Right Triangle, tiiem together and extract the square root

of this sum.

Example. What is the length of the hypotenuse of a

right triangle whose base is 18" and whose altitude is 12" ?

Fig. 37.

144

1|2

324

144

V4'68'.00'00(21.63"

144
324

468

Arts.

426

4323

68
41

2700

2556
14400
12969

POWERS

131

Explanation. We square the value
get 324 and the value of the altitude, 12"
these two together gives 468 for their
root of 468 is 21.63. The hypotenuse
therefore, 21.63" long.

We may also find the length of the
angled triangle if we know the length
and base by turning this rule around.

of the base 18" to

to get 144. Adding

sum. The square

of the triangle is,

altitude of a right-
of the hypotenuse
The altitude of a

Fir.. 38.

right triangle is equal to the square root of the difference
of the squares of the hypotenuse and the base. We may
also say that the base is equal to the square root of the
difference of the squares of the hypotenuse and the alti-
tude.

Example. In Fig. 38 the length of the rafter from the
ridge center to the plate is 14.42 ft. and one-half the span
of the roof is 12 ft. What is the vertical distance from the
plate to the ridge?

132 ARITHMETIC FOR CARPENTERS AND BUILDERS

14.42

2884
5 768
57 68

24
12

144 2

= 208

144

207.9364=

208

144

V64 = 8'

•

Ans.

Explanation. We first find the square of the hypote-
nuse, which is 208, very nearly, and then the square of the
base, which is 144. The difference between them is 64.
The square root of this difference is 8. The required dis-
tance is, therefore, 8 ft.

119. The Circle. The processes of involution and evo-
lution are quite common in connection with calculations
of the circle. If we wish to find the number of square
inches in a circle, for example, we square the radius in
inches and multiply this result by ir or 3.1416. Also if we
wish to find the radius of a circle when the area is given
we divide the area by 3.1416 and extract the square root
of this quotient.

Example. Find the area of a circle in square inches, the
radius of which is 30".

302 = 900

900X3.1416 = 2827.44 sq.ins. Ans.

POWERS 133

Explanation. Multiplying 30 by itself to get its square
gives 900 and multiplying this by ir or 3.1416 gives 2827.44
sq.in. in the circle.

Example. What is the radius of a circle whose area is
1017.87 sq. in.?

3.1416)1017.8700(324

942 48

75 390
62 832

12 5580
12 5664

V3'24
1

28 224
224

(18"

Ans.

Explanation. This problem is just the reverse of the
one given above. Divide 1017.87 by tt to get 324, nearly.
Extract the square root of 324 to get 18", which is the
required radius.

If we work with the diameter instead of the radius we
use the value .7854 in place of 3.1416. The radius of a
circle is one-half the diameter; therefore the square of the
diameter will be four times the square of the radius. This
accounts for the use of the value .7854 instead of 3.1416
when using the diameter. .7854 = J of 3.1416.

120. Capacities of Circular Tanks and Cisterns. The
builder often finds it necessary to determine the number of
gallons or barrels that a cistern will hold. This information
may usually be found in tables, but it is convenient to kno\»'

134 ARITHMETIC FOR CARPENTERS AND BUILDERS

how to figure it so that if you do not have a table at hand
you can figure it out for yourself.

Find the area of the bottom of the tank in square feet
and multiply this by the height in feet. This product gives
the contents in cubic feet. There are 7^ gallons of water
to every cubic foot, so, in order to get the contents in gallons,
we multiply the total cubic feet by 7^. There are 31^
gallons to a barrel; therefore, to get the capacity in barrels
we divide the total number of gallons by 31^. Multipl5dng
by 7^ and dividing by 31^ is the same as multipljdng by
.238, because 7^ -i- 31^ = .238. Hence to find the capacity
in barrels we may multiply the contents in cubic feet by 238.

Example: Find the number of gallons a circular tank
will contain if the diameter is 6 ft. and the height is 10 ft.

62 = 36

.7854

144
180
2 88
25 2

28.2744

28.27
10

sq.ft.

282.70

282.7
7.5

cu.ft.

141 35
1978 9

21?0. 25 = 21201 gallons. Ans.

POWERS 135

Explanation. The square of the diameter 6 is 36 and
the product of this multiplied by .7854 gives 28.27 sq.ft.
in the bottom of the cistern. It is 10 ft. high and 10 times
28.27 gives 282.7 cu.ft. as the contents of the cistern. Mul-
tiplying this by 7.5 gives 2120j gallons as the capacity of
the cistern.

Example. How many barrels will this tank contain?

31.5) 2120.25 (67.3 barrels. Ans.

1890

2302
2205

975
945

Explanation. We have just figured the number of
gallons the tank will contain. There are 31 1 gallons to
the barrel. If we divide the number of gallons 2120.25
by 31.5 the result, 67.3, will give the answer in barrels.

Example. How many barrels will a cistern hold the
cubic contents of which is 282.7 cu.ft.?

282.7 cu.ft.
.238

2 2616

8 481

56 54

67.2826 = 67.3 barrels. Ans.

Explanation. This example is used to show that if we
multiply the cubic contents of a cistern expressed in cubic

136 ARITHMETIC FOR CARrENTERS AND BUILDERS

feet by .238 we will get the same result approximately as
to multiply by 7 5 and divide by 31 5. The value given here,
282.7 cu.ft., is the same as that found in the second example
above. Multiplying this by .238 gives 67.2826 barrels, which
is very nearly 67.3 barrels as found above.

POWERS 137

Summary of Chapter XI

64. To find the power of a number, use it as many times
as a factor as is indicated by its exponent: that is, to find
the second power use the number twice as a factor; to find
the third power use the number three times as a factor,
(Sec. 109.)

65. To raise a fraction to a given power, raise the nu-
merator to the required power and also the denominator.
(Sec. 110.)

66. To find the square root of a number, follow the
explicit directions given in Sec. 113.

67. To prove a square root, multiply the root by itself
to get the square. (Sec. 114.)

68. To find the side of a square when its area is given,
find the square root of the area. The result will be the
length of the side of the square. (Sec. 117.)

69. To find the length of the hypotenuse of a right
triangle when the base and the altitude are given, square
the base and the altitude and extract the square root of
their sum. (Sec. 118.)

70. To find the length of the altitude of a right triangle,
square both the hypotenuse and the base and extract the
square root of their difference. (Sec. 118.)

71. To find the length of the base of a right triangle,
square the hypotenuse and altitude and extract the square
root of their difference. (Sec. 118.)

72. To find the area of a circle, square the radius and
multiply by 3.1416 or square the diameter and multiply
by .7854. (Sec. 119.)

73. To find the diameter of a circle when the area is
given, divide the area by .7854 and extract the square root

138 ARITHMETIC FOR CARPENTERS AND BUILDERS

of the quotient. If the radius is wanted, divide by 3.1416
and extract the square root of the quotient. (Sec. 119.)

74. To determine the contents of a circular tank in
gallons, find the area of the bottom in square feet, multiply
by the height of the tank in feet and multiply this by
7i (Sec. 120.)

75. To find the capacity of a circular tank in barrels,
multiply the contents in cubic feet by .238. (Sec. 120.)

Fig. 39. — Roof for a Leau-to.

PROBLEMS

162. Find the value of the following:

(o) V178084 422

Ans

(b) V198.1369 14.070

Ans

163. Find the value of the following:

(a) 14» (c) 9^X3.1416

(6) 286* (rf) (f)^

POWERS

139

164. A pergon standing requires a space 20"X20". What are
the dimensions of a square platform large enough to hold 75 people,
allowing 1 ft. extra all around the edge?

165. What will be the length of a brace cut for 6 ft. run and
6 ft. rise?

166. Fig. 39 shows a roof for a lean-to. The run is 13 ft. and
the rise is 6 ft. Find the necessary length of a 2X4 rafter allowing
12" for overhang and 6" for trimming. What length of 2X4
would you order?

Fig. 40. — Scissors Roof Truss.

167. What will be the length of a rafter for 12 ft. run and 10
ft. rise allowing about 4" for trimming and 2 ft. for overhang?

168. Fig. 40 shows a scissors roof truss and the diagram to the
right gives the dimensions. What is the length of the main rafter
A and what is the length of the strut B?

169. Fig. 41 shows the stringer for a flight of stairs. What is
the length of the stringer?

170. Fig. 42 shows a queen-post truss and the diagram to the
right gives the dimensions on the center lines. Find the missing
dimensions which are indicated by letters.

171. A circular platform has an area of 240^ sq.ft. What is
its diameter?

172. What is the area in square inches of a circle two feet in
diameter?

140 ARITHMETIC FOR CARPENTERS AND BUILDERS

173. A circular window opening is 54" in diameter. How
many square feet of glass surface does it contain?

174. What is the capacity of a circular cistern 5 ft. in diameter
and 8 ft. deep? Give the answer in barrels.

Fig. 41. — Stringer for a Flight of Stairs.

175. A hot- water tank for an apartment house is said to hold
100 gallons. It is 22" in diameter and 60" high. Does it hold
100 gallons as stated? If not, what is its capacity?

Fig. 42. — Queen-post Truss.

176. A circular cistern is 5' 6" in diameter and 10' deep. How
large must a square cistern be built of the .same height to contain
the same amount of water? Why are cisterns usually built round
instead of square?

POWERS 141

177. A 40-gallon tank is 18" in diameter. How high must
it be?

178. A concrete silo is 16 ft. in diameter and 30 ft. high. What
is its capacity in cubic feet?

179. Silage weighs about 40 lbs. per cubic foot. How many
tons of silage will the above silo hold?

180. What must be the height of a circular silo to hold 10,000
cu.ft. if its diameter is 18 ft.?

181. What must be the diameter of a drain tile into which one
4" and one 6" tile empty? The large tile must have an area
equivalent to the two small ones.

CHAPTER XII

LUMBER TERMS AND DIMENSIONS. FLOORING.
SHINGLES.

121. Lumber Trade Customs. Our methods of grading
and measuring lumber are the result of custom and these
customs vary somewhat in different localities. The sizing
and grading of lumber is regulated by the various lumber
manufacturers' associations. Some of the terms and cus-
toms are of such wide usage that they may be explained here.

122. Ltmiber Terms. The term timber is generally
applied to sticks more than 4" in thickness the smaller way,
and the term liunber is applied to sticks less than 4" in
thickness. A plank is a piece from 1|" to 4" thick; while
a board is a piece less than 1^" in thickness. A scantling
is a piece of lumber 4"X6" in size or less. The term is
most frequently applied to a 2" X4" stick, though it may
refer to a 4"X4" or 2"X6" stick equally well.

Rough stock means lumber that is sawn so as to admit
of dressing but that is not dressed. Dressed stock means
lumber that has been planed or surfaced on one or more
sides. S. 1 S is an abbreviation for " surfaced on one side."
S. 2 S. means " surfaced on two sides," S. 1 S. 1 E. means
" surfaced on one side and one edge " and S. 4 S. means
" surfaced on four sides."

Dimension Lumber is usually dressed on one side and
one edge in order to size it. Finishing lumber is lumbei

142

LUMBER TERMS AND DIMENSIONS

143

intended for interior or exterior finishing. It is dressed
on one or both sides or as directed.

123. Allowance for Dressing. In planing or dressing a
board about h" is removed for each surface dressed. Lumber

2-Door8

Fig. 43. — Garage.

which has not been thoroughly dried and seasoned before
being dressed will shrink sufficiently to cause considerable
reduction in the width of a board. A board which is nom-
inally one inch in thickness which has been dressed on one
or both sides will measure actually about xi" in thickness.
If lumber is wanted to full dimensions after it is dressed,

144 ARITHMETIC FOR CARPENTERS AND BUILDERS

it is necessary to specify that the dimensions must be full
size after the stock is dressed. Dressed stock is measured
strip count; that is, at the full size of the rough material
used to manufacture the piece.

Fig. 44. — Floor Plan of a Small Cottage.

124. Width of Lumber. Boards run in multiples of one
inch in width. Owing to the processes used in manufacture
there may be considerable actual variation from the rule.
If a board is more than j" scant on the 8" width or of less
width it is counted in the next lower width. A board may
be I" scant on the 9" or 10" widths or 5" scant on the 11"

LUMBER TERMS AND DIMENSIONS

145

or 12" widths or wider. If the boards of any specified
nominal width do not meet these requirements they are
counted in the next lower width.

PLAN

END ELEVATION

Fig. 45.— Plan of Roof.

125. Standard Lengths. Rough stock or finish lumber
is usually cut in multiples of two feet in length. Standard
lengths for finishing lumber are usually in multiples of one
foot.

126. Sizes of Common Boards. Boards of 1" nominal

146 ARITHMETIC FOR CARPENTERS AND BUILDERS

size S. 1 S. or S. 2 S. as previously explained, are dressed to
\l" thick; ly nominal size S. 1 S. or S. 2 S. are dressed
to liV"; H" nominal size S. 1 S, or S. 2 S. are dressed to

127. Sizes of Dimension Lumber. The sizes of dressed
scantling and dimension lumber are given in the following
table:

li^zSShipUp
I'x «-3'o.c.

z 1* braces

iSfUlong

SECTION O^
1 Vt"/. S'SHIP LAP

ELEVATION

Fig. 46. — Forms for Concrete.

TABLE XIII
SIZES OF DIMENSION LUMBBB

Nominal Size.

Drcssod.

Actual Size.

2X4

2X6

2X8

2X10

2X12

S. 1 S. 1 E.
S. 1 S. 1 E.
S. 1 S. 1 E.
S. 1 S. 1 E.
S. 1 S. 1 E.

if"x.3r

li"X5|"

ii"x7r'

1|"X9|"

irxiu"

If the lumber is dressed on all four sides, the pieces may
be I" less in thickness and width than S. 1 S. 1 E.

128. Sizes of Dressed Finishing Ltunber. Finish lum-
ber is dressed to the same thickness that common boards

LUMBER TERMS AND DIMENSIONS

147

are. Thus, 1" nominal size S. 1 S. or S. 2 S. is dressed to
if" in thickness; U" S. 1 S. or S. 2 S. is dressed to 1^" in
thickness; U" S. 1 S. or S. 2 S. is dressed to 1 j^" in thick-
ness. Stock which is nominally 4" in width is dressed to
3^" when finished. Other widths are also dressed to Y'
less than nominal widths.

129. Flooring. To allow for the tongue and groove
matching 1"X3", 1"X4" and 1"X6"
thick and show 2^", 3i" and 5i" face.

flooring will run jf"

'Ceiling line

Drawers

^Floor line
ELEVATION

SECTION

Fig. 47. — Cupboard.

130. Sizes of Framing Lumber. Joists are reduced j"
on the side and |" on the edge if surfaced on one side and
one edge when green and unseasoned. If they are surfaced
on four sides, they are reduced |" for each side surfaced.
When rough and green, they should not be more than
j" scant in width or thickness. Joists may be had in
the following nominal sizes: 2", 2^" and 3"X10", 12"
and 14".

131. Measurement of Lumber. The unit of measure-
ment of lumber is the board foot. The board foot is the

148 ARITHMETIC FOR CARPENTERS AND BUILDERS

measure of a board an inch thick and a foot square. Boards
less than one inch thick are counted the same as though
they were one inch thick. If the boards are more than an
inch in thickness they are counted at their nominal size.
Finish and dimension lumber is counted at the size neces-
sary to manufacture the piece as previously explained.

For boards one inch or less in thickness and of various
widths the following table gives a convenient method of
figuring board feet.

TABLE XIV

BOARD FEET IN LUMBER OF DIFFERENT WIDTHS

Boards 3" wide will contain i as many board feet as they are ft. long.

4" J

6" J

9" f

12" wide contain as many board ft. as they are ft. long.
15" wide will contain 1 J as many board ft. as they are ft. long.
16" U

In general the following rule may be used for finding
the board measure of any board, plank or timber. Multi-
ply the width in feet by the length in feet by the thickness in
inches. If the width is given in inches, divide by 12 to gel
it in feet.

Example. How many board feet are there in a scant-
ling 2" X4"X 16 ft.?

2><|><i^ = ^ = lof f t . B.M. Ans.

Explanation. To change the width to feet we must
divide the 4" by 12 but this may be done by expressing the
operation and canceling. Thus 4 divided by 12 is \ and
the result is ^ or 10| ft. B.M.

LUMBER TERMS AND DIMENSIONS

149

132. Counting Lumber. In the yard, at the car or on
the dock, the tallyman is expected to count the lumber. His
record is kept in a small book called the tally book. The

J2
O

tally is recorded in different ways, depending upon the kind
of material. Framing lumber and other lumber of specified
lengths is generally tallied by the piece. Shiplap, sheathing,

150 ARITHMETIC FOR CARPENTERS AND BUILDERS

ceiling and flooring is generally tied in bundles containing
a certain number of feet B.M. and tallied by bundles.

133. Flooring. In figuring quantities of lumber required
to cover surfaces, we must allow for matching and waste.
A 1"X4" flooring board will show only 3j" face. We must,
therefore, allow sufficient quantity to make up for this
difference. It is necessary to add about 20% in board
measure to the surface measure for this purpose and from
5% to 10% for waste.

Example. How many board feet of 1"X4" flooring will
be required to lay a floor 28 ft. wide by 62 ft. long, allowing
25% for matching and waste?

62'
28'

496
124

1736 sq.ft.
.25 %

86 80
347 2

434.00 Add.
1736

2170 ft. B.M. Ans.

Explanation. The number of square feet floor surface
is found by multiplying the length 62 ft. by the width 28 ft.
to get 1736 sq.ft. 25% of this is 434 sq.ft. which when
added to the actual area gives 2170 ft. B.M. of flooring
required.

LUMBER TERMS AND DIMENSIONS

151

134. Shingles. Shingles are figured by the thousand.
A thousand common shingles are equal to 1000 shingles 4"
wide. They are put up in bundles of different quantities
according to the thickness of the butts. Shingles 18" long
having 5 butts to 2^' are packed five bundles to the thou-
sand. Shingles 16" and 18" having 5 butts to 2" are packed
four bundles to the thousand.

-58}i-

A-

-40— — *^5-

2 X 8' 4^

88"

ELEVATION

Fig. 49.— Work Bench.

-18' 'I I

END VIEW

1 X 10 i

The common unit for measuring roof surface is the
square. A square is a surface containing 100 sq.ft., or
it is equivalent to a surface 10 feet square.

Shingles are laid so that from 4" to 5" of the butts are
exposed to the weather. The table given below indicates
the number of shingles required to cover a square of roof
surface when laid with various exposures allowing some
waste.

152 ARITHMETIC FOR CARPENTERS AND BUILDERS

TABLE XV
AREA COVERED BY SHINGLES*

Laid to the weather.

1000 will cover square feet.

No. shingles to a square.

44"

41"
5"

100
110
120
133

1000
910
833
752

* Kidder's Pocket Book.

Example: How many shingles laid 4^" to the weather
will be required to cover a roof 12 ft. by 24 ft.? How many
bundles if the shingles are 5 butts to 2"?

24'
12'

48
24

288 sq.ft. = 2.88 squares.

2.88
833

8 64
86 4
2304

2399.04 = 2.4 thousand shingles.

2.4

9 . 6 or 10 bundles of shingles. Ana.

LUMBER TERMS AND DIMENSIONS

153

Explanation. From the above table we find that when
shingles are laid 4|" to the weather 833 are required per
square. We must first find the number of squares of roof
surface to be covered. There are 288 sq.ft. or 2.88 squares.
At 833 shingles per square, it would require about 2400
shingles. Counting 4 bundles to the thousand, it would
require 10 bundles to cover the roof.

Fig. 50.— Roof.

135. Material Lists. The builder is frequently called
upon to make out material lists or lumber bills for various
structures. In doing this work it is imperative that it
should be done in an orderly manner. The different kinds
of material should be grouped and their use in the structure
designated. A reference mark on the drawing and a cor-
responding mark on the computations will often serve to
connect the two. The results should be recorded in such
a way that they may be readily understood at anv time.

154 ARITHMETIC FOR CARPENTERS AND BUILDERS

If the habit of neat and orderly work in making out lumber
bills and purchasing lists is formed early it will be of great
value during the entire business career of the builder. The
following example will show a convenient method of making
out lumber bills.

Example: Make out the lumber bill for the private
garage shown in Fig. 43.

A =

■■ Ipc 4X4X6' S. IS. IE.

•

Ipc 4X4X12' **

Sill

2 pes 4X4X18' "

B =

30 pes 2X4X10' S. IS. IE.

Posts

C =

= 2 pes 2X4X12' S. IS. IE.

1 Plate

2 pes 2X4X18' "

D =

= 28 pes 2X4X10' S. 1 S. 1 E.

Rafters

E =

= 1 pc 2X6X22' S. 2 S.

Ridge

F =

= 3 pes 2X4X14' S. IS. IE.

Belt

G =

= 2 pes 2X4X10' S. IS. IE.

Headers

H =

= 375 ft. B.M. 1X6 S. IS.

Sheathing

1 =

= 14 bundles shingles

J =

= 700 ft. B.M. f"X5i" drop siding

K =

= 4 pes 2X6X10' S. 1 S. 1 E.
2 pes 2X6X12

Stiles and rails

Explanation. For most of the material in this list
it is only necessary to take the quantities off of the plan,
adjusting the lengths to the proper stock sizes. Some
judgment must be exercised in making this adjustment,
but one soon becomes accustomed to the work and can tell
at a glance the proper lengths to specify. Since so much
does depend upon judgment one man's list will not be
exactly like another's on the same structure. The sheathing
is figured as previously explained, allowing 10% for waste

LUMBER TERMS AND DIMENSIONS

155

and adding sufficient material for the ridge board. Twenty
per cent is allowed for lapping and waste in figuring the
siding. It is always better to allow for liberal quantities
rather than to run short of material on the job.

This is not a purchasing list. In sending the list to

Fig. 51. — Garage,

the lumber dealer the items should be arranged so that
all the material of a given kind and length is grouped
together. This material list should be carefully preserved
and placed in the hands of the foreman when the material
is delivered on the job so that he will know without ques-
tion just where every stick belongs in the structure.

156 ARITHMETIC FOR CARPENTERS AND BUILDERS

;:0.8-

LUMBER TERMS AND DIMENSIONS 157

Summary of Chapter XII

76. To find the number of feet board measure in a
piece of lumber, multiply the width in feet by the length
in feet by the thickness in inches. (Sec. 131.)

77. To find the number of feet board measure of fioor-
mg to cover a given floor, find the area of the floor in square
feet and add 25% to 30% for matching and waste. (Sec.
133.)

78. To find the mmiber of shingles required to cover
a given roof surface, find the size of the surface in squares
and multiply by the number of shingles required per
square as given in the table. Multiply the number of
thousand of shingles thus obtained by 4 or 5 according
to the number of bundles to the thousand. (Sec, 134.)

PROBLEMS

182. How many board feet in a piece of lumber 1"X8", 14'
long?

183. Six pieces of finish lumber l5"X8", 12' long are required
for a certain job. How many board feet are there in the lot?

184. A contractor orders the following bill of dimension stuff
for concrete forms. How many feet B.M. are there in each lot?

24 pes. 4X4 S. 1 S. 1 E. 16' long
180 pes. 2X4 S. 1 S. 1 E. 10' long
200 pes. UXS S. 1 S. 1 E. 12' long.

185. How many board feet are there in 200 pieces 2"X14"
joists 16' long?

186. How many board feet of sheathing 1"X6" are required
to cover the side of a barn 12' high and 20' long, allowing 10%
for waste?

187. How many feet B.M. of sheathing 1"X6" are required
to sheath a roof surface 16' wide and 32' long, allowing a 2" space
between boards and 10% for waste?

158 ARITHMETIC FOB CAKPENTERS AND BUILDERS

l.A.

LUMBER TERMS AND DIMENSIONS 159

188. A man wishes to build a sidewalk 6' wide and 150' long.
The planks are 1§"X8" S. 1 S. 1 E. He iJUts in 3 sleepers 4"X4"
under the planks for nailing. How many board feet must he order
of each lot, allowing 10% waste?

189. How many board feet of 1"X3" tongue-and-groove floor-
ing will be required to lay a floor 14'X16', allowing for face measure
and 5% for waste?

190. Fig. 44 shows the first-floor plan of a small cottage. How
many feet B.M. of l"x4" flooring will be required, including the
porches, allowing 5% for waste?

191. Fig. 45 shows the plan of an ordinary ridge roof. How
many feet B.M. will be required to sheath the roof using 1"X6"
sheathing spaced 2" apart?

192. Fig. 46 shows the forms for a concrete job in which 1 j"X8"
shiplap is used. How many feet B.M. of shiplap will be required?

193. How much material will be required to make the cupboard
shown in Fig. 47?

194. Figure the lumber required to make the kitchen cabinet
shown in Fig. 48. Give the number of pieces and the B.M.

195. How much material will be required to make the work-
bench shown in Fig. 49. Give the number of pieces and the B.M.

196. How many thousand shingles will be required for the roof
shown in Fig. 50, if they are laid 4j" to the weather?

197. How many bundles of shingles will be required for the
garage shown in Fig. 51, if they are laid 4^" to the weather?

198. Make out the complete lumber bill in pieces for the shed
shown in Fig. 52. Give the number of pieces of each different
length.

199. Make out the complete lumber bill for the barn shown in
Fig. 53.

CHAPTER XIII

BUILDERS' GEOMETRY. GEOMETRIC CONSTRUC-
TIONS. ANGLES AND ANGULAR MEASURE.
ELLIPSE. POLYGONS.

136. Use of Geometry. The carpenter and wood-
worker has to deal largely with points, straight and curved
lines, plane and curved surfaces and various kinds of
solids. Geometry is a study of the properties, construction
and measurement of lines, surfaces and solids. A thorough
knowledge of geometry is very useful.

137. Geometric Points and Lines. A point has no
dimensions; it merely has position. In marking points
with a pencil or scriber, we must give them some size in
order to see them, but theoretically they are only imaginary.

A line has only one dimension, that of length. A
line drawn with a pencil or chalk has some width, but in
theory it should have no width. Lines may be straight
or curved. A straight line is one that does not change its
direction. A curved line is one which changes its direction
at every point. A broken line consists of a series of straight
lines variously directed and joined together.

138. Geometric Surfaces and Solids. A surface has
two dimensions — length and breadth. A plane surface is
one which will wholly contain a straight line no matter in
which direction the line is laid in the plane. A curved
surface is one which changes its direction in accordance

160

BUILDERS' GEOMETRY 161

with a given law. The surface of a cyhnder or sphere is a
curved surface.

The definition of a plane surface gives a practical
method of testing for a true plane. If a surface has a
warp or wind, it may easily be detected by laying a straight-
edge on the surface in different positions and sighting for
unevenness. If the plane is true, the straightedge will
lie wholly in the surface in whatever position it may be
placed. This method of testing is illustrated in Fig. 54,

Fig. 54. — Testing a Surface.

A solid is space completely surrounded by surfaces.
A solid has length, breadth, and thickness.

139. Geometric Angles. An angle is formed by two
straight lines which meet at a point. The point is called
the vertex of the angle. A right angle is one in which the
two intersecting lines are perpendicular to each other.
The angle between the two edges of a steel square is a
right angle. An acute angle is one in which the lines
make less than a right angle. An obtuse angle is one in
which the lines make more than a right angle.

An angle is designated by letters placed at the point
or vertex and on the sides or legs. The letter at the vertex
is always written between the letters representing the sides.

162 ARITHMETIC FOR CARPENTERS AND BUILDERS

In Fig. 55 the angle DAB is a right angle, as is also
the angle BAC, for the reason that the line AB is -per-:
pendicular to the line DC at the point A. The angles
EOF and GOH are both acute angles because each is less
than a right angle. The angles EOG and FOH are both
obtuse angles because each is greater than a right angle.

140. Circular or Angular Measure. In Chapter X the
circle was studied in part and some of its properties were
investigated. In this chapter additional definitions and
properties will be considered. The arc of a circle is a

Riffht
Angle

Right
Angle

Fig. 55.

part of its circumference. The part of the circumference
between the points A and C in Fig. 56 is called the arc ABC.
A chord is a straight line connecting two points on the
circumference of a circle. The straight line joining the
points A and C in Fig. 56 is spoken of as the chord ADC.

Circular arcs are measured in degrees. In a complete
circle there are 360 degrees. In a fourth of a circum-
ference or quadrant, there are 90 degrees of arc. An angle
at the center is measured by the number of degrees of
arc intercepted between the sides of the angle; the num-
ber of degrees in the arc gives the number of degrees in
the angle. For precise measurements degrees are divided
into minutes and minutes are again divided into seconds.

BUILDERS' GEOMETRY 163

There are 60 seconds in one minute and 60 minutes in
one degree. Degrees, minutes, and seconds are designated
by the symbols °, ', ", respectively. Thus 34 degrees,
42 minutes and 30 seconds are written 34° 42' 30''.

141. Sectors and Segments. The area included be-
tween two radii and the arc is called a sector. In Fig.
57 the area AOBC is a sector of the circle. A segment
is that part of a circle which is included between an arc

Fig. 56.— Chord and Arc. Fig. 57. — Segment.

and its chord. Thus in Fig. 57 the shaded portion ACBD
is a segment.

142. To Bisect a Line and to Erect a Perpendicular.

To bisect a line means to divide it into two equal parts.
Ordinarily it would only be necessary to measure the
total length of the Une, divide this measure by two and
lay off the half distance on the Hne. By the methods of
geometry we not only can bisect a line without actually
measuring it, but we can also erect a perpendicular line,
or a line making a right angle with the first, at the middle
point.

164 ARITHMETIC FOR CARPENTERS AND BUILDERS

Let AC, Fig. 58, be the line of which we wish to find
the middle point. Using the ordinary carpenter's compass,

set the points a Uttle
more than half the length
of the line AC apart. With
this distance as a radius
and the points A and C as
centers, strike two arcs so
that they cross above and
below the line ylC at D
and E. Through these
points draw the line DE.
The line DE will bisect
AC at the point F and is
also perpendicular to AC.

Fig. 68. — Perpendicular Bisector.

DE is called the perpendicular bisector of AC

143. To Bisect an Angle. Let AOB in Fig. 59 be an
angle which is to be divided into two equal angles. With

Fig. 59. — Bisecting an Angle.

O as a center and any convenient distance as a radius,
draw an arc which cuts the legs of the angle at C and D.

BUILDERS' GEOMETRY

165

Then, with C and D as centers, and any convenient radius
strike two arcs. These will intersect at the point E. Join
0 and E. The line OE divides the angle into two equal
parts because it is everwhere equally distant from the
lines OA and OB.

144. To Erect a Perpendicular at any Point on a Line.
We have already learned how to erect a perpendicular
bisector to any line, but suppose we wish to erect a per-
pendicular to a line at a given point on the line. Let
DB, Fig. 60, be the line on which we desire to erect a per-

FiG. 60. — Erecting a Perpendicular.

pendicular at the point C. From any convenient point
above the line, such as 0, strike an arc passing through
C, the given point. It will also cut DB at another point
F. From F draw a line through the center 0 and extend
it to cut the arc at A. Join A and C. This line AC will
be perpendicular to DB.

145. To Construct a Right Angle. The following
method, known as the " 3, 4, 5 " method, is frequently used
to lay out a right angle.

On the line AC in Fig. 61 measure three units, say
three feet, from A to E. With A as a center strike an arc
four units long. With E as a, center strike an arc five units

IQQ ARITHMETIC FOR CARPENTERS AND BUILDERS

long. They will intersect at D. Join D and A. AD
will make a right angle with AC. In other words the line
AB, which is a continuation of AD, will be square with AC.
The reason for the above is that a right triangle is
formed. We have learned that in a right triangle the
square of the hypotenuse is equal to the sum of the squares
of the other two sides. If this is true, then 3^+4^ must be
equal to 5^, 3^+42 = 25. Also, 5^ equals 25, which estab-

Fin. 61. — Constructing a Right Angle.

lishes the truth of the proposition. Notice that any
multiple of the units 3, 4, and 5 may be used, such as 9,
12, and 15, or 12, 16, and 20.

In laying out foundations, batter boards are set up
and lines estabUshed as shown in Fig. 62. The accuracy
of the right angle at the corner may be tested by using
the " 3, 4, 5 " method. A cloth measuring-tape is held
with the zero mark at A, the intersection of the lines.
Measure out 9 ft. to the point E and let out 27 ft. of tape.
Grasp the tape at a point 15 ft. from E and secure the
36 ft. mark at A. Draw the tape taut and a right angle

BUILDERS' GEOMETRY

167

will be formed the sides of which are 9', 12', and 15' long.
The lines may then be estabUshed square with each other.

Fig. 62. — Laying out Foundations.

146. To Construct Various Angles. A right angle may
be constructed as directed above or it may be laid out
by means of the steel square. An angle of 45° may be
obtained by bisecting a right angle. It may also be con-
structed by drawing a diagonal of a

square. To lay out a 60° angle,

draw a horizontal line, as AB in Fig.

63. With AB as a, radius and A and

B as centers, scribe arcs intersecting

at C. Draw the line AC and the

angle BAC will be 60°. If BC is

drawn, each of the angles will be ^i^ b

60°. A 30° angle may be found by fig. 63.— A 60° Angle.

bisecting a 60° angle.

147. To Find the Center of an Arc or Circle. The
carpenter often finds practical cases in which it is necessary

168 ARITHMETIC FOR CARPENTERS AND BUILDERS

to find the radius of a circle when the segment or arc is
given. In Fig. 64 it is desired to find the radius of the
circle of which the arc is a part. From any point on the
arc, such as A, draw the two chords interescting the arc
in any two points as B and C. With the points A, B, and
C as centers and any convenient radius, draw the short
intersecting arcs dd', ee', ff, and gg'. Through the points

Fig. 64. — Finding the Radius of a Circle

in which these arcs intersect, draw the lines MO and NO.
These lines will be the perpendicular bisectors of the chords
AC and AB and they will intersect each other at the
point 0, which is the center of the required circle. The
distance from 0 to the given arc is the required radius.

148. The Ellipse. The practical man sometimes has
occasion to use the ellipse in laying out work. Such work
is most usual in the construction of concrete forms for

BUILDERS' GEOMETRY

169

Fig. 65. — Constructing an Ellipse.

sewers and arches. There are several methods of construct-
ing an ellipse, but the most convenient is by means of the
trammel.

To construct an ellipse by the trammel method, having
given the length and width of the ellipse, lay out the long
and short axes at right angles to each other as shown in
Fig. 65. The short axis
CD is'called the minor axis d

and the long axis AB is
the major axis. With a
straightedge or the tram-
mel, using three trammel
points, lay off the distance
from 0 to D, or one-half
the minor axis between
the points a and b on the
straightedge. From a, again
lay off one-half the major axis or the distance from 0 to
B. This will locate the point c on the straightedge. Now
keep the point h on the major axis and the point c on the
minor axis and locate several positions of the point a by
making a mark in the wood or paper as shown in the figure.
Join the points thus found with a smooth curve and the
result will be an ellipse. The points should be located
fairly close together in order to make the curve a smooth one.

149.* Polygons. A polygon is a plane figure bounded
by any number of straight lines. The side of a polygon
is any one of these lines. The point where two lines meet
is called a vertex. A polygon is designated by reading the
letters at the vertices. A triangle is a polygon which has
three sides. A quadrilateral is a polygon which has four
sides. A parallelogram is a quadrilateral whose opposite
sides are parallel to each other. Fig. 66 shows a parallel-

170 ARITHMETIC FOR CAHPENTERS AND BUILDERS

ogram. When all the angles are right angles and the
sides are all equal the figure is a square. When all of the

angles are right angles but
when the opposite sides
only are equal to each
other, the figure is a rect-
angle. A hexagon is a
polygon having six equal
sides and six equal angles.
An octagon is a figure
having eight equal sides and eight equal angles.

160. To Construct a Hexagon. To construct a hexagon
in a given square, proceed as follows: First, find the center
of the square by drawing a diagonal from comer to coiner
as shown in Fig. 67. Then inscribe a circle within the square

D C

Fig. 66. — ^A Parallelogram.

Fig. 67.
Constructing a Hexagon.

C d

Fig. 68. '
Constructing an Octagon.

SO that the circumference of the circle just touches the sides
of the square. With the center of one side of the square,
the point A for example, as a starting point, and with the
length of the radius of the circle as a measure, step off
around the circle. This should divide the circle exactly

BUILDERS' GEOMETRY 171

into six equal parts. Now join these points just found with
a straight hne to form the hexagon ABCDEF.

151. To Construct an Octagon. The octagon or eight-
sided figure may be constructed as shown in Fig. 68. Find
the center of the square as explained above. Then with
the distance from 0 to J. as a radius and with the points
A, B, C, and D used one after the other as centers, draw
arcs to cut the edge of the square in the points a, a', b,
b', etc. Join these points to form the octagon.

172 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter XIII

79. Geometry is a study of the properties of lines, sur-
faces, and solids. (Sec. 136).

80. A point is a geometric representation which has
position only. (Sec. 137).

81. A line is a geometric representation which has
direction and length. Lines may be straight or curved.
(Sec. 137).

82. A surface is a geometric representation which
has length and breadth, but no thickness. Surfaces may
be either plane or curved. (Sec. 138).

83. A solid is a geometric representation which has
length, breadth, and thickness. (Sec. 138).

PROBLEMS

Note. To lay out the construction asked for in the following prob-
lems, use a small steel square or wooden or celluloid triangle and a
compass or dividers. A cheap compass which may be attached to
a lead pencil may be obtained at any book store. Use a fairly hard
lead pencil so that your work will not smear and get dirty.

200. Construct a square exactly 4" on a side and square the
corners by the method given in Section 144.

201. Two lines are stretched for the corner of a concrete
foundation. Show how you can test them by the "3, 4, 5," method.
Make a diagram.

202. Construct angles of 60°, 45°, and 30°.

203. Construct the largest hexagon you can draw in a 4"
square.

204. Construct the largest octagon you can draw in a 4"
square.

205. Construct a hexagon each side of which is 2|" long.

Hint. The radius of the circle in which it is inscribed must be
2i".

BUILDERS' GEOMETRY 173

206. Draw an ellipse by the trammel method, using a straight
piece of paper as a straightedge. Make the eUipse 3" wide and
5" long.

207. Lay out a circle 4' in diameter in six segments so that it
may be built from 8" boards. Make your drawing to a scale
of l|" = r. How long must the pieces be cut to make each
segment?

208. Draw a circle 4" in diameter and lay out in it a sector
of which the angle at the center is 60°.

209. A window is to be constructed in the form of a half
circle or semicircle with a radiuo of 1| ft. The glass is to set in
lead. At the center is a small semicircle with a 3" radius and
the rest of the glass area consists of six sectors. Make a drawing
showing the size and shape of all the pieces in the window.

CHAPTER XIV

AREAS OF SURFACES AND VOLUMES OF SOLIDS.
AREA OF TRIANGLE, PYRAMID, CYLINDER AND
CONE. VOLUME OF CYLINDER, PRISM, PYRAMID
AND CONE. MEASURING SURFACES AND VOLUMES

We have already learned how to find the area of the
square and rectangle and also how to find the length of
the hypotenuse of a right-angled triangle. The student
should review these subjects carefully before taking up this
chapter.

152. Area of a Triangle. Fig. 69
shows how two triangles are formed
by drawing a diagonal line from one
corner to the other of a rectangle.
You can readily see that the area of
each of these triangles will be one-half
of the area of the rectangle. If the
letter S stands for the area of the
rectangle, L for its length, and W for
its width; then.

y_.

Fig. 69.

S = LXW.

The area of each triangle is one-half the area of the
rectangle. Let the letter s stand for the area of the tri-
angle, b for its j^ase and a for its altitude; then

s = iS = ^bXa.

174

AREAS OF SURFACES 175

If you look at the rectangle you will now see that the
length of the rectangle is the same as the altitude of the
triangle and that the width of the rectangle is equal to the
base of the triangle. We may now formulate the rule that
the area of a triangle is equal to one-half the product of the
base hy the altitude. Please note that, as in the case of finding
the area of the square and of the rectangle, the altitude
and the base must he expressed in the same units.

Example. Find the area of one triangle in Fig. 69 if
7. = 30"andTf = 20".

20"
30"

2)600 sq.in.

300 sq.in. area of triangle. Ans.

Explanation. The area of a rectangle 30" in length and
20" in width equals 20" X 30'-' = 600 sq.in. The area of
the triangle formed on these sides is one-half the area of
the corresponding rectangle. Hence, we divide this product
by 2 and get 300 sq.in as the area of the triangle.

153. Area of a Pyramid. A sohd whose base is a poly-
gon and whose sides are all triangles which meet in a com-
mon point above the base is called a pyramid. The common
point in which the sides meet is called the vertex or the
apex. When the vertex is perpendicularly above the center
of the base the pyramid is a right pyramid. If the upper
portion of a pyramid is removed, as by a saw cut parallel
to the base, the portion that remains is called the frustimi
of the pyramid. Pyramids are named according to the
number of sides there are in the base. If the base is a
square, it is called a square pyramid. If the base is a
hexagon, it is called a hexagonal pyramid. Various forms

176 ARITHMETIC FOR CARPENTERS AND BUILDERS

of the pyramid are often found in roofs, walls, bins, hoppers,
etc. Fig. 70 is an example of the hexagonal pyramid.

Since the sides of a pyramid are triangles, we can use
the information given above to find the surface area of a

Fig. 70.— Hexagonal Pyramid. Fig. 71.— Square Pyramidal Rpof.

pyramid. An example applying to a pyramidal roof is given
below.

Example. Find the number of feet B.M. necessary to
cover the square pyramidal roof shown in Fig. 71.

72 = 49
72 = 49

9.9'
• T

AREAS OF SURFACES
V^ = 9.9' = length of a.

177

69 . 3 sq.ft. in one side
4 sides

277.2 sq.ft. in roof
20% for waste

55.440

277.2
55.4

332.6 sq.ft. or say 335 ft. B.M. of lumber required.

Ans.

Explanation. Since the surface we are measuring is
inclined, the altitude of the triangle the area of which we
require is the line a. This length is found by solving for
the hypotenuse of a triangle
when the base and the alti-
tude are given. The altitude
in this case is the line EO = 7'
and the base is AE = \^- = 7'.
Solving the triangle we get
9.9' as the length of the line
a. When we lay the side of
the roof out flat, it appears as
shown in Fig. 72. The alti-
tude is 9.9' and the base is 14'. If we divide 14 by 2 before
multiplying it will amount to the same thing as dividing
the product by 2. Multiplying 9.9' by 7' gives 69.3 sq.ft.
for the area of the triangle. Since there are four sides, we

Fig. 72.

178 ARITHMETIC FOR CARPENTERS AND BUILDERS

multiply by 4 to get the total area of the roof. Allowing
20% for waste gives 332.6 sq.ft or say 335 ft. B.M. of lum-
ber required.

154. Area of a Cylindrical Surface. A cylinder is a
solid whose ends or bases are two parallel circles and the
side of which is a curved surface. A cylinder may be

formed by revolving a rectangle
about one of its sides as an
axis, in which event the cylinder
is a right cylinder. Fig. 73
shows a right cylinder. We have
many familiar examples of the
use of cylinders in practical
building construction.

If we should roll a cylinder
out on a flat surface as shown
in Fig. 74, it would form a
rectangle the width of which
would equal the height of the
cylinder and the length of which
would be equal to the circum-
ference of the base of the cylin-
der. Therefore, to find the area of the surface cf a cylinder,
find the circumference of the base and multiply by the height
of the cylinder. To find the total area add to this the
area of the two bases.

We have already learned that the circumference of a
circle is equal to t times the diameter and that the area of
a circle is equal to the radius squared times tt. Let h equal
the height of the cylinder and r equal the radius. Then
the diameter d equals 2r. If s represents the area of the
surface; then

s=2irrh=irdh/

Fig. 73.— Cylinder.

AREAS OF SURFACES 179

If S represents the total area, then,

S = 2'irrh-\-2irr^ = Tdh+^Td^.

Example. Find the number of feet B.M. required to
build the walls of a wooden stave silo 12' in diameter and
30' high. Plank 2" thick are to be used. Add 20% for
matching and cutting.

3.1416

12' d = 2r

6 2832
31 416

37.6992 ft.

circumference.

37.7'

30' high

1131.0 sq.ft

. in surface.

1131

2 for 2"

plank.

2262

20% for matching and

cutting.

452.40

2262

452

2714 ft. B.M. of lumber required. Ans.

Explanation. Multiply the diameter by x to get the
circumference. This times the. height in feet will give the
number of square feet in the surface. If 2" plank are used

180 ARITHMETIC FOR CARPENTERS AND BUILDERS

and 20% allowed for matching and cutting, the total number
of feet B.M. required will be about 2700.

155. Area of the Stirface of a Cone. A cone is a solid
the base of which is a circle and the side of which is curved
surface which closes in to a point called the vertex. The
distance from the vertex to a point on the circumference
of the base is called the slant height. A cone may be formed
by revolving a right triangle about its altitude as an axis.

Fig. 74.

When the upper portion of a cone is cut away as with a
saw cut, the part that remains is called a frusttun of a cone.
Fig. 75 shows a cone and a frustum. This surface is found
in practical construction in the roofs of round buildings
and towers.

If we should roll a cone out upon a flat surface as we
did for the cylinder, we would get a part or sector of a circle.
This is shown in Fig. 76. The length of the arc AB, Fig. 76,
is equal to the circumference of the base of the cone. The
radius of the sector is equal to the length of the slant height
or the line a on the cone. This length may be found by

AREAS OF SURFACES

181

solving for the hypotenuse of the triangle OEK, Fig. 75.
The area of this sector is equivalent to the area of a triangle
which has a base equal to the length of the arc AB and an
altitude equal to the slant height of the cone. If we let s
equal the lateral area of the cone, r the radius of the base
and a the slant height, then

s = §-2-7rra=7rra.

Fig. 75. — Cone and Frustum.

Just as in the triangle it is equal to one-half the circum-
ference of the base times the slant height. The circum-
ference of the base is equal to 27rr and canceling the 2 with
I the formula reads irra. In other words this is: The lateral
area of a cone is equal to the product of ir by the radius of the
base and by the slant height. If the total area S is wanted,
it is equal to S=irra-\-jrr^, and is found by adding the area
of the base circle to the lateral area.

182 ARITHMETIC FOR CARPENTERS AND BUILDERS

Example. How many shingles laid i^'' will be required
to cover the roof of the silo shown in Fig. 77?

72-1-72 = 98

V98 = 9.9' slant height the roof.
3.1416

21.9912= 22'
9.9

198
198

217.8 sq.ft. in roof.
217.8

10% for waste

21 add for waste
217.8

238.8 sq.ft. or say 2§ squares.
833
2i

416
1666

2082 shingles required.

4 bundles to the thousand

8.328, say 9 bundles of shingles required. Ans.

Explanation. Solve the triangle to find the slant height
of the roof. Find the product of w times the radius of the
base and times the slant height. Add 10% for waste and
count 4 bundles of shingles to the thousand.

AREAS OF SURFACES

183

156. Volume. We have already learned something
about cubic measure and how to find the volume of a cube.
It is sometimes necessary to know how to find the cubic
contents of bins and other structures, having various shapes.
It will usually be possible to separate an irregularly formed
structure into parts which are, or very nearly are, classed
under one of the regular forms. We should, therefore, know
how to figure the volume of the cylinder, the pyramid and
its frustum and the cone and its frustum.

157. Volume of a Cylinder. We have already learned
that the volume of a cylinder is equal to the product of the
area of the base by the height of the cylinder. We can now
express this as a formula. Let V equal the volume of the
cylinder, r the radius of the base and h the height of the
cylinder; then

Example. How many cubic feet in a concrete silo
baving 12 ft. inside diameter m^ 30 ft. high?

184 ARITHMETIC FOR CARPENTERS AND BUILDERS

62 = 36
36X3.1416 = 113 sq.ft. in bottom.
113 X 30 = 3390 cu.f t. in silo. Ans.

Fig. 77.— SUo.

Explanation. Find the area of the bottom of the silo
by squaring the radius and multiplying by tt. Multiply this
by the height to get the cubic contents.

158. Volume of a Prism. A solid whose ends are parallel
polygons and whose sides are parallelograms is called a

AREAS OF SURFACES

185

prism. When the edges of the sides make right angles with
the edges of the base, the prism is a right prism. Our most
famihar example of a right prism is simply a square or
rectangular block of wood. The volume of a prism is equal
to the product of the area of the base by the height.

A F

Fig. 78. — Three Pyramids Cut from a Cube.

159. Volume of a Pyramid. Fig. 78 shows how three
pyramids may be cut out of a cube. This demonstrates
that a pyramid contains one-third as much as the corre-
sponding cube or prism. Therefore, the volume of a pyramid
is equal to one-third the product of the area of the base by the
altitude.

186 ARITHMETIC FOR CARPENTEKA ^ND BUILDERS

If a represents the length of one side of the base, h the
length of the other side, h the altitude, then if V represents
the volume,

V = \ahh

Fig. 79. — Bin with Hopper.

Fig. 80. — Gas Container.

Example. Fig. 79 shows a bin with an inverted pyramid
for a hopper at the bottom. Find the total contents of
the bin in bushels.

For the volume of the upper prism :

10' X 10' X 20' = 2000 cu.f t.

For the volume of the hopper:

|XlO'X10'X8' = 266cu.ft.

A^EAS OF SURFACES 187

For the total volume :

2000+266 = 2266 cu.ft.
There are 1.25 cu.ft. in a bushel,

2266-^1.25 = 1813 bu. Ans.

Explanation. To find the contents of the upper bin,
multiply the length by the width and again by the height.
To find the contents of the hopper take one-third of the
product of the area of the base by the altitude. Adding
this to the quantity for the upper portion we have 2266 cu.ft.
for the total contents of the bin. There are Ij cu.ft. in
each bushel; hence we divide the contents in cubic feet by
11 to get 1813 bushels.

160. Volume of a Cone. The method of finding the
volume of a cone is the same as that for finding the volume
of a pyramid. It is equal to one-third the product of the area
of the base by the height. The volume is

V = iT7^h,

in which r is the radius of the base and h is the height or
altitude.

Example. A gas container is in the form of a cyhnder
topped by a cone. The dimensions of the container are
given in Fig. 80. How many cubic feet of gas will the con-
tainer hold?

Note. Squaring the diameter and multiplying by .7854 is the same
as squaring the radius and multiplying by 3.1416.

52 = 25
25 X. 7854 = 19.63 sq.ft.
19.63X8=157 cu.ft. in the cylinder.

188 ARITHMETIC FOR CARPENTERS AND BUILDERS

19.63X2.5 = 49.07.

^X49.07 = 16.35 cu.ft. in cone.

157+16.35 = 173.35 cu.ft. total. Ans.

Explanation. Find the contents of the cylinder in the
usual way. Find the contents of the cone by taking | of

\ ^

A /

A \^

/ ^

f a' b

A y^

\. A

/ ^

A \^

-3*-

FiG. 81.— A Deck Roof

the product of the area of the base, 19.63 cu.ft., by the
altitude, 1\ ft., to get 16.35 cu.ft. Add the contents of
the cylinder to the contents of the cone to get 173.35 cu.ft.
as the total contents of the container.

161. Volume of the Frustum of a Cone or Pyramid.
A frustum of a cone or pyramid is figured by first finding
what the contents would be if a part were not cut away.

AREAS OF SURFACES

189

Then find the volume of the part cut away and subtract
from the whole volume to get the volume of the frustum.

162. Measuring Surfaces and Volumes. Many of the
cases which one finds in practical work do not fit these regu-

FiG. 82.— Building with Cambrel Roof.

lar forms precisely. It is necessary, then, to divide the whole
into a number of parts which will come into one of these
classes. The accuracy of the result will depend largely upon
the judgment used in making these divisions and skill in
carrying out the operation.

190 ARITHMETIC FOR CARPENTERS AND BUILDERS

Example. Fig. 81 shows a deck roof and gives the
dimensions. Figure the total area of the roof.

Slant height of roof

82=64
82 = 64

128 = 11.31'
Area of 8 small triangles A, A, A, etc.:

— —^ =45. 24 sq.ft. in one triangle.

45.24X8 = 361.92 sq.ft. in 8 triangles.
Area of 2 rectangles B, B:

11.31X18X2 = 407.16 sq.ft.
Area of 2 rectangles C, C:

11.31X4X2=90.48 sq. ft.
Area of rectangle D:

18X4 = 72 sq.ft.

Adding:

361.

407.

90.48

931.56 sq.ft. total. Ans.

Explanation. Divide the roof surface up into triangles
and rectangles. Figure the area of each of these separately
and then add them all together to get the total area.

AREAS OF SURFACES

191

Summary of Chapter XIV

84. The area of a triangle is equal to one-half the prod-
uct of the base by the altitude. S = ^ ha. (Sec. 152.)

85. The lateral area of a cylinder equals two times t
times the radius of the base times the height. S = 2Trrh.
(Sec. 154.)

86. To get the total area of a cylinder add to the lateral
area the area of both bases. S = 2rrh-\-2Tn^. (Sec. 154.)

Fig. 83.

87. The lateral area of a cone equals w times the radius
of the base times the slant height. S = Trra. (Sec. 155.)

88. To get the total area of a cone add to the lateral
area the area of the base. S = irra-{-7rr^. (Sec. 155.)

89. The volume of a cylinder is equal to r times the
square of the radius of the base times the height. V=irr^h.
(Sec. 157.)

90. The volume of a prism is equal to the product of
the area of the base by the height. (Sec. 158.)

192 ARITHMETIC FOR CARPENTERS AND BUILDERS

I I

AREAS OF SURFACES

193

91. The volume of a pyramid is equal to one-third the
product of the area of the base by the height. V = lahh.
(Sec. 159.)

92, The volimie of a cone is equal to one-third the
area of the base by the height. V'=lirr^h. (Sec. 160.)

Fig. 85.

PROBLEMS

210. A triaogle has a base of 20' and an altitude of 15'. What
will be the side of a square having the same area?

211. Fig. 82 shows the end view of a building with a gambrel
roof. Figure the area of the walls and area of the roof.

194 ARITHMETIC FOR CARPENTERS AND BUILDERS

212. In the problem above how many feet B.M. of T'XS"
shiplap will be required to sheath the walls? How many feet B.M.
of r'X4" laid 2" apart will be required to sheath the roof? How
many shingles will be required for the roof?

213. In Fig. 83 find the length of rafters, struts and other
parts not given.

214. Fig. 84 shows the framing for a hip roof. How many
feet B.M. will be required if 2"X4" are used for jack rafters and
2" X 6" for hip and ridge?

Roof for a Pavilion.

215. An octagonal tower is 20' across from side to side. The
height of the apex is 11 ft. above the eaves. How many shingles
will be required for the roof? Draw the octagon to scale to find
the length of a side.

216. How many cubic feet in a monoUthic concrete silo 16'
in diameter on the outside and 45' high, if the walls are 9" thick?

217. Find the number of shingles required for the roof shown
in Fig. 85.

218. How many square feet of surface in a conical roof 15'
in diameter at the base and 8' high?

219. Fig. 86 shows a roof for a pavilion. If t^e shingles cost
$5.00 per thousand and they are laid 5" to the weather what will
be the cost of shingling the roof? Figure $3.00 per M shingles
for labor and nails.

CHAPTER XV

USE OF THE CARPENTER'S SQUARE. BRACE MEASURE.
OCTAGONAL SCALE. ESSEX BOARD MEASURE.
DIAGONAL SCALE. POLYGON SCALE. GEOMETRIC
APPLICATIONS

163. Carpenter's Square. The steel square or car-
penter's square is probably the most useful tool which the
carpenter has in his kit. In the usual form it consists of
blade or " body " 24 in. long and 2 in. wide and a," tongue "
16 in. long and 1| in. wide. The tongue makes an exact
right angle or an angle of 90° with the blade. Both the
blade and the tongue are graduated in inches and fractions
of an inch. These graduations begin at the corner and run
out on both edges of the tongue and blade. The fractional
graduations are usually in thirty-seconds, sixteenths,
twelfths, tenths and eighths. The square is also supplied
with a brace measure, an octagonal scale, the Essex board
measure and a diagonal scale, the uses of each of which
will be explained. Some squares give a rafter table instead
of the Essex board measure scale. The side of the square
which has the maker's name stamped upon it is called the
face and the opposite side is called the back. Fig. 87 shows
the face of a steel square and Fig. 88 shows the back.

164. The Brace Measure. The brace measure is used
to determine the length of a diagonal brace. When the
length of two sides of a square are known, the length of the
diagonal may be found by using the brace measure. The

195

196 ARITHMETIC FOR CARPENTERS AND BUILDERS

ITI'I'I'I'I'I'ITI'I'I'I'ITI'I'ITITI

figures along the cen-
ter of the back of the
tongue as shown in
Fig. 88 constitute the
brace measure. The
equal figures, one
above another, repre-
sent the side of the square and the figure to
the right and between them represents the
length of the diagonal.

Example. What is the length of a diag-
onal brace for a square 48" on a side?

Explanation. Look on the brace meas-
ure to find If. Between these two figures
and the right you will find 67.88. This the
length of the required brace in inches and
decimals.

165. The Octagonal Scale. The octag-
onal scale is sometimes called the " eight
square scale." It is shown along the center
of the face of the tongue in Fig. 87. To
use the scale you take as many divisions
from the scale as there are inches in the
width of the stick. Lay this distance off
on each side of the center line of the end
of the stick after it has been carefully cen-
tered. Connect these points diagonally
across the corners of the square and an
octagon or " eight square " figure will be the
result. This method is nearly correct but is
not exact.

Example. Lay out an octagon on a
stick that is 12" square.

ill

— i>_:z

— M -=

-!>> .:i.-*-

_fc -» ~=

« _ fcj — =

-<g ~=

-» — =

I'lG. 87.

USE OF THE CARPENTER'S SQUARE

197

v> «* —

— 1^ lO

00 V> —

0\ t* —

Explanation. This
example is illustrated
in Fig. 89. Find the
center of the stick
by drawing diagonals
from corner to corner
and scribe the center
line, using a try-square. This will give the
lines A B and C D shown in the figure. The
stick is 12" square; therefore, you should
take twelve divisions off of the scale. Lay
this distance off in each direction to the
right and left of the points A, B, C and D.

Fig. 88.

J B I
Fig. 89. — Laying out an Octagon.

Join these points with straight lines across
the corners. The result will be an octagon.

166. The Essex Board Measure. The
Essex board measure is used to determine
the number of board feet contained in a
stick without figuring it out. This scale is

198 ARITHMETIC FOR CARPENTERS AND BUILDERS

shown along the cen-
ter of the body in
Fig. 90. The start-
ing point for each
determination is the
figure 12 on the outer
edge. This is the number of board feet
in a 1" board 12" wide and 12 feet long.
The small figures in a line under the
12 mark represent other lengths of
boards. To find the number of board
feet in a piece which is more or less than
12" wide and from 8 to 15 ft. long, run
down the column under the 12 mark
until the proper length is found. Then
run along either to the right or left un-
til under a mark on the outer edge
which corresponds to the proper width
of the board and the small figure will
be the required value in board feet.

Since a board 12 feet long would
have as many board feet as it is inches
wide, the figures are omitted for this
length. Since a board 6 feet long would
have one-half as many board feet as it is
inches wide, this length is also omitted.
If a board is less than 8 feet long its
board measure may be found by divid-
ing its double by two. If a board is
longer than the lengths given on the
scale, divide the length into two parts,
find the board measure for each of these
and add them together. Don't try to

« wo*

*.mm at —

^■»»

t0 ^ 9 -^
^ ^ 99

itf «» —

cp».» A =

9 99

CO 00^ a>

~PI

»9i* ^ —

fe-

C-

o • n vj

O *" ~

W-»

fe-

«..o.0B r

* CD *

-. O « OD

« * _

w-^

s-,

>i>a«0 —

E-

9 9 99

0.^.5 =

P"C9

E-

a »2 M
to C ^ C

~«o

...-"^

E-

* o — "*

3*" w =

'~o

E-

E-r:

Fig. 90.

USE OF THE CARPENTER'S SQUARE

199

work out a problem on the Essex board measure if you can
do it quicker in your head.

Example. How many board feet in a board 1" thick,
15 feet long and 8" wide?

Explanation. Referring to Fig. 90 under the figure 12
on the outer edge look down until you find 15, which is the
length of the board. Then run between the lines to the left
until you are under the figure 8 on the outer edge of the
scale. This represents the width of the board. Here you
will find the figure 10. There are 10 feet B.M. in the board.

Example. How many board feet in a board 2" thick,
22 feet long and 10" wide?

Explanation. Divide the length into two parts, one 10
ft. and one 12 ft. We know at once that a board 10" wide
and 12 ft. long will have 10 ft. B.M. Since this board is
2" thick this part will contain 20 ft. B.M. For the part
10 ft. long look down under the figure 12 to find 10 and to
the left until you come to the figure 10 on the outer edge of
the body. Here read 8-4 which means Sy'V or 8| ft. B.M.
Since the board is 2" thick double this to get 16|. Adding
this to the value 20, previously
found in the first part, we get 36|
ft. B.M. in the piece.

167. The Diagonal Scale. Fig.
91 shows the diagonal scale used
on some steel squares. Its purpose
is to enable the workman to make
measurements to the hundredth of
an inch. A square one inch on a
side is divided into ten equal parts
on each side. In one direction

these lines are joined with lines parallel to the sides of
the square. In the other direction the point marked 0 is

„0 1 2 3 4 5 6 7 8 9 10

— r

0123466789 10

Fig. 91.

200 ARITHMETIC FOR CARPENTERS AND BUILDERS

joined to the point
1 on the opposite
side. You will now
see that the dis-
tance on the line
61 from the zero
vertical line to the

diagonal intersection is ^^jfth^ of an inch.
In like manner the distance from the zero
vertical line to the diagonal intersection
on hi is x^^ths of an inch. To read
.78" on the scale, using a pair of dividers
you place one point on the intersection of
the zero vertical line with the horizontal
line iS. The other divider point should
be extended to touch the diagonal in-
tersection on iS just beyond the vertical
line marked 7. The distance between the
divider points will then be iVV+ifir =
■^% = .78, which is the required distance.
Read tenths on the horizontal scale and
hundredths on the vertical scale.

168. The Polygon Scale. This scale
is useful for finding the correct angle to
cut sticks of lumber so that they will
form regular polygons when fitted to-
gether. It is illustrated in Fig. 92. On
some squares this scale is placed along the
center of the face of the tongue. The
first figure indicates the number of sides
in the polygon. The second figure indi-
cates the point on the body, and the third
figure the point on the tongue that is

- hj

- -» «* JW-E

Ci>

0) ^

H ^

y ~z

Fig. 92.

USE OF THE CARPENTER'S SQUARE

201

made to coincide with the edge of the stick in order to get
the correct angle as shown in Fig. 93. If it is desired to
cut a polygon of eight sides you would take 18 on the
body and 7| on the tongue of the square. Mark the
stick along the edge of the tongue as illustrated in the
figure. Eight pieces of equal length cut to this angle will
fit together to make an eight-sided figure.

169. Geometric Applications. If the workman under-
stands fully the properties of the right triangle and the
applications of geometry, he will have but little trouble

^2^

Mark here

Fig. 93. — Getting Correct Angle for a Polygon Frame.

in making his steel square serve him as a very useful tool.
A few of the common applications are given here. Others
will occur to the practical man in connection with his daily
work.

170. Hypotenuse of a Right-angled Triangle. To find
the length of the hypotenuse of a right-angled triangle
when the base and altitude are unequal, lay a two-foot
rule across the square with the zero mark of the rule at the
point on the body which corresponds to one leg of the tri-
angle and the edge of the rule crossing the tongue at a point
corresponding to the other leg of the triangle. The
reading of the rule will give the length of the hypotenuse.

Example. What is the length of the hypotenuse of a
triangle the base of which is 8 ft. and the altitude 14 ft.?

202 ARITHMETIC FOR CARPENTERS AND BUILDERS

Explanation. In this case it is convenient to allow 1"
on the rule and on the square to represent a foot in the
triangle. Referring to Fig. 94, we place the zero mark of
the rule on the 8" mark on the body and lay the edge of
the rule across the 14" mark on the tongue. At this point
the rule reads 16|". This means that the hypotenuse is
16J ft. long or, expressing this in feet and inches, 16' Ij".

' I 1 'I I ' I ' 'I I 1 1 I

Fig. 94. — Finding Length of Hypotenuse.

171. To Divide a Line. Suppose we wish to divide the
width of a board into ten equal parts for ripping into strips.
Lay the square diagonally across the board so that the
zero mark is at one edge of the board and the 10" mark is
at the other edge as shown in Fig. 95. Then make a mark
exactly at each one of the inch divisions on the edge of the
square. Drawing a line parallel to the edge of the board
through each one of these points will divide the board into
ten equal strips.

172. To Bisect an Angle. Measure out equal distances

USE OF THE CARPENTER'S SQUARE

203

on each leg of the angle, OA and OB in Fig. 96. Hold the
square so that the same figure on each side of the square

Fig. 95.— Finding Width of Strips.

coincides with the points A and B. The vertex or point
of the square will then be at the exact middle of the angle.

Fig. 96. — Bisecting an Angle.

Mark this point D and join it with a straight line to the
vertex of the angle 0. The line OD is the required bisector
of the angle.

204 ARITHMETIC FOR CARPENTERS AND BUILDERS

Summary of Chapter XV

93. The brace measure is used to find the length of a
diagonal brace. (Sec. 164.)

94. The octagonal scale is used to find the correct angle
for dressing a square stick into an octagon. (Sec. 165.)

95. The Essex board measure scale is used to find the
number of feet B.M. in any board. (Sec. 166.)

96. The diagonal or hundredth scale is used to measure
hundredths of an inch. (Sec. 167.)

97. The polygon scale is used to find the correct angle to
cut sticks of equal length to form polygons having from five
to twelve sides. (Sec. 168.)

98. The steel square may be used to find the length of
the hypotenuse of any right-angled triangle. (Sec. 170.)

99. The steel square may be used to divide a given Une
into any number of equal parts. (Sec. 171.)

100. The steel square may be used to bisect an angle.
(Sec. 172.)

PROBLEMS

220. Fiild the length of a diagonal brace for a square 54" on
a side.

221. Find the length of a diagonal brace for a square 4' 9"
on a side.

222. Find the length of a diagonal brace for a square 10' on
a side.

223. Draw a diagram showing how to lay out an octagon on
a stick 6" square, using the octagon scale. If you take your work
from Fig. 87 you will have to make your drawing half size.

224. Using the Essex board measure scale find how many
board feet there are in a board 2" thick, 9" wide and 16' long.

225. How many feet B.M. are there in 16 pieces li"XlO",
26' long?

226. How many feet B.M. are there in a board 1"X14", 18'
long?

USE OF THE CARPENTER'S SQUARE" 205

227. A six-sided figure each side of which is 18" long is required.
Tell how to find the correct angles to cut the boards.

228. What is the length of the hypotenuse of a triangle the
base of which is 6 ft. long and the altitude 10 ft.?

229. What is the length of the hypotenuse of a right-angled
triangle 16" on a side?

230. Divide a line which is 6" long accurately into 13 equal
parts.

231. Construct a diagonal scale similar to the one shown in
Fig. 91. Show the distance .78" on your diagram.

232. Bisect a 45° angle, using a steel square.

CHAPTER XVI

USE OF THE CARPENTER'S SQUARE IN FRAMING.
FLOOR BRIDGING. STAIR BEAMS. RAFTER
FRAMING. ROOF PITCHES. RAFTER TABLE.
ROOF FRAMING. HIP OR VALLEY RAFTERS. JACK
RAFTERS

Not only may the steel square be used for the purposes
indicated in the last chapter, but it may also be used very
conveniently in the processes required for framing a building.
As everyone knows, it may readily be used for determining

Fig. 97. — Cuts for Floor Bridging.

a right angle. It may also be used to determine many of
the other angles to which it is necessary to cut the lumber
in framing. These angles are called cuts or bevels.

173. To Find the Cuts for Floor Bridging. Take the
width of the joist on the tongue of the square and the dis-
tance between the joist on the body. Take the measure-
ments on opposite sides of the stick as shown in Fig. 97.
The tongue will give the cut and the body will give the
length.

206

THE CARPENTER'S SQUARE IN FRAMING 207

174. To Find the Cuts on a Stair Beam. Suppose the
stair is to be laid out with 7|" risers and 9" treads. Then
it is simply necessary to lay the square on the beam in such
a manner that the reading on the tongue from the edge of
the beam is 7^" and a similar reading on the body is 9", as
shown in Fig. 98. Mark the line and repeat for each

Fig. 98. — Laying out Cuts on a Stair Beam.

succeeding step. Remember that the thickness of the floor
must be taken into account when laying out the top and
bottom steps.

175. Rafter Framing. Before taking up the subject of
roof framing it is necessary to explain some of the terms
and customs commonly used. The span of the roof is the
same as the width of the building as shown in Fig. 99. The
run of the rafter is one-half of the span and the rise is the
vertical distance from the top of the plate to the top of

208 ARITHMETIC FOR CARPENTERS AND BUILDERS

the ridge. The pitch of the rafter is the ratio of the rise to
the span. The rafter length is the distance from the outside
comer of the plate to the center of the ridge. The exten-
sion is the distance from the outside corner of the plate to
the end of the rafter. When finding the actual length of a
rafter, the distance from the outer corner of the plate to the
center of the ridge is first found. The square is then set for

Fig. 99.— Span of a Roof.

the ridge cut as explained in Sec. 177 and one-half the width
of the ridge subtracted in the direction of the body of the
square. Sufficient length must then be added for the exten-
sion. The result will be the exact length of the rafter.

176. Pitch. The most common pitches are indicated
in the table. They can be found on the steel square by
laying a straightedge across the square with one end on
the figure 12 on the body and the other end on that figure
of the tongue which corresponds to the rise as illustrated in
Fig. 100.

THE CARPENTER'S SQUARE IN FRAMING 209

TABLE XVI
COMMON ROOF PITCHES
i pitch IS 12 on the body and 4 on the tongue.

12
12
12
12
12
12

8
10
12
15
18

' ' ' I ' ' ' I I I I '^ I r I I I I

12
Fig. 100.— Finding Pitch

177. To Find the Cuts for a Common Rafter. The cut

at the top of the rafter is called the ridge cut or the plumb
cut. The cut at the lower end of the rafter is called the
bottom cut or heel cut.

First, to find the plumb cut place the square as shown
in the upper position of Fig. 101. Use 12 on the body and
other figures on the tongue, depending upon the pitch of
the roof. For example, if the rafter is to be set at one-third
pitch use 12 on the body and 8 on the tongue as indicated
in the table above. For pitches not given in the table

210 ARITHMETIC FOR CARPENTERS AND BUILDERS

multiply the pitch by 2, express as a proportion and solve
for the tongue reading: Thus, for one-fifth pitch

iX2 = |

2: 5::x: 12

a: = 244-5=4|.

Ans.

Next, to find the heel cut place the square as shown in
the lower portion of Fig. 101. If the rafter is to extend

- Mark here

X Pitch

Repeat Plumb cut here

Fig. 101. — Finding Rafter Cuts.

beyond the plate, the square is again set so as to give the
plumb cut indicated in the figure. The process is repeated
again for the end cut.

178. The Rafter Table. The rafter table found on the
back of the body of rafter squares or framing squares is used
to find the length of a common rafter for various pitches.
This table is shown in Fig. 102. The graduations on the
outside edge of the body are used in connection with the

THE CARPENTER'S SQUARE IN FRAMING 211

^NJ

^0)

table. These grad-
u a t i o n s are in
twelfths and the
inches may be used
to represent either
inches or feet while

the twelfth marks represent either
twelfths of a foot (inches) or twelfths
of an inch. These figures on the edge
when making use of the table corre-
spond to the run of the rafter. To use
the rafter table, look at the left end of
the table to find the required pitch.
Then under the mark on the outer
edge which corresponds to the run of
the rafter in feet will be found the
required length in feet, inches and
twelfths of an inch. Thus a rafter set
at one-half pitch with a run of ten feet
is 14' ly\" long. Again a rafter set at
quarter pitch with a run of 12 ft. is 13'
5" long.

When the run is given in inches the
length will be in inches. Thus, a rafter
set at quarter pitch with a run of 8"
would be 8i|"4-i\ of -^-^" , which, neg-
lecting the added fraction, is approxi- pi
mately Sji". If the run is given in both
feet and inches find each separately and
add the lengths together.

Example. What is the length of a
common rafter set at one-third pitch
with a run of 10 ft. 6 in.?

— RAFTERS

1 FEET-INCHES

--* AND TWELFTHS

^0)

-co

0) =

<0 =

o =

Fig. 102.

212 ARITHMETIC FOR CARPENTERS AND GUILDERS

For run of 10 ft. the rafter length is 12' 0/^"
For run of 6" the rafter length is 1^"

Total length 12' 7^\" Ans.

Explanation. Looking for the run in feet, we first find
that the length for a 10 ft. run will be 12' 0^\". Now finding
the length for the run in inches we get 7^^", disregarding
the last figure in the table. Adding these figures together
gives 12' 1 ^^" for the distance from the outer edge of the
plate to the center of the ridge. For the actual length
deduct one-half the thickness of the ridge and add sufficient
for the extension.

179. Roof Framing. Up to this point we have taken
up only the lengths and cuts for common rafters. Any roof
which is framed for gables, hips and valleys will require
rafters of various lengths having both plumb cuts and side
cuts. Fig. 103 shows the framing for such a roof. The
ridge board marked A is the piece at the center to which
the rafters are nailed. The plate marked B is the board
placed on top of the studs to which the foot of each rafter is
secured. The common rafters, C, are those which set square
against the ridge board and which have no side cut. Hip
rafters are those marked D. Valley rafters are those
marked E. Jack rafters, marked F, are those the upper end
of which set against the hip rafter. Cripple rafters — rafters
having no foot — span the space between the ridge board
and valley rafter or between hip and valley rafters. They
are marked G in the illustration.

180. Length of Hip or Valley Rafters. The length of
the hip rafter is the hypotenuse of a right-angled triangle,
one side of which is the common rafter and the other side
is the distance on the plate from the foot of the hip rafter to

tat

214 ARITHMETIC FOR CARPENTERS AND BUILDERS

the foot of the common rafter. The length of the valley
rafter is also the hypotenuse of a similar triangle, and for
this reason the following example and explanation will apply
to a valley rafter as well.

4 .

' '

— 4! >!

a .

Fia. 104.— Hip Rafter.

Example. What is the length of the hip rafter shown
in Fig. 104?

5' 8" = 5.66'

5.662 = 32
42 = 16
16+32=48
V48 = 6.93

6.93' = 6' 11 A" Ans.

THE CARPENTER'S SQUARE IN FRAMING 215

Explanation. The length of the common rafter having
4 ft. run for one-half pitch roof is 5' 8". The distance from
the foot of the hip rafter to the foot of the common rafter
is 4'. The hip rafter is the hypotenuse of this triangle.
Solving by finding the square root of the sum of the squares
of the base and altitude gives 6' 11^" for the length of this
rafter. Notice that the hip rafter is not shown in its true
length in the figure.

181. Cuts for Hip or Valley Rafters. The rise for a
hip or valley rafter is the same as the rise for the common
rafter. The run of a hip rafter is the horizontal distance
from the outside comer of the plate at the foot to a vertical
line from the center of the ridge against which it rests.
This is equivalent to the length of the corresponding com-
mon rafter. In every case the ratio of the run for the hip
or valley rafter to the run of the common rafter is as 17 is to
12. To find the plumb and heel cuts for a hip or valley
rafter for a given roof pitch we proceed exactly as for a
common rafter, but we use the 17 mark on the body instead
of the 12 mark.

Example. What figures should be used in marking the
plumb and heel cuts for the hip rafter of a roof of one-third
pitch when the run of the common rafter is 8 ft.?

Explanation. To find the plumb and heel cut of a com-
mon rafter having one-third pitch use 12 on the body and
8 on the tongue as explained in Sec. 177. For the hip
rafter use 17 on the body and 8 on the tongue.

182. Hip Rafter Table. The necessary information for
finding the side cuts for hip and valley rafters and the length
and side cuts for jack rafters may be found in the hip rafter
table shown in Fig. 105.

In the table under the heading PITCH is given the read-
ing for the pitches in common use. The fractions represent

216 ARITHMETIC FOR CARPENTERS AND BUILDERS

i|i|i|'|T|'i'|'i'|'i'|'i'|'i'|'i'l''T''l''T''

=- PAID. APPO. FOR

— o

;;;

—

— ^

- Z

:- o

=- *
= — *
=- o
u

0>~

e
e

«— _

- ,3

- a

- o

«0

t- _

Fia. 105.

the pitches and the
figures to the right
are the ones used
for determining the
plumb and heel cuts
of common rafters.
The next set of figures under the
heading H I P indicate the length of the
hip and valley rafters for each foot of
run for the common rafter of the differ-
ent pitches. To show how this is used
let us take the same example that is
worked out above.

Example. What is the length of the
hip rafter shown in Fig. 104?

1-8-10x4 = 4-32-40

4-32-40 = 4' 32f|" = 6' 11,V' = 6' 11^"

Ans.

Explanation. From the table for
every foot run of the common rafter the
hip rafter is 1' 8j|" long. Since the run
of the common rafter is 4' we multiply
this by 4 and reducing the product to its
simplest form find the length of the hip
rafter to be 6' 11 3", which is approxi-
mately the same value that we got by
solving the triangle.

183. Jack Rafters. The next set of
figures on the hip rafter table gives the
length of the shortest jack rafter when
they are spaced 16" on center. Since

THE CARPJi.iNTER'S SQUARE IN FRAMING 217

Mark here

these rafters are evenly spaced the second will be twice as
long as the first and the third three times as long as the
first. The lengths for 24" spacing are also given.

Example. Find the length of the third jack rafter set
at one-half pitch.

22f"X3 = 67r = 5' 7|" Ans.

Explanation. The first
jack rafter set 16" on centers
at one-half pitch will be 22|"
long as indicated in the table.
The third rafter will be three
times as long as this or 5' 7|".

The plumb and heel cuts
for jack rafters are the same
as for common rafters of the
same pitch. To find the side
cut use the next set of figures
to the left in the table. For
one-half pitch take 17 on the
body and 12 on the tongue
and and mark the rafter as
shown in Fig. 106.

184. Side Cut for Hip
Rafters. To find the side cut
for hip or valley rafters use
the figures under the heading

HIP marked SIDE-CUT. For one-half pitch the setting
on the square will be 8 on the body and 11 on the tongue.
Mark the cut along the tongue.

185. Length of Hip or Valley Rafters. The remainder
of the table indicates the length of the hip or valley rafters
for the run of the corresponding common rafter for the

Fig. 106.— Side Cut for Rafter.

218 ARITHMETIC FOR CARPENTERS AND BUILDERS

various pitches. Find the run of the common rafter on the
outside edge of the scale. Under this figure will be found
the length of the hip or valley rafter for the corresponding
pitch.

Example. Find the length of hip rafter required for
one-half pitch roof when the building is 24 ft. wide.

Explanation. The run of the common rafter is one-half
the width of the building. Under the figure 12 on the outer
edge and to the left between the lines which represent one-
half pitch find the figures 20—9—5. This means that the
hip rafter is 20' 9^%" long.

186. Cuts for Cripple Rafters. The plumb and side cuts
for cripple rafters are the same as for jack rafters. The
cuts at the bottom of a cripple rafter are the same as those
at the top.

187. Actual Lengths. In all of the above tables the
lengths given are from the outside edge of the plate to the
center of the ridge or hip. To find the actual length deduct
one-half the width of the ridge or hip and add sufficient
material for the extension at the eaves.

THE CARPENTER'S SQUARE IN FRAMING 219

Summary of Chapter XVI

101. The steel square may be used to find the cuts for
floor bridging as explained in Sec. 173.

102. It may be used to find the cuts on a stair beam.
(Sec. 174.)

103. It may be used to find the cuts on a common
rafter. (Sec. 177.)

Fig. 107.— Struts.

104. The rafter table may be used to find the length of
common rafters required for various runs and pitches.
(Sec. 178.)

105. The length of a hip or valley rafter is equal to the
hypotenuse of a right-angled triangle, the base of which is
the distance from the foot of the hip rafter to the foot of
the common rafter and the altitude of which is equal to the
length of the common rafter. (Sec. 180.)

106. The plumb and heel cuts for hip and valley rafters
may be found by using 17 on the body instead of 12 as
explained in Sec. 181.

220 ARITHMETIC FOR CARPENTERS AND BUILDERS

107. The hip rafter table may be used to find the plumb
and heel cuts of common, hip and jack rafters; to find the
side cuts of these rafters, and to find the length of hip,
valley and jack rafters. (Sec. 182.)

1
\

"^ N

/ '

-26-

,2x6 Ridge

.2z6^It>

Jack Bafteis

Fig. 108.— Roof Framing.

108. The cuts for cripple rafters are the same as for jack
rafters except that the bottom cuts are like the top cuts.
(Sec. 186.)

109. The actual length of the rafters is found by
deducting one-half of the ridge or hip and adding sufficient
material for the overhang. (Sec. 187.)

THE CARPENTER'S SQUARE IN FRAMING 221

PROBLEMS

233. Ten inch floor joists are placed 16" on centers. Draw a
diagram to scale showing how to find the top and bottom cuts
for the bridging.

Fig. 109.— Plan of Roof.

234. A stair joins two floors, the second of which is 10' 3" above
the first. Twice the riser plus the tread should about equal 24".
Make a sketch showing how to lay out the cuts on the stair beam.

235. A building is 24' wide. The roof ridge is 8' above the
plate. What is the pitch of the roof?

222 ARITHMETIC FOR CARPENTERS AND BUILDERS

236. A building is 32' wide. The roof ridge is 16' above the
plate. How would you determine the plumb and heel cuts for the
common rafters?

237. The roof of a building has a pitch of three-eighths. How
would you find the plumb and heel cuts of the common rafters?

238. What is the length of a common rafter set at five-twelfths
pitch when the run is 12' 6"?

239. What would be the length of the hip rafters for the roof
in the problem above?

240. What is the length of a valley rafter for a roof of one-half
pitch when the span is 16 ft.

241. Show how to find the cuts for the struts marked a and I
in Fig. 107.

Hint. — Struts b are perpendicular to the rafters.

242. By using the hip rafter table wherever possible find the
number of lineal feet 2x6, 2x4 and 4x4 required for the roof
framing in Fig. 108.

Note. — ^The pitch of the lower section is not to be found in
the rafter table.

243. Show how to find the side cuts for the jack rafters shown
in Fig. 108.

244. Show how to find the side cuts for the hip rafters shown
in Fig. 108.

245. Using the hip rafter table, find the length of a hip rafter
for a roof having one-fourth pitch when the building is 18 ft. wide.

246. Find the actual lengths of all the rafters in the roof shown
in Fig. 109. Make a roof framing plan to accompany your solution.
Space the rafters 16" on centers.

INDEX

Abbreviations, 2
Accounts, wage, 55
Accuracy of results, 66
Actual length of rafters, 218
Acute angle. An, 161
Adding feet and inches, 2
Allowance for dressing, 143
Altitude of right triangle, to find,

131
Altitude of triangle, 130
Analysis of problems in percent-
age, 83
Angle, An, 161

acute, 161

obtuse, 161

right, 161
Angle, designating an, 161
Angle, to bisect, 164
Angles, to construct, 167
Angular measure, 162
Apex of a pyramid, 175
Arc of a circle, 162
Area, 108

of a circle, 132

of a cylindrical surface, 178

of a pyramid, 175

of a rectangle, 108

of a square, 108

of surface of a cone, 180

of a triangle, 174

Area covered by shingles, table of,
152

Assets, 54

Avoirdupois system of weight, 107

Bank accounts, 50

balance, 51

check, 51

deposit, 50

pass book, 50
Barrel, number of gallons in, 134
Base of triangle, 130
Batter Boards, 166
Bearing power of soils, 115
Belts, rule for the width of, 74
Bisect an angle, to, 164

with the steel square, 202
Bisector, perpendicular, to find,

163
Board, 142
Board feet in lumber of different

widths, table of, 148
Board feet, rule for, 76
Board foot, 147
Board measure, Essex, 197
Board measure, to find, 148
Boards, sizes of, 145
Body of steel square, 195
Bottom cuts for rafters, 209
Brace measure, 195

223

224

INDEX

British system of weights and

measures, 105
Broken line, A, 160
Building estimates, 57
Building material, weight of, 114

Cancellation, 33
Capacity, measures of, 106
Capacity of cisterns and circular

tanks, 133
Carpenters' square, 195
Cement and concrete mixtures, 92
Center of an arc or circle, to find,

167
Changing a common fraction to a

decimal, 67
Changing a decimal to a common

fraction, 66
Changing a fraction to a per cent,

82
Changing improper fractions, 15
Changing mixed numbers, 14
Changing whole numbers to im-
proper fractions, 14
Check for addition, 78

for division, 79

for multipUcation, 79

for subtraction, 78
Chord, A, 162
Circle, 109

arc of, 162

circumference, 109

diameter, 109

radius, 109
Circle, to find area of, 132
Circle, to find radius when area is

given, 133
Circular measure, 162

Circumference of a circle, 109
Circumference, to find, 110
Cisterns and circular tanks, ca-
pacities of, 133
Common rafters, 212

to find cuts for, 209
Comparative values, liquid and

dry measure, 107
Compound fractions, 41
Compound interest, 88
Concrete, quantities of materials

per cubic yard, 117
Cone, A, 180

frustum of, 180

slant height of, 180

vertex, 180

volume of, 187

volume of frustum of, 188
Construct a right angle to, 165
Counting lumber, 149
Cripple rafters, 212

cuts for, 218
Crow-bar, 100
Cube root, 124

to find, 129
Cube, volume of, 112
Cubic foot, 112
Cubic inch, 112
Cubic measure, 106
Curved fine. A, 160
Curved surface. A, 160
Cylinder, A, 178

right, 178
Cylinder, volume of, 183
Cylindrical surface, area of, 178

Decimal equivalents of the frac-
tions of an inch, 69

INDEX

225

Decimal fractions, 60

addition, 62

division, 64

multiplication, 63

reading, 61

subtraction, 63

writing, 60
Decimal point. The, 48
Degrees, 162

Designating an angle, 161
Diagonal scale, 199
Diameter of a circle, 109

to find. 111
Difference, 4
Dimension lumber, 142

sizes of, 146
Discounts and list prices, 86
Divide a line, to, with the steel

square, 202
Dividend, 5
Dividing a whole number by a

fraction, 41
Division of feet and inches, 5
Divisor, 5
Dressed finishing lumber, sizes of,

146
Dressed stock, 142
Dressing, allowance for, 143
Dry measure, table of, 107

Eight square scale, 196
EUipse, The, 168

major axis, 169

minor axis, 169
Essex board measure scale, 197
Estimates, building, 57
Evolution, 124

practical applications, 129

Extension of rafter, 208
Extracting square root, directions

for, 127
Extremes of a proportion, 94

Factors, 32

Feet board measure, 76

Finishing lumber, 142

Floor bridging, to find cuts for,

206
Flooring, 107

to find amount required, 150
Foot, 1

Footings, to find size of, 116
Force arm, 98
Formulas, 76

how to use, 76
Fraction, terms of a, 10
Fractions, 10

adding, 20

changing to a per cent, 82

compound, 41

denominator, 11

division of, 39

improper, 12

multiplication of mixed num-
bers, and, 29

multiplying a group of whole
numbers and, 31

numerator, 11

of an inch, 12

powers of, 123

product of two, 30

product of whole numbers and,
29

proper, 11

reading, 11

reduction to higher terms, 13

226

INDEX

Fractions, reduction to lower
terms, 13

subtracting, 20

writing, 11
Framing lumber, sizes of, 147
Framing, rafter, 207

roof, 212
Framing squares, 210
Frustum of a cone, 180

volume of, 188
Frustum of a pyramid, 175

volume of, 188
Fulcrum, 98

Gallons of water per cubic foot,

134
Gallons of water to a barrel, 134
Geometric applications of the steel

square, 201
Geometry, 160
uses of, 160

Heel cut for rafter, 209
Hexagon, 170

to construct, 170
Hexagonal pyramid, 176
Hip rafters, 212

side cut for, 217
Hip rafter table, 215
Hip or valley rafters, cuts for, 215

length of, 212, 217
Hypotenuse of a right triangle,
130

to find, 130
Hypotenuse of a right triangle,
steel square method, 201

Inch, 1
Interest, 87

compound, 88
Inverse proportion, 94
Inverse ratio, 91
Involution, 122

practical apphcations, 129

Jack rafters, 212
length of, 216

Least common denominator, 15

to find, 17

to reduce to, 19
Length, measures of, 105
Length of lumber, standard, 145
Levers, 98

arrangement of, 99

force arm, 98

fulcrum, 98

types, 99

weight arm, 98
LiabiUties, 54 .
Line, A, 160

broken, 160

curved, 160

straight, 160
Liquid measure, table of, 106
List prices and discounts, 86
Lumber, counting, 149

measurement of, 147

weight of, 113

width of, 144
Lumber tally book, 149
Lumber terms, 142

INDEX

227

Lumber, board, 142

dressed stock, 142

plank, 142

rough stock, 142

scantling, 142

timber, 142
Lumber trade customs, 142

Major axis of ellipse, 169
Material lists, 153
Means of a proportion, 94
Measurement of lumber, 147
Measures of capacity, 106

of surface, 106

of volume, 106

of weight, 107
Measuring materials for concrete,

92
Measuring surface and volumes,

189
Metric system of weights and

measures, 105
Mile, 2

Minor axis of ellipse, 169
Minuend, 4
Minutes, 162
Mixed numbers, 12

adding, 26

dividing, 41

multiplication of fractions and
29

multiplying, 32

subtracting, 27

working with, 26
Mixtures, cement and concrete,

92
Money, 47

addition, 48

Money division, 50

multiplication, 49

subtraction, 49
Money sums, writing and reading,

47
Multiplicand, 5
Multiplier, 5
Multiplying feet and inches, 5

Numbers, mixed, 12
Numerator, 11

Obtuse angle, An, 161
Octagon, 170

to construct, 171
Octagonal scale, 196
Of means times, 30

Parallelogram, 169

Pay roll, 55

Percentage, 82, 83 T

amount, 83

analysis of problems, 83

base, 83

difference, 83

meaning of term's, 83

rate, 83
Per centum, meaning of, 82
Perpendicular bisector, to find,
163

to erect at any point on a line,
165
Pi=x = 3.1416, 110
Pitch of a roof, 208
Plane surface. A, 160

228

INDEX

Plank, 142
Plate, 212
Plumb cut, 209
Point, A, 160
Polygon scale, 200
Polygons, 169

hexagon, 170

octagon, 170

parallelogram, 169

quadrilateral, 169

rectangle, 170

side, 169

square, 170

triangle, 169

vertex of, 169
Powers, 122

base, 122

exponents, 122

of common fractions, 124
Principal, 87
Prism, 185

right, 185

volume of, 185
Product, 5
Profits, 87

rroof of square root, 128
Proportion, 94

extremes, 94

inverse, 94

levers, 99

meaning of, 94

means, 94

solving a, 94

statement of a, 96

statements about a, 95
Problems, 43

classes of, 42

conclusion, 42

operation, 42

Problems, soluticHi of, 42

statement of, 42
Problems, analysis of in percent-
age, 83
Problems in percentage, solution

of, 85
Pulley sizes, rule for, 73
Pyramid, 175

apex, 175

frustum, 175

right, 175

vertex, 175

volume of, 185

volume of frustum of, 188
Pyramidal roof, 176

Quadrant, 162

Quadrilateral, 169

Quantity of materials per cubic

yard of concrete, 117
Quotient, 5

Radical sign, 124
Radius of a circle, 109

to find. 111

to find when area is given, 133
Rafter, common, to find cuts, for,

209
Rafter extension, 208
Rafter framing, 207
Rafter, hip, side cuts for, 217
Rafter, hip, table, 215
Rafter, length, 208
Rafter, pitch of, 208
Rafter rise, 207
Rafter run, 207

INDEX

229

Rafter squares, 210
Rafter table, 210
Rafters, 212

common, 212

cripple, 212

hip, 212

jack, 212

valley, 212
Rafters, actual length of, 218
Rafters, cripple, cuts for, 218
Rafters, hip, side cuts for, 217
Rafters, jack, length of, 216
Rafters, length of hip or valley,

212
Ratio, 91

inverse, 91

meaning of, 92
Receipt, 54
Rectangle, 170

area of, 108
Rectangular block, volume of, 112
Ridge board, 212
Ridge cut, 209
Right angle. A, 161

to construct, 165
Eight triangle, 130

altitude, 130

base, 130

hypotenuse, 130
Right prism, 185
Rise of a rafter, 207
Rod, 2

Roof framing, 212
Roots, 124

imperfect, 124

index, 124

radical sign, 124
Rough stock, 142
Rules, 73

Rules for pulley sizes, 73
for the width of belts, 74
using letters in, 76

Run of a rafter, 207

Scantling, 142
Seconds, 162
Sector, A, 163
Segment, A, 163

Shingles, area covered by, table
of, 152

to find amount of, 151
Short methods, 78

addition, 78

division, 79

multiphcation, 79

subtraction, 78
Side of a square, to find, 129
Sign of division, 10
Significance of position, 61
Sizes of common boards, 145
Sizes of dimension lumber, 146
Sizes of dressed finishing lumber,

146
Sizes of framing lumber, 147
Slant height of a cone, 180
Slopes, 97

Soils, bearing power of, 115
Solid, A, 161

Solution of problems in percent-
age, 85
Solving a proportion, 94
Span of a roof, 207
Square, 170

area of, 108

to find side of, 129
Square of roof surface, 151
Square or surface measi re, 108

230

INDEX

Square root, 124

directions for extracting, 127

periods, 125

proof of, 128

to find, 125

to locate decimal point, 126
Square, steel, 195

back, 195

body, 195

face, 195

tongue, 195
Stair beam, to find cuts on, 207
Standard lengths of lumber, 145
Statement, 52

Statement of a proportion, 96
Statements about a proportion, 95
Steel square, geometric applica-
tions, 201
Stone measure, 106
Straight line. A, 160
Strip count, 144
Subtracting feet and inches, 4
Subtrahend, 4
Surface, A, 160

curved, 160

cylindrical, area of, 178

plane, 160
Surface, measures of, 106
Surface, testing a plane, 161
Surfaces and volumes, measuring,
189

Table of area covered by shingles,
152
avoirdupois weight, 107
bearing power of soils, 116
board feet in lumber in differ-
ent widths, 148

Table of common cubic measure,
106
common linear measure, 105
common liquid measure, 106
comparative values of liquid

and dry measure, 107
decimals of an inch for each ^,

68
dry measure, 107
quantities of material per cubic

yard of concrete, 117
sizes of dimension lumber, 146
weight of lumber per cubic foot,

114
weight of various building ma-
terials per cubic foot, 114
Tally book, lumber, 149
Terms of a ratio, 92
Testing a plane surface, 161
Timber, 142
Time book, 55
Tongue of steel square, 195
Trammel method of constructing

ellipse, 169
Triangle, 130, 169
area of, 174

Units of length, 1
Using letters in rules, 76

Valley rafters, 212
cuts for, 215
length of, 212, 217
Vertex of a cone, 180
Vertex of a pyramid, 175

INDEX

231

Volume, 183

measures of, 106

of a cone, 187

of a cylinder, 183

of a prism, 184

of a pyramid, 185
Volumes and surfaces, measuring,
189

Wage accounts, 55
Weight arm, 98

Weight, avoirdupois, table of, 107
Weight of building material, 114
Weight of lumber, 113

per cubic foot, table of, 114
Weight measures of, 107
Weights and measures, 105
Width of belts, rule for, 74
Width of lumber, 144

Yard, 1

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