b
1. c. s.
REFERENCE LIBRARY
A SERIES OF TEXTBOOKS PREPARED FOR THE STUDENTS OF THE
INTERNATIONAL CORRESPONDENCE SCHOOLS AND CONTAINING
IN PERMANENT FORM THE INSTRUCTION PAPERS.
EXAMINATION QUESTIONS, AND KEYS-USED
IN THEIR VARIOUS COURSES
ARITHMETIC
MENSURATION
MECHANICAL DEFINITIONS
MECHANICAL CALCULATIONS'
YARN CALCULATIONS, COTTON
YARN CALCULATIONS, WOOLEN
AND WORSTED
YARN CALCULATIONS, GENERAL
CLOTH CALCULATIONS, COTTON
CLOTH CALCULATIONS, WOOLEN
AND WORSTED
DRAFT CALCULATIONS
READING TEXTILE DRAWINGS
983 E
SCRANTON
INTERNATIONAL TEXTBOOK COMPANY
86
Copyright, 1906, by International Textbook Company.
Entered at Stationers’ Hall, London.
Arithmetic: Copyright, 1901, by Christopher Parkinson Brooks. Copyright, 1905,
by International Textbook Company. Entered at Stationers’ Hall, London.
Mensuration: Copyright, 1905, by International Textbook Company. Entered
at Stationers’ Hall, London.
Mechanical Definitions: Copyright, 1901, by Christopher Parkinson Brooks.
Copyright, 1905, by International Textbook Company. Entered at Stationers’
Hall, London.
Mechanical Calculations: Copyright, 1901, by Christopher Parkinson Brooks.
Copyright, 1905, by International Textbook Company. Entered at Stationers’
Hall, London.
Yarn Calculations, Cotton: Copyright, 1901, by Christopher Parkinson Brooks.
Copyright, 1905, by International Textbook Company. Entered at Stationers'
Hall, London.
Yarn Calculations, Woolen and Worsted: Copyright, 1901, by Christopher
Parkinson Brooks. Copyright, 1905, by International Textbook Com¬
pany. Entered at Stationers’ Hall, London.
Yarn Calculations, General: Copyright, 1901, by Christopher Parkinson Brooks.
Copyright, 1905, by International Textbook Company. Entered at Stationers’
Hall, London.
Cloth Calculations, Cotton: Copyright, 1901, by Christopher Parkinson Brooks.
Copyright, 1905, by International Textbook Company. Entered at Stationers’
Hall, London.
Cloth Calculations, Woolen and Worsted: Copyright, 1901, by Christopher
Parkinson Brooks. Copyright, 1905, by International Textbook Com¬
pany. Entered at Stationers’ Hall, London.
Draft Calculations: Copyright, 1903, by International Textbook Company.
Entered at Stationers’ Hall, London.
Reading Textile Drawings: Copyright, 1905, by International Textbook Com¬
pany. Entered at Stationers' Hall, London.
All rights reserved.
86
Printed in the United States.
16651
THE GETTY RESEARCH
INSTITUTE LIBRARY
PREFACE
Formerly it was our practice to send to each student
entitled to receive them a set of volumes printed and bound
especially for the Course for which the student enrolled.
In consequence of the vast increase in the enrolment, this
plan became no longer practicable and we therefore con¬
cluded to issue a single set of volumes, comprising all our
textbooks, under the general title of I. C. S. Reference
Library. The students receive such volumes of this
Library as contain the instruction to which they are entitled.
Under this plan some volumes contain one or more Papers
not included in the particular Course for which the student
enrolled, but in no case are any subjects omitted that form
a part of such Course. This plan is particularly advan¬
tageous to those students who enroll for more than one
Course, since they no longer receive volumes that are, in
some cases, practically duplicates of those they already
have. This arrangement also renders it much easier to
revise a volume and keep each subject up to date.
Each volume in the Library contains, in addition to the
text proper, the Examination Questions and (for those
subjects in which they are issued) the Answers to the
Examination Questions.
In preparing these textbooks, it has been our constant
endeavor to view the matter from the student’s standpoint,
and try to anticipate everything that would cause him
trouble^ The utmost pains have been taken to avoid and
correct any and all ambiguous expressions—both those due
to faulty rhetoric and those due to insufficiency of state¬
ment or explanation. As the best way to make a statement,
explanation, or description clear is to give a picture or a
iii
IV
PREFACE
diagram in connection with it, illustrations have been used
almost without limit. The illustrations have in all cases
been adapted to the requirements of the text, and projections
and sections or outline, partially shaded, or full-shaded
perspectives have been used, according to which will best
produce the desired results.
The method of numbering pages and articles is such that
each part is complete in itself; hence, in order to make the
indexes intelligible, it was necessary to give each part a
number. This number is placed at the top of each page, on
the headline, opposite the page number; and to distinguish
it from the page number, it is preceded by a section
mark (§). Consequently, a reference, such as §3, page 10,
can be readily found by looking along the inside edges of
the headlines until §3 is found, and then through §3 until
page 10 is found.
International Correspondence Schools
CONTENTS
Arithmetic Section Page
Definitions. 1 1
Notation and Numeration . 1 2
Addition. 1 6
Subtraction. 1 12
Multiplication. 1 16
Division . 1 22
Factors. 1 27
Greatest Common Divisor. 1 29
Least Common Multiple. 1 31
Cancelation. 1 33
Fractions. 2 1
Decimals. 2 18
Percentage. 3 1
Interest. 3 7
Ratio. 3 10
Proportion . 3 12
Signs of Aggregation. 3 19
Involution . 4 1
Evolution. 4 3
Square Root . 4 4
Cube Root. 4 10
Roots of Fractions. 4 15
Table Method of Extracting Square and
Cube Root. 4 16
Denominate Numbers. 5 1
Measures. 5 2
Addition of Compound Quantities .... 5 12
Subtraction of Compound Quantities ... 5 13
Multiplication of Compound Quantities . . 5 14
Division of Compound Quantities .... 5 16
v
VI
CONTENTS
Mensuration Section
Mensuration of Surfaces. 6
Triangles. 6
Quadrilaterals. 6
Polygons. 6
The Circle . 6
Mensuration of Solids. 6
The Prism . . . 6
The Cylinder. 6
The Pyramid and Cone. 6
The Sphere. 6
Mensuration of Lumber.• . . 6
Mechanical Definitions
Machine Elements. 7
Shafting. 7
Power Transmission. 7
Belting. 7
Pulleys. 7
Rope Transmission. 7
Gearing. 7
Cams. 7
Levers. 7
Mechanical Calculations
Rules Pertaining to the Transmission of
Power. 8
Rules Applying to Shafting. 8
Rules Applying to Speeds. 8
Rules Applying to Belts. 8
Rope Transmission. 8
Rules Applying to Gears. 8
Rules Applying to Levers. 8
Yarn Calculations, Cotton
Single Yarns .. 9
Sizing Roving and Yarn.9
Ply Yarns. 9
Beam Calculations. 9
CONTENTS vii
Yarn Calculations, Cotton —Continued Section Page
Average Numbers. 9 15
Fancy Warps. 9 16
Twist in Yarns . 9 18
Breaking Weight of Cotton Warp Yarn . . 9 19
Metric System of Numbering Yarns ... 9 21
Yarn Calculations, Woolen and Worsted
Worsted Yarns.10 1
Sizing Worsted Yarns.10 3
Sizing Worsted Roving .10 6
Ply Yarns.10 7
Woolen Yarns.. ... 10 13
Sizing Woolen Roving and Yarn .... 10 15
Ply Yarns.10 17
Beam Calculations.10 18
Fancy Warps. .10 21
Average Numbers.10 23
Metric System of Numbering Yarns ... 10 25
Yarn Calculations, General
Single Yarns.11 2
Cotton Yarns.11 2
Numbering System.11 2
Sizing Cotton Roving and Yarn.11 4
Woolen Yarns.11 7
Run System of Numbering.11 7
Cut System of Numbering.11 9
Sizing Woolen Roving and Yarn .... 11 9
Worsted Yarns.11 11
Sizing Worsted Yarns.11 12
Sizing Worsted Roving .11 13
Silk Yarns . 11 14
Jute, Linen, and Ramie. .11 16
Equivalent Counts.• . . 11 18
Ply Yarns. 11 20
Woolen and Worsted Ply Yarns.11 26
Ply Yarns of Different Materials .... 11 27
Diameter of Yarns ..11 29
Vlll
CONTENTS
Yarn Calculations, General —Continued Section Page
Beamed Yarns.11 30
Average Numbers .11 33
Fancy Warps.11 35
Metric System of Numbering Yarns . . 11 38
Cloth Calculations, Cotton
Calculations Necessary for Cloth Produc¬
tion . 12 1
Harness Calculations.12 3
Reeds .12 4
Calculations for Warp Yarns.12 5
Calculations for Filling Yarn.12 10
Finding Yards per Pound.12 13
Figuring Particulars From Cloth Samples . 12 16
Finding Hanks of Warp Yarn.12 18
Finding Hanks of Filling.12 20
Counts of Filling to Preserve Yards per
Pound.12 22
Average Sley.12 23
Short Rules Used in Obtaining Particu¬
lars of Cloth Samples . ■.12 24
Cloth Calculations, Woolen and Worsted
Calculations Necessary for Cloth Produc¬
tion .13 1
Construction of Fabrics.13 2
Harness Calculations.13 4
Reed Calculations.13 5
Contraction in Weaving .13 9
Shrinkage in Finishing.13 11
Fancy Patterns. .... 13 14
Figuring Particulars From Cloth Samples 13 21
Draft Calculations
Drafting .15 1
Drafting With Common Rolls.15 5
Gearing of Rolls.15 9
Driving and Driven Gears . •.15 11
CONTENTS
IX
Draft Calculations— Continued Section
Calculating Draft for Common Rolls . . 15
Break Draft.15
Metallic Rolls.15
Drafting With Special Reference to the Mill 15
Reading Textile Drawings
Representation of Objects.91
Kinds of Drawings.91
Mechanical Drawings.91
Elevations or Side Views.91
Plans.91
Sectional Views.91
Detail and Assembly Drawings.91
Perspective Views . . . •.91
Diagrammatic Views.91
Lines Used on Drawings.91
Conventional Methods of Representing
Objects.91
Page
12
17
20
25
1
2
3
6
7
11
17
18
20
22
28
ARITHMETIC
(PART 1)
DEFINITIONS
1. Arithmetic may be defined as the science of numbers
and the art of computing with them.
2. A unit is a single thing, or one, as one mill, one loom,
or it may be used collectively, as one dozen, which really
means twelve articles, or one hank, which is a definite number
of yards.
3. A- number is a unit or a collection of units, as one,
three, five, seven.
4. A concrete number is a number that refers to some
particular article or quantity, as four spindles, seven boilers,
three gallons.
5. An abstract number is a number that has no refer¬
ence to any particular article or quantity, as three, five, seven,
nine.
6. The unit of a number is one of the same kind as
the collection of units represented by the number. The unit
of twelve is one , of fifty looms is one loom, of ten dollars is
one dollar.
7 . TAke numbers are numbers that refer to units of the
same kind, as three yards and six yards, five quarts and eight
quarts.
8. Unlike numbers are numbers that refer to units of
different kinds, as two miles and ten minutes , four spools and
six shuttles.
For notice of copyright. see page immediately following the title page
?r
2
ARITHMETIC
§1
NOTATION AND NUMERATION
9. Numbers are expressed in three ways: (1) by words;
(2) by figures; (3) by letters.
10 . Notation is the art of expressing numbers by
figures or letters.
11. Numeration is the art of expressing in words num¬
bers that have been previously expressed by figures or letters.
12 . Arabic notation, which is commonly used for
arithmetical calculations, employs ten characters , or signs ,
called figures for expressing numbers. These, with their
names, are as follows:
Figures 1234567 89 0
Names one two three four five six seven eight nine naught ,
cipher ,
or zero
The last character (0) is called naught, cipher, or zero,
and when standing alone has no value.
The other nine figures are called digits, and each has an
independent value of its own. The cipher is not a digit;
hence, whenever the word digit is used, it refers to one of
the first nine figures above named.
Any whole number is called an integer, or integral
number.
13 . As there are only ten figures, it is evident that in
expressing numbers each figure may have a different value
at different times. In other words, figures have simple values
and local , or relative , values, according to their position in
relation to other figures.
14 . The simple value of a figure is its value as a digit
when standing alone or, in a collection of figures, when stand¬
ing at the right of the other figures.
§1
ARITHMETIC
3
15 . The local, or relative, value of a figure is the value
it expresses when placed at the left of other figures.
16 . The particular position that a figure occupies with
relation to other figures is called its place; thus, in 45
(forty-five) the 5 occupies the first place counting from the
right, and the 4.the second place.
17 . The local size, or value, of the units of a figure differs
according to the place they occupy; thus, in the number 566
[five hundred (and) sixty-six], each of the figures without
regard to its place represents units; but the 6 occupying the
first place, counting from right to left, represents only
6 single units, while the 6 occupying the second place repre¬
sents 6 tens, or 6 units each ten times the size, or value, of
a unit of the first place, and the 5 occupying the third place
from the right represents 5 hundreds, or 5 units each one
hundred times the size, or value, of a unit of the first place,
and ten times the value of a unit of the second place.
An illustration of the increasing value of figures due to their
local, or relative, place in relation to the unit place is as follows:
The figure 7 standing alone represents seven
units; thus.... v 7
If another 7 is placed at the left of the previous
figure, its value will not be seven units, but seven
tens; i. e., ten times seven units, or seventy units,
and the whole number will be the sum of seventy
and seven, or seventy-seven; thus.. . 77
If another 7 is placed at the left of the previous
one, its value will be seven hundreds; i. e., one
hundred times the value of the seven in the first
place or ten times the value of the seven in the
second place, and the number will be the sum of
seven hundred, seventy, and seven, or seven
hundred (and) seventy-seven; thus. 777
If still another 7 is added its value will be seven
thojisands; i. e., ten times the seven in the third
place, one hundred times the seven in the second
place, and one thousand times the value of the
4
ARITHMETIC
§1
seven in the first, or units, place, and the number
will be the sum of seven thousand, seven hundred,
seventy, and seven, or seven thousand seven
hundred (and) seventy-seven; thus. 7777
Another 7 will increase the number seven tens of
thousands , and it will read seventy-seven thousand
seven hundred (and) seventy-seven; thus .... 77777
Another 7 will increase the number seven
hundreds of thousands , and it will read seven hun¬
dred (and) seventy-seven thousand seven hundred
(and) seventy-seven; thus .... ....... 777777
Another 7 added in the seventh place, beginning
with the unit and counting to the left, will increase
the number seven millions , and it will then read
seven million seven hundred (and) seventy-seven-
thousand seven hundred (and) seventy-seven; thus 7777777
18. In general, the following law may be drawn from
the above:
Law.— The value expressed by any figure is always increased
tenfold each time it is removed one place to the left.
19. The cipher (0) has no value in itself but when used
in conjunction with other figures it is useful in determining the
place of such other figures. To represent the number six hun¬
dred seven only two digits (see Art. 12) are necessary, one to
represent six hundred units by being placed in the third , or
hundreds , place, and one to represent the seven units. The tens ,
or second , place is represented by a cipher; thus, 607. If the
two digits were placed together, thus 67, the 6 would be in
the tens , or second, place and the number would be sixty-seven.
If there were tivo ciphers between the 6 and 7, the 6 would
be thrown into the thousands place and the number would be
composed of thousands and units , reading six thousand seven ,
and written 6,007. If the number were six hundred seventy ,
it would have the cipher at the right-hand side to show that
there were no units. In this case, the number would be
composed of hundreds and tens , and would be read six hun¬
dred seventy , which is written 670.
§1
ARITHMETIC
5
20. In writing numbers it is customary to separate
them by means of commas into periods of three figures
each, beginning at the right. These groups of three figures
are called, respectively, the periods of units, of thousands , of
millions , of billions , etc. . •
A clear understanding of the method of separating a
number into periods, the names of the periods, and the
names of the respective places of each period may be
obtained from the following example, showing how the
number 466,735,438,894,278 is divided into periods.
Period
of
Period of
Period
of
Period of
Period of
Trillions
Billions
Millions
Thousands
Units
in
XJ
G
in
g
o
C/3
a
C/3
G
o
Gj
in
G
in
• H
o
• H
o
TJ
\ -<
C/3
r—H
r-H
in
a
o
.G
G
u
EH
c
_o
• rH
m
C/3
a
o
• r-t
§
Eh
G
in
M—1
M—1
• rH
• rH
F <
M—1
o
o
• rH
u
o
• t-H
o
• rH
o
.G
in
C/3
EH
C/3
PQ
n
in
Eh
in
C/3
H3
TD
e
r O
).
Note. —It frequently happens, when adding a long column of figures,
that the sum of two numbers, one of which does not occur in the
addition table, is required. Thus, in the first column of example 2 ( a)
the sum of 16 and 6 was required. We know from the table that
6 + 6 = 12; hence, the first figure of the sum is 2. Now, the sum of any
number less than 20 and of any number less than 10 must be less than 30,
since 20 + 10 = 30; therefore, the sum is 22. Consequently, in cases of
this kind, add the first figure of the larger number to the smaller number,
and if the result is greater than 9, increase the second figure of the
larger number by 1. Thus, 44 + 7 — ? 4 + 7 = 11; hence, 44 + 7 = 51.
The addition may also be performed as follows:
4 2 5
3 6
9 2 15
4
9 0 7
sum 1 0 5 8 7 Ans.
Explanation.— The sum of the numbers in the units
column is 27 units, or 2 tens and 7 units. Write the 7 units as
the first, or right-hand figure, in the sum. Reserve the 2 tens
and add them to the figures in the tens column. The sum of
the figures in the tens column, plus the 2 tens reserved and
carried from the units column is 8, which is written down as
the second figure in the sum. There is nothing to carry to
the next column, because 8 is less than 10. The sum of the
numbers in the next column is 15 hundreds, or 1 thousand
and 5 hundreds. Write down the 5 as the third, or hundreds,
figure in the sum and carry the 1 to the next column. 1 + 9
= 10, which is written down at the left of the other figures.
10
ARITHMETIC
1
This method saves space and figures, but that shown in
example 2 is to be preferred when adding a long column.
Example 3. — Add the numbers in the column below:
.Solution . — 8 9 0
82
90
393
281
80
770
83
492
80
383
84
191
sunt 3 8 9 9 Ans.
Explanation.— The sum of the digits in the first column
equals 19 units, or 1 ten and 9 units. Write down the 9 and
carry 1 to the next column. The sum of the digits in the
second column + 1 = 109 tens, or 10 hundreds and 9 tens.
Write down the 9 and carry the 10 to the next column.
The sum of the digits in this column plus the 10 carried is 38.
The entire sum is 3,899.
28 . The process of mentally adding the second figure of
the sum of a column of figures in with the next column is
' i . . r
called carrying.
In some cases, where long columns of figures are being
added, it is necessary to add tzvo or more figures in with
the next column. To illustrate this point, suppose that in
a problem in addition the sum of the units column was' 347.
In this case, the 7 would be placed under the units column
and the 34 mentally added in with the tens column.
29 . Rule.—I. Write the nunibers so that all the figures
of the same order will be in the same column.
II. Add all the figures in the right-hand column , or
column of units , and if their sum is less than ten , write it
underneath. If the sum is ten or more , write down the unit
1
ARITHMETIC
11
figure only and add the figure or figures denoting the tens
in with the next column.
III. Proceed in like manner with each column until all are
added, taking care to write down the entire sum of the last , or
left-hand , column.
30 . Proof. —There is no complete proof possible of
addition. The usual method is to repeat the operation in
the reverse order; that is, add each column from top to
bottom. If the same result is obtained as by adding from
bottom to top, the work is probably correct.
Another method is to add the columns of figures by the
method of addition explained in example 2, Art. 27 .
EXAMPLES FOR PRACTICE
Find the sum of:
(а) 24 + 100 + 365.
(б) 310 + 985 + 67 + 546.
(c) 444 + 1,362 + 4,678 + 5,003.
(d) 7,891 + 338 + 467 + 2,136/
(e) 9 + 90 + 900 + 9,000 + 90,000.
(/) 1,234 +-2,341 + 3,412 + 4,123. 1
\g) 34,567 + 8,934 + 303+4,212.
'(a) 489
(5) 1,908
Ans. •
(c) 11,487
(if) 10,832
(e) 99,999
(f) 11,110
-(g) 48,016
1. What is the sum of five million three hundred seventy thousand
four hundred seven plus seven hundred seventy-four thousand three
hundred forty-four pluk five thousand three hundred ninety-five?
:r ^ V» - Ans. 6,150,146
2. A contractor was employed 1 weeks in erecting a chimney.
The first week he built 30 feet, the seeqpd week 40 feet, the third and
fourth weeks each 50 feet; what was the height of the chimney?
Ans. 170 ft.
3. A large corporation operates three weave sheds; the production
of No. 1 weave room was 236,450 yards of sheeting for a certain week;
during the same week, weave rooms Nos. 2 and 3 produehd, respect¬
ively, 125,325 and 133,650 yards of fancy goods; what was the total
production in yards for the week? Ans. 495,425 yd.
4. A mill receives machinery, stock, and supplies during the first
week of the month valued at $3,475; during the second week, $2,950
worth; during the third week, $4,380 Worth; and during the rest of the
month, $4,895 worth; what was the total expense of these materials
for the month? Ans. $15,700
12
ARITHMETIC
1
SUBTRACTION
31. Subtraction is the process of finding what part of
a given quantity remains when a certain part is taken from
it, and is the reverse of the process of addition.
The minuend is the given quantity that is the larger of
the two numbers involved in the process of subtraction.
The subtrahend is the given part that is taken from the
minuend and thus is necessarily the smaller of the two
numbers.
The remainder, or difference, is the part of the minu¬
end that is left after the subtrahend has been taken from it.
32. The sign of subtraction is a short horizontal
line, thus — . It is read minus, and means less. Thus,
8 — 5 is read 8 minus 5, and means that 5 is to be subtracted ,
or taken from 8.
The whole example would be written 8 — 5 = 3, and would
be read 8 minus 5 equals 3. In this case, the number 8 is
the minuend, 5 is the subtrahend, and 3 is the remainder.
33. Only like numbers can be subtracted, as it is evidently
impossible to subtract feet from dollars or pounds from yards.
34. When one number is to be subtracted from another,
the subtrahend is placed tmder the minuend in such a position
that all the figures of the same order will be in the same
column, or, in other words, units must be placed under units ,
tens under tens, and so on.
After the subtrahend is placed under the minuend, a line
is drawn below it and the result of the subtraction, or the
remainder , placed beneath this line.
Example.— From 7,849 subtract 5,323.
Solution.— minuend 7 8 4 9
subtrahend 5 3 2 3
remainder 2 5 2 6 Ans.
§1
ARITHMETIC
13
Explanation. —Begin at the right-hand, or units, column
and subtract in succession each figure in the subtrahend from
the one directly above it in the minuend, and write the remain¬
ders below the line. The result is the entire remainder.
35. It will be noticed in the example just given that in
no case does the value of the subtrahend figure of any order
exceed in value the minuend figure of the same order. If in
any case the minuend figure of any order (excepting the order
on the extreme left) has a value less than the corresponding
subtrahend figure, then another operation is involved, namely,
that of borrowing. If the minuend figure of the left-hand
order is of a less value than the subtrahend figure of the same
order, then subtraction is impossible, since the whole minuend
is smaller than the subtrahend, and in arithmetic a larger
value cannot be taken from a smaller.
Example 1.—From 8,453 take 844.
Solution. — minuend 8 4 5 3
subtrahend 8 4 4
remahider 7 6 0 9 Ans.
Explanation. —Begin at the right-hand, or units, column
to subtract. We cannot take 4 from 3, and must, therefore,
borrow 1 from the 5 in the tens column and annex it to the 3
in the units column. The 1 ten = 10 units, which added to
the 3 in the units column = 13 units. 4 from 13 = 9, the
first, or units, figure in the remainder.
Since we borrowed 1 from the 5, only 4 remains; 4 from 4
= 0, the second, or tens, figure. We cannot take 8 from 4,
and must, therefore, borrow 1 from 8 in the thousands column.
Since 1 thousand = 10 hundreds, 10 hundreds + 4 hundreds
= 14 hundreds, and 8 from 14 = 6, the third, or hundreds,
figure in the remainder.
Since we borrowed 1 from 8, only 7 remains, from which
there is nothing to subtract; therefore, 7 is the next figure in
the remainder, or answer.
The operation of borrowing is performed by mentally
placing 1 before the figure following the one from which it is
borrowed. In the above example the 1 borrowed from 5 is
14
ARITHMETIC
§1
placed before 3, making it 13, from which we subtract 4.
The 1 borrowed from 8 is placed before 4, making 14, from
which 8 is taken.
Example 2.—Find the difference between 10,000 and 8,763.
Solution. — minuend 100 00
subtrahend 8 7 6 3
remainder 12 3 7 Ans.
Explanation. —In the above example, we borrow 1 from
the second column and place it before 0, making 10;
3 from 10 = 7. In the same way we borrow 1 and place
it before the next cipher, making 10; but as we have
borrowed 1 from this column and have taken it to the
units column, only 9 remains from which to subtract 6;
6 from 9 = 3. For the same reason we subtract 7 from 9
and 8 from 9 for the next two figures, and obtain a total
remainder of 1,237.
36 . Rule.—I. Place the subtrahend (or smaller number)
under the minuend {or larger number) so that the figures of the
same order will be in the same column. Commencing at the
right, subtract each order of the subtrahend from the correspond-
ing order of the minuend and write the result in the same order
of the remainder.
II. If any order of the minuend has a less vahie than the
corresponding order of the subtrahend , increase its value by 10
and subtract; then diminish by 1 the figure in the next higher
order of the minuend and subtract.
37 . There are two methods of proving subtraction, as
follows:
Proof 1. — Add the remainder and the subtrahend, and, if
the work is correct, their sum will equal the minuend.
Example 1. — Prove that the difference between 1,000 and 987 is 13.
Solution. — subtrahend 98 7
remainder 13
minuend 10 00 Ans.
This method of proof depends on the principle that sub¬
traction is the reverse of addition, and vice versa.
1
ARITHMETIC
15
Proof 2.— Subtract the remainder from the minuend and
if the work is correct the result will equal the subtrahend.
Example 2.—Prove that 1,265 minus 823 equals 442.
Solution.— minuend 12 65
remainder 4 4 2
subtrahend 8 23 Ans.
This method of proof depends on the principle that the
smaller of any two numbers is equal to the larger less the
difference of the numbers.
EXAMPLES FOR PRACTICE
Subtract:
(a) 31 from 58.
(b) 92 from 99.
(c) 328 from 649.
{d) 837 from 979.
(e) 345 from 568.
(/) 799 from 843.
(g) 876 from 1,123.
(/;) 7,346 from 9,258.
(i) 8,675 from (175 + 8,700 + 98).
(/) 3,258 from (2,001 + 7,890).
f(«)
27
-(*)
7
(c)
321
id)
142
\(e)
223
(/)
44
(g)
247
W
1,912
(i)
298
(/)
6,633
Note—I n the last two examples, it will be noticed that a portion of the example
is included in parentheses, thus ( ). In both cases this indicates that the operations
indicated within the parentheses are to be performed first; and after the value of the
part within the parentheses is obtained as a single number, the subtraction is per¬
formed as in the preceding examples.
i
16
ARITHMETIC
1
MULTIPLICATION
38. Multiplication is a short process of adding a
number to itself a definite number of times.
The multiplicand is the number that is added, or
multiplied.
The multiplier is the number that indicates the number
of times the multiplicand is to be added to itself, or it is the
number by which we multiply.
The product is the result obtained by multiplication, or
the number obtained by adding the multiplicand to itself the
number of times indicated by the multiplier.
39. The multiplier always signifies a number of times and
is therefore an abstract number.
The multiplicand may be either an abstract or a concrete
number.
The product and the multiplicand are always like numbers.
40. The sign of multiplication is a small oblique
cross, thus X, and is placed between the multiplicand and
the multiplier.
When the multiplier precedes the multiplicand the sign of
multiplication is read times; thus 6 X $5 = $30 is read six
times five dollars equals thirty dollars. In this example the
figure 6, being an abstract number, must be the multiplier.
When the multiplicand precedes the multiplier, the sign of
multiplication is read multiplied by; thus, $5x6 = $30 is
read five dollars multiplied by six equals thirty dollars.
41. In the table on the following page, the product of
any two numbers (neither of which exceeds 12) may be
found.
§1
ARITHMETIC
17
I
times
i
is
i
2
times
1
is
2
3
times
I
is
3
I
times
2
is
2
2
times
2
is
4
3
times
2
is
6
I
times
3
is
3
2
times
3
is
6
3
times
3
is
9
I
times
4
is
4
2
times
4
is
8
3
times
4
is
12
I
times
5
is
5
2
times
5
is
10
3
times
5
is
15
I
times
6
is
6
2
times
6
is
12
3
times
6
is
18
I
times
7
is
7
2
times
7
is
14
3
times
7
is
21
I
times
8
is
8
2
times
8
is
16
3
times
8
is
24
I
times
9
is
9
2
times
9
is
18
3
times
9
is
27
I
times
IO
is
IO
2
times
IO
is
20
3
times
10
is
30
I
times
11
is
11
2
times
11
is
22
3
times
11
is
33
I
times
12
is
12
2
times
12
is
24
3
times
12
is
36
4
times
I
is
4
5
times
1
is
5
6
times
1
is
6
4
times
2
is
8
5
times
2
is
10
6
times
2
is
12
4
times
3
is
12
5
times
3
is
15
6
times
3
is
18
4
times
4
is
16
5
times
4
is
20
6
times
4
is
24
4
times
5
is
20
5
times
5
is
25
6
times
5
is
30
4
times
6
is
24
5
times
6
is
30
6
times
6
is
36
4
times
7
is
28
5
times
7
is
35
6
times
7
is
42
4
times
8
is
32
5
times
8
is
40
6
times
8
is
48
4
times
9
is
36
5
times
9
is
45
6
times
9
is
54
4
times
IO
is
40
5
times
IO
is
50
6
times
10
is
60
4
times
11
is
44
5
times
11
is
55
6
times
11
is
66
4
times
12
is
48
5
times
12
is
60
6
times
12
is
72
7
times
I
is
7
8
times
1
is
8
9
times
1
is
9
7
times
2
fs
14
8
times
2
is
16
9
times
2
is
18
7
times
3
is
21
8
times
3
is
24
9
times
3
is
27
7
times
4
is
28
8
times
4
is
32
9
times
4
is
36
7
times
5
is
35
8
times
5
is
40
9
times
5
is
45
7
times
6
is
42
8
times
6
is
48
9
times
6
is
54
7
times
7
is
49
8
times
7
is
56
9
times
7
is
63
7
times
8
is
56
8
times
8
is
64
9
times
8
is
72
7
times
9
is
63
8
times
9
is
72
9
times
9
is
81
7
times
IO
is
70
8
times
IO
is
80
9
times
10
is
90
7
times
ii
is
77
8
times
11
is
88
9
times
11
is
99
7
times
12
is
84
8
times
12
is
96
9
times
12
is
108
IO
times
I
is
IO
11
times
1
is
11
12
times
1
is
12
IO
times
2
is
20
11
times
2
is
22
12
times
2
is
24
IO
times
3
is
30
11
times
3
is
33
12
times
3
is
36
IO
times
4
is
40
11
times
4
is
44
12
times
4
is
48
IO
times
5
is
50
11
times
5
is
55
12
times
5
is
60
IO
times
6
is
60
11
times
6
is
66
12
times
6
is
72
IO
times
7
is
70
11
times
7
is
77
12
times
7
is
84
IO
times
8
is
80
11
times
8
is
88
12
times
8
is
96
IO
times
9
is
90
11
times
9
is
99
12
times
9
is
108
TO
times
IO
is
100
11
times
IO
is
110
12
times
10
is
120
IO
times
ii
is
I IO
11
times
11
is
121
12
times
11
is
132
IO
times
12
is
120
11
times
12
is
132
12
times
12
is
144
This table should be carefully committed to memory.
Since 0 has no value, the product of 0 and any number is 0.
18
ARITHMETIC
1
42, To multiply a number by one figure only:
Example. —Multiply 425 by 5.
Solution. — multiplicand 4 25
multiplier 5
product 212 5 Ans.
Explanation. —For convenience, the multiplier is gener¬
ally written under the right-hand figure of the multiplicand.
On looking in the multiplication table, we see that 5 X 5 is 25.
Multiplying the first figure at the right of the multiplicand,
or 5, by the multipler, 5, it is seen that 5 times 5 units is
25 units, or 2 tens and 5 units. Write the 5 units in the units
place in the product and reserve the 2 tens to add to the
product of tens. Looking in the multiplication table again,
we see that 5 X 2 is 10. Multiplying the second figure of
the multiplicand by the multiplier 5, we see that 5 times
2 tens is 10 tens, and 10 tens plus the 2 tens reserved are
12 tens, or 1 hundred plus 2 tens. Write the 2 tens in the
tens place, and reserve the 1 hundred to add to the product
of hundreds. Again, we see by the multiplication table that
5 X 4 is 20. Multiplying the third, or last, figure of the
multiplicand by the multiplier 5, we see that 5 times 4 hun¬
dreds is 20 hundreds, and 20 hundreds plus the 1 hundred
reserved are 21 hundreds, or 2 thousands and 1 hundred,
which we write in the thousands and the hundreds places,
respectively. Hence, the product is 2,125. This result is
the same as adding 425 five times. Thus,
425
425
425
425
425
sum 2 12 5 Ans.
§1
ARITHMETIC
19
EXAMPLES FOR PRACTICE
Find the product of:
(a)
61,483 X 6.
(«)
368,898
(6)
12,375 X 5.
(b)
61,875
(c)
10,426 X 7.
(c)
72,982
(d)
10,835 X 3.
Ans.-
(d)
32,505
(e)
98,376 X 4.
4
(e)
393,504
(f)
10,873 X 8.
( f)
86,984
(g)
71,543 X 9.
(g)
643,887
(h)
218,734 X 2.
(h)
437,468
43. To multiply a number by two or more figures:
Example.— Multiply 475 by 234.
Solution. — multiplicand 4 7 5
multiplier 2 3 4
19 0 0
14 2 5
9 5 0
product 11115 0 Ans.
Explanation.— For convenience, the multiplier is gener¬
ally written under the multiplicand, placing units under units,
tens under tens, etc.
We cannot multiply by 234 at one operation; we must,
therefore, multiply by the parts and then add the partial
products.
The parts by which we are to multiply are 4 units, 3 tens,
and 2 hundreds. 4 times 475 = 1,900, the first partial
product; 3 times 475 = 1,425, the second partial product , the
right-hand figure of which is written directly under the figure
multiplied by , or 3; 2 times 475 = 950, the third partial
product , the right-hand figure of which is written directly
under the figure multiplied by, or 2.
The sum of these three partial products is 111,150, which
is the entire product.
44. Rule.—I. Write the m^dtiplier under the multipli¬
cand , so that units are under units, tens under tens , etc.
II. Begin at the right and multiply each figure of the mul¬
tiplicand by each successive figure of the multiplier , placing the
20 ARITHMETIC § 1
right-hand figure of each partial product directly under the
figure Jised as a multiplier.
III. The sum of the partial products will equal the required
Product.
45. Proof. —Review the work carefully; or multiply the
multiplier by the multiplicand, afid if the results agree, the
work is correct.
46. When there is a cipher in the multiplier , multiply
by it the same as with the other figures. Since the product
of any number and 0 is 0, the work may be greatly short¬
ened by writing the first cipher of the partial product, then
multiplying by the next figure of the multiplier and writing
the partial product beside the cipher, as shown in the follow¬
ing examples:
(a)
(■ b )
(c)
(d)
0
2
1 5
7 0 8
X
0
X 0
X
0
X 0
0
Ans.
0
Ans.
0 0 Ans.
0 0 0
Ans.
(e)
(f)
(g)
3
1 1
4
4
0 0
8
3 1 2
6 4
2 0
3
3 0
5
1 0
0 2
9
3 4
2
2 0
0 4
0
6 2 5
2 8
2 2
8 0
12 0 2
4 0
312640
0
3 2
1 4
2
Ans.
12 2 2
4 4
0
Ans.
313265
2 8 Ans.
47.
If
the
multiplier
or
multiplicand,
or both,
ends in
one or more ciphers, write the multiplier so that the right-
hand digit of the multiplier is under the right-hand digit of
the multiplicand; then, disregarding the ciphers on the right
of either multiplier or multiplicand, or both, multiply as
previously directed and write as many ciphers on the right
of the product as there are ciphers on the right of multiplier
and multiplicand. Thus:
Example. —Find the product of: (a) 78,392 X 248,000; (b) 7,839,200
X 248; (<-) 7,839,200 X 248,000.
§1
ARITHMETIC
21
Solution.—
(a)
7 8 3 9 2
248000
627136
313568
156784
194412160 0 0 Ans.
(c)
7839200
248000
6 2 7 1 3 6
3 1 3 5 6 8
156784
(*)
7839200
2 4 8
627136
313568
156784
1 944121600 Ans.
1944121600000 Ans.
EXAMPLES FOR PRACTICE
Find the product of:
(a)
13 X 3.
(*)
2,679 X 5.
(c)
2,893 X 14.
(d)
6,734 X 30.
(e)
39,465 X 89.
(/)
14,674 X 103.
U)
367,801 X 2,705.
w
567,893 X 1,003.
'(a)
39
(6)
13,395
(c)
40,502
(d)
202,020
(e)
3,512,385
(/)
1,511,422
Or)
994,901,705
(h)
569,596,679
1. A certain mill produces on an average 68,725 yards of cloth
per day; what is the production in a year of 296 working days?
Ans. 20,342,600 yd.
2. A reservoir is fed by two pumps each having a capacity of
875 gallons per hour. During a certain week one pump was in
operation 39 hours and the other 55 hours; during the same week
75,000 gallons of water was drawn from the reservoir; what was the
excess of water supplied to the reservoir by the pumps during the
week? Ans. 7,250 gal.
3. A certain mill contains 1,830 looms, each of which produces
175 yards of cloth per week; what is the total production?
Ans. 320,250 yd.
4. During a snow storm, snow fell to such an extent that the pres¬
sure on the roof of a mill was 4 pounds per square foot. The area
of the roof is 23,625 square feet; what was the total weight of snow
supported by the roof? Ans. 94,500 lb.
22
ARITHMETIC
1
DIVISION
48. Division is a process of finding how many times
one number is contained in another.
The number that is divided is called the dividend.
The number by which the dividend is divided is called the
divisor.
The number that denotes the number of times the divisor
is contained in the dividend is called the quotient.
49. When the dividend does not contain the divisor an
exact or even number of times, the excess is called the
remainder. The remainder, being part of the dividend, is
always of the same kind as the dividend, and must always be
less than the divisor.
50. If the divisor and dividend denote the same kind of
units, then the quotient is an abstract number; thus, to divide
20 looms by 5 looms is to find the number of times 5 looms
must be taken to obtain 20 looms, or 4 times.
51. If the divisor is an abstract number, then the quo¬
tient denotes units of the same kind as the dividend; thus, to
divide 20 looms by 4 is to find the number of looms in each
part when the 20 looms are divided into 4 equal parts.
52. The sign of division is a short horizontal line with
a dot above and a dot below it, thus - 4 - , and indicates that
the number preceding the sign is to be divided by the num¬
ber following it, the sign being read divided by. Thus, the
expression 12 -4- 6 = 2 is read 12 divided by 6 equals 2, or
G into 12 equals 2.
Division is also indicated by writing the dividend above a
short horizontal line and the divisor below; thus, f = 2, and
is read 8 divided by 4 equals 2 or 8 over 4 equals 2.
Division is the reverse of multiplication.
1
ARITHMETIC
23
53. Dividing or multiplying both the divisor and divi¬
dend by the same number does not alter the value of the
quotient.
54. To divide wlien the divisor consists of but one
figure:
Example. —What is the quotient of 875 4- 7?
divisor dividend Quotient
Solution. — 7)875(125 Ans.
7
T 7
1 4
~3 5
3 5
remainder 0
Explanation. — 7 is contained in 8 once. Place the 1
as the first, or left-hand, figure of the quotient. Multiply
the divisor 7 by the 1 of the quotient, and place the prod¬
uct 7 under the 8 in the dividend, and subtract. Beside the
remainder 1, bring down the next figure of the dividend, in
this case 7, making 17; 7 is contained in 17, 2 times. Write
the 2 as the second figure of the quotient. Multiply the
divisor 7 by the 2 in the quotient, and subtract the product
from 17. Beside the remainder 3, bring down the next
figure of the dividend, in this case 5, making 35. 7 is con¬
tained in 35, 5 times, which is placed in the quotient. Mul¬
tiplying the divisor by the last figure of the quotient,
5 times 7 = 35, which, subtracted from 35 under which it
is placed, leaves 0. Therefore, the quotient is 125. This
method is called long division.
55. In sliort division, only the divisor, dividend, and
quotient are written, the operations being performed mentally.
dividend
divisor 7 ) 8 1 7 3 5
quotient 12 5 Ans.
The mental operation is as follows: 7 is contained in 8,
once and 1 remainder; imagine 1 to be placed before 7,
making 17; 7 is contained in 17, 2 times and 3 over; imagine
24
ARITHMETIC
§1
3 to be placed before 5, making 35; 7 is contained in 35,
5 times. These partial quotients, placed in order as they
are found, make the entire quotient 125.
56. To divide when the divisor consists of two
or more figures:
Example. —Divide 2,702,836 by 63.
divisor dividend quotient
Solution.— 63)2702836(42902 Ans.
2 5 2
18 2
12 6
5 6 8
5 6 7
1 3 6
12 6
1 0
Explanation. —As 63 is not contained in the first two
figures, 27, we must use the first three figures, 270. Now,
by trial we must find how many times 63 is contained in 270.
6 is contained in the first two figures of 270, 4 times. Place
the 4 as the first, or left-hand, figure in the quotient. Mul¬
tiply the divisor, 63, by 4, and subtract the product, 252, from
270. The remainder is 18, beside which we write the next
figure of the dividend, 2, making 182. Now, 6 is contained
in the first two figures of 182, 3 times, but on multiplying
63 by 3, we see that the product, 189, is too great, so we
try 2 as the second figure of the quotient. Multiplying the
divisor, 63, by 2 and subtracting the product, 126, from 182,
the remainder is 56, beside which we bring down the next
figure of the dividend, making 568. 6 is contained in 56
9 times. Multiply the divisor, 63, by 9 and subtraQt the
product, 567, from 568. The remainder is 1, and bring¬
ing down the next figure of the dividend, 3, gives 13. As 13
is smaller than 63, we write 0 in the quotient and bring down
the next figure, 6, making 136. 63 is contained in 136,
2 times, with a remainder of 10. Therefore, 42,902e3 is
the quotient.
1
ARITHMETIC
25
57. The proper remainder is always smaller than the
divisor. If, while dividing at any time, the remainder is
found to be larger than the divisor, it indicates that the
quotient figure is not large enough and it should be increased.
58. Rule. — I. Write the divisor at the leit of the divi¬
dend , with a line between them.
II. Find how many times the divisor is contained in the
lowest number of the left-hand figures of the dividend that will
contain it, and write the result at the right of the dividend, with
a line between, for the first figure of the quotient.
III. Multiply the divisor by this quotient figure; subtract
the product from the figures of the dividend used to obtain the
quotient figure, and to the remainder annex the next figure of
the dividend: Divide as before, and thus continue until all the
figures of the dividend have been used.
IV. If, when a figure is brought doivn from the dividend ,
the number formed is smaller than the divisor, place a cipher in
the quotient and bring dozv?i the next figure of the dividend, and
so ozi until the number is laige enough to co?itain the divisor.
V. If there is finally a remainder, write it after the quotie7it
with the divisor underneath.
VI. If at a7iy time the divisor multiplied by the quotient
figure is larger than the poi'tion of the dividend bemg used, it
shows that the qziotient figure is too large a?id should be
diminished.
59. Proof. —Multiply the quotient by the divisor, and
to the product add the remainder, if any, and the result
obtained will equal the dividend if the work is correct. Thus,
divisor dividend quotient
23)1383(60^ Ans.
1 3 8
3
0
retnainder 3
26
ARITHMETIC
§1
Proof.— quotient 6 0
divisor 2 3
180
1 2 0
13 8 0
remainder 3
dividejid 13 8 3
Or, divide the dividend by the quotient, and, if the work
is correct, the result will equal the original divisor. If there
is any remainder in the original example omit it when divi¬
ding the quotient into the dividend and the same remainder
will occur in the proof. Thus,
quotient dividend divisor
60 ) 1 3 8 3( 2 3
1 2 0
183
1 8 0
remainder 3
EXAMPLES FOR PRACTICE
Solve the following expressions:
(a) 541,926 4- 7,854.
(t>) 1,460,883 4 - 83.
( e) 350,096 4 - 8.
{d) 962,251 4 - 107.
(e) 417,789 4- 549.
(/) 190,850 4 - 25.
(g) 99,999,999 4 - 3,333.
(//) 22,407 4 - 462.
'(«)
69
(*)
17,601
(C)
43,762
id)
8,993
(e)
761
if)
7,634
Or)
30,003
l(^)
48|
1. A mill purchased cotton to the value of $12,600, averaging $42
per bale; how many bales were purchased? Ans. 300 bales
2. A certain chimney contains 1,120,000 bricks; five men w r ere
employed in its construction, each man laying 2,000 bricks per day on
an average; how long did it take to complete the brickwork?
Ans. 112 da.
§1
ARITHMETIC
27
FACTORS
60. The factors of a number are those numbers that,
when multiplied together, produce the number.
61. A prime number is a number that has no integral
factor except itself and one; or in other words, a prime
number is one that cannot be divided, without a remainder,
except by itself and one.
The following table includes all of the prime numbers up
to and including 499:
I
37
89
151
223
281
359
433
2
4i
97
157
227
283
367
439
3
43
IOI
163
229
293
373
443
5
47
103
167
233
307
379
449
7
53
107
173
239
311
383
457
11
59
109
179
241
3i3
389
461
13
6i
113
181
251
3i7
397
463
1 7
67
127
191
257
33i
401
467
19
7 1
131
193
263
337
409
479
23
73
137
197
269
347
419
487
2 y
79
139
199
271
349
421
491
3i
83
149
211
277
353
43i
499
62. A prime factor is a factor that is a prime number.
63. A composite number is a number that is the
product of two or more integral factors; thus, 15, 35, or 117
are composite numbers since 3x5 = 15, 7x5 = 35, and
13 X 9 = 117. ,
When referring to integral factors, the number itself and
one are not considered.
A composite number can have only one set of prime
factors; thus, the number 16 cannot be expressed as the
product of any set of prime factors except 2 X 2 X 2 X 2.
28
ARITHMETIC
1
It is the product of 4 X 4 and of 8x2, but in the former
case both of the factors are composite, and in the latter case
one of them is composite.
64. An even number is a number that is exactly
divisible by two.
65. An odd number is a number that is not divisible
by two, without a remainder.
A number that can be divided by another, without a
remainder, is said to be exactly divisible, and the divisor is
called an exact divisor.
The process of finding the prime factors of a given number
is one of repeated divisions, using for divisors the smallest
prime numbers that are exactly divisible into the number,
or into the quotients resulting from previous divisions.
66. To find the prime factors of a number:
Example.— Find the prime factors of 576.
Solution. — 2)576
2 ) 2 8 8
2 )144
2 )_ 72
2 )_ 36
2 )_ 18
3 )_9
3
That is: 2 X 2 X 2 X 2 X 2 X 2 X 3 X 3 = 576. Ans.
Explanation.— In this example the division is accom¬
plished by the short method of division, using the smallest
prime number possible for each divisor; these divisors and
the quotient of the last division are the prime factors of the
number 576, since they are all prime numbers, and when
multiplied together produce the number 576.
Rule. —Divide the given number by the smallest prime number
that will exactly divide it. If the quotient is a composite number ,
proceed in the same manner; and continue until a quotient is
1
ARITHMETIC
29
obtained that is a prime number. The several divisors and the
last quotient are the prime factors required.
If no prime factor is found before the quotient becomes
equal to, or less than, the divisor, the number is a prime
number.
EXAMPLES FOR PRACTICE
Find the prime factors of the following numbers:
( a ) 24.
'(«) 2, 2, 2, 3
CO
00
(b) 2, 19
(c) 108.
(c) 2, 2, 3, 3, 3
(. d) 2,160.
. AnsJ
(d) 2, 2, 2, 2, 3, 3, 3, 5
(e) 4,224.
(e) 2,2,2,2,2,2,2,3,11
(/) 36,960.
(f) 2,2,2,2,2,3,5,7,11
(g) 59,049.
(g) 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
(h) 11,760.
[(h) 2, 2, 2, 2, 3, 5, 7, 7
GREATEST COMMON DIVISOR
67. A common divisor of two or more numbers is any
divisor that will exactly divide them; thus, 4 is a common
divisor of 8, 12, 16, and 20, because
8-4 = 2
12 - 4 = 3
16 - 4 = 4
20 - 4 = 5
68. To find a common divisor of two or more
numbers:
Rule. —Resolve the given numbers into two factors , one of
which is common to all; this common factor is a common divisor
of the given numbers.
69. The greatest common divisor of two or more
numbers is the greatest number that will divide each of them
without a remainder.
The divisors of 36 are 2, 3, 4, 6, 9, 12, and 18, and the
divisors of 24 are 2, 3, 4, 6, 8, and 12. The only common
divisors of 36 and 24 are 2, 3, 4, 6, and 12, of which 12 is the
greatest; therefore, 12 is the greatest common divisor of 36
and 24.
30
ARITHMETIC
§1
If two integral numbers have no common divisor, they are
said to be prime to each other; thus, 9 and 16 are prime to
each other, although both are composite numbers.
The letters G. C. D. stand for greatest common divisor,
or in some instances, where the term greatest common
measure is used, the abbreviation G. C. M. is adopted.
In finding the greatest common divisor of two or more
numbers, first reduce the numbers to their prime factors, and
then find the product of those factors that are common to all
the numbers.
Example. —What is the G. C. D. of 72 and 168?
Solution.—
2)72
2)168
2)36
2 ) 8 4
2)18
2) 4 2
3 ) 9
3 ) 2 1
3
7
2 is a common divisor of 72 and 168 three times, and 3 is a
common divisor once, so the greatest common divisor of 72 and 168
is 2 X 2 X 2 X 3 = 24. Ans.
70. To find the greatest common divisor of two
or more numbers:
Rule. —Resolve the given numbers i?ito their prime factors;
the product of those factors that are common to all the given
numbers is the greatest common divisor.
EXAMPLES FOR PRACTICE
Find the G. C. D. of:
(a) 45 and 75.
(b) 120 and 168.
(c) 60 and 220.
{d) 630 and 2,730.
( e) 63, 171, and 207.
(/) 1,890, 4,410, and 10,710.
(a)
15
(b)
24
(c)
20
id)
210
( •
ARITHMETIC
(PART 2)
FRACTIONS
1. A fraction is one or more parts of a unit. One-half,
three-fourths, two-fifths are fractions.
If a weave room contains 1,000 looms and has 500 of this
number on fancy work and the other 500 on plain work, it
might be stated that one-half of the looms were on one class
of work and one-half on another class of work. Instead of
writing out the expression one-half in words, it could be
shown by means of figures, thus i, and would be read one-
half. When any part of a number or unit is expressed in
this manner, it is said to be expressed fractionally , and the
expression (in this case 2 ) is known as a fraction.
2. Two numbers are required to express a fraction; one
is called the numerator, and the other, the denominator.
The numerator is placed above the denominator, with a
line between them, as f. Here 3 is the denominator, and
shows into how many equal parts the unit, or o?ie, is divided.
The numerator 2 shows how many of these equal parts are
taken or considered. The denominator also indicates the
names of the parts.
2 is read one-half
I is read three-fourths
f is read three-eighths
te is read five-sixteenths
is read twenty-nine forty-sevenths
For notice of copyright, see page immediately following the title page
22
2
ARITHMETIC
§2
3. In the expression “f of an apple,” the denominator 4
shows that the apple is to be (or has been) cut into 4 equal
Parts, and the numerator 3 shows that three of these parts, or
fourths, have been taken or considered.
If each of the parts, or fourths, of the apple were cut into
two equal pieces, there would then be twice as many pieces as
before, or 4 X 2 = 8 pieces in all; one of these pieces would
be called one-eighth, and would be expressed in figures as i.
Three of these pieces would be called three-eighths, and
written f. The words three-fourths, three-eighths, five-six¬
teenths, etc. are abbreviations of three one-fourths, three
one-eighths, five one-sixteenths, etc. It is evident that the
larger the denominator, the greater is the number of parts
into which anything is divided; consequently, the parts
themselves are smaller, and the value of the fraction is less
for the same number of parts taken. In other words, 9 , for
example, is smaller than i, because if an object be divided
into 9 parts, the parts are smaller than if the same object
had been divided into 8 parts; and, since i is smaller than Jr,
it is clear that 7 one-ninths is a smaller amount than 7 one-
eighths. Hence, also, I is less than I.
The line between the numerator and denominator means
divided by, or -f-. t is equivalent to 3 -h 4. 1 is equivalent
to 5 -r- 8.
4. The value of a fraction is equal to the numerator
divided by the denominator, as 4 = 2, f = 3.
The value of a fraction whose numerator and denominator
are equal is 1 .
7 , or four-fourths = 1
I, or eight-eighths = 1
ti, or sixty-four sixty-fourths = 1
5. The numerator and denominator of a fraction are
called the terms of a fraction.
6 . A proper fraction is a fraction whose numerator is
less than its denominator. Its value is less than 1. i, t, 9
are all proper fractions, since the number above the line, or
2
ARITHMETIC
3
the numerator, in each case is less than the number below
the line, or the denominator.
7. An improper fraction is a fraction whose numerator
is egjial to , ox greater than , its denominator. Its value is one
or more than one. i are -all improper fractions, since
in each case the numerator is equal to, or greater than, the
denominator.
8. A simple fraction is a fraction with but one number
for its numerator and one number for its denominator. Such
a fraction may be either proper or improper. All the illus¬
trations so far given are simple fractions.
9. A mixed number is a whole number and a fraction
combined. 8| (read eight and three-fourths) is a mixed
number, as it is composed of the whole number 8 and the
fraction f.
10. A complex fraction is one that has a fraction or a
mixed number for either its numerator or denominator, or
for both.
— 6 , and ^ are examples of complex fractions.
24 5|
11. To read a simple fraction, read the numerator as
though it were a whole number and then read the denomi¬
nator adding th or ths to the name of this figure, with the
few exceptions noted in the following:
\ is read one-half
-iV is read one-twelfth
2 h is read one twenty-second
All other denominators of fractions ending with a 2
(except those ending with 12) are read similar to the last
example given above.
i is read one-third
iV is read one-thirteenth
2 V is read one twenty-third
All other denominators of fractions ending with a 3
(except those ending with 13) are read similar to the last
example given above.
4
ARITHMETIC
2
i is read one-fifth
tV is read one-fifteenth
2 V is read one twenty-fifth
All other denominators of fractions ending with a 5
(except those ending with 15) are read similar to the last
example given above.
-To is read one-tenth
2 V is read one-twentieth
All other denominators of fractions ending in ty are spelled
similar to the last example.
The fraction 4 is read one-fourth and also one-quarter.
All denominators of fractions having two figures or more
and ending with a 4 are read by adding th or ths to the name
of the figure. _
REDUCTION OF FRACTIONS
12 . The process of changing the terms of a fraction
without changing the value of the fraction is known as
reduction.
13. Multiplyivg both the numerator and denominator of
a fraction by the same number does not change the value of the
fraction. Thus, multiplying both terms of the fraction 4 by 2,
3 X 2 = 6
4X2 8
The value is not changed, since f = f. Suppose that
an object, say an apple, is divided into 8 equal parts. If
these parts be arranged in 4 piles, each containing 2 parts,
it is evident that each pile will be composed of the same
amount of the entire apple as would have been the case
had the apple been originally cut into 4 equal parts. Now,
if one of these piles (containing 2 parts) be removed,
there will be 3 piles left, each containing 2 equal parts,
or 6 equal parts in all; i. e., six-eighths. But, since one
pile, or one-quarter, was removed, there are three-quarters
left. Hence, 4=1. The same course of reasoning may
be applied to any similar case. Therefore, multiplying
2
ARITHMETIC
5
both terms of a fraction by the same number does not alter
its value.
14. Dividing both the numerator and denominator of a
fraction by the same number does not change the value of the
fraction. Thus, dividing both terms of the fraction i%- by 2,
_8_ -7- 2 _ 4
10 -s- 2 5
That = f is readily seen from the explanation given in
Art. 13; for, multiplying both terms of the fraction 1 by 2,
4 X 2 _ 8
5x2 10
and, if i = must equal 1. Hence, dividing both terms
of a fraction by the same number does not alter its value.
15. A fraction is reduced to lower terms by dividing both
terms by the same number.
A fraction is reduced to its lowest terms when its numer¬
ator and denominator cannot both be divided by the same
number, except 1, without a remainder; for example, f, t,
1 1 _ 8 _
2 4 , 15 .
16. Suppose that it is desired to reduce the fraction fo to
its lowest terms; that is, to have the smallest possible numbers
for the numerator and denominator.
Both numerator and denominator can be divided by 2 and
the result is if. Both the numerator and denominator of
this new fraction, if, cannot be divided by 2 without a
remainder, but both can be divided by 3, giving the result f.
The fraction f cannot be reduced any lower, since there
is no number greater than 1 that will exactly divide both
terms of the fraction. Therefore, f is the lowest terms in
which the fraction ff can be expressed.
17. To reduce a fraction to its lowest terms:
Rule .—Divide both the numerator and denominator by any
number greater than 1 that will divide into both without a
remainder, a?id continue this operation until there is no number
greater than 1 that will exactly divide both terms.
6
ARITHMETIC
§2
EXAMPLES FOR PRACTICE
Reduce the following fractions to their lowest terms:
(a) fi
( b) wi.
(c) A.
(d) n.
Ans.-
(a)
( b)
{c)
(d)
1 _
2
X
8
3.
8
69
7 O
(*)
2 16
1 2 9 6*
(e) i
(0
94
5 64*
l(/)
JL
6
18. A fraction is reduced to higher terms by multiplying
both the numerator and denominator by the same number.
Thus, if both terms of the fraction I are multiplied by 2,
the result is the fraction f, which is of the same value as i.
19. To reduce a fraction to an equal fraction hav¬
ing a given denominator:
Example. —Reduce f to an equal fraction having96fora denominator.
Solution. —Both the numerator and the denominator must be multi¬
plied by the same number in order not to change the value of the frac¬
tion. The denominator must be multiplied by some number which
will, in this case, make the product 96; this number is evidently 96 -f- 8
7 V 12 84
= 12, since 8 X 12 = 96. Hence, - x ^ = 95 - Ans -
Rule. — Divide the denominator of the nezv fraction by the
denominator of the given fraction and multiply both terms of
the fraction by this result.
20. To reduce a whole number to an improper
fraction:
Example. —How many fifths in 7?
Solution.— Since there are 5 fifths in 1 (f = 1), in 7 there will be
7X5 fifths, or 35 fifths; i. e., 7xf = s g L . Ans.
Rule. — Multiply the whole number by the denominator of
the desired fraction and use this result as the numerator.
21. To reduce a mixed number to an improper
fraction:
Example. —Reduce 4| to eighths.
Solution. —Since there are 8 eighths in 1 (f = 1), in 4 there will be
4X8 eighths, or 32 eighths, and 32 eighths plus 5 eighths equals
37 eighths, or Therefore, 4| = nr. Ans.
§2
ARITHMETIC
7
Rule .—Multiply the whole number by the denominator of the
fraction, and to this result add the numerator of the fraction.
Use the result tlms obtained for the numerator of the improper
fraction, and for the denominator take the denominator of the
fraction in the mixed number.
22. To reduce an improper fraction to a whole or
a mixed number:
Example 1.—Reduce ^ to a whole or a mixed number.
Solution. —Since one unit contains 6 sixths, there areas many units
in or 48 sixths, as 6 is contained times in 48, or 8 . Therefore, the
whole number 8 is equal to the fraction Ans.
Example 2.—Reduce x to a whole or mixed number.
Solution. —Since in one unit there are 8 eighths, the number of
units in ^, or 21 eighths, must equal 21 eighths divided by 8 or 2 plus
a remainder of 5 eighths. Therefore, ^ = 2 + f = 2f. Ans.
Rule .—Divide the numerator by the denominator. If there
is a remainder write it over the denominator of the fraction
and place the fraction thus formed with the whole number,
making the result a mixed number.
EXAMPLES FOR PRACTICE
(a) Reduce § to a fraction having 12 for a denominator. Ans.
(b) Reduce xc to a fraction having 90 fora denominator. Ans. iH>
{,c ) Reduce ^ to a fraction having 96 as a denominator. Ans. trt
(d) Change 6 to an improper fraction having 24 as its denominator.
Ans. ^
( e) Change 12 to an improper fraction having 5 as its denominator.
Ans. ^
(/) Reduce 14 to an improper fraction with 3 as its denominator.
Ans. ^
(e)
Reduce to an improper fraction.
Ans. ff
(h
Reduce 12f to an improper fraction.
Ans.
(i)
Reduce 4} to an improper fraction.
Ans. ^
Change the following to whole or mixed numbers:
(/)
25
6 •
Ans. 5
(h
4 7
7 •
Ans. 6 f
(l)
1 9
4 •
Ans. 4\
(m)
2 7
3 •
Ans. 9
8
ARITHMETIC
§2
23. A common denominator of two or more fractions
is a number that will contain (i. e., that may be divided by)
the denominator of each of the given fractions without a
remainder. The least common denominator is the least
number that will contain each denominator of the given frac¬
tions without a remainder.
24. To find the least common denominators
Example 1. — Find the least common denominator of 1, and re.
Solution. —First place the denominators in a row, separated by
commas.
2 ) 4, 3, 9, 16
2 ) 2, 3, 9, 8
3 ) 1, 3, 9, 4
3 ) 1, 1, 3, 4
4 ) 1, 1, 1, 4
1, 1, 1, 1
2X2X3X3X4 = 144, the least common denominator. Ans.
Explanation. —Divide each of the numbers by some
prime number that will divide at least two of them without a
remainder (if possible), bringing down to the row below those
denominators that will not contain the divisor without a
remainder. Dividing each of the numbers by 2, the second
row becomes 2, 3, 9, 8, since 2 will not divide 3 and 9 without
a remainder. Dividing again by 2, the result is 1, 3, 9, 4.
Dividing the third row by 3, the result is 1, 1, 3, 4. So con¬
tinue until the last row contains only l’s. The product of
all the divisors, or 2 X 2 X 3 X 3 X 4, is 144, and is the least
common denominator.
Example 2. —Find the least common denominator of £, •&, A-
Solution.—
3)9,
12,
18
3)3,
4,
6
2)1,
4,
2
2)1,
2,
1
1 , 1 , 1
3 X 3 X 2 X 2 = 36. Ans.
25. To reduce two or more fractions to fractions
having a common denominator:
§2
ARITHMETIC
9
Example. —Reduce f, f, and £ to fractions having a common
denominator.
Solution. —The common denominator is a number that will contain
3, 4, and 2. The least common denominator is 12, because it is the
smallest number that can be divided by 3, 4, and 2 without a
remainder.
2 _8 3 _ _9 1 _ _6
3 " 12’ 4 ” 12’ 2 ~ 12
Reducing f (see Art. 19), 3 is contained in 12, 4 times. By multi-
2 4 8
plying both numerator and denominator of f by 4, we find - X = ys-
In the same way, we find f = A, and | = A- Ans.
Rule. —For each fraction to be reduced, divide the common
detiominator by the denominator of the fraction , and multiply
both terms ot the fraction by the quotient.
EXAMPLES FOR PRACTICE
Reduce the following to fractions having the least common denom
inator:
(a)
'•fit to and +
(a) A,
8 16 12
24) 24» 24
(b)
H and A •
(b) «,
8
2 4
(0
t> vr> and A"■
Ans.-
(c) «,
10 8
4 2) 42
(d)
■§■> ts 1 and 5 +
id) f§,
28 12 24
60) 60) 60
(*)
4|, 3-f, and 5*.
(p\ 4iO Ol_2 C_3_
\C) ^16) ^16» ^16
(0
-Z- 3 pn/1 4*.
16) 4 > «*UU 24 •
1(0 H,
36 10
48) 48
ADDITION OF FRACTIONS
26. Fractions cannot be added unless they have a common
denominator. We cannot add 1 to -g- as they now stand, since
the denominators represent parts of different sizes. Fourths
cannot be added to eighths.
Suppose that we divide an apple into 4 equal parts, and
then divide 2 of these parts into two equal parts. It is
evident that we shall have 2 one-fourths and 4 one-eighths.
Now, if we add these parts, the result is 2 + 4 = 6 some¬
thing. But what is the something? It is not fourths, for
six fourths are li, and we had only 1 apple to begin with;
neither is it eighths, for six eighths are f, which is less than
1 apple. By reducing the quarters to eighths, we have
<- = I, and adding the other 4 eighths, 4 + 4 = 8 eighths.
10
ARITHMETIC
§2
This result is correct, since 1=1. Or we can, in this case,
reduce the eighths to quarters. Thus, i = f; whence, adding,
2 + 2 = 4 quarters, a correct result, since 1=1.
Before adding, fractions should be reduced to a common
denominator, preferably the least common denominator.
Example 1.—Find the sum of 1, f, and f.
Solution. —The least common denominator, or the least number
that will contain all the denominators, is 8.
14 3 _ 6 5 _ 5
2 8’ 4 8’ aUd 8 8
Explanation. —As the denominator tells, or indicates,
the names of the parts, the numerators only are added, to
obtain the total number of parts indicated by the denominator.
Thus, 4 one-eighths plus 6 one-eighths plus 5 one-eighths =
4 6 5 4 + 6 + 515
8 + 8 + 8 8 8
li. Ans.
Example 2.—What is the sum of 121, 14f, and 7A?
Solution.— The least common denominator in this case is 16.
12 ^ = 1241
14| = 141§
7A = T&
sum 33 + 44 = 33 + 144 = 344+ Ans.
The sum of the fractions is 44 or 144, which added to the sum of the
whole numbers is 3444-
Example 3.—What is the sum of 17, 13-&, and 34?
Solution. —The least common denominator is 32. 13^ = 13^,
34 = 3A- 17
13A
9
32
3O0 .
16 X 3 10 X 0 6 ' A
This brings the same quotient as in the first case.
Example 3.—Divide f by j.
Solution. —We cannot divide f byi, as in the first case of example
2 , for the denominators are not the same; therefore, we must solve as
in the second case.
3^1 = 3 4 = 3X4 3
8 ’ 4 8*1
8X1
2
Ans.
16
ARITHMETIC
§2
Example 4.—Divide 5 by fi.
Solution.—
1 0
1 6
inverted becomes
8
_ i6 9 x n
5 x lo = ~vr
t
1 6
1 O •
= 8 .
Ans.
36. Whenever a mixed number has to be dealt with in
the division of fractions, the best method is to change the
mixed number to an improper fraction and then proceed
according to the foregoing rules.
Example 1. —How many times is 3f contained in 7re?
Solution.-
0 3 1 5 .
“ 0 4 — 4 ,
119 4 _
16 15
7A =
119
1 6 •
119 X 4
16 X 15
4
inverted equals A-
119 e
60 8
Ans.
Example 2. —Divide 5| by 2|.
Solution.— 5| = V; = f.
Then,
26^_5 _ 26 2
5 ' 2 ~ 5*5
52
25
— 9-A
— ^26*
Ans.
EXAMPLES FOR PRACTICE
Divide:
(a) 15 by 6 f.
(£) 30 by!.
(c) 172 by |.
(d) by lAu
(e) ^byl4f.
(/) W by 17|
Ans. •
(а) 2|
(б) 40
(c) 215
(d) m
(e) Hi
1(0
37. As explained in Art. 10, when a fraction contains a
fractional number for its numerator or denominator, or both,
i i 4
it is known as a complex fraction. Thus. A and - are all
8 s "3
complex fractions.
38. It will be noticed in the above illustrations of com¬
plex fractions that a heavier line is used to separate parts of
the fractions. This heavy line always separates the numer¬
ator and denominator of complex fractions. Thus, in the
first case, f is the numerator and 8 the denominator. In
§2
ARITHMETIC
17
the second fraction, £ is the numerator and i the denomi¬
nator. In the third fraction, 4 is the numerator and f, the
denominator.
A line drawn between two numbers in this manner indi¬
cates that the one above the line is to be divided by the
one below the line. Consequently, in order to reduce a
complex fraction to a common one, divide the numerator
by the denominator.
3.
Example. —Reduce £ to a common fraction.
O
c i 3 _ 3 1 3 .
Solution.— g=| + 8 = |Xg=g 2 . Ans.
39. Whenever an expression like one of the three fol¬
lowing is obtained, it may always be simplified by trans¬
posing the denominator from above to below the line, or
from below to above, as the case may be, taking care, how¬
ever, to indicate that the denominator when so transferred is
a multiplier.
3 1
1. - — - = — = —: for, regarding the fraction above
9 9 X 4 36 12
the heavy line as the numerator of a fraction whose denom¬
inator is 9, - X ^ as before.
9X4 9X4’
2. | = ^ — 12. The proof is the same as in the
4 O
first case.
O I = 5X 4
'I 3X9
20
27’
For, regarding f as the numerator
5 x 9
of a fraction whose denominator is 1, 4
t X 9
3X9’
and
5 X4 5X4 20 ,
- = as above.
3 X 9 x 4 3 X 9 27’
This principle may be used to great advantage in cases like
X 310 X n X 72
40 X 4i X 5i
Reducing the mixed numbers to fractions,
18
ARITHMETIC
§2
. , i X 310 X fi X 72 , T , f .
the expression becomes . —. Now, transferring
40xfx¥
the denominators of the fractions and canceling,
3
X0 3 0 3
1 X 310 X 27 X 72 X 2 X 6 = 1 X 340 X X U X % X 0
40 X 9 X 31 X 4 X 12 40 X 0 X 34 X 4 X 42
4 2
= — = 13i ^
Greater exactness in results can usually be obtained by
using this principle than by reducing the fractions to deci¬
mals. The principle, however, should not be employed if a
sign of addition or subtraction occurs either above or below
the dividing line.
DECIMALS
40. Decimals are tenth fractions; that is, the parts of a
unit are expressed on the scale of ten, as tenths, hundredths ,
thousandths , etc.
41. The denominator, which is always ten or a multiple
of ten, as 10, 100, 1,000, etc., is not expressed, as it would be
in common fractions, by writing it under the numerator with
a line between them, as mf, tot, toVo, but is expressed by
placing a period (.), which is called a decimal point, to
the left of the figures of the numerator, so as to indicate
that the number on the right is the numerator of a fraction
whose denominator is 10, 100, 1,000, etc.
42. The reading of a decimal number depends on the
number of decimal places in it, or the number of figures
to the right of the decimal point.
One decimal place expresses tenths
Two decimal places express hundredths
Three decimal places express thousandths
Four decimal places express ten-thousandths
Five decimal places express hundred-thousandths
Six decimal places express millionths
2
ARITHMETIC
19
Thus:
.3
3
1 0
= 3 tenths
.03
3
10 0
= 3 hundredths
.003
s
10 0 0
= 3 thousandths
.0003 =
3
1 0 0 0 0
= 3 ten-thousandths
.00003 =
3
100000
= 3 hundred-thousandths
.000003 =
3
1000000
= 3 millionths
We see in the above that the number of decimal places in a
decimal equals the number of ciphers to the right of the figure 1 in
the denominator of its equivalent fraction. This fact kept in mind
will be of much assistance in reading and writing decimals.
43. Whatever may be written to the left of a decimal
point is a whole number. The decimal point merely separates
the fraction on the right from the whole number on the left.
44. When a whole number and decimal are written
together, the expression is a mixed number. Thus, 8.12 and
17.25 are mixed numbers.
45. The relation of decimals and whole numbers to each
other is clearly shown by the following table:
m
G
m
G
o
• T-(
m
G
C /3
• i-H
tn
O
C
a
£
o
X
4-1
a
C /3
• 4-4
• r-i
> 4-4
3
O
• T—C
O
o
m
a
in
m
XI
-*—>
TG
G
aJ
in
G
O
X
m
X
-i->
r a
G
rt
m
G
O
X
4-1
I
G
CD
CO
X
4J
TG
G
a]
in
G
O
X
4-1
I
"O
CD
U
TG
G
G
X
m
X
4->
G
O
2 3 4 5
a
6 7
in
X
4-1
G
O
G
sa ^ 12
g
Hence, .3917 = —, nearly. Ans.
1Z
Rule. —Reduce 1 to a fractio?i having the given denominator.
Multiply the given decimal by the fraction so obtained, and the
result will be the fraction required.
Express:
(a)
(b)
(c)
(d)
(e)
(f)
EXAMPLES FOR PRACTICE
.625 in 8 ths.
.3125 in 16ths.
.15625 in 32ds
.77 in 64ths.
.81 in 48ths.
Ans.
'(«)
(b)
(c)
(■ d)
5
1 6
5
32
49
64
(e)
39
48
.923 in 96ths.
(0
89
9 6
62. The sign for dollars is $. It is read dollars. $25 is
read 25 dollars.
Since there are 100 cents in a dollar, 1 cent is 1 one-
hundredth of a dollar; the first two figures of a decimal part
of a dollar represent cents. Since a mill is iV of a cent, or
ToVo of a dollar, the third figure represents mills.
Thus, $25.16 is read twenty-five dollars and sixteen cents;
$25,168 is read twenty-five dollars sixteen cents and eight mills.
■
ARITHMETIC
(PART 3)
PERCENTAGE
DEFINITIONS AND PRINCIPLES
1. In certain operations it is very convenient to regard a
quantity as being divided into 100 equal parts; thus, instead
of using the ordinary fractions i, f, f, the equivalent frac¬
tions or their equivalent decimals .25, .60, .28-7
100 100 100
are used. This practice is a very convenient one in all com¬
putations involving United States money, because, since $1
equals 100 cents, it is easier to comprehend what part of the
whole i^o is than some other equivalent fraction, as t 4 ^-; it
is also much easier to compute with fractions whose denom¬
inators are 100 than it is to compute with fractions whose
denominators are composed of other figures.
2. Percentage is a term applied to those arithmetical
operations in which the number or quantity to be operated on
is supposed to be divided into 100 equal parts.
/
3. The term per cent, means by the hundred. Thus,
8 per cent, of a number means 8 hundredths; i. e., tTo, or
.08, of that number; 8 per cent, of 250 is 250 X tTo, or 250
X .08 = 20; 47 per cent, of 75 bushels is 75 X to = 75
X .47 = 35.25 bushels. The statement that the population
of a city has increased 22 per cent, in a given time, say from
For notice of copyright , see page immediately following the title page
23
2
ARITHMETIC
§3
1890 to 1900, is equivalent to saying that the increase is 22
in every hundred; that is, for every 100 in 1890, there are
22 more, or 122, in 1900.
4. The sign of per cent, is %, and is read per cent.
Thus, 6% is read six per cent.; 12i% is read twelve and one-
half per cent., etc.
5. When expressing the per cent, of a number to use in
calculations, it is customary to express it decimally instead
of fractionally. Thus, instead of expressing 6%, 25%,
and 43% as tot, i^o, and it is usual to express them
as .06, .25, and .43.
6. The following table will show how per cent, can be
expressed either as a decimal or as a fraction:
Per Cent.
Decimal
Fraction
Per Cent.
Decimal
Fraction
1% . .
.01
1
100
i%. .
.0025
i _JL_
10 0 OT 4 0 0
2 % . .
.02
2
100
or bV
\° fo . .
.005
i r , r . 1...
100 ° f 2 00
s % . .
•05
5
100
or to
ii% . .
.015
li nr 3
100 01 200
10% . .
.IO
1 0
10 0
or to
6i% .' .
■ .064
nr _1_
100 or 1 6
25 % • •
.25
2 B
10 0
or i
8i% . .
.083
8i 1
Too or IT
50% . .
•50
B 0
10 0
or 2
122% . .
.125
I2i nr J-
100 or 8
75 % • •
•75
7 B
1 0 0
or f
i6f%. .
.i6f
163 X
100 or 6
100% . .
I.OO
1 0 0
10 0
or 1
33 a% . .
• 333
33 * X
100 or 3
125% . .
1.25
1 2 B
10 0
or 1 i
37 *% • •
• 372
37 i a
100 or 8
150% . .
1.50
1 B 0
10 0
or ii
62!% . .
.625
62 * nr £
100 or 8
500% . .
5.00
5 0 0
10 0
or 5
87 ¥% . .
.875
87* nr i
10 0 or 8
7. The names of the terms used in percentage are:
the base , the rate or rate Per cent., the percentage, the amount ,
and the difference.
8. The base is the number or quantity that is supposed
to be divided into 100 equal parts.
§3
ARITHMETIC
3
9. The rate per cent, is that number of the 100 equal
parts into which the base is supposed to be divided that is
taken or considered. The rate is the number of hundredths
of the base that is taken or considered. The distinction
between the rate per cent, and the rate is this: the rate per
cent, is always 100 times the rate. Thus, 7% of 125 and .07
of 125 amount in the end to the same thing; the former, 7,
is the rate per cent.—the number of hundredths of 125
intended; the latter, .07, is the rate, the part of 125 that is
to be found; 7% is used in speech, .07 is the form used in
computation. So, also, 12= .125, = .005, If °Io = .0175.
10. The percentage is the result obtained by multiply¬
ing the base by the rate. Thus, 7% of 125 = 125 X .07
= 8.75, the percentage.
11. The amount is the sum of the base and the per¬
centage.
12. The difference is the remainder obtained when the
percentage is subtracted from the base.
13. The terms amount and difference are ordinarily used
when there is an increase or a decrease in the base. For
example, suppose the population of a village is 1,500 and it
increases 25 per cent. This means that for every 100 of the
original 1,500 there is an increase of 25, or a total increase of
15 X 25 = 375. This increase added to the original population
gives the amount , or the population after the increase. If the
population had decreased 375, the final population would
have been 1,500 — 375 = 1,125, and this would be the differ¬
ence. The original population, 1,500, is the base on which
the percentage is computed; the 25 is the rate per cent., and
the increase or decrease, 375, is the percentage. If the base
increases, the final value is the amount, and if it decreases,
its final value is the difference.
4
ARITHMETIC
3
BASE, RATE, AND PERCENTAGE
14 . Rule.— To find, the percentage , multiply the base by
the rate.
Example 1.—A farmer raised 650 bushels of wheat and sold 64%;
how many bushels did he sell?
Solution. —The base is 650 bu. Out of every 100 bu. raised 64 were
sold; that is, the number of bushels sold was nf 0 or .64 of the number
raised. The rate is, then, .64, and
650 X .64 = 416 bu., the percentage. Ans.
Example 2. —If cotton is selling at 8 cents per pound and the price
goes up 121%, what will be the new selling price?
Solution. —It is evident that the new selling price is equal to the
original selling price plus the percentage of increase. The base is
8 ct.; the rate is Vl\% , or .125, and the percentage is .125 X 8 ct. = let.
Hence, the new selling price is 8 ct. -f 1 ct. = 9 ct. Ans.
15 . In the last example, the result sought, 9 cents, was
the amount. In all cases where the amount is involved, it
can be obtained directly by multiplying the base by 1 plus
the rate. Thus, in the last example 1 plus the rate is
1 + .125 = 1.125, and 1.125 X 8 cents = 9 cents, the same
result as before.
Similarly, the difference may be found directly by multi-
%
plying the base by 1 minus the rate. An example will
illustrate this.
Example. —If wool is selling for 20 cents per pound and the price
drops 10%, what will be the new selling price?
Solution. —The new selling price is evidently equal to the original
selling price less the percentage of decrease. Here 20 ct. is the base
the rate .10, and the percentage is .10 X 20 ct. = 2 ct. Hence, the
new selling price is 20 ct. — 2 ct. = 18 ct. Ans.
Or, the difference expressed as a per cent, is 1 — .10 = .90, and
.90 X 20 ct. = 18 ct. Ans.
16 . When the percentage and rate are known, the base
may be found by dividing the percentage by the rate. Thus,
suppose that 12 is 6%, or too, of some number; then 1%, or
Too', of the number, is 12 -t- 6, or 2. Consequently, if 2 = 1%
or too - , 100%, or too = 2 X 100 = 200. But, since the same
§3
ARITHMETIC
5
result may be arrived at by dividing 12 by .06, for 12 -r- .06
= 200, it follows that:
Rule.— When the percentage and rate are known, to find the
base , divide the percentage by the rate , expressed decimally.
Example 1.—Bought a certain number of bushels of apples and
sold 76% of them; if I sold 228 bushels, how many bushels did I buy?
Solution. —Here 228 is the percentage, and 76%, or .76, is the rate;
hence, applying the rule,
228 .76 = 300 bu. Ans.
Example 2. —A corporation is operating 500 looms, which is 25%
of the number of looms in the mill; how many looms are in the mill?
Solution. —Here 500 is 25% of some number; in other words, 500
is the percentage and .25 is the rate. Therefore, applying the rule,
the number of looms in the mill equals the base, which equals 500
-p .25 = 2,000. Ans.
Example 3.—A mill sells cloth for 12 cents per yard, which is 20%
more than the cost to manufacture; what is the cost per yard to
manufacture?
Solution. —The cost to manufacture is the base, or 100%. The
selling price is 20% greater, or 100% + 20% = 120%. The rate is
therefore 1.20. Applying the rule,
12 ct. -r- 1.20 = 10 ct. Ans.
Example 4.—Suppose that in the last example, the selling price,
12 cents, is 20% less than the cost to manufacture; what is the cost to
manufacture?
Solution. —The cost to manufacture is the base, 12 ct. is the per¬
centage, and 1 — .20 = .80 is the rate. Applying the rule,
12 ct. -r- .80 = 15 ct. Ans.
17. When the base and percentage are given, the rate
(expressed decimally) may be found by dividing the per¬
centage by the base. Thus, suppose that it is desired
to find what per cent. 12 is of 200. 1% of 200 is 200 X .01
= 2. Now, if 1% is 2, 12 is evidently as many per cent, as
the number of times that 2 is contained in 12, or 12 -4- 2 = 6%.
But the same result maybe obtained by dividing 12, the per¬
centage, by 200, the base, since 12 4- 200 = .06 = 6%. Hence,
Rule.— When the percentage and base are given, to find the
rate , divide the percentage by the base , and the result will be the
rate , expressed decimally.
6 ARITHMETIC §3
Example 1.—Bought 300 bushels of apples and sold 228 bushels.
What per cent, of the total number of bushels was sold?
Solution.—H ere 300 is the base and 228 is the percentage; hence,
applying rule,
rate = 228 -^ 300 = .76 = 76%. Ans.
Example 2. —What per cent, of 875 is 25?
Solution. —Here 875 is the base, and 25 is the percentage; hence,
applying rule,
25 -T- 875 = .02f = 2f%. Ans.
Example 3.—A mill contains 300 looms, which are capable of pro¬
ducing 200 cuts per day of a certain kind of cloth, if running all the
time. The actual production is 180 cuts per day. What per cent, of
the possible total production per day are the looms turning off?
Solution. —Here 200 is the base, because the question is, What per
cent, of the total production is 180? Hence, 180 is the percentage and
200 is the base. Therefore, applying the rule,
180 -f- 200 = .90 = 90%. Ans.
Example 4.—Cloth costing 75 cents per yard to manufacture is sold
at 60 cents per yard; what is the loss per cent.?
Solution. —Here 75 ct. is the base, because it is desired to know
the loss per cent, on the cost. The loss was 75 ct. — 60 ct. = 15 ct.
Hence,
15 ct. -5- 75 ct. = .20 = 20%. Ans.
Example 5.—Suppose that, in the last example, the cloth had been
sold for 80 cents per yard; what would have been the gain per cent.?
Solution. —As before, 75 ct. is the base; the gain is 80 ct. — 75 ct.
= 5 ct.; and the gain per cent, is 5ct. -5- 75 ct. = ,06f = 6|%. Ans.
EXAMPLES FOR PRACTICE
1. Cloth costing 30 cents per yard to manufacture was sold at a
loss of 16|%; what was the selling price per yard? Ans. 25 ct.
2. Cloth costing 60 cents per yard to manufacture is sold at 70
cents per yard; what is the gain per cent.? Ans. 16f%
3. Cloth is sold at 70 cents per yard, which gives a profit of 20%;
what is the cost per yard to manufacture? Ans. 58£ ct.
4. An agent receives $50 for selling a machine. The money he
receives is 20% of the cost of the machine; what is the cost of the
machine? Ans. $250
§3
ARITHMETIC
7
5. A certain amount of yarn weighs 1,800 pounds before being
sized, and 2,000 pounds after being sized; what per cent, of this 2,000
pounds is yarn, and what per cent, is size? Ang j90% yarn
’\10% size
6. A mixing of wool has to weigh 730 pounds, of which 9% is
red, 17% black, 69% natural, and the balance olive; find the weight
of each color required. 165.7 lb. red
A 124.1 lb. black
'503.71b. natural
36.5 lb. olive
7. A cotton broker sold 100 bales of cotton for $4,000, and charged
$200 commission; what rate per cent, did he charge for selling?
Ans. 5%
8. The production of a machine has to be increased 8%. It now
turns off 14 ounces per spindle in a given time; how many ounces
will a spindle produce in the same time after the change?
Ans. 15.12 oz.
9. A comber turns off 50 pounds of cotton in a day, which is 84
per cent, of the cotton fed in: (a) What is the weight of the cotton fed
in? (A) How many pounds of waste are there in one day?
Ans.
{
(a) 59.51b.
(b) 9.51b.
INTEREST
18. Interest as a general term may be defined as a
form of percentage dealing particularly with the use of money.
The rules and explanations previously given will be found
to apply equally well when dealing with interest.
One point, however, that should be carefully noted is that
whenever the use of money is being considered the element
of time enters into the calculation.
19. The terms commonly used in connection with this
subject are as follows:
Interest, which is the sum of money paid for the use of
money for a certain time.
Principal, which is the money for the use of which the
interest is paid.
Rate of interest, which is a certain per cent, and which is
also known as the rate per cent.
Amount, which is the principal and the interest added
together.
8
ARITHMETIC
3
20. To illustrate these terms suppose that a person bor¬
rows $100 for a year, at the end of which time he pays
the $100 and also the interest at the rate of 6% per year.
6% of $100 is $6. Therefore, to cancel his debt he will
pay $100 + $6, or $106.
In the above case $100 is the principal , or the money
loaned by one person and borrowed by another, $6 is the
interest , or the amount of money the borrower pays for the
use of the principal, 6% per year is the rate per cent., and
$106 is the amount.
21. The unit of time when figuring interest is generally
one year. Very frequently the time is not stated and in all
such cases should be considered as one year.
Example.— Find the interest on $500 for 4 years at 2%.
Solution. —In such a case the time is understood to be one year;
that is, 2% per year. Thus,
$ 5 0 0
.0 2
$ 1 0.0 0 int. for 1 yr.
$10 X 4 = $40 int. for 4 yr. Ans.
22. Interest, however, may be computed for shorter
periods, as semiannually (6 mo.) or quarterly (3 mo.),
although the general rule is 1 year.
Example. —A man borrows $400 for 1 year agreeing to pay
2% quarterly; how much interest will he have paid at the end of
the year?
Solution. — $ 4 0 0
.0 2
$ 8.0 0 int. for 3 mo.
$8X4 = $32 int. for 1 yr. Ans.
23. Another form of interest is that known as compound
interest. In this case the interest is added to the principal
at certain intervals and this amount is taken as a new
principal.
Interest may be compounded, or added to the principal,
annually, semiannually, or quarterly, but if the time is not
stated it is understood to be 1 year.
§3
ARITHMETIC
9
Example. —Find the amount to be paid at the end of 1 year
6 months on $500 at 4% compounded semiannually.
Solution. —In this example, the rate 4% is for 1 yr., but the
interest must be added at the end of every 6 mo.; consequently, the
rate must be considered as 2% every 6 mo., which equals 4% for
1 yr. The method of performing the example is as follows:
$500
.0 2
$ 1 0.0 0
5 0 0
$ 5 1 0
.0 2
$ 1 0.2 0
5 1 0
$ 5 2 0.2 0
.0 2
$ 1 0.4 0 4 0
5 2 0.2 0
$ 5 3 0.6 0
int. at end of 6 mo.
first principal
new principal
int. at end of 1 yr.
second principal
new principal
int. at end of 1 yr. 6 mo.
third principal
amount at end of 1 yr. 6 mo. Ans.
EXAMPLES FOR PRACTICE
1. What is the interest on $800 for 5 years at 6%? Ans. $240
2. What is the amount to be paid at the end of 3 years for $300
at 4% ? Ans. $336
3. A person pays $40 per year for the use of a certain sum of
money, the interest being 4%; what is the principal? Ans. $1,000
4. If $36 is paid for 2 years’ interest on a sum of money, the rate
per cent, being 3, what is the principal? Ans. $600
5. What is 6 months’ interest on $5,000 at 4%. Ans. $100
10
ARITHMETIC
§3
RATIO
24. Suppose that it is desired to compare two numbers,
say 20 and 4. If we wish to know how many times larger
20 is than 4, we divide 20 by 4 and obtain 5 for the quotient;
thus, 20 -T- 4 = 5. Hence, we say that 20 is 5 times as large
as 4; i. e., 20 contains 5 times as many.units as 4. Again,
suppose that we desire to know what part of 20 is 4. We
then divide 4 by 20 and obtain i; thus, 4 -f- 20 = i, or .2.
Hence, 4 is i, or .2, of 20. This operation of comparing two
numbers is termed finding the ratio of the two numbers.
Ratio, then, is a comparison. It is evident that the two
numbers to be compared must be expressed in the same unit;
in other words, the two numbers must both be abstract num-
bers or concrete numbers of the same kind. For example,
it would be absurd to compare 20 horses with 4 birds, or
20 horses with 4. Hence, ratio may be defined as a com¬
parison between two numbers of the same kind.
25. . A ratio may be expressed in three ways; thus, if it is
desired to compare 20 and 4 and express this comparison as
20
a ratio, it may be done as follows: 20 -f- 4; 20 : 4, or -—. All
4
three express the ratio of 20 to 4. The ratio of 4 to 20 would
4
be expressed thus: 4 -4- 20; 4 : 20, or —. The first method of
expressing a ratio, although correct, is seldom or never
used; the second form is the one most often met with, while
the third form, called the fractional form, possesses great
advantages to students of algebra and of higher mathe¬
matical subjects. The second form is better adapted to
arithmetical subjects and is the one ordinarily adopted in
this Course.
26. The terms of a ratio are the two numbers to be
compared; thus, in the above ratio, 20 and 4 are the terms.
§3
ARITHMETIC
11
When both terms are considered together, they are called a
couplet; when considered separately, the first term is called
the antecedent and the second term, the consequent.
Thus, in the ratio 20 : 4, 20 and 4 form a couplet, and 20 is
the antecedent and 4 the consequent.
27. A ratio may be direct or inverse. The direct ratio
of 20 to 4 is 20 : 4, while the inverse ratio of 20 to 4 is 4 : 20.
The direct ratio of 4 to 20 is 4 : 20, and the inverse ratio is
20 : 4. An inverse ratio is sometimes called a reciprocal
ratio. The reciprocal of a number is 1 divided by the
number. Thus, the reciprocal of 17 is iV; of f is 1 -r- f = I;
i. e., the reciprocal of a fraction is the fraction inverted.
Hence, the inverse ratio of 20 to 4 may be expressed as 4 : 20,
or as to : i. Both have equal values; for, 4 -f- 20 = i, and
_J_.i__l_w4._A %
20 “ 4 — 2 0 X 1 — 5 •
28. The term vary implies a ratio. When we say that
two numbers vary as some other two numbers, we mean that
the ratio between the first two numbers is the same as the
ratio between the other two numbers.
29. The value of a ratio is the result obtained by per¬
forming the division indicated. Thus, the value of the ratio
20 : 4 is 5; it is the quotient obtained by dividing the ante¬
cedent by the consequent. The value of a ratio is always
an abstract number, regardless of whether the terms are
abstract or concrete numbers.
30. When a ratio is expressed in words, as the ratio of
20 to 4, the first number named is always regarded as the
antecedent and the second as the consequent, without regard
to whether the ratio itself is direct or inverse. When not
otherwise specified , all ratios are understood to be direct. To
express an inverse ratio, the simplest way of doing it is to
express it as if it were a direct ratio, with the first number
named as the antecedent, and then transpose the antecedent
to the place occupied by the consequent, and the consequent
to the place occupied by the antecedent; or, if expressed in
the fractional form, invert the fraction. Thus, to express the
12
ARITHMETIC
3
inverse ratio of 20 to 4, first write it 20 : 4, and then, trans¬
posing the terms, as 4 : 20; or as • S i a , and then inverting,
as 2 *d. Or, the reciprocals of the numbers may be taken, as
explained above. To invert a ratio is to transpose its terms.
31. Instead of expressing the value of a ratio by a single
number, it is more convenient to express it by means of
another ratio in which the consequent is 1. Thus, suppose
that it is desired to find the ratio of the weights of two pieces
of iron, one weighing 45 pounds and the other weighing
30 pounds. The ratio of the heavier to the lighter is then
45 : 30, an inconvenient expression. Using the fractional
form, we have
45
30
quent, we obtain
U
Dividing both terms by 30,* the conse-
, or I 2 : 1. This is the same result as
obtained above, for li -f- 1 = li, and 45 -f- 30 = I 2 .
PROPORTION
32. Proportion is an equality of ratios, the equality
being indicated by the double colon ( :; ) or by the sign of
equality ( = ). Thus, to write in the form of a proportion the
two equal ratios, 8 : 4 and 6 : 3, which both have the same
value, 2, we may employ any one of the three following forms:
8 : 4 :: 6 : 3 (1)
8 : 4 = 6 : 3 (2)
33. The first form is the one most extensively used, by
reason of its having been exclusively employed in all the
older works on mathematics. The second and third forms
are being adopted by all modern writers on mathematical
subjects, and in time will probably entirely supersede the
first form. In this Course, the second form will be adopted,
*This evidently does not alter the value of the ratio, since by the
laws or fractions, both numerator and denominator may be divided by
the same number without changing the value of the fraction.
§3
ARITHMETIC
13
unless some statement can be made clearer by using the
third form.
34. A proportion may read in two ways. The old
way to read the preceding proportion was: 8 is to A as 6 is to 3;
the new way is: the ratio of 8 to 4 equals the ratio of 6 to 3.
The student may read it either way, but we recommend the
latter.
35. Each ratio of a proportion is termed a couplet. In
the above proportion, 8 : 4 is a couplet, and so is 6 : 3.
36. The numbers forming the proportion are called
terms; and they are numbered consecutively from left to
right, thus:
first second third fourth
8 : 4.= 6 : 3
Hence, in any proportion, the ratio of the first term to the
second term equals the ratio of the third term to the fourth
term.
37. The first and fourth terms of a proportion are called
the extremes, and the second and third terms the means.
Thus, in the foregoing proportion, 8 and 3 are the extremes
and 4 and 6 are the means.
38. A direct proportion is one in which both couplets
are direct ratios.
39. An inverse proportion is one that requires one
of the couplets to be expressed as an inverse ratio. Thus,
8 is to 4 inversely as 3 is to 6 must be written 8 : 4 = 6 : 3;
i. e., the second ratio (couplet) must be inverted.
40 . Proportion forms one of the most useful sections of
arithmetic. In our grandfathers’ arithmetics, it was called
“The rule of three.”
41 . Rule.— In any proportion , the product of the extremes
equals the product of the means.
Thus, in the proportion
17 : 51 = 14 : 42
17 X 42 = 51 X 14, since both products equal 714.
14 ARITHMETIC §3
42. Rule.— The product of the extremes divided by either
mean gives the other mean.
Example. —What is the third term of the proportion 17 : 51 = : 42?
Solution. —Applying the rule,
17 X 42 = 714, and 714 -r 51 = 14. Ans.
43. Rule.— The product of the means divided by either
extreme gives the other extreme.
Example. —What is the first term of the proportion : 51 = 14 : 42?
Solution. —Applying the rule,
51 X 14 = 714, and 714 -f- 42 = 17. Ans.
44. When stating a proportion in which one of the terms
is unknown, represent the missing term by a letter, as x.
Thus, the last example would be written
x : 51 = 14 : 42
and for the value of x, we have x = ^ ^ = 17.
45. The principle of all calculations in proportion is
this: Three of the ter?ns are always given a?id the remaining
one is to be fou?id.
Example 1.—If 4 men can earn $25 in 1 week, how much can
12 men earn in the same time?
Solution. —The required term must bear the same relation to the
given term of the same kind as one of the remaining terms bears to
the other remaining term. We can then form a proportion by which
the required term may be found.
The first question the student must ask himself in every calculation
in proportion is: “What is it I want to find?” In this case it is
dollars. We have two sets of men, one set earning $25, and we want
to know how many dollars the other set earns. It is evident that the
amount 12 men earn bears the same relation to the amount that 4 men
earn as 12 men bears to 4 men. Hence, we have the proportion, the
amount 12 men earn is to $25 as 12 men is to 4 men; or, x : $25 = 12
men : 4 men. Since either extreme equals the product of the means
divided by the other extreme, we have,
Since it matters not which place x, or the required term, occupies,
the problem could be stated in any of the following forms, the value of
x being the same in each:
3
ARITHMETIC
15
(a) $25 : x = 4 men : 12 men; or the amount 12 men earn
$25 X 12
= - 1 or $75, since either mean equals the product of the
extremes divided by the other mean.
(b) 4 men : 12 men = $25 : x\ or the amount that 12 men earn
$25 X 12
= ---, or $75, since either extreme equals the product of the
means divided by the other extreme.
t
(c ) 12 men : 4 men = x : $25; or the amount that 12 men earn
$25 X 12
= - - -, or $75, since either mean equals the product of the
extremes divided by the other mean.
Example 2.—If 200 looms produce 300 cuts in a given time, how
many cuts will 400 looms produce in the same time, when working
under similar conditions?
Solution. —Comparing the number of looms in the two cases, the
ratio 200 : 400 is obtained; comparing the number of cuts, the ratio is
300 : x. Hence, the proportion is
200 : 400 = 300 : x
400 X 300
from which
x =
200
= 600 cuts. Ans.
Another way of stating- the proportion in examples like
the one last given is to write the first number, in this case
200 looms; then write the next number of the same kind, in
this case 400 looms, with the ratio sign between; write the
sign of equality and then the number mentioned in connec¬
tion with the first and which results from it, in this case
300 cuts; lastly, write the number of the same kind as the
third number. An example will illustrate this.
Example 3.—If 60 bales of cotton are used in 5 days, how many
bales will be used in 18 days at the same rate?
Solution. —The first number is 60 bales; the other number
expressed in bales is to be found and is represented by x. The num¬
ber mentioned in connection with the first and resulting from it is
5 da.; that is, it took 5 da. to use 60 bales; hence, 5 da. is the third
term and 18 da. is the last, or fourth, term. The proportion is, therefore,
60 bales : x bales = 5 da. : 18 da.
or, 60 : x = 5 : 18
from which x = ^ ^5 —- = 216 bales. Ans.
16
ARITHMETIC
3
EXAMPLES FOR PRACTICE
Find the value of x in each of the following:
(a)
$16
: $64
x : $4.
(a)
x =
$1
(*)
x :
85 =
10
: 17.
(b)
X —
50
24
x —
15
: 40.
(c)
X =
64
(d)
18
94 =
2
x. Ans.<
(d)
X =
10|
(e)
$75
: $100
—
x : 100.
(e)
X =
75
(f)
15
x =
21
10.
(0
X =
7y
Or)
x :
75 yd.
=
$15 : $5.
(g)
X =
225 yd.
1. If 75 pounds of lead costs $2.10, what will 125 pounds cost at the
same rate? Ans. $3.50
2. If A does a piece of work in 4 days and B does it in 7 days, how
long will it take A to do what B does in 63 days? Ans. 36 da.
3. How long will it take 60 cards to card a certain amount of
cotton if 40 cards can do the work in 3 days? Ans. 2 da.
4. A mill is running 10 hours a day for 6 days in the week and is
turning off 2,400 cuts a week, but they wish to run on reduced time so
that their production will be but 1,600 cuts per week; how many hours
per day will the mill run? Ans. 6f hr.
5. If 4 combers turn off a certain weight of cotton in 6 days, how
many combers will have to be procured to do the same work in
2 days? Ans. 12 combers
INVERSE PROPORTION
46. In Art. 39, an inverse proportion was defined as one
that required one of the couplets to be expressed as an
inverse ratio. Sometimes the word inverse occurs in the
statement of the example; in such cases the proportion can
be written directly, merely inverting one of the couplets.
But it frequently happens that only by carefully studying
the conditions of the example can it be ascertained whether
the proportion is direct or inverse. When in doubt, the
student can always satisfy himself as to whether the propor¬
tion is direct or inverse by first ascertaining what is required,
and stating the proportion as a direct proportion. Then, in
order that the proportion may be true, if the first term is
smaller than the second term , the third term must be smaller
than the fourth; or if the first term is larger tha?i the second
3
ARITHMETIC
17
term , the third term must be larger than the fourth term.
Keeping this 'in mind, the student can always tell whether
the required term will be larger or smaller than the other term
of the couplet to which the required term belongs. Having
determined this, the student then refers to the example and
ascertains from its conditions whether the required term is
to be larger or smaller than the other term of the same
kind. If the two determinations agree, the proportion is
direct; otherwise, it is inverse, and one of the couplets must
be inverted.
Example. —A’s rate of doing work is to B’s as 5 : 7; if A does a
piece of work in 42 days, in what time will B do it?
Solution. —The required term is the number of days it will take B
to do the work. Hence, stating as a direct proportion,
5:7 — 42 : x
Now, since 7 is greater than 5, at will be greater than 42. But, refer¬
ring to the statement of the example, it is easy to see that B works
faster than A; hence, it will take B a less number of days to do the
work than A. Therefore, the proportion is an inverse one, and should
be stated
5:7 — x 42
5 X 42
from which x = —=— - 30 da. Ans.
Had the example been stated thus: The time that A requires to do
a piece of work is to the time that B requires, as 5 : 7; A can do it in
42 da., in what time can B do it? it is evident that it would take B
a longer time to do the work than it would A; hence, x would be
greater than 42, and the proportion would be direct, the value of x being
—=■— = 58.8 da.
o
EXAMPLES FOR PRACTICE
1. If a pump that discharges 4 gallons of water per minute can fill
a tank in 20 hours, how long will it take a pump discharging 12 gal¬
lons per minute to fill it? Ans. 6f hr.
2. The circular seam of a boiler requires 50 rivets when the pitch
is 2i inches; how many would be required if the pitch were 3| inches?
Ans. 40
3. The, spring hangers on a certain locomotive are 2i inches wide
and -f inch thick; those on another engine are of the same sectional
area, but are 3 inches wide; how thick are they? Ans. f in.
18
ARITHMETIC
§3
4. A locomotive with driving wheels 16 feet in circumference runs
a certain distance in 5,000 revolutions; how many revolutions would it
make in going the same distance if the wheels were 22 feet in circum¬
ference (no allowance for slip being made in either case) ?
Ans. 3,636 a rev.
UNIT METHOD
47. In the older books on arithmetic, a large number of
problems were solved by compound proportion; but these
problems can be solved much more easily by the unit
method, which will now be explained by means of examples.
Example 1.—If a pump discharging 4 gallons of water per minute
can fill a tank in 20 hours, how long will it take a pump discharging
12 gallons per minute to fill the tank?
Solution.— -A pump discharging 4 gal. per min. fills the tank
in 20 hr. Therefore, a pump discharging 1 gal. per min. fills it
in 4 X 20 hr. Hence, a pump discharging 12 gal. per min. fills it in
4 X 20 hr. _ 20 hr.
12 " ~ 3
- = 6| hr.
Ans.
Example 2.—If 4 men earn $65.80 in 7 days, how much can 14 men,
paid at the same rate, earn in 12 days?
Solution.—
Therefore,
Therefore,
Therefore,
Therefore, 14
4 men in 7 da. earn $65.80
, • - . $65.80
1 man in 7 da. earns —-—
4
, .' . . $65.80
1 man in 1 da. earns ——
» X »
1 man in 12 da. earns
men in 12 da. earn
$65.80 X 12
4X7
$65.80 X 12 X 14
4X7
Canceling,
14 men in 12 da. earn $65.80 X 3 X 2 = $65.80 X 6 = $394.80.
Ans.
48. The student will notice that in the solution of these
examples, in the successive steps, the operations of mul¬
tiplication and division were merely indicated, and no
multiplication or division was performed until the very last,
and then the answer was obtained easily by cancelation.
In arithmetical calculations, the student should make it an
invariable habit to indicate the multiplications and divisions
that occur in the successive steps of a solution, and not to
§3
ARITHMETIC
19
perform these operations until the very last. Then, he will
probably be able to use the principle of cancelation.
Example.— If a block of granite 8 feet long, 5 feet wide, and 3 feet
thick weighs 7,200 pounds, what is the weight of a block of granite
12 feet long, 8 feet wide, and 5 feet thick?
Solution. —If a block 8 ft. long, 5 ft. wide, and 3 ft. thick weighs
7,200 lb., a block 1 ft. long, 5 ft. wide, and 3 ft. thick weighs
7,200
lb.;
a block 1 ft. long, 1 ft.' wide, and 3 ft. thick weighs
7,200
8X5
lb.; and a block
7 200
1 ft. long, 1 ft. wide, and 1 ft. thick weighs ’ lb. Therefore,
8X5 X o
by the same reasoning, a block 12 ft. long, 8 ft. wide, and 5 ft. thick
weighs
7,200 X 12 X 8 X 5
8X5X3
4
7,200 X It X $ X 5
8x5x3
28,800 lb. Ans.
EXAMPLES FOR PRACTICE
1. If a pump discharges 90,000 gallons of water in 20 hours, in
what time will it discharge 144,000 gallons? Ans. 32 hr.
2. If 15 looms produce 450 yards of cloth in a certain length of
time, how many yards will 75 looms produce in the same time?
Ans. 2,250 yd.
SIGNS OF AGGREGATION
49. The vinculum-, parentheses (), brackets [ ] ,
and braces { } are called symbols of aggregation, and are
used to include numbers that are to be considered together.
Thus, 13 X 8 — 3, or 13 X (8 — 3), shows that 3 is to be
taken from 8 before multiplying by 13.
13 X 8^3 = 13 X 5 = 65
13 X (8 - 3) = 13 X 5 = 65
When the vinculum or parentheses are not used, we have
13 X 8 - 3 = 104 - 3 = 101
50. In any series of numbers connected by the signs + ,
— „ X, and -s-, the operations indicated by the signs must be
performed in order from left to right, except that no addition
20
ARITHMETIC
§3
or subtraction may be performed if a sign of multiplication
or division follows the number on the right of a sign of addi¬
tion or subtraction, until the indicated multiplication or
division has been performed. In all cases the sign of multi¬
plication takes the precedence, the reason being that when
two or more numbers or expressions are connected by the
sign of multiplication, the numbers thus connected are
regarded as factors of the product indicated, and not as
separate numbers.
Example 1.—What is the value of 4 X 24 — 8 + 17?
Solution. —Performing the operations in order from left to right,
4 X 24 = 96; 96 — 8 = 88; 88 + 17 = 105. Ans.
Example 2.—What is the value of the following expression: 1,296
-e- 12 + 160 - 22 X 3i = ?
Solution. — 1,296 4- 12 = 108; 108 + 160 = 268; here 22 cannot be
subtracted from 268 because the sign of multiplication follows 22;
hence, multiplying 22 by 31 gives 77, and 268 — 77 = 191. Ans.
Had the above expression been written 1,296 4- 12 + 160
— 22 X 3i r 7 + 25, it would have been necessary to
divide 22 X 3| by 7 before subtracting, and the final result
would have been 22 X 3a = 77; 77 ■— 7 = 11; 268 — 11
= 257; 257 + 25 = 282. In other words, it is necessary to
perform all the indicated multiplication or division included
between the signs + and —, or — and + , before adding or
subtracting. Also,, had the expression been written 1,296
-4- 12 + 160 — 24| -4- 7 x 32 + 25, it would have been neces¬
sary to multiply 3i by 7 before dividing 241, since the sign
of multiplication takes the precedence, and the. final result
would have been 31 X 7 = 24l; 24l 4- 24l = 1; 268 — 1
= 267; 267 + 25 = 292.
It likewise follows that if a succession of multiplication
and division signs occur, the indicated operations must not
be performed in order, from left to right—the multiplication
must be performed first. Thus, 24 x34-4x24-9x5 = 1.
In order to obtain the same result that would be obtained by
performing the indicated operations in order, from left to
right, symbols of aggregation must be used. Thus, by using
§3
ARITHMETIC
21
two vinculmns, the last expression becomes 24 X 3 -r- 4
X 2 -r 9 X 5 = 20, the same result that would be obtained
by performing the indicated operations in order, from
left to right.
EXAMPLES FOR PRACTICE
Find the values of the following expressions:
(a)
(8 + 5 — 1) 4-4.
f(«)
3
(b)
5 X 24 - 32.
(b)
88
(c)
5 X 24 -r 15.
(c)
8
(d)
144 - 5 X 24.
A ns
(d)
24
(e)
(1,691 - 540 + 559) -r
- 3 X 57.
(e)
10
(O
2,080 + 120 - 80 X 4
- 1,670.
if)
210
(g)
(90 + 60 25) X 5 -
29.
(g)
1
W
90 + 60 25 X 5.
iw
1.2
ARITHMETIC
(PART 4)
INVOLUTION
1. Involution is the process of multiplying a number by
itself one or more times. The product obtained by this
process is called a power of that number.
Thus, the second power of 3 is 9, since 3 X 3 is 9.
The third power of 3 is 27, since 3 X 3 X 3 is 27.
The fifth power of 2 is 32, since 2x2x2x2X2is 32.
2. An exponent is a small figure placed to the right and
a little above a number to show to what power it is to be
raised, or how many times the number is to be used as a
factor, as the small figures *, 3 , and 5 in the following:
Thus, 3’ = 3 X 3 = 9
3 3 = 3x3x3 = 27
2 5 = 2x2x2x2x2 = 32
3 . The root of a number is that number which, used the
required number of times as a factor, produces the number.
In the above cases 3 is a root of 9, since 3 X 3 is 9. It is
also a root of 27, since 3 X 3 X 3 is 27. Also, 2 is a root of
32, since 2 X 2 X 2 X 2 X 2 is 32.
4. The second poiver of a number is called its square.
Thus, 5* is called the square of 5, or 5 squared , and its value
is 5 X 5 = 25.
5. The third power of a number is called its cube. Thus,
5 3 is called the cube of 5, or 5 cubed, and its value is
5x5x5= 125.
For notice of copyright , see page immediately following the title page
2
ARITHMETIC
§4
6. To find any power of a number:
Example 1.—What is the third power, or cube, of 35?
Solution.— 3 5
3 5
* 1 7 5
10 5
12 2 5
3 5
6 12 5
3 6 7 5
cube 4 2 8 7 5 Ans.
Example 2.—What is the fourth power of 15?
Solution.— 1 5
1 5
7 5
1 5
2 2 5
1 5
112 5
2 2 5
3 3 7 5
1 5
1 6 8 7 5
3 3 7 5
fourth power 5 0 6 2 5 Ans.
Example 3.— 1.2 3 — what?
Solution.— 1.2
1.2
1.4 4
1.2
2~88
14 4
cube 1.7 2 8 Ans.
§4
ARITHMETIC
3
Example 4.—What is the third power, or cube, of §?
Solution.—
5 _33 v 3_3X 3 X 3 _ 27
8 X 8 X 8 8X8X8 512’ AnS ‘
Rule. — I. To raise a whole number or a decimal to any
power , use it as a factor as many times as indicated by the exponent.
II. To raise a fraction to any power , raise both the numer¬
ator and denominator to the power indicated by the exponent.
EXAMPLES FOR PRACTICE
Raise the following to the powers indicated:
(a) 85 2
(b) (If)’
(r) 6.5 2
. (d) 14*
(e) (l ) 3
(0 ( t ) 3
Or) (i) 3
(h) 1.4 s
EVOLUTION
7. Evolution is the reverse of involution. It is the
process of finding the root of a number that is considered
as a power.
8. The square root of a number is that number which,
when used twice as a factor, produces the number.
Thus, 2 is the square root of 4, since 2 X 2, or 2\ = 4.
9. The cube root of a number is that number which,
when used three times as a factor, produces the number.
Thus, 3 is the cube root of 27, since 3 X 3 X 3, or 3 3 , = 27.
10. The radical sign V, when placed before a number,
indicates that some root of that number is to be found.
11. The index of the root is a small figure placed over
and to the left of the radical sign, to show what root is to be
found. -
Ans. •
(a)
(b)
(c)
(d)
(e)
(f)
(g)
iw
7,225
144
169
42.25
38,416
2 7
64
12 5
2 16
3 4 3
8
5.37824
4
ARITHMETIC
§4
Thus, ^100 denotes the square root of 100.
\125 denotes the cube root of 125.
^256 denotes the fourth root of 256, and so on.
When the square root is to be extracted, the index is
generally omitted. Thus, VlOO indicates the square root
of 100. Also, V225 indicates the square root of 225.
SQUARE ROOT
12. The largest number that can be written with one figure
is 9, and 9 s = 81; the largest number that can be written with
two figures is 99, and 99“ = 9,801; with three figures 999, and
999“ = 998,001; with four figures 9,999, and 9,999“ = 99,980,-
001, etc. In each of the above it will be noticed that the
square of the number contains just twice as many figures as
the number itself.
In order to find the square root of a number, the first step
is to find how many figures there will be in the root. This is
done by pointing of! the number into periods of two figures
each, beginning at the right. The number of periods will
indicate the number of figures in the root. Thus, the square
root of 83,740,801 must contain 4 figures, since, pointing
off'the periods, we get 83 / 74'08'01, or 4 periods; consequently,
there must be 4 figures in the root. In like manner, the
square root of 50,625 must contain 3 figures, since we have
S'06'25, 3 periods.
13 . Suppose that it is desired to find the square root
of 49,729. Applying the method of pointing off to the
number in the example, gives 4'97'29. It will be noticed
that the left-hand period contains but one figure. This must
necessarily result whenever the number contains an odd
number of figures.
Next draw a line at the right of the number and find the
largest number whose square is less than, or equal to, 4, the
first period. This number is evidently 2, since 2x2 = 4.
4
ARITHMETIC
5
Place this number at the right of the line as the first figure
of the answer. The process up to this point is as follows:
4'9 7'2 9 ( 2
Subtract the square of the first figure in the answer from
the first period and bring down the next period, which gives
4'9 7'2 9 ( 2
4
9 7
Multiply the first figure of the answer by 2, and use this
result as a trial divisor to divide into the remainder not con¬
sidering its last figure.
In this example, 2x2 = 4, which, as a trial divisor, gives
2 as a result when divided into 9, which is the remainder
without the last figure. Place this 2 as the second figure of
the answer.
Place the second figure, 2, of the answer at the right of the
trial divisor, making it the unit figure of the new divisor,
which gives 42 as the new divisor. This new divisor multi¬
plied by the second figure of the answer gives 84, which is to
be subtracted from the remainder.
The process thus far is as follows:
4'9 7'2 9 ( 2 2
4
4 2) 97
8 4
1 3
Note. —It frequently happens that the result obtained by multiplying
the complete trial divisor by the last quotient figure obtained by
dividing the remainder, exclusive of its right-hand figure, by the first
trial divisor is larger than the remainder, in which case the next
number smaller must be tried. Thus, in this example, suppose that
the remainder had been 80 instead of 97; then the result obtained by
multiplying 42 by 2 would have been larger than the remainder and 1
would have been tried. The new divisor would therefore have been
41 and the second figure of the answer 1.
The rest of the process is simply a repetition of that
already described. Consequently, the next period is now
brought down, which gives 1,329 as the complete remainder.
6
ARITHMETIC
4
The two figures in the answer are multiplied by 2, which
gives 44 as a trial divisor.
The trial divisor divided into the remainder without the
last figure, or 132, gives 3 as the third figure of the answer.
Place the third figure, 3, of the answer at the right of the
trial divisor used to obtain it, making it the unit figure of
the new divisor, which gives 443 as the new divisor; this
multiplied by 3, gives 1,329, the complete operation being
as follows:
4'9 7'2 9 ( 2 2 3 Ans.
4
42 m
84
443)1329
13 2 9
It will be seen that there is no remainder; 223 is conse¬
quently the square root of 49,729.
14. If, in any case, the trial divisor is larger than the
remainder omitting the last figure, a cipher must be placed
in the answer and also at the right of the trial divisor, making
a new trial d'visor. Bring down the next period for a new
remainder and divide by the trial divisor to obtain the next
figure of the answer.
Example. —Find the square root of 255,025.
Solution. — 2 5'5 0'2 5 ( 5
2 5
1 0) 5 0
It will be seen that when the example has been performed to this
point, the trial divisor 10 will be found to be larger than 5, which is
the remainder, omitting the last figure. Therefore, place a cipher as
the next figure of the answer; bring down the next period for a new
remainder, and multiply the figures in the answer by 2 in order to
obtain a trial divisor. This process is as follows:
2 5'5 0'2 5 ( 5 0
2 5
1 0 ) 5 0
0 0
100 ) 5025
§4
ARITHMETIC
7
Dividing the trial divisor 100 into the remainder without the last
figure, or 502, 5 is obtained. Place this number as the next figure of
the answer and also as the last figure of the trial divisor, which gives
1,005 as the complete divisor.
Multiply the complete divisor by the last figure of the answer and
place the result under the remainder. The process is as follows:
2 5'5 O'2 5 ( 5 0 5 Ans.
2 5
10 ) 50
0 0
1005 ) 5025
5 0 2 5
Since the product obtained by multiplying the complete divisor by
the last figure of the answer exactly equals the remainder, then 505
must be the square root of 255,025.
15 . The square of a decimal always contains twice as
many figures as the number squared. For example, .1*
= .01, .13 2 = .0169, .751 2 = .564001, etc. If it be required
to find the square root of a decimal and the decimal has not
an even number of figures in it, annex a cipher. The best
way to determine the number of figures in the root of a
decimal is to begin at the decimal point, and, going toward
the right, point off the decimal into periods of two figures
each. Then, if the last period contains but one figure,
annex a cipher. Thus, the decimal .625 is pointed off as
follows: .62'50.
16 . If the number is a whole number and decimal, the
whole number is pointed off to the left and the decimal to
the right. That is, commence at the decimal point in both
cases and point off. Thus, the number 126.423 is pointed
off as follows: 1'26.42'30.
17 . The operation of finding the square root in all these
cases is similar to that previously described, except that
when the decimal point is reached, a decimal point is placed
in the answer. This gives as many decimal places in the
answer as there are periods of decimals in the number. An
example without explanation will be given here.
8
ARITHMETIC
§4
Find the square root of 606.6369.
6'0 6.6 3'6 9 ( 2 4.6 3 Ans.
4
44)206
1 7 6
486 ) 3063
2 9 16
4923) 14769
1 4 7 6 9
18. There are comparatively few numbers that are exact
squares and, consequently, it is possible to find the exact
square root in only a small number of cases. Numbers
that have an exact root are called perfect powers and the
factors of the perfect powers are called rational factors.
Numbers that have no exact root are called surds and the
factors are called irrational factors. In the numbers
from 1 to 1,000, inclusive, there are only 42 perfect powers,
not counting 1, and of these only 30 are perfect squares and
9 perfect cubes.
20 is an example of a surd. This number lies between
16 (= 4 2 ) and 25 (= 5 2 ); hence, the square root of 20, or
V20 is greater than 4 and less than 5 and is, therefore, equal
to 4 plus an interminable decimal. In other words, no
matter to how many figures the square root of 20 may be
calculated, the root will never be found exactly.
Although the root of a surd cannot be found exactly, as
close an approximate may be obtained as is desired, since
the result may be carried to any required number of decimal
places by annexing periods of two ciphers each to the number.
Example.—
Solution.—
§4
ARITHMETIC
9
Example. --What is the square root of 3? Find the result to five
decimal places.
Solution— 3.0 O'O O'O O'O O'O 0 ( 1.7 3 2 0 5 Ans.
1
27)200
1 8 9
343 ) 1100
10 2 9
3462 ) 7100
6 9 2 4
3464) 17600
0 0 0 0 0
346405) 1760000
1732025
2 7 9 7 5
Explanation. —Annexing five periods of ciphers to the
right of the decimal point, the first figure of the root is
found to be 1. Multiply 1 by 2 in order to obtain a trial
divisor and find how many times it is contained in the
remainder without the last figure, or 20. It is contained
10 times, but this is evidently too large. Trying 9 it is
found that this is also too large, since 29 X 9 = 261 and
261 is greater than the remainder 200. In the same way it
is found that 8 is also too large. Trying 7, 7 times 27 is
189, a result smaller than 200; therefore, 7 is the second
figure of the root. The next two figures, 3 and 2, are
easily found. The fifth figure in the root is a cipher, since
the trial divisor 3,464 is greater than the remainder with¬
out the last figure, or 1,760. Bringing down the next period
and multiplying the number in the answer by 2 to obtain
the trial divisor, it is found that the next figure in the
answer is 5. This gives the required number of decimal
places in the answer.
Proof. —To prove square root, square the result obtained.
If the number is an exact power, the square of the root will
equal it; if it is not an exact power, the square of the root,
plus the remainder, will equal it.
10
ARITHMETIC
4
19 . To find the square root of a number:
Rule. —Divide the number into periods of two figures each ,
commencing at the right , or if the number contains a decimal ,
at the decimal point and working both ways.
Find the largest number that multiplied by itself will not be
greater than the first left-hand period and zvrite this number as
the first figure of the answer.
Square the first figure of the answer and subtract this square
from the left-hand period of the number. To this result annex
the next period of the number for a complete remainder.
Multiply the first figure of the answer by 2 and use the result
as a trial divisor.
Divide the complete remainder , leaving off the last figure , by
the trial divisor and write the result , or the result diminished ,
as the second figure of the answer.
To the trial divisor , annex this second figure of the answer
and use the result as a complete divisor.
Multiply the complete divisor by the second figure of the root
and subtract this result from the complete remainder.
To the result thus obtained annex the next period for a com¬
plete remainder.
Proceed in this manner until all the periods of the number
have been brought down.
EXAMPLES FOR PRACTICE
Find the square root of the following:
(а) 1,936
(б) 110,889
(c) 4.9729
(d) .0196
(e) 9,604
(/) 163,216
Ans.
(a) 44
{b) 333
( c ) 2.23
(d) .14
(e) 98
( f) 404
CUBE ROOT
20 . Since cube and higher roots are seldom
required in mill work, the student may, if he so
desires, omit the following articles. No examples
relating to them are included among the Exami¬
nation Questions at the end of this Section;
4
ARITHMETIC
11
nevertheless, the student is advised to read care¬
fully the following articles, as they contain much
that may be of value to him.
21. In the same manner as in square root, it can be
shown that the periods into which a number, whose cube root
is to be extracted, is divided must contain three figures,
except that the first, or left-hand, period of a whole or mixed
number may contain one, two, or three figures.
Suppose that it is desired to find the cube root of 11,390,625.
First separate the number into periods of three figures each,
commencing at the right. This gives H'390'625. Draw a
line at the right of the number as was done when extract¬
ing the square root and find the largest number whose cube
will not be greater than the left-hand period. Place this
number as the first figure of the answer.
In the example given it will be seen that the left-hand
period is 11 and the largest number whose cube will not be
greater than 11 is 2, since 3 3 —■ 27.
Place the 2 as the first figure of the answer and subtract
the cube of this number from the left-hand period of the
number. To the remainder thus obtained annex the next
period of the number. The process thus far is as follows:
1 1'3 9 0'6 2 5 ( 2
8
3 3 9 0
It is next necessary to find a trial divisor. This is obtained
by considering the number in the answer as tens, squaring
it, and multiplying the result by 3. 2 considered as tens'is 20,
which squared gives 400. 400 multiplied by 3 gives 1,200 as
a trial divisor.
Dividing the trial divisor into the remainder 3,390, it is
seen that it is contained 2 times. Place the 2 as the second
figure of the answer. This gives the following:
1 1'3 9 0'6 2 5 ( 2 2
8
20* X 3 = 1200 |3 3 90
12 ARITHMETIC §4
In order to obtain a complete divisor , it is necessary to add
together three products.
The first product is that obtained for the trial divisor, or
in this case 1,200.
The second product is obtained by considering the first
figure of the answer as tens , and multiplying it by the second
figure of the answer, and by 3. In the example being worked,
the first figure of the answer considered as tens gives 20,
which multiplied by the second figure of the answer, or 2,
gives 40, which result multiplied by 3 gives 120 as the
second product.
The third product is obtained by squaring the second
figure of the answer, which in this case is 2 2 = 4.
Adding these three products together gives, for the com¬
plete divisor, 1,200 + 120 + 4 = 1,324. The example worked
to this point is as follows:
1 1'3 9 0'6 2 5 ( 2 2
8
20* X 3 = 1 2 0 0
(20 X 2) X 3 = 120
2 2 = 4
3 39 0
13 2 4
Next multiply the divisor by the second figure of the
answer and subtract the result from the remainder previ¬
ously obtained. To the result annex the next period for
a new remainder. The process is as follows:
1 1'3 9 0'6 2 5 ( 2 2
8
20 2 X 3 = 1 2 0 0
(20 X 2) X 3 = 120
2 2 = 4
33 9 0
13 2 4
2 6 4 8
742625
The further processes are but repetitions of those previ¬
ously described and, consequently, will not need much
explanation.
§4
ARITHMETIC
13
It is next necessary to find a trial divisor for the
remainder. This is done in the same manner as in the
first case, the number in the answer being considered as
tens, and squared, after which it is multiplied by 3. In
the example given this will be 220 2 = 48,400, which, multi¬
plied by 3, gives 145,200 for a trial divisor.
The trial divisor divided into the remainder gives 5 as
the next figure of the answer.
Next, find the complete divisor by adding together the
three products, taking the trial divisor as the first product;
3 times the number already in the answer considered as tens
multiplied by the next figure of the answer for the second
product, and the square of the third figure of the answer for
the third product. Multiply the complete divisor by the
third figure of the answer and subtract the result thus
obtained from the remainder previously obtained. The
complete process is as follows:
20 2 X 3 = 12 0 0
(20 X 2) X 3 = 120
T = 4
132 4
220 2 X 3 =14 5 2 0 0
(220 X 5) X 3 = 3300
5 2 =_25
148525
1 1'3 9 0'6 2 5 ( 2 2 5 Ans.
8
3 39 0
26 4 8
742625
742625
Since the product obtained by multiplying the complete
divisor by the third term of the answer is just equal to the
remainder, the process is complete, and the answer, or 225,
is the cube root of the number 11,390,625.
22. In extracting the cube root of a decimal, proceed as
above, taking care that each period contains three figures.
Begin the pointing off at the decimal point, going toward
the right. If the last period does not contain three figures,
annex ciphers until it does.
14
ARITHMETIC
§ 4
The number of decimal places in the answer will equal
the periods in the decimal whose root is to be extracted.
Example.— What is the cube root of .009129329?
Solution. — .0 0 9'1 2 9 r 3 2 9 (.2 0 9 Ans.
8
20* X 3 =
12 0 0
112 9
0 0 0 0
200* X 3 = 1 2 0 0 0 0
(200 X 9) X 3 = 5400
9 5 = 8 1
1129329
125481
1129329
Explanation. —In this example it will be noticed that in
the first case the trial divisor 1,200 is greater than the
remainder 1,129; consequently, the next figure of the answer
is 0 and it is necessary to bring down the next period for a
new remainder, after which the regular process is followed.
23. One example of extracting the cube root of a mixed
number will be given here without any further explanation.
Example.— What is the cube root of 47.832147?
Solution.—
30 a X 3 = 2 7 0 0
(30 X 6) X 3 = 5 4 0
6 2 = 3 6
3 2 7 6
360 2 X 3 = 388800
(360 X 3) X 3 = 3240
3 2 = 9
392049
4 7.8 3 2'1 4 7 ( 3.6 3 Ans.
2 7
JOSS 2
1 9 6 5 6
117 6 1 4 7
1176147
Proof.— To prove cube root, cube the result obtained^ If
the given number is an exact power, the cube of the root
will equal it; if not an exact power, the cube of the root, plus
the remainder, will equal it.
24. To find tlie cube l'oot of a number:
Rule. — Sep a ra te the number into periods of three figures
each , commencing at the right , or if the number contains a
decimal, at the decimal point and working both ways.
§4
ARITHMETIC
15
Find the largest number whose cube is not greater than the
left-hand period and write this number as the first figure of the
answer.
Subtract the cube from the left-hand period and to this result
annex the next period of the number.
Square the member in the answer considered as tens , and mul¬
tiply this result by 3.
Use the result thus obtained for a trial divisor to divide into
the remainder , and place the number resulting from this division ,
or the number diminished , as the next figure of the answer.
To the trial divisor , add the result obtained by multiplying
the first figure of the answer considered as tens by the second
figure of the answer and by 3; also add the square of -the second
figure. The sum thus obtained is the complete divisor.
Multiply the complete divisor by the second figure of the
answer and subtract the result thus obtained from the remainder.
To this result annex the next period of the number and proceed
in the manner described until all the periods have been tised.
EXAMPLES FOR PRACTICE
Find the cube root of the following numbers:
(«)
1,860,867
»
123
(0
185,193
(b)
57
(0
636,056
Ans. •
{c)
86
(d)
87,528.384
id)
44.4
(e)
74,088
{e)
42
(f)
257,259,456
1(0
636
ROOTS
OF FRACTIONS
25. •If the given number is in the form of a fraction, and
it is required to find some root of it, the simplest and most
exact method is to reduce the fraction to a decimal and extract
the required root of the decimal. If, however, the numerator
and denominator of the fraction are perfect powers, extract
the required root of each separately, and write the root of
the numerator for a new numerator, and the root of the
denominator for a new denominator.
16
ARITHMETIC
§4
Example
Solution.—
1.—What is the square root
IT = = 3
\64 _ yl64 8'
of A?
Ans.
Example 2.—What is the square root of f ?
Solution. —Since f = .625, Vf = V. 625 = .7906. Ans.
Example 3.—What is the cube root of ff ?
Solution.—
Example 4.—What is the cube root of i?
Solution. —Since \ = .25 = C25 = .62996.
Ans.
Rule. —Extract the required root of the numerator and
denominator separately; or, reduce the fraction to a decimal,
and extract the root of the decimal.
TABLE METHOD OF EXTRACTING SQUARE
AND CUBE ROOTS
26. In any number, the figures beginning with the first
digit* at the left and ending with the last digit at the right,
are called the significant figures of the number. Thus,
the number 405,800 has the four significant figures 4, 0, 5, 8;
and the number .000090067 has the five significant figures
9, 0, 0, 6, and 7.
The part of a number consisting of its significant figures
is called the significant pai't of the number. Thus, in the
number 28,070, the significant part is 2807; in the number
.00812, the significant part is 812; and in the number 170.3,
the significant part is 1703.
In speaking of the significant figures or of the significant
part of a number, we consider the figures, in their proper
order, from the first digit at the left to the last digit at the
right, but we pay no attention to the position of the decimal
point. Hence, all numbers that differ only in the position ot
the decimal point have the same significant part. For example,
.002103, 21.03, 21,030, and 210,300 have the same significant
figures 2, 1,0, and 3, and the.same significant part 2103.
*A cipher is not a digit.
§4
ARITHMETIC
17
SQUARES AND CUBES
No.
Square
Cube
1.0
I .OO
1.000
1.1
1.21
I- 33 I
1.2
1.44
1.728
1.3
1.69
2.197
i -4
1.96
2-744
i -5
2.25
3-375
1.6
2.56
4.096
i -7
2.89
4-913
i.8
3-24
5-832
1-9
3.61
6.859
2.0
4.00
8.000
2.1
4.41
9.261
2.2
4.84
10.648
2-3
5-29
12.167
2.4
5-76
13.824
2-5
6.25
i 5-625
2.6
6.76
I 7.576
2.7
7.29
19.683
2.8
7.84
21.952
2.9
8.41
24.389
3-0
9.00
27.000
3 -i
9 , 6 1
29.791
3-2
10.24
32.768
3-3
10.89
35-937
3-4
11.56
39-304
3-5
12.25
42.875
3-6
12.96
46.656
3-7
13.69
50.653
3-8
14.44
54-872
3-9
15.21
59-319
4.0
16.00
64.000
4.1
16.81
68.921
4.2
17.64
74.088
4-3
18.49
79.507
4-4
19.36
85.184
4-5
20.25
91.125
4.6
21.16
97-336
4-7
22.09
103.823
4-8
23.04
no.592
4-9
24.01
117.649
5-0
25.00
125.000
5 -i
26.01
132.651
5-2
27.04
140.608
5-3
28.09
148.877
5-4
29.16
157.464
No.
Square
Cube
5-5
30.25
166.375
5-6
3 I .36
175.616
5-7
32.49
185.193
5-8
33-64
I 95 -II 2
5-9
34 - 8 i
205.379
6.0
36.00
2x6.000
6.1
37.21
226.981
6.2
38.44
238.328
6.3
39.69
250.047
6.4
40.96
262.144
6.5
42.25
274.625
6.6
43-56
287.496
6-7
44.89
300.763
6.8
46.24
314.432
6.9
47 . 6 i
328.509
7-0
49.00
343.000
7 -i
50.41
357.911
7.2
51.84
373.248
7-3
53-29
389.017
7-4
54-76
405.224
7-5
56.25
421.875
7.6
57-76
438.976
7-7
59-29
456.533
7-8
60.84
474.552
7-9
62.41
493.039
8.0
64.00
512.000
8.1
65.61
531.441
8.2
67.24
551.368
8-3
68.89
571.787
8.4
70.56
592.704
8-5
72.25
614.125
8.6
73.96
636.056
8-7
75.69
658.503
8.8
77-44
681.472
8.9
79.21
704.969
9.0
81.00
729.000
9.1
82.81
753.571
9.2
84.64
778.688
9-3
86.49
804.357
9.4
88.36
830.584
9-5
90.25
857.375
9.6
92.16
884.736
9-7
94.09
912.673
9.8
96.04
941.192
9.9
98.01
970.299
18
ARITHMETIC
4
27. Figuring’ the root of a whole number to one decimal
place is often accurate enough for practical purposes in mill
work and the following method of obtaining the square and
cube roots of numbers, which will in almost every case give
sufficiently accurate results, will be found to be of advantage.
28. By means of the table which contains the squares and
cubes of numbers from 1 to 10, varying by tenths, the first
three, and frequently the first four, significant figures of the
square root or cube root of any number can be readily
determined.
29. By the aid of this table, the first two significant
figures of the root can be obtained directly and one more by
a slight calculation. For example, suppose that it is desired
to find the first three significant figures of V5,269.73. Point¬
ing off into periods and moving the decimal point so that
it falls between the first and second periods, the number
becomes 52.69 / 73; in other words, the significant figures of
V5,269.73 are the same as of V52.6973. Since four figures
only are given in the table, reduce the given nifmber to four
figures. The problem then becomes: find the first three fig¬
ures of V52.70. Referring to the table, 52.70 lies between
51.84 ( = 7.2 2 ) and 53.29 ( = 7.3 2 ); hence, the first two figures
of the root are 7.2. Find the difference between the two
numbers in the table between which the given number falls
and call it the first difference; thus, 53.29 — 51.84 = 1.45 =
the first difference. Find the difference between the smaller
number in the table and the given number and call it the
second difference; thus, 52.70 — 51.84 = .86 = the second
difference. Divide the second difference by the first differ¬
ence, and the first figure of the quotient, if the quotient
is .05 or greater, will be the third figure of the root, when
reduced to one figure. If the quotient is less than .05, the
third figure of the root is a cipher. Thus, .86 -r- 1.45 => .59,
or .6 when reduced to one figure. Therefore, the first three
figures of V52.70 are 7.26. Since the integral part of the
given number contains two periods, there are two figures in
§4
ARITHMETIC
19
the integral part of the root; therefore, V5.269.73 = 72.6 to
three figures.
30. The cube root is found to three significant figures in
exactly the same way, as shown in the following example:
Example. —Find the first three figures of V~0625.
Solution. —Pointing off and placing the decimal point between the
first and second significant periods, the result is 62.500. Referring to
the table, the first two figures of the root are 3.9; the first difference is
64.000 - 59.319 = 4.681; the second difference is 62.500 - 59.319
= 3.181; 3.181 -r- 4.681 = .67, or .7 to one figure. Therefore, V62.5
= 3.97, and V.0625 = .397 to three significant figures. Ans.
TO EXTRACT OTHER ROOTS THAN THE SQUARE
AND CUBE ROOTS
31. By two or more operations certain roots other than
square and cube roots may be extracted; thus, to find the
fourth root of a number it is only necessary to find the
square root of the number and then extract the square root
of the result thus obtained, or to find the sixth root first
extract the cube root and then the square root, etc.
Example 1. —What is the fourth root of 256?
Solution. — V256 = 16; Vl6 = 4
Therefore, V256 = 4. Ans.
In this example, V256, the index is 4, which equals 2X2. The root
indicated by 2 is the square root; therefore, the square root is extracted
twice.
Example 2.—What is the sixth root of 64?
Solution.— >/64 =4; Vi = 2
Therefore, V64 = 2. Ans.
In this example, V64, the index is 6, which equals 3 X 2. The root
indicated by 3 is the cube root; therefore, the cube and square roots
are extracted in succession.
Example 3.—What is the sixth root of 92,873,580 to two decimal
places?
Solution. — 6 = 3X2. Hence, extract the cube root, and then
extract the square root of the result. V92,873,580 = 4.52.8601, and
V452.8601 = 21.28. Ans.
20 ARITHMETIC §4
It matters not which root is extracted first, but it is probably easier
and more exact to extract the cube root first.
Rule. —Separate the index of the required root into its factors
(3's and 2’s), and extract , successively , the roots indicated by the
several factors obtained. The fuial result will be the required
root.
ARITHMETIC
(PART 5)
DENOMINATE NUMBERS
1. A simple number expresses units, either abstract or
of a single denomination; thus, 2 dollars or 7 gallons are
simple numbers.
2. A compound number is a collection of concrete
units of several denominations; thus, 10 pounds and 4 ounces
or 5 gallons and 2 quarts are compound numbers.
3. A denominate number is a concrete number that
may be changed to a different denomination without chan¬
ging its value; thus, 2 gallons equal 8 quarts, both numbers
expressing the same quantity.
4. Reduction, as applied to arithmetic, is the changing
or reducing of a number into one of a different denomination
but having an equal value.
5. Reduction descending; is the process of changing a
number of one denomination into a number of a lower
denomination, as, for instance, gallons to quarts or dollars to
cents. Reduction descending involves the process of multi¬
plication.
6. Reduction ascending is the process of changing a
number of one denomination into a number of a higher
denomination, as, for instance, quarts to gallons or cents to
dollars. Reduction ascending is the reverse of reduction
descending and involves the process of division.
For notice of copyright, see page immediately following the title page
i 5
2
ARITHMETIC
§5
MEASURES
7. A measure is a standard unit , established by law or
custom, by which quantity of any kind is measured.
MEASURE OF MONEY
UNITED STATES MONEY
8. The following table is the one used for the currency
of
the United States
•
Table
10
mills (ra.).
. . = 1
cent.
. ct
10
cents.
, . . = 1
dime.
. d.
10
dimes.
, . . = 1
dollar.
. $
10
dollars.
. . . = 1 eagle.
. E
Mills
Cents
Dimes
Dollars Eagle
10 =
1
100 =
10
- 1
1,000 =
100
= 10
= 1
10,000 =
1,000
= 100
= 10 = 1
9. The various denominations of United States currency
are based on a decimal system, the unit being 1 dollar; thus,
one-tenth of 1 dollar is 1 dime and ten times 1 dollar is 1 eagle.
Dollars are separated from cents and mills by a decimal
point, the cents occupying the first two, and the mills, the
third place to the right of the point, since cents represent
hundredth parts of a dollar and mills, thousandth parts; thus,
$25,487 is read twenty-five dollars forty-eight cents and seven
mills.
When the number of cents in an expression of dollars and
decimal parts of a dollar is less than ten, a cipher is inserted
between the decimal point and the figure denoting the num¬
ber of cents, since cents represent hundredth parts of a
dollar, thus $14.06.
§5
ARITHMETIC
3
MEASURES OF WEIGHT
AVOIRDUPOIS WEIGHT
10. Avoirdupois weight is commonly used for weigh¬
ing goods of all descriptions, except in compounding
medicines, and for all metals except gold, silver, etc.
. Table
16 drams (dr.).= 1 ounce .oz.
16 ounces.= 1 pound.lb.
100 pounds.= 1 hundredweight . . . cwt.
20 hundredweight.= 1 ton.T.
Drams Ounces Pounds Ton
W r.KiH 1
16 = 1
256 = 16 1
25,600 - 1,600 = 100 = 1
512,000 = 32,000 = 2,000 = 20 = 1
11. A long ton is equal to 2,240 pounds and is used
in connection with large lots of merchandise, notably iron
and coal, when bought and sold by the wholesale. A
long hundredweight is 112 pounds. The long ton and
long hundredweight are used in the United States Custom
Houses. Unless otherwise stated, the short ton (2,000
pounds) and short hundredweight (100 pounds) are always
referred to.
12. An adaptation of the avoirdupois weight that is
used in mill work for weighing yarn, roving, etc., is as
follows:
Table
27.34+ grains = 1 dram
437.50 grains = 16 drams = 1 ounce
7,000.00 grains = 256 drams = 16 ounces = 1 pound
TROY WEIGHT
13. Troy weight is used in weighing gold and silver
and in mints and jewelers’ shops generally.
Table
24 grains (gr.)
20 pennyweights
12 ounces . . .
1 pennyweight .... pwt.
1 ounce.oz.
1 pound.lb.
4
ARITHMETIC
§5
Grains
Penny¬
weights
Ounces
Pound
24
480
= 1
= 20 =
= 1
5,760
= 240 =
= 12
= 1
APOTHECARIES’ WEIGHT
14. Apothecaries’ weight is used in mixing medicines.
Table
20 grains (gr.) = 1 scruple ....
3 scruples.= 1 dram.
8 drams.= 1 ounce.
12 ounces.= 1 pound ....
Grains Scruples Drams Ounces Pound
20 = 1
60 = 3 = 1
480 = 24 = 8 = 1
5,760 = 288 - 96 = 12 = 1
sc.
dr.
. oz.
. lb.
MEASURES OF QUANTITY
LIQUID MEASURE
15. Liquid measure is commonly employed for meas¬
uring the more common liquids.
4 gills (gi.) . .
2 pints . . . .
4 quarts . . .
31| gallons . . .
63 gallons . . .
Gills Pints
4 = 1
8 = 2
32 = 8
1,008 = 252
2,016 = 504
Table
. . = 1 pint.pt.
. . = 1 quart.qt.
. . = 1 gallon.gal.
. . = 1 barrel.bbl.
= 1 hogshead.hhd.
Quarts
1
4
Gallons
= 1
Barrels
Hogshead
126
= 311
1
252
= 63
2 =
1
APOTHECARIES’ FLUID MEASURE
16. Apothecaries’ fluid measure is used for measuring
medicines and other liquids that are used in small quantities.
Table
60 minims (min.).
8 fluid drachms :.
16 fluid ounces .
8 pints.
1 fluid drachm
1 fluid ounce
1 pint
1 gallon
§5
ARITHMETIC
5
DRY MEASURE
17. Dry measure is used for measuring grains, fruits,
vegetables, salt, coal, and similar substances.
Table
2 pints (pt.)
=
1 quart.
. qt.
8 quarts . .
=
1 peck .
. pk.
4 pecks . . .
1 bushel .
bu.
8 bushels . .
=
1 quarter.
. qr.
Pints
Quarts
Pecks
Bushels Quarter
2
= 1
16
= 8
= 1
64
= 32
= 4
= 1
512
= 256
= 32
= 8 = 1
MEASURES OF LENGTH
LINEAR, OR LONG, MEASURE
18. Linear, or long, measure is used for measuring
distances of any magnitude and in any direction.
12 inches (in.) or (") . . .
Table
1 foot . . .
... ft. or (')
3 feet .
=
1 yard . . .
. . . yd.
51 yards, or 161 feet ....
=
1 rod . . .
. . . rd.
40 rods .
=
1 furlong
. . . fur.
8 furlongs, or 320 rods . .
=
1 mile . . .
. . . m.
Inches Feet
Yards
Rods Furlongs Mile
12 = 1
36 = 3 =
198 = 161 =
1
51
1
7,920 = 660 =
220
=
40 =
1
63,360 = 5,280 =
1,760
=
320 =
8 = 1
SURVEYORS’
MEASURE
19. Surveyors’ measure
is
used by
surveyors in
measuring land, roads, etc.
Table
7.92 inches.=
1 link . . .
.I.
25 links .
=
1 pole . . .
.P-
100 links, 4 poles, or 66 feet .
=
1 chain . . .
.cha.
10 chains.
=
1 furlong . .
.fur.
8 furlongs, or 80 chains . .
=
1 mile . . .
.m.
6
ARITHMETIC
5
Gunter’s cliain is 66 feet in length and divided into 100
links and is used in ordinary land surveys, but for locating
roads and laying out public works an engineer’s chain 100'
feet in length and containing 100 links is used.
MEASURES OF AREA
SQUARE MEASURE
20. Square, or surface, measure is used in measuring
surfaces and areas of all kinds.
Note.—A square is a figure having four equal sides and all of its
angles equal: a square inch , therefore, is the amount of surface
enclosed by a square the sides of which are 1 inch long; similarly a
square foot or square mile is the amount of surface or the area enclosed
in squares the sides of which are each 1 foot or 1 mile long.
Table
144 square inches (sq. in.) . . . = 1 square foot.sq. ft.
9 square feet.= 1 square yard.sq. yd.
301 square yards, or 1 , , ,
, ^ > .= 1 square rod.sq. rd.
2721 square feet j
160 square rods.= 1 acre.A.
640 acres.= 1 square mile.sq. m.
Square Square Square Square , Square
Inches Feet Yards Rods Mile
144 = 1
1,296 = 9 = 1
39,204 = 2721 = 301 = 1
6,272,640 = 43,560 = 4,840 = 160 = 1
4,014,489,600 = 27,878,400 = 3,097,600 = 102,400 = 640 = 1
MEASURES OF SOLIDITY OR VOLUME
CUBIC MEASURE
21. Cubic, or solid, measure is used for measuring
such materials as have length, breadth, and thickness, as
timber, stone, and other similar materials.
Note.—A cube is a body bounded by six square and equal sides;
therefore, a cubic inch is the volume contained in a cube having six
faces each 1 inch square, and similarly a cubic foot or a cubic yard is
the volume of cubes with sides 1 foot or 1 yard square.
§5 ARITHMETIC 7
Table
1,728 cubic inches (cu. in.) . . . = 1 cubic foot. cu. ft.
27 cubic feet.= 1 cubic yard.cu. yd.
Cubic Inches Cubic Feet Cubic Yard
1,728 = 1
46,656 = 27 = 1
22. A table that is used for measuring wood for fuel is
as follows:
Table
16 cubic feet.= 1 cord foot .c. ft.
8 cord feet, or1 . ,
>.= 1 cord .c.
128 cubic feet j
MEASURE OF TIME
23. Time is measured as follows:
Table
60 seconds (sec.)
60 minutes . .
1 minute.min.
1 hour.hr.
24 hours . . .
7 days ....
365i days, or
52 weeks H days
1 day.d.
1 week.wk.
1 year.yr.
Seconds
Minutes
Hours
Days
Weeks
60
= 1
3,600
60
= 1
86,400
= 1,440
= 24
= 1
604,800
= 10,080
= 168
= 7
= 1
31,557,600
= 525,960
= 8,766
= 365-1
= 52^
Note.— For convenience it is customary to reckon 365 days as a year and call
every fourth year 366 days, placing the extra day in the month of February, which
then has 29 days. This is known as a leap year. A year is equal to 12 months
(mo.) and for convenience a month is considered as 30 days.
ANGULAlt MEASURE
24. Angular, or circular, measure is employed for
measuring the angle between two lines or surfaces and also
for measuring circles, latitude, longitude, etc.
Table
60 seconds (") .= 1 minute. '
60 minutes .= 1 degree.°
90 degrees.= 1 right angle, or quadrant L
360 degrees, or 4 L .= 1 circumference .... cir.
8
ARITHMETIC
§5
Seconds
Minutes
Degrees
Quadrants Circumference
60
= 1
3,600
= 60
= 1
324,000
- 5,400
= 90
= 1
1,296,000
= 21,600
= 360
= 4 1
MISCELLANEOUS MEASURES
1 pound sterling (^).
1 league (lea.).
1 fathom .
1 knot, or nautical mile.
1 meter.
1 decimeter.
1 centimeter.
1 millimeter.
1 dozen (doz.).
1 gross .
1 great gross .
1 quire .
1 ream .
1 large ream.
1 perch.
1 gallon.
1 tierce.
1 puncheon .
1 carat .
1 butt.
1 bushel.
1 palm.
1 hand .
1 span .
1 gallon of water (U. S. Standard)
1 gallon of water (British Imperial
gallon) .
1 cubic foot.
= $4.8665
= 3 miles
= 6 feet
- 1H miles
= 39.37 inches
= 3.937 inches
= .3937 inch
= .03937 inch
= 12 articles
= 12 dozen
= 12 gross
= 24 sheets of paper
= 20 quires
= 500 sheets
= 24f cubic feet
= 231 cubic inches
— 42 gallons
= 2 tierces
= 3| grains (troy)
= 108 gallons
= 2,150.42 cubic inches
= 3 inches
= 4 inches
= 9 inches
= 231 cubic inches = 8.355 pounds
= 277 cubic inches = 10 pounds
= 7.481 gallons
§5
ARITHMETIC
9
REDUCTION OF COMPOUND QUANTITIES
25. Reducing a quantity to a lower denomination, or
reduction descending:
Example 1.—Reduce 1 T., 3 cwt., 45 lb. and 11 oz. to ounces.
Solution.—
1 T. 3 cwt. 45 lb. 11 oz.
20 cwt. in 1 T.
2 0 cwt.
3 cwt.
2 3 cwt.
10 0 lb. in 1 cwt.
2 3 0 0 lb.
_4 5 lb.
2 3 4 5 lb.
1 6 oz. in 1 lb.
1 4 0 7 0
2 3 4 5
3 7 5 2 0 oz.
1 1 oz.
3 7 5 3 1 oz. Ans.
Explanation.— One ton multiplied by the number of
hundredweight in 1 ton (20) = 20 hundredweight. 20 hun¬
dredweight plus 3 hundredweight = 23 hundredweight,
which when multiplied by the number of pounds in one
hundredweight (100) = 2,300 pounds; adding 45 pounds
= 2,345 pounds. This product multiplied by the number
of ounces in 1 pound (16) = 37,520 ounces; adding 11 ounces
= 37,531 ounces, the number of ounces in 1 ton 3 hun¬
dredweight 45 pounds 11 ounces.
Example 2. —Reduce 4 m. 25 rd. 10 ft. to ft.
Solution. — 4 m. 25 rd. 10 ft.
3 2 0 rd. in 1 m.
1 2 8 0 rd.
2 5 rd.
13 0 5
1 6 i ft. in 1 rd.
6 5 2 4
7 8 3 0
13 0 5
2 1 5 3 2 4 ft.
1 0 ft.
2 1 5 4 2 4 ft. Ans.
10
ARITHMETIC
§5
26. To reduce a compound quantity to a lower
denomination:
Rule. —Multiply the given number of units of the highest
denomination by the member of emits of the next lower denomi¬
nation required to make a unit of the highest denomination , and
to the product add the corresponding denomination of the multi¬
plicand if there be any. Proceed in like manner until the required
denomination is reached.
EXAMPLES FOR PRACTICE
Reduce:
(а) 2 lb. (avoirdupois) to ounces.
(б) 7 T. 17 cwt. 48 lb. 8 oz. 14 dr. (avoir¬
dupois) to drams.
(c) 1 lb. 11 oz. to grains. (See Art. 12.)
( d ) 31b. 6 oz. to grains. (See Art. 12.)
[(a)
(b)
Ans. •
32 oz.
4,031,630 dr.
(c) 11,812.5 gr.
{d) 23,625 gr.
27. Reducing- a compound quantity to a higher denomina¬
tion, or reduction ascending:
Example 1. —Reduce 35,678 drams to higher denominations.
Solution.— 1 6 )3 5 6 7 8 dr.
1 6) 2 2 2 9 oz. and 14 dr.
1 0 0 ) 1 3 9 lb. and 5 oz.
1 cwt. and 39 lb.
1 cwt. 39 lb. 5 oz. 14 dr. Ans.
Explanation.— Dividing the total number of drams by
the drams in 1 ounce gives 2,229 ounces and 14 drams left
over. Dividing the ounces by the ounces in 1 pound,
139 pounds is obtained and 5 ounces left over; dividing
the pounds by the pounds in 1 hundredweight, 1 hundred¬
weight and 39 pounds is obtained. Thus, 35,678 drams
equals 1 hundredweight 39 pounds 5 ounces 14 drams.
Example 2.—Reduce 454,621 pt. (dry measure) to highest denomina¬
tions.
2 HA 4 HA
8 ) 2 2 7 3 1 0 qt. and 1 pt.
4 )2 8 4 1 3 pk. and 6 qt.
8)7103 bu. and 1 pk.
8 8 7 qr. and 7 bu.
887 qr. 7 bu. 1 pk. 6 qt. 1 pt. Ans.
Solution.—
§5
ARITHMETIC
11
28. To reduce a quantity to a higher denomina¬
tion:
Rule. —Divide by the number of units required to make one
unit of the next higher denomination.
Divide the quotient and each successive quotient thus acquired
in like manner until the required denomination is reached.
The last quotient and the sevetal remainders arranged in the
order of their successive denominations is the answer required.
EXAMPLES FOR PRACTICE
Reduce the following numbers to highest denominations:
( a ) 2,789 oz. (avoirdupois).
(b) 348 dr. (avoirdupois).
( c ) 16,001 oz. (avoirdupois).
(d) 32,645 dr. (avoirdupois).
Ans.
' (a) 1 cwt. 74 lb. 5 oz.
(b) 1 lb. 5 oz. 12 dr.
(c) 10 cwt. 1 oz.
.(d) 1 cwt. 27 lb. 8 oz. 5 dr.
REDUCTION OE UNITED STATES MONEY
29. Since United States currency varies as a decimal
system, the following rules will enable the student to reduce
it to a lower denomination:
Rule. — I. To reduce dollars to cents, annex two ciphers.
II. To reduce dollars to mills, annex three ciphers.
III. To reduce cents to mills, annex one cipher.
IV. Dollars, cents, and mills expressed as a single number
arc reduced to mills by removing the decimal point.
V. Dollars and cents are reduced to mills by removing the
point and annexing one cipher.
30. To reduce United States currency to a higher
denomination:
Rule. — I. To reduce mills to cents, divide by ten or point off
with a decimal point one place on the right.
II. To reduce cents to dollars, divide by one hmidred or point
off two places.
III. To reduce mills to dollars, point off three places or
divide by one thousand.
12
ARITHMETIC
5
ADDITION OF COMPOUND QUANTITIES
31. Addition of compound numbers is the process of
finding the sum of two or more denominate numbers when
one or more of them is compound.
When compound numbers are to be added, they must be so
placed that numbers of the same denomination shall stand in the
same vertical column.
Example 1.—Find the sum of 4 lb. 13 oz. 14 dr., 2 cwt. 93 lb. 15 oz.
9 dr., and 3 T. 19 cwt. 35 lb. 12 oz. 11 dr. (avoirdupois).
Solution. —Arranging the numbers as explained above,
T.
cwt.
lb.
OZ.
dr.
4
13
14
2
93
15
9
3
19
35
12
11
sum 4
2
34
10
2 Ans.
Explanation.
— The
column
containing
the smallest
denomination is first added and equals 34 drams, or 2 ounces
and 2 drams. Placing the 2 drams under the dram column,
add the 2 ounces in with the ounce column, the sum of
which equals 42 ounces, or 2 pounds and 10 ounces. Placing
the 10 ounces under the ounce column, add the 2 pounds in
with the pound column, which equals 134 pounds, or 1 hun¬
dredweight and 34 pounds. Placing the 34 pounds under
the pound column, add the 1 hundredweight in with the hun¬
dredweight column, which equals 22 hundredweight, or 1 ton
and 2 hundredweight. Placing the 2 hundredweight under
the hundredweight column and adding the 1 ton in with the
ton column, which equals 4 tons, the final sum is 4 tons
2 hundredweight 34 pounds 10 ounces 2 drams.
Example 2.—Add 78 lb. 6 dr., 85 lb. 9 oz., and 39 lb. 15 oz. 11 dr.
(avoirdupois).
Solution. —Arranging like denominations under each other,
cwt. lb. oz. dr.
78 6
85 9
39 15 11
sum 2
3
9
1 Ans.
§5
ARITHMETIC
13
32. To add compound numbers:
Rule.— Write the given numbers so that numbers denoting
the same denomination will sta?id in the same column.
Add as in simple addition and carry from the sum of each
denomination one for as many units as it takes of that denomi¬
nation to make a unit of the next higher denomination.
EXAMPLES FOR PRACTICE
Add the following compound numbers:
(a) 3 lb. 24 oz. 3 dr. and 7 lb. 11 oz. 14 dr. (avoirdupois).
( b ) 3 gal. 2 qt. 1 pt. 3 gi. and 21 gal. 3 qt. 1 pt. 3 gi. (liquid
measure).
( c ) 56 T. 4 cwt. 75 lb. 14 oz. and 48 T. 18 cwt. 37 lb. 8 dr.
(avoirdupois).
( d) 12 oz. 13 pwt. 19 gr. and 11 oz. 18 pwt. 21 gr. (troy).
( a ) 12 lb. 4 oz. 1 dr.
A (6) 25 gal. 2 qt. 1 pt. 2 gi.
(c) 105 T. 3 cwt. 12 lb. 14 oz. 8 dr.
( d) 2 lb. 12 pwt. 16 gr.
SUBTRACTION OF COMPOUND QUANTITIES
33. Subtraction of compound numbers is the process of
finding the remainder, or difference, between two denominate
numbers when one or both of them are compound.
Example 1.—From 2 lb. 7 oz. 5 dr. 2 sc. 12 gr. subtract 1 lb. 9 oz.
7 dr. 2 sc. 16 gr. (apothecaries’ weight).
Solution. — lb. oz. dr. sc. gr.
minuend 2 7 5 2 12
subtrahend 1 9 7 2 16
remainder 0 9 5 2 16 Ans.
Explanation. —It is impossible to subtract 16 grains from
12 grains, so 1 scruple, or 20 grains, is borrowed from the
scruple column making 32 grains, from which 16 grains is
subtracted, leaving 16 grains as a remainder. In the same
manner, as 2 scruples cannot be subtracted from 1 scruple,
1 dram, or 3 scruples, is borrowed from the dram column
making 4 scruples, from which 2 scruples is subtracted,
leaving 2 scruples. Similarly, as 7 drams cannot be sub¬
tracted from 4 drams, 1 ounce, or 8 drams, is borrowed from
the ounce column making 12 drams, from which 7 drams is
14
ARITHMETIC
§5
subtracted, leaving 5 drams as a remainder. The same
method is followed out in subtracting the ounces and the
final remainder is found to be 9 ounces 5 drams 2 scruples
16 grains.
Example 2.—Subtract 3 bu. 3 pk. (5 qt. from 7 bu. 3 pk. 3 qt. (dry-
measure) .
Solution.— bu
minuend 7
subtrahend 3
remainder 3
pk.
3
3
3
qt.
3
J)
5 Ans.
34. To subtract compound numbers:
Rule.— Write the lesser compound number under the greater
so that numbers of the same denomination zvill be in the same
column and proceed as in the subtraction of simple numbers.
If any number of the subtrahend is larger than the correspond¬
ing minuend number , add to the minuend number as many units
as make one of the next higher denomination and take one from
the minuend number of such higher denomination.
EXAMPLES FOR PRACTICE
Subtract:
(«)
ib)
(c)
(d)
15 oz. from 1 lb. 3 oz. (avoirdupois).
7 lb. 5 oz. 3 dr. from 14 lb. 3 oz. 2 dr. (apothecaries’).
1 lb. 10 oz. 16 pwt. from 3 lb. 8 oz. 4 pwt. (troy).
44 lb. 3 oz. 10 dr. from 11 cwt. 34 lb. 2 oz. 6 dr.
Ans.
(a) 4 oz.
(b) 6 lb. 9 oz. 7 dr.
(c) 1 lb. 9 oz. 8 pwt.
(d) 10 cwt. 89 lb. 14 oz. 12 dr.
MULTIPLICATION OF COMPOUND QUANTITIES
35. Multiplication of compound quantities is the process
of adding a compound number to itself a definite number of
times.
Example 1.—Multiply 3 gal. 2 qt. 1 pt. 3 gi. (liquid measure) by 6.
Solution.—
gal.
qt.
pt.
gi-
multiplicand
3
2
1
3
multiplier
6
product
22
1
0
2 Ans.
§5
ARITHMETIC
15
Explanation. —In this example multiplying 3 gills by 6,
the product obtained is 18 gills, which is equal to 4 pints and
2 gills. Write the 2 gills in the gill place of the final
product and add the 4 pints to the product of 1 pint multi¬
plied by 6, thus 1x6 = 6 and 6 -f 4 = 10. The product of
the pints is equal to exactly 5 quarts, so a cipher is written
in the pint place of the final product and the 5 quarts is
added to the product of 2 quarts multiplied by 6, thus 2x6
= 12 and 12 + 5 = 17. The product of the quarts is equal
to 4 gallons and 1 quart. Write the 1 quart in the quart
place of the final product and add the 4 gallons to the product
of 3 gallons multiplied by 6, thus 3x6 = 18 and 18 + 4
= 22. As there is no higher denomination than gallons in
the example, it is complete and the final product is 22 gallons
1 quart 2 gills.
Example 2.—Multiply 35 bu. 3 pk. 6 qt. (dry measure) by 9.
Solution.— bu. pk. qt.
multiplicand 35 3 6
multiplier 9
product 323 1 6 A ns.
36. To multiply compound numbers:
liule. —Multiply each denomination of the multiplicand by
the multiplier , as in the multiplication of simple numbers , and
carry as in the addition of compound numbers.
EXAMPLES FOR PRACTICE
Find the product of the following compound numbers:
( a ) 5 gal. 2 qt. 1 pt. (liquid measure) multiplied by 3.
(b) 13 cwt. 3 lb. 8 oz. (avoirdupois) multiplied by 7.
(c) 3 lb. 7 oz. 16 pwt. (troy) multiplied by 4
id) 25 T. 18 cwt. 66 lb. 12 oz. (avoirdupois) multiplied by 6.
' {a) 16 gal. 3 qt. 1 pt.
(b) 91 cwt. 24 lb. 8 oz.
(c) 14 lb. 7 oz. 4 pwt.
.(d) 155 T. 12 cwt. 8 oz.
Ans.
16
ARITHMETIC
§5
DIVISION OF COMPOUND QUANTITIES
37. Division of compound numbers is the process of
dividing a compound number into a definite number of equal
parts.
Example 1.—Divide 143 sq. yd. 4 sq. ft. 81 sq. in. (square measure)
by 5.
Solution. — sq. yd. sq. ft. sq. in.
divisor 5 ) 143 4 81 dividend
quotient 28 6 45 Ans.
Explanation.—I n this example, 143 square yards divided
by 5 equals 28 square yards and 3 square yards left over.
Write the 28 square yards in the quotient and reduce the
3 square yards to square feet, adding them to the 4 square
feet in the dividend; thus, 3 square yards = 27 square feet,
adding 4 square feet = 31 square feet, which when divided
by 5 equals 6 square feet and 1 square foot as a remainder.
Reduce the 1 square foot to square inches and add them to
the 81 square inches in the dividend; thus, 1 square foot
= 144 square inches, adding 81 square inches = 225 square
inches, which when divided by 5 equals 45 square inches.
The final quotient obtained is 28 square yards 6 square feet
45 square inches.
Example 2. —Divide 323 bu. 1 pk. 6 qt. (dry measure) by 9.
Solution.— bu. pk. qt.
divisor 9 ) 323 1 6 dividend
quotient 35 3 6 Ans.
38. To divide compound numbers:
Rule. —Proceed as in the division of simple numbers , dividing
each denomination in order beginning with the highest. If there
is a remainder after dividing any denomination , reduce it to the
next lower denomination, adding in the number of this denom¬
ination in the dividend if any , and proceed as before.
§5
ARITHMETIC
17
EXAMPLES FOR PRACTICE
Divide:
(a) 3 T. 13 cwt. 55 lb. 10 oz. (avoirdupois) by 6.
(b) 72 sq. yd. 4 sq. ft. 31 sq. in. (square measure) by 7.
(c) 36 cu. yd. 1 cu. ft. 960 cu. in. (cubic measure) by 8.
( d ) 21 T. 5 cwt. 52 lb. (a oirdupois) by 8.
(a) 12 cwt. 25 lb. 15 oz.
Ans.
(b) 10 sq. yd. 3 sq. ft. 25 sq. in.
\c) 4 cu. yd. 13 cu. ft. 1,200 cu.
(d) 2 T. 13 cwt. 19 lb.
in.
MENSURATION
MENSURATION OF SURFACES
DEFINITIONS
1. Mensuration treats of the measurement of lines ,
angles, surfaces, and solids.
2. A line expresses length or distance without breadth
or thickness.
3. A straight line, Fig. 1, is one that
does not change its direction throughout its fx G . i
whole length.
4. A curved line, Fig. 2, changes its
direction at every point.
Fig. 2
5. Parallel lines, Fig. 3, are those
that are equally distant from each other at
all points.
6. A line is perpendicular to another,
Fig. 4, when it meets that line so as not to
incline toward it on either side. _
Fig. 4
7. A vertical line, Fig. 5, is one that
points toward the center of the earth; it is
also known as a plumb-line.
8. A horizontal line, Fig. 5, is one
that makes a right angle with a vertical line.
8
f
g
Horizontal
Fig. 5
9. A surface is that which has length and breadth
without thickness.
For notice of copyright, see page immediately following the title page
$6
2
MENSURATION
§6
10. A plane surface is one in which if two points be
taken, a straight line connecting them will be wholly in the
surface.
11. A curved surface is one no part of which is
plane.
12. An angle is the inclination of two lines one to the
other and is measured in degrees (°).
13. A riglit angle, Figs. 4 and 5, is formed by two
lines that are perpendicular to each other, and is 90° in
magnitude.
14. An acute angle, Fig. 6, is an angle of less than 90°.
Fig. 6 Fig. 7
15. An obtuse angle, Fig. 7, is an angle of more
than 90°.
16. A plane figure is any part of a plane surface
bounded by straight or curved lines.
17. The ai’ea of a plane figure is its surface contents.
TRIANGLES
18. A triangle is a plane figure bounded by thrde
straight lines and having three angles.
19. The altitude of a triangle is the
distance from its apex to base measured
perpendicularly to the base. In the tri¬
angle a be, Fig. 8, the dotted line b d repre¬
sents the altitude of the triangle, while
the base of the triangle is represented by
the line ac.
20. An equilateral triangle, Fig. 9, is one that has
all of its sides equal and each of its angles of 60° magnitude.
§6
MENSURATION
3
21. An isosceles triangle,
equal sides and two equal
angles.
22. A scalene triangle,
Fig. 11, is one that has all of
its sides and all of its angles
unequal.
Fig. 10, is one that has two
23. A right tri¬
angle, Fig. 12, is one that
has one angle a right
angle.
24. To find the area
of a triangle:
Fig. 11
Rule .—Multiply the base by the altitude and divide the
product by 2.
Example.— The base of a triangle is 14 inches in length and the
altitude is 12 inches; what is the area?
14 in. X 12 in. .
Solution.— - 2 - = °4 sq. in. Ans.
Note.— In the above example it will be noticed that by multiplying: inches by
inches the product obtained is square inches; similarly, feet multiplied by feet or
rods by rods equals square feet or square rods, etc. It must be remembered that
only like numbers can be multiplied together and that feet can never be multiplied
by inches, nor rods by feet; consequently, in all problems dealing with mensuration,
all dimensions must be reduced to like terms before multiplying.
EXAMPLES FOR PRACTICE
Find the areas of the following triangles, the length of the base and
altitude being respectively:
(a) 10 inches and 8 inches. (a)
(b) 34 feet and 42 feet.
( c ) 114 inches and 212 inches.
(d) 34 miles and 18 miles.
Ans.
W
(c)
(d)
40 sq. in.
714 sq. ft.
12,084 sq. in.
306 sq. mi.
25. To find tlie area of a triangle when the alti¬
tude is unknown but the length of each side is given:
Rule .—From one-half the sum of the three sides , subtract
each of the sides separately and multiply the remainders together
and by one-half the sum of the sides; the square root of the
Product will be the area of the triangle.
4
MENSURATION
6
Example. —What is the area of a triangle the sides of which are,
respectively, 16, 16, and 12 feet in length?
Solution.— 16 + 16 + 12 = 44; 44 -5- 2 - 22; 22 - 16 = 6;
22 — 16 = 6; 22 - 12 = 10;
6 X 6 X 10 X 22 = 7,920; VL920 = 88.99 sq. ft. Ans.
EXAMPJ.ES for practice
1. Find the area of a triangle the sides of which are, respectively,
32, 32, and 24 inches in length. Ans. 355.977 sq. in.
2. Find the area of a triangle the sides of which are, respectively,
18, 24, and 22 feet in length. Ans. 189.314 sq. ft.
QUADRILATERALS
26. A quadrilateral is a plane figure bounded by four
straight lines.
27. A parallelogram is
sides of which are parallel.
a quadrilateral the opposite
28. A rectangle,
Fig. 18, is a parallelogram
having all of its angles
right angles.
29. A square. Fig.
14, is a parallelogram hav¬
ing all of its angles right angles and all of its sides of equal
length.
30. A rhomboid, Fig. 15, is a parallelogram having
none of its angles right angles.
Fig. 15
31. A rhombus, Fig. 16, is a parallelogram having all
of its sides of equal length but none of its angles right
angles.
§6
MENSURATION
5
32. The altitude of a parallelogram is the distance
between two opposite sides measured perpendicularly, as
indicated by the dotted lines in Figs. 15 and 16.
33. To fiud the area of a parallelogram:
Rule .—Multiply the altitude by the base and the product will
be the area.
Example. —Find the area of a parallelogram the base of which is
345 inches and the altitude 423 inches.
Solution.— 423 in. X 345 in. = 145,935 sq. in. Ans.
EXAMPLES FOR PRACTICE
Find the areas of the following parallelograms, the lengths of the
bases and altitudes being respectively:
(a) 145 inches and 136 inches.
(b) 2,034 feet and 23 feet.
(c) 135 rods and 4| rods.
(d) 39 feet and 14^ feet.
Ans.<
' (a) 19,720 sq. in.
(b) 46,782 sq. ft.
( c ) 567 sq. rd.
. ( d ) 559 sq. ft.
34. A trapezoid, Fig. 17, is a
quadrilateral having only two of
its sides parallel. fig. 17
35. The altitude of a trapezoid is always measured
perpendicularly between the parallel sides, as shown by the
dotted line in Fig. 17.
36. To find the area of a trapezoid:
Rule.— Multiply one-half the sum of the parallel sides by the
altitude.
Example. —The parallel sides of a trapezoid are, respectively, 12
and 28 feet in length, and the altitude is 30 feet; what is the area of
the figure?
Solution. — 12 ft. + 28 ft. = 40 ft.; 40 ft. -p 2 = 20 ft.
20 ft. X 30 ft. = 600 sq. ft. Ans.
EXAMPLES FOR PRACTICE
1. What is the area of a trapezoid the parallel sides of which are,
respectively, 54 and 78 feet in length, and the altitude of which is
64 feet? Ans. 4,224 sq. ft.
6
MENSURATION
§6
2. What is the area of a trapezoid the parallel sides of which are,
respectively, 8 and 18 inches in length, and the altitude of which is
23 inches? Ans. 299 sq. in.
37. A trapezium, Fig. 18, is a quadrilateral that has
no two sides parallel.
38. A line joining two oppo¬
site corners of a quadrilateral, as
for instance the line a b, Fig. 18,
is known as a diagonal.
39. To find the area of a
trapezium:
b Rule. —Divide the figure into
tzvo triangles by ?neans of a diagonal;
the sum of the areas of these triangles equals the area of the
trapezium.
Example. —What is the area of a trapezium whose diagonal is
43 inches long, the length of the perpendicular lines dropped on the
diagonal from the opposite corners being 22 and 26 inches, respectively?
Note.— The perpendicular lines drawn from opposite corners of a quadrilateral
to its diagonal constitute the altitudes of the two triangles into which the diagonal
divides the quadrilateral. Thus, in Fig. 18, the line f d represents the altitude of the
triangle a db, and the line ec the altitude of the triangle ac b.
Solution. — 43 in. X 22 in. = 946 sq. in.; 946 sq. in. -p 2 = 473 sq.
in., area of one triangle; 43 in. X 26 in. = 1,118 sq. in.; 1,118 sq. in.
ri- 2 = 659 sq. in., area of other triangle.
473 sq. in. + 559 sq. in. = 1,032 sq. in., area of trapezium. Ans.
a
EXAMPLES FOR PRACTICE
1. The diagonal of a trapezium is 26 feet in length and the per¬
pendiculars from the opposite corners to the diagonal are 8 and 14 feet,
respectively, in length; what is the area? Ans. 286 sq. ft.
2. The diagonal of a trapezium is 78 inches and the perpendicu¬
lars 42 and 48 inches, respectively, in length; what is the area of the
figure? Ans. 3,510 sq. in.
§6
MENSURATION
7
POLYGONS
40. A polygon is a plane figure bounded by straight
lines. The term is usually applied to a figure having more
than four sides. The bounding lines are called the sides,
and the sum of the lengths of all the sides is called the per¬
imeter of the polygon.
41. A regular polygon is one in which all the sides
and all the angles are equal.
42. A polygon of five sides is called a pentagon; one
of six sides, a hexagon; one of seven sides, a heptagon,
etc. Regular polygons having from five to twelve sides
are shown in Fig. 19.
Pentagon Hexagon Heptagon Octagon Decagon Dodecagon
Fig. 19
43. To find the area of a regular polygon:
Rule .—Multiply the perimeter by one-half the length of the
perpendicular from its center to one of its sides.
Example. —The perimeter of a regular polygon is 28 inches in length
and the perpendicular distance from its center to one side is 8 inches;
what is its area?
Solution. — 8 in. -f- 2 = 4 in.; 28 in. X 4 in. = 112 sq. in. Ans.
EXAMPLES FOR PRACTICE
1. If the perimeter of a regular polygon is 78 feet in length and the
distance from its center to one side measured perpendicularly is 21 feet,
what is its area? Ans. 819 sq. ft.
2. The perimeter of a regular polygon is 112 inches in length and
the perpendicular distance from the center to one side is 32 inches;
what is the area? Ans. 1,792 sq. in.
8
MENSURATION
§6
THE CIRCLE
44. A circle, Fig. 20, is a plane figure bounded by a
curved line, called the circumference, every portion of
which is equally distant from a point within called the center.
45. The diameter of a circle is any straight line drawn
through its center and terminating at each end in the cir¬
cumference. Thus the line a b, Fig. 21, is a diameter of
the circle.
46. If a circle is divided into halves, each half is called
a semi-circle, and each half of the circumference is called a
semi-circumference.
47. Any straight line terminating at each end in the
circumference but not passing through the center is called
a chord, as for instance
the line a e. Fig. 22.
Fig. 23 Fig. 24
49. An arc of a circle (see
its circumference.
48. A straight line
drawn from the center to
the circumference of a
circle (as a c, Fig. 23) is
called a radius.
a d e, Fig. 24) is any part of
50. To find the circumference of a circle:
Rule.— Multiply the diameter by 3.1416.
Note. — 3.1416 is the approximate length of the circumference of
a circle whose diameter is 1.
Example.—W hat is the circumference of a circle the diameter of
which is 48 inches?
Solution. — 48 in. X 3.1416 = 150.7968 in. Ans.
6
MENSURATION
9
51. To find the diameter of a circle with a given
length of circumference:
Rule.— Divide the circumference by 3.1416.
Example. —What is the diameter of a circle the length of circum¬
ference of which is 8 feet?
Solution. — 8 ft. -r- 3.1416 = 2.5465 ft. Ans.
EXAMPLES FOR PRACTICE
Find the circumferences of circles having the following diameters:
(а) 24 inches.
(б) 35 feet.
( c ) 27 rods.
(a) 79 yards.
Ans.
f (a) 75.3984 in.
(b) 109.956 ft.
(0 84.8232 rd.
l(rf) 248.1864 yd.
52. To find the area of a circle:
Rule.— Multiply the square of the diameter by .7854.
Note. — .7854 is the area of a circle whose diameter is 1.
Example. —What is the area of a circle the diameter of which is
75 inches?
Solution. — 75 in. X 75 in. X .7854 = 4,417.875 sq. in. Ans.
EXAMPLES FOR PRACTICE
Find the areas of circles of the following diameters:
(a) 22 inches.
(b) 47 yards.
( c ) 768 rods.
(d) 176 inches.
Ans.
(a) 380.1336 sq. in.
{d) 1,734.9486 sq. yd.
' (c) 463,247.7696 sq. rd.
.(d) 24,328.5504 sq. in.
53. To find the length of one side of a square
inscribed in a given circle:
Rule.— Multiply the diameter of the circle by .707107'.
Note. —A square is said to be inscribed in a circle when the vertices
of all its angles lie in the circumference of the circle. .707107 is the
length of the side of a square inscribed in a circle whose diameter is 1.
Example. —How thick is the largest square stick that can be inserted
in a pipe 5 inches in diameter on the inside?
Solution. — 5 in. X .707107 = 3.5355 in. Ans.
10
MENSURATION
§6
EXAMPLES FOR PRACTICE
1. What is the thickness of the largest square iron rod that will just
fit in a round hole li inches in diameter? Ans. 1.06 in.
2. What is the thickness of a square stick of timber that may be
cut from a log 28 inches in diameter? Ans. 19.7989 in.
54. To find the length of one side of a square
equal in area to a given circle:
Rule.— Multiply the diameter of the circle by .886227.
Note. — .886227 is the length of the side of a square equal in area
to a circle whose diameter is 1.
Example. —What is the length of one side of a square that is equal
in area to a circle 15 inches in diameter?
Solution. — 15 in. X .886227 = 13.293 in. Ans.
EXAMPLES FOR PRACTICE
1. What is the length of one side of a square that will have an area
equal to that of a piston 20 inches in diameter? Ans. 17.7245 in.
2. What is the length of one side of a square field that will contain
the same number of acres as a circular field 700 feet in diameter?
Ans. 620.3589 ft.
MENSURATION OF SOLIDS
THE PRISM
55. A solid, or solid body, is one that has three
dimensions; viz., length, breadth, and thickness.
56. A prism is a solid body the ends of which are
formed by two similar plane figures that are
equal and parallel to each other, and whose
sides are parallelograms. Prisms are triangular,
rectangular, square, etc. according to the char¬
acter of the figure forming the ends.
57. A parallelopipedon, Fig. 25, is a
prism whose bases (ends) are parallelograms.
A
/
!
i
j
i
i
i
i
■
7
Fig. 25
6
MENSURATION
11
58. A cube, Fig. 26, is a prism whose faces and ends
are squares. All the faces of a cube are equal.
59. In the case of plane figures, perimeters
and areas must be considered. In the case of
solids, the areas of their outside surfaces and
their contents or volumes must be considered.
Fig. 2f>
60. The base of a prism is either end, and of solids in
general, the ends on which they are supposed to rest.
61. To find the surface area of a prism:
Rule. —Multiply the length of the perimeter of the base by the
altitude, and to the product add the area of both ends.
Example. —What is the surface area of a square prism the base of
which is 14 inches square and the altitude 25 inches in length?
Solution. — 14 in. X 4 = 56 in., perimeter of base
56 in. X 25 in. = 1,400 sq. in., area of sides
14 in. X 14 in. = 196 sq. in., area of one base
196 sq. in. X 2 = 392 sq. in., area of both bases
1,400 sq. in. + 392 sq. in. = 1,792 sq. in., total surface area. Ans.
EXAMPLES FOR PRACTICE
1. Find the surface area of a rectangular prism whose base is
10 inches long and 12 inches wide and which is 13 inches high.
Ans. 812 sq. in.
2. What is the surface area of a square prism the base of which
is 175 inches square and whose altitude is 342 inches?
Ans. 300,650 sq. in.
62. To find tlie contents or volume of a prism
or rectangular box:
Rule. —Multiply the width by the depth and by the length;
or find the area of the base according to the rule previously
given , which when multiplied by the height equals the contents
or solidity of the prism.
Example.— What is the capacity of a box 36 inches long, the
ends being 14 inches by 28 inches?
Solution. — 28 in. X Min. X 36 in. = 14,112 cu. in. Ans.
12
MENSURATION
§6
Note.— It lias been stated that inches multiplied by inches equals square inches
or, similarly, yards multiplied by yards equals square yards. Continuing: still further,
as is necessary in finding: the contents, volume, solidity, or capacity of solids: square
inches or square yards multiplied by inches or yards equals cubic inches or cubic
yards, etc.
From this it will be seen that by multiplying: together the two dimensions of a
surface, such as a rectangle, the area of the figure will be expressed in square units,
while if the three dimensions of a solid, as for instance, a parallelopipedon. are multi¬
plied together the contents, or solidity, of the solid is expressed in cubical units.
EXAMPLES FOR PRACTICE
1. What is the capacity of a box the ends of which are 24 inches
square and the length of which is 44 inches? Ans. 25,344 cu. in.
2. What is the capacity, in cubic feet, of a freight car 34 feet long,
8 feet wide, and 7 feet high? Ans. 1,904 cu. ft.
3. How many cubic inches are there in a stick of timber 8^ feet
long, 7 inches wide, and 4^ inches thick? Ans. 3,213 cu. in.
Suggestion.— Reduce the 8i feet to inches.
4. What is the contents of a tank 13 feet square and 12 feet high?
Ans. 2,028 cu. ft.
5. How many cubic feet are there in a cube whose sides are 12 feet
in length? Ans. 1,728 cu. ft.
THE CYLINDER
A cylinder, Fig. 27, is a body of uniform diameter
the ends, or bases, of which are equal parallel
circles.
64. To find the surface area of a cylinder:
Rule.— Multiply the circumference of the base by
the height of the cylinder and to this product add the
area of the ends.
Example. —What is the surface area of a cylinder 6 inches in
diameter and 13 inches high?
Solution.—
6 2 X .7854 = 28.2744 sq. in., area of one end
28.2744 sq. in. X 2 = 56.5488 sq. in., area of both ends
6 in. X 3.1416 = 18.8496 in., length of circumference
18.8496 X 13 = 245.0448 sq. in., area of convex surface
245.0448 + 56.5488 = 301.5936 sq. in., total surface area. Ans.
Note.—T he convex surface of a solid is the curved surface; thus, the area of the
convex surface of a cylinder is its total surface area less the area of the ends.
63.
Fig. 27
6
MENSURATION
13
EXAMPLES FOR PRACTICE
1. What is the surface area of a cylinder 7 feet long and 21 inches
in diameter? Ans. 6,234.5052 sq. in.
2. What is the surface area of a cylinder 21 feet long and 27 inches
in diameter? Ans. 22,520.5596 sq. in.
3. What is the surface area of a 21-foot boiler, 6 feet in diameter?
Neglect the curvature of the heads. Ans. 452.3904 sq. ft.
65. To find the contents or volume of a cylinder:
Rule .—First find the area of the base according to the rule in
Art. 52, and then multiply the area of the base by the altitude.
Example. —How many cubic feet of water will a cylindrical tank
12 feet in diameter and 14 feet high hold?
Solution. — 12 2 X .7854 = 113.0976 sq. ft., area of base; 113.0976
sq. ft. X 14 ft. = 1,583.3664 cu. ft. Ans.
EXAMPLES FOR PRACTICE
1. What is the contents of a cylinder 35 inches in diameter and
4| feet high? Ans. 51,954.21 cu. in.
2. What is the capacity of a cylindrical tank 8 feet in diameter and
9 feet deep? Ans. 452.3904 cu. ft.
3. If the water in a circular tank 44 inches in diameter is 27 inches
deep, how many cubic feet of water are there in the tank?
Ans. 23.7583 cu. ft.
4. How many cubic inches in a cylinder 19 inches long and 11 inches
in diameter? Ans. 1,805.6346 cu. in.
THE PYRAMID AND CONE
66. A pyramid, Fig. 28, is a solid the base of which
is a polygon and the sides of
which taper uniformly to a
point called the apex, or
vertex.
67. A cone, Fig. 29, is
a solid having a circle as a
base and a convex surface
tapering uniformly to a point called the apex, or vertex.
14 MENSURATION §6
68. The altitude of a pyramid or cone is the perpen¬
dicular distance from the vertex to the base.
69. To find tlie contents ox* volume of a cone or
pyramid:
Rule .—Multiply the area of the base by one-third the altitude.
Example. —What is the solid contents of a cone 30 feet high and
5 feet in diameter at the base?
Solution. — 5 s X .7854 = 19.635 sq. ft. area of base;
i of 30 ft. = 10 ft.
19.635 sq. ft. X 10 ft. = 196.35 cu. ft. Ans.
EXAMPLES FOR PRACTICE
1. Find the contents of a cone 39 inches high and 12 inches in
diameter at the base. Ans. 1,470.2688 cu. in.
2. Find the contents of a square pyramid 300 feet high and 325 feet
square at the base. Ans. 10,562,500 cu. ft.
70.
THE FRUSTUM OF A PYRAMID OR CONE
If a pyramid be cut by a plane parallel to the base,
so as to form two parts, as in
Fig. 30, the lower part is called
the frustum of the pyramid.
If a cone be cut in a similar man¬
ner, as in Fig. 31, the lower part
is called the frustum of the cone.
71. To find tlie contents or
volume of the frustum of a
pyramid oi* cone:
Rule. —Find the areas of the two ends of the frustum; multiply
them together and extract the square root of the product. To
the result thus obtained add the two areas and multiply the sum
by one-third of the altitude.
Example. —What is the capacity of a tank shaped like the frustum
of a cone, the inside diameter of the top being 10 feet and of the
bottom 14 feet, and the depth of the tank being 12 feet?
§6
MENSURATION
15
Solution. — 10 ft. X 10 ft. X .7854 = 78.54 sq. ft., area of small
end; 14 ft. X 14 ft. X .7854 = 153.9384 sq. ft., area of large end;
153.9384 X 78.54 = 12,090.321936; Vl2, 090.321936 = 109.956 sq. ft.;
109.956 + 153.9384 + 78.54 = 342.4344 sq. ft.; 12 ft. -p 3 = 4 ft.
342.4344 sq. ft. X 4 ft. = 1,369.7376 cu. ft. Ans.
EXAMPLES FOR PRACTICE
1. What is the volume of the frustum of a square pyramid the
length of which is 30 feet, the top being 10 feet square and the bottom
20 feet square? Ans. 7,000 cu. ft.
2. What is the contents of a round stick of timber 20 feet long,
1 foot in diameter at the larger end, and \ foot at the small end?
Ans. 9.162 cu. ft.
THE SPHERE
72. A sphere, Fig. 32, is a solid bounded by a continuous
convex surface, every part of which is
equally distant from a point within called
the center.
73. The diameter, or axis, of a sphere
is a line passing through its center and ter¬
minating at each end at the surface.
74. To find the surface area of a sphere:
Rule.— Square the diameter and multiply the result by 3.1416.
Example. —What is the surface area of a sphere 14 inches in
diameter?
Solution.—
14 3 X 3.1416 = 14 X 14 X 3.1416 = 615.75 sq. in. Ans.
EXAMPLES FOR PRACTICE
1. What is the surface area of a sphere 10 inches in diameter?
Ans. 314.16 sq. in.
2. The diameter of the earth is about 8,000 miles; what is its
approximate surface area? Ans. 201,062,400 sq. mi.
75. To find the contents or volume of a sphere:
Rule.— Multiply the ciibe of the diameter by .5236.
Fig. 32
16 MENSURATION §6
Example. —How many cubic inches of ivory in a billiard ball
2 inches in diameter?
Solution.— 2 3 X .5236 = 4.1888 cu. in. Ans.
EXAMPLES FOR PRACTICE
1. How many cubic inches are there in a sphere 20 inches in
diameter? Ans. 4,188.8 cu. in.
2. What is the contents of a spherical float 9 inches in diameter?
Ans. 381.7044 cu. in.
MENSURATION OF LUMBER
76. Lumber is measured by board measure, which is
an adaptation of square measure.
77. A board foot is considered as 1 square foot of board
1 inch thick: therefore 1,000 feet of lumber is equal to
1,000 square feet of boards 1 inch thick.
78. To find the number of feet of lumber in 1-incli
boards:
Rule.— Multiply the length of the board , in feet, by the width ,
in inches , and divide the product by 12.
Example.— How many feet of lumber are there in a 1-inch board
18 feet long and 8 inches wide?
Solution. — — — = 12 ft. Ans.
EXAMPLES FOR PRACTICE
1. How many board feet in a 1-inch board 21 feet long and
18 inches wide? Ans. 31J ft.
2. How many board feet in a 1-inch board 12 feet long and
24 inches wide? Ans. 24 ft.
79. To find the number of feet of lumber in
joists, beams, etc.:
Rule. —Multiply the tvidth, in inches , by the thickness , in
inches , and by the length, in feet. Divide this product by 12
and the quotient is the number of feet of lumber in the stick.
§6
MENSURATION
17
Example. —How many feet of lumber in a joist 4 inches wide,
3 inches thick, and 12 feet long?
„ 4X3X12 10£l .
Solution.— -- = 12 ft. Ans.
EXAMPLES FOR PRACTICE
1. How many feet of lumber in a beam 8 inches wide, 3 inches
thick, and 16 feet long? Ans. 32 ft.
2. How many feet of lumber in a timber 12 inches wide, 16 inches
deep, and 24 feet in length? Ans. 384 ft.
MACHINE ELEMENTS
SHAFTING
CLASSES OF SHAFTING
1. A shaft may be defined as a rod, generally of iron or
steel, that may be rotated for the purpose of transmitting
power in a certain direction and quantity, depending largely
on the connections of the driven pieces with the shaft and its
size. The word' shafting, as used in the mill, generally
includes only that which is employed in the main transmis¬
sion of power to various machines, although the term actually
includes parts of a machine that are of similar construction.
The size of a shaft is determined by its diameter, which is
expressed in inches and fractional parts of an inch. It is
always better to use a shaft of sufficient diameter to transmit
the required horsepower easily than to place the strain on a
very small shaft and run the risk of its breaking.
Formerly wrought-iron shafts were largely used, but these
are being replaced by turned or cold-rolled steel shafting.
Large shafts, such as are employed for flywheels and shafts
where a considerable amount of power is transmitted, are
generally forged from the best quality of steel.
Shafting of almost any diameter and of nearly any desired
length may be purchased to suit circumstances; the freight,
however, is higher on pieces over a certain length.
For notice of copyright , see page immediately following the title page
27
2
MECHANICAL DEFINITIONS
§7
2. The shafting used in a mill may be divided into three
general classes as follows: (1) The main, or head, shaft,
which is driven directly from the engine or other source of
power; this shaft is sometimes called the first, or prime,
mover. (2) The second movers, or (as they are gener¬
ally known) line shafts; these are the main driving shafts
of each room and derive their power from the prime mover.
(3) Countershafts for simply transmitting power to differ¬
ent parts of the room or for making changes in the speed
for driving some particular machine or machines; these are
located with reference to the positions of different machines
in order to supply them with power as economically as pos¬
sible. Long countershafts are classed as second movers.
SHAFT COUPLINGS
v
3. Where a long line of shafting is required to economic¬
ally distribute the power it becomes necessary to couple the
several sections of the shaft in order to secure the transmis¬
sion of the power. Three types of couplings are in general
use: (1) Fast, or permanent, coupIi?igs; (2) loose couplings ,
or clutches, by means of which shafts may be connected or
disconnected as required; (3) friction clutches, which are loose
couplings that hold by friction. The couplings are connected
to the shafts by keys—tapered pieces of metal that fit into
slots in both the shaft and the coupling. These slots are
called keyways, or key seats.
Fig. 1
4. Box, or muff, couplings, shown in Fig. 1, consist
of a cast-iron cylinder a that fits over the ends of the shafts;
the ends are prevented from moving relative to each other
by the sunken key b. The keyway is cut half into the box
and half into the shaft ends; quite commonly the ends of
§7
MECHANICAL DEFINITIONS
3
the shafts are enlarged to allow the keyway to be cut with¬
out weakening the shafts.
5. A clamp coupling is shown in Fig. 2. To make
this coupling the face of each half for the joint is first
planed off, the holes for the bolts drilled, and the two
halves bolted together with pieces of paper between them,
after which the coupling is bored out to the exact size of
Fig. 2
the shaft. The pieces of paper, on being removed, leave
a slight space between the halves, and the coupling when
bolted to the shaft grips it firmly. This form of coupling is
very easily removed or put on; it has no projecting parts,
and may be used as a driving pulley, if desired. The key in
this coupling is straight; i. e., not tapered.
Fig. 3
6. The flange coupling, shown in Fig. 3, consists of
cast-iron flanges a , a keyed to the ends of the shafts. To
insure a perfect joint, the flange after being keyed to the
shaft, is usually faced in a lathe. The two flanges are then
4
MECHANICAL DEFINITIONS
7
brought face to face and bolted together. Sometimes the
ends of the shafts are enlarged to allow for the keyway.
7. Sometimes a clutch, coupling, shown in Fig. 4, is
used. By this means the two sections of the shaft may be
readily disengaged; so that if any part of the room is shut
Fig. 4
down, the shafting may be stopped without affecting the rest
of the plant, thus saving wear on the equipment and a great
amount of power. This coupling consists of a piece a fast¬
ened securely to the end of one shaft, and a movable piece b
Fig. 5
sliding on a key let into the end of the other shaft. The
two halves of the coupling may be connected or discon¬
nected, at pleasure, by means of a lever having a yoke at
one end that engages with the groove b t .
§7
MECHANICAL DEFINITIONS
5
8. Friction couplings, or clutches, are used as loose,
or disengaging, couplings on shafts running at high speeds;
they are often used to couple wheels or pulleys to shafts.
The form shown in Fig. 5 is simple in construction but it is
difficult to put 'in action; besides, it is necessary to exert
a pressure in a horizontal direction in order to keep the
clutch in action and this causes an end thrust on the shaft.
A better form is shown in Fig. 6. The shaft n carries a
flange or cylinder a , while the shaft m has a ring b keyed to
it; this ring is split and fits inside the flange or cylinder a.
The split ends are connected by a screw with right-hand
and left-hand threads; the screw is turned by the lever c,
Fig. 6
which is connected by the link d to the sleeve e. When the
sleeve is pushed toward the clutch, the rotation of the screw
throws the ends /, / of the ring b apart, and thereby causes
the ring b to grip the flange a tightly. A clutch of this form
is easy to operate, and produces no end thrust on the shaft.
Friction clutches may be placed in contact while the shaft
is running at high speed, but to do this with the ordinary
form of clutch coupling would break the coupling.
SHAFT HANGERS
9. For convenience and safety, the shafting is generally
supported from the ceiling by hangers. A type that is
often used in textile-mill equipments is shown in Fig. 7.
This hanger consists of an iron frame a provided with
shoulders through which holes are bored so that it can be
6
MECHANICAL DEFINITIONS
§7
securely fastened to the ceiling by means of lagscrews.
This frame supports a box, or bearing, b in which the shaft
rotates, and below which there is a drip pan c that catches
the oil that works out of the box.
Though all hangers are not pro¬
vided with drip pans, they are val¬
uable adjuncts to them in weave
rooms where the oil is liable to
drop on fabrics or warps, or in
other places where cleanliness is
desired.
The box, or bearing, of a hanger
should have a free lateral move¬
ment and also a vertical rocking
action, in order that it may con¬
form, or aline, with the shaft.
There should also be a positive
vertical screw adjustment, in order
that the height of the bearing may be regulated when lining up
the shafts. The length of the box, or bearing surface, within
which the shaft turns, varies from 3| to 5 times the diameter
of the shaft, according to the ideas of different makers. It is
safe to use one from 4 to 5 times the diameter of the shaft.
The inner portion of the box, which comes in contact with
the shaft, is made of Babbitt metal, brass, or some similar
antifriction metal in order that the shaft may run with the
least possible friction and heating.
Shafting may be easily lined up by adjusting the boxes of
the hangers by means of screws with which they are equipped.
Shafting should be lined up at least once a year as consider¬
able damage may be caused by imperfect alinement. Many
mills attend to this every 6 months, thereby saving wear on
the boxes of the hangers and guarding against the liability
of a broken shaft.
Hangers should preferably be attached to the supporting
beams of the floor above and as near to a line of posts as
possible so that varying weights on the floor above will not
throw the shaft out of line to any great extent.
Fig. 7
7
MECHANICAL DEFINITIONS
7
10. Distance Between Hangers. —When hangers are
put up they should be lined perfectly true, both laterally and
vertically, and should not be placed too far apart. The
distance between the bearings should not be great enough
to permit a deflection of the shaft of more than .01 inch per
foot of length. Hence, when the shaft is heavily loaded with
pulleys, the bearings must be closer than when it carries
only a few. Pulleys that transmit a large amount of power
should be placed as near a hanger as possible.
The following table gives the maximum distances between
the bearings of different sizes of continuous shafts that are
used for the transmission of power:
Diameter of Shaft
Inches
Distance Between Bearings
Feet
Wrought-Iron Shaft
Steel Shaft
2
I I
id5
3
1 3
13-75
4
15
15-75
5
1 7
18.25
6
19
20.5
7
2 1 \
22.25
8
23
24
9
25
26
11. Types of Hangers. —Some hangers are designed
to be attached to posts or columns instead of to the ceiling;
in this case they are called post hangers. A hanger of
this type is shown in Fig. 8.
Occasionally supports for shafting are constructed to be
attached to the walls of the mill and are known as wall
brackets. A common type of wall bracket is shown in
Fig. 9.
When the support, or bearing, of a shaft is designed to
rest on the upper side of a horizontal surface, it is called a
8
MECHANICAL DEFINITIONS
7
. Fig. 8 Fig. 9
Fig. 10
Fig. 11
Fig.12
§7
MECHANICAL DEFINITIONS
9
pillow-block. When the same bearing is fastened on the
under side of a horizontal surface, it is termed an inverted
pillow-block. A pillow-block bearing is shown in Fig. 10.
When it is necessary to pass a shaft through a wall, a
special bearing known as a wall box is used, a view of
which is shown in Fig. 11. This bearing is designed to be
built into the brickwork of the wall.
A floor stand is a support for a shaft running near the
floor. This is often used to support the end of a long
shaft that projects from a machine. An ordinary type of
floor stand is shown in Fig. 12.
POWER TRANSMISSION
FRICTIONAL CONTACT
12. Power transmission in mill work is accomplished
by different methods according to the needs of the individual
mill; but where a great amount of power is to be transmitted,
it generally necessitates the employment of shafting, pul¬
leys, belts, ropes, gearing, etc.
Frictional contact is sometimes used for transmitting
power where it is desired to apply the power quickly or to
disconnect it with as little loss of time as possible. The
simplest method of doing this is
by means of the frictional con¬
tact between the circumferences
of circular disks or pulleys, as
shown in Fig. 13. Two pul¬
leys d, / are shown attached to
shafts with their circumferences
in contact at a. If d imparts
motion to /, then d is the driver and / the driven, and vice
versa; in any case the motion that is imparted is in the
opposite direction to that of the driver.
In order that an excessive amount of slippage may not take
place, the face of the driving pulley is usually made of wood
or is covered with leather or paper, while the driven pulley
10
MECHANICAL DEFINITIONS
§7
is made of cast iron; if any degree of efficiency is to be
obtained it is imperative that the driving and driven pulleys
be pressed together with considerable force. The amount
of power that may be transmitted by a friction drive depends
on the magnitude of this force together with the coefficient
of friction between the pulleys. Frictional contact is only
feasible when small amounts of power are to be transmitted.
It is not often used for power transmission in textile mills.
BELTING
13 . Belts running over pulleys form a convenient means
for transmitting power, but they are not suited to transmit a
precise velocity ratio, owing to their tendency to stretch and
to slip. For driving machinery, however, this freedom to
stretch and to slip is an advantage, since it prevents shocks
that are liable to occur when a machine is suddenly
started, or when there is a sudden fluctuation in the load.
Three kinds of belts are frequently used in textile-mill work:
leather belts, cotton belts, and rubber belts.
14 . Leather. —The material most commonly used for
belts is leather tanned from ox hides. It averages -ft inch
in thickness and is obtained in strips up to 5 feet long.
Belts of any required length are made by joining these strips
together. The wider belts are made from the thicker por¬
tions of the hide. Leather belts should be oak-tanned, as
those tanned with hemlock bark or extracts do not have the
same strength nor do they last as long.
15 . Cotton helts can be made very wide and without
as many joints as leather belts. The necessary thickness
is obtained either by sewing together from four to ten plies
of cotton duck, or weaving a belt as a ply cloth so as to form
a fabric of several plies securely interwoven and with good
selvages. Cotton belting is cheaper than leather.
16 . Rubber belts, which are made by cementing plies
of cotton duck with india rubber, are more adhesive than
leather belts and, therefore, have greater driving capacity.
§7
MECHANICAL DEFINITIONS
11
They are largely used in dye houses, or other places where
the air is damp or full of steam, as under these conditions
leather will quickly stretch and rot.
17. Single and Double Belts. —A single belt is one
that consists of only one thickness of hide. A double belt
consists of two thicknesses of the hide cemented, stitched, or
pegged together with the grain, or hair, side out. They are
used for heavy drives where a large amount of power is
transmitted. All belts over 6 inches in width should be
double. The ends of a double belt should be tapered , or
skived down , and then cemented together. Double belts are
always made of leather taken from the center of the hide
along the backbone, or from strips taken from each side of
the center; these latter make the best belts as the hide is apt
to be thick along the backbone. The strength of a double
belt is to that of a single belt as 10 is to 7.
Occasionally belts are made three-ply, that is, they consist
of three thicknesses of hide.
18. Belt Fastenings. —There are many good methods
of fastening the ends of belts together, but lacing is gener¬
ally used, as it is flexible like the belt itself, and runs noise¬
lessly over the pulleys. The ends to be laced should be cut
squarely across and the holes
in each end for the lacings
should be exactly opposite
each other when the ends are
brought together. Very nar¬
row belts, or belts having only
a small amount of power to
transmit, usually have only one
row of holes punched in each
end, as in Fig. 14; a is the out¬
side of the belt, and b the side running next to the pulley.
To lace, the lacing should be drawn half way through one
of the middle holes, from the under side, as for instance
through 1 ; the upper end should then be passed through 2,
under the belt and up through 3, back again through 2 and 3,
12
MECHANICAL DEFINITIONS
§7
through 4 and up through 5, where an incision is made in
one side of the lacing, forming a barb that will prevent the
end from pulling through. The other side of the belt is
laced with the other end, it first passing up through 4.
Unless the belt is very narrow, the lacing of both sides
should be carried on at once.
Fig. 15 shows a method of lacing where double lace holes
are used, b being the side to run next to the pulley. The
lacing for the left side is begun at 1, and continues through
2, 3, 4, 5 , 6, 7, 6, 7, 4, 5, etc. A
6-inch belt should have seven
holes, four in the row nearest
the end, and a 10-inch belt, nine
holes. The edges of the holes
should not be nearer than f inch
from the sides; and the holes
should not be nearer than
1 inch from the ends of the belt.
The second row should be at
least If inches from the end.
Another method is to begin the lacing at one side instead of in
the middle. This method will give the rows of lacing on the
under side of the belt the same thickness all the way across.
19 . Although a laced belt with the holes punched as
small as possible is the best, a good copper-riveted belt will
give good service. Belt hooks are not desirable except in
cases of emergency or for narrow belts or those that have
to be frequently altered, as they cut the belt and are also
liable to catch in the clothing of the operatives and thus
cause accidents.
20. Care of Belts.—1. Belts should be run with the
smooth, or grain, side next to the pulley for the following
reasons: ( a ) There is more friction of the belt on the pul¬
ley and, therefore, less slipping and consequent loss of
power, {b) The center of strength in a belt is located one-
third of the distance through the belt from the flesh side
and it is better to crimp the grain, or weak, side around the
§7
MECHANICAL DEFINITIONS
13
pulley than to strain it. ( c ) The stronger side of the belt
receives the least wear when run in this manner.
Some authorities recommend that the flesh side of a belt
be run next to the pulleys, and while this is contrary to the
general practice, in some cases it gives good results.
2. The lower part of a horizontal or inclined belt should
be the driving part; then the slack part will run from
the top of the driving pulley. The sag of the belt will
then cause it to encompass a greater length of the cir¬
cumference of both pulleys. Long belts, running in any
direction other than the vertical, work better than short
ones, as their weight holds them more firmly to their
work. There is, however, a disadvantage in long belts,
since they greatly increase the strain on the bearings of
the shaft.
3. The accumulations of grease and gummy matter should
be frequently removed and the belts dressed with castor oil
or some other suitable dressing on the side of contact, in
order to keep them moist and pliable. It is bad practice to
use rosin to prevent slipping; it gums the belt, causes it
to crack, and prevents slipping for only a short time.
4. If a belt properly cared for persists in slipping, a wider
belt or larger pulleys should be used; the latter to increase
the belt speed. Belts should not be run tight, as the strain
thus produced will wear out both the belt and the bearings
of the shaft.
PULLEYS
21. Pulleys are wheels that are provided with wide
faces, or rims, in order to give sufficient frictional contact
with the belts that run around them to transmit the required
amount of power. They are designed to be attached to
shafts for the purpose of transmitting their power or for
communicating power t® them, and are known, respectively,
as driving and driven pulleys.
The size of a pulley is gauged by its diameter, the width
of the face, and the bore, or diameter, of the hole designed
for the reception of the shaft. Split pulleys are made with
MECHANICAL DEFINITIONS
14
§7
a large bore, the bushings (see Art. 24) being made to suit
any diameter of shafting.
22. A flat, or straiglit-faced, pulley is one whose
face, on which the belt runs, is parallel to the shaft to which
it is fastened. A section of one is shown in Fig. 16.
-
23. A crown, or crown-faced, pulley is one with a
greater diameter in the center of its face than at either edge,
a section of its rim showing a
curve on "the exterior side, as seen
in Fig. 17.
The object of crowning a pulley
is to keep the belt moving on the
center of its face, even if the aline-
ment of the two pulleys on which
the belt runs is not perfect. The
crown of the pulley accomplishes
this because the tendency of a
running belt is to run on the high¬
est part of the pulley, which in this
case is in the center of its face.
If a belt is placed to run on a
conical pulley, it tends to move
up to the largest diameter. To
prevent the belt from running off, two short, conical pulleys
might be used, placed on the same shaft with the large diam¬
eters together. From this the crowned, or rounded, form
of the face of pulleys has been developed.
For driving shaft pulleys, the crown is usually from tV to
h inch per foot of width for high-speed pulleys and i inch
per foot of width for slow-speed pulleys. Machine pulleys
are frequently crowned more than this, sometimes as much
as 4 inch on a 3-inch-face pulley. As a general rule, a high¬
speed pulley requires less crown than a slow-speed pulley,
and a pulley of large diameter less than one of small diam¬
eter. Pulleys are sometimes crowned by drilling holes
through their faces and riveting leather laps around them
with copper rivets.
a»Mai
Fig. 16
Fig. 17
§7
MECHANICAL DEFINITIONS
15
24. Split and Solid Pulleys. —Pulleys are either split
or solid. A split pulley is made in halves and is attached
to the shaft by clamping the hub of the pulley to the shaft or
to segments of metal or wood. These segments are known
as a bushing and are inserted between the shaft and the
inside of the hub.
A solid pulley is attached to the shaft either by a key or
by a setscrew, which is threaded in the hub of the pulley and,
when turned in, binds the hub to the shaft. Sometimes two
or more setscrews are used for fastening a solid pulley on
a shaft, in which case the screws are placed at an angle of 90°
to one another. In this posi¬
tion the pulley is held firmer
than with the screws oppo¬
site each other in the hub.
Pulleys may be made of
cast iron, wrought iron, or of
wood. The latter are not
generally constructed entirely
of wood, the rim only being
made of this material; it is
built of blocks nailed and
glued together. Sometimes
pulleys are made of wrought
sheet steel or iron, the pulley
being stamped out and built
up with rivets. A solid iron pulley is shown in Fig. 18; it
will be noticed that the face of the pulley is crowned. This
illustration also shows the method of placing the setscrews
for attaching the pulley to the shaft. The hub of this pulley
is key-seated, so that it may be attached to the shaft by
means of a key if desired.
Solid pulleys are placed on the shaft before it is raised
into position and placed in the hangers, while split pulleys
may be attached or removed at any time without disturbing
the shaft.
A wood-rim split pulley is shown in Fig. 19; it will
be noticed that this is a straight-faced pulley.
Fm. 18
16
MECHANICAL DEFINITIONS
7
25. A drum is a long, cylindrical pulley, the term being
generally used when more than one belt, or band, is driven
by the same pulley.
Fig. 19 Fig. 20
26. A slieave, or slieave pulley, is one with a groove
or grooves cut or cast in its face for the reception of driving
ropes, or bands. Sheave pulleys are sometimes known as
grooved pulleys. An ordinary type of sheave pulley is shown
in Fig. 20.
Fig. 21
27. Flange pulleys are often used on various machines
and are sometimes found on countershafts. Different types
are shown in Fig. 21, the flanges being designed to pre¬
vent the belt’s running off or coming in contact with another
belt running on the same pulley.
7
MECHANICAL DEFINITIONS
17
28. An idle pulley, or idler, is a pulley around which
a belt runs without imparting motion to any particular
mechanism other than the pulley itself, which is usually
loose on a stud.
29. A guide pulley is an idler placed in such a position
and at such an angle that it guides the belt on to a pulley on
which it could not otherwise run.
30. A binder pulley is an idler so arranged between
driving and driven pulleys that by moving it in a certain
direction the slack of the belt will be taken up.
a f
31. Tight and Loose Pulleys. —The customary method
of communicating power to a machine is to have the machine
connected, by means of a belt, with a pulley on a driving
shaft, which is usually located near the ceil¬
ing, but sometimes arranged underneath the
floor, the driving belt passing through holes
in the floor. As it is obviously impossible
to control the machine economically by means
of the driving shaft, there are two driven pul¬
leys on the machine, one of which is fast to the
shaft of the machine and the other loose, so
that it may rotate without imparting motion
to the machine. Since the pulley on the dri¬
ving shaft has a sufficient width of face to admit
of shifting the belt, it will be seen that the
machine may be controlled independently of
the driving shaft by shifting the belt on to the
tight or the loose pulley, according to whether
it is desired to start or stop the machine.
Such an arrangement is spoken of as tight
and loose pulleys and is shown in Fig. 22.
In this illustration the overhead driving
shaft is marked a and has the driving pulley b
attached to it. This pulley has a face of sufficient width so
that the belt may be moved sidewise a trifle more than its
own width. The motion may be transmitted by the belt from
the driving pulley b to the loose pulley c, in which case the
Pig. 22
18
MECHANICAL DEFINITIONS
§7
machine will not be driven, as the loose pulley will simply
turn on the driving shaft e of the machine. If, however, the
belt is shifted so that the power is transmitted from the dri¬
ving pulley b to the tight pulley d, which is keyed to the shaft e
with the key /, motion will be imparted to the machine.
32. The belt is moved, or shipped, from the loose to the
tight pulley, or vice versa, either by hand or by means of a
belt shipper. The usual type of belt shipper consists of a
rod moving in guides in a direction parallel to the shaft that
carries the tight and loose pulleys, and provided with a fork,
between the tines of which the belt runs. When the rod is
moved in the direction of the loose pulley, the fork pressing on
the side of the belt ships it on to that pulley, and vice versa.
33. Speed cones, or step pulleys, are used in machines
where different speeds are required; this variation in speed
is obtained by means of pulleys of different diameters. The
7
MECHANICAL DEFINITIONS
19
usual method of making step pulleys, or speed cones, is to
have pulleys of different diameters cast in one piece or built
up with wood so that a series of faces is formed for the
belt to run on. A similar pulley is used for the driven
pulley, but is placed on the shaft in a reversed position, so
that when the belt is on the largest diameter of one pulley,
it is on the smallest diameter of the other pulley, and vice
versa. The diameters of the intermediate steps are of course
proportionally changed because the sum of the diameters of
any two opposite steps must be constant so that the belt will
be of the right length for any position and will have the
same tension when on any pair of pulleys.
Fig. 23 shows an arrangement of speed cones that is very
often adopted. In this case the driving speed cone e is placed
on a countershaft, which is also provided with a tight pulleys
and a loose pulley d, by means of which its motion may be
controlled independently of the main shaft of the room from
which it is driven. A belt k connects the driving speed cone
with the driven cone /, which is fastened to the shaft h of the
machine. The belt may be shifted from one step of the
driving and driven cones to another in order to secure differ¬
ent diameters of driving and driven pulleys, thus producing
a wide variation of possible speeds of the shaft h.
With the belt on the largest diameter of the driving cone e
and the smallest diameter of the driven cone / the maximum
speed of the machine is attained. The minimum speed of
the machine is produced with the belt on the smallest diameter
of e and the largest diameter of /. With the belt in interme¬
diate positions, speeds ranging between the maximum and
minimum are obtained. It should be noted that, when speak¬
ing of a cone pulley as having a certain number of steps, the
number of steps is one less than the number of pulleys on the
cone. Thus, the cones in Fig. 23 are four-step cones and
have five pulleys. Consequently, if a cone pulley (or cone)
is spoken of as having five steps, there are six pulleys and
six changes of speed.
In Fig. 23, the tight and loose pulleys are on the counter¬
shaft instead of the driving shaft of the machine. The
20
MECHANICAL DEFINITIONS
7
machine is stopped or started by moving the lever n } which
operates the bar in. Attached to m is the fork / between the
tines of which the countershaft belt passes. By moving the
lever to the left the belt is shipped ofi to the tight pulley c,
thus imparting motion to the countershaft and then to the
cones and the machine driven by them. Moving the lever to
the right ships the belt from c on to the loose pulley d, thus
stopping the machine.
34. From step pulleys, or speed cones, the tapered, or
true, cones, which have no steps for the belt to rest on,
have originated, the belt being automatically shifted along
their surfaces by a belt guide, thus securing the gradual
variation in speed that is necessary in certain classes of
textile machinery.
35. Dii*ection of Rotation.—The connection between
two pulleys by a belt or rope can be made in two ways;
namely, by an open belt, as shown in Fig. 24, or by
a crossed belt, Fig. 25. When a pulley is driven by an
open belt, its direction of rotation is the same as that of the
driving pulley, as indicated by the arrows in Fig. 24. When
a pulley is driven by a crossed belt, its direction of rotation
§7
MECHANICAL DEFINITIONS
21
is opposite to that of the driving pulley, as shown by the
arrows in Fig. 25.
36. Countershafts.—In practice there is a limit to the
size of pulleys that can be used, and therefore to the ratio
of speeds that can be obtained from two shafts. The size of
driving pulleys is limited by the distance between the ceiling
and the shaft and also by the size of the driven pulley. For
instance, an extremely large pulley cannot be used to drive
an extremely small one to obtain a desired speed, because
the belt will slip on the smaller pulley owing to the lack of
sufficient surface contact.
If the belt is applied with
sufficient tension to over¬
come this, an undue strain
will be placed on the bear¬
ings of the shaft and they
will become heated. To
overcome this difficulty an
additional shaft, called a
countershaft, is used and.
in some cases, more than
one is necessary before the
desired speed can be ob¬
tained with an economical
transmission of power.
The transmission of
power in this manner re¬
quires one pulley on the
first, or driving, shaft and one on the driven shaft, while
each intermediate shaft, or countershaft, requires two pul¬
leys of different diameters. An arrangement of this kind
is shown in Fig. 26. If a is the driving pulley and d the
driven, the speed of d is greater than that of a\ but if d is
the driver and a the driven, the speed of a is less than that
of d. If pulleys b and c are of the same size, the power will
be transmitted through the countershaft without any change
of speed being given by it to d.
22
MECHANICAL DEFINITIONS
§ 7
37. Speed of Pulleys. —Pulleys over 4 feet in diameter
and flywheels, especially cast-iron ones, should never be
speeded so fast that their surface velocity exceeds 5,000 feet
per minute, since there will be a danger of their bursting.
Many authorities give 3,750 feet per minute as a limit to
the surface speed of large pulleys. Smaller pulleys may
have a higher surface velocity, but excesses should be
avoided.
NON-PARALLEL SHAFTS
c
J3
U
38. In the drives previously mentioned the driving and
driven shafts have been parallel to each other. But condi¬
tions that frequently arise where power is required to be
transmitted by belts or ropes are:
(1) when the driving and driven
shafts are at right angles to each
other and are not in the same plane;
(2) when the driving and driven
shafts are non-parallel and are both
in the same plane.
39. Quarter-Turn. —When
shafts are arranged as in the first
instance, the pulleys must be so
placed that the belt is delivered from
one pulley into a plane passing
through the center of the face of
the other pulley. The arrangement
shown in Fig. 27 is known as a
quarter-turn because the belt is
given a quarter twist. A connec¬
tion of this kind can only be driven
in the direction indicated by the
If the direction of the belt is reversed
it will run off the pulleys unless a guide pulley is used.
The easiest and most convenient way of fixing the posi¬
tion of quarter-turn pulleys is to plumb the leaving sides of
each pulley; that is, drop a plumb-line from the center of the
Fig. 27
arrows on the belt.
§7
MECHANICAL DEFINITIONS
23
face of the leaving side, where the belt leaves the driving
pulley, and arrange the driven pulley so that the plumb-
line shall just touch the center of its face on the side from
which the belt leaves it. This is shown by the two pul¬
leys at the top of Fig. 27, which represents a plan of two
quarter-turn pulleys as seen from above.
The objection to a quarter-turn belt is that, when the angle
at which the belt is drawn off the pulleys is large, the belt
is strained, especially at
the edges, and it does
not hug the pulleys well.
Small pulleys placed
quite a distance apart,
with narrow belts give
the best results, from
which it follows that
quarter-turn belts are
not well suited to trans¬
mit much power.
Fig. 28 shows how
the arrangement can be
improved by placing a
guide pulley against
the loose side of the
belt. The driver d re¬
volves in a left-hand
direction, making a b the
driving, or tight, side of the belt. To determine the posi¬
tion of the guide pulley, select some point in the line a b,
as g. When the pulleys differ in diameters this point should
be somewhat nearer the smaller pulley. Draw lines eg and
eg; the middle plane of the guide pulley should then pass
through the two lines. Looked at from a direction at right
angles to pulley /, line eg coincides with a b; looking at
right angles to pulley d, line eg also coincides with a b.
Fig. 28
40. If two shafts are not parallel to each other and are
both in the same plane, they may be driven by a belt by
24
MECHANICAL DEFINITIONS
§7
means of guide pulleys as shown in Fig. 29. In a case like
this the guide pulleys must be placed so that the active part
of their surfaces will be tangent to planes passing through
the centers of the faces of the driving and driven pulleys at
the intersection of these planes, as shown in the upper, or
plan, view in Fig. 29. They must also be placed so that
their center planes are tangent to the surfaces of the dri¬
ving and driven pulleys, as shown in the elevation, or lower,
view. A pair of guide pulleys and their shaft, as illustrated
in Fig. 29, are known as a mule pulley stand.
ROPE TRANSMISSION
41. Many American mills are introducing rope drives
for transmitting power, especially for the main drives from
the engine, for which this method is particularly adapted.
The distance to which power can be transmitted by means of
sheave pulleys and ropes is practically unlimited, as is also
the amount of power. Except for very short distances, rope
driving is the cheapest method of transmitting power, being
economical not only in the first cost, but in the maintenance.
This in itself is an important item. An evenness of motion
that cannot • be obtained by any other system of power
§7 MECHANICAL DEFINITIONS 25
transmission is obtained by transmitting power in this man¬
ner; this is due to the lightness, elasticity, and slackness of the
rope, which takes up all inequalities between the power and
the load. Rope drives are noiseless because of the flexibility
and lubrication of the rope and because of the air passage
underneath the rope, owing to the Y-shaped groove in which
it runs. An exact alinement of the driving and driven
pulleys is not necessary when ropes are used, and by properly
placing idle pulleys power may be transmitted in any desired
direction. The security that a rope drive affords against
shut-downs due to the crippling of the drive is one of the
great advantages of this system. This is due to the fact
that before breaking, the rope stretches excessively, though
gradually, thus giving warning that it should be replaced.
The absence of electrical disturbances and the almost total
immunity from slip are among the many advantages that
may well be claimed by this system for power transmission.
42. There are two systems of rope transmission in
common use. In the first, the transmission is effected by
several parallel, independent ropes that pass around the
26
MECHANICAL DEFINITIONS
flywheel of the engine and the pulley or pulleys to be driven.
Each rope is made quite taut at first, but stretches until it
slips, after which it is respliced. A good example of this
system is shown in outline in Fig. 30. The flywheel m
carries thirty-five parallel ropes that distribute power to the
pulleys a , b, c, d , e, /, located on the five floors of the mill.
The ropes are distributed as follows: a, four ropes; b and c,
five ropes each; d, e, and /, seven ropes each. A secondary
system of ropes drives the pulleys^, h, k , /.
In the second system of rope transmission, a single rope
is carried around the pulleys as many times as is necessary,
to transmit the required power; the necessary tension is
obtained by passing a loop of the rope around a weighted
pulley. An example of this system is shown in Fig. 31.
The rope is wrapped continuously around the flywheel d and
the driven pulley e. From the last groove of e , the rope is
led over the idlers f,g, which are set at such an angle as to
lead it back to the first groove in d. The weight w is
attached to the pulley /, which is movable along the rod Ji.
The movement of the pulley /, therefore, takes up the
stretch of the rope, and keeps it always at the same tension.
Pulleys may be attached to the shaft of the pulley e , and the
§7
MECHANICAL DEFINITIONS
27
power received by e may thus be transmitted to any desired
points.
The first of the above systems of transmission is used
chiefly in Europe; the second, in the United States. The
ropes generally used are of Manila hemp or cotton, some¬
times with a wire core. For transmitting power long dis¬
tances, especially where the rope is exposed to the weather,
a wire rope is used. For inside drives the cotton rope
without a wire core is suitable.
43. Next in importance to the rope are the grooved
pulleys, or sheaves, on which the rope runs. The grooves
are made of metal or wood and must be smooth, in order to
prevent the rope from wearing, and true, to keep it from
swaying. These grooves are made Y-shaped so that they
may grip, or bind, the rope and not allow it to slip; the
best forms are those in which the sides approach each
other at an angle of from 45° to 60°. The rope does not
touch the bottom of the groove, but is wedged in between
the sides.
An idle pulley for a rope drive may have half-round
grooves, as there is no necessity for gripping the rope at
this point.
GEARING
44. Gearing. —The transmission of power for short
distances at slow speeds, as between the driving and driven
shaft of a machine, is generally accomplished by means of
gears. A gear is a disk with a hole in its center, so that it
can be readily attached to a shaft. Its face is cut, or cast,
with teeth that are so designed as to mesh with the teeth
of another gear, thus producing motion by a direct pressure
of the teeth of one gear on the teeth of the other.
Gearing is ordinarily made of cast iron; if great strength
is required, steel may be used. Gears that are called on
to resist shocks may be made of gun metal or phosphor
bronze. Fast-running gears are sometimes made with
wooden cogs, or of rawhide or fiber, instead of metal.
28
MECHANICAL DEFINITIONS
§7
45. Teetli is the name applied to the projections around
the circumference of a gear when they are made from the
same piece as the body of the gear; while cogs is the name
used when they are made of separate material and fastened
to the rim of the gear. Gears with cogs inserted in their rims
are called mortise gears. Toothed gears are in gear when
their teeth are engaged, and out of gear when separated.
Fig. 32 shows two gears, either of which may be the driver
and communicate motion to the driven gear by direct contact
of its teeth. It will be noticed that the direction in which
the driven gear rotates is opposite to that of the driver.
46. The pitch circle of a gear is an imaginary circle
described, with the axis of rotation for a center, through the
point of contact of the teeth of one gear with those of another
gear. It is its effective circumference and determines its ratio
of velocity when working with other gears. In Fig. 32 the
dot-and-dash lines represent the pitch circles of the two gears.
Suppose that two shafts 10 inches apart are to be connected
by gears and are required to revolve in the ratio of 3 to 2;
then the diameters of the pitch circles of the gears will be 8
7
MECHANICAL DEFINITIONS
29
and 12 inches. Two circles centered on the shafts and having-
radii of 4 and 6 inches will give the required ratio of speeds
by rolling contact; thus, this regulates the diameters of the
pitch circles of two gears to give the required relative
velocity.
47. The circular pitch of a gear is the distance
between the centers of two consecutive teeth measured on
the pitch circle or, as it is sometimes called, the pitch line.
48. The diametral pitch of a gear is equal to the
number of teeth on its circumference divided by the number
of inches in the diameter of the pitch circle. In order to
mesh and run together, gears must be of the same pitch.
49. The part of the gear-tooth outside of the pitch circle
is called the point, or addendum, of the tooth and the
part inside the pitch circle is called the root of the tooth.
The root of a tooth is usually about four-sevenths, and the
point about three-sevenths, the length of the tooth. This
allows the point of the tooth to clear the bottom of the space
between the teeth of the other gear by one-seventh of
its length.
50. The amount of space by which the width between
the teeth is greater than the thickness of the tooth is called
the backlasli. The dimensions of gears are given by the
diameter of the bore, by the number of teeth that the gear
contains, and by the pitch.
51. Trains of Gears. —When the driving and driven
shafts are so far apart as to necessitate the use of gears of
too large diameters, or when the desired ratio of velocity is
such that one gear must have an excessive diameter, it is
necessary to interpose gears between the driving and driven
shafts. When gears are combined in this manner they are
known as a train of gears. Fig. 33 shows a train of gears
thus arranged, the motion being transferred from the shaft e
to the shaft / by means of the gears a , d , c, b. The gears d , c
are used to transfer the motion of a to b and to give the
latter a velocity different from that of the gear a.
30
MECHANICAL DEFINITIONS
§7
52. Spur gears have the teeth arranged on the outer
rim parallel to the shaft on which the gear is placed. The
action of a pair of these gears may be regarded as that of
two rolling cylinders minus any slip. This type of gear is
one that is largely used for transmitting power between
parallel shafts and is shown in Fig. 33.
Fig. 34
53. Bevel gears are designed to transmit power from
one shaft to another when the shafts are in the same plane
but not parallel. The action of a pair of bevel gears may be
regarded as that of two rolling cones whose axes intersect
§7
MECHANICAL DEFINITIONS
31
at their apexes. Fig. 34 shows an ordinary type of bevel
gears. When two bevel gears of the same size are designed
for shafts that are at right angles to each other, as in Fig. 34,
they are also known as miter gears.
54. A ratchet gear, as shown in Fig. 35, is designed to
engage with a small piece of metal called a pawl. The
teeth are so shaped that motion can be imparted to the gear
in one direction only by a pawl similar to b. It is held from
turning in the other direction
by the action of the pawl b,
which engages with the
Fig. 36
teeth. Owing to the inclination of the teeth, when the ratchet
a is turning in the proper direction, the pawl slips over the
backs of the teeth.
Sometimes a ratchet gear is driven a definite number of
teeth at a time by the action of an oscillating pawl, and is
then held by a stationary pawl while the driving pawl is work¬
ing backwards. At other times the ratchet is attached to a
shaft having an intermittent motion, the pawl preventing the
shaft from turning backwards during the periods of rest, but
offering no resistance to its forward motion.
55. Sprocket gears, or wheels, are designed to be
driven by and to drive chains. A common type of this gear
is shown in Fig. 36.
32
MECHANICAL DEFINITIONS'
§7
56. Spiral gears are those that have the teeth cut
obliquely on their surfaces and are used to connect shafts
that form an angle with each other.
57. Star gears are those having recesses cut in them at
wide intervals and are driven by special contrivances, often
by a pin. This pin is often attached to a disk and engages
with the star gear at each rotation of the shaft. A star gear
is always a driven gear. *
58. Mangle gears are used to reverse their own direc¬
tion of rotation and are always driven gears. They are either
eccentric or concentric. When concentric the center of the
pitch circle coincides with the axis of rotation, and when
eccentric the center of the pitch circle is removed from the
axis of rotation.
§7
MECHANICAL DEFINITIONS
33
An eccentric mangle gear / is shown in Fig. 37. The gear
is of somewhat peculiar construction, having instead of the
ordinary gear-teeth, a number of pegs, or pins, fastened in
its rim. These are designed to mesh with the driving pinion e
fastened to the shaft d, by means of which motion is imparted
to the mangle. The shaft d is carried in a wide guide, or
bearing, d x so that the pinion e may accommodate itself to
the eccentricity of the mangle and engage with either the
outside or the inside of the row of pins, thus enabling the
mangle to be driven in either direction. The shaft d projects
beyond the pinion e a slight distance and engages with a
flange A extending entirely around the gear on the inside of
the pins and part way around the outside. By this means
the pinion is guided from the outside to the inside of the
row of pins, and vice versa.
Suppose that in Fig. 37 the shaft d is rotating in the direc¬
tion shown by the arrow; then the pinion e will work around
on the inside of the pins on the mangle until the last pin A
is reached, when the flange A will guide the gear to the out¬
side of the row of pins and the direction of the rotation of
the mangle will be reversed. This motion will continue
until the other end of the row of pins is reached, whereupon
the flange will guide the pinion back to the inside of the row
of pins and the direction of rotation of the mangle will again
be reversed. It is not necessary to have the outside flange
pass entirely round the mangle gear, since when the teeth
are engaging with the outside of the pins on the portion of
the gear that is farthest from its axis, the pinion is held in
contact with the teeth of the mangle by means of the shaft d,
which in this case is pressed against the end of the wide
bearing d x .
A mangle gear is said to perform a complete cycle of its
movements in making a double revolution; that is, one revo¬
lution in one direction and one in the opposite direction.
59. A worm is an endless screw designed to mesh with
a gear called a worm-gear. A worm and a worm-gear are
shown in Fig. 38. Worms are single-threaded when they
34
MECHANICAL DEFINITIONS
§7
have a single thread cut on them and double-threaded when
they have two threads. A worm is always a driver, a single-
threaded worm driving the worm-gear one tooth for each
revolution and a double-threaded worm moving it two teeth.
Occasionally worms are met with that have three threads
cut on them. Worms furnish a means of reducing a great
velocity of a shaft to the slow speed of the worm-gear. In
calculations a worm is counted
as a one-tooth gear if single-
threaded and as a two-tooth
gear if double-threaded.
60. An annular gear is
one whose teeth are cut on the
inside of its rim, or circum¬
ference.
61. A rack is a straight bar
with teeth cut on its surface.
62. A pinion is a small
gear and is in most cases a spur
gear.
63. Eccentric gears are
gears that rotate about points
not centrally located with regard to their pitch circles. A
pair of eccentric gears require setting in the proper posi¬
tion before they will run; otherwise, they will bind and spring
the shafts to which they are attached.
A pair of eccentric gears is always marked—in some cases,
by nicking with a cold chisel two teeth on one gear and one
tooth on the other. When setting the gears, place them
together so that the single nicked tooth on one gear will
mesh between the two nicked teeth on the other gear.
64. Elliptic gears are made in the form of an ellipse
and are marked and set after the manner of eccentric gears.
Elliptic and eccentric gears are used for producing a variable
speed of the driven shaft during each rotation of the pair
of gears.
7
MECHANICAL DEFINITIONS
35
65. A carrier, intermediate, or idle, gear is a gear
running on a stud used either to change the direction of
rotation of a driven gear or to connect a driving and driven
shaft when gears of sufficiently large diameter cannot be
used to make the connection directly. When one inter¬
mediate is used, the direction of rotation of the driven gear
is the same as that of the driver; but when two intermediates
are used, the direction of rotation is the reverse, as though
the driver meshed directly with the driven gear.
SCREWS
66. A screw is a straight shaft with a thread cut
spirally around it. It may have a right-hand or a left-hand
thread, depending on whether it advances when turned to
the right or to the left.
The pitch of a screw is expressed in threads per inch for
small screws, but for large screws it is usually expressed as
the distance between the centers of consecutive turns of the
thread measured parallel to the axis of the screw.
67. A multiple-threaded screw is one with two or
more threads cut on it.
68. A reciprocating screw is one with both a right-
hand and a left-hand thread cut on it, the threads being con¬
nected at each end and designed to engage with a pin, or
some form of traveler, for the purpose of imparting a
reciprocating motion.
69. A differential screw is one with a varying pitch
and is designed to engage with a pin, as it is obviously
impossible to fit such a screw with a threaded nut.
CAMS
70 . A cam is a turning or sliding piece that, by the shape
of its curved edge or face, or a groove in its surface, imparts a
variable or intermittent motion to a roller, lever, rod, or other
moving part, the character of the imparted motion depending
on the shape of the cam. This motion, where a small amount
of power is involved, is often transmitted by sliding contact;
MECHANICAL DEFINITIONS
36
§7
but in cases where much force is employed, the contact is
generally rolling.
The lieel of a cam is that part of its - effective circumfer¬
ence that is nearest the axis of rotation. The toe of a cam
is the part of its effective circumference that is farthest from
the axis of rotation. The difference in the distance between
the toe of the cam from the axis of rotation and the heel of
the cam from the axis of rotation is
called the throw of the cam.
71. A positive-motion cam is
one that actuates the part that is in
direct contact with it equally well in all
directions without the aid of a spring
or weight; a common type is shown
in Fig. 39. It consists of a plate a fast¬
ened to a rotating shaft and having a
groove b cut in one face. A roller c
attached to a three-armed lever d
engages with the groove; thus, as the cam rotates, it imparts
to the lever a motion that depends on the shape of the guid¬
ing groove. The lever d is moved positively in both direc¬
tions and nothing less than actual breakage of some part can
prevent its working.
72. A non-positive cam is one that depends on either
a spring or a weight to keep the part on which it acts in
direct contact with it at all times; a non-positive cam is
§7
MECHANICAL DEFINITIONS
37
shown in Fig. 40. Like the positive-motion cam, it consists
of a plate a fastened on a rotating shaft, but the roller c,
instead of being guided by a groove, rests on the circum¬
ference of the cam. Motion will be positively imparted to
the rod e only while the cam-bowl, or roller, is traversing
from the heel to the toe of the cam. While passing from the
toe to the heel, the roller is held in contact with the cam by
the spring b attached to the lever d. If the rod e or lever d
is caught on the return motion so that the tension of the
spring is overcome, the roller c will be held away from the
cam.
73. A cam-bowl is a small roller that runs either in the
groove of a positive cam or on the circumference of a non¬
positive cam, in the latter case being held in position by a
spring or weight.
74. If the action of a cam is intermittent, it is called
a wiper. Occasionally the term wiper is applied to any
non-positive cam. _
LEVERS
75. A lever is an inflexible bar capable of being freely
moved about a fixed point, or line, called the fulcrum.
The bar is acted on at two points by two forces that tend to
rotate it in opposite directions about its fulcrum. Of these
two forces, the one that is applied with the purpose of
Fig. 41
imparting motion is termed the power , while the force that
is to be overcome is the weight , or resistance. The parts a f
and b /, Fig. 41, are the arms of the lever.
There are three classes, or varieties, of levers depending
on the relative positions of the power p , weight w, and ful¬
crum /.
38
MECHANICAL DEFINITIONS
§7
If the fulcrum is between the power and the weight ( p , /, w ),
as shown in Fig. 41, the lever is of the first class. In this
combination equilibrium exists if the product of the force p
times arm a f equals the product of the force w times arm b /.
l-/ _
/«}
Fig. 42
If the weight is between the power and the fulcrum ( p , w, /)
as shown in Fig. 42, the lever is of the second class.
If the power is between the weight and the fulcrum ( w,p, /),
the lever is of the third class, Fig. 43.
Fig.43
76 . Sometimes it is not convenient to use a lever suffi¬
ciently long to make a given power support a given weight.
In this case, combinations of levers known as compound
level’s are used.
MECHANICAL CALCU¬
LATIONS
RULES PERTAINING TO THE TRANS¬
MISSION OF POWER
RULES APPLYING TO SHAFTING
1. Cold-Rolled Shafting. —The following rules will be
found useful in finding the required size of a cold-rolled
shaft necessary to transmit a given horsepower.
To find the required diameter of a main shaft:
Rule I. —Find the cube root of 100 times the required horse¬
power divided by the desired number of revolutions of the shaft
per minute.
To find the required diameter of line shafts to transmit a
given horsepower with the power taken off at intervals and
the bearings of the shaft not more than 8 feet apart:
Rule II. —Find the cube root of 50 times the required horsc-
pozver divided by the desired number of revolutions per minute.
To find the required diameter of short countershafts for
transmitting a given horsepower:
Rule III. —Find the cube root of 30 times the required
horsepower divided by the desired number of revolutions per
minute.
For notice of copyright, see page immediately following the title page
2
MECHANICAL CALCULATIONS
§8
Example. —Suppose that it is desired to purchase a line shaft for a
weave room requiring 350 horsepower; it is desired to have the shaft
make 300 revolutions per minute and a cold-rolled shaft is to be used.
What diameter of shafting is required?
50 X H P
Solution. —Diameter of shaft equals cube root of
rev. oer mm.
50X 350
300
= 3.87+ in. Ans.
Note.—I n a case like this a 4-inch cold-rolled shaft would probably be ordered' as
this would allow for the extra power required to overcome the friction of the shaft in
its bearings.
2. Turned Shafting.— The following rules give the
methods of finding the required size of turned shafting to
transmit a required horsepower.
To find the required diameter of a main shaft:
Rule I .—Find the cube root of 125 times the required horse¬
power divided by the desired number of revolutions per minute.
To find the required diameter of line shafts with the
power taken off at intervals and the bearings not more than
8 feet apart:
Rule II .—Find the cube root of 90 times the reqziired horse-
pozver divided by the desired number of revolutions per minute.
To find the required diameter of short countershafts:
Rule III. — Find the cube root of 50 times the required horse¬
power divided by the desired number of revolutions per minute.
Example.— What diameter of turned shafting is capable of trans¬
mitting 45 horsepower, the shaft to be the main driving, or line, shaft
of the room and the bearings not more than 8 feet apart. It is desired
that the shaft make 150 revolutions per minute.
90 X H P
Solution. —Diameter of shaft equals cube root of - : ——•
rev. per mm.
J 90 X 45 = 3 in Ans
\ 150
Note.—W hen the hangers are placed far apart, a larger shaft is necessary in
order that it may have stiffness to withstand the bending strain due to its lack of sup¬
port and to its own weight.
8
MECHANICAL CALCULATIONS
3
RULES APPLYING TO SPEEDS
3. Speeds of Pulleys.—In connection with pulleys it
must be remembered that the driving pulley is the one that
furnishes the power to the driven pulley. The tight side of
the belt always travels toward the driving pulley and the
slack side toward the driven pulley.
To find the number of revolutions per minute of the
driven pulley, the diameter and number of revolutions of
the driving pulley and the diameter of the driven pulley
being known:
Rule I.— Multiply the diameter of the driving pulley by its
revolutions and divide the product by the diameter of the driven
pulley.
Example 1.—A driving shaft making 350 revolutions per minute
carries a 24-inch pulley that drives a 14-inch pulley on the main shaft
of a machine; find the revolutions of the main shaft of the machine.
24 in. X 3.50 .
Solution.— -——-= 600 rev. per mm. Ans.
14 m. v
To find the revolutions of the driving shaft, the diameter
and revolutions of the driven pulley and the size of the dri¬
ving pulley being known:
Rule II. — Multiply the diameter of the driven pulley by its
speed and divide by the diameter of the driving pulley.
Example 2.—The shaft of a machine makes 700 revolutions per
minute. The size of the driven pulley is 8 inches and the driving pul¬
ley on the main shaft is 14 inches; find the revolutions of the main
driving shaft.
8 in. X 700
Solution.— — 7 —; -= 400 rev. per mm. Ans.
14 in. ^
To find the diameter of a driven pulley to give any desired
number of revolutions per minute, the diameter of the driving
pulley and its speed being known:
Rule III. — Multiply the diameter of the driving pulley by
its speed and divide the product by the desired number of revolu¬
tions of the driven pulley.
Example 3.—The main shaft of a room makes 225 revolutions per
minute and carries a 20 -inch pulley from which it is desired to drive
4
MECHANICAL CALCULATIONS
§8
a countershaft 300 revolutions per minute; what size pulley must be
ordered for the countershaft?
20 v 225
Solution. — . — = 15-in. pulley. Ans.
oUU
To find the diameter of a driving pulley necessary on a
shaft making a definite number of revolutions in order to
drive a driven pulley of a given diameter a certain speed;
Rule IV .—Multiply the diameter of the driven pulley by the
desired speed and divide the product by the speed of the driving
shaft.
Example 4.—Find the size of the pulley required on a driving shaft
making 360 revolutions per minute in order to drive a machine 600 revo¬
lutions per minute. The size
of the driven pulley on the
machine is 12 inches.
Solution.—
12 X 600 „
360— = 20-in. pulley. Ans.
4. Effect of Coun¬
ter s li a f t s on Speed.
It often happens that
the power is transmitted
through a countershaft,
or countershafts, carrying
different-sized pulleys be¬
fore being applied to the
pulley whose speed it is
desired to find.
To find the speed of a
driven pulley when the
power is transmitted through one or more countershafts
carrying different-sized pulleys:
Rule .—Multiply the speed of the driving shaft by the product
of the diameters of all the driving pulleys and divide the result
by the product of the diameters of all the driven pulleys.
Example. —Referring to Fig. 1, suppose that the driving shaft
makes 375 revolutions per minute and that the main driving pulley a
is 18 inches in diameter and drives a 12-inch pulley £ on a countershaft.
MECHANICAL CALCULATIONS
5
On this countershaft a 22-inch pulley c drives the 10-inch pulley d of
a machine. Find the number of revolutions of the pulley d.
Solution.—
375 X 18 in. X 22 in.
12 in. X 10 in.
1,237.5 rev. per min.
Ans.
5. Ratios of the Speeds and the Diameters of
Pulleys.—From these rules it will be seen that speeds and
variations of speeds and the diameters of driving and driven
pulleys should be treated as ratios and that the following
equality of ratios, or proportion, is always true; namely, that
the diameter of the driving pulley is to the diameter of the
driven pulley as the speed of the driven pulley is to the speed
of the driving pulley. As the product of the extremes is
always equal to the product of the means in a proportion, so
the diameter of the driving pulley multiplied by its speed
always equals the diameter of the driven pulley multiplied
by its speed.
6. To find the surface velocity of a rotating pulley or
cylinder or the speed of a belt passing around it, in feet
per minute (slip neglected):
Rule.—I. Multiply the diameter of the pulley or cylinder
in feet by 3.1416 and by the number of revolutions per minute.
II. If the diameter of the pulley or other cylinder is expressed
in inches , multiply its diameter by 3.1416 and by the member of
revolutions per minute that it makes and divide the product by 12.
Example.— -Find the surface velocity, in feet per minute, of a
50-inch cylinder making 160 revolutions per minute.
c 50 X 3.1416 X 160 onni4Ii . .
Solution. — -^ - = 2,094.4 ft. per mm. Ans.
J -Pj
EXAMPLES FOR PRACTICE
1. Required the speed of a driven shaft on which there is a 14-inch
pulley, the power being transmitted by a 22-inch pulley making
280 revolutions per minute. Ans. 440 rev. per min.
2. What size pulley should be placed on a driving shaft making
320 revolutions per minute in order to drive a machine 500 revolutions
per minute, the size of the driven pulley being 12 inches?
Ans. 18f-in. pulley
6
MECHANICAL CALCULATIONS
§8
3. What size pulley is required on a countershaft that is to be
driven 160 revolutions per minute from a 24-inch pulley on a main
shaft making 220 revolutions per minute? Ans. 33-in. pulley
4. A main shaft carries a 28-inch pulley and drives a 21-inch pulley
700 revolutions per minute; what is the speed of the main shaft?
Ans. 525 rev. per min.
5. A 20-foot flywheel makes 60 revolutions per minute and drives
a 6-foot pulley on the head-shaft. A 5-foot pulley on the head-shaft
drives a 4-foot pulley on the main shaft of a weave room. On the
main shaft of the weave room a 2-foot pulley drives a 2^-foot pulley
on a countershaft. Find the speed of the countershaft.
Ans. 200 rev. per min.
Note.—I n these Examples for Practice it will be found an advantage in order to
better understand any example to make a rough sketch of the arrangement of pulleys
and then designate the driving and driven pulleys.
RULES APPLYING TO BELTS
7. Length of Belts.—The following rules will enable
the student to perform calculations in connection with belts.
To find the length of an open belt, the distance between
the centers of the shafts and the diameters of the driving and
driven pulleys being known:
Rule I .—Multiply half the sum of the diameters of the
driving and driven pulleys by 3.1416 and to this product add
twice the distance between centers.
Example 1.—A countershaft is to be driven from the main shaft
with an open belt; the distance between the centers of the shafts is
12 feet, and the diameters of the driving and driven pulleys are,
respectively, 2 and 3 feet; how long a belt is required?
Solution.— ~1 = 2.5; 2.5 X 3.1416 = 7.854
12 X 2 = 24; 24 + 7.854 = 31.854 ft. of belt. Ans.
Note—I n case one pulley is much larger than the other, it is well to cut the belt
2 or 3 inches longer than calculated by the above rule.
To find the length of a crossed belt, the size of the driving
and driven pulleys and the distance between the centers of
the shafts being known:
Rule II .—To one-half the Product of the sum of the diameters
of the driving and driven pulleys and 3.1416 add twice the
§8
MECHANICAL CALCULATIONS
7
square root of the sum of the square of the distance between the
centers of the shafts and the square of one-half the sum of
the diameters of the driving and driven pulleys.
Example 2.—A countershaft is to be driven from the main shaft
with a crossed belt, the distance between the centers of the shafts is
12 feet, and the diameters of the driving and driven pulleys are,
respectively, 2 and 3 feet; how long a belt is required?
Solution.—
(2 + 3) X 3.1416
2
2 X Vl44 +X+ =
2 X Vl44 + 6.25 =
2 X VloO.25 =
2 X 12.25 = 24.5
24.5 + 7.854 = 32.354 ft. of belt. Ans.
Note.—T hese rules, although not absolutely accurate, are near enough for prac¬
tical purposes when it is impossible to measure the length of belt required.
8 . Horsepower Transmitted by Belts. —As the width
of a belt required to transmit a given horsepower depends
on the speed and tension of the belt, the size of the smaller
pulley, and the relative amount of its surface touched by the
belt, no rule can be given that will apply to all cases.
A belt that is being constantly shifted from a tight to a
loose pulley, or vice versa, must be wider than one running
on the same pulley all the time, while innumerable other
conditions govern the horsepower capable of being trans¬
mitted by a given belt and the life of the belt.
It has been found by exhaustive experiments that a single
belt traveling 900 feet per minute will transmit approxi¬
mately 1 horsepower per inch of width when the arc of con¬
tact on the smaller pulley does not vary much from 180°.
This is used by many engineers as a general law for belting
and is applied in all cases. From this fact the following
rules in connection with belting are obtained.
To find the horsepower transmitted by a given belt:
Kule I .—Divide the product of the width of the belt , in
inches , and the speed , in feet per minute, by 900.
8
MECHANICAL CALCULATIONS
8
To find the required width of a belt to transmit a given
horsepower:
llule II. —Divide the horsepower multiplied by 900 by the
speed of the belt , in feet per minute.
Example.—T wo 48-inch pulleys are to be connected by a single
belt and make 200 revolutions per minute; if 40 horsepower is to be
transmitted what must be the width of the belt?
200 X 48 X 3.1416
12
= 2,513 ft. per
Solution. — The belt speed = —
Ans.
Note.—A 14-inch belt might safely be used, since the rule gives a liberal width
when the pulleys are of equal size.
9. In these rules it has been assumed that the belt is
open and also that the driving and driven pulleys are of the
same diameter, the belt consequently being in contact with
half of the circumference of each pulley. But when one
pulley is larger than the other, the horsepower transmitted
is reduced as the arc of the smaller pulley that is in contact
with the belt is reduced. With a crossed belt the amount of
horsepower that can be transmitted by a given width of belt
is increased, as there is then more of the surface of the
pulleys in contact with the belt.
As the rules for single belts are based on the strength
at the lace holes, a double belt, which is twice as thick,
/■
should be able to transmit twice as much power as a single
belt and, in fact, more than this where, as is quite common,
the ends of the belt are cemented together instead of being
laced. Where double belts are used on small pulleys, how¬
ever, the contact with the pulley face is less nearly perfect than
it would be if a single belt were used, owing to the greater
rigidity of the former. More work is also required to bend
the belt as it runs over the pulley than in the case of the
thinner and more pliable belt, and the centrifugal force
tending to throw the belt from the pulley also increases with
the thickness. For these reasons, the width of a double
belt required to transmit a given horsepower is generally
assumed to be seven-tenths the width of a single belt
§8
MECHANICAL CALCULATIONS
9
required to transmit the same power. Therefore, in order
to find the width of a double belt required to transmit a
given horsepower, proceed as with a single belt and multiply
the result by tV; and in order to find the horsepower trans¬
mitted by a given width of a double belt, proceed as with a
single belt and multiply the result by - 7 °.
ROPE TRANSMISSION
10. Rope drives give the most satisfactory results when
the ropes run in grooves the sides of which have an angle
of about 45°.
To find the horsepower transmitted by a single rope run¬
ning under favorable conditions in a 45° groove:
Rule. —Multiply the speed of the rope, in feet per second , by
the square of its diameter , in inches , and divide the product
by 825. This quotient multiplied by the result obtained by sub¬
tracting from 200 the speed of the rope per second , squared , and
divided by 107.2 equals the horsepower that can be transmitted
with a single rope.
Example. —A flywheel designed for a rope drive is 22 feet in diam¬
eter and is equipped with 30 grooves; the diameter of the rope is
li inches and the flywheel makes 50 revolutions per minute; what
horsepower can be safely transmitted?
Solution.—
22 X 3.1416 X 50
60 (sec.)
57.596 X (li)
825
57.596 X 1.5625
825
= 57.596, speed of rope in ft. per sec.
3317.299216^ _
X
^200 -
107.2
.109083 X (200 - 30.9449)
.109083 X 169.0551 = 18.441, H. P. for one rope. Since there are
30 grooves in the flywheel, in each of which there is one rope, the
total power transmitted will be 18.441 X 30 = 553.23 H. P. Ans.
Fig. 2 shows the horsepower transmitted by 1-inch, li-inch,
li-inoh, lf-inch, and 2-inch ropes for various velocities. The
horizontal distances represent velocities in feet per second,
10
MECHANICAL CALCULATIONS
§8
and the vertical distances the horsepower transmitted by a
single rope. It shows that the maximum power is obtained
at a speed of about 84 feet per second. For higher veloci¬
ties, the centrifugal force becomes so great that the power is
decreased, and when the speed reaches 145 feet per second.
Pig. 2
the centrifugal force just balances the tension, so that no
power at all is transmitted. Consequently, a rope should not
run faster than about 5,000 feet per minute, while it is prefer¬
able, on the score of durability, to limit the velocity to 3,500
feet per minute.
RULES APPLYING TO GEARS
11 . The rules that apply to pulleys are also applicable to
gears with the exception that the number of teeth on the
gears are taken instead of the diameters.
To find the speed of a driven gear:
Rule. —Multiply the speed of the first driving gear by its
number of teeth and by the number of teeth on each driving gear
in the train, if there is more than one , and divide the Product by
§8
MECHANICAL CALCULATIONS
II
the number of teeth on the driven gear or by the product of the
teeth on the driven gears.
Example. —Suppose that the shaft e in Fig. 3 is the driving shaft
and makes 40 revolutions per minute; find the number of revolutions
of the driven shaft f when a and c have each 24 teeth and d and b
11 teeth.
c 40 X 24 X 24
Solution.— — 11 1 —- = 190.413 rev. per min. Ans.
11 y\ 11
A good method of determining whether a gear is a driver
or driven gear is to notice which side of its teeth are worn,
or polished smooth. The driving gear always has its teeth
polished on the side facing in the direction in which the gear
is moving; a driven gear has the opposite side of the teeth
polished; while on an intermediate gear both sides of the
teeth are worn.
For solving problems that deal with gears, use the same
rules as are given for pulleys, remembering that the number
of teeth on the driving gear or gears multiplied together and
by the speed of the first driver equals the number of teeth on
the driven gear or gears times the speed of the last driven
gear. Speeds and sizes of gears, like pulleys, should be
treated by proportion. Intermediate gears should not be
used when finding speeds or sizes of gears.
Where two gears are fastened together and interposed
between driving and driven gears after the manner of an
intermediate, they are said to be compounded or to constitute
12
MECHANICAL CALCULATIONS
§8
a compound. If the two gears vary in size, such an arrange¬
ment will affect the value of the train and they should be
figured as a driver and a driven gear, as they really are.
12 . Sizing: Gear-Blanks. —Many mills are equipped
with gear-cutting machines for cutting gears to replace
broken or worn ones, making change gears, etc. When it is
desired to cut a gear, it is necessary to select a gear-blank of
the correct diameter for the desired number of teeth and pitch.
To find the desired diameter of blank for any number of
teeth and any diametral pitch:
Rule I .—Add 2 to the required number of teeth and divide by
the desired pitch; the quotient is the diameter , expressed in
inches , of the blank required.
Example 1.—A change gear with 33 teeth and 10-pitch is desired.
To what diameter must the gear-blank be turned?
33 + 2
Solution.— ~To” = ^ * n - ^ ns -
To find the number of teeth the gear-cutter must space to
cut a given blank a required pitch:
Rule II. —Multiply the diameter of the blank by the required
pitch and from the product thus obtained subtract 2; the answer
is the number of teeth required.
Example 2.—How many teeth must the gear-cutter space to cut
a gear-blank 21 inches in diameter, 12-pitch?
Solution.— 2 \ X 12 = 30; 30 — 2 = 28 teeth. Aus.
13. Speed of Worm-Gears.—To find the speed of a
worm-gear driven by a single- or double-threaded worm:
Rule I .—If the worm is single-threaded divide its speed by
the number of teeth in the worm-gear.
II. If the worm is double-threaded multiply its speed by 2
and divide the product by the number of teeth in the worm-gear.
Example 1.—A 00-tooth worm-gear is driven by a single-threaded
worm making 300 revolutions per minute; find its number of revolu¬
tions per minute.
Solution.—
—- = 5 rev. per min. Ans.
60
8
MECHANICAL CALCULATIONS
13
Example 2.—An 80-tooth worm-gear is driven by a double-threaded
worm making 160 revolutions per minute; find the number of revolu¬
tions per minute of the worm-gear.
c 160 X 2 A
Solution. — —--— = 4 rev. per mm. Ans.
oU
A worm is always a driver and reckoned as a one-tooth
gear if single-threaded and a two-tooth gear if double-
threaded.
14. A mangle gear is always a driven gear and its speed
is found in the same manner as that of an ordinary gear
except that its size is reckoned as somewhat larger than
it really is, because the driving pinion, while rounding each
end of the row of pegs, makes a half-revolution, which
moves the mangle but one peg.
To find how many complete cycles, or double revolutions,
a mangle gear will make per minute:
Rule. — Divide the product of the number of revolutions per
minute of the driving pinion and the number of teeth that it con¬
tains by the sum of twice the number of pegs in the mangle and
the number of teeth in the driving pinion diminished by 2.
Example. —A 10-tooth pinion gear making 216 revolutions per
minute drives a mangle gear with 176 teeth, or pegs; how many com¬
plete cycles per minute does the mangle gear make?
0 216 X 10 216 X 10 216 X 10
{solution. (176 X 2) + (10 - 2) ~ 352 + 8 ~ 360
complete cycles. Ans.
That is, the mangle gear would make 12 revolutions, 6 in one direc¬
tion and 6 in the other.
EXAMPLES FOR PRACTICE
1. A shaft making 30 revolutions per minute must drive a shaft
carrying a 40-tooth gear 21 revolutions per minute; what gear is
required as a driver? Ans. 28-tooth gear
2. A driving gear with 60 teeth makes 70 revolutions per minute
and drives a pinion gear of 18 teeth; find the speed of the pinion gear.
Ans. 233^ rev. per min.
3. The speed of the first driving shaft is 40 revolutions per minute;
the driving gears contain 48, 60, and 40 teeth, respectively; the driven
gears, 40, 48, and 40 teeth, respectively; find the speed of the last
driven shaft. Ans. 60 rev. per min.
14
MECHANICAL CALCULATIONS
§8
4. A double-threaded worm making 300 revolutions per minute
drives an 80-tooth worm-gear; how many revolutions per minute does
the worm-gear make? Ans. 7i rev. per min.
5. A mangle gear having 116 teeth is driven by a 10-tooth pinion
gear making 120 revolutions per minute; how many double revolutions
per minute does the mangle gear make? Ans. 5 double revolutions
6. A train of gears is composed of four drivers containing 32, 24,
40, and 16 teeth, respectively, and four driven gears containing 24, 16,
32, and 8 teeth, respectively; the first driver makes 16 revolutions per
minute; find the speed of the last driven shaft. Ans. 80 rev. per min.
7. A 72-tooth worm-gear is driven by a double-threaded worm
making 198 revolutions per minute; find the number of revolutions per
minute of the worm-gear. Ans. 5i rev. per min.
8. A 6-tooth pinion making 265 revolutions per minute drives an
83-tooth mangle gear; how many complete cycles does the mangle
gear make? Ans. 9 cycles
CONSTANTS
15. It often happens that though a machine contains a
more or less complicated train of gears, only one of them is
changed for alterations in the speed of the driven gear. This
gear is known as a change gear, and the arrangement of
the train is such that its size may readily be changed without
disturbing the other members of the train. If the change
gear is a driven gear, an increase in its size will diminish
the speed of the driven gear or shaft. If it is a driver, an
increase in its size will increase the speed of the driven gear
or shaft.
Where a train of gears is employed, the calculation of the
required size of change gear to produce a given speed of the
driven gear becomes rather long unless some method of
shortening the operation is adopted. This may be accom¬
plished by partly performing the operation and securing a
number that expresses the value of the train, excluding the
change gear, and that needs but one multiplication or divi¬
sion to obtain the desired speed or the desired size of change
gear; such a number is called a constant.
16. A constant number may be either a constant factor
or a co?istant dividend. A constant factor is a number
§8
MECHANICAL CALCULATIONS
15
that, when multiplied by the change gear, gives the speed
of the driven shaft of a train of gears and, when divided into
the speed of the driven shaft, gives the number of teeth in
the change gear. A constant dividend is a number that,
when divided by the number of teeth in a change gear,
gives the speed of the driven shaft of a train of gears and,
when divided by the speed of the driven shaft, gives the
number of teeth in the change gear.
For all speed calculations the constant number for a train
of gears is a constant factor if the change gear is a driver
and a constant dividend if the change gear is a driven gear.
Where a constant number is used in connection with some
result dependent on the action of a train of gears, it may be
either a constant factor or dividend, depending on whether
the value of the said result is increased or decreased by an
increase in the size of the change gear.
To find a constant:
Rule. —Perform the calculation of the train of gears in the
ordinary manner , using a theoretical change gear of one tooth.
Example. —A certain roll is driven as follows: The first driving
gear has 40 teeth and makes 390 revolutions per minute; this gear
drives a 39-tooth gear attached to a shaft on which there is also a
64-tooth gear driving a 32-tooth gear on a shaft. On this latter shaft
is fastened a 40-tooth gear that drives a 40-tooth gear on another shaft;
on this shaft a change gear drives a 128-tooth gear on the shaft of the
roll. What is the constant for finding the speed of the roll with various
change gears?
Solution. —In this case the constant number must be a constant
factor, as the change gear is a driver and an increase in its size increases
the speed of the roll.
390 X 40 X 64 X 40 X 1
39 X 32 X 40 X 128
6.25, constant factor. Ans.
Note.— If any change gear is used, its size multiplied by 6.25 in this case, will give
the speed of the roll; also, if any speed is desired, the required change gear can
be found by dividing the desired speed by 6.25.
EXAMPLES FOR PRACTICE
1. The constant factor for a certain train of gears is 7.5; what
change gear will give 150 turns of the driven shaft per minute?
Ans. 20-tooth gear
16
MECHANICAL CALCULATIONS
§8
2. Find the constant for the following train of gears, the first driver
making 220 revolutions per minute: drivers 24 , 42, 30, respectively,
and change gear; driven 18, 36, 24, and 42 teeth, respectively.
Ans. 10.185, constant
3. The constant dividend for a train of gears is 768; what will be
the speed of the driven gear when a 24-tooth change gear is used?
Ans. 32 rev. per min.
4. Find the constant for the following train of gears in which the
first driving gear makes 120 revolutions per minute: drivers, 30, 45, 90,
and 40 teeth, respectively, driven, 30, 60, 50, respectively, and change
gear. Ans. 6,480, constant dividend
RULES APPLYING TO LEVERS
17. To find the weight supported or the pressure exerted
at the weight end of a lever, the length of the weight arm,
the length of the power arm, and the power applied being-
known:
Rule. —Multiply the power by the length of the power arm
and divide the product by the length of the zveight arm.
Example. —A 25-pound weight is placed on a lever that is so con¬
nected as to exert a pressure on a pair of rolls; the weight is 4 feet
from the fulcrum of the lever and the rod connecting the lever with the
rolls is 1| feet from the fulcrum of the lever; find the pressure exerted.
25 X 4
Solution.— —— j — = 66f lb. pressure. Ans.
18. Any problem of levers may be solved by treating it
as a proportion in which the power is to the weight as the
weight arm is to the power arm; and, as in proportion
the product of the extremes is equal to the product of the
means, so the power times the power arm is equal to
the weight times the weight arm.
When dealing with compound levers it must be remembered
that the continued product of the power multiplied by the
power arms is equal to the continued product of the weight
multiplied by all the weight arms, every alternate arm of
the combination of levers, starting with the power arm,
being a power arm and every alternate arm, starting with
the weight arm, being a weight arm.
§8
MECHANICAL CALCULATIONS
17
Example. —A 40-pound weight (power) acts through the following
power arms: 4 feet, 3 feet, and 3 feet, respectively; the corresponding
weight arms being 3 feet, 2 feet, and 2 feet, respectively; what weight
is supported, or pressure exerted, at the extremity of the last
weight arm ?
40 X 4 X 3 X 3
120 lb. Ans.
Solution.—
3X2X2
YARN CALCULATIONS,
COTTON
SINGLE YARNS
NUMBERING SYSTEM
1. From the earliest times of the manufacture of yarns
it has been found necessary to adopt some system by means
of which the different sizes of yarns may be designated,
since, otherwise, the manufacturer would have no means of
readily distinguishing one yarn from another.
In the cotton system of numbering yarns, all counts are
based on the standard length of 840 yards, which is known
as 1 hank; in other words, a hank of cotton yarn contains
840 yards. That this is a constant number, never varying,
is a fact that should always be remembered. For example,
1 hank of a certain number of cotton yarn contains 840 yards,
and a single hank of any yarn either finer or coarser will
contain the same number of yards.
The method of numbering the yarns is that of calling a
yarn that contains 1 hank, or 840 yards, in 1 pound a No. 1
yarn. If the yarn contains 2 hanks, or 1,680 yards in
1 pound, it is known as a No. 2 yarn; if it contains 3 hanks,
or 2,520 yards, in 1 pound, it is known as a No. 3 yarn. Thus
it will be seen that the number of hanks (840 yards) that it
takes to weigh 1 pound determines the counts of the yarn.
The counts of a yarn are generally indicated by placing a
letter s after the figure representing the number of the yarn.
Thus, 26s shows the counts of a yarn and indicates that the
yarn contains 26 hanks (26 X 840 yards) in 1 pound.
For notice of copyright, see page immediately following the title page
§9
2
YARN CALCULATIONS, COTTON
§9
CALCULATIONS OF SINGLE YARNS
2. There are several calculations that become necessary
when dealing with single yarns, but all are based on one
equation, namely,
length of yarn, in yards _ ^
weight, in pounds X counts X standard
The standard will always be the same when dealing with
cotton (840 yards), and any one of the other items can be
found by simply substituting the values of the rest in the
above equation.
3. To find the counts of a yarn when the length and
weight are given:
Rule.— Divide the total length of yarn, expressed in yards,
by the weight, expressed in pounds, times the standard length.
Example.— If 168,000 yards of yarn weighs 5 pounds, what are
the counts?
Solution.—
168,000 (length of yarn, in yards)
■=—, —r-rr —, . w Q , , —y—yv = 40s, counts. Ans.
5 (weight, in pounds) X 840 (standard)
4. To find the weight of yarn when the length and
counts are known:
Rule. —Divide the length , in yards, by the counts times the
standard length.
Example. —What is the weight of 42,000 yards of number 5s yarn?
42,000 (length, in yards)
Solution.-
5 (counts) X 840 (standard)
= 10 lb. Ans.
5. To find the length of yarn when the weight and
counts are known:
Rule. —Multiply the weight, in pou?ids, counts, a?id standard
length together.
Example. —What is the length of yarn contained in a bundle that
weighs 8 pounds, the counts of the yarn being 26s?
Solution. — 8 (weight, in pounds) X 26 (counts) X 840 (standard)
= 174,720 yd. Ans.
$9
YARN CALCULATIONS, COTTON
3
6 . When numbering yarns several different weights and
names are used, for which reason it will be well to memorize
the two following tables. These tables include only those
terms used in yarn numbering.
TABLE I
The table of lengths is as follows:
li yards = 1 thread, or circumference of wrap reel.
120 yards = 80 threads = 1 skein, or lea.
840 yards = -560 threads = 7 skeins, or leas = 1 hank.
TABLE II
The table of weights is as follows:
27.34 grains = 1 dram.
437.5 grains = 16 drams = 1 ounce.
7,000 grains = 256 drams = 16 ounces = 1 pound.
The part of this table that is most frequently used and
which, therefore, should be carefully noted is that which
states that there are 7,000 grains in 1 pound.
SIZING ROVING AND YARN
7. From the time the cotton is placed in the form of
roving until it becomes the finished yarn, some method has
to be adopted in every well-regulated mill by means of which
the exact counts of the roving and yarn being run may be
known.
Roving is a term used to designate a strand of loosely
twisted fibers from which a yarn may be spun, while the
term yai*n is applied to a thread composed of fibers uniformly
disposed throughout its structure and having a certain
amount of twist for the purpose of enhancing its strength.
It is generally the custom to weigh a certain quantity from
each machine, at least once a day, and by this means ascer¬
tain whether the roving or yarn is being kept at the
required weight. This process is known as sizing, and is
a matter that should always be carefully attended to.
From the rules and explanations previously given it will
be plain that if 840 yards (1 hank) was always the length
4 YARN CALCULATIONS, COTTON §9
weighed, in order to learn the counts of the yarn, it would
simply be necessary to divide the weight, expressed in
Fig. 1
pounds, into 1 pound, or if expressed in grains, into 7,000
(the number of grains in 1 pound). It will readily be seen
Fig. 2
that to measure off 840 yards of yarn would not only require
considerable time, but would also produce an unnecessary
§9
YARN CALCULATIONS, COTTON
5
waste of material. To overcome these difficulties, when
sizing yarn, it is customary to measure off one-seventh of
840 yards, or 120 yards; to weigh this amount; and divide its
weight, in grains, into one-seventh of 7,000, or 1,000. The
result obtained in this manner will be the same as if 840 yards
was taken and the weight, in grains, divided into 7,000.
8 . The Wrap Reel. —When sizing yarns, an instrument
known as a wrap reel is used to measure the yarn. As its
name indicates, this instrument consists of a reel, generally
I 2 yards in circumference. The yarn is wound on this
reel and a finger indicates on a disk the number of yards
reeled. Fig. 1 shows an ordinary type of wrap reel, while
Fig. 2 shows yarn and roving scales. These scales are suit¬
able for weighing by tenths of grains.
Example. — 120 yards of yarn is reeled and found to weigh
40 grains; what are the counts?
Solution. — 1,000 -p 40 = 25s. Ans.
9. This basis for indicating the size of the yarn is also
used to designate the size of roving, although when sizing
roving a shorter length is used. It is customary in this case
to measure off one-seventieth of 840 yards, or 12 yards, and
divide the weight, in grains, of this length of roving into
one-seventieth of 7,000, or 100.
Example. — 12 yards of roving is found to weigh 20 grains; what
are the counts?
Solution. — 100 -p 20 = 5-hank. Ans.
The counts of roving are indicated in a somewhat different
manner from the counts of yarn. Thus, in this case, the
roving would be known as 5-hank ?vving, indicating that
5 hanks weigh 1 pound.
10. From the foregoing, it will be readily understood
that the counts of a yarn may be obtained by finding the
weight of any length of yarn and dividing the weight into a
certain number of grains, this number having the same pro¬
portion to 7,000 that the length of yarn weighed has to 840.
6
YARN CALCULATIONS, COTTON
9
As an aid to the student in any calculations that he may
have to deal with, the following lengths and dividends
are given:
TABLE III
If 480 yards is weighed, divide
If 240 yards is weighed, divide
If 120 yards is weighed, divide
If 60 yards is weighed, divide
If 40 yards is weighed, divide
If 30 yards is weighed, divide
If 20 yards is weighed, divide
If 15 yards is weighed, divide
If 12 yards is weighed, divide
If 10 yards is weighed, divide
If 8 yards is weighed, divide
If 6 yards is weighed, divide
If 4 yards is weighed, divide
If 3 yards is weighed, divide
If 2 yards is weighed, divide
If 1 yard is weighed, divide
weight in grains into 4,000.
weight in grains into 2,000,/
weight in grains into 1,000.
weight in grains into 500.
weight in grains into 333#.
weight in grains into 250.
weight in grains into 166f.
weight in grains into 125.
weight in grains into 100.
weight in grains into 83#.
weight in grains into 66#.
weight in grains into 50.
weight in grains into 33#.
weight in grains into 25.
weight in grains into 16#.
weight in grains into 8#.
EXAMPLES FOR PRACTICE
1. 50,400 yards of cotton yarn weighs 3 pounds; what are the counts?
Ans. 20s
2. 100,800 yards of cotton yarn weighs 10 pounds; what are the
counts? Ans. 12s
3. 100 skeins of cotton yarn, each containing 4,200 yards, weighs
20 pounds; what are the counts? Ans. 25s
4. What are the counts of cotton yarns 120 yards of which weighs,
respectively, 17, 21, 26, and 32 grains? *
Ans.
58.82s
47.61s
38.46s
131.25s
5. If a cop is known to contain 960 yards of cotton yarn and 35 of
these cops weigh 1 pound, what are the counts? Ans. 40s
6. What is the weight of 37,000 yards of Is cotton? Ans. 44.04 lb.
7. What is the weight, in ounces, of 10,000 yards of 36s cotton?
Ans. 5.29 oz. (practically)
8. How manyyards of 60s cotton in 24 pounds? Ans. 1,209,600 yd.
§9
YARN CALCULATIONS, COTTON
7
9. What is the length of cotton yarn in 50 pounds of 20s?
Ans. 840,000 yd.
10. A spool of 28s cotton yarn weighs 28 ounces. Find the length
of yarn on the spool if the spool itself weighs 8 ounces. Ans. 29,400 yd.
11. What is the weight of 336,000 yards of 40s cotton? Ans. 10 lb.
12. What is the weight of 151,200 yards of 60s cotton? Ans. 3 lb.
PLY YARNS
METHOD OF NUMBERING
11. Definition. —Very frequently during the process of
manufacturing yarns, it becomes necessary to twist two or
more threads together to form one coarser thread. Such
yarns are commonly known as ply yarns, also sometimes
called folded, or twisted, yarns. It should be understood
that when yarns are folded in this manner it does not change
their length, with the exception of a slight percentage of
contraction, which takes place during the twisting, but it does
change their weight.
The method of numbering cotton ply yarns is that of
giving the counts of the single yarns that are folded and
placing before these counts the number that indicates the
number of threads folded; thus, 2/40s indicates that two
threads of 40s single yarn are folded together, causing the
ply yarn to be equal in weight to a single 20s. As previ¬
ously stated, during the process of twisting, a slight con¬
traction takes place. Consequently, to make the resultant
counts 20s, the single yarns folded would have to be slightly
finer than 40s. However, this contraction will not be con¬
sidered in the rules and examples that follow, since it is so
slight that it is more a matter of experience than one of
mathematics.
8
YARN CALCULATIONS, COTTON
§9
CALCULATIONS OF PLY YARNS
FOLDED YARNS OF THE SAME COUNTS
12. It is not often the custom in mills to fold yarns of
different counts, since single yarns of the same number
make the best double, or ply, yarns. Consequently, when
yarns of the same counts are folded, in order to find the
counts of the resulting ply yarn, it is simply necessary to
divide the counts of the yarns folded by the number of
threads that constitute the ply yarn. For example, if three
threads of 90s cotton are folded to form a ply yarn, the
resultant yarn will be equivalent in weight to a single 30s
(90 4- 3 = 30).
The method of indicating the counts of this ply yarn is
3/90s. The rules for finding the counts, weight, and length
of ply yarns are similar to those given for finding the same
particulars of single yarns, with the exception that the
counts of the ply yarn do not indicate the actual counts of
the yarn, but instead indicate the counts of the yarns folded.
Consequently, when figuring to find these particulars, the
actual counts of the folded yarn must be taken. The
following examples will make this subject clearer:
Example 1.—What is the weight of 642,000 yards of 2-ply 40s
cotton yarn?
Solution.—
2-ply 40s cotton yarn is equal in weight to a single 20s.
642,000
20 X 840
= 38.21 lb. Ans.
Example 2.—What is the length of 20 pounds of 2-ply 60s cotton?
Solution. —The 2-ply 60s cotton is equal in weight to a single 30s.
20 X 30 X 840 = 504,000 yd. Ans.
Example 3.—What are the counts of a 2-ply yarn 352,800 yards of
which weighs 10 pounds?
Solution.—
_352,800
10 X 840
= 42s
This answer is the counts of the ply yarn considered as a single
thread, but since it is made from two threads of the same counts it will
be a 2-ply 84s. Therefore, 352,800 yd. of the 2-ply 84s cotton yarn
will weigh 10 lb. Ans.
§9
YARN CALCULATIONS, COTTON
9
FOLDED YARNS OF DIFFERENT COUNTS
13 . As was previously stated, it does not often occur that
different counts are folded together to form a ply yarn, since
singles of the same number make the best double, or ply,
yarns. However, this will sometimes happen, especially
when it is desired to make a fancy yarn, in which case
two yarns of different counts are often folded.
For an illustration, suppose that it is desired to find the
resultant counts of a 40s folded with a 20s. Take as a basis
840 yards of each yarn; then 840 yards of the 40s weighs
tV pound. 840 yards of the 20s weighs 2 V pound. Conse¬
quently, after these yarns are folded there will be 840 yards
of a ply yarn the weight of which is i 1 ? + 2 V = A pound.
The example now resolves itself into the following: What
are the counts of a yarn 840 yards of which weighs impound?
Since length divided by (weight times standard) equals
counts, then - 3 — ^0 —- = 13.33s, counts of the ply yarn.
To X 840
This example has been worked out to some length in
order to enable the student to thoroughly understand the
method of numbering ply yarns. A somewhat shorter
method is as follows:
14 . To find the resultant counts when two threads of
different numbers are folded:
Rule. —Multiply the tzvo counts together and divide the result
thus obtained by the sum of the counts.
Example. —What will be the counts of a yarn obtained by folding
a 40s with a 20s?
Solution. — ^ = 13.33s, counts. Ans.
40 -f 20
PLY YARNS COMPOSED OF MORE THAN TWO THREADS
15 . In many cases it will be found necessary to find the
counts of a ply yarn that is made from more than two single
threads. Under such circumstances a somewhat different
process must be followed. For example, suppose that
10
YARN CALCULATIONS, COTTON
9
one thread each of 24s, 36s, and 72s are folded to form a
ply yarn and it is required to ascertain the counts of the
resultant yarn. This may be found by following the rule
given and performing two operations as follows:
First find the counts of the yarn that will result from fold¬
ing the 24s with the 36s. — * -- = 14.4s. The example
24 + 36
then resolves itself into the following: What are the counts
of a ply yarn made from one thread of 14.4s and one of 72s.
14.4 X 72
14.4 + 72
12s. Ans.
16 . A somewhat shorter method than this, however, may
be applied to 3 or more ply yarns made from different counts.
Rule. — Take the highest counts and divide it by itself and by
each of the other counts. Add the results thus obtamed a?id
divide this result into the highest counts.
Example. —Same as in the preceding article.
Solution. — 72 -r- 72 = 1
72 -I- 36 = 2
72 -I- 24 = 3
6
72 -I- 6 = 12s. Ans.
17 . To find the resultant counts when more than one
end of the different counts are folded:
Rule .—Divide the highest counts by itself and by each of the
other counts. Multiply the result in each case by the number of
ends of that counts. Add the results thus obtained and divide
this result into the highest counts.
Example. — 4 ends of 80s and 3 ends of 60s are folded to form a
ply yarn; what are the resultant counts?
Solution. — 80 = 80 = 1; 1X4 = 4
80 = 60 = H; H X 3 = 4
8
80 -7- 8 = 10s, resultant counts. Ans.
18. Another calculation that will frequently occur when
dealing with ply yarns is that of finding the counts of a yarn
that must be folded with another to produce the given counts.
§9
YARN CALCULATIONS, COTTON
11
Rule.— To find what counts must be folded with another to
produce a given counts, nniltiply the two counts together and
divide by their difference.
Example. —What counts of yarn must be folded with a 50s to pro¬
duce a 30s ply yarn?
Solution.—
50 X 30
50 - 30
75s. Ans.
The student may prove this example by taking the two
single yarns, 50s and 75s, and finding the ply yarn resulting
from folding them, following the rule given for finding the
counts of ply yarn resulting from the folding of single yarns.
19 . Still another calculation that enters into ply yarns is
that of finding the weight of single yarns of different counts
that are to be twisted together to produce a given weight of
ply yarn.
Rule.— When tzvisting together two or more threads of dif¬
ferent counts , to find the weight of each required to produce
the giveti weight , find the counts resulting from folding the
two or more threads; then as the counts of one thread is to
the resulting counts so is the total weight to the weight required
of that thread.
Example.—I t is desired to produce 100 pounds of a ply yarn from
an 80s and a 32s; what weight of each single thread must be used in
order to produce the 100 pounds of ply yarn?
Solution.—
80 X 32
80 + 32
22.85s
32 : 22.85 = 100 : 71.40
Therefore, 71.40 lb. of 32s would be required.
Ans.
It will readily be seen that in order to find the weight of the other
thread it is only necessary to subtract the weight of the 32s from the
total weight, or 100 lb. But in order that this may be fully under¬
stood, the required weight of 80s will be worked out similarly to that
of 32s.
Resulting counts = 22.85; 80 : 22.85 = 100 : 28.56 lb. of 80s. Ans.
Note.— In the previous example the weight of the 80s yarn plus the weight of
the 32s yarn should equal the weight of the ply yarn, but owing to the use of decimals,
examples of this kind seldom give exact results. Thus, 71.40 pounds + 28.56 pounds.
= 99.96 pounds; whereas the total weight should be 100 pounds.
12
YARN CALCULATIONS, COTTON
§9
exampi.es for practice
1. If a thread of 40s and one of 60s are twisted together, what are
the counts of the resultant ply yarn? Ans. 24s
2. A 20s, 30s, and 60s are twisted together; what are the counts of
the ply yarn? Ans. 10s
3. What are the counts of a yarn that must be twisted with a 44s
to produce a ply yarn equal to a 20s? Ans. 36.66s
4. What weight of 40s must be twisted with a 20s to produce
200 pounds of ply yarn? Ans. 66.65 lb.
5. What weight of 100s, 80s, and 60s would be used in producing
500 pounds of ply yarn? r 127.65 lb. of 100s
Ans. { 159.56 lb. of 80s
1212.75 lb. of 60s
6. What counts of ply yarn will result if 18s, 24s, and 36s are
twisted together, and what weight of each will there be in 240 pounds
of the ply yarn?
Ans.
8s
106| lb. of 18s
80 lb. of 24s
53i lb. of 36s
7. A ply yarn is made from 1 end each of 10s, 12s, and 15s; what
are the counts of the ply yarn? Ans. 4s
8. A 3-ply yarn is made from 80s, 40s, and 30s and weighs
100 pounds; what weight does it contain of each counts of single yarn
and what are the counts of the ply yarn?
Ans.
14.117s
17.64 lb. of 80s
35.29 lb. of 40s
147.05 lb. of 30s
9. What will be the resultant counts if 2 ends of 40s and 1 end of
60s are twisted to make a ply yarn? Ans. 15s
10. What counts result from twisting a 60s with a 36s? Ans. 22.5s
9
YARN CALCULATIONS, COTTON
13
BEAM CALCULATIONS
20 . The yarn that forms the warp of a piece of cloth
after being spun is placed on what are known as section
beams, and is then run from these section beams on to the
loom beam, which is placed at the back of the loom. The
yarn is then ready to be made into cloth. Whether this
warp yarn is on the section beam or loom beam, there are
several calculations that it is necessary to understand and
that apply equally well to both cases.
21 . To find the counts of the yarn on a beam that con¬
tains one size of yarn, the weight, length, and number of
ends being given:
Rule .—Multiply the length , in yards , by the mimber of ends
on the beam and divide the result thus obtained by the weight ,
in pounds, times the standard mimber of yards to the pound.
Example.— A warp beam contains 2,400 ends of cotton, each
200 yards long; the weight of this yarn is 15 pounds. What are
the counts?
c 200 X 2,400 OQ .
Solution. — --rz - _ = 38.095s. Ans.
15 X 840
Note.—I t has been shown how it is possible to obtain the counts of a yarn when
its weight and length are given. The same rule is adopted in this case with the
exception that the length of only one end is given and, as there are several on a
beam, it is necessary to multiply the length of each end by the number of ends on the
beam in order to obtain the total length of yarn.
It should be carefully noted in connection with these
examples that in many cases the weight given will be found
to include not only the weight of the yarn, but also that of
the beam on which it is placed. Under such circumstances
it becomes necessary to deduct the weight of the beam from
the weight given, in order to find the true weight of the yarn.
22 . To find the number of ends on a beam when the
weight, length, and counts are given:
14
YARN CALCULATIONS, COTTON
§9
Rule. —Multiply the weight, in pounds, by the standard num¬
ber of yards and also by the counts. Divide this result by the
length of the warp, in yards.
Example. —The yarn on a beam weighs 200 pounds; the counts are
20s cotton, and the length of each end 1,400 yards. How many ends
does the beam contain?
Solution.—
200 X 840 X 20
1,400
2,400 ends. Ans.
23. To find the weight of yarn on a beam when the
length, number of ends, and counts are given:
Rule. —Multiply the length, in yards, by the number of ends
on the beam and divide the result thus obtained by the standard
number of yards times the counts of the yarn.
Example. —A beam contains 2,400 ends of 20s cotton, each end
being 500 yards long; find the weight of the yarn.
_ 500 X 2,400
Solution. — — = 71.428 lb. Ans.
o4U X
24. To find the length of warp on a beam when the
weight, number of ends, and counts are given:
Rule. —Multiply the weight, in pounds, by the standard
length and by the counts. Divide the result thus obtained by
the number of ends on the beam.
Example. —It is required to make a warp of 1,600 ends of 20s from
30 pounds of raw stock; what will be the length of the warp?
Solution.—
30 X 840 X 20
1,600
315 yd. Ans.
25. To find the length of warp that can be placed on
a beam:
Rule. —Find the weight of yarn that the beam will contain,
by weighing a beam of the same size when filled with yarn and
deducting the weight of the beam itself. Then apply the rule
in Art. 24.
Example 1.—A certain size beam when filled with yarn weighs
140 pounds, the beam itself weighing 50 pounds. What length of a
warp composed of 1,800 ends of 20s cotton can be placed on it?
§9
YARN CALCULATIONS, COTTON
15
Solution.—
140 — 50 = 90 lb. of yarn.
90 X 840 X 20
1,800
= 840 yd.
Ans.
Example 2.—Same beam as in example 1. How many yards of a
warp containing 3,000 ends of 40s cotton can be placed on it?
Solution.—
140 — 50 = 90 lb. of yarn.
90 X 840 X 40
3,000
1,008 yd.
Ans.
AVERAGE NUMBERS
26 . Very frequently different counts of yarn will be
found on the same beam or in the same warp. When this
is the case, it becomes necessary to find the average number ,
or average counts , of the different yarns before other calcula¬
tions can be made. By the term average number, or
average counts, of tlie warp, a count of yarn is meant
that will give the same weight as the warp yarn, considering
that an equal number of ends are used.
27 . To find the average number of the warp yarn when
more than one count is used:
Rule. — Divide the total number of ends of each count by its
own count. Add these results together and divide the result thus
obtained into the total number of ends in the warp.
Example. —Suppose that on the same beam there are 1,800 ends
of 60s and 800 ends of 40s; what counts of yarn will weigh the same as
the total weight of warp yarn, using the same number of ends in
both cases?
Solution. — 1800-5-60 = 30
8 0 0 -s- 40 = 20
2600 50
2,600 -T- 50 = 52s, average counts. Ans.
Proof. —That the above example is correct can be shown
by finding the weight of the warp yarn and also the weight
of 2,600 ends of 52s, considering the length to be 1 yard.
Applying the rule given for finding weight when ends,
length, and counts are g.ven, — —^ = .0357 lb.,
16
YARN CALCULATIONS, COTTON
9
= .0238 lb.; .0357 + .0238 = .0595 lb. = the total weight
of 1 yard of warp yarn.
Applying the same rule for finding the weight of 2,600
ends of 52s, X 2,600 _ ^ Ans.
840 X 52
28 . The above rule will be found to apply equally well
when more than two counts are used in the same warp.
Example.—W hat will be the average number of a warp that con¬
tains 200 ends of 20s, 1,000 ends of 40s, and 900 ends of 45s?
Solution. —Applying the rule just given,
200 -v- 20 = 10
1,000 -p 40 = 25
900 -v- 45 = 2 0
5 5
Total ends = 2,100; 2,100 -f- 55 = 38.18s, average number. Ans.
29 . In cases where the order of arranging the different
counts of yarn in the warp is known, but the total number of
ends is not given, the same rule will be found to apply if the
number of ends in the arrangement is considered as the total
number of ends.
Example. —A warp is arranged 48 ends of 36s and 2 ends of 10s.
Find the average number.
Solution.— 48 -f- 36 = 1.3 3 3
2 -s- 10 = .2
1.5 3 3
Total number of ends = 50; 50 1.533 = 32.615s, average number.
Ans.
FANCY WARPS
30. When more than one color of yarn is placed on the
same beam, it frequently becomes necessary to find the total
number of ends of each color, and also the weight of each
particular yarn.
31. To find the number of ends of each color of yarn on
a beam when the total number of ends are given:
Rule .—Multiply the number of ends of any one color in one
pattern by the total number of ends on the beam and divide by
the total ends in one pattern.
9
YARN CALCULATIONS, COTTON
17
Note. —One repeat of the colors in a warp is termed the pattern
of tlie warp. For an illustration, suppose that the ends on a beam
are arrauged 16 ends black, 8 ends white, 16 ends black, 8 ends gray.
This is known as the pattern of the warp, since all the ends are only
repeats of this arrangement.
Example. —Suppose there are 2,400 ends on a beam arranged as
stated in the note; how many ends of each color will there be?
Solution. — 16 ends black, 8 ends white, 16 ends black, and
8 ends gray, give 48 as the total number of ends in 1 pattern.
There are 32 ends of black in 1 pattern.
32 X 2,400 = 76,800; 76,800 -t- 48 = 1,600 ends of black on the beam.
Aus.
There are 8 ends of white in 1 pattern.
8 X 2,400 = 19,200; 19,200 h- 48 = 400 ends of white on the beam.
Ans.
There are 8 ends of gray in 1 pattern.
8 X 2,400 = 19,200; 19,200 -s- 48 = 400 ends of gray on the beam.
Ans.
After the total number of ends of each color is found in
this manner, in order to learn the weight of each color of yarn
it is simply necessary to apply the rule given for finding the
weight when the length, counts, and number of ends are
given.
EXAMPLES FOR PRACTICE
1. What is the weight of a warp 200 yards long, that contains 2,400
ends of 60s cotton? Ans. 9.52 lb.
2. What are the average counts in a warp made from 24 ends of
40s and 1 end of 8s cotton? Ans. 34.48s
3. A warp is arranged 3 ends black; 1, white; 7, black; 1, white;
counts, 24s cotton; 154 yards long; 2,400 ends; find the weight of each
color and the total weight of the warp. ( 15.27 lb. of black
Ans. <3.05 lb. of white
118.32 lb., total weight
4. A warp is arranged 24 ends of 48s and 1 end of 2-ply 10s cotton;
what are the average counts? Ans. 35.71s
5. What is the weight of warp yarn in a warp 400 yards long,
that contains 3,360 ends of 40s cotton? Ans. 40 lb.
6. 4 pounds of 25s cotton is made into a warp containing 875 ends;
what is the length? Ans. 96 yd.
18
YARN CALCULATIONS, COTTON
§9
7. What is the length of a warp made of 30s cotton and con¬
taining 1,800 ends, if the weight is 1 pound? Ans. 14 yd.
8. A warp is arranged 3 ends black, 1 end white, 1 black, 3 white,
and weighs 9 pounds; what is the weight of each kind of yarn if the
counts are 36s cotton? Ans. 4.5 lb.
9. What are the counts of warp yarn in a cotton warp 200 yards
long that contains 2,400 ends, and weighs 15 pounds? Ans. 38.09s
10. A cotton warp 120 yards long contains 4,480 ends, and weighs
40 pounds; what are the counts? Ans. 16s
11. How many ends are there in a warp 1,400 yards long and
weighing 200 pounds, the counts of the yarn being 20s cotton?
Ans. 2,400 ends
12. What will be the cost of a warp made of 30s cotton at 16 cents
per pound, if the warp is 500 yards long, and contains 1,800 ends?
Ans. $5.71
13. How many ends are there in a warp 700 yards long and
weighing 100 pounds, the counts being 40s cotton? Ans. 4,800 ends
TWIST IN YARNS
32. To enable the yarn, whether warp or filling, to with¬
stand the strain that is brought on it, it is necessary to insert
what is known as twist. Warp yarn contains more twist
per inch than filling yarn, since it has to withstand a greater
strain during the process of weaving. The common practice
as to the number of turns per inch to be placed in yarn varies
in different mills, but is always based on multiplying the
square root of the number being spun by a constant number.
In American mills, for ordinary warp yarns this constant
number is 4.75; for filling yarn, it is 3.25.
33. To find the twist to be inserted in any counts of
yarn:
Rule. —Extract the square root of the coiuits and multiply
the result by the standard adopted.
Example. —Find the turns per inch in number 25s ordinary warp.
Solution. — V25 = 5; 5 X 4.75 = 23.75, turns per in. Ans.
It should be understood that the constants given above
are not necessarily adopted in all cases, since varying
§9
YARN CALCULATIONS, COTTON
t9
circumstances will require different amounts of twist in the
yarn. These constants, however, form a basis on which to
determine the amount of twist that is to be placed in any
counts being spun. _
BREAKING WEIGHT OF COTTON WARP YARN
34 . An instrument that is commonly adopted for finding
the strength of warp yarns is shown in Fig. 3. In test¬
ing the strength of the yarn it
is the custom to wrap, or reel,
120 yards and place this yarn
in the form of a skein on the
hooks a,b, Fig. 3. By turn¬
ing the handle until the yarn
breaks, the breaking weight
is shown by means of the
finger and dial. When finding
the breaking weight of any
counts of warp yarn, the fol¬
lowing rule will be found to be
of advantage.
35 . To find the standard
breaking weight of warp yarns:
Rule. —Multiply the known
coiaits of the yarn by 5. Add the
constant 1,600. Divide the sum
by the counts of the yarn. The
answer is the approximate stand¬
ard breaking weight, in pounds.
Example. —What is the standard
breaking weight of 40s warp yarn?
40 X 5 + 1,600
Fig. 3
Solution.—
40
= 45 lb. Ans.
The same number of yarn from the same mill or from
the same mixing of cotton will not show an absolutely
uniform breaking strength. This being the case it can
readily be understood that yarns from different mills or
20
YARN CALCULATIONS, COTTON
§9
from different mixings of cotton will always show a vari¬
ation, which should, however, be reduced as much as
possible.
TABLE IV
BREAKING WEIGHT OF AMERICAN WARP YARNS, PER
SKEIN. WEIGHT GIVEN IN POUNDS AND TENTHS
Num-
Breaking
Num-
Breaking
Num-
Breaking
Num-
Breaking
ber
Weight
ber
Weight
ber
Weight
ber
Weight
I
26
66.3
51
36.6
76
25.8
2
27
63.6
52
36.1
77
25-5
3
530.0
28
61.3
53
35-5
78
25-3
4
410.0
29
59-2
54
34-9
79
24.9
5
330.0
30
57-3
55
34-4
80
24 .6
6
275.0
31
55-6
56
33-8
81
24 -3
7
237.6
32
54-o
57
33-4
82
24.0
8
209.0
33
52.6
58
32.8
83
23-7
9
186.5
34
51.2
59
32-3
84
23-4
10
168.7
35
50.0
60
3i-7
85
23.2
11
1 54 -1
36
48.7
61
3i-3
86
22.8
12
142.0
37
47.6
62
30.8
87
22.6
13
I3I-5
38
46.5
63
30.4
88
22.4
14
122.8
39
45-5
64
30.0
89
22.2
15
1 15 -1
40
44.6
65
29.6
90
22.0
16
108.4
4i
43-8
66
29.2
91
21.7
1 7
102.5
42
43-o
6 7
28.8
92
21.5
18
97-3
43
42.2
68
28.5
93
21-3
i9
92.6
44
41.4
69
28.2
94
21.2
20
88.3
45
40.7
70
27.8
95
21.0
2 I
83.8
46
40.0
7 1
27.4
96
20.7
22
79-7
47
39-3
72
27.1
97
20.5
23
75-9
48
38.6
73
26.8
98
20.4
24
72.4
49
37-9
74
26.5
99
20.2
25
69.2
50
37-3
75
26.2
100
20.0
Table IV gives what is known as Draper’s standard
for the breaking weight of American warp yarns and is
9
YARN CALCULATIONS, COTTON
21
generally used in the United States as a standard, being based
on breaking weights of samples of yarn taken from a large
number of mills and over extended periods. This table
may be used for reference, although the rule given should
be memorized for use when it is not convenient to refer
to the table.
METRIC SYSTEM OF NUMBERING
YARNS
36 . In recent years there has been from time to time
considerable agitation along the lines of adopting one sys¬
tem of numbering all classes of yarns used in textile manu¬
facturing. The objects of these movements are: first, to
bring about the unification of the method of stating the
degrees of fineness of the yarns from all varieties of fibers
used in the textile industry in the whole world; second, to
adopt a method that would be practical in every way.
The advantages of such a system as this are apparent.
The chief objection to it is that, from long usage, the
methods at present adopted are too well developed for a
single corporation or a single country to take on itself such
a reform, without being assured that its neighbors and
competitors would simultaneously do the same thing, aL
a mill instituting such a reform would be at a disadvantage
in the markets.
The method usually set forth is that of numbering all
classes of yarns by what is known as the metric system,
and consists of considering 1 meter of No. 1 yarn to weigh 1
gram, the meter being the unit of length in the metric sys¬
tem and the gram being the unit of weight. The only parts
of the metric system that it is necessary to understand in
order to convert one system into the other, are the equiva¬
lents of the meter and the gram, which are as follows:
1 yard = .914 meter; 1 pound = 453.59 grams
37 . To find the number of yarn in any present standard
system that corresponds to the number of yarn in the metric
standard system:
22
YARN CALCULATIONS, COTTON
9
Rule. —Multiply the counts , given in the metric system , by
453.59 (grams in 1 pound) and divide by the standard number
of yards to the pound in the present system multiplied by .914
(meter in 1 yard).
Example. —A cotton yarn numbered according to the metric system
is marked 40s. Find the counts in the present system.
0 40 X 453.59 00 ao .
Solution.— n ,„ s „ — 23.631s. Ans.
840 X .914
38. To find the number of yarn in the metric standard
system that corresponds to the number of yarn in any
present standard system:
Rule. —Miiltiply the counts , given in the present system , by
the preseiit standard number of yards to the pound and by .914
{meter in 1 yard) and divide by 453.59 {grams in 1 pound).
Example. —A cotton yarn numbered according to the present sys¬
tem is marked 46s. Find the counts in the metric system.
46 X 840 X .914
77.86s. Ans.
Solution.—
453.59
YARN CALCULATIONS,
WOOLEN AND WORSTED
WORSTED YARNS
SYSTEM OF NUMBERING SINGEE YARNS
1. From the earliest times of the manufacture of yarns
it has been found necessary to adopt some system by means
of which the different sizes of yarns may be designated,
since, otherwise, the manufacturer would have no means of
readily distinguishing one yarn from another. The sizes
of different yarns are known as the numbers, or counts, of
the yarns. The words number and counis in all cases repre¬
sent a certain length for a certain weight.
In the worsted system of numbering yarns, all counts are
based on the standard of 560 yards, which is known as
1 hank. In other words, a hank of worsted yarn contains
560 yards. That this is a standard number and never varies
is a fact that should always be remembered. For example,
1 hank of a certain number of worsted yarn contains 560
yards and a hank of finer or coarser yarn also contains the
same number of yards.
The method of numbering the yarns is that of calling a yarn
that contains 1 hank, or 560 yards, in 1 pound a No. 1 yarn.
If the yarn contains 2 hanks, or 1,120 yards, in 1 pound, it is
known as a No. 2 yarn; if it contains 3 hanks, or 1,680 yards,
in 1 pound, it is known as a No. 3 yarn. Thus it will be seen
For notice of copyright, see page immediately following the title page
2 YARN CALCULATIONS §10
that the number of hanks (560 yards) that it takes to weigh
1 pound determines the counts of the yarn.
The counts of a yarn are generally indicated by placing
the letter s after the figure representing the number of the
yarn. Thus, 26s shows the counts of a No. 26 yarn and
indicates that the yarn contains 26 hanks (26 X 560 yards)
in 1 pound.
CALCULATIONS OF SINGLE YARNS
2. There are several calculations that must be considered
when dealing with single worsted yarns, but all are based on
one equation, namely,
_ length of yarn, in y ards _ _ ^
weight, in pounds X counts X standard
The standard will always be the same when dealing with
worsted (560 yards), and any one of the other items can be
found by simply substituting the values of the rest in the
above equation.
3. To find the counts of a yarn when the length and
weight are given:
Rule .—Divide the total length of yarn , expressed in yards ,
by the product of the weight , expressed in pounds,and the standard
length .
Example. —If 168,000 yards of yarn weighs 10 pounds, what are
the counts?
Solution.—
168,000 (length of yarn, in yards)
10 (weight, in pounds) X 560 (standard)
= 30s, counts.
Ans.
4. To find the weight of yarn when the length and counts
are known:
Rule .—Divide the length , in yards , by the product of the
counts a?id the standard length.
Example. —What is the weight of 42,000 yards of number 5s yarn?
42,000 (length, in yards)
5 (counts) X 560 (standard)
15 lb. Ans.
Solution.—
§10
WOOLEN AND WORSTED
3
5 . To find the length of yarn when the weight and counts
are given:
Rule. —Multiply the weight, in pounds, the counts, and the
standard length together.
Example. —What is the length of yarn contained in a bundle that
weighs 8 pounds, the counts of the yarn being 26s?
Solution. — 8 (weight, in pounds) X 26 (counts) X 560 (standard)
= 116,480 yd. Ans.
6. When numbering yarns several different weights and
also names for lengths will be found to be used, for which
reason it will be well for the student to memorize the two
following tables. These tables include only those terms
used in yarn numbering.
TABLE I
The table of lengths is as follows:
1 yard = 1 thread (circumference of reel).
80 yards = 80 threads = 1 lea.
560 yards = 560 threads = 7 leas = 1 hank.
TABLE II
The table of weights is as follows:
27.34 grains = 1 dram.
437.5 grains = 16 drams = 1 ounce.
7,000. grains = 256 drams = 16 ounces = 1 pound.
The part of this table that is most frequently used and
which, therefore, should be carefully noted is that which
states that there are 7,000 grains in 1 pound.
SIZING WORSTED YARNS
7. It is generally the custom in mills to weigh a certain
quantity of roving or yarn from each machine at least once
a day and by this means ascertain whether it is being kept
at the required weight. This process is known as sizing
and is a matter that should always be very carefully
attended to.
From the rules and explanations previously given it will
be plain that if 560 yards, or 1 hank, were always the length
4
YARN CALCULATIONS,
§10
weighed, in order to learn the counts of the yarn it would
simply be necessary to divide the weight, expressed in
pounds, into 1 pound, or if expressed in grains, into 7,000
(the number of grains in 1 pound). It will readily be seen
that to measure off 560 yards of yarn would not only require
considerable time but would also produce an unnecessary
waste of yarn. To overcome these difficulties, when sizing
yarn it is the custom to measure off one-seventh of 560
yards, or 80 yards; to weigh this amount; and divide its
weight, in grains, into one-seventh of 7,000, or 1,000.
The result obtained in this manner will be the same as
though 560 yards were taken and its weight, in grains,
divided into 7,000.
8. The Wrap Reel. —When sizing yarns, an instrument
known as a wrap reel is used to measure the yarn. As its
name indicates, this instrument consists of a reel, generally
1 yard in circumference. The yarn is wound on this reel
and a finger indicates on a disk the number of yards reeled.
Fig. 1 shows an ordinary type of wrap reel, while Fig. 2
shows yarn and roving scales. These scales are suitable for
weighing by tenths of grains.
§10
WOOLEN AND WORSTED
5
Example. — 80 yards of yarn is reeled and found to weigh
20 grains; what are the counts?
Solution.— 1,000 4 - 20 = 50s. Ans.
9 . From the foregoing the student will readily under¬
stand that the counts of a yarn may be obtained by finding
the weight of any length of yarn and dividing the weight into
a certain number of grains, this number having the same
proportion to 7,000 that the length of yarn-weighed has to 560.
Fig. 2
As an aid to the student in any calculations that he may have
to deal with, the following lengths and dividends are given:
TABLE III
If 320 yards
If 160 yards
If 80 yards
If 60 yards
If 40 yards
If 30 yards
If 20 yards
If 15 yards
If 10 yards
If 8 yards
If 6 yards
If 4 yards
If 3 yards
If 2 yards
If 1 yard
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
is weighed,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
divide weight,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
in grains,
into 4,000.
into 2 , 000 .
into 1 , 000 .
into
750.
into
500.
into
375.
into
250.
into
187*.
into
125.
into
100 .
into
75.
into
50.
into
CO
into
25.
into
12 *.
6 YARN CALCULATIONS, §10
SIZING WORSTED ROVING
10 . Roving is a term used to designate a strand of
loosely twisted fibers from which a yarn may be spun, while
the term yarn is applied to a thread composed of fibers uni¬
formly disposed throughout its structure and having a certain
amount of twist for the purpose of enhancing its strength.
It is usually customary to designate the size of worsted
roving by the number of drams that 40 yards weighs; thus,
if 40 yards of a certain roving weighs 3.2 drams, it is known
as a 3.2-dram roving. When very fine yarns are being made,
however, the roving is generally designated by the weight,
in drams, of 80 yards, this length being taken because of the
greater accuracy obtained by weighing a greater length. In
sizing the roving, a scale capable of weighing drams and
tenths of drams is commonly used, but should it be neces¬
sary to use a grain scale, the measurements given in Table II
would prove convenient.
Since worsted yarn is numbered on the basis of the number
of hanks of 560 yards in 1 pound and roving by the number
of drams that 40 or 80 yards weighs, it is sometimes desir¬
able to find the counts of a roving.
11 . To find the counts of a roving when its weight, in
drams, is known:
Rule.— If the weight , in drams, of 40 yards is known, divide
this weight into IS.3. If the weight , in drams, of 80 yards is
known, divide this weight into 36.6.
Example. —What are the counts of worsted roving, 40 yards of
which weighs 6.1 drams?
Solution.— 18.3 4- 6.1 = 3s. Ans.
Note.— 18.3 and 36.6 are known as jrauge points and are obtained by multi¬
plying the drams in 1 pound by the number of yards weighed (40 or 80) and dividing
by the number of yards in 1 hank, thus:
266 V 40
-= 18.285. practically 18.3
560
256 V 80
= 36.57, practically 36.6
560
From this it will be seen that 40 vards of number Is weighs 18.3 drams and
80 yards of number Is weighs 36.6 drams, which is the true significance of the
gauge points.
10
WOOLEN AND WORSTED
7
EXAMPLES FOR PRACTICE
1 . 16,800 yards of worsted yarn weighs 1 pound; what are the
counts? Ans. 30s
2. 33,600 yards of worsted yarn weighs 5 pounds; what are the
counts? Ans. 12s
3. 80 skeins of worsted yarn, each containing 2,100 yards, weighs
20 pounds; what are the counts? Ans. 15s
4. What are the counts of worsted yarns 80 yards of which weighs,
respectively, 17, 21, 26, and 32 grains?
Ans. ■
58.82s
47.61s
38.46s
31.25s
5. If a cop is known to contain 960 yards of worsted yarn
and 35 of these cops weigh 1 pound, what are the counts? Ans. 60s
6 . What is the weight of 37,000 yards of Is worsted? Ans. 66.07 lb.
7. What is the weight, in ounces, of 10,000 yards of 36s worsted?
Ans. 7.93 oz.
8 . How many yards of 40s worsted are there in 24 pounds?
Ans. 537,600 yd.
9. What is the length of yarn in 25 pounds of 35s worsted?
Ans. 490,000 yd.
10. What is the weight of 168,000 yards of 20s worsted?
Ans. 15 lb.
PLY YARNS
METHOD OF NUMBERING
12. Definition. —Very frequently during the process of
manufacturing yarns, it becomes necessary to twist two or
more threads together to form one coarser thread. Such
yarns are commonly known as ply yarns, also as folded, or
twisted, yarns, and sometimes as double and twist yarns.
It should be understood that when yarns are folded in this
manner it does not change their length, with the exception
of a slight percentage of contraction, which takes place
during the twisting, but it does change their weight.
The method of numbering worsted ply yarns is that of
giving the counts of the single yarns that are folded and
placing before these counts the number that indicates the
8
YARN CALCULATIONS,
§10
number of threads folded. Thus, 2 '40s indicates that two
threads of 40s single yarn are folded together, causing the
ply yarn to be equal in weight to a single 20s. As previ¬
ously stated, a slight contraction takes place during the
process of twisting; consequently, to make the resultant
counts 20s, the single yarns folded must be slightly finer than
40s. This contraction, however, will not be considered in the
rules and examples that follow, since it is so slight that it is
more a matter of experience than one of mathematics.
CALCULATIONS OF PLY YARNS
13. Folded Yarns of tlie Same Counts.—It is not
often the custom in mills to fold yarns of different counts,
since single yarns of the same number make the best double,
or ply, yarns. Consequently, when yarns of the same counts
are folded, in order to find the counts of the resultant ply
yarn, it is simply necessary to divide the counts of the
yarns folded by the number of threads that constitute the
ply yarn. Thus, for example, if three threads of a 45s are
folded to form a ply yarn, the resultant yarn will be equiv¬
alent in weight to a single 15s (45 -4- 3 = 15).
The method of indicating the counts of this ply yarn
is 3/45s. The rules for finding the counts, weight, and
length of ply yarns are similar to those given for finding the
same particulars of single yarns, with the exception that the
counts of the ply yarn do not indicate the actual counts of
the yarn, but instead indicate the counts of the yarns folded;
consequently, when figuring to find these particulars, the
actual counts of the ply yarn must be taken.
Example 1.—What is the weight of 642,000 yards of 2-ply 40s?
Solution. —A 2-ply 40s is equal in weight to a single 20s.
642,000
20 X 560
57.32 lb. Ans.
Example 2.—What is the length of 20 pounds of 2-ply 60s?
Solution.— 2-ply 60s is equal in weight to a single 30s.
20 X 30 X 560 = 336,000 yd. Ans.
§10 WOOLEN AND WORSTED 9
Example 3.—What are the counts of a 2-ply yarn, 176;400 yards of
which weighs 10 pounds?
o 176,400
Solution.— - ■ ’ - = 31.5s. Ans.
10 X 5b0
This answer is the counts of the ply yarn considered as a single
thread, but since it is made from two threads of the same counts it
will be a 2-ply 63s. Therefore, 176,400 yd. of the 2-ply 63s will weigh
10 lb.
14 . Folded Yarns of Different Counts. —As was
stated, it does not often occur that different counts are folded
together to form a ply yarn, since singles of the same num¬
ber make the best double, or ply, yarns. However, this will
sometimes occur, especially when it is desired to make a
fancy yarn, in which case two yarns of different counts are
often folded.
Suppose, for illustration, that it is desired to find the result¬
ant counts of a 40s worsted folded with a 20s worsted. Take
as a basis 560 yards of each yarn; then 560 yards of the 40s
weighs iV pound, and 560 yards of the 20s weighs 2 V pound.
Consequently, after these yarns are folded, there will be
560 yards of a ply yarn the weight of which is to + 2 V = 1 \
pound.
The example now resolves itself into the following: What
are the counts of a yarn 560 yards of which weighs -40 pound?
Since length divided by (weight times standard) equals
560
counts, then -5 -= 13.33s, counts of the ply yarn.
it) X 560
This example has been worked out to some length in
order to enable the student to understand thoroughly the
method of numbering ply yarns. The following, however,
is a somewhat shorter method.
15 . To find the resultant counts when two yarns of dif¬
ferent numbers are folded:
Rule. — Multiply the two counts together and divide the result
thus obtained by the sum of the counts.
Example.— Same as in Art. 14.
40 v 90
Solution.— i/TroTi = 13.33s. Ans.
40 + 20
10
YARN CALCULATIONS,
10
16. Ply Yarns Composed of More Than Two
Threads. —In many cases it will be found necessary to find
the counts of a ply yarn that is made from more than two
single threads. Under such circumstances it will be neces¬
sary to follow a somewhat different process. For example,
suppose that three single threads of 24s, 36s, and 72s,
respectively, are folded to form a ply yarn and that it is
required to ascertain the counts of the resultant yarn. This
may be found by following the rule previously given and
performing two operation's as follows:
First find the counts of the yarn that will result from
= 14.4s. The
folding the 24s with the 36s. Thus:
example now resolves itself into the following: What are
the counts of a ply yarn made from one thread of 14.4s
and one of 72s? . 14-4 X 72 _ ^ Ans.
14.4 + 72
A somewhat shorter method than this, however, may be
applied to 3 or more ply yarns made from different counts.
Rule. —Take the highest counts and divide it by itself and
by each qf the other counts. Add the results thus obtained and
divide this result into the highest cojcnts.
Example.— Same as above.
Solution. — 72 h- 72 = 1
72 -s- 36 = 2
72 -j- 24 = 3
6
72 4- 6 = 12s. Ans.
17. To find the resultant counts when more than one end
of the different counts are folded:
Rule. —Divide the highest counts by itself and by each of the
other counts. Multiply the result in each case by the number of
ends of that counts. Add the results thus obtained and divide
this result into the highest counts.
Example. — 4 ends of 80s and 3 ends of 60s are folded to form a
ply yarn. What are the resultant counts?
§10
WOOLEN AND WORSTED
11
Solution.— 80 -s- 80 = 1; 1 X 4 = 4
80 4- 60 = H; 1£ X 3 = 4
8
80 -f- 8 = 10s, resultant counts. Ans.
Another calculation that will frequently occur when dealing
with ply yarns is that of finding the counts of a yarn that
must be folded with another to produce the given counts.
18. To find what counts must be folded with another to
produce a given counts:
Rule. —Multiply the two counts together and divide by their
difference.
Example.— What counts of yarn must be folded with a 50s to
produce a 30s ply yarn?
Solution.—
50 X 30
50 - 30
= 75s.
Ans.
This example may be proved by taking the two single yarns
50s and 75s and finding the ply yarn resulting from folding
them, following the rule previously given for finding the
counts of ply yarn resulting from the folding of single yarns.
Still another calculation that enters into ply yarns is that
of finding the weight of single yarns to produce a given
weight of ply yarn when twisting together two threads of
different counts.
19. When twisting together two or more threads of dif¬
ferent counts, to find the weight of each required to produce
the given weight:
Rule. —Find the counts resulting from folding the two or
more threads; then as the counts of one thread is to the resultant
counts so is the total weight to the weight required of that thread.
Example. —It is desired to produce 100 pounds of a ply yarn from
an 80s and a 32s; what weight of each single thread must be used?
o!! = 22.85s, resultant counts.
oU -f- oZ
32 : 22.85 = 100 : at
22.85 X 100
Solution.—
x =
32
= 71.40 lb. of 32s. Ans.
It will readily be seen that in order to find the weight of the other
thread it is only necessary to subtract the weight of the 32s from the
12
YARN CALCULATIONS,
§10
total weight, or 100 lb.; but, in order to enable the student to fully
understand this rule, the required weight of the 80s will be worked out
similarly to that of the 32s. Thus:
x =
Resultant counts = 22.85s
80 : 22.85 = 100 : x
22.85 X 100
80
= 28.56 lb. of 80s.
Ans.
Note.—I n the previous example the weight of the 80s yarn plus the weight of the
32s yarn should equal the weight of the ply yarn, but owing to the use of decimals,
examples of this kind seldom give exact results. Thus. 71.40 pounds + 28.56 pounds
= 99.96 pounds: whereas the total weight should be 100 pounds.
EXAMPLES FOR PRACTICE
1. If a thread of 40s and one of 60s are twisted together, what are
the counts of the resultant ply yarn? Ans. 24s
2. A 20s, 30s, and 60s are twisted together; what are the counts of
the ply yarn? Ans. 10s
3. What are the counts of a thread that must be twisted with a 44s
to produce a ply yarn equal to a 20s? Ans. 36.66s
4. What weight of 40s must be twisted with a 20s to produce
200 pounds of ply yarn? Ans. 66.65 lb.
5. What weight of 100s, 80s, and 60s would be used in producing
500 pounds of ply yarn? r 127.65 lb. of 100s
Ans.< 159.56 lb. of 80s
(212.75 lb. of 60s
6. What counts of ply yarn will result if 18s, 24s, and 36s are
twisted together, and what weight of each will there be in 240 pounds
of the folded yarn? r 106f lb. of 18s
Ans. 8s< 80 lb. of 24s
I 53^ lb. of 36s
7. A ply yarn is made from 1 end each of 10s, 12s, and 15s; what
are the counts of the ply yarn? Ans. 4s
8. A 3-ply yarn is made from 80s, 40s, and 30s, and weighs
100 pounds; what weight does it contain of each count of yarn and
what are the counts of the ply yarn? f 17.64 lb. of 80s
Ans. 14.117s{ 35.29 lb. of 40s
147.05 lb. of 30s
9. What will the resultant counts be if two ends of 40s and one of
60s are twisted to make a ply yarn? Ans. 15s
10. What counts result from twisting a 60s with a 36s? Ans. 22.5s
§10
WOOLEN AND WORSTED
13
WOOLEN YARNS
SYSTEMS OF NUMBERING SINGLE YARNS
RUN SYSTEM
20. The common method of numbering woolen yarns is
by what is known as the run system. With this system
«
the number of runs that weigh 1 pound indicates the size of
the yarn, a run being 1,600 yards. For example, if a woolen
yarn is a 4-run yarn, it will contain 4 X 1,600 yards to
1 pound.
It will be seen that this method of numbering the yarn is
very similar to that explained in connection with worsted
yarn, the only difference being that 1,600 yards in this case
is taken as the standard instead of 560, which is the standard
for worsted. Consequently, the rules given for finding the
length, weight, and counts of worsted yarns will be found to
apply equally well in this case with the exception that 1,600 will
be used in place of 560. However, in order that the student
may thoroughly understand the application of these rules,
an example of each will be given.
Example 1.—A woolen yarn 72,000 yards in length is found to weigh
10 pounds; what run is it?
.Solution.- Io VX600 = 4i ' ru “- Ans '
Example 2.—What is the weight of 64,000 yards of 2-run woolen
yarn?
e 64,000 on .
SOLUTION.- 210^00 = 20 ,b ' Ans -
Example 3.—What is the length of 5 pounds of a 6-run yarn?
Solution.— 5 X 6 X 1,600 = 48,000 yd. Ans.
14
YARN CALCULATIONS,
§10
21. A very convenient method of finding the weight,
run, or length of a woolen yarn when the weight is
expressed in ounces may be explained as follows:
Example 1.—What is the weight, in ounces,
51-run yarn?
Solution.—
8,800
5.5 X 1,600
1 lb.
of 8,800 yards of a
But it will be noticed that it is desired to obtain the weight
in ounces; therefore, the result in pounds must be multiplied
by 16, which gives 16 ounces as the answer. If, instead of
multiplying the answer by 16, a new standard is obtained by
dividing 1,600 by 16, the result will be the same but the
process will be considerably shortened. Therefore, when¬
ever the weight is expressed in ounces the standard
employed should be 100.
Example 2.—What is the weight, in ounces, of 5,250 yards of a
3-run yarn?
5,250 1(7l
Solution. — ^ ^ = 17| oz. Ans.
Example 3.—
run is it?
Solution.—
4,000 yards of woolen yarn weighs 5 ounces.
4,000
5 X 100
8-run. Ans.
What
Example 4.—How many yards are there in 10 ounces of a 3-run
yarn ?
Solution. — 10 X 3 X 100 = 3,000 yd. Ans.
CUT SYSTEM
22. In the vicinity of Philadelphia the counts of woolen
yarns are based on what is termed the cut system, which
has for its standard length 300 yards. In all other respects
the calculations are similar; consequently, the rules given
will apply in this case with the exception that 300 is used as
the standard.
Example 1.—What is the size, in the cut system, of a woolen yarn
36,000 yards of which weighs 10 pounds?
36,000
10 X 300
Solution.—
= 12-cut. Ans.
WOOLEN AND WORSTED
15
810
Example 2.—What is the weight of 66,000 yards of a 20-cut yarn?
c 66,000
Solution. 20 X 300 = U lb ' Ans -
Example 3.—What is the length of 10 pounds of a 24-cut woolen
yarn?
Solution.— 10 X 24 X 300 = 72,000 yd. Ans.
SIZING WOOLEN ROVING ANII YARN
23. Both woolen roving (in some mills and districts this
term is corrupted into roping) and yarn are sized according
to the number of runs or cuts in 1 pound. In sizing both
the yarn and roving a scale known as a run, or cut, scale is
commonly used. This scale, shown in Fig. 3, consists sim¬
ply of a brass beam that is carried by an upright standard
and so arranged that when a given number of yards of
roving or yarn is placed on the beam in such a position as
to balance (the beam being out of balance when not in
use), the position of the yarn will indicate its size on a
graduated scale on the beam. In order that the reading
may be as accurate as possible, the yarn or roving to be
weighed is twisted into a small bunch and suspended by
a single thread. Run or cut scales are usually made to
indicate the correct size from 50 yards of yarn or roving,
but they may be made to indicate the size with any desired
number of yards.
16
YARN CALCULATIONS,
§10
If other than the number of yards for which the scale is
graduated is used in weighing, the reading of the scale
must be corrected. In order to do accurate work, the
equilibrium of a run or cut scale should be adjusted with
a known weight or with a known size of yarn. This can be
accomplished by turning to the right or to the left the set¬
screw in the heavy end of the beam.
When a mill is not equipped with a run scale, the size of
the yarn must usually be determined by means of a grain
scale. In most mills 20 yards of yarn or roving is measured
off for sizing, but in other mills 50 yards, 100 yards, or
various other lengths may be used.
TABLE IV
Runs
Grains
Runs
Grains
1
Runs
Grains
JL
2
175.00
5
17.50
10
8.75
e
8
140.00
5 T
16.66
1O4
8-53
3
4
116.66
52
15.90
1O2
8-33
7
8
100.00
5 f
15.21
107
8.13
i
87.50
6
14.58
11
7-95
14
70.00
67
14.00
I I 4
7.77
1-2
58.33
62
13.46
1 ii
7.60
if
50.00
67
12.96
Ilf
7-44
2
43.75
7
12.50
12
7.29
27
38.88
-X
7 4
12.06
124
7.14
27
35-00
72
11.66
122
7.00
2!
31.81
7 f
11.29
I2f
6.86
3
29.16
8
10.93
13
6.73
3T
26.92
8|
10.60
I 3 T
6.60
32
25.00
8i
10.29
132
6.48
34
23.33
8f
10.00
1 3 f
6.36
4
21.87
9
9.72
14
6.25
44
20.58
9 i
9-45
142
6.03
42
19.44
92
I ^
9.21
15
5.83
4f
18.42
3
94
8.97
1
§10
WOOLEN AND WORSTED
17
24. To find the size of a woolen roving or yarn when
the weight of a given number of yards is known:
Rule. — Divide the number of yards weighed, multiplied by
7,000 , by the standard number ( 1,600 or 300), mtiltiplied by the
weight, in grains.
Example. —If 20 yards of woolen yarn weighs 25 grains, what is
the size of the yarn?
c 20 X 7,000 Q1
Solution.— ■ gn _ w og = 3 i-run. Ans.
I,b00 X 25
The table on page 16 gives the size, in runs, and the weight,
in grains, of 20 yards of woolen roving or yarn. If any num¬
ber of yards other than 20 is weighed, the reading of the
table will have to be corrected. For instance, if 100 yards
of yarn is found to weigh 175 grains, the size of the yarn is
not l-run but five times (100 20 = 5) i-run, or 2i-run.
PLY YARNS
25. Ply yarns made from woolen threads are num¬
bered exactly the same as worsted ply yarns; consequently,
the rules given in connection with worsted ply yarns will be
found to apply equally well when dealing with ply yarns
made of woolen with the exception, of course, that in each
case the standard number of yards to the pound used in the
woolen system must be adopted.
EXAMPLES FOR PRACTICE
1. Woolen yarn weighing 1 pound contains 18,000 yards; what
is the number in both run and cut systems? . „ f 11.25-run
Ans ' 160-cut
2. What is the weight, in ounces, of 10,000 yards of woolen yarn,
the yarn being 4-run? Ans. 25 oz.
3. What is the length of 75 pounds of 2-run woolen yarn?
Ans. 240,000 yd.
4. Find the weight of 60,000 yards of 4-cut woolen yarn.
Ans. 50 lb.
18
YARN CALCULATIONS,
§ 10
5. What is the length of 100 pounds of 3-cut woolen yarn?
Ans. 90,000 yd.
6. What are the cut counts of a woolen yarn when 1,200 yards
weighs 2 pounds? Ans. 2-cut
7. What are the run counts of a woolen yarn when 100,000 yards
weighs pounds? Ans. 8.33-run
BEAM CALCULATIONS
26. The yarn that forms the warp of a fabric after being
spun is wound on long wooden spools known as jack, or
dresser, spools. The yarn from a number of these spools
is then passed through several intermediate processes and
ultimately wound on the loom beam, from which it is
slowly unwound in the loom as the fabric is woven. Whether
the yarn is on the jack-spool or the loom beam, there are
several important rules that apply equally well in either case.
27. To find the size of yarn on a jack-spool or beam con¬
taining only one size of yarn, the weight of the yarn, its
length, and the number of ends being known:
Rule. — Multiply the length, in yards, of the warp on the
beam or length on the jack-spool by the number of ends on the
beam or spool, as the case may be. Divide the result thus
obtained by the weight of the yarn, in pounds, multiplied by
the standard number.
Example 1 .—A worsted warp 350 yards long contains 1,600 ends
and weighs 50 pounds; what is the number of the yarn?
0 350 X 1,600 on
Solution. — .. w = 20 s. Ans.
50 X obO
Example 2.— A jack-spool contains 200 yards of woolen yarn that
weighs 2 pounds, there being 40 ends on the spool; what is the size of
the yarn?
0 200 X 40 o1
Solution.— _ w , = 2 ^-run. Ans.
2 X 1,600
Note — The method of finding 1 the size of a given yarn when its length and weight
are known was explained in Art. 3, and it will be noticed that the same method is
adopted in the above rule, the only difference being that in the former case the length
of only one thread was given, while in this case there are a number of threads, or
ends, together, thus necessitating the multiplication of the length of each end by the
number of ends in order to find the total length of yarn. The rules that follow are
also similar to those employed for single threads except that a number of ends are
dealt with instead of a single one.
§10
WOOLEN AND WORSTED
19
It should be noted in connection with these examples that
the weight of the spool or beam has not been taken into con¬
sideration, the weight of the yarn only being given. In the
mill the yarn and the spool, or beam, on which it is wound
are usually, of necessity, weighed together; in which case it
becomes necessary to deduct the weight of the beam or
spool in order to obtain the true weight of the yarn.
Example 3.—A woolen warp contains 1,200 ends and is 400 yards
in length; the warp and beam together weigh 140 pounds and the
beam alone weighs 65 pounds. What is the size of the yarn?
Solution.—
140 - 65 = 75 lb.
400 X 1,200
75 X 1,600
4-ruu. Ans.
28. To find the number of ends on a beam when the
weight and length of the warp and the size of the yarn are
known:
Rule. —Multiply the weight , in pounds , by the standard
number and by the size of the yarn. Divide the result thus
obtained by the length of the warp , in yards.
Example 1.—A worsted warp is 800 yards long and weighs
200 pounds exclusive of the beam. If the warp is composed of
20s yarn, how many ends does it contain?
Solution.—
200 X 560 X 20
~ 800
2,800 ends. Ans.
Example 2.—A woolen warp is 600 yards long and weighs
120 pounds exclusive of the beam. If the warp is composed of
4^-run yarn, how many ends does it contain?
Solution.—
120 X 1,600 X 4|
600
1,440 ends.
29. To find the weight of a warp when the length, num¬
ber of ends, and size of the yarn are known:
Rule. —Multiply the length of the warp , in yards , by the
number of ends that it contains and divide the result thus
obtamed by the standard number multiplied by the size of
the yarn.
20
YARN CALCULATIONS,
10
Example 1.—A woolen warp 475 yards long contains 960 ends of
3-run yarn; what is the weight of the yarn?
Solution.—
475 X 960
1,600 X 3
95 lb. Ans.
Example 2. —A worsted warp 650 yards long contains 1,860 ends of
16s yarn; what is the weight of the yarn?
650 X 1,860 00
Solution. — . = 134.933 lb. Ans.
560 X 16
30. To find the length of a warp when the weight, num¬
ber of ends, and the size of the yarn are known;
Rule. —Multiply the weight of the warp , in pounds , by the
standard number and by the size of the yarn , and divide the
result thus obtained by the member of ends in the warp.
Example 1.—A worsted warp contains 2,400 ends of 12s yarn and
weighs 200 pounds; how long is it?
Solution.—
200 X 560 X 12
2,400
560 yd. Ans.
Example 2.—A woolen warp contains 1,200 ends of 2-run yarn and
weighs 231 pounds; how long is it?
Solution.—
231 X 1,600 X 2
1,200
616 yd. Ans.
31. To find the length of warp that can be placed on a
beam:
Rule. —Find the weight of yam that the beam will contain ,
by weighing a beam of the same size wheel filled with yarn and
deducting the weight of the beam itself. Then apply the rule in
Art. 30.
Example 1.—A certain size beam when filled with yarn weighs
140 pounds, the beam itself weighing 50 pounds. What length of a
warp composed of 1,800 ends of 28s worsted can be placed on it?
Solution.—
140 — 50 = 90 lb. of yarn.
90 X 560 X 28
1,800
= 784 yd. Ans.
Example 2.—Same beam as in Example 1. How many yards of a
warp containing 3,000 ends of 40s worsted can be placed on it?
Solution.— 140 — 50 = 90 lb. of yarn.
90 X 560 X 40
§10
WOOLEN AND WORSTED
21
EXAMPLES FOR PRACTICE
1. A worsted warp and beam weigh 170 pounds, the beam alone
weighing 70 pounds. If the warp is 350 yards long and contains
2,240 ends, what is the size of the yarn? Ans. 14s.
2. A woolen warp is 320 yards long and weighs 124 pounds, exclu¬
sive of the beam. If the warp is composed of 2|-run yarn, how many
ends does it contain? Ans. 1,550 ends
3. A worsted warp 780 yards long contains 2,480 ends of 20s yarn;
what is the weight of the yarn? Ans. 172.714 lb.
4. A woolen warp contains 824 ends of 2i-run yarn and weighs
103 pounds. How long is the warp? Ans. 450 yd.
FANCY WARPS
32. It frequently happens that in the production of fancy
goods there are two or more colors of yarn used in a single
warp in order to obtain a certain effect in the fabric. When
this is the case, it often necessitates the finding of the total
number of ends of each color in the warp or the weight of
each particular yarn, etc. When the yarn is on the loom
beam the following rule will be found of value.
33. To find the number of ends of each color of yarn on
a beam when the pattern of the warp and the total number of
ends in the warp are known:
Rule. —Multiply the number of ends of any one color in the
pattern by the total number of ends on the beam and divide by
the total number of ends in the pattern.
Example 1.—A warp is arranged, or dressed, 16 ends of black,
8 ends of slate, 20 ends of black, and 4 ends of white, and contains
2,400 ends; how many ends of each color are there?
Solution. —In this example there are 16 -f- 8 -f 20 + 4 = 48 ends
in one repeat of the pattern, of which 36 ends are black; 8, slate; and
4, white.
36 X 2,400
48
8 X 2,400
48
4 X 2,400
1,800 ends of black
400 ends of slate
200 ends of white
■Ans.
48
22
YARN CALCULATIONS
Proof. — 1,800 + 400 + 200 = 2,400 ends in warp.
Note. —One repeat of the colors or counts in a warp is known as the
pattern of the warp. For an illustration, suppose the ends on a
beam are arranged 8 black, 2 white, 8 black, 2 slate. This would be
the pattern of the warp and all the other ends arranged in the same
manner would simply form repeats of the pattern.
After the total number of ends of each color are found, it
is only necessary to apply the rule in Art. 29 in order to find
the weight of each color in the warp, the length of the warp
and the size of the yarn being known.
Example 2. —If the warp in example 1 is 640 yards long and is
composed of 20s worsted yarn, how many pounds of each color of
yarn does it contain?
Solution.—
640 X 1,800
560 X 20
640 X 400
560 X 20
640 X 200
560 X 20
= 102.857 lb. of black yarn
= 22.857 lb. of slate yarn >Ans.
= 11.428 lb. of white yarn
34. When warps are composed of yarns of different
counts arranged in a pattern, the same method may be
employed to find the number of ends of each count, after
which the rule in Art. 29 may be applied to find the weight
of yarn of each count.
Example. —A worsted warp is to be made containing 3,000 ends
and 432 yards long. The yarn used is of two counts arranged 12 ends
of 28s, 2 ends of 6s, 4 ends of 28s, 2 ends of 6s. How many pounds of
each yarn is necessary for the warp?
Solution. —In this example there are 12 + 2 + 4 + 2 = 20 ends in
each pattern, of which 16 are of 28s yarn and 4 of 6s yarn.
16 X 3,000
20
= 2,400 ends of 28s
4 X 3,000
20
= 600 ends of 6s
432 X 2,400
560 X 28
432 X 600 '
-660W = 77
= 66.122 lb. of 28s
>Ans.
= 77.142 lb. of 6s
§10
WOOLEN AND WORSTED
23
AVERAGE NUMBERS
35. When a warp contains yarn of more than one size, it
is often of advantage to find the average size of the warp in
order that other calculations may be more easily performed.
By the term average number of the warp, which is often
used in this connection, a number of yarn is meant that will
be of the same weight as the yarns of unequal counts that are
used in the warp, if the same number of ends are used.
36. To find the average size of warp yarn in a warp that
contains two or more sizes:
Rule. — Divide the total number of ends of each counts by its
size. Add the results thus obtained arid divide the sum into the
total number of ends in the warp.
Example. —Suppose that on the same beam there are 1,600 ends of
40s worsted and 800 ends of 10s. It is desired to know what number
of yarn will make a warp of the same weight per yard with the same
number of ends.
Solution. — 1,600 4 40 = 40
800 4 10 = 80
2+>0 120
2,400 4 120 = 20s, average number. Ans.
Proof. —That the above example is correct may be readily
shown by finding the weight of 1 yard of warp composed of
1,600 ends of 40s and 800 ends of 10s worsted and also the
weight of 1 yard of a warp composed of 2,400 ends of 20s
worsted.
= .0714 lb., weight of 40s yarn in first warp
= .1428 lb., weight of 10s yarn in first warp
= .2142 lb., weight of 1 yard of first warp
= .2142 lb., weight of 1 yard of second warp
From this it will be seen that the weight of the original
warp and that of the one composed of the same number of
ends of 20s is identical; therefore, the example must be correct.
1X1,600
560 X 40
1 X 800
560 X 10
.0714 + .1428
IX 2,400
560 X 20
24
YARN CALCULATIONS, §10
37. The rule just given will be found to apply equally
well if more than two sizes of warp yarn are used.
Example. —A warp is composed of 1,200 ends of 40s worsted,
600 ends of 20s, and 120 ends of 10s. What is the average number?
Solution. — 1,200 40 = 30
600 -f- 20 = 30
120 -p 10 = 12
L920 72
1,920 -f- 72 = 26.66s, average number. Ans.
38. In cases where the order of arranging the different
counts of yarn in the warp is known but the total number of
ends is not given, the same rule will be found to apply if
the number of ends in one pattern of the warp is considered
as the total number of ends.
Example. —A warp is arranged 48 ends of 36s and 2 ends of 10s.
Find the average number.
Solution. — 48 -s- 36 = 1.333
2 - 5 - 10 = .2
50 1.533
50 -r* 1.533 = 32.615s, average number. Ans.
EXAMPLES FOR PRACTICE
1. A worsted warp contains 1,620 ends of 45s yarn and 840 ends of
20s yarn; find the average counts of the warp. Ans. 31.53s
2. A woolen warp contains 900 ends of 44-run yarn and 450 ends
of 24-run; find the average size of the yarn. Ans. 3f-run
§10
WOOLEN AND WORSTED
25
METRIC SYSTEM OF NUMBERING
YARNS
39. In recent years there has been from time to time
considerable agitation along the lines of adopting one
system of numbering all classes of yarns used in textile
manufacturing. The objects of these movements are: first,
to bring about the unification of the method of stating the
degree of fineness of the yarns for all varieties of fibers used
in the textile industry in the whole world; second, to adopt
a method that would be practical in every way.
The advantages of such a system as this would be many.
The chief objection to it is that, from long usage, the methods
at present adopted are too well developed for a single corpo¬
ration or a single country to take on itself such a reform,
without being assured that its neighbors and competitors
would simultaneously do the same thing, as a mill instituting
such a method would be at a disadvantage as compared with
its competitors.
The method usually set forth is that of numbering all
classes of yarns by what is known as the metric system
and consists of considering 1 meter of number Is yarn to
weigh 1 gram, the meter being the unit of length in the
metric system and the gram being the unit of weight. The
only parts of the metric system that it is necessary to under¬
stand in order to convert one system into the other are the
equivalents of the meter and the gram, which are as follows:
1 yard = .914 meter, 1 pound = 453.59 grams
40. To find the number of yarn in any present standard
system that corresponds to the number of yarn in the metric
standard system:
Rule. —Multiply the counts, given in the metric system , by
453.59 (grams in 1 pound ) and divide by the standard number
26
YARN CALCULATIONS,
§10
of yards to the pound in the present system multiplied by .914
(meter in 1 yard).
Example. —A worsted yarn numbered according to the metric
system is marked 40s. Find the counts in the present system.
40 X 453.59
560 X .914
= 35.447s. Ans.
Solution.—
41. To find the number of yarn in the metric standard
system that corresponds to the number of yarn in any pres¬
ent standard system:
Rule. —Multiply the counts , given in the present system , by
the present standard number of yards to the pound and by .914
(meter in 1 yard) and divide by 453.59 (grams in 1 pound).
Example. —A worsted yarn numbered according to the present
system is marked 46s. Find the counts in the metric system.
Solution.—
46 X 560 X .914
453.59
= 51.907s. Ans.
INTRODUCTION
1. From the earliest times of the manufacture of yarns it
has been found necessary to adopt some system by means of
which the different sizes of yarns can be designated, since
otherwise the manufacturer has no means of readily distin¬
guishing one yarn from another. As no uniform standard
has been adopted, the methods used in the various manufac¬
turing centers of the world have become too numerous to
allow a detailed description being given here; therefore,
only those found in common practice, with particular refer¬
ence to the American methods, will be dealt with.
The different yarns employed in textile manufactures may
be divided into several classes: For example, they may be
divided into (1) vegetable fibers , which include cotton, linen,
jute, etc.; (2) animal fibers , which include woolen, worsted,
silk, etc. Again, these yarns may be divided with reference
to the system employed when numbering them, as, for
example, one class may be said to be that which includes all
yarns based on the system of having higher numbers for
coarser threads; in this class are included coarse linen, coarse
jute, and raw silk. The second class includes all yarns based
on the system of having higher numbers for finer threads.
This method will be found to be the more common and
includes cotton, woolen, worsted, mohair, spun silk, fine
linen, and fine jute.
For notice of copyright , see page immediately following the title page
l 11
2
YARN CALCULATIONS, GENERAL
§11
2. The words member and counts when applied to yarns
represent in all cases a certain length for a certain weight.
Other terms frequently used are run, cut , skein , spindle , grist ,
etc. A person figuring counts in any mill, especially when
dealing with cloth calculations, will find it necessary to
understand the systems employed when numbering different
yarns, such as wool, cotton, worsted, and silk, for which
reason the numbering of these will be thoroughly dealt with,
each one being considered by itself.
SINGLE YARNS
COTTON YARNS
NUMBERING SYSTEM
3. Method of Determining tlie Counts. — In the
cotton system of numbering yarns all counts are based on
the standard of 840 yards, which is known as 1 hank; in
other words, a hank of cotton yarn contains 840 yards, with¬
out regard to the fineness or coarseness of the yarn. This
is a constant number, and the fact that it never varies should
always be remembered.
The method of numbering the yarns is that of calling a
yarn that contains 1 hank, or 840 yards, in 1 pound a No. 1
yarn. If the yarn contains 2 hanks, or 1,680 yards, in
1 pound, it is known as a No. 2 yarn; if it contains 3 hanks,
or 2,520 yards, in 1 pound, it is known as a No. 3 yarn.
Thus it will be seen that the number of hanks (840 yards) it
takes to weigh 1 pound determines the counts of the yarn.
The counts of a yarn are generally indicated by placing a
letter s after the figure representing the number of the yarn;
thus, 26s shows the counts of a yarn and indicates that the
yarn contains 26 hanks (26 X 840 yards) in 1 pound.
11
YARN CALCULATIONS, GENERAL
3
CALCULATIONS OF SINGLE COTTON YARNS
4. Several calculations become necessary when dealing
with single yarns, but all are based on one equation; namely,
length of yarn, in yards _ ^
weight, in pounds X counts X standard
The standard will always be the same when dealing with
cotton (840 yards), and any one of the other items can be
found by simply substituting the values of the rest in the
above equation.
5. To find the counts of a yarn when the length and
weight are given:
Rule. —Divide the total length of yam, expressed in yards,
by the product of the weight, expressed in pounds, and the stand¬
ard length.
Example.— 168,000 yards of yarn is found to weigh 5 pounds;
what are the counts?
Solution.—
168,000 (length of yarn, in yar ds) _
5 (weight, in pounds) X 840 (standard)
= 40s counts.
Ans.
6. To find the weight of yarn when the length and counts
are known:
Rule. —Divide the length, in yards, by the product of the
couiits and the standard length.
Example. —What is the weight of 42,000 yards of number 5s yarn?
.Solution.—
42,000 (length, in yards)
5 (counts) X 840 (standard)
10 lb. Ans.
7. To find the length of yarn when the weight and counts
are given:
Rule. —Multiply the weight, in pounds, the cowits, and the
standard length together.
Example. —What length of yarn is contained in a bundle weigh¬
ing 8 pounds, the counts of the yarn being 26s?
Solution.— 8 (weight, in pounds) X 26 (counts) X 840 (standard) =
174,720 yd. Ans.
4
YARN CALCULATIONS, GENERAL
§11
8. From the rules and examples given, the student will
readily see that there is one primary fact that should con¬
stantly be kept in mind when trying to understand the system
of cotton-yarn numbering-; viz., in all cases the number of
hanks (840 yards) of any yarn that it takes to weigh 1 pound
is the counts of that yarn.
When numbering yarns, it will be found that several
different weights and names of lengths are used, for which
reason it will be well to memorize the two following tables.
These tables include only those terms used in yarn numbering.
TABLE I
The table of lengths is as follows:
1 | yards = 1 thread, or circumference of wrap reel.
120 yards = 80 threads = 1 skein, or lea.
840 yards = 500 threads = 7 skeins, or leas = 1 hank.
TABLE II
The table of weights is as follows:
27.34 grains = 1 dram.
437.5 grains = 16 drams = 1 ounce.
7,000 grains = 256 drams = 16 ounces = 1 pound.
The part of this table most frequently used and which,
therefore, should be carefully noted is that which states
that there are 7,000 grains in 1 pound.
SIZING COTTON ROVING AND YARN
9. Roving is a term used to designate a strand of loosely
twisted fibers from which a yarn may be spun, while the term
yarn is applied to a thread composed of fibers uniformly
disposed throughout its structure and having a certain
amount of twist for the purpose of enhancing its strength.
Some method must be adopted in every well-regulated
mill by means of which the exact counts of the roving and
yarn being run may be known. It is generally the custom
to weigh a certain quantity from each machine at least once
a day and by this means ascertain whether the roving or
yarn is being kept at the required weight. This process is
known as sizing and is a matter that should always be very
carefully attended to.
§11
YARN CALCULATIONS, GENERAL
5
10. The Wrap Reel.—When sizing yarns, an instru¬
ment known as a wrap reel is used to measure the yarn.
As its name indicates, this instrument consists of a reel,
generally I 2 yards in circumference. The yarn is wound
on this reel and a finger indicates on a disk the number
of yards reeled. Fig. 1 shows an ordinary type of wrap
6
YARN CALCULATIONS, GENERAL
§11
reel, while Fig. 2 shows yarn and roving scales. These
scales are suitable for weighing by tenths of grains.
In case the standard length of the yarn being sized was
reeled off, all that would be necessary, in order to find the
counts, would be to divide the weight into 1 pound. This
would, however, require considerable time, in addition to
causing an unnecessary amount of waste yarn. It is, there¬
fore, customary to reel off a certain number of yards and
divide the weight of this amount, expressed in grains, into
the number of grains that bears the same relation to 7,000
(grains in 1 pound) that the number of yards weighed bears
to the standard length. For example, if it is desired.to size
the yarn running on a cotton spinning frame, the bobbin of
yarn is removed from the frame and taken to the wrap reel.
120 yards is reeled off and the weight, in grains, of this
120 yards divided into 1,000, the answer being the counts of
the yarn. This must be true, because 120 yards bears the
same relation to 840 yards (the standard) that 1,000 bears
to 7,000 (the number of grains in 1 pound).
Example.— 120 yards of cotton yarn is reeled and found to
weigh 40 grains; what are the counts?
Solution.— 1,000 -p 40 = 25s. Ans.
11. This basis of indicating the size of the yarn is also
used to designate the size of roving, although when sizing
roving, a shorter length is used. It is customary in this
case to measure off one-seventieth of 840 yards, or 12 yards,
and divide the weight, in grains, of this length of roving
into one-seventieth of 7,000, or 100.
Example.— 12 yards of roving is found to weigh 20 grains; what
are the counts?
Solution.— 100 -p 20 = 5-hank. Ans.
The counts of roving are indicated in a somewhat different
manner from the counts of yarn. Thus, in this case, the
roving would be known as 5-hank roving , indicating that
5 hanks weigh 1 pound.
12. From the foregoing, the student will readily under¬
stand that the counts of a cotton yarn may be obtained by
11
YARN CALCULATIONS, GENERAL
7
finding the weight of any length of yarn, and dividing the
weight into a number of grains that has the same propor¬
tion to 7,000 that the length of yarn weighed has to 840.
As an aid to the student the following lengths and dividends
are given:
TABLE III
If 480 yards is weighed, divide weight, in grains, into 4,000.
If 240 yards is weighed, divide weight, in grains, into 2,000.
If 120 yards is weighed, divide weight, in grains, into 1,000.
If 60 yards is weighed, divide weight, in grains, into 500.
If 40 yards is weighed, divide weight, in grains, into 3334-
If 30 yards is weighed, divide weight, in grains, into 250.
If 20 yards is weighed, divide weight, in grains, into 166f.
If 15 yards is weighed, divide weight, in grains, into 125.
If 12 yards is weighed, divide weight, in grains, into 100.
If 10 yards is weighed, divide weight, in grains, into 834-
If 8 yards is weighed, divide weight, in grains, into 66f.
If 6 yards is weighed, divide weight, in grains, into 50.
If 4 yards is weighed, divide weight, in grains, into 334-
If 3 yards is weighed, divide weight, in grains, into 25.
If 2 yards is weighed, divide weight, in grains, into 16f.
If 1 yard is weighed, divide weight, in grains, into 84.
WOOLEN YARNS
RUN SYSTEM OF NUMBERING
13. The common method of numbering woolen yarns is
by what is known as the run system. With this system
the number of runs that weigh 1 pound indicates the size of
the yarn, a run being 1,600 yards. Thus, for example, if a
woolen yarn is a 4-run yarn, it will contain 4 X 1,600 yards
to 1 pound.
It will be seen that this method of numbering the yarn is
similar to that explained in connection with cotton yarn, the
only difference being that 1,600 yards in this case is taken as
the standard in place of 840, which is the standard for cotton.
Consequently, the rules given for finding length, weight,
and counts of cotton yarns will be found to apply equally
well in this case, with the exception that 1,600 must be used
in place of 840. However, in order that the student may
8
YARN CALCULATIONS, GENERAL
§11
thoroughly understand the application of these rules, an
example of each will be given.
Example 1.—A woolen yarn 72,000 yards in length is found to
weigh 10 pounds; what run is it?
Solution.—
72,0 00
10 x 1,600
— 4^-run.
Ans.
Example 2.—What is the weight of 64,000 yards of 2-run woolen
yarn ?
Solution.—
64,000
2 X 1,600
20 lb. Ans.
Example 3.—What is the length of 5 pounds of a 6-run yarn?
Solution.— 5 X 6 X 1,600 = 48,000 yd. Ans.
14. A very convenient method of finding the weight,
run, or length of a woolen yarn when the weight is expressed
in ounces may be explained as follows;
Example 1.—What is the weight, in ounces, of 8,800 yards of
a 5i-run yarn?
e 8 « 800
Solution.— - w , ,, nn = 1 lb.
5.5 X 1,600
16 (oz. in 1 lb.) X 1 = 16 oz. Ans.
It will be noticed that as it is desired to obtain the weight
in ounces, the result in pounds is multiplied by 16, which
gives 16 ounces as the answer. If, instead of multiplying
the answer by 16, a new standard should be obtained
by dividing 1,600 by 16, the result would be the same, but
the process would be considerably shortened. Therefore,
whenever the weight is expressed in ounces the standard
employed should be 100.
Example 2.—What is the weight, in ounces, of 5,250 yards of
a 3-run yarn?
5,250
Solution.
3 X 100
= 17^ oz. Ans.
Example 3.— 4,000 yards of woolen yarn weighs 5 ounces; what
run is it?
4,000
Solution.—
5 X 100
= 8-ruu. Ans.
Example 4.—How many yards are there in 10 ounces of a 3-run
yarn?
Solution. — 10 X 3 X 100 = 3,000 yd. Ans.
§11
YARN CALCULATIONS, GENERAL
9
CUT SYSTEM OF NUMBERING
15 . In the vicinity of Philadelphia the size of woolen
yarns is based on what is termed the cut system, which
has 300 yards for its standard length of yarn. In all other
respects the calculations are similar; consequently, the rules
previously given will apply in this case, with the exception
r
that 300 is used as the standard.
Example 1.—What is the size in the cut system of a woolen yarn
36,000 yards of which weighs 10 pounds?
„ 36,000 10 . .
Solution. — = 12-cut. Ans.
10 X 300
Example 2.—What is the weight of 66,000 yards of a 20-cut yarn?
Solution. — 20 X 300 ~ ^ us '
Example 3.—What is the length of 10 pounds of a 24-cut woolen
yarn ?
Solution. — 24 X 10 X 300 = 72,000 yd. Ans.
SIZING WOOLEN ROVING AND YARN
16 . Both woolen roving (in some mills and districts the
term roving is corrupted into roping ) and yarn are sized
according to the number of runs or cuts in 1 pound. In sizing
both the yarn and roving, a scale known as a run or cut
scale is commonly used. This scale, shown in Fig. 3, con¬
sists of a brass beam carried by an upright standard, and so
10
YARN CALCULATIONS, GENERAL
§11
arranged that when a given number of yards of roving or
yarn is placed on the beam in such a position as to balance
(the beam being out of balance when not in use), the posi¬
tion of the yarn will indicate its size on a graduated scale
on the beam. In order that the reading may be as accurate
as possible, the yarn or roving to be weighed is twisted into
a small bunch and suspended by a single thread. Run or
cut scales are usually made to indicate the correct size from
50 yards of yarn or roving, but they may be made to indicate
the size with any desired number of yards. If other than
the number of yards for which the scale is graduated is
used in weighing, the reading of the scale must be corrected.
In order to do accurate work the equilibrium of a run or
cut scale should be adjusted with a known weight or a
known size of yarn. This can be accomplished by turning
to the right or the left the setscrew in the heavy end of
the beam.
When a mill is not equipped with a run scale, the size of
the yarn must usually be determined by means of a grain
scale. In most mills 20 yards of yarn or roving is measured
off for sizing, but in other mills 50 yards, 100 yards, and
various other lengths are used.
17 . To find the size of a woolen roving or yarn when
the weight of a given number of yards is known:
Rule. —Divide the number of ya?'ds zveighed, multiplied by
7,000, by the standard number ( 1,600 or 300), multiplied by
the weight, in grains.
Example. —If 20 yards of woolen yarn weighs 25 grains, what is
the run of the yarn?
c 20 X 7,000 o1
Solution. 1|600 x g, = Si-run. Ans.
The following table gives the size, in runs, and the weight,
in grains, of 20 yards of woolen roving or yarn. If any
number of yards other than 20 is weighed, the reading of the
table will have to be corrected. For instance, if 100 yards
of yarn is found to weigh 175 grains, the size of the yarn is
not 2 -run, but 5 times (100 -f- 20 = 5) i-run, or 22-run.
11
YARN CALCULATIONS, GENERAL
11
TABLE IV
Runs
Grains
Runs
Grains
Runs
Grains
1
2
175.00
5
17.50
10
8-75
5
8
140.00
5 4
16.66
ioi
8-53
3
4
I l6.66
5 *
15.90
ioi
8-33
7
8
100.00
5 f
15.21
1 of
8.13
i
87.50
6
14.58
11
7-95
I 4
70.00
67
14.00
1 if
7-77
i!
58-33
6i
13.46
1 is
7.60
if
50.00
6f
12.96
1 if
7-44
2
43-75
7
12.50
12
7.29
21
38.88
74
12.06
I2f
7.14
2|
35-00
7 s
11.66
12!
7.00
2T
31.81
7 f
. 11.29
I2f
6.86
3
29.16
8
10.93
13
6-73
37
26.92
r-t|^
OO
10.60
137
6.60
32
25.00
8i
10.29
132
6.48
34
23-33
8f
10.00
1 3 f
6.36
4
21.87
9
9.72
14
6.25
44
20.58
94
9-45
142^
6.03
42
19.44
92
9.2 1
15
5-83
44
18.42
97
8.97
WORSTED YARNS
18 . The basis on which worsted yarns are numbered is
that of having 560 yards constitute 1 hank, and numbering
the yarns according to the number of hanks that weigh
1 pound; consequently, the student will understand that the
rules given for numbering cotton yarns apply equally well
when numbering worsted yarns, with the difference, how¬
ever, that 560 yards is taken as the standard instead of 840.
Examples for finding counts, weight, and length of worsted
yarns will be given in order further to enable the student to
understand the system employed.
Example 1.— 268,800 yards of worsted yarn weighs 10 pounds.
What are the counts?
12
YARN CALCULATIONS, GENERAL
11
0 268,800
Solution.— .. w = 48s. Ans.
10 X 5b0
Example 2.—What is the weight of 161,280 yards of a 36s worsted
yarn?
161,280 0 .
Solution.— = 8 lb. Ans.
36 X 560
Example 3.—What is the length of 20 pounds of a 28s worsted
yarn ?
Solution. — 20 X 28 X 560 = 313,600 yd. Ans.
Besides the table of weights given in connection with
cotton yarns, the following table of lengths is used for
worsted yarns:
TABLE V
1 yard = 1 thread (circumference of reel).
80 yards = 80 threads = 1 lea.
560 yards = 560 threads = 7 leas = 1 hank.
SIZING WORSTED YARNS
19 . When finding the counts of worsted yarns it is the
usual custom to reel off 80 yards and divide the weight, in
grains, into 1,000. It will be seen that this method is similar
to that of reeling 120 yards of cotton and dividing its weight
into 1,000, as previously explained, since one-seventh of 560
(standard length for worsted) is 80. Other lengths and
dividends for worsted yarns are as follows:
TABLE VI
If 320 yards is weighed, divide
If 160 yards is weighed, divide
If 80 yards is weighed, divide
If 60 yards is weighed, divide
If 40 yards is weighed, divide
If 30 yards is weighed, divide
If 20 yards is weighed, divide
If 15 yards is weighed, divide
If 10 yards is weighed, divide
If 8 yards is weighed, divide
If 6 yards is weighed, divide
If 4 yards is weighed, divide
If 3 yards is weighed, divide
If 2 yards is weighed, divide
If 1 yard is weighed, divide
weight,
in
grains,
into
4,000.
weight,
in
grains,
into
2,000.
weight,
in
grains,
into
1,000.
weight,
in
grains,
into
750.
weight,
in
grains,
into
500.
weight,
in
grains,
into
375.
weight,
in
grains,
into
250.
weight,
in
grains,
into
187|.
weight,
in
grains,
into
125.
weight,
in
grains,
into
100.
weight,
in
grains,
into
75.
weight,
in
grains,
into
50.
weight,
in
grains,
into
371.
weight,
in
grains,
into
25.
weight,
in
grains,
into
12*.
§11
YARN CALCULATIONS, GENERAL
13
SIZING WORSTED ROVING
20. It is usually customary to designate the size of
worsted roving by the number of drams that 40 yards weighs;
thus, if 40 yards of a certain roving weighs 3.2 drams it is
known as a 3.2-dram roving. When very fine yarns are
being made, however, the roving is generally designated by
the weight, in drams, of 80 yards, this length being taken
because of the greater accuracy obtained by weighing a
greater length. In sizing the roving, a scale capable of
weighing drams and tenths of drams is commonly used, but
should it be necessary to use a grain scale the measure¬
ments given in Table II should be employed.
Since worsted yarn is numbered on the basis of the
number of hanks of 560 yards in 1 pound and roving by the
number of drams that 40 or 80 yards weighs, it is sometimes
desirable to find the counts of a roving.
21 . To find the counts of a roving when its weight in
drams is known:
Rule .—If the weight , in drams , of 40 yards is known , divide
this zveight fnto 18.3. If the zveight, i?i drams , of 80 yards is
known , divide this weight into 36.6.
Example. —What are the counts of worsted roving 40 yards of which
weighs 6.1 drams?
Solution. — 18.3 -f- 6.1 = 3s. Ans.
Note. — 18.3 and 36.6 are known as ftauge points and are obtained by multi¬
plying the drams in 1 pound by the number of yards weighed (40 or 80) and dividing
by the number of yards in 1 hank; thus,
266 V 40
——j— = 18.285, practically 18.3
256| 360 " = 36 ‘ 57, practically 36 - 6
From this it will be seen that 40 yards o£ number Is weighs 18.3 drams, and
80 yards of number Is weighs 36.6 drams, which is the true significance of the
gauge points.
14
YARN CALCULATIONS, GENERAL
811
SILK YARNS
22. Two classes of silk are used for manufacturing
purposes —spun and raw silk. As these are numbered in an
entirely different manner, they will be considered separately.
23. Single spun silk is numbered in a manner exactly
similar to cotton yarns, containing 840 yards to the hank and
basing the counts on the number of hanks that weigh 1 pound.
Example 1.—What are the counts of a spun-silk thread, if 756,000
yards weighs 15 pounds?
Solution.—
756,000
15 X 840
= 60s.
Ans.
Example 2.—What is the weight of 403,200 yards of a 40s spun-
silk thread?
Solution.—
403,200
40 X 840
= 12 lb.
Ans.
Example 3.—What is the length of 4 pounds of a 48s spun-silk
yarn?
Solution. — 4 X 48 X 840 = 161,280 yd. Ans.
24. There are two distinct systems employed for num¬
bering raw silk, one of which is used for designating the
size of raw silk as it comes from the cocoon, while the other
is used for numbering thrown silk , i. e., raw silk that has been
doubled and reeled.
Raw silk as it comes from the cocoon is numbered accord¬
ing to the weight of one skein 476 meters in length. This
weight is expressed in units equal to -gVth part of a denier , or
.0531 gram. Although .0531 gram is really equal to only a
fractional part of a denier, it is commonly spoken of as
a denier. The so-called denier is actually equal to
.053115 gram, but in practice .0531 gram is used, while
476 meters is equal to 520.57 yards, although in practice
520 yards is used as the standard length of a skein. If
therefore 520 yards of raw silk weighs .7965 gram (.0531
X 15 = .7965), it is spoken of as a 15-denier silk. From
this explanation it will be noted that the system of numbering
raw silk differs materially from that employed for numbering
cotton, woolen, worsted, spun silk, etc., since the higher the
11
YARN CALCULATIONS, GENERAL
15
number, the coarser is the silk; whereas, in the systems
previously explained, higher counts indicate finer yarns.
Example 1.—If 520 yards of raw silk weighs .6372 gram, what is
the size of the silk?
Solution. — .6372 -=- .0531 (grams in 1 denier) = 12-denier silk.
Ans.
Example 2.—What is the weight, in grams, of 15,600 yards of
20 -denier raw silk?
Solution. — 520 yards weighs 20 deniers, or 1.062 grams (.0531
X 20 = 1.062); 15,600 yards, therefore, since 520 is contained in 15,600
exactly 30 times, will weigh 30 X 1.062 grams, or 31.860 grams. Ans.
Example 3.—How many yards are there in 4.248 grams of 26-denier
raw silk?
Solution. —Since 520 yards weighs 26 deniers, 20 yards (520 -=- 26
= 20) weighs 1 denier, or .0531 gram. If 20 yards weighs .0531 gram,
there will be as many times 20 yards in 4.248 grams as .0531 is contained
in 4.248, or 80 times (4.248 -f- .0531 = 80). Therefore, 80 times 20 yards
= 1,600 yards. Ans.
Note.—T here are practically 28j grams, or 533i deniers, in 1 ounce (avoirdupois).
25. Although in Europe thrown silk is numbered the
same as raw silk, in America a different system is employed;
namely, the dram system, in which the method of speci¬
fying the size of such yarns is that of giving the weight of
1,000 yards in drams (avoirdupois). For example, if
1,000 yards of a certain thrown-silk thread weighs 8 drams,
it is known as an 8-dram silk. It will be seen from this that
the system employed in thrown-silk numbering also differs
materially from that employed with cotton, woolen, etc., since
in this case, as well as in the case of raw silk as it comes
from the cocoon, the higher the count of the thread, the
coarser it will be.
When figuring the counts of thrown silk, Table II, given
in connection with cotton yarns, must be employed.
<*
Example 1.— How many yards are there in 1 pound of a 5-dram
silk?
Solution. — 1,000 -5- 5 = 200 yd. in 1 dram
200 X 256 (drams to the pound) = 51,200 yd. in 1 lb.
Ans.
16 YARN CALCULATIONS, GENERAL §11
Example 2.—How many yards are therein 10 pounds of a 2^-dram
silk thread?
Solution. — 1,000 -f- 2\ = 400 yd. in 1 dram
400 X 256 - 102,400 yd. in 1 lb.
102,400 X 10 = 1,024,000 yd. in 10 lb. Ans.
Example 3. —What are the counts of a thrown-silk thread if 1 pound
contains 25,600 yards?
Solution. — 25,600 -f- 256 = 100 yd. in 1 dram
1,000 4- 100 = 10 drams — weight of 1,000 yd.
Therefore, the thread is a 10-dram silk. Ans.
Example 4. —What are the counts of a thrown-silk thread that con¬
tains 800 yards in 1 ounce?
Solution.— 800 4- 16 = 50 yd. in 1 dram
1,000 4- 50 = 20 drams = weight of 1,000 yd.
Therefore, the thread is a 20-dram silk. Ans.
Example 5.—What is the weight of 1,000 yards of a 10-dram silk
thread?
Solution. —The weight is 10 drams, since thrown silk is numbered
by the dram weight of 1,000 yd. Ans.
Example 6 .—What is the weight of 48,000 yards of an 8-dram silk?
Solution. — 1,000 yd. weighs 8 drams. Therefore, 48,000 yd. will
weigh 48 X 8 drams = 384 drams. Ans.
JUTE, LINEN, ANTI RAMIE
26. Fine jute yarns, fine linen, and ramie, or China grass,
are numbered in a manner exactly like the woolen cut sys¬
tem; that is, 300 yards to the hank is taken as the standard
length, and the number of hanks that weigh 1 pound deter¬
mines the counts of the yarn. Consequently, all calculations
given in connection with the cut system of numbering
woolen yarns will apply equally well when dealing with
these materials.
The counts of coarse jute and coarse linen yarns are deter¬
mined by the weight, in pounds, of a spyndle (14,400 yards).
Thus, if 14,400 yards weighs 4 pounds, it is known as
4-pound yarn. This system will be seen to compare with
the system of numbering raw silk, in that the coarser the
yarn, the higher will be the counts; but, as will be noticed
§11
YARN CALCULATIONS, GENERAL
17
from the preceding- explanations, this method of numbering
yarns is the exception rather than the rule, since generally
yarns are numbered on the basis of having the higher counts
for the finer yarns.
SUMMARY
27 . If the student carefully memorizes the following few
facts, he will be saved a great deal of difficulty that might
otherwise occur when figuring any single yarns having higher
counts for finer threads. Such yarns are those made from
cotton, wool, worsted, and spun silk, and are the yarns that
will be most frequently met with.
1. If the length of yarn is given, always divide this length
by the product of the other item given and the standard to
find the required item.
2. If it is desired to find the length of yarn, always multi¬
ply the given items and the standard together.
3. Cotton has for its standard length 840 yards; worsted,
560 yards; spun silk, 840 yards; woolen (run system),
1.600 yards; woolen (cut system), 300 yards.
EXAMPLES FOR PRACTICE
1. 50,400 yards of cotton yarn weighs 3 pounds; what are the
counts? Ans. 20s
2 . 100,800 yards of cotton yarn weighs 10 pounds; what are the
counts? Ans. 12s
3. 100 skeins of cotton yarn, each containing 4,200 yards, weighs
20 pounds; what are the counts? Ans. 25s
4. What are the cotton counts of yarns 120 yards of which weighs
respectively 17, 21, 26, and 32 grains?
Ans.
58.82s
47.61s
38.46s
31.25s
5. If a cop is known to contain 960 yards and 35 of these cops
weigh 1 pound, what are the cotton counts of the yarn? Ans. 40s
6 . Woolen yarn weighing 1 pound contains 18,000 yards; what is
the number in both run and cut system? f 11.25-run
Ans ' 160-cut
7. What is the weight of 37,000 yards of Is worsted yarn?
Ans. 66.07 lb.
18
YARN CALCULATIONS, GENERAL
§11
8 . What is the weight of 10,000 yards of woolen yarn, the yarn
being 4-run? Ans. 25 oz.
9. How many yards of 60s spun-silk yarn are there in 24 pounds?
Ans. 1,209,600 yd.
10. If 200 yards of thrown silk weighs 6 drams, what number is it?
Ans. 30-dram silk
11. What is the length of yarn in 50 pounds of 20s cotton?
Ans. 840,000 yd.
12. A spool of 28s cotton yarn weighs 28 ounces. Find the length of
yarn on the spool if the spool itself weighs 8 ounces. Ans. 29,400 yd.
13. What is the weight of 336,000 yards of 40s spun silk?
Ans. 10 lb.
14. What is the weight of 151,200 yards of 60s cotton? Ans. 3 lb.
15. How many yards are there in 4 ounces of 5i-run woolen yarn?
Ans. 2,100 yd.
EQUIVALENT COUNTS
28. Many calculations are met with in which it becomes
necessary to place the count of one yarn in the system of
another. That is, if a cotton thread is a certain count in the
cotton system, it may be necessary to learn what its counts
would be if it were numbered similarly to a worsted thread.
When two, three, or more threads made from different raw
stock and numbered according to different methods are
placed in the same system, they are said to be reduced to
equivalent counts.
Example. —Suppose that it is desired to learn what the counts of a
15s cotton would be if numbered similarly to worsted yarn.
Solution. —A 15s cotton has 15 X 840 = 12,600 yd. in 1 lb. The
question then resolves itself into the following: What are the worsted
counts of a yarn containing 12,600 yd. to the lb.?
12,600 -f- 560 = 22.5s, worsted counts. Ans.
29. To find the count of one system that is equivalent
to that of another:
Rule. —Multiply the given counts by the member of yards hi
the standard length of the specified system and divide by the
number of yards in the standard length of the system required .
11
YARN CALCULATIONS, GENERAL
19
Example 1.—Find the equivalent of a 40s cotton in worsted
counts.
Solution. — 840 X 40 = 33,600
33,600 4 - 560 = 60s. Ans.
Explanation.— Since there are 840 yards of yarn in
1 pound of Is cotton, there will be 40 X 840, or 33,600, yards
in 1 pound of 40s. The question then is to find the worsted
counts of a yarn containing 33,600 yards to the pound. Since
length divided by (standard multiplied by weight) equals
counts, then 33,600 -4- (560 X 1) must equal the counts.
Example 2. —Find the equivalent of a 16s cotton yarn in the woolen
run system.
Solution. — 840 X 16 = 13,440
13,440 4 - 1,600 = 8.4-run, woolen. Ans.
Example 3.—Place in worsted counts, a 60s cotton, an
woolen, and a ? 0 s worsted.
Solution. —The cotton thread equals = 90s worsted.
obO
• 8 x 1 600
The 8 -run woolen thread equals-= 7 ^— = 22.85s worsted.
560
The 20s worsted would, of course, be the same.
8 -run
Ans.
Ans.
EXAMPLES FOR PRACTICE
1. Find the equivalent of a 13s cotton in worsted counts.
Ans. 19.5s
2. Find the worsted counts of a 48s cotton. Ans. 72s
3. What are the cotton counts of a 40s worsted? Ans. 26.66s
4. What counts of cotton are equivalent to 90s worsted? Ans. 60s
5. Convert 7-run to cut system. Ans. 37.33-cut
6 . What are the equivalent counts of a 12s silk in woolen runs?
Ans. 6.3-run
7. Find the equivalent of a 65s worsted in cotton counts.
Ans. 43.33s
20
YARN CALCULATIONS, GENERAL
§11
PLY YARNS
METHOD OF NUMBERING
30. Definition. —Very frequently during the process
of manufacturing yarns, two or more threads are twisted
together to form one coarser thread. Such yarns are com¬
monly known as ply yarns, also sometimes called folded,
or twisted, yarns.
31. The method of numbering cotton ply yarns is that
of giving the counts of the single yarns that are folded and
placing before these counts the number that indicates the
number of threads folded; thus, 2/40s indicates that two
threads of 40s single yarn are folded together, the folded
yarn being equal, in weight, to a single 20s thread. During
the process of twisting a slight contraction takes place.
Consequently, to make the resultant counts 20s, the single
yarns that are folded must necessarily be slightly finer than
40s. However, this contraction will not be considered in
the rules and examples to be given, since it is so slight that
it is more a matter of experience than one of mathematics.
CALCULATIONS OF COTTON PLY YARNS
FOLDED YARNS OF THE SAME COUNTS
32. It is not customary in mills to fold yarns of different
counts, since single yarns of the same number make the best
double, or ply, yarns. Consequently, when yarns of the same
counts are folded, in order to find the counts of the result¬
ing ply yarn, it is simply necessary to divide the counts
of the yarns folded by the number of threads that constitute
the ply yarn. For example, if three threads of 90s cotton
are folded to form a ply yarn, the resultant yarn will be
equivalent in weight to a single 80s (90 -f- 3 = 30). The
YARN CALCULATIONS, GENERAL
21
§11
student should be careful to distinguish between the counts
of the ply yarn and the counts of the single yarn that equal
in weight the ply yarn. Thus, if three threads of 90s are
twisted together, the yarn that results from folding these
three yarns, or in other words the ply yarn, is equal in
weight to a single 30s, although when indicating the count
of the ply yarn it is spoken of as a 3-ply 90s.
The method of finding the counts, weight, and length of
ply yarns is similar to that explained in connection with
single yarns, with the exception that the counts of the ply
yarn do not indicate the actual counts of the thread but
instead indicate the counts of the yarns folded. Conse¬
quently, when figuring to find these particulars, the actual
weight of the ply yarn must be taken into consideration
and, on this account, the counts of the single yarn that
the ply yarn equals are considered and not the counts of
the single yarns that are folded.
Example 1.—What is the weight of 642,000 yards of 2-ply 40s cotton
yarn ?
c 042,000 QO 01 n .
Solution. — ■■■ = 08.21 lb. Ans.
LJu X o4U
Explanation. —To make a 2-ply 40s, two ends of 40s are
twisted together; consequently, a yard of the ply yarn will
weigh just twice as much as a yard of one of the single
yarns folded, which will make the ply yarn equal in weight
to a 20s single yarn. Therefore, 20, which is the actual
counts of the ply yarn, is used in the calculation. Since
length divided by (counts multiplied by standard) equals
weight, then 642,000 -f- (20 X 840) must equal the weight of
the yarn.
Example 2.—What is the length of 20 pounds of 2-ply 36s cotton?
Solution. — 20 X 18 X 840 = 302,400 yd. Ans.
Explanation. — 2-ply 36s is composed of two threads
of 36s folded together; consequently, the weight of a yard
of the ply yarn must be just twice that of a yard of one of
the ends folded to make the ply yarn. This will make the
ply yarn equal in weight to an 18s single yarn and 18s must
22
YARN CALCULATIONS, GENERA!
§11
be used as the counts of the ply yarn in the calculation.
Since weight times counts times standard equals length,
then 20 X 18 X 840 must equal the number of yards in
20 pounds of 2-ply 36s.
Example 3.—What are the counts of a 2-ply cotton yarn, 352,800
yards of which weighs 10 pounds?
0 352,800 , 00,01
Solution. — v , nAr . = 42 = 2-ply 84s. Ans.
10 X 840 r 7
Explanation. —Since length divided by (weight times
standard) equals counts, then 352,800 -s (10 X 840) must give
the actual counts of the ply yarn; that is, this result gives the
counts of the ply yarn considered as a single yarn, but since
two single yarns are folded and each of these is just half as
heavy as the folded yarn, then two ends of 84s must be
folded to make the ply yarn, which, consequently, will be
known as a 2-ply 84s.
FOLDED YARNS OF DIFFERENT COUNTS
33. Although not a common practice, in some cases,
especially when it is desired to make a fancy yarn, two
yarns of different counts are folded and sometimes two
yarns of different materials.
Suppose, for illustration, that it is desired to find the
resultant counts of a 40s cotton folded with a 20s cotton.
Take as a basis 840 yards of each yarn; then 840 yards of
the 40s weighs pound; 840 yards of the 20s weighs
To pound. Consequently, after these yarns are folded, there
will be 840 yards of a ply yarn the weight of which is
4~o -f- 2 V = 4at pound.
The example now resolves itself into the following: What
are the counts of a yarn 840 yards of which weighs -40 pound?
Since length divided by (weight times standard) equals
counts, then, 3 ^0 = 13.33s, counts of the ply yarn.
To X 840
This example has been worked out to some length in order
to enable the student thoroughly to understand the method
of numbering ply yarns. A somewhat shorter method, how¬
ever, is as follows:
§11 YARN CALCULATIONS, GENERAL 23
34. To find the resultant count when two threads of
different numbers are folded:
Rule. —Multiply the two counts together and divide the
result thus obtained by the siwi of the counts.
Example. —Same as previous example.
Solution.—
40 X 20
40 + 20
13.33s, counts.
Ans.
PLY YARNS COMPOSED OF MORE THAN TWO THREADS
35 . In many cases it will be found necessary to find the
counts of a ply yarn made from more than two single threads.
Under such circumstances it will be necessary to follow a
somewhat different process. For example, suppose that
three single threads—24s, 36s, and 72s, respectively—are
folded to form a ply yarn and it is required to ascertain the
counts of the resultant yarn. This may be done by follow¬
ing the rule previously given and performing two operations
as follows:
First find the counts of the yarn that would result from
folding the 24s with the 36s as follows: ^ — ^ _ 14 4 S
24 + 36
The example then resolves itself into the following:
What are the counts of a ply yarn made from one thread
of 14.4s and one of 72s? 14.4 X 72 _ ^s. Ans.
14.4 + 72
A somewhat shorter method than this, however, may be
applied to 3 or more ply yarns made from different counts.
Example. —Same as given above.
Solution. — 72 -p 72 = 1
72 -f- 36 = 2
72 -p 24 = 3
6
72 -f- 6 = 12s. Ans.
Rule. — Take the highest counts and divide it by itself and by
each of the other counts. Add the results thus obtained and
divide this result into the highest counts.
24 YARN CALCULATIONS, GENERAL §11
36. To find the resultant counts when more than one
end of the different counts are folded:
Rule. —Divide the highest counts by itself and by each of the
other counts. Multiply the result in each case by the number of
ends of that counts. Add the results thus obtained and divide
this result into the highest counts.
Example. — 4 ends of 80s and 3 ends of 60s are folded to form a
ply yarn; what are the resultant counts?
Solution. — ' 80-5-80 = 1; 1X4 = 4
80 - 5 - 60 = 1|; \\ X 3 = 4
8
80 -5- 8 = 10s, resultant counts. Ans.
37 . When dealing with ply yarns it often becomes neces¬
sary to find the counts of a yarn to be folded with another
to produce the given counts.
Rule. —Miiltiply the two counts together and divide by their
difference.
Example. —What counts must be folded with a 50s to produce
a ply yarn equal in weight to a 30s?
Solution.—
50 X 30
50 - 30
75s. Ans.
Proof.— What are the counts of a ply yarn made by
twisting a 50s with a 75s? = 30s. Ans.
50 + 75
38. Another calculation that must frequently be made
when dealing with ply yarns is that of finding the required
weight of each thread folded in order to produce a required
weight of the ply yarn.
Rule. —Find the counts resulting from folding the two or
more threads; then , as the counts of one thread is to the result¬
ing counts, so is the total weight to the weight required of that
thread.
Example. —It is desired to produce 100 pounds of a ply yarn com¬
posed of an 80s and a 32s twisted together; what will be the required
weight of the 80s and also of the 32s?
11
YARN CALCULATIONS, GENERAL
25
80 X 32
Solution.— ^ = 22.85s, counts of the ply yarn expressed in
single numbers.
x =
80 + 32
32 : 22.85 = 100 : x
100 X 22.85
32
= 71.40 lb. of 32s. Ans.
x —
80 : 22.85 = 100 : x
100 X 22.85
80
= 28.56 lb. of 80s. Ans.
In a case similar to the example given above, after the
weight of one thread has been obtained, it is of course only
necessary to subtract that weight from the total weight in
order to obtain the weight of the other thread; or, in case
more than two threads are folded, then the weight of one of
these threads may always be obtained by subtracting the
combined weight of the other threads from the total weight
of the ply yarn.
Note. —In the previous example the weight of the 80s yarn plus the
weight of the 32s yarn should equal the weight of the ply yarn, but
owing to the use of decimals, examples of this kind seldom give exact
results. Thus, 71.40 pounds + 28.56 pounds = 99.96 pounds; whereas
the total weight should be 100 pounds.
CALCULATION OF COST OF PLY YARNS
39. If the price of each yarn is given and it is required
to find the price per pound of the resultant yarn, it becomes
necessary to multiply the weight of each count of yarn by its
price, add the results, and divide by the total weight. The
answer will be the price per pound of the ply yarn.
Example. —If in the example given in Art. 38, the 80s yarn is
worth 72 cents per pound and the 32s is worth 48 cents per pound, what
will be the cost per pound of the ply yarn?
Solution.—
71.40 lb. of 32s at 48 cents per lb. = $34.27, cost of the 32s yarn
28.56 lb. of 80s at 72 cents per lb. = $20.56, cost of the 80s yarn
$34.27 + $20.56 = $54.83, total cost of ply yarn
$54.83 -f- 100 = 54.8 cents per lb., cost of the ply yarn. Ans.
40 . Another rule for finding the price of 2-ply yarns when
the threads to be twisted together are of different values and
different counts is as follows:
26
YARN CALCULATIONS, GENERAL
§11
Rule. —Multiply the highest coiints by the Price of the lowest
counts and the lowest counts by the price of the highest. Add the
results thus obtained and divide this result by the sum of the
counts. The ansiver will be the price of the ply yarn.
Example. —A 32s yarn costs 42 cents per pound and a 16s yarn
costs 18 cents per pound; what will be the cost per pound of a ply
yarn resulting from twisting these two?
Solution. — 32 X $.18 = $5.76; 16 X $.42 = $6.72
$5.76 + $6.72 = $12.48; 32 + 16 = 48
$12.48 4- 48 = 26 cts. Ans.
WOOLEN AND WORSTED PLY YARNS
41. Ply yarns made from worsted or woolen yarns are
numbered exactly the same as cotton ply yarns. Therefore,
the rules given will be found to apply equally well when
dealing with ply yarns made of worsted and woolen, with
the exception that in each case the standard number of yards
to the pound of the system being dealt with must be used.
Consequently, ply yarns of these systems need no further
explanation here.
PLY YARNS OF SPUN SILK
42. The numbering of ply yarns made from spun silk
will be found to differ somewhat from the methods pre¬
viously explained. Thus, when numbering silk ply yarns,
the counts resulting after folding the yarns is given and this
number is followed by the number that indicates how many
threads are folded.
For example, 60/2 spun silk indicates that two threads
of 120s have been folded together. Thus, it will be seen
that the actual counts of the ply yarn are given instead of
the counts of the single yarn, as is the case in cotton, woolen,
and worsted ply yarns.
Example 1. —What is the weight of 642,000 yards of a 40s 2-ply
spun silk?
„ 642,000 in in _ ,, .
Solution. — — — 19.10/ lb. Ans.
40 X 840
11
YARN CALCULATIONS, GENERAL
27
Explanation.— 40 s 2-ply spun silk is equal in weight
to a single thread of 40s. Consequently, 40 should be con¬
sidered as the counts of the ply yarn when finding weight or
length. Since length divided by (counts times standard)
642,000
equals weight, 40 x g40 mus t equal the weight of the yarn.
Example 2.—What is the length of 20 pounds of a 30s 2-ply spun
silk?
Solution. — 840 X 30 X 20 = 504,000 yd. Ans.
Explanation. —A 30s 2-ply spun silk is equal in weight
to a single 30s; consequently, 30 should be considered as the
counts of the ply yarn. Since standard times counts times
weight equals length, 840 X 30 X 20 must equal the length
of the yarn.
Example 3.—What are the counts of a 2-ply silk yarn if 352,800 yards
weighs 10 pounds?
o 352,800 _ ,
Solution.— 1^X840 = Ans -
Explanation. —The counts of the 2-ply yarn would be
indicated as follows: 42/2 spun silk, which shows that
two ends of 84s have been twisted to make the ply yarn.
PLY YARNS OF DIFFERENT MATERIALS
43 . In all cases where threads of different materials are
twisted together, in order to perform any of the calculations
previously explained, it becomes necessary first to place
these counts in the same system of numbering yarns.
Example. —A 36s cotton and a 48s worsted are twisted to form a
ply yarn; what are the counts of the resultant yarn?
Solution. —It is first necessary to ascertain in which system the
resultant yarn should be placed. In this case the counts of the ply
yarn will be found in both the worsted and cotton systems. In the
first case then, to find the worsted counts of the ply yarn resulting
from twisting these two yarns it is necessary to find the equivalent
36 X 840
counts of the 36s cotton in the worsted system.
560
2.S
YARN CALCULATIONS, GENERA!
11
The 36s cotton is found to equal a 54s worsted so that the question
resolves itself into the following: What are the counts of a ply yarn
resulting from twisting a 54s worsted and a 48s worsted?
54 X 48
1-- -= 25.41, worsted counts of the ply yarn. Ans.
54 + 48
Since in this example it is also required to find the counts of the
ply yarn in the cotton system, it is therefore necessary first to find
48 X 560
the equivalent counts of the 48s worsted in cotton.
840
= 32s.
Having placed the 48s worsted in the cotton system, treat the
worsted as though it were cotton and find the counts of a ply yarn
that will result from folding a 32s and a 36s cotton.
32 X 36
. rr = 16.94, cotton counts of the ply yarn. Ans.
oZ «jo
From this it is seen that if a 36s cotton and a 48s worsted are
twisted together, the counts of the resultant ply yarn will be either
25.41s worsted or 16.94s cotton.
EXAMPLES FOR PRACTICE
1. If a thread of 40s cotton and one of 60s cotton are twisted
together, what are the counts of the resultant ply yarn? Ans. 24s
2. A 20s, 30s, and 60s cotton are twisted together; what are the
counts of the ply yarn? Ans. 10s
3. What are the counts of a yarn that must be twisted with a
44s cotton to produce a ply yarn equal to a 20s? Ans. 36.66s
4. What weight of 40s cotton yarn must be twisted with a 20s to
produce 200 pounds of ply yarn? Ans. 66.65 lb. of 40s
5. What weight of 100s, 80s, and 60s cotton would be used in pro¬
ducing 500 pounds of ply yarn? f 127.65 lb. of 100s
Ans.< 159.56 lb. of 80s
[212.75 lb. of 60s
6 . What counts of ply yarn would result if 18s, 24s, and 30s cotton
are twisted together, and what weight of each would there be in
240 pounds of the folded yarn? f 106.66f lb. of 18s
Ans. 8sI 80.00 lb. of 24s
[53.33ilb. of 36s
7. A ply yarn is made from one end each of 10s, 12s, and 15s; what
are the counts of the ply yarn? Ans. 4s
8 . Find the price per pound and the weight of each yarn in
100 pounds of a ply yarn made from one thread of 40s 2-fold silk at
$2.52 per pound and one thread of 4-run woolen at 40 cents per pound.
173.92 ct. per pound
Ans. < 16 lb. of 40s-2 silk
[84 lb. of 4-run woolen
§11
YARN CALCULATIONS, GENERAL
29
9. A 3-ply yarn is made from 80s, 40s, and 30s worsted and weighs
100 pounds; what weight does it contain of each count of yarn and
what are the counts of the ply yarn? (17.64 lb. of 80s
Ans. 14.117s< 35.29 lb. of 40s
[47.05 lb. of 30s
10. What is the cost per pound of a ply yarn composed of one
thread of 22s yarn at 40 cents per pound and one thread of 40s at
84 cents per pound? Ans. 55.6 ct.
11. What is the price per pound of a ply yarn composed of one
thread of 40s worsted at 96 cents per pound and one thread of 80s 2-fold
silk at $5.28 per pound? Ans. $2.04
12. A ply yarn is composed of one thread of 10s cotton and one
thread of 26-cut woolen; what are the counts in the cut system?
Ans. 13.481-cut
13. What would be the resultant counts in the cotton system of one
end of 40s cotton, one of 40s worsted, and one of 60s silk twisted
together? Ans. 12.631s cotton
14. What counts result from twisting a 60s cotton with a 36s
worsted? Give answers in both cotton and worsted systems.
, f 17.14s in cotton system
s ‘ 125.71s in worsted system
DIAMETER OF YARNS
METHODS OF DETERMINING THE DIAMETER
OF YARNS
44 . Definition.— By the term diameter of yarns is
meant their thickness. Thus* if a thread is said to be
gV inch in diameter, it means that it is -gV inch in thickness,
or in other words, that sixty of these threads can be placed
side by side in 1 inch. The diameters of threads become an
important matter when considering the number of threads
that should be placed in a given space of cloth, since if too
small a number of ends are put in a cloth, it will not in most
cases give good results, while, on the other hand, if more
ends are crowded into a given space than can lie side by
side in that space, it will give the cloth a cockled appear¬
ance. There are two universal methods of finding the
diameters of yarns—by means of the microscope and a
30
YARN CALCULATIONS, GENERAL
§11
micrometer, and by means of the following rules; but it
should be understood that these rules give only approximate
results, since the diameter of a thread is influenced very
largely by the class of material used and by the twist, per
inch placed in the thread.
45. A rule for finding the diameter of yarn that is based
very largely on experience is as follows:
Rule. —Find the number of yards per pound in the counts
under consideration a?id extract the square root of this number.
This result less a deductio)i gives the denominator of a fraction
having 1 as its numerator that expresses the diameter of the yarn
in inches.
The deductions generally made are 16 per cent, for
woolen, 10 per cent, for worsted, and 8 per cent, for cotton.
Example. —Find the diameter of a 40s worsted thread.
Solution. — 40 X 560 = 22,400 yd. per lb.
>122,400 = 150, nearly
150 less 10 per cent. = 135
Therefore, the diameter of a 40s worsted is approximately rss in-,
or in other words, 135 ends of a 40s worsted can be placed side by
side in 1 in. Ans.
BEAMED YARNS
46. A part of the yarn before being woven into cloth is
placed on what are known as loom beams, a large number
of ends of the same length being placed on one beam. The
calculations necessary in connection with the yarn on a beam
will be found to be very similar to those used in connection with
the length, weight, and counts of single ends with this differ¬
ence; viz., that whereas formerly only a single end was being
dealt with, in the present case a large number of ends must
be taken into consideration. Thus, for example, if each end
on a beam is 1,000 yards long and there are 2,000 ends, then
there must be 2,000 X 1,000 = 2,000,000 yards of yarn.
This point should always be taken into consideration when
dealing with yarn placed on a beam. The yarn on a beam
is generally spoken of as the warp yarn, or the warp.
11
YARN CALCULATIONS, GENERAL
31
Note. —Since the examples and rules for finding different data in
regard to the yarn on a beam apply equally as well to one system of
numbering yarns as to another, no particular system will be dealt with,
the examples being taken from different classes of yarn.
47 . To find the counts of the yarn on a beam containing
only one size of yarn, the weight, length, and number of
ends being given:
Rule. — Multiply the length , expressed in yards , by the num¬
ber oh ends on the beam and divide the result thus obtained by the
weight , expressed in pounds , times the sta?idard number oh yards
to the pound.
Example. —A warp beam contains 2,400 ends of cotton each 200
yards long. The weight of this yarn is 15 pounds; what are the counts?
o 200 X 2,400 oc .
Solution.— - = 38.095s. Ans.
lo X o4u
Explanation. —Since there are 2,400 ends and each end
is 200 yards long, there must be 2,400 X 200 = 480,000 yards
in all. The question then resolves itself into finding the
counts of a yarn 480,000 yards of which weighs 15 pounds.
Since length, in yards, divided by (weight, in pounds, times
standard) always equals counts, 480,000 divided by (15 X 840)
must give the counts.
In some cases the weight given will be found to include
not only the weight of the yarn, but also that of the beam on
which the yarn is placed. When this occurs, it is necessary
first to deduct the weight of the beam from the weight given,
in order to obtain the true weight of the yarn.
48. To find the number of ends on a beam when weight,
length of the warp, and size of the yarn are known:
Rule .—Multiply the weight , in pounds , by the standard
number and by the size oh the yarn. Divide the result thus
obtained by the length oh the warp , in yards.
Example 1.—A worsted warp is 800 yards long and weighs
200 pounds exclusive of the beam. If the warp is composed of
20 s yarn, how many ends does it contain?
200 X 560 X 20
2,800 ends. Ans.
Solution.
800
32
YARN CALCULATIONS, GENERAL
§11
Example 2.—A woolen warp is 600 yards long and weighs
120 pounds exclusive of the beam. If the warp is composed of
4|-run yarn, how many ends does it contain?
120 X 1,600 X 41
Solution.—
600
-= 1,440 ends. Ans.
49. To find the weight of yarn on a beam when length,
number of ends, and counts are given:
Rule. —Multiply the length , expressed in yards , by the num¬
ber oh ends on the beam and divide the result thus obtained by
the standard number of yards times the counts of the yarn.
Example.— A beam contains 2,400 ends of 20s cotton, each end
being 500 yards long; find the weight of the yarn.
Solution.—
500 X 2,400
840 X 20
71.428 lb. Ans.
Explanation.— By multiplying the length of one end by
the total number of ends on the beam the total length of
yarn on the beam is obtained; and since the length, expressed
in yards, divided by the standard times the counts equals
the weight, in pounds (2,400 X 500) -4- (840 X 20), will give
the weight of the yarn on the beam.
50. To find the length of a warp when weight, number
of ends, and size of the yarn are known:
Rule.— Multiply the weight of the warp , in pounds , by the
standard number and by the size of the yarn , and divide the
result thus obtained by the number of ends in the warp.
Example 1. — A worsted warp contains 2,400 ends of 12s yarn and
weighs 200 pounds; how long is it?
„ 200 X 560 X 12 C/VA J A
Solution.— - 2~40b- = •^ ns>
Example 2. — A woolen warp contains 1,200 ends of 2-run yarn and
weighs 231 pounds. How long is it?
„ 231 X 1,600 X 2 J A
Solution. —- — *>00 - = 616 yd. Ans.
51. To find the length of warp that can be placed on
a beam:
Rule. —Find the weight of yarn that the beam will contain ,
by weighing a beam of the same size when filled with yarn and
§11
YARN CALCULATIONS, GENERAL
33
deducting the weight of the bea?n itself. Then apply the rule in
Art. 50.
Example 1.—A certain size beam when filled with yarn weighs
140 pounds, the beam itself weighing 50 pounds. What length of a
warp composed of 1,800 ends of 20s cotton can be placed on it?
Solution.—
140 — 50 = 90 lb. of yarn
90 X 840 X 20
1,800
840 yd. Ans.
Example 2. —Same beam as in example 1 . How many yards of
a warp containing 3,000 ends of 40s cotton can be placed on it?
Solution.—
140 — 50 = 90 lb. of yarn
90 X 840 X 40
3,000
1,008 yd. Ans.
AVERAGE NUMBERS
52. In case different counts of yarn are placed on the
same beam, as very frequently occurs, it will be found neces¬
sary to first find the average number , or average counts, of the
different yarns before making other calculations. By the
term average number, or average counts, is meant a
count of yarn that will give the same weight, provided that
the same number of ends and the same length occur in both
cases. Thus, if 400 ends of 10s and 800 ends of 20s weigh a
certain number of pounds, then 1,200 (400 + 800) ends of the
average counts will weigh the same, provided that the ends
are the same length in both cases.
53. To find the average counts of the ends on a beam
when the ends are of different counts:
Rule. —Divide the total number of ends of each count by its
own count. Add these results together and divide the result thus
obtained into the total number of ends in the warp.
Example. —There are placed on the same beam 1,800 ends of 60s
cotton and 800 ends of 40s cotton; what are the average counts?
Solution. — 1 8 0 0 = 60 = 30
800 = 40 = 20
2600 50
2,600 -r- 50 = 52s, average counts. Ans.
34
YARN CALCULATIONS, GENERAL
§11
Explanation. —Suppose that each end on the beam is
, 840 yards long, then (1,800 X 840) 4- (60 X 840) = 30 pounds,
. weight of the 60s yarn. Also (800 X 840) 4- (40 X 840)
= 20 pounds, weight of the 40s. The combined weight of
the 60s and 40s will, of course, be 30 + 20 = 50 pounds.
There are 2,600 ends each 840 yards long and this length of
yarn weighs 50 pounds. The question, therefore, is if all the
yarn is the same counts, what must be the counts when
2,600 X 840 yards weighs 50 pounds? Since length, expressed
in yards, divided by standard times weight, expressed in
pounds, equals counts, then (2,600 X 840) 4- (840 X 50) must
equal the average counts. Since the length in this case, or
840 yards, cancels in every calculation with the standard
number of yards, which is also 840, these two terms may be
omitted and the example stated as in the solution. This
holds good with any length.
54 . In case more than two different counts are placed
on the same beam, the same rule will be found to apply.
Example. —What are the average counts in case 200 ends of 20s,
1,000 ends of 40s, and 900 ends of 45s are placed on the same beam?
Solution. — 2 0 0 4- 20 = 10
10004-40 = 25
9004-45 - 20
2T(To 5~5
2,100 4-55 = 38.18s, average counts. Ans.
55 . In cases where the order of arranging the different
counts of yarn in the warp is given, the total number of
ends in the warp not being known, the same rule will be
found to apply by considering the number of ends in the
arrangement as the total number of ends.
Example. —A warp is arranged 48 ends of 36s and 2 ends of 10s;
find the average number.
Solution. — 4 8 4- 36 = 1.3 3 3
2 4-10 = .2
5 0 1.5 3 3
Total number of ends = 50. 50 4- 1.533 = 32.615, average num¬
ber. Ans.
§11
YARN CALCULATIONS, GENERAL
35
56. If the yarn is of different material, such as cotton
and worsted, then it is necessary first to place the different
counts in the same system before applying the rule for
finding the average number.
Example. —There are placed on a beam 2,000 ends of 40s cotton
and 450 ends of 45s worsted; what are the average counts in the
cotton system?
Solution. —First find the equivalent cotton counts of 45s worsted.
45 X 560
—r— = 30s cotton
o4(J
This example then resolves itself into finding the average counts
of 2,000 ends of 40s and 450 ends of 30s.
2 0 0 0 -=- 40 = 50
450-T-30 = 15
2450 65
2,450 -5- 65 = 37.69s, average counts. Ans.
FANCY WARPS
57. When more than one color of yarn is placed on the
same beam, it frequently becomes necessary to find the total
number of ends of each color and the weight of each par¬
ticular yarn.
In order fully to understand the explanations given in this
connection it will be necessary first to consider a few terms
that will frequently be met with. The yarn that is placed on
the loom beam is known as the warp, or warp yarn. It is
this yarn that forms the threads running lengthwise in the
cloth and is thus distinguished from the yarn running across
.the cloth, which is known as the fillingr- In case the warp
yarn is composed of different colors or different counts, the
order in which the different counts or colors are placed on
the beam is known as the pattern of tlie warp. Thus, if
the warp is arranged 4 ends of black, 4 ends of white, 4 ends
of black, 4 ends of white, and so on across the cloth, the
warp pattern is said to be 4 black, 4 white.
58. To find the number of ends of each color of yarn on
a beam when the warp pattern and total number of ends
are given:
36
YARN CALCULATIONS, GENERAL
§11
Rule .—As the number of ends in one pattern is to the number
of ends of any one color in the pattern , so is the total number of
ends in the warp to the total manber of ends of that color.
Example. —The yarn on a beam is arranged 16 ends black, 8 ends
white, 16 ends black, 8 ends gray; how many ends of each color are
there if there are 2,400 ends on the beam?
Solution.— 1 6 ends black
8 ends white
1 6 ends black
8 ends gray
4 8 = total number of ends in one pattern.
There are 32 ends of black in one pattern.
Therefore, 48 : 32 = 2,400 : ;tr
32 X 2,400 , AAA ^ t U1 ,
x = - -r^ -= 1,600 ends of black.
4o
There are 8 ends of white in one pattern.
Therefore, 48 : 8 = 2,400 : x
8 X 2,400
Ans.
x —
48
= 400 ends of w T hite. Ans.
There are 8 ends of gray in one pattern.
Therefore, 48 : 8 = 2,400 : x
8 X 2,400
x —
48
400 ends of gray. Ans.
Explanation.— The total number of ends in one pattern
must always bear the same relation to the number of ends of
any one color in the pattern that the total number of ends in
the warp bears to the total number of ends of that color.
59. If it is desired to find the weight of the ends of each
color, after having obtained the total number of ends of each
color, apply the rule for finding weight when length, counts,
and number of ends are given.
EXAMPLES FOR PRACTICE
1. What is the weight of a cotton warp 200 yards long that con¬
tains 2,400 ends of 60s? Ans. 9.52 lb.
2. A cotton warp is arranged 3 ends black, 1 white, 7 black,
1 white, counts 24s, 154 yards long, 2,400 ends; find the weight of
each color and the total weight of the warp.
f 15.27 lb., black
Ans.< 3.05 lb., white
118.32 lb., total weight
§11
YARN CALCULATIONS, GENERAL
37
3. A worsted warp 780 yards long contains 2,480 ends of 20s yarn;
what is the weight of the yarn? Ans. 172.714 lb.
4. What are the counts of warp yarn in a cotton warp 200 yards
long that contains 2,400 ends and weighs 15 pounds? Ans. 38.09s
5. A cotton warp is 120 yards long, contains 4,480 ends, and
weighs 40 pounds; what are the counts? Ans. 16s
6. A worsted warp and beam weigh 170 pounds, the beam alone
weighing 70 pounds; if the warp is 350 yards long and contains
2,240 ends, what is the size of the yarn? Ans. 14s
7. 40 pounds of 25s cotton is made into a warp containing 875 ends;
what is the length? Ans. 960 yd.
8. A woolen warp contains 1,648 ends of 4j-run yarn and weighs
103 pounds; how long is the warp? Ans. 450 yd.
9. A woolen warp is 320 yards long and weighs 124 pounds, exclu¬
sive of the beam; if the warp is composed of 2^-run yarn, how many
ends does it contain? Ans. 1,550 ends
10. How many ends are there in a cotton warp 1,400 yards long
weighing 200 pounds, the counts of the yarn being 20s?
Ans. 2,400 ends
11. What are the average counts in a cotton warp arranged 24 ends
of 40s and 1 of 8s? Ans. 34.48s
12. A worsted warp contains 1,620 ends of 45s yarn and 840 ends
of 20s yarn; find the average counts of the warp. Ans. 31.53s
13. A woolen warp contains 900 ends of 4^-run yarn and 450 ends
of 2i-run; find the average size of the yarn. Ans. 3f-run
14. A cotton warp is arranged 40 ends of 20s and 20 ends of 10s;
what are the average counts? Ans.' 15s
38
YARN CALCULATIONS, GENERAL
§11
METRIC SYSTEM OF NUMBERING
YARNS
60. In recent years there has been from time to time
considerable agitation along the lines of adopting one system
of numbering all classes of yarns used in textile manufac¬
turing. The objects of these movements are: first, to bring
about the unification of the method of stating the degree of
fineness of the yarns for all varieties of fibers used in the
textile industry in the whole world; second, to adopt a
method that would be practical in every way.
The advantages of such a system as this would be many.
The chief objection to it is that, from long usage, the methods
at present adopted are too well developed for a single cor¬
poration or a single country to take on itself such a reform,
without being assured that its neighbors and competitors
will simultaneously do the same thing, as a mill instituting
such a method would be at a disadvantage as compared with
its competitors.
The method usually set forth is that of numbering all
classes of yarns by what is known as the metric system,
in which 1 meter of No. 1 yarn weighs 1 gram, the meter
being the unit of length in the metric system and the gram
the unit of weight. The only parts of the metric system
that it is necessary to understand, in order to convert one
system into the other, are the equivalents of the meter and
the gram, which are as follows:
1 yard = .914 meter, 1 pound = 453.59 grams
61 . To find the number of yarn in any present standard
system that corresponds to the number of yarn in the metric
standard system:
Rule .—Multiply the counts, given in the metric system , by
453.59 (grams in 1 lb.) and divide by the standard number of
§11
YARN CALCULATIONS, GENERAL
39
yards to the pound in the present system multiplied by .914
(meter in 1 yard).
Example. —A cotton yarn numbered according to the metric system
is marked 40s. Find the counts in the present system.
40 X 453.59
Solution.— 840 X 914 = 23 ‘ 631s - Ans -
62. To find the number of yarn in the metric standard
system that corresponds to the number of yarn in any present
standard system:
Rule. —Multiply the counts, given in the present system, by
the present standard number of yards to the pound and by .914
(meter in 1 yd.) and divide by 453.59 {grams in 1 pound).
Example. —A worsted yarn numbered according to the present
system is marked 46s. Find the counts in the metric system.
46 X 560 X .914
51.907s. Ans.
Solution.—
453.59
CLOTH CALCULATIONS,
COTTON
CALCULATIONS NECESSARY FOR CLOTH
PRODUCTION
INTRODUCTION
1. Importance of Cloth Calculations. —The calcula¬
tions that come under the head of cloth are by no means the
least essential of the many dealing with cotton-mill processes.
Important alike to the designer, superintendent, and treas¬
urer, they play a leading part in the manufacture of the raw
stock into the finished product. Not only is it necessary to
ascertain the many details regarding the structure of a piece
of cloth before it can be made, but it must also be learned
whether it is possible for the mill to make such a cloth.
2 . Definitions. —After the warp yarn has been wound
on the loom beam, the separate ends are drawn through the
harnesses and afterwards through the reed. The warp is
then ready to be placed in the loom. The harnesses are
attached to mechanisms that raise and lower them; and,
since some of the harnesses are up while others are down,
a division of the warp yarn must necessarily take place.
It is through the space formed by this division that the
filling passes and, by this manner of interlacing, the cloth
is formed.
In the language of the mill, the threads of a cloth that run
lengthwise of the piece, or the warp, are always spoken of as
For notice oi copyright, see page immediately following the title page
2 12
2
CLOTH CALCULATIONS, COTTON
§12
the ends, while those that run from side to side are known
as the picks. A cloth is said to be so many sley, which
means that it contains so many ends per inch. It is also
spoken of as being such a pick cloth, by which it is meant
that the cloth has so many picks per inch. Thus, regular
print cloth is said to be 64-sley and 64-pick, which means
that the cloth contains 64 ends and 64 picks per inch; this is
known as the counts of the cloth. When cloth contains the
same number of ends per inch as picks it is spoken of as
being so many square. Thus, the print cloth just referred
to is known as 64 square.
When specifying the counts of a cloth in writing, the
number of ends per inch is always placed first and is fol¬
lowed by the multiplication sign after which the number
of picks per inch is placed. Thus, if a cloth contains 80 ends
and 60 picks per inch, it is written 80 X 60 and, in speaking
of the counts of this cloth, it is said to be eighty by sixty.
In speaking of the weight of cotton cloth, the number
of yards in a pound is considered and the cloth is said to
be a so many yard cloth. Thus, ordinary print cloth is
spoken of as being a 7-yard cloth , which means that it takes
7 yards of the cloth to weigh 1 pound. This method differs
very materially from that in practice in the woolen and
worsted trades, where a cloth is said to be a so many
ounce cloth; that is, if a piece of cloth weighs 12 ounces
to the yard it is said to be a 12-otince cloth. The weight
of heavy cotton goods, such as duck, is indicated by the
weight of a square yard; that is, a piece of duck weighing
7 ounces to the square yard, is spoken of as 7-ounce duck.
The other specifications necessary in connection with a
piece of cloth are the width, the counts of the warp yarn,
and the counts of the filling. In giving these specifica¬
tions they are shown as follows: 48 X 52 — 36" — 4.15 yards
— 18s warp — 22s filling. The counts of the warp and filling
are sometimes written in the following form: 18s/22s. These
specifications show that the cloth is 48-sley, 52-pick, 36 inches
wide, 4.15 yards to the pound, the warp being 18s, and the
filling 22s.
12
CLOTH CALCULATIONS, COTTON
3
■8 , ..
« -.. ' '
K|
1 ^©»«*5^»<4 = 20s reed '
Ans.
12. When cloth is unevenly reeded it is more difficult to
find the required reed than with evenly reeded cloth. In
cloth of this description, the reeding of the warp is usually
arranged in a definite order in conjunction with the pattern
of the warp, which is generally, although not always, com¬
posed of yarns of different counts.
13. To find the reed to use for an unevenly reeded cloth:
Rule. —Multiply the dents in one pattern by the patterns in
the warp and divide by the width in the loom.
WOOLEN AND WORSTED
9
§13
Example. —A warp contains 3,920 ends and is to be reeded 62 inches
wide in the loom; what number of reed is required if the warp is
dressed and reeded as follows?
4
white
drawn in
1 dent
1
white
1
1
tan
white
> drawn in
1 dent
1
tan
4
white
drawn in
1 dent
12
tan
drawn in
6 dents
4
white
drawn in
1 dent
1
white 1
1
1
tan
white
* drawn in
2 dents
1
tan
12
white
drawn in
6 dents
4
blue
drawn in
1 dent
1
tan
2
white
drawn in
2 dents
1
tan
144
tan
drawn in
7 2 dents
19 6 ends in pattern 9 3 dents in pattern
Note.—T he above is known as a reed draft.
Solution.— 3,920 -p 196 = 20 patterns in warp
= 30s reed. Ans.
CONTRACTION IN WEAVING
14 . Fabrics contract in two ways during the weaving
process. They contract in length; that is, to weave one cut
of 50 yards of cloth, the warp must be longer than 50 yards
because of the contraction in interlacing with the filling. As
already explained, all cloths are formed by the interlacing of
two series of yarns—the warp and the filling—and, since in
interlacing in weaving, the two series of yarns bend around
each other to a greater or less extent, it naturally follows
that a piece of cloth will not be so long as the warp from
which it was woven. There will be found many causes that
affect the amount of contraction of the warp in weaving.
The tension of the warp during weaving will affect it to a
10
CLOTH CALCULATIONS,
§13
slight extent. The comparative counts of the warp and
filling also have an influence, since it will readily be seen
that, if the warp is made of coarser yarn than the filling, it
will not be deflected to so great an extent and therefore
will not contract so much. The contraction of the warp in
weaving is known as take-iip in weaving.
The cloth also contracts in the width, or in the direction of
the filling; that is, the woven cloth is narrower than the cloth
is reeded. This latter depends largely on the character of
the filling, whether coarse or fine, or hard or soft twist.
The contraction of goods in weaving, in general, is largely
influenced by the class of weave that is used, or in other
words the manner of interlacing the warp and filling; thus, it
will be seen that a weave that interlaces only once in 5 ends
or 5 picks, like a satin weave, will not contract the same as a
weave that interlaces at every end and pick, like a plain
weave. Often certain weaves will greatly increase the con¬
traction of one system of yarn, either warp or filling, and
correspondingly decrease the contraction of the other. The
contraction of- cloth in weaving varies from 4 to 15 per cent.
No definite rule can be given for this contraction with woolen
and worsted goods, and the value of minute records of each
piece of cloth that the mill makes cannot be overestimated.
The contraction in weaving must not be confounded by
the student with the shrinkage of the cloth in finishing. This
latter varies with the class of goods from 10 to 15 per cent,
for cassimeres up to from 25 to 30 per cent, for beavers
and kerseys.
15. To find the length of warp required to weave a
required length of cloth, the take-up of the warp in weaving
being known:
Rule. —Divide the required length of cloth by 100 per cent,
minus the percentage of take-up in weaving.
Example. —How many yards of warn are necessary to weave
72 yards of cloth if the take-up in weaving is 4 per cent.?
Solution. — 100 per cent. — 4 per cent. = 96 per cent.
72 -p .96 = 75 yd. Ans.
13
WOOLEN AND WORSTED
11
Explanation. —It would seem in the above example that
all that is necessary is to find 4 per cent, of 72 yards and
add it, but it must be remembered that it is the warp
that takes up 4 per cent., not the cloth. The length of the
cloth (72 yards) is the length of the warp minus 4 per cent.;
therefore, it is 96 per cent, of the original length of warp.
That this example is correct may be seen by finding 4 per
cent, of the length of the warp (75 yards) and subtracting
the length obtained (3 yards) from it.
16 . In making allowances for various percentages when
figuring any contraction or shrinkage of cloth or yarn or
other results dependent on such contraction or shrinkage, in
case of any doubt as to the proper method of applying the
percentage the following rules should be borne in mind:
Rule I. — When figuring forwards in the direction of the process
of manufacture, as for instance, from yarn to cloth, or from woven
cloth to finished cloth , if the result is to be increased, multiply by
100 per cent, plus the percentage to be taken; or if the result is to
be decreased, by 100 per cent, minus the percentage to be taken.
Rule II. — When figuring in the opposite direction to the proc¬
ess of manufacture, as for instance, from finished cloth to woven
cloth, or from woven cloth to yarn, divide, if the resjilt is to be
increased, by 100 per cent, mums the percentage to be taken; or
if the result is to be decreased, by 100 per cent, plus the per¬
centage taken. _
SHRINKAGE IN FINISHING
17 . The shrinkage of a cloth in finishing is entirely dis¬
tinct from the contraction of the cloth in weaving, and is an
item that can only be determined by the experience and
judgment that is obtained by a familiarity with different
fabrics, the methods employed in finishing them, and their
peculiarities in finishing. After close observation it will be
possible to tell about how much each fabric made in the
mill will shrink in the finishing process. The shrinkage, of
course, varies with the stock entering into the goods, the
kind of cloth desired, and the finishing process. Goods
12
CLOTH CALCULATIONS,
§13
should be reeded wide enough to allow for the shrinkage
during finishing. It should be understood, however, that
although due consideration of the probable shrinkage is nec¬
essary, there is some leeway, and the finisher will finish the
goods to the width and weight required, within reasonable
limits.
Woolen goods shrink more than worsted, and consequently
should be reeded wider and warped longer for the same width
and length of cloth. Goods that are fulled shrink more than
goods that are not fulled. Heavy woolen goods usually
are heavily fulled, often being triple milled, and will shrink
as high as 30 per cent., averaging from 25 to 30 per cent.
For light-weight, fulled woolens an allowance for shrinkage
in width of from 12^ to 18 per cent, is necessary, while if
not fulled, a smaller allowance, say from 10 to 15 per cent.,
is sufficient. For light-weight worsted goods with a clear
finish, from 8 to 12i per cent, is a sufficient allowance for
shrinkage from reed to finished cloth, while if fulled (which
is rarely done) from 12i to 15 per cent, should be allowed.
For heavy-weight worsted goods with a clear finish, from
12i to 15 per cent, should be allowed for the shrinkage in
width, while if the goods are fulled, an allowance of from
15 to 20 per cent., or even more in some cases, should
be made.
As a general rule, fabrics do not shrink as much in length as
as in width, especially those that are not fulled, the action in
passing through the finishing machinery being to keep the
goods stretched in length. This, however, is not a universal
rule. When a fabric shrinks in length, the counts, or size, of
ihe warp yarn is made coarser and the weight of the cloth per
yard is increased. A shrinkage in width does not result in
an increase in weight, but does make the counts of the filling
coarser, since the width of the cloth is made narrower and
the length of the filling consequently shorter, while the
weight remains the same. The gain in weight that a fabric
obtains in shrinking in length, and also the increase in the
counts of the warp and filling, is somewhat altered by the
loss of weight in finishing, which occurs from scouring out
13
WOOLEN AND WORSTED
13
the oil and grease in the cloth, as well as the loss in certain
processes, such as napping, shearing, etc. In the case of
worsted goods, this loss will about counterbalance the
weight gained by the shrinkage of the cloth in length, so
that the counts of warp and filling and the weight per yard
of the finished goods are practically the same as from the
loom. Woolen goods if shrunk to a considerable extent in
length will weigh more per yard after finishing than before,
but if not shrunk greatly in length, will be lighter when
finished, owing to the loss in weight in finishing, due to the
cleansing of the cloth from grease, etc.
18. To find the original size of a yarn when the counts
of the yarn in the finished cloth, the total shrinkage, and the
loss in finishing are known:
Rule. —Divide the finished cozints of the yarn , multiplied
by 100, by 100 minus the difference between the total shrinkage
and the loss in finishing.
Example 1.—If the warp in a piece of cloth has shrunk 20 percent,
in length from the beam to the finished cloth, and the goods have
lost 10 per cent, in finishing, what were the original counts of the
warp yarn if the counts in the cloth are found to be 3.15-run?
Solution. — 20 per cent. — 10 per cent. = 10 per cent.
3.15 X 100
100 - 10
3.5-run. Ans.
Note.—I f in this example the shrinkage in finishing and the contraction in
weaving were considered separately, the example would have to be solved in two
steps, first finding the counts of the yarn in the woven cloth and then the counts of
the yarn on the beam.
Example 2.—A piece of cheviot-finish worsted cloth has shrunk
18 per cent, in width from the reed to the finished cloth. If the
counts of the filling in the finished cloth are 31.68s and the cloth has
lost 6 per cent, during finishing, what were the original counts of the
filling?
Solution. — 18 per cent. — 6 per cent. = 12 per cent.
31.68 X 100
100 - 12
36s. Ans.
In a manner similar to that explained for finding the
original counts of the yarn, the weight of the cloth from the
14
CLOTH CALCULATIONS,
13
loom may be found if the finished weight, the shrinkage in
length, and the loss in finishing are known. The shrinkage
of the cloth in width must be taken into consideration in cal¬
culations for finding the width of the cloth in the reed, etc.
FANCY PATTERNS
19. When it is desired to find the number of ends of each
color or count in the warp from a piece of cloth containing a
warp pattern, the following method may be used to advan¬
tage in some instances, especially if the cloth is a stripe
pattern.
20. To find the number of ends of each count or color m
a cloth when different counts, colors, or materials are used:
Rule .—Find the number of patterns in the cloth by dividing
the total width by the width of one pattern and multiply the
result thus obtained by the number of ends of each color in the
pattern. The result will in each instance be the number of ends
of that particular counts or color in the warp.
Example. —A small sample of worsted cloth is found to be dressed
12 black, 2 slate, 12 black, and 2 white, and it is desired to make a
similar cloth 56 inches wide; how many ends of each color will there
be in the warp?
Solution. —With a small steel rule and a pair of dividers it is found
that there are exactly three patterns in 1 inch of the sample. Then,
one pattern occupies i inch. In each pattern there are 24 ends of
black, 2 ends of slate, and 2 ends of white.
56 -p §■ = 168 patterns in the cloth
168 X 24 = 4,032 ends of black
168 X 2 =
168 X 2 =
336 ends of slate
336 ends of white
Ans.
This gives a total of 4,032 + 336 + 336=4,704 ends in the
warp, but as in this example no allowance has been made
for selvages, it will be assumed that 36 ends are drawn
on each side of the warp for a selvage, or 72 ends for both
selvages. The total number of ends in the above warp will
then be 4,704 + 72 = 4,776 ends.
13
WOOLEN AND WORSTED
15
Continuing still further with this same example, it will be
supposed that it is desired to find the reed to use and the width
that this cloth must be set in the loom to finish to the desired
width. Suppose that this particular kind of cloth shrinks
15 per cent, from reed to finished cloth; that is, the total
shrinkage, including the contraction from reed to cloth and
the shrinkage in finishing, is 15 per cent, in width. In order
to find the reed to use, the ends per inch in the loom must
be found. It must be thoroughly understood in finding this
item that, as the cloth shrinks 15 per cent, from reed to
finished cloth, the width in the reed represents 100 per cent,
and the width of the finished cloth is 85 per cent, of this width.
21 . To find the width in the loom inside selvages:
Rule. —Multiply the finished width inside of selvages by 100
and divide the result thus obtained by 100 minus the percentage
of total shrinkage from reed to finished cloth.
Example. —The same as in Art. 20.
Solution.—
56 X 100
100-15
65.88 in., width in reed inside selvages. Ans.
22 . To find the number of ends per inch in the reed:
Rule. —Divide the total number of ends in the warp exchisive
of the selvage ends by the width in the reed inside of the selvages.
Example. —The same as in Art. 20.
Solution. — 4,704 -5- 65.88 = 71.40 ends per in. in reed. Ans.
In a case like this it is customary to take the nearest
number that will divide evenly, in order to obtain a reed that
is in stock if possible, so that in this case it will be assumed
that there are 72 ends per inch in the reed.
23 . To find the reed when the ends per inch in the loom
are known:
Rule. —Divide the ends per inch in the loom by the mimber of
ends per de?it that are suitable for the warp yarn and the fabric
being woven.
16 ' CLOTH CALCULATIONS, §13
Example. —Suppose that in the sample considered in Art. 22 the
fabric is such that it should be reeded four in a dent; find the reed.
Solution. — 72 h- 4 = 18s reed. Ans.
In the above fabric, the calculated number of ends per inch in
the loom, 71.40, was not used, owing to the reed; so in order
to find the actual width in the reed, it will now be necessary
to use an 18s reed, four in a dent, or 72 ends per inch in
the loom.
4,70 4
4 X 18
65i inches, width in loom inside of selvages
It was stated that 72 ends of selvage, or 36 ends on each
side, would be added to this warp. These, if reeded six in
a dent with an 18s reed, will make exactly a 3 -inch selvage
on each side of the cloth; therefore, to find the width of the
cloth in the reed over selvages, it is simply necessary to
add f inch to the width in the reed inside selvages; thus,
65i -f- f = 66 inches.
24. Although rules have been given for finding the
counts, or size, of a yarn when the weight of a given
number of yards is known, it is well to give here the rule for
finding the size of the yarn under other conditions, such as
occur when analyzing a sample of cloth.
25. To find the size of a given yarn when the weight of
a definite number of inches is known:
Rule. —Multiply the number of inches weighed by 7 ,000
{grams in 1 pound ) and divide the result thus obtained by
the weight , in grains , times the standard number times 36
(inches in 1 yard.)
Example 1.—It is found that 29 inches of woolen yarn weighs
.7 grain; what is the size of the yarn?
c 29 X 7.000 _ _ 0 .
Solution. 7x 1,600 X 36 = 6 03 - run ' Ans '
Example 2. —It is found that 58 inches of worsted yarn weighs
1.4 grains; what are the counts of the yarn?
58 X 7,000
1.4 X 560 X 36
Solution.—
= 14.38s. Ans.
13
WOOLEN AND WORSTED
17
26. Nothing has yet been said of the weight of cloth,
which is in itself an important item and deserves the careful
attention of the student.
27. To find the weight per yard of finished cloth, in
ounces, when the weight of 1 square inch, in grains, is
known:
Rule. — Multiply the weight of 1 square inch, in grains, by
the width of the cloth and by 36 ( inches i?i 1 yard), and divide
by 437.5 (grains in 1 ounce).
Example.— A square inch of worsted dress goods weighs 2.2 grains.
If the cloth is 44 inches wide when finished, what is the weight per yard?
0 2.2 X 44 X 36 „
Solution. — - . - = /.96oz. Ans.
46 / .5
28. To find the weight of the warp in 1 yard of finished
cloth:
Rule. —Multiply the ends per inch by the width of the cloth,
thus obtaining the ends in the warp, and by 16 (ounces in
1 pound), and divide the result thus obtained by the size of the
yarn in the finished cloth times its standard number.
Example.— In the cloth mentioned in the previous example there are
44 warp threads per inch, and the counts of the yarn in the finished cloth
are 13.88s; what is the weight of the warp in 1 yard of finished cloth?
„ 44 X 44 X 16
Solution. i 3 .88 x 6W = 3 98 °*' Ans '
29. To find the weight of filling in 1 yard of finished
cloth:
Rule. — Multiply the picks per inch in the finished cloth by
the width and by 16 (ounces in 1 pound), and divide the result
thus obtahied by the counts of the filling in the finished cloth
multiplied by the sta?idard number.
Example.— Suppose that in the same piece of cloth as in the
previous examples there are 41 picks per inch and the size of the
filling in the finished cloth is 12.94s; find the weight of filling in 1 yard
of finished cloth.
Solution.- ~ 12 , 94 x 660 = 3.98 os. Ans.
In this piece of cloth it will be noticed that the weight of
warp and filling in a yard of cloth is the same. To find the
18
CLOTH CALCULATIONS,
13
weight of 1 yard of finished cloth it is now only necessary to
add the weights of the warp and filling, in this case 3.98 and
3.98 ounces, which, when added, equal 7.96 ounces. It will
be seen that this is a method of proving the weight of 1 yard
of finished cloth as found from the weight of 1 square inch.
Note. —In Arts. 27, 28, and 29 the width of the finished cloth is
considered inside of the selvages, and the calculations and rules for¬
mulated on this basis. The reason for this is that when analyzing a
small sample of cloth, the required finished width is usually given
inside of the selvages, and by ignoring the selvages at this point the
weight per yard of the cloth, as found from the weight in grains of
1 square inch, can be readily proved, as explained. The desired num¬
ber of ends for selvages can afterwards be added to the warp and
width in reed in finding other particulars.
30. The weight of the finished cloth is easily obtained
by means of these rules, but in cases where it is desired to
know the weight of the cloth from the loom the following
method is pursued.
31. To find the weight of the warp yarn in a yard of
woven cloth from the loom:
Rule.— Multiply the number of ends in the warp , including
the selvage e?ids, by 16 ( dunces in 1 pound ), a?id divide the result
thus obtained by the original size of the warp yarn multiplied
by the standard number. To the result tlms obtained , add the
percentage of take-up in weaving.
Example.— A woolen warp in a piece of cloth contains 840 ends and
30 selvage ends; the original size of the yarn is 2.14-run and the warp
takes up 6 per cent, in weaving; what is the weight of warp in the cloth
from the loom?
Solution.—
4.06 oz. + 6 per cent. = 4.30 oz. Ans.
While it is true that the cloth takes up, or shrinks, from
the reed to the cloth, or in width, and this is take-up, or
shrinkage, in weaving, it does not affect the weight of the
cloth, as cloth is measured and weighed by the linear yard;
therefore, it is only the shrinkage of the cloth in length that
affects its weight.
32. To find the weight of filling in a yard of woven
cloth from the loom:
13
WOOLEN AND WORSTED
19
Rule .—Multiply the width of the cloth in the reed in inches
by the picks per inch and by 16 (ounces in 1 pound) and divide
the result thus obtained by the size of the filling multiplied
by the standard number .
Example.— The cloth mentioned in the example in Art. 31 is
34.15 inches in the reed and contains 27 picks per inch woven; the size
of the filling is 2.14-run; what is the weight of filling in 1 yard of
cloth from the loom?
Solution.—
34.15 X 27 X 16
2.14 X 1,600
4.30 oz. Ans.
33. To find the weight of the cloth from the loom it is
simply necessary to add the weights of the warp and filling,
which in the above examples, it will be noticed, are equal.
Therefore, the weight of the above cloth from the loom is
equal to 4.30 ounces + 4.30 ounces, or 8.60 ounces.
EXAMPLES FOR PRACTICE , 7 ^
1. A warp contains 2,924 ends and is dressed as follows: 8 black,
2 slate, 12 black, 4 slate, 4 black, 3 slate, and 1 fancy; how many ends
of each color are there in the warp? r2,064 black
Ans. < 774 slate
186 fancy
2. A warp containing 1,800 ends is drawn in according to the draft,
shown in Fig. 1; how many heddles are required for each harness?
’540 heddles on first harness
A 720 heddles on second harness
' 360 heddles on third harness
180 heddles on fourth harness
7
7
7
6
6
6
6
5
5
5
5
4
4
4
4
3
3
3
3
2
2
2
2
1
1
1
Fig. 2
3. A warp containing 3,900 ends is drawn in according to the draft
shown in Fig. 2; how many heddles will be required on each harness?
! 450 heddles on first and seventh harnesses
600 heddles on second, third, fourth, fifth,
and sixth harnesses
20
CLOTH CALCULATIONS,
13
4. A woolen warp is drawn three per dent in an 8s reed and con¬
tains 912 ends; how much reed space does the warp occupy?
Ans. 38 in.
5. A worsted warp is reeded four per dent in a 15s reed and set
34 inches wide; how many ends are there in the warp?
Ans. 2,040 ends
6. A warp containing 2,700 ends is set 30 inches wide in the reed
and is reeded five per dent; what is the number of the reed used?
Ans. 18s reed
7. A finished cloth is 28 inches wide inside of selvages and has
shrunk 12 per cent, from reed to finished cloth; if the selvage occupies
1 inch in the reed, what is the total width of the cloth in the loom?
Ans. 32.81 in.
8. How many yards of cloth from the loom will be obtained from
240 yards of warp, if the contraction in weaving is 8 per cent.?
Ans. 220.8 yd.
9. A warp contains 2,592 ends and is reeded four per dent in an
18s reed; how wide is the cloth in the loom? Ans. 36 in.
10. A finished piece of cloth is 60 yards in length and has shrunk
20 per cent, in finishing; how long was the unfinished piece?
Ans. 75 yd.
§ 13
WOOLEN AND WORSTED
21
FIGURING PARTICULARS FROM CLOTH
SAMPLES
34. In order that the method of analyzing’ a sample of
cloth may be understood, a sample of cloth is enclosed and
the method of procedure will be explained with reference to
this sample. It is not intended at this stage to teach the
method of picking out the weave and of obtaining the draw¬
ing-in and chain drafts, but simply to acquaint the student
with the method of figuring the given sample with a view to
its reproduction.
The sample of cloth that has been taken to illustrate the
process of cloth analysis is a piece of worsted dress goods
finished 48 inches wide inside the selvages.
WEIGHT OF CLOTH
35. The first operation in analyzing a sample of cloth
is to find the weight, in grains, of 1 square inch of the
cloth. This is usually accomplished by cutting out a square
inch with a steel die and weighing it with a grain scale.
A steel rule and a sharp knife are also often used for cut¬
ting out a given area of cloth. Quite often 2 or 3 square
inches are cut if the sample of cloth is large enough to
allow it, and the resulting weight divided by 2 or 3, as the
case may be. In the cloth under consideration, it has been
found that 6 square inches of cloth weigh exactly 8 grains.
The first calculation is to find the weight of the cloth in
ounces per yard from the data that are now possessed con¬
cerning this sample, namely, the weight of 6 square inches
and the finished width. = 5-26 ounces, finished
6 X 437.5
weight exclusive of selvages.
22
CLOTH CALCULATIONS,
§13
COUNTS OF THE WARP AND FILLING
36. Next, it is desired to find the size, or counts, of the
yarns in this sample of cloth. It is found that the brown
and the white warp yarns are exactly the same size, and
that 100 inches of the warp yarn weighs exactly 1 grain.
Then, as this is worsted yarn, the standard of which is
560 yards to the pound, this number must be used in finding
the counts of the yarn.-—^—— = 34./2s, counts of
1 X 560 X 36
warp yarn.
Proceeding in the same manner, it is ascertained that the
counts of the brown and white filling are the same, and that
104 inches of filling yarn is required to weigh 1 grain.
104 X 7,000
1 X 560 X 36
36.11s, counts of filling yarn.
Ans.
NUMBER OF ENDS IN THE WARP
37. It is next desired to find the number of ends in the
warp. By counting with a pick glass or by measuring off
a given space and counting the ends it is found that there
are 74 ends per inch, and, as the cloth is 48 inches wide,
it would naturally be assumed that there are 48 X 74, or
3,552 ends, in the warp. Before the number of ends in the
entire warp is finally determined on, however, it is best to
see if the number of ends in the warp is divisible by the
number of ends in the pattern, and, if not, to increase or
decrease the ends in the warp, as the case may be, in order
to make even patterns. In this case it is found that there
are 24 ends in one pattern of the warp; there are therefore
148 repeats in the total width of 48 inches.
PROOF OF WEIGHT AND COUNTS
38. Having found the number of ends in the warp
and the size of the yarns used, the next operation will
be to find the weight of the cloth, as figured from the
yarn, in order to prove that these items have been found
correctly.
§13
WOOLEN AND WORSTED
23
39. Weight of the Warp. —To accomplish this it is
first necessary to find the weight of the warp per yard of
cloth.
3,552 X 16
34.72 X 560
2.92 ounces, weight of warp yarn.
Had there been two or more counts of yarn in the warp in
this sample of cloth, it would have been necessary to find
the number of ends and weight of each and add the results,
or to find the average counts of the warp yarn, whichever
would have been most convenient.
40. Weight of the Filling. —It is next necessary to
find the weight of the filling per yard of cloth and, in order
to accomplish this, the number of picks per inch must be
determined. This can be done in the same manner as when
finding the ends per inch. The sample under considera¬
tion will be found to have 62 picks per inch.
48 X 62 X 16
36.11 X 560
2.35 ounces, weight of filling yarn.
If the filling in the sample under consideration had been
composed of two or more different counts of yarn, it would
have been necessary to find either the weight of each or
the average counts of the filling.
The weight of the warp as previously found was 2.92 ounces,
which, when added to 2.35 ounces, the weight of the filling,
equals 5.27 ounces, which is the weight of 1 yard of the
finished cloth, exclusive of the selvage. It will be noticed
that this result proves within .01 the weight of 1 yard as found
from the weight of 6 square inches; therefore, the sizes of
the yarn and weights of warp and filling must be correct. It
is not always possible to prove the work as close as this,
but it is advisable to prove as close as .05, which allows for
small errors occurring through not carrying out the decimals.
If the error is greater than .05, it is better to go over the
work again, although some designers only prove within .1.
24
CLOTH CALCULATIONS,
13
REED AND ENDS PER DENT
41. The next item of importance to decide on with
regard to this piece of cloth is the reed, in order to find
which it is necessary to find the ends per inch in the reed.
This being a clear-finished worsted, the total shrinkage from
the reed to the finished cloth is about 10 per cent.
In a case like this it is customary to drop the fraction and
make a whole number of ends per inch in the loom, in order
that a reed may be readily obtained; in this case say
66 ends per inch in the reed, drawn 3 ends in 1 dent.
66 -4- 3 = 22s reed. Ans.
WIDTH IN REED
42. It is next desired to find the width of the cloth in the
reed. 3,552 4- 66 = 53.81 inches in the reed. This width in
the reed is of course inside of the selvages.
Suppose that in this fabric 7 dents are used for each sel¬
vage, or 14 dents for both selvages. Then the width of the
selvages in the reed will equal 14 divided by 22, or .63 inch,
and the total width of the fabric in the reed will be 53.81 inches
plus .63 inch, or 54.44 inches.
WEIGHT OF CLOTH FROM LOOM
43. It is next desired to find the weight of the woven cloth
from the loom. If 6 ends per dent are drawn for selvages
of the same yarn as the warp, there will be 14 X 6 = 84 sel¬
vage ends, which, added to the ends in the body of the cloth,
will make 3,636 ends in the whole warp. The sample of
cloth being a clear-finished worsted, it will be assumed that
the shrinkage from warp yarn to finished cloth and from reed
to finished cloth is 10 per cent., and that the yarn will lose
10 per cent, in finishing, so that the original counts of the
yarn in warp and filling will be the same as found in the
finished cloth.
§13
WOOLEN AND WORSTED
25
Had this cloth been woolen and shrunk very much, it would
have been necessary to find the original size of the warp and
filling yarns, but generally speaking, in worsted goods, the
loss in finishing will counterbalance the gain in weight of
the yarn due to the shrinkage.
44. Weiglit of Warp. —The first step in finding the
weight of the cloth from the loom is to find the weight of
1 yard of the warp yarn, including the selvage yarn.
3,636 X 16 _ 2.99 ounces, weight of 1 yard of warp. As this
34.72 X 560 ^
cloth will take up in weaving about 4 per cent., it is necessary to
add this to the weight of the warp, in order to find the weight of
the warp in 1 yard of the cloth from the loom. 2.99 + 4 per cent.
= 3.10 ounces, weight of warp in 1 yard of cloth from loom.
45. Weight of Filling. —It is next necessary to find the
weight of the filling from the loom, in order to find the
weight of the cloth from the loom. It was found that there
were 62 picks per inch in the finished cloth, but as the finished
cloth has undergone a shrinkage of 10 per cent, in the length,
the number of picks must have been increased 10 per cent., so
that 62 picks represent 110 per cent, of the original number of
picks per inch.
62 X 100
= 56.36. This number represents
100 + 10
the picks put in by the loom, but as the warp takes up
4 per cent, in weaving, and contracts after being taken from
the loom there must be 4 per cent, more picks in the cloth
from the loom, which must be taken into account in finding
the weight of the cloth from the loom. 56.36 + 4 per
cent. = 58.61 picks per inch in cloth. Having found the
picks per inch in the woven cloth and knowing the width
in the loom, the weight of the filling is easily obtained.
54.44 X 58.61 X 16
36.11 X 560
2.52 ounces, weight of filling in 1 yard
of cloth from the loom.
Adding the weight of warp and filling, the weight of the
cloth from the loom is obtained. 3.10 + 2.52 =p 5.62 ounces
from loom.
26
CLOTH CALCULATIONS,
§13
SUMMARY
The data obtained from the sample of cloth and the entire
figuring for this sample of cloth are as follows:
Data
6 square inches weighs 8 grains.
100 inches of warp yarn weighs 1 grain.
104 inches of filling yarn weighs 1 grain.
10-per-cent, shrinkage from reed to finished cloth.
10-per-cent, shrinkage from warp yarn to finished cloth.
10-per-cent, loss of weight in finishing.
4-per-cent, take-up of warp in weaving.
48 inches, finished width inside of selvages.
74 ends per inch, finished.
62 picks per inch, finished.
24 ends in a pattern (warp and filling).
Analysis
48 X 36 X 8 =
6 X 437.5
100 X 7,000
1 X 560 X 36
1 04 X 7,00 0
1 X 560 X 36
5.26 ounces, finished weight, inside selvages.
= 34.72s, counts of warp yarn.
= 36.11s, counts of filling yarn.
74 X 48 = 3,552 ends in warp.
3,552 -i- 24 = 148 patterns in warp.
3,552 X 16
34.72 X 560 '
48 X 62 X 16
36.11 X 560
- 2.92 ounces, weight of warp in finished cloth.
= 2.35 ounces, weight of filling in finished
cloth.
2.92 ounces + 2.35 ounces = 5.27 ounces, proof of finished
weight as found from the weight of 6 square inches.
3,552 -r-
/48 X 100 \
Vioo -10/
66+, say 66 ends per inch in reed.
66 -h 3 = 22s reed.
3,552 -j- 66 = 53.81 inches* width in reed inside selvages.
13
WOOLEN AND WORSTED
27
14 -f- 22 = .63 inch for selvages.
53.81 + .63 = 54.44 inches, total width in reed.
14 X 6 = 84 ends for selvages.
3,552 + 84 = 3,636, total ends in warp.
3,636 X 16 0 on • u*. £ i ^ t
- nn -= 2.99 ounces, weight of 1 yard of warp.
34.72 X 560
2.99 ounces + 4 per cent. = 3.10 ounces, weight of warp
in 1 yard of woven cloth.
g*™ _ 5 g 30 . 50,30 _f_ 4 p er cent. = 58.61 picks in the
cloth from loom.
54,44 X 58 .-61 X 16
36.11 X 560
= 2.52 ounces, weight of filling in
1 yard of cloth from loom.
3.10 ounces + 2.52 ounces = 5.62 ounces, weight of cloth
from loom.
It is always best to figure a sample through carefully, using
the exact counts of yarn as found in the cloth and other items,
in general, of the exact value found, and afterwards make
any allowances that may be deemed necessary to facilitate
the process of manufacture. For instance, if it were desired
to reproduce this sample, it would be best to spin 36s yarn for
both warp and filling, as it would be inadvisable to spin 34s
for the warp and 36s for the filling. It mignt also be neces¬
sary to alter the picks per inch, since it might be found that
60 picks would produce a better looking fabric, etc.
EXAMPLES FOR PRACTICE
1. What weight of number 20s worsted yarn will be required for
filling in a piece of cloth 35 inches wide at the reed, 50 yards long, and
containing 60 picks per inch? Ans. 9-f lb.
2. If 44 inches of worsted yarn weighs .7 grain, what is the size of
the yarn? Ans. 21.82s yarn
3. What is the weight of 1 yard of cloth 35 inches wide if 1 square
inch weighs 1.7 grains? Ans. 4.89 oz.
4. What is the weight of 1 yard of warp containing 1,008 ends
of 3-run yarn? Ans. 3.36 oz.
28
CLOTH CALCULATIONS
13
5. The warp yarn in a piece of cloth is found to be 17s; if this
cloth has shrunk 25 per cent, in length and lost 10 per cent, in weight
in finishing, what was the original number of the yarn? Ans. 20s
6. A certain cloth is made from 3-run yarn, both warp and filling.
The warp contains 1,044 ends, including the selvages, and is reeded
35 inches wide in the loom. There are 29 picks per inch in the cloth
from the loom. The take-up of the warp in weaving is 4 per cent.
What is the weight of this cloth from the loom? Ans. 7 oz.
7. One square inch of a 28-inch cloth weighs 2.75 grains; what is
the weight of the cloth in ounces per yard? Ang. 6.33 oz.
8. If 40 inches of woolen yarn weighs 1.2 grains, what run is
the yarn? Ans. 4.05-run
9. A certain warp contains 1,920 ends of 2/40s worsted; how many
pounds of warp will it take to weave a 50-yard cut of cloth if the warp
contraction in weaving is 6 per cent.? Ans. 9.118 lb.
10. What is the weight of filling in 1 yard of cloth 32 inches wide
containing 30 picks per inch of 4-run woolen yarn? Ans. 2.4 oz.
DRAFT CALCULATIONS
DRAFTING
INTRODUCTION
1. In the manufacture of cotton yarn a principle is
adopted that has to be considered in connection with almost
every process between the opening- of the raw cotton and the
spinning of the yarn —that known as drafting. This prin¬
ciple is one of great importance to any one desiring to
obtain a thorough knowledge of cotton-yarn-mill machinery,
and should be the subject of careful study and constant
watchfulness in order that the best results may be obtained.
As the principle of drafting cannot be considered a special
feature of any one machine, and as it must be considered so
frequently in connection with many machines, a general
explanation of it is here given.
The word drafting in the cotton-mill business is applied
to the principle of attenuating, or drawing out, a compara¬
tively large mass of cotton fibers to the condition of a
thinner but longer mass. The fibers found in the cotton
bale possess no regular order or arrangement. The
machines through which the cotton passes in the early
stages of its manufacture into cotton yarn arrange these
fibers into some sort of order until they assume the form of
the sliver , which is a rope of loose cotton in which the fibers
lie side by side, or as nearly so as it is possible to arrange
them. Some fibers overlap, while others are end to end, but
the object is to arrange them all approximately parallel so
For notice of copyright, see page immediately following the title page
§15
2
DRAFT CALCULATIONS
15
that the ribbon or rope of fibers can be handled without being
broken, and so that the weight of 1 yard of the sliver may
be approximately the same as that of another yard, varying
only to the extent of a few grains.
Previous to this form, the cotton has been arranged in a
sheet 36 or 40 inches wide, and made into a roll, but without
any attempt having been made at parallelizing the fibers; in
this condition it is known as a lap. The expression draft
is applied to indicate the extent to which these laps are
drawn out or elongated to form other laps; to the resultant
lap drawn out to make a sliver; to the slivers drawn out to
make still smaller slivers, known as rovings; or to the extent
to which these rovings are drawn out to make still smaller
rovings, or to make yarn.
The principle of attenuating the mass of cotton fibers by
various means is one of the fundamental principles of cotton-
yarn preparation. It may be done by means of air-currents,
by which the fibers are separated one from the other and
carried along by a current of air and deposited on rotating
screens delivering the sheet of cotton at a higher speed than
that at which it is fed into the machine; it may be performed
by rapidly rotating cylinders and rolls covered with wire
teeth, which elongate the mass of fibers even to the extent
of separation, depositing them again at a given rate on a
condenser, or doffer; or it may be, and most frequently is,
performed by means of revolving rolls. It is to the prin¬
ciples of drafting by means of successive pairs of revolving
rolls that most frequent reference will be made.
2. Objects of Drafting.—In attenuating, or drawing
out, the mass of cotton, there are three principal objects:
the first is to reduce the lap, sliver, or roving to a less weight
per yard, that is, attenuating it gradually to the desired
degree of fineness; the second object is that of arranging
and improving the arrangement of the fibers in a parallel
order so that they may lie side by side and overlap one
another; the third object is that of evening the strand of
fibers to eliminate thick or thin places, which is done by a
§15
DRAFT CALCULATIONS
3
combination of drafting and doubling. The use of succes¬
sive pairs of drawing rolls is largely adopted to arrive at
these results.
The action of the drawing rolls can be imitated in the fol¬
lowing manner: Grasping in the left hand a tuft of loose
cotton that has been partially cleansed at one of the earlier
processes, with the right hand pull the extreme projecting
ends forwards; then, after obtaining a fresh grip farther back
with the left hand, and a fresh grip farther back with the
right hand, pull still more fibers forwards. By repeating
this, what was a loose tuft of cotton will gradually be drawn
out to a thin film of partially straightened fibers.
This crude method illustrates that the process of drawing
a number of cotton fibers past one another, always holding the
front ends firmly and the rear ends of the same fibers loosely,
results in the whole mass of fibers being gradually straight¬
ened out. This principle is made use of in most cotton-yarn-
preparation machines by having carefully constructed and
adjusted rolls, the rear ones holding the mass of fibers and
running at a slow speed, the forward ones tightly gripping a
portion of the fibers and revolving at a greater speed. This
arrangement is duplicated again and again, until in some
machines there are as many as four pair of rolls successively
acting on the fibers in this manner. The quickly rotating
pair of rolls draws the fibers away from the slowly rotating
rolls, and as the fibers are gripped by their fore ends and
pulled forwards, the loose rear ends trail behind and tend to
become straightened out as they are drawn from the tuft held
by the slowly rotating rolls.
In early days, when cotton yarns were made by hand, the
deftness of the spinster’s fingers was relied on to obtain
this parallelizing of fibers, but over 100 years ago the
principle of drafting by means of rolls was discovered by Sir
Richard Arkwright, and has been in use ever since.
3. Doubling. —The explanation just given shows that
attenuating and parallelizing the mass of fibers tends to
reduce its thickness and make a thin sheet where there was
4
DRAFT CALCULATIONS
15
formerly a thick one, and if continued indefinitely would
result in destroying the continuity of the sliver or roving.
To prevent this, doubling is resorted to in most of the cotton-
yarn-preparation machines. Briefly explained, this means
>
that instead of feeding only one lap, sliver, or roving at the
back of each machine, two or more are fed together, making
one at the front; this not only helps to compensate for the
excessive attenuation, but has the great advantage of helping
to neutralize unevenness in the original mass of fiber fed to
the machine. By feeding several together, the thick or thin
places of any one are combined with other slivers of normal
size, or thick places with thin ones, and the combination of
two, three, four, five, or six independent slivers or rovings,
which are drawn out into one, results in an evenness not
attainable in any other manner.
4. The general operation or application of these principles
is spoken of as attenuation, or drawing out, and sometimes
as drafting , but strictly speaking, a more limited meaning
attaches to the word drafting. Draft as applied to cotton-
yarn machinery should only be considered as referring to the
ratio of attenuation, and drafting should refer to the attenu¬
ation only, without being understood to include the parallel¬
izing or evening features that have just been referred to,
although these follow as a matter of course. A full consid¬
eration of all these features cannot be made until the con¬
struction and operation of the machines have been dealt with
in their proper places, but the subject of draft, pure and
simple, can now be considered and the calculations connected
therewith explained, so that this information can be applied
to each machine in which draft rolls are used.
The word draft as used here must always be considered as
referring to the ratio of attenuation, and must not be con¬
fused with the word draft that indicates a draft of air, as
used in connection with the pneumatic drafts found in some
cotton-yarn-preparation machines.
15
DRAFT CALCULATIONS
5
DRAFTING WITH COMMON ROLLS
5. The study of draft calculations is of great value for
many reasons, and its importance cannot be exaggerated.
A thorough knowledge of this subject enables the amount of
draft, or attenuation, that is taking place in each machine to
be determined; it enables the right size of lap, sliver, roving,
or yarn to be produced, or changes to be made from one size
to another; it provides a means of calculating how errors of
size in the resulting product of any machine can be corrected,
and to anticipate beforehand what resultant size of lap,
sliver, roving, or yarn can be made from a machine or a
series of machines.
The subject will be dealt with here only so far as the cal¬
culations are concerned, but before considering these calcu¬
lations a brief description of the passage of cotton through
drafting rolls, according to one method, will be given.
Fig. 1 represents a section though four pair of rolls, the
Fig. i
lower rolls a , b , c , d being constructed of steel and fluted
longitudinally. These flutes are cut in such a manner that
the outer edge of the projections ends in almost a flat surface;
strictly speaking, they are arcs of a circle which has the center
of the roll as a center. The upper rolls a lt b lt f,, d y are con¬
structed of iron with a covering of flannel immediately
around them, and a thin leather covering outside of the
flannel. These rolls are absolutely smooth, and are pressed
against the bottom rolls by means of weights.
6
DRAFT CALCULATIONS
§15
The rolls d, d 1 into which the material is fed should always
be spoken of a.s the feed-rolls or back rolls , the roll d x being
distinguished from the roll d by the term back top roll. The
rolls delivering the material, represented by a and a ly should
always be spoken of as the delivery rolls or front rolls , the
roll a , being called the front top roll. The first pair of inter¬
mediate rolls, counting from the front, is spoken of as the
second pair of rolls; and the third pair, counting from the
front, as the third pair of rolls. Thus, the roll a would be
the front, or delivery, roll; b would be the second roll; c, the
third roll; and d, the back roll, or feed-roll.
The circumferential speed of the upper and lower roll in
each pair should be the same; that is, a point on the surface
of d should move at the same speed as a point on the sur¬
face of d ly because d x is driven by frictional contact with d.
The same remarks apply to any other pair in the series. In
practice, however, it is found that there is a slight slippage
between the bottom and top rolls, but in case the top -
roll is properly constructed, weighted, and lubricated, this
slippage is very slight and will be ignored in the following
calculations.
The back roll, which is the feed-roll, always rotates at the
slowest speed and the front roll at the highest, the speed of
the other rolls being so arranged that c revolves a little
more quickly than d, and b still more quickly than c, but at a
less speed than a. The speed of these rolls is usually
indicated by the number of revolutions per minute, often
abbreviated to R. P. M. The direction of rotation of the
rolls is shown by a small arrow within the section of each.
Between d and d t a ribbon of cotton is fed and is carried
forwards, as shown, between each pair of rolls, until it
emerges at the front. The direction of movement of this
ribbon of fibers is shown by the arrows. The upper rolls
are weighted in such a manner as to firmly grip the fibers
that pass below them, and thus if the spaces between the
centers of each pair of rolls are properly adjusted and the
relative speeds of the rolls accurately arranged, the principle
of drawing the fibers past one another by means of a firm
15
DRAFT CALCULATIONS
7
grip of their fore ends, the rear ends trailing behind, is
satisfactorily adopted. The same conditions continuously
exist in the machine, because as the forward rolls pass fibers
forwards, the rear rolls are supplying new ones, and the
results are thus comparatively even and regular.
The illustration shows the gradual attenuation or reduction
in size of the mass of cotton, owing to the increased speed
of each pair of rolls over the preceding pair. It will be
seen that if the surface speed of the back roll is 60 inches
per minute and that of the front roll 360 inches, the sliver
emerging from the front roll will be six times as long and
consequently six times as fine as when entering the back roll.
The arrangement just described is only one of many found
in cotton-yarn-preparation machinery and is merely given as
an example. Draft could be produced between only two
pair of rolls almost contiguous; again, these two rolls,
known as the feed-roll and delivery roll, respectively, might
have between them a large number of other rolls, or a num¬
ber of cylinders or rollers, or other means of producing
draft, but the draft would be computed between the feed-rolls
and the delivery rolls if the total draft were desired.
6. Methods of Finding Draft. —Draft is the ratio of
the speed of the delivery to that of the feed part of a
machine. It indicates the ratio between the surface speed of
the front, or delivery, roll and the surface speed of the back,
or feed, roll, and may be found in different ways.
1. Draft may be found by dividing the space moved
through in a given time by a point on the surface of the
feed-roll, into the space moved through in the same time by
a point on the surface of the delivery roll.
Example 1.—Referring to Fig. 1, if a point on the surface of d
moves 3 inches while a point on the circumference of the roll a moves
18 inches, what is the draft?
Solution. — 18 -=- 3 = 6, draft. Ans.
2. Draft may be found by dividing the weight per yard
of the product delivered, into the weight per yard of the
material fed into the feed-rolls.
8
DRAFT CALCULATIONS
§15
Example 2.—If 1 yard of cotton fed into the machine weighs
72 grains, and 1 yard delivered at the front of the machine weighs
12 grains, what is the draft?
Solution. — 72 -p 12 = 6, draft. Ans.
3. Draft may be found by dividing the number of yards
delivered by the delivery roll in a certain time, by the num¬
ber of yards fed into the feed-roll in the same time.
Example 3.—If 2 yards of cotton is fed into a machine during the
same time that 12 yards is delivered, what is the draft?
Solution.— 12 -p 2 = 6, draft. Ans.
It will be observed that these three methods of finding the
draft deal with the ratio between the length, weight, or speed
of the material fed and the corresponding condition of the
material delivered; and from these examples will be deduced
the facts that while the length of material fed into the
machine is increased by drafting, the weight per unit of
length is always decreased in the same proportion. The
problems just presented are comparatively easy to under¬
stand, but the data necessary to solve the problem of draft
in any of the methods just shown is not always available.
It may be necessary to calculate the draft before a machine
is put into operation or without being able to ascertain the
weight or length of material fed or delivered, and it is then
necessary to find the draft and perform various other draft
calculations by a consideration of the gearing that connects
the rolls and also the sizes of the rolls themselves. This
is the most common method of providing data for draft
calculations.
Draft may therefore be defined in various ways, with ref¬
erence to the machine or part of a machine in which draft is
being produced: (1) The ratio between the length delivered
and the length fed in a certain time; (2) the ratio of speed
between a point on the delivery roll and a point on the
feed-roll; (3) the number of times that a certain length of
material is increased while being operated on; (4) the ratio
between the weight of a certain length of material fed and
the weight of the same length of material delivered; (5) the
§15
DRAFT CALCULATIONS
9
number of times that the weight of a certain length of
material is decreased while being operated on.
Each of these definitions has the same meaning, the same
facts being stated in different ways. The different methods
of finding draft are each based on one of these definitions,
according to the data provided or available.
GEARING OF ROLES
7. Figs. 2 and 3 represent different views of four pair of
rolls and their gearing, Fig. 2 showing the principal driving
end. The figures show the front, second, third, and back
Fig. 2
top rolls, and the front bottom roll. The front rolls are
marked a and a x ; the second top roll, d 1 ; the third, c and the
back top roll, d x . The bottom roll a drives the back bottom
roll by a train of gears e, /, g, h; e is on the roll a; h is on
the back roll; / and g are compounded and revolve on a stud.
The third bottom roll is driven from the back roll by means
of three gears j, k , /, Fig. 3; j is on the back roll; l is on the
third roll; while k is an idler, or carrier, gear revolving on a
stud. The second bottom roll is driven from the roll a by
10
DRAFT CALCULATIONS
§15
means of three gears m, n , o; m is on the second roll; o is on
the front roll a ; while n is a carrier gear revolving on
a stud.
A carrier gear is usually placed between a driver and a
driven gear when it is not convenient to make them large
enough to mesh with each other, or where it is necessary to
change the direction of motion of the driven gear without
changing its speed. It is important, in connection with
draft calculations, to notice which gears are merely carrier
gears, as a carrier gear does not affect the speed, and can
be left out of all calculations of trains of gears of which it
forms a unit.
The sizes of the rolls shown in Figs. 2 and 3 are as
follows: Front roll a. If inches; second roll, If inches; third
roll, li inches; fourth roll, li inches. These dimensions
represent the diameter of the roll in each case.
Although Figs. 2 and 3 show a view of four pair of draft
rolls mounted on stands as they would appear in actual
operation, it would be unnecessary and inconvenient in mill
work to make a sketch of this kind in order to illustrate the
various methods of combining and driving draft rolls. The
simplest method of showing draft rolls and their gearing,
15
DRAFT CALCULATIONS
11
and the one usually adopted, is to make a diagram in which
horizontal lines are drawn to show the lines of rolls, and short
lines drawn at right angles to these to indicate the gears
connecting the rolls. It is perhaps better to show the out¬
line of the gears, instead of using a line.
Fig. 4 shows a diagram that would represent the rolls and
gearing shown in both Figs. 2 and 3. This indicates that
there are four lines of rolls and that the power is received by
the tight and loose pulley shown on the front-roll shaft.
It further shows that motion is conveyed to the back roll
from the front roll by means of the gears )
Fig. 16
tails of its various
parts. Fig. 17 would
be known as an assembly drawing, and Fig. 18 as the detail
drawings.
It frequently happens that in large assembly drawings
certain parts are necessarily so small as to render them
indistinct; in such cases, detail drawings of these parts are
18
READING TEXTILE DRAWINGS
§91
often given. These detail drawings may not necessarily
be drawings of a single part, but may be enlarged draw¬
ings of a group of
parts or of the assem¬
bled parts of some par¬
ticular mechanism.
PERSPECTIVE VIEWS
21. A perspec¬
tive view may be said
to combine several
views of an object into
one. For this reason
it gives at one glance
a correct idea as to the
shape of an object, while in a regular mechanical drawing it
may be necessary to study several views before the mind
h
©
©
i
«
o 2
J
too
w
Fig. 18
will be able to imagine its true shape. A perspective view
is drawn so as to represent the object as it will appear to the
eye when viewed under ordinary conditions, while the views
\
§91
READING TEXTILE DRAWINGS
19
given in mechanical drawings are, for the sake of convenience,
drawn under assumed conditions, as previously described.
Fig. 1 is an example of a perspective view. It may be
said in this instance to combine a front elevation, a right-
side elevation, and a plan view in one, and gives at once
a good idea of the shape of the object.
20
READING TEXTILE DRAWINGS
§91
As a further illustration of perspective views, Fig. 19 is
given. This represents a loom as it appears when viewed
from a point in front and to the right of the machine. It will
be seen that this view resembles a photograph with its lights
and shadows omitted. Perspective views are much more
valuable in many cases for illustrating purposes than photo¬
graphic reproductions, since they show clearly parts that in
a photograph would be in the shadow and, hence, indistinctly
perceived. _
DIAGRAMMATIC VIEWS
22. Diagrammatic views are those that do not show
the exact structure of an object, but are simply used for indi-
Dra/t Change 6ear—g8/~
£/£ |
e20
Fig. 20 L_L_
eating it, or indicating its position or some other special
feature. It is difficult to lay down any definite definition of
a diagrammatic drawing, since a great many types of these
drawings and many methods of indication are used. In
general, the purpose of a diagrammatic drawing is more that
of indicating the principles on which the action of a mech¬
anism are based than its construction. This class of draw¬
ings indicates therefore the elements of the mechanism
mostly by means of conventional symbols instead of giving
the correct outlines and proportions of the parts. It is also
allowable to change the positions of the parts and even to
eliminate some parts altogether, if this tends to make the
§91
READING TEXTILE DRAWINGS
21
drawing more clear. Fig. 20 shows one form. It is a rep¬
resentation of four pairs of drawing rolls a,b,c,d, together
with the method of driving them by gears, and the sizes of
the gears used. The front roll a receives motion through
the tight pulley; the gear e on this roll drives the third roll c
by means of the gears f,g, h\ the fourth roll is driven from
the third roll by the gears/, k, l\ the gear k is an idler, or
carrier, gear. The second roll b is driven from the third roll
by the gears /, m , n\ the gear rn in this case is an idler, or
carrier, gear. It will be seen that while this drawing does
not give an exact representation either of the rolls or of
the gears used to connect them, it does give an indication of
the rolls and of the gears that drive them. Fig. 21 gives
another illustration of a diagrammatic drawing. In this case
a representation is given of the method of driving a section
of twelve spindles by means of one driving band, to which
motion is imparted by the rotating drum c. This drawing
does not give an exact representation of the parts, but
simply shows the method of passing the band around the
drum and around the spindles in order to drive the twelve
spindles with one band.
22
READING TEXTILE DRAWINGS
§91
(a)-
(b)
LINES USED ON DRAWINGS
23. In general, there are but six kinds of lines used on
drawings; they are shown in Fig. 22. The light, full line ( a )
is used more than any of the others and is used for drawing
the outlines of all parts of an object that can be seen by the
eye. The dotted line ( b) consists of a series of very short
dashes. It is used for showing the position and shape of an
object or part of an object that is concealed from the eye in
the view shown, on account of other parts of the object or
__ other objects intervening.
The use of the dotted
line is shown in Fig. 1.
_The broken-and-dotted
lines (c) and (d ), Fig. 22,
' sometimes called dot-
_and-dasli lines, consist
of a long dash followed
by either one or two dots,
repeated regularly. It is
very seldom that both these lines appear in the same
drawing, being used either to represent center lines or indi¬
cate where a section has been taken when a sectional view
of an object is shown in another drawing. The broken
line (e) is used chiefly for dimension lines; it consists of a
series of long dashes. This line appears in Figs. 28 and 29.
The heavy, full line (/) is made not less than twice as thick
as ordinary full lines on a drawing, and is used only for
shade lines.
(c)-
(d)-
(e)-
(f)-
Fig. 22
SHADE LINES
24. On some mechanical drawings, it will be observed
that a number of the lines are made very heavy. These
heavy lines are put on in accordance with a certain almost
universally adopted system; they show by their location
whether the part looked at is a hole in the object, or is raised
beyond the surface. They are called shade lines. Thus,
in Fig. 23, by means of the shade lines a person can
91
READING TEXTILE DRAWINGS
23
determine, without looking at any other view of the object,
that the rectangles 1 and 4 represent square holes, and 2
and 3 square bosses. A
Likewise, the heavy
lines CD and D B show
that the object has a
material thickness.
25. In the almost
universally adopted
system of shade lining,
the light is assumed to
come in one invariable
direction, in such a man¬
ner as to be parallel
with the plane of the
paper; to make an angle
of 45° with all horizontal and vertical lines of the drawing,
and to come from the upper left-hand corner of the draw¬
ing. Each view of the
object represented is
shaded independently of
any of the others; and,
when shading, the ob¬
ject is always supposed
to stand so that the view
that is being shaded
will be a top view.
Any surface that can
be touched by any one
of a series of parallel
straight lines, drawn at
an angle of 45° with the
horizontal and vertical
lines of the drawing, is
called a light surface;
a surface that cannot be touched by lines having this angle
is called a dark surface. All the edges produced by the
Fig. 23
24
READING TEXTILE DRAWINGS
91
intersection of a light and a dark surface are usually shade-
lined.
26. An application of these rules is shown in Fig. 24,
where a top view of a series of triangular wedges radiating
from the common center O is given. The top surfaces are,
of course, light surfaces, but some of the perpendicular sur¬
faces, which are indicated by the outlines of the wedges, are
light and some dark. Observing the arrows, which here
represent the direction of the light, it is noticed that they
strike the edge O A in
the wedge R O A; this
side will therefore be a
light surface, and as
the top surface is also
a light surface, the
line O A is drawn as a
light line. OR, on the
contrary, is a heavy
line, since the light
cannot strike this
side without passing
through the wedge;
hence, this is a dark
surface, and its junc¬
tion OR with the light
surface OAR requires
a shade line. For the
same reason AR is also
shaded. The same reasoning applies to the lines OB, O D,
OG, 01, OK, and O M; also to Q N, ML, and K J. CB is
not shaded, because the light strikes the surface of which C B
is the edge. O N makes an angle of exactly 45° with a
horizontal line, and is treated as if it were the edge of a
light surface; this is done in every case in which the line
considered makes an angle of 45° with a horizontal line.
27. The benefit to be derived from shade lines is well
illustrated by Fig. 25. Let ( a) be a front view, and ( b) a
Fig. 25
§91
READING TEXTILE DRAWINGS
25
side view of an object. Looking only at the front view, it is
impossible to determine whether the part A is a hole, as
shown by the side view given in Fig. 25 (b), or a boss,
as shown by the side view given in Fig. 25 (c ). Now, if the
front view of the object be drawn with heavy shade lines
placed as in Fig. 25 (d ), the shade lines by their position
immediately show that the part A is a hole, without having
first to refer to the side view shown at (b). If the shade
lines are placed as in Fig. 25 (e ), they show that the part A
is a boss, without any reference to the side view.
BREAKS
28. When a long and comparatively slender object is to
be drawn, it often happens that, when drawn to a sufficiently
large scale to make it intelligible, it will extend beyond the
space available. In such a case, part of the object is shown
Fig. 26
broken out and the remaining ends are close to each other.
The fact that part of the object is broken away for the sake
of convenience is indicated by a so-called break. It is
always understood that the part broken away and not shown
is of the same transverse dimensions and shape as the parts
contiguous to the break. In some cases, one end of the
object is broken away.
Fig. 26 shows a common method of drawing a rod or any
similar object too long for the space available. It being-
essential in the particular case shown to know the shape and
size of the ends, the rod is therefore shown with its central
part broken away, as indicated by the breaks at a and b. It
is customary always to section the break to suit the material
of which the object is made.
29. Breaks may be indicated in various ways; most com¬
monly, the break is given an outline that will reveal the.
26
READING TEXTILE DRAWINGS
§91
shape of the object. Conventional methods of indicating
breaks are shown in Fig. 27. Wood is usually shown broken
in the manner illustrated at (a); angle
irons, as at ( b)\ T irons, as at ( c);
Z bars, as at (d ). Cylindrical objects
are occasionally broken as shown at (e ),
but most frequently in the manner
shown at (/). Pipes and similar hol¬
low cylindrical objects may be broken
as shown at (g), but more frequently
the break is made as shown at (/i).
Rectangular objects may be broken in
the manner shown at (z); plates and
objects other than those included
between views (a) and (z), are often
shown broken off by drawing a wavy
freehand line as in (/) and (k).
(b)
(o)
(d)
(e)
(f)
(O)
(h)
d)
(j)
(k)
Fig. 27
-I
3 .
□
~1
5
]
SCALES
30. Drawings of an object may be
made the actual size of the object, may
be enlarged so as to be larger than the
object itself, or, as is generally the case,
may be made smaller than the object
itself. When a drawing of an object is
not made the exact size of the object,
it is said to be drawn to a scale. For
instance, suppose that it is required to
make a drawing one-quarter the size
of the object to be represented, so that
3 inches on the drawing will represent
1 foot on the object. Then, if 3 inches
is laid off and divided into twelve equal
parts, each part, which is actually i inch
in length, will represent 1 inch on the object. If each part is
divided into eight parts, each part will represent i inch on the
object. A scale of this kind is called a quarter scale, or a scale
§91
READING TEXTILE DRAWINGS
27
of 3 inches to the foot. An eighth scale , or a scale of 1 k inches
to the foot , would be constructed in a similar manner, except
that lv inches would be laid off and subdivided, instead of
3 inches. In cases where small objects are drawn to an
enlarged scale, the scale might be 18 inches to the foot, in
which case the object would be drawn li times its real size,
or if the scale had 24 inches to the foot, the drawing would
be twice its actual size.
DIMENSION LINES
31. Dimension lines are used on drawings to show the
size, or dimensions, of the object; they are especially neces¬
sary when objects are drawn larger or smaller than they
actually are. Dimension lines may be placed directly on an
object, or they may be placed outside of the
object, as, for instance, the dimension line
in Fig. 28 marked inches, which shows *
that the length of the cop there shown is
62 inches. In the latter case, extension,
or limiting, lines, as a and b, are drawn.
The dimension line is then placed between
them, and arrowheads are placed on its
ends in contact with the limiting lines, as
shown. This indicates that the measure¬
ment is made between the points from
which the limiting lines have been drawn.
Limiting lines c , d, e, f are also drawn in
Fig. 28, and the dimension lines drawn to
them show that the length of taper of the
cop bottom is f inch, the length of the body ^
of the cop 4i inches, and the length of the
taper at the nose li inches.
As a further illustration of the use of FlG ‘ 28
dimension lines, Fig. 29 ( a ) and ( b ) is given, (a) showing a
plan view and (b) a side elevation of a shuttle. According to
the dimensions given, the entire length of the shuttle is
17 inches; its greatest width measured across the top is
lit inches; the thickness of its sides is i inch; the entire
Fig. 28
28
READING TEXTILE DRAWINGS
§91
length of the hollow portion is 9H inches, and this hollow is
located 4 inches from the heel of the shuttle. The length of
the spindle in the shuttle is 7 inches, and its diameter varies
from f inch at the base to iV inch at the tip. The height of
the shuttle is 1| inches and the groove on the side is inch in
width; the eye of the shuttle is located 3f inches from the tip.
CONVENTIONAL METHODS OF REPRESENTING
OBJECTS
32. Owing to the fact that certain parts occur with great
frequency in mechanical drawings and that considerable time
is required to draw them in the regular manner, it is prefer¬
able to represent them by certain conventional methods.
Take for instance screw threads and tapped holes, as shown
in Fig. 30.
Referring to the figure, probably the clearest representa¬
tion of a screw thread is shown in Fig. 30 (a) and (e).
Fig. 30 (a) is a full view of a screw (a bolt in this instance);
Fig. 30 («?) is a sectional view of a tapped hole. Fig. 30 ( b)
and (/) shows a representation of a screw thread and tapped
hole that may be drawn more easily and still give a good
representation of the thread. Fig. 30 ( c ) and (g) shows a
conventional method of representing a screw thread and
tapped hole that is very largely used. While it does not by
any means show the exact appearance of the thread, it readily
§91
READING TEXTILE DRAWINGS
29
conveys to the mind what the
Fig. 30 ( d ) and (k) illustrates
drawing is intended to show,
another and still more simple
(c) (d)
method of representing a screw thread and tapped hole,
but this method is so crude that it is little used except
in drawings where the appearance must be sacrificed to
(a)
(t»
(c)
(d)
Fir,. 31
rapidity of execution on the part of the draftsman. It will
be noticed that in each case the slant of the thread in the hole
is in the opposite direction to that on the screw.
30
READING TEXTILE DRAWINGS
§91
33. When the screw thread is hidden by part of the
object, and it is deemed necessary to show it in dotted lines,
it is usually drawn in one of the four ways illustrated in
Fig-. 31. Of these, the method shown in Fig. 31 (a) is prob¬
ably the clearest; that shown in Fig. 31 (b) is fairly good; and
the one shown in Fig. 31 ( c) may be drawn quickly but is not
a good representation. The method illustrated in Fig. 31 ( d )
is practically no representation of a screw thread at all.
34. All the threads shown in Figs. 30 and 31 are riglit-
handed; that is, when looking along the axis of the screw,
the thread advances in the direction in which the hands of a
watch move. When the thread advances
in an opposite direction, the thread is
left-handed; it is then drawn with
the thread slanting the other way from
that shown in Figs. 30 and 31. Thus,
in Fig. 32 a left-handed screw thread is
shown in full, and a left-handed tapped
hole in section beneath it. It will be
noticed that the thread in the hole
slants the opposite way to that on the
screw.
In case of doubt, a right-handed
thread can always be told from a left-
handed thread by holding the screw,
or the drawing of it, so that the axis
is vertical. Then, if the thread slants
upwards to the right in case of a screw shown in full, or
upwards to the left in case of a tapped hole in section, the
thread is right-handed; if otherwise, it is left-handed.
35. Springs also are frequently represented by con¬
ventional methods; for instance, Fig. 33 (a) shows a method
of representing a spring that practically gives its exact
appearance, but as this method of showing springs is difficult
to draw, conventional methods, as shown at Fig. 33 (b) and
(c), are frequently used. In addition, chains are frequently
represented on a drawing in some conventional manner.
§91
READING TEXTILE DRAWINGS
31
Fig. 34 (a) shows a method of representing a chain that gives
its exact appearance; Fig. 34 (b) shows a method of indi¬
cating a small chain that may be much more readily drawn
Fig. 33
0 ! <
r)
'o'
v ) i,i
n f (
OJ
&
P i ,
n 1 1
o)
'O'
in | «
O)
&
a <
6)
'o'
I | ,
o )
&
OJ
'o'
U A (
o)
V&
(a) (b) (c) (d)
Fig. 34
by the draftsman. Fig. 34 (c) shows the exact appearance
of a sprocket chain, while Fig. 34 (d) is a conventional repre¬
sentation that is frequently used. Conventional representa¬
tions of ropes are often used. Fig. 35 (a) shows the exact
(a)
(b)
Fig. 35
appearance of a rope, while Fig. 35 {b) is a conventional repre¬
sentation that is frequently used. The object of using all
these conventional methods of representing various objects is,
of course, to save the draftsman’s time in making drawings.
Fig. 36
§91
READING TEXTILE DRAWINGS
33
MOVING PARTS SHOWN IN TWO OR MORE
POSITIONS
36. Dotted lines, as previously explained, are used to
show those parts of an object that are concealed from the
eye in the view of the object given, by other parts interve¬
ning between them and the eye. In addition, they are used to
indicate parts of mechanism when it is desired to show them
in more than one position. This use of dotted lines is
merely for the purpose of indicating on the drawing the
extent of the movement of certain parts or the positions that
they assume under certain conditions; as an illustration of
this use of dotted lines Fig. 36 is given. This is a view of the
shedding mechanism of a cam-loom. The harness m and the
jack l are shown in one position in full lines, while in order
to show the extent of their movement, they are shown in
dotted lines in the position that they assume when the
harness is lowered, at which time the cam-bowl / 2 on the
jack is in its position nearest to the cam-shaft k x : The full
lines represent the position of the harness and the jack when
the harness is at its highest position, while the dotted lines
show their position when the harness is in its lowest position.
REFERENCE TETTERS
37. Reference letters are placed on the various parts
shown in a drawing so that these parts can be definitely
referred to in the accompanying text. There are, in general,
three ways of placing reference letters on a drawing.
The letter may be placed directly on the object to which
reference is to be made, as for instance in the case of the
letter h, Fig. 19, where the reference letter is placed directly
on the take-up roll of the loom. A reference letter may be
placed to one side of the object to be referred to and a
line drawn to the object, as for instance in the case of the
letter k 2 , Fig. 19, where the reference letter refers to the reed
cap of the loom. Or, where confusion will not result, the
letters are placed beside the object to which reference is
READING TEXTILE DRAWINGS
34
§91
to be made and the line omitted, as in the case of the
letter v at the shipper handle in Fig. 19.
When a machine has a series of parts that cooperate in
such a manner as to constitute a definite mechanism, more or
less complete in itself, each part of such a combination is
referred to by the same letter, distinction being made
between the several parts by giving the letter different sub-
scripts. For instance, suppose that a certain mechanism
contains a train of three gears; these gears would be let¬
tered a,a 1} a a , or if more gears were contained in the same
mechanism, the subscripts would run still higher until each
part referred to in the text was lettered.
Where several drawings are used to illustrate the same
machine or mechanism, the same part, if shown in more than
one of these drawings, always has the same reference letter,
except in special cases. Thus, a gear that is marked a in
one drawing will be marked a in every other drawing in which
it may appear.
When reference letters are enclosed in parentheses
thus (a), (b ), etc., the reference may be to a separate view
of some particular element of a mechanism, but is mostly to
different views of the entire mechanism. For instance, if
reference is made to Fig. 29 (a), it is desired to call atten¬
tion to that view of the object marked (a) in Fig. 29.
A SERIES OF QUESTIONS
Relating to the Subjects
Treated of in This Volume.
It will be noticed that the questions contained in the fol¬
lowing pages are divided into sections corresponding to the
sections of the text of the preceding pages, so that each
section has a headline that is the same as the headline of
the section to which the questions refer. No attempt should
be made to answer any of the questions until the corre¬
sponding part of the text has been carefully studied.
ARITHMETIC
(PART 1)
EXAMINATION QUESTIONS
(1) In a certain year, Great Britain used 4,141,000 bales
of cotton, while the United States used 2,812,000 bales; how
many bales of cotton were used by both of these countries in
that year? Ans. 6,953,000 bales
(2) In 12 months, Russia used 778,000 bales of cotton,
France 650,000, Germany 1,170,000, Austria 530,000, Switzer¬
land 150,000, Sweden 80,000, Holland 65,000, Belgium
145,000, Spain, Portugal, and Greece together 360,000, and
Italy 390,000; what was the total number of bales used by
✓
these countries? Ans. 4,318,000 bales
(3) In a certain year, the number of bales of cotton
raised by North Carolina was 336,261, while South Carolina
produced 747,190; how many more bales were produced by
South Carolina than by North Carolina? Ans. 410,929 bales
(4) In a certain year, Georgia produced 1,191,864 bales
of cotton and Mississippi 1,154,725; what was the total
number of bales produced by Georgia and Mississippi in this
year? Ans. 2,346,589 bales
(5) Solve the following problems:
(a) 436 + 14 + 89 + 1,075; (b) 100,173 + 19 + 11 + 8 + 53.
A ns J(«) 1-614
Ans T( 17.618 lb. black
(19) («) A cotton warp contains 4 pounds of 18s and
has 1,200 ends; how long is it? (b) 15 pounds of 2-ply
60s cotton is made into a warp 80 yards long; how many
ends does it contain? A \ (a) 50.4 yd.
Ans 'l {b) 4,725 ends
(20) Give a rule for finding the counts of a ply yarn
composed of more than two threads of single yarn.
(21) State how to find the weight of each thread in a ply
yarn when the total weight and the counts of the separate
threads are given.
(22) Two threads of 60s and one of 40s are twisted; what
are the counts of the ply yarn? Ans. 17.142s
(23) 52^ pounds of 20s cotton is used to make a warp
700 yards long; how many ends are there in the warp?
Ans. 1,260 ends
§9
YARN CALCULATIONS, COTTON
3
(24) What are the average counts of the yarn in a warp,
the ends arranged 34 ends of 48s and 1 end of 2-ply 20s?
Ans. 43.316s
(25) What twist would be inserted in an ordinary 56s
cotton filling yarn? Ans. 24.319 turns per inch
EXAMINATION QUESTIONS
(1) What counts of yarn should be twisted with a 24s to
make a ply yarn equal in weight to an 8s single? Ans. 12s
(2) An 80s, 40s, and 30s are twisted together; what are
the counts of the ply yarn? Ans. 14.117s
(3) How many yards are there in 2 pounds of 50s
worsted? Ans. 56,000 yd.
(4) Explain what is meant by the word counts.
(5) State what is meant by ply yarns.
(6) State how many yards there are in 1 pound of yarn,
the counts being 5s: (a) in the woolen (run and cut) systems;
(b) in the worsted system.
A f (a) 8,000 yd. (woolen run), 1,500 yd. (woolen cut)
ns 'i(£) 2,800 yd. (worsted)
(7) If 368,000 yards of single worsted yarn weighs 16
pounds, what are the counts? Ans. 41.071s
(8) What counts of yarn are twisted to make a 2-ply 38s
worsted yarn?
(9) What is the weight of 360,000 yards of: (a) 2-ply
36s worsted? (b) 5-run woolen?
A f (a) 35.714 lb. worsted
ns, \(£) 45 lb. woolen run
§10
2
YARN CALCULATIONS,
§10
(10) How many ounces of yarn are there in 14,000 yards
of 2-ply 36s worsted? Ans. 22.208 oz.
(11) What is the length of 18 pounds of 8s yarn: (a) in
the worsted system? ( b) in the woolen systems (run and
cut)? A \ (a) 80,640 yd.
ns ‘t(£) 230,400 yd. (run), 43,200 yd. (cut)
(12) What counts must be twisted with a single 14s to
produce a 2-ply thread equal in weight to a single 10s?
Ans. 35s
(13) What weight of 60s worsted will be required to
produce 120 pounds of 2-ply yarn when twisted with a 38s
worsted? Ans. 46.530 lb.
(14) A jack-spool containing 40 ends is spooled with 300
yards of woolen yarn that weighs 3 pounds; what run is the
yarn? Ans. 2.5-run
(15) A worsted warp containing 2,800 ends is 800 yards
long and weighs 200 pounds; what is the size of the yarn?
Ans. 20s
(16) A woolen warp is 400 yards long and weighs
80 pounds. If the warp is composed of 42"-run yarn, how
many ends does it contain? Ans. 1,440-ends
(17) A worsted warp 740 yards long contains 1,600 ends
of 2-ply 36s; what is the weight of the yarn in the warp?
Ans. 117.460 lb.
(18) A warp containing 2,772 ends is dressed 6 black,
2 slate, 3 black, 1 slate, 12 black, 4 slate; how many ends
of each color are there in the warp?
Ans. 2,079 ends of black, 693 ends of slate
(19) A woolen warp contains 864 ends of 3-run yarn and
weighs 85i pounds; what is the length of the warp?
Ans. 475 yd.
(20) A warp contains 1,400 ends of 2-ply 40s worsted
and 700 ends of 10s; what is the average number of the
warp? Ans. 15s
§10
WOOLEN AND WORSTED
3
(21) If 2 yards of worsted yarn weighs 1 grain, what
is the size of the yarn? Ans. 25s
(22) If 1 yard of woolen yarn weighs 1.2 grains, what
run is the yarn? Ans. 3.645-run
(23) 300,000 yards of woolen yarn weighs 50 pounds;
what are the counts in both the run and cut systems?
Ans. 3.75-run, 20-cut
(24) Give a rule for finding the counts of a ply yarn
composed of more than two threads of single yarns.
(25) State how to find the weight of each thread in a ply
yarn when the total weight and the counts of the separate
threads are given.
YARN CALCULATIONS,
GENERAL
EXAMINATION QUESTIONS
(1) What counts of yarn should be twisted with a 24s
cotton to make a ply yarn equal in weight to an 8s single?
Ans. 12s
(2) An 80s, 40s, and 30s worsted are twisted together;
what are the counts of the ply yarn? Ans. 14.117s
(3) How many yards are there in 2 pounds of 50s cotton
yarn? Ans. 84,000 yd.
*
(4) How many yards are there in 13 pounds of 3l-run
woolen yarn? Ans. 70,200 yd.
(5) Explain what is meant by the word counts.
(6) State how many yards there are in 1 pound of yarn
in each of the following systems, the counts being 5s single:
woolen (run and cut), cotton, spun silk, and worsted.
8,000 yd. (woolen run), 1,500 yd. (woolen cut)
Ans.(4,200 yd. (cotton), 4,200 yd. (spun silk)
.2,800 yd. (worsted)
(7) If 368,000 yards of single worsted yarn weighs
16 pounds, what are the counts? Ans. 41.071s
(8) What counts of single yarns are twisted to make a
2-ply 38s worsted yarn?
2
YARN CALCULATIONS, GENERAL
11
(9) Find the counts in the woolen (run and cut),
worsted, and cotton systems equal to 36s spun silk.
« j 18.9s (woolen run), 100.8s (woolen cut)
ns 'l54s (worsted), 36s (cotton)
(10) What is the weight of 360,000 yards of: ( a) 2-ply
36s worsted? ( b ) 2-ply 48s cotton? ( c) 5-run woolen?
(a) 35.714 lb.
Ans.] (b) 17.857 lb.
Ac) 45 lb.
(11) How many ounces of yarn are there in 14,000 yards
of 2-ply 36s worsted? Ans. 22.208 oz.
(12) Change 2-ply 40s worsted to 2-ply cotton counts.
Ans. 26fs
(13) What is the length of 18 pounds of 8 s yarn: (a) in
the cotton system? ( b) in the worsted system? (c) in the
spun-silk system?
Ans.
(a) 120,960 yd.
(b) 80,640 yd.
Ac) 120,960 yd.
(14) What counts must be twisted with single 14s
worsted to produce a 2 -ply yarn equal to a single 10 s?
Ans. 35s
(15) What weight of 60s worsted will be required to pro¬
duce 120 pounds of 2-ply yarn when twisted with a 38s
worsted? Ans. 46.530 lb.
(16) What is the price per pound of a ply yarn composed
of one thread of 36s cotton at I 4 cents per ounce and one
thread of 54s worsted at 6 cents per ounce? Ans. 58 ct.
(17) What counts of cotton yarn must be twisted with
a 40s cotton to make a 2-ply yarn equal to a 24s single?
Ans. 60s
(18) If one end of 30s and one of 10s are twisted, what
will be the weight of each yarn in 50 pounds of the ply yarn?
A {12.5 lb. of 30s
Ans -137.5 lb. of 10s
11
YARN CALCULATIONS, GENERAL
3
(19) • A given length of 2-ply cotton yarn weighs
70 pounds and is composed of one end of 24s and one end
of 9s; what weight of each is there in the ply yarn?
, f 19.089 lb. of 24s
Ans -(50.905 lb. of 9s
(20) 300,000 yards of woolen yarn weighs 50 pounds;
what are the numbers in both the run and cut systems?
Ans. 3.75s (run), 20s (cut)
(21) A 60s cotton, 40s worsted, and 60s silk are twisted;
what are the counts of the ply thread in the worsted system?
Ans. 21.176s
(22) How many ends are there in a 32s cotton warp
200 yards long and weighing 15 pounds? Ans. 2,016 ends
(23) A worsted warp containing 2,800 ends is 800 yards
long and weighs 200 pounds; what is the size of the yarn?
Ans. 20s
(24) If 76 pounds of 30s cotton warp yarn is used in a
warp of 3,600 ends, what is the length of the warp?
Ans. 532 yd.
(25) A woolen warp is 400 yaids long and weighs
80 pounds; if the warp is composed of 42-run yarn, how
many ends does it contain? Ans. 1,440 ends
CLOTH CALCULATIONS,
COTTON
EXAMINATION QUESTIONS
(1) How many ends are there in each of the following-
warps, allowing 48 ends for selvages: ( a) Width at reed 80
inches, reed 20 dents per inch, 2 ends per dent? (b) Width
at reed 32 inches, reed 40 dents per inch, 2 ends per dent?
a f (a) 3,248 ends
s ‘l(£) 2,608 ends
(2) A warp contains 3,800 ends and is woven in a reed
containing 30 dents per inch, the ends being drawn in 4 per
dent; what is the width at the reed, neglecting selvages?
Ans. 31f in.
(3) What is the weight of yarn in a warp 28 inches wide
at the reed including selvages, drawn 6 per dent in a reed
having 12 dents per inch, the length of the warp being
50 yards and the counts of the yarn 60s? Ans. 2 lb.
(4) What is the weight of yarn in a warp containing
2,800 ends including selvages, the warp being 60 yards long
and the counts of the yarn 24s? Ans. 8i lb.
(5) A cut of cloth made with 36s filling is 50 yards long,
32 inches wide at the reed, including selvages, and contains
56 picks per inch; what is the weight of the filling?
Ans. 2.962 lb.
(6) (a) A warp is arranged 8 ends of black and 4 of
white. It is 160 yards long and contains 1,800 ends of 30s;
§ 12
2
CLOTH CALCULATIONS, COTTON
12
what is the weight of each color in the warp? (b) What
weight of filling will be required if the above piece is woven
30 inches wide at the reed, including selvages, with 64 picks
per inch, the warp yielding 150 yards of cloth? The counts
of the filling are the same as the warp. ( c ) What is the
weight per yard in ounces of the above cloth, allowing 10 per
cent, for size on warp yarns?
(a) 3.809 lb. of white, 7.618 lb. of black
Ans.{ {b) 11.428 lb.
(c) 2.5 oz.
(7) (a) A warp contains 4 pounds of 18s and has
1,200 ends; how long is it? (b) 15 pounds of 2-ply 60s is
made into a warp 80 yards long; how many ends does it
contain? . f (a) 50.4 yd.
Ans ’t(£) 4,725 ends
(8) What is the cost of the yarn in a piece of cloth made
as follows: 70 ends per inch, in the reed, of 35s at 20 cents
per pound, 72 picks per inch of 40s at 22 cents per pound?
The piece is woven 55 yards long from 58 yards of warp and
is 30 inches wide at the reed, including selvages.
Ans. $1,606
(9) 52i pounds of 20s is used to make a warp 700 yards
long: (a) How many ends are there in the warp? (b) If
the yarn is 21 inches wide at the reed, including selvages,
and the reed contains 20 dents per inch, the selvages being
reeded like the body of the warp, how many ends are there
to the dent? A \ (a) 1,260 ends
11S ’1(£) 3 ends per dent
(10) A warp pattern is as follows: 4 ends of blue, 16 ends
of white, 4 ends of red. The width at the reed is 32 inches,
not including selvages, and the warp is drawn 2 ends per
dent in a reed containing 30 dents per inch. How many
ends of each color will be required for the body of the
warp? Ans. 320 ends red, 320 ends blue, 1,280 ends white
(11) What will be the weight of filling in: («) 56 yards
of cloth 29 inches wide at the reed, including selvages, 78
picks per inch, using 60s filling? {b) 100 yards of cloth
§12
CLOTH CALCULATIONS, COTTON
3
32 inches wide at the reed including selvages, 36 picks per
inch, using 16s filling? A f (a) 2.513 lb.
Ans ‘l(£) 8.5711b.
(12) A cloth is woven 100 yards long from 104 yards of
warp 29 inches wide at the reed, including selvages, 64 picks
per inch of 36s filling, and 64 ends per inch of 27s warp;
what will be the weight per yard in ounces of above cloth,
allowing 10 per cent, for size on warp yarn?'
Ans. 2.47 oz.
(13) What length of cloth will 100 pounds of 30s supply
with filling if the cloth is to be 29 inches wide at the reed
including selvages, and to contain 72 picks per inch?
Ans. 1,206.896 yd.
(14) What is the weight per yard, in ounces, of a cloth
56 inches wide including selvages, if 1 square inch weighs
2.625 grains? Ans. 12.096 oz.
(15) A cloth 28 inches wide at the reed including sel¬
vages is made with 56 ends per inch in the reed, and 56
picks per inch of 30s. What is the weight per yard in
ounces, adding 7 per cent, to warp yarn for size and con¬
traction in weaving? Ans. 2.048 oz.
(16) A warp pattern is drawn in as follows:
Dents
16
4
2
4
26
Ends
32
16
4
16
68
(a) How many patterns will there be in a warp 30 inches
wide at the reed excluding selvages, using a 34s reed?
( b) What will be the total number of ends required for the
body of the warp? A \(a) 39 patterns, 6 dents over
Ans, l(£) 2,664 ends
(17) A piece of cloth 4 inches square weighs 14 grains;
if the cloth is 32 inches wide including selvages, what are
the yards per pound? Ans. 6.944 yd.
4
CLOTH CALCULATIONS, COTTON
§12
(18) What would be the width in the reed of a 64-sley
cloth containing 1,800 ends in the body of the cloth drawn
2 per dent, and 48 ends drawn 4 per dent for selvages?
Ans. 30.476 in.
(19) What reed will be required to weave a 50-sley cloth,
if drawn 2 ends per dent? Ans. 23.275s
(20) A cloth is of the following construction: 70 X 76
— 38 in. wide including selvages — 70s warp — 12 yd.; what
counts of filling must be used? Ans. 118s (about)
CLOTH CALCULATIONS,
WOOLEN AND WORSTED
EXAMINATION QUESTIONS
(1) A worsted cloth is 28 inches wide and 1 square inch
weighs 2.6 grains; what is the weight of the cloth per yard?
Ans. 5.99 oz.
(2) A warp containing 2,600 ends is to be drawn through
the harnesses according to the draft shown in Fig. 1; how
many heddles are required for each harness?
Ans.
780 on first harness
1,040 on second harness
520 on third harness
260 on fourth harness
(3) A warp containing 2,772 ends is dressed 6 black,
2 slate, 3 black, 1 slate, 12 black, 4 slate; how many ends of
each color are there in the warp?
* , [2,079 ends of black
ns * 1693 ends of slate
(4) The above warp is composed of 2/40s worsted yarn
and is 560 yards long; what is the weight of each color of
yarn?
A [103.95 lb. black
134.65 lb. slate
(5) A warp contains 1,400 ends of 2/40s worsted and
700 ends of 10s; what is the average number of the warp?
Ans. 15s
2
CLOTH CALCULATIONS,
13
(6) A warp contains 2,520 ends including selvage ends
and is reeded 5 per dent; if there are 70 ends per inch in the
reed: (a) what reed should be used? (b) what is the width
of the goods in the loom?
a f (a) 14s reed
Ans -1(*) 36 in.
(7) If 50 inches of worsted yarn weighs 1 grain, what is
the size of the yarn? Ans. 17.361s
(8) A piece of cloth contains 1,960 ends of 20s worsted
in the warp; the cloth is 30 inches wide and contains 44 picks
per inch of 4-run woolen yarn; what is the weight of the
cloth per yard? Ans. 6.1 oz.
(9) If 30 inches of woolen yarn weighs 1.2 grains, what
run is the yarn? Ans. 3.038-run
(10) A cloth is woven 62 inches wide in the reed; the
woven cloth has 28 picks per inch of 3i-run yarn; in a cer¬
tain length of cloth there are 14 pounds of filling; what is the
length? Ans. 45.16 yd.
(11) One square inch of cloth weighs 2.8 grains; what is
the weight of a yard if the cloth is 56 inches wide?
Ans. 12.902 oz.
(12) A certain piece of cloth has shrunk 20 per cent, from
reed to finished goods and lost 10 per cent, in finishing; the
filling in the finished cloth is 14.4s worsted; what was the
original size? Ans. 16s
(13) A piece of cloth is 36.96 inches wide finished and
has shrunk 16 per cent, from reed to finished cloth; what was
the width in the loom? Ans. 44 in-
(14) A warp contains 1,120 ends of 22-run yarn and takes
up 4 per cent, in weaving; what is the weight of warp in
1 yard of cloth from the loom? Ans. 4.65 oz.
(15) If a piece of cloth is 34 inches wide and contains
28 picks per inch of 10s worsted, what is the weight of filling
in 48 yards of cloth? Ans. 8.160 lb.
§13 WOOLEN AND WORSTED 3
(16) What is the weight of a yard of cloth composed of
960 ends and 32 picks per inch of 3.2-run yarn if it is 30 inches
wide? Ans. 6 oz.
(17) If the above cloth has shrunk 20 per cent., both warp
and filling ways, what was the original size of the yarn and
what was the width in the reed, supposing the finished width
given in the previous example to be outside of selvages?
Neglect loss in finishing.
A J 4-run
Ans * 137.5 in.
(18) A warp contains 2,592 ends and is reeded 2 per dent,
the reed being an 18s; what is the width in the loom?
Ans. 72 in.
(19) What is the weight of the filling in a yard of goods
44 inches wide and containing 44 picks per inch of 14.65s
worsted filling? Ans. 3.775 oz.
(20) A warp is reeded 3 per dent and is 37 inches wide
in the loom; what is the number of ends in the warp if an
11s reed is used? Ans. 1,221 ends
EXAMINATION QUESTIONS
(1) What are the objects of drafting?
(2) Why is more draft produced by a set of metallic
rolls than by a set of common rolls that correspond in
diameter with the metallic rolls and that are geared in the
same manner?
(o) If the number of ends put up at the back of a certain
machine and drawn into one, the weight per yard of each
end at the back, and the weight per yard of the end
delivered at the front are known, how could the draft of the
machine be obtained?
(4) Explain the terms: (a) draft change gear; (b) crown
gear.
(5) Give the rule for finding the required draft gear to
give a certain draft when the constant is known, the constant
being a constant dividend.
(6) What is the effect of placing a larger draft gear on
machines geared similarly to Figs. 4, 5, and 6?
(7) Explain the terms: (a) constant factor; (b) constant
dividend.
(8) What is meant by saying that a machine has a
draft of 6?
(9) Explain the term break draft.
(10) Give the rule for finding the draft constant of a set
of rolls connected by gears.
2
DRAFT CALCULATIONS
§15
(11) Find the break draft for the rolls and gears shown
in Fig. 5. Ans. 2.915
(12) Find the total draft of a machine that has a draft of
1.4 between third and back rolls; a draft of 1.5 between
second and third rolls, and a draft of 3.2 between front and
second rolls. Ans. 6.72
(13) If the feed-roll and delivery roll of a machine make
the same number of revolutions per minute, what will be the
draft if the feed-roll is If inches in diameter and the delivery
roll If inches in diameter? Ans. 1.555
(14) Find the total draft for Fig. 5, supposing that
metallic rolls are used and a 60-grain sliver is passing into
the machine. Ans. 7.04
(15) Find the total draft for the set of rolls shown in
Fig. 6. Ans. 6.006
(16) Find the total draft for the set of rolls shown in
Fig. 5. Ans. 6.459
(17) Find the constant for Fig. 5, the gear,g being the
change gear. Ans. 523.25
(18) Find the constant for Fig. 6, the gear^' being the
change gear. Ans. 468.518
(19) Using the constant found in answer to question 18,
what change gear will be required to give a draft of 7?
Ans. 67, gear
(20) If a draft gear of 68 teeth gives a draft of 6.04 on a
certain machine, what draft gear will be required to give a
draft of 6.4, the draft gear being a driver? Ans. 64, gear
READING TEXTILE
DRAWINGS
EXAMINATION QUESTIONS
(1) Name two uses of dotted lines.
(2) What is an assembly drawing?
(3) (a) What is a perspective drawing? ( b) What is the
principal difference between a perspective drawing and a
photographic reproduction of the same object from the same
point of view?
(4) What is meant when it is said that a certain drawing
is a right-side elevation of an object?
(5) For what are broken lines chiefly used?
(6) In a drawing of a section, what is done by the drafts¬
man to enable it to be recognized as a sectional view and not
one of the outside surfaces of the object?
(7) Show, by sketches, how cast iron, wrought iron, steel,
and brass are indicated by section lines.
(8) When it is desired to draw a long piece, such as a
shaft, in a limited space, how can the drawing be made with¬
out reducing the size of the object to such an extent as to
make the diameter appear so small that details cannot be
shown plainly?
(9) Give a sketch showing the conventional method of
representing a screw thread that is largely used.
I 91
2 READING TEXTILE DRAWINGS §91
(10) How are sections of an object shown that is too
small or thin to be readily sectioned?
(11) Why is Fig. 15 (c) preferable to Fig. 15 (£)?
(12) What is a diagrammatic drawing?
(13) State briefly the use of shade lines.
(14) In the object shown in Fig. I, (a) how many separate
pieces are shown? (5) what
different materials are indi¬
cated?
(15) Make a sketch show¬
ing a conventional method
of representing a spring.
(16) How can it be deter¬
mined if a screw has a right-
hand thread?
(17) Why are reference
letters placed on drawings?
(18) What can be said in
regard to reference letters referring to a piece that is shown
in several drawings?
(19) What is meant by a partial section?
(20) If a drawing of an object is made to an eighth scale
and the object is 36 inches long, what will be the length of
the drawing?
INDEX
Note.— All items in this index refer first to the section and then to the page of the sec¬
tion. Thus, “Addition 1 6” means that addition will be found on page 6 of section 1.
A
Sec. Page
Sec. Page
Abstract number.
1
1
Average number of the warp . . .
10
23
Acute angle.
6
2
yarns.
9
15
Addition.
1
6
“ “ ** “
11
33
of compound quantities .
5
12
sley.
12
23
“ decimals.
2
21
Avoirdupois weight.
5
3
“ fractions.
2
9
Axis of a sphere.
6
15
“ Proof of.
1
11
Rule for.
1
10
B
Sign of.
1
6
Back rolls.
15
6
Aggregation, Signs of.
3
19
Backlash.
7
29
Altitude of a parallelogram ....
6
5
Base.
3
2
“ pyramid or cone . . .
6
14
“ of a prism.
6
11
“ “trapezoid.
6
5
rate, and percentage.
3
4
“ “ triangle.
6
2
Beam calculations ,.
9
13
American warp yarns. Breaking
“ “
10
18
weight of.
9
20
Beamed yarns.
11
30
Amount.
S
3
Beams, Loom.
9
13
tl
3
7
“ Section.
9
13
Angles.
6
2
Belt, Crossed.
7
20
Angular measure.
5
7
" fastenings.
7
11
Animal fibers.
11
1
“ Open.
7
20
Annular gear.
7
34
“ Quarter-turn.
7
22
Antecedent.
3
11
“ shipper.
7
18
Apothecaries’fluid measure . . . .
5
4
Belting.
7
10
weight.
5
4
Belts, Care of.
7
12
Arabic notation.
1
2
“ Cotton.
7
10
Arc of a circle.
6
8
“ Horsepower transmitted by
8
7
Area.
6
2
“ Leather.
7
10
“ Measures of.
5
6
" Length of.
8
6
“ of a circle.
6
9
Rubber.
7
10
“ “ “ cylinder.
6
12
“ Rules applying to.
8
6
“ M prism.
6
11
“ Single and double.
7
11
“ “ “ regular polygon ....
6
7
Bevel gears.
7
30
. sphere .
6
15
Binder pulley.
7
17
“ “trapezium..
6
6
Board measure.
6
16
“ “ “ trapezoid.
6
5
Body, Solid.
6
10
“ “ triangle.
6
3
Borrowing numbers.
1
13
Arithmetic.
1
1
Box couplings.
7
2
Ascending, Reduction.
5
1
Braces .
3
19
Assembly and detail drawings . .
91
17
Brackets.
3
19
Average counts.
11
33
“ Wall.
7
7
“ “
12
21
Break draft.
15
17
XI
Xll
INDEX
Sec. Page
Sec. Page
Breaking- weight of warp yarns . .
9
19
Circle, Radius of a.
6
8
Breaks in drawings.
91
25
“ Rule to find area of.
6
9
Broken-and-dotted lines.
91
22
“ “ “ circumference
“ line.
91
22
of.
6
8
Bushings, Pulley.
7
15
Circular measure.
5
7
“ pitch of a gear.
7
29
C
Circumference of a circle.
6
8
Calculating draft for common rolls
15
12
Clamp coupling.
7
3
production and draft .
15
22
Classes of shafting.
7
1
Calculation of cost of ply yarns . .
11
25
Cloth calculations, Cotton ....
12
1
Calculations, Beam.
9
13
Importance of
12
1
It i i
10
18
“ “ ** “
13
1
Cotton-cloth.
12
1
Woolen and
“ yarn .
9 *
1
worsted . .
13
1
“ Draft.
15
1
“ Counts of.
12
2
for filling yarn . . .
12
10
“ Double-width.
13
3
warp yarns . . .
12
5
“ from loom. Weight of. . . .
13
24
General yarn ....
11
1
“ Ounce.
12
2
Harness.
12
3
“ Pick.
12
2
(1 “
13
4
“ production, C al cul a ti on s
Importance of cloth .
12
1
necessary for.
12
1
“ “ “ “
13
1
“ production, C alcul at ion s
Mechanical.
8
1
necessary for.
13
1
necessary for cloth
“ samples, Figuring particu-
production ....
12
1
lars from ....
12
16
necessary for cloth
F i g u r ing particu-
production ....
13
1
lars from ....
13
21
of cotton ply yarns .
11
20
Obtaining particu-
“ ply yarns.
9
8
lars of .
12
24
it it it it
10
8
“ Single-width.
13
3
“single cotton
“ Weight of.
13
3
yarns.
11
3
“ “ “
13
21
" single worsted
“ Yard.
12
2
yarns.
10
2
Clutch coupling.
7
4
“ single yarns . . .
9
2
Clutches.
7
2
“ Reed.
13
5
Friction.
7
5
Woolen and worsted
Cogs.
7
28
cloth .
13
1
Cold-rolled shafting, Rule to find
Woolen and worsted
the size of.
8
1
yarn.
10
1
Colors, Repeat of.
9
17
Cam-bowl.
7
37
Common denominator of fractions
2
8
Cams.
7
35
divisor. Greatest ....
1
29
Cancelation.
1
33
multiple.
1
31
Canceling.
1
33
rolls. Calculating draft
Care of belts.
7
12
for.
15
12
Carrier gear.
7
35
Drafting with . .
15
5
Center of a circle.
6
8
Complex fraction.
2
3
Chain, Engineer’s.
5
6
Composite number.
i
27
“ Gunter’s.
5
6
Compound interest.
3
8
Change gear.
8
14
levers.
7
38
<1 <4
15
25
numbers .
5
1
Characters.
1
2
“ “ To add ...
5
13
Chord of a circle.
6
8
quantities, Addition of
5
12
Cipher.
1
2
Division of
5
16
Circle.
6
8
Multiplica-
“ of a gear, Pitch.
7
28
tion of . .
5
14
INDEX
xm
Sec.
Compound quantities, Reduction
of ... . 5
Subtraction
of ... . 5
Concentric gears . 7
Concrete number. 1
Cone and pyramid. 6
“ or pyramid, Altitude of a . . 6
Frustum of a . . 6
Cones, Speed. 7
Consequent . 3
Constant dividend. 8
dividends.15
factor . 8
factors.15
Rule to find a. 8
Constants. 8
“ 15
Construction of fabrics.13
Contact, Frictional. 7
Contraction during weaving ... 12
in weaving.13
Cost of ply yarns, Calculation of . 11
Cotton belts. 7
“ cloth calculations.12
ply yarns. Calculations of 11
“ roving and yarn, Sizing . . 11
warp yarn. Breaking
weight of. 9
“ yarn calculations. 9
“ yarns .11
Countershafts. 7
“ 7
on speed, Effect of 8
Counts and weight. Proof of . . . 13
“ Average .11
“ “ .. 12
“ Equivalent.11
“ Folded yarns of different . 9
“ . . 10
“ . . 11
the same 9
“ .. “ 10
“ “ “ “ “ 11
Method of determining the 11
“ of cloth.12
“ filling, to preserve yards
per pound.12
“ the warp and filling . . 13
Couplet. 3
Couplings. 7
Cross-section.91
Crossed belt. 7
Crown-faced pulley. 7
Cube, A. 6
“ Definition of a. 5
Sec. Page
Cube of a number. 4 1
“ root. 4 3
“ “ 4 10
“ “ Table method of ex¬
tracting . 4 16
Cubes and squares. 4 17
Cubic measure. 5 6
Curved line, A. 6 1
“ surface, A. 6 2
Cut scale.11 9
“ system of numbering woolen
yarns.10 14
numbering woolen
yarns.11 9
Cylinder, Surface area of a ... . 6 12
“ The. 6 12
Volume of a. 6 13
1 )
Dark surface.91 23
Decagon. 6 7
Decimal, Reducing a fraction to a 2 29
to a fraction, Reducing a 2 30
Decimals. 2 18
Addition of. 2 21
Division of. 2 24
Multiplication of ... . 2 23
Rule for the division of . 2 27
Subtraction of. 2 22
Definitions. 1 1
Mechanical. 7 1
of mensuration .... 6 1
“ percentage. 3 1
Delivery rolls.15 6
Denominate numbers. 5 1
Denominator. 2 1
of fractions. Common 2 8
Dent.13 6
Descending, Reduction. 5 1
Detail and assembly drawings . . 91 17
Determining the counts, Method of 11 2
“ diameter of yarns 11 29
Diagrammatic views.91 29
Diameter of a circle. 6 8
“ sphere ....... 6 15
“ gear-blank. 8 12
“ yarns.11 29
Diameters of pulleys, Ratio of
speeds to ... . 8 5
“ rolls on draft, Ef¬
fect of the .... 15 12
Diametral pitch of a gear. 7 29
Difference. 1 12
“ 3 3
Different counts. Folded yarns of 9 9
Differential screw. 7 35
Page
9
13
32
1
13
14
14
18
11
15
28
14
28
15
14
28
2
9
8
9
25
10
1
20
4
19
1
2
2
21
4
22
33
21
18
9
9
22
8
8
20
2
2
22
22
11
2
15
20
14
11
6
XIV
INDEX
Sec.
Page
Sec.
Page
Digits.
i
2
Drawings, Kinds of.
91
2
Dimension lines.
91
27
Lines used on.
91
22
Direct proportion.
3
13
Mechanical.
91
3
“ ratio.
3
11
Reading textile ....
91
1
Direction of pulley rotation ....
7
20
Driven and driving gears.
15
11
Distance between hangers ....
7
7
Drives, Rope .
7
24*
Dividend.
1
22
Driving and driven gears.
15
11
Constant .
8
15
Drum, A.
7
16
Dividends, Constant .
15
28
Dry measure.
5
5
Division.
1
22
of compound quantities
5
16
E
“ decimals.
2
24
Eccentric gears.
7
32
Rule for the
2
27
Effect of countershafts on speed
8
4
“ fractions.
2
15
“ diameter of rolls on
Proof of.
1
25
draft.
15
12
Rule for.
1
25
“ gears and rolls on
Sign of.
1
22
draft.
15
14
Divisor .
1
22
size of gears on draft . .
15
13
“ Greatest common ....
1
29
Elements, Machine.
7
1
Dodecagon .
6
7
Elevations.
91
6
Dot-and dash lines .
91
22
Elliptic gears .
7
34
Dotted line.
91
22
End.
13
2
Double and single belts.
7
11
Ends and reed per dent.
13
24
" width cloth.
13
3
“ Definition of.
12
2
yarns .
10
7
“ in a warp, Finding number
Doubling.
15
3
of.
12
5
* 4
15
30
“ the warp, Number of . .
13
22
Draft and production, Calculating
15
22
Engineer’s chain.
5
6
“ Break.
15
17
Equality, Sign of.
1
6
“ calculations.
1
Equilateral triangle.
6
2
“ Combined effect of gears and
Equivalent counts.
11
18
rolls on.
15
14
Even number.
1
28
“ Drawing-in.
13
4
Evolution.
4
3
“ Effect of diameters of rolls
Exponent of a number.
4
1
on.
15
12
Extension lines.
91
27
“ Effect of size of gears on . .
15
13
Extracting square and cube
“ for common rolls, Calcu-
roots.
4
16
lating.
15
12
gears .
15
25
F
“ Harness .
12
3
Fabrics, Construction of.
13
2
13
4
Factor, Constant.
8
14
“ Methods of finding.
15
7
Prime.
1
27
Drafting.
15
1
Factors.
1
27
with common rolls . . .
15
5
“ Constant.
15
28
“ special reference to
“ Rational and irrational . .
4
8
the mill.
15
25
Rule to find prime ....
1
28
Drafts, Irregular reed.
12
12
Fancy patterns.
13
14
“ Table of.
15
24
“ warps .
9
16
Drawing, Definition of .
91
1
“ “
10
21
Fundamental principles
4 ( 4 4
11
35
of.
91
3
Fast couplings.
7
2
Scales for.
91
26
Fastenings, Belt.
7
11
in draft.
4
Feed-rolls.
15
G
Drawings, Breaks in.
91
25
Fibers, Animal .
11
1
Detail and assembly . .
91
17
“ Vegetable.
11
1
Freehand .
91
2
Figure, A plane.
6
2
INDEX
xv
Sec. Page
G
Sec.
Page
Figures .
1
2
Gauge points.
10
6
Significant .
4
16
Cl Cl
11
13
Figuring particulars from cloth
Gear, Annular.
7
34
samples.
12
16
11 blank, Diameter of.
8
12:
particulars from cloth
“ blanks, Sizing.
8
12:
samples.
13
21
“ Carrier .
7
35
Filling.
11
35
“ Change .
8
14
(«
13
2
c c c c
15
25
“ and warp. Counts of the , .
13
22
“ Idle.
7
35
“ Finding hanks of.
12
12
“ Intermediate.
7
35
II “ ti <<
12
20
Pitch circle of a.
7
28
“ to preserve yards per
“ “ of a .
7
29
pound, Counts of . . . .
12
22
Ratchet.
7
31
“ Weight of.
13
23
Gearing.
7
27
“ yarn. Calculations for . . .
12
10
of rolls.
15
9
Finishing, Shrinkage in.
13
U
Gears and rolls on draft, Effect of .
15
14
First or prime mover.
7
2
“ Bevel.
7
30
Flange coupling.
7
3
“ Concentric.
7
32
“ pulleys.
7
16
“ Draft.
15
25
Flat pulley.
7
14
“ Driving and driven.
15
11
Floor stand.
7
9
“ Eccentric.
7
32
Fluid measure. Apothecaries’ . ,
5
4
Effect of size of, on draft . .
15
13
Folded yarns.
9
7
“ Elliptical.
7
34
i < c c
10
7
Mangle.
7
32
11
20
“ Miter.
7
31
of different counts . .
9
9
Mortise.
7
28
<« l« Cl II II
10
9
Rule to find teeth required in
8
12
II <1 II Cl II
11
22
“ Rules applying to.
8
10
“ “ the same counts .
9
8
for draft.
15
26
II cc cc cc <* c*
10
8
“ Spiral.
7
32
Cl Cl CC CC Cl Cl
11
20
“ Sprocket .
7
31
Fraction, Complex.
2
3
“ Spur.
7
30
Improper.
2
3
“ Star .
7
32
Proper .
2
2
“ Trains of.
7
29
Simple.
2
3
Greatest common divisor.
1
29
Terms of a.
2
2
Groove pulleys.
7
16
to a decimal, Reducing a
2
29
Guide pulley.
7
17
Value of a.
2
2
Gunter’s chain.
5
6
Fractions.
2
1
II
Addition of.
2
9
Common denomina-
Hanger, Wall-box.
7
9
tor of .
2
8
Hangers, Distance between . . .
7
7
Division of.
2
15
Shaft .
7
5
Multiplication of ... .
2
13
Types of .
7
7
Reduction of .
2
4
Hanks of filling, Finding.
12
12
Roots of .
4
15
CC CC Cl II
12
20
Subtraction of .
2
11
“ '! warp yarn. Finding . . .
12
18
Freehand drawings . . . • .
91
2
“ Finding num-
Friction clutches .
7
2
ber of . . .
12
7
II II
7
5
Harness .
12
i
“ couplings .
7
5
c c
13
2
Frictional contact ..
7
9
calculations .
12
3
Front rolls.
15
6
II 4 1
13
4
Frustum of a pyramid or cone . .
6
14
“ draft.
12
3
Full line.
91
22
Cl cc
13
4
Fundamental principles of drawing
91
3
Head, or main, shaft.
7
2
XVI
INDEX
Sec.
Page
Sec. Page
Heavy full line.
91
22
Line, Light full.
. 91
22
Heddles.
12
3
“ shafts.
. 7
2
Heel of a cam.
7
36
Linear or long measure.
. 5
5
Heptagon.
6
7
Linen, jute, and ramie.
. 11
16
Hexagon.
6
7
Lines.
6
1
Horizontal line, A.
6
1
“ Broken-and-dotted . . . .
. 91
22
Horsepower transmitted by belts .
8
7
Dimension .
. 91
27
Hundreds.
1
3
Dot-and-dash.
. 91
22
f
Extension.
. 91
27
“ Limiting.
. 91
27
Idle gear.
7
35
“ Parallel.
6
1
“ pulley.
7
17
“ Shade .
. 91
22
Idler.
7
17
used on drawings.
. 91
22
Improper fraction.
2
3
Liquid measure.
5
4
Index of the root.
4
3
Listing.
. 13
7
Integer.
1
2
Local value.
1
3
Integral number.
1
2
Long division.
1
23
Interest.
3
7
“ or linear measure.
. 5
5
Intermediate gear.
7
35
Longitudinal sections.
. 91
15
Inverse proportion.
3
13
Loom beams.
9
13
3
16
4 4 4 4
. 11
30
ratio.
3
11
“ Weight of cloth from . . .
. 13
24
Inverted pillow-block.
7
9
Loose and tight pulleys.
. 7
17
plan.
91
7
“ couplings.
. 7
2
Involution.
4
1
Lumber, Mensuration of ....
6
16
Irrational factors.
4
8
Isosceles triangle.
6
3
M
J
Machine elements.
. 7
1
Main, or head, shaft.
7
2
Jute, linen, and ramie.
11
16
Mangle gears.
7
32
K
Measure, Angular.
. 5
7
Key seats.
7
2
Apothecaries’ fluid . .
. 5
4
Keys.
7
2
Board.
6
16
Keyways.
7
2
Circular.
. 5
7
Cubic.
. 5
6
L
“ Dry.
. 5
5
Law of value expressed by figures
1
4
Linear or long.
. 5
5
Lay.
13
2
Liquid.
. 5
4
Least common denominator of
of money.
. 5
2
fractions . . .
2
8
“ time.
. 5
7
multiple ....
1
31
Square.
. 5
6
Leather belts.
7
10
Surface.
. 5
6
Left-handed threads.
91
30
Surveyors' .
. 5
5
Length, Measures of.
5
5
Measures.
. 5
2
“ of belts.
8
6
Miscellaneous ...
. 5
8
Letters, Reference.
91
33
of area.
. 5
6
Lever, Pressure exerted by a . . .
8
16
“ length.
. 5
5
Levers .
7
37
“ quantity.
. 5
4
“ Rules applying to.
8
16
“ solidity or volume .
. 5
6
Light full line.
91
22
“ weight.
. 5
3
“ surface.
91
23
Mechanical calculations.
. 8
1
Like numbers.
1
1
definitions.
. 7
1
Limiting lines.
91
27
drawings .
. 91
3
Line, Broken .
91
22
Mensuration.
. 6
1
“ Dotted.
91
22
of lumber.
6
16
“ Heavy full.
91
22
“ solids.
6
10
INDEX
XVII
Sec. Page
Sec. Page
Mensuration of surfaces.
6
1
Number of the reed.
13
6
Metallic rolls .
15
20
“ “ warp, Average . .
10
23
Method of numbering- ply yarns . .
9
7
“ yarns. Average . . . .
9
15
<« 44 14 44 14
10
7
44 44 44 44
11
33
44 44 44 44 44
11
20
Power of a.
4
1
Methods of finding draft.
15
7
Prime.
1
27
Metric system of numbering yarns
9
21
Reciprocal of a.
S
11
44 44 44 44 44
10
25
Root of a.
4
1
44 44 44 44 44
11
38
Square of a.
4
1
Minuend.
1
12
Unit of a.
1
1
Minus sign.
1
12
Numbering ply yarns.
9
7
Miscellaneous measures.
5
8
44 44 44
10
7
Miter gears.
7
31
44 44 44
11
20
Mixed number.
2
3
single worsted yarns .
10
1
Money, Measure of.
5
2
yarns .
10
13
“ Reduction of United States
6
11
system.
9
1
“ United States.
5
2
“ Yarn.
11
2
Mortise gears.
7
28
woolen yarns, Cut sys-
Mover, First, or prime.
7
2
tem of.
10
14
Movers, Second.
7
2
woolen yarns. Cut sys-
Moving parts shpwn in two or
tem of.
11
9
more positions.
91
33
woolen yarns. Run
Muff couplings .
7
2
system of.
10
13
Mule pulley stand.
7
24
woolen yarns, Run
Multiple, Common .
1
31
system of.
11
7
threaded screw.
7
35
yarns, Metric system
Multiplicand.
i
16
of.
9
21
Multiplication.
i
16
“ Metric system
of compound quan-
of.
10
25
tities.
5
14
“ Metric system
“ decimals ....
2
23
of.
11
38
“ fractions . . . .
2
13
Numbers, Borrowing.
1
13
Proof of.
1
20
Compound.
5
1
Rule for.
1
19
Denominate.
5
1
Sign of.
1
16
“ Like.
1
1
“ table.
1
17
Simple.
5
1
Multiplier.
1
16
To add compound . . .
5
13
N
Unlike.
1
1
Numeration.
1
5
Naught .
1
2
and notation.
1
2
Non-parallel shafts.
7
22
Numerator
2
1
“ positive cam.
7
36
O
Notation.
i
5
and numeration.
i
2
Objects, Conventional methods of
Arabic.
i
2
representing.
91
28
Number, A .
i
1
“ of drafting.
15
2
Composite.
i
27
Representation of ... .
91
1
Cube of a.
4
1
Obtuse angle .
6
2
Even.
1
28
Octagon.
6
7
Exponent of a.
4
1
Odd number.
1
28
Integral.
1
2
Open belt.
7
20
Mixed.
2
3
Ounce cloth.
12
2
" Odd.
1
28
of ends in a warp . . . .
12
5
P
“ “ “ the warp . . .
13
22
Parallel lines.
6
1
“ hanks of warp yarn .
12
7
Parallelogram.
6
4
XV111
INDEX
Sec.
Page
Sec.
Page
Parallelopipedon .
6
10
Polygon, Area of a regular ....
6
7
Parenthesis.
19
Regular.
6
7
Part, Significant.
4
16
Polygons .
6
7
Partial products.
1
19
Positive-motion cam .
7
36
“ sections.
91
14
Post hangers .
7
7
Parts shown in two or more posi-
Pound, Yards of cotton cloth per .
12
21
tions, Moving.
91
33
Power of a number.
4
1
Pattern of the warp.
9
17
“ transmission.
7
9
it < < < ( 11
10
22
of.
8
1
“ “ “ “
11
35
Powers, Perfect.
4
8
Patterns, Fancy.
13
14
Prime factor.
1
27
Pentagon .
6
7
“ number.
1
27
Per cent., Definition of .
3
1
“ or first mover.
7
2
" “ Sign of.
3
2
Principal.
3
7
Percentage .
3
1
Principles of drawing, Funda-
base, and rate.
3
4
mental.
91
3
Rule to find.
3
4
“ percentages . . . .
3
1
Perfect powers .
4
8
Prism, The.
6
10
Permanent couplings.
7
2
Product .
1
16
Perpendicular line, A.
6
1
Production and draft, Calculating
15
22
Perspective views.
91
18
Products, Partial .
1
19
Pick.
13
2
Proof of addition .
1
11
“ cloth .
12
2
“ “ division.
1
25
Picks, Definition of.
12
2
“ “ multiplication.
1
20
Pillow-block.
7
9
“ “ square root .
4
9
Pinion, A.
7
34
“ “ subtraction.
1
14
Pitch circle of a gear.
7
28
" “ weight and counts . . . .
13
22
“ of a gear.
7
29
Proper fraction.
2
2
“ 44 44 screw.
7
35
Proportion.
3
12
Plan, Inverted.
91
7
Inverse.
3
16
“ Top.
91
7
Unit method of ....
3
18
Plane figure, A.
6
2
Pulley, Binder.
7
17
“ surface, A.
6
2
“ bushings.
7
15
Plans.
91
7
“ Crown-faced.
7
14
Plumb-line.
6
1
Flat.
7
14
Ply yarns .
9
7
“ Guide .
7
17
10
7
Idle .
7
17
10
17
“ Sheave .
7
16
11
20
“ stand, Mule.
7
24
" Calculation of cost of . .
11
25
“ Straight-faced.
7
14
“ Calculations of.
9
8
Pulleys.
7
13
44 II II II
10
8
“ Flange and groove ....
7
16
cotton
11
20
“ Ratios of speeds to diam-
“ composed of more than
eters of .
8
5
two threads.
9
9
“ Solid.
7
15
" composed of more than
“ Speed of.
7
22
two threads.
10
10
“ Speeds of.
8
2
" composed of more than
“ Split.
7
15
two threads.
11
23
“ Step.
7
18
Method of numbering . .
9
7
Surface velocity of rotating
8
5
II 14 It II II
10
7
“ Tight and loose.
7
17
14 II II II II
11
20
Pyramid and cone.
6
13
" of different materials . .
11
27
“ “ spun silk.
11
26
Q
“ “ Woolen and worsted . .
11
26
Quadrilaterals.
6
4
Points, Gauge.
10
6
Quantities, Addition of compound
5
12
INDEX
xix
Sec.
Page
Sec. Page
Quantities, Division of compound
5
16
Repeat of colors.
9
17
Multiplication of com-
Representation of objects.
91
1
pound .
5
14
Rhomboid.
6
4
Reduction of com-
Rhombus.
6
4
pound .
5
9
Ribs.
12
4
Subtraction of com-
“
13
5
pound .
5
13
Right angle.
6
2
Quantity, Measures of.
5
4
“ handed threads.
91
30
Quarter-turn belt
7
22
triangle.
6
3
Quotient.
1
22
Rolls and gears on draft, Effect of
15
14
Back.
15
6
R
“ Calculating draft for com-
Rack, A.
7
34
mon.
15
12
Radical sign.
4
3
“ Delivery.
15
6
Radius of a circle.
6
8
“ Drafting with common . . .
15
5
Ramie, jute, and linen.
11
16
“ Front.
15
6
Ratchet gear.
7
31
“ Gearing of.
15
9
Rate, base, and percentage ....
3
4
" Metallic.
15
20
“ of interest.
7
“ on draft, Effect of diameters
“ per cent.
3
3
of.
15
12
Ratio.
10
“ Top.
15
6
Rational factors.
4
8
Root, Cube.
4
3
Ratios of speeds to diameters of
“ “
4
10
pulleys.
8
5
“ Index of the .
4
3-
Raw silk.
11
14
“ of a number.
4
1
Reading textile drawings.
91
1
“ Square .
4
3
Reciprocal of a number.
3
11
Roots, Extracting square and cube
4
16
ratio.
8
11
“ of fractions.
4
15
Reciprocating screw.
7
35
Rope drives.
7
24
Rectangle.
6
4
“ transmission.
7
24
Reducing a decimal to a fraction,
4< <1
8
9-
Rule for.
2
30
Rotation, Direction of pulley . . .
7
20
fraction to a decimal
2
29
Roving and yarn, Sizing.
9
3
“ its lowest
cotton . .
11
4
terms .
2
5
“ woolen
10
15-
Reduction.
5
1
44 44 44 44 44
11
9-
of compound quantities
5
9
" Definition of.
9
3
“ United States money
5
11
“ Sizing worsted.
10
6
Reductions of fractions.
2
4
“ “ “
11
13
Reeds . .
12
4
Rubber belts.
7
10-
“ and ends per dent.
13
24
Rule for addition.
1
10
“ calculations.
13
5
“ for addition of compound
“ drafts, Irregular.
12
12
numbers.
5
13
“ Number of the.
13
6
“ for addition of decimals . . .
2
21
“ table.
12
17
" for addition of fractions . . .
2
10
“ Width at.
12
10
“ for allowing for shrinkage
41 in.
13
24
of cloth or yarn.
13
11
Reeds .
12
4
“ for cancelation.
1
35
Reel, Wrap.
9
5
“ for division.
1
25
<1 4 4 >
10
4
“ for division of compound
4 4 4 1
11
5
numbers.
5
16
Reference letters .
91
33
for division of decimals . . .
2
27
Regular polygon.
C
7
“ for division of fraction by
Relative value.
1
3
whole number.
2
15
Remainder.\
1
12
“ for division of whole number
“
1
22
or fraction by fraction . . .
2
15-
XX
INDEX
Sec. Page Sec. Page
for extracting other roots than
Rule to find base when percentage
square and cube.
4
19
and rate are known ....
3
5
for extracting roots of frac-
to findcircumferenceof a
tions.
4
16
circle .
6
8
for filling required to preserve
“ to find common divisor . . .
1
29
the weight of cloth.
12
25
to find cost of two-ply yarns
for finding cube root.
4
14
when threads are of differ-
for finding square root . . . .
4
10
ent values.
11
26
for length of belts.
8
6
to find count of one system
for multiplication.
1
19
equivalent to that of an-
for multiplication of com-
other.
11
18
pound quantities.
5
15
to find counts in ply yarns . .
9
10
for multiplication of decimals
2
24
to find counts of a roving . .
10
6
for multiplication of fractions
2
14
to find counts of a roving . .
11
13
for raising a number or deci-
to find counts of a yarn to
mal to any power.
4
3
be folded with another . . .
11
24
for reducing decimal to frac-
to find counts of a yarn when
tion.
. 2
30
length and weight are given
9
2
for reducing fraction to deci-
to find counts of a yarn when
mal.
2
29
length and weight are givbn
10
2
for reducing fraction to low-
to find counts of a yarn when
est terms.
2
5
length and weight are given
11
3
for reducing fractions to frac-
to find counts of filling to pre-
tions with common denom-
serve weight.
12
23
inator.
2
9
to find counts of yarn on a
for reducing inches to deci-
beam containing only one
mal part of foot.
2
30
size of yarn.
11
31
for reduction of compound
to find counts of yarn on a
quantities to lower denomi-
beam that contains one
nations.
5
10
size of yarn.
9
13
for reduction of a quantity to
“ to find dents per inch in a
higher denominations . . .
5
11
reed .
12
10
for rope transmission ....
8
9
“ to find diameter of circle with
for speed of worm-gears . . .
8
12
given circumference ....
6
9
for subtraction.
1
14
“ to find diameter of a driven
for subtraction of compound
pulley.
8
3
quantities.
5
14
“ to find diameter of a driving
for subtraction of fractions .
2
12
pulley.
8
4
for subtraction of decimals .
2
22
“ to find diameter of counter-
to find a constant.
8
15
shafts.
8
1
to find area of circle.
6
9
“ to find diameter of gear-blank
8
12
to find area of parallelogram
6
5
“ to find diameter of line shafts
8
1
to find area of regular polygon
6
7
to find greatest common di-
to find area of trapezium . .
6
6
visor.
1
30
to find area of trapezoid . . .
6
5
“ to find least common multiple
1
32
to find area of triangle ....
6
3
“ to find percentage.
3
4
to find average counts of ends
“ to find prime factors.
1
28
on a beam.
11
33
to find rate when percentage
to find average counts of yarn
and base are known ....
3
5
in a piece of cloth.
12
21
to find size of cold-rolled
to find average number of
shafting.
8
1
warp yarn.
9
15
to find the diameter of yarns .
11
30
to find average number of
“ to find the draft between two
yarn in a cloth.
12
24
pair of rolls.
15
14
to find average size of warp
to find the draft constant of a
yarn.
10
23
machine.
15
28
INDEX
xxi
Sec. Page
Sec.
Page
Rule to find the draft gear.
15
29
Rule to find the number of feet of
“ to find the draft of a machine
15
31
lumber in 1-inch boards . .
6
16
to find the hank of a roving
15
32
( «
to find the number of heddles
to find the hanks of filling
required on any harness . .
12
3
contained in a cloth ....
12
12
4 4
to find the number of heddles
“ to find the hanks of warp yarn
required on each harness .
13
4
contained in a cloth ....
12
9
( <
to find the number of revolu-
“ to find the horsepower trans-
tions per minute of a driven
mitted by a belt.
8
7
pulley.
8
3
“ to find the length of a square
6
10
il
to find the number of teeth re-
to find the length of an in-
quired in gears.
8
12
scribed square.
6
9
4 1
to find the number of yards
“ to find the length of a warp .
10
20
per pound in a sample of
to find the length of a warp .
11
32
cloth.
12
14
to find the length of warp on
44
to find the number of yarn in
a beam.
9
14
any system corresponding
“ to find the length of warp re-
to number in metric sys-
quired for a certain length
tern.
9
22
of cloth.
13
10
4 1
to find the number of yarn in
“ to find the length of warp that
any system corresponding
can be placed on a beam .
9
14
to number in metric sys-
“ to find the length of waip that
tern.
10
25
can be placed on a beam •
10
20
“
to find the number of yarn in
“ to find the length of warp that
any system corresponding
can be # placed on a beam
11
32
to number in metric sys-
“ to find the length of yarn . . .
9
2
tem.
11
38
“ to find the length of yarn . . .
10
3
4 4
to find the number of yarn in
“ to find the length of yarn . . .
11
3
metric system correspond-
“ to find the number of counts
ing to number of yarn in
'
in a ply yarn.
10
10
another system.
9
22
“ to find the number of ends of
4 4
to find the number of yarn in
each color of yarn on a
metric system correspond-
beam .
10
21
ing to number of yarn in
“ to find the number of ends of
another system.
10
26
each color of yarn on a
to find the number of yarn in
beam.
9
16
metric system correspond-
“ to find the number of ends of
ing to number of yarn in
each color of yarn on a
another system.
11
39
beam.
11
36
44
to find the original size of a
“ to find the number of ends of
yarn.
13
13
each count or color in a
44
to find the percentage of warp
cloth .
13
14
contraction during weaving
12
9
“ to find the number of ends of
“
to find the pressure exerted
each kind in the body of the
by a lever.
8
16
cloth.
12
7
“
to find the reed to use for an
“ to find the number of ends on
unevenly reeded cloth . . .
13
8
a beam.
10
19
11
to find the reed when the ends
“ to find the number of ends on
per inch in the loom are
a beam..
9
14
known.
13
15
“ to find the number of ends on
4 4
to find the reed when the width
a beam.
11
31
in the loom is known . . .
13
8
“ to find the number of ends per
“
to find the required weight of
inch in the reed.
13
15
each thread folded to pro-
“ to find the number of feet of
duce a ply yarn.
11
24
lumber in joists, beams.
4 4
to find the required width of
etc.
6
16
belt.
8
8
XXII
INDEX
Sec.
Page
Sec. Page
Rule to find the resultant count of
Rule
to find the volume of the
ply yarns.
ii
23
frustum of a pyramid or
4 4
to find the resultant counts
cone.
6
14
when different counts are
44
to find the weight of a sliver
folded.
9
10
or roving produced by a
4 4
to find the resultant counts
machine.
15
31
when different counts are
to find the weight of a warp
10
19
folded.
10
10
44
to find the weight of filling in
< 4
to find the resultant counts
a yard of cloth.
13
17
when different counts are
4 4
to find the weight of single
folded...
11
24
yarns required to produce a
4 4
to find the resultant counts
ply yarn.
9
11
when two threads are
4 4
to find the weight of the warp
folded.
9
9
in a yard of cloth.
13
17
(4
to find the resultant counts
4 4
to find the weight of threads
when two yarns are folded
10
9
required to produce a given
4 4
to find the revolutions of the
count.
10
11
driving: shaft.
8
3
4 4
to find the weight of yarn on
4 4
to find the size of a given yarn
12
22
a beam .
9
14
4 4
to find the size of a given
4 4
to find the weight of yarn on
yarn.
13
16
a beam .
11
32
4 4
to find the size of a woolen
41
to find the weight of yarn
roving: or yarn.
10
17
when length and counts are
4 4
to find the size of a woolen
given.
9
2
roving: or yarn.
11
10
4 4
to find the weight of yarn
4 4
to find the size of yarn on
when the length and counts
jack spool or beam con-
are known.
11
3
taining only one size of
4 4
to find the weight per yard of
yarn.
10
18
finished cloth.
13
17
4 4
to find the speed of a driven
44
to find the width in the loom
g:ear.
8
10
inside selvages.
13
15
4 4
to find the speed of a driven
4 4
to find the width occupied by
pulley.
8
4
the warp yarn in the reed
12
11
4 4
to find the standard breaking
4 4
to find the width of cloth in
weight of warp yarns . . .
9
19
the reed.
13
7
4 4
to find the surface area of a
4 4
to find the weight of a yarn .
10
2
cylinder.
6
12
4 4
to find what counts must be
4 4
to find the surface area of a
folded to produce a given
prism .
6
11
count.
10
11
4 4
to find the surface area of a
4 4
to find what counts must be
sphere .
6
15
folded to produce given
4 4
to find the surface velocity of
counts .
9
11
rotating pulleys .
8
5
4 4
to reduce a fraction to a
4 4
to find the total ends in a
given denominator ....
2
6
cloth.
12
6
4 4
to reduce a mixed number to
4 4
to find the total number of
an improper fraction . . .
2
6
ends in a warp.
13
8
“
to reduce an improper frac-
4 4
to find the twist to be inserted
tion to a whole or mixed
in any counts of yarn . . .
9
18
number .
2
7
4 4
to find the volume of a cone
4 4
to reduce a whole number to
or pyramid .
6
14
an improper fraction . . .
2
6
41
to find the volume of a cylin-
Rules applying to belts .
8
6
der .
6
13
44
“ gears .
8
10
to find the volume of a prism
6
11
4 4
“ levers .
8
16
to find the volume of a
44
shafting ....
8
1
sphere ..
6
15
44
“ speeds .
8
2
INDEX xxiii
Sec. Page
Sec. Page
Rules for draft gears.
15
26
Shrinkage in finishing.
13
11
“ “ proportion .
3
13
Side views.
91
6
“ “ the reduction of United
Significant figures.
4
16
States money.
5
11
part.
4
16
“ pertaining to the transmis-
Sign, Minus.
1
12
sion of power.
8
1
“ of addition.
1
6
“ to find size of turned shaft-
“ division.
1
22
ing for given horsepower .
8
2
equality.
1
6
used in obtaining particu-
multiplication ....
1
16
lars of cloth samples . .
12
24
“ “ per cent.
3
2
Run scale.
11
9
“ “ subtraction.
1
12
“ system of numbering woolen
Radical.
4
3
yarns .
10
13
Signs.
1
2
“ numbering woolen
“ of aggregation .
3
19
yarns .
11
7
Silk, Ply yarns of spun ....
11
26
Runs, Table of.
11
11
“ Varieties of .
11
14
“ yarns .
11
14
S
Simple fraction.
2
3
Samples, Figuring particulars
“ numbers.
5
1
from cloth.
12
16
“ value.
1
2
Figuring particulars
Single and double belts ....
7
11
from cloth.
13
21
“ cotton yarns.
n
3
Rules used in obtaining
“ spun silk.
ii
14
particulars of cloth . .
12
24
“ width cloth.
13
3
Scale, Cut.
11
9
“ worsted yarns ....
10
1
“ Run.
11
9
“ yarns .
9
1
Scalene triangle.
6
3
11 n
11
2
Scales for drawing.
91
26
“ Calculations of .
9
2
Screw, Differential.
7
35
“ Systems of number-
“ Multiple-threaded.
7
35
ing.
10
13
“ Pitch of a.
7
35
Size, Allowance for weight of .
12
19
“ Reciprocating.
7
35
“ of gears on draft. Effect
of
15
13
Screws.
7
35
Sizing cotton roving and yarn
11
4
Seats, Key.
7
2
“ gear-blanks.
8
12
Second movers.
7
2
“ roving and yarn ....
9
3
Section beams.
9
13
“ woolen roving and yarn
10
15
Sectional views.
91
11
<< it a a n
11
9
Sections, Longitudinal.
91
15
“ worsted roving.
10
6
Partial.
91
14
“ “ “
11
13
Selvage ends .
12
6
“ yarns .
10
3
Selvages.
13
7
(4 44 44
11
12
Semicircle.
6
8
Sley.
12
16
Semi-circumference.
6
8
“ Average.
12
23
Shade lines .
91
22
“ Definition of.
12
2
Shaft, A.
7
1
Solid, A.
6
10
“ couplings.
7
2
“ pulleys.
7
15
“ hangers .
7
5
Solidity, or volume. Measures of
5
6
“ Main or head.
7
2
Solids, Mensuration of ....
6
10
Shafting.
7
1
Speed cones.
7
18
Rules applying to ....
8
1
“ Effect of countershafts
on
8
4
Shafts, Line.
7
2
“ of pulleys ......
7
22
“ Non-parallel.
7
22
“ “ worm-gears.
8
12
Sheave pulley.
7
16
Speeds of pulleys.
8
2
Shed.
13
2
“ Ratios of, to diameters
of
Shipper, Belt.. . . . .
7
18
pulleys .
8
5
Short division.
1
23
“ Rules applying to . . .
8
2
XXIV
INDEX
Sec. Page
Sec.
Page
Sphere .
6
15
Table of liquid measure.
5
4
Spiral gears.. . .
J
32
“ “ mill weights.
5
3
Split.
13
6
runs.
11
11
“ pulleys.
7
15
“ “ sizing for cotton roving
Sprocket gears .
7
31
or yarn .
11
7
wheels.
7
31
“ “ sizing for worsted yarns .
11
12
Spun silk, Ply yarns of.
ii
26
“ “ square measure.
5
6
“ Single.
ii
14
“ squares and cubes . . . .
4
17
Spur gears.
7
30
surveyors’ measure . . .
5
5
Square .
6
4
“ time measure.
5
7
< <
12
2
“ “ Troy weight.
5
3
“ Definition of a.
5
6
“ “ United States money. . .
5
2
“ roots. Extracting.
4
16
“ Reed.
12
17
measure.
5
6
Teeth.
7
28
“ of a number.
4
1
“ required in gears, Rule to
“ root .
4
3
find.
8
12
“ Proof of.
4
9
Tens.
1
3
Squares and cubes.
4
17
Terms of a fraction.
2
2
Star gears.
7
32
“ “ “ ratio.
3
10
Step pulleys.
7
18
Textile drawings, Reading . . .
91
1
Straight-faced pulley.
7
14
Threads.
91
30
line, A.
6
1
Throw of a cam.
7
36
Subtraction.
1
12
Thrown silk.
11
15
of compound quanti-
Tight and loose pulleys.
7
17
ties.
5
13
Time, Measure of.
5
7
“ decimals.
2
22
Toe of a cam.
7
36
11 fractions.
2
11
Top plan.
91
7
Proof of .
1
14
“ roll.
15
6
Rule for.
1
14
Trains of gears.
7
29
Sign of.
1
12
Transmission of power.
8
1
Subtrahend.
1
12
Power.
7
9
Surds.
4
8
Rope.
7
24
Surface.
6
1
“ “
8
9
Dark.
91
23
Trapezium.
6
6
“ Light.
91
23
Trapezoid .
6
5
measure.
5
6
Triangle.
6
2
Surfaces, Mensuration of.
6
1
Troy weight.
5
3
Surveyors’ measure.
5
5
Twist in yarns.
9
18
System of numbering yarns, Metric
11
38
Twisted yarns.
9
7
Systems of numbering single yarns
10
13
“ “
10
7
H t<
11
20
T
Types of hangers.
7
‘ 7
Table method of extracting roots .
4
16
U
“ Multiplication.
1
17
of angular measure.
5
7
Unit, A.
1
1
apothecaries’ fluid meas-
“ method of proportion ....
3
18
ure. . . .
5
4
“ of a number.
1
1
weight. . .
5
4
United States money.
5
2
“ avoirdupois weight. . . .
5
3
“ Reduction of
5
11
breaking weight of warp
Unlike numbers.
1
1
yarns.
9
20
cubic measure.
5
7
V
“ distance between hangers
7
7
Value, Local.
1
3
“ “ drafts.
15
24
“ of a fraction.
2
2
“ dry measure.
5
5
“ “ “ ratio.
3
11
“ “ linear measure.
5
5
“ Simple.
1
2
INDEX
XXV
Sec.
Vegetable fibers.11
Velocity of rotating pulleys .... 8
Vertical line, A. 6
Views and their arrangement ... 91
Diagrammatic.91
“ Perspective. . . 91
“ Sectional.91
“ Side.91
Vinculum. 3
Volume of a cone or pyramid ... 6
“ cylinder. 6
“ prism. 6
“ “ sphere. 6
“ the frustum of a pyra¬
mid or cone. 6
“ or solidity. Measures of 5
W
Wall-box hanger. 7
“ brackets. 7
W arp .11
“ .11
.13
“ and filling, Counts of the . . 13
“ Average number of the ... 10
“ Finding the number of ends
in a.12
“ Number of ends in the ... 13
“ Pattern of the. 9
“ “ " “ .10
..11
“ Weight of the.13
“ “ “ “ 13
“ yarn.11
“ “ .11
“ Breaking weight of cot¬
ton . 9
“ Finding hanks of ... 12
“ Finding number of
hanks of.12
“ yarns, Breaking weight of
American. 9
“ Calculations for . . 12
Warps, Fancy. 9
“ “ .10
“ “ .11
Weaving, Contraction during ... 12
“ “ in.13
Weight and counts. Proof of ... 13
“ Apothecaries’. 5
“ Avoirdupois . . ,. 5
“ Measures of. 5
“ of cloth.13
“ cotton cloth.12
“ cloth from loom .... 13
“ “ filling.13
Sec. Page
Weight of the filling.
13
23
warp.
13
23
warp.
woolen and worsted
13
25
cloth .
13
3
“ Troy .
5
3
Wheels, Sprocket.
Woolen and worsted cloth calcula-
7
31
tions . . .
13
1
“ ply yarns . .
“ yarn calcula-
11
26
tions . . .
10
1
“ roving and yarn, Sizing . .
10
15
“ “ “ 44 4 4
11
9
yarns.
“ Cut system of num-
11
7
bering.
“ Cut system of num-
10
14
bering.
“ Run system of num-
11
9
bering.
“ Run system of num-
10
13
bering.
11
7
Woolly yarns.
10
13
Worm, A.
7
33
“ gear .
7
33
“ gears. Speed of.
Worsted and woolen cloth calcula-
8
12
tions . . .
13
1
“ ply yarns . .
“ yarn calcula-
11
26
tions . . .
10
1
roving, Sizing.
10
6
“ “ “
11
13
yarns .
10
1
II 14
11
11
“ Sizing.
10
3
44 44 44
11
12
Wrap reel.
9
5
4 4 4 4
10
4
Y
11
5
Yard cloth.
12
2
Yards per pound .
12
21
“ Finding.
12
13
Yarn and roving. Sizing.
9
3
.. “ cotton . .
11
4
“ woolen . .
10
15
44 44 44 44 44
11
9
“ Breaking weight of cotton .
9
19
“ calculations, Cotton.
9
1
for filling....
12
10
General ....
“ “ Woolen and
11
1
worsted . . .
10
1
Page
1
5
1
3
20
18
11
6
19
14
13
11
15
14
6
9
7
30
35
2
22
23
5
22
17
22
35
23
25
30
35
19
18
7
20
5
16
21
35
8
9
22
4
3
3
21
2
24
25
XXVI
INDEX
Sec. Page
Sec. Page
Yarn, Finding: hanks of warp . . .
12
18
Yarns,
Metric system of number-
“
number of hanks of
ing.
10
25
warp.
12
7
44
Metric system of number-
“
numbering: system.
11
2
ing.
11
38
“
Warp.
11
30
i 4
of different counts, Folded .
9
9
“
4 1
11
35
11
44 44 44 44
10
9
Yarns, Average number of . . . .
9
15
« (
44 44 44 44
11
22
4 t
44 44 “
11
33
« t
materials. Ply .
•11
27
“
Beamed.
11
30
< <
“ spun silk, Ply.
11
26
Breaking: weight of ... .
9
20
« (
“ the same counts, Folded
9
8
“
Calculation of cost of ply .
11
25
< (
4444 44 44 44
10
8
44
Calculations for warp .
12
5
t <
44 44 44 44 44
11
20
“
of cotton ply .
11
20
4 4
Ply.
9
7
“ply.
9
8
4 4
44
10
7
“
“ " “
10
8
4 4
‘ ‘
10
17
single . . .
9
2
4 4
44
11
20
44
“ single cot-
4 4
Run system of numbering
ton . . .
11
3
woolen.
10
13
4 l
“ single wors-
4 4
Run system of numbering
ted . . .
10
2
woolen.
11
7
composed of more than two
4 4
Silk.
11
14
threads, Ply. .
9
9
4 4
Single.
9
1
of more than two
4 4
44
11
2
threads, Ply. .
10
10
4 4
Sizing worsted .
10
3
41
of more than two
4 4
4 4 4 4
11
12
threads, Ply . .
11
23
4 4
System of numbering single
1 1
Cotton.
11
2
worsted .
10
1
4i
Cut system of numbering
4 4
Systems of numbering sin-
woolen .
10
14
gle .
10
13
41
“ numbering
4 4
Twisted .
9
7
woolen .
11
9
4 4
4 4
10
7
Diameter of .
11
29
4 4
“
11
20
Folded .
9
7
4 4
Twist in .
9
18
“
( i
10
7
4 4
Woolen .
11
7
1 ‘
11
20
4 4
“ and worsted ply . .
11
26
Method of numbering ply
9
7
4 4
Woolly .
10
13
44 “ 4 4 4 4
10
7
4 4
Worsted .
10
1
n il a il
11
20
4 4
4 4
11
11
Methods of determining the
diameter of.
11
29
z
4 ‘
Metric systemof numbering
9
21
Zero
1
2