40001820 textile mathematics part 1

8/3/2019 40001820 Textile Mathematics Part 1

1/63

TEXTILEMATHEMATICS

PART IBY

THOMAS WOODHOUSEHead of the Weaving and Designing Department,Dundee Technical College and School of Art

AND

ALEXANDER BRANDChief Draughtsman, Messrs. Douglas Fraser

& Sons, Ltd., Textile Engineers andIronfounders, Arbroath

BLACKIE AND SON LIMITED50 OLD BAILEY LONDONGLASGOW AND BOMBAY

19Z0 PRINTED A\l' 'ou:-m IN GREAT i a lT ! l! '.


2/63

PREFACEThe recent Education Act, in its relation to instruc-

tion in Continuation Schools and Classes, makes itdesirable, if not absolutely necessary, that textbooksof an elementary nature should appear, such booksdealing with the Science subjects most closely corre-lated to each large industry. Some knowledge ofElementary Science is essential, not only to the stu-dents of Continuation Schools and Classes, but topractically all who wish to prosecute intelligently thestudy of the technology of any particular branch ofindustry.. There are very few science textbooks which referparticularly to the textile industry, and it is hopedthat this book - the first of two parts on TextileMathematics-as well as other books which are tobe published shortly, will supply, in part at least,this want in textile literature.The general principles involved in all preliminary

books on Mathematics are identical, although theseprinciples may be explained in different ways. Thepresent book, while embodying these common prin-ciples, differs from the usual books on the subjectin that the principles are explained so as to appealto textile students, while practically all the examplesand exercises bear directly on the problems of textilework.

CONTENTSCHAP.

I. INTRODUCTORYII. SIGNS AND DEFINITIONSIII. REMOVAL OF BRACKETSIV. SYMBOLIC NOTATIONV. RECTANGLES AND SQUARESVI. TRIANGLES -VII. QU,,"DRILATERALS-VI 11 . POLYGONS

Page

2

7II16233037

IX. CIRCLE - 41X. ARCS, CHORDS, SECTORS, AND SEGMENTS 56XI. GEAR 'WHEELS 65XII. RECTANGULAR SOLIDS, RELATIVE DENSITY, AND

SPECJFIC GRAVITY - 69

XIII. PRISMS AND CYLINDERS 76XIV. PYR.~MID. CONE, AND SPHERE 84EXERCISES - 107T. WOODHOUSE.A. BRAND.

July, Ir)30.


3/63

2 TEXTILE MATHEMATICSwith little trouble, two or more yarns of differentnumbers. The number 8 actually indicates in the firsttwo cases the relation between the lengths of yarncontained in a fixed weight, while in the case ofjute,the same number shows the relation between a certainweight and a fixed length. All yarn tables arefounded on one or other of these two bases; if it werepracticable to number yarns according to their dia-meters, such a basis would possess obvious advantagesover the others when dealing with questions affectingcloth structure.Practically every person has at least an elementary

knowledge of mathematics, although the knowledgemay not be recognized by that name. In many in-stances the mathematical principle involved is appliedonly to particular cases, and no notion of its generalapplication is entertained. It is the purpose of thefollowing pages to show the mathematical principlesunderlying many kinds of textile calculations, and toshow the application of general mathematical ideas towhat are apparently intrinsically different problems.

TEXTILEMATHEMATICS-J

CHAPTER IINTRODUCTORY

Mathematics is the science of quantities, not merelyof quantities in themselves, but also of the exact anddefinite relation between simple quantities and sets ofquantities.In textile work of all kinds, the s~lution of practical

problems demands exact and definite relations betweencertain sets of quantities. To take a simple example,one speaks, say, of a No.8 yarn; the significance ofthis number varies greatly according to the fibre ofwhich the yarn is composed, and according to thedistrict in which it is spun and woven. If the yarn ismade of cotton fibres, No.8, or simply 88, will meanthat in 1 lb. of the yarn there should be 8 hanks,and each hank should measure 840 yd. in length. Ifthe yarn is linen or flax, 8 leas of 300 yd. each shouldweigh 1 lb., while if the yarn be jute, 14,400 yd.should weigh 8 lb.In each case the figure 8 expresses a definite re-

lation between the length arid the weight of a certainquantity ofyarn, and it is usually possible tocompare,

1

CHAPTER IISIGNS AND DEFINITIONS

In the manipulation of quantities there are, in realitvJ two elementary ideas involved-that of addition and

that of subtraction; for multiplication may be re-garded as a quick method of continued addition, whiledivision may be regarded as a process of continuedsubtraction. In every case all quantities may bemanipulated to obtain the desired results by one or


4/63

SIGNS AND DEFINITIONS 3 4 TEXTILE MATHEMATICSmore of these four operations, viz. addition, sub-traction, multiplication, and division.To save time and trouble these operations are de-noted by signs, as under:

The sign

means that 6 multiplied by 4 is equivalent to 24;while

24 -;-4 = 6

" minusmultiplied bydivided by

" subtraction.multiplication.division.

indicates that 24 divided by 4 is equivalent to 6.It is important to note that however many numbers

appear on each side of the equality sign (=), those onone side are exactly of the same value as the simplenumber or group of numbers on the other side; thisstatement is true for all the different kinds of signsattended by numbers or the like which are connectedby the sign (=) of equality.Many simple numerical equations need all the first

five signs which appear in the above four examples.Thus

+ known as plus stands for addition.x " "" "Again:

The sign =" ."" 00"

known as equal to stands for equality.is used for approximately equal to." to imply difference between.means infinity." is to. (6 + 4) - (3 x 2) + { (IS -;- 9) x 4} = 12

6 x 4 = 24

means that the so-called "expression" on the left ofthe mark = would, when simplified, have the valueI 2.On the left-hand side of the above equation are

three pairs of numbers, 6 and 4, 3and 2, and 18 and 9,enclosed in what are known as brackets ( ). Thelast pair, 18 and 9, and also their brackets and thenumber 4, are enclosed in a larger bracket { },whilea still larger group of bracketed sets may appearwithin brackets marked []. It will be seen that itis necessary to have differently-marked brackets inorder to avoid mistakes.In the above equation the brackets, and indeed all

the signs used, have exactly the same meaning as inordinary arithmetic; arithmetic is simply a branch ofthe parent subject of mathematics. The brackets, inparticular, are used to group together quantities whichhave to be treated as single simple quantities. Thus,

The undermentioned numbers and signs,6 + 4 = 10,

are read, 6 added to 4 equals 10,or, 6 plus 4 equals 10.

This isa" numerical equation", although a very simpleone, and implies that the terms on the left-hand sideof the sign = are of the same value as the single termon the right of the same sign. In a somewhat similarmanner we may introduce other simple numericalequations as under:

10 - 6 = 4,and read, 10minus 6 equals 4.

Again,


5/63

SIGNS AND DEFINITIONS 5 6 TEXTILE MATHEMATICSthe above equation means that 6 is to be added to 4,giving 10; that 3 is to be multiplied by 2, giving 6;that 18 is to be divided by 9, giving 2, and this 2multiplied by 4, giving 8 j and further, that the pro-duct of3 and 2, or 6, is to be subtracted from 10, giving4, and, finally, 8 is to be added to this 4, resulting in 12,the answer or single value on the right. All theseoperations may be set down briefly in the followingmanner:

(6 + 4) - (3 X 2) + {(18 -;- 9) X 4) 1210 6 + (2 X 4) 1210 6 + 8 12"-,.--or 4 + 8 12.

Then all the brackets may be removed, and the equa-tion stated as under: eba + b - cd + f = x.Note that cd means c X d, the product of c and d;

and that f means e -;.I, the quotient o'f e and f.The simplified form, eba + b - cd + f = x,

The foregoing example of a numerical equationdeals essentially with a particular case; in many in-stances general results are required, and these generalresults can only be obtained by using symbols insteadof numbers. The most convenient symbols are theletters of the alphabet, although any kind of markscould be made to serve the purpose. When a groupor groups of letters appear on the left side of theequality mark (=), and another set appears on theright side of the same mark, the arrangement is saidto be an "algebraical equation ".An algebraical equation of the same form as theabove numerical equation may appear as follows:

is at present unsolved, and cannot be solved unlessvalues are given to all the letters but one. Supposethat a = 6, b = 4, c = 3, d = 2, e = 18, and f = 9Then we can find the value of x by replacing thevarious letters by their numerical values. Thus, wechange eba + b - cd + f

18 X 4to 6 +4 - (3 X 2) +~9or 6 + 4 - 6 + 8 = x,whence 12 x

xx ,

or x 12.

(a + b) - (c x d) + {(e -;-I) X b} = x.The principal point to notice at this stage is the

fundamental difference between the two examples.The arithmetical equation is a particular case, and theanswer must always be 12, because the equation dealsonly with definite and concrete quantities. On theother hand, the algebraical equation is a general case,because x may be of different values, according as wechange the values of the remaining letters, and theequation is correct irrespective of whatever concretevalues may be assigned to the symbols. It is not

The bracket containing e and Imay be removed first,and the equation simplified:

(a + b) - (c x d) + { j x b} = x.


6/63

REMOVAL OF BRACKETS 7 8 TEXTILE MATHEMATICSdifficult to see that if, in the above example, a = 5instead of 6, the value of x must be II.Exercises, with answers, on p. 107.

latter the length CD. The length now remaining isclearly ABD. The result is evidently AD, and ADis therefore equal to x +y - e. Now AB =x, andBD = (y - z), so that x + (y - z) = x +y - z,2. When the sign immediately preceding the first

bracket of a pair is negative (-), the brackets mayalso be removed without altering the value of theexpression, provided that all the positive signs withinthe bracket are made negative, and all the negativesigns within the bracket made positive. That is to

CHAPTER IIIREMOVAL OF BRACKETS

The use of brackets is attended with one difficulty,that of their removal when an expression or an equa-tion requires to be simplified. The following instruc-tions should be carefully noted:-I. When the sign immediately preceding the firstbracket of a pair is positive (+). the brackets lTlaybe

removed and the signs unchanged without alteringthe value of the expression. That is to say,

x + (y - z) = x +y - z.This process may be proved as follows : -Let the

lengths of the straight lines AB, BC, and CD or DC

say, x - (y - z) =x - y + z.x - (y - z) really means x - (+y - z)= x -y + z.

This process may also be demonstrated in a similarmanner to the above. Again, let the line AB =x,Be = y, and CD = z.Draw AB or x (fig. 2) to the right, BC or y to the

%- ZA B B C 0 c

A I :.t:- 'j - i : I ex+(y-z) ZFig. I

x Y zA B C B C 0zX-I I 6C A 0

YFig...

(fig. I), represent to any scale the values of x, y, and zrespectively. Lines AB and BC from left to rightmay be considered positive, while CD, being fromright to left, may be considered negative. To thelength AB join the length BC, and cut off from the

left, and CD or z to the right; the result of thesethree is again AD.Now, x - (y - z) = AB - DB

,c-A._= AB - BC + CD=x -y + z.

Example I.-Prove that7 - (4 + 2) + (3 - 8) - (- 24 + 3) = 17


7/63

REMOVAL OF BRACKETS 9 10 TEXTILE MATHEMATICS

34 = 17

2. Now multiply each term within the brackets bythe numberoutside the brackets, and we obtain(27x + Io~t- (2~_-:-_4~):-(81-=_I_3_6~) = 139X+ 236 36 3. Remove all the brackets as explained, and then

obtain27x + 108- 24X+ 48 - 84 + I36x = !J9~ + 2.36 364. For convenience only, collect all terms which

contain x, and arrange as follows:_2 7x - 24 X + J3 6x t_!o8 + 48 - 84: = I39X + 2.

36 365. Find the value of the terms involving x, as wellas those which do not contain x.

J39~ + 72 = I39x + 2.36 366. Cancel out 36 and 72, and we have

!~~X + 2 = I~~X + 2.The left-hand side of the original equation has now

been reduced to its simplest form, and the resultproves that the two sides of the equation were equal,as indicated by the mark (=) of equality.Exercises, with answers, on p. 108.

This becomes 7 - 4 - 2 + 3 - 8 + 24 - 3 = 17after removing all the brackets as explained. Now,collect all the positive quantities and all the negativequantities from the lower line.

7 + 3 + 24 - 4 - 2 - 8 - 3 = 17'"--y--"Example 2.-Simplify the expression

3X - 9 + 4X - 12 _ 8x + 12.3 " 2 4

rst step: x - 3 + (2X - 6) - (2X + 3).znd step: x - 3 + 2X - 6 - 2X - 3.3rd step: x - 3 - 6 - 3.Final step: X- 12.It is worth noting that the division sign, i.e. the

line separating the numerator from the denominator,in J~ - 9 acts in the same way as a bracket, because3it compels one to work out each of the terms in thegroup as if each were enclosed in brackets.Example J.-Prove by removing the brackets that

3X + 12 2X - 4 2 I - 34X _ I39X'-4 -- 3- - ---9~ - 36 + 2.I. Find the least common multiple of 4, 3, and 9,

as in arithmetic; the L.C.M. is 36. Then say 4 into36 = 9 for the first group, and we have 9(3x + 12)for the numerator. Do this for each of the remainingtwo groups, and the result will be9(3X + 12) - I2(2X - 4) - 4(21 - 34x)36 ., I~~X + 2. (DtI4)


8/63

SYMBOLIC NOTATION II J2 TEXTILE MATHEMATICS

CHAPTER IV of a; t is the reciprocal of ~.or 1t; and:: and ~ arereciprocals, and so on. YIn Example 4, T\ is the reciprocal of 16; but T\ in.is the pitch of the screw, and 16 is the number ofthreads per inch. These particulars may be embodiedin a mathematical law:-The pitch of a screw ininches is the reciprocal of the number of threadsper inch. In symbols,

if p the pitch in inches,then ~ = the threads per inch.

Further, if t = the threads per inch, the pitch willbe 7 ' because p and } and t and 7 ' are two pairs ofreciprocals.Example 5.- The fly-wheel of a mill engine makes

240 revolutions per minute (often abbreviated to240 r.p.m.). This is equivalent to 2640 revs. = 4 revs.o sees.per second. How long does it take to make onerevolution?

SYMBOLIC NOTATIONIt is most useful, and sometimes absolutely essen-

tial, to be able to convert a mathematical expressionor law or rule into words, and also to be able torepresent any law or rule by means of symbols. Thefollowing examples illustrate various forms, and indi-cate how a rule may be expressed by means ofsymbols ; they also show how mathematical expres-sions may be put into words.Example 4.- The pitch of a screw is lif in. How

many threads are there per inch?The pitch of a screw is the distance, centre tocentre, of two neighbouring threads, as indicated Intwo places in fig. 3.

Fig . 3 Time taken.evolutions made.

In the above example there is I thread in Y if in.,1 In -;-TI6 in. or 1 X 16 = 16 threads in I in.I

Imin. or 60 sec.I sec.~ sec.i sec.hence

The pitch is therefore the reciprocal of the threadsper inch. Any two numbers of which the product isunity (I) are called reciprocal numbers. Thus 5 and tare reciprocals; so are l2 and 12; ~ is the reciprocal

(D64) a 2

Note that i is the reciprocal of 4, and I sec. is aninterval of time.~_- is the reciprocal of a time interval, briefly4 sec.termed a frequency. If the fly-wheel revolves once


9/63

SYMBOLIC NOTATION 13 TEXTILE MATHEMATICSin t sec., it will revolve 4 times in I sec.: In otherwords, its frequency is 4-Example 6.-Express V ft. per minute, in feet persecond.Since I sec. is { f 1 r part of Imin., it follows that

V ft. per minute = ~ ft. per second.Example 7.-A mill-engine uses S lb. of steam

per hour when its load is Hhorse-power. How manypounds of steam will it use in D days if it runs t hr.per day? What are the meanings of the following

S Hterms: D x t; HiS?The engine uses S lb. of steam in r hr.It will use S x t St lb. of steam in t hr. per day.

" St x D = StD lb. of steam in D days oft hr. each.D days of t hr. eachthe total time the engine is running.

~ is the reciprocal of ~, so that the lb. of steam perhorse-power hour is the reciprocal of the horse-power(h.p.) developed per pound of steam per hour.Example 8.-A 100m runs at 200 picks per minute

on cloth requiring 40 shots per inch. If the loomloses 15 per cent of its possible picks through un-avoidable stoppages, find its production in yards perday of 9 hr.

200 picks per minute x 60 min. per hour= (200 X 60) picks per hour.(200 X 60) picks per hour X 9 hr. per day

= (200 X 60 X 9) picks per day.If 15 per cent picks are lost, then 85 per cent ofpicks are effective, and we have

8~%200 X60 x 9 X ~ % 0 = total effective picks per day.roo 0D X t =DtSH

pounds of steam per hourhorse-power developed

There are 40 shots per inch, so that200 X 60 X 9 X 85 .~-- IOO---~ -;-40

= 2~_~~ X 85 = inches of doth per day.100 X 40There are 36 in. per yard, therefore200 X 60 X 9 X 85..!.. 6

100 X 40 3200 X 60 )( 9 X685 =63.75 yd. per day.

100 X 40 X 3In the above arithmetical example there are alto-

gether 7 terms, 3 of which are variable for any par-ticular cloth, the remaining 4 being constant for anyparticular weaving factory. The constant terms are

= the pounds of steam used per hour for each horse-power developed

= the pounds of steam per horse-power hour.This ratio ~ is commonly used to compare the

relative efficiencies of different classes and types ofsteam-engines.

H horse-power developedS - pounds of steam per hour= the horse-power developed per pound of steam

per hour.


10/63

SYMBOLIC NOTATION IS 16 TEXTILE MATHEMATICS60, g, and 36, representing, respectively, minutesper hour, hours per day, and inches per yard, and 100the percentage basis.Lety = yards of cloth per loom per day;e percentage of effective working time;p = picks per minute, the 100mspeed;and s shots of weft per inch.Then, expx60xgy= . IOOxsx- j6-

60xgxexp100 x 36 x s60xg x ep

100 x 36 sl_ xep20 s3 ep . 15 ep-- or ---..----.20S S

the second group of symbols is to be subtracted fromthe value of the first group contained in the brackets.Example IO. - The picks or shots in any cloth,multiplied by the reed width in inches, multiplied by

the cloth length in yards, and the product of thesethree terms divided by 840, is equal to the hanks ofweft in a piece of cotton cloth. Express this value insymbols.

Let SRCH

shots (or picks) per inch;reed width in inches;cloth length in yards;hanks of weft in the cloth.

Then,H = S x _ ~ ~ _ C : ; = ?_R C hanks.840 840

Exercises, with answers, on p. IOg.

The above numerical example can be checked bythe latter expression, thus,

Y = "1 5 eps. [5 x 85 x 100

4063'75 yd. per day.

CHAPTER VRECTANGLES AND SQUARES

Example 9.-Express in words,a(b x a c) - b c'

The two lines in a compass which cross each otherand which point respectively to North and South,and to East and West, are perpendicular to eachother, or at right angles to each other, because eachpair of adjacent lines from the crossing-point C, infig. 4, embraces one-quarter of the 360 degrees,or = go degrees. It will be evident that godegrees (one right angle) will be enclosed betweentwo perpendicular lines whether such lines be ofequal length or of different lengths. Thus the lineEB, an extension of the line WE, is longer than

The above signifies that b is to be multiplied bythe product of a and c; that a is to be divided by theproduct of band c; and that the value or quotient of


11/63

RECTANGLES AND SQUARES 18 TEXTILE MATHEMATICSthe line NS, and the line WB is considerably longerthan the line NS, but all the angles in the figureare right angles. The four heavy lines which cir-cumscribe the ring or circle are all the same length;these four heavy lines constitute a rectangle as wellas a square. The dotted extension in conjunctionwith the east side, E, is also a rectangle, but not asquare, and the complete outline of the figure in theillustration is stil l another but larger rectangle.

equal to 90. If AB is divided into 6 equal partsby the points I, 2, 3, &c., and BC into 4 similarequal parts by the points 6, 7, 8, &c., and linesdrawn from these points parallel to two adjacent

o 13 12 II 10 9 c14

4-32I 2 :; " ' 1 - 5 6

815 7

r---~;';'_:::::::----'--------- --------------------~IIIIIII~ ~ ~ ~ ~ : B16 6

A I 3 5 BFig. 5

Fig. 4

sides, the rectangle is divided into 4 rows of 6 littlesquares each, or altogether 24 little .squares. Eachsquare is Iunit long and Iunit broad, and is called asquare unit. The whole area of the rectangle thuscontains 24 sq. units.But 6 units X 4 units

length X breadth 24 sq. unitsj hence

area of rectangle.

,,_ _.;;::~-::::,.__~ --J

THE RECTANGLE.-A rectangle is thus a 4-sidedfigure in which all the angles are right angles, andin which two parallel sides are perpendicular to theother two parallel sides.THE SQUARE.-A square is a 4-sided figure in

which all the sides are of the same length, and all theangles are right angles.Fig. 5 is a rectangle 6 units long and 4 unitsbroad. It has four sides-AB, BC, CD, and DA.The side AB = the side CD, and the side BC = theside AD. The words "the angle enclosed betweentwo adjacent sides AB and BC" may be expressed

A A A Aby ABC. The four angles ABC, BCD, CDA, andADAB are all right angles, that is, each angle is Fig . 6

If the linear unit is I in.,o the unit of area will be "r sq.in.; similarly, if the linearunit be I yd., the unit ofarea will be I sq. yd., andso on.Again, with a square, such

as that in fig. 6, the sidesEF, FG, GH, and HE are

F all equal, and the 4 anglesare right angles. It is clear

H-t32.I 2 . 3 -4

E


12/63

RECTANGLES AND SQUARES 19 20 TEXTILE MATHEMATICSthat each side is 4 units long, and that there are16 sq. units in the area.Area of square length of side X length of side= side squared, or side",

By a similar method it may be shown thatA

b = TAgain, let A the area of a square,

and s = the side of same square.ide" means side multiplied by itself. In the sameway we have 42 = 16; 82 = 64; a2 = a x ajx2 =x x x, &c.Symbolically, let

A area of rectangle jI -- length of long side;b breadth of short side.Then, A I x b,

or A = lb.

Then, as indicated above,A S2,

or . 1 ' 2 = A.If the square root of each equal quantity is taken,

we have "ji2 "jA,or s "jA.

In such an equation, A is said to be given explicitlyin terms of I and b, and that I and b are only' givenimplicitly. It is most important to note, however,that if any equation is stated in the form A = Ib,that one can also find the values of I and b. Thus,if we divide each side of the equation A = Ib by b,we obtain A lbb -- Ii'

In words, the side of a square is equal to the squareroot of its area.In making use of these rules, it is essential that the

units of area and length should correspond. Thus,if the square or rectangle be measured in inches, thearea will be in square inches; if the area be in squareyards, the length will be in yards.A knowledge of the following tables must be

acquired:--Then, by cancelling the two b's on the right handside of the equality mark (=), we get

I. LINEAR MEASURE

AIi I,12 inches3 feetst yards40 poles8 furlongs

I foot.Iyard.Ipole or perch.I furlong, or 220 yards.

= Imile, or 1760yards, or5280 feet.

so that I A=7iIn words, the length of a rectangle is equal to its areadivided by its breadth.


13/63

RECTANGLES AND SQUARES 21 22 TEXTILE MATHEMATICS

144 square inches9 square feet30t square yards40 square poles4 roods, or 4840 sq. yd.

= Isquare foot.Isquare yard.1 square pole.

= I rood.Iacre.

Example IJ. -A cotton carding - room measures108 ft. long by 48 ft. wide. It contains 36 carding-engines each occupying a floor space of 10 ft. 4 in.by 6 ft. Find the area of the floor, the area actuallyoccupied by the machinery, and the area taken up forpasses and the like.

Area of room = length x breadth= 108 ft. x 48 ft.= 5184 sq. ft.Floor space for I engine IO! ft. x 6 ft.62 sq. ft.

Area occupied by 36 engines (62 x 36) sq. ft.= 2232 sq. ft.

Area left for passes (5184 - 2232) sq. ft.= 2952 sq. ft.

II. SQUARE MEASURE

It is also useful to remember the length known as achain (Gunter's chain). This is extensively used inland surveying, and is equal to 22 yards, or -.lu-thof amile.Example II.-Apiece of canvas is 24 in. wide and50 yd. long. What is its area in square yards?Length 50 yd.Breadth 24 in. -;-36 in. per yard 2436 yd. Example I4.-Find the cost of paving a passage

in a factory, 88 ft. long by 9 ft. wide, at the rate of47S. per square yard.Area of passage length x breadth

88 ft. x 9 ft.(88 x 9) sq. ft.88 x 9f d = 88 sq. yd.9 sq. t. per sq. y .Cost of paving = 88 sq. yd. x 47s. per sq. yd.

4136s., or 206, I6s.

Area I x b24 100 1 d50 x -=- =333 sq. Y .36 3

Example I2.-A small sample of cloth measures2 in. square and weighs 28 gr. Find the weight ofthe cloth in ounces per square yard. (7000 gr. inI lb., 4371 gr. in Ioz.)

Area of sampleArea of Iyd.

28gr,_4 sq. in,:. Weight of I sq. yd.

or weight of cloth

== side"= 2 X 2 4 sq. in.36 X 36 = 1296 sq. in.weight of Isq. in.28 X 1296~---- gr.428 X 1296

= 4 x 437~ oz.20'736 oz.,20'736 oz. per square yard.

Example Is.-Find the cost of a carpet, 27 in. wide,at lOS. 6d. per yard, for the complete covering of aroom 22 ft. long by 18 ft. wide.

Area of carpet = area of floor= 22 ft. X 18 ft.= (22 x 18) sq. ft.Width of carpet 27 in.= 21 ft.


14/63

TRIANGLESLength of carpet (22 X 18) sq. ft. -;- 2t ft.

17 6 ft.58* yd.Cost of carpet = S8t yd., at lOS. 6d. per yard,176 21 hilli-- X - SIlO (YS3 2 to.616 shillings, or 30, 16.1'.

23 TEXTILE MATHEMATICS. fig. 7. It may be proved (Euc. I, 5) that the anglesopposite the equal sides are equal; hence, if one angleis known, the others may easily be found.SCALENETRIANGLE.-A scalene triangle has threeunequal sides, as GHK in fig 7.All the above types are named with reference to the

Area of square =and side

sides,,.jarea

= ,.j336458 ft.

c FExample I6.-A wool-store is to be erected on asquare plot of ground known to contain 3364 sq. ft.Find the length of its sides.

Exercises. with answers, on p. 110.

CHAPTER VITRIANGLES

As the name implies, a triangle is a plane figurebounded by three straight lines and including threeangles. The definitions regarding the various typesof triangles appear below.EQUILATERALTRIANGLE.-An equilateral triangle

has its three sides equal, as ABC in fig. 7. It maybe proved (Euc. I, 5) that the three angles of thistriangle are equal, and since the three angles of anytriangle are equal to 180, each angle of an equilateraltriangle is 60.ISOSCELESTRIANGLE.-An isosceles triangle has

two equal sides, as DF and FE in triangle DEF in

Fig. 7

lengths of the sides. Triangles may also be classifiedaccording to the size of the angles they contain, andas follows:-RIGHT-ANGLEDTRIANGLE.-A right-angled triangle

has one of its angles a right angle (oo"), as in thetriangle LMN in fig 7. The angle M is a right angle,and the side LN opposite the right angle is called thehypotenuse. It is proved in Euc. I, 47, that the


15/63

TRIANGLES TEXTILE MATHEMATICSsquare on the hypotenuse is equal to the sum of thesquares on the other two sides, i.e, m2 = /2 + n2OBTUSE ANGLEDTRIANGLE.-An obtuse - angledtriangle has one of its angles obtuse, i.e, greater than

goo, as angle K in triangle GKH, fig. 7.ACUTE-ANGLEDTRIANGLE. An acute-angled tri-

angle has all its angles acute, i.e, each is less than900, as ABC or DEF, fig. 7.The right-angled triangle is a very important one

in all branches of work; its properties enable a certainclass of problem to be easily solved. The word tri-angle may be conveniently represented bythe sign L3..Example I7.-Find the length of the hypotenuse

of a right-angled triangle when the other two sidesare respectively 9 ft. and 12 ft.Referring to fig. 7, in the dLMN,

Consequently, it follows that a right-angled trianglemay always be drawn if the three sides are kept in theproportion of 3, 4, and 5. Many textile machines arein two or more separate parts; to take a simpleexample, a beaming- machine consists, say, of abobbin creel or bank and the winding-on mechanism.

~III

1 2 + n2,92 + 12281 + 144225,./225r y ft., the length of hypotenuse.

It should also be noted that if

A Bandm

RFig'. 8nz 2 = /2 + n2 ,

and if /2 be subtracted from each side of the equation,then m2 - /2 =n2,

whence n = .Jm2 - 1 2 .In the same way it may be shown that

/ = .../m 2 _ n2Notice that 9 + 16 25,

or 32 + 42 =5 2 .

In order that each part may work properly with theother, it is essential that they should have a commoncentre-line, i.e. the centre-line of each part should liein one straight line. Thus, in fig. 8, AB representsthe back line of the beaming-machine proper trans-ferred to the floor, and C is the middle point. FromC mark off to left and to right the points D and E,each 3 units from C. The units would probably beI ft., but any other suitable unit may be utilized.


16/63

TRIANGLES TEXTILE MATHEMATICSWith C as centre, and a radius of 4 units, draw thearc FG. With D as centre, and a radius of 5 units,draw the arc HK, and with the same radius and Eas centre, draw the arc LM. Ifcorrectly drawn, thethree arcs will meet at a common point P. Join Cto P, and extend the line in the required direction toSand R.In the triangles PCD and PCE, CD and CE are

each 3 units, PD and PE are each 5 units, and thecommon side CP is 4 units. The triangles are there-

/\fore right-angled, the right angles being PCD and/\PCE. The line RS will thus coincide with the centre-

line of all the parts of the machine.The following three statements concerning tri-

angles are proved by Euclid, and are worth keepingin mind:-I. Any two sides of a triangle are together greater

than the third side. (Euc. I, 20.)2. The three angles of a triangle are together equal

to two right angles. 2 x 90 = 180. (Euc. I, 32.)3. The area of a triangle is half that of the rectangle

on the same base and having the same altitude.(Euc. I, 41.)Any of the three points or apices of a triangle may

be called the vertex; the side opposite the vertex thenbecomes the base. In fig. 9, the point A is the vertexof the 4ABC, and BC is the base; similarly, G isthe vertex of the 4 GBC. The perpendicular AF isdrawn from the vertex A of the first triangle to the

/\ /\base, so that BFA = AFC =90, and the line AFmeasures the height or altitude of the triangle. Theline ED is parallel to the line BC, and hence therectangle EBCD is on the same base, BC, as the

(D64) 3

aABC, and the altitude AF = the side BE. Itis not difficult to see, and it may easily be proved,that the two shaded triangles are equal in area to

Fig. 9

the two unshaded triangles, and therefore 4ABC= t reet. EBCD. Consequently,

Area of 4ABC Q area of rect. EBCDt (BC x EB)BC x AF

2ab"2'

where a the altitude, and b = the base.Let A = the area of the triangle,abthen A. = - .2

MUltiplyboth sides of the equation by 2, then 2A =ab oDivide both sides of the equation by b, then 2~ = a,

2Aor a = T;


17/63

TRIANGLES 30 TEXTILE MATHEMATICSthat is to say, the altitude of a triangle is equal totwice the area divided by the base.In a similar way, it may be shown that

2Ab =-,a

Then, 2.1'and .I'

a + b + c,a+b+c------.2

= II in.

It can be proved that the area A of a triangle isA =../s(s-a) (s-b) (s-c)~

Example I9.-Find the area of a triangular plot ofground intended for drying bleached hanks of yarn,the sides of the plot measuring 40 yd., 46 yd. , and50 yd. respectively.

Area = ../s(s-a) (s-b) (s-c),and .I' = k (a + b + c)

(40 + 46 + 50) = 68r. A . ./ 68 (6 8 - 4of( 68-- 46)(68 - -50)sq~yd.../68x28x--2Z-XI8Sq. yd."/753984, or 868'3 sq. yd.

i.e. the base of a triangle is equal to twice the areadivided by the altitude.Example I8.-Find the weight of a triangular sheet

of metal intended for a sliver conductor if the base6 in"easures 6 ft. II in. and the altitude is 3 ft.the sheet weighing 8 oz. per square foot.

abArea

2=42 i n~_~ . ._ .~~1_~~~q. ft .

2 x 14412' 104 sq. ft .

Weight 12' 104 sq. ft. x 8 oz.= 12'104 x 816 oz. per pound6.052 lb.

In practice, it is not always possible to obtain thevertical height i the following method illustrates acase where it is more convenient to measure the threesides only. It is a common and excellent method toindicate the lengths of the sides of a triangle by smallletters which correspond to the similar larger lettersat the opposite vertices. This is done in two of thetriangles in fig. 7.

Exercises, with answers, on p. 1 I r.

CHAPTER VIIQU ADRI LATERALS

Let s the semi-perimeter of a triangle,i.e. half the sum of its 3 sides.the 3 sides,

QUADRILATERAL.-Theterm quadrilateral may beapplied to any plane figure bounded by four straightlines. The square and rectangle are thus special kindsof quadrilaterals. Other special forms of quadrilateralare the parallelogram, the rhombus, the trapezium,and the trapezoid.PARALLELOGRAM.-Aparallelogram is a quadri-

lateral figure in which both pairs of opposite sidesare para1tel. The square and rectangle are thereforequadrilateral figures, or rather parallelograms; in, b, and c


18/63

QUADRILATERALS TEXTILE MATHEMATICStion of the two upper figures, and is proved inEuc. I, 34, that:-I. The opposite sides of a parallelogram or rhombusare equal.2. The opposite angles of a parallelogram or

rhombus are equal.3. The diagonals of a parallelogram or rhombus

bisect one another, i.e, they divide each other into twoequal parts, and also divide the figure into two tri-angles which are equal in every respect.

general, a parallelogram is considered to have angleswhich are not right angles. Examples appear atABCD and EFGH in fig. ro.RHoMBus.-A rhombus is a quadrilateral figure

all the four sides of which are equal, and all theangles differing from right angles. E FGH, fig. IOjis therefore a rhombus.The straight lines AC, BD, KM, LN, &c., drawn

Figv r r

Fig. 10It may also be shown that the diagonals of a

rhombus bisect each other at right angles.All the rules which are applicable to the solution of

triangles may also be applied to the solution of quadri-laterals, since the latter can be divided into two tri-angles. Special rules are, however, often much moreconvenient than the above method.Fig. I I illustrates a parallelogram ABCD on

a base AB j this line AB is also the base of therectangle ABEF.It may be proved (Euc. I, 35) that the area of a

parallelogram is equal to that of a rectangle on thesame base and having the same height. If ~ BCEis cut off 'and placed in a position so as to form

from corner to corner in fig. 10 are called diagonals.Note that each diagonal divides the quadrilateral intotwo equal or unequal triangles.TRAPEZIUM.-A trapezium is a quadrilateral which

has one pair of opposite sides parallel, as KL and NMin the diagram KLMN, fig. 10.TRAPEZOID. A trapezoid is a quadrilateral in

which no two sides are parallel, as PQRS, fig. 10.N.B.-These two definitions are often reversed.Referring to fig. 10, it is evident from the construe-


19/63

QUADRILATERALS 33 34 TEXTILE MATHEMATICS..::\ADF (shown shaded in fig. 11), a rectangle ABEFis formed. Then,Area of parallelogram ABeD = area of rectangleABEF

=AB x AF= base x perpendicular height.

TRAPEzIUM.-In fig. 12, KLMN represents atrapezium with the base KL parallel to the opposite

M

If Abhthen A

area of parallelogram,= base,perpendicular height,

bit,K

Fig. Y2

bAh'

and It A7 } '

side MN. Join LN, and draw NT perpendicular to/\ /\KL; then NTK and NTL are right angles. Now,

Area of trapezium KLMN = area of aLN +area of L3.LMN= KL NT + J Y I N x_~T.

2 2

from which it may be deduced that,

Example 2o.-Find the area of a parallelogram thebase of which is 4 ft. 3 in. and the perpendicularheight (the height at right angles to the base) is6 ft. 2 in.Taking the common factor NT outside, we have2

Area A biz.51 in. X 74 in.3774 sq. in.

.:._;-26 2 sq. ft.

A = NT (KL + MN)z(KL + MN)= NT ------------------------.

2

Example 2I.-Find the height of a rhombusarea of which is 18 sq. ft., and the side 4 ft. 6 in.Area A = b x h,

AIi

theBut KL and MN are the two parallel sides of thetrapezium, so thatKL + MN = half the sum of the parallel sides,

2

= 18 sq. ft.4ft.6fil---.IS- = 4ft.4!

and NT is the perpendicular distance between them,so thatArea of trapezium = !um of parallel sides X theperpendicular distance between them.

whence h


20/63

QUADRILATERALS 35 TEXTILE MATHEMATICSIfx and yare the parallel sides, and d is the per-

pendicular distance between x andy, then, sinceA = the area of the trapezium,A =d ( X ; Y ) ,

may be termed the off-set method; it is explainedbelow.Example 23.-Find the rent, at 5 per acre, of a

quadrilateral plot of ground of which one diagonalmeasures 15 chains, and the off-sets from it to theother two vertices are 8 chains and 12 chains respec-tively.Referring to fig 13, ABCD is a quadrilateral repre-

senting the plot; the diagonal AC is IS chains; FDo

from which it may be deduced thatd = _ ~ _ r ~ _ .x+y

Example 22.-A factory or mill site is in the formof a trapezium; the two parallel sides measure 6 chainsand 8 chains respectively, and the perpendicular dis-tance between them is 5 chains. Find the cost of thesite at the rate of 750 per acre.Note: 22 yd. = I chain.

222 or 484 sq. yd. = T \r ac.Area of site = area of trapezium

= d (x +y)2

(6 + 8)= 5 2A

BFig. '3= 5 X 735 sq. chains

3'5 ac.Price of site 750 X 3'5 ac.2625.

IRREGULARQUADRILATERALS.-The area of anirregular quadrilateral or trapezoid, such as PQRSin fig. 10, is found by drawing a diagonal, and cal-culating the areas of the two triangles thus formed;any measurements which enable the areas of the twotriangles to be found will also be necessary to calcu-late the area of the quadrilateral. The most con-venient way involves only three measurements, and

and EB are the two off-sets of 8 and 12 shains, bothbeing perpendicular to the diagonal AC.

Let AC =d;FD = aEB =h.Area of quadrilateral ABCD =area of ..dACD+ area of ..dABC.

A = (!AC x FD) + (tAC x EB)= (~ X a ) + (~X h )= ~(a + b)= ~-diagonal X the sum of the off-sets.


21/63

POLYGONS 37 TEXTILE MATHEMATICSWith the values given in Example 23, we have:

A ~ (8 + 12)2

= 15 x 202ISO sq. .chains

= IS ac.Rent 15 x S =7S

Exercises, with answers, on p. 113.

ABCDEFGH; this outer circle is called the circum-scribed circle.The dotted lines which bisect the angles of the 8

isosceles triangles also bisect the 8 sides of the octagon.Ifa circle is drawn with a radius equal to the distancefrom the common central to point the middle of anyof the sides of the octagon, this circle will pass through,

F

CHAPTER VIIIPOLYGONS

POLYGON.-A polygon is a plane figure bounded bymore than 4 straight lines. A regular polygon is onein which all the sides are equal and all the anglesequal.A polygon with S sides is called a Pentagon.

" " 6 " " Hexagon." " 7 " " Heptagon." " 8 " " an Octagon." " 9 " "a Nonagon." " 10 " " Decagon.ABCDEFGH in fig. 14 is a regular octagon, since

all its 8 sides are equal and all its angles are equal.Ifthe angles are bisected, the bisectors (shown in thinlines from the centre of the circle to the outer circle)will all meet at one point, the centre, as stated.With a radius equal to half the diameter, i.e. half ofAE, and rotating about the common central point, acircle may be drawn which will touch all the points

Fig. '4

or touch, all the points K, L, M, &c. Such an innercircle is called the inscribed circle.Since the octagon is divided up into 8 equal isosceles

triangles, the area of the octagon may be found bycalculating one of the triangles and multiplying theresult by 8. The general case may be stated asfollows. Let the polygon have n sides, then: Areaof polygon = the area of n equal triangles.A abn. X - (where a = altitude, and b = base)2G x b ) It.

But since (n x b) = the number of sides multiplied


22/63

POLYGONS 39 TEXTILE MATHEMATICSby the length of each, the result is the perimeter ofthe polygon. Hence

(i X b ) = ~the perimeter,like, or into a number of each kind. Such a case isdemonstrated in the next example.Example 25.-A plan of a dining-room is repro-

o Fso that the rule deduced may be stated as follows:-To find the area A of a regular polygon of n sides,multiply half the perimeter by the perpendicular drawnfrom the centre of a side to the centre of the circum-scribed and inscribed circles.Example 24.-Find the cost of a hexagonal sheet of

metal, 16 in. side, at the rate of 4S. per square foot;the perpendicular which bisects the side and passes tothe centre is 13~ in.

nabA

--,26 X 13k X 16 sq. m, Iio'

Area2

6 X 13.875 X 16 sq. ft.

It is not always convenient in practice to measurethe perpendicular; difficulty may be experienced infinding the exact centre of the circumscribed and in-scribed circles. Rules can be deduced, however,which are independent of the perpendicular, but theseinvolve trigonometrical ratios of the angles of thetriangle; the discussion of these ratios is outside thescope of the present chapter.When the areas of irregular polygons are required,

it is always possible to divide such figures into anumber of triangles, rectangles, trapeziums, or the

IL ._R ~o

.~

III!I--------_js

Cost2 X 1444.625 sq. ft.

= 4.625 sq. ft. X 4s. per square footISS. 6d.

A BFig . 15

duced in fig. 15. Part of the floor is covered with acarpet RSTV, 4 yd. by 4t yd., the remainder beingcovered with linoleum. Find the whole area of theroom, that of the carpet, and that covered bylinoleum.


23/63

CIRCLE TEXTILE MATHEMATICSArea of floor =2 unequal rectangles ABCL and

NPFG,+ 2 equal triangles KMH and DQE,+ 2 equal trapeziums MNGH andPQEF.

AI (r6ft. X IS ft. 6 in.) + (4ft. X 3ft. 6 in.)( 1 2 in. .) ( 2 ft.6in. + 3ft. 6in. .)+ 2 -2-x2ft.6ll1. + 2 -------2---X 21111.(16 X r5~)+(4 X 3k)+2 (~X 2k)+2 (3 X IV(248 + 14 + 2~ + 10k) sq. ft.275 sq. ft.

shows three points on the circumference of a circle,and C indicates the central point. It will thus beseen that the lines CA, CB, and CD are three radiiof this circle. Moreover, two of these radii, CA andCD, are in one straight line; the two radii, so dis-posed, form a diameter of the circle; hence, AD isa diameter, and all other straight lines which touchtwo points on the circumference and pass through the

Area of carpetAc

I3 ft. 6 in. X 12 ft.(I 3~ X 12) sq. ft.162 sq. ft. of carpet.275 sq. ft. - 162 sq. ft.I 13 sq. ft.

Area of linoleum:. Az

NOTE.-The subscript letters /, I, and c in AfAl,and A, must not be confused with such expressions asN, At,and lV. In the former cases, the small lettersare distinguishing signs; in the latter cases, the smallletters are indices. Indices are fully explained inChap. XX.Exercises, with answers, on p. 113.

Fig . , 6

CIRCLE

centre C are diameters. Circles are often measuredin terms of their radii, but in practice it is perhapsmore common to measure them in terms of theirdiameters.

It is found that the ratio between the circumferenceand the diameter of a circle is constant. Thus,

Circumference . 020' = approximately 3+or ~1-'iameterThe exact value of the ratio between the circumferenceand the diameter cannot be expressed in figures,although it may, of course, be written with any re-quired degree of accuracy. The above value, 3t. isroughly correct; the value 3' 14 is more correct, andis often used; 3' 142 is correct to three places of

CHAPTER IX

CIRCLE.-A circle is a plane figure enclosed by acurved line called the circumference j all points onthis circumference are equidistant from the centre,and this distance is equal to a radius. ABO, fig. 16,


24/63

CIRCLE 43 44 TEXTILE MATHEMATICSdecimals; 3 '1416 is also often used, and is correctto four places of decimals; still more correct valuesare: 3'14159, 3.141593, and 3'1415926. This ratioin mathematics is generally expressed by the Greekletter ?r (pronounced Pi); hence, one may write

Circumference of circleDiameter of circle = . 7 1 " ,

Example 26.-Find the circumference of the take-up roller of a loom if the effective diameter of theroller is 5t in. (Use 71 " = 3'14')

c

where 71 " = ' 4 1 , 3'14, 3,1416, &c., according to thedegree of accuracy required.Let c the circumference of a circle;

r the radius;d the diameter;7r the above-mentioned ratio.

Example 27.-Two shafts are connected by means

Then, cd14-8

or c7 1 " ( I )

7 1 " d (2)Fig. r

whence d c

of two pulleys and a belt. See fig. 17. Find the mini-mum length of belt required.Length =2(~-pulley circumference)+ 2(distance between centres)

pulley circumference + (2 X 14ft. 8 in.)7 1 " d + 29 ft. 4 in.(3'14 x 4) + 29 ft. 4 in.12' 56 ft. + 29 33 ft.4189 ft. '. 41 ft. IO~ in.

Dividing each side of the latter equation by 7 1 " , wehave

By definition, a diameter equals 2 radii, hence,d = 21 (4)

It is occasionally convenient to use the radiusinstead of the diameter, so, substituting zr for d inequations (I), (2), and (3), it is found that:I. c2 r ?r (5)

7 1 " 2 r , usually stated 27r1' (6)c c-, or r = (7)7 1 " 2 7 1 "

In this example and the next no allowance has beenmade for the thickness of the belt.Example 28.-A double-beater scutcher is drivenby means of a IO-in.-diameter pulley running at1200 revolutions per minute. Find the speed of thebelt in feet per minute.

2. C3. 21

(D 64) 4


25/63

CIRCLE 45 TEXTILE lVlATHEIVATICSIn 1 revolution of the pulley the belt passes through

a distance equal to the circumference of the pulley,Circumference of pulley 7 T ' d

3,14 x ro in.= 3' 14 x 10 ft.12

In 1 min, the pulley makes 1200 revolutions;

The fly-wheel is therefore 0'03S in, (fully ill) toolarge in diameter.Example 3I.-An emery-wheel, ro-in, diameter,

is to have a grinding speed of 5000 ft. per minute.How many r.p.m. should it make?Surface speed 7 T ' d x r.p.m.,

or S 7 T ' d X r.p.rn.jSr.p.m. 7 T ' d= 3.14 x 10 X 1200123140 ft. per minute.

Example 29,-The cylinder of a carding-engine isso-in. diameter over the wire, and runs at I6S revo-lutions per minute. Find the surface speed of thepins in feet per minute.Surface speed = circumference of roller in feet

X revolutions per minute.

:. speed of belt = 3'14 X 10 in.SOOO

7 T '37 ft. S! in.3'I4I6= 144'035 in.

= 500() X I~3'14 X IO 111.

AREAOF A CIRCLE.-Referring to fig. 14, wherea regular octagon (S-sided polygon) is shown, let itbe -assumed that the number of sides within the cir-cumscribed circle be increased to 16; then it is clearthat the length of each side would become muchshorter than those shown, and that the lines wouldappear nearer to the circumscribed circle; as thenumber of sides is further increased, they approachnearer and nearer to the outer circle, while if thenumber of sides is increased to a sufficiently greatdegree, it will become impossible to distinguishbetween the polygon and the circumscribed circle.A circle may thus be regarded as a polygon with aninfinitely large number of sides, and hence the rulefor finding the area of a regular polygon may also beapplied to the circle.

Area of circleA

191O'S r.p.rn.

S 7 T ' d X r.p.m.[(3' 14 x 50) X 165] in. per minute314XSOXI6Sf .= t. per minute12

= 215S'75 ft. per minute.Example so.-It is desired to check the diameter

of a fly-wheel by measuring its circumference. Asteel tape is passed round the wheel and indicates37 ft. 8 k in. If the diameter is nominally 12 ft.,what is the error?

c452' S .6m3'14112 ft. 0'035 in.

area of polygon,iperimeter x perpendicular= !circumference x radius

H 2 7 T ' r ) X r= 7 T ' r 2 ,

d


26/63

CIRCLE 47 TEXTILE MATHEMATICS

A .. d " 1 di dgam, smce a iameter IS twice t ie ra lUS, Y = -.2:. Area of circle 7 r y 2= 7 r ( ~ ) 2

Example 32.-Find the area of the ro-in.ediameterpiston of a gas-engine.

a27r-'i2"

d27 1 ' 4 'or, as it is commonly expressed, !!_d24

Area A = !!_d24= 7854d2= '7854 X IO X IO sq. in.= 78. 54 sq. in.

Example 33.- The total pressure on the face of thepiston of a steam-engine at a certain point of itsstroke is 16,000 lb. If the piston is zo-in. diameter,find the pressure per square inch.Area of face of piston

or A =The area of a circle may therefore be expressed bythe formula'A = .78S4 d2

Dividing each side of the equation by '7854, wehave Pressure per sq. in.

' ! ! . d 24'7854 X 20 X 20 sq. in.314' 16 sq. in.Total pressure in poundsArea in square inches .16000 _ 50'93 lb. per314'16 - square inch.

A :. Phence d Example 34.- The plunger of a pump used in size

or starch mixing has a lifting area of 12' 57 sq. in.What is its diameter?The equation may also be expressed in terms of the

radius. Thus,Area A = 7 r r ,whence y2 A ,7 r

Y = ~= ~3.~i6

d =~'7~54= ~~;~; I

= ~16'004 = 4 in.Certain problems may be solved by finding the

difference between the areas of two circles. In thisconnection special methods are employed which oftensave time and trouble.


27/63

The above is true in every case, and the processmay be stated generally as under :--The difference of the squares of any two quantitiesis equal to the product of the sum and the difference

of the two quantities. Let the two quantities be re-presented symbolically by the letters A and Bithen

A2 - B 2 = (A + B) (A - B).Ifx and yare the quantities, then

x2 _ y2 = (x +y)(x - y);(x +y) and (x - y) are said to be the factors of theexpression x2 _ y2.To prove that the two factors are as stated. wemight multiply them together, and the result shouldbe x2 - y2. The operation of multiplication appearsbelow, and the description follows immediately:

x+y_.x:~Lx2 + xy= _ x )'_ _ _ )'2x2 _y2

Notice first that 16and 9

Now, (16 - 9):. (42 - 32)Also, (4 + 3) (4 - 3)since 7 X I

Again, 144and 81144 - 81

:. 122 - 92Now, (12 + 9) (12 - 9)

21 X 3

CIRCLE TEXTILE MATHEMATICS9 50

= 16,9

Place the two expressions x + y and x - y asshown. Multiply x +y by x, then multiply x +yby -y, and add the two results. The mental pro-cess is somewhat as follows: x times x =x X x orx2iput down x2 under the x. x times y = x X y or

42, since 4 X 432, since 3 X 37 7 7 .

= 7 122, since 12 X 12 =92, since 9 X 9 = 81.63-63-

= 63,63

I#

Fig . , 8

XYi put down . : x y under the y_ Now, multiply by-Yi -y times x = -)'X or -.:Xi.Yi put down -xyunder the + xy already found. -y timesy = _y2;put down - y2 a little to the right. (Note: when thesigns are the same, both + or both -, the productis + or positive i and when the signs are different,one - and one +, the product is - or negative.)


28/63

CIRCLE 5' TEXTILE MATHEMATICSNow add together all the similar terms: there is onlyone x2; put down x2 Add +xy to -XYi there isone of each, so the result is 0 or zero; leave the placeblank as indicated. Finally, there is only onej/', andit is negative, i.e. _y2; write down - y2. The resultis thus x2 _ y2.The proof may also be shown graphically, as in

fig. 18. ACEG is any square, and ABKH is asmaller square. The difference between the two isrepresented by a small shaded square KDEF, andtwo equal rectangles BCDK and HKFG.

Let x AC, the side of the large square,andy = AB, the side of the small square.

Then BC = x -y,and GH = x -yo

Area of ring area of outer circle - area of innercircle

'7854 D2 - -78S4d 2= .7854 (D2 _ d2)= -7854 (D + d) (D - d).

Area of difference of squares =But side BCand side CL

: . ( X 2 _ y2)

rectangle BCLM.(x - y),(x +y).(x +y) (x - y).

The two circles need not necessarily have the samecentre, i.e. they need not be concentric.Example 3S.-A vertical boiling'-kier 6 ft. 6 in.

diameter is provided with a man-hole 3 ft. diameter.Find the weight of the top plate of the kier if it ismade from metal weighing 18lb. per square foot.

Area of plate area of upper surface - area ofman-hole

= area of 6 ft. 6 in.-diameter circle- area of 3 ft.-diameter circle

= .7854 (0 + d) C D - d)= '7854 (6 ft. 6 in. + 3 ft.) (6 ft.6 in. - 3 ft.)= '7854 x 9! ft. x 3t ft.= '7854 X 33'25 sq. ft.= 26IlS sq. ft.

Weight of plate = 18 lb. per sq. ft. x 26. lIS sq. ft.= 470 lb.

Suppose the upper rectangle HKFG be cut off andplaced endways on the top of the shaded small square;it would appeal' as indicated by the stippled rect-angle ELMF, so that

This important deduction may be applied to findingthe area of a circular ring of any description. Thus,

Let d = the inside diameter of a plane cir-cular ring,

and D the outside diameter of same planecircular ring.

VARIATIONOF AREA OF CIRCLES.- The area ofa circle = .78S4d 2 In this, '7854 is constant, there-fore the area A varies directly as the square of thediameter, i.e, as d2 d thus varies as the . . . / A .

If the diameter is doubled, the area is increasedfourfold. The diameter of a circle, for example, is3 in.i the area = 7854 X 32 = 7' 07 sq. ins. Thearea of a 6-in.-diameter circle is '7854 X 62 = 28'28sq. in. The area of the large circle is, as stated,


29/63

4 times the area of the small circle, although theirdiameters are but 2 to I. This fact can be deducedfrom the statement given above that the area of acircle varies as the square of the diameter, just as theareas of squares vary as the squares of their sides.Consider, for example, the diagrams in fig. 19j ineach case there is a circle enclosed in a square,the side of which is equal to the diameter of thecircle.

CIRCLE

Upper diagram square I X IMiddle" ,,2 X 2Lower "

53 54 TEXTILE MATHEMATICS

I circle 12 X 7854 .22 " 22 X 7854 .

, .. . . .') Nl e t S ~ x . . o cl~ 00 \0 .-C l 00 "> N !2 ~.~


30/63

CIRCLE 5 5 TEXTILE MATHEMATICSExample 37.- The approximate diameter of a jute

yarn, expressed as a fraction of an inch, is obtainedby the formula: d = ,.je, where c is the count in

120pounds per spindle of 14,400yd. Find the diameterof a 16-Ib. yarn, and compare it with the diameterof a g-Ib. yarn. .r : ,.j16 3_ 1ro-lb, yarn: d = -in.120 120 120 30,.jc ,.j9_ . . . , 1g-Ib. yarn: d = = ~ = - in.120 120 120 40

r-e-- ~ ~ ~

o c

Fig. 20

s A

It should be particularly noticed that the aboverefers to jute; the diameters of cotton, woollen, spunsilk, worsted, and linen yarns vary inversely as thesquare root of the count.Exercises, with answers, on p. 114.

CHAPTER XARCS, CHORDS, SECTORS, AND SEGMENTSAacs.s--An arc of a circle is the name given to any

part of its circumference. Thus, in fig. 2I, a circle isB

E

Fig.2tNotice that the count of a yarn depends on its area,

which, for comparative purposes, and as alreadydemonstrated, may be taken as a circle. The area ofa circle varies as the square of the diameter, or, inother words, the diameter varies as the square rootof the area. In the case of jute yarn, since the areacorresponds to the count, the diameters vary as thesquare roots of the counts. Thus, if 16-lb. yarn hasa diameter of , ; ; ' 0 in., the diameter of 9-Ib. yarnwill be , . j9T6:30 x ,.j 16 lb. 30 x J =4

I - tn.40

illustrated with its centre at C. The heavy part in thecircumference between A and B is an arc; similarly,DE is an arc, but a smaller one than AB; again, thetwo remaining parts of the circle in lighter lines arealso arcs; hence, a circle may contain any number ofarcs according to the size of the-latter. If CA andCB are joined, the angle ACB is formed at the centreof the complete circle, and is termed the central angle.It is obvious that if this angle ACB is doubled, thelength of the arc will also be doubled. The lengthsof arcs of any particular circle are thus proportionalto the central angle.


31/63

ARCS, CHORDS, SECTORS, SEGMENTS 57Suppose the radius CA rotates round the point C

in the direction shown by the ar~ow, i.e. counter-clockwise.. By the time it again reaches its presentposition it will have passed through an angle of360, or 4 right angles. The angular division knownas a degree may be further divided into minutes andseconds in order to make more accurate measurementspossible. Thus,

58 TEXTILE MATHEMATICSIn the expression given there are thus three factors

which may vary; these are a, d, and D; if any twoare known, the third may readily be found by a suit-able transposition of the formula. Thus,

60" or 60 seconds =60'" 60 minutes

3600 " 360 degrees =Iminute.Idegree.1 complete circle =4 rightangles.

a = .00872d D (I)ad = '00872D (2)

D = . O O S 7 2d , (3 )Example 38.-The diameter of a circle is 42 in.,

find the length of the arc which subtends an angleof 60 at the centre,If the angle ACB, fig. 21, measures 72, then thearc AB will be 3 " - , ; \ of the complete circumference;and if the circumference is 5 in., the arc will be i-hof5 in. = 1 in. tong.

Let a = length of arc;c = circumference of circle;

and D = the number of degrees.DThena -c360Dc3 6 0 '

a oo872dD= -00872 X 42 X 60= 21 '9744 in.

Example 39. - The length of an arc of a circle42 in. diameter is 21 '9744 in., what angle does itsubtend at the centre?

It is most convenient, as already indicated, tomeasure circles by their diameters; since the cir-cumference = 7rd , the value trd may be substitutedfor c in the above expression, giving

7rdD3bo= 3'1416dD360= .00872dD.

aD = ~'00872d21 '9744:0087 ; - > < - ' 4 2 .

= 60 degrees.

a

CHORDS.-If any two points on the circumferenceof a circle are joined by a straight line, this straightline is termed a chord. Thus, AB in fig. 22 is achord of the circle ADBE. The centre of the circleis at C, and the radius CD bisects the chord AB, thetwo lines being perpendicular to each other. AB isalso the chord of the arc AD B, as well as the chordofthe arc AEB.


32/63

ARCS, CHORDS, SECTORS, SEGMENTS ~Since the radius CD is drawn at right angles to the

chord AB, the point F is the centre of the chord, andthe point D is the centre of the arc ADB. IfCA andCB are joined, right-angled triangles CF A and CFBare formed. The distance FD, on the radius CD, iscalled the height of the arc ADB.It has already been shown that in a right-angled

triangle the square on the hypotenuse is equal to the

60 TEXTILE MATHEMATICSIt is also possible to deduce the following rule,

which should be specially used to find the diameteronly when the chord and the height of the arc aregiven: c2 + 1 1 2d =-7z~'The expressions for lz in terms of d and c, and for c

10 terms of Iz and d are rather unwieldy and seldomo

E.Fig. a2

sum of the squares on the other two sides. This factmay be used to solve many questions regardingchords. Thus, in the triangle AFC,/2 = c2 + a2,/ being the radi us of the circle, c being half the chord,and a being the radius minus the height of the arc.

Now, if/2 = c2 + a2 , ./ = '" &-+a 2Similarly, by transposition, we have

c2 = /2 _ a2 or c = " ' J 2 _ a2 ,and a2 =p- c2 or a = "'f2= -C 2

Note that there are three variables involved, and ifany two are given, the third may be found.

(D64) 5

Fig . 23

used. If, however, the student desires practice, hecan find them from the following data, which illus-trates a method of finding d:

h(d - h)hd - J z2 =

hd =. d

c2,c2,c2 + h2c2 + h2--Iz-' as already stated.

Example 40.-In a ao-in, circle, fig. 23, there is achord drawn 4 in. from the centre; find the length ofthe chord.


33/63

ARCS, CHORDS, SECTORS, SEGMENTS 61 TEXTILE MATHEMATICSThe solution of this example appears in two ways: Area A of sectorp=a2 + C2, C2 It(d - h)c2 =2 - a2, = (10-4)(2 X 10 - 6)~f2 - a2, 6 X 14 AC2C = chord S4,= 2 ~f2 - a2 c = ../S+= 2002 - .f :.2C = 2 "'84= 2~JOO-I6 = 2X9'165

= 2 ~S4 = IS'33 in.= 2X9'I6S= IS'33 in.

Example 4I.- The chord of an arc is 40 ft., and itslength 13 ft. Find the diameter of the circle. '

Fig. 24

d c2 + 1t2=r:= ()2 + 13 2

13400+ 169

13= 569 = 43'77 ft.13

Example 42.-ln a 20-in.-diameter circle, find thearea of a sector the arc of which subtends an angle of80at the centre.

Area A of sector '00218Dd2-00218 X 80 X 202'0021S X 80 x 20 X 20.69'76 sq. in.

SECTORs.-A sector of a circle is that part which isbounded by an arc and 2 radii drawn from its endsto the centre, as shown in fig. 24, where ACB isa sector of the circle AEB.The angle ACB is termed the sector angle, and in

any circle the area of a sector is proportional to the.sector angle. If the angle is 360, the sector becomesa circle, and its area = ' ! ! . t f 2 or 7 r r . If the sector4 '

It is not always convenient to measure the sectorangle; when this is the case, the arc may be measuredinstead. The formula then becomes:

Area A of sector = arc X radius.. 2

angle be D degrees, then

Example 4J.-Find the area of a sector of a zo-in.vdiameter circle, the arc of which is 40 in. in length.

Area A of sector ~ arc X radius40 X 10.2A zo X 10 =200sq. in.


34/63

ARCS, CHORDS, SECTORSi SBGMENTS 63 TEXTILE MATHEMATICSThis rule is based on the law for the area of a circle.

If a sector angle be made near enough to 360, thesector approaches more and more closely to a com-plete circle. When the sector angle reaches 360, thefigure is a circle. Now,

Area A of circle 7rr2= ~X r=!ircumference X radius.'I the case of a sector less than a circle, the

circumference in theabove is replaced bythe value of the arc.:. Area A ofsector =

B

The area of the sector is found by the rules justdiscussed, viz. half the product of the arc and theradius, or 00218Dd2 The area of the triangle maybe found by any of the methods indicated in ChapterVI, or by any of the trigonometrical methods whichare explained in a later chapter.Example N.-Find the area of the segment, thechord of which is 20 ins. long, the height 5 in., andthe arc 26 in. long.Referring to Example 41, we see that

c2 + h2d = It (where c = !hord)_ I ~ _ _ S :5arc X radius. A 0SEGMENTS.-A seg-ment of a circle is that

part bounded byachordand its arc.Thus, in fig. 25,ABDE is a Circle with

centre C.ABO is an Earc with AD the cor- Fig. 25responding chord. Theshaded portion is a segment of the circle, while theunshaded portion is another segment.

If the radius CFB cuts AD at right angles, thepoint F is the centre of the chord AD, and the pointB is the centre of the arc ABO.Note that ABDCA is a sector of the circle, and

that AFDC is an isosceles triangle of which' thetwo sides CA and CD are radii of the circle, and thethird side is the chord AD of the arc ABD.Area A of segment ABD = area of sector ABDC

- area of triangle ACD.

= 100+ 255 = ~S = 25 in. diameter.5If d 25 in., r = I2~ in.

:. Area A of sector =!rc X radius= t X 26 X 12~= 162'5 sq. in.Area of triangle = !base X altitude= !chord X (radius-height)=! 20 X (I2~ - 5)= 10 X 7t = 75 sq. in.Area of segment area of sector - area of

triangle= 162'5 - 75= 87' 5 sq. in.

Exercises, with answers, on p. u6.


35/63

GEAR WHEELS 66 TEXTILE MATHEMATICS

GEAR WHEELS

by two considerations: (a) the space available, and(h ) the strength of tooth required. It is not proposedto deal with the strength question in this work, butmerely to discuss the relations between pitch anddiameter..The diameter of the pitch circle is called the pitchdiameter, and the pitch itself may be measured inat least three different ways.I. The number of teeth in the rim may bear a

certain relation to the pitch diameter of the wheel.Thus, in certain trains of gear, each wheel has6 teeth in the rim for each inch in the pitch diameter,i.e. 6 teeth in 3' 1416 in. of the pitch circle. For

CHAPTER XI

GEAR WHEELs.-It may be an advantage at thisstage to refer briefly to the relations between the cir-cumferences and diameters of toothed wheels, soextensively used in textile machinery. Fig. 26 is adrawing of a small spur pinion containing 18 teeth.

= 6 teeth per inch of pitch diam.3 in. pitch diameterexample:

18 teeth

Fig. 26

The number 6 expresses the ratio between the numberof teeth in the wheel and the pitch diameter in inches.This ratio is briefly termed No.6, and such a wheelis described as a spur pinion of 18 teeth, No.6 pitch.This kind of pitch is termed Dianzetral Pitch. Ingeneral:

Let DP dand N

Then N

diameter of wheel in terms of pitch,diametral pitch,number of teeth in the wheel.D x P d ~N

D =P d 'N=I)'

These teeth are situated at regular intervals on acircle known as the" pitch circle ", and indicated bythe outer circle which passes through every tooth.The distance between any point on the pitch circleof one tooth and the corresponding point on a neigh-bouring tooth is called the "pitch". Each pitchis obviously a certain definite measurement, and allwheels which are to gear together, or to form thesame unbroken straight train, must have the samepitch. In any machine wheel the pitch is determined

2. Another common method of expressing the pitchis to state the actual distance in inches, or in a fractionof an inch, between corresponding points on two


36/63

GEAR WHEELS 68 TEXTILE MATHEMATICSadjacent teeth, measured along the pitch circle. Forexample, in the I8-tooth pinion illustrated in fig. 26,suppose the pitch circle is 18in. in circumference, thedistance between each pair of teeth, or the pitch ofthe teeth, will be I in. This kind of pitch is calledCircular Pitch (see the arc AB in fig. 26).In general:

Let DP c ~and N

o s:N x P c ,Then N

the diameter of the pitch circle,the circular pitch,the number of teeth in the wheel.7rD

and P c Fig. 273. When making complete patterns of gear wheels,

it is not always easy to mark out the pitch along thepitch circle, and successive trials, with the necessaryadjustments, are made until the correct division isobtained. Note that if any instrument of the dividertype is used, the measurement marked off will be thestraight line between the centres oftwo adjoining teeth.This straight-line measurement is really a chord ofthe pitch circle, such as chord CD in fig. 26. Thiskind of pitch is termed the Chordal Pitch.The chordal pitch is not commonly used to express

the relation between the pitch and the diameter j thecalculation involved is rather cumbersome, and furtherdiscussion would be out of place in an elementarystudy of mathematical principles.Example 45. - Two gear wheels (fig. 27) which

mesh directly with one another contain 50 and 25teeth respectively. The 25-tooth wheel is keyed to a

loom crank-shaft, and the 50-tooth wheel is keyed tothe bottom or low shaft (called wyper-shaft or cam-shaft). If the pitch is NO.5, find the exact distancebetween the centres of the two shafts.Distance 0

-2 - diam. of wheel A + 2 - diam. of wheel Bl(diam. of wheel A + diam. of wheel B)t (25 teeth 5 pitch + 50teeth 5 pitch)l75 teeth 5 pitch)_ !. x 752 57~in. between the centres.

Example 46.-A 72-tooth wheel on a factory main-:shaft is to have the teeth set at z-in.ccircular pitch.Find the pitch diameter, and calculate the mean speed-of the wheel rim in feet per minute if the shaft makes;200.revolutions per minute.


37/63

RECTANGULAR SOLIDSD=NxPc

6g TEXTILE MATHEMATICSsmall one on the left is a special form, known as acube, each of its six faces being a square.The rectangular solid on the right has six rectan-gular faces, each pair of opposite faces, ABeD and

EFGH, ABGF and DCHE, ADEF and BCHG, beingequal and parallelrectangles. Thelength, breadth,andheight of this par-ticular solid areall different j whenthese three factors


38/63

RECTANGULAR SOLIDS 71 TEXTILE MATHEMATICSrather a similar one, appearing in the lower diagram.Each of the slices may be divided into four longstrips, each strip containing 6 units of I unit side.Suppose the unit to be 1 in., then each little cube isI cu. in. In each slice there will be four strips of6 cu. in. = 24 cu. in., and in the three slices ofthe complete solid there will be 3 X 24 = 72 cu. in.The volume of the solid in the upper diagram infig. 29 is therefore

6 X 4 X 3 = 72 cu. in.In general terms, the volume V of any rectangular

solid is equal to the product of its length, breadth, andheight; that is to say,

V = l X b X h,or V = lbh,

Exarnple 48.-How many bales of the above cottoncould be stored in a warehouse 120 ft. long, 80 ft.wide, and 40 ft. available height. (No reduction foreartways, &c.)Volume V =

No. of bales =

I X b X h= (120 X 80 X 40) cu. ft.= 384,000 cu. ft.total volume of warehouse~-~voiume of Ibale. } ~ 1 f ~ : ~ t t ~ _ f ~ ~38,400 bales.

From this general formula, which contains fourvariables, any unknown value of a term may befound when the remaining three are known.

V lblz: ~ 1and h ~ J

RELATIVE DENSITY.-It is often required to knowhow the weight of a given volume of one material com-pares with the weight of an equal volume of someother material. The weight of a substance measuredin this way, i.e. weight per unit volume, is known asthe density of the substance.Example 19.-Find the relative density of a bale of

Indian cotton, 10 cu. ft. in volume, and weighing396 lb., and a Brazilian bale measuring 49 in. X 20 in.x 18 in . , and weighing 220 lb.

Note that biz, Ill, and lbare the areas of the variousfaces.Density O f Indian bale

_ 396 lb.-i-o CtL ft.Density of Brazilian bale

220 lb.

39.6 lb. per cu. ft.

Example 47.-A bale of Indian cotton is 48 in.long, 20 in. wide, and 18 in. thick; find its volumein cubic feet.

V lxbxh(48 X 20 X 18) cu. in.48 x 20 x 18 ficu. t.1728ro cu. ft.

(49 X 20 X 18) cu. in.220 lb.= 49 ~__() ~,~~ cu. ft.1728

220 lb.w.2i cu. ft.;: 2 I 55 lb. per cubic foot.


39/63

RECTANGULAR SOLIDS 7 3 74 TEXTILE MATHEMATICSRelative Density

Indian _ 39.6 _ 1.84- Brazilian - '21~55 - -1-' -.In other words, baled Indian cotton is, bulk for

bulk, 1.84 times heavier than baled Brazilian cotton.If Brazilian cotton is taken as the standard of com-parison, thenRelative Density

for all solids and liquids, while hydrogen is thestandard for all gases.)Ifit is known, for example, that the specific gravityof aluminium is 258, then it is understood that[ cu. ft. of aluminium weighs 258 times as much asIcu. ft. of water.The following particulars concerning the weight

and volume of water should be committed to memory,as they are very important:

Brazilian - -Indian 2155 I .3 g - : ( ) = .547 = 1.84' I cu. ft. of water = 1000 oz. = 62~- lb.I gall. of water = 277~ cu. in. = 10 lb.Icu. ft. of water = 6! gall.that is to say. baled Brazilian cotton is only. 547 timesas heavy as baled Indian cotton.SPECIFIC GRAvITY.-In Example 49, baled Indian

and baled Brazilian cotton were in turn used asstandards of comparison, and the relative densities ofeach with respect to the other were found. It is,however, much more convenient to have one invari-able standard of comparison, and this standard ofcomparison is pure water. The density of water isregarded as unity, i.e. its numerical value is taken tobe I,and all other densities thus become:(a) fractions of I if the substances are lighter than

water, bulk for bulk, Of,(b ) multiples of I if the substances are heavier than

water, bulk for bulk.When water is used as the standard of comparison,

the ratio is given a special name, viz. SPecificGravity; that is to say, the number given to thematerial expresses the ratio of the weight of a givenvolume of the substance to the weight of an equalvolume of water, when the weight of the latter isrepresented by I. (Water is used as the standard

The above values are only approximate ones, butthey are sufficiently accurate for all practical purposes.(If very accurate results are required, then Icu. ft. ofwater may be taken at 996'46 oz.; I gall. at 277'46cu. in.; and Icu. ft. at 622 8 gall.)Example 50.- The guide-bars of a packing-pressmeasure II ft. 3 in. long by 6 in. wide and It in.thick. Find their weight if the specific gravity ofsteel is 7.8.

V lxbxhI I ft. 3 in. X 6 in. X 11 in.

= I35~6_x 1.25 cu. ft.1728= 0'586 cu. ft.Icu. ft. of water =

'586 cu. ft.1000 oz._ , - 586 X 1000 lb.16 oz. per lb.36.6 lb.

Specific gravity of steel is 7.8, thereforeWeight of bar 7.8 x 36.6 lb.= 28S481b.


40/63

RECTANGULAR SOLIDS 75 TEXTILE MATHEMATICSAREA OF SURFACES.-If a rectangular solid has

length I, breadth 0, and height h, the whole outer sur-face of the solid will be2(1 x It) + 2(1 x b) + 2(h x b)= 2(lh + Ib + hh) sq. units.

Ifa cube has a side s, the area of the outer surfacewill be

CHAPTER XIIIPRISMS AND CYLINDERS

PRISMS.-A prism is a solid bounded by plane faces,of which two, called ends, are parallel, equal in area,and exactly similar, while the other faces are parallelo-grams. Fig. 30 shows two forms of prism; in each2( S X s) + 2( S X s) + 2( S X s)

= 6(s x s) = 6s2 sq. units.Example sr. - An acid-tank in a bleachfield

measures 10ft. long by 6 ft. wide by 4 ft. deep. Findthe cost of lining it with sheet lead at IS. per squarefoot.Area of tank

(I x b) + 2( 1 x It ) + 2( h X b)N.B.-There are 4 sides and bottom (no top).

Ib + 2(llz + hb )(10 x 6) + 2{(IO x 4) + (4 x 6)}

= 60 + 2(40 + 24)60+80+48188 sq. ft.

Cost = 188 sq. ft. x IS. per square foot188s. or 9, 8s.

The thickness of the wood has been ignored.Exercises, with answers, on p. 119.

I!

i.I....,,i;I~t

--~\

A BFig. 30

the ends are equal in area, parallel to each other, andof the same size and shape. The figure marked A iscalled an oblique prism, because the sides are notperpendicular to the lower end or base; the otherfigure, marked B, is termed a right prism, because itslateral edges are at right angles to the base.The examples shown are hexagonal prisms, because

the two ends are hexagons; the base of a prism may,however, have any shape, provided that the area ofthe end be bounded by straight lines.AREA OF SURFACEOF PRISM.-The area of the

surface of the prism is the area of the two ends added(:0 64) 6


41/63

PRISMS AND CYLINDERS 77 TEXTILE MATHEMATICSArea of surface= Area of 2 ends + (perirneter X height)= 2(12 ft. X 3 in.) + {(I2 ft.+ 3 in. + rz ft. +3 in.) X ro ft.]= (2 X 12 X t) + (24! X 10)= (6 + 245) sq. ft.= 251 sq. ft.VOLUME OF A PRIsM.-The rectangular solid illus-

trated in fig. 29, p. 70, is evidently only a specialform of prism. It was shown that its volume wasobtained from the formula

V = I X b X hibut (I X b) = area of base.:. V = area of base X height.

The same reasoning applies to any prism. Thevolume of a prism is, therefore,V =area of base X height.Example .5J.-Find the weight of a hexagonal bar

of iron 12 ft. long by2 in. side, it beinggiven that I cu. ft. ofiron weighs 484 lb.NOTE. - A regular

hexagon is made up of6 equilateral triangles(see fig. 32). The areaof an equilateral tri-angle may be found inthe usual way if itsaltitude be known, but

if there is difficulty in finding the altitude, use may bemade of a special rule, based on the trigonometricalratios of its angles (discussed in Chapter VII, Part II);in this easel if the length of side be I units,

to the area of the lateral sides. In the majority ofcases which occur in practice, the sides are rectangles,so that the finding of the area should present littledifficulty. ,There is, however, a short method of finding thearea of the lateral sides. It will be evident that if apiece of paper be cut to the size necessary to coverexactly the lateral sides of a right prism, and thepaper then removed and laid flat, that it will forma rectangle in which the length equals the perimeterof the prism, and the breadth equals the height of theprism. Then,

Area of surface of prism = area of 2 ends +(perimeter of base X height of prism).

0"0. . .

12'-0"

'========~Example 52.-A plain rectangular warehouse door,

IO ft. high by ia ft. wide by 3 in. thick, is to be fire-proofed by covering it with sheets of tinned steel.Find the area of the sheet required, neglecting waste.The dimensions are shown in fig. 31.

Fig. 32


42/63

so that in the example under discussion we haveVolume V 2'59812 X L (where L= 12 ft.)

2'598 X 22 X 12 ft.(2'598 X 4 X 144) cu. in.

= 2'598 X 4 X 144 f8 cu. t,172o- 866 cu. ft.Weightofbar = 866 x 484 lb.

419' 144 lb.CYLINDER.-A cylinder may be roughly described

as a special form of prism with circu-lar ends. A right circular cylinder ismore correctly defined as the soliddescribed by one complete revolutionof a rectangle, one side of which actsas the fulcrum. Thus, in fig. 33,ABCD is a rectangle of which theside AB is the line about which therectangle may be assumed to revolvein order to sweep out in space avolume of the shape known as a rightcylinder. It is one of the commonestshapes adopted in textile machinery,and indeed in all mechanical work,and its mathematical properties are Fig. 33therefore of extreme importance.The complete surface of the cylinder consists of the

PRISMS A ND CYL IN DE RSP.. /3.Area of equilateral triangle = 4

Area of hexagon 6 X 12../34= 6xf2x 1'732 =

4

79 80 TEXTILE MATHEMATICScurved surface swept out by the side CD, fig 33, andof the two circular surfaces or ends swept out by thesides BC and AD.The dimensions ofa cylinder are generally expressedby stating the diameter and the length or height, and

practical mathematical laws should therefore be statedin terms of these dimensions.SURFACEOFACYLINDER.- The area of the curved

surface of a cylinder is that traced out by a line, equalto the height or length, moving through a distanceequal to the circumference of the end, so thatArea of curved surface = circumference of base

X height or length=7r d X 1,or surface S =7rd l .The whole surface area will be equal to the above

added to the two circular ends, or, stated symboli-cally,

A = 7 rd l + 2 X Z ! : X tJ 247rd(1 + ~)

= 7rd(1 + y) , where y = radius.The volume of the cylinder will beV area of base or section X length or height: ! ! . . t J 2 X I4

= 7!{[2/, or '7854d21.4If preferable, the equation may be stated in terms

of the radius. Thus,V = 7 r y 21.


43/63

PRISMS AND CYLINDERS 81 TEXTILE MATHEMATICS

drum, or S=7rdl3' 1416 X 6 ft. X 60 in.(3'1416 X 6 X 5) sq. ft.94'248 sq. ft. in Irev.Area printed in I min. = (94'248 X 5 revs.) sq. ft.

II 1 hour = (94' 248 X 5 X 60) sq. ft.94.248 X 5 X 60 sq. yd.93141.6 sq. yd.

HOLLOW CVLINDERs.-In practice, it is customary,whenever feasible, to use hollow cylindrical objects inpreference to solid cylinders. In such a case, thevolume of the hollow cylinder only is required, andsuitable rules for this calculation can be deduced fromthe general statement given above.Example 57.-Fig. 34 illustrates an extensively-

used hollow cylinder, that of a delivery pressing-roller

Example 51.- The colour-drum of a calico-printingmachine is 6 ft . diameter, and runs at 5 r.p.m., print-ing cloth 60 in. wide. Find the number of squareyards of cloth capable of being printed in one hour.Curved surface of

" "

Example 55.- The starch or sowbox roller of acylinder dressing-, slashing-, or sizing-machine isIS in. diameter and 66 in. long; find the area insquare feet of the copper sheet required to cover itcompletely.

Whole area A 7r d ( I + r){3' 1416 X IS(66 + v sq. in.3' 1416 X IS(66 + 9) sq. in.J'1416 X 18 X 75 sq. ft .

14429'45 sq. ft.Example 56.-A roller weight used in a drawing-

frame is 3 k in. diameter and S in. long i find itsweight if 1 cu. in. of cast iron weigh '263 lb.V '!!.d2l4('7854 X 3z X 3~ X 8) cu. in.76'97 cu. in.

Weight 76'97 cu. in. X 263 lb. per cubic inch.= 2o.241b., say 201 lb.

Fig. 34

suitable for a flax or jute drawing-frame. The rollerconsists of three hollow cylinders joined together-two end-pieces and one centre-piece.Volume of centre-}_ volume of solid cylinder-piece - volume of hole

= !:D21 _ ? !d 214 4

= :!l(D2 - d2)4= ~l(D + d)(D - d)4= ' ! ! . X 9H6 + 3t)(6 - 3!)4= .7854 X9! X9! x 2!= 177.21 cu. in.


44/63


45/63

PYRAMID, CONE, AND SPHERETHE PVRAMID.- The name pyramid is applied to

any solid which is bounded by plane surfaces, one ofwhich, termed the base, may be any rectilinear figure,while the others are all triangles. The triangularfaces have a common vertex, which must of necessitylie outside the plane of the base. The point may benear the base, or a distance removed from it, andfig. 36 illustrates one formof pyramid in which thedistance of the point fromthe base is greater thanthe width of the baseABCDEF. The triangu-lar sides are represented byABV, BCV FAV,ali of which taper to thepoint V, the common ver-tex.The pyramid illustrated

is fully described as a righthexagonal prism. It hasalready been indicated whyit may be termed a prism.It is a hexagonal prismbecause its base has 6sides, and is therefore a hexagon j it is a right prismbecause the line VP, a perpendicular dropped fromthe vertex V to the base, meets the base at its centreP. Point P is the centre of the inscribed circle indi-cated j it is also the centre of the circumscribed circlewhich is not shown, but which would pass outside thehexagon and touch all points A, B, C, D, E, F. Ifthe base had happened to be a rectangle, the-point Pwould have coincided with the crossing or intersectionof the diagonals.

85 TEXTILE MATHEMATICS6

v

The distance VP is called the Iwight of the pyra-mid, while the distance VR is termed, for the sake ofdistinction, the slant heiglzt. The combined area. ofthe triangular faces is the sum of the slant surfaces.The total surface area of the pyramid is, of course,the area of the slant surfaces + the area of the base.VOLUMEOF PVRAl\lID.-It may be proved that the

volume of a pyramid is one-third of the volume ofa prism which has the same base and the same per-pendicular height.

IfVh

and Athe volume,the height,the area of the base, thenAh3V

With this formula as a starting-point, it will bea useful exercise for the student to deduce an expres-sion for A in terms of V and h, and then one for h interms of V and A.SLANTSURFACEOF PVRAMID.-Since all the slant

surfaces of the pyramid are triangles, the area ofeach of which is} base X altitude, it is evident thati g. 36

Surface Swhere S

pand s

ps-,2the area of the slant surfaces,the perimeter of the base,the slant height.

The above formula applies only to pyramids withregular bases, as this type is the only form of pyramidin which the slant heights of the various triangularfaces are uniform.WHOLE SURFACE OF PVRAMID.-As previously


46/63

PYRAMID, CONE, AND SPHERE 88 TEXTILE MATHEMATICSindicated, the whole surface area of the pyramid isrepresented by

Area of slant surfaces + area of base.The method of finding the area of the slant sur-faces has just been discussed; the area of the base is

obtained by applying the rules relating to rectilinealfigures given in previous chapters.

300 - (2 X IS) = 270 ft. long; and two halves (oneat each end) of a rectangular pyramid KMNCB, eachwith a base 33 ft. by IS ft., and a height of 8 ft.Cubical contents of one bay= Rect. prism + triang. prism + rect. pyramid

= 1M + Al + l~__3 ab lbh= (300 X 33 x 14) + 2 I +3= (30~ X 33 x 14)+ (8:33 X 270) + (32 X (2 ~ 15)X)= 138,600 + 35640 + 2640= 176,880 cu. ft.

Cubical contents of 6 bays= 176,880 X 6 = 1,061,280 cu. ft .Maximum number of employees

1,061,280= --- -- =4245 persons.250

Example 59.-Fig. 37 represents part of one bay ofa weaving-shed, the inside dimensions being as indi-cated. Find the maximum number of persons whichmay be employed in a shed consisting of 6 similarbays, given that the Factory Acts require 250 cu. ft.of space per person.An examination of the figure shows that each

bay consists of a rectangular prism ABCDEFGH,300 ft. long, 33 ft . broad, and 14 ft. high; a triangu-lar prism DNMALK, 33 ft. wide, 8 ft. high, and

NOTE.- The number of workers actually employedin a factory of such dimensions would be considerablyless than the above, since the floor space for eachoperative would be less than 14 sq. ft.; actually,

3=--3=----X-=3()Q_x_6= 13' 99 sq. ft.4245

The calculation shows, however, the generous pro-portions of modern mills and factories compared withthe requirements of t

40001820 textile mathematics part 1

Documents