An interesting series of calculations useful to the knitting industry can be built up in connection with the methods of finding the resultant single counts of two or more yarns folded together. The counts of twist yarns is in the first instance of special application to the spinner, but there are many circumstances in which they may be useful in the yarn store of a hosiery factory. When two yarns of the same counts are twisted, the resultant equivalent single counts is obtained by dividing by 2, but when the yarns vary in size the proceeding is rather different. A common fault is to add the two counts together and divide by 2, but this gives a result which is entirely wrong. For purposes of comparison it is useful to have the equivalent single counts when two or more yarns are folded together, but the special interest to this branch lies in using two single yarns to take the place of one, in cases where supplies of the first material have run short. It is then of practical importance to have the two substitute yarns chosen from those in stock so that they approach as nearly as possible the size of the original.

Example 27.—Find the counts of 60's and 40's folded together. This example can be used to build up the formula from first principles. Take for convenience 60 hanks of the highest counts and twist this with an equal length of 60 hanks of the second counts stated thus—

60 hanks of 60's counts weigh	1	lb.
60 hanks of 40's counts weigh	1½	lb.
60 hanks of folded thread =	2½	lb.

therefore, 60 divided by 2½ gives 24 hanks of folded yarn in 1 lb. which is the counts 24's.

To find out a shorter rule for estimating the counts of a two-fold thread, let the first counts be represented by A, and the second counts by B, and let R be the resultant counts of yarns A and B folded together. Following the concrete example 27, it may be stated generally thus—

Example 28.—

A hanks of A counts	=	1	lb.
A hanks of B counts	=	A/B	lb.
A hanks of folded yarn	=	1 + A/B	lb.

The resultant counts R is obtained thus—

A
1 + A/B	= R,

but

	A		A + B
1 +		=
	B		B

and making the fraction proper we obtain the rule—

AB
	= R.
A + B

Stated in words the rule is "To find the resultant counts of two threads folded together, multiply the two counts together and divide by their sum."

It often happens that a counts of a given size is required from two single yarns as in the frequent case of yarns running down before the contract for goods has been delivered. In such instances the resultant counts required is known and given one of the constituent singles, the other can be obtained by the rule: "Multiply the two counts together and divide by the difference." This can be proved in a general way from the last-found formula—

Example 29.—

A × B
	= R
A + B

in this equation the following also holds good

AB =	R (A + B)
AB =	AR + BR
A B - BR =	AR
B (A - R) =	AR
	AR
B =
	A - R

Similarly, if A is the missing counts of the two-fold yarn, the rule for A can be proved thus—

AB =	R (A + B)
AB =	AR + BR
AB - AR =	BR
A (B - R) =	BR
	BR
A =
	B - R

Examples in folded yarns.

Example 30.—Find the counts of 64's, 48's and 32's yarns folded together, and also give average when they are used separately one thread of each size in a garment. From the formula proved in Example 28 we have the following, taking the highest counts as starting-point—

64 hanks of 64's counts weigh	1	lb.
64 hanks of 48's counts weigh	1?	lb.
64 hanks of 32's counts weigh	2	lb.
64 hanks folded yarn weigh	4?	lb.

therefore—

64
	= 1410/13 × 3 = 444/13 counts.
4?

In the hosiery trade such yarns are more often used separately than folded together, when the more useful problem is to find the average counts of the three threads which is obtained by multiplying this result by the number of threads in the set, in this case 3.

Example 31.—Give the resultant equivalent single counts of a silk and wool yarn composed of one thread of 2/40's worsted folded with a thread of 60/2 spun silk, also give the weight of each material in 110 lb. of garments and state the price ratios of worsted and silk respectively, taking the silk at 12s. 6d. per lb. and the worsted at 5s.

For the counts calculations both yarns have to be expressed in the same denomination and let the worsted system be taken. Transferring 60/2 silk, the counts is 60's as it stands and the transfer is made thus—

60 × 840
	= 90's.
560

To find the counts formula 28 can be used, that is, multiply the two counts together and divide by their sum—

90 × 20		90 × 20		180
	=		=		= 164/11.
90 + 20		110		11

To obtain the weight ratio the calculation had better be worked thus—

90 hanks of 90's	= 1	lb.
90 hanks of 20's	= 4½	lb.
90 hanks twist	= 5½	lb.

For the weight proportion the total is given as 110 lb., which has to be divided in the ratio of the weight column, that is, of a total of 5½ lb., 1 lb. is silk and the remainder worsted—

1
	× 110 = 20 lb. silk.
5½
4½
	× 110 = 90 lb. worsted.
5½

For the price let the weight column be again used—

1	lb. silk @ 12/6	= 12/6
4½	lb. wor. @ 5/-	= 22/6
5½	lb. twist	= 35/-

also ⁵/₁₄ of cost is due to silk and ⁹/₁₄ due to worsted.

It is quite evident that this method of procedure yields results of the utmost interest and value to the maker of knitted goods.

Example 32.—Find the counts of 48's cashmere folded with 30's merino. Answer in merino system.

Change 48's cashmere into merino—

48 × 560
	= 32's.
840

Using the rule as in (28)—

32 × 30		32 × 30
	=		= 1515/31.
32 + 30		62

Example 33.—Find the resultant counts of 2/32's cashmere or worsted folded with 21's skeins counts. State answer in skeins system.

Transfer worsted to skeins—

16 × 560
	= 35 skeins.
256

By one method the counts is given by—

35 × 21		35 × 21
	=		= 13? counts skeins.
35 + 21		56

In this connection it might be useful to have the weight proportion of each, also the average counts if the two threads are not folded but placed side by side. Let the price of the 2/32's be 7/6 per lb., and that of the 21's skeins, 3/-per lb., when the cost per lb. of the combination is obtained as follows—

35 hanks of 35's	= 1	lb. @ 7/6	=	7/6
35 hanks of 21's	= 1?	lb. @ 3/-	=	5/-
35 hanks of folded	= 2?	lb.	=	12/6

For the counts ³⁵/_2? = 13? if folded, but if separate, number × 2 will give the average counts 13? × 2 = 26¼ average counts.

The weight percentage of each constituent in the garment is stated thus—

1		1?
	× 100 = 37½ % worsted.		= 62½ % skeins.
2?		2?

For the average price 2? lb. = 12/6 = 4/8¼ per lb.

Example 34.—Find in Scotch system the resultant and average counts of 36/36 Scotch with 2/40's cotton.

Bringing the cotton to Scotch counts—

20 × 840
	= 84 cut,
200

by the rule—

84 × 18
	= 14·82 counts, or × 2 = 29·6 average counts.
84 + 18

This rule can be applied twice in succession to find the counts of a three-fold yarn.

Example 35.—Find the counts of 80's, 40's and 20's folded together and give the average counts if they are used separately in a garment, one thread of each.

Taking 80's and 40's the resultant counts of these two is secured by the usual method—

80 × 40		80 × 40		80
	=		=		or 26? counts.
80 + 40		120		3

This is now folded with the remaining counts 20, in a similar operation—

26? × 20
	= 113/7 resultant counts or × 3 = 342/7 average counts.
26? + 20

This result can be verified by the original method—

80 ÷ 80	= 1
80 ÷ 40	= 2
80 ÷ 20	= 4
80 hanks	= 7 lb.

Example 36.—Give all useful particulars for the following combination of yarns in a garment—

1 - 2/42 wor.	@	5/6	per lb.
1 - 12's cotton	@	3/6	per lb.
1 - 80/2 silk	@	15/-	per lb.

Bring the worsted to cotton counts and the three yarns will then be in a like denomination—

21 × 560
	= 14's cotton.
840

Arranging these for the resultant counts, price and weight relationships—

80 ÷ 80	=	1	lb. @	15/-	= 15/-
80 ÷ 14	=	5·7	lb. @	5/6	= 31/4
80 ÷ 12	=	6·6	lb. @	3/6	= 23/1
80 hanks	=	13·3	lb.		= 69/5

80 ÷ 13·3 = 6·1 counts if twisted, and × 3 = 18·3 counts average.

By computation from the table of prices it is found that

For the weight percentage the proportions are taken from the weight column—

1			5·7
	× 100 = 7·52 % silk.			× 100 = 42·8 % worsted.
13·3			13·3

The remaining figure 49·68 % cotton is obtained by subtraction.

Example 37.—Find the useful particulars for a plated fabric composed of 2/20's worsted @ 8/6 per lb., 10's cotton @ 4/- per lb. The cotton yarn appears on the back and has 50 per cent. more take-up than the face.

10 × 840
	= 15's worsted.
560

15 plus 50 % of 15	= 22½	÷ 15	=	1½	lb. @ 4/-	= 6/-
15 plus 0	= 15	÷ 10	=	1½	lb. @ 8/6	= 12/9
				3	lb.	= 18/9

From the column of weights it is seen that this is divided equally between the two yarns; although the cotton is finer in diameter, yet the weights are made equal by extra take-up in loop formation.

Example 38.—A garment order is being executed from 2/42's yarn which runs short, 2/60's is in sufficient stock, and it is required to find what counts used with this will give the same weight of garment on the frame. Applying the difference rule—

2/60's	= 30's		30 × 21		30 × 21
				=		= 70's.
2/42's	= 21's		30 - 21		9

Example 39.—Given two counts 48's and 24's in stock, find the third yarn needed to twist with these to obtain a counts equivalent to 10's single. The first step is to fold the two yarns together by the addition rule—

48 × 24
	= 16's.
48 + 24

Then by the difference rule—

16 × 10		160
	=		= 26? Ans.
16 - 10		6

Example 40.—40's and 30's worsted are yarns in stock, find a third thread to fold with these to make a three-fold counts = 10's.

40 × 30		40 × 30
	=		= 17? counts of the two yarns folded.
40 + 30		70

Using the difference rule—

17? × 10
	= 24's Ans.
17? - 10

To provide proof use the ordinary method of folding—

40 ÷ 40	= 1	lb.
40 ÷ 30	= 1?	lb.
40 ÷ 24	= 1?	lb.
40 hanks	= 4	lb.

If the third thread were wanted in another quality such as cotton, the answer would be secured by transfer—

24 × 560
	= 16's cotton
840

or-

24 × 560
	= 52½ skeins.
256

Example 41.—Find the particulars for a fabric containing three yarns used, one of each in succession, where the take-up of each thread varies in the fabric as shown. This take-up of different yarns in knitting is frequently neglected in calculations, but obviously should be carefully taken into account for it has a very direct influence on the weight.

We proceed by stating the length of hank in the ratio of the various take-ups thus—

225 hanks of 12's	=	18·75	lb.
150 hanks of 30's	=	5	lb.
125 hanks of 40's	=	3·12	lb.
125 hanks normal	=	26·87	lb.

Working out percentages in the usual way from the column of weights we find that—

40's yarn = 11·4	% of weight.
30's yarn = 18·6	% of weight.
12's yarn = 70	% of weight.

Example 42.—2/24's worsted yarn has been used for an order and runs out before completion of the weight required; sufficient weight of 16's is in stock, find the nearest counts which must be used along with this to give a fabric of the same weight as the original; 2/24's = 12's.

Using the rule of differences—

16 × 12
	= 48's.
16 - 12

The ratio of weights can be found thus, 48's and 16's—

48 hanks of 48's =	1 lb.
48 hanks of 16's =	3 lb.
	4 lb.

That is, out of a total of 4 lb. one-quarter is of the finer counts 48, whilst the remaining three-quarters is of the second counts 16's. By making this calculation it will be seen whether there is a sufficient weight of each yarn to complete the order.

Example 43.—As a substitute for solid 2/22's yarn in a garment it is proposed to use one thread of 18's and a second yarn of suitable size to give the same weight. Find this yarn and state the ratio in which the two will occur in the garment.

By the rule of differences—

18 × 11		18 × 11
	=		= 282/7 counts.
18 - 11		7

For the weight of each, this can be done quite simply by adding the two counts and taking the relative proportions inversely—

18	+ 282/7	= 462/7.
18
	× 100	= 388/9 % of 282/7 counts.
462/7
282/7
	× 100	= 61? % of 18's counts.
462/7

This last example shows clearly that the weight of the constituents varies inversely according to the yarn counts, the higher the counts the lower the weight and vice versÂ. The matter is more prominent in problems where a resultant counts is given with two weight ratios, the counts to produce these being sought by calculation.

Example 44.—A garment is required equal to 12's counts composed of two yarns where one-third of weight is on the face and two-thirds on the back. Find two counts which will fulfil these conditions.

The counts are inverse to the weights; if the proportion had been direct we should have stated: ? of 12's, but seeing that the ratio is inverse we state: ³/₁ of 12's = 36's for one yarn.

The other thread is ? of 12, which inversely gives ³/₂ of 12 = 18. For proof—

36 × 18
	= 12's counts.
54

Example 45.—Find two yarns one having one-fifth of the weight and the other four-fifths to give a resultant counts = 12's.

? of 12, inverse	= 5/1 × 12	= 60's counts.
? of 12, inverse	= 5/4 × 12	= 15's counts.

These counts are 60's and 15's and they together produce a thread = 12's.

Example 46.—A three-fold yarn is equal to a counts of 8's, the first thread gives one-seventh of the weight, the next three-sevenths and the next four-sevenths, find each counts in the folded thread.

Example 47.—A fabric with an average counts of 20's is required in two materials, one giving two-fifths of weight and the other three-fifths, find the counts of each.