COUNTS OF YARN
Within recent years a great improvement has been effected in the matter
of yarn numbering for the hosiery trade. Formerly a number of systems
were in vogue which were distinctly local in their character and
application, but these now tend to confine themselves to the standards
common to other branches of textiles. Most yarns can be classed under
the worsted, cotton or silk systems; woollen yarns spun on the Borders
of Scotland are based on the Galashiels counts, whilst those from
Yorkshire are counted on the skein system. Artificial silk yarns are
numbered on the denier system which has come into greater prominence
recently in connection with the growth of artificial silk goods on the
market. The more irrational and arbitrary methods of numbering yarns are
rapidly declining in use and the great majority of yarns now supplied
are given in one or other of the systems named.
_Worsted Yarns_, including those coming under the term cashmere, botany
and mohair are numbered on the worsted system which has its basis in the
number of 560-yd. hanks which weigh 1 lb. of 16 oz.
_Cotton Yarns_, including those spun from a mixture of cotton and wool
under the term merino, and spun silk are estimated on the number of
840-yd. hanks which weigh 1 lb. of 16 oz. There is a reservation in the
case of two-fold silk yarns, the counts giving the exact number whether
single, two-fold, or three-fold.
_Woollen Yarns._—Alloa is an important centre of hosiery yarns spinning,
and a system used in this locality is based on the equivalent of the
number of 240-yd. hanks which weigh 1 lb.
Woollen yarns spun on the Scottish Borders are calculated on the
Galashiels method which is equivalent to the number of 200-yd. hanks in
1 lb. The Yorkshire woollen skein system is based on the number of yards
per oz. which, brought into line with others, gives a basis of the
number of 256-yd. hanks which weigh 1 lb. Leicester lamb's wool system
is equivalent to the number of hanks of 176 yd. each in 1 lb.
_Silk Yarns._—In addition to the spun silk yarns mentioned as being
counted on the basis of the cotton hank of 840 yd., what is known as the
Organzine silk system is given by the number of yards per oz.
Tram silk is calculated on the weight in drams of 1,000 yd., and in the
case of artificial silk yarns the counts are gauged by the weight in
deniers of 520 yd. There are 1,600/3 or 533⅓ deniers in 1 oz.
_Yarn Testing for Counts._—This subject has been taken up with greater
interest by hosiery manufacturers, who are now installing suitable
apparatus for conducting the important test as to whether a yarn is up
to standard in regard to counts. Variation in the yarn size at once
reflects itself in the weight per dozen garments, the usual trade basis,
and with increased prices of yarns these tests are likely to become more
prevalent in the future. Compared with the weaving trade the question of
gauging the size of a yarn by the method of inspection and handling is
by no means effective on account of the loose nature of many knitting
yarns; in general they appear to have much less weight than their
diameter would lead the observer to suppose. In weaving yarns the twist
is much more decided in effect, but in hosiery materials accurate
estimation can only be performed by making a calculation based on the
weight of a given length of the sample. A number of simple devices are
on the market whereby the counts of yarn can be accurately determined by
weighing a given number of threads cut to a certain template according
to the yarn system, and these instruments are being largely employed in
cases where the overseer is too busy to give the matter personal
attention. An intelligent yarn foreman, however, prefers the method of
weighing off a given length of the thread and finding the counts by
direct calculation or by the aid of an assimilating table. He can also
devise short ways of making the calculation of counts adapted to the
class of yarns being handled in greatest numbers, and these do not
depend for their accuracy on any accidental mechanical factors.
Let it be supposed that the custom in a factory is to test a yarn by
unravelling a length of 20 yd. and finding the weight of this in grains.
In each calculation the proportion will be repeated of finding the
number of yards in 1 lb. or 7,000 grains. Again, if the prevailing
counts be worsted, then this will involve 560 in each calculation as in
the following example.
_Example 1._—On unreeling a yarn it is found that a length of 20 yd.
weighs 10 grains, find the counts in worsted.
By proportion, if there are 20 yd. in 10 grains, the yards in 1 lb. or
7,000 grains will give the yards per lb. This obtained, we divide by 560
the length of the worsted hank to obtain the counts thus—
20 × 7,000
—————————— = 25's worsted counts of yarn.
10 × 560
_Example 2._—A worsted hosiery yarn is tested and 20 yd. are found to
weigh 35 grains, find the counts.
20 × 7,000
—————————— = 7⅐ worsted counts.
35 × 560
If these two examples be observed it will be noted that for every
calculation of this type such as a yarnman might be expected to make
frequently, the common numbers are (20 × 7,000)/560 = 250. These will
occur in every calculation of this kind and this gives a short method of
getting the result, for in place of using these three factors we take
the resultant 250 as shown and divide the weight of grains into it.
_Example 3._—Find the counts of a cashmere hosiery yarn, 20 yd. of which
weigh 24 grains.
Following the method indicated we can obtain this result at once by
dividing 250 by 24 = 10·4 counts cashmere.
The other counts met with frequently is the cotton or merino system
where the hank number is 840 and the value in all such calculations is
given by the numbers—
20 × 7,000 500
—————————— = ——— or 166⅔.
840 3
_Example 4._—Find the counts of a merino yarn of which 20 yd. weigh 14
grains.
Taking the value 500/3 divide it by 14, thus—
500
—————— = 11·9 counts.
3 × 14
Similarly, if working with dram weights and a standard length of 20 yd.
we should devise a value for the figures constantly recurring and this
would greatly simplify the calculation of the counts.
_Example 5._—Find the counts in Galashiels or Scotch woollen system of a
yarn, 20 yd. of which weigh 2 drams.
The first step is to find the yards in 1 lb. = 16 × 16 = 256 drams, and
then divide by the hank length of 200 yd.—
20 × 256
———————— = 12·8 cut Scotch woollen.
2 × 200
In all calculations of this character the numbers (20 × 256)/200 will
occur and these reduced give a value of 25·6 which is taken as the
constant and for all similar calculations the weight in drams is simply
divided into this value to obtain the result.
_Example 6._—Find the counts in Scotch woollen of a yarn, 20 yd. of
which weigh 1·25 grains.
Taking the value as 25·6, this is divided by 1·25 = 20·4 cut woollen.
Sufficient has been given to show that it is comparatively easy to
calculate counts of yarns regularly coming into the yarn store where we
have a few standard hank lengths to consider along with the values
obtained for each type of calculation met with in practice. Tables in
each can be constructed from which the counts of yarn may be seen at a
glance, the only work being to find the weight of the test length.
_Yarn Conversion._—When two or more yarn classes are used in the same
garment it is necessary for purposes of calculation to translate the
counts into one or other of the systems, the most common system for
choice. Thus, in cotton and wool, or silk and wool twist yarns, it may
be necessary to make a calculation for counts and this cannot be
affected unless both yarns are in the same denomination. The rule is to
multiply the given counts by its own hank length and divide by the hank
length of the yarn required.
_Example 7._—Change 2/40's merino counts into worsted and Yorkshire
skeins. 2/40's merino = 20's single, and by the rule—
20 × 840
———————— = 30's worsted.
560
To convert into skeins counts the hank length is 256, and the formula is
given by—
20 × 840
———————— = 65·6 skeins.
256
_Example 8._—Change 24/24 Scotch woollen into Alloa and skein systems.
24/24 = 12 cut single.
12 × 200
———————— = 10's counts Alloa.
240
12 × 200
———————— = 9·4 skeins counts.
256
_Example 9._—Find the equivalent of 40/2 spun silk in worsted and skeins
counts. In silk the number is always the exact counts whether it be
folded two or more ply. Thus, we have it stated as—
40 × 840
———————— = 60's counts in worsted.
560
40 × 840
———————— = 131·25 counts skeins.
256
_Example 10._—Find the yards of yarn in 3 lb. of 2/48's worsted and ¼
oz. of 60/2 spun silk respectively.
In 2/48's worsted 24 × 560 = yards in 1 lb., ∴ 24 × 560 × 3 = 40,320 yd.
in 3 lb. 60/2 spun silk = 60 × 840 = yards in 1 lb. or 16 oz., and to
obtain the length in ¼ oz. divide by 4 × 16—
60 × 860
———————— = 787·5 yd.
4 × 16
_Weight of Knitted Fabrics._—These calculations lead to examples where
the weight of knitted fabric has to be found. The ordinary plain knitted
loop in which the bulk of textures is worked consists really of a weft
structure, that is, the yarns run predominantly crosswise, and are
intersected with the preceding loops in the manner already described. In
determining the weight of a given length of plain knitted fabric we
require various factors, these being taken as they are on the frame. It
is essential in the first place to know the counts of yarn employed, and
the number of courses inserted per inch into the fabric, and again it is
essential to know the width at which the fabric is being worked on the
machine. Finally, it is necessary to estimate what is known as the
"take-up," for the yarn is pushed into curved formation which "takes up"
yarn about twice the width of the fabric by the intersecting of the yarn
over the needles and this has a very definite influence on the weight.
_Example 11._—Find the weight of 10 yd. of knitted fabric made from
2/20's worsted yarn with 18 courses per inch at a width of 32 in. on the
needles. The take-up is 2, that is, to form one course of loops, a
length of yarn equal to twice the width is required.
If the question of take-up be ignored for the moment, let the yarns be
inserted as weft threads crosswise in the fabric and we shall have in 1
in. of cloth 18 courses or threads each 32 in. wide. This gives (18 ×
32)/36 = yards of yarn in 1 in. of cloth or (18 × 32 × 36)/36 yards of
yarn in 1 yd. of cloth × 10 for 10 yd., but from the yarn counts we know
that the size is such that 2/20's worsted = 10 × 60 = yards in 1 lb., so
that dividing the latter by this number of yards will give the weight of
the fabric in pounds, thus—
18 × 32 × 36 × 10
————————————————— = 1·03 lb. as the weight of 10 yd. of fabric.
36 × 10 × 560
But this is the weight if the yarns are straight in the fabric, which
they are not, for there is a take-up of 2, that is, the weight has to be
doubled—
1·03 × 2 = 2·06 lb. weight.
From this concrete example may be derived a formula which can be applied
to all cases where the sufficient particulars are given, and following
the above example we obtain—
_Example 12._—
Courses per in. × width × length × take-up
—————————————————————————————————————————— = weight in lb.
counts × basis
In this statement let the courses per inch or the sett be represented by
_S_, the width by _w_, and the weight by _W_, counts = _C_, basis = _B_,
length = _L_, from which we obtain the following equation—
_Example 13._—
_S_ × _w_ × _L_ × _T_
————————————————————— = _W_,
_C_ × _B_
or-
_S_ × _w_ × _L_ × _T_ = _W_ × _C_ × _B_.
From this it follows that given any six of the seven factors we may
obtain the seventh by substitution of values. Some of these
possibilities are of academic interest only and are seldom required in
practice, but a few examples may be given of the use of this formula.
_Example 14._—Find the weight of 200 yd. of knitted webbing worked from
2/40's cashmere yarn with 28 courses per inch to a width of 48 in. with
a take-up of 1·75.
Substituting as in formula 12—
28 × 48 × 200 × 1·75
———————————————————— = 42 lb. weight.
20 × 560
_Example 15._—Calculate the weight of fleecy fabric worked one thread
2/30's worsted yarn on face with 12 skeins yarn on back; the worsted has
a take-up of 1·75, whilst the back yarn take-up is 2. There are 24
courses per inch of each thread, the length is 150 yd. and the width
equivalent to 60 in.
For this example it is most expeditious to work out each yarn separately
according to formula 12. For the face yarn the items will be stated
thus—
24 × 60 × 150 × 1·75
———————————————————— = 45 lb. weight,
15 × 560
for the woollen—
24 × 60 × 150 × 2
————————————————— = 140·6 lb. woollen.
12 × 256
These added give 45 + 140·6 = 185·6 for the total weight in pounds.
_Example 16._—Find the weight of a fabric plated as follows—
1-40/2 spun silk with take-up of 1·75
1-2/32's merino with take-up of 2·25
Length 320 yd., 24 courses per inch, 60 in. wide.
For the silk the counts 40/2 are taken as 40's single—
24 × 60 × 320 × 1·75
———————————————————— = 24 lb.
40 × 840
For merino—
24 × 60 × 320 × 2·25
———————————————————— = 77·14 lb. Total 101·14 lb.
16 × 840
_Example 17._—Find the merino counts of yarn to produce 72 yd. of fabric
56 in. wide, 18 courses per inch, 24 lb. of material with a take-up in
knitting of 2½.
The formula for this type of calculation can be derived from that given
for finding the weight, all items being the same except that the weight
is substituted for the counts.
Following formula 12—
18 × 56 × 72 × 2½
————————————————— = 9's counts single or 2/18's.
24 × 840
_Example 18._—Calculate the yarn counts in the Alloa system for a fabric
180 yd. long with 14 courses per inch, 66 in. wide, the take-up is 2 and
the weight 24 lb.
14 × 66 × 180 × 2
————————————————— = 57¾ Alloa.
24 × 240
_Example 19._—Estimate the skein counts for a fabric 200 yd. long, 80
lb. in weight, 60 in. wide, 12 courses per inch, with a take-up of 2·25.
12 × 60 × 200 × 2·25
———————————————————— = 15·8 skeins counts.
80 × 256
_Example 20._—Give the worsted counts to reduce the weight to 60 lb. for
example 19.
12 × 60 × 200 × 2·25
———————————————————— = 9·6 worsted or 2/19's approx.
60 × 560
_Example 21._—Calculate the length of knitted fabric which can be
secured from 30 lb. of 2/42's cashmere, 22 courses per inch, 63 in.
wide, take-up 1½.
This is obtained from formula 13 and may be stated thus—
weight × counts × basis
————————————————————————— = length.
courses × width × take-up
30 × 21 × 560
————————————— = 170 yd. approx.
22 × 63 × 1½
_Example 22._—Find the length per 100 lb. obtained in 2/30's merino,
take-up 1·75, 21 courses per inch, 64 in. wide.
100 × 15 × 840
—————————————— = 535·7 yd.
21 × 64 × 1·75
_Example 23._—Find the length of rib fabric with a take-up of 3 obtained
from 100 lb. of 20/20 Scotch woollen counts, 15 courses per inch, 60 in.
wide.
100 × 10 × 200
—————————————— = 74 yd.
15 × 60 × 3
_Example 24._—Find the width at which a fabric will require to be worked
to use 80 lb. of 2/16's worsted counts, take-up 2½, length 240 yd., 18
courses per inch.
This formula is identical to that used to find the length, except that
the length is substituted for the width—
weight × counts × basis
—————————————————————————— = width.
courses × length × take-up
80 × 8 × 560
————————————— = 33·19 in. wide.
18 × 240 × 2½
_Example 25._—Find the width for a cotton fabric weighing 40 lb., 120
yd., 2/32's counts, 21 courses per inch, take-up is 2½.
40 × 16 × 840
————————————— = 85·3 in. wide.
21 × 120 × 2½
_Example 26._—Find the courses per inch required for a fabric worked in
18 skeins, 60 in. wide, take-up 175, 120 yd., 50 lb. weight. This is
obtained from the same formula as Examples 24 and 25 except that the
width is substituted for courses per inch.
50 × 18 × 256
——————————————— = 18·3 courses per inch.
60 × 120 × 1·75