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The Expressions For F1,
F2, F3 , and F4: A Recapitulation |
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In the general Equation
F1 quantifies the distinctive contribution
to yarn tenacity of yarn irregularity that
is inevitable in roller drafting. F2, F3,
and F4 quantify the distinctive contribution
of twist that has necessarily to be introduced
into the drafted fleece to convert it into
formed yarn. |
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We have already noted
that for any one given system of fibre and
yarn tests, the set of numerical values
in the expressions for is the same for all
ring-spun yarns; also is unique, and the
same for all ringspun yarns. The values
of l and m, the numerical constants in F1,
the expression for the irregularity fraction,
are, understandably, specific to the drafting
system used to spin the yarn. Therefore,
any mill that wishes to use the General
Equation needs to determine the constants
in the expressions for F2 and F3 only once;
they can then keep on using these constants
for any ringspun yarn; they need to estimate
afresh only the numerical constants in F1
for each drafting system. |
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In the ATIRA investigations
all the cotton fibre length tests were carried
out by Baer sorter; the fibre bundle tenacity
tests were carried out on tufts held by
a pair of clamps, in such a way that all
the fibres in the tuft bridge the distance
between the two clamps. The yarns were all
tested for tenacity on the conventional
lea tester. The same set of numerical constants
in the expressions for F2 and F3 could,
therefore, be used to estimate the CS of
yarns from any one of the many spinnings
that were carried out in the investigations.
Only the numerical constants in F1 were
determined afresh for each drafting system. |
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Estimation of Yarn CS
Using the General Equation: a Worked-out Example |
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We denote by |
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L effective length |
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S, % fibres shorter than 12-mm |
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B % fibres longer than 24-mm |
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H fibre-fineness, millitex,
not micronaire |
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Z zero-gauge fibre-bundle tenacity with
the Stelo level calibration cotton |
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F 1/8-in. tenacity with the Stelo level
calibration cotton. |
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Then the steps in the calculation for estimating
the CSP are: |
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where l and m are specific
to a drafting system, and have, therefore,
to be determined afresh for each drafting
system, |
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Table I – 6 –
i, gives the numerical values of each step
in the calculation. In this exercise we
take l=4.9613 and m=-1.2583, the values
from one of the ATIRA spinnings. |
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Practical Application
of the General Equation |
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We can
take advantage of the fact that the expressions
for F2 and F3 are unique for any system
of fibre and yarn tests, and are applicable
to ring yarns from any different drafting
system. For any one system of cotton and
yarn testing, we need to derive the numerical
constants in the expressions for F2 and
F3 only once; we can then use them for any
ring-spun yarn; only the numerical constants
in the expression for F1 need to be determined
anew for every different drafting system.
The determination of the constants in F1
requires only the spinning of each of five
different cottons, each to three different
counts, each count at any one convenient
T.M. |
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The
procedure for determining the constants
in F2 and F3 is explained in Part II; that
for determining the constants in F1 is explained
in the next section, Table – 6 - ii. |
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We can use the general
equation to generate a wealth of information
for use in cotton selection. Immediately
after carrying out the cotton fibre tests,
the laboratory can trace out the CS –
T.M. curves for a range counts of yarn of
likely interest to the spinner. All that
we require is the storage in a computer
of the requisite programme for the numerical
evaluation of the expressions for F1, F2,
f3 and F4 and for curve tracing. |
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|
cotton |
S4 |
F414 |
cotton |
S4 |
F414 |
| L |
32.3 |
27.1 |
H |
130 |
150 |
| H |
130 |
150 |
f(H) |
0.0169 |
0.0225 |
| Z |
35.4 |
43.4 |
lm |
0.169 |
0.225 |
| F |
19.3 |
21.4 |
fmax |
0.94273 |
0.91822 |
| S |
19.8 |
19.7 |
E0 |
0.572423 |
0.745453 |
| B |
56.1 |
37.8 |
E |
0.772558 |
1.107395 |
| G |
0.377658 |
0.438537 |
lm+LM |
0.9416 |
1.3324 |
| V |
0.145259 |
0.147055 |
F(3) |
0.77835 |
0.70045 |
| W |
0.314682 |
0.233531 |
f(3) |
0.82564 |
0.76283 |
| A0 |
2.665567 |
1.994458 |
DEN |
6956.49 |
8303.46 |
| A |
36.4878 |
30.69601 |
Q |
4.26080 |
4.37552 |
| L_1 |
2.987192 |
3.515494 |
N |
90.854 |
78.740 |
| Q |
4.260801 |
4.375517 |
F(1) |
0.42863 |
0.35431 |
| M |
3.25 |
3.25 |
X |
2982 |
2942 |
| C |
50 |
50 |
F(4) |
0.871175 |
0.871175 |
| N |
90.854 |
78.74013 |
EST CS |
2024 |
1375 |
| nl |
1.508391 |
1.476093 |
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| U0 |
1.533742 |
3.052998 |
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| U |
5.115905 |
23.37368 |
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| F2 |
0.943691 |
0.703154 |
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Table I – 6
- i: Illustrative Calculation Of CSP from
The General Equation |
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Note: In this exercise
we take l =4.9613 and m=-1.2583, the values
from one of the ATIRA spinnings. |
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More decimals than
are really required are given to help you
verify your computer output. |
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Determining the Constants
in the Expression for F1 |
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Given a system of fibre
and yarn tests, for every new drafting system
we have to determine the numerical constants
in F1. In such a situation, to get accurate
enough values of the constants, we have
to spin each of five disparate cottons to
three different counts; we need to spin
each count at any one convenient twist multiplier
only. We determine the CSP of all the yarns.
We then proceed with the calculation upto
the stage, |
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in Table – 6
– ii. We calculate the value of Y
for each of the three counts of yarn spun
from each of the five cottons. We then prepare
a plot of the fifteen Y-values against the
corresponding Q –values. We will have
three points over each of the five Q’s.
We draw the best fitting straight line through
the points. The intercept of the line gives
the numerical value of l; the slope of the
line gives the value of m. Alternatively,
we can use the ‘solver’ of the
computer to get l and m. |
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|
cotton |
S4 |
F414 |
cotton |
S4 |
F414 |
| L |
32.3 |
27.1 |
U |
5.115905 |
23.37368 |
| H |
130 |
150 |
F2 |
0.94369 |
0.70315 |
| P0 |
40.8 |
50.0 |
H |
130 |
150 |
| P1/8 |
24.3 |
27.0 |
f(H) |
0.0169 |
0.0225 |
| Z |
35.4 |
43.4 |
lm |
0.169 |
0.225 |
| F |
19.3 |
21.4 |
fmax |
0.94273 |
0.91822 |
| S |
19.8 |
19.7 |
E0 |
0.572423 |
0.745453 |
| B |
56.1 |
37.8 |
E |
0.772558 |
1.107395 |
| G |
0.377658 |
0.438537 |
lm+LM |
0.941558 |
1.332395 |
| V |
0.145259 |
0.147055 |
F(3) |
0.77835 |
0.70045 |
| W |
0.314682 |
0.233531 |
f(3) |
0.825635 |
0.762834 |
| A0 |
2.665567 |
1.994458 |
F(4) |
0.871175 |
0.871175 |
| A |
36.4878 |
30.69601 |
OBS X |
3409.1 |
4705.8 |
| I |
165.175 |
485.262 |
Q |
4.260801 |
4.375517 |
| L_1 |
2.987192 |
3.515494 |
f(Q) |
4.260801 |
4.375517 |
| Q |
4.260801 |
4.375517 |
N |
90.854 |
78.740 |
| C |
50 |
50 |
DEN |
6956.49 |
8303.46 |
| N |
90.854 |
78.740 |
Y |
-0.2662 |
-0.0748 |
| nl |
1.508359 |
1.476093 |
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| M |
3.25 |
3.25 |
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| U0 |
1.533742 |
3.052998 |
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Table – 6 –
ii Determination of the Numerical Values of
the Constants In the Expression for F1 |
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We calculate the value
of Y for each of the three counts of yarn
spun from each of the five cottons. We then
prepare a plot of the 15 Y-values against
the corresponding Q-values. We will have
three points over each Q. We draw the best
fitting straight line through the points.
The intercept of the line gives the numerical
value of l; the slope of the line gives
the value of m. Alternatively, we can use
the ‘solver’ of the computer
to get l and m. |
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