Illustrative example |
L effective length |
| S, % fibres shorter than 12-mm. |
| B % fibres longer than 24-mm |
| H fibre-fineness, millitex, not micronaire |
| Z Stelometer zero-gauge fibre-bundle tenacity, and |
| F Stelometer 1/8-in. tenacity by. |
| Then |
 |
Table II - A– 3 - i gives, by way of illustration, the numerical values of each step in the calculation. |
|
Table II – A - 3 -i: Illustrative Calculation Of CSP from The General Equation |
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 |
|
|
|
U0 |
1.533742 |
3.052998 |
|
|
|
U |
5.115905 |
23.37368 |
|
|
|
F2 |
0.943691 |
0.703154 |
|
|
|
|
| Note TO Table II – A - 3 -i: Illustrative Calculation Of CSP from The General Equation |
More decimals than are really required are given to help you verify your computer output. |
***When we have to deal with a new spinning set-up, we have to estimate the numerical constants in F(1). In such a situation to get accurate enough values we have to spin each of five disparate cottons to three different counts, and each count at one convenient twist multiplier. We then proceed with the calculation upto the stage, |
DEN =208.35xZxfmax. |
Then, |
OBS X=(OBS CS)/(F2xf3xF4)
|
 |
We calculate the value of Y for each count of yarn spun from each cotton. We then prepare a plot of the 15 Y-values against the corresponding Q – we will have three points over each Q. We draw the best fitting straight line through the points. The intercept of the line =l; the slope of the line is = m. As an alternative method we can use the ‘solver’ of the computer to get l and m |
