Question

If $W$ and $S$ be the wales per inch and fraction shrinkage of a knitted fabric then the machine gauge $G$ will be

• A. $W(1+S)$
• B. $W(1-S)$
• C. $\dfrac{W}{(1-S)}$
• D. $\dfrac{W}{(1+S)}$

Hint:

In knitting calculations, always remember that shrinkage increases fabric density. Machine settings must account for this increase to achieve the desired final fabric structure.

Answer 1

The Correct Option is C

Step 1: Understanding the terms.
Wales per inch ($W$) represents the number of wales measured in the finished fabric after shrinkage. Fractional shrinkage ($S$) indicates the proportion by which the fabric dimensions reduce during finishing. Machine gauge ($G$) refers to the number of needles per inch on the knitting machine before shrinkage occurs.
Step 2: Relating fabric shrinkage to machine gauge.
Due to shrinkage, the wale density in the finished fabric increases. Therefore, the wale density on the machine must be lower than the final measured value. The relationship between machine gauge and finished wale density is:
\[ W = G(1-S) \] Step 3: Rearranging the formula.
To find the machine gauge, rearrange the equation:
\[ G = \frac{W}{(1-S)} \] Step 4: Conclusion.
Hence, the correct expression for machine gauge in terms of wales per inch and shrinkage is $\dfrac{W}{(1-S)}$.

Answer 2

To relate machine gauge to the wales measured on the finished fabric, start from the physical picture of what happens as a knitted fabric relaxes after coming off the machine.

On the machine, the needle spacing sets the "as-knitted" wale density, equal to the machine gauge \( G \) (needles, and therefore wales, per inch). Once the fabric is taken off the machine and allowed to relax, the loops reorganize into a more balanced, rounder shape. This relaxation causes the fabric to contract lengthwise but spread out widthwise, which pulls the wales apart and reduces the number of wales counted per inch in the finished fabric compared to the machine setting.

If the fractional shrinkage (contraction) associated with this relaxation is \( S \), the finished wales per inch \( W \) works out to a fraction \( (1-S) \) of the original machine gauge:
\[ W = G(1-S) \]

Rearranging this to express the machine gauge needed to obtain a target finished wales-per-inch value:
\[ G = \frac{W}{1-S} \]

Since \( (1-S) \) is less than 1, this correctly predicts that the required machine gauge \( G \) must be somewhat higher than the wales per inch eventually measured on the relaxed, finished fabric - consistent with the widening/relaxation behaviour described above.

Comparing to the other options: \( W(1+S) \) and \( \dfrac{W}{(1+S)} \) would imply the machine gauge relates to \( W \) through an increase in \( S \)'s effect in the wrong direction for a contraction, and \( W(1-S) \) would make \( G \) smaller than \( W \), which is the reverse of what the relaxation behaviour requires.

Therefore, the correct answer is \( \dfrac{W}{(1-S)} \).