Knitting is one of the most used fabric production techniques, where yarns are interlooped to form the flat fabric. Two types of knitting exist, namely, warp knitting and weft knitting. A weft-knitted fabric can be produced using one single strand of yarn Figure-1(a). while a warp-knitted fabric or a woven fabric requires the same number of yarns as required through the width of the fabric Figure-1(b). Knit fabric production calculation has been explained in detail as below.

Knit Fabric Production Calculation:

👉1. Production in length (meter/hour):

$= frac{{Course/hour}}{{Course/meter*Feeder/course}}$

$= frac{{Machine RPM*60*No.of.feeders*Efficiency% }}{{Course/inch*39.37*Feeder/course}}$

$= frac{{Machine.RPM*No.of.feeders*Eff% }}{{Course/inch*Feeder/course}}x1.524$

$= frac{{Machine.RPM*No.of.feeders*Eff% }}{{Course/cm*2.54*Feeder/course}}x1.524$

$= frac{{Machine.RPM*No.of.feeders*Eff% }}{{Course/cm*Feeder/course}}x0.60$

Example: a circular knitting machine has 30” diameter, 24 gauge and 96 feeders. It is producing a single jersey fabric with 15 course per cm. If the machine runs at 22 rpm and 85% efficiency, calculate the length of the fabric produced in one hour.

Solution:

Here, Machine RPM= 22, Course/cm= 15, No of feeders= 96, Feeder/course for single jersey fabric= 1, Efficiency= 85%

Therefore,

$Pr oduction = frac{{Machine.RPM*No.of.feeders*Eff% }}{{Course/cm*Feeder/course}}x0.60$

$= frac{{22*96*0.85}}{{15*1}}x0.60$

$= 71.81meter/hour$

👉2. Fabric width (meter):

$= frac{{pi *Machine.dia.in.inch*Machine.gauge}}{{Wales/cm*100}}$

Example: a circular knitting machine has 30” diameter, 28 gauge and 96 feeders. It is producing a single jersey fabric with 15 course per cm and 20 wales per cm. If the machine runs at 20 rpm and 85% efficiency, calculate the width and length of the fabric produced in one hour.

Solution: Here,

Machine RPM= 20, Course/cm= 15, Machine dia= 30 inch, Wales/cm= 20, Machine gauge= 28, Feeder/course for single jersey fabric= 1, No of feeders= 96, and Efficiency= 85%.

Therefore,

$(i).Fabric.production = frac{{Machine.RPM*no.of.feeders*Eff% }}{{Course/cm*feeder/course}}0.60meter/hour$