Some Essential Calculations for Knitting
Introduction:
Knitting is completely calculation based production. Before production any item knitter must calculate first. If calculation is done perfectly then production efficiency will increase. Now I’m giving some problems of knitting for helping purpose novice knitting muster. In previous article I have written about different necessary formula for knitting. You can read it to understand following problems.
Calculations for Knitting
Problem-1: Calculate the production per hour (in gauge) of a circular knitting machine from the following data –Cylinder dia – 30”
Cylinder speed – 20 rpm
No. of feeder – 36
Efficiency – 80%
Course/inch – 30
Course/rpm – 36
Solution
Given,
Cylinder dia – 30”
Cylinder speed – 20 rpm
No. of feeder – 36
Efficiency – 80%
Course/inch – 30
Course/rpm – 36
Production/8 hour =?
We know that,
Course/min = No. of feeder X Cylinder speed
= 36 X 20
= 720
Again,
Course/min
Production = ……………………… X Efficiency
Course/inch
= (720/30) X (80/100) inch/min
= (720/30) X (60/30) X (80/100) gauge/hr
= 32 gauge (ANS)
Problem-2: Calculate the production per 8hr (in meter) of a circular knitting machine from the following data –
No. of feeder – 48
Fabric width – 264 cm
Stitch density – 15
Machine speed – 20 rpm
Gauge – 14
Efficiency – 75%
Solution:
Given,
No. of feeder – 48
Fabric width – 264 cm
Stitch density – 15
Machine speed – 20 rpm
Gauge – 14
Efficiency – 75%
Production/8hr =?
We know that,
No. of Wales in fabric Gauge X π X Dia
Wales/cm = …………………………………... = ……………………………….
Width of fabric Width of fabric
= (14 X 3.14 X 30)/264 = 4.995 = 5
Again,
Course/cm = (Stitch density)/(Wales/cm) = 15/5 = 3
Course/min No. of feeder X m/c speed
Production = ……………………. X Efficiency = ………………………… X Efficiency
Course/cm Course/cm
= (48 X 20)/3 X .75 cm/min
= (48 X 20)/3 X (60 X 8)/100 X .75 m/8hr
= 1152 m/8hr (ANS)
Problem-3: Calculate the production per day (in gauge) of a circular knitting machine from the following data –
Machine speed – 25 rpm
Machine gauge – 28
Machine dia – 24”
No. of feeder – 96
Course/cm – 16
Efficiency – 90%
Solution:
Given,
Machine speed – 25 rpm
Machine gauge – 28
Machine dia – 24”
No. of feeder – 96
Course/cm – 16
Efficiency – 90%
Production/24hr =?
We know,
Course/min = No. of feeder X Machine speed = 96 X 25 = 2400
And, Course/inch = Course/cm X 2.54 = 16 X 2.54 [2.54 cm = 1 inch]
= 40.64
Again,
Course/min
Production = …………………….. X Efficiency
Course/inch
= (2400/40.64) X 0.90 inch/min
= (2400/40.64) X (60 X 24)/36 X 0/90 Gauge/24 hr
= 2126 gauge/24 hr (ANS)
Problem-4: Calculate the production per shift (in kg) of a single jersey circular knitting machine from the following data –
Cylinder speed – 35 rpm
No. of feeder – 96
Cylinder dia – 30 inch
Machine gauge- 28
Efficiency – 85%
Course density – 18 course/ cm
Wales density – 13 Wales/cm
Fabric wt. – 125 gm/m2
We know that,
Fabric length per shift
No. of feeder X Machine speed
= …………………………………………………….. X 60 X 8 X (85/100)
Course/cm
= (96 X 35)/18 X 60 X 8 X (85/100) m/shift
= 761.6 m/shift
Total amount of walse
Fabric width = …………………………………….
Wales/cm
Π X Machine dia X Machine gauge
= …………………………………………………
Wales/cm
= (3.14 X 30 X 28)/13 = 202.89 cm
= 2.03 m
Total wt. of fabric per shift
= Total length of fabric X fabric width X Wt. of fabric
= 761.6 X 2.03 X 125 gm/m2
= (761.6 X 2.03 X 125)/1000 kg/m2
= 193.26 kg/m2 (ANS)
Problem-5: In a circular knitting machine, machine speed = 135 rpm, no. of feeder = 18, Produced fabrics course/cm = 28, Wales/cm = 24. Calculate the production per 8 hr and fabric density.
Solution:
Given,
Machine speed = 135 rpm
No. of feeder = 18
Course/cm = 28
Wales/cm = 24
Production/8 hr =?
Stitch density =?
We know that,
Stitch density = Course/cm X Wales/cm = 28 X 24 = 672 (ANS)
No. of feeder X Machine speed
Production = ……………………………………………………
Course/cm
= (18 X 135)/28 cm/min
= (18 X 135)/28 X (60 X 8)/100 m/8hr
= 416.57 m/8hr (ANS)
