# Knitting Production Formulas, Problems and Solutions

Introduction:
The production of a circular knitting machine is done by both length and weight. There are several types of formulas for the determination of production. Some knitting production formulas, problems, and solutions are given below– ### Production Related Formulas, Problems and Its Solutions of Knitting

Knitting Production Formulas for the determination in length:

………………………………Course/Minute
1. Fabric length = …………………………………
…………………………………Course/cm

2. Course/Min = No. of feeder X Cylinder speed

3. Stitch density = Course/cm X Wales/cm

No. of Wales
4. Fabric width = ………………………….
Wales/cm

……No. of Needleπ X Cylinder Dia X Gauge
= …………………………… = ………………………………………..
……Wales/cmWales/cm

Formulas for the determination of production in weight:

Course/min X Yarn length per course
1. Fabric weight = …………………………………………………………
Yarn count

2. Yarn length per course = Total needle no. of machine X Loop length

3. Total needle no. m/c = π X Cylinder dia X Gauge

Some mathematical problems and their solutions:

1. Calculate the production per day in kg of a plain single jersey knitted at 30-inch diameter, 24 gauge circular machine having 96 feeders and 0.25 cm stitch length produced by 30/1’s. The machine operates at 25 rpm at 70% efficiency.

Solution:
Here data given,

• Machine dia = 30”
• Machine gauge = 24
• No. of feeders = 96
• Stitch length = 0.25 cm
• Yarn count = 30/1’
• Machine rpm = 25
• Efficiency = 70%
• Now, length of yarn in a loop = 0.25 cm

So, Length of yarn in full course = 0.25 X π X G X d cm

= 0.25 X π X 24 X 30 cm

So, Length of yarn used in a minute for producing course,

= 0.25 X π X 24 X 30 X 96 X 25 cm

We get, Production per day at 70% efficiency,

0.25 X π X 24 X 30 X 96 X 25 X 60 X 24 X 70
= ………………………………………………………………………
2.54 X 36 X 840 X 30 X 100 X 2.2

= 269.85 kg (ANS)

Or,

RPM of cylinder X No. of feeder X π X Cylinder dia (inch)
Production in weight = ………………………………………………………………………………………….
……1000 X 1000

…………Gauge X Loop length (mm) X tex X 60 X 24 X Efficiency
X ………………………………………………………………………………………….
………………………………………1000

………..25 X 96 X π X 30 X 24 X 0.25 X 10 X 590.5/30 X 60 X 24 X 0.70
= ………………………………………………………………………………………………………
………..………..………..………..1000 X 1000 X 1000

= 269 kg/day (ANS)

2. Calculate the production of a single jersey circular knitting m/c per shift from the following data –
Here,

• Cylinder dai = 30”
• Cylinder speed = 20 rpm
• No. of feed = 36
• No. of course per inch = 30
• Machine eff = 80%

Solution:

Here the given data,

• Cylinder dai = 30”
• Cylinder speed = 20 rpm
• No. of feed = 36
• No. of course per inch = 30
• Machine eff = 80%
• Production/8 hr =?

We know,
No. of course per min = No. of feeder X Cylinder speed

= 36 X 20 = 720

Again,
………………………No. of course per min
Production = ………………………………….. X Machine efficiency
……………………..No. of course per inch

= 720/30 X 80/100 inch/min

= 720/30 X (60 X 8)/36 X 80/100 gauge/8hr

= 256 gauge/shift (ANS)

3. Determine the no. of course per cm of fabric from the following data. This fabric is produced 1152 m per shift in a single jersey circular knitting machine.

Here,

• Knitting m/c speed = 20 rpm
• No. of feeder = 48
• Machine efficiency = 75%

Solution:

Given,

• Knitting m/c speed = 20 rpm
• No. of feeder = 48
• Machine efficiency = 75%
• Length of produced fabric per shift = 1152 m
• No. of course per cm =?

We know,

Fabric production per shift,

……Course per min
= …………………………. X 60 X 8 X Efficiency
…..Course per cm

……….No. of feeder X Machine speed X 60 X 8
= ………………………………………………………………….. X Efficiency
……………………………..Course per cm

……………………………..48 X 20 X 60 X 8…………75
Or, 1152 X 100 = ………………………………… X ……………
………………………………..Course/cm……………….100

…………………………….48 X 20 X 60 X 8 X 75
Or, Course/cm = ………………………………………. = 3 (ANS)
…………………………….1152 X 100 X 100

```Author of this Article:
Md. Abu Sayed
Founder of Textile Apex
Pabna Textile engineering college, Pabna, Bangladesh
Email: [email protected]
Cell : +8801745214773```

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