**Production Related Formulas, Problems and Its Solutions of Knitting**

**Md. Abu Sayed**

Founder of Textile Apex

Pabna Textile engineering college, Pabna, Bangladesh

**Email:**sayed1952@facebook.com

**Cell :**+8801745214773

**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 of them are given below –

**Formulas for the determination of production 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 Needle π X Cylinder Dia X Gauge

=

**……………………………**=

**………………………………………..**

Wales/cm Wales/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 its 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 =?

**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 a 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 =?

**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