**Some Essential Calculations for Knitting**

**Md. Abu Sayed**

Founder of Textile Apex

Pabna Textile engineering college, Pabna, Bangladesh

**Email:**sayed1952@facebook.com

**Cell :**+8801745214773

**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 =?

**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 Wales in fabric Gauge X π X Dia

Width of fabric Width of fabric

= (14 X 3.14 X 30)/264 = 4.995 = 5

Again,

Course/min No. of feeder X m/c speed

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

- No. of feeder – 48
- Fabric width – 264 cm
- Stitch density – 15
- Machine speed – 20 rpm
- Gauge – 14
- Efficiency – 75%
- Production/8hr =?

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 = 3Course/min No. of feeder X m/c speed

**Production**=**…………………….**X Efficiency =**…………………………**X EfficiencyCourse/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 =?

**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

**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 =?

**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)**