Calculations in weft knitting machines

CALCULATIONS IN WEFT KNITTING MACHINE

A.MURALIKRISHNAN

LECTURER – TEXTILES

P.A.C.RAMASAMYRAJA POLYTECHNIC COLLGE,

RAJAPALAYAM, TAMILNADU, INDIA

Email: [email protected]

Compiled by A.Muralikrishnan, Lecturer, PACR polytechnic college, Rajapalayam, TN, India

Courses Courses : It is a horizontal rows of loops across the width of Fabric produced by the adjacent needles

- No of courses determines the length of fabrics

- Measured in Courses / cm or Courses/Inch

• No. of courses in 1 machine revolution = number of feeders

• No. of loops in course = no. of needles in machine


Courses

• No of courses produced in 1 rev. = No. of feeders

Example 1: A knitting machine is running at a speed of 20 rpm with 30 feeders. Calculate the total no. of courses produced in 1 hr.

Ans:

Total number of courses = 20 x 30 x 60

= 36000 courses/hr


Wales • It is a vertical column of loops produced by the Same

needles.

– Number of wales determines the width of fabric and

– measured in wales /cm or wales/Inch


Machine parametersGauge: Number of needles per 1 inch.

Ex: 18 , 20, 22 ….

M/c Diameter : Diameter of cylinder Ex: 18 “, 20 “…..

Total no. of needles in a machine :3.142 x machine cylinder diameter x gauge

Example 2: Find out the total number of needles in a machine having cylinder diameter of 20” and 26 gauge

Ans: Total number of needles = 3.142 x 20 x 26 = 1634 needles


Circumference of fabric

• Total no. of wales in a fabric

= total number of needles

Wales / inch = Machine gauge/ (1 - Shrinkage/100)

Example 3: Find the approximate circumference of fabric and WPI in a knitted fabric having 24 gauge, 20 inch diameter (assume fabric shrinkage is 35%)

Ans: Circumference of fabric = 3.142 x 20x (1-0.35) = 40.8 “

WPI = machine gauge / (1-shrinkage%) = 24/(1-0.35) = 36.92


Stitch length or Loop length: Length of yarn in loop

• ( ranges from 0.28 “ to 0.35” or 7 to 10 mm)

Course length: Length of yarn consumed for full course

during knitting.

= Total no. of loops in 1 course x Loop length

(No. of loops in course = no. of needles in m/c )

• Total no. of courses for 1 rev = Total no. of feeder

Total no. of courses for n rev = n x feeders

Total no. of loops for n rev = n x feeders x total no. of needles.

Example 4: A knitting machine having following specifications: Machine Speed– 32 rpm, 24 gauge, 30 “ cylinder diameter, 40 feeders, 85% efficiency, CPI – 45, Loop lenth – 0.28” and yarn tex – 30. Calculate

1. Total no. of courses produced per shift

2. Total length of fabric produced per shift

3. Total weight of yarn required per shift.


• Loop shape factor: CPI / WPI =1.3

• Stitch density : Total no. of loops in a given area

S.D = CPI x WPI or CPCM x WPCM

Example 5: Calculate stitch density for a knitting machine having 24 gauge, 20 “ cylinder diameter, shrinkage – 30%, Loop shape factor – 1.3

Ans: WPI = 24/0.75 = 34.28

CPI = WPI x 1.3 = 44.57

Stitch density = 1527.9 loops / sq. inch

Loop shape factor and stitch density


Tightness factor

• It is a ratio of area covered by yarn in one loop to the area occupied by the loop.

• Indicate the relative tightness or looseness of knitted fabric.

• T.F = √Tex / l where l = loop length in cm

Example 6: Find out the tightness factor of knitted fabric made of 36s count with loop length of 0.12”.


Important formulas based on constants• CPI = Course constant / LL (Kc / LL)

• WPI = Wale constant / LL (Kw / LL)

• S.D = CPI x WPI or (Course const x wale const)/ LL2

= Kc x Kw / LL2

• Fabric GSM = CPCM x WPCM x LL in cm x Tex / 10

– Or Ks x Tex / 10 x LL in cm

Example 7: Given data : Kc – 5.5 , Kw – 4.2, 32 Ne, Tightnessfactor =15, Loop length- 0.28”.

Calculate 1. CPCM , WPCM

2. Stitch density

3. GSM


Example 8: A knitting machine is running at 25 rpm with 60feeders and the fabric density is measured with CPI = 36.

WPI = 27 (14.2 c/cm; 10.6 w/cm) and the fabric weight is 173g/M2 and width -70”.

Calculate 1. Production in length per hour

2. Production in weight per hour

[1] Production in length per hour :

= (60 feeders X 25 rpm X 60 min)/(36 CPI)

= 2500 “/hr or (63.4 m/hr)

[2] The production in weight per hour :

given the finished width of fabric is 70 inches (1.78 meter)

= (63.4 m/hr X 1. 76 m X 173 g/m2 )/(1000g)

= 19.5 kg/hr (43 lb/hr)


Example 9:

• A knitting machine have 14 gauge, 20” dia, 40feeders operating at 25 rpm with 90% efficiency.Calculate the length, width and weight per hour forthe following fabrics using two fold resultant yarncount of 12s Ne.

• 1x1 rib, loop length – 0.2” , CPI = 27, WPI = 20

• Interlock, loop length – 0.12” , CPI = 37, WPI = 30


References

• Spencer D J, “Knitting Technology”, Pergamon Press, UK, 1998.

• Anbumani N , “Knitting- Fundamentals, Machines, Structures and Developments”, New Age International Private Limited, India, 2006


Engineering

Calculations in weft knitting machines