Knitting Calculations
Knitting speed
π×𝑑×𝑛
Knitting speed, V = 39.37 × 60
Where,
π = 3.1416
d = cylinder diameter in inch
n = cylinder rpm
39.37 =conversion factor from inch to meter
60 = conversion factor from minutes to hour
Example:
If the cylinder diameter is 30 inch and the cylinder rpm is 35 then find out the knitting speed.
3.1416 × 30 × 35
Knitting speed, V = 39.37 × 60
= 1.396 m/hour
The above equation can be simplified:
3.1416
Factor = 39.37 × 60
= 0.00133
Thus, V = 0.00133 x d x n m/hour
For calculating the machine rpm from the knitting speed, the above equation can be transposed as follows:
V
The machine rpm, n =
0.00133 × d
Example:
If the knitting speed is 1.63 m/hour and the cylinder diameter is 26 inch, then find out the machine rpm.
1.63
The machine rpm, n = 0.00133 × 26
= 47.137 ~ 47 rpm
Speed factor or performance number
With the aid of the speed factor (SF) or the performance number (L), circular knitting machines with different
systems/feeders and operating speeds can be compared in productivity terms. The performance number (L) is
calculated in accordance with the equation as follows:
L = S x n courses/min
A circular knitting machine with 72 systems (no. of feeders) and an rpm of 25 per min. has therefore a
performance number of:
L = 72 x 25 = 1800 courses/min
The performance number 'L‘ indicates therefore the theoretical number of courses/min. produced by the
circular knitting machine.
SF = L = S x n courses/min
SF
n= per minute
S
Example:
If the speed factor (SF) is 2520 courses/min and system count i.e. no. of feeder is 84, then find out the
machine rpm.
SF 2520
n = = = 30 rpm
S 84
Performance and efficiency
Performance or efficiency plays a decisive role in perfect productivity calculations and cost accounting data
for all machines used in warp and weft knitting. By performance, we understand actual output expressed as a
% of theoretical output.
Actual output
Performance = × 100 %
Theoretical output
Due to various influencing variables, performance is always less than 100% or expressed as a decimal, less
than 1.
Production Calculations
The following are the important parameters which decide the production calculations of circular weft
knitting:
Machine Parameters:
Machine speed (rpm)
Machine diameter (inch)
Machine gauge
No of feeders
Machine efficiency
Number of needles
Yarn and fabric parameters:
The construction (e.g. single-jersey, rib, I/L, purl etc.)
The course density or courses/cm, and
The weight per unit area in gm/m²
Machine output
The machine capacity or performance in running m/hr is calculated in accordance with the following equation:
Speed of machine in rpm x No. of feeders. x Efficiency x 60 minutes
Machine capacity, L = No of courses per cm. x 100
m/hr
Example
Calculate the length in meters of a plain, single sided or single-jersey fabric knitted at 20 courses/cm. on a 30
inch diameter 22 gauge circular machine having 108 feeds. The machine operates for 8 hours at 36 rpm at 87%
efficiency.
36 x 108 x 87 x 60 x 8
L= 20 x 100 x 100
= 811.82 meters
Example:
Calculate the length in meters of a plain, single sided or single-jersey fabric knitted at 16 courses/cm. on a 26
inch diameter 28 gauge circular machine having 104 feeds. The machine operates for 8 hours at 29 rpm at 95%
efficiency. [Machine capacity i.e. the total length of the fabric in metres].
Speed of machine in rpm x No. of feeders x Efficiency x 60 minutes
L=
No. of courses per cm x 100
29 x 104 x 95 x 60 x 8
= 16 x 100 x 100
= 859.56 meters
Fabric width
The fabric width in meter is calculated in accordance with the following equation:
3.1416 x Cylinder diameter in inch x Machine gauge
Fabric width = No. of wales per cm. x 100
Example:
If the cylinder diameter is 30 inch, machine gauge is 32 and the wales per cm is 14, then
3.1416 x 30 x 32
Fabric width = = 2.153 metres
14 x 100
Production capacity
If the production capacity (P) of a circular knitting machine is to be calculated in kg/hr., it can be
calculated in accordance with the following equation:
Running length in metre per hour x Fabric width in metre x Weight in GSM
Production capacity, P = kg/hr
1000
Example:
If the production in running metres per hour is 63.76 meters, fabric width is 1.76 meters and the fabric
weight is 160 gm/m² , then
63.76 x 1.76 x 160
Production capacity, P = = 17.95 kg/hr.
1000
PRODUCTION EXAMPLE
Plain circular knitting machine:
Values of circular knitting machine: Values of article:
Machine diameter 30 inch Structure: plain (Single-jersey)
Gauge E 28 Yarn: cotton Nm 50/1 (Ne 29.6/1)
Number of feeders S = 96 Course density 18 courses/cm.
Machine speed n = 35 rpm Wales density 13 wales/cm.
Machine efficiency = 85% Fabric weight 125 gm/m2
n x S x Efficiency x 60 x hour 35 x 96 x 0.85 x 60
Machine performance L in meter per hour = = = 95.2 m/hr
Courses/cm x 100 18 x 100
3.1416 x Cylinder diameter in inch x Machine gauge 3.1416 x 30 x 28
Fabric width in meter = = = 2.03 m
Wales per cm.x 100 13 x 100
Length in meter per hour x Fabric width in metre x Weight in GSM
Machine performance in kg per hour = kg/hr
1000
95.2 x 2.03 x 125
= kg/hr = 24.157 kg/hr
1000
Interlock circular knitting machine:
An interlock fabric comprising, in the simplest case, two part courses. These part courses complement each other
to make a full course, and therefore two systems or feeders are required for producing one course.
The following data were assumed for the interlock fabric production:
Example 1
Values of circular knitting machine: Values of article:
Machine diameter 30 inch Structure: plain interlock
Gauge E 28 Yarn: polyester D 76/1
Number of feeders 96 Course density 17 courses/cm
Machine speed 31 rpm Wales density 14 wales/cm
Machine efficiency 85% Fabric weight 100 gm/m2
n x S x Efficinecy x 60 x hour 31 x 96 x 0.85 x 60
Machine performance L in meter per hour = feeders/course x courses/cm x 100 = = 44.64 m/hr
2 x 17 x 100
3.1416 x Cylinder diameter in inch x Machine gauge 3.1416 x 30 x 28
Fabric width in meter = = = 1.88 m
Wales per cm. x 100 14 x 100
Length in metre per hour x Fabric width in metre x Weight in GSM
Machine performance in kg per hour = kg/hr
1000
44.64 x 1.88 x 100
= kg/hr = 8.39 kg/hr
1000
Example 2
Values of circular knitting machine: Values of article:
Machine diameter 30 inch Structure: plain interlock
Gauge E 42 Yarn: polyester filament yarn dtex 50
Number of feeders 108 Course density 19 courses/cm
Machine speed 31 rpm Wales density 23 wales/cm
Machine efficiency 87% Fabric weight 100 gm/m2
n x S x Efficiency x 60 x hour 31 x 108 x 0.87 x 60
Machine performance L in meter per hour = feeders/course x courses/cm x 100 = = 45.99 m/hr
2 x 19 x 100
3.1416 x Cylinder diameter in inch x Machine gauge 3.1416 x 30 x 42
Fabric width in meter = = = 1.72 m
Wales per cm. x 100 23 x 100
Length in meter per hour x Fabric width in meter x Weight in GSM
Machine performance in kg per hour = kg/hr
1000
44.99 x 1.72 x 100
= kg/hr = 7.91 kg/hr
1000
Areal Density/Weight per unit area
By knowing the stitch density (CPI × WPI), loop length (l), and yarn count, the areal density of the knitted
fabric can be calculated as,
CPI × WPI × l (mm)
Areal density/GSM = × 0.9158
Ne
CPI × WPI × l (cm)
Areal density/GSM = × 9.158
Ne
CPI × WPI × l (mm)
Areal density/GSM = × 1.55
Nm
Areal density/GSM = CPI × WPI × l mm × Tex × 0.00155
Areal density/GSM = CPI × WPI × l mm × Denier × 0.00017
Thank you!
sonjitkumar@buft.edu.bd
