Shedding Mechanism and Gearing calculation
Shedding Mechanism and Gearing of Loom calculation
Introduction:
Shedding is the first primary motion of the loom and Dividing of warp threads into two parts for insertion of weft thread is called shed, and the mechanism to form shed is called shedding.
There are two types of shedding – 1) Positive shedding and 2) Negative shedding.
Basic shedding mechanisms are Paddle/treadle, Tappet, Dobby and Jacquard. calculations play an important role in shedding process. Some important calculation (Crank Shaft RPM, Counter Shaft RPM, Bottom Shaft RPM, PPM, Machine RPM, Production of Loom, Driver or Driven Shaft RPM etc.) of gearing and shedding mechanism of a loom are discussed below.
Calculation of Gearing and Shedding Mechanism of Loom:
- Find the RPM of bottom shaft, counter shaft and PPM from the following data-
RPM of Crank Shaft = 800
Picks to the round = 4
Solution:
We know,
PPM (Picks Per Minute) = RPM of Crank Shaft
PPM = 800 (Ans)
…………………………………………PPM
RPM of Counter Shaft = ————————
…………………………………No of Tappets
Given, Picks to the round = No of tappets = 4
………………………………………800
So, RPM of counter shaft = ———- = 200(Ans)
………………………………………..4
Again, we know that,
…RPM of Crank Shaft 2
———————————– = ———–
…RPM of Bottom Shaft 1
………………………………………RPM of Crank Shaft
Or, RPM of Bottom shaft = —————————————
……………………………………………………2
….800
= ————-
…..2
=400 (Ans)
- Find out the PPM of the loom from following information-
- Motor RPM = 800
- Motor Pulley Dia = 3”
- M/C Pulley Dia = 8”
- No of Tappet = 4
- RPM of Counter Shaft = 78
- Crank Shaft Wheel = 74T
Solution:
We Know,
…………………………………………..PPM
RPM of Counter Shaft = —————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=78x 4
= 312
RPM of Crank Shaft = 312 (Ans)
- Find out the production per hour of a loom in which the auxiliary shaft is running at 40 rpm and there are 4 picks to the round.
Assume,
PPI = 40 and
Efficiency = 90%
Solution:
We Know that,
…………………………………………..PPM
RPM of Counter Shaft = —————————
……………………………………No of Tappets
So, PPM = RPM of Counter Shaft x No of Tappets
= 40 x 4
= 160
……………………………………….PPM
So, Production of loom = ————— x Efficiency
……………………………………….PPI
….160 90
= ——– x ————– inch/min
….40 100
….160 x 90 x 60
= —————————- yds/hr
….40 x 100 x 36
= 7.3 yds/hr (Ans)
- Find RPM of bottom shaft, crank shaft and PPM from the following data-
No of teeth in crank shaft = 48T
RPM of counter shaft = 80
No of tappets = 4
Solution:
We know,
……………………………………………PPM
RPM of Counter Shaft = ————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
= 80 x 4
= 320
RPM of crank shaft = 320 (Ans)
…RPM of Crank Shaft 2
———————————- = ————-
…RPM of Bottom Shaft 1
……………………………………..RPM of Crank Shaft
Or, RPM of Bottom shaft = ——————————
………………………………………………….2
….320
= ————-
…..2
= 160 (Ans)
- Tabulate the speed of crank, bottom and auxiliary shafts for a loom running at 400 PPM having four tappets on the auxiliary shaft.
Solution:
Here Given,
PPM =400
Crank shaft speed =400 rpm (Ans)
……………………………………..RPM of Crank Shaft
Or, RPM of Bottom shaft = ——————————–
………………………………………………….2
…400
= ————-
….2
=200(Ans)
………………………………………..PPM
RPM of Counter Shaft = ————————–[Counter Shaft = Auxiliary Shaft]
…………………………………No of Tappets
…400
= ———-
….4
RPM of auxiliary shaft = 100 (Ans)
- Find out the rpm of the loom from the following information-
- Motor RPM = 950
- Motor Pulley Dia = 3”
- Machine Pulley Dia = 8”
- Crank Shaft Wheel = 37 T
- Bottom Shaft Wheel = 74T
Solution:
We know that,
………………………………………………Motor Pulley Dia No of Teeth of Crank Shaft Wheel
Loom RPM = PPM = Motor RPM x —————————— x —————————————————
…………………………………………….Machine Pulley Dia No of Teeth of Bottom Shaft Wheel
…………….3 37
= 950 x ——– x ———
……………8 74
= 178.125 (Ans)
- Find out the PPM of loom from the following information –
- Motor RPM = 1440
- Motor Pulley Dia = 3”
- Machine Pulley Dia = 8”
Solution:
…………………………………Motor Pulley Dia
PPM = Motor RPM x ———————————-
………………………………Machine Pulley Dia
………………3
= 1440 x ———-
………………8
= 540 (Ans)
- A wheel of 80 teeth fixed on a shaft of 120 rpm drives a wheel of 50 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 120
- Driver Shaft Wheel Teeth = 80 T
- Driven Shaft Wheel Teeth = 50 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………80
= 120 x ——- = 192 (Ans)
9.A wheel of 90 teeth fixed on a shaft of 130 rpm drives a wheel of 60 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 130
- Driver Shaft Wheel Teeth = 90 T
- Driven Shaft Wheel Teeth = 60 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………90
= 130 x ——- = 195 (Ans)
60
10 Find out the PPM of the loom from following information-
- Motor RPM = 800
- Motor Pulley Dia = 3”
- M/C Pulley Dia = 8”
- No of Tappet = 8
- RPM of Counter Shaft =96
- Crank Shaft Wheel = 74T
Solution:
We Know,
…………………………………………..PPM
RPM of Counter Shaft = —————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=95x 8
= 760
RPM of Crank Shaft = 760 (Ans)
11.Find out the PPM of loom from the following information –
- Motor RPM = 2440
- Motor Pulley Dia = 5”
- Machine Pulley Dia = 10”
Solution:
…………………………………Motor Pulley Dia
PPM = Motor RPM x ———————————-
………………………………Machine Pulley Dia
………………5
= 2440 x ———-
………………10
= 1220 (Ans)
12.Find the RPM of bottom shaft, counter shaft and PPM from the following data-
RPM of Crank Shaft = 600
Picks to the round = 8
Solution:
We know,
PPM (Picks Per Minute) = RPM of Crank Shaft
PPM = 600 (Ans)
…………………………………………PPM
RPM of Counter Shaft = ————————
…………………………………No of Tappets
Given, Picks to the round = No of tappets = 8
………………………………………600
So, RPM of counter shaft = ———- = 75(Ans)
……………………………………….8
Again, we know that,
…RPM of Crank Shaft 2
———————————– = ———–
…RPM of Bottom Shaft 1
………………………………………RPM of Crank Shaft
Or, RPM of Bottom shaft = —————————————
……………………………………………………2
….600
= ————-
…..2
=300 (Ans
13.Find out the PPM of the loom from following information-
- Motor RPM = 800
- Motor Pulley Dia = 3”
- M/C Pulley Dia = 8”
- No of Tappet = 6
- RPM of Counter Shaft = 86
- Crank Shaft Wheel = 74T
Solution:
We Know,
…………………………………………..PPM
RPM of Counter Shaft = —————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=86x 6
= 516
RPM of Crank Shaft =516 (Ans
14.Find out the production per hour of a loom in which the auxiliary shaft is running at 60 rpm and there are 4 picks to the round.
Assume,
PPI = 60 and
Efficiency = 90%
Solution:
We Know that,
…………………………………………..PPM
RPM of Counter Shaft = —————————
……………………………………No of Tappets
So, PPM = RPM of Counter Shaft x No of Tappets
=60 x 4
= 240
……………………………………….PPM
So, Production of loom = ————— x Efficiency
……………………………………….PPI
….240 90
= ——– x ————– inch/min
….60 100
….240 x 90 x 60
= —————————- yds/hr
….60 x 100 x 36
= 7.7 yds/hr (Ans)
15.Find RPM of bottom shaft, crank shaft and PPM from the following data-
No of teeth in crank shaft = 48T
RPM of counter shaft = 90
No of tappets = 5
Solution:
We know,
……………………………………………PPM
RPM of Counter Shaft = ————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=90 x 5
= 450
RPM of crank shaft =450 (Ans)
…RPM of Crank Shaft 2
———————————- = ————-
…RPM of Bottom Shaft 1
……………………………………..RPM of Crank Shaft
Or, RPM of Bottom shaft = ——————————
………………………………………………….2
….450
= ————-
…..2
= 225 (Ans)
16.Tabulate the speed of crank, bottom and auxiliary shafts for a loom running at 400 PPM having four tappets on the auxiliary shaft.
Solution:
Here Given,
PPM =500
Crank shaft speed =500 rpm (Ans)
……………………………………..RPM of Crank Shaft
Or, RPM of Bottom shaft = ——————————–
………………………………………………….2
…500
= ————-
….2
=250(Ans)
………………………………………..PPM
RPM of Counter Shaft = ————————–[Counter Shaft = Auxiliary Shaft]
…………………………………No of Tappets
…500
= ———-
….4
RPM of auxiliary shaft = 125 (Ans)
17.Find out the rpm of the loom from the following information-
- Motor RPM = 650
- Motor Pulley Dia = 3”
- Machine Pulley Dia = 8”
- Crank Shaft Wheel = 37 T
- Bottom Shaft Wheel = 74T
Solution:
We know that,
………………………………………………Motor Pulley Dia No of Teeth of Crank Shaft Wheel
Loom RPM = PPM = Motor RPM x —————————— x —————————————————
…………………………………………….Machine Pulley Dia No of Teeth of Bottom Shaft Wheel
…………….3 37
= 650 x ——– x ———
……………8 74
= 121.87 (Ans)
18.Find out the PPM of loom from the following information –
- Motor RPM = 2000
- Motor Pulley Dia = 3”
- Machine Pulley Dia = 8”
Solution:
…………………………………Motor Pulley Dia
PPM = Motor RPM x ———————————-
………………………………Machine Pulley Dia
………………3
=2000 x ———-
………………8
= 750 (Ans)
19.A wheel of 80 teeth fixed on a shaft of 240 rpm drives a wheel of 50 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 240
- Driver Shaft Wheel Teeth = 80 T
- Driven Shaft Wheel Teeth = 50 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………80
= 240 x ——- = 384(Ans)
20.A wheel of 90 teeth fixed on a shaft of230 rpm drives a wheel of 60 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 230
- Driver Shaft Wheel Teeth = 90 T
- Driven Shaft Wheel Teeth = 60 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………90
= 230 x ——- = 345 (Ans)
21. Find out the PPM of the loom from following information-
- Motor RPM =1000
- Motor Pulley Dia = 3”
- M/C Pulley Dia = 8”
- No of Tappet = 8
- RPM of Counter Shaft =80
- Crank Shaft Wheel = 74T
Solution:
We Know,
…………………………………………..PPM
RPM of Counter Shaft = —————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=80x 8
= 640
RPM of Crank Shaft =640(Ans)
22.Find out the PPM of loom from the following information –
- Motor RPM = 2000
- Motor Pulley Dia = 5”
- Machine Pulley Dia = 10”
Solution:
…………………………………Motor Pulley Dia
PPM = Motor RPM x ———————————-
………………………………Machine Pulley Dia
………………5
= 2000 x ———-
………………10
=1000(Ans)
23.A wheel of 90 teeth fixed on a shaft of 240 rpm drives a wheel of 50 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 240
- Driver Shaft Wheel Teeth = 90T
- Driven Shaft Wheel Teeth = 60 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………90
= 240 x ——- = 360(Ans)
60
24.A wheel of 90 teeth fixed on a shaft of230 rpm drives a wheel of 60 teeth. Find the speed of the shaft on which the driven wheel is fixed.
Solution:
Here given,
- Driver Shaft RPM = 230
- Driver Shaft Wheel Teeth = 70 T
- Driven Shaft Wheel Teeth = 70 T
- Driven Shaft RPM =?
…………………………………………………………..Driver Wheel Teeth
Driven Shaft RPM = Driver Shaft RPM x —————————————
………………………………………………………….Driven Wheel Teeth
……………70
= 230 x ——- =230 (Ans)
70
25. Find out the PPM of the loom from following information-
- Motor RPM =1000
- Motor Pulley Dia = 3”
- M/C Pulley Dia = 8”
- No of Tappet =10
- RPM of Counter Shaft =90
- Crank Shaft Wheel = 74T
Solution:
We Know,
…………………………………………..PPM
RPM of Counter Shaft = —————————–
…………………………………….No of Tappets
PPM = RPM of Counter Shaft x No of Tappets
=90x 10
= 900
RPM of Crank Shaft =640(Ans)
26.Find out the PPM of loom from the following information –
- Motor RPM = 2000
- Motor Pulley Dia = 15”
- Machine Pulley Dia = 10”
Solution:
…………………………………Motor Pulley Dia
PPM = Motor RPM x ———————————-
………………………………Machine Pulley Dia
………………15
= 2000 x ———-
………………10
=3000(Ans)
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