Torque and power requirement for roller

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SUMMARY

The forum discussion centers on calculating the torque and power requirements for a roller system designed to drive a fabric roll. The roller has a diameter of 90 mm, a length of 400 mm, and is connected to a 25 mm diameter shaft. The total weight of the roller and shaft is 6 kg, while the fabric roll weighs 20 kg. The motor must operate at 10 RPM, and calculations indicate that the required torque is 0.18 N·m, with further considerations needed for steady-state power requirements due to opposing torques from friction and load.

PREREQUISITES
  • Understanding of basic physics principles, including torque and power calculations.
  • Familiarity with angular velocity and its relationship to RPM.
  • Knowledge of moment of inertia and its impact on torque requirements.
  • Experience with mechanical systems involving rollers and fabric handling.
NEXT STEPS
  • Research the calculation of moment of inertia for different shapes and materials.
  • Learn about the effects of friction and opposing torques in mechanical systems.
  • Investigate motor selection criteria based on torque and power requirements.
  • Explore existing designs of fabric handling machines for comparative analysis.
USEFUL FOR

Mechanical engineers, design engineers, and anyone involved in the design and operation of fabric handling systems will benefit from this discussion.

saruvanan
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We have roller setup which is to be driven by a motor. The roller dia is 90 mm. A shaft which is connected to its end is 25 mm of dia. The roller weight is around 10 kg. On the periphery of the roller there are some 40 kgs of mass is acting. The roller rpm is 10. What is the torque required to run this roller and power required by motor. Thanks in advance
 
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Once the system you describe is operating it will take little energy to keep it going.
What is the purpose of the roller ? What does it do ?
Where will the motor energy go when the system is operating ?
Is there a belt ?
 
The roller is used to drive another roller by surface contact. The driven roller is placed directly over the driver and it sits by means of gravity. The driven roller weight is 40 kg. The driving roller weight is 10 kg. The motor is to be directly connected at the driver,s roller end. No belt. Thanks for replying baluncore. If posible i need calculation steps
 
Where does the energy go?
You have still not identified why this is being done.
 
Baluncore, its difficult application, hard to explain words. Can you help me with calculations?
 
Last edited by a moderator:
Power is energy flow per second. That requires work to be done on something.
You have not specified where the energy will go and so there can be no power calculation.
What is between the rollers?
 
saruvanan said:
Baluncore, its difficult application, hard to explain words. Can you help me with calculations?

Why are you refusing to show any work on your part? Is this for schoolwork? We do not do your schoolwork for you here...
 
saruvanan said:
The roller is used to drive another roller by surface contact. The driven roller is placed directly over the driver and it sits by means of gravity. The driven roller weight is 40 kg. The driving roller weight is 10 kg. The motor is to be directly connected at the driver,s roller end. No belt. Thanks for replying baluncore. If posible i need calculation steps

P=Tω
power = torque . angular velocity

To calc power required (for a constant speed) you need to figure out/measure the opposing torque due to whatever loads are acting and know what speed the roller needs to run at.
 
Let me rephrase my question

Dimensions
We’ve a roller of length 400mm & Dia of 90mm. At both ends of the roller a shaft is fitted. Shaft dia is 25 mm & length is 100mm each side. On the 25mm Dia shaft, two flanged bearings are fitted. The total roller & shaft weight is 6 kgs.

Working Principle:

Over the roller, a fabric in a roll form is loaded. The fabric roll sits over this roller by means of gravity. The roller has to drive the fabric roll on its circumference & feed the fabric end to another machine.

Problem:

1. What is the torque required to drive the roller at the shaft end.

2. What HP of motor, we should select to drive the shaft, if the motor is directly connected (without any gear box) to the 25mm dia shaft
 
  • #10
saruvanan said:
Let me rephrase my question

Dimensions
We’ve a roller of length 400mm & Dia of 90mm. At both ends of the roller a shaft is fitted. Shaft dia is 25 mm & length is 100mm each side. On the 25mm Dia shaft, two flanged bearings are fitted. The total roller & shaft weight is 6 kgs.

Working Principle:

Over the roller, a fabric in a roll form is loaded. The fabric roll sits over this roller by means of gravity. The roller has to drive the fabric roll on its circumference & feed the fabric end to another machine.

Problem:

1. What is the torque required to drive the roller at the shaft end.

2. What HP of motor, we should select to drive the shaft, if the motor is directly connected (without any gear box) to the 25mm dia shaft

Thread is locked for Moderation. Please PM me with the answers to my questions.
 
Last edited:
  • #11
Thread reopened for now.

@saruvanan -- do you understand what billy_joule wrote in post #8? Can you post a picture of your setup or a similar one?
 
  • #12
@Billy_joule
Power = torque. Angular velocity
Torque = radius . Force
Force = mass. Acceleration
Acceleration= velocity/time
Velocity= radius. Angular velocity
Angular velocity = (rpm.2.pi )/ 60
The roller needs to run @ 10 rpm.
Using the above formulas I've calculated this.
Hence angular velocity= 1.05 rad/secs
Velocity = 0.045 x 1.05= 0.05 m/secs
Acceleration= 0.05/3 = 0.02 m/sq.secs (0 to 10 rpm in 3 secs)
Force = (20+6).0.02x 9.8= 4.0 N ( 20 kg fabric roll, 6 kg roller self weight )
Now torque= 4.0 x 0.045 = 0.18 N.m

This is the calculation I've done. Not sure its right. I cannot attach a picture of the setup. Perhaps a drawing i can attach
 
  • #13
attached a conceptual drawing

This is a cinceptual design
 

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  • #14
saruvanan said:
@Billy_joule
Power = torque. Angular velocity
Torque = radius . Force
Force = mass. Acceleration
Acceleration= velocity/time
Velocity= radius. Angular velocity
Angular velocity = (rpm.2.pi )/ 60
The roller needs to run @ 10 rpm.
Using the above formulas I've calculated this.
Hence angular velocity= 1.05 rad/secs
Tangential Velocity = 0.045 x 1.05= 0.05 m/secs
Acceleration= 0.05/3 = 0.02 m/sq.secs (0 to 10 rpm in 3 secs)
Bolded word mine.

These calculations are all correct.

It's a little strange that you specified the rpm and not the tangential velocity. I would've expected you to need the fabric to be fed at a certain rate as that is the entire purpose of the machine. 5 cm of fabric per second seems mighty slow to me, but then again you haven't told us the purpose of the machine... Why did you select 10 rpm? Why did you select 3 seconds? why not 6? That would half the motor cost/size/energy use..


Force = (20+6).0.02x 9.8= 4.0 N ( 20 kg fabric roll, 6 kg roller self weight )
Now torque= 4.0 x 0.045 = 0.18 N.m

These are wrong.
The weight of rolls acts through the center of rotation so the 'r' in the T=Fr equation is zero. That is to say they make no contribution to torque about the shaft.

You need to calculate the moment of inertia ('I') of the rotating masses.
http://en.wikipedia.org/wiki/Moment_of_inertia

Then you can calculate the torque required for the acceleration:
http://en.wikipedia.org/wiki/Angular_acceleration


But what you are really need to know is the steady state operation- how much power you need to keep the motor turning at your target rpm. for that you need to know where energy is going (eg the opposing torques from friction etc); not a trivial task.

EDIT: I now see your design. How does it compare with other similar machines?

Many people have invested millions of dollars and man hours in designing machines just like that...Do you think you'll compete with those designs? Why do you want to reinvent the wheel? Why not get (or copy...) an off the shelf design with years of R&D already done for you?
 
  • #15
billy_joule said:
... Why do you want to reinvent the wheel? Why not get (or copy...) an off the shelf design with years of R&D already done for you?


Ummm - do they give these designs away for free ?
 
  • #16
Re; post #13.
As the fabric unrolls at a set speed determined by the bottom roller, it rises up between two other pinch rollers.
Fabric cannot be pushed uphill, so it must be pulled up by the two pinch rollers. All the work of moving the fabric must be done by the upper pinch rollers. The bottom "driving" roller is not driving and will therefore not require power.
 
  • #17
Dnomyar said:
Ummm - do they give these designs away for free ?


google image search "*machine you want to know about*" and look at the images.
I bet every single image returned will have a better design than the OP's.

You don't need the original engineering drawings to copy a simple design. a single photo or a 20 second explanation will do. This isn't cutting edge confidential technology. They have probably been made the same for 50 years!
 
  • #18
billy_joule said:
google image search "*machine you want to know about*" and look at the images.
I bet every single image returned will have a better design than the OP's...!

Nope, not a one looking like the concept put forward - and Zero computations.
 
  • #19
Dnomyar said:
Nope, not a one looking like the concept put forward - and Zero computations.

Of course none look like the OP's, his idea won't work at all!

Like Baluncore explained, fabric cannot be pushed off a roll.
And generally you won't find calculations/computations with an image search.
 
  • #20
"And generally you won't find calculations/computations with an image search."
Duh - Really ? You didn't comment on the lack of suitable designs either, so you suggestion of Googling the solution wasn't so helpful.
So, since there were no computations, perhaps you don't mind the OP trying to establish the calculations then ? Perhaps he is allowed to expand his understanding a bit ?"Like Baluncore explained, fabric cannot be pushed off a roll."
Fabric 'can' be pushed off a roll. The easiest way would be to put two rollers under the fabric roll, and power them, making them rotate the fabric roll - the material will be pushed off the roll of fabric.
 
  • #21
Dnomyar said:
"And generally you won't find calculations/computations with an image search."
Duh - Really ? You didn't comment on the lack of suitable designs either, so you suggestion of Googling the solution wasn't so helpful

What lack of designs?
Mechanisms to pull sheet or tape off rolls can be found easily. Powered pinch rollers are used. a single roller can be used if the wrap angle and friction are sufficient.

here are some example I found with a "fabric roller machine" search term, better terms (like the name of the machine he wants to build) will net better results.

Hand cranked pinch rollers
http://gannetaus.com.au/UserFiles/hw11.png

powered pinch rollers:
http://i00.i.aliimg.com/photo/106970453/Fabric_Rolling_Machine.jpg

The bottom line is that you need to pull the sheet off the reel.

So, since there were no computations, perhaps you don't mind the OP trying to establish the calculations then ? Perhaps he is allowed to expand his understanding a bit ?

The calculations required appear earlier in this thread, namely P = Tω. see post #14 for further advice. Without further info from the OP little more can be said.

"Like Baluncore explained, fabric cannot be pushed off a roll."
Fabric 'can' be pushed off a roll. The easiest way would be to put two rollers under the fabric roll, and power them, making them rotate the fabric roll - the material will be pushed off the roll of fabric.

Fabric cannot be pushed up through unpowered pinch rollers. See post #16
 

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