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conveyor torque calculation
- Thread starter barron
- Start date
MikeHalloran
Mechanical
Rent an engineer.
Mike Halloran
Pembroke Pines, FL, USA
Mike Halloran
Pembroke Pines, FL, USA
diskullman
Industrial
Or call your local power transmission distributor. They will assist you.
Russell Giuliano
Russell Giuliano
PeterCharles
Mechanical
F (N) = mu x N (N)
T (Nm)= F (N) x R (m)
Power (W) = F (N) x V (m/s)
This should get you started, OR HIRE SOMEONE WHO CAN DO THE JOB FOR YOU.
T (Nm)= F (N) x R (m)
Power (W) = F (N) x V (m/s)
This should get you started, OR HIRE SOMEONE WHO CAN DO THE JOB FOR YOU.
- Thread starter
- #5
Thanks to the guys who replied.
I am working in Bangkok and we are the only company in Thailand building this type of equipment also this drive is a first off so 'renting an engineer' would be like finding a virgin in Pattaya !!!
I am asking for a professional confirmation that my calculations are correct theory-wise.
The pallets are fixed to an aluminium articulating block chain and are supported by precision bearing wheels (mu = 0.075).The loop is 60 mtr. The weight of moving parts and load is 140kg/mtr. and the velocity 24mtr/min.
From this I calculate the torque to be 6.2 kN and using an efficiency factor of 0.85 the power needed would be 3.0 kW.
The driving force is provided by forcing the smooth surface of a grooved transmission belt, driven by the motor, against the surface of the chain block over a distance of 1.2mtr. This force is supplied by a number of "Rosta" tensioning arms.
Laboratory tests show the static coefficient of friction between the block and belt to be 0.58.
My assumption is that 6200N x 0.58 = 3596N would be taken up by friction.
My question is do I have to provide an additional 6200 - 3596 = 2604N by the tension units to prevent slippage ???
Again here's hoping
Roy
I am working in Bangkok and we are the only company in Thailand building this type of equipment also this drive is a first off so 'renting an engineer' would be like finding a virgin in Pattaya !!!
I am asking for a professional confirmation that my calculations are correct theory-wise.
The pallets are fixed to an aluminium articulating block chain and are supported by precision bearing wheels (mu = 0.075).The loop is 60 mtr. The weight of moving parts and load is 140kg/mtr. and the velocity 24mtr/min.
From this I calculate the torque to be 6.2 kN and using an efficiency factor of 0.85 the power needed would be 3.0 kW.
The driving force is provided by forcing the smooth surface of a grooved transmission belt, driven by the motor, against the surface of the chain block over a distance of 1.2mtr. This force is supplied by a number of "Rosta" tensioning arms.
Laboratory tests show the static coefficient of friction between the block and belt to be 0.58.
My assumption is that 6200N x 0.58 = 3596N would be taken up by friction.
My question is do I have to provide an additional 6200 - 3596 = 2604N by the tension units to prevent slippage ???
Again here's hoping
Roy
PeterCharles
Mechanical
Oooh, Pattaya sounds an interesting place to visit.
Anyway, all the parts are supported on wheels with bearings. Mu at 0.075 may be OK but I'd try and do a practical check, load up a pallet and pull it with a spring balance.
Torque is measured in Nm, so you can't have a torque in newtons. So is your 6,200 a force in Newtons or a torque in Newton Metres?? I might think it's a force in Newtons.
You are going to use a friction drive to produce the necessary linear force and the coefficient of friction in the force transfer area is 0.58. Lets say the linear transmission force is horizontal, then to generate this will require a vertical force of 10,690 N (6,200 / 0.58).
This vertical force will be transferred through the pallet wheels thus increasing the driving force by 801 N (10,690 * 0.075).
Of course, now you need a bit more vertical force to create the extra transmission force. But you get the idea.
Now, regarding the directions to Pattaya ................
Anyway, all the parts are supported on wheels with bearings. Mu at 0.075 may be OK but I'd try and do a practical check, load up a pallet and pull it with a spring balance.
Torque is measured in Nm, so you can't have a torque in newtons. So is your 6,200 a force in Newtons or a torque in Newton Metres?? I might think it's a force in Newtons.
You are going to use a friction drive to produce the necessary linear force and the coefficient of friction in the force transfer area is 0.58. Lets say the linear transmission force is horizontal, then to generate this will require a vertical force of 10,690 N (6,200 / 0.58).
This vertical force will be transferred through the pallet wheels thus increasing the driving force by 801 N (10,690 * 0.075).
Of course, now you need a bit more vertical force to create the extra transmission force. But you get the idea.
Now, regarding the directions to Pattaya ................
- Thread starter
- #7
Hi Peter
Thanks for your input but I am still not absolutely clear.
The objective here is to prevent slippage when the conveyor starts up (sometimes fully loaded)and when running.
I have seen this type of drive overseas for the same load conditions using 10 tension arms each providing about 200N force.
However I would like to determine mathematically what lateral force must be applied to the belt and block.
I forgot to mention that on the opposite side of the chain block there is another set of tension arms fitted with an idler roller comprising a nylon tyred precision bearing so that the block cannot move laterally.
In the attached photo the pallet is fixed to the block by the bolt seen in the centre of the block.
Best regards
Roy
Thanks for your input but I am still not absolutely clear.
The objective here is to prevent slippage when the conveyor starts up (sometimes fully loaded)and when running.
I have seen this type of drive overseas for the same load conditions using 10 tension arms each providing about 200N force.
However I would like to determine mathematically what lateral force must be applied to the belt and block.
I forgot to mention that on the opposite side of the chain block there is another set of tension arms fitted with an idler roller comprising a nylon tyred precision bearing so that the block cannot move laterally.
In the attached photo the pallet is fixed to the block by the bolt seen in the centre of the block.
Best regards
Roy
PeterCharles
Mechanical
Must admit I haven’t seen one of these drives before, but correctly engineered there’s no reason why it shouldn’t work.
We’ve already established how much pull is needed to move the pallets. If you’re starting fully loaded then in addition to the force to overcome friction there will need to be a bit more to overcome the inertia of the loaded pallets. How much depends on the acceleration necessary for the application, low acceleration low additional force.
From the photo I understand a bit more about the drive to the pallets. You drive a belt, then use the belt to drive the pallets. It’s a friction drive from the belt to the pallets so to get the normal load the pallets are clamped against the belt using the Rosta arms, the lower set running against the belt, the upper set running on the top of the pallets and providing the reaction against the lower arms. All seems OK to me.
I would be concerned that the exact loading from the arms may be a bit hit and miss. To give adjustment I’d have adjusting screws between the lower and upper frames. By moving the frames closer together or further apart the clamping (normal) force will be varied so varying the drive force between the belt and pallets. It’s all just F = mu x N. However you may need to make a bit of extra allowance for friction in the top as well as bottom rollers, and an allowance for compression effects in the drive belt due to the lower rollers.
With no background data I’d certainly be carrying out a few practical tests to verify any paper calculations (which will never be 100% accurate).
We’ve already established how much pull is needed to move the pallets. If you’re starting fully loaded then in addition to the force to overcome friction there will need to be a bit more to overcome the inertia of the loaded pallets. How much depends on the acceleration necessary for the application, low acceleration low additional force.
From the photo I understand a bit more about the drive to the pallets. You drive a belt, then use the belt to drive the pallets. It’s a friction drive from the belt to the pallets so to get the normal load the pallets are clamped against the belt using the Rosta arms, the lower set running against the belt, the upper set running on the top of the pallets and providing the reaction against the lower arms. All seems OK to me.
I would be concerned that the exact loading from the arms may be a bit hit and miss. To give adjustment I’d have adjusting screws between the lower and upper frames. By moving the frames closer together or further apart the clamping (normal) force will be varied so varying the drive force between the belt and pallets. It’s all just F = mu x N. However you may need to make a bit of extra allowance for friction in the top as well as bottom rollers, and an allowance for compression effects in the drive belt due to the lower rollers.
With no background data I’d certainly be carrying out a few practical tests to verify any paper calculations (which will never be 100% accurate).
- Thread starter
- #9
Hi Peter
Took a couple of days off for Xmas - hope you had a merry one.
I think we are about done with this topic. Just a couple of comments - the angle of the photo may be misleading.
The whole drive is horizontal with the pallet fixed above the chain. There is no upper and lower only left and right side of the chain.
The force of the tension arm can be increased by changing the angle of the arm to the chain (from 10 thro 30 degrees)
Further all tensioners are mounted on a sliding plate which can be adjusted in or out.
I would like summarise what you have advised :
Friction force => pulling force = mu x (N left + N right)
if this is so then the required force from the tensions arms would be much higher than they can provide.
The photo is a conveyor currently in operation and yesterday I checked with the people who installed it.
They set total tension force (N) = Pulling force x mu and it works.
How this reconciles to the laws of dynamics I don't know.
Back to Pattaya - check out and take a flight to Bangkok then a taxi to Pattaya but beware nothing is free and everything !! is for sale.
Best regards and Happy New Year
Roy
Took a couple of days off for Xmas - hope you had a merry one.
I think we are about done with this topic. Just a couple of comments - the angle of the photo may be misleading.
The whole drive is horizontal with the pallet fixed above the chain. There is no upper and lower only left and right side of the chain.
The force of the tension arm can be increased by changing the angle of the arm to the chain (from 10 thro 30 degrees)
Further all tensioners are mounted on a sliding plate which can be adjusted in or out.
I would like summarise what you have advised :
Friction force => pulling force = mu x (N left + N right)
if this is so then the required force from the tensions arms would be much higher than they can provide.
The photo is a conveyor currently in operation and yesterday I checked with the people who installed it.
They set total tension force (N) = Pulling force x mu and it works.
How this reconciles to the laws of dynamics I don't know.
Back to Pattaya - check out and take a flight to Bangkok then a taxi to Pattaya but beware nothing is free and everything !! is for sale.
Best regards and Happy New Year
Roy
PeterCharles
Mechanical
As I said, I'd never seen this arrangement before but I'll tuck it away for the future. I've also looked at the Rosta web site where they have full details of these mounts complete with tensioner arms.
Combining all the features together -
1) add the weight of all the moving pallets and anything they are carrying together to get a total weight (8,400 kg).
2) multiply this by mu1 (0.075), a suitable value according to the bearings used in the wheels.
3) this gives the drive force that needs to be provided by the belt drive unit (8,400*9.81*0.075 = 6,180 N).
4) divide the drive force by mu2 (0.58), a suitable value according to the belt and surface it is in contact with.
5) this gives the total clamping force to be provided by the Rosta tensioning units. (6,180/0.58 = 10,655 N)
6) divide this total force by the number of pairs of tensioning units (i.e. one above and one below the belt/pallet contact). The photo shows 7 pairs on one side, I'm presuming there are 7 on the opposite side (?) totalling 14
7) this gives the load per pair. (10,655/14 = 761 N)
8) IMPORTANT, this is a vertical load clamping the belt to the pallets, not a tangential load as shown in the Rosta catalogues!
9) not knowing the Rosta units installed that's as far as I can go, but if you know the unit reference then the catalogues will enable you to estimate the angle of the arms to give the necessary vertical clamping load.
As you say, we've probably flogged the topic to death now.
Oh, and "her indoors" won't let me fly to Bangkok :-((
Combining all the features together -
1) add the weight of all the moving pallets and anything they are carrying together to get a total weight (8,400 kg).
2) multiply this by mu1 (0.075), a suitable value according to the bearings used in the wheels.
3) this gives the drive force that needs to be provided by the belt drive unit (8,400*9.81*0.075 = 6,180 N).
4) divide the drive force by mu2 (0.58), a suitable value according to the belt and surface it is in contact with.
5) this gives the total clamping force to be provided by the Rosta tensioning units. (6,180/0.58 = 10,655 N)
6) divide this total force by the number of pairs of tensioning units (i.e. one above and one below the belt/pallet contact). The photo shows 7 pairs on one side, I'm presuming there are 7 on the opposite side (?) totalling 14
7) this gives the load per pair. (10,655/14 = 761 N)
8) IMPORTANT, this is a vertical load clamping the belt to the pallets, not a tangential load as shown in the Rosta catalogues!
9) not knowing the Rosta units installed that's as far as I can go, but if you know the unit reference then the catalogues will enable you to estimate the angle of the arms to give the necessary vertical clamping load.
As you say, we've probably flogged the topic to death now.
Oh, and "her indoors" won't let me fly to Bangkok :-((
- Thread starter
- #11
Hello Peter Charles
I was not satisfied with the outcome of our previous calculations as the result did not fit the parameters of existing conveyors.
Hence I went back to first principles where I believe we made a simple error.
FRICTION DRIVE CALCULATIONS
Using actual load test method determine Coefficient of Friction
from mu = N/F where N = W (total weight of parts)
Nylon wheel with precision bearing to steel mu = 0.075
Applying the result to the conveyor parameters F = ~6.2 KN
Using ASTM method determine Coefficient of Friction
Rubber to Aluminium mu = 0.57
Drive is effected by forcing a rubber belt (driven by the motor) against the aluminium side of the load by means of tension springs without slippage.
For calculation purposes imagine the drive rotated so that the belt is below the load and apply F = mu x N (=W)
In this case W = 6.2 KN thus F = 6200 x 0.57 = 3534 Newtons
which would be the point of slippage.
Therefore a force >F should be applied by the tension arms.
This result supports existing conveyor parameters.
A diagramatic file is attached hereto.
What do you think.
Roy
W friction force ( F )
rubber belt
> F
In this case : W = N = ~6200 Newtons mu = 0.57
F = 6200 x 0.57 = 3534 Newtons
Thus when F = 3534 then slip will occur
Hence a force greater than F must be applied by the tension springs
I was not satisfied with the outcome of our previous calculations as the result did not fit the parameters of existing conveyors.
Hence I went back to first principles where I believe we made a simple error.
FRICTION DRIVE CALCULATIONS
Using actual load test method determine Coefficient of Friction
from mu = N/F where N = W (total weight of parts)
Nylon wheel with precision bearing to steel mu = 0.075
Applying the result to the conveyor parameters F = ~6.2 KN
Using ASTM method determine Coefficient of Friction
Rubber to Aluminium mu = 0.57
Drive is effected by forcing a rubber belt (driven by the motor) against the aluminium side of the load by means of tension springs without slippage.
For calculation purposes imagine the drive rotated so that the belt is below the load and apply F = mu x N (=W)
In this case W = 6.2 KN thus F = 6200 x 0.57 = 3534 Newtons
which would be the point of slippage.
Therefore a force >F should be applied by the tension arms.
This result supports existing conveyor parameters.
A diagramatic file is attached hereto.
What do you think.
Roy
W friction force ( F )
rubber belt
> F
In this case : W = N = ~6200 Newtons mu = 0.57
F = 6200 x 0.57 = 3534 Newtons
Thus when F = 3534 then slip will occur
Hence a force greater than F must be applied by the tension springs
- Thread starter
- #12
Hello Peter Charles
(ignore preious post - incorrect file attached)
I was not satisfied with the outcome of our previous calculations as the result did not fit the parameters of existing conveyors.
Hence I went back to first principles where I believe we made a simple error.
FRICTION DRIVE CALCULATIONS
Using actual load test method determine Coefficient of Friction
from mu = N/F where N = W (total weight of parts)
Nylon wheel with precision bearing to steel mu = 0.075
Applying the result to the conveyor parameters F = ~6.2 KN
Using ASTM method determine Coefficient of Friction
Rubber to Aluminium mu = 0.57
Drive is effected by forcing a rubber belt (driven by the motor) against the aluminium side of the load by means of tension springs without slippage.
For calculation purposes imagine the drive rotated so that the belt is below the load and apply F = mu x N (=W)
In this case W = 6.2 KN thus F = 6200 x 0.57 = 3534 Newtons
which would be the point of slippage.
Therefore a force >F should be applied by the tension arms.
This result supports existing conveyor parameters.
A diagramatic file is attached hereto.
What do you think.
Roy
(ignore preious post - incorrect file attached)
I was not satisfied with the outcome of our previous calculations as the result did not fit the parameters of existing conveyors.
Hence I went back to first principles where I believe we made a simple error.
FRICTION DRIVE CALCULATIONS
Using actual load test method determine Coefficient of Friction
from mu = N/F where N = W (total weight of parts)
Nylon wheel with precision bearing to steel mu = 0.075
Applying the result to the conveyor parameters F = ~6.2 KN
Using ASTM method determine Coefficient of Friction
Rubber to Aluminium mu = 0.57
Drive is effected by forcing a rubber belt (driven by the motor) against the aluminium side of the load by means of tension springs without slippage.
For calculation purposes imagine the drive rotated so that the belt is below the load and apply F = mu x N (=W)
In this case W = 6.2 KN thus F = 6200 x 0.57 = 3534 Newtons
which would be the point of slippage.
Therefore a force >F should be applied by the tension arms.
This result supports existing conveyor parameters.
A diagramatic file is attached hereto.
What do you think.
Roy
PeterCharles
Mechanical
I'm happy with parts A & B, down to the 3kW.
But I disagree with your part C.
Your friction force induced between the rubber belt and aluminium needs to be 6,180 N, i.e. equal to the friction force calculated in part B.
To achieve this the normal load between the rubber belt and aluminium has to be 6,180 divided by 0.57, NOT 6,180 times 0.57 giving 10,842 N as the vertical load to be imparted by the Rosta spring units.
But I disagree with your part C.
Your friction force induced between the rubber belt and aluminium needs to be 6,180 N, i.e. equal to the friction force calculated in part B.
To achieve this the normal load between the rubber belt and aluminium has to be 6,180 divided by 0.57, NOT 6,180 times 0.57 giving 10,842 N as the vertical load to be imparted by the Rosta spring units.
- Thread starter
- #14
Hi Peter
You are correct of course but I was just looking for a solution to fit an existing case.
However I believe I have now solved the anomalies - thanks to a comment you made in an earlier reply.
Like you I had no experience in this type of drive but I did have access to an existing conveyor overseas and took photographs and recorded the operating parameters.
Having no idea of 'mu' for the carrier wheels but knowing the total load/mtr, overall length of conveyor, speed in mtr/min and the use of a 3 KW motor I back calculated to a coefficient of 0.075. I then had a testing laboratory, using actual sample materials, provide 'mu' for the rubber/aluminium.
I am now setting up the conveyor in Factory for testing and yesterday, remembering your comment " mu at 0.075 may be OK but I'd try and do a practical check, load up a pallet and pull it with a spring balance", I loaded up 5 pallets (= 40 kg = 1 mtr) with a load of 100 kg. I pulled it horizontally for a distance of 10 mtrs along the track and the result was constant at 1.8 kg.
This makes 'mu' 0.013 in which case the force required by the 'Rosta' units is well within their capability but a 3 KW motor is ridiculous.
There is a possible answer to this - redundancy - which has to do with the conveyor usage. I do not want to get into specifics in an open forum but I can say that this conveyor ( with various types of drive) has a specific usage and operates at frequent intervals 24 hrs per day, 7 days per week, 52 weeks per year and is used at the same locations all over the world. The big 'no no' is stoppage or downtime so prevention cost is not a factor.
One way is to provide two drive units diametrically opposite to each other, each one capable of taking the full load. The friction drive is ideal for this as there is no need for motor synchronisation.
I would like to say thank-you for your comments and advice - it has been good to have someone to talk to and you certainly have educated me in F=mu*N.
Best regards
Roy
You are correct of course but I was just looking for a solution to fit an existing case.
However I believe I have now solved the anomalies - thanks to a comment you made in an earlier reply.
Like you I had no experience in this type of drive but I did have access to an existing conveyor overseas and took photographs and recorded the operating parameters.
Having no idea of 'mu' for the carrier wheels but knowing the total load/mtr, overall length of conveyor, speed in mtr/min and the use of a 3 KW motor I back calculated to a coefficient of 0.075. I then had a testing laboratory, using actual sample materials, provide 'mu' for the rubber/aluminium.
I am now setting up the conveyor in Factory for testing and yesterday, remembering your comment " mu at 0.075 may be OK but I'd try and do a practical check, load up a pallet and pull it with a spring balance", I loaded up 5 pallets (= 40 kg = 1 mtr) with a load of 100 kg. I pulled it horizontally for a distance of 10 mtrs along the track and the result was constant at 1.8 kg.
This makes 'mu' 0.013 in which case the force required by the 'Rosta' units is well within their capability but a 3 KW motor is ridiculous.
There is a possible answer to this - redundancy - which has to do with the conveyor usage. I do not want to get into specifics in an open forum but I can say that this conveyor ( with various types of drive) has a specific usage and operates at frequent intervals 24 hrs per day, 7 days per week, 52 weeks per year and is used at the same locations all over the world. The big 'no no' is stoppage or downtime so prevention cost is not a factor.
One way is to provide two drive units diametrically opposite to each other, each one capable of taking the full load. The friction drive is ideal for this as there is no need for motor synchronisation.
I would like to say thank-you for your comments and advice - it has been good to have someone to talk to and you certainly have educated me in F=mu*N.
Best regards
Roy
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