# Conveyor Torque

1. Oct 30, 2016

### SevenToFive

A company called the other day with a question about their conveyor. My company did not build the conveyor but did supply the gearbox. The conveyor is 8 feet long, and 4 feet wide, it is level and transports castings to the grinding department of a foundry. The motor they are using is a 4 pole 5HP motor, assuming at 60Hz. I was told the castings get "dumped" onto this conveyor which is always turning, so there is a bit of a shock load. The foundry wants to be able to move 3000lbs of castings on this conveyor, however when they put the load on the conveyor the motor trips out. The gearbox that is on their conveyor is rated for 9000in-lbs with a 5HP motor and 29 rpm. The conveyor shaft is 11 inches in diameter and coupled directly to the output shaft of the gearbox.

I used an equation that I found but had to convert the values to metric and then back to standard.
The force is MU * Mass* Gravity, where MU is 0.5, the 3000lbs is converted to Newtons:13344.66 Newtons, gravity is 9.81
So F=0.5*13344.66*9.81
Force = 65455.6Newtons

Torque = 65455.6N * 0.2794m : the 0.2794 is the 11 inch shaft converted to meters
Torque = 18288.3Nm; 161865.2in-lbs

Am I on the right path with this? If so the current gearbox is greatly undersized.

Thanks for the help it is greatly appreciated.

2. Oct 30, 2016

### TomHart

What does MU stand for?

You converted 3000 pounds to 13344.66 Newtons. That is correct. But then in your F = MU*Mass*Gravity equation, you are not using the mass; you are using the weight of 13344.66 Newtons. So you end up multiplying the mass by gravity 2 times. To find the mass of a 13344.66 Newton weight, you have to divide by 9.81 to yield 1360.3 kg. If I use that as the mass in your equation, I get:
F = 0.5 * 1360.3 * 9.81 = 6672.3 Newtons.

So it appears that your "Force" result is almost 10 times higher than it should be - unless I am misunderstanding something.

3. Oct 30, 2016

### SevenToFive

MU was a coefficient of friction. Thanks for the help.

4. Nov 1, 2016

### SevenToFive

Well the customer called, it seems that many of the original things have changed. Now the castings will be sliding onto this conveyor that will be 4 feet wide and 10 feet long and level with a 14 inch roller. Can my same calculation as above used to determine the torque requirement?

Thanks to all of those who reply.

5. Nov 1, 2016

### tygerdawg

You're working with a distributed load. It's been several years, but I recall that this is a standard model of most gearmotor manufacturer's Engineering Guides. I know SEW Eurodrive has their online sizing tool PTPilot, but probably the others have something similar (Dodge, etc.). Worst case, this is a standard calculation to be found in PDF Engineering Guides.

Offhand, 5HP seems a little light for this load situation. These calculations all deal with Peak Torque, which accounts for starting from zero velocity and accelerating to max speed. Which is a lot different from steady state torque. In any case, you must take into account all of the inertias, both translational & rotational.

6. Nov 3, 2016

### CWatters

If they dump/drop the castings onto the belt the normal force could be quite a bit higher than it is "at rest". Could they visually estimate how far the belt moves before the castings are up to speed/not sliding? That would allow you to calculate the acceleration of the casting and hence the force needed top accelerate them.

7. Nov 3, 2016

### CWatters

I take it the output of the gearbox is doing 29rpm? In which case I make the belt speed 0.42m/s (1.4ft/s). Perhaps I'm wrong but that sounds a bit fast for heavy castings. Do they really want a 3000lb casting coming off the end at that speed or do they mean there are 3000lb in total of smaller castings on the belt at once?

What exactly are they doing when it trips? Are they dropping one 3000lb casting? Are they putting 3000lb of smaller castings on the belt and then turning it on? Are they putting 9 * 300lb castings on the belt and dropping another one on?

8. Nov 3, 2016

### jack action

In addition to @TomHart 's comment, the torque is obtained by multiplying the force with the radius of the shaft, not the diameter. Thus your number is not 10 times, but 20 times larger than expected.