# Formula to calculate start up torque

1. Aug 8, 2016

### tpa

How to calculate the start up torque in inch pounds to start a plastic tube of 1.90" OD x 1.50" Id that has a shaft and ball bearings on each end. The shaft will be supported on each end, and will not rotate. The tube is horizontal. It would be similar to putting a conveyor roller in a frame, taping a string over the center so it hangs down both side equally with buckets on each side of equal weight, then add weight to one side only until the roller moves.. How much force did that take in inch pounds? That is what I want to measure and need a formula for it. Then I want to be able to build a machine to measure same, if you have any ideas on that too. Something repeatable and reliable.
Thanks

2. Aug 9, 2016

### Bystander

Tell you what ... You've described the measurement approach exactly ... build it and go from there.

3. Aug 9, 2016

### tpa

The problems I have are these:
1. I still need a formula to that shows how to calculate the value.
2. The machine I am talking about needs to be able to measure the value of at minimum 3500 different tubes per day. On a production basis. I don't know if that could be accomplished by using a wheel on an electric motor that touches the tube and then we measure resistance of the motor and convert to inch pounds or by using a Stair gauge somehow, but its the volume of measurements per day that presents the challenge.
3. what formula would one use to calculate the breakaway torque (per roller) required to start a 5 pound box that sits on 6 - 1.90" OD rollers going down a 3.5 degree downslope.

4. Aug 9, 2016

5. Aug 9, 2016

### Nidum

(1) You can get an estimate of bearing friction torque from manufacturers data . For small bearings with light loads and using oil lubricants it is usually next to nothing .

(2) If the rollers are designed properly then you are not going to get any large differences between them - so why test them all ?

6. Aug 9, 2016

### Nidum

To give a more complete answer to this question we need to see a drawing of the roller assembly .

7. Aug 9, 2016

### tpa

Nidum,
Customer is demanding values for start up torque for every roller produced. Even though they are all the same product. My way of thinking is that first we need to establish how much start up torque is required to get that 5 pound box to free start when it sits on 4 to 6 rollers, then divide that value by the number of rollers to get a "per roller value". We do not have the formula that will give us that answer, hence the request for how to calculate that value.

Then, simplistically, we need to be able to test every to verify the roller was under the value required or if it's over the value will need to rework or scrap.
For the formula of how many inch pounds of force to start a 3 pound box on 4.0 degree slope on 1.9"OD rollers I have seen this data, maybe you can verify if it is correct.
F start = W * Sin 4.0 degrees
= 3 lb * Sin 4.0 degrees
= .157lbs
F Start roll = .157/6 rollers = .026 lbs
F Start roll = Torque start/radius of roller
= To/.95
To = .026 * .95 = .025 inch pounds

8. Aug 9, 2016

### tpa

Attached is the only information I have ever seen on start up torque. Included is a sample roller drawing of a 1.9" OD roller with ball bearings each end and a shaft through the bore of the bearings. The shaft ends mount in a frame and the roller tube turns.

can you verify the calculation is correct for a 3 pound box on 6 rollers on a 4 degree slope is correct?

#### Attached Files:

• ###### Start up torque and roller drawing.pdf
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107.2 KB
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9. Aug 10, 2016

### Tom.G

A bit of a problem there. You seem to have used the Sin of 3o instead of 4o.
= 0.209lbs
0.209/6 = 0.0349
0.0349 *0.95 = 0.033 inch lbs

10. Aug 10, 2016

### tpa

Thanks for the correction, as I mentioned it is the only formula for this calculation I could find and it was printed that way. The math I can correct, the real question is, is the formula correct??

11. Aug 10, 2016

### Tom.G

Formula/approach looks good to me.

I do miss things on occassion, so responses from others are encouraged.

What say folks, is the math procedure correct for the stated problem?

Have fun in a production environment, that's only 1/2 Ounce of torque on the roller periphery.

12. Aug 11, 2016

### tkyoung75

That calculation is for conveyor starting torque, where as problem definition appears to suggest quality control.

13. Aug 11, 2016

### Nidum

@tpa Please clarify whether the box is being pushed to start it up the slope or down the slope .

The calculation method and the answers given are in any case not correct .

I'll come back later but in the meantime see if you can find the source of the error .

Clue : If it takes a tangential force of Ftr to start one roller then what is the sum of the tangential forces needed to start six rollers ?

Last edited: Aug 11, 2016
14. Aug 11, 2016

### tpa

The box is placed on the conveyor rollers and must "self start" with no assist down the conveyor.
That is the formula I am looking for. How do you calculate the start up torque required for a self start in this scenario "per roller" under the box.