Turning torque

[Juggernaut]

At work I am working on designing a swerve drive for a fairly large vehicle. I am attempting to calculate the torque needed to rotate the wheel of a swerve drive to make sure I have enough torque from my steering motor.

I have a 4" wide 8" diameter wheel. There is 1500lbs of load on this wheel. The total contact area under this load is .002in^2. The coefficient of friction for the wheel and the surface it will be running on is. 0.5.

If anyone could point me in the right direction to be able to figure this out it would be greatly appreciated.

[Ether]

Originally Posted by Juggernaut

I have a 4" wide 8" diameter wheel. There is 1500lbs of load on this wheel. The total contact area under this load is .002in^2

How did you figure the .002 square inch number?

[IKE]

Something is not making sense. A 1500 lb load with 0.002 in^2 is 750KSI which is about 10X the yield strength of many quality steels.

In general the calcualtion for torque would be the force of a specific point, times its coefficient of friction, times the radius from the axis of rotation.

[Aren Siekmeier]

A little integral calculus helps you out here. You can add up small contributions to the total required torque from small portions of the wheel's width depending on the frictional force and the radius, take the limit, and compute the integral.

Spoiler for I guess I need a title:

We'll split the wheel in half to make it easier, starting from distance r=0 to r=W/2 from the wheel pivot point, with W the wheel width (assuming wheel pivot is in the middle, it's easy to compensate otherwise).

At each dr chunk, the normal force there is the total normal force (equal to the load, unless you're moving in and out of the ground...) on the wheel scaled to the size of dr, or L*dr/W, where L is the load on the wheel, dr is the width of the small, and W is the total wheel width.

The frictional force then is the normal force times u, the coefficient of friction, or uLdr/W. The torque is the force crossed with the radius, so we have the torque for each chunk as uL/W * rdr, which we integrate from 0 to W/2 and multiply by 2 to get our final expression, T = uLW/2.

This assumes the contact area between the wheel and the floor can be described as a line. You could integrate over a rectangular contact if you knew roughly how big that was and would find a slightly larger answer, due to the larger moment. This also assumes uniform distribution of the wheel load over the contact area.

[Juggernaut]

Im not sure what I was thinking with the .002 inches. The wheel manufacturer gave me a calculated surface area for this load of .051inches.

Thanks for all the help.

[Ether]

Im not sure what I was thinking with the .002 inches. The wheel manufacturer gave me a calculated surface area for this load of .051inches.

Surface area is not measured in inches.

What material is the wheel made of, and what is the cross-sectional shape of the wheel? (i.e. like a forklift wheel shape or a motorcycle wheel shape)

[Juggernaut]

Sorry, .051in^2.

It is a Shore A durometer material call Vulkosoft. The wheel has a flat surface like a forklift wheel.

[jspatz1]

Any calculation you might do on this has a good chance of being off by such a large margin that it is not practical to actually use it in sizing your system. You would probably get a far more usable result by measuring a physical test if you can arrange it. Build a pivot, load the wheel, torque it with a lever, measure the force, calc the torque.