In designing a swerve module, I’ve come across a bit of a problem; my basic knowledge of physics leaves me without a way of calculating how much torque is needed to turn a wheel under load normal to its standard axis of rotation, as is done in swerve drive, both while stationary and while moving. Could anyone please assist me in figuring it out, and/or direct me to a relevant paper?
For an ‘order of magnitude’ calculation (usually written on a napkin), just take the width of the wheel, divide it in half. That’s your ‘torque arm’.
Then take the force needed to slide the wheel (say, in Pounds) over the surface.
Multiply the force by the torque arm, that’s a close estimate. So, if you need 24 pounds of force to slide the wheel sideways, and it is a 1" wheel, you get a torque of about 1 Ft-lb.
You can measure the sliding force, which depends on the coefficient of friction with the surface and considering the weight bearing on the wheel, with a pull-type scale.
This is purely an estimate, since it does not take into account several factors, including:
- The friction in the shaft & bearing assembly
- The force resisting the wheel from turning varies by the distance from the axis, we are assuming a point-load at the furthest edge only.
- we are considering the static case only (the robot is not moving). With movement the forces are reduced, so we can consider only the worst case.
- We assume the axis of the wheel turn is the centerline of the wheel. The force is significantly reduced if the axis is offset - look at a lawn tractor front wheel to see the idea.
No, it’s not a scientific or engineering calculation, but it’ll get you close. You probably would need about 2 or 3 FtLb for a 1" wheel, double that for a 2" wheel, 160 lb robot with wedgetop on carpet. But in the engineering world, this kind of guesstimate is usually sufficient.
You CAN also build a model and measure the torque required! That the “empirical” method.
I’d like to see someone answer the original question though with a more rigorous treatment.
I realized I never got a chance to thank you for the answer, so let me properly do that now.
The other reason for necroing this thread is to search for what Don and I are asking for - namely, the more rigorous method of doing it.
I worked myself through the equation I believe you are looking for late one night. As far as I know it is correct:
Coefficient of friction * tread width * weight supported by the wheel = force required to turn the wheel
If you want to get fancy:
Coefficient of friction * √((tread width)² + (tread length in contact with the carpet)² ) * weight supported by the wheel = force required to turn the wheel
tread length in contact with the carpet - if you were to look at a side profile of the wheel and the carpet so that you are seeing the wheel as a circle and the carpet as a long, thin rectangle, this would be the distance of a straight line from the first point that the wheel and carpet intersect to the second point that the wheel and carpet intersect.
When calculating the weight supported by each wheel for non-crab, 4+ wheel swerves remember the wobbly four legged chair/stool in your kitchen. At any given time there are only three legs of it bearing its weight, and the same can be true for your swerve. This may not always be the case, and even if it is you may be able to dig deep enough into the motors’ torque to compensate without stalling them. Either way it’s at least something to consider.
Some other things you may also want to consider are additional weight you may pick up during the course of a match (i.e.: game pieces), what happens if another robot gets under your bumper or crossing the bump in 2012 and you are left with fewer wheels than you designed for in contact with the ground, as well as any other scenario you can think of that may compromise the turning ability of the wheels in your swerve. While easily over looked, not thinking about things like these could end up costing you a match or two through the course of a season or worse yet robot functionality, but that’s just my two cents.