Re: The physics of pushing
 

Assuming you're pushing against an identical robot, this limits pushing force F to approximately (W*k-tau)/h, where h is the height (above the horizontal witness line through the center of the wheels) of the effective contact point of the pushing force and tau is the torque on the rear wheels

Note that if F is below the horizontal witness line through the center of the rear wheels, then "h" is negative.

corrected diagram

original diagram

Re: The physics of pushing
 

Awesome diagram! I hadn't ever really tried to figure out the solution before, but with your diagram, it's pretty easy! Thanks, Ether!

However, I think there's an extra variable (tau) in your solution. From looking at your diagram, I get the maximum pushing force as being (W*k)/h, with F=tau. However, I think I must be missing the reason that you have included "tau" in the calculation. Can you explain? I set the rotational torques about the ground contact point equal to one another (F*h = W*k) and then solved for F. However, either I'm missing something (the reason for tau being in the answer) or you made a minor mistake.

When coming up with physical solutions like this, I like considering each of the independent variables and how making them bigger / smaller effects the result, particularly at the extremes. (W*k)/h seems to make sense from a number of practical ways:

• The lower you can get the pushing point ("h"), the harder the robot can push without tipping over. If the height can be made to be zero, then this constraint (tipping over) is never a limiting factor. Very tall robots can't push well at all, as they'll simply tip over instead.
  
• The heavier you can make the robot ("W"), the harder the robot can push. A weightless robot can't push at all, and an infinitely heavy robot won't have it's pushing power limited by tipping over.
  
• The further you can move the CG forward of the rear wheels ("k"), the harder the robot can push. If the CG is directly over the rear wheels (k=0), then the slightest push will tip the robot over. If the CG is infinitely far in front of the rear wheels, then the robot isn't limited in pushing by tipping over, either.

In any case, thanks, Ether! I'll have to encourage our team to consider this factor when designing gear ratios for pushing next year -- we've definitely had more pushing power than we could use during these past few years! Accordingly, we probably could have geared higher to get a little more speed without compromising our pushing ability (barring electrical energy considerations.)

PS: "You must spread some Reputation around before giving it to Ether again."

Re: The physics of pushing
 

Nice catch. In my haste I labelled "h" as the vertical distance from the force to the ground. It should be the vertical distance from the force to the horizontal witness line through the center of the rear wheels.

See corrected diagram in original post.

In the corrected diagram, you get F*h + tau = W*k,

so F = (W*k -tau)/h (as originally posted).

If you then assume tau = F*r in the above equation (where "r" is wheel radius),

you get F = k*W/(h+r), where "h+r" in the new diagram is the "h" in the old one.

So yes, the formula for the original diagram should have been F = W*k/h.

For the corrected diagram, it is F=W*k/(h+r), which amounts to the same thing.

Sorry for the confusion.

Re: The physics of pushing
 

Oops! I actually liked where you had it the first time -- changing the diagram has now gone and made my earlier post invalid -- except it now agrees with my calculations! Whew!

Re: The physics of pushing
 

Ah, I didn't consider that. I re-attached the original diagram to provide context for your post.

Yes, the rotational reaction torque on the robot due to the torque on the wheels.

Yes. I think you posted this while I was busy wordsmithing my previous post:) It now explains the role of "r", the wheel radius.