Quote:
Originally Posted by pogenwurst
We've had a lot of debate about this on my team, and this seems like an appropriate thread to ask for clarification:
What sort of difference, exactly, is there between adding extra driven wheels versus adding extra undriven to our robot, and why?
I've seen snatches of conversation on the issue, but nothing really substantial enough for me to draw a conclusion from.
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Good question. The robot's weight is supported by every wheel that touches the ground, driven or undriven. This weight corresponds to the normal force, which represents how much frictional force each wheel could generate (mu*Fn=Ff). However, it can only generate this force if it's being driven. Make sense? So say you had a 120lb robot and 4 wheels evenly supporting the weight--not likely, as robots usually aren't perfectly weight-symmetrical, but nonetheless. Each wheel supports 30lbs, so each can generate a maximum force of mu*30. So if all the wheels are driven, it's 4(mu*30). If only 2 are driven, it's only 2(mu*30), significantly less.
EDIT: For those wondering, "mu" is the coefficient of friction (I imply it's static in this case), which is basically a measure of how difficult it is to move two materials against each other, with larger numbers (usually around 1) being more difficult than smaller ones. As you can imagine, the mu's in this game a very low, between .15 and .05, depending on who you ask. Also, the reason spinning wheels generates a frictional force in the first place basically goes back to Newton's third law, the old action-reaction one.
Quote:
Originally Posted by Gdeaver
Keep in mind that most of the discussions concerning friction and traction assume a flat surface contact. The FRP material is going over carpet. If the FRP was placed on a hard surface like concrete or a gym floor the 2 D assumptions would be valid. Because of the carpet the FRP may be deformed changing the contact area. This puts the physics into a 3 D problem. Does this change things? Do fewer wheels cause more defection and better performance or does the deflection make it worse? Is the deflection enough to have a significant effect? I don't know the answers. Not sure about the math either, but it needs to be looked at. This is a case where one needs to consider the assumptions of a math model to determine if the model is valid for the problem.
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Very true. In fact, if you've seen the FRP, it itself is bumpy. I can't explain the math, but I think what it comes down to is that contact area is actually important. Because of the contact area changes, the more surface area you have, the more likely you are to hit a higher mu. (Note this conclusion is based solely on dynamic contact issues, I'm not sure about deformation.) To at least quasi-demonstrate this to your team, I might recommend an incline test. That is, slap together a chassis, put the wheels on the FRP and tilt the plastic until the chassis starts to slip. The terminal angle can also give you static mu values if you run the force-sum equations, but it's worth it just to see wheels get stuck on a bump, causing the angle to change slightly.