[tennispro9911]

Those look like huge wheels. CIM wheels have 64 oz in torque under "normal load" which is 27 A and 12 V. I've found that at a high current, the battery cannot sustain 12 V even if it is fully charged, so you could use the figure for 40 A, 12 V but I personally doubt you will get that ideal figure.

64 oz in * 12.75 (toughbox speed reduction, torque multiplier) = 816 oz in, although the toughbox is only 80% or 90% effecient so multiply by .8 or .9 depending on your assumptions.

816 oz in * 1lb/ 16 oz = 51 lb in

I don't know the diameter of your wheel but my guess is a 16" bike tire so 8 in radius and 8in lever arm.

51 lb in / 8 in = 6.375 lb force from each wheel/motor. If you had a sprocket drive then you can multiply this number by whatever speed reduction you would use. Of course, you would go slower but thats a small price to pay for being able to turn.

To quote the possibly overused but completely accurate words of an announcer from last year "And this is why we do the math."

So I don't know how much force you need. That depends on the radius that the frictional force is acting on the turning of your robot and at what angle, and at what angle and radius the force of your wheels are acting on the turning of the robot and the frictional force of your caster wheels. I don't even know how you would go about figuring out some of those things, but just a does this number make sense check would tell you that 12.75 lbs from all of your wheels will not sufficiently power a 150 lb robot (15 lb battery and 15 lb bumpers).