Robot wheel torque to weight calculation help

It's been a while since I've had a physics class and I'm scratching my head to remember how to figure out the following problem.  If I have a pair of motors for a robot wheeled base that have a torque of 6 lb/in each and, lets say, a 5 inch wheel, how heavy of a robot can I build?  I'm starting to build a robot base that I would like to future proof so I can add extra electronics and hardware and start with a battery with plenty of power (which of course means more weight).  I'd like to avoid building the robot and find out later that the motors don't deliver enough force.  Any ideas?

Grosseinator (author) 5 years ago
Thanks. That has dusted some cobwebs off a long unused part of my brain. I'm getting back into electronics and robotics via means of a home made atomatic garden irrigation system and have forgotten some things over that last 16 years.
The torque divided by the wheel radius will give you the amount of forward force that will be produced.

F_wheel = Torque / Radius_of_Wheel

The force that needs to be overcome will be primarily the friction between the wheels and the ground. Neglecting any other sources of friction and air resistance, and assuming you are travelling over a level surface, the frictional force will be equal to the weight of the vehicle multiplied by whatever coefficient of friction is appropriate for the road surface and the material the wheel is made of:

F_friction = Total_vehicle_weight * u_friction

You'll have to find what the friction coefficient is. I assume your road surface is probably concrete and your wheels are rubber or plastic? You may want to find the worst case coefficent you will encounter, just to be conservative in your design.

If you are driving with two motors (one on each wheel?) then 2 times F_wheel will need to be greater than F_friction.

There will be a different (and greater) coefficient of friction necessary to get moving from a standstill. Also, if you intend to climb any hills you will have to recompute. The frictional force will be less, as the normal force will be less, but part of your weight force will be acting against you.