Minibot Math + Physics Example Calculations - An Educational Read

[sanddrag]

I see several teams with large minibots, with large wheels. To me, it just doesn't make sense. I'd like to show a simple physics calculation that says why. I'll present it in an easy-to-read format, in case pages full of equations confuse you (like they confuse me sometimes).

As Mark Leon says, "Do the math." In this exercise, be aware that I'll jump back and forth between English and Metric units. Anyone who has completed high-school physics should be able to follow along. For clarity, I don't present unit conversions. I like using www.convert-me.com when I need one quick.

In Physics, Work is the product of Force and Distance. In equation form: Work=Force*Distance. Furthermore, Power = Work/Time. So, to minimuze the time, you want to minimize the work. Make sense? The distance is fixed (90some inches to the top of the pole), so you have to minimize the force, which in this case, is the force of gravity (weight) pulling down on the minibot. Every gram of mass you add that isn't necessary is slowing you down!

So, for an example calculation, let's assume a minimal mass (and weight) minibot, that weighs 2.5 pounds (lbf). I think this is possible to achieve, if the gear-heads are removed from the motors, small rollers are used instead of wheels, and light weight construction techniques are employed. Why remove the gear-heads? Well, if we had a higher angular (rotational) speed, we could use a smaller wheel and get the same liner speed. Also, the gear-heads have internal frictional inefficiencies, which is basically robbing the mini bot of power. Think of it this way. Would you put 20W50 oil made for a diesel truck into a sport bike? Would you drive a Corvette with half the spark plug wires removed?

Each motor, with the gearhead removed, has a peak power of approximately 16 Watts, as indicated by this dynomometer test CD User Richard performed. http://www.chiefdelphi.com/forums/sh...8&postcount=15

We are allowed two motors. This means, the minibot, as a whole, has a peak power of 32 Watts (assuming no inefficiencies). If we convert this to English units, it equates to 23.6 Foot-Pounds Force per second (ft-lbf/s) of peak power.

Remember from above, that Power=Work/Time. Substituting in the equation for Work, we get Power=Force*Distance/Time. Let's rearrange to solve for Distance/Time, or in other words, speed. Distance/Time = Power/Force. Let's calculate: Distance/Time =Speed = (23.6 ft-lbf/s) / 2.5 lb. The result, after division, is 9.44 feet per second (ft/sec).

So what does the above calculation and result mean? It means that a minibot weighing 2.5 lbs could theoretically travel at a constant upward speed of 9.44 feet per second, reaching the top of the pole in approximately (or just longer than) 1 second! However, this is IF and ONLY IF it is designed to operate at the PEAK POWER point of the motors, as we assumed.

So how do we design a minibot to operate at the peak power point?

From the motor dyno curve, we can read off the graph that the torque one motor puts out at the peak power point, is approximately 45 milliNewton-meters, or 0.045 N-m, which is equivalent to approximately 0.4 lbf-in in the English system.

Now, Newton's laws tell us that if the minibot is moving up the pole at constant velocity, the friction force of its tires on the pole (and oppositely, of the pole on the tires) is equal to that of its weight. Let's make the assumption that each tire is directly driven by one motor (no gearhead), and the two tires share the friction (force) equally. Thus, the force on each tire is 1.25 pounds.

Let's recall the equation Torque=Force*Radius. Now, let's rearrange the torque equation to solve for Radius. Thus, Radius = Torque/Force. Substitute in our known values. Radius = (0.4 lbf-in) / (1.25 lb). The result, after division, is 0.32 inches.

So what does this mean?

It means that in the above calculation, we have optimized the design, by designing around the peak power point of the motor curve. In theory, if you have a minibot weighing 2.5 pounds, with two motors directly driving 0.64 inch diameter rollers, it will climb the pole at a speed of 9.44 feet per second after it has accelerated. In practice, we may see significant losses due to friction and whatnot. So, to be conservative, you should use a smaller diameter roller.

If anyone else would like to add to the above, please do. Special thanks to Ether and a couple other users whose names escape me for posting some similar example calculations in other threads.