# Minibot Math + Physics Example Calculations - An Educational Read

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/showpost.php?p=1005008&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.

This is an awesome introduction to the physics behind robot design! I tried to explain the P=W/t to the minibot freshman at the beginning of the season, and I just got blank stares :(.

We did this calculation and everyone on the team was really excited, but we spent a week trying to get our “wheels” mounted directly onto the motors, and decided we would rather just get a minibot going up the pole and try to make the fast version once we had that working. We never did get our shafts mounted on the motors, we just didn’t have the resources to do it properly. I wonder if this was true for any other teams?

Here’s a rudimetary explanation:

If you are holding a stack of school books, even though they are heavy, you are doing no work. Although a force is being applied to hold them, they are not being moved at any distance.

Now say you take that stack of school books, and walk them upstairs to the second floor. You have now done work.

Now say you take that stack of school books, and run them upstairs to the second floor. You’re out of breath. Why? You just did the same amount of work, but in half the time.

How did you do it? You used twice the power.

What if you had only a certain amount of power? Well, to get the books up there twice as fast, you’d need to carry only half as many.

Now, keep in mind that this example neglects a lot of things, but it’s a basic way to describe it, that seems to get the point across.

Of course, you’ll always get one kid who says “but the library is on the third floor.”

I think a better analogy would be, would you put an overdrive transmission in a car, instead of changing the rearend gear ratio and making low gear in the transmission a steeper ratio? One way has two sets of gears working against each other when cruising down the highway, the other makes the transmission act like a straight thru shaft.

oh…yeah…the first way is how all modern cars work…ok, forget that idea.

Anways, back to the topic, the physics of it is not really the problem. It’s the building of it. FTC is an “erector set” type of thing, where the normal design method is to start screwing kit parts together, and see how it works.

FRC is completely different, you know going in that you’re gonna have to design and fabricate parts to make something work at all.

I think there are some teams who look at the minibot as an FTC thing, and other teams who look at it as an FRC thing. The FTC mentality leaves the stock gearboxes on, and the minibot is big and heavy and slow. The FRC mentality starts with just the essential parts (motors and battery), and builds up from there as needed to make it work.

But thanks for the lesson, I hope teams will work to understand and implement it.

OK I’m sold. Just one question, What do you use to attach traction material to the miniature rollers? Just to test, I attatched a 3 inch wheel directly to a motor with the gear head removed. It spun so fast, the rubber actually expanded and separated from the wheel.

Thanks,
–Al

3 inch wheel is way too big.
You should probably ask some questions:

1. under load, what will be the approximate RPM of the motor?
2. what is the speed necessary reach the top in under 2 seconds?
3. what is the diameter of the wheel needed to reach that speed?

Those will get you in the ballpark.
You can get more precise by:

1. trying different tests
2. using explanations of the math/physics- there may even be some spreadsheets you can use:

There are several variables that need to be controlled to make a direct drive minibot work.

I like it! Now to grab a stack of A.P. Physics books and get the freshman running!

Several weeks ago our team put together a spreadsheet that made it easy to determine the optimum wheel diameter once we knew the robot’s mass.
(This exercise is left to the student. )

After a couple of iterations, we achieved a climb rate of 1.17 seconds, just about what the laws of physics had predicted.

The only variable in our equation that we didn’t know for sure was the overall efficiency of the drive train. We assumed ~ 90%, as we had tossed out the gearbox. The timed result suggests that it was more like 85%.

Thanks,

Yes, I didn’t think the 3"inch wheels would be small enough for direct drive, It was just what I had on hand that I could quickly attach to the motor to see how fast it drove with the Gearhead removed. Since it was fast enough to drive the rubber tread off the 3" wheels, I was wondering what would be needed to keep traction material attached to fabricated small diameter wheels.

Elasticity, and friction.

We use a 3/8" aluminum shaft, drilled out to 0.097" to fit the motor shaft, and press a 1" piece of 1/2" OD surgical tubing over it.