# Thrust in a drivetrain?

I’ve learned a lot about drivetrain calculations and construction over the past year, much thanks to the CD community. I have one question on how thrust (torque/radius) relates to acceleration, and other qualities.

It seems you forgot to post the question

Depends on how big your jet engine is: The bigger the engine, the more thrust it generates, and the faster you can go. Also works for rocket engines and rocket motors.

And what you’re ACTUALLY looking for is what might better be called “force to floor”. Which…
F=ma, but in this case you’re looking at the applied force, which we’ll call Ff.

Ff = (Torque/radius)*mu, where mu is the coefficient of friction between your wheel(s) and the floor in question. Generally this will be some number between 0 and 1, but can be slightly higher than 1 depending on how “grippy” your tread is (and significantly lower than 1 if you’ve broken traction and are spinning your wheels against the wall). For the sake of this post, we’ll call it 0.8–just so it’s not 1.

Also, we’ll call m as being 150 lbm.

So…
(Torque/radius)0.8=150a. You can do the math to get acceleration from there, I think.

In brushed DC electric motors, torque is proportional to current.

The more you want your drivetrain to put torque to the ground, the more current will be drawn.

The more current is drawn, the lower your battery voltage will go (non-ideal voltage source).

The lower your battery voltage is, the lower the voltage you are able to apply to the motor is. This will limit the overall acceleration possible.

Additionally, more current means more power dissipated as heat. The more you try to make your motors push, the more they will heat up.

Motors lose their ability to push as they start to spin faster. You can push hardest when the motor is stalled. This is why motors don’t accelerate infinitely. This is also part of the reason your drivetrain has a maximum “runout” speed.

Hot wires conduct worse than cold wires (usually). A hot motor is less efficent at converting electrical energy to mechanical energy than a cold motor.

Friction is usually pretty constant. No matter how hard you try to push, there will always be kinetic friction opposing the motion proportional to the normal forces applied to the body in motion.

No. This is… this is just wrong.

There are two limitations on Ff. One is motor, one is traction.

The traction limit is Ff <= N*mu (N being the normal force on your tires, er, wheels). This normal force resists the robot’s weight- so Ff <= Weight * mu.

The other limit is from your motor and is more complicated (as the torque your motor can produce is a function of its speed)- but fundamentally, it is Ff <= (Torque at wheel)/(radius of wheel) = (Torque at motor)*(Gearbox reduction)/(radius of wheel).

Ff will be the smaller of these two, and if the traction is the limit, your tire will likely start spinning. Usually this isn’t the limit except maybe at low speeds with a high reduction gearbox.

If you like math, not to toot my own horn, but maybe look at these resources: