Permanent Magnet DC Motor Calcuations

Question

I have to determine speed and torque suitable for my combat robot. I've done some calculations and I need to know whether they're right or not (because they don't seem to be right).

Suppose I have 10kg robot, I want to push other 10kg robot to the arena walls. Let's first assume opposite robot is not opposing my push. Suppose I want it to move it with acceleration of 10cm/sec/sec (also lemme know if acceleration assumed is suitable for robowar or not). Then force required will be,

F = (total mass to be moved) * (acceleration) = (10+10)*(10e-2) = 2 N. Assume opposite robot is also pushing with me same force, in that case I will need another 2 N for opposing his push, Now taking into account frictional retardation, suppose I need another 2 N so total would be 6 N.

Assuming velocity = 10cm/sec, then Power required will be 6 * velocity = 6*10e-2 = 600mW. For 12V motor even with 20% efficiency it would cause current of 250mA only. Also output power = speed X torque.

Now I've heard that robowars usually require motor with high full load current otherwise they'll be damaged.

So it looks like I am somewhere wrong in my calculations. If so, lemme know where,am I wrong where I assumed that opposing robot will cause just 2 N, because generally they'd be fitted with more high torque motors. But is it the only thing which goes wrong in above calculations? If opposite bot had to just exert 2N on my bot, are calculations right?

PS: By side-shaft geared motors, it is meant that it is internally geared with horizontal shaft, right? Or "side-shaft" means something more?

Answer 1

The logic for the calculation seems ok, you can also check the results here and here

When building robots that would oviously benefit from as much force (motor torque) as possible, I would suggest the following approach.

Rather than defining speed and acceleration requirements, I would approximate what is the maximum amount of force the wheels can apply to the ground without causing the wheels to slip. This would lead to the maximum torque of the motor by calculating backwards from the wheel force to the motor torque. If you consider that this amount of torque has to be applied to the ground also when the maximum velocity of the robot is reached, you have defined the motor. You can then check what acceleration will the motor provide. Since it is limited by wheel friction, even if you would increase motor torqe, the acceleration would not incears, since wheels would start to spin.

This way you exploint your wheels to their maximum amount and so deliver the maximum force possible to the oponent robot.

Answer 2

I affraid you are very wrong with your assumptions. I'm not exactly sure what competition are you going to start in, but take a look at some videos with 3kg sumo robots:

• Vienna: https://youtu.be/7T0en-RcljE
• Riga: https://youtu.be/Z_WE-ppaE5A
• Japan: https://youtu.be/lUpUQf16qzQ

On all of these videos you will see robots that reach speeds of couple m/s and accelerate in less than 1s. Some of these competitions may be played on metal rings and magnets on the robots to increase friction and in result maximum possible accelerations. Still, I think the smallest reasonable speed is 1-1.5m/s with acceleration at least $1m/s^2$.

Now, let's find out how much force do you need to push your opponent. For a moment let's assume it just a 10kg block of rubber (because a robot will likely have rubber wheels). From a table of friction coefficients you select one between rubber and whatever material the arena is made of - let's say dry concrete. That means that friction coefficient $\mu = 0.85$. Now the friction force $F_f = \mu * m * g$, so for 10kg block of rubber it will be around 85N and that's under the assumption your opponent is doing nothing!

To sum up aim for the biggest motors you can pack in your robot. Also, keep in mind that the more friction you have, the more push you will get. That means that good tires for your wheels are as important as the motors.