I am building a robot, a battlebot to be precise, and I cannot, for the life of me, find out how to calculate how strong my motors need to be. PLEASE!!!! NOBODY PUBLISH FORCE TIMES DISTANCE!!! ITS NOT WHAT I AM ASKING!!!! I have 4 wheels, each with one motor. The back wheels are 3 times the size of the front ones (I will fix that with gears). BUT, IGNORING THE GEARS, HOW DO I CALCULATE THE TORQUE NEEDED FOR EACH MOTOR(ASSUME THEY ARE EQUAL) TO MOVE THE ROBOT, WEIGHING x NUMBER OF POUNDS, y NUMBER OF FEET?

OHHH, so i doesn't matter the distance. I didn't really understand all the math. I'm pretty good at math, but going into 8th grade. (so i dont have an engineering text)

I'm still confused. Can you just give me a single formula, please? You know, with wheel radius, weight of robot, and distance, and time, using variables like r(radius), w(weight), d(distance), and t(time)

Break the problem up into several fairly simple physics problems.

Start with the force that you would need to simply push the robot.

If it ever needs to go up a slope, you will have the force working against the force of gravity to raise it to a higher level.

To accelerate the mass, you have F=MA.

You need to overcome the rolling friction of the wheels.

If you will need to push another robot, you may have a lot of sliding friction assuming its wheels may not be cooperating with your robot.

Sum up all of those forces. Divide the force by 4 and multiply by the wheel radius. That is the torque per wheel. The smaller wheels will require less torque.

Then you need to add the bearing friction and the friction in the gears. The friction in the gears could be a major part of the total. If you buy a gear box, the manufacturer may supply that information.

The torque at the motor will be (motor torque) = (wheel torque) x (wheel RPM) / (motor RPM).

Edit 1

For a battlebot, you will need all the torque that you can get to push head to head against the torque of the opponent. If you push the opponent sideways, you don't have opposing torque, but you have maximum friction to skid the opponents wheels sideways.

Since reducing the speed gives more wheel torque for a given motor torque, you may want to consider carefully the ability to go fast vs the ability to push hard.

Edit 2

For your level of experience, it is probably best to just buy the motor or motors that you can afford and remember if you gear down to a lower speed, you will be able to push harder.

### Related Reading:

#### Incoming search terms:

Uncategorized

Well yes, I’m afraid the answer is force times perpendicular distance ðŸ™‚

I’m guessing that you have a mass that you want to accelerate. For example you might want a 1Kg robot to race across a play area in x seconds. You can calculate the acceleration you need from a standing start. Or the swing of some horrible cleaver, &c.

If the torque of the motor must provide this acceleration, then the torque of the motor is exerted at the radius of the wheel. It is equivalent to a force at that radius, ie torque = force times distance.

That force is the one that accelerates the mass.

So the force (backwards) on the ground likely equals the force forwards on the craft (mass) you have F = ma.

If you assume sort-of constant acceleration you can likely get the rest.

a = dv/dt. so velocity grows linearly with time. v = 0 + at. area of the triangle is distance. Want to cross x distance in some time: Substitute what you know, you’ll likely have a torque.

I guess you will try to get all wheel drive? 1/4 torque at each wheel. Likely if the drive is electrical, the motor will share load well anyway.

I’m guessing that in the acceleration calculation, you can likely ignore the rotational inertial of the wheels, (& therefore the need to accelerate them, rotationally) but likely you will need a friction term to discount the available acceleration.

Hope this helps.

References :Your mechanics text will have the kinematics equations.

It depends on all the friction losses in the gears, and how fact you want to accelerate.

Once you are moving the y number of feet at some constant speed, all the torque is going into overcoming the friction and resistance in the gears of your drive-chain. Motor-gearbox-diff-axle.

Go for the highest torque you can afford.

References :you can calculate the torque by analysing the free body diagram of your robot as a mechanism ,then you can calculate the maximum torque -using either analytical or graphical methods- thereafter you can choose your motor according to your need.

References :Break the problem up into several fairly simple physics problems.

Start with the force that you would need to simply push the robot.

If it ever needs to go up a slope, you will have the force working against the force of gravity to raise it to a higher level.

To accelerate the mass, you have F=MA.

You need to overcome the rolling friction of the wheels.

If you will need to push another robot, you may have a lot of sliding friction assuming its wheels may not be cooperating with your robot.

Sum up all of those forces. Divide the force by 4 and multiply by the wheel radius. That is the torque per wheel. The smaller wheels will require less torque.

Then you need to add the bearing friction and the friction in the gears. The friction in the gears could be a major part of the total. If you buy a gear box, the manufacturer may supply that information.

The torque at the motor will be (motor torque) = (wheel torque) x (wheel RPM) / (motor RPM).

Edit 1

For a battlebot, you will need all the torque that you can get to push head to head against the torque of the opponent. If you push the opponent sideways, you don't have opposing torque, but you have maximum friction to skid the opponents wheels sideways.

Since reducing the speed gives more wheel torque for a given motor torque, you may want to consider carefully the ability to go fast vs the ability to push hard.

Edit 2

For your level of experience, it is probably best to just buy the motor or motors that you can afford and remember if you gear down to a lower speed, you will be able to push harder.

References :