Calculate torque for a wheel to move

can someone double check my calculations for the required torque to move a 1" wheel on a 60 Kg bot?

Here are my calculations:

I have a NEO motor:
Max RPMs: 5600
stall torque: 2.6 Nm

The gravitational force on the robot is 60Kg * 9.8 m/s/s which is just under 600 N pushing down on the robot.

Because there are four wheels, the force is split 4 ways, 150 N per wheel.

I am trying to drive a 1" diameter wheel on a gear train. The gear ratio is 43:7 from the driver to the idler, and 7:5 from the idler to the output. The wheel is connected to the output gear.

load torque = force * wheel radius so the load torque for each wheel is 150 N * 0.5" or 0.0127m = 1.91 Nm

because the gear train reduces the torque of the motor, a 43:7 and 7:5 increase the speed to 48160 RPMs and reduce the stall torque of the motor acting on the output to 0.3 Nm, not enough to overcome the 1.91 Nm required.

Therefore to make the motor have enough torque to make the robot move, putting the motor on an 8:1 gearbox increases the torque on the output to 2.4 Nm and decreases the speed to 6040 RPMs, definitely enough to drive the wheel.

Can someone let me know whether my calculations make sense and if they are correct? I’d really appreciate it! Thanks everyone!!!

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You may want to draw a free body diagram. the motors do not need to overcome gravity to move the robot on the floor, you only need to overcome friction in the system which is usually pretty small. Also not sure I follow but you usually do not want to speed up the motor to drive the drivetrain, are you sure your gear train is setup to reduce torque? This would seem non optimal. I think your best bet is to consult the JVN design worksheet to figure what is your optimal gear ratio for the given current draw and speed/acceleration you want to achieve.

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You’re on the right track, but I believe you’re a little misunderstood about “load torque”.

Each wheel supports 150N. This doesn’t dictate how much force it takes to move the robot, though; it limits how much torque you can put into the ground before the wheel starts to slip. What you’re actually solving for here is “how much torque do I need to spin the wheels if my robot was driving against a wall”.

Here’s a great site explaining the concept in greater detail: Combat Robot Drive Train - Optimum Gear Ratio Selection

The torque required to move the robot is less than that, and much harder to calculate - it relies on a bunch of physical properties. In an ideal world, it would take almost no torque to begin moving your robot.

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the friction force would be the coefficient of friction of the carpet we’re driving on * the normal force right?

No this is not correct, that would be the maximum amount of force that could be applied to the floor by the wheel before the wheel slips (IE breaks friction), wheels actually make use of static friction to rotate, the force of friction applied by the wheel to the surface will only be as much as is needed to transfer the energy to accelerate the robot, as long as this does not exceed the max static friction you will not encounter wheel slip.

The friction i mention is the resistance to motion within the gearbox including, but not limited to, bearing resistance from grease, gear mesh friction, etc. This is usually quite low.

So I shouldn’t be worried about the robot moving with this gear orientation?

What should fall out from the math:

Assuming the motor system can overcome static friction, the mass of the robot influences the acceleration of the robot, not whether it can or cannot move at all.

More specifically:

The question is probably ill defined - In a frictionless environment, any non-zero torque will cause the robot to move by some small amount.

This is a direct outcome of Newton’s 2nd law, applied in the direction of travel of the robot:

\sum F = ma

where:

\sum F is the sum of all forces acting on the robot, in the direction of travel
m is the robot’s mass - positive and nonzero
a is the acceleration of the robot in the direction of travel.

Assuming the robot starts stationary, “move” would be defined by having non-zero a. Any non-zero a over time would imply nonzero velocity, hence “motion”

In a frictionless environment, the only F in the sum of forces is the force from the motor. For any m, as long as F > 0, a will also be > 0. Therefor, motion will occur.

The question should probably be - Calculate the required torque/forces to cause some specific amount of motion.

In FRC, this is often defined as “sprint time” - amount of torque required to cause a robot to move from Point A to Point B within a certain time limit.

All this stated another way: the calculation you performed is for lifting a 60kg load upward, against gravity. However, the FRC case involves moving a 60kg load move perpendicular to gravity. These aren’t the same physical setup, and some the math and intuition should be different.

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The robot will most likely move; I suggest you dig a bit deeper, though, and define your goal more specifically. Do you want your robot to move a certain speed? Is there a certain distance it’s repeatedly travelling?

If you can narrow down those specifics, you can end up with a gear ratio much more optimal for what your robot is designed to do.

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Spreadsheets are great, but you should use them after you fully understand how the calculation works. The OP is looking for help understanding the calculation, not just for correct numbers, so this isn’t helpful.

Re: the OP - posters above have covered how, in an idealized environment, the amount of torque it takes to “move” any robot is “anything more than no torque at all.” In the presence of friction, though, there is some minimal torque required to overcome the “static” friction in the drive. In FRC, we usually quantify this with a characteristic voltage term (typically called kStatic or kS) - it can be easily measured by slowly ramping the voltage to the drive motors and recording the maximal voltage applied before motion begins. Once you have kS, it is not difficult to back-calculate (an estimate of) the corresponding torque from the motor specifications.

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