Calculating Linear Force from Motor Torque?

[ksafin]

We've been wondering how one goes about calculating linear force from motor torque.

For example, if you have a motor of X torque, with a spindle of radius R attached to it, how can you determine the linear force (ie tension in the string being rolled onto the spindle)?

That's just an example.

In reality, our setup is such that the motor has an 40 tooth gear to a 80 tooth gear, which is then on the same axle as a pinion gear which moves across a rack for our linear slide.

We're trying to find out how much this rack and pinion can lift; the motor's torque is 283 oz-in.

[AdamHeard]

I'll give a half answer, as I wager you can figure it out from there (and you gain more insight that way!).

Torque = Force x Distance

This is the perpendicular distance from force to the point of rotation.

The next step of course is to determine whether or not it's appropriate to use the motor's stall Torque for such calculations. And if not, what Torque would you use?

[ksafin]

Originally Posted by AdamHeard

I'll give a half answer, as I wager you can figure it out from there (and you gain more insight that way!).

Torque = Force x Distance

This is the perpendicular distance from force to the point of rotation.

The next step of course is to determine whether or not it's appropriate to use the motor's stall Torque for such calculations. And if not, what Torque would you use?

I'm actually in IB Higher Level Physics and have been in AP Physics B and all, so I'm familiar with torques, but this in particular has been a confusing question for me.

So assuming the motor has torque of 283 oz-in, and a shaft diameter of .25 in (thus radius of .125), if you were to simply wrap a string around the shaft and attach a mass to the end of it, would the rotation of the motor shaft produce a linear force of (283 / .125)? It just sounds unreasonable because the result is 141 lbs of force.

[AdamHeard]

Originally Posted by ksafin

I'm actually in IB Higher Level Physics and have been in AP Physics B and all, so I'm familiar with torques, but this in particular has been a confusing question for me.

So assuming the motor has torque of 283 oz-in, and a shaft diameter of .25 in (thus radius of .125), if you were to simply wrap a string around the shaft and attach a mass to the end of it, would the rotation of the motor shaft produce a linear force of (283 / .125)? It just sounds unreasonable because the result is 141 lbs of force.

That number is correct. That's the effect of small lever arms. Also, electric motors have peak torque at stall.

So your reasoning is valid thus far, but what is the consequence of using stall Torque? What is the current draw? How fast does it move?

[ksafin]

Originally Posted by AdamHeard

That number is correct. That's the effect of small lever arms. Also, electric motors have peak torque at stall.

So your reasoning is valid thus far, but what is the consequence of using stall Torque? What is the current draw? How fast does it move?

Well, it wouldn't move

By what means can I calculate a given torque at a given speed, then?

The motor is 152 rpm.

If I wanted, say, 80 rpm, could I find the percentage of total RPM (80/152) and multiply that by the stall torque to find corresponding torque at the given speed?

I'm assuming speed and torque change by the same factor, proportionally to the drop in voltage.

[AdamHeard]

Originally Posted by ksafin

Well, it wouldn't move

By what means can I calculate a given torque at a given speed, then?

The motor is 152 rpm.

If I wanted, say, 80 rpm, could I find the percentage of total RPM (80/152) and multiply that by the stall torque to find corresponding torque at the given speed?

I'm assuming speed and torque change by the same factor, proportionally to the drop in voltage.

Yes, Good.

Assuming constant voltage, T = -Ts/wf * w + Ts.

T = torque
w = speed (in the same units as wf)

These are both variables.

Ts = Stall Torque
wf = Free Speed

These are both constants (and linearly scale with voltage).

So, to give the last tidbit. An accepted practice is to determine the max load the lift must lift and/or hold. You'd then want to choose your gear ratio and drum diameter such that this corresponds to some percentage of stall torque. The lower the percentage, the less heat you'll make and the longer the motor will last (but the slower you'll go).

Depending on the time it needs to run, and how much it has to hold still (Versus just lift up, then down quickly) anywhere from 10%-25% are valid numbers for FRC. Be careful trusting the word of others (including me) too much on that %age though. It varies motor to motor and is a function of how much heat is made to how much heat the motor is able to dissipate from the armature.

[ksafin]

Gotcha, thanks so much!

Just a quick question.

Right now we have it set up such that the motor has an 80 tooth gear, turning a 40 tooth gear, which is on the same axle as a rack and pinion gear (32 tooth I believe).

I feel like the calculation is a bit different in this case.

My assumption is that I can say the motor torque "doubles" from the gear ratio, lending to a 283 * 2 oz-in torque on the axle w/ the 40 tooth and pinion gear.

At this point, do I use the radius of the pinion gear to determine linear force on the rack and pinion?

[AdamHeard]

Originally Posted by ksafin

Gotcha, thanks so much!

Just a quick question.

Right now we have it set up such that the motor has an 80 tooth gear, turning a 40 tooth gear, which is on the same axle as a rack and pinion gear (32 tooth I believe).

I feel like the calculation is a bit different in this case.

My assumption is that I can say the motor torque "doubles" from the gear ratio, lending to a 283 * 2 oz-in torque on the axle w/ the 40 tooth and pinion gear.

At this point, do I use the radius of the pinion gear to determine linear force on the rack and pinion?

Correct. Don't forget that free speed will be halved by that ratio as well.

For a Given gear ratio N, To = N*Ti and wo = wi/N.

To = output Torque
Ti = input Torque

wo = output speed
wi = input speed

I'd wager you already knew that, but I figured others reading might not.

Also, for multiple stages of gear reduction, lets say N1... N4. The total reduction Ntot = N1*N1*N2*N4.

EDIT: Magnets is correct that each stage of gearing will cause some energy loss. However I do disagree that a significant safety factor is always necessary.

[ksafin]

OK, so from the gear ratio, we calculate a new torque of 35.375 lbs-in.

The radius of the pinion gear is ~ 0.4 in, lending to a linear force of 88.4375 lbs at stall Torque.

Are there any calculations that need be taken into account regarding the pinion gear teeth number or the length of the rack or whatnot?

[AdamHeard]

Originally Posted by ksafin

OK, so from the gear ratio, we calculate a new torque of 35.375 lbs-in.

The radius of the pinion gear is ~ 0.4 in, lending to a linear force of 88.4375 lbs at stall Torque.

Are there any calculations that need be taken into account regarding the pinion gear teeth number or the length of the rack or whatnot?

The pinion teeth number would be used to find it's radius, the length of the rack wouldn't matter for the current calculations (but could be used for calcs like how much energy is required to traverse the entire rack, etc...).

[BBray_T1296]

To find torque at a specific RPM/Amp draw (at least, for a CIM motor), Direct your attention to the bottom left of the AndyMark CIM specs Try searching your motor number and see what you can find

[KrazyCarl92]

Originally Posted by ksafin

Right now we have it set up such that the motor has an 80 tooth gear, turning a 40 tooth gear, which is on the same axle as a rack and pinion gear (32 tooth I believe).

Originally Posted by AdamHeard

Correct. Don't forget that free speed will be halved by that ratio as well.

For a Given gear ratio N, To = N*Ti and wo = wi/N.

To = output Torque
Ti = input Torque

wo = output speed
wi = input speed

If I am understanding the set up properly, I believe this would have the opposite effect. The set up described has an 80 tooth gear on the motor and a 40 tooth gear being driven, which results in double the speed but half the torque. Using the equations above, isn't N = 0.5?

[safiq10]

Completley off topic.. OMG so this is how you use math in FRC???? Sorry our team measures and dose basic math but nothing like this.

[KrazyCarl92]

Originally Posted by safiq10

Completley off topic.. OMG so this is how you use math in FRC???? Sorry our team measures and dose basic math but nothing like this.

Utilizing motor curves and calculations from physics are necessary conditions for optimization of an FRC robot, or any system with electric motors.

JVN's Mechanical Design Calculator and Ether's Simple Motor Calculator are great tools available on these forums that can help out.

[cgmv123]

Originally Posted by safiq10

Completley off topic.. OMG so this is how you use math in FRC???? Sorry our team measures and dose basic math but nothing like this.

You won't ever need to do math like the above examples to succeed in FRC.

That said, understanding physics principles (like the ones above) can be helpful and are often a part of the engineering process, which is one of the key parts of FRC.

Keep in mind that the goal of FIRST is to inspire future engineers. You aren't expected to know anything coming in. Learning engineering principles is also a part of the process.