Belt Drive System Prob.


by Hiro Kazayama
Tags: belt, drive, prob
Hiro Kazayama
Hiro Kazayama is offline
#1
Mar22-13, 02:52 AM
P: 1
Ok. Say you have a belt drive system that is run by a motor at one end. You have a load F located somewhere in the middle of the belt and the belt does not deflect (it is rigid).

How would I determine the torque required by the motor to move this load F?

Thanks in advance!
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Travis_King
Travis_King is offline
#2
Mar22-13, 10:31 AM
P: 763
Do you know how to calculate the mass moment of inertia of an object at a distance from the primary axis?
tygerdawg
tygerdawg is offline
#3
Mar22-13, 11:01 AM
P: 138
You must accelerate the mass.
F=ma.
Torque-pulley = F x Diameter-pulley.

You must also keep it moving, which is an inertia issue.
T=Jα

Then add all the other torques that you can account for. Size the motor for that, plus safety factor & efficiencies.

Travis_King
Travis_King is offline
#4
Mar22-13, 11:59 AM
P: 763

Belt Drive System Prob.


Well...It's more in depth than that leads one to think.

It all comes down to inertia and, as tygerdawg said, acceleration.

But you've got to consider them together, so instead of using m*a*D, as tygerdawg suggested, you should use T=J*(Angular accel).

So you have the following things to consider:
-- The belt is considered to be rigid, so if you assume that the belt/pulley system transfers torque with perfect efficiency (there's no losses from slip at the pulleys or stretch in the belt (and if there is, just add the efficiency factor in)), the mass is effectively acting on the motor pulley.
-- For masses that are relatively low in profile (that is, the COM isn't incredibly high above the belt) you can consider the effective inertial radius as the radius that of the pulley diameter for quick calculations.
-- If the mass has to be moved upward (i.e. at an incline) you have to account for that as well.

Inertia of a point mass (COM) at a distance from the primary axis is:
J(load) = m(load) * R(pulley)^2

m(load) is going to be:
W(load) / a(gravity)

Make sure you divide by the correct units for gravity. Inertia deals with masses, not forces. (which is why torque calculations are much easier on the brain in SI than in Imperial)

When you find your inertia, J(load), you can then multiply it by the angular acceleration:

(Angular accel) = (wi - wf) / (time)

Remember though, this will only be the torque required to accelerate the object. If you wanted to be more precise, you would add in the inertia of the belt (which can be approximated as m(belt)*R(pulley)^2), the inertia of the motor (as in the rotor inerta, which is found using the moment of inertia formula for a hollow cylinder), and the inertia of the pulleys, J(pulley)= .5 * m(pulley) * (Router^2 + Rinner^2).

Then you combine them all to get the total system inertia and multiply that by the angular acceleration to get your actual required torque.

There are other formulas for finding moments of inertia (using densities and what have you), and systems can get very complex. For instance, if you have any gearing, you have to translate the inertia through the gearing (which goes as a function of the ratio of the angular velocities squared), and if you are considering losses due to friction or slip or stretch, or if your object has a strange geometry that must be considered in the inertia calcs, or if it moves on the conveyor, etc. etc.

But this should give you a back of the envelope type of answer.


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