Load Inertia through gear reduction 3

[Tagger]

I was at the understanding that the load inertia (lbft^2) is reduced or increased based on the gear ratio. Example:
10,000 lbft^2 at 756 RPM, the reflected load inertia on the low speed side of 282 (2.681 gear reduction) will give 26,810 lbft^2.
But I have seen an equation where the reduction ratio is squared. Reflected inertia = Load inertia/(reduction ratio^2).
Which is correct?

 

[scarecrow55]

I think the suared forml is correct - see in here

http://uk.geocities.com/mikebull2001/veryfirstpage.htm

 

[PeterCharles]

I was always taught it was r^2, where r is the gearbox ratio.

 

[gaufridus]

Torque varies as the ratio of the gears (units of torque = lbft)
Inertia varies as the square of the ratio (units of inertia = lbft^2)

 

[Tagger]

So what ya'll are saying is if there is 803,000 lbft^2 inertia acting on the shaft where I want to place a brake that is rotating at 756 RPM, then if I place the brake on the low speed side (282 RPM) then the inertia would increase to 803,000/((282/756)^2) = 5,771,130 lbft^2? That can't be right....or can it. The brake would be impossibly hugh to apply the torque to stop that inertia in an emergency stop time!

 

[Warpspeed]

The total braked energy (frictional heat generated) will be the same, but the torque required to do the same job will be far higher. It all depends on how quickly you need to pull the speed down.

 

[zekeman]

Yes. Keep in mind that the energy that you have to stop is
1/2*I*w^2.and it remains the same no matter where you choose to absorb it. If you absorb it at the low speed end, the I is increased by the gear ratio squared; However, the speed is reduced by that same factor, so that energy is the same. NO FREE LUNCH.

 

[tygerdawg]

Perhaps it may be more like this:
the mass moment of inertia of your rotational loads (gears in the gearbox, your turning load, and the motor rotor) never changes. Look at the formula for MMI for proof: it is not a function of any speed, but only geometry and mass.

A couple of things DO CHANGE, however. But it ain't MMI.

Torque required to accelerate or decelerate the load will change with speed. Remember that fella Newton and his F = Ma and T = J(alpha) formulas? "a" and "alpha" is what changes, not inertia or mass.

Also, the torque at each point in the drive train is what changes. Take for example the case of a standard industrial gearmotor. You have an AC induction motor connected to a gearbox with the output shaft turning a load. The motor produces 1750 RPM and, say, 50 lb-in of torque. The output shaft will produce RPMout = (1750/ratio) with torque Tout = (Tin X ratio). The motor rotor shaft may only be 12mm in diameter, but the gearbox output shaft will be 50mm. It must carry the extra torque produced by the gears.

And to stop a large-inertia output load, all that is necessary is to put a brake on the motor sufficient to apply work in the opposite direction of rotation to convert the motor rotor speed & torque to 0 RPM and heat in a certain period of time. The applied torque will be multiplied through the gear ratio. When braking torque is applied, then it is used to decelerate the rotational inertia of the drive train. That's T=Ja and the mounting bolts on the motor will see that.

Perhaps you are getting confused about the concept of "inertia matching" of the load to the motor rotor. This is a function of the square of the ratio. This type of analysis is done during sizing studies to help eliminate resonance or motion ringing of the motor rotor RPM when it tries to accelerate a large load inertia. What is sought is a certain ratio of load-to-rotor inertia through a gear train. There are several rules of thumb used in industry for various light- and heavy-duty applications.

Do a Google search for "Smart Motion Cheat Sheet". It's a PDF file for free and has all this stuff in it. Too bad we were all too busy having fun in college to actually learn this stuff in Sophomore Dynamics, eh?

TygerDawg