Simple gear pair: efficiency and output torque 1

[liauw]

Hi. I have a question about output torque and efficiency of a simple gear pair as shown on the picture. So, I have a pinion and a gear. I give an input torque Tp in the clockwise direction. Therefore, the pinion will rotate with ωp angular velocity in clockwise and the gear ωg in counter-clockwise. Then, I apply a load Tl in the opposite direction of the gear motion (clockwise direction). The gear is still rotating in the counter-clockwise. I neglect the friction between the gear mesh, but I consider the bearing friction both in pinion cp*ωp and gear cg*ωg. The moments of inertia of the pinion and the gear are Ip and Ig, respectively. My question is what the output torque To is because I want to find the efficiency of this gear pair. Thank you very much.

 

[GregLocock]

Tl=wg/wo*(Tp-cp*wp)-cg*wg at constant speed

obviously.

However by ignoring the friction at the teeth you are ignoring the main source of inefficiency in a typical gear pair, so your calculation of efficiency is probably useless.

Cheers

Greg Locock


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[liauw]

Hi GregLocock! Thanks for your reply. Sorry, I should've said that I'm interested in non-steady state condition (non-zero acceleration). But, I assume you think To equals Tl, right? I have thought about it, but I don't think it's correct. Here's my explanation.

If To = Tl, I would not get a correct efficiency η.

η = Po/Pi = To ωg/(Tp ωp) = Tl ωg/(Tp ωp)

we know that rg ωg = rp ωp and ωg/ωp = rp/rg, so

η = Tl rp/(Tp rg)

In this equation, we can see that Tl/Tp has to be rg/rp in order to have η = 1. If I don't have Tl/Tp close to rg/rp, the efficiency could be much less than 1, for example 0.2, which doesn't make sense.

 

[GregLocock]

Tl+Ig*wgdot=wg/wo*(Tp-cp*wp-Ip*wpdot)-cg*wg

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[liauw]

Thanks for the equation GregLocock! But, I still don't think that the output torque equals the load (if that's what you meant). what do you think about my explanation on the efficiency? Thanks again

 

[GregLocock]

"what do you think about my explanation on the efficiency? "

Incomprehensible.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[3DDave]

Greg - you are far more patient than most.

 

[GregLocock]

Nah, just waiting for the the sun to power up the solar before I go outside for another exciting session of electric wood splitting. Very glamorous lifestyle round here!

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

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[BrianPetersen]

Overcomplexification.

 

[3DDave]

I guess the real question is one about transmissibility, which is frequency dependent and very touchy with backlash. At a certain input frequency the input gear teeth don't rotate far enough to touch the mating teeth before reversing.

It's weird to refer to it as efficiency because there is no loss of energy due to the mesh.

 

[liauw]

Hi 3DDave. Thanks for your reply. I agree with you that the efficiency should be 1 because the loss of energy due to mesh is neglected. But, let's say I consider it now. Still, I don't know what the output torque is. My objective is to develop a mathematical model for this system. I can use a spring and damper to account for the gear-mesh. Thanks a lot!

 

[gruntguru]

Some things don't deserve to be mathematically modeled.

je suis charlie

 

[BrianPetersen]

Skip all the greek letters. The output torque is the input torque multiplied by the tooth ratio multiplied by the mechanical efficiency. The inertia of the gears isn't worth worrying about in most applications.

 

[SwinnyGG]

I don't know what the output torque is

The torque applied to the gear by the pinion is unaffected by load.