Gear Ratios and Efficiency

[knji]

Assuming that we have a gear ratio g < 1 from the driving to the driven gear in terms of the number of teeth. Say g = 0.01 and assume the gear train efficiency e is 0.8.

To me, this means that the effective output torque, Tout, will be reduced so that

Tout = e x Tin/gr .

Also, if one uses energy conservation

[Tout x Omega_out] / [Tin x Omega_in] = e,

this also tells me that the effective speed at the shaft will reduce.

This is where I get confused. Motors have a inversely proportional torque-speed relationship. Doesn't this say that if the effective torque reduces, then the effective speed increases?

TIA

Klaus

 

[knji]

I must be missing something in the above equations. Does a reduction in efficiency translate to both a reduction in output torque and speed or just output torque alone?

i.e

Tout = e x Tin/gr and
Vout = gr*Vin

=> Tout x Vout = e x Tin x Vin

Is this valid?

TIA

Klaus

 

[GregLocock]

The average tooth speed changeth not.

with a gear.

Cheers

Greg Locock

 

[ivymike]

...unless the gear is made of something verrrrrry stretchy, right?

 

[knji]

So are we saying the a reduction in drive train efficiency translates to a reduction in output torque ONLY?

TIA

 

[GregLocock]

Yes. Unless you have rubber gears.

Cheers

Greg Locock

 

[knji]

Thank you Greg.

Klaus

 

[Sparweb]

It might be helpful to look at the lost torque more carefully:

T_out * Omega_out = (T_in - T_friction)* Omega_in

If your goal is to apply a specific output torque, then you need to use more input power than you would expect if you didn't take friction (and other losses) into account.


Steven Fahey, CET
"Simplicate, and add more lightness" - Bill Stout