Total machine system efficiency

[Pinon1977]

TL;DR Summary

Trying to determine the total efficiency of a system of gearboxes within a larger machine system

Please see the attached sketch. Basically I have a system of three gear boxes, each with their own respective efficiencies. I'm trying to determine, at the end of this string of gearboxes, what the overall efficiency is. How might one go about determining this? Do you just take the average? 70 + 80 + 90 / 3?

 

[berkeman]

I'm no expert on gearboxes, but for most systems you would multiply the efficiencies of the series systems to get the overall efficiency. So 0.7 x 0.8 x 0.9 = ?

 

[Baluncore]

Consider a line of 16 gearboxes, each with an efficiency of 90%. The average efficiency would be 90%. But energy must pass through each gearbox in turn to reach the next, with a loss at each step.
In reality, the efficiency would be;
0.90¹⁶ = 0.1853 = 18.5%.

 

[Pinon1977]

Consider a line of 16 gearboxes, each with an efficiency of 90%. The average efficiency would be 90%. But energy must pass through each gearbox in turn to reach the next, with a loss at each step.
In reality, the efficiency would be;
0.90¹⁶ = 0.1853 = 18.5%.

Wow!!!! That's not the explanation I was expecting, but it make sense to a certain degree.

Does it matter what kind of gearbox it is? Planetary vs worm gear vs helical, etc? I was loosely presuming that there would be some sort of gearbox constant or multiplier (depending upon the type of gearbox being used).

 

[jack action]

Do a Free Body Diagram (FBD) on each gearbox to get the answer.

 

[Baluncore]

Does it matter what kind of gearbox it is? Planetary vs worm gear vs helical, etc?

Different types of gearboxes have different energy efficiencies.

Generally, a two-step reduction box is less efficient than a one-step reduction, but the one-step reduction weighs more for the same ratio and power.

The bearings used inside the gearbox make a big difference as they are subjected to significant side forces on the shafts.

Planetary gears can cancel side forces on the shafts, so are often more efficient.

The ease of driving a gearbox backwards has efficiency implications. A worm gear is very inefficient when driven backwards.