Chief Delphi - Gearbox efficiencies

I have searched the white papers, but can not find the one I use regularly for calculations. It was originally posted by Joe Johnson. I will copy it in it's entirety here, hopefully someone can locale the link to it's posting location.

Quote:

Originally Posted by Joe Johnson
Yes, efficiency is a huge factor and very important for sizing gearboxes and designing mechanisms.

I use the following for designing using torque*:
• Straight Spur Gear with good bearing condition 95%
• Straight Spur Gear with funky bearing conditions 90% (look at the final stage of the FP transmission for an example of a funky bearing condition)
• Planetary
o Low ratio (<5:1) 85-90% (depends on how good the bearings are, the grade of the gears, the size of the planets w.r.t. their axles... things like that)
o High ratio (>7:1) 50%
o Very High Ratio (>20:1 -- "hunting tooth" stages) 15% <<these are great for speed reduction by division, but lousy for torque increases by multiplication
• Helical Gear, parallel axis 80%
• Helical Gear, cross axis see Worm Gear
• Worm Gear -- totally depends on lead angle
o Best case 50-60% (high lead angles of 40 deg, good bearings, etc.),
o Worst case 5-15% (lead angles of 10 deg, bad thrust management, etc.)
• Conical Gears / Bevel Gears depends on bear arrangement and alignment 60-90%
• Chain 90% (assuming good alignment and tension)
This is PER STAGE.

Example: If you have a 4 stage 4:1 per stage spur gear gearbox with good bearings it would be .95^4 = 81% efficient. So... ...instead of getting a ratio of 256:1 your "effective ratio" (from a torque point of view) would be 207:1.

Continuing with the example, if you put a FP in with a stall torque of .45N-m then you would get 93N-m out of this gearbox, not 115N-m. Now suppose you are trying to lift your robot with this gearbox and you have the output connected to a .17m arm (and assume your robot weight is 600N, then you need 100N-m to lift your robot.

NOTE: You are not going to lift that robot, all you are going to do is turn a lot of electrons into heat.

Continuing, if you put a 3:1 chain stage between the arm and the gearbox, the effective ratio would be 560:1 (207X3*.9). You could put 250N-m of torque on your arm. Now your motor would be loaded at 40% of its stall during your lift (and the motor would be running at 60% of its free speed or the arm would be turning under load conditions at 12RPM = 16,000RPM *.6/(256*3)<<Note: Actual Ratio used for SPEED, Effective Ratio used for Torque).

Now you'd lift in a heartbeat (1/2 turn in 6 seconds -- well... ...kind of a LONG heartbeat ;-) and you have extra torque should another robot get in your way on the way up.

Life is good.... ...always.

Joe J.

*Some say I am too conservative but my experience with FIRST and with automotive actuators tells me that these numbers are not far from the right ones.