Fisher Price Motor Power

[billbo911]

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
  • 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)
  • High ratio (>7:1) 50%
  • 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
  • Best case 50-60% (high lead angles of 40 deg, good bearings, etc.),
  • 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.

Joe,

WOW, this is one of the best and easiest to follow posts on this subject that I have ever read! Being an Electronics Engineer, this mechanical example goes a long way to explain to me how to approach this type of design.

I do have have one question though. The following comment lost me, I would like to know where you came up with the 100N/m value?

...and assume your robot weight is 600N, then you need 100N-m to lift your robot.

I'm sure the answer is obvious to some, but for some reason I just can't seem to see how you came up with this value.