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Unread 11-03-2011, 01:02
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Ether Ether is offline
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Re: Motor for arm: what should be used when?

Quote:
Originally Posted by clayman View Post
Ballpark, I'd say 40 ft-ibs.
I assume that's 40 ft-lbs before you balance it with springs :-)

Just for fun, let's do a little engineering and see what happens at 40 ft-lbs
and why the gear ratio is so important.

Let's say you want to move the arm at 90 degrees per second. That's 15 rpm.

You said your total gear ratio is 12.75*2:1 = 25.5:1.

So your speed at the motor is 15*25.5 = 383 rpm.

Your torque at the motor is (40/25.5)*12*16 = 301 ozin.

Using the motor curves for the CIM and a bit of algebra, 11.4 volts will give you 301 rpm at 383 ozin. The motor efficiency at this operating point will be 6.4% and you'll be pulling 117 amps (until your circuit breaker trips).


Now let's see what happens if you change the gear ratio to 250:1

The rpm at the motor is now 3750 and the torque at the motor is 31 ozin.

Using the CIM motor curves again, we see that 9.6 volts will produce 31 ozin of torque at 3750 rpm. The motor efficiency will now be 65% and you'll be drawing only 14 amps.


Of course, the problems with high gear ratios are

- some loss of efficiency (which I haven't accounted for here)

- weight, size, expense

- problems with overloading the gears if they are too small


But there are problems with using balancing springs instead of gear-down:

- you can't balance over the entire operating range

- if you don't gear down enough, you lose fine control of the arm


So, no surprise, as with everything in engineering it's a matter of finding the right compromise to meet the task at hand.




Last edited by Ether : 11-03-2011 at 01:48.
 


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