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necessary torque for an arm
We've calculated that there are 51 ft-lbs of torque resisting our arm's movement. To move the arm we were thinking of running two RS-550's at peak efficiency. They would be run through a CIM-U-LATOR (2.7:1) and then a p80 (256:1) this gives us 63 ft-lbs of torque as an output.
The question is, do we then do an additional reduction to increase the torque? The current reductions have us running at 25 RPM and we would like to stay above 15 RPM. |
Re: necessary torque for an arm
51 ft-lbs is a pretty heavy arm...
that's 63 ft lbs at peak efficiency, right? Also, 25 RPM is 150 degrees per second - a QUITE high arm speed. |
Re: necessary torque for an arm
The arm needs to travel about 90 degrees (it's a 4-bar linkage) so at 15 fps we're looking at one second total for arm travel. I've tried to convince the others that that's a bit too fast (I use "a bit" loosely) but to no avail. Also, the arm is still in the design stage, so I guessed high on the weight.
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Re: necessary torque for an arm
The main thing would be that it's a lot more difficult to precisely control an arm that travels over its entire reach in half a second than it would to control one going up in 2 seconds.
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Re: necessary torque for an arm
Agreed. 25 rpm is way too high. We're looking at something like 10 rpm
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Re: necessary torque for an arm
Okay, so if I add a 2:1 reduction, we'll be at 126 ft-lbs and 12.5 RPM at peak efficiency.
There's a potential problem. The p80's are rated for 85 ft-lbs max. Is this for the torque inside the gearbox, or the torque of the system? |
Re: necessary torque for an arm
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Re: necessary torque for an arm
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And so that system would be fine? (mind-numbing speed aside) |
Re: necessary torque for an arm
If it's worth anything to help you judge speed:
We will be lifting a 3.5 pound arm that travels at 100 deg/sec (100 deg is full travel). We also have options to slow it down via programming or different sprockets incase we find it is either too quick or too slow. |
Re: necessary torque for an arm
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http://www.chiefdelphi.com/forums/sh...php?p=1006598# |
Re: necessary torque for an arm
The way I got to the 51 ft-lbs was by figuring out the reaction forces on the two pivot points for the top bar in our four-bar linkage. The reaction force was 76 pounds, multiplied by the length of our driven bar gives us 51 ft-lbs.
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Re: necessary torque for an arm
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The tests you linked to demonstrate 40 lbs on a 2" pulley which is 80 inch-pounds of torque. The original poster calculated that they need up to 51 foot-pounds of torque for their arm. That's 7.65 times more torque than the scenario in your video. Quote:
Without doing the math myself, and without any formal mechanical training, the numbers you have posted seem plausible. In addition, you may be able to lower the required torque by partially balancing the arm with surgical tubing or a gas spring. |
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Re: necessary torque for an arm
I'd suggest some possible safety issues in addition to what has already been said. That heavy of an arm going at those speed could do serious damage to other robots. I'm all for an aggressive game, but lets not crack each others lexan.
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Re: necessary torque for an arm
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