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.

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.

Re: necessary torque for an arm
 

Agreed. 25 rpm is way too high. We're looking at something like 10 rpm

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
 

That rating is the max recommended torque at the gearbox output.

Re: necessary torque for an arm
 

Happy Birthday!

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
 

I feel like this needs to be reduced further, perhaps with a chain. If it is that fast it will be hard to control. Anyway nice to know some other people know the Cim-u-lator exists, we were thinking of using one also because weve reached our CIM limit because of mecanum drive. But unless your arm weighs more then 40 pounds, one rs550 with a 64:1 transmission will do just fine. I've posted some videos of our team testing an rs550 the thread is here:
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.

Re: necessary torque for an arm
 

Please be careful giving advice like this. This statement is absolutely false.

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.

Thanks!

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.

Re: necessary torque for an arm
 

Great advice, thanks! My main concern was with the theory being employed. I understand all the math we're using and can tweak the pivot distances to minimize the torque.

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.

Re: necessary torque for an arm
 

This is assuming a 10 lb manipulator approximately five feet away from the closest pivot point on our four-bar linkage.

Re: necessary torque for an arm
 

10 lbs isn't bad considering that. Is that including the weight of the arm itself? or is that just the actual manipulator on the end? If your considering the weight of just the manipulator, your numbers are going to be a bit low. If your considering the weight of the arm too and are placing it at the end, they will be a bit high. You should be placing the weight of the arm at the midpoint of the arm and it should give a pretty good approximation. There will be a bit of error because of various things, but you should be pretty close.

PS: I'm intrigued by the thought of a 5 foot long manipulator.

EDIT: Also, don't forget rotational inertia. Its easy to focus on gravity being your resistance that is fighting you, but it isn't the only one. You also need to consider rotational inertia. You should be able to do some research on it, but if you continue to have problems after that...give me a PM and I'll help you through the math.

Re: necessary torque for an arm
 

That pesky Tnet=Ia equation?

Not too concerned about that, especially since it looks like we'll be going with 126 ft-lbs of total torque.

Re: necessary torque for an arm
 

This year there are going to be some heavy arms with heavy manipulators hang way out there. As noted the forces on joints, motors and gear boxes are substantial. We are going to use a passive device to balance the arm so that the gear train only has to hold position not lift the entire load. Fiber glass pultrustions can reduce weight over AL.