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Unread 23-01-2011, 21:36
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Re: necessary torque for an arm

Quote:
Originally Posted by 548swimmer View Post
This is assuming a 10 lb manipulator approximately five feet away from the closest pivot point on our four-bar linkage.
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.
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Last edited by Molten : 23-01-2011 at 21:46.
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Unread 23-01-2011, 21:58
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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.
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Unread 24-01-2011, 08:06
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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.
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