Mechanics and Construction > Mechanics and Construction

Robotic arm

**alehandro**:

Hi new here, I'm making a project that need's a robotic arm. My base will rotate 180 and the elbow will have a length of 15cm and weight 1kg. Which should the torque of the elbow servo be? Is this a good one?? http://www.ebay.co.uk/itm/1-2-4x-Digital-MG996R-High-Torque-Servo-Metal-Gear-for-RC-Car-Truck-Boat-Model-/400669823345?pt=UK_ToysGames_RadioControlled_JN&var=&hash=item5d49c85171 Thanks..

**bdeuell**:

torque = force*distance (distance being the radial distance between the axis of rotation and the load vector)

I'm not sure I understand your design ... you talk about the base and then the elbow. Is there more than one joint? Perhaps a better description or ideally a picture.

If you are designing a joint that will move a 15cm long arm in a vertical plane then the max load would be when the arm is in the horizontal position (where the radial distance the force vector acts on is longest). This would give a torque of 15 kg-cm not accounting for any acceleration, frictional losses, or the mass of the arm. As your servo can only achieve this torque at stall it is probably a little too small especially given the other loads not accounted for in the calculation.

Note N-cm are the correct units to express torque but I used kg-cm because that is how your servo is speced. To convert mass to weight (force) use Force = mass*acceleration (acceleration in this case is acceleration due to gravity = 9.8 m/s^2)

**alehandro**:

Thank you very much for your answer!! To be more precise my robotic arm will have a base that will only rotate 180 degrees and then 10cm arm L1 until the elbow ant then another 15cm long arm L2. That's it only that. My L2 will be 1Kg with the arm mass. The L2 will be at a vertical position and I want it to move 45degrees right and left. I mean that the center will be at the vertical position and wont go at a horizontal position ever.

Also what does at stall mean? It's probably a stupid question but my inglish is a little rusty..

And to find the exact torque for my servo can you explain it one more time :) . You wrote me that "torque = force*distance" so my distance is 15cm and my force is "mass*acceleration" which mass in this case is 1kg and acceleration 9.8m/s^2?? Thank you..

**jwatte**:

You will need to post a drawing or sketch of some kind -- I don't understand what you intend to convey with "L2 at vertical."

If the longest distance your gripper will be from the center is 25 centimeters, and the weight of your load plus the arm is 1 kg, then the holding torque of the L1 servo needs to be at least 25 kg.cm.

"stall torque" is the rating for an actuator/motor at which the motor stops and cannot produce any more torque. A motor that is held in stall for more than one or a few seconds is likely to destroy itself. When you see only one rating for a motor, it's usually the "stall torque," if you intend to apply continuous torque, the "working torque" of a motor will typically be 0.20 or 0.25 times the stall torque.

Thus, for a holding torque of 25 kg.cm, you may need a motor with stall torque rating of 125 kg.cm.

**alehandro**:

Thank you for replying. Here's a sketch, if it's to sloppy tell me to do a better one.. I will have a sun panel instead of a gripper and some other stuff too and that's why the weight of the arm and load will be 1kg.. It will go only 45degrees left and 45degrees right and the center of the L2 arm 15cm will be facing up. Which should the torque of the elbow servo be?

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