How does gravity torque affect the position control system of a robot arm?

In summary, the robot arm experiences a downwards force due to gravity. This force needs to be linearised so that it does not disturb the system's dynamics.
  • #1
remz
9
0
I'm working on a developing a position control system for a robotic shoulder joint driven by a DC motor. To simplify, the robot arm can be viewed as being in a fully extended state.

I have now reached a stage where I can no longer ignore the affect gravity has on the torque generated by the motor.

The robot arm, in a fully extended position slighty above the horizontal, will experience a downwards force of F=mg sin(theta), where theta is measured from the vertical equilibrium point to the centre of gravity of the arm. As the arm is connected to a pivot, a moment is created with magnitude, T=mgr sin(theta), where r is the length of the arm measured from the pivot to the centre of gravity.

The presence of sin(theta) is non-linear and thus has to be linearised around an operating point. If we were to assume that the robot shoulder arm would not pivot at more than +-45deg from the horizontal axis, then, taking theta from the horizontal line, rather than from the vertical as we did previously, we can linearise the torque caused by gravity on the arm, as T = mgr cos(theta), where cos(theta) ~= 1 and so becomes, T = mgr.

The problem with this is that in developing the transfer function for my robot arm, that is (dw/dt) / i, where i is the current fed to the shoulder motor, the torque caused by gravity now appears as a disturbance. i.e., it is not a coefficent of either my input or output. Which is ideally what I would like.

Actually, I've just realized this isn't a problem as I have found a reasonable way to represent this disturbance in my block diagram and hence, my system transfer function. But, having spent all this time writing this I think i'll just post it all the same. Maybe you can criticize the logic of the above.

Rem
 
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  • #2
Heh, well, for what it's worth, except for not seeing why you care if it is a linear function or not, the logic looks fine.
 
  • #3
The linearisation will be to analyse the system in the s-domain.
 

What is gravity torque on a robot arm?

Gravity torque on a robot arm refers to the force exerted by the weight of the arm and any attached objects, such as tools or parts, that act on the joints of the arm. This force can cause the arm to move or resist movement, and must be accounted for in the design and control of the robot arm.

Why is gravity torque important for a robot arm?

Gravity torque is important for a robot arm because it affects the stability, accuracy, and performance of the arm. If not properly accounted for, the arm may not be able to accurately complete tasks or may become unstable, potentially causing damage to itself or its surroundings.

How is gravity torque calculated on a robot arm?

Gravity torque is calculated by multiplying the mass of each part of the arm by the distance from the center of gravity to the joint and then by the acceleration due to gravity. This calculation must be done for each joint of the arm and must also take into account any external forces acting on the arm.

How can gravity torque be minimized on a robot arm?

Gravity torque can be minimized on a robot arm by using lightweight materials, reducing the length of the arm, or redistributing the weight of the arm and any attached objects. Additionally, advanced control techniques can be used to compensate for the effects of gravity torque.

What are the potential consequences of not accounting for gravity torque on a robot arm?

If gravity torque is not properly accounted for, the robot arm may not be able to accurately perform its intended tasks. This could result in errors, damage to the arm or its surroundings, or even safety hazards. In extreme cases, the arm may become unstable and unable to function properly, requiring costly repairs or replacements.

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