Joint torques coupled to motor torque in cable-actuated robotic finger

[Yorik]

I am working on creating a robotic finger with the design choice as in the picture above. The green cable should be attached to a motor/pulley, which is the driving actuator. The cable is then wound around three passive joint pulleys in the finger and each joint has a counteracting spring for returning to its rest position. Before going to a real setup, I am modeling the dynamic equations of this system in Matlab. I derived dynamic equations using the Euler-Lagrange method on a triple pendulum, which of course does not yet include the cable mechanism. Now, I have equations for all three joint torques [tau1, tau2, tau3].

From here, I do not know how to model the cable transmission mechanism and relate the motor torque (tau_m) to the individual passive joint torques. It is obvious that in a kinematic chain like this, the first joint experiences a higher torque than the second and third due to gravity, inertia and external forces further on in the kinematic chain. However, since the joint are coupled through the cable and to the motor torque and the latter is the only variable I can control, how do the dynamic equations relate to the individual joint torques?

Because, in a cable (I know there are some losses due to friction, but leave them out for now) the tension force should be the same along its entire length. But if the tension is the same everywhere, then also should the induced torque in all joints, right? This seems like a contradiction to me.

Can someone try to explain what happens in a mechanism like this and how to model the dynamics properly?

 

[BrianE22]

Interesting - no help to you but how do you preload the cable? I'm guessing you put a bar of a certain diameter in the grip?

 

[waross]

tension force should be the same along its entire length

Try: tension force should be the same along the entire length of each section.
Consider the finger working horizontally to avoid complications due to gravity.
Ignore friction.
Foe explanation assume that each return spring exerts an equal force.
Now visualize a weight equal to the force of the last spring suspended on a rope.
Above that fasten a weight representing the second last joint.
Above that another weight representing the first joint.
When you get that working you may calculate and add in the gravity contribution.
Hint: Gravity will act with the return spring, for or against depending on the orientation.


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[Yorik]

@waross. Thankyou for your reply!

I still do not entirely get it, though. The way I see it is that tau1 is larger than tau2, which is larger than tau3. This seems logical, since the first joint also feels the inertia, combined spring forces and a larger moment due to external forces.
However, in a cable the tension should be the same throughout its length (assuming that the cable is massless). This would mean that the tension and thus the torque exerted on each joint is the same, so tau1 = tau2 = tau3. In an ideal frictionless cable transmission system, this is also what you would expect. Which way of thinking is true? And if the tensions and therefore the joint torques are the same, what is the torque the motor should then deliver? Can I just only consider the dynamic equation for the first joint (tau1), and say that the other two joint torques are equal to tau1?

 

[BrianE22]

If the pulley drums have rotational inertia then the cable loads going in and out of it would be different.

 

[waross]

You have four separate cable sections. Each pulley adds tension. Like adding weights to a vertical cable.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[BrianE22]

Ignoring pulley rotational inertia and friction, the cable loads (T1, T2) on each pulley would be the same. That's assuming the pulleys are idlers.

 

[Yorik]

This still holds true even though it is the same cable?

If I understand it correctly, it would then mean that the tension is highest between the motor and the first joint, creating a joint torque tau1 which is able to withstand all counteracting torques of the finger. The tension becomes lower after one pulley and thus exerts a lower torque on the second and eventually even lower on the third joint.

Then, is the motor torque needed to withstand all induced torques equal to the sum of tau1 + tau2 + tau3?

 

[waross]

If the pulleys are not anchored but are free turning, the cable tension will be balanced by the combined tensions of the return springs.
The same torque will be exerted at each joint.
That torque will be the directed sum of return spring torque, gravity and gripping force.
The difference is whether the pulleys are anchored or free turning.
With free turning pulleys, and neglecting the weight of the cable and friction,
Suggest that you compare gravity force with intended gripping force and decide whether gravity forces may be ignored for practical purposes.
Note that if the model is inverted so that the fingers are below horizontal, the gravity forces will be reversed.

If I understand it correctly, it would then mean that the tension is highest between the motor and the first joint,

Rather than more tension, think, slightly different deflection of the joint.
You are correct that with free turning pulleys the tension is equal over the length of the cable.
The tension is opposed at each joint by the directed sum of the return spring, gravity, and friction.
In the first joint, gravity will oppose and deflection will be less.
In the second joint gravity will be neutral.
In the third joint gravity will aid and deflection will be greater.
Friction:
Friction will restrain the pulleys and the tension may be greater or less closer to the operating motor, depending on the direction of movement.
The action of small changes may be erratic due to the difference between sticktion and friction.
This may be most noticeable when tension is reduced.
(Work on reducing the friction to negligible.)


--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

Friction is your enemy.
Friction will produce an effect that is an indeterminate combination of free turning pulleys and fixed pulleys.
As the cable tension is increased, stiction will resist movement.
When the stiction is overcome in one joint, that joint will move, relieving some cable tension and thus some torque on the other joints.
With significant stiction, the joints may move in small jerks and independently.
When evaluating friction you may be best to consider both static friction (stiction) and sliding or rolling friction.
The worst case that I know of when stiction was neglected was during the design and construction of the TRIUMF accelerator. (Tri Univesities Meson Facility) in Vancouver Canada.
This was a very large electromagnet enclosed in a vacuum tight vessel.
The vessel split horizontally and the top was lifted by a number of screw jacks.
The top weight several tons.
When it came time to open the vessel for the first time, the lift was stopped part way and then could not continue. The screw jacks would not lift it. They had been designed for sliding friction and could not overcome static friction.
The vessel had to be closed so that the screw jacks got a running start and then opened all the way without stopping.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

Then, is the motor torque needed to withstand all induced torques equal to the sum of tau1 + tau2 + tau3?

Try: The motor torque needed to withstand all induced torques equal to tau1 = tau2 = tau3 plus indeterminate friction.
In each case (neglecting friction) tau = spring force +- gravity.
(correction made)
--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[BrianE22]

The motor torque needed to withstand all induced torques equal to the sum of tau1 = tau2 = tau3

I disagree. The motor supplied force, F, equals tau1 = tau2 = tau3 (not the sum of each).

 

[waross]

Thanks for the heads up BrianE22. Cut and paste mistake. Correction made.

Is this clearer?
(tau 1 minus friction effects) = (tau 2 minus friction effects) = (tau 3 minus friction effects)
Friction may cause different tension in different sections of the cable.



--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[waross]

The joints will find an equilibrium position where opposing forces such as gripping force, spring tension and stiction balance the tension in the cable.
Friction and stiction in the pulley axles complicates things.

--------------------
Ohm's law
Not just a good idea;
It's the LAW!

 

[GregLocock]

This would be much easier to understand if you drew a free body diagram for each segment of the finger separately. Are the pulleys anchored to the segment, or are they free to rotate?

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?

 

[prex]

1) Think of a common belt drive: you have two pulleys with a tight side under tension and a slack side with nearly zero tension, excluding pretension. This explains how is it that the different sections of your cable have different tension forces: the moment applied to each pulley determines the difference of tensions in the intervening cable sections
2) Think, as suggested above, of a different but similar situation: a vertcal cable with masses attached to it at different heights: the force required to retain it is the sum of the individual weights. You can add a spring, but in this example, the spring is attached between each mass and a fixed point: the force to retain the cable will be the sum of the weights and of the spring forces. And of course, if you want to move the cable suspension point, you'll need to add inertia forces.
3) Lagrangian mechanics is the best way to handle this problem. My suggestion is that you start from a simpler system, then adding elements and complexity to it after you've mastered the simpler version.

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[Yorik]

Thanks for all the replies!

Let me try to summarize and see if this is the right way of thinking.

Right now, I derived the complete dynamic equations for tau1, tau2 and tau3 in the case of a planar manipulator with actuators at the joints. Of course, tau1 is larger than tau2, which is larger than tau3, which is to be expected. However, as you guys mention, using a cable transmission system basically gives the following constraints: tau_motor = (tau 1 minus friction effects) = (tau 2 minus friction effects) = (tau 3 minus friction effects), joint angles theta_motor = theta1 = theta2 = theta3, angular velocities theta_dot_motor = theta_dot1 = theta_dot2 = theta_dot3 and angular accelerations theta_dotdot_motor = theta_dotdot1 = theta_dotdot2 = theta_dotdot3. This under the assumption of equal pulley sizes (I know about the gear ratio, but just for simplicity) and that the cable does not slip on the pulley (friction effects are significant enough).

My equations have the following form right now, using Euler-Lagrange method: tau = M*thetadotdot + C*thetadot + S*theta + G, where tau = [tau1,tau2,tau3], thetadotdot = [thetadotdot1,thetadotdot2,thetadotdot3], thetadot = [thetadot1,thetadot2,thetadot3], theta = [theta1,theta2,theta3], M is the mass/inertia matrix, C the Coriolis matrix, S the matrix including spring forces and G the gravity vector.
But, since tau_motor = (tau 1 minus friction effects) = (tau 2 minus friction effects) = (tau 3 minus friction effects), does this reduce to only one dynamic equation, which is the first row of the equation that I have right now? Meaning: using only the equation derived for tau1 and relate it to tau_motor, since tau1 obviously includes all the forces that act on the finger and thus the tension of the cable? And if so, then tau2 and tau3 must equal this same torque (minus friction effects), since the cable transmits the same tension over the cable?

Please correct me if I am mistaken.

Furthermore, suppose that on each link of the finger an external force is working. Do I then only relate these forces to the first joint, so they can be added to the equation of tau1?
Let's say, external force F3 (at the finger tip) creating a large moment, F2 creating a smaller moment and F1 and even smaller moment? Or do I also still need to consider the moments with respect to tau2 and tau3?