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The parallel mechanism is composed of a moving platform and a base (or frame) connected by four legs. Each leg is composed of five revolute joints numbered 1, 2, … 5 from the base to the moving platform. The axes of the revolute joints 3, 4 and 5 in leg 1, the revolute joints 1 and 5 in leg 2, as well as the revolute joints 1, 2 and 3 in leg 3 are all parallel to the Z-axis. The axes of the remaining revolute joints in each leg are parallel

I want to know the DOF. Can someone help?

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So I guess you want to know the DOF of the moving platform. First of all keep in mind that for a parallel mechanism with a fixed base and one moving platform the maximum degrees of freedom for the moving platform are 6, e.g. 3 rotational and 3 translational DOF.

In this case it is kind of hard to see intuitively what DOF the moving platform could reach. So I would suggest following procedure:

  1. Imagine the legs without the revolute joints, linked to the moving platform in a way they can provide the 6 DOF to the moving platform. In this case the link should be able to change its length and be able to rotate around the joints at the base and the moving platform freely. The first example that comes to my mind would be telescopic legs with ball joints at the base and the moving platform.

  2. Now we know what kind of motion the legs need to be able to achieve to have 6 DOF at the moving platform. This means you can analyze the legs individually to determine the actual DOF of the platform. Take the actual leg 2 for example and forget the moving platform. Think only about the leg which is attached to the base. The leg can achieve all translations and rotations, at least inside a very small workspace. (Take a minute to think about it)

  3. Do the same for leg 1 and 3. Looking at leg 3 we can see that we could remove the revolute joints 1 and 2 and the moving platform would have the same DOF, so we only have to consider 3, 4 and 5. Where it is clear that the leg can achieve all rotations but is not variable in length, e.g. the translational motion can't be achieved and therefore has only 3 DOF. And since leg one is the same as leg 3 only in reverse, it also has only 3 DOF.

  4. The moving platform can only achieve the DOF of the leg with the minimum DOF. This is because all legs have to provide the same 'amount' of DOF in order to achieve the DOF. This sounds a bit weird but again take a minute and think about it. For example if wanted the platform to move towards us in an only translational motion, leg 2 would try to pull the platform towards us but leg 1 and 3 would only come towards us including some rotational motion. Therefore an only translational motion would not be possible. Try to think about more restrictions. The final conclusion would therefore be that the moving platform has only 3 DOF.

Maybe somebody else knows about some special formula you can apply, but I hope this helps a bit in terms of intuitive understanding about DOF.

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  • $\begingroup$ Your description is very intuitive and it is an interesting take on the DOF analysis. I do not agree with every detail you wrote (e.g. I do not think the rotational joints can be neglected if they are not co-axial). Generally if you find the kinematic chain (leg) with the minimum amount of DOF that does not always equal the DOF of the structure, it is only an upper bound for the DOF count on the structure level. $\endgroup$
    – 50k4
    Sep 7 at 11:49
  • $\begingroup$ @50k4, thank you for your comment. I agree that that in general the 'appendix' of a mechanical structure with minimum DOF does not imply a global restriction in DOF of the whole mechanical system. However in case of simple parallel mechanism with a base and only one moving platform, I suggest that this assumption holds true for every combination of joints, which is a direct consequence of the legs being parallel. Please correct me if I'm wrong! $\endgroup$
    – Matthias
    Sep 8 at 18:26
  • $\begingroup$ It holds true in all normal/usual cases, but one can think of somewhat forces examples, where the DOFs of the other kinematic chains are in directions which are not allowed by the chain with the minimum DOF (let us call this chain of interest). So even if the other chains have more DOF they can be in directions which would cancel out at least 1 DOF of the chain of interest. $\endgroup$
    – 50k4
    Sep 8 at 18:51
  • $\begingroup$ You are right I just thought of an example where all legs have one DOF, but at the end the MP has zero. So the statement is indeed not true in general, even in terms of this simple parallel mechanism. I thought about adding the constraint that the minimal DOF have to be also shared with the other legs. But then it is not as intuitive anymore. So I guess for an intuitive approach there is no straightforward solution. There is a linked post to this discussion, which may lighten up the theory a bit. $\endgroup$
    – Matthias
    Sep 21 at 14:32

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