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look at the above image, the red coordinates is attached to the robot end-effector, and the other one is the base.coordinates.

Now, As we know, the end-effector can do any rigid motion, so if a point P is attached on the robot end-effector. we can calculate the velocity of the point P.

The general formula is just like: [tex]\dot{P}=\omega\times a+b[/tex]

Here, the [itex]\omega[/itex] is just the angular velocity, and b is the translation velocity.

But when reading some papers, I found the general formula has different expression.

One is:

[tex]\dot{P}=\Omega\times P+T[/tex]

here, P has the coordinates in base frame. T and [itex]\Omega[/itex] is defined as the velocity screw.

Another expression:

[tex]\dot{P}=\omega\times(R\cdot^{1}P)+v[/tex]

here, Here, the R is the rotation matrix of end-effector frame wrt base frame, and [itex]^{1}P[/itex] is the vector wrt end-effector frame. [itex]\omega v[/itex] is also some kind of velocity description about the end-effector. In-fact, this is the most convenient way I can understand.

I'm quite confusing that it seems both expression is valid, but what exactly does the velocity screw means?

For more details about the paper, I have also wrote another post, see here:

https://www.physicsforums.com/showthread.php?t=512697

Hope someone can give me some explanation on why there have two different formulas.

thanks very much!

zyh

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# Velocity screw description of a robot end-effector question

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