# Screw Axes in the Base Frame

I'm trying to find screw axes for joint (on pic) with angle θ. It's on zero position. As I understood:

• joint has an angular velocity on the Z axis, clockwise, so the sign is positive and ω=(0,0,1);
• joint is located at height L along the X axis, so v=(L,0,0).

But right answer is [0,0,1,0,-L,0]. Why?

If I'm understanding you correctly, you're attempting to put the position of the joint in for the translational velocity component of the screw axis. What you actually want to put in there is the velocity of a point rigidly attached to the rotating part of the joint, but currently located at the origin of the base frame, $$v = \begin{bmatrix} \omega_{x} \\ \omega_{y} \\ \omega_{z} \end{bmatrix} \times \begin{bmatrix} -x \\ -y \\ -z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \times \begin{bmatrix} -L \\ \phantom{-}0 \\ \phantom{-}0 \end{bmatrix} = \begin{bmatrix} \phantom{-}0 \\ -L \\ \phantom{-}0 \end{bmatrix}.$$ You can interpret $$\begin{bmatrix} \omega \\ v \end{bmatrix} = \begin{bmatrix} \phantom{-}0 \\ \phantom{-}0 \\ \phantom{-}1 \\ \phantom{-}0 \\ -L \\ \phantom{-}0 \end{bmatrix}$$ as the translational velocity you would have to give an object at the origin to have it act as if it were rotating about the $$z$$ axis passing through $$x=L, y=0$$. This way of treating velocities is called the "spatial" approach, and I discuss it further in my answer here: Robotic manipulator Jacobian by product of exponentials.
• $\times$ is the vector cross product.