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I am studying robotic kinematics , and given the following manipulator with its D-H frames assigned:

enter image description here

I have to find the rotation matrix $R_{e}^{2}$, which is the rotation matrix of frame $e$ with respect to frame $2$.

I know that this matrix is :

$\begin{pmatrix} 0 &sin\beta & cos\beta \\ 0& cos\beta & -sin\beta\\ -1 &0 &0 \end{pmatrix}$

but I cannot understand why.

Can somebody please help me understand how does this reotation matrix comes out?

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It looks like both frames are attached to the end effector at point P, and are offset by a constant angle $\beta$. So you don't really have to consider the D-H parameters to answer this question - just do a simple rotation matrix evaluation.

So how are the frames, $F_2$ = $[\hat{x_2} \; \hat{y_2} \; \hat{z_2}]^T$ and $F_e$ = $[\hat{x_e} \; \hat{y_e} \; \hat{z_e}]^T$, related? Well, from the diagram you have above, we have the following relationships:

$$\hat{x_2} = \cos\beta \hat{y_e} + sin\beta \hat{z_e}$$ $$\hat{y_2} = \cos\beta \hat{y_e} - sin\beta \hat{z_e}$$ $$\hat{z_2} = -\hat{x_e}$$

Note that $\hat{x_e}$ points into the page and $\hat{z_2}$ points out of the page.

Then do you see how your rotation matrix above falls out of that? Clearly,

$$F_2 = R_e^2F_e$$

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A rotation matrix is also called a director cosine matrix. The elements of the rotation matrix are the cosines of the unit vectors of two coordinate systems involved. You can find a more generic explanation here.

Let $\angle (e_{2,i}, e_{e,j})$ denote the angle between the angle between unit vector on the i axis of the fixed reference frame and the unit vector of the j axis of the rotated frame.

$$ R_{2e} = \\ \begin{pmatrix} cos(\angle (e_{2,x}, e_{e,x},)) & cos(\angle (e_{2,x}, e_{e,y},)) & cos(\angle (e_{2,x}, e_{e,z}))\\ cos(\angle ( e_{2,y}, e_{e,x},))) & cos(\angle (e_{2,y}, e_{e,y},)) & cos(\angle (e_{2,y}, e_{e,z}))\\ cos(\angle (e_{2,z}, e_{e,x},)) & cos(\angle (e_{2,z}, e_{e,y},)) & cos(\angle (e_{2,z}, e_{e,z})) \end{pmatrix} = \begin{pmatrix} 0 &sin\beta & cos\beta \\ 0& cos\beta & -sin\beta\\ -1 &0 &0 \end{pmatrix} $$

In your case we identify that:

$cos(\angle (e_{2,x}, e_{e,x})) = 0$

This translates to a $90^\circ $ angle between the unit vectors of the two x axes.

$cos(\angle (e_{2,y}, e_{e,x})) = 0$

This translates to a $90 \deg $ angle between the unit vectors of the y axes of the reference frame and the x axis of the rotated frame.

$cos(\angle (e_{2,z}, e_{e,x})) = -1$

This translates to a $180^\circ $ angle between the unit vectors of the z axes of the reference frame and the x axis of the rotated frame.

Similarly for all other elements of the matrix, but here is the most interesting ones:

$cos(\angle (e_{2,y}, e_{e,y},)) = cos(\beta)$

This translates to a $\beta^\circ $ angle between the unit vectors of the y axes of the reference frame and the y axis of the rotated frame.

$cos(\angle (e_{2,z}, e_{e,x})) = cos(\beta)$

This translates to a $\beta^\circ $ angle between the unit vectors of the z axes of the reference frame and the x axis of the rotated frame.

$cos(\angle ( (e_{2,y}, e_{e,z}) ) = -sin(\beta)$

Using the quarter period phase shift property $cos(x + \frac{\pi}{2}) = -sin(x)$

$cos(\angle ((e_{2,y}, e_{e,z})) = cos(\beta + \frac{pi}{2})$

This translates to a $\beta + 90^\circ $ angle between the unit vectors of the y axes of the reference frame and the z axis of the rotated frame.

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