# Problem defining rotation matrix

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

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.

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,

$$F_2 = R_e^2F_e$$

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.