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