# Wrong forward-kinematic after calculating from DH-parameters

I'm working on an exercise about DH-parameters for my robotics course in university and I ran into a problem, because my solution is different from the one given by my professor. I calculated the DH-parameters $$\theta=\theta_1-90˚, d=20m, a=0, \alpha=-90˚$$. So far my solution is identical with the one of my professor. Now the next task is to calculate the transformation matrix $$T_{0,1}$$ using the calculated DH-parameters. Now I calculated: $$\begin{equation} T_{0,1}= \left(\begin{matrix} \cos(\theta_1-90) & 0 & -sin(\theta_1-90) & 0 \\ sin(\theta_1-90) & 0 & cos(\theta_1-90) & 0 \\ 0 & -1 & 0 & 20\\ 0 & 0 & 0 & 1 \end{matrix}\right) \end{equation}$$ Now the solution of my professor states: $$\begin{equation} T_{0,1}= \left(\begin{matrix} \cos(\theta_1-90) & 0 & -sin(\theta_1-90) & 0 \\ sin(\theta_1-90) & 1 & cos(\theta_1-90) & 0 \\ 0 & -1 & 0 & 20\\ 0 & 0 & 0 & 1 \end{matrix}\right) \end{equation}$$ Which is almost what I got, but not exactly since they put a 1 where I put a 0 in position (2,2). After multiplicating the transformation matrixes of the single DH-parameters I had found that the position (2,2) evaluates to $$\cos(\theta_i)\cdot\cos(\alpha_i)$$ and since since $$cos(-90˚)$$ evaluates to 0 it should be 0, or not? Where am I wrong?

• Your professor's solution seems fishy. I would check if the two matrices are actually rotation matrices by verifying that M*M^T = I (the matrix multiplied by its transpose should give you the identity). Dec 27, 2018 at 21:16

The upper-left 3x3 matrix must be an orthonormal rotation matrix. Every column of that must have a unit norm. The second column $$[0, 1, -1]^T$$ has a norm of $$\sqrt{2}$$ which makes the rotation matrix invalid.