# Question on Denavit–Hartenberg method for forward kinematics

This is a rather simple question, but I'm new to the field so I need some help.

I am trying to compute the forward kinematics using the DH method as shown below:

I extract the rotation and translation matrices. Then I try to check if they are correct. I take the base vector $$[1,0,0]^T$$ in the $$(x_0, y_0, z_0)$$ system and using the $$A^0_{0'}$$ transformation matrix I want to see what vector I get in the $$(x_0', y_0', z_0')$$ axis.

As you can see, I get the base vector $$[0,1,0]^T$$ which is the $$y_0'$$ vector, when I should be getting $$[0,0,1]^T$$ i.e the $$z_0'$$ vector.

What am I missing?

Your transformation matrix should written as $$A_{0}^{0'}$$. In your case, say we have a vector $$q_0$$.

And for the formula:

$$q_{0} = A_{0}^{0'}q_{0'}$$

The vector $$q_{0'}$$ is originally represented on frame $$0'$$, and you want to find the value $$q_{0}$$, which is the value of the vector $$q_{0'}$$ represented on frame $$0$$.

From your image, the vector $$[1,0,0]$$ is already on frame $$0$$. So the vector $$[1,0,0]$$ is the result ($$q_{0}$$), not the transformation matrix multiplier ($$q_{0'}$$).

$$[1,0,0]^T = A_{0}^{0'}q_{0'}$$

Where $$q_{0'}$$ here is $$[0,0,1]^T$$, the value that you are looking for.