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An example problem in the textbook I'm reading from has the spherical wrist system and the DH parameter table as shown below: enter image description here enter image description here

giving us the transformation matrices as below

$$ A_4=\left(\begin{array}{cccc} c_{4} & 0 & -s_{4} & 0\\ s_{4} & 0 & c_{4} & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right) $$ $$ A_5=\left(\begin{array}{cccc} c_{5} & 0 & s_{5} & 0\\ s_{5} & 0 & -c_{5} & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right) $$ $$ A_6=\left(\begin{array}{cccc} c_{6} & -s_{6} & 0 & 0\\ s_{6} & c_{6} & 0 & 0\\ 0 & 0 & 1 & d_{6}\\ 0 & 0 & 0 & 1 \end{array}\right) $$ giving us the final transformation matrix as $$ T_6^3=A_4A_5A_6=\left(\begin{array}{cccc} c_{4}\,c_{5}\,c_{6}-s_{4}\,s_{6} & -c_{6}\,s_{4}-c_{4}\,c_{5}\,s_{6} & c_{4}\,s_{5} & c_{4}\,d_{6}\,s_{5}\\ c_{4}\,s_{6}+c_{5}\,c_{6}\,s_{4} & c_{4}\,c_{6}-c_{5}\,s_{4}\,s_{6} & s_{4}\,s_{5} & d_{6}\,s_{4}\,s_{5}\\ -c_{6}\,s_{5} & s_{5}\,s_{6} & c_{5} & c_{5}\,d_{6}\\ 0 & 0 & 0 & 1 \end{array}\right) $$ per the textbook.

I think $A_5$ given in the text is wrong. Shouldn't the $\theta$ parameter against link 5 be $\theta_5-90^\circ$? That would give us $$ A_5=\left(\begin{array}{cccc} s_{5} & 0 & -c_{5} & 0\\ -c_{5} & 0 & -s_{5} & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right) $$ and the final transformation matrix as $$ T_6^3=\left(\begin{array}{cccc} c_{4}\,c_{6}\,s_{5}-s_{4}\,s_{6} & -c_{6}\,s_{4}-c_{4}\,s_{5}\,s_{6} & -c_{4}\,c_{5} & -c_{4}\,c_{5}\,d_{6}\\ c_{4}\,s_{6}+c_{6}\,s_{4}\,s_{5} & c_{4}\,c_{6}-s_{4}\,s_{5}\,s_{6} & -c_{5}\,s_{4} & -c_{5}\,d_{6}\,s_{4}\\ c_{5}\,c_{6} & -c_{5}\,s_{6} & s_{5} & d_{6}\,s_{5}\\ 0 & 0 & 0 & 1 \end{array}\right) $$ which I think is the right one.

PS:
I did tried my method with different orientation for frame 4 (x-axis flipped 180) and frame 5 (x-axis flipped 180) and I get the same final transformation matrix. If I try the one give by the textbook, I end up getting different answer.

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If the only problem you see in the DH parameters presented is with the $\theta$ values, it is common practice to drop any offset associated with this value. In other words, $\theta = 0$ is not necessarily the same pose as what is presented in diagrams.

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  • $\begingroup$ "it is common practice to drop any offset associated with this value" How can the offset be dropped without affecting the transformation matrix? $\endgroup$ Commented Dec 2, 2022 at 4:14
  • $\begingroup$ It does affect it! You are correct, the transformation matrix you have solved for is the correct solution for the given pose - but remember that the pose in the figure is not necessarily at a configuration where all the joint angles are zero! So in reality, the pose appears to represent the case where $\theta_{5} = -90$. $\endgroup$ Commented Dec 2, 2022 at 16:04
  • $\begingroup$ Gotcha! Thanks man! $\endgroup$ Commented Dec 3, 2022 at 9:52

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