Given a desired transform matrix of the end effector relevant to the base frame of the P560: transform

John J. Craig, in his book, Introduction to Robotics Mechanics and Control, computes the inverse kinematic solutions of a Puma 560, with (correct me if wrong) Modified DH parameters and gets the following set of equations for theta angles: transform

and I noticed that there are no alpha angles in these calculations.

So, my question is why aren't the alpha angle values not used in the calculation for the desired pose with the given end effector transform? Why is it independent of the axes twist angles of the robot?


Check out Equation 3.6:

$$ ^{i-1}_iT = \left[ \begin{array}{c} c\theta_i & -s\theta_i & 0 & r_{i-1} \\ s\theta_ic\alpha_{i-1} & c\theta_ic\alpha_{i-1} & -s\alpha_{i-1} & -s\alpha_{i-1}d_i \\ s\theta_is\alpha_{i-1} & c\theta_is\alpha_{i-1} & c\alpha_{i-1} & c\alpha_{i-1}d_i \\ 0 & 0 & 0 & 1 \end{array} \right] $$

and Figure 3.21, "Link parameters of the PUMA 560":

$$ \begin{array}{c} i & \alpha_{i-1} & r_{i-1} & d_i & \theta_i \\ \\ 1 & 0 & 0 & 0 & \theta_1 \\ 2 & -90^{\circ} & 0 & 0 & \theta_2 \\ 3 & 0 & a_2 & d_3 & \theta_3 \\ 4 & -90^{\circ} & a_3 & d_4 & \theta_4 \\ 5 & 90^{\circ} & 0 & 0 & \theta_5 \\ 6 & -90^{\circ} & 0 & 0 & \theta_6 \\ \end{array} $$

Here I've used $r$ instead of $a$ to prevent any confusion between $\alpha$ and $a$. In Equation 3.6, $s$ means $\sin$ and $c$ means $\cos$.

To your question, $\alpha$ is more related to the orientation of two joints relative to each other as opposed to the motion of the joints relative to each other. That is, $\alpha$ is generally fixed by physical construction and thus, as a parameter, isn't time-dependent.

So hopefully you can see that, from Figure 3.21, all of the $\alpha$ values are either +/- 90 degrees or zero. Equation 3.6 then uses it only in sine and cosine terms, meaning that they can all be reduced to +/-1 or zero.

I really, really hate when authors jump through simplifications like this without a heads up, so I totally understand the confusion.

  • $\begingroup$ yup, thats it, i got it, thanks! If you could also suggest a book that is detailed, as correct as possible and goes from less than intermediadiate level of knowledge of robotics to a good enough depth to deal with the majority of the issues that arise generally and maybe a bit more specifically(if anything like that exists). $\endgroup$
    – hey
    Aug 16 '16 at 12:06
  • $\begingroup$ @hey - It's difficult to say what would be a good resource for someone (relatively) new, as it depends on what you want to focus on. That said, the Springer Handbook of Robotics is a great resource in general. Lots of topics are covered - if you follow the link, under "Table of Contents" (after the list of chapters) there's a "Show the Next 20" link. You can click that repeatedly - there are 65 chapters on a very large range of topics. $\endgroup$
    – Chuck
    Aug 16 '16 at 12:14
  • $\begingroup$ It's an expensive book, but if you have access to a university library you can probably get an electronic copy of the book for free. I will warn you that the Dynamics section by Featherstone can be pretty hard to follow. He basically invented his own math (it was his PhD dissertation, I believe) to simplify dynamic chains, and he abuses a bit of notation - the symbols are familiar but they don't quite mean what you're used to them meaning. That said, it's the basis for the Rigid Body Dynamics Library and, though I don't have proof, I think Simulink Multibody, too. $\endgroup$
    – Chuck
    Aug 16 '16 at 12:17
  • $\begingroup$ @hey - If you've got any other questions or would like to chat, please join us in Robotics Chat! Comments are for comments, so I generally try to avoid conversations here, but I figured I'd try to at least point you in the right direction. Happy reading! $\endgroup$
    – Chuck
    Aug 16 '16 at 12:18
  • $\begingroup$ Well, ,thanks again, I will join next time when ill need something b4 asking. Have a nice day $\endgroup$
    – hey
    Aug 16 '16 at 12:22

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