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I'm an Electronics student taking a module in Robotics.

From the example,

Example from lecture notes

I understand line 1 as the Jacobian is found from the time derivative of the kinematics equation and such relates joint angles to velocity.

I do not understand why the transpose has been taken on line 3 and how line 4 is produced.

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  • $\begingroup$ Can you provide more detail? Is $J$ the Jacobian or is it moment of inertia? Is $m$ mass or something different? What is $q$? $\endgroup$ – Chuck Apr 30 '16 at 23:54
  • $\begingroup$ Unfortunately this is my problem, this is the lecture slide as given, so this is as much information as I have sorry! $\endgroup$ – LazyMoggy May 2 '16 at 11:04
  • $\begingroup$ Since this came from a slide, your best bet might be to ask the person who presented the slide. $\endgroup$ – Ian May 3 '16 at 21:12
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How the third equation is used in the derivation would be helpful in order to explain why $\delta x$ is being transposed. But when you take $$\delta x = J \delta q$$ and transpose it, you must reverse the order of the terms of the right hand side. This makes sense if you consider the dimensions of $J$. If $x \in \Re^n$ and $q \in \Re^m$ then $J$ is $n$ x $m$. To postmultiply $q^T$ (which is 1 x $m$) you must transpose $J$.

The fourth equation is just a statement that the seesaw is in equilibrium - the total moment is zero.

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