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It is well known that the dynamics equation for a mechanical system (e.g. a robotics manipulator) is given be the Euler-Lagrange equation which takes the particular form (in the simplified version),

$$ M(q)\ddot{q}+N(q,\dot{q})=u$$

where $M$ is the inertia tensor, $N$ the Coriolis/centripetal vector and $u$ the input (torque). In a given coordinate change of the state e.g.

$$q=h(y)$$

following a similar procedure as in "Robot manipulator control: Theory and Practice" pp 148-149, by Frank Lewis, after some algebra we get,

$$M(q)J\ddot y+(N(q,\dot q)+M(q)\dot J \dot y)= u$$

Apparently this can be shorthanded as,

$$\bar{M}\ddot{y} + \bar{N} =u$$

by collecting terms. The problem is that $M$ and $N$ are still functions of $q$. Given that $M$ is a tensor and $N$ a vector, they transform as such. Thus

$$\hat{M}(y)=J^{T}M(h(y))J$$ and

$$\hat{N}(y)=JN(h(y))$$

Substituting in the transformed equation we get,

$$J^{T}M(h(y))JJ\ddot{y}+(JN(h(y))+J^{T}M(h(y))J\dot{J}\dot{y})= u$$

The question is, are the last three equations correct? i.e. does the inertia tensor and the Coriolis vector transform this way?

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  • $\begingroup$ Do you have a motivation? From the way you've written your question, it appears you're meandering around in a textbook and manipulating expressions, looking for people to tell you whether or not you've made a mistake. MO is generally for more focused questions, rather than "please check my argument for errors" type requests. $\endgroup$
    – Ryan Budney
    Commented Jan 3, 2012 at 21:50
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    $\begingroup$ Ryan i'm not sure what you mean by motivation. From my end, my motivation is very clear: i'm doing research in nonlinear control & robotics and i'm investigating a very specific thing (the application of a specific map $q=h(y)$). However, tensorial analysis is not strong point and i'm not really sure about the last three equations. I could rephrase the question with a very specific argument: does the inertia tensor and the Coriolis vector transform this way? However in order to provide a background, i've elaborated on the whole thing! $\endgroup$
    – Jorge
    Commented Jan 3, 2012 at 22:51
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    $\begingroup$ There's quite a bit wrong here. My suggestion would be to write everything in coordinates: $q^i=h^i(y),\ \dot{q}^i = \frac{\partial h^i}{\partial y^m}\dot{y}^m,\ \ddot{q}^i=\frac{\partial h^i}{\partial y^m}\ddot{y}^m+\frac{\partial^2 h^i}{\partial y^n\partial y^n}\dot{y}^n\dot{y}^m$. Substitute this into $M(q)\ddot{q}+N(q,\dot{q})=u$, multiply everything by $J$, and massage what you get into the form $\hat{M}(y)\ddot{y}+\hat{N}(y,\dot{y}) = \hat{u}$. Your assumed transformation of $\hat{M}$ is correct, but $\hat{N}$ is way off (and $\hat{u}=J u$). $\endgroup$
    – user17945
    Commented Jan 3, 2012 at 22:52
  • $\begingroup$ I agree with Ryan, though, this isn't really an appropriate question for this site. $\endgroup$
    – user17945
    Commented Jan 3, 2012 at 22:53

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