Is it possible to express the jacobi derivative of position with respect to angular displacement easier?

Question

Let

$$ \vec{\omega} = \frac{\vec{s} \times \vec{v}}{\left\lvert s\right\rvert^2} $$

and define

$$ \vec{\theta} = \int \vec{\omega} \,\mathrm{d} t $$ $$ R = \frac{\mathrm{d} \vec{s}}{\mathrm{d} \vec{\theta}} $$ $$ D = \frac{\mathrm{d} \vec{\theta}}{\mathrm{d} \vec{s}} $$

I think one can obtain the relations:

$$ \vec{v} = R \vec{\omega} $$

and

$$ D \vec{v} = \vec{\omega} $$

but how does one solve for $R$ and $D$ themselves?

I'm also not sure about the meaning of the two and when each one is well defined.

Answer 1

From $\vec{v} = R\vec{\omega}$

$$ D \vec{v} = \frac{\vec{s} \times \vec{v}}{\left\lvert s\right\rvert^2} $$

$$ D \hat{e}_k = \frac{\vec{s} \times \hat{e}_k}{\left\lvert s\right\rvert^2} $$

$R$ is then the inverse of that but I'm not sure it is always valid to take the inverse.

Answer 2

Notice from your first equation that $w$ is perpendicular to $v$, so $$w\cdot v=0$$

Define a new vector that is perpendicular to both $w$ and $v$ $$b=w\times v$$

Use the "BAC-CAB" rule evaluate the cross product $$\eqalign{ b\times w &= (w\times v)\times w \cr &= w^2v - (w\cdot v)w \cr &= w^2v \cr }$$ Now use the Levi-Civita tensor to define the cross-matrix for $b$ $$\eqalign{ B &= -\varepsilon\cdot b \cr Bw &= b\times w &= w^2v \cr \Big(\frac{B}{w^2}\Big)w &= v \cr\cr }$$ So we have found an expression for the $R$ matrix $$R=\frac{B}{w^2}$$ This result can be simplified to $$\eqalign{R=\frac{vw^T-wv^T}{w^Tw}\cr\cr}$$ It is worth noting that $R$ is singular, so you won't get anywhere with an approach based on $R^{-1}$.