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I have done some research on vector alignment here on math exchange and in some technical papers. I seem to find some discrepancies in the formulae presented and I am not sure if these are errors or if they present slightly different concepts.

So the question in general is vector alignment via the Rodriguez Formula. The mathexchange post in question is this one mathexchange

The formula for R (from the post) is as follows:

$\mathit{R} = \mathit{I} + [v]_x + [v]_x^2\frac{1-c}{s^2}$

where $s = \sin(\theta)$ and $c = \cos(\theta)$

On wikipedia the formulae is as follows:

$\mathit{R} = \mathit{I} + s[v]_x + [v]_x^2(1-c)$

Using the notes of Ethan Eade we get:

$\mathit{R} = \mathit{I} + \frac{s}{\theta}[v]_x + [v]_x^2\frac{1-c}{\theta^2}$

Equation 82 on page 10.

So which one is correct? Or do they express slightly different things? I am inclined to believe Ethan Eade's notes as he has a track record of publications.

Community
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Marc HPunkt
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1 Answers1

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In Eade's article, the vector $\omega$ (which you've changed to $v$) is arbitrary, whereas in the wikipedia article, $\mathbf k$ (your $v$) is a unit vector. If you define $\theta = \|\omega\|$ (as Eade does in formula (10)), then $\omega = \theta\mathbf k$, and $[\omega]_\times = \theta \mathbf K$,so his formula now matches the one in wikipedia (see also this derivation).

Similarly, in the other Math Exchange thread, $v = a \times b$ is also not a unit vector. It's norm is $\sin \theta$. Thus $[v]_\times = \sin \theta \mathbf K$, and with this we see why that version of the equation drops the $s$ from the $[v]_\times$ term, and divides by $s^2$ in the $[v]_\times^2$ term.

Paul Sinclair
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  • A big thanks to you. So the difference between Wikipedia and Eade was the unit vector. The difference between mathexchange and the rest was the choice of the vector norm? Why does the post choose the norm to be $\sin(\theta)$? – Marc HPunkt Jun 29 '19 at 08:57
  • Edit: I think I "understand" a difference, such that In eade the $\omega$ is in lie algebra space while in the mathexchange post its a normal vector. So maybe the $\sin$ as norm is a result of going from lie algebra space to lie group space. But im just guesssing. – Marc HPunkt Jun 29 '19 at 09:54
  • Wikipedia is doing "rotate by angle $\theta$ about the unit axis vector $\mathbf k$". Jur van den Berg is doing "rotate unit vector $a$ to unit vector $b$". He gets an axis vector using the cross-product, whose magnitude is $\sin \theta$. Rather than divide that out to get a unit vector, he just carries it. Ethan Eade is relating Jur van den Berg's rotation with its linearized version in the tangent space. The linear approximation to $\sin \theta$ at $0$ is just $\theta$, so his vector has magnitude $\theta$ (an oversimplification, but that is the jist of why he has $\theta$). – Paul Sinclair Jun 29 '19 at 15:09
  • I see. So its an approximation around 0. Thank you! – Marc HPunkt Jun 29 '19 at 16:05
  • Perhaps "differential equation" would be a closer description than "approximation". – Paul Sinclair Jun 29 '19 at 17:31