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I am attempting to derive the following formula for rotation of a vector $\mathbf{u}$, undergoing a left-handed rotation $\mu$, to achieve a new vector $\mathbf{v}$. The expression is as follows:

$\mathbf{v} = (1-\cos{\mu})\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n} + (\cos{\mu})\mathbf{u} - \sin{\mu}(\mathbf{n}\times\mathbf{u})$

Here is a diagram of what is going on:

Rotation from u to v

The textbook writes the following:

$\mathbf{v} = \vec{ON} + \vec{NW} + \vec{WV}$

$\mathbf{v} = \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n} + \frac{\mathbf{u} - \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}}{|\mathbf{u}-\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}|}NV\cos{\mu} + \frac{\mathbf{u}\times\mathbf{n}}{|\mathbf{u}|\sin{\phi}}NV\sin{\mu}$

Note that $\phi$ was never defined in the text, which makes this slightly frustrating.

I am trying to understand these three terms; they will simplify to the formula I want. I get that the first term is the projection of $\vec{ON}$ onto $\mathbf{u}$, but I am not sure of what the other two terms mean. Any help would be much appreciated.

(for more context, see page 8 of the pdf file here - this gives the derivation, but no explanation of what each term is. This page is an excerpt from my textbook.)

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1 Answer 1

up vote 3 down vote accepted

$\phi$ is the angle between $\bf{n}$ and $\bf{u}$. The factor $\sin \phi$ arises from the identity $| \mathbf{u} \times \mathbf{n}| = |\mathbf{u}| \sin \phi.$

What is happening here: first, we project onto $\bf{n}$, which yields the vector $\vec{ON}.$ The 'rest', which is the vector $\vec{NV}$, lies on a plane perpendicular to $\bf{n}$; this is the plane on which the circle drawn in the picture lies. The book then chooses a basis on this plane. As a first basis vector, they choose $$\bf{e_1} = \frac{\mathbf{u} - \langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}}{|\mathbf{u}-\langle\mathbf{u},\mathbf{n}\rangle\mathbf{n}|},$$ which points from $N$ towards $U$, as a second basis vector they take take the unit vector perpendicular to the $\bf{u,n}$-plane, i.e. along $$\bf{e_2} = \frac{\mathbf{u}\times\mathbf{n}}{|\mathbf{u}\times\mathbf{n}|}.$$ $\mu$ is the angle between the points $V$ and $W$ on the circle. Note that $(\mathbf{e_1},\mathbf{e_2}) = 0$, so indeed the two basis vectors are orthogonal.

Hope this helps!

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2  
Note that $|\mathbf u-\langle\mathbf u,\mathbf n\rangle\mathbf n|=|\mathbf u\times\mathbf n|=NU=NV$, so all those factors actually cancel (leading to the result in the first displayed equation in the question). See also the Wikipedia article (which doesn't mention "Goldstein's rotation formula" as an alternative name; is that how the book calls it?). –  joriki Oct 9 '11 at 23:12
    
@joriki: I don't believe the name 'Rodrigues formula' is well-known in the physics community; I had actually never seen this result before... –  Gerben Oct 10 '11 at 12:53
    
@joriki - My textbook and professor both referred to this as a result derived by Goldstein; actually, I've seen it referred to as the Rodrigues formula before, but I've never really used it (typically used quaternions or DCM, the formula was only mentioned in passing).. –  Dang Khoa Oct 10 '11 at 16:40

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