Talk:Rotation matrix

30 March 2007 - Leonid

Hi everyone!

I just added the bit about the Rotation Tensor representation and Rotation Matrix invariance with respect to change of coordinate frame. I think this invariance note is important in "Rotation Matrix" talks. From the other side, dealing with invariant objects is much more convinient especially if you start doing some advanced stuff like Elasticity Theory and Mechanics of Beams and Shells.

Does rotation tensor thing deserve a page on it's own?

There is some similar bit on "Representation of Rotation" page, but the guy whom wrote that page is cleary in love with quaternions. I am in love with quaternions as well the moment some simulation of rigid body dynamics need to be done, but theory of rotation tensors is very old and deserve it's place (I think). The whole Classical Ellasticity Theory is built on it.

Regards, Dr.Leonid Paramonov PDRA, Imperial College London, UK

17:41, 24 August 2006 - Alanic

I did the changes I talked about and I'm ready to defend them if you don't agree. The big matrices in "three dimensions" section may need an update, too. This was my very first contribution to Wikipedia and I didn't read any help pages, so feel free to cancel my changes if I'm not abiding any rules.

13:25, 22 August 2006 - Alanic

${\begin{pmatrix}\cos {\alpha }&\sin {\alpha }\\-\sin {\alpha }&\cos {\alpha }\\\end{pmatrix}}.{\begin{pmatrix}1\\0\\\end{pmatrix}}={\begin{pmatrix}\cos {\alpha }\\-\sin {\alpha }\\\end{pmatrix}}$ ${\begin{pmatrix}\cos {\alpha }&\sin {\alpha }&0\\-\sin {\alpha }&\cos {\alpha }&0\\0&0&1\end{pmatrix}}.{\begin{pmatrix}1\\0\\0\\\end{pmatrix}}={\begin{pmatrix}\cos {\alpha }\\-\sin {\alpha }\\0\\\end{pmatrix}}$

The first equation shows that the matrix rotated the vector clockwise. This conflicts with the definition that rotation matrices rotate vectors counter-clockwise.

According to the right hand rule, the second equation shows that this matrix rotated this vector around the -z axis. This conflicts with the definitions on the page that a rotation matrix is a matrix that when multiplied with a vector it rotates the vector counter clockwise and that this matrix is a rotation around the z axis. I think all matrices in this page need to be transposed.

67.122.123.121 21:19, 21 August 2006 (UTC) - Geoff Dolan[reply]

Hopefully not stepping on any feet, but I added a minus sign on the 21 term of the three dimensional rotation matrix. Besides disagreeing with my text (Sidi, Spacecraft Dynamics and Control, 1997), it's easy to program the matrix into matlab and see that it isn't a rotation matrix without the minus sign.

I think the signs for the rotation about the z-axis on the 12 and the 21 entries are wrong. Can someone confirm this?

I haven't looked at the other matrices to see if they're wrong too. Thanks

I don't think they're wrong. They look exactly how I've learnt them on the University a few months ago :)

Torzsmokus 20:36, 3 January 2006 (UTC)[reply]

It depends on how you define your positive angle. Conventions differ.

I've added a yaw-pitch-roll system today and I've adapted the signs to the current status (out of respect to the authors). However I must say, that at my university the yaw-pitch-roll system was defined like this:

${\mathcal {R}}(\gamma ):={\begin{pmatrix}1&0&0\\0&\cos {\gamma }&\sin {\gamma }\\0&-\sin {\gamma }&\cos {\gamma }\end{pmatrix}}$ , ${\mathcal {P}}(\beta ):={\begin{pmatrix}\cos {\beta }&0&-\sin {\beta }\\0&1&0\\\sin {\beta }&0&\cos {\beta }\end{pmatrix}}$ , ${\mathcal {Y}}(\alpha ):={\begin{pmatrix}\cos {\alpha }&\sin {\alpha }&0\\-\sin {\alpha }&\cos {\alpha }&0\\0&0&1\end{pmatrix}}$ . For the 2-dimenional matrix I agree that it usually is defined like this: ${\begin{pmatrix}\cos {\theta }&-\sin {\theta }\\\sin {\theta }&\cos {\theta }\end{pmatrix}}$ Wedesoft 21 Mar 2006 22:12 BST

All given matrices are correct. But it should read "with the equivalent counter-clockwise rotation in \mathbb{R}^2". I'll correct this immediately. (AK, 2006-04-05)

rewrite by TomViza

I just completed a major rewrite, this article is now much more generalised and also more easy to read. I moved some stuff in from Rotation (mathematics) which is very long.

I also added the formula to find the matrix in terms of the Euler angles.

TomViza 16:41, 21 May 2006 (UTC)[reply]

The following is the derivation of the second function in the 3D section. When I moved the equation from Rotation (mathematics), I thought that the derivation was not very encyclopdic, but have kept it here for thouroughness. TomViza 16:41, 21 May 2006 (UTC)[reply]

Derivation. This matrix is derived from the following vector algebraic equation (see dot product, cross product, and matrix multiplication):

\mathbf {u'} =(\cos \theta )\mathbf {u} +(1-\cos \theta )\mathbf {v} (\mathbf {v} \cdot \mathbf {u} )+\sin \theta (\mathbf {v} \times \mathbf {u} ),\qquad \qquad (1)

which in turn is derived from

\mathbf {u'} =\mathbf {u_{\|}} +(\cos \theta )\mathbf {u_{\perp }} +\sin \theta (\mathbf {v} \times \mathbf {u_{\perp }} ).

Here

\mathbf {u_{\|}} =\mathbf {v} (\mathbf {v} \cdot \mathbf {u} ),

\mathbf {u_{\perp }} =\mathbf {u} -\mathbf {u_{\|}} ,

\mathbf {v} \times \mathbf {u_{\perp }} =\mathbf {v} \times \mathbf {u} ,

which shows that u is resolved (see Gram-Schmidt process) into a parallel and a perpendicular component (to v). The parallel component does not rotate, only the perpendicular component does rotate. This rotation is similar to a two dimensional rotation, except that instead of x and y axes, there are $\mathbf {u_{\perp }}$ and $\mathbf {v} \times \mathbf {u_{\perp }}$ axes, both of which are perpendicular to v.

Quaternions

It would be nice with a link to quaternions

Rotation matrix vs orthogonal matrix

Hi, just came across this article and have a a couple of question.

First, it says in the definition part that a rotation matrix is equivalent to an orthogonal matrix. If this is so, why is there a separate article on rotation matrices when there already is one on orthogonal matrices. What additional information is provided here which does not fit in the article on orthogonal matrices.

Second, is the equivalence of rotation and orthogonal matrices established in the literature? Personally, I would suggest that a rotation matrix is a special case of an orthogonal matrix (for the n-dim case) which only has two eigenvalues not equal to one [my correction]. Such a matrix always appears as a generalization of a 2D rotation for the the n-dim case in the sense that it has one well-defined rotation space in which it rotates with one well-defined angle. Also, in the 2D and 3D cases such a matrix is equivalent to an orthogonal matrix, but in 4 and higher dimensions a general orthogonal matrix is the product of two or more such matrices. Don't know if this is an established way of defining rotation matrices. --KYN 18:48, 12 July 2006 (UTC)[reply]

No, a rotation matrix is an orthogonal matrix with the additional restriction that the determinant is +1. Thus

{\begin{bmatrix}0&1\\1&0\end{bmatrix}}

is an orthogonal matrix, but since its determinant is −1 it is not a rotation matrix. The eigenvalues of a rotation matrix are guaranteed to include a single +1 in odd dimensions, but otherwise may have as many repetitions of −1 or complex conjugate pairs with magnitude +1 as the dimension allows. Thus

{\begin{bmatrix}0&-1&0&0\\1&0&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix}}

is a rotation matrix with four eigenvalues, none of which are +1. We can state the definition as

An n×n matrix M is a rotation matrix if M^TM = I and if det(M) = +1.

Planar rotations are a very special case, and only 3D rotations have a rotation axis.

We can derive the algebraic conditions from the geometric statement that a rotation is a direct isometry leaving one point fixed. Take the fixed point to be the origin so we are working with a Euclidean inner product space.

Then isometry means preservation of distances (and by implication, angles), which is equivalent to preservation of the inner product. In vector form the inner product of a vector with itself is v^Tv. Therefore an isometry satisfies (Mv)^T(Mv) = v^Tv, for all vectors v. Rewrite the left-hand side as v^T(M^TM)v, and rewrite the right-hand side as v^TIv; then to obtain equality for all v we must have M^TM = I, as stated.
For an isometry to be direct it must not reverse "handedness". The identity transformation is obviously direct, and the identity matrix is a (null) rotation, with determinant +1. Furthermore, in the Lie group of Euclidean isometries there must be a connected path from the identity to any direct isometry. Thus in the orthogonal group of n×n matrices there are two disjoint connected components: the special orthogonal group, which contains the identity and all of whose members have determinant +1; and the remaining component (which is only a coset, not a group), all of whose members have determinant −1.

Apparently the article could use some work. --KSmrq^T 13:10, 21 August 2006 (UTC)[reply]

Use of homogenous Coords?

Hello,

This may be confusing for people who have to implement this with translation compounded, as there is no section on homogenous coords, should this be added? 129.78.208.4 02:51, 7 November 2006 (UTC)[reply]

Added generators, reformat

I have added a description of the generators of the group, and reorganized things so that the roll, pitch, and yaw matrices come first. That way, the other representations are more easily understood as compositions of these basic rotations. I also added expressions for the generators, except the Euler angle matrix, because I don't have an expression for that generator. Nothing has been deleted, but some things have been rephrased. PAR 20:52, 5 January 2007 (UTC)[reply]

How to find the generator A?

I'm reading the section about the properties of the rotation matrix...if I have M, but I want to find the generator A, what is the best way to do this? --HappyCamper 22:27, 26 April 2007 (UTC)[reply]

Transpose as inverse

Hmm...it doesn't seem as obvious to me that the transpose of a rotation matrix is its inverse as it perhaps was to the writer of this article. Might this section perhaps be altered? I'm not actually certain of the best clear way of showing this - perhaps showing that under the appropriate orthogonal basis, the leading diagonal is 1's except for 2 Cos(theta)s, and everywhere else in 0 except the sin thetas, then simply subbing in -theta for theta? Wrayal 15:17, 13 May 2007 (UTC)[reply]

The rotation matrices about the x, y, and z axis do not seem to be equate to the "generator" written to their right (the rotation matrices with the matrix exponentials). It seems that one is the transpose of the other. Can someone please explain?

Thanks- James