Coordinate System Rotation Matrix (global to local)

dhume878
Messages
6
Reaction score
1
I feel I may have improperly posted this thread https://www.physicsforums.com/showthread.php?t=469331" but am just not as knowledgeable in my matrix math as I need to be. One (me) would think that somehow you should be able to get a rotation matrix from these two systems.

Homework Statement


So I have two matrices composed of 3 orthogonal vectors

G = [1 0 0;
0 1 0;
0 0 1]
and
L = [0.96247 -0.03259 -0.266524;
0.02676 0.99932 -0.025578;
0.26718 0.018486 -0.962682]

I have a point in the global system, which i would like to rotate in the same manner one matrix is rotated from the other. (I think .. R(alpha) * R(beta) * R(gama))

It can be assumed the two systems have the same origin.

Homework Equations


Possibly relevant

x'_vector = R_matrix * x_vector

The Attempt at a Solution



Here is my MATLAB attempt. I feel I'm way off (no laughing!)

Code:
x = [1 0 0];
x_p = [0.96247 -0.0325928 -0.266524];

A = x_p \ x

y = [0 1 0];
y_p = [0.0267575 0.999315 -0.0255778];

B = y_p \ y

z = [0 0 1];
z_p = [0.267175 0.0174863 0.962682];

C = z_p \ z

Again I hope I'm not being too hasty with this repost, but i didn't want to be scorned for my incorrect post location.

Thanks for any help,

Cheers!
 
Last edited by a moderator:
Physics news on Phys.org
The matrix L is not a rotation of the matrix G. |det(L)| = 0.85, while |det(G)| = 1. Furthermore the vectors composing L are not orthogonal, e.g., {0.96247, -0.03259, -0.266524}.{ 0.26718, 0.018486, -0.962682} = 0.51. There is a linear transformation between the bases, but it is not an orthogonal transformation (rotation).
 
:rolleyes:

Actually it was my mistake I typed in the numbers incorrectly.

[PLAIN]http://j.drhu.me/2011-02-03_1426.png

dot = 0 as well.

Annnnddd, I believe the matrix I was referring to as L .. is just my rotation matrix and I should be able to multiply it by a point, and get that point in the local system.

Does this make sense at all?
 
Last edited by a moderator:
dhume878 said:
:rolleyes:

Actually it was my mistake I typed in the numbers incorrectly.


dot = 0 as well.

Annnnddd, I believe the matrix I was referring to as L .. is just my rotation matrix and I should be able to multiply it by a point, and get that point in the local system.

Does this make sense at all?

I think it makes sense. If two coordinate systems are linearly related and have the same origin, then they will be related by a rotation matrix. So if you indeed computed R(alpha) * R(beta) * R(gamma) correctly, you should be ok. The determinant and orthogonality of rows are good sanity checks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top