# How to use Euler's angle theorem in rotation of a coordinate

1. Dec 4, 2012

### hei

If i have a point at (0,0,5) in x,y,z system, then i make 2 rotation on the point with center at origin.
i)the first rotation is on y-axis with angle P in clockwise direction.
ii) the second rotation is on the point's new x-axis rotate in angle Q in clockwise direction.

How can i find the new coordinates after these 2 rotation with Euler's rotation theorem?

2. Dec 4, 2012

### haruspex

That theorem only tells you the net transformation is a rotation. Rodrigues' rotation formula might be more use.

3. Dec 5, 2012

### hei

using the point (0,0,5) , i make a rotation of 30 degree in clockwise direction about y axis,

so i should get the rotation matrix of

| cos 30 0 -sin 30 | |x| | cos 30 0 sin 30 | |x|
| 0 1 0 | |y| or | 0 1 0 | |y|
| sin 30 0 cos 30| |z| | -sin 30 0 cos 30| |z|

which one is the right one or i got both wrong?

Last edited: Dec 5, 2012
4. Dec 5, 2012

### BobG

Second one is correct (I think, they don't line up very well).

One rotation matrix for positive (direction of a right hand screw, or counter-clockwise as viewed from the outside) rotations:

Code (Text):

1        0        0
0     cos x     sin x
0    -sin x     cos x

With columns x, y, z and rows x, y, z, shift the matrix so the 1 lies at the intersection of whichever axis you're rotating around.

For negative (or clockwise), reverse the sign of the sines.

Since you're dealing with orthogonal matrices, the safer way is that clockwise is the inverse of counter-clockwise and the inverse of an orthogonal right hand matrix (which your rotation matrices are) is the transpose. In other words, take the transpose of the counter-clockwise matrix to get the clockwise rotation matrix.

The latter helps you when you start combining rotations into a single matrix and you don't want to have to reaccomplish the work for the opposite direction.

Additionally, sometimes the known angle won't be the rotation angle, but the complement of the rotation angle. Instead of building an ugly rotation matrix consisting of cos (90 - x), sine will substituted for cosine and cosine for sine to make a cleaner matrix consisting of just x.

5. Dec 6, 2012

### hei

lets say if i have a Z-axis which perpendicularly point into the screen, X-axis which point to its right side and Y-axis which point upward and a point with coordinate ( 0 , 0 , 5 ).

then i rotate the point's y-axis 30 degree in clockwise direction.
so the rotation operation matrix should be
Code (Text):

|cos 30      0    sin 30   |
|    0         1      0    |
|-sin 30      0    cos 30  |

Sorry about the disoriented matrix in previous post...

then on the point's new x-axis, i rotate the point 60 degree counter clockwise, so the point will move upward. And the matrix is:
Code (Text):

|    1        0        0       |
|    0       cos 60   sin 60   |
|    0      -sin 60   cos 60   |

so to get the point final position, i just multiply all the stuff together like:

Code (Text):

|cos 30      0    sin 30   | |    1        0        0       | | 0 |     |1.25   |
|    0        1     0      | |    0       cos 60   sin 60   | | 0 | =   |4.3301 |
|-sin 30      0    cos 30  | |    0      -sin 60   cos 60   | | 5 |     |2.165  |

Is that the right way?

6. Dec 6, 2012

### hei

Last edited: Dec 6, 2012
7. Dec 7, 2012

### hei

I get it now, thanks everyone for teaching me.