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How to rotate Cartesian coordinate system? 
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#1
Jul3013, 12:36 PM

P: 5

Hello,
I would like to rotate the Cartesian coordinate system ( i=(1,0,0); j=(0,1,0); k=(0,0,1) ) so that angles between new and the old axes be equal to α, β and γ, respectively. Is any simple way similar to the Euler transformations to accomplish that? 


#2
Jul3013, 01:47 PM

P: 615




#3
Jul3013, 03:58 PM

P: 5




#4
Jul3013, 04:06 PM

P: 615

How to rotate Cartesian coordinate system?
It's very helpful. Wikipedia is always the first place I start looking when I have a question. If you don't like Wikipedia, though, there's always this alternative. 


#5
Jul3013, 04:38 PM

P: 5




#6
Jul3013, 04:55 PM

P: 327

You're basically looking for a matrix A such that the vectors Ai, Aj and Ak form a basis for 3D space and are in angles α, β and γ relative to the vectors i, j and k.
You can form a system of equations from which you can solve the elements of matrix A by requiring that: 1. A must be an orthogonal matrix, i.e. vectors Ai, Aj and Ak form an orthonormal set, too, when i, j and k do. 2. The dot products are [itex] {\bf i} \cdot(A {\bf i} )=cos(\alpha)[/itex], [itex] {\bf j} \cdot(A {\bf j} )=cos(\beta)[/itex], [itex] {\bf k} \cdot(A {\bf k} )=cos(\gamma)[/itex] These conditions are enough to determine the matrix elements. 


#7
Jul3013, 05:00 PM

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PF Gold
P: 6,344

Don't give up. I think this fits your bill: http://inside.mines.edu/~gmurray/ArbitraryAxisRotation/



#8
Jul3013, 06:06 PM

P: 5

From the initial conditions (2.) I can define immediately three components of new basis vectors ([itex] A{\bf i}_x = acos(\alpha)[/itex], [itex] A{\bf j}_y = acos(\beta)[/itex], [itex] A{\bf k}_z = acos(\gamma)[/itex]). For the rest of components I can write the orthogonality (using scalar or/and vector products) and the normalization conditions for new basis vectors. However, this gives system of six quadratic equations with six unknowns which is quite ugly to solve generally. Bedsides that, I would get 8 or 16 different solutions of that system and I doubt existence of more than two solutions for the original problem. 


#9
Jul3013, 06:11 PM

P: 5

The page looks interesting but it's not clear for me how to correlate the Rotation About an Arbitrary Axis to my problem. 


#10
Nov1713, 01:21 PM

P: 19

Could someone help me figure out how to rotate a coordinate system about the zaxis such that the the line y = mx + c coincides with the xaxis?
Shouldn't a simple projection of all the coordinates i.e xj = xicos(theta), yj =yicos(theta) and zj = zj cos(theta) work? 


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