Euler angles equivalence with a single rotation

AI Thread Summary
The discussion centers on the possibility of representing a rotation defined by Euler angles (alpha, beta, gamma) using a single rotation around a unit vector N and an angle zeta. It is established that N is not unique due to the transformation properties of rotation vectors, although it can be uniquely defined under certain conditions. The conversation emphasizes the use of rotation matrices and geometric algebra to derive N and zeta as functions of the Euler angles, independent of the specific vector being rotated. The method involves chaining rotations and using rotors to express the overall rotation in a manageable form. Ultimately, the goal is to find a clear relationship between the Euler angles and the parameters of a single rotation.
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imagine we rotate a vector centered at the origin with euler angles alpha,beta,gamma.
now the question is, can we do this rotation by the means of defining a vector N(which its length is 1).and rotating the vector zeta radians counter clockwise around N?
I think it must be possible and I want to know if N is unique(we have only one N for the given euler angles) and I want to find N and zeta in terms of alpha beta and gamma.
thank you.(complete answers really apreciated!GOD bless those who give complete anwers!)
P.S: what I've done to find the answer:I used cartesian coordinates and found the rotatian matrix correspending to a rotation zeta over a given N.>>RA=A'
I found R.A is a column matrix (x,y,z) and A'(x',y',z').and R is a 3x3 matrix which I found in terms of N(nx,ny,nz) and zeta.
 
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Do you want to transform a specific, known vector? If you want to express a general rotation with just N and zeta, those will depend on the rotated vector. You can look how this vector transforms under the euler rotations and compare this to a rotation around an arbitrary N. Solving the equations will give an appropriate N.

The transformation \vec{N} \to -\vec{N} and \zeta \to - \zeta +2 \pi k always works, so N is not unique. You can restrict ζ to [0,pi], and N will be unique apart from special cases (ζ=0 and ζ=pi).
 
thank you sir.
of course we know the vector we want to rotate.its say;A(Ax,Ay,Az) but it doesn't matter I think.for what ever vector, we use we want to know N and zeta which must be functions of only alpha beta gamma(the euler angles correspending to rotation).
I mean imagine we have the rotation matrix (possibly NOT a function of A and only function of alpha beta gamma) and also we know the rotation matrix for the rotaion zeta around an N.(also NOT a function of A) so if we simply put this two matrix equal we should get N and zeta as a function of the euler angles.A is not involved in the answer.
now I want some angel! to write N and zeta as function of the euler angles.
it means around which axis and how much do we need to do a single rotation to get an euler rotation of alpha beta gamma.
 
Rotations are very easy to deal with in geometric algebra and/or with quaternions.

The rotation is entirely characterized by a rotor: R = e^{- i \theta /2} where i is some bivector, or a geometric object, that obeys i^2 = -1. There are three such objects in 3D space: we'll call them e_{xy}, e_{yz}, e_{zx}. They describe rotations in the xy, yz, and zx planes, respectively.

We can chain three such rotations together:

R = \exp(-e_{xy}\alpha/2) \exp(- e_{yz}\beta/2) \exp(- e_{zx} \gamma/2)

(You should already know that this is a perfectly valid means of parameterizing rotations; you can always go from fixed axes to rotated axes with Euler angles.)

When you multiply all three rotors out, you'll get something of the form

R = a + b e_{xy} + c e_{yz} + d e_{zx} = \cos \frac{\theta}{2} - u \sin \frac{\theta}{2}

where u is a unit bivector. Taking \cos^{-1} a = \theta/2 gives the angle. Dividing out \sin \theta/2 gives the plane of rotation.

Example: consider a rotation composed of a \pi/3 rotation in the xy plane, \pi/2 in the yz plane, and \pi/3 in the zx plane. The rotor is then

R = \left(\frac{\sqrt{3}}{2} - e_{xy} \frac{1}{2} \right) \left(\frac{\sqrt{2}}{2} - e_{yz} \frac{\sqrt{2}}{2} \right) \left(\frac{\sqrt{3}}{2} - e_{zx} \frac{1}{2} \right)

Right away, just by multiplying all the first terms and all the last terms, we can find that \cos \theta/2 = \sqrt{2}/4.

The rest of the multiplication is best done by selecting for the other terms:

R = \frac{\sqrt{2}}{4} - e_{xy} \frac{\sqrt{6}}{4} - e_{yz} \frac{\sqrt{2}}{4} - e_{zx} \frac{\sqrt{6}}{4}

(May have done arithmetic wrong, but it should give the idea that this is not difficult--only tedious--to do.)

From here, the answer is within sight. The angle is 2\cos^{-1} (\sqrt{2}/4) \approx 2.42 = 139^\circ. Written in terms of the unit bivector, the rotor is

R = \frac{\sqrt{2}}{4} - \frac{\sqrt{14}}{4} \left(\sqrt{\frac{3}{7}} e_{xy} +\sqrt{\frac{1}{7}} e_{yz} + \sqrt{\frac{3}{7}} e_{zx} \right)

Or, more plainly, we have a 139 degree rotation about e_x + 3 e_y + 3 e_z. And this was all possible through geometric algebra/quaternions.
 
mfb said:
If you want to express a general rotation with just N and zeta, those will depend on the rotated vector.
This is wrong - sorry, I was confused.
See Muphrid.
 
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