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Euler angles equivalence with a single rotation

  1. Sep 8, 2012 #1
    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 lenght 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.
     
  2. jcsd
  3. Sep 8, 2012 #2

    mfb

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    Staff: Mentor

    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 [itex]\vec{N} \to -\vec{N}[/itex] and [itex]\zeta \to - \zeta +2 \pi k[/itex] 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).
     
  4. Sep 8, 2012 #3
    thank you sir.
    of course we know the vector we want to rotate.its say;A(Ax,Ay,Az) but it doesnt 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.
     
  5. Sep 8, 2012 #4
    Rotations are very easy to deal with in geometric algebra and/or with quaternions.

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

    We can chain three such rotations together:

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

    (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

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

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

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

    [tex]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)[/tex]

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

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

    [tex]R = \frac{\sqrt{2}}{4} - e_{xy} \frac{\sqrt{6}}{4} - e_{yz} \frac{\sqrt{2}}{4} - e_{zx} \frac{\sqrt{6}}{4}[/tex]

    (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 [itex]2\cos^{-1} (\sqrt{2}/4) \approx 2.42 = 139^\circ[/itex]. Written in terms of the unit bivector, the rotor is

    [tex]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)[/tex]

    Or, more plainly, we have a 139 degree rotation about [itex]e_x + 3 e_y + 3 e_z[/itex]. And this was all possible through geometric algebra/quaternions.
     
  6. Sep 8, 2012 #5

    mfb

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    This is wrong - sorry, I was confused.
    See Muphrid.
     
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