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Eliminating the Euler Angle singularity without quaternions?

  1. Aug 23, 2011 #1
    Hi all,

    I've formulated using Lagrangian formalism the equations of motion for a spinning top. I know about the gimbal lock/singularity that occurs at theta=0 and I was wondering if there was any other way to do it without dwelving into quaternions.

    Yogi published a paper "A Motion of Top by Numerical Calculation" suggesting a replacement: [itex]\dot{\beta} = \dot{\varphi} cos \vartheta[/itex] and [itex]\dot{\alpha} = \dot{\varphi} sin \vartheta[/itex], but this hasn't worked for me (I find myself getting [itex]\ddot{\alpha} = 0[/itex], which isn't true)

    I've had a look at quaternions but I'm not inclined to understand it or be able to simply 'convert' my equations over into quaternion calculus, neither (I believe) can I use Lagrangian mechanics on quaternions.
     
  2. jcsd
  3. Aug 23, 2011 #2
    One way to avoid the problem without using quaternions is to lock the possible values of theta to a given range that doesn't include the singularities. For example, theta cannot be smaller than 1E-6 degrees and larger than 179.99999 degrees. But the simplest way is using quaternions.

    M.
     
  4. Aug 23, 2011 #3
    Thanks Mbert, unfortunately the case I am looking at causes the top to rise to the steady position.. which unfortunately is the theta 0 position.

    Looks like I'll have to dwelve into quaternions. Is it possible to 'convert' my equations in Euler angles into quaternions if I have the equations in Euler's (rigid body) equations? Or must everything be redefined again?
     
  5. Aug 23, 2011 #4
    To convert from Euler angles to quaternions take a look at this website:

    http://www.flipcode.com/documents/matrfaq.html#Q60"

    Especially the part:

    M.
     
    Last edited by a moderator: Apr 26, 2017
  6. Aug 23, 2011 #5
    Thanks Mbert, I came across several articles on how to convert the angles themselves to quaternion, however the equations of motion are in the forms of: [itex]\ddot{\phi},\ddot{\psi},\ddot{\vartheta} = f(\phi,\psi,\vartheta,\dot{\phi},\dot{\psi},\dot{\\theta})[/itex], so I cannot apply the conversions in that manner.

    Does Lagrange formalism work on quaternions?
     
  7. Aug 23, 2011 #6
    Last edited by a moderator: Apr 26, 2017
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