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I Making Eulers eqs. comply with Lagrange eqs.

  1. May 26, 2016 #1
    Lately when doing a simulation for a quadrocopters most reports I've come across regarding modeling use Eulers equation of motion. That makes sense, as the quadrocopter is a body rotating in 3 dimensions.

    Then I tried to model the system using Lagrange equations instead but I don't get the same result. I wrote the Lagrangian as

    L = T-U = T = 0.5*(Ixωx2 +Iyωy2+Izωz2 )

    But applying Lagrange equations to L doesn't give Eulers equation of motions. Actually though, it gives the linearized version of Eulers, when linearized around the angular velocities zero.

    Why is this? I must have missed some crucial condition that must be fulfiled for Lagrange equations to hold true? Did I get the expression for the lagrangian L wrong? Is it because Eulers model a continous mass distribution, while Newton/Lagrange rather deal with a system of masses? I assumed the lagrangian to be in the rotating frame, as are Eulers equations in their standard version.
     
  2. jcsd
  3. May 26, 2016 #2
    The Lagrangian L must be expressed in terms of generalized coordinates and generalized velocities. This is only I can answer on such a level of detailing
     
  4. May 26, 2016 #3
    But we have three general forces that are the respective input torques for each axis, and three general coordinates that are the euler angles, right?
     
  5. May 27, 2016 #4
    it is hard to find your mistake without formulas
     
    Last edited: May 27, 2016
  6. May 27, 2016 #5

    vanhees71

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    You must set up your Lagrangian in terms of generalized coordinates and the corresponding generalized velocities, i.e., you always work in a holonomous basis on the tangent space of your manifold, but the Cartesian coordinates of the momentaneous angular velocity in the body-fixed system of reference, as used in Euler's equations, are not such holonomous coordinates.

    For the rigid-body description you can use Euler angles in the Lagrangian formalism.
     
  7. May 28, 2016 #6
    vanhees71: Hmm. Is there some way of seeing this algebraically?
     
  8. May 28, 2016 #7

    vanhees71

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    You can evaluate the Cartesian components in terms of Euler angles. I can point only to my German mechanics manuscript, but it's in any textbook on analytical mechanics treating the spinning top:

    It starts here (unprimed coordinates refer to the inertial space-fixed and primed ones to the non-inertial/rotating body-fixed reference frame)

    http://theory.gsi.de/~vanhees/faq/mech/node74.html

    The relations between Euler angles and Cartesian coordinates of the angular velocity is given at

    http://theory.gsi.de/~vanhees/faq/mech/node78.html

    The Euler angles are depicted here (note that the meaning of the angles in different textbooks differ!):

    http://theory.gsi.de/~vanhees/faq/mech/node22.html

    I follow the convention in

    A. Sommerfeld, Lectures on Theoretical Physics, vol. 1 (Mechanics)

    which has a very good treatment of the top (which is no surprise, because Sommerfeld wrote a comprehensive mathematical treatment of the spinning top together with Felix Klein).
     
  9. May 28, 2016 #8
    there is a simple way to obtain formulas for angular velocity in terms of Euler angles. This way is based upon the addition theorem for angular velocities (without calculating of the matrices
     
  10. Jun 7, 2016 #9
    Hmm. Food for thoughts.

    I should try to understand the issue of rotation in a rotating coordinate system first. Because, if the coordinate system is rotating exactly the same as some solid object, the rotation of the object in this coordinate system must be zero. But, it seems like rotation like this is defined such that the object is making an infinitely small rotation before the coordinate system follows? So that the rotation of the coordinate system is always lagging behind an infinitely small angle? I can't see how this could happen otherwise. Am I on the right track?
     
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