Making Eulers eqs. comply with Lagrange eqs.

• I
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.

wrobel
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

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?

wrobel
it is hard to find your mistake without formulas

Last edited:
vanhees71
Gold Member
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.
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.

vanhees71: Hmm. Is there some way of seeing this algebraically?

vanhees71
Gold Member
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

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).

wrobel