What Are the Correct Euler's Angles for Rotating a Cube?

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SUMMARY

The correct Euler's angles for rotating a cube around its principal axes are defined as follows: for rotation around the x-axis, use θ = θ, φ = 0, ψ = 0; for the z-axis, set ψ = 0, θ = 0, φ = θ; and for the y-axis, apply φ = θ, θ = θ, ψ = -θ. The rotation matrices involved are R_x(α), R_y(α), and R_z(α), with the overall transformation represented by R(θ, φ, ψ) = R_z(φ) R_x(θ) R_z(ψ). The condition for the y-axis rotation leads to θ = -π/2 - α and φ = ψ = -π/2.

PREREQUISITES
  • Understanding of Euler's angles and their application in 3D rotations
  • Familiarity with rotation matrices R_x(α), R_y(α), and R_z(α)
  • Basic knowledge of trigonometric functions and their role in rotation
  • Concept of principal axes in three-dimensional geometry
NEXT STEPS
  • Study the derivation and application of rotation matrices in 3D graphics
  • Learn about quaternion representations for 3D rotations
  • Explore the implications of Euler's angles in robotics and aerospace engineering
  • Investigate the limitations and ambiguities associated with Euler's angles
USEFUL FOR

Students in physics or engineering, 3D graphics developers, and anyone involved in robotics or simulations requiring precise rotational transformations.

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Homework Statement


What are the Euler's angles corresponding to the rotations of a cube in [tex]\theta[/tex] radians around each of its principal axes



around the x: [tex]\theta[/tex]=[tex]\theta[/tex]
[tex]\phi[/tex]=[tex]\psi[/tex]=0

around z:[tex]\psi[/tex]=[tex]\theta[/tex]=0
[tex]\phi[/tex]=[tex]\theta[/tex]

around y:[tex]\phi[/tex]=[tex]\theta[/tex]=[tex]\theta[/tex]
[tex]\psi[/tex]=-[tex]\theta[/tex]


is it correct? how can I make it more clear? It's very confusing...
 
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Help me please...
 
In order to formalize the solution you can use the rotation matrices: [tex]\hat{R}_x(\alpha)[/tex], [tex]\hat{R}_y(\alpha)[/tex], [tex]\hat{R}_z(\alpha)[/tex] and the matrix which describes the whole Euler's transform:

[tex] \hat{R}(\theta, \phi, \psi) =<br /> \hat{R}_z(\phi) \hat{R}_x(\theta) \hat{R}_z(\psi) \quad (1)[/tex]

It's easy to get rotation around the x and z axes from (1) and your answers for these cases are right.

As for y-axis the condition

[tex] \hat{R}(\theta, \phi, \psi) = \hat{R}_y(\alpha)[/tex]

yields

[tex] \theta = -\frac{\pi}{2} - \alpha;[/tex]

[tex] \phi = \psi = -\frac{\pi}{2}.[/tex]
 

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