Coordinate system transformation problem (about Euler angle)

Question

This question comes from the work that constructing the general cylinder by the NURBS. I have implemented the point-set $P_1,P_2,\cdots,P_n$ that in the $O-xyz$ plane, now I want to implement the point-set that in the plane $O_1-x_1y_1z_1$. So I need to transform the coordinate of $P_1,P_2,\cdots,P_n$ from the CS $O-xyz$ to CS $O_1-x_1y_1z_1$ firstly.

---

Given that I have a set of points $P_1,P_2,\cdots,P_n$ in the $O-xyz$ plane. Please see the following schematic diagram. Here, $P_i=\{x_i,y_i,0\}$

Now I would like to transform these points to other coordinate system($O_1-x_1y_1z_1$).

My trial

I believe that the key step is solving the transformation matrix between coordinate system ($O-xyz$) and coordinate system ($O_1-x_1y_1z_1$).

• Step1: Using the translational transformation $ \begin{pmatrix} 1 && 0 && 0 && O_{1x}\\ 0 && 1 && 0 && O_{1y}\\ 0 && 0 && 1 && O_{1z}\\ 0 && 0 && 0 && 1 \end{pmatrix} $
• Step2: Using the composite rotational transformation.

For the second step, I discovered the Euler-angle reference in Wiki-Encyclopedia.

However, I didn't know how to confirm the angles like $\alpha$ and $\gamma$ via the $\vec{O_1z_1}=\{n_x,n_y,n_z\}$owing to that I am struggling to understand $\vec{N}$ axis.

For the angle $\beta$, it is easy to compute. Namely,

$$\cos\beta=\frac{n_z}{\sqrt{n_x^2+n_y^2+n_z^2}}$$

---

Question

I have searched the DOC of Wolfram Mathematica by the keyword Euler in V9. Unfortunately, I cannot find the help info.

• Is there related built-in about Euler-angle transformation in Mathematica?
• Or how to do the Euler-angle transformation?

Answer 1

To address your actual problem:

If you're just looking to re-orient your B-spline cylinder, there's no need to go through the Euler angles. Here's one way.

Consider the following cylinder:

myCyl = BSplineSurface[{{{0, 0, 0}, {0, 1, 0}, {1, 1, 0}, {1, 0, 0}},
                        {{0, 0, 1}, {0, 1, 1}, {1, 1, 1}, {1, 0, 1}}}, 
                       SplineClosed -> {False, True}, SplineDegree -> {1, 3}, 
                       SplineWeights -> {{1, 3, 3, 1}, {1, 3, 3, 1}}];

Graphics3D[myCyl, Axes -> True]

Note that this unit-height cylinder is pointing towards the positive $z$-axis. To re-orient this cylinder to point in a different direction, you can use RotationTransform[]. Here's how to re-orient it to point at the direction of $(3,1,2)$:

Graphics3D[{MapAt[Map[RotationTransform[{{0, 0, 1}, {3, 1, 2}}], #] &, myCyl, 1],
            {Black, Arrow[Tube[{{0, 0, 0}, {3, 1, 2} // Normalize}]]}},
           Axes -> True]

You can of course add a translation:

Graphics3D[{MapAt[Map[Composition[TranslationTransform[{-1, -1, 1}], 
                                  RotationTransform[{{0, 0, 1}, {3, 1, 2}}]], #] &,
                  myCyl, 1],
            {Black, Arrow[Tube[{{0, 0, 0}, {3, 1, 2} // Normalize}]]}},
           Axes -> True]

Answer 2

EulerMatrix is available in MMA 10. To obtain the matrix for the transformation shown in your sketch, apply

EulerMatrix[{α,β,γ},{3,1,3}]

This transformation is known as the x-convention, because the second rotation is about x'-axis. The Wikipedia designates this by ZXZ.

Those who do not have MMA 10 can obtain the same x-convention transformation using RotationMatrix applied 3 times. The code is

Transpose@
  Dot[
    RotationMatrix[γ, {0, 0, -1}], 
    RotationMatrix[β, {-1, 0, 0}], 
    RotationMatrix[α, {0, 0, -1}]]

As you know from the Wikipedia, there is more than one set of Euler angles. The y-convention is more common in quantum mechanics, so let me give that transformation here as well. In MMA 10, the y-convention transformation is

 EulerMatrix[{α,β,γ},{3,2,3}]

or

 EulerMatrix[{α,β,γ}]

since the y-convention is the default when the axes are not specified. The Wikipedia designates this by ZYZ. In terms of the RotationMatrix function, the y-convention transformation is

Transpose@
  Dot[
   RotationMatrix[γ, {0, 0, -1}],
   RotationMatrix[β, {0, -1, 0}], 
   RotationMatrix[α, {0, 0, -1}]]

A reference for these transformations is Classical Mechanics by Herbert Goldstein, 2nd Edition, Section 4-4 and Appendix B.

Answer 3

Suppose you have a vector p expressed as

p = p1 e1 + p2 e2 + p3 e3

where e1, e2, e3 are a dextral triad of unit vectors and p1, p2 and p3 are scalars.

Let n1, n2, n3 be another dextral triad of unit vectors.

Then p = p1 (e1.n1 n1 + e1.n2 n2 + e1.n3 n3) + p2 (e2.n1 n1 + e2.n2 n2 + e2.n3 n3) + p3 (e3.n1 n1 + e3.n2 n2 + e3.n3 n3)

You can collect terms to put this in the form p =()n1 + ()n2 + ()n3. The () contain scalars involving dot products.

e1.n1 is the cosine of the angle between e1 and n1. And similarly for the other 8.