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Let's suppose we have two rotations about two different axes represented by vectors $v_1$ and $v_2$: $R_1(v_1, \theta_1)$, $R_2(v_2,\theta_2)$.

It's relatively easy to prove that composition of these two rotations gives rotation about axis $v_3$ distinct from axes $v_1$ and $v_2$ .

Indeed
if for example $v_3=v_1$ then
$R_1(v_1, \theta_1) R_2(v_2,\theta_2)=R_3(v_1,\theta_3)$ leads to $R_2(v_2,\theta_2)=R_1^T(v_1, \theta_1)R_3(v_1,\theta_3)=R(v_1,\theta_3 -\theta_1)$ what gives $v_1=v_2$. ... Contradiction...
We see that composition of two rotations about different axes always generates a new axis of rotation.

The problem can be extended for condition of the plane generated by the axes.

Question:

  • Is it true that composition of two rotations generates the axis which doesn't belong to the plane which is constructed by the original axes of rotations ?

  • How to prove it ?

  • If the statement is not however true what are conditions for not changing a plane during the composition of rotations $ ^{[1]}$ ?

$ ^{[1]}$ It can be observed that even in the case of quite regular rotations the above statement is true

Let's take $Rot(z,\dfrac{\pi}{2})Rot(x,\dfrac{\pi}{2})= \begin{bmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{bmatrix} = Rot([1,1,1]^T, \dfrac{2}{3}\pi)$

or

$Rot(x, \pi )Rot(z, \pi )= \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix} = Rot( y, \pi)$

So I suppose it is generally true but how to prove it ?

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2 Answers 2

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This is easily seen if we assume familiarity with the use of unit quaternions in representing rotations. A rotation $R$ about the axis given $\vec{v}=v_1\bf{i}+v_2\bf{j}+v_3\bf{k}$ by the angle $\theta$ is represented by the quaternion $$ q=\cos\frac\theta2+\sin\frac\theta2\vec{v}. $$ Here it is essential that $\vec{v}$ is a unit vector. The connection is that the rotated version $R\vec{u}$ of a vector $\vec{u}$ is then given by the quaternion product $$ R\vec{u}=q\vec{u}\overline{q}, $$ where $\overline{q}=\cos\frac\theta2-\sin\frac\theta2\vec{v}$ is the conjugate quaternion.

The composition of two such rotations is then faithfully reproduced as a product of the representing quaternions. So if another rotation $R'$ is represented by $q'=\cos\frac\alpha2+\sin\frac\alpha2\vec{v}'$, the composition $R'\circ R$ is represented by the product $$ \begin{aligned} qq'&=\left(\cos\frac\alpha2\cos\frac\theta2-\sin\frac\alpha2\sin\frac\theta2\,\vec{v}'\cdot\vec{v}\right)+\\ &+\cos\frac\alpha2\sin\frac\theta2\vec{v}+\cos\frac\theta2\sin\frac\alpha2\vec{v}'+\sin\frac\alpha2\sin\frac\theta2\,\vec{v}'\times\vec{v}. \end{aligned} $$ From the second row we can read the axis of the composition - it is the unit vector parallel to that linear combination of $\vec{v}$, $\vec{v}'$ and their cross product. The first two terms are in the plane $T$ spanned by $\vec{v}$ and $\vec{v}'$, but the cross product is perpendicular to $T$. Therefore the axis of the combined rotation is in the plane $T$ if and only if that cross product term is zero. Either of the sines vanishes only when the rotation is trivial ($\alpha=0$ or $\theta=0$). The cross product vanishes iff $\vec{v}$ and $\vec{v}'$ are parallel.


In other words, your hunch is correct.

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  • $\begingroup$ I wonder if starting from $R=\exp{(\theta K)}$, with $K$ as in the matrix version of Rodriguez’ formula leads to anything interesting. $\endgroup$
    – amd
    Jul 18, 2016 at 20:23
  • $\begingroup$ Nah... seems like a dead end unless I can somehow reach the requirement that $\theta_1K_1$ and $\theta_2K_2$ must commute. $\endgroup$
    – amd
    Jul 19, 2016 at 4:51
  • $\begingroup$ @amd In what case they commute? Orthogonal axes ? Probably not.. $\endgroup$
    – Widawensen
    Jul 19, 2016 at 6:23
  • $\begingroup$ @Widawensen: Those $K$ are antisymmetric 3x3 matrices with zeros on the diagonal. Their commutators are essentially the cross product. So they commute, iff the rotations are about the same axis. $\endgroup$ Jul 19, 2016 at 6:27
  • $\begingroup$ @JyrkiLahtonen Right. If you multiply out their commutator and set it to zero, the three equations that you get say that the projections of the axes onto any of the coordinate planes are colinear, i.e., that the two rotation axes coincide. That was the easy part. I gave up trying to end up with this a necessary condition for the composite axis to be coplanar with the two original rotation axes. $\endgroup$
    – amd
    Jul 19, 2016 at 6:35
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Using quaternions, as in Jyrki’s answer, is straightforward and provides some useful insights into the nature of the resultant rotation. As he suspects, the result can also be found another way.

Suppose that the new rotation axis is coplanar with the original axes. Since vectors along the axis of rotation are eigenvectors of $1$, we must have $$R_1R_2(av_1+bv_2)=av_1+bv_2$$ for some $a$ and $b$ not both zero. Rearranging and again using the fact that $v_1$ and $v_2$ are eigenvectors of the respective rotations gives $$a(R_2-I)v_1=b(R_1^{-1}-I)v_2.$$ Wlog we can take $v_1$ and $v_2$ to be unit vectors. Substituting Rodrigues’ formula for the rotations then makes this $$a[(1-\cos{\theta_2})v_2\times(v_2\times v_1)+\sin{\theta_2}(v_2\times v_1)] = b[(1-\cos{\theta_1})v_1\times(v_1\times v_2)-\sin{\theta_1}(v_1\times v_2)]$$ which can be rearranged as $$[a(1-\cos{\theta_2})v_2+b(1-\cos{\theta_1})v_1]\times(v_2\times v_1) + (a\sin{\theta_2}-b\sin{\theta_1})(v_2\times v_1) = 0.\tag{*}$$ This clearly holds if $v_1$ and $v_2$ are colinear. If they aren’t, then the two vectors being added together in (*) are orthogonal, so they must both be zero. Any linear combination of $v_1$ and $v_2$ is orthogonal to $v_1\times v_2$, so the expression in square brackets must be zero, and since by hypothesis $v_1$ and $v_2$ are not colinear, we have $a(1-\cos{\theta_2})=0$ and $b(1-\cos{\theta_1})=0$, therefore at least one of the rotation angles must be zero.

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  • $\begingroup$ Nice! I made it as far as your second equation but, not knowing my Rodriguez, could not really do much with it :-) $\endgroup$ Jul 18, 2016 at 20:29
  • $\begingroup$ @amd So we have also proof without quaternions, it's interesting that in both answers cross product plays important role.... $\endgroup$
    – Widawensen
    Jul 18, 2016 at 20:45
  • $\begingroup$ @Widawensen The cross products are an artifact of working in $\mathbb R^3$. A rotation operates on the orthogonal rejection of a vector $v$ relative to the axis $u$, which in $\mathbb R^3$ can be expressed as $-u\times(u\times v)$. In Rodrigues’ construction, this pair is extended to an orthogonal basis, which involves another cross product. $R(x\times y)=Rx\times Ry$, so these cross products survive the second rotation. Also, since we’re interested here in the span of $v_1$ and $v_2$, I’m not terribly surprised that a normal to that plane crops up somewhere. $\endgroup$
    – amd
    Jul 19, 2016 at 4:49
  • $\begingroup$ Yup. The cross product is also the Lie bracket of the Lie algebra of $SO_3(\Bbb{R})$ (or of the group of unit quaternions $SU_2(\Bbb{C})$), when we identify it with $\Bbb{R}^3$. In other dimensions it is more complicated. $\endgroup$ Jul 19, 2016 at 5:24
  • $\begingroup$ @Jyrki Jyrki, Could I add in the separate answer which would be a kind of supplement to yours a discussion of the basic most interesting cases derived from your formula? I mean case for orthogonal axes and rotation angles 90 and 180 degrees.. or if you want you can describe them..I'm really astonished that it is so straightforward ... only using these half-angles in it is some kind mysterious... $\endgroup$
    – Widawensen
    Jul 19, 2016 at 5:54

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