How to calculate instantaneous rotation axis of rigid body from point velocity vectors?

Question

When we have the angular velocity vector and point positions, calculation velocity vectors for point is not a problem. We use formula $$ \vec v = \vec \Omega \times \vec r $$ When I started writing algorithms for rigid body rotation I encountered the problem, how to calculate angular velocity from point velocity vectors using formulas, without the use of geometric relationships.

Flat example

We have two points of rigid body $m_1(x_1,0,0) ; m_2 (0,y_2,0) $ and velocity vectors for these points $ V_1(0,0,v_z1) ; V_2 (0,0,v_z2)$. How to calculate the instantaneous angular velocity for this example?

A three-dimensional example

We have three points of rigid body $m_1(x_1,0,0) ; m_2 (0,y_2,0); m_3 (0,0,z_3) $ and angular velocity vectors $\Omega (\omega_x, \omega_y, \omega_z) $. So point velocity vectors are

$ V_1(0,\omega_z x_1,-\omega_y x_1) ; V_2 (v_x2,0,v_z2) ; V_3 (v_x3,v_y3,0)$.

How to calculate the instantaneous angular velocity for this example using formulas not geometry?

Answer 1

Here's an answer that generalizes the question that was asked. The answer I originally posted here (which is preserved at the bottom of this answer) addressed only the very limited set of circumstances in the question in a limited way.

It is implicitly assumed in the question that the axis of the rotation passes through the point $(0,0,0)$ and that this point has zero velocity. Without such an assumption we cannot really conclude anything when the velocities of only two points are given; what if $v_{z1}=v_{z2}=1$, that is, both of the specified points have velocity $(0,0,1)$, and the point at $(0,0,0)$ also has velocity $(0,0,1)$? Then the motion of the object has no rotation at all.

But if we assume that the point at $(0,0,0)$ is within the rigid object and that this point within the object does not move, it is sufficient to have the velocities of just two other points, and these can be any two points as long as they do not both lie on a single line through $(0,0,0)$.

Let $\vec r_1$ and $\vec r_2$ be position vectors of two points of the rigid object, and let these points have velocities $\vec v_1$ and $\vec v_2$, respectively. Further assume that $\vec r_1$ and $\vec r_2$ are linearly independent.

Since the point at the origin is fixed, we know that \begin{align} \vec v_1 &= \vec\Omega \times \vec r_1, \\ \vec v_2 &= \vec\Omega \times \vec r_2 \end{align} where $\vec\Omega$ is the vector representing the (initially unknown) rotational velocity.

If $\vec v_1 = 0$, then $\vec\Omega = q_1 \vec r_1$ for some real-valued scalar $a_1$. Otherwise, since the body is rigid, we know that the radial component of $\vec v_1$ is zero. That means $\vec v_1$ is orthogonal to $\vec r_1$. Let

$$ \hat r_1 = \frac{\vec r_1}{\lVert\vec r_1\rVert}, \quad \hat v_1 = \frac{\vec v_1}{\lVert\vec v_1\rVert}, \quad \text{and} \quad \hat u_1 = \hat r_1 \times \hat v_1. $$ That is, $\hat r_1, \hat v_1, \hat u_1$ is an orthonormal basis. Write $\vec\Omega$ as a linear combination of this basis:

$$ \vec\Omega = a_1 \hat r_1 + b_1 \hat u_1 + c_1 \hat v_1. $$

Then $0 = \vec\Omega \cdot \vec v_1 = c_1 \lVert v_1\rVert$ implies that $c_1 = 0$. Moreover, \begin{align} \vec v_1 &= \vec\Omega \times \vec r_1 \\ &= (a_1 \hat r_1 + b_1 \hat u_1) \times \lVert\vec r_1\rVert \hat r_1 \\ &= a_1\lVert\vec r_1\rVert (\hat r_1 \times \vec r_1) + b_1\lVert\vec r_1\rVert (\hat u_1 \times \hat r_1) \\ &= b\lVert\vec r_1\rVert \hat v_1 \\ &= b_1\frac{\lVert\vec r_1\rVert}{\lVert\vec v_1\rVert} \vec v_1, \end{align} so $ b_1 = \dfrac{\lVert\vec v_1\rVert}{\lVert\vec r_1\rVert} $ and we have

$$ \vec\Omega = a_1 \hat r_1 + \frac{\lVert\vec v_1\rVert}{\lVert\vec r_1\rVert}(\hat r_1\times\hat v_1) = q_1 \vec r_1 + \frac{1}{\lVert\vec r_1\rVert^2} (\vec r_1 \times \vec v_1) \tag1 $$ where $q_1 = \dfrac{a_1}{\lVert\vec r_1\rVert}$ is a real-valued scalar.

Note that this is true also in the case where $\vec v_1 = 0$, so we can remove the condition that $\vec v_1 \neq 0$.

The same argument applies to $\vec v_2$ and $\vec r_2$, that is, $$ \vec\Omega = q_2 \vec r_2 + \frac{1}{\lVert\vec r_2\rVert^2} (\vec r_2 \times \vec v_2) \tag2 $$ where $q_2$ is a real-valued scalar.

Two vector expressions that are equal to $\vec\Omega$ are equal to each other, so $$ q_1 \vec r_1 + \frac{1}{\lVert\vec r_1\rVert^2} (\vec r_1 \times \vec v_1) = q_2 \vec r_2 + \frac{1}{\lVert\vec r_2\rVert^2} (\vec r_2 \times \vec v_2). $$

Rearranging the terms, $$ q_1 \vec r_1 - q_2 \vec r_2 = \frac{1}{\lVert\vec r_2\rVert^2} (\vec r_2 \times \vec v_2) - \frac{1}{\lVert\vec r_1\rVert^2} (\vec r_1 \times \vec v_1). $$ On the right side of this equation is a vector that can be computed directly from the given information in the problem. On the left side is a linear combination of two known vectors, in which only the two scalars $q_1$ and $q_2$ are unknown. If we write the equations componentwise in $x$, $y$, and $z$ components then we have a system of three linear equations in two unknowns. At least one of these equations is redundant, but since $\vec r_1$ and $\vec r_2$ are linearly independent, the solution of this system of equations is unique. Solve for $q_1$ or $q_2$ and then use either Equation $(1)$ or Equation $(2)$ to compute $\Omega$.

If you happen to choose $\vec r_1$ and $\vec r_2$ so that each has only one non-zero coordinate, that is, each one lies on a different axis, then the equation at the end becomes very easy to solve. But you do not need to use points on the axes, or even points whose position vectors are orthogonal. It is sufficient that the position vectors do not both lie in the same line.

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The previous contents of this answer

Since you have been working with rigid body rotations you are probably familiar with some of the unpleasant properties of those rotations, such as the fact that they are not commutative (the order in which you do the rotations matters).

Fortunately, angular velocity is not like that. It adds like an ordinary vector. (See here or here.)

Note, however, that for vector addition of angular velocities to be physically meaningful, the angular velocities must be physically composed in some way. In one of the examples in the link there is a platform rotating around one axis with a pendulum swinging around another axis that rotates with the platform. These angular velocities add as vectors. You don't need a physical linkage like that; it is enough to identify coordinate frames rotating relative to the inertial frame or to each other.

To reconstruct an angular velocity from components, however, the components must actually be components of the angular velocity relative to a single set of vectors, properly computed for that geometry. We face the same problem with linear velocity: if you want to decompose a linear velocity in a plane into one component parallel to the $x$ axis and one component parallel to the line $y=x,$ the calculation that gives the correct $x$ component for the usual orthogonal decomposition will generally give wrong answers for this non-orthogonal decomposition. It can be tricky to get this right, and it is no easier for decomposition of angular velocity than for decomposition of linear velocity.

But since you are working with orthogonal vectors, the decomposition is relatively straightforward. For your three-dimensional example, first we can establish that the linear velocity of a point at $(0,0,0)$ on this rigid object would be zero, so any rotation occurs around an axis through that point. No angular velocity applied to either the $y$ or $z$ axes can give the linear velocity of the point $(0,y_2,0)$ a non-zero $z$ component, so any $z$ component of that point must be due entirely to the component of angular velocity around the $x$ axis. Likewise the $x$ component of the linear velocity of $(0,y_2,0)$ must be due entirely to the component of angular velocity around the $z$ axis, and the $z$ component of the linear velocity of $(x_1,0,0)$ must be due entirely to the component of angular velocity around the $y$ axis.

That is, \begin{align} v_{z2} &= y_2 \omega_x,\\ v_{x2} &= -y_2 \omega_z,\\ v_{z1} &= -x_1 \omega_y,\\ \end{align} and the vectors $(\omega_x, 0, 0),$ $(0, \omega_y, 0),$ and $(0, 0, \omega_z)$ are components of the angular velocity of the body. The total angular velocity is $(\omega_x, \omega_y, \omega_z),$ that is, an angular velocity of $\sqrt{\omega_x^2 + \omega_y^2 + \omega_z^2}$ around the axis parallel to the vector $(\omega_x, \omega_y, \omega_z).$

Answer 2

It is not true that you never can commutative operation vector product. For the equation we are interested $$\vec v = \vec \Omega \times \vec r \tag 1$$ this can be easily done for a single point rotation (where vectors $\vec v, \vec \Omega, \vec r$ are perpendicular) and the value of the position vector is one $| \vec r|=1$. Then $$\vec \Omega = \vec r \times \vec v \tag 2$$ However, if the position vector is different from one $| \vec r|\neq1$, we must upgrades the pattern $$\vec \Omega = \frac {\vec r \times \vec v}{| \vec r|^2} \tag 3$$ When I understood what is temporary instantaneous angular velocity then I found a way to commutative the pattern (1) based on formula (3). I will now use a controversial record a inverse vector. I have already asked about the correctness of such a record here Can I use a fraction of vector to reverse cross product? and because so far no one has proved that such a record is wrong I will still use it.

I write down as a fragment of formula (3) as follows $$ \frac {\vec r} {|\vec r|^2}=( \frac {r_x} {|\vec r|^2},\frac {r_y} {|\vec r|^2},\frac {r_z} {|\vec r|^2})= \vec {\frac 1 r} \tag 4$$ And the current pattern (3) record is as follows $$\vec \Omega = \vec{\frac 1 r} \times \vec v \tag 5$$

Now we can proceed to the solution of the first example that was in the question. To be correct, the conditions must be met, we must have two perpendicular point location vectors $ \vec r_1 (r_x,0,0), \vec r_y (0,r_y,0)$. And the angular velocity must be on surface of these vectors $\vec \Omega (\omega_x, \omega_y,0)$. We know from properties the rigid body that all points must move according to the formula (1) hence we learn what point velocity vectors are $$\vec v_1= \vec \Omega \times \vec r_1 = (0,0,- \omega_y r_x) \tag 6 $$ $$\vec v_2= (0,0, \omega_x r_y) $$ The points rotate only at an angular velocity perpendicular to the position vector. So we count vectors of angular velocity of points from the formula (5)

$$\vec \Omega_1 = \vec{\frac 1 r_1} \times \vec v_1 = (0, \frac 1 r_x r_x \omega_y,0) \tag 7$$ $$\vec \Omega_2 = \vec{\frac 1 r_2} \times \vec v_2 = ( \frac 1 r_y r_y \omega_x,0,0) $$ So we see that $$\vec \Omega = \vec \Omega_1 + \vec \Omega_2= \vec{\frac 1 r_1} \times \vec v_1 + \vec{\frac 1 r_2} \times \vec v_2 \tag 8 $$ This is the answer to the first example. Everything works in a flat arrangement. To calculate any instantaneous angular velocity in three-dimensional space need some more calculations.

Answer to the second example

Important! The points must be perpendicular to each other.

Again, we are counting velocity vectors for points $$\vec v_1= \vec \Omega \times \vec r_1 = (0,\omega_z r_x,- \omega_y r_x) \tag 9 $$ $$\vec v_2= (-\omega_z r_y,0, \omega_x r_y) $$ $$\vec v_3= (\omega_y r_z, -\omega_x r_z,0) $$ In a freely rotating rigid body in the reference system where the center of mass is stationary point velocity vectors are perpendicular to their position vectors.

We now calculate the angular velocities of the points from the formula (5)

$$\vec \Omega_1 = \vec{\frac 1 r_1} \times \vec v_1 = (0, \omega_y, \omega_z) \tag {10}$$ $$\vec \Omega_2 = \vec{\frac 1 r_2} \times \vec v_2 = (\omega_x,0,\omega_z) $$ $$\vec \Omega_3 = \vec{\frac 1 r_3} \times \vec v_3 = (\omega_x,\omega_y,0) $$ When these vectors are added together, the angular velocity values are doubled, it means that you have to divide the result by two $$\vec \Omega = \frac {\vec{\frac 1 r_1} \times \vec v_1 + \vec{\frac 1 r_2} \times \vec v_2 + \vec{\frac 1 r_3} \times \vec v_3 } 2 \tag {11} $$

But we don't have to use such a long formula, observing the results (10) we see the information on the coordinates of the instantaneous angular velocity can be drawn using the formula (5) any two points whose position vectors are perpendicular.

If someone claims that it is impossible to perform the inverse operation of a vector product. In the sense that it is impossible in the case $\vec a \times \vec b = \vec c $ perform the reverse operation, i.e. having only vectors $\vec b $ and $ \vec c$ get the whole vector $ \vec a$ or having only vectors $\vec a $ and $ \vec c$ get the whole vector $ \vec a$ . It seems like I've done something impossible:)