Rotating reference frames

Question

I'm trying to understand the equations that govern velocity in a rotating reference frame... \begin{equation} v_i = (\frac{dr}{dt})_r + \Omega \times r . \end{equation}

I'd like to build a simple simulation of a rocket taking off from earth with some constant inertial velocity, say: $v_i=[1,0,0]^T$.

I assume some $\Omega$ value to represent the rotation of the earth about z, say $\Omega=[0,0,1]$.

And then solve for the percieved velocity in the rotating frame: \begin{equation} (\frac{dr}{dt})_r = v_i - \Omega \times r . \end{equation}

What I expect, after integrating velocity into position, would be an outwardly rotating spiral showing the relative position of the "rocket" to an observer in the rotating earth frame. What I see, from a simple simulink sim, is quite different.

My sim:

The output:

Thoughts?

Answer 1

$\def\m{\mathbf}$ Coordinate vector of a point in static frame: $r^s$

Coordinate vector of the same point in rotating frame: $r^r$

(Pure rotation, both frames have the same origin.)

Coordinate transformation (rotation matrix): $R$

The matrix is orthogonal, i.e., $R^TR=RR^T=\m1$ (the unit matrix)

Important property: $\m0=\frac{d}{dt} \m1 = \frac{d}{dt} (R^TR) = \dot R^T R + R^T \dot R$

That means the matrix $\m\Omega := R^T\dot R$ is anti-symmetric $\m\Omega = -\m\Omega^T$ (with only three relevant components $\Omega_1 := \m\Omega_{32}, \Omega_2 := \m\Omega_{13}, \Omega_3:=\m\Omega_{21}$) and the products $\m\Omega v$ can be expressed with the vector $\Omega=(\Omega_1,\Omega_2,\Omega_3)$ as $\Omega\times v$.

Coordinate vector in rotating frame:

$$ r^s = R\cdot r^r $$

Velocity, time-derivative in the static frame: $$ v_s^s := \frac{d r^s}{dt} = \dot R\;r^r + R\;\dot r^r $$

Apply $R^T$ to this equation:

$$ R^T v_s^s = R^T\dot R\;r^r + \dot r^r $$

You see you transform the velocity $v_s^s$ into the rotating frame (the same where also $r^r$ lives). The right name in our nomenclature for the time derivative calculated in the static frame and transformed into the rotating one would be $v_s^r = R^T v_s^s$.

With this you get your formula $$ v_s^r = \Omega \times r^r + \dot r^r. $$

Answer 2

Let me first show you the math: Your position in the rotating frame be $\vec x_\text r$ where in the static frame it be $\vec x_\text s$. Now the velocity is: $$ \vec v_\text s = \frac{\mathrm d}{\mathrm dt} \vec x_\text s. $$

This should be fine. Now to go to the rotating frame, we will have to expand the notation a little bit. We have to take a look at the unit vectors and decompose the position vector into components: $$ \vec x_\text s = \sum_{i=1}^3 {x_\text s}_i \vec e_i, $$ where $\vec e_i$ are the unit vectors of the reference frame. In the static frame, they are: $$ \vec e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\quad \vec e_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\quad \vec e_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} $$

They do not depend on the time, so you do not have to take that into account, when doing a time derivative.

The rotating frame will have unit vectors that are time dependant, you will have the following product rule: $$ \vec v_\text r = \frac{\mathrm d}{\mathrm dt} \sum_{i=1}^3 {x_\text r}_i \vec e_i = \sum_{i=1}^3 \left( {\dot x_\text r}_i \vec e_i + {x_\text r}_i \dot{\vec e}_i \right)$$

The last term is where your $\Omega \times$ comes into play. See Derivation of the centrifugal and coriolis force for remaining the derivation.

Answer 3

If you want an alternative way to look at this problem, you may switch to complex numbers notation (since this is a planar problem).

If in the non rotating system the position of the rocket is $z=\gamma (t)$, then in a system that rotates with costant angular velocity $-\omega$, you have $z=e^{i\omega t}\gamma(t)$. If $\gamma (t)=vt$, then $$z=e^{i\omega t}vt=k\theta \,e^{i\theta},\qquad \text{with }k=\frac{v}{\omega},$$ that is, as you said, a spiral.