Introduction

In typical problems in physics, we tend to have a fixed coordinate system that we can reference back to throughout the problem. However, with planes, where do we make the coordinate system originate from? From takeoff or the destination? What about the plane itself?

Typically, for aircraft flying at low speeds, they use the Earth-fixed Earth-centric (EFEC) coordinate system for their reference frame while spacecraft use the Earth centric inertial (ECI) coordinate system. ECI is an inertial system so it does not accelerate, however, EFEC is not an inertial frame because it rotates with the Earth. Inertial reference frames are ideal, but if we were in a rotating frame, how would we account for the changing movement of the axes? Well that is what this tutorial is about, so let's dive in!

Pure Translation

Let's start off by picturing a vector $ \vec{A} $ translating with no rotation in an inertial frame. It can be written as

$$ \vec{A} = A_x \hat{i} + A_y \hat{j} + A_z \hat{k} $$

Since $ \vec{A} $ is not rotating, the rotation vector $ \vec{\omega} = [p;q;r] = [0;0;0] $. Now if we were to take the time rate of change of $ \vec{A} $, that would look something like this.

$$ \frac{d}{dt} \vec{A} = \frac{d}{dt} (A_x \hat{i} + A_y \hat{j} + A_z \hat{k}) $$
$$ = \frac{d}{dt} (A_x \hat{i}) + \frac{d}{dt} (A_y \hat{j}) + \frac{d}{dt} (A_z \hat{k}) $$

Now the first term can be written out using chain rule as

$$ \frac{d}{dt} (A_x \hat{i}) = (\frac{d}{dt} A_x) \hat{i} + A_x (\frac{d}{dt} \hat{i}) $$

This can be continued to get the full expression for $ \frac{d}{dt} \vec{A} $.

This is a decent sized expression, and there is in fact a simplification we can make because $ \vec{A} $ is not rotating. Because $ \vec{A} $ is in pure translation, we can say that $ \frac{d}{dt} \hat{i} = \frac{d}{dt} \hat{j} = \frac{d}{dt} \hat{k} = 0 $.
This leaves us with

$$ \frac{d}{dt} \vec{A} = \dot{A_x} \hat{i} + \dot{A_y} \hat{j} + \dot{A_z} \hat{k} $$

Now what would the expression be if $ \vec{A} $ were rotating?

Add Rotation

Now let's say the vector $ \vec{A} $ from before was in a rotating as well as translation. We would follow the same process as before to get Eq. (1).

$$ \frac{d}{dt} \vec{A} = (\frac{d}{dt} A_x) \hat{i} + A_x (\frac{d}{dt} \hat{i}) + (\frac{d}{dt} A_y) \hat{j} + A_y (\frac{d}{dt} \hat{j}) + (\frac{d}{dt} A_z) \hat{k} + A_z (\frac{d}{dt} \hat{k}) $$

However, now the rotation vector $ \vec{\omega} = [p;q;r] \neq [0;0;0] $. What we want to find now is what the time rate of change of $ \hat{i} $, $ \hat{j} $, $ \hat{k} $ will be. Let's start with $ \frac{d}{dt} \hat{i} $.

The way we find the time rate of change of $ \hat{i} $ is by looking at how $ \hat{i} $ rotates given rotation about the y and z-axes.

Fig. 1 - Time rate of change of $ \hat{i} $ graphic.

In Figure 1, the top-most axes shows a typical xyz coordinate system with the positive rotation directions marked in blue. There are two other axes below, each of which finds a component of $ \frac{d}{dt} \hat{i} $ because we need to see how $ \hat{i} $ changes as the y and z axis rotate individually.

Looking at the bottom left set of axes, we are looking down the y axis at the xz plane. Now what we are doing is looking to see in which direction $ \hat{i} $ moves as the y axis rotates positively at angular speed q, hence, $ \hat{i} $ moves in the negative $ \hat{k} $ direction at angular speed q, so the first component of $ \frac{d}{dt} \hat{i} $ is $ -qk $. Looking at the other set of axes on the bottom right, we are now looking down the z axis at the xy plane. As the z axis rotates positively at a rate r, we see $ \hat{i} $ moves in the positive $ \hat{j} $ direction at a rate of r, so the full expression for $ \frac{d}{dt} \hat{i} $ is

We can do the same process to derive the equation for $ \frac{d}{dt} \hat{j} $.

And repeat for $ \frac{d}{dt} \hat{k} $ as well.

Now if we write Eq. (1) as

$$ \frac{d}{dt} \vec{A} = \dot{A_x} \hat{i} + \dot{A_y} \hat{j} + \dot{A_z} \hat{k} + A_x (\frac{d}{dt} \hat{i}) + A_y (\frac{d}{dt} \hat{j}) + A_z (\frac{d}{dt} \hat{k}) $$

We can take the last three terms and make some substitutions and modifications.

$$ A_x (\frac{d}{dt} \hat{i}) + A_y (\frac{d}{dt} \hat{j}) + A_z (\frac{d}{dt} \hat{k}) $$
$$ = A_x(r\hat{j} - q\hat{k}) + A_y(p\hat{k} - r\hat{i}) + A_z(q\hat{i} - p\hat{j}) $$

Now, remember that $ \omega = [p;q;r] $, then we can write Eq. (5) as $ \omega \times \vec{A} $, where

$$ \vec{\omega} \times \vec{A} = \begin{bmatrix} \hat{i} & \hat{j} & \hat{k} \\ p & q & r \\ A_x & A_y & A_z \end{bmatrix} $$

So now we can write for a rotating and translating vector $ \vec{A} $, its time rate of change is

In Eq. (6), $ (\frac{d}{dt} \vec{A})_{xyz} $ is the derivative with respect to the body frame and $ \vec{\omega} \times \vec{A} $ is due to angular velocity in the instantaneous rotating frame. This full expression is called Chasle's Theorem. This theorem is often used in kinematics!