Introduction
In dynamics, one frequently needs a derivative of a vector in an inertial frame of reference in order to apply Newton’s laws when it is much more convenient to express that vector in a different frame. The Basic Kinematic Equation (also known by many other, less-informative names) relates the derivatives in two frames to each other:
$$\displaystyle \frac{^e d\vec{q}}{dt} = \frac{^f d\vec{q}}{dt} + (^e \vec{\omega}^f)\times \vec{q} $$
where $\vec{q}$ is some arbitrary vector and $^e \vec{\omega}^f$ is the angular velocity of the $f$ frame relative to the $e$ frame. A left-superscript on a derivative indicates the frame in which the derivative is taken, and vector quantities are denoted by arrows.
It’s important to keep in mind that there is a difference between a vector and it’s representation in a coordinate system (i.e. in a frame). The vector from London to Tokyo is still the vector from London to Tokyo regardless of whether one observes it while fixed on Earth or from orbit.
Our goal is to derive the Basic Kinematic Equation (BKE) from first principles. This is the derivation I would have wanted to see as a Purdue AAE student, so to any fellow Boilermakers out there reading this — you’re welcome.
Deriving the BKE
For each $i = 1,2,3$, let $\hat{e}_i$ and $\hat{f}_i$ be the unit vectors pointing in the $i$-th coordinate direction of frames $e$ and $f$, respectively. We’ll start by writing our vector $\vec{q}$ in the $f$ frame:
$$\displaystyle \vec{q} = \sum_i q^{f}_i \hat{f}_i$$
Taking the time derivative and using the product rule on each term in the sum:
$$\displaystyle \begin{aligned} \frac{d\vec{q}}{dt} &= \sum_i \left( \dot{q}^{f}_i\hat{f}_i + q^{f}_i \dot{\hat{f}}_i \right) \\ &= \left( \sum_i \dot{q}^{f}_i\hat{f}_i \right) + \left( \sum_i q^{f}_i \dot{\hat{f}}_i \right) \end{aligned} $$
Crucially, the left-hand side does not involve any particular frame whereas the right-hand side involves quantities expressed in the $f$ frame. The term in the first set of parentheses is the rate of change of the vector of $f$-frame coordinates of $\vec{q}$, which is another way of saying that it is the derivative of $\vec{q}$ in the $f$ frame. Rewriting that term, we get:
$$\displaystyle \frac{d\vec{q}}{dt} = \frac{^f d\vec{q}}{dt} + \left( \sum_i q^{f}_i \dot{\hat{f}}_i \right) $$
Now we need to figure out the time derivatives of each of the $\hat{f}_i$ vectors. These are always unit vectors, so each one’s rate of change must be perpendicular to the vector itself. In three dimensions, we can therefore write the rates of change as cross products with some other vectors:
$$\displaystyle \begin{aligned} \dot{\hat{f}}_1 &= \vec{v}_1\times\hat{f}_1 \\ \dot{\hat{f}}_2 &= \vec{v}_1\times\hat{f}_2 \\ \dot{\hat{f}}_3 &= \vec{v}_3\times\hat{f}_3 \end{aligned} $$
But since the three unit vectors all comprise the the same frame, they undergo the same rotation according to the same angular velocity $ \vec{\omega}^f = \vec{v}_1 = \vec{v}_2 = \vec{v}_3$. Therefore:
$$\displaystyle \dot{\hat{f}}_i = \vec{\omega}^f \times \hat{f}_i $$
The angular velocity $\vec{\omega}^f$ of the $f$ frame has to be defined relative to (but not necessarily expressed in) some frame. We will not bother specifying which frame that is because in a moment, it will be seen not to matter. Now we can rewrite $\frac{d\vec{q}}{dt}$ as:
$$\displaystyle \begin{aligned} \frac{d\vec{q}}{dt} &= \frac{^f d\vec{q}}{dt} + \left( \sum_i q^{f}_i (\vec{\omega}^f\times\hat{f}_i) \right) \\ &= \frac{^f d\vec{q}}{dt} + \vec{\omega}^f\times \left(\sum_i q^f_i \hat{f}_i \right) \\ &= \frac{^f d\vec{q}}{dt} + \vec{\omega}^f \times \vec{q} \end{aligned} $$
We can do the same process over again for the $e$ frame to obtain:
$$\displaystyle \frac{d\vec{q}}{dt} = \frac{^e d\vec{q}}{dt} + \vec{\omega}^e \times \vec{q} $$
Again not specifying the frame relative to which $\vec{\omega}^e$ is defined. We now have two different ways of expressing $\frac{d\vec{q}}{dt}$ which must be equal to each other:
$$\displaystyle \frac{^e d\vec{q}}{dt} + \vec{\omega}^e \times \vec{q} = \frac{d\vec{q}}{dt} = \frac{^f d\vec{q}}{dt} + \vec{\omega}^f \times \vec{q} $$
Subtracting the $\vec{\omega}^e$ term from both sides:
$$\displaystyle \frac{^e d\vec{q}}{dt} = \frac{^f d\vec{q}}{dt} + (\vec{\omega}^f – \vec{\omega}^e) \times \vec{q} $$
As long as $\vec{\omega}^f$ and $\vec{\omega}^e$ are defined relative to the same frame, the difference $\vec{\omega}^f – \vec{\omega}^e$ is the angular velocity of the $f$ frame relative to the $e$ frame: $\vec{\omega}^f – \vec{\omega}^e = {^e}\vec{\omega}^f$. So we have the BKE:
$$\displaystyle \frac{^e d\vec{q}}{dt} = \frac{^f d\vec{q}}{dt} + {^e}\vec{\omega}^f \times \vec{q} $$