# Equation governing relative motion between aircraft and store

I am reading a paper titled Conditions for safe separation of external stores by EE Covert [Link]. In Section 1 Introduction of the paper, the author has given an expression for $$d$$, the relative distance between some point $$r_{p}$$ on the store and $$r_{q}$$ on the aircraft as:

$$d = d \left( 0 \right) + \int\limits_{0}^{t} \left\{ \left[ R \right] \cdot \left( U + \omega \times r_{p}\right) - \left( V_{A/C} + \Omega \times r_{q} \right)\right\} \mathrm{d}t$$

where,

1. $$V_{A/C}$$: aircraft translational velocity in aircraft body coordinates.

2. $$\Omega$$: aircraft rotational velocity in aircraft body coordinates.

3. $$r_{q}$$: position vector between aircraft center of gravity, and point of interest in aircraft.

4. $$r_{p}$$: position vector between origin and point on store.

5. $$\left[ R \right]$$: rotation matrix to transfer store velocities to aircraft coordinate frame.

6. $$U$$: store linear velocity in store coordinates.

7. $$\omega$$: store angular velocity in store coordinates.

I want to derive this expression, but I don't know where to start.

## 1 Answer

To derive these EOM, you have to understand that there are two coordinate systems defined here, namely the store coordinate system ($$X,Y,Z$$) and the aircraft coordinate system ($$X',Y',Z'$$). Suppose the store has a linear velocity $$U$$ in its coordinate system, meaning $$U := U_x\hat{i}' + U_y\hat{j}' + U_z\hat{k}'$$. In the aircraft coordinate system, the velocity will be different, because the aircraft is translating and rotating as well. In general, the velocity at a point $$Q$$ in some reference frame, given the velocity at point $$P$$ is given by

$$\vec{v}^Q = \vec{v}^P + \vec{\omega} \times \vec{r}^P,$$ where $$\omega$$ denotes the angular velocity of that reference frame. This just comes from regular dynamics. Applied to this problem, the velocity of the origin (see Figure 1 in the paper) given the velocity of the store is

$$U_0 = U + \omega\times r_p$$ However, this is still in the reference frame of the store. In order toconvert this to the reference frame of the aircraft, you have to change reference frames. To do this, you have to multiple by the direction cosine matrix $$R := R_{A/S}$$, which rotates the store frame to the aircraft frame, so in total we get

$$^AU_0 = R \ ^SU_0 = R(U + \omega\times r_p).$$ This is just the first term in the equation in the paper, because it only considers the velocity of the store. However, the aircraft is moving too! So, you apply the same procedure to get the second term, except you have to add in a negative sign. This gives you the full equation that you are trying to understand.