Drag separation occurs when the drag of the rocket booster (body) increases relative to the drag of the upper airframe of the rocket (nose cone). This difference in drag results in the body falling behind and separating from the rocket nose which can lead to a premature deployment of the recovery harness. The theory behind this suggests that since the drag on the nose cone is much lower than the drag on the body, the momentum of the nose cone carries it forward faster (alternatively, as the body’s drag is larger it decelerates quicker in comparison to the nose cone) which creates tension on the joint where the nose and body connect. Separation pins are utilized as a result, to prevent early deployment and prevent drag separation. This can be observed at the orange line (which would typically be higher up).
This concept is illustrated in the diagrams below. Here, the drag forces and momentum corresponding to both the nose cone and body are displayed outlining the difference between the drag in both these components which causes drag separation.
To determine whether or not drag separation would be an issue, the following calculations were performed:
Complete Rocket:
First, to analyze the rocket as a whole, compute only the external forces (Gravity and Drag) and ignore the internal forces (Normal forces).
| (1) | F_{net}=M_{1}g+M_{2}g+D_{1}+D_{2}=(M_{1}+M_{2})a_{net}\\ a_{net}=\frac{D_{1}+D_{2}}{M_{1}+M_{2}}+g |
Here, N is the normal force, which represents the interaction between the two sections. If N is negative, there will be drag separation (nothing will be holding the sections together).
System 1:
| (2) | M_{1}a_{net}=N+D_{1}+M_{1}g\\ N=M_{1}(\frac{D_{1}+D_{2}}{M_{1}+M{2}}+g)-D_{1}-M_{1}g\\ N=\frac{M_{1}D_{2}-M_{2}D_{1}}{M_{1}+M_{2}} |
System 2:
| (3) | M_{2}a_{net}=D_{2}+M_{2}g-N\\ N=D_{2}+M_{2}g-M_{2}a_{net}\\ N=\frac{M_{1}D_{2}-M_{2}D_{1}}{M_{1}+M_{2}} |
It is reassuring to see that the results for N are the same when solving via System 1 as when solving by means of System 2. This gives us confidence in our answer.
These results tell us that drag separation will occur (N<0) if M2 is sufficiently large and/or D1 is sufficiently large.