
Ejection Charge CalculationsBy Darrell Mobley This is an article posted on the WEB by Darrell Mobley. The theory behind the numbers. If you've never had a rocket crash because the parachute didn't come out right, go on to read something else. But if you know what the term "core sampling" is from experience, here is some good information. Parachute failure often comes from an incorrect or insufficient amount of ejection charge. This is caused by either having too large a parachute compartment inside the rocket, or not using the right amount of black powder in a reloadable motor. Ejection charge weight can be calculated based on the desired ejection pressure and the internal "freevolume" of the rocket airframe. Normally the volume of the parachute and rigging inside is neglected. This approach is used in industry for closed bomb calculations and pulsar (pressure cartridge) applications. First you need to determine the required pressure to separate and deploy the recovery system. This depends on the area of the bulkhead, hence body diameter, and the mass of the nose section. The force from the pressure must be enough to overcome the inertia and drive the mass of the nose section the length of the coupler inside the tube to the point of separation, plus a little more for momentum to fully deploy everything. If you consider the nose having to deploy into a wind, or not near apogee, you need a little more push again. Assume that the gas expands and the pressure occurs instantly and uniformly throughout the volume. The pressure exerts an instant force on the forward bulkhead intended for extension. Neglect any change in pressure and temperature from the change in volume as the nose moves forward, (unless you just like calculus.) This is the simplest case for a single set of variables and adequate for most ejection systems. The ejection charge equation is: Wp = dP * V / R * Twhere

Here's an example calculation. Suppose you want to generate 15 psi inside a 4" diameter rocket in a parachute compartment 18" long. That makes a volume of 226 in3. The amount of powder you need will be:
Wp = 15 * 226(454) / 12 (22.16) 3307
Wp = 1.75 grams
The equation can be turned around to find what pressure is produced by a given charge mass. So that you won't have to think, it's just:
dP = Wp * R * T / V
Now, so you will have to think, given the ejection charge mass in a D12 motor is .85 grams, what pressure is generated inside an Estes Phoenix model, with a 2.4" diameter and 8" long chute compartment?
From this relationship of parameters, you can better design the amount of recovery space in your rocket, or customize the amount of powder to successfully deploy the system.
Of course, since you're being this careful to determine the correct charge amount, you'll want to verify it by a ground test. That way you can get a better handle on other less precise variables like nose coupler friction inside the tube and necessary shock cord length. But how to do that is another article!