Shock! (Part 2)

A little advancement in the business of how strong and long a shock cord should be.

I asked some questions over on YORF. After some confusion the following information came to light, from an article, “Ejection Charge Measurements”, by Ed Brown in the January/February 2007 Sport Rocketry magazine: For Estes 13mm motors, 0.4 gram; for 18mm, 0.6 gram; and for 24mm, 1.0 gram. I want to get a look at that article but NARTS doesn’t have that back issue available individually, only as a CD of all the 2007 issues. I may have to buy that, though \$20 for an article that may have no further useful information in it is a little hard to swallow. Wonder if I can get a copy through Interlibrary Loan?

On various web pages you can find information or Javascript calculators for calculating ejection charges, intended for those doing e.g. altimeter ejection. Such a calculator may be found here along with a fair amount of text on the subject. Near the bottom it says:

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?

So, first, this is a similar number for the ejection charge mass in a 24 mm motor as that given above, though a little smaller. Where it came from, I don’t know.

Second, the implication is that the formula given is valid for BP motor ejection.

Third, the answer to the question, using the formula given, is: 45.5 PSI! And for a Big Bertha, using 1.6″ diameter, 15″ length, and 0.6 g charge, it’s 38.6 PSI. A factor of about 2.5 times the 15 PSI I was assuming earlier (and was hoping was too large). For a Baby Bertha, with about 3.5″ length, it’s 165 PSI!

That, of course, shoots the nose and body apart at a much higher speed than I was estimating.

So is this at all reasonable?

The formula used by the calculator, as it says in the text, comes simply from computing how much gas is produced when black powder burns. The assumption is “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 [ejection].” Of course in reality nothing is instantaneous, but I have a feeling it’s close enough. Another assumption is that all the gas goes into pressurizing the volume. For a  motor ejection, there’s another way the gas can go: out the back of the motor. But that aperture is constricted (by the nozzle), so maybe that correction is small. Can you really get any factors like 5 out of that?

There is also the clay cap on the ejection charge; some energy has to go into breaking that. Could that be significant?

On another aspect of the subject, there’s this business about the effects of the parachute during ejection. When does it inflate, and how fast? Questions like that. I went looking on YouTube for some videos of model rockets, but not surprisingly, very few of them give you any useful view at all of the ejection. The ones that do often do so because ejection was late and close enough to the ground to be captured by the video camera, but that’s not the kind of typical ejection I was looking for! I was thinking to myself, “It’s too bad no one ever mounts a keychain camera on a rocket pointing forward”… and then I found these:

Awesome. Not sure how much there is to be learned from these, but worth closer examination than I’ve given them so far.