Gold dropping rapidly

In retrospect the snap decision to go with a streamer in the Au-198 should have been given more careful thought. With a C11 casing and an altimeter on board it’s fairly heavy, about 106 grams, and the fins do sweep back so are pretty much guaranteed to hit the ground first.

Tim Van Milligan gives us a formula for computing streamer dimensions:

A = \frac{2mg}{\rho C_d v^2}

where A is the streamer area, m is the rocket mass, g = 9.8 m/s2 is the acceleration of gravity, \rho = 1225 g/m3 is the density of air, C_d is the drag coefficient, and v is the desired terminal velocity. Of course the problem is you don’t know the drag coefficient, which depends not only on the material but on the geometry of the streamer (and not in a simple way), but Sunward tells us the range is from 0.14 to 0.40; if you take 0.3 that’s probably good for a rough estimate. As for velocity, let’s say 20 ft/s = 6.1 m/s. Then A = 2 * 106 * 9.8 / (1225 * 0.3 * 6.1*6.1) = 0.15 square meters. For a streamer length to width ratio of 10:1 (which is, contrary to popular belief (and Sunward), not shown to be optimal for all streamers, but is reasonable) the size should be 1.2 meters by 0.12 meters, or 49 inches long by 4.9 inches wide. Yikes.

Well, 20 ft/s is pretty gentle, really. How about more like 35 ft/s = 10.7 m/s? The streamer dimensions are inversely proportional to the velocity so this 175% higher velocity requires dimensions (49, 4.9)/0.175 = 28 by 2.8 inches.

My streamer was about that long, more like 29 inches, but only an inch wide. 29 square inches is 0.0187 m2. Turning the above formula around:

v = \sqrt{\frac{2mg}{\rho C_dA}}

gives a velocity estimate of 57 ft/s.

It’d be nice to know what the velocity really was. It’d be nice if I’d had a device aboard the rocket that could measure its altitude as a function of time, so I could get the velocity. OH WAIT I DID.

au-198 flightIt reached its apogee of 543 feet in about five or six seconds (blue curve), then took about 15 seconds to come down. 543/15 = 36 feet per second average. The red curve is velocity obtained by dividing changes in altitude by changes in time and averaging over 11 measurements; it says the downward velocity increased, then decreased, and then went a little crazy — probably all that’s due to wind. But 35 ft/s looks like not a bad estimate. Which evidently was too fast. But also, it contradicts the above calculation! Why? Was the drag coefficient higher? Using

C_d = \frac{2mg}{\rho A v^2}

I get the drag coefficient was 0.8! Nope.Van Milligan says the coefficient gets smaller for length to width ratios much above 10:1, so my 29:1 can’t be the problem.

What’s gone wrong? I don’t know. A drag coefficient of 0.8 is crazy. Air density varies, but 1225 g/mis at sea level and above sea level it’ll be lower, making the required streamer dimensions larger; anyway, that value won’t be anywhere near wrong enough to explain things. The 106 g mass is correct, I just checked. I’m pretty sure our launch site is on Earth with 9.8 m/s2 acceleration of gravity.

So the input numbers are right, and the answer is wrong; is the formula wrong? But the derivation looks reasonable.

I’ll think about it some more. While I’m fixing the fin, putting more dog barf in, and picking out a Mylar or thin mil chute for the next flight.

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