"pushing it to the limit"
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Bowling Ball Design Considerations Retention There has been a lot of discussion on various forums about the dangers of launching bowling ball rockets. Obviously, any 12 pound rocket weighs the same whether it contains a bowling ball or not. But, if something goes wrong with the flight and the ball gets separated from the recovery device (parachute) there's a big problem. Once again, it's no worse than a separated nose cone weighing 8 pounds (and would actually fall slower) but the perception to others is quite frightening. I feel design considerations should be taken in the connection between the ball and the parachute to handle the worst case scenario. I've been guilty of underestimating the potential forces that may apply, or more to the point, underestimating the likelihood of a non optimal deployment. As can be seen with my discussion under electronics it's easy with bowling ball rocket designs to get in a situation where the electronics deploy the parachute under extreme conditions. The ideal and least stressful deployment is obviously at apogee where the rocket basically stalls in the air without any horizontal velocity. The worst deployment is either during descent once the rocket has reached terminal velocity, or at maximum velocity during ascent which would be at motor burnout. Descent: Terminal velocity can be calculated as follows: Vt = sqrt( 2W/(Cd * rho * A))A light rocket with a spent motor, parachute, electronics and an 8 pound ball weighs around 12 pounds. Based on simulations and comparisons to actual altitudes Cd values as low as .08 should be used when basing the diameter at 8.5 inches. Assuming the fins add very little, the surface area is just the area of the ball, pi x r2 = 56.7 sq. inches = 0.39 sq. feet If you plug all this in the formula you get 566 feet/sec If you do the same calculation with a 16 pound ball based on a total of 20 pounds you get 731 feet/sec. Ascent: Based on simulations, actual altitudes and flight data from AltAcc I can offer the following:
In my opinion the most likely failed deployment situation would be during ascent, primarily because of incorrect use of electronics, or possibly from underestimating the forces on the rocket at very high speeds. So, what forces are exerted on a rocket deploying a parachute at 1100 feet/sec ? Just so you know, 1100 feet/sec is 750 miles per hour, or 0.95 the speed of sound. Think about what happens in this situation (based on a design where the ball is the nose cone): The rocket would separate...the ball would continue to go up . . . the rest of the rocket would slow down quickly because without the ball it catches the wind and relatively speaking doesn't weigh much . . . the chute would be pulled from the rocket very quickly . . . the chute would open upside down trying to slow the ball. In the first second or two the parachute and the body of the rocket would be trying to slow down the ball, eventually the combination of the ball and chute would have a lower speed than the rocket body and the chute would begin slowing that too. The biggest force is the fully deployed parachute slowing the ball at the instant it's still traveling at 1100 feet/sec. The force the parachute exerts on the ball can be calculated by using descent calculations. It' equivalent to the weight that would be required for the parachute to descend at that speed. The larger the parachute the more force. For this example if we use the recommended Rocketman chute size for a 12 pound rocket, a R9C.... With a 12 pound rocket the R9C is listed at a descent of 20 feet/sec. This can be used as the terminal velocity in the equation above to back out a Cd for the parachute. The surface area isn't known but if the Cd is determined from a given area, and the same area/Cd used to calculate the force at a different value then the calculation is valid. So, taking an estimate of the area at 50 sq. feet, this gives a Cd of exactly 0.5. If this is then used to calculate weight (W) needed to get a terminal velocity of 1100 feet/sec, we get 35,000 pounds If this is done with the example above where it might be traveling at 800 feet per second this drops to 20,000 pounds This doesn't make sense! Perhaps my math is wrong? or maybe you can't use the terminal velocity formula at these extremes? Or maybe it does make sense when you consider that it's an absolute maximum force. This assumes the ball doesn't slow down at all before the parachute fully opens, and it assumes there is no 'give' in the shock cord, the parachute lines and the chute itself. Another way to look at it: I've had 2 bowling ball rockets deploy right after motor burnout (oops). Both had the same size and make of parachute, the same harness for retention and the same basic design and weight. The first time it came down safely on the chute, the second broke the kevlar shock cord and shredded the chute. The only difference was the motor being used. I've concluded from this the first flight was just barely strong enough to deal with the forces involved. According to the AltAcc accelerometer at the time of deployment it was traveling at 850 feet/sec, and when the chute opened it was subjected to a maximum deceleration of -35g for a very brief time. The shock cord used was flat braided Kevlar listed with a breaking strength of 1000 pounds. 35g on a 12 pound rocket comes out to 420 pounds of force. The accelerometer may not sample quickly enough to catch the maximum deceleration in a case like this. It should have required -80g to break the Kevlar. The second early deployment was with a K700 (as opposed to a K250) which would have resulted in considerably higher speed as listed above, and so much higher potential -g forces. Obviously, it's difficult to accurately determine the strength required in such an extreme situation. But, based on the above I would recommend the retention between the ball and the parachute be capable of handling at least 1g for every 10 feet per second. In the example above this would be 85g for the rocket that survived the early deployment and 110g for the one that didn't. So if the table from above were redone:
Connection to the ball On the AHPRA site it states " The ball must be drilled through to retain it. Weld the eyebolt." I don't agree with this, I think that in many cases this method is not strong enough. I think it's best to tie the shock cord through a section of the ball. If the retention needs to hold a force of at least 1400 pounds consider the following:
Unfortunately, I'm guilty of not knowing how strong this connection is. This is the method used to attach the ball in the early deployment examples above. In the example with the K700, the kevlar didn't break right at the ball, but a few inches away. In both cases there were no signs of damage to the ball where it was connected and no signs that the connection to the ball was about to fail. Shock Cord The 2 main choices here are Kevlar shock cord or Nylon. Kevlar has the advantage of being extremely strong for it's size/weight. It's also resistant to heat so protection from ejection charges isn't needed. Readily available options are: 1/4" flat braided 1000 pound (I won't use this again!) 1/4" tubular 3000 pound (my new shock cord of choice for Bowling Ball rockets) 5/8" tubular 7000 pound Some Nylon has the distinct advantage that it stretches slightly. Ask anyone who has fallen while climbing and they'll tell you the advantage of stretch. It reduces the 'shock' of deceleration by absorbing some of the highest initial force. If a climbing rope didn't have any stretch it would need to be thicker to handle the extra load, and the person that fell would probably be hurt by the extra g forces. If size and weight (and heat resistance) weren't a factor in rocketry Nylon would probably always be the best choice. But, be careful, not all nylon cord stretches! For a detailed discussion on nylon as a shock cord and the choices and strengths see the article at info central . Parachute If everything else has been designed to handle over 1000 pounds during deceleration, the parachute becomes the weak link. Not many parachutes are designed to deploy at 750 miles per hour! All this discussion is based on dealing with the worst case scenario (other than forgetting to turn electronics in which case none of this even matters). Not many rockets are designed to handle this, and that's why not many people use main chutes that are ballistic grade. I think the parachute should be the weakest link! When parachutes fail because of excessive force they usually either break a few of the lines to the canopy or blow out one or more panels. If either of these happen and the ball is still connected it will still fall too quickly. The good news is it will have a shock cord and a mess of fabric flailing behind it making it easy to spot, and therefore avoid. In the failure I had at LDRS 20 I was using a custom light weight parachute (1.1 ounce rip stop nylon) and it survived high speed deployment without any damage! So, if a graceful descent is needed no matter what, all these factors will need to be considered and combined with a ballistic grade parachute. Obviously, the real goal is to make a design that will deploy only at apogee. Conclusion I feel that given the public perception of rockets containing bowling balls, precautions should be taken so that, at the very least, in the hopefully unlikely situation where early deployment occurs, the ball will not fall to the ground completely unattached. In addition to this I would support, if determined to be necessary, the use of ballistic grade parachutes to ensure an intact canopy in this extreme situation. |