Eureka! I think.

I recently decided against, and scrubbed my Level 2 cert. attempt this past Monday at NARAM 57. There were just too many "little things" that came up the morning of the launch. Anyways I wasn't 100% ready and sure, so no launch. Based on recommendations from more experienced flyers...more recovery harness (length) is more of your friend when it comes to reducing load on the recovery system, rather than using less of ultra high strength material. Right? Makes total sense now...if my weakest "link" was a 500# rated U-bolt...there is no need for harness material rated higher than 1750#. My rocket is a 5.5" diameter 77" tall Nike Smoke, weighing 18.6 #'s prepped for flight without motor, probably a bit more with more harness length added (there's plenty of room in there!) I just need to add more harness length, 4 to 5 times the length of the rocket. As the parts separate further apart, they decelerate more. Slower parts means less load on the entire system. Force = mass x velocity squared.

The opening forces can be quite high on a chute. You might want to reconsider your choice of u bolt- a bit more will not hurt you.

Thanks for the advise Mark, I truly appreciate any helpful suggestions...I'll double check that figure...it was the most heavy duty that I saw available from Rocketry Warehouse. Will check McMaster and Grainger sites too.

AZRxocet said:

View attachment 268767

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I thought that the rocket looked like something from a Van Halen album... Turns it's more like Van Hagar.

Looks Good!

Yeah, as it turns out I really can't fly 55...at least until I get my recovery issues worked out.

Having a long harness helps prevent "bounce back" -- where your nose cone bounces back into the airframe, either tangling the chute or doing damage. Also, ensures you can use enough BP to get a good ejection and pull out the chute without bounce back.

Just make sure you don't use so much ejection charge that the pieces stretch the harness full length and hit the ends hard.

You didn't mention using DD so I assume motor ejection. Either way, long shock cords can reduce the shock load on the system when using energetic ejection charges, but they can also provide a long accelerating drop for one part of the rocket after the chute opens. If you really watch some of the DD flight, sometimes you'll see the main deployed from the payload while pointing down and the fin can well above the payload. The main will open and almost stop mid air since there is no weight on it. The payload will reach the end of it's cord first and put some shock loading on the system. When the fin can falls past the payload and hits the end of its cord it will really put a huge shock load on the system. The longer the cord, the larger the shock load. Like you said, Force = mass x velocity squared. That can happen with motor deploy too. Usually when it's pointing down and deploys after apogee.

AZRxocet said:

Force = mass x velocity squared.

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Force = mass times acceleration. Kinetic Energy equals mass times velocity squared divided by two. Work is the difference in Kinetic Energy. Force is Work divided by distance.

If the difference in Kinetic Energy is created over a short distance (i.e. the parts are still moving apart quickly when they reach the length of the shock cord) then the force is great. If the relative velocities of the parts decrease slower, over a long distance, the force is less.

Apologies for being picky ..... but this is rocket science!

-- Roger

Thanks to everyone. I've got a much better understanding now.