How many shear pins does it take...

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g zilla

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Im building a 3" 2 stage rocket about 10' tall fiberglass rocket. How many 2-56 shear pins do I need to keep it from drag sepperating.
 
It will depend on the retained weight and deceleration drag of your flight. Figure that out from simulation, ground test accordingly. You'll probably end up with 3x 2-56.
 
I would be blown away if you need more than (3) 2-56 pins but if you are worried about it, SIM away.
 
I was going to pin the interstage coupler and the top coupler.I was hoping to use 1 or 2 but it looks like 3 is the way to go.
 
I was going to pin the interstage coupler and the top coupler.I was hoping to use 1 or 2 but it looks like 3 is the way to go.

I have used 1 and 2 often. As long as your coupler isn't sloppy there is no problem.
 
Yea the common safe practice is to use 3+ to avoid an oft-cited cocked/stuck coupler, but anecdotes seem to support the safe use of 1 or 2 under the right circumstances. Some kit manufacturers even recommend it on certain birds. Never seen an actual report of someone proving that their rocket didn't separate because of one or two shear pins. I've never done it but if it passes ground testing and everything is tight enough it should be fine.
 
You have two concerns: 1) drag separation at the coupler when the motor burns out, and 2) separation of the nose cone at the apogee charge due to the ejection force.

Do a sim. Look at your max acceleration. If it is in f/s2, convert it to G force. Weigh the rocket above the booster. To calculate F (force), use F = ma. 'm' is mass in pounds; 'a' is acceleration in G force. Assuming a sudden dead stop of the booster at motor burnout, how much force is pushing against the payload bay?

Ex.: if the upper section of the rocket weighs 3 pounds, and the max acceleration is 17 G, assume a sudden stop to 1 G. So, 3 * 16 = 48. A #2-56 nylon screw has a sheer strength of somewhere between 30 - 45 psi. So you can see that one pin is not enough. Take the lowest number, 30. How many do you need to retain 48 pounds? At least two.

Now, go to an ejection charge calculator. Most will tell you the maximum number of pins the given charge can sheer. As long as you remain >= 2 and <= the maximum number, you are good to go.

Repeat for the nose cone, but this time the acceleration needs to be taken from the sim at the apogee (drogue) charge. Often this is quite high. The mass is the weight of the nose cone.

So, if the nose weighs 0.5 pounds, and the acceleration at apogee event is 42 G, then the F is 21 pounds. One pin would do.

However, there are those who will argue that any less than three may cause the coupler (or shoulder) to jam in the tube, or at least struggle to slide out. So many people set three as their minimum. As long as you stay above the maximum number of sheerable pins for the size of the charge, you are golden.

Do the math and come back and let us know what you come up with.
 
[rant] So I'm going to talk about getting into the weeds. That is fine if it's what you want to do. But...

If you are flying a 3FNC or 4FNC rocket without a large draggy bottom end, friction fit of the lower part to the upper will work find. As long as you can pick the rocket up by the nose cone, give it a good shake and nothing comes apart, you are good to go. Just don't over do your drogue ejection charge. Remember, you only need to separate the two halves enough so the drogue chute gets into the air stream. Reasonable charges, short shock cords and a successful shake test will do you well.

I know, that doesn't follow all the math, calculations, goes with the "shear pins or fail" trend or falls into to being a "rocket scientist" mode, but it works and works well and consistently.

[/rant]

If doing all the calculations and following the latest ideas is what you enjoy, Go For It! Enjoy, it's your hobby!
 
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