Where Do The (8-16 psi) Numbers Come From For Pressure to Eject Parachute?

Going through the ideal gas law calculations to calculate how many grams of black powder I'll need to eject my main chute. I keep seeing this number of 100-200 lbf for a 4 inch as the suggested value for my pressure (8-16psi). They say these are typical values, but I have no idea where they are drawing this conclusion from.

Usually I'd wouldn't be that much of a stickler about it, but as I'm doing a formal academic writeup on it I can't quite just say "well, these are typical values" without anything backing it up. Any ideas about what formula's they're using to calculate these numbers?

The shear force of one shear pin x number of pins x a safety factor is the force needed. Divide that by the area of the coupler to find pressure needed.

Super helpful, thank you.

If you're ejecting an unpinned segment, you could also document aiming for an ejection force that'll accelerate the two halves apart with sufficient energy so that they don't get hung up.

(for bonus points, do some dynamics and try and estimate the shocks when the pieces reach the extent of the cord!)

Nytrunner said:

If you're ejecting an unpinned segment, you could also document aiming for an ejection force that'll accelerate the two halves apart with sufficient energy so that they don't get hung up.

(for bonus points, do some dynamics and try and estimate the shocks when the pieces reach the extent of the cord!)

Click to expand...

I think the bonus points is the most important part. Having too strong of ejection leads to compensating by using longer cords, heavier components, etc. Most cause more problems then what they solve. On 4" and smaller non-fiberglass rockets, I've only been friction fitting and using small apogee charges. It takes some dialing in, but it works surprisingly well and has a lot less wear and tear on the rockets then using shear pins.