jimzcatz
Boss, Carolina Rocket Mafia
I use 4-40 screws on 7.5" fiberglass I use 4 of them.
The force trying to separate the rocket at burnout is given below. If you just consider inertia forces you can set R=1
This force can be - or + depending on the mass ratios. If the mass of the lower section is more than 1/2 the mass of the total rocket then the joint is in compression after burnout. No shear pins necessary.
Fsep = a [ M / (1+R) - M1 ]
Where:
a = max deceleration [just prior to motor burnout]
M = total mass of rocket (mass NOT weight)
M1 = mass of lower section
R = drag ratio
Yes, that is the right calculation.A very naive question about deceleration (or inertial) separation force.
I started with this:
So,
Fsep = a [ M / (1+R) - M1 ]
Fsep = a [ M / (1+1) - M1 ] => inertial separation force only
Fsep = a [ M/2 - M1 ]
Given:
M1=24 lb, M=35 lb, a=-80 ft/sec^2
Fsep = -80 * [ 35/2 - 24 ] = -80 * [ 17.5 -24 ] = -80 * -6.5 = 520 lb ft/sec^2
Fsep lbf = [ 520 lb ft/sec^2 ] / [ 32.1705 ft/sec^2 ]
Fsep lbf = 16.1621
Given:
Coupler friction constant = 5 lbf
Net Fsep lbf = 16.1621 - 5 = 11.1621 [if >0 then shear pins needed]
Does this look right?
Yes, that is the right calculation.
I believe you are right. Let me check my original derivation to confirm.One more clarification. If drag is being considered, which is the correct ratio for R:
#1 CdA upper / CdA lower
OR
#2 CdA lower / CdA upper
To me, the calculation only make sense if the correct ratio is #2. Is that correct?
Disregard last answer. R = CdA upper / CdA lower. Note the direction of a (positive a is decelleration). If using imperial units, express a in g's, mass in lb and units for Fsep will be in lb.One more clarification. If drag is being considered, which is the correct ratio for R:
#1 CdA upper / CdA lower
OR
#2 CdA lower / CdA upper
To me, the calculation only make sense if the correct ratio is #2. Is that correct?
Most normal rockets with heavy motor cases in the lower section will not require shear pins for the apogee break. None of the rockets I fly do. What you can do with the Fsep equations is you can reality check what the drag ratio needs to be to be in danger of drag separation. Then judge for yourself if that R value could possibly happen.For what it is worth, calculating the different drag coefficients for different pieces of a single body is not easy. Most drag coefficients from look-up tables account for the entire object tested in a wind tunnel. You have to separate out pressure drag on the nose, shear drag on the nose, body and fins and base drag from flow separation. I have crude estimates in Rocksim for some of my higher performance rockets.
It is worth noting that over-pinning the joint for the drogue can have pretty big consequences. Also adding pins for that joint means larger ejection charges which adds another stress point if the shock cord gets pulled hard. Unless you are flying past Mach, I recommend a decent friction fit for the drogue joint and shear pins for the main. By decent, have the joint slightly stronger than the weight of the booster. This is just a rule of thumb and different rockets will necessitate different joints.
And even still that may not be adequate, be sure, config pins for worst case pressure differential and use a larger charge if need be.Can't remember if it was mentioned: but do put a vent hole in the BT so as the exterior air pressure drops the interior pressure does not push the ebay off.
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