Shear pin retention stregth. Check my math.

[jderimig]

I your case, it is more likely to have been early ejection than drag separation, if you are sure that the friction between your nose cone and sustainer was sufficient to provide some resistance to separation.

Hello BBS,

Rocksim routinely simulates altitude generally within 10% of measured values which means that it probably can get the drag and acceleration modeling correct within 10% as well. Using those results Tim estimates ~41lbs of force attempting to separate the nosecone from the airframe shortly after burnout in a "normal" flight of that rocket. Tim used no shearpins.

I am curious why do you doubt Tim's Airfest incident couldn't be a case of drag separation?

Is it because you disagree that ~41lbs of spearation force would be generated by his flight profile?
OR
Is because you think ~41lb of force is easy reliably retained by a friction fit?

 

[BBS]

doubts are based on missing evidence. i didn't check the math, i trust what you say. there wasn't 41 lbs more force trying to seperate the nosecone, there was 41 pounds less force that was not slowing down the sustainer.

added thought: newton says that the two masses shouldn't sererate based on their individual masses since the nosecone is half of the weight of the sustainer. drag is acting on the two masses as a unit, even if there was no friction between the airframe and nosecone, the two parts will not separate. the fact that there is a boat tail effectively decreases the drag of the sustainer, that is the purpose of a boat tail. so the drag seperation theory is becoming less likely as data points are added.

of course i could be wrong, but there is no evidence that would support that conclusion either.

 

[BBS]

wow, calm down Dixon

john's post :
...
2. Consider a rocket with a single joint, the drag force above and below the joint are the same but the masses for each section are different. If the mass above the joint is less than the mass below the joint, F=ma says it will need less net force than the bottom section if both sections have the same "a". Since the drag force on both sections are the same the only way that the top section can have less net force is if the bottom section pushes (compression) the top section. No drag separation can occur in this mode. If the mass of the upper section is greater, then the opposite happens, the bottom section would have to pull the top section to keep the accelerations the same.
...

my post, 8 posts earlier:
...
if the sustainer inertia is > the payload inertia then your answer is that you need to impart >0 lbs of resistance in the payload/sustainer interface to insure the two masses won't separate. On the other hand, if the payload inertia > the sustainer inertia then you need to impart a greater resistance than the differential of inertia to insure the two sections won't separate.
...

how are these statements in conflict? there was no flip flopping, i have been consistent.

 

[pythonrock]

Dixon,

from an engineering standpoint, the only force that you need to consider when designing a retention method is how much more potential inertia the payload will have, than the sustainer, at burnout. if the sustainer inertia is > the payload inertia then your answer is that you need to impart >0 lbs of resistance in the payload/sustainer interface to insure the two masses won't separate. On the other hand, if the payload inertia > the sustainer inertia then you need to impart a greater resistance than the differential of inertia to insure the two sections won't separate.

The maximum necessary resistance, to keep the two together, is >0 lbs more than the weight of the sustainer at motor burnout. A shear pin that has 30lbs of shear strength, has an excess of 2X more strength than necessary to retain a sustainer that weighs 14lbs. That is not not a condition to worry about, it is an indicator of how much minimum force you will need the ejection charge to impart upon the payload bulkhead to overcome the shear strength of the shear pin. If you are curious to demonstrate this to your satisfaction, you can do simple experiments that use a compressor, a cylinder, a piston and a pressure gauge to determine the nominal shear strength of the shear pins you're testing.

Regardless of the differential of inertia, the sustainer is along for the ride if it cannot overcome its own weight in order to separate from the payload. 30lbs of shear strength may not be adequate to insure retention of a 31lb sustainer if the differential of inertia is > 1lb, then I would add another pin.

It is just physics not rocket science. In my world, when I want to figure out if my gun mount will not kill the occupants of a helicopter in a 20g crash, I figure out what is necessary to keep the mount and all of the components of it, and all accessories attached to it, in place. We (the company) use FEA to analyze and verify the design and if the customer desires it, we will have a pull test performed to validate the design. Trust me, our rockets are not complex devices to understand. The difference between an aircraft crash and a rocket going into coast phase is that the earth is affecting sudden and unforgiving force upon the vehicle at impact. The transition from boost phase to coast phase is only changing the force from +g's to -g's. It is not a violent event from the perspective of the rocket. A moving mass smashing into a stationary mass imparts a known quantity of force, F=ma. A mass set to free-fall will continue happily going along its trajectory until it has an impact with another mass.

The problem is that you continue to deny drag as a factor.

A metal gun mount in a helicopter has no joints designed to separate and is never exposed to aerodynamic forces that will shear a grade 8 steel bolt. In a helicopter crash, the deceleration forces produced by impact are orders of magnitude greater than any aerodynamic drag any part of the helicopter would ever see. In those situations, yes, you can ignore drag. ( it is insignificant, but not 0 )

We are talking about a rocket in flight that does have a joint designed to separate and hopefully is not impacting the earth. As soon as the motor burns out,

the force acting differentially on the rocket across the joint is aerodynamic drag.

(the pressure from an insufficiently vented bay also impacts the required strength of the shear pins)

In this situation your post is incorrect and highly misleading to those who want to learn something about drag separation. If you are an engineer, you should be able to understand the free body diagram. Please study it again before you post any more nonsense.

 

[BBS]

The problem is that you continue to deny drag as a factor.

A metal gun mount in a helicopter has no joints designed to separate and is never exposed to aerodynamic forces that will shear a grade 8 steel bolt. In a helicopter crash, the deceleration forces produced by impact are orders of magnitude greater than any aerodynamic drag any part of the helicopter would ever see. In those situations, yes, you can ignore drag. ( it is insignificant, but not 0 )

We are talking about a rocket in flight that does have a joint designed to separate and hopefully is not impacting the earth. As soon as the motor burns out,

the force acting differentially on the rocket across the joint is aerodynamic drag.

(the pressure from an insufficiently vented bay also impacts the required strength of the shear pins)

In this situation your post is incorrect and highly misleading to those who want to learn something about drag separation. If you are an engineer, you should be able to understand the free body diagram. Please study it again before you post any more nonsense.

No, you’re right, I am ignoring drag. This it is not because I refuse to acknowledge it or don’t believe in it or don’t understand it. I am just sidelining it for the sake of discussion so that the other forces at play can be considered. Most, probably all, of the other forces affecting the flight have more impact on the flight dynamics than drag.

I guess my primary objection to the concept of ‘Drag Separation’ as a term is that the aerodynamic ‘drag’ is rarely the effect being observed. it is a fictional term given to a failure mode that has no wide acceptance outside hobby rocketry.

I will admit, If all things were otherwise equal, between the two sections, then drag may be the remaining force causing the separation.

In Mr. Dixon’s submission, The rocket, as a complete assembly has less drag than each if its parts or the sum of its parts/2. Only after nosecone separation does the sustainer Cd increase significantly, as does the nosecone Cd, and for similar reasons. The timing is important. The rocket design is essentially an airfoil, albeit clipped at the trailing end, since it has an ogive nosecone and an ogive tail cone as shown in the rocksim profile.

The nose cone Cd increases once it separates from the sustainer because it suddenly changes form from an airfoil to a flat end ogive. The sustainer Cd also increases once it separates from the nosecone transforming from an airfoil to an open ended cylinder with an interior bulkhead to inhibit airflow through it and an ogive tail cone.

If the sustainer did pass the nosecone, in flight, as described, and I have no doubt that it did, then we know that the sustainer had the larger potential energy stored in it. Conversely, the nosecone was able to shed its stored energy faster, because it was lighter, the Cd changed the effective drag making the loss of energy more efficient.

So why did the nose cone feel it necessary to leave the sustainer if the sustainer was more energetic? That is the essence of the question we need to answer in the example given by Mr. Dixon.

 

[pythonrock]

OK, lets go back farther than the free body diagram, how about high school physics.
The forces on an airfoil in flight are thrust, drag, lift and gravity.
Lift BTW is what turns rockets, not drag, as you claimed in your first post.
If there is no thrust, then the only force acting on the airfoil in the direction of motion is DRAG.
Inertia is not a force.
A force acts on a mass producing an acceleration.
A mass in motion has inertia.
The resistance of the air applies a force to the airfoil.
That force is called drag.
The inertia tends to keep the airfoil moving and the drag tends to slow it down (that is an acceleration).
If it was in a vacuum it would continue moving at the same rate, (and would not separate regardless of the masses of it's parts.)

For a rocket moving through air, drag acts on all exposed parts of the rocket.

If the drag on the parts of the rocket behind the joint, relative to that mass is greater than the drag on the front parts, relative to their mass, THEN THERE WILL BE A DIFFERENCE OF FORCE THAT WILL TRY TO SEPARATE THE PARTS

I hope that is simple enough for you to follow.

Otherwise, I repeat the conclusion of my last post. with !!!

 

[dixontj93060]

So why did the nose cone feel it necessary to leave the sustainer if the sustainer was more energetic? That is the essence of the question we need to answer in the example given by Mr. Dixon.

I already explained this, but will calmly do it (only) one more time. Once the nose cone separated it continued forward for a while due to mass. A nose cone by itself is unstable (you agree with this right?). Due to instability it flipped and the open back of nose cone took it's Cd to 1.0. This additional drag slowed its movement compared to the fin can/body which was, in relative terms, more stable, given the corrective force from the fins, and continued to move forward until the end of the shock cord was reached zippering the body.

...'drag'...is a fictional term given to a failure mode that has no wide acceptance outside hobby rocketry.

This statement is ludicrous. Read NASA manuals and reports that analyze flight dynamics. I just picked one such report at random linked here: https://fas.org/spp/guide/usa/launch/sr_handbook.pdf. This report, the NASA Sounding Rocket Program Handbook, discusses multi-stage sounding rocket design. Please read it (if you don't read it at least use your browser "Find" function and search for drag). You will see that it points to drag separation as a basic design principle utilized to conduct the staging, i.e., stage separation for sounding rocket design. So is drag or drag separation "fictional"?--not according to NASA or, to any other rocket or aerodynamic engineer in the world for that matter.

 

[dixontj93060]

I haven't been following this thread, but was curious, did this rocket have shear pins?

It did not. See post #48.

 

[jderimig]

I guess my primary objection to the concept of ‘Drag Separation’ as a term is that the aerodynamic ‘drag’ is rarely the effect being observed.

BBC,

I think the point you are making is that although differential drag force may exist across the rocket sections the dominant cause of separation in most cases is the difference in inertial forces across the sections after engine burnout. It may or not be but for the sake of argument lets assume it is....

This inertial force difference is a(M2 - M1) where a is the deceleration of the rocket after motor burnout.

But if there is no drag, then there is no "a" (a=0) and no inertial force difference. The larger the drag, the larger the a, the larger the differential inertial force. So drag IS the root cause of separation events whether it be differential drag force or differential inertial forces. "Drag Separation" is then a totally appropriate term for this type of incident.

 

[BBS]

thanks John.

sorry for the heat everyone.

i'm really not trying to be an ass.

 

[ChrisAttebery]

This thread got my curiosity peaked so I looked at the sims for my Punisher 4 last night. The worst case I could find was 14G deceleration on a M3700. For my 4.25 lb. nose cone that equates to about 55 lbs. on the shear pins. I'm using 3x nylon rivets that need about 80lbs each to shear. I think I'm good. :wink:

 

[jderimig]

This thread got my curiosity peaked so I looked at the sims for my Punisher 4 last night. The worst case I could find was 14G deceleration on a M3700. For my 4.25 lb. nose cone that equates to about 55 lbs. on the shear pins.

How did you come up with 55lbs on the nosecone shearpins?