SHEAR PIN CALCULATION, THE ENGINEERING APPROACH

BBS

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Introduction

Last week I felt like a mailman being attacked by a wild pack of wiener dogs after trying to suggest that “Drag Separation” was not really hard to understand or overcome. Some members here took exception to my simplistic view of physics and the way that things work, in my experience. It was a weird place for me to find myself in. Rather than argue the validity of a myth, or that of a methodology, to overcome the belief in a misnamed phenomenon, let me begin fresh. I really have no desire to respond to attacks on either my intelligence or my beliefs. Faith is what we humans revert to, in the absence of evidence, to support our beliefs. I do not ask for your faith, I ask for your open mind and the benefit of the doubt.

Background

I am an Engineer. I have been making a living as an Engineer of one type or another, primarily a Mechanical Engineer, for 40 years. I have a degree in art.

Motivation

Having had such a lengthy and varied career, there are some things that have become second nature in my approach to problem solving in my work, and by extension, in my hobby. I thought that I could help someone else struggling to grasp the concepts and forces that impact my favorite past time, high power rocketry.

I like designing. I like building. I like learning. I like improving my skills. I like passing on knowledge.

A member, Bat-mite, who began a thread that I referenced in my introduction, asked for his peers to check his math as he was trying to understand how many shear pins he needed to use to prevent drag separation.

Simple question, I thought, I have something to offer. It didn’t go well for me.

So I thought; how do I convince someone who has such faith in their own hypothesis that they will never permit anyone to question their beliefs and are willing to defend their logic by swearing and making an outrageous claim that I have nothing to contribute to a hobby forum?

So contemplating my new found heaping helping of humility, I knew that it was futile to even try. So I decided to create the following, a narrative about how real engineers come up with real solutions to real problems using real data. And start my own thread.

I will start slow and break it down into manageable elements that can be used as building blocks so that non-engineers can also learn from the process. It can be dreadfully boring to write, and to read, to all but those who are interested in learning the process. Part of what you do, as an Engineer, is to relate your solution, or an approach to a solution, to someone in a way in which they do not need to understand the minutia it takes to arrive at the solution, but that they are confident that you know how you found the solution and that the solution is valid. This is my goal here.

I make no claim that all of you will get it, or that all will agree, or that I won’t make illogical assumptions, or make errors in math, or even offer conflicting information or even sound like a fool. In fact I would be very happy if some of those things do happen and some of them do not happen. I am writing this in a vacuum with only myself to review my work. My philosophy is that success feeds your pride and failures feed your experience. I don’t remember the source of a quote, probably not me but it could have been; “experience is what you get when you don’t get what you want”. My advice to everyone is: “Never start investigating a failure with the assumption that the failure is not your fault.” Start your investigation by verifying your work is not the cause of the failure. Once you know this, the rest should be easy to prove

Procedure: A practical guide of determining the number of shear pins you need.

Finding a solution to a problem begins with a statement. For this article I will use the following statement: “Drag separation will keep my rocket to from flying successfully.” Hear is how I find the solution:

I first rephrase the statement into the form of a question:

Question 1: How do I prevent my rocket from suffering from drag separation?

Answer 1: Make the rocket so that it cannot drag separate.

You are now done, it is just that simple…….
Okay, lets consider how our solution can be accomplished. Start again.

Q2: How do I keep the rocket from (drag) separating?

A2a: Design it as a single part that can never separate.
A2b: Design it so that it remains a single part until you no longer want it to be a single part, so that you can deploy a recovery device and fly it again.

I will assume that most of you prefer answer A2b over answer A2a.

Analogy: Answer A2a is what many munitions designers would prefer. Answer A2b, minus the recovery device, is what Bendix adopted in 1948 when they replaced the warhead on the Loki with a dart that would separate to leave the drag of the booster behind.

Q3: So how do I insure that my rocket does what I want it to?

A3: You must make the top stay attached to the bottom with a retention system.

Q4: What are my options for retention?

A4a: You could pin them together by adding something such as screws or pins, in shear, that will hold the two halves together until you want them to separate
A4b: You could tape them together
A4c: You could temporarily glue them together
A4c: You could design a mechanism that locks the two halves together that can also release them at the appropriate time.
A4d: You could use Builders prerogative: you invent or adopt another way to retain the two halves

So lets decide that we want to take the most direct path that we think fits our needs with the least complexity.

To me A4a: “pin them together” is my preferred approach. However any of the answers are just as valid as my choice.

So lets start with a question we can find an answer to:

Q5: How many screws or pins do I need to prevent separation?

I recognize this question! It is the question that Bat-mite asked in another thread!

Q6: When does it matter?

A6: let's figure out that answer

Let us begin with the data that we know, and apply logic to our assumptions in order to come to a conclusion.
Data:
1. Two rockets named Rocket A and Rocket B
2. Rocket A is a symmetrical airfoil. See the drawing of “airfoil”.
3. Rocket B is the same symmetrical airfoil but it has a joint to allow the payload to separate from the sustainer in order to deploy a parachute.
4. Rocket A mass = Rocket B mass
airfoils.jpg rockets.jpg

The data, so far, should lead us to some logical conclusions:
1. As long as Rocket B remains undivided, Rocket A drag = Rocket B drag.
2. Since the mass of Rocket A and Rocket B are equal, and so is their coefficient of drag, the velocity of Rocket A is equal to the velocity of Rocket B, through the entire flight , and therefore,
3. Both Rocket A and Rocket B are equal in every way based on the available data. This remains a true statement as long as rocket B remains undivided.

Now the science
1. One force present to keep Rocket B undivided, while in motion, is Newton’s 1st law of motion. (copied from Wikipedia) “…an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force”
2. One force acting to divide Rocket B into its two subassemblies subC and subD is the differential of inertia, or stored energy, acting on subC and subD. This, we find, can be shown using Newton’s 2nd law of motion, F=ma, where Force equals mass times acceleration.

So we know the mass (weight) of each subassembly, at rest, because I assigned them weight, and we find that subC weights 1.35 kilograms (3 pounds) and that subD weighs 2.27 kilograms (5 pounds). For the purposes of demonstration, let us assume that the masses given are the masses at motor burnout.

So now we need to know the acceleration to use in our formula.

The easy button, for us at this point, is to define our rocket in simulation software and apply a specific motor profile to simulate flight. We discover that the rocket motor we have chosen will accelerate our rocket, as the fuel is consumed, to 304 meters per second (997 feet per second) when the fuel is depleted. Now, we have the necessary data to perform the following math:

Relevant Math
1. The quantity of stored energy in subC can be found by applying our known values. F=ma which is F=1.35 X 304 which results in 410.4 Newtons.
2. We also know that the quantity of stored energy in subD can be found by applying it’s known values. F=ma or 2.27 X 304 which results in 690.1 Newtons.

Scenario One:
1. The force of subC = 410.4, while the force of subD, 690.1.
2. 410.4 minus 690.1 equals -279.7 Newtons, a negative 62.8 pounds.
3. SubD has more stored energy than subC and the rocket will continue in motion until both have expended their stored energies and the negative G’s become 1 G.
4. There will be no drag separation.
5. There could be separation, from pressure differential, but that is another consideration.

Scenario Two: Let’s reverse the weight of subC and subD to see the inverse result.
1. If the force of subC is 690.1 Newtons and the force of subD, 410.4 Newtons, we have a difference of 279.7 Newtons, a positive 62.8 pounds.
2. SubC now has more stored energy than subD so what happens at burn out? Separation?
3. Since the result is positive, then you assume that subD will expend it’s stored energy before subC. And you can expect the two subassemblies to separate (drag separate?).

So now you have your results
1. We see, in Senario One, that we can ignore the negative result as irrelevant to the question because the answer is less than zero. Any resistance subC and subD have to separate from each other is enough to insure the stay together as well.
2. We need only be concerned with a positive result from our findings in Scenario Two,

So the practical application is this:

When we subtract subC from subD we get 279.7 Newtons, this is our ‘Delta’.

Since the results are: Delta = 62.8 pounds, the shear pin strength required to overcome 62.8 is obviously greater than 62.8 pounds, right?

More Math

The equation to find the number of shear pins necessary is: N=62.8/Eforce, where 62.8 = the force we need to overcome and Eforce = the strength of one shear pin.

More data is needed
We now have to find the strength of the shear pins. This number is based on the cross sectional area of the pin times the strength of the material.

I will use data found here: https://www.feretich.com/rocketry/Resources/shearPins.html : a single 2-56 nylon 6/6 screw has a minimum of 31 pounds of shear strength.

Finally we have all of the data necessary.
N=62.8/31 which equals 2.025 pins
Rounding up tells you to use three 2-56 screws

At this point engineers have all of the information they need to make an informed decision. A prudent Engineer should also include some margin of safety to account for variations in fastener strength and motor performance so as not to risk failure. The answer did not require math to calculate drag, drag was neutralized by not allowing it to be a factor.

If you worked for me, and it took you longer to solve the problem than it takes to read this post, then you would probably not last very long in my employ.

END of Process.

Your next assignment is to figure out how much black powder it takes to break your shear pins.

It really isn’t rocket science, it is engineering.

Now, admittedly, I have no idea who of you are engineers, who are physicist’s, who are students, who are rocket scientist’s (I only know one real one of those), nor who among you are just thirsty for knowledge. I don’t know, nor do I care, who agrees with me and who does not agree with me. Everyone is entitled to his or her own opinion and everyone is entitled to my opinion as well.

You are bound to succeed just by keeping an open mind and being humble enough to question a conclusion when the evidence is not compelling enough to support the conclusion. No matter how violent the reaction is, come to your own conclusions and don’t let people belittle you.

brandy
 
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les

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Introduction



The easy button, for us at this point, is to define our rocket in simulation software and apply a specific motor profile to simulate flight. We discover that the rocket motor we have chosen will accelerate our rocket, as the fuel is consumed, to 304 meters per second (997 feet per second) when the fuel is depleted. Now, we have the necessary data to perform the following math:

Relevant Math
1. The quantity of stored energy in subC can be found by applying our known values. F=ma which is F=1.35 X 304 which results in 410.4 Newtons.
2. We also know that the quantity of stored energy in subD can be found by applying it’s known values. F=ma or 2.27 X 304 which results in 690.1 Newtons.

I have a question - and please, I am not trying to attack but to understand the 304 m/s. This value was then used in the F = ma equation. But isn't the 304 m/s the velocity, not acceleration?

So we launched our rocket and it achieved a speed at motor burnout of 304 m/s. I would think we would need to use this speed number and the Cd of Sub C and the Cd of Sub D to determine the actual deceleration force now being applied to each section. Then we need to determine if the force on Sub C is less than or greater than the force on Sub D. If the drag force on Sub C is equal to or greater than the drag force on Sub D, the rocket will stay together. But if the drag force on Sub D is greater than Sub C it will want to decelerate faster while the Sub C part will want to continue at the higher speed (Newton's Law as you pointed out).
Now we need to calculate those differences and determine how to keep them together - which could be the number of shear pins.
 

BBS

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les,
yes you are correct. One assumption that I make, in order to extract data points, was that I froze the moment of motor burnout. it is neither accelerating nor decelerating I froze the data at an assumed maximum velocity.
thanks for the question
 

dixontj93060

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Brandy, mass is one component but in your analysis you are missing another critical factor. Consider scenario 2 in your original post (SubC having greater mass than SubD), but instead SubC is now shaped as below. Will it separate? How do you know or not know without considering the Cd and frontal area of each subassembly (in addition to it's mass)?

NewShape.PNG
 
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BBS

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... How do you know or not know without considering the Cd and frontal area of each subassembly (in addition to it's mass)?

View attachment 283106

Because I am capable of neutralizing drag, as a factor, by not allowing the two components to separate in the first place, regardless of their form. I have no beef with physics or people who want to teach it. As an Engineer is not particularly relevant to consider frontal area in retaining one mass to another mass at a specific velocity.

thanks Tim,
brandy
 

dixontj93060

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Because I am capable of neutralizing drag, as a factor, by not allowing the two components to separate in the first place, regardless of their form. I have no beef with physics or people who want to teach it. As an Engineer is not particularly relevant to consider frontal area in retaining one mass to another mass at a specific velocity.

thanks Tim,
brandy

But Brandy, you say in scenario 2, point 3 with SubC mass > SubD mass that separation does in fact occur as you state below.

"Since the result is positive, then you assume that subD will expend it’s stored energy before subC. And you can expect the two subassemblies to separate (drag separate?)."

So, as I stated in the sketch, same mass distribution that you specify in your original post but SubC is an inverted cone with the forward diameter ~3X the body of SubD... Taking it further, what if SubC was a flat disc of same mass but disc diameter is 10X the body of SubD... Will the inverted cone separate only due to mass regardless of drag? How about the flat disc? These solutions cannot be attained by considering mass only. Drag and area are a fundamental part of the physics in this scenario.
 
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jderimig

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Because I am capable of neutralizing drag, as a factor, by not allowing the two components to separate in the first place, regardless of their form. I have no beef with physics or people who want to teach it. As an Engineer is not particularly relevant to consider frontal area in retaining one mass to another mass at a specific velocity.

thanks Tim,
brandy

What if I deploy a chute out the rear of the rocket after burnout? Won't the sections separate regardless of the inertia ratio?

Or if I deploy an air brake on the upper section after burnout won't the sections stay together even if interia upper is a little higher than inertia lower?
 
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dixontj93060

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What if I deploy a chute out the rear of the rocket after burnout? Won't the sections separate regardless of the inertia ratio?

John, you are right, changing the shape on SubD under Brandy's scenario 1 (greater mass than SubC) is another way to illustrate it. With the right size/shape drag plate (or chute) deployed on SubD the rocket could be made to separate regardless of the mass distribution.
 
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Worsaer

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I have been following this interesting exchange over the two threads, and it occurs to me that BBS may have figured out the math, but the calcs only work in a vacuum. That's the only way I can imagine that one can "neutralize drag", as the coefficient of drag is not a constant.

I am reminded of a NASA publication I recently read entitled, "Breaking the Mishap Chain", in which the author describes a number of failed test flights. and what was learned from them. One of those flights involved the F22a, which had an ejection seat rated for 600 knots per hour.

Note the effect of drag on the pilot, compared to the plane:

"The aircraft remained inverted with the dive steepening from 65 to 72 degrees nose low. At 14,880 feet and Mach 1.49, with an 83-degree nose-low attitude, he finally rolled the aircraft upright and pulled full aft on the stick; however, the F-22 continued to descend rapidly. Cooley ejected just 3,900 feet above the ground while traveling at 765 knots equivalent airspeed. Unfortunately, this speed was 165 knots above the maximum ejection speed of his ACES II ejection system, and he sustained fatal injuries due to the blunt force trauma of the resultant windblast."

Why did the pilot perish? It was not a mechanical failure of the ejection seat, nor was it impact with the ground. If the OP's calcs were complete, the pilot would have continued at the same speed as the plane after ejecting. It's a sad outcome, but NASA learned from it and uses those lessons to reduce the chances of other pilots suffering the same outcome.
 

jderimig

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I have been following this interesting exchange over the two threads, and it occurs to me that BBS may have figured out the math, but the calcs only work in a vacuum.

In a vacuum there would be no deceleration and hence no separation forces.
 

dixontj93060

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Worsaer, certainly a graphic :eyepop:, yet good example.

But, come to think of it, maybe you are on to something... Instead of using the word drag, maybe the term "air resistance" and the concept of molecular interference is more palatable. Below is an excerpt from high school level exam prep notes defining Air Resistance.

Air Resistance.jpg
 
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jderimig

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What? There's no gravity in a vacuum?

In the rocket inertial frame of reference (the one that generates the inertial separation forces) the acceleration is zero. The rocket is weightless.
 

Bat-mite

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In the rocket inertial frame of reference (the one that generates the inertial separation forces) the acceleration is zero. The rocket is weightless.

I think that's over my stubby head, but I don't want to derail the thread. Carry on!
 

dixontj93060

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Here is another primer on air resistance / drag. Part of the external ballistics discussion (starting at 8:54) in this old video posted by TopRamen a few days ago...

[YOUTUBE]avoqxyyX42w?[/YOUTUBE]
 

BBS

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seriously?
you guys are kidding, right?
man, you had me going, I was buying it.
now I feel really stupid.
you're all fired!
 
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les

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I agree with the approach of stating the problem and the known facts, but I do disagree with some steps and comments.

First, while calling F=ma, yet in the calculations it shows using the mass time VELOCITY. Not acceleration. The product of mass and velocity is momentum.

Introduction

Newton’s 1st law of motion. (copied from Wikipedia) “…an object either remains at rest or continues to move at a constant velocity, unless acted upon by a force”
2. One force acting to divide Rocket B into its two subassemblies subC and subD is the differential of inertia, or stored energy, acting on subC and subD. This, we find, can be shown using Newton’s 2nd law of motion, F=ma, where Force equals mass times acceleration.

So we know the mass (weight) of each subassembly, at rest, because I assigned them weight, and we find that subC weights 1.35 kilograms (3 pounds) and that subD weighs 2.27 kilograms (5 pounds). For the purposes of demonstration, let us assume that the masses given are the masses at motor burnout.

So now we need to know the acceleration to use in our formula.

The easy button, for us at this point, is to define our rocket in simulation software and apply a specific motor profile to simulate flight. We discover that the rocket motor we have chosen will accelerate our rocket, as the fuel is consumed, to 304 meters per second (997 feet per second) when the fuel is depleted. Now, we have the necessary data to perform the following math:

Relevant Math
1. The quantity of stored energy in subC can be found by applying our known values. F=ma which is F=1.35 X 304 which results in 410.4 Newtons.
2. We also know that the quantity of stored energy in subD can be found by applying it’s known values. F=ma or 2.27 X 304 which results in 690.1 Newtons.

Scenario One:
1. The force of subC = 410.4, while the force of subD, 690.1.
2. 410.4 minus 690.1 equals -279.7 Newtons, a negative 62.8 pounds.
3. SubD has more stored energy than subC and the rocket will continue in motion until both have expended their stored energies and the negative G’s become 1 G.
4. There will be no drag separation.
5. There could be separation, from pressure differential, but that is another consideration.

Scenario Two: Let’s reverse the weight of subC and subD to see the inverse result.
1. If the force of subC is 690.1 Newtons and the force of subD, 410.4 Newtons, we have a difference of 279.7 Newtons, a positive 62.8 pounds.
2. SubC now has more stored energy than subD so what happens at burn out? Separation?
3. Since the result is positive, then you assume that subD will expend it’s stored energy before subC. And you can expect the two subassemblies to separate (drag separate?).

Now given that at motor burnout both sections (Sub C and Sub D) have the same velocity, the different masses will have different momentum, but what will keep them together or make them separate? As previously pointed out "an object in motion with remain in motion unless acted upon by a force". What forces are in play?

The 4 forces on an object in flight are thrust, lift, gravity, and drag.
Since we re looking at when the motor burned out there is no thrust.
Assume the fins are symmetrical and the rocket is pointing straight up with no lateral wind, so there is no lift.
Gravity exists, but it will operate on both pieces equally (consider Galileo's experiment of dropping 2 balls of different masses and them landing at the same time)
That leaves DRAG.

Brandy stated in another post,
"Because I am capable of neutralizing drag, as a factor, by not allowing the two components to separate in the first place, regardless of their form".

Since the initial question was how many shear pins are needed to prevent drag separation, we are trying to find that answer so as to "not allowing the two components to separate in the first place".

SO back to these 2 sections. Each is now being acted on by drag. Is the drag the same on both pieces?

Drag force is proportional to the medium density, area, velocity, and Coefficient of drag (Cd).

At this point we need to look at both pieces.

As a thought exercise - imagine the 2 pieces are separate and flying in formation side by side and the motor has burned out. In a vacuum the two pieces would stay in formation. But for our flights we are still in atmosphere so there is drag. If the drag is greater on one piece over the other, that piece will slow down faster.
Case 1: If the higher drag is on Sub C, then it wants to slow down faster and would fall behind Sub D.
Case 2: If the higher drag is on Sub D, then it wants to slow down faster and would fall behind Sub C.

Now instead of the formation being side by side, sub C is directly in front of Sub D. In Case 1, Sub C is slowing down faster so Sub D will "crash" into the rear of Sub C and try to push it along. So, if they had been attached there would be no drag separation,
In Case 2, Sub D is slowing down faster. Sub C wants to keep going at its higher velocity. Without any means of retention, there would be drag separation.

SO the critical factor is the Cd and this drag force that may or may not cause separation. This becomes the hard part as we may not know all of the information. We can measure the area and look up the density of air at the expected altitude of motor burnout. We will also know the velocity of the rocket. But we then need to know the Cd for each piece. Frankly, determining the Cd will probably be an estimate. And I will be honest - I am not sure how to get accurate numbers for the Cd of the 2 sections (and also consider the base drag for the Sub D section.)

EDIT - It was brought to my attention that OR & RS can provide the individual Cd for each section. One can then calculate the drag on each portion. Once that force is known then Brandy's equations for determining the number of shear pins can be used


In terms of the original conclusion that the separation is based on the mass, and the drag can be ignored, let's do another thought experiment.

For this case our rocket has a large flat drag plate at the base of Sub D - sort of like a spool rocket. And the nose section (Sub C) is very loose but stream lined.
At motor burnout the drag plate will have a very large drag force and want to rapidly slow down Sub D. Sub C, regardless of having more or less mass, will want to keep going at its original speed - oops - drag separation.

Again, the critical item is the drag forces acting on each section. These forces cannot be ignored.

As a side note, if the body is sealed and reaches sufficient altitude, then the trapped air will want to expand and will also add an internal force trying to separate the two sections as well.... So as a good engineer one would over-design the retention (like number of shear pins) to account for this.

Unfortunately, I did not solve the original question. How many shear pins are needed. But the presented approach I believe would provide an incorrect answer.
 
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Igotnothing

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Truth is, the pins are likely to encounter much worse forces on their way to apogee.
I have a slightly larger rocket than Brandy's - 9 pounds, 3 inch dia, 6 feet long, 6 fins. I find that 1 2-56 nylon shear pin is sufficient on the drogue bay, 2 on the main bay, but have not experimented with 0 - relying just on coupler friction and air resistance to keep things together. I also have a 23 pound rocket with HED that relies on exactly 0 shear pins. Coupler friction is enough.


If you have watched James Jarvis Three Carb Yen video, you saw the on-board camera view of a high-alpha, dramatic buckling load displace the outer tube at a coupler - carbon fiber acted like rubber under the loads, and the rocket survived to get the final stage to 130000'. If a shear pin had been at that location it would have failed in tension.

I think "drag separation" is more insidious than just air resistance difference on the booster versus the payload. Likely culprits blamed on "drag separation" are: the above buckling forces, ejection charge delay grain failure, case leaks (I've seen that one), loose chutes striking a fwd bulkhead at burnout, insufficient volume venting to ambient,
 

JimJarvis50

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If you have watched James Jarvis Three Carb Yen video, you saw the on-board camera view of a high-alpha, dramatic buckling load displace the outer tube at a coupler - carbon fiber acted like rubber under the loads, and the rocket survived to get the final stage to 130000'. If a shear pin had been at that location it would have failed in tension.

Actually, if you're referencing the separation at 37 seconds in the video, that was the point where the sections were supposed to separate but didn't. But, this was indeed the result of forces on the airframe.

[video=youtube;eHloNCGlYz4]https://www.youtube.com/watch?v=eHloNCGlYz4[/video]

On the other hand, the buckling that occurred for the third stage just before that (at 35 seconds in the video), likely did break the shear pins that I used to hold the third stage to the second (they are there because you don't want the third stage to drag separate before the second stage motor ignites - that would be bad). Fortunately, I don't think there is much of a drag separation force on the third stage. I have previously approximated the tendency for my 4x4 and 4x3 two-stagers to drag separate. I used the following equation, which has been posted on TRF by others in the past:

Sep force, N = (Mass Booster / Mass Total x Drag Sust) - (Mass Sust / Mass Total x Drag Booster)

I approximated the drag forces for the sustainer and for the entire rocket (and then the booster by difference) from RockSim and RasAero. This is not rigorous, but I was just trying to get a rough idea of the forces at the time I did the analysis. The calculation was done at the estimated burnout velocity, and the results are in the attached picture. If the Sep force is positive, the parts will not separate, if negative, they will. I was surprised to find that the FCY (4" x 4") is more likely to drag separate than the TCY (4" x 3"). The upper two stages of the three stager are essentially the TCY (except that the second stage in the three stager has a little more mass and a little less drag than the TCY). So, if the pins were broken during the flight, perhaps this is why the stages didn't separate.

Jim

Drag Sep Chart.png
 

Hal8472

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I wanted to add an additional thanks for posting. I was pointed to this thread for asking a similar question. I found the discussions very helpful, particularly including the details on atmospheric drag.
 
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