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
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
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
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
Last edited:
