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  1. #1
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    Calculating Nose Cone Drag Separation?

    I am trying to wrap my head around drag separation, particularly as it applies to nose cones.

    Definition: Drag Separation - The unintended (or in a few cases, intended) separation of bodies while in flight that occur after motor burnout, due to the inertia of one section having more force to separate the bodies than another section has force to keep the bodies together.

    As I understand it, the heavier a nose cone is, the more prone it will be to drag separate with high thrust motors.

    Can someone show me the way to calculate out these forces? I did a cursory search and didn't find much.

    With this knowledge, it would be easier to make a determination on when to think about the use of shear pins.

    Greg


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    ALWAYS use shear pins. The simple solution. You could find the aerodynamic drag on all the components, nosecone, rocket with fins and then force exerted on the nosecone shoulder by the airframe and figure out if you need them also. Or you could just always use them.

    Edward


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    "Definition: Drag Separation - The unintended (or in a few cases, intended) separation of bodies while in flight that occur after motor burnout, due to the inertia of one section having more force to separate the bodies than another section has force to keep the bodies together"

    It's not inertia that causes drag separation, it's ... drag.

    After burnout, drag and gravity are working to slow the rocket. Gravity affects all parts of the rocket equally, but drag doesn't. The nose of a rocket is more streamlined than the fins. So, the drag force on the back end of the rocket is greater than on the front. It's as if the air is trying to pull the rocket apart. If the friction (or shear pins) holding the parts together is less than the difference in drag, the parts can be pulled apart.

    If you can calculate the drag on the nose section and the drag on the fin section then compare the difference to the force required to separate the sections of the rockets, then you can determine if drag separation will occur.

    -- Roger

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    Sometimes, just using shear pins is not enough.

    If you trust all of its calculations sufficiently, you can use OpenRocket's component analysis to compute what percent of the drag on the rocket comes from forward of the break. Then, figure out what percentage of the mass is forward of the break.

    The drag separation force (disregarding internal pressure) will be:
    (deceleration of rocket(t)*mass of front) - (% of drag from front(velocity(t))*total drag force(t))

    For a given deceleration of the rocket, the total force needed to keep the front part attached will come from two sources: the friction at the break and the drag on the front. Subtracting the drag at the front will yield the amount of friction at the break.

    Note that for supersonic rockets, the distribution of drag force will vary greatly with Mach number, so be sure to record a few values: the highest Mach numbers will be when the magnitude of the drag force is highest, but it might not be the point when the drag separation force peaks.

    I wish you could set a separation point in OpenRocket, so that it could output drag from forward and aft of the separation point in graphs. Then I could set up a formula to plot drag separation force versus time.
    Several rocket motors burned this year.

  5. #5
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    Quote Originally Posted by jadebox View Post
    It's not inertia that causes drag separation, it's ... drag.
    Not completely true. There was a nice derivation of this on RP before it went down, but in a nutshell you can calculate the separation force as follows:

    Fsep = a [ M / (1+R) - M1 ]

    Where:
    a = max deceleration
    M = total mass of rocket (mass NOT weight)
    M1 = mass of lower section
    R = drag ratio

    If Fsep is +, then there is a drag separation force, if - then there isn't.

    Thus we can define the Derimiggio-Newton 1st law of Drag Separation:
    If M/(1+R) < M1, then the sections won't drag separate period.

    I've computed this for typical rockets and it convinced me that in many cases you could have a totally loose coupler and the sections just couldn't drag-separate. This is one reason why I have never bothered to use shear pins at the booster-payload joint.

    Of course, at the nose cone the larger forces are not from drag but from shock cord snapback, so shear pins there are usually needed.

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    Quote Originally Posted by mikec View Post
    Not completely true.
    I realized that what I said wasn't exactly right as soon as I clicked submit. Gravity accelerates all the parts of the rocket equally, but the force is proportional to the mass of each object (otherwise, we'd all weight the same and I wouldn't need to go on a diet).

    But, I was right in the sense that, absent drag, there would be no "drag separation." In a vacuum, for example, the parts wouldn't separate after burnout regardless of the difference in mass.

    -- Roger
    Last edited by jadebox; 1st February 2013 at 10:59 PM. Reason: spelinge

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    Quote Originally Posted by mikec View Post

    ...
    Fsep = a [ M / (1+R) - M1 ]

    Where:
    a = max deceleration
    M = total mass of rocket (mass NOT weight)
    M1 = mass of lower section
    R = drag ratio

    If Fsep is +, then there is a drag separation force, if - then there isn't.
    ...
    Ok, this is the kind of stuff I was looking for: the math.

    Is [a] measured in m/s/s?

    Is [Mx] = W/g? (where W is in kg and g 9.81 m/s/s)

    Is the [R] the Cd or is it something else?

    Do you have a handy sample example?

    Greg

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    Quote Originally Posted by GregGleason View Post
    Ok, this is the kind of stuff I was looking for: the math.

    Is [a] measured in m/s/s?

    Is [Mx] = W/g? (where W is in kg and g 9.81 m/s/s)


    Is the [R] the Cd or is it something else?

    Do you have a handy sample example?

    Greg
    First off, to give credit where it's due, this derivation is by John Derimiggio, not me. He had it written up well (with ASCII free-body diagrams)
    but that site is no longer available (it was a club website, not RP, IIRC).

    Units are in any consistent format, like all MKS. Mass is kg, acceleration in m/s2, Cd is dimensionless.

    The drag ratio is the ratio between the Cds of the upper and lower sections by themselves; you can look at the Cd breakdown in OpenRocket to get a handle on this.

    An example: for my L3 rocket, upper airframe drag is 0.12 and lower is 0.38. M1 is mass of
    lower section, total mass of rocket at burnout is 9 kg, let's say,
    and upper mass is 1 kg. So R is 0.32, M is 9 kg, M1 is 8 kg, and max
    drag decel on an M1315 is 50 m/s2. So fsep is 50*(9/(1+0.32) - 8) = -59.1
    and there is no drag separation.
    Last edited by mikec; 31st January 2013 at 12:33 AM. Reason: fixed mass values

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    Thanks mikec.

    I'll see if I can plug in the values and come up with something.

    Greg

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    Update to an old thread. Tim Dixon and I have been sharing thoughts on this derivation (which is nearly 7 years old) and a refresh was done.

    Below is a rederivation of the problem. I switched the coordinate system + direction this time so the signs are different but the theory is the same.

    Note that the Drag Force ratio R is really the ratio of Cd*A of the top and bottom airframe sections. Usually the fin area is negligible so the Cd ratio is sometimes a good approximation, but to be totally accurate the Cd*A should be used for R. Both easily obtained from Rocksim or OR.

    Also "a" is the deceleration from drag only at motor burnout. So if your sim says 5g's of deceleration, 1g of that is from gravity. So subtract that 1g out and use 4g. Note in the derivation below + is pointing "down", so deceleration is entered as a positive value in the equation. Also in the below derivation Fsep is defined as a "compression" force. So that if Fsep is positive then drag separation will not occur. Sorry for flipping the sign convention this time.....

    John Derimiggio NAR/TRA L3
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    Quote Originally Posted by mikec View Post
    I've computed this for typical rockets and it convinced me that in many cases you could have a totally loose coupler and the sections just couldn't drag-separate. This is one reason why I have never bothered to use shear pins at the booster-payload joint.
    Only thing is, in most cases (that many don't realise) most drag separation events will occur from deficient venting of the internal volume of the airframe. A post I posted to ARocket some time ago explains how this occurs:

    A rocket airframe on the ground (ready for launch) generally contains an internal gas pressure of 1 atm – that’s ~14.5 Psi which roughly translates to 14.5 pounds per every square inch of internal cross-sectional area x2 trying to separate the rocket. Fortunately, there’s also the same amount of pressure acting on the outside of the airframe to equalize the forces and keep the airframe together.
    Now launch the rocket and accelerate to a touch over M1 before engine burnout. Let’s also assume the rocket has a standard streamlined nose cone and flat based aft and is unvented. At burnout, that flat based aft is now effectively seeing a complete vacuum as the gas molecules in the atmosphere no longer can impart their momentum on it (the rocket is travelling too fast in the opposite direction). This phenomenon is what we refer to as “base drag”. The motor has finished its thrust phase, so it no longer is providing any force either.
    If we assume the internal volume of the rocket is still at 1 atm or 14.5Psi, we again investigate the force distribution within the airframe structure: we still have 14.5 psi on the inside trying to separate the rocket, and we have an even greater force on the NC that’s effectively pushing it down on the airframe, however, there’s no longer any external force on the base pushing it together. The base is effectively being pulled off the rocket, but actually being pushed by the internal gas within the airframe. For a 6 inch diameter rocket, the amount of force trying to push the aft away from gas pressure alone is (6/2)^2 * pi * 14.5 = ~410 pounds of force. As you can see, not trivial by any means! This is irrespective of what altitude you’re at, so as you can see, at these velocities, the *static* pressure differential between the inside of the airframe and out is considerably less important than intuition would suggest. Some might suggest it’s irrelevant, but that’s not at all correct and here’s why:
    There are forces counterbalancing this like that from the momentum of the decelerated mass of the aft end that effectively helps keep the rocket together. The magnitude of deceleration is very much determined by the density of the atmosphere acting on the frontal areas of the rocket hence altitude and static pressure does have a significant influence, but not as much as the mechanism of base drag.
    In extreme examples where the frontal area of the fins is significant, the mechanism causing drag separation can be more intuitive to current perceptions of the mechanism ie. what is more commonly considered the mechanism causing drag separation as per the discussion above.

    Troy
    Last edited by rocket_troy; 7th June 2016 at 12:03 AM.

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    Tim, his analysis doesn't show a increase in pressure ,it shows an increase in pressure difference. That's a result of the drop in external static pressure with altitude and counteracted by equalization through the vents.

    Troy's post points out that due to the near vacuum of the base the pressure differential is actually much higher than the static differential alone.
    Last edited by Ravenex; 18th February 2016 at 03:44 AM.
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    Calculating Nose Cone Drag Separation?

    Quote Originally Posted by Ravenex View Post
    Tim, his analysis doesn't show a increase in pressure ,it shows an increase in pressure difference. That's a result of the drop in external static pressure with altitude and counteracted by equalization through the vents.

    Troy's post points out that due to the near vacuum of the base the pressure differential is actually much higher than the static differential alone.
    I see that now. I think I was reading Troy's post wrong. Let me re-look. Plus, I attached the wrong spreadsheet so I'm sure that was utterly confusing. I deleted the wrong one but only on my phone now so can't drag/drop the right on as an attachment. I'll fix in the morning and look at what pressures Troy was referring to.\

    Note: Deleted my post as I misunderstood Troy's point. I believe I understand it and am now looking at the vacuum impact at nozzle exit as he has pointed out.
    Last edited by dixontj93060; 18th February 2016 at 02:51 PM.
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    Quote Originally Posted by rocket_troy View Post
    Only thing is, in most cases (that many don't realise) most drag separation events will occur from deficient venting of the internal volume of the airframe. A post I posted to ARocket some time ago explains how this occurs:

    A rocket airframe on the ground (ready for launch) generally contains an internal gas pressure of 1 atm Ė thatís ~14.5 Psi which is translates to 14.5 pounds per every square inch of internal cross-sectional area x2 trying to separate the rocket. Fortunately, thereís also the same amount of pressure acting on the outside of the airframe to equalize the forces and keep the airframe together.
    Now launch the rocket and accelerate to a touch over M1 before engine burnout. Letís also assume the rocket has a standard streamlined nose cone and flat based aft and is unvented. At burnout, that flat based aft is now effectively seeing a complete vacuum as the gas molecules in the atmosphere no longer can impart their momentum on it (the rocket is travelling too fast in the opposite direction). This phenomenon is what we refer to as ďbase dragĒ. The motor has finished its thrust phase, so it no longer is providing any force either.
    If we assume the internal volume of the rocket is still at 1 atm or 14.5Psi, we again investigate the force distribution within the airframe structure: we still have 14.5 psi on the inside trying to separate the rocket, and we have an even greater force on the NC thatís effectively pushing it down on the airframe, however, thereís no longer any external force on the base pushing it together. The base is effectively being pulled off the rocket, but actually being pushed by the internal gas within the airframe. For a 6 inch diameter rocket, the amount of force trying to push the aft away from gas pressure alone is (6/2)^2 * pi * 14.5 = ~410 pounds of force. As you can see, not trivial by any means! This is irrespective of what altitude youíre at, so as you can see, at these velocities, the *static* pressure differential between the inside of the airframe and out is considerably less important than intuition would suggest. Some might suggest itís irrelevant, but thatís not at all correct and hereís why:
    There are forces counterbalancing this like that from the momentum of the decelerated mass of the aft end that effectively helps keep the rocket together. The magnitude of deceleration is very much determined by the density of the atmosphere acting on the frontal areas of the rocket hence altitude and static pressure does have a significant influence, but not as much as the mechanism of base drag.
    In extreme examples where the frontal area of the fins is significant, the mechanism causing drag separation can be more intuitive to current perceptions of the mechanism.

    Troy
    Troy do you mind posting a diagram of the forces you are referring to? Below is my interpretation of what I believe you are saying and what I think happens at burnout, but I may be way off.

    Click image for larger version. 

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    Last edited by dixontj93060; 18th February 2016 at 07:07 PM.
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    I think Troy is saying that the entire base area of the rocket has no atmospheric pressure on it because of the rocket's velocity. I'd agree that there could be some reduction in pressure dependent on the rocket's velocity, but it seems implausible that it would go all the way to zero.

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    Quote Originally Posted by mikec View Post
    I think Troy is saying that the entire base area of the rocket has no atmospheric pressure on it because of the rocket's velocity. I'd agree that there could be some reduction in pressure dependent on the rocket's velocity, but it seems implausible that it would go all the way to zero.
    Mike, although I drew it as different sizes as I thought the main effect would be from engine burnout and the transient immediately at the nozzle, but there really is no definition of the shape/size of volume #2. Even if the top of the cylinder were the airframe width leading to an inverted cut off cone, the "path of least resistance" to filling up that vacuum (assuming it is a vacuum) whatever its size/shape, is still from the all sides and bottom. So, yes, the "vent hole size" is not infinity, but it is very large. If you look at the model, any large vent will equalize the pressure immediately with a resultant spike function in force that is infinitesimal in time as attached (note: force is shown negative model versus positive direction in sketch above but magnitude is calculated for 6" dia. airframe, burnout at 1s). So although there may be a momentum change from an impulse spike (impulse=F*t) pulling on the base of the rocket (likely part of the overall drag force, D, and included in Cdrag), the pressure transient happens so fast, I don't believe the internal pressure inside volume #1 would be affected much, if at all.

    Click image for larger version. 

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    Last edited by dixontj93060; 18th February 2016 at 07:46 PM.
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  17. #17
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    BTW, a pretty good discussion of base drag, or it's estimation/prediction isolated from other drag factors is found starting on page 10 in this doc: http://web.aeromech.usyd.edu.au/AERO...Prediction.pdf.
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    I'm going to do some diagramming, but I believe counting the pressure differential from the tail vacuum and counting the base drag causes you to count the same force twice. I think if you've already counted the force from the vacuum behind the tail in the drag on the tail, then you should only count static pressure in the pressure forces.
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    Quote Originally Posted by Ravenex View Post
    I'm going to do some diagramming, but I believe counting the pressure differential from the tail vacuum and counting the base drag causes you to count the same force twice. I think if you've already counted the force from the vacuum behind the tail in the drag on the tail, then you should only count static pressure in the pressure forces.
    ^+1 Correct! Base drag is built into the Cd. (Just like the pressure differential X nozzle throat is already accounted for in motor thrust).
    John Derimiggio NAR/TRA L3
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    Calculating Nose Cone Drag Separation?

    Quote Originally Posted by Ravenex View Post
    I'm going to do some diagramming, but I believe counting the pressure differential from the tail vacuum and counting the base drag causes you to count the same force twice. I think if you've already counted the force from the vacuum behind the tail in the drag on the tail, then you should only count static pressure in the pressure forces.
    I agree. I think base pressure change is one component of the base drag.

    I'm trying to resolve Troy's comment regarding base pressure effecting internal pressures...
    Last edited by dixontj93060; 18th February 2016 at 10:05 PM.
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    Quote Originally Posted by dixontj93060 View Post
    I agree. I think they are one in the same.

    I'm trying to resolve Troy's comment regarding base pressure effecting internal pressures...
    Reminds me of an idea I had a while ago to use a venturi to create a negative pressure in the booster section to retain the sections instead of shear pins. If you have a rocket with alot of base drag and vent the lower section through the rear you can accomplish the same thing.
    John Derimiggio NAR/TRA L3
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    Quote Originally Posted by jderimig View Post
    Reminds me of an idea I had a while ago to use a venturi to create a negative pressure in the booster section to retain the sections instead of shear pins. If you have a rocket with alot of base drag and vent the lower section through the rear you can accomplish the same thing.
    This might be a good starting point...

    Click image for larger version. 

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    Quote Originally Posted by mikec View Post
    I think Troy is saying that the entire base area of the rocket has no atmospheric pressure on it because of the rocket's velocity. I'd agree that there could be some reduction in pressure dependent on the rocket's velocity, but it seems implausible that it would go all the way to zero.
    Well, it all depends how fast you're going and the geometry of your base. If you're travelling faster than 99% of the molecules in the air along a given vector, how do you expect them to impart any of their momentum onto your base if it's flat? If you have (effectively) no momentum transfer from gas molecules onto a surface, it's effectively seeing a vacuum.
    Yes (responding to later posters) the base drag is included into the Cd - in fact, it's likely to be the major component of Cd at velocities under M1.5 because of its typical geometry (flat) but my response was strictly relating to *drag separation* and like primary mechanism that causes it - base drag.

    Think in terms of gas dynamics here: Just say we have a rocket with no nose cone ie. a flat front and a flat back - a typical closed cylinder. As we increase in velocity from zero, we experience a certain amount of drag from the forward face because that face is effectively imparting some momentum onto the gas molecules it collides with in the atmosphere. However, what's also happening is we're experiencing just as much drag from the rear of the rocket because the gas molecules in the air are imparting less of their momentum onto the aft face because the rocket is moving away. The drag will continue to rise on both front and back faces until you reach the velocity (a bit over M1) where there are (effectively) no longer any molecular collisions with the aft face. Beyond that point, the base drag doesn't change (it's effectively already seeing a full vacuum) so the only increase in drag comes from the front face. At that point, you see the Cd of your rocket change (it reduces) because there's only one end providing additional drag with an increase in speed.
    Adding a streamlined nose cone, will significantly reduce the change in momentum of the gas molecules your rocket collides with thereby reducing drag - so if you still maintained a flat aft, most of your drag up until approx M1+(M1xCd of your NC) will come from your aft face. Add a boat tail to your aft and you're effectively giving the gas molecules in the air more time to collide with your base allowing the rise in base drag to be a much slower one (like a Nose Cone).
    Last edited by rocket_troy; 7th June 2016 at 12:06 AM.

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    Quote Originally Posted by jderimig View Post
    Below is a rederivation of the problem. I switched the coordinate system + direction this time so the signs are different but the theory is the same.
    I don't think this is quite correct. You're using a in the equations for both the upper and lower sections, but a1 and a2 must actually be different if they are separating.

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    Quote Originally Posted by micro View Post
    I don't think this is quite correct. You're using a in the equations for both the upper and lower sections, but a1 and a2 must actually be different if they are separating.
    At the point in time that John is analyzing a=a1=a2. Only after separating would things begin to deviate.
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    Quote Originally Posted by micro View Post
    I don't think this is quite correct. You're using a in the equations for both the upper and lower sections, but a1 and a2 must actually be different if they are separating.
    The equation calculated the force needed at the joint for them to stay together (a1=a2). If the force is negative then the joint is in tension and will separate.
    Last edited by Ravenex; 4th June 2016 at 02:20 PM.
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    Quote Originally Posted by jderimig View Post
    Reminds me of an idea I had a while ago to use a venturi to create a negative pressure in the booster section to retain the sections instead of shear pins. If you have a rocket with alot of base drag and vent the lower section through the rear you can accomplish the same thing.
    Interesting ides. On our upscale Dragonfly (which has an internal volume of about 4500 cubic inches) I ran two half-inch vent tubes out the aft end. I could have simply drilled holes in the airframe, but I thought that the low-pressure zone behind the rocket would be helpful. It also looks a lot better...

    A friend/local club member that is also a retired aircraft engineer came over just yesterday to go over drag separation forces in preparation for installing our shear pins. He actually had two different formulas to calculate drag - one was a spreadsheet created by a friend and the other was a slightly more complex formula that he had dug up online. I will try to puzzle over the math some more (not my strong point) and see how it all compares to what has been posted here.

    Great thread! Except it makes me wish I would have given more attention to my math studies...

    Dan Feller
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    Instantaneous velocity will be equal for both bodies, not necessarily instantaneous acceleration. If you propose they are accelerating together, then they will obviously remain together. Fsep is not an actual force acting on the bodies, it's a quantification of the difdifferential between Fd1 and Fd2 resulting in a difference in acceleration.

    Take the case m1=m2=1kg, Fd1=2N, Fd2=1N. One would expect Fsep=1N since the forces are known. But the equation yeilds Fsep=-.3a.
    Last edited by micro; 5th June 2016 at 08:19 AM. Reason: typo decimal point

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    Quote Originally Posted by micro View Post
    Instantaneous velocity will be equal for both bodies, not necessarily instantaneous acceleration. If you propose they are accelerating together, then they will obviously remain together. Fsep is not an actual force acting on the bodies, it's a quantification of the difdifferential between Fd1 and Fd2 resulting in a difference in acceleration.

    Take the case m1=m2=1kg, Fd1=2N, Fd2=1N. One would expect Fsep=1N since the forces are known. But the equation yeilds Fsep=-3a.
    Fsep IS an actual force acting on the bodies. It is the inter-stage coupling force. As drawn in the free body diagram a positive Fsep means the coupling is in compression which means the sections have the same acceleration which means they are not drag separating.

    If Fsep calculates to be (-) then conditions for drag separation exist and yes in that case each section will not have the same a. But that point is moot, the purpose of the derivation is to indicate if the conditions for drag separation exist, not to determine the dynamics in that drag separation condition.

    In other words, equal acceleration at each section is the condition where no drag separation occurs. Fsep is the intercoupler force required enforce this condition. If Fsep calculates (+), then this force can be supplied by the butt joint at the coupler. If Fsep calculates (-), then you have to supply this force either through coupler friction or shear pins that can support the force of Fsep.
    Last edited by jderimig; 4th June 2016 at 10:19 PM.
    John Derimiggio NAR/TRA L3
    MarsaSystems

  30. #30
    Join Date
    20th January 2009
    Location
    Salem, MA
    Posts
    8,264
    Note to Troy

    Drag Separation and Internal Pressurization Separation are 2 different processes that are caused by very different mechanisms.

    1. Drag separation is due to differential retained momentum. It does not require internal pressurization.
    2. Internal pressurization is due to the pressure differential between the interior and exterior of the rocket. Drag is not relevant. There is no difference between the differential pressure created by an ejection charge or altitude induced internal pressurization as the net forces are the same.

    Bob Krech
    CMASS, MMMSC, Tech Officer, NAR S&T
    BOD URRG, MRSI, MMMSC
    L2 NAR 78096 TRA 14588
    CAP 536940 NER-MA-019 CDC, AEO
    KC1FVI General

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