Shear pin retention stregth. Check my math.

[pythonrock]

Tim,
Exactly. It's the difference in drag, which will be maximum when the velocity is highest - immediately after motor burnout.
The weight of the parts, their inertia, will try to make the parts continue, and the aerodynamic drag, trying to slow them down, will determine the force difference on the two parts. If that force is high enough to overcome the friction fit (or pins) they will separate.

My point was that Gs of acceleration of the motor burn is irrelevant.

 

[dixontj93060]

My point was that Gs of acceleration of the motor burn is irrelevant.

I agree, that is why I referenced the links in post #26 (but trying to get through this without providing the answer outright so some learning occurs).

 

[pythonrock]

Tim,
Exactly.
You do have to consider the weights to determine the net forces on the two parts, basically drag/mass for each part. If the drag/mass for the front is less than the back, there will be a net force trying to separate the parts. Other than for something like boosted darts, that difference is generally less than a few Gs. It might separate friction fit parts that aren't tight enough, but I can't imagine a situation that could break shear pins.

The forces will be maximum when velocity is highest - immediately after burnout, so if it doesn't separate then, it won't drag separate up near apogee.

My point was that the acceleration Gs of the motor burn is totally irrelevant to drag separation.

sorry missed prev post and reposted with corrections

 

[dixontj93060]

Tim,
Exactly.
You do have to consider the weights to determine the net forces on the two parts, basically drag/mass for each part. If the drag/mass for the front is less than the back, there will be a net force trying to separate the parts. Other than for something like boosted darts, that difference is generally less than a few Gs. It might separate friction fit parts that aren't tight enough, but I can't imagine a situation that could break shear pins.

The forces will be maximum when velocity is highest - immediately after burnout, so if it doesn't separate then, it won't drag separate up near apogee.

My point was that the acceleration Gs of the motor burn is totally irrelevant to drag separation.

sorry missed prev post and reposted with corrections

Dan, I believe you are missing the time component. Look at the equations, F = momentum / time = mass*velocity/time = m*a, so highest (upward vertical) forces are when acceleration (change in velocity) is the highest, not when velocity the highest (for any given mass being the same).

 

[pythonrock]

I believe I misused the term "force" in parts of my posts. It is the difference in deceleration ( which includes both the drag force and the mass) that is related to "drag separation".

 

[FredA]

You are right - the acceleration G's of the motor BURN are irrelevant....100% correct.
What you need is to know the peak dV/dT which is at burnout.
This dV/dT is caused by change in THRUST from burning to not-burning.

How are you going to get this number?
It's the peak of the slope of the burn-out phase of the thrust-curve.
For a moon-burn it's insignificant. For Warp-9 it's a big number.

For a worst case, if you have a motor with a crisp shutoff, use 100% of the thrust.

Don't believe me....
Give us a better way to calculate the FORCE on the retention at burnout.

 

[pythonrock]

Again, think of a rocket in space. At motor burn out the rocket will coast at whatever velocity it had before shutdown. dv/dt = 0 regardless of the thrust the motor was producing before. The dv/dt at burnout in the atmosphere is caused by aerodynamic drag. It has nothing to do with motor shutdown.
The FORCE at the joint is the difference between the deceleration of the front and back of the rocket. That deceleration is a function of the aerodynamic drag of the respective part and it's mass.

 

[dixontj93060]

The FORCE at the joint is the difference between the deceleration of the front and back of the rocket. That deceleration is a function of the aerodynamic drag of the respective part and it's mass.

I have contacted Mr. Derimiggio and working now on re-deriving old physics equations from decades past... Be patient, we may have something of substance soon. Older minds take a bit of revving to get up to "velocity"

 

[BBS]

Wow, I was sure that g's were the effect of gravity upon mass, that is what the g stands for in "g's" after all. Drag is insignificant as a force in deceleration of a rocket. The only significant consideration in separation of a payload from a booster is IF the mass differential that is present between the two parts can overcome the resistance (friction or shear force) to separate.

Newton's first law says that the payload and the booster will continue to move at a constant velocity unless a force acts upon them. Gravity is the force we try to overcome with rockets, not drag. We use drag to steer and Newtonseconds of force to overcome gravity.

Remember, shear pins are just the new masking tape. All of the same principals apply today that many of us learned 50 years ago flying our Estes and Centuri models. Gravity has remained constant. I use shear pins on fiberglass rockets because it is cooler than tape, fiberglass can withstand the localized shear forces with little or no damage and fiberglass allows us such close fit tolerances that there is no room for tape between the coupler and airframe. I use tape on cardboard rockets because it works, is simple, is adjustable and it does not damage the airframe. Both, shear pins and tape, do the same thing.

if you are puking your laundry early when you use shear pins, you have something else going on besides drag separation. The rule of thumb that you should be able to hold up the payload without the sustainer falling off is simply checking the resistance to separate, it is not a hard number unless your sustainer is tethered to the earth.

 

[pythonrock]

BBS,
I'm sorry you think that way.

 

[pythonrock]

Gravity is the curvature of spacetime as a result of mass.
"G" is the acceleration equivalent to gravity at earths surface and is approx 32 ft/s/s.

 

[FredA]

I give...time to sit back and watch the experts....

op:

 

[BBS]

Math is great, i like math, but sometimes over thinking the problem obscures the logic necessary to solve the problem.

 

[dixontj93060]

Just wanted to refer to John Dermiggio's repost over here: https://www.rocketryforum.com/showt...se-Cone-Drag-Separation&p=1550813#post1550813.

Interestingly math shows that if you want to leave it up to the shake test, most people are doing it wrong. Instead of holding the nose/payload and shaking, the math shows you'd be better off holding the fin can and shaking upside down as the momentum/mass of the aft section works to keep the sections from separating but not the other way around (but everyone knew that already, right :wink.

 

[Ravenex]

I am working on a complete write-up on forces on the joint of a rocket. I will be expanding it eventually to a complete explanation of the basic physics/mechanics of rocketry. Right now it's almost 3 pages in word and I am working on formatting my equations into images to make them more readable. It will take a little long to finalize, please hold tight.

 

[BBS]

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.

 

[rharshberger]

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 the designs 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, that we call g's. A mass set to free-fall will continue happily going along its trajectory until it has an impact with another mass.

So basically BBS if I understand correctly we have two forces affecting whether the rockets seperates early or not: the forward section inertia and the aft sections inertia+/-drag. If the drag is high enough or the tail section light enough that it losses velocity faster than the forward section we get a seperation. However if those two sections are pinned together then they are one piece and it is only necessary for the pins to overcome the forces placed on them by drag (for the most part, EDIT: I forgot about positive airpressure in the airframe too).

I was having a hard time accepting the theory of a fast burnout causing enough deceleration to cause separation, "as an object in motion tends to stay in motion", drag was the main cause of the separation to me, and with Vmax motors the problem with early separation is possibly insufficient airframe venting and separation due to positive airframe pressure vs the external air pressure.

Before people get to excited about what I may or may not understand properly based on the above statements, please realize A) I am not an engineer or a physicist B) I try and understand the how and why without the math, I am sure I can figure the math out but have no desire or time to do so. Feel free to pick the above statements apart, but I am trying to understand what exactly we are dealing with with terms I am more comfortable with.

 

[dixontj93060]

BBS, I beg to differ as drag is a fundamental part of rocket physics/dynamics and, more practically, the simple shake test or differential in weight type of calculations have proven inaccurate for me in more than one flight. One example of an actual flight, its outcome and the overall physics is shown below. Right now I am rebuilding this rocket and that is one of the reasons I am interested in this subject--want to get it right this second time around.

In general I thought it might be useful to other TRF members to show an example of John D.’s math and provide a tool for future use. This example was a drag separation event I had at Airfest a few years back. It was a Public Enemy Ultra Fatboy on a K1100T. The separation point was the nose cone where there was located a Defy Gravity holding a deployment bag for main parachute release. Before the flight I did the typical test by shaking from the nose and it was on tight (using wraps of blue tape inside on shoulder, not shear pins). From what I gather now shaking provides ~2 to 3 gee impact (looking at Android and iPhone reported readings). So even in my shaking in preparation for launch with a 5.5lb aft section I was really only measuring around 11lbs to 17lbs of holding force in that test. As I look closer and in hindsight, I see from Rocksim and RasAero II they show -13 and -15 gee (-482 ft/s*s) burnout decel, respectively. As shown in the attached Excel tool the separation force experienced in the flight was closer to 41lbs. Even if I had used shear pins, I probably would have only used a single 2-56 nylon screw which probably still wouldn't have been enough to hold the nose cone in this case.

Attached are:

1. Rocksim image of loaded rocket. Before others mention it, yes, margin shown is 0.81 but once adding “zero mass” aft cone to account for stubby design, margin is actually 1.25 and as such, rocket flew straight as an arrow.
2. Two simulations—one from Rocksim, one from RasAero II to show burnout negative deceleration excursion.
3. Excel Separation Analysis spreadsheet with data for this flight/rocket entered (but should be appropriate for general FNC rockets).

View attachment Separation Analysis.xlsx

 

[BBS]

I know what you are saying. I experienced the same phenomenon with my Dyacom Scorpion. If I had a cardboard airframe it would definitely zippered as well.

There were three altimeters onboard. A Raven, an SL100 and a Multitronics ‘Kate’. I also had onboard radio control where I could deploy the main chute on command.

The following is my flight report:

1. The rocket weighing 20lbs at ignition was overweight for the k250.
2. The light breeze caused weathercocking.
3. The k250 burned out at 2453ft AGL and then coasted to 6Kft AGL.
4. The Angle of attack was ballistic at apogee.
5. Drogue deployed while the rocket was traveling at about 187ft/sec.
6. Inertia of the payload section when the drogue deployed caused the stitching on the attached third loop of the shock cord to fail. The third loop of the shock cord was attached to the electronics bay.
7. The forces that tore the stitching also broke away two of the three anchors holding the aft bulkhead onto the electronics bay.
8. The same shock event allowed the nose cone to break the shear pins and pull out the main chute.
9. The main chute charge fired at 700ft AGL according to the Stratologger data.
10. The Stratologger altimeter worked as intended and deployed the drogue chute as soon as it detected that the rocket had begun decent.
11. The Raven Altimeter data is totally worthless. It indicated that launch altitude was negative 5600ft AGL and apogee was +400ft AGL. It also indicates that the temperature was 110degrees and we know it was in the 70’s.
12. Conclusion: Root cause failure can only be attributed to flyer error. The thrust to weight ratio was too low. No damage was obvious to the parachute.
13. The Kate Altimeter is the most telling. It reported the following data:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
¦ Flight Card ¦
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Launch date: Sat Sep 26 2015
Launch site: BLACK ROCK PLAYA NEVADA
Rocket name: SCORPION
Motor: K250
Expected alt: 8000 feet
Dual deploy: Yes, main parachute set at 700 feet
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
¦ Flight Results ¦
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Max GPS altitude: 5837 feet AGL
Max Acc altitude: 6496 feet AGL

Max GPS velocity: 485 feet/sec 330.7 mph 0.444 Mach
Max Acc velocity: 529 feet/sec 360.7 mph 0.485 Mach
Max GPS Mach num: 0.490 Mach

Max Acceleration: 6.8 G's 218.1 feet/sec/sec

Motor burnout alt: 2453 feet AGL
Motor burn time: 7.9 sec (approx)

Distance to apogee: 3728 feet 0.71 miles
Bearing to apogee: 248 degrees (from true north)
Ground speed at apogee: 187 feet/sec 127.5 mph
Heading at apogee: 248 degrees (from true north)

Descent rate on drogue: 12 feet/sec 8.4 mph
Descent rate on main: 11 feet/sec 7.5 mph
Touch down velocity: 12 feet/sec 8.2 mph

Main chute deployed at: ? feet AGL

Time to apogee: 22.1 sec
Time on drogue: 8 min 41.7 sec
Time on main: ? sec
Total flight time: 9 min 3.7 sec

Lift off: 12:05:45.0 19:05:45.0 UTC
Apogee: 12:06:07.2 19:06:07.2 UTC
Touch down: 12:14:48.8 19:14:48.8 UTC

Launch pad coordinates: N 40° 50.0955' W 119° 8.0476'
Landing site coordinates: N 40° 51.5796' W 119° 6.8134'
Distance to landing site: 10668 feet 2.02 miles
Bearing to landing site: 32 degrees (from true north)
Ground level: 3921 feet MSL

Liftoff Apogee Landing
Battery: 4.12v 4.11v 4.09v
Temperature: 76°F 76°F 79°F

Transmitter: Mx110A0219 Firmware: 2.0.0.0

 

[BBS]

bear with me,

let us exclude fins for a moment, and only consider drag on the airframe itself. The payload has a drag coefficient that, by itself, is equal to the sustainer+payload drag coefficient when they are a single unit. So the difference in drag between the two sections can only be attributed to the fins.

so what is the drag that the fins impart?

Force(ns)= .5*Cd*p(air density)* A(frontal cross section area of the fin semi-span X number of fins)*V2(velocity squared)

For my EZI-65 with beveled 3/16” thick fins @ 700f/s the theoretical numbers for the drag imparted by the fins is:

F=.5*.05= .025*1.204kg/m3=.0301*.0067m2=.0002*213.4m/s2=9.18ns/4.45= 2.064lbs

2 pounds is insignificant on a 4” dia x 58” long rocket but it is close to the dry weight of the sustainer in this case. A light press fit using tape between the coupler and sustainer airframe tube is sufficient to keep the rocket together at burnout. I have validated this with empirical testing so I’m confident that my numbers and deductions are viable. if you double the coefficient of drag the result is only 4lbs.

 

[dixontj93060]

I know what you are saying. I experienced the same phenomenon with my Dyacom Scorpion.

I guess you don't know what I'm saying as my flight profile was nothing like the Scorpion flight you described.

In my case there was no weathercocking, the flight was not underpowered, I had no arching apogee event leading to high velocity apogee deployment; in fact, no early apogee event per se and the main event was triggered on decent per altimeter programming. Instead the Ultra Fatboy flight was perfectly vertical and upon engine burnout it visibly separated (with no charge report or smoke, also born out on altimeter data) as the nose cone hurtled ahead of the body/fin can immediately, i.e., drag separation.

 

[dixontj93060]

bear with me,

let us exclude fins for a moment, and only consider drag on the airframe itself. The payload has a drag coefficient that, by itself, is equal to the sustainer+payload drag coefficient when they are a single unit. So the difference in drag between the two sections can only be attributed to the fins.

so what is the drag that the fins impart?

Force(ns)= .5*Cd*p(air density)* A(frontal cross section area of the fin semi-span X number of fins)*V2(velocity squared)

For my EZI-65 with beveled 3/16” thick fins @ 700f/s the theoretical numbers for the drag imparted by the fins is:

F=.5*.05= .025*1.204kg/m3=.0301*.0067m2=.0002*213.4m/s2=9.18ns/4.45= 2.064lbs

2 pounds is insignificant on a 4” dia x 58” long rocket but it is close to the dry weight of the sustainer in this case. A light press fit using tape between the coupler and sustainer airframe tube is sufficient to keep the rocket together at burnout. I have validated this with empirical testing so I’m confident that my numbers and deductions are viable. if you double the coefficient of drag the result is only 4lbs.

Incorrect. What happens is that the rocket as a whole is impacted by high deceleration at burnout, and that creates a force (F=m*a, i.e., rocket mass * burnout deceleration). That force is distributed across both sections of the rocket in an equal/opposite way proportional to the drag coefficient of each section (no thrust, so free object only impacted by drag). If this force distribution is divided per the complex ratio of mass and Cd*A differences (per allocation shown in John Dermiggio's derivation) you will get a separation force that is in no way directly related to, and only by happenstance may equate to, the drag of the fins.

Further, your assumption that drag differential is only affected by the addition of fins is wrong. Drag is a combination effect of frontal area, shape, base shape eddies, skin friction, amount of skin being effected, etc.--so no, the fins' area and Cd is only a small piece of what is going on and thus the combined drag of each section needs to be considered as in the equation/relationship already referenced.

 

[BBS]

in your case, have you considered pressure differential at altitude? did you have a vent hole in the parachute compartment?

the air pressure at 7000' is ~3psi lower than at sea level and remains constant in AGL relationships. you got there pretty fast. so your 4" flat plate cross section has ~12.5 square inches X 3 psi = 37.5lbs trying to separate the nose cone from the sustainer. this is the reason we use vents in all compartments. if your rocket had no vents, i would expect the results you got. I learned vent is always a good practice, the hard way.

so what do you suppose caused the zippering on your rocket? the drag of two flat plates of the sustainer and nose cone applying the irresistible force of the shock cord accross the body tube from drag, or from the weight of the nose cone transitioning from a Cd of .05 to a Cd of .1 instantaneously while the heavier sustainer didn't want to stop? been there, done that.

 

[jderimig]

Thank goodness for Isaac Newton. His laws make what we are discussing much less complicated than some are making it.

At the joint between two rocket sections the airframe will either be in compression or tension. If its under compression then the design is drag separation safe, no retention is needed, if the joint is in tension then some form of retention is needed. Fortunately the law of dynamic motion (F=ma) are well know and can be applied to provide a simple solution. Only three types of quantities need to be known, forces, masses and acceleration. Those are the only terms I see in F=ma.

Lets do two simple thought experiments in rockets that are decelerating.
1. Consider a rocket with a single coupling, mass of the 2 sections are the same. If the drag force above the joint is greater than the drag force below the joint, then the joint will be in compression, no drag separation will occur. If the drag force below the joint is greater than above the joint, then the joint will be in tension and drag separation will occur unless there is retention that can support this tensile force.

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.

Now unfortunately we will rarely design rockets where the section masses are matched or the drag forces are matched. But fortunately Newton F=ma and a free-body-diagram provides a simple solution.

Fjoint = a [ m1 - M /(R+1) ] (Note the equation is in the form F = m a )
where m1 is the mass of the lower section and R is the drag-force ratio of the upper to lower section.

If Fjoint is positive then the joint is in compression and no special retention is needed to prevent drag separation.
If Fjoint is negative then the joint is in tension and retention must be provided to withstand the estimated value calculated from the above equation.

Derivation of the above formula is here: https://www.rocketryforum.com/showt...se-Cone-Drag-Separation&p=1550813#post1550813
Props go to Newton.

PS. Thanks Tim for reinvigorating this.

 

[BBS]

thanks John, i couldn't seem to say it as succinctly as you put it.

 

[dixontj93060]

in your case, have you considered pressure differential at altitude? did you have a vent hole in the parachute compartment?

the air pressure at 7000' is ~3psi lower than at sea level and remains constant in AGL relationships. you got there pretty fast. so your 4" flat plate cross section has ~12.5 square inches X 3 psi = 37.5lbs trying to separate the nose cone from the sustainer. this is the reason we use vents in all compartments. if your rocket had no vents, i would expect the results you got. I learned vent is always a good practice, the hard way.

so what do you suppose caused the zippering on your rocket? the drag of two flat plates of the sustainer and nose cone applying the irresistible force of the shock cord accross the body tube from drag, or from the weight of the nose cone transitioning from a Cd of .05 to a Cd of .1 instantaneously while the heavier sustainer didn't want to stop? been there, done that.

There was a vent hole in the body. I already modeled that with Dave Schultz's simple analysis (https://www.rocketryforum.com/showt...ar-forward-as-practical&p=1549085#post1549085). The body only had about 18" free space and the pressure buildup is/was minimal even with no venting. The rocket didn't get close to the sim'd 7,000 ft, more like 3,000 ft.

The nose cone Cd was/is ~0.09. Once the nose drag separated it is clearly unstable. After a bit it flipped and it had an open shoulder (which allowed access for addition of required/adjustable nose weight). Once that happened, sure the Cd was then essentially 1.0 (see bottom right illustration analysis on page 2 of this Apogee Rockets report: https://www.apogeerockets.com/education/downloads/Newsletter346.pdf). Upon flipping, the accel-based altimeter in the nose fired the apogee charge. The stable fin can continued upward until it reached the end of the shock cord and zippering ensued and the whole assembly began to descend. When down to 800 feet, the main charge fired.

 

[BBS]

So basically BBS if I understand correctly we have two forces affecting whether the rockets seperates early or not: the forward section inertia and the aft sections inertia+/-drag. If the drag is high enough or the tail section light enough that it losses velocity faster than the forward section we get a seperation. However if those two sections are pinned together then they are one piece and it is only necessary for the pins to overcome the forces placed on them by drag (for the most part, EDIT: I forgot about positive airpressure in the airframe too).

I was having a hard time accepting the theory of a fast burnout causing enough deceleration to cause separation, "as an object in motion tends to stay in motion", drag was the main cause of the separation to me, and with Vmax motors the problem with early separation is possibly insufficient airframe venting and separation due to positive airframe pressure vs the external air pressure.

Before people get to excited about what I may or may not understand properly based on the above statements, please realize A) I am not an engineer or a physicist B) I try and understand the how and why without the math, I am sure I can figure the math out but have no desire or time to do so. Feel free to pick the above statements apart, but I am trying to understand what exactly we are dealing with with terms I am more comfortable with.

Rich, you get it. you don't have to be an engineer to understand Newton you only have to be an engineer to drive a train.

 

[BBS]

There was a vent hole in the body. I already modeled that with Dave Schultz's simple analysis (https://www.rocketryforum.com/showt...ar-forward-as-practical&p=1549085#post1549085). The body only had about 18" free space and the pressure buildup is/was minimal even with no venting. The rocket didn't get close to the sim'd 7,000 ft, more like 3,000 ft.

The nose cone Cd was/is ~0.09. Once the nose drag separated it is clearly unstable. After a bit it flipped and it had an open shoulder (which allowed access for addition of required/adjustable nose weight). Once that happened, sure the Cd was then essentially 1.0 (see bottom right illustration analysis on page 2 of this Apogee Rockets report: https://www.apogeerockets.com/education/downloads/Newsletter346.pdf). Upon flipping, the accel-based altimeter in the nose fired the apogee charge. The stable fin can continued upward until it reached the end of the shock cord and zippering ensued and the whole assembly began to descend. When down to 800 feet, the main charge fired.

pressure (psi) has nothing to do with volume it has to do with surface area as in: pounds per square inch. So if your rocket went to 3000' the force from within was ~19 lbs if no effective venting was present

so maybe the vent was obstructed? it happens. without recorded data you really don't know when your separation charge fired, assuming it fired, or at what velocity, do you? it could have been before separation, and certainly could have been after separation. it is not a foregone conclusion that your early separation was due to drag separation . . . unless you are sure that the sustainer somehow ignored Newton's laws of motion (your sustaner weighed 3 lbs more than your nosecone according to your own data). a boat tail on your rocket lowers the rocket Cd and moves the CP forward. so your own conclusion that your rocket was more stable than the simulation claimed is irrational.

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. evidence would be that the parachute has traces of black powder residue from the separation charge. that would indicate that it was inside the airframe when the charge when off. a lack of this evidence is not evidence to the contrary it is simply not evidence at all.

by reading John's narrative and following the link to Tim's update, and reading his post, you should see that three of us agree on this. My understanding comes from the work i do and the hobby I pursue. Simulations are not real data they are calculations. it doesn't matter what your simulations are telling you if the reality conflicts. and it does in your example. simulations are only telling you what you can expect to happen in a perfect world given nominal, baseline, data. your actual flight was not within the parameters of the perfect world in which your simulations were run.

We all want an explanation for a result we didn't anticipate. That is why engineers can't assume that we understand what happened by {edit} 'simple' observation, but should only base our conclusions on physical evidence. observation is not evidence it is perception, ask any lawyer. none of us want to be wrong in our conclusions but we often are, in the absence of actual data. interpretation of data is an argument that all engineers have with other engineers sooner or later.

 

[BBS]

good exchange, i enjoyed that

 

[dixontj93060]

pressure (psi) has nothing to do with volume it has to do with surface area as in: pounds per square inch. So if your rocket went to 3000' the force from within was ~19 lbs if no effective venting was present.

Clearly you don't know what you are talking about. The enclosed compartment, it's volume, builds up pressure as the rocket ascends (see: https://home.earthlink.net/~david.schultz/rnd/pressurelag/parachute.html). That pressure then applies a psi to the bulkhead--but maybe you are too dense to follow real math.

AND THEN WORSE you flip-flop your whole viewpoint once John D. posts his drag force analysis but before that you were spewing crap that was completely off base and incorrect.

You only have 14 posts and clearly have no background in any kind in science. Stop posting your B.S. You aren't providing any value.

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.

This was already explained--can you READ? I said it was tested by "your golden shake test" and it was at least 11 to 17 lbs of holding force. That was not sufficient based on the real drag separation force of 41 lbs which was also posted--can you read a spreadsheet?

...without recorded data you really don't know when your separation charge fired, assuming it fired, or at what velocity, do you?

Again, not reading... Post #56 provided detailed sequencing of events and it was noted I was using an accel-based altimeter. Where did that come from?; the altimeter (DUH).