Shear pin retention stregth. Check my math.

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Bat-mite

Rocketeer in MD
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Okay, please tell me if I am on track, here.

My Radial Flyer's burnout weight with an M1101 is 10.5#. The max acceleration simmed is 17.8G. Taking a worst case scenario, the maximum separation force on the rocket would be F = MA, or 186.9.

I have a ton of #2-56 nylon screws already, and would prefer to use them. These screws have a shear strength of 30 - 45. If I assume the worst case 30, then 186.9 / 30 = 6.23, meaning I would need roughly six #2-56 pins. In the best case, it comes to 186.9 / 45 = 4.15.

So I am looking at anywhere from four to six #2-56 screws. Am I right so far?

Now, here is another question. My RW X-Celerator has a burnout weight of 13.6 on a J1520. It pulls a max of 25.3G. That gives me an F of 343.1. 343.1 / 45 (best case) is 7.6 shear pins. But I have never used more than two, and never gotten early separation. Since this is a VMAX, that means that I pass through the heaviest G forces during motor burn, and by the time of burnout, the G Force is very low.

So if we say ~40 per #2-56 pin, at two pins, I have 80# of shear strength. Since mass is constant at 13.6, then my G force after burnout must really be about 5.9. Right?

And with that in mind, how many pins do I really need for my Radial Flyer? The M1101W is a longer burn-time motor, so I need to look at my simulations and see what the actual maximum acceleration is after burnout, then use that as my A factor to get the real number of pins.

Make sense?
 
Not exactly. You want to look at the mass of the part of the rocket you are trying to retain. For instance if your NC is only 1.5lbs of the 10.5lbs, then that is what you are trying to secure in a nominal recovery event. Problem is it is hard to determine nominal and in fact most design to worst case anomaly events. So instead of your 17.8gees, I'd use 3X to 5X that, say 90gees. So your retention needs for the NC in the example scenario above would be no more than 145lbf.

I'll leave it as a homework assignment to figure out the retention needs for the booster :). Hint: it again isn't the full weight of the rocket and max forces there happen at a different time in the flight profile and are generally more deterministic.
 
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I was actually thinking booster. The rocket has a very light polystyrene NC, and I doubt needs more than one pin, although I will use two.

So, by your reasoning above, I need the weight of everything thing above the booster separation point. And then, would I be looking at the G force at the apogee event?
 
The shear pins only need to secure the weight of the components above the break line. The other thing is that they are only needed to prevent drag separation. You're not going to get 90g of drag on anything we fly. The rule I learned is that you should be able to pick up the rocket by the nose cone and not have it come apart. I'd use 2 or 3 2-56 pins on your nose cones and be done with it.
 
The shear pins only need to secure the weight of the components above the break line. The other thing is that they are only needed to prevent drag separation. You're not going to get 90g of drag on anything we fly. The rule I learned is that you should be able to pick up the rocket by the nose cone and not have it come apart. I'd use 2 or 3 2-56 pins on your nose cones and be done with it.

I hear ya! But if my L3CC asks me how I arrived at the number of pins, I want to be able to give an intelligent-sounding answer.
 
I WISH IT WAS REQURIED THAT EVERY TAP/L3CC ASKS ABOUT SHEER PINS!

This is an L2 skill that should be mastered before starting an L3 project.
It kills me to see these threads.
Instant FAIL, IMHO.
 
The shear pins only need to secure the weight of the components above the break line. The other thing is that they are only needed to prevent drag separation. You're not going to get 90g of drag on anything we fly. The rule I learned is that you should be able to pick up the rocket by the nose cone and not have it come apart. I'd use 2 or 3 2-56 pins on your nose cones and be done with it.

OK, well if you are going to give him the answer...

The 90gees is for a violent recovery event at apogee and making sure you don't loose the nosecone and deploy the main (and a lot of people I notice loose their main laundry because of not understanding this).

Drag separation forces apply mostly at burnout and affect the booster retention point.
 
Fred,

This is your most common reply on TRF. Why don't you spend some time teaching people how to do this instead of bashing them? Make yourself useful instead of annoying. I'm all ears if you want to help me.
 
Okay, please tell me if I am on track, here.

My Radial Flyer's burnout weight with an M1101 is 10.5#. The max acceleration simmed is 17.8G. Taking a worst case scenario, the maximum separation force on the rocket would be F = MA, or 186.9.

I have a ton of #2-56 nylon screws already, and would prefer to use them. These screws have a shear strength of 30 - 45. If I assume the worst case 30, then 186.9 / 30 = 6.23, meaning I would need roughly six #2-56 pins. In the best case, it comes to 186.9 / 45 = 4.15.

So I am looking at anywhere from four to six #2-56 screws. Am I right so far?

Now, here is another question. My RW X-Celerator has a burnout weight of 13.6 on a J1520. It pulls a max of 25.3G. That gives me an F of 343.1. 343.1 / 45 (best case) is 7.6 shear pins. But I have never used more than two, and never gotten early separation. Since this is a VMAX, that means that I pass through the heaviest G forces during motor burn, and by the time of burnout, the G Force is very low.

So if we say ~40 per #2-56 pin, at two pins, I have 80# of shear strength. Since mass is constant at 13.6, then my G force after burnout must really be about 5.9. Right?

And with that in mind, how many pins do I really need for my Radial Flyer? The M1101W is a longer burn-time motor, so I need to look at my simulations and see what the actual maximum acceleration is after burnout, then use that as my A factor to get the real number of pins.

Make sense?

John,,
Be careful here..
The way you worded this it's incorrect..
Especially if the load is a "V Max"...
You will sustain the heaviest positive G's at motor ignition...
And then will sustain the heaviest negative G's on motor burnout.....
In your minds eye envision an extremely light rocket flying on a "V Max" load...
The negative G's from motor burnout can be quite substantial to say the least.....

Teddy
 
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Without doing the math, six #2-56 pins for a 10 lb rocket sounds like way too many. I think three would be plenty.

...My RW X-Celerator has a burnout weight of 13.6 on a J1520. It pulls a max of 25.3G. That gives me an F of 343.1. 343.1 / 45 (best case) is 7.6 shear pins. But I have never used more than two, and never gotten early separation. Since this is a VMAX, that means that I pass through the heaviest G forces during motor burn, and by the time of burnout, the G Force is very low...

Really, flight data from my Raven altimeters always show that max G force is from the ejection charge at either apogee or at main deployment. I have seen over 50 Gs acceleration at ejection charge. Once when I had a CATO and the main deployed at over 300 mph the max acceleration was about 90 Gs. Didn't hurt the chute but the Blue Tube airframe got a really bad zipper.
 
I have a ton of #2-56 nylon screws already, and would prefer to use them. These screws have a shear strength of 30 - 45.
Make sense?

FYI:

Years ago I saved this reference from Doc's strength of materials website & keep it for my own reference.

Screen Shot 2016-02-03 at 3.24.12 PM.png

The finding are significantly different than yours. These pins were from Mac-Carr. I don't know where you got your figures,but if they are confirmed, you should be OK.
You may wish to double check the stats on yours to be safe.

On another note.... after saving data from my altimeters over the years, I discovered the most force applied during my flights was from ejection charges.
Some over 75 G's......since tabulating. Far higher then any from high thrust motors.
It helped me lower my charge weight considerably & still maintain "healthy" separations.
It has entirely eliminated issues at apogee , where I was reaching full shock cord extension and occasionally had main come out at the top.
Previously I adhered to the "blow it up or blow it out" camp. A bit of data collecting "wised" me up & saved me from long walks.....LOL
 
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John, if you have not yet done so, review the comments on the other current shear pin thread - no need to repeat it here. Overall there is a good discussion of why we use shear pins, what to consider, how to measure the factors, etc. It is important to understand the 'Why', not just the how. Ultimately the responsibility for a safe flight is on the flier's shoulders.

Have fun, and be safe.
 
I have read the other threads. I want to understand the math and do it correctly. But there seems to be some kind of secret club where the people who know how to do it won't share their knowledge with others.

I have many flights with shear pins under my belt. I know that they are needed to prevent drag separation of the booster, and prevent premature ejection of the nose cone at the apogee event. I have never had drag separation, and I only once had a premature NC ejection (and that was because I left my shear pins home and thought I could get away with friction fit).

I am 100% confident that three pins in the booster is sufficient, and two in the NC is sufficient. I am 75% confident that I could use two in the booster.

I can follow the equations presented here for NC retention, no problem. What I cannot figure out how to do is calculate drag separation forces. I have looked at many web sites, posts here on TRF, and in books. I have seen equations that require a drag coefficient, but I also know that the drag coefficient is constantly changing.

If no one "in the know" wants to teach me or share his information, I am 100% confident that shear pins will not be the thing that causes me to fail my L3 attempt.

I hope the fact that I am even asking is proof that I want to know how to do this mathematically instead of just feeling my way through. But in the end, feeling my way through hasn't let me down yet.
 
I WISH IT WAS REQURIED THAT EVERY TAP/L3CC ASKS ABOUT SHEER PINS!

This is an L2 skill that should be mastered before starting an L3 project.
It kills me to see these threads.
Instant FAIL, IMHO.


I agree that it should be asked why by TAP/L3CC in the construction process before the build.
I almost think it should be a part of the level 2 cert as well.

However, from everything I have been researching, I can't find stead fast answer.
I have found all kinds of charts, and black powder calculators out there.
But nothing say why.

I have been told, if you can pick up the rocket by the nose cone then it shouldn't drag separate.
which if you put a vmax motor in it, the neg g force on burn out can be crazy which will void out the statement above.

I guess the real answer is trial and error on your level 2 flight, and make a good guess on your level 3.
 
Drag separation is the hardest to model since you must know the drag on the parts of the rocket on each side of the separation point.
Drag sep occurs [usually] because the drag force on the fins is more than the drag on the NC by an amount large enough to overcome the retention (if any).

How to get a handle on drag-sep?
First - rule of thumb is that 1/3 of the drag is on the NC, one third on the fins and the rest on the body, especially the rail buttons.
Now, this is a ROT -- if you have giant fins, they will have more drag....and get you in trouble faster.
What you really need to get a handle on is the "differential drag" across you separation point. If the difference is zero, you are done.

If there is non-zero drag difference, then you need to understand the forces involved.
In most rockets, the bulk of the body stays with the fins so the NC section usually has less drag than the remainder of the rocket.
Drag sep occurs at burn out. Where the drag is highest and the deceleration peaks.

The [de] acceleration at burn out is a function of how fast the motor shuts down - again, hard to model.
Assume a hard shutoff from full thrust and you can't go wrong.
So, for example, if you motor is pulling 10 G's at burnout. Has a crisp shutdown. The deceleration is assumed to be -9G's.

Those 9G's are ALL DRAG.
If the drag is equalized front to back, no separating forces are generated.
If the fins produce twice the drag of the NC, then the fins will generate about 3G's more deceleration force than the nose.
Take that 3G's and the mass of the forward section and using F=MA, figure the force pulling the rocket apart.
Provide retention for that force.

So -- as you can see - hanging a rocket upside down and seeing if the NC falls off is NOT SUFFICIENT by any means.

To summarize:
Assume 1/3:2/3 ratio for front-to-back drag as worst case [unless you have huge fins]
Assume a crisp shutdown from 100% thrust for maximum deceleration.
Calculate separation force using ~1/3 of the deceleration force (drag difference) times the mass retained.
Calculate the proper retention to survive this force.
Find the proper separation charge to remove the retention.
 
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So -- as you can see - hanging a rocket upside down and seeing if the NC falls off is NOT SUFFICIENT by any means.

Especially on rockets like this one... Where thick, squarish fins went up against a nosecone with nearly 3lbs of weight. The subsequent drag separation and zipper at LDRS30 gave me more impetus to validate my shear pins on every design.

8250220470_81a1992ef9_z.jpg
 
John, if you look at the RocketMaterials link (via the internet archive), you will notice the data and argument for not using only two shear pins. I can vouch for the "3 or more" argument, having had inconsistent deployment with two shear pins.
 
This thread has been very informative. Nothing I need yet, but for when I do, thanks!
 
To summarize:
Assume 1/3:2/3 ratio for front-to-back drag as worst case [unless you have huge fins]
Assume a crisp shutdown from 100% thrust for maximum deceleration.
Calculate separation force using ~1/3 of the deceleration force (drag difference) times the mass retained.
Calculate the proper retention to survive this force.
Find the proper separation charge to remove the retention.

Okay, so ... my maximum acceleration shown in a simulation is 17.8 G. That gives me a -16.8 G deceleration force. One third of that is 5.6 G. The mass retained is my booster to aft, which comes in at about 9.8 pounds. F = 9.8 * 5.6 = 54.88, or ~55.

A #2-56 shear pin will retain anywhere from 36 to 45 pounds. Thus two shear pins is sufficient for retention of the booster after burnout. That is consistent with what I have seen in a similarly sized rocket.

Using calculators online, it appears that 1.9 grams of BP is required for the volume of the booster above the top CR. Calculators also reveal that this is sufficient to break up to five #2-56 pins.

Since there is some question as to whether or not one gets an even separation with one or two pins, I can step up to three pins without upgrading my ejection charge size. Ground testing will demonstrate the efficiency of this charge. The back-up charge will be sized 15% larger than the primary.

As for nose cone retention, it is a light, polystyrene cone with a 3/8" diameter eye bolt. It is less than a pound. I have no data to calculate the number of Gs at apogee ejection since this is an unflown rocket. But if I use 50 G as a middle-of-the-road figure, and one pound for the NC, then F = 49, and two pins are sufficient. Again, I could step up to three if I want to ensure a more even breakage.

Thanks, Fred. If I have misunderstood or misrepresented you, please let me know.
 
Okay, so ... my maximum acceleration shown in a simulation is 17.8 G. That gives me a -16.8 G deceleration force. One third of that is 5.6 G. The mass retained is my booster to aft, which comes in at about 9.8 pounds. F = 9.8 * 5.6 = 54.88, or ~55.

A #2-56 shear pin will retain anywhere from 36 to 45 pounds. Thus two shear pins is sufficient for retention of the booster after burnout. That is consistent with what I have seen in a similarly sized rocket.

Using calculators online, it appears that 1.9 grams of BP is required for the volume of the booster above the top CR. Calculators also reveal that this is sufficient to break up to five #2-56 pins.

Since there is some question as to whether or not one gets an even separation with one or two pins, I can step up to three pins without upgrading my ejection charge size. Ground testing will demonstrate the efficiency of this charge. The back-up charge will be sized 15% larger than the primary.

As for nose cone retention, it is a light, polystyrene cone with a 3/8" diameter eye bolt. It is less than a pound. I have no data to calculate the number of Gs at apogee ejection since this is an unflown rocket. But if I use 50 G as a middle-of-the-road figure, and one pound for the NC, then F = 49, and two pins are sufficient. Again, I could step up to three if I want to ensure a more even breakage.

Thanks, Fred. If I have misunderstood or misrepresented you, please let me know.

Tally-ho :horse:
 
Your "retained mass" is [usually] the mass FORWARD of the separation point.
If I read correctly, you used the AFT mass.
But otherwise you got the idea.

Also - don't forget to ADD any pressure imbalance in un-vented bays as more force against your retainers.
For DD rockets, check both separation points the same way with the adjusted "forward" mass.


And a note about the "hang upside down" check.
This works for many cases because:
Flipping over the rocket tests the retention to 1G.
If you go with the 1/3:2/3 ratio of drag, then you need -3G of drag force to generate 1G of separation force.
If you have a soft-shutdown motor, as many are, odds are you won't get a 3G shutdown impulse.
A Warp-9 shutdown....goodbye NC.....
[This is an approximation....but you get the idea, I hope.]
 
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Yes, the KEY is to figure the drag ratio - then you need to USE that ratio to estimate the force on the retainer. Which is a function of the mass ahead of the separation point (where the drag differential is applied) and the de-acceleration peak impulse due to motor turn-off.
 
OK, I'll stick my neck out. Can't help myself.

When a motor burns out, there is NO deceleration related to the motors thrust. ( OK a slight change related to the increase base drag without the motor exhaust) The motor burn itself has NOTHING to do with the deceleration after burnout. Think of a rocket in space. As soon as the engine shuts down you are weightless, instant zero G. No backwards "deceleration" or "de-acceleration" No unattached parts flying apart.
The drag is only aerodynamic, related to the velocity, the air density and the drag coefficients of the rocket parts.

There is also no force on the shear pins during boost the way most rockets are built. The nose cone and each body tube is resting on the tube below it. The shear pin holes are drilled with the tubes in contact, so each tube supports the next and no force is on the pins at all.

The differential drag on the front vs rear parts of the rocket can cause drag separation as above.
 
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OK, I'll stick my neck out. Can't help myself.

When a motor burns out, there is NO deceleration related to the motors thrust. ( OK a slight change related to the increase base drag without the motor exhaust) The motor burn itself has NOTHING to do with the deceleration after burnout. Think of a rocket in space. As soon as the engine shuts down you are weightless, instant zero G. No backwards "deceleration" or "de-acceleration" No unattached parts flying apart.
The drag is only aerodynamic, related to the velocity, the air density and the drag coefficients of the rocket parts.

There is also no force on the shear pins during boost the way most rockets are built. The nose cone and each body tube is resting on the tube below it. The shear pin holes are drilled with the tubes in contact, so each tube supports the next and no force is on the pins at all.

The differential drag on the front vs rear parts of the rocket can cause drag separation as above.

Good luck with this Dan,,,
Let me know how this works out for ya,,, lol...

Teddy
 
OK, I'll stick my neck out. Can't help myself.

When a motor burns out, there is NO deceleration related to the motors thrust. ( OK a slight change related to the increase base drag without the motor exhaust) The motor burn itself has NOTHING to do with the deceleration after burnout. Think of a rocket in space. As soon as the engine shuts down you are weightless, instant zero G. No backwards "deceleration" or "de-acceleration" No unattached parts flying apart.
The drag is only aerodynamic, related to the velocity, the air density and the drag coefficients of the rocket parts.

There is also no force on the shear pins during boost the way most rockets are built. The nose cone and each body tube is resting on the tube below it. The shear pin holes are drilled with the tubes in contact, so each tube supports the next and no force is on the pins at all.

The differential drag on the front vs rear parts of the rocket can cause drag separation as above.

Dan, it is forces derived from momentum and drag,

Force = Momentum / Time and,​
Momentum = Mass * Velocity​

So you have both an upward force (from momentum during acceleration change) and downward force (from drag moving through air) acting on each section of the rocket. Ignoring gravity, drag separation occurs when you have a heavy nose section with less drag which will want to pull away from a lighter, more draggy boost section which happens in the short time that the acceleration changes.
 
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