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Shear pin sizing on a NON supersonic single event rocket.

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jahall4

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I’m sure this has been hashed over before, but since there is always neat little new products coming out I’d thought I ask.

In a 4” all G12/G10 rocket, is there any reason to use more than 3 #2 nylon shear pins? 3 #4 maybe?

Note the rocket is around 5lbs wet and will not be flown supersonic also it is NOT Dual deploy, single event only.

Also, anyone sell some type of “shear plate”
 

dixontj93060

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Really?! This kind of question belies your inexperience.
 

DavidMcCann

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single event?

No plate, no pins. If you pick it up by the nose and it comes right off, tape on the shoulder until it doesn't
 
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jahall4

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single event?

No plate, no pins. If you pick it up by the nose and it comes right off, tape on the shoulder until it doesn't
Yep, normally I would, the "slit" scotch tape technique has worked well for me even on 4" rockets, but this one is different in that aft section is smaller (and may end up lighter) than the forward section and my first all FG rocket. So I'm understandably cautious.
 

cerving

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Are you flying with electronics (i.e. something in the nose)? If not, there's no easy and reliable way to test the ejection charge, so I vote for the taped-nose method, with one small vent hole in the body tube. If you ARE flying with electronics, you can drill the tube/NC for two #2 shear pins and do a ground test. If it doesn't work, then you're out nothing, and you can always go back and use tape.
 

jahall4

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Are you flying with electronics (i.e. something in the nose)? If not, there's no easy and reliable way to test the ejection charge, so I vote for the taped-nose method, with one small vent hole in the body tube. If you ARE flying with electronics, you can drill the tube/NC for two #2 shear pins and do a ground test. If it doesn't work, then you're out nothing, and you can always go back and use tape.
My fault, I wasn't clear. My concern is drag separation. Yes, there is an avbay in the nose which contributes to my separation concerns.

BTW... I have simulated motor "ejection charges" it was actually very easy and I think reliable. If you're interested I'd be glad to start another thread?
 

smapdiage9

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Shear pins have a relatively known breaking force, documented on a few websites. By simulating the rocket's flight, and using the mass of all the retained components (NC/Chute/Laundry/Avionics?) multiplied by the maximum deceleration force (most people overestimate by using the flight's maximum acceleration) you can come up with how many pins would be required to prevent premature separation. Then using the volume of your tube to be pressurized by ejection charge (again, a few websites to calculate this) and the area of your nosecone end or bulkplate you can figure out the size of charge necessary to overcome that many pins. That said I'm sure most people's 4" or smaller rockets use 3x 2-56 nylon screws and function just fine.
 

jahall4

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Yes. Thanks James! I was just trying to get in the ball park and also ask about any neat little products that might be available. I suspect that the reason most use #2s is that you can't get any smaller. That's why I often just use tape, but as I mentioned this rocket is unusual.

You lost me on “mass … multiplied by the maximum deceleration force” since the force of deceleration on the aft section of the rocket is what we are trying to determine, is it not? Why would we multiply it by the mass which is already part of the equation?
 

Bat-mite

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I needed help with this some time ago, and it resulted in a massive debate that even, in some cases, degenerated into name-calling. Here is a summary. I hope it is helpful.

Normally, when doing dual deployment, you have to calculate for drag separation of the booster, and nose cone retention when the drogue charge fires. In your case, it is much simpler.

You need to know the weight of the NC (or the half above the separation point). You are trying to retain the nose cone after motor burnout, so that the nose doesn't continue upward while the booster or airframe starts to get pulled downward by gravity.

You also need to know the rocket's acceleration at burnout. You can get this from a sim. If you are not using OR or Rocksim, do a basic sim on ThrustCurve.org and use the maximum acceleration given for your motor.

Then take a worst case scenario in which, at burnout, the booster immediate decelerates to zero and begins to fall. The question is: how much force is required to retain the nose cone so that its momentum doesn't keep it going upward?

The equation to use is Force = mass (or we'll use weight) X acceleration; or, F = ma. You know the 'm' from weighing the NC, and the 'a' from your sim.

For our example, let's say the NC weighs 1.2 pounds, and the maximum acceleration is 8G. Subtract 1G from the acceleration, since the acceleration of a static velocity is 1. So, we get F = 1.2 x 7, or F = 8.4.

This means you need to apply 8.4 pounds of force to retain the NC. If you look online at the shear strength of whatever you are choosing to use for shear pins, take the minimum value, since you can't afford to assume that you will get a higher value.

6/6 nylon #2-56 screws have a minimum shear strength of 31 pounds. Since you only need to retain 8.4 pounds, you can see that one #2-56 nylon screw is way more than sufficient to prevent drag sep. of your NC.

With one screw, however, there is the chance of binding, since the pressure is applied unevenly. IOW, one side of the NC is free to fly, while the other is momentarily bound, potentially causing the NC shoulder (or coupler) to get cock-eyed in there.

There are many people who will simply go with one screw. Personally, I am comfortable going with two, while others will insist on three.

Thus, the next question is: when I do want the NC to separate (at apogee), how many pins can my BP charge break?

To answer that, use one of the online BP charge size calculators. Put in your diameter, tube length, and desired pressure, and it will calculate the required charge size and also tell you how many shear pins that size of charge can break.

Try the equation with your actual values. If you are using a high thrust motor like a VMAX, maybe your max acceleration is much higher, like 22 G. Even so, if your NC is 1.2 lb., you only need one pin.

Unless your shoulder is super loose, binding is very unlikely, and you can get away with one pin easily. But #2-56 screws are cheap, and if you are more comfortable with two or three, go for it.

A final note. FG is more than capable of cleanly breaking nylon screws, so you don't need any shim or anything. You will have to decide whether or not to tap them, or just drill clean holes and shove them in. What I do is tap the inner tube (in this case, your NC shoulder), and drill the outer hole slight larger. Then I screw them in. Many others will tell you this is a pain in the butt, and they like to tap theirs in with a tack hammer, or something.

Come back and let us know how it goes.
 

mpitfield

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I needed help with this some time ago, and it resulted in a massive debate that even, in some cases, degenerated into name-calling. Here is a summary. I hope it is helpful.

Normally, when doing dual deployment, you have to calculate for drag separation of the booster, and nose cone retention when the drogue charge fires. In your case, it is much simpler.

You need to know the weight of the NC (or the half above the separation point). You are trying to retain the nose cone after motor burnout, so that the nose doesn't continue upward while the booster or airframe starts to get pulled downward by gravity.

You also need to know the rocket's acceleration at burnout. You can get this from a sim. If you are not using OR or Rocksim, do a basic sim on ThrustCurve.org and use the maximum acceleration given for your motor.

Then take a worst case scenario in which, at burnout, the booster immediate decelerates to zero and begins to fall. The question is: how much force is required to retain the nose cone so that its momentum doesn't keep it going upward?

The equation to use is Force = mass (or we'll use weight) X acceleration; or, F = ma. You know the 'm' from weighing the NC, and the 'a' from your sim.

For our example, let's say the NC weighs 1.2 pounds, and the maximum acceleration is 8G. Subtract 1G from the acceleration, since the acceleration of a static velocity is 1. So, we get F = 1.2 x 7, or F = 8.4.

This means you need to apply 8.4 pounds of force to retain the NC. If you look online at the shear strength of whatever you are choosing to use for shear pins, take the minimum value, since you can't afford to assume that you will get a higher value.

6/6 nylon #2-56 screws have a minimum shear strength of 31 pounds. Since you only need to retain 8.4 pounds, you can see that one #2-56 nylon screw is way more than sufficient to prevent drag sep. of your NC.

With one screw, however, there is the chance of binding, since the pressure is applied unevenly. IOW, one side of the NC is free to fly, while the other is momentarily bound, potentially causing the NC shoulder (or coupler) to get cock-eyed in there.

There are many people who will simply go with one screw. Personally, I am comfortable going with two, while others will insist on three.

Thus, the next question is: when I do want the NC to separate (at apogee), how many pins can my BP charge break?

To answer that, use one of the online BP charge size calculators. Put in your diameter, tube length, and desired pressure, and it will calculate the required charge size and also tell you how many shear pins that size of charge can break.

Try the equation with your actual values. If you are using a high thrust motor like a VMAX, maybe your max acceleration is much higher, like 22 G. Even so, if your NC is 1.2 lb., you only need one pin.

Unless your shoulder is super loose, binding is very unlikely, and you can get away with one pin easily. But #2-56 screws are cheap, and if you are more comfortable with two or three, go for it.

A final note. FG is more than capable of cleanly breaking nylon screws, so you don't need any shim or anything. You will have to decide whether or not to tap them, or just drill clean holes and shove them in. What I do is tap the inner tube (in this case, your NC shoulder), and drill the outer hole slight larger. Then I screw them in. Many others will tell you this is a pain in the butt, and they like to tap theirs in with a tack hammer, or something.

Come back and let us know how it goes.
Very helpful reply...I would also add that you should do ground testing if possible.
 

jahall4

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Then take a worst case scenario in which, at burnout, the booster immediate decelerates to zero and begins to fall. The question is: how much force is required to retain the nose cone so that its momentum doesn't keep it going upward? The equation to use is Force = mass (or we'll use weight) X acceleration; or, F = ma. You know the 'm' from weighing the NC, and the 'a' from your sim. For our example, let's say the NC weighs 1.2 pounds, and the maximum acceleration is 8G. Subtract 1G from the acceleration, since the acceleration of a static velocity is 1. So, we get F = 1.2 x 7, or F = 8.4. This means you need to apply 8.4 pounds of force to retain the NC. If you look online at the shear strength of whatever you are choosing to use for shear pins, take the minimum value, since you can't afford to assume that you will get a higher value.
So... in your calculation you're assuming the aft section's acceleration is 0 zero as opposed to the acceleration (or rather deceleration) delta (and resulting force) between fore and aft sections?

Regardless, based on your explanation, more than couple of #2 is overkill.
 
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Bat-mite

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You're taking a worst-case scenario so that you are covered under any unforeseen conditions, and all types of motors. There is obviously a maximum number of pins that can be shreared by a deployment charge, and you don't want to exceed that. And, again, there is a minimum number of pins you need to prevent separation at burnout, if the booster stops dead or burnout. Is that likely? Of course not. Will it ever happen? Probably not. But by finding your end points, you can pick a comfortable number and size of pins that will cover you in all scenarios.
 

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Great explanation, Bat-mite.

Here is how I rigged a cutting plate for my sheer pins. The NC was pretty loose, so there was plenty of room for a wrap of redneck aluminum that was held down with epoxy. (Note that the large hole in the BT below is for a plastic rivet that was used for a previous single deploy flight. The small hole for the shear pin is hard to see.)

IMG_1351.jpg
 

Bat-mite

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Let me add, too, that in a dual deployment situation, you are calculating for drag separation of the upper half of the rocket (payload and NC), and thus the 'm' will be the weight of the entire upper half at burnout.

The other critical calculation is for nose cone retention, not at burnout, but at apogee deployment. In this case, the NC needs to be retained in the payload tube when the drogue charge goes off. Now, for the 'a' value, you are not using the rocket's motor acceleration, but the acceleration created by the apogee event.

You may get something extremely high like 110 G at apogee. So if we had the 1.2 pound NC from the previous example, we're now coming up with 132 pounds force. Now you are needing four #2-56 pins in the nose.
 

jahall4

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Great explanation, Bat-mite.

Here is how I rigged a cutting plate for my sheer pins. The NC was pretty loose, so there was plenty of room for a wrap of redneck aluminum that was held down with epoxy. (Note that the large hole in the BT below is for a plastic rivet that was used for a previous single deploy flight. The small hole for the shear pin is hard to see.)

View attachment 297764
Mountain Dew? :)
 

jahall4

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Let me add, too, that in a dual deployment situation, you are calculating for drag separation of the upper half of the rocket (payload and NC), and thus the 'm' will be the weight of the entire upper half at burnout.

The other critical calculation is for nose cone retention, not at burnout, but at apogee deployment. In this case, the NC needs to be retained in the payload tube when the drogue charge goes off. Now, for the 'a' value, you are not using the rocket's motor acceleration, but the acceleration created by the apogee event.

You may get something extremely high like 110 G at apogee. So if we had the 1.2 pound NC from the previous example, we're now coming up with 132 pounds force. Now you are needing four #2-56 pins in the nose.
Certainly, but as I mentioned single event. Really all I'm trying to do is ensure the aft section won't separate when engine completes its burn. Tape would probably be adequate, but I thought I would give shear pins a try since it was a FG air-frame. I'll probably drill 3 #2 and test with just the 1 pin

Thanks for you help, good common sense explanation.
 

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You may get something extremely high like 110 G at apogee. So if we had the 1.2 pound NC from the previous example, we're now coming up with 132 pounds force. Now you are needing four #2-56 pins in the nose.
No, that is not the separation force. The separation force due to inertia is the acceleration times the difference in mass between the NC and lower section of the rocket. If the lower section of the rocket has higher mass than the NC then there will be no separation force (which is the most common case).

Just light friction tape is all that is necessary. Why risk a ballistic recovery? Don't use shear pins.
 

Bat-mite

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No, that is not the separation force. The separation force due to inertia is the acceleration times the difference in mass between the NC and lower section of the rocket. If the lower section of the rocket has higher mass than the NC then there will be no separation force (which is the most common case).

Just light friction tape is all that is necessary. Why risk a ballistic recovery? Don't use shear pins.
You lost me, John. Are you going back to the OP using single deployment? The post of mine that you quoted was regarding DD, in which case in a FG rocket, the NC is almost always heavier than the payload tube.
 

jderimig

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You lost me, John. Are you going back to the OP using single deployment? The post of mine that you quoted was regarding DD, in which case in a FG rocket, the NC is almost always heavier than the payload tube.
Ah, yes I was just thinking NC and rest of the rocket. Yes indeed, for a nosecone on an av-bay that shock cord jerk can be quite high and that is a different calculation (like yours....). Sorry for the confusion.
 

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if you're trying to get super technical with separation forces you probably also want to consider the drag difference between the retained and retaining halves of the rocket since we're not coasting in a vacuum, and the volume of ground level air inside the airframe versus the vent hole area and the magnitude and rate of altitude gain/pressure diff.
 

jahall4

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The separation force due to inertia is the acceleration times the difference in mass between the NC and lower section of the rocket.
... but that does not account for the additional drag force. Seems to me it's all about the delta in deceleration. If the aft section decelerates faster than to forward section you risk separation. The trick obviously is determining the max deceleration of the fore and aft sections independent of each other. Keep in mind that may not be at the end of the burn depending on the thrust curve :).
 

jahall4

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if you're trying to get super technical with separation forces you probably also want to consider the drag difference between the retained and retaining halves of the rocket since we're not coasting in a vacuum
Exactly. like I said already, my concern is a smaller lighter tail section.
 

Bat-mite

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Exactly. like I said already, my concern is a smaller lighter tail section.
I hope you find the exact numbers you are looking for. I simply tried to show that any number of shear pins between the worst-case minimum and the maximum that can be sheared for a given charge are sufficient.

There is the old method for a taped friction fit NC that if you hold the rocket by the NC and the airframe doesn't fall off, it is secure. One could do the same with shear pins.

Any science beyond that is way over my head.
 

jderimig

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... but that does not account for the additional drag force. Seems to me it's all about the delta in deceleration. If the aft section decelerates faster than to forward section you risk separation. The trick obviously is determining the max deceleration of the fore and aft sections independent of each other. Keep in mind that may not be at the end of the burn depending on the thrust curve :).
Its the drag force that creates the deceleration. There could be a delta drag force difference between the sections across the coupler but usually this is small compared to the mass differences. The fore and aft sections do not have different decelerations unless they are separating.

Here is the math and analysis to calculate the force required to keep the sections together. If Fsep is calcuated to be >0 then they will not separate.

http://www.rocketryforum.com/showth...se-Cone-Drag-Separation&p=1550813#post1550813
 

jahall4

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Its the drag force that creates the deceleration. There could be a delta drag force difference between the sections across the coupler but usually this is small compared to the mass differences. The fore and aft sections do not have different decelerations unless they are separating.
Exactly, which is my concern. You make mention in your analysis (which was the direction my thinking was heading) that: "... Cd*A ... Both easily obtained from Rocksim".... where/how do I get the Cd for the fore and aft sections separately?
 

jderimig

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Exactly, which is my concern. You make mention in your analysis (which was the direction my thinking was heading) that: "... Cd*A ... Both easily obtained from Rocksim".... where/how do I get the Cd for the fore and aft sections separately?
In OpenRocket its in the "Component Analysis" menu. In Rocksim its in Cd Analysis somewhere, its been awhile since I have used RS.
 

jahall4

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In OpenRocket its in the "Component Analysis" menu. In Rocksim its in Cd Analysis somewhere, its been awhile since I have used RS.
Yep, Problem is it groups the nose and body together so you don't know what % is just the nose.
2016-07-26_16-49-34.jpg
 
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