# Thread: Three 4-40 nylon shear pin (screws) right amount for 10 pound fiberglass rocket?

1. Originally Posted by r66astro
what we need is exploding shear pins! why not?
Sounds like a Mythbusters episode! I'm guessing your screw holes would get a bit bigger after the first boom.

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Originally Posted by Steve Shannon
Disregarding the difference between static friction and dynamic friction, it's exactly the same.
It isn't just static vs dynamic friction I'm pretty sure. It takes force to stop a moving object (the nose cone). The amount of course depends on the nose cone mass, velocity when the shear pin hits the edge, and shear pin stiffness, but I could imagine it being a meaningful effect since ejection charges accelerate the nose cone at 20g to 100g in many rockets so the velocity rises quickly, and that force is strictly added to the force applied by the ejection charge toward shearing the pin.

3. Originally Posted by rocketsam2016
It isn't just static vs dynamic friction I'm pretty sure. It takes force to stop a moving object (the nose cone). The amount of course depends on the nose cone mass, velocity when the shear pin hits the edge, and shear pin stiffness, but I could imagine it being a meaningful effect since ejection charges accelerate the nose cone at 20g to 100g in many rockets so the velocity rises quickly, and that force is strictly added to the force applied by the ejection charge toward shearing the pin.
Nope, F = mass * acceleration, no matter which direction.

Steve Shannon

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Originally Posted by Steve Shannon
Nope, F = mass * acceleration, no matter which direction.

Steve Shannon
I'm pretty confident.... I'm at work so replying with math will have to wait until tonight :-)

As a thought experiment, imagine a hypothetical ejection charge that provides a very small and constant force (not nearly enough on it's own to shear the pins), but that the slack before the pins engage is very very long. As the nose cone moves and eats up that slack, it builds up velocity. When the pins finally engage, the force from the ejection charge itself pushes against them (obviously) the same amount as it would have if there was no slack. What's different though is that for the pins not to shear, they must provide enough force to fully stop the moving nose cone. This force is trivial if the velocity is tiny, and massive if the velocity is large.

This is the same reason that if you pulled back the string on a bow-and-arrow and rested the tip of the arrow against a tree, the tree could fully prevent the arrow from traveling forward when you release the bow string. If you step back a foot so that the arrow accelerates before hitting the tree, the arrow will penetrate the tree.

5. Originally Posted by rocketsam2016
I'm pretty confident.... I'm at work so replying with math will have to wait until tonight :-)

As a thought experiment, imagine a hypothetical ejection charge that provides a very small and constant force (not nearly enough on it's own to shear the pins), but that the slack before the pins engage is very very long. As the nose cone moves and eats up that slack, it builds up velocity. When the pins finally engage, the force from the ejection charge itself pushes against them (obviously) the same amount as it would have if there was no slack. What's different though is that for the pins not to shear, they must provide enough force to fully stop the moving nose cone. This force is trivial if the velocity is tiny, and massive if the velocity is large.

This is the same reason that if you pulled back the string on a bow-and-arrow and rested the tip of the arrow against a tree, the tree could fully prevent the arrow from traveling forward when you release the bow string. If you step back a foot so that the arrow accelerates before hitting the tree, the arrow will penetrate the tree.
Your first statement was simply that it requires less force to keep an object from moving than it does to stop it. That's a fairly hypothetical statement which I responded to by saying that with the exception of the difference between static friction and dynamic friction the force would be the same. That's the basis of Newtonian physics, Newton's first law. It takes the same amount of force to accelerate something as it does to decelerate something. I was talking about the instantaneous application of force
Your latest example is something different from how I understood your original statement. You're describing impulse; some force applied over time to get it moving which must be countered by some force applied over time in order to cause it to stop. If that time is shorter, such as when the arrowhead strikes the tree, the force must be greater in order to return the velocity to zero.
I agree completely.

Steve Shannon

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Originally Posted by Steve Shannon
Your first statement was simply that it requires less force to keep an object from moving than it does to stop it. That's a fairly hypothetical statement which I responded to by saying that with the exception of the difference between static friction and dynamic friction the force would be the same. That's the basis of Newtonian physics, Newton's first law. It takes the same amount of force to accelerate something as it does to decelerate something. I was talking about the instantaneous application of force
Your latest example is something different from how I understood your original statement. You're describing impulse; some force applied over time to get it moving which must be countered by some force applied over time in order to cause it to stop. If that time is shorter, such as when the arrowhead strikes the tree, the force must be greater in order to return the velocity to zero.
I agree completely.

Steve Shannon
Sorry for confusing reply then, my bad! FYI however I wasn't OP, I had made no comments at all on this thread until I responded to your comment about static and dynamic friction, which I mistakenly interpreted as saying that static vs dynamic friction was the only reason that widened shear pin holes would change things.

So I think we agree on the following? Even with a hypothetical frictionless system, if one widens the shear pin hole (and thus allow the nose cone to move before the shear pin engages) it can decrease the amount of bp needed to shear the pins. The static vs dynamic friction differences also contribute to possibly less BP being needed since dynamic friction is less than static friction. Both effects together explain the observation that widening the shear pin holes decreased the amount of BP needed.

Finally, I definitely I agree with other folks that I'd rather have tight holes and not rely on the accelerated nose cone and/or decrease in friction (from static to dynamic) to shear the pins. That dynamic behavior seems more prone to inconsistent behavior than the simple static case. It also seems best to have the friction between nose and tube be as low as possible, since friction could easily change with things like temperature, whereas shear pin shear force should be much more consistent. If it takes too much pressure to shear the pins, it seems best to just use fewer and/or smaller pins.

7. Originally Posted by rocketsam2016
Sorry for confusing reply then, my bad! FYI however I wasn't OP, I had made no comments at all on this thread until I responded to your comment about static and dynamic friction, which I mistakenly interpreted as saying that static vs dynamic friction was the only reason that widened shear pin holes would change things.

So I think we agree on the following? Even with a hypothetical frictionless system, if one widens the shear pin hole (and thus allow the nose cone to move before the shear pin engages) it can decrease the amount of bp needed to shear the pins. The static vs dynamic friction differences also contribute to possibly less BP being needed since dynamic friction is less than static friction. Both effects together explain the observation that widening the shear pin holes decreased the amount of BP needed.

Finally, I definitely I agree with other folks that I'd rather have tight holes and not rely on the accelerated nose cone and/or decrease in friction (from static to dynamic) to shear the pins. That dynamic behavior seems more prone to inconsistent behavior than the simple static case. It also seems best to have the friction between nose and tube be as low as possible, since friction could easily change with things like temperature, whereas shear pin shear force should be much more consistent. If it takes too much pressure to shear the pins, it seems best to just use fewer and/or smaller pins.
I've had too weak of shear pins shear from the force of a compressed chute expanding and then walking it out to the pad, bouncing. Also my first L3 attempt was DQed because my main came out at apogee because I used too weak of shear pins. My preference would be heavier pins and more BP.
Yes, having a larger hole will decrease the force needed to shear/cut the pin for a couple reasons but I think by doing so you reduce the predictability which is one of the greatest advantages of using shear pins in the first place.
My recommendation would be to drill as close to the size of the pin as possible, then calculate your BP requirements, add a little to give yourself a margin of safety, then ground test.

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Originally Posted by Steve Shannon
I've had too weak of shear pins shear from the force of a compressed chute expanding and then walking it out to the pad, bouncing. Also my first L3 attempt was DQed because my main came out at apogee because I used too weak of shear pins. My preference would be heavier pins and more BP.
Yes, having a larger hole will decrease the force needed to shear/cut the pin for a couple reasons but I think by doing so you reduce the predictability which is one of the greatest advantages of using shear pins in the first place.
My recommendation would be to drill as close to the size of the pin as possible, then calculate your BP requirements, add a little to give yourself a margin of safety, then ground test.
I agree with all of that. I'm not the OP, but on my current build that I may use for L3 I'm choosing shear pins to hold against a 50G force trying to pull the nose cone off. That means intrinsically that when the pins do separate, the nose cone will accelerate at 75G (50% safety factor), so hopefully my tracker stays intact :-)

9. Originally Posted by rocketsam2016
So I think we agree on the following? Even with a hypothetical frictionless system, if one widens the shear pin hole (and thus allow the nose cone to move before the shear pin engages) it can decrease the amount of bp needed to shear the pins. The static vs dynamic friction differences also contribute to possibly less BP being needed since dynamic friction is less than static friction. Both effects together explain the observation that widening the shear pin holes decreased the amount of BP needed.

Finally, I definitely I agree with other folks that I'd rather have tight holes and not rely on the accelerated nose cone and/or decrease in friction (from static to dynamic) to shear the pins. That dynamic behavior seems more prone to inconsistent behavior than the simple static case. It also seems best to have the friction between nose and tube be as low as possible, since friction could easily change with things like temperature, whereas shear pin shear force should be much more consistent. If it takes too much pressure to shear the pins, it seems best to just use fewer and/or smaller pins.
I disagree with this. The pressure generated by the BP charge is NOT constant through the whole shear processes. I believe it would be dropping very quickly by the time the nose cone begins to move. The energy of the BP charge is changed into momentum in the nose cone, but is reduced by the same amount imparted to the nose cone. The only difference in this set up as I see it would be the difference between static and dynamic friction and I'm not sure that would cause that much difference since the charge energy has to overcome static friction and shear force of the pin, or static friction and have enough force left to add to the momentum of the nose cone to overcome the shear force of the pins.

This is a fun thought problem, but just ground test.

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Originally Posted by Handeman
I disagree with this. The pressure generated by the BP charge is NOT constant through the whole shear processes. I believe it would be dropping very quickly by the time the nose cone begins to move. The energy of the BP charge is changed into momentum in the nose cone, but is reduced by the same amount imparted to the nose cone. The only difference in this set up as I see it would be the difference between static and dynamic friction and I'm not sure that would cause that much difference since the charge energy has to overcome static friction and shear force of the pin, or static friction and have enough force left to add to the momentum of the nose cone to overcome the shear force of the pins.

This is a fun thought problem, but just ground test.
I agree with: (a) just ground test (b) don't use loose pins, though I’m a relative amateur and (c) this is a fun thought problem, so I hope other folks see it that way too.

I'm not sure I agree with your analysis though. It's been a long time since physics for me but let's give this a shot., Consider a chamber that is 100mm long pressurized to 10psi with a nose cone with 10in^2 of surface area that weighs 1lb. Let's assume no friction since we all agree on the effect that has. In this case, a static setup will exert 100lbs of force (10psi * 10in^2) on the nose cone and therefore on the shear pins. For the scenario with slack in the setup, let's say there 1mm of slack before the shear pin engages. Once the nose cone has moved this 1mm, we have a 1% increase in volume and a corresponding 1% decrease in pressure, so the force applied by the pressure alone when the shear pins engage is approximately 99lbs. In that 1mm of travel, however, we'll have exerted between 99lbs and 100lbs of force along a mm of travel, so conservatively let's call it 99lbs of force applied along 1mm. F=ma, so acceleration (in Gs) is 99Gs. Using vf^2=vi^2 + 2*a*d, we see that the final velocity when we hit the shear pin is sqrt(2* 99G * 1mm) = sqrt(2*99G*9.8m/s^2/G * 1mm) = 1.4 m/s.

Let's say the shear pin can deflect 1mm before breaking. Let's also imagine the force applied during this deflection is constant. This isn't realistic, but if it weren't constant then it must at least at some point exceed the force we would calculate if it were, so assuming it constant is a conservative assumption. To stop the nose cone in 1mm will take exactly the same force it took to accelerate to 1.4 m/s in 1mm, so somewhere between 99lbs and 100lbs. This is in addition though to the 99lbs of force still being applied by the pressure, so we end up with 198lbs on the shear pin.

One thing I'm obviously skipping out on here is that the expansion of the volume causes a temperature drop which also decreases the pressure, but this effect should be miniscule I would think.

Another more practical thing I'm skipping out on is that the pressure leaks out of other places as the nose cone moves. I have no way to model this, but I'd imagine that if your chamber leaks even close to as fast as the nose cone is able to move out there are pretty big problems...

Yet another thing: this effect is obviously smaller if the slack distance before the shear pins engage is much smaller than the amount the shear pins can deflect. My point is to prove the effect rather than calculate an accurate prediction of the effect magnitude on a real system.

Finally, regarding your point about energy, I don’t think there is any sort of energy equality observed here. The fact is that a lot of the BP energy will get dissipated after the nose cone is pushed off, so there is nothing to say we can’t extract a larger fraction of it before the nose cone leaves.

I’d say there’s at least a 75% chance my math is broken somewhere, looking forward to finding out where :-)

11. It's way to late to really go through the math right now, but where did you get 99G of acceleration for the nose cone of what weight/mass? You should be calculating the acceleration of the nose cone based on the force and mass of the nose cone over the 1mm to get the speed. That will give you the amount of momentum in the nose cone when it hits the sheer pin. You also need to keep the units the same. You need to convert everything to metric or standard first. Trying to do it piecemeal during the process is just mistake prone.

I got a launch tomorrow so if I think about it, I'll come back and look over the math.

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Originally Posted by Handeman
It's way to late to really go through the math right now, but where did you get 99G of acceleration for the nose cone of what weight/mass? You should be calculating the acceleration of the nose cone based on the force and mass of the nose cone over the 1mm to get the speed. That will give you the amount of momentum in the nose cone when it hits the sheer pin. You also need to keep the units the same. You need to convert everything to metric or standard first. Trying to do it piecemeal during the process is just mistake prone.

I got a launch tomorrow so if I think about it, I'll come back and look over the math.
99lbs of force applied to a 1lb object will cause 99Gs of acceleration right? If you prefer metric units:
F = 99lbs = 440.37N (at sea level)
m = 1lb = 0.454kg
a = F/m = 440.37N/0.454kg = 970 m/s^2
1G = 9.8m/s^2

so accel in Gs = (970m/s^2) / (9.8m/s^2/1G) = 99G

Regarding the rest of the math, I actually did it out with unit conversions on paper but felt the text was more readable with units we all know and I didn't need to clutter the post with a bunch of unit conversions. Regardless, you can type an expression like sqrt(2*99G*9.8m/s^2/G * 1mm) into google and it will do the unit conversion for you (for that expression google gives me 1.39m/s). It's pretty convenient for quick-and-dirty rocket math :-) It's still possible I messed units up somewhere, but not any place I see. Let me know if you do the proper units on your own and get a different result though! Like I said it's been a while since physics class....

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Originally Posted by rocketsam2016
99lbs of force applied to a 1lb object will cause 99Gs of acceleration right? If you prefer metric units:
F = 99lbs = 440.37N (at sea level)
m = 1lb = 0.454kg
a = F/m = 440.37N/0.454kg = 970 m/s^2
1G = 9.8m/s^2

so accel in Gs = (970m/s^2) / (9.8m/s^2/1G) = 99G

Regarding the rest of the math, I actually did it out with unit conversions on paper but felt the text was more readable with units we all know and I didn't need to clutter the post with a bunch of unit conversions. Regardless, you can type an expression like sqrt(2*99G*9.8m/s^2/G * 1mm) into google and it will do the unit conversion for you (for that expression google gives me 1.39m/s). It's pretty convenient for quick-and-dirty rocket math :-) It's still possible I messed units up somewhere, but not any place I see. Let me know if you do the proper units on your own and get a different result though! Like I said it's been a while since physics class....
I feel kinda bad that I started this whole rigmarole with my comment that I found a clearance hole through the airframe improved my odds, demonstrated empirically through several ground tests, but this is really educational! I'm not even the 'Original' OP, but this is making me remember why I ditched Chemistry as a major and went to Biology...MUCH less math! The argument laid out by RocketSam is logical to me, and NOT just because it fits with my experience! (let's not even GO there! WAY too much of that in American politics lately). Then again the counterpoints from Handeman are valid too!

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Originally Posted by rocketsam2016
if one widens the shear pin hole (and thus allow the nose cone to move before the shear pin engages) it can decrease the amount of bp needed to shear the pins.
Against my better judgement I will weigh in here.

When you talk about breaking pins an energy perspective is an easier framework than a force perspective. It will take a given amount of energy in the form of work (F * d) to shear the pins. That energy comes from the BP charge less any frictional loses in the nosecone motion.

So if it takes a fixed amount of energy to break the pin, the required BP charge would be the same whether there is no motion before the pin is hit or if the nosecone had a "running" start before hitting the pin.

But the amount of energy to break a pin isn't a constant. It takes less energy to break plastic if you apply the force faster than if the force was applied more slowly. So yes the shear pin may require less BP to break if you hit the pin with a running start, but not because more force is applied to the pin but because the pin is weaker in that deformation case.

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Originally Posted by jderimig
Against my better judgement I will weigh in here.

When you talk about breaking pins an energy perspective is an easier framework than a force perspective. It will take a given amount of energy in the form of work (F * d) to shear the pins. That energy comes from the BP charge less any frictional loses in the nosecone motion.

So if it takes a fixed amount of energy to break the pin, the required BP charge would be the same whether there is no motion before the pin is hit or if the nosecone had a "running" start before hitting the pin.

But the amount of energy to break a pin isn't a constant. It takes less energy to break plastic if you apply the force faster than if the force was applied more slowly. So yes the shear pin may require less BP to break if you hit the pin with a running start, but not because more force is applied to the pin but because the pin is weaker in that deformation case.
Agreed it's often easier to think about these things in terms of energy, though I'd be genuinely curious to know what you think is wrong with my math before.

Your point is compelling that applying the energy more quickly might mean that less energy is needed. I maintain that having a running start results in more force being applied to the pin (I think my math from before demonstrates it, and intuitively it takes force to stop a moving object, and we know that the force from the pressure won't have decreased much). If we think about it in energy terms: when the shear pins engage most of the energy from the BP won't have been extracted yet since the gas is still highly pressurized. The pressurized chamber only loses potential energy IIUC when the volume expands (or through heat exchange), so by getting a running start before hitting the pins we've extracted more energy from the pressurized chamber before engaging the shear pins that can then be applied to breaking them. The longer the running start, the more energy has been extracted from the original pressurized system.

Anyway, we can let this lie - there are a few theories to support the "running start" making it easier to shear pins, we have experimental evidence confirming the effect, and I think there is consensus that it's best not to rely on this effect. Interesting stuff, but sorry for dragging things into the weeds.

16. Just off the top off my head, if it's easier to shear pins with a "running start" wouldn't that "running start" be applied when the payload section hit the end of the apogee shock cord at apogee. The whole assembly compresses when the apogee charge goes off and the nose cone gets its running start when the tube hits the end of the shock cord and the nose cone inertia keeps it moving forward.

This would seem to make the shear pins less effective if you set up with the "running start" design.

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Originally Posted by rocketsam2016
Agreed it's often easier to think about these things in terms of energy, though I'd be genuinely curious to know what you think is wrong with my math before.
I don't think anything is wrong with your math. And you have good point that more potential energy may be converted into kinetic energy by allowing the section some motion before it encounters the shear pins. But we agree its best to size the ejection charge to break the pins in the worst case (no previous motion), we do not need to economize BP usage.

18. I'd like to chime in on shear behavior. John makes a good point that the energy required to break a pin isn't constant. You want the pin to undergo brittle failure (fracture) rather than rather than tear. If the pin tears, then there is an enormous amount of energy goes into polymer deformation, which of course requires more energy in the ejection charge. That also means more force will need to be applied to the shear pin holes, all of which is not desirable. So the way to ensure brittle failure is to ensure that the shear rate is high enough that polymer acts glassy rather than viscoelastic. (I won't bore everyone here with a discussion of polymer rheology and time-temperature superposition. If you are curious, PM me). In order to best achieve that, you do not want the pin to have any wiggle room at all. If it moves, it lowers the shear rate and leads to deformation rather than fracture. I've actually seen this on one of my rockets - one pin was a bit loose, and you could see that the nylon was deformed and actually tore, whereas the other two fractured cleanly. I had to drill a new hole to fix that. Another key is that there should be minimal gap between the NC and wall. So for NCs that have rings on them it becomes important to put the shear pins through the rings, not the gaps in between (Madcow NCs are a common example). I have used 3 2-56 nylon screws with plastic NC/cardboard tubes for >15 flights on the same holes. They key is I drill the holes with a #50 bit, harden the cardboard with CA, then I actually tap the holes for 2-56 and thread the screws in. A little extra effort, but well worth it for reliability. I haven't had the holes elongate at all in either the cardboard or the plastic NC, and I have never used any sort of added cutting device (brass shims, etc.). I use 3 2-56 pins for 4" NCs, and 3 4-40 for 5.5" and up (for 3" and under, unless it is a weighted or heavy NC, I just use friction fit). I've used this approach in blue tube and fiberglass as well, and I've never had a shear-pin deployment issue in over 50 flights with this setup.

David

19. Originally Posted by Wizard
I'd like to chime in on shear behavior. John makes a good point that the energy required to break a pin isn't constant. You want the pin to undergo brittle failure (fracture) rather than rather than tear. If the pin tears, then there is an enormous amount of energy goes into polymer deformation, which of course requires more energy in the ejection charge. That also means more force will need to be applied to the shear pin holes, all of which is not desirable. So the way to ensure brittle failure is to ensure that the shear rate is high enough that polymer acts glassy rather than viscoelastic. (I won't bore everyone here with a discussion of polymer rheology and time-temperature superposition. If you are curious, PM me). In order to best achieve that, you do not want the pin to have any wiggle room at all. If it moves, it lowers the shear rate and leads to deformation rather than fracture. I've actually seen this on one of my rockets - one pin was a bit loose, and you could see that the nylon was deformed and actually tore, whereas the other two fractured cleanly. I had to drill a new hole to fix that. Another key is that there should be minimal gap between the NC and wall. So for NCs that have rings on them it becomes important to put the shear pins through the rings, not the gaps in between (Madcow NCs are a common example). I have used 3 2-56 nylon screws with plastic NC/cardboard tubes for >15 flights on the same holes. They key is I drill the holes with a #50 bit, harden the cardboard with CA, then I actually tap the holes for 2-56 and thread the screws in. A little extra effort, but well worth it for reliability. I haven't had the holes elongate at all in either the cardboard or the plastic NC, and I have never used any sort of added cutting device (brass shims, etc.). I use 3 2-56 pins for 4" NCs, and 3 4-40 for 5.5" and up (for 3" and under, unless it is a weighted or heavy NC, I just use friction fit). I've used this approach in blue tube and fiberglass as well, and I've never had a shear-pin deployment issue in over 50 flights with this setup.

David
Great description! One suggestion. I tap the outer hole, the one in the body tube, because the screw head remains attached to that portion of the screw. For the hole in the nosecone shoulder or coupler I drill the hole for a slight interference fit to the screw threads. That way, when prepping for the next flight I can just push that portion of the screw in, instead of trying to unscrew it.

Steve Shannon

20. What is with all the necrobumping lately????

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Originally Posted by Steve Shannon
Great description! One suggestion. I tap the outer hole, the one in the body tube, because the screw head remains attached to that portion of the screw. For the hole in the nosecone shoulder or coupler I drill the hole for a slight interference fit to the screw threads. That way, when prepping for the next flight I can just push that portion of the screw in, instead of trying to unscrew it.

Steve Shannon
This is a great idea. I was going to do the opposite on my current build and wasn't looking forward to unscrewing the headless stump post flight, so your way is much better

22. Is there an advantage to threading either side? Why not just drill a slight interference fit hole and push it in both sides? Especially in fiberglass.

23. Originally Posted by Handeman
Is there an advantage to threading either side? Why not just drill a slight interference fit hole and push it in both sides? Especially in fiberglass.
That would work just fine. For smaller rockets where I've used styrene rods that's all I've done. There have been a few times I've had to remove screw type shear pins from an unlaunched rocket and having threads on the outside has made that very slightly easier. Without those threads I would have use an Exacto knife to lift the head enough to grab.

Steve Shannon

24. I don't think there is a performance advantage, I just find it makes it easier to install. Plus, as Steve said, its much easier to remove if the rocket needs disassembly before launch. For removal after recovery, I find a very small flathead screwdriver can be gently pushed into the stub of the pin in the NC and screwed out. In a plastic NC, it usually can just be popped out. Surprisingly, it doesn't seem to damage the threads in plastic NC, at least for 2-56 pins.

David

25. Originally Posted by Wizard
I don't think there is a performance advantage, I just find it makes it easier to install. Plus, as Steve said, its much easier to remove if the rocket needs disassembly before launch. For removal after recovery, I find a very small flathead screwdriver can be gently pushed into the stub of the pin in the NC and screwed out. In a plastic NC, it usually can just be popped out. Surprisingly, it doesn't seem to damage the threads in plastic NC, at least for 2-56 pins.

David
I kind of found it to be opposite. I just have drilled holes and push the 4-40 screws right through both holes. When the two aren't lined up quite right, I can't imagine how much hard it would be if I had to screw the 4-40 pin in while trying to get that first one lined up. With just holes, you can just push on it and wiggle until it drops in.
As for removal, I have long fingernails and have been able to grab the head and pull them out when I needed to. I also carry a Leatherman Mini so I have the blade if I ever do need it.

Just my opinion, YMMV.

26. Originally Posted by Handeman
I kind of found it to be opposite. I just have drilled holes and push the 4-40 screws right through both holes. When the two aren't lined up quite right, I can't imagine how much hard it would be if I had to screw the 4-40 pin in while trying to get that first one lined up. With just holes, you can just push on it and wiggle until it drops in.
As for removal, I have long fingernails and have been able to grab the head and pull them out when I needed to. I also carry a Leatherman Mini so I have the blade if I ever do need it.

Just my opinion, YMMV.
Absolutely nothing wrong with your method! Fortunately we have a hobby that allows many correct ways to do things!

Steve Shannon

27. Agreed!

28. Originally Posted by Steve Shannon
Absolutely nothing wrong with your method! Fortunately we have a hobby that allows many correct ways to do things!

Steve Shannon
Absolutely! What I like about this forum is I can learn other ways of doing things. I certainly can't think of them all....

29. Originally Posted by Handeman
Absolutely! What I like about this forum is I can learn other ways of doing things. I certainly can't think of them all....
Exactly! My favorite group activity is learning.

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Originally Posted by DAllen
What is with all the necrobumping lately????
Because sometimes you never stop learning, as highlighted by the posts quoted below:

Originally Posted by Steve Shannon
Absolutely nothing wrong with your method! Fortunately we have a hobby that allows many correct ways to do things!

Steve Shannon
Originally Posted by Handeman
Absolutely! What I like about this forum is I can learn other ways of doing things. I certainly can't think of them all....
Originally Posted by Steve Shannon
Exactly! My favorite group activity is learning.

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