Quick links and swivel strength

[highflyer1968]

I have been looking for quick links and swivels for my 4 inch frenzy with av bay and have noticed that there are a variety of strengths. Any where from 250 lb up to 1500 lb strength. How do you know what size or strength you would need for a rocket? Is bigger always better?

 

[rharshberger]

What does the rocket weigh RTF (ready to fly)? What are the maximun simulated G's? How much safety factor do you want to have for the hardware?

 

[Tractionengines]

WATCH for Ultimate Strength versus Allowable Load.

Some things are rated at "WLL" (Working Load Limit). Usually your better Quick Links, and SOME Heavy Duty swivels from name brand manufacturers are advertised at WLL limits. Then a data sheet will give the safety factor used. 3:1, 5:1, 10:1, are all common factors for items for differnt types of rigging. (Securing a load, Lifting a load, Lifting a person.)

Other items like fishing swivels, and low cost "no-name" links, rings, swivels, and misc. hardware, are advertised at ultimate strength. This is point were damage to the device occurs. (Stretched, bent, or breaking.)

So having a quick link with a WLL rating of 400lbs and a 3:1 safety factor; connected to a fishing swivel with an ultimate strength of 1200lbs is actually connecting equal components.

BUT having the same 400lbs quick link connected to a 450lb fishing swivel, is very miss applied (one way or the other.)

Mike

 

[mtnmanak]

Rich and Mike have talked about the two main issues you need to consider when rating your overall recovery system for the proper amount of stress weight load limit. Those considerations are the amount of “G” load your rocket will be subject to and the load limits of the components in your system.


First, let’s discuss the G-forces you need to consider.

Normally, the highest recovery load limit your rocket will encounter is when the main parachute deploys. To calculate this, we need to think about the “deceleration” encountered. I put deceleration in quotes because the physicists among us will balk at the use of the term since deceleration is merely acceleration vectored backwards, but, for simplicity’s sake, we will use the term here.

To calculate deceleration, we use the simple formula:

S _i ² – S _f ² / 2 x d

Where:

S _i = Initial speed (fps)

S _f = Final speed (fps)

d = distance (ft) required to go from Si to Sf

Once we have the deceleration, we divide by 32 (gravity’s acceleration on earth) and that will give us the G’s

Multiply that by the weight of your rocket and you then know how much “shock load” your recovery system needs to handle.

Let’s first look at a nominal flight of a rocket weighing 20 pounds.

Let’s assume the flier has chosen a drogue that will result in a descent rate of 70 fps and main that results in 20 fps. Let’s also assume the main parachute deploys quickly and achieves that speed in 50 feet. Thus, we get:

Deceleration = 70² – 20² / 2 x 50 = 4900 – 400 / 100 = 45 ft/sec/sec

G’s: 45 / 32 = 1.4 G

Load limit = 20 lbs x 1.4 = 28 lbs

So, on a nominal flight, a 20 pound rocket in this scenario would need a recovery system that could handle 28 pounds of shock loading.

Let’s look at the absolute worst case scenario, which would be the main coming out right after motor burnout (unlikely, but, for the purpose of looking at worst case, this would be the rocket’s highest velocity during flight). This worst case scenario should be calculated using the largest motor the rocket would be expected to fly on.

Let’s use 750 fps (about 500 MPH) for this calculation. That is a fairly common speed achieved by Level 2 rockets.

Let’s assume it takes a little longer for the main to slow the rocket down at this speed and use 100 feet for the stopping distance (and assume the main parachute doesn’t shred when it deploys at this speed), we get the following:

Deceleration = 750² – 20² / 2 x 100 = 562,500 – 400 / 200 = 2810.5 ft/sec/sec

G’s: 2810.5/32 = 87.8 G

Load limit = 87.8 x 20 = 1756 lbs

It is unlikely that a main would deploy right at motor burnout, but, if you were trying to calculate the absolute worst case of a 20 pound rocket with a max velocity of 500 MPH, you would need to ensure your recovery system is capable of handling a shock load of 1756 lbs.

Once you know your worst case, you have to do some risk analysis to determine how you want to proceed. For a 20 pound rocket, it isn't very difficult to design a recovery system to handle 2000 lbs of shock loading, so you can probably do that without significant financial or technical difficulty. For a 100 pound rocket, though, you will probably need to limit your risk acceptance to something lower than absolute worst case. In our example above, a 100 pound rocket would need a recovery system capable of handling a shock load close to 9000 pounds. This is probably not realistic on a hobby budget, so you would probably want to look at "most likely worst case" scenarios, like a drogue that fails to deploy.

To address the load limits of the hardware, Mike is absolutely right – you need to be careful of what the advertised load limits for each component are. Frankly, if you buy the components on Amazon, it is a crap shoot. You often have no idea if the limits listed are working load, shock loads, etc. Further, you have no real confidence they are even telling the truth. If I buy on Amazon, I personally target a number that is about double the limit I think I need, just to be safe. If you buy from someplace like McMaster-Carr, you can be more than reasonably sure they are giving you accurate information about the hardware, but you will probably pay a premium. In my lighter rockets, I just go with the cheaper stuff on Amazon and assume the load ratings are low. For my big expensive rockets, I only use hardware I get from places like Mcmaster-Carr.

Also, don’t forget that you need to rate the load limit of the entire recovery system from the eye-bolt in the nosecone tip right down to your motor mount. If you get a bunch of swivels and quicklinks with a high load limit rating and then the non-forged eye-bolt opens up or a cheap piece of all-thread strips out or the nylon shock cord burns through, your rocket will still come burning in. Start at the nosecone tip and follow the recovery system all the way to the aft end and look for any weak points.

(side rant – I wish this forum had the capability to do subscripts and superscripts!)

(edited side rant – I wish the text formatting on this forum allowed subscript and superscript without using HTML codes!)

 

[tsmith1315]

(side rant – I wish this forum had the capability to do subscripts and superscripts!)

(sub) (/sub) and (sup) (/sup) in proper brackets for formatting

S_i²

 

[mtnmanak]

(sub) (/sub) and (sup) (/sup) in proper brackets for formatting

Thanks! that looks cleaner and I edited my side rant

 

[mtnmanak]

Another option to the deceleration calculation above, is to calculate the deceleration as a function of time rather than distance in terms of the interval between the initial and final speeds. This is usually simpler since the equation is easier and the time to deploy the parachute is easier to figure out from empirical data rather than distance traveled.

I generally use three main techniques to pack my parachutes:

• “Open” – burrito folded in a nomex blanket
• “Pouched” – using a semi-closed pouch such as the “parachute liners” from Fruity Chutes
• “Bagged” – using a fully closed (flap) deployment bag with a pilot chute

I have studied many of my rocket videos and determined that the amount of time it takes from the deployment event to the time I have a full main parachute varies depending on how I pack the parachute. The average times to full main I get are:

• “Open” – 1.5 seconds
• “Pouched” – 2.5 seconds
• “Bagged” – 4 seconds

The calculation using a time interval is much simpler since you don’t have to square anything.

S _i – S _f / t

Where:

S _i = Initial speed (fps)

S _f = Final speed (fps)

t – time (sec) required to go from Si to Sf


So, for a 20 pound rocket using an “open” parachute pack on a nominal flight:


Deceleration = 70 – 20 / 1.5 = 33.33 ft/sec/sec

G’s: 33.3 / 32 = 1.04 G

Load limit = 20 lbs x 1.04 = 20.8 lbs

 

[Johnly]

Decades ago Drake Damereau tested quick links to the point of failure on a calibrated Instron.
Even the small quicklinks endured 1000 pounds or more before they pulled apart and opened.
It's likely that after experiencing 1/2 their point of failure force they would could be opened and reused, but the recovery system component wouldn't have failed at that point.

 

[waltr]

Great information on deployment stresses.

In addition to having all components properly rated having a way to absorb forces and reduce 'shock' goes a long way to prevent failures.
For larger rockets with nylon straps I braid the strap. This helps it to cleanly deploy without tangling and, most important, slows down the separation speed that reduces the 'shock' when the strap becomes fully extended. Smaller rockets (mid power) with I do three bundles taped.
Both the braid and taped bundles take energy to pull apart slowing the separation speed.

I did the braid on 1/2 inch nylon in my LOC 4" Goblin. Motor eject near apogee and the chute slowly opened as the booster mass pulled apart the braid. I would say there was NO shock to any of the recovery components.
I other rockets/flights with non-idea ejection times this has keep the chute from shredding and without zippering the tube.

 

[highflyer1968]

Rich and Mike have talked about the two main issues you need to consider when rating your overall recovery system for the proper amount of stress weight load limit. Those considerations are the amount of “G” load your rocket will be subject to and the load limits of the components in your system.


First, let’s discuss the G-forces you need to consider.

Normally, the highest recovery load limit your rocket will encounter is when the main parachute deploys. To calculate this, we need to think about the “deceleration” encountered. I put deceleration in quotes because the physicists among us will balk at the use of the term since deceleration is merely acceleration vectored backwards, but, for simplicity’s sake, we will use the term here.

To calculate deceleration, we use the simple formula:

S _i ² – S _f ² / 2 x d

Where:

S _i = Initial speed (fps)

S _f = Final speed (fps)

d = distance (ft) required to go from Si to Sf

Once we have the deceleration, we divide by 32 (gravity’s acceleration on earth) and that will give us the G’s

Multiply that by the weight of your rocket and you then know how much “shock load” your recovery system needs to handle.

Let’s first look at a nominal flight of a rocket weighing 20 pounds.

Let’s assume the flier has chosen a drogue that will result in a descent rate of 70 fps and main that results in 20 fps. Let’s also assume the main parachute deploys quickly and achieves that speed in 50 feet. Thus, we get:

Deceleration = 70² – 20² / 2 x 50 = 4900 – 400 / 100 = 45 ft/sec/sec

G’s: 45 / 32 = 1.4 G

Load limit = 20 lbs x 1.4 = 28 lbs

So, on a nominal flight, a 20 pound rocket in this scenario would need a recovery system that could handle 28 pounds of shock loading.

Let’s look at the absolute worst case scenario, which would be the main coming out right after motor burnout (unlikely, but, for the purpose of looking at worst case, this would be the rocket’s highest velocity during flight). This worst case scenario should be calculated using the largest motor the rocket would be expected to fly on.

Let’s use 750 fps (about 500 MPH) for this calculation. That is a fairly common speed achieved by Level 2 rockets.

Let’s assume it takes a little longer for the main to slow the rocket down at this speed and use 100 feet for the stopping distance (and assume the main parachute doesn’t shred when it deploys at this speed), we get the following:

Deceleration = 750² – 20² / 2 x 100 = 562,500 – 400 / 200 = 2810.5 ft/sec/sec

G’s: 2810.5/32 = 87.8 G

Load limit = 87.8 x 20 = 1756 lbs

It is unlikely that a main would deploy right at motor burnout, but, if you were trying to calculate the absolute worst case of a 20 pound rocket with a max velocity of 500 MPH, you would need to ensure your recovery system is capable of handling a shock load of 1756 lbs.

Once you know your worst case, you have to do some risk analysis to determine how you want to proceed. For a 20 pound rocket, it isn't very difficult to design a recovery system to handle 2000 lbs of shock loading, so you can probably do that without significant financial or technical difficulty. For a 100 pound rocket, though, you will probably need to limit your risk acceptance to something lower than absolute worst case. In our example above, a 100 pound rocket would need a recovery system capable of handling a shock load close to 9000 pounds. This is probably not realistic on a hobby budget, so you would probably want to look at "most likely worst case" scenarios, like a drogue that fails to deploy.

To address the load limits of the hardware, Mike is absolutely right – you need to be careful of what the advertised load limits for each component are. Frankly, if you buy the components on Amazon, it is a crap shoot. You often have no idea if the limits listed are working load, shock loads, etc. Further, you have no real confidence they are even telling the truth. If I buy on Amazon, I personally target a number that is about double the limit I think I need, just to be safe. If you buy from someplace like McMaster-Carr, you can be more than reasonably sure they are giving you accurate information about the hardware, but you will probably pay a premium. In my lighter rockets, I just go with the cheaper stuff on Amazon and assume the load ratings are low. For my big expensive rockets, I only use hardware I get from places like Mcmaster-Carr.

Also, don’t forget that you need to rate the load limit of the entire recovery system from the eye-bolt in the nosecone tip right down to your motor mount. If you get a bunch of swivels and quicklinks with a high load limit rating and then the non-forged eye-bolt opens up or a cheap piece of all-thread strips out or the nylon shock cord burns through, your rocket will still come burning in. Start at the nosecone tip and follow the recovery system all the way to the aft end and look for any weak points.

(side rant – I wish this forum had the capability to do subscripts and superscripts!)

(edited side rant – I wish the text formatting on this forum allowed subscript and superscript without using HTML codes!)

Wow thanks for all the information. I will be using this for calculations as I build my rocket. This is what I love about the forum. There is always somebody to answer and help with questions.

 

[rharshberger]

Wow thanks for all the information. I will be using this for calculations as I build my rocket. This is what I love about the forum. There is always somebody to answer and help with questions.

One thing to remember a recovery system is only as strong as its weakest link, you can intentionally design a specific failure point into the system. For example if you use 3 loop harnesses the drogue harness could be designed that IF a failure occurs (due to shock) it breaks away at the avbay side allowing the fin can section to at least still have a drogue chute to help it survive the fall and not come in ballistic, then the main chute section and hopefully still functioning electronics will deploy the main and save the upper section of the rocket and any payload it may contain. Parachutes can fail too, I have seen drogues ripped away (swivel failures, shroud line failures, and panels ripped out) though it is less common than main chute failures.

 

[Onebadhawk]

That’s cool. I should not have connected the moonburner to the M2020 which does use Gorilla or Elmers. Any reason for those vs epoxy?

Rich and Mike have talked about the two main issues you need to consider when rating your overall recovery system for the proper amount of stress weight load limit. Those considerations are the amount of “G” load your rocket will be subject to and the load limits of the components in your system.


First, let’s discuss the G-forces you need to consider.

Normally, the highest recovery load limit your rocket will encounter is when the main parachute deploys. To calculate this, we need to think about the “deceleration” encountered. I put deceleration in quotes because the physicists among us will balk at the use of the term since deceleration is merely acceleration vectored backwards, but, for simplicity’s sake, we will use the term here.

To calculate deceleration, we use the simple formula:

S _i ² – S _f ² / 2 x d

Where:

S _i = Initial speed (fps)

S _f = Final speed (fps)

d = distance (ft) required to go from Si to Sf

Once we have the deceleration, we divide by 32 (gravity’s acceleration on earth) and that will give us the G’s

Multiply that by the weight of your rocket and you then know how much “shock load” your recovery system needs to handle.

Let’s first look at a nominal flight of a rocket weighing 20 pounds.

Let’s assume the flier has chosen a drogue that will result in a descent rate of 70 fps and main that results in 20 fps. Let’s also assume the main parachute deploys quickly and achieves that speed in 50 feet. Thus, we get:

Deceleration = 70² – 20² / 2 x 50 = 4900 – 400 / 100 = 45 ft/sec/sec

G’s: 45 / 32 = 1.4 G

Load limit = 20 lbs x 1.4 = 28 lbs

So, on a nominal flight, a 20 pound rocket in this scenario would need a recovery system that could handle 28 pounds of shock loading.

Let’s look at the absolute worst case scenario, which would be the main coming out right after motor burnout (unlikely, but, for the purpose of looking at worst case, this would be the rocket’s highest velocity during flight). This worst case scenario should be calculated using the largest motor the rocket would be expected to fly on.

Let’s use 750 fps (about 500 MPH) for this calculation. That is a fairly common speed achieved by Level 2 rockets.

Let’s assume it takes a little longer for the main to slow the rocket down at this speed and use 100 feet for the stopping distance (and assume the main parachute doesn’t shred when it deploys at this speed), we get the following:

Deceleration = 750² – 20² / 2 x 100 = 562,500 – 400 / 200 = 2810.5 ft/sec/sec

G’s: 2810.5/32 = 87.8 G

Load limit = 87.8 x 20 = 1756 lbs

It is unlikely that a main would deploy right at motor burnout, but, if you were trying to calculate the absolute worst case of a 20 pound rocket with a max velocity of 500 MPH, you would need to ensure your recovery system is capable of handling a shock load of 1756 lbs.

Once you know your worst case, you have to do some risk analysis to determine how you want to proceed. For a 20 pound rocket, it isn't very difficult to design a recovery system to handle 2000 lbs of shock loading, so you can probably do that without significant financial or technical difficulty. For a 100 pound rocket, though, you will probably need to limit your risk acceptance to something lower than absolute worst case. In our example above, a 100 pound rocket would need a recovery system capable of handling a shock load close to 9000 pounds. This is probably not realistic on a hobby budget, so you would probably want to look at "most likely worst case" scenarios, like a drogue that fails to deploy.

To address the load limits of the hardware, Mike is absolutely right – you need to be careful of what the advertised load limits for each component are. Frankly, if you buy the components on Amazon, it is a crap shoot. You often have no idea if the limits listed are working load, shock loads, etc. Further, you have no real confidence they are even telling the truth. If I buy on Amazon, I personally target a number that is about double the limit I think I need, just to be safe. If you buy from someplace like McMaster-Carr, you can be more than reasonably sure they are giving you accurate information about the hardware, but you will probably pay a premium. In my lighter rockets, I just go with the cheaper stuff on Amazon and assume the load ratings are low. For my big expensive rockets, I only use hardware I get from places like Mcmaster-Carr.

Also, don’t forget that you need to rate the load limit of the entire recovery system from the eye-bolt in the nosecone tip right down to your motor mount. If you get a bunch of swivels and quicklinks with a high load limit rating and then the non-forged eye-bolt opens up or a cheap piece of all-thread strips out or the nylon shock cord burns through, your rocket will still come burning in. Start at the nosecone tip and follow the recovery system all the way to the aft end and look for any weak points.

(side rant – I wish this forum had the capability to do subscripts and superscripts!)

(edited side rant – I wish the text formatting on this forum allowed subscript and superscript without using HTML codes!)

I don't agree that a deployment at motor burnout is rare or unlikely..
We grow in our rocketry journey..
It is exactly this scenario ( high thrust, short burn motor ) that brings us to shear pins..
It is these occurrences that we learn from..

Teddy

 

[mtnmanak]

I don't agree that a deployment at motor burnout is rare or unlikely..
We grow in our rocketry journey..
It is exactly this scenario ( high thrust, short burn motor ) that brings us to shear pins..
It is these occurrences that we learn from..

Teddy

I was referring to a main parachute deployment at motor burnout, which is rare compared to a drogue deployment at motor burnout, which happens somewhat frequently due to drag separation. A main deployed at motor burnout is much more likely to happen in a single deploy situation, but, usually, single deploy flights employ smaller motors and lower velocities.

A drogue deployed at motor burnout still inflicts a massive stress on the recovery system, but deploying a main at burnout on a high velocity flight would be considerably more stress. In all likelihood, the hardware would not be your issue here - the airframe and parachute would mostly likely be destroyed before your hardware gave out. This was just an exercise in the considerations of choosing quicklinks and swivels that will meet the most likely stress demands of a flight.

As for shock cords, harnesses and attach points, that is why I only use Onebadhawk kevlar - everything else in the system will fail before those parts will

 

[JohnCoker]

(sub) (/sub) and (sup) (/sup) in proper brackets for formatting

S_i²

It's also possible to use Unicode super- and subscript characters directly for simple cases: x² y₀, but I agree it's simpler to use bbcode since they will handle arbitrary content.
https://en.wikipedia.org/wiki/Unicode_subscripts_and_superscripts

 

[bjphoenix]

The rocket is pulling against a string connected to a parachute, it's not pulling against the bumper of a big truck. What is the highest velocity it will have when the parachute deploys? What is the drag of the parachute at that velocity? How can it have much more force on the system than that?
If you have a rocket with a very long delay and it is on its way down at good velocity when the parachute ejects- the parachute will open and immediately start pulling on the shock cord and try to zipper the tube. The pull of the parachute against the front of the tube is whatever drag the parachute can generate at that velocity.
It seems that there could be some inertia if pieces separate- say the nose cone is blown off at deployment, catches the air stream and starts slowing down as the heavier rocket body passes it by. When the nose cone reaches the end of the shock cord there is a high shock load depending on the weight of the nose cone and the elastic stiffness of the shock cord. But in this scenario the nose cone wouldn't have much mass. If the nose cone had a lot of mass it wouldn't slow down very fast in the slipstream, it would fall at the same speed as the main part of the rocket and wouldn't create high shock loads when it reaches the end of the cord.
A rocket with properly functioning dual deployment system probably doesn't generate much inertia load in the system but you want to design for more than that because sometimes things don't work out as planned.

 

[rharshberger]

The rocket is pulling against a string connected to a parachute, it's not pulling against the bumper of a big truck. What is the highest velocity it will have when the parachute deploys? What is the drag of the parachute at that velocity? How can it have much more force on the system than that?
If you have a rocket with a very long delay and it is on its way down at good velocity when the parachute ejects- the parachute will open and immediately start pulling on the shock cord and try to zipper the tube. The pull of the parachute against the front of the tube is whatever drag the parachute can generate at that velocity.
It seems that there could be some inertia if pieces separate- say the nose cone is blown off at deployment, catches the air stream and starts slowing down as the heavier rocket body passes it by. When the nose cone reaches the end of the shock cord there is a high shock load depending on the weight of the nose cone and the elastic stiffness of the shock cord. But in this scenario the nose cone wouldn't have much mass. If the nose cone had a lot of mass it wouldn't slow down very fast in the slipstream, it would fall at the same speed as the main part of the rocket and wouldn't create high shock loads when it reaches the end of the cord.
A rocket with properly functioning dual deployment system probably doesn't generate much inertia load in the system but you want to design for more than that because sometimes things don't work out as planned.

Drag separation is one of those scenarios, things get shock loaded quickly even though they a nominally moving in the same direction.

 

[Chad]

When the nose cone reaches the end of the shock cord there is a high shock load depending on the weight of the nose cone and the elastic stiffness of the shock cord.

I like to put my ebays in the nose cone. I 3dprint the parts needed to do that, basically a cap and some kind of mount that attaches to the inside of the nosecone, but never worked out the math on the forces involved. I haven't had a failure but wonder how much in the ballpark i am.

If you imagine the ideal case, a rocket just reaching apogee and horizontal then would there would be two calculations per @mtnmanak formulas above? One for the nose cone ebay cap and one for the body tube bulkhead since they'd both be moving at different speeds? In the formula, given the change in speed has a square on it that seems to be what contributes the most to the resulting forces.

The ejection charge goes off and the nose cone accelerates away from the body tube but the body tube is also going to accelerate away from the nose cone. The difference in velocity would be due to the difference in mass between the two. Given momentum is linear with respect to velocity and mass (p=mv) do you think i could just measure the velocity of the nosecone with a camera during ground testing and then base the body tube velocity off its mass proportional to the nosecone? I feel like the body tube accelerating away would rob the nose cone of some energy though which wouldn't be the case in ground testing since the body tube is typically braced against something (i.e. the ground).

I hope that made sense i am not a mathmagician

 

[Handeman]

I like to put my ebays in the nose cone. I 3dprint the parts needed to do that, basically a cap and some kind of mount that attaches to the inside of the nosecone, but never worked out the math on the forces involved. I haven't had a failure but wonder how much in the ballpark i am.

If you imagine the ideal case, a rocket just reaching apogee and horizontal then would there would be two calculations per @mtnmanak formulas above? One for the nose cone ebay cap and one for the body tube bulkhead since they'd both be moving at different speeds? In the formula, given the change in speed has a square on it that seems to be what contributes the most to the resulting forces.

The ejection charge goes off and the nose cone accelerates away from the body tube but the body tube is also going to accelerate away from the nose cone. The difference in velocity would be due to the difference in mass between the two. Given momentum is linear with respect to velocity and mass (p=mv) do you think i could just measure the velocity of the nosecone with a camera during ground testing and then base the body tube velocity off its mass proportional to the nosecone? I feel like the body tube accelerating away would rob the nose cone of some energy though which wouldn't be the case in ground testing since the body tube is typically braced against something (i.e. the ground).

I hope that made sense i am not a mathmagician

I think you have it exactly right. In flight when both move away from the point of ejection, the nose cone would have less energy than it does when the BT is braced on the ground during ground test.
If the BT doesn't move on the ground test, all the energy is imparted to the nose cone. The shock load should be the same as and in-air deployment. The energy is just coming from the nose cone alone instead of coming from both ends of the shock cord.

 

[mtnmanak]

I like to put my ebays in the nose cone. I 3dprint the parts needed to do that, basically a cap and some kind of mount that attaches to the inside of the nosecone, but never worked out the math on the forces involved. I haven't had a failure but wonder how much in the ballpark i am.

If you imagine the ideal case, a rocket just reaching apogee and horizontal then would there would be two calculations per @mtnmanak formulas above? One for the nose cone ebay cap and one for the body tube bulkhead since they'd both be moving at different speeds? In the formula, given the change in speed has a square on it that seems to be what contributes the most to the resulting forces.

The ejection charge goes off and the nose cone accelerates away from the body tube but the body tube is also going to accelerate away from the nose cone. The difference in velocity would be due to the difference in mass between the two. Given momentum is linear with respect to velocity and mass (p=mv) do you think i could just measure the velocity of the nosecone with a camera during ground testing and then base the body tube velocity off its mass proportional to the nosecone? I feel like the body tube accelerating away would rob the nose cone of some energy though which wouldn't be the case in ground testing since the body tube is typically braced against something (i.e. the ground).

I hope that made sense i am not a mathmagician

I wouldn't over think the nosecone-BT separation too much - you aren't getting huge velocities on the nosecone ejection that cause it to have a huge impact when it reaches the end of the shock cord. You also get into a more complicated equation if you start trying to compute multiple acceleration vectors (i.e - horizontal acceleration from the black powder charge and vertical acceleration from gravity). The equation isn't all that complicated, but I don't think it is worth the time it would take to compute it.

I don't think it is really necessary. If you want to take a stab at it, just use the length of the shock cord. If you have a 2 pound nose cone and 30 foot shock cord, we could estimate that the nosecone would be falling at around 62 fps after a 30 ft fall and, if we assume the nosecone "snaps" to "stop" at the end of the shock cord, using the equations above, you would end up with a 4G impact, so about an 8 pound load. Not a huge deal for most hardware or internal components.

I have used a lot of 3d printed sleds in nosecones and the only time I have broken them is when a nosecone/rocket came in ballistic. I have not had any issues with 3d printed parts in the nosecones when an event happened - even when I had a drag separation on an L motor launch of a 4" tube fin rocket (which almost destroyed half the booster BT - but no issues to the 3d printed sled in my NC).

 

[waltr]

Here is some real data from Ejection and then parts hitting end of cord and/or Chute opening.



Data is Accelerometer data of x, y & z (Series 1, 2, 3) with y-axis (series2) in line with rocket.
This is from an instrument package in the payload section of a BMS 3" school rocket flying an F67W.
Ejection change at ~6.6seconds and peaks at 116m/s/s (11.8Gs). At ~6.9seconds is the peak of -140m/s/s (14Gs) when cord reaches end and/or chute opens.

Note the acceleration from Motor peaked at ~80m/s/s (8Gs).
I think these forces (accelerations) are under estimated at times.

 

[Hobie1dog]

I never worry about the stainless steel ball bearing snap swivels or the quick links, as every failure that I've had is when the parachute shroud lines break/tear in pieces

 

[FredA]

Oh, on the contrary, I've seen many stretched swivels and snapped quick-links.
Build rockets big enough and use 9/16th TN [or larger] and you'll start losing hardware.
Check your quick-links - the HomeCheapo variety stainless are only rated for something like 250 pounds - the zinc-steel are stronger.

I've seen tons of flight traces with 30G shocks at apogee.
I've seen too many with over 50G's.
Fly a big rocket and have a 30G deploy and it's amazing the "ease" which stuff gets yanked apart.

 

[mtnmanak]

The rule of thumb I was taught was to design the entire system to handle 50Gs. That is pretty easy to do with smaller diameter rockets, but it gets really difficult when the rockets start getting big. The current rocket I am working on is going to be about 250 lbs on the pad with an O motor. Figure a recovery weight somewhere in the 225+ range. Designing the system for a 50G event would mean everything would need to be rated for 11K-12K lbs. Not impossible, but hardware rated to those levels is very expensive (for example, McMaster-Carr has some quick links rated to 10,800 lbs - at $115 a piece). You have to do your risk analysis and, at some point, make some calculated compromises or engineering changes.

 

[rharshberger]

The rule of thumb I was taught was to design the entire system to handle 50Gs. That is pretty easy to do with smaller diameter rockets, but it gets really difficult when the rockets start getting big. The current rocket I am working on is going to be about 250 lbs on the pad with an O motor. Figure a recovery weight somewhere in the 225+ range. Designing the system for a 50G event would mean everything would need to be rated for 11K-12K lbs. Not impossible, but hardware rated to those levels is very expensive (for example, McMaster-Carr has some quick links rated to 10,800 lbs - at $115 a piece). You have to do your risk analysis and, at some point, make some calculated compromises or engineering changes.

At that point you ditch quick links and start using rigging and lifting shackles, heavy but much cheaper, shackles also have the advantage of being able to use a safety wire through the pin to ensure they will not loosen, and there is no strain on the threads like a quick link.

 

[FredA]

Use Lifting Eyebolts - thread on the end of your central all-thread.
Dill a hole at the end of the all-thread for a cotter pin to prohibit unscrewing.

 

[kalsow]

Dill a hole at the end of the all-thread for a cotter pin

Over the years, I've tried a few times. Each time it cost me a drill bit. I guess I don't have the right mojo.