Aluminum U-bolts

[m85476585]

I have some from McMaster Carr, and they are rated for 218lbs. Are they strong enough for a 3-4lb rocket?

 

[11bravo]

Static load?
Sure, more than enough.
WAY more than enough.
Dynamic/shock load?
I wouldn't bet on it.

 

[bobkrech]

Probably. It all depends on your recovery loads and how they get distributed to the rest of the airframe.

The allowable G loading is simply the part rating divided by the rocket weight: 218/4= 54.5 G and 218/3= 72.7 G. Many chutes will shread before you reach these G levels so it's unlikely the U-bolt will break under normal and under most abnormal situations.

What is of more concern is how you spread out the load on the bulkhead. If the U-bolt pulls through the bulkhead, it doesn't matter what the U-Bolt is rated for.

Fortunately it's easy to test it. Simply perform a static load test on the assembly. Take a strong platform and drill a hole in it for your shockcord and stand your rocket upside down on the board. Hang 220 pounds (100 kg) from the shockcord. If nothing breaks, you have proven your design.

Bob

 

[m85476585]

When does most of the shock occur? At separation, or when the chute opens?

 

[cjl]

If your cord is long enough, there shouldn't be much shock at separation.

Looking at some pics of deployment my dad caught, the largest shock by far on my rockets is at main opening.

 

[m85476585]

200lbs doesn't seem like much, especially compared to 600lb-rated quicklinks and 2000lb rated shock cords, but it sounds like it will work. Is everything else we use for recovery overkill?

 

[bobkrech]

Originally posted by m85476585
200lbs doesn't seem like much, especially compared to 600lb-rated quicklinks and 2000lb rated shock cords, but it sounds like it will work. Is everything else we use for recovery overkill?

Yes, for the most part folks overbuild their rockets.

You wouldn't put the Queen Mary 2's anchor on your 17' runabout, but that effectively what would would be doing with those recovery components on a 3-4 pound rocket.

Many chutes go south at > 50G shock load, and most are gone by 100G. A 600 pound link can take a 200G -> 150G shock load and the 2000 pound kevlar is good for 667G -> 500G shock loads.

IMO using 600 pound rated quick links and 2000 pound kevlar rated on a run of the mill 3-4 pound rocket is a waste of money. It simply adds weight and little else.

Most folks don't understand what a G-load is. A 100G deceleration is a 3200 fps2 velocity change rate.

For example if you are +/-3 seconds of of apogee when you deploy, your rocket is traveling at 96 fps. To generate a 100G shock load with a standard chute, the rocket has to slow down from 96 fps to ~16 fps in ~(96-16)/3200 = 80/3200 = 0.025 seconds. In that time the rocket will have dropped only 96 fps * 0.025 s = 2.4 ft. Now you typical Estes rocket can possibly do this with a light parachute and a short shock cord (how many times have we seen a modroc strip a chute), but a high power rocket with 10-20 ft of shock cord won't slow down that quickly. It will take 0.1 to 0.2 seconds just to extend the shock cord, and since you have destroyed the aerodynamics of the rocket by blowing the NC or airframe apart, it's already slowing down all by itself due to the configurational change before the chute fully deploys.

If you really are concerned, add a slider to your chute to slow the opening time. The average 3-4 pound rocket with a reasonable length shock cord and a folded chute is unlikely to fully deploy in less than 0.1 seconds. That alone would reduce the shock load to 25 G or less. Add a slider to double the opening time and you have reduced the shock load to 10-15G which is easily handled by any chute on the market.

By doing a little math ahead of time, you can come up with a matched set of components that will simplify your rocket design and make your rockets lighter and less expensive provided you use a reliable and reasonably timed ejection method.

Bob

 

[fumoffu]

Originally posted by bobkrech
If you really are concerned, add a slider to your chute to slow the opening time. The average 3-4 pound rocket with a reasonable length shock cord and a folded chute is unlikely to fully deploy in less than 0.1 seconds. That alone would reduce the shock load to 25 G or less. Add a slider to double the opening time and you have reduced the shock load to 10-15G which is easily handled by any chute on the market.
Bob [/B]

What is this slider you speak of it interests me!

I have also been confused about the terms "working load" and "test" when used to describe these items. And many times the store just gives a lbs rating. I have heard there is like a 4x difference between the two?

I have been meaning to look up this physics stuff for the forces at work during deployment, any one have a more formal descrition (with the icky equations) of this stuff?

Thanks!

 

[m85476585]

I don't know for sure, but I would assume the "working load" is what is safe to use it at, and the "test" strength is where it will physically break.

 

[falingtrea]

Originally posted by fumoffu
What is this slider you speak of it interests me!

A slider is a small disk with holes around the rim that you feed the shroud lines through. You then slide the disk up next to the canopy before you fold up the parachute. When the parachute is deployed the disk slides down the shroud line, and the chute opens more the farther the disk travels down the lines.

 

[brianc]

Originally posted by fumoffu
What is this slider you speak of it interests me!

GiantLeap has some pics and movies of their's in action
https://giantleaprocketry.com/products_recovery.asp#slider













OT-- Props to GL! I placed an order the other day, fully expecting it
to arrive after the holiday rush. It was on my doorstep early this morning!
Always great service for Ed and Kent.

 

[bobkrech]

Originally posted by fumoffu
What is this slider you speak of it interests me!

I have also been confused about the terms "working load" and "test" when used to describe these items. And many times the store just gives a lbs rating. I have heard there is like a 4x difference between the two?

I have been meaning to look up this physics stuff for the forces at work during deployment, any one have a more formal descrition (with the icky equations) of this stuff?

Thanks!

Working load is the rated load capacity of a device.

Proof Test load is a laboratory test typically at a somewhat higher load than the working load (x2 is common) for quality control purposes. A part may not deform under a proof test load.

https://www.certifiedslings.com/loadlimits/loadlimits.html

The physics of load testing is really not that difficult to understand.

1 G is 32 ft/s/s acceleration

Load is a force, weight is a force. G load = load force/weight

delta v / delta t = a a in Gs is simply delta v /(delta a * 32)

Conservative testing for shock load is simply using a dead weight equivalent to the shock load.

Bob

 

[Ted Cochran]

Originally posted by bobkrech
Yes, for the most part folks overbuild their rockets.
...
For example if you are +/-3 seconds of of apogee when you deploy, your rocket is traveling at 96 fps. To generate a 100G shock load with a standard chute, the rocket has to slow down from 96 fps to ~16 fps in ~(96-16)/3200 = 80/3200 = 0.025 seconds. In that time the rocket will have dropped only 96 fps * 0.025 s = 2.4 ft.

I generally agree with this analysis, but be careful here. If you use long shock cord, especially long kevlar shock cord (say, 30' for the purposes of the example that follows), what might happen is that

--the nose cone comes off,
--a 30' shock cord fully extends (remember, all parts are continuing to accelerate downward; air resistance just reduces the rate of acceleration and does not slow anything down unless it's already above its terminal velocity; not likely for LMR and HPR),
--the parachute (attached near the nose cone) opens, stopping the nose cone abruptly
--the fin can continues to sail along in trail until it either
a) crashes into the parachute, collapsing it and initiating an overly speedy recovery, or
b) the fin can misses the parachute, travels another 30' until it reaches the end of its travel, and then is abruptly brought to a stop. Falling against an open parachute means you're trying to accelerate the column of air under the parachute, which compresses, and the forces thereby generated are much more significant than if you expose a load to a gradually opening parachute.

If something like this scenario happens, 200G's are entirely possible, because the parachute will already be fully opened, and the fin can will have traveled an additional 60' in free fall. Even without an initial velocity caused by a nonoptimum delay, you're basically asking the attachment to hold after tying one end of the shock cord to the roof and then throwing the fin can off of a six story building.

Originally posted by bobkrech
... Now you typical Estes rocket can possibly do this with a light parachute and a short shock cord (how many times have we seen a modroc strip a chute)

That's a function of the speed with which a small parachute can open, and the frailty of those chutes--especially the attachments.

Originally posted by bobkrech
... but a high power rocket with 10-20 ft of shock cord won't slow down that quickly. It will take 0.1 to 0.2 seconds just to extend the shock cord, and since you have destroyed the aerodynamics of the rocket by blowing the NC or airframe apart, it's already slowing down all by itself due to the configurational change before the chute fully deploys.

Unless we are talking about early deployment (at above terminal velocities), the airframe is always accelerating. Long shock cords only have value for absorbing the energy caused by the relative motion of the parts of the rocket created by the ejection charge.

Cheers,

--tc

 

[Stymye]

whats funny is the question required a simple yes or no answer.

 

[Ted Cochran]

Ah, then the original question should have been accompanied by a specific design, with a load analysis, and a flight profile with a six-sigma uncertainty envelope.

P.S. And then the answer would be "no" for anything other than lazy flights to 2000' with good delay choices and consistent chute packing.

 

[Sailorbill]

Originally posted by Ted Cochran
I generally agree with this analysis, but be careful here. If you use long shock cord, especially long kevlar shock cord (say, 30' for the purposes of the example that follows), what might happen is that

--the nose cone comes off,
--a 30' shock cord fully extends (remember, all parts are continuing to accelerate downward; air resistance just reduces the rate of acceleration and does not slow anything down unless it's already above its terminal velocity; not likely for LMR and HPR),
--the parachute (attached near the nose cone) opens, stopping the nose cone abruptly
--the fin can continues to sail along in trail until it either
a) crashes into the parachute, collapsing it and initiating an overly speedy recovery, or
b) the fin can misses the parachute, travels another 30' until it reaches the end of its travel, and then is abruptly brought to a stop. Falling against an open parachute means you're trying to accelerate the column of air under the parachute, which compresses, and the forces thereby generated are much more significant than if you expose a load to a gradually opening parachute.

If something like this scenario happens, 200G's are entirely possible, because the parachute will already be fully opened, and the fin can will have traveled an additional 60' in free fall. Even without an initial velocity caused by a nonoptimum delay, you're basically asking the attachment to hold after tying one end of the shock cord to the roof and then throwing the fin can off of a six story building.

This looks like a good reason to put a chute near the fin can and stop/slow the heavy part first.
Both my DD Endeavour and my Grandson's motor ejection Endeavour have a small drogue on the fin can. My grandson's rocket is of the zipper proof design so the small drogue helps pull the main chute from the forward tube.

 

[Ted Cochran]

Originally posted by Sailorbill
This looks like a good reason to put a chute near the fin can and stop/slow the heavy part first.

I did an article for Sport Rocketry in 2000 or so that went through a bunch of data to get to that very conclusion.

The big benefit is to avoid having the fin can collide with an open parachute. It happens quite a lot--if you start looking for it, you'll see it 5-10% of the time if the chute is attached to the nose cone and the shock cord is long.

--tc

 

[m85476585]

It will be dual-deployment, probably drogueless. It seems like that will cause more stress on the recovery system.

 

[bobkrech]

Originally posted by Ted Cochran
I did an article for Sport Rocketry in 2000 or so that went through a bunch of data to get to that very conclusion.

The big benefit is to avoid having the fin can collide with an open parachute. It happens quite a lot--if you start looking for it, you'll see it 5-10% of the time if the chute is attached to the nose cone and the shock cord is long.

--tc

Ted

That was one of the better documented technical articles in SR in the recent past and something I have done for sometime.

While I agree that the scenerio you propose can happen, it happens because of a poor recovery system design.

1.) Tubular products are a poor shock cord materials. They have little stretch so they doesn't absorb shock, it transmits it. If you use a tubular shock cord, you should also have a loop more elastic material to absorb shock.

To illustrate this, let's use your 6 story drop analogy. For convenience I'll use a 64 ft drop which represents a 2 second freefall. A body in freefall for 2 seconds has attained a 64 fps velocity and has fallen 64'. All lines have some stretch. If the the line has a 0.5% stretch at the load, the object decelerates to 0 velocity in 4" at 192G. If the rope has a 1.5% stretch, the object stops in 1' and the G-loading drops to 64G. A shock cord with 3% stretch drops the G-loading to 32G in 2' distance, a 6% stretch drops it to 16G in 4' distance, and 12% stretch drops the G-loading to only 8G in a 8' distance. The lesson to be learned is that a shock cord with a reasonable amount of stretch won't strip a chute whereas an overly strong shock cord with little or no stretch will strip a chute most of the time in a less than optimal deployment senerio.

2.) The terminal velocity of a rocket depends on its square root of its sectional density which is the weight divided by the cross-section. A typical rocket with a 10:1 side-on to head-on cross-scetion will have a terminal velocity more than 3 times slower if it descends horizontally instead of vertically. This is illlustrated in dual deployment applications, a rocket that breaks into 2 equal drag pieces at apogee will usually descend in a horizontal orientation at a slower descent rate than if a drogue were deployed from the NC and the rocket maintained a vertical descent orientation. Similarily if a main chute is deployed from center of the rocket you get some aerodynamic slowing due to the conformational change before the chute deploys. This can make a difference in the final outcome since shedding even a few 10s of fps velocity makes a difference in the G-loading.

3.) I took a long look at this problem a few years ago for safe recoveries of bowling balls in accidental deployments at max V. It was clear that simply attaching a ballistic drogue directly to a bowling ball drilled to accept a tubular nylon harness would result in a safe recovery of a 16 pound bowling ball at velocities in excess of 700 mph using parachutes similar to https://aeroconsystems.com/chutes/p73_xform.htm A similar design could be used to safely decelerate large motor casing from L3 rocket shreds.

4.) It's pretty easy to assemble a lead sled made from oversized components, throw in a large motor and fly it. If all goes well in the flight, you have a nominal recovery, but if the recovery system hicups, you more than likely have a major recovery safety issue.

It doesn't take a lot more effort to otimize a design by selecting stong but light components, and figure out how to mitigate stresses from less than optimal recovery circumstances using simple techniques such as parachute sliders to slow chute opening time in the event of a less than optimal high speed deployment, adding stretch to shock cord to minimize G-loading, during such events, and investigating configurational options to aerodynamically limit the terminal velocity in case of a chute failure.

The devil's in the detail.

Bob

 

[Ted Cochran]

Originally posted by bobkrech


[We must avoid] poor recovery system design.

The devil's in the detail.

[I hope that's a reasonable paraphrase ]

[Aside to folks who thought there ought to be a simple answer--nope ]

I completely agree with the idea that recovery systems should be designed. The whole point of drogues, pilot chutes, and deployment bags is to make the behavior of the recovery system predictable. This doesn't matter as much on itty bitty rockets, or on vastly over-engineered rockets, but eventually it matters because the forces get too large to over-engineer for.

Lest folks think this is all there is to it, consider impact effects of stainless steel quick links on aluminum u-bolts, the deleterious effects of knots in the shock cord, and the strength of the bulkhead to which the u-bolt is mounted (I've seen perfectly adequate u-bolts pulled through 1/2" marine plywood bulkheads).

It's called rocket science for a reason!

Cheers,

--tc

 

[m85476585]

What about something completely different, like epoxying a Kevlar strap through a slot in the bulkhead, and attaching the TN to it with a quicklink? It seems like it would be strong enough and as light as aluminum.

 

[Brian62]

When we build rockets, and overbuild, that is not done for all the near perfect flights......rather for the one nightmare flight when you really need it.
Brian

 

[Stymye]

overbuilding is overbuilding..no matter how you phrase it

In a catostrophic situation the rocket should be frangeable to some extent for safety reasons alone.

Mabey more emphasis should be put on avoiding those situations in the first place than just building a tank that can handle anything.

 

[denverdoc]

Originally posted by stymye
overbuilding is overbuilding..no matter how you phrase it

In a catostrophic situation the rocket should be frangeable to some extent for safety reasons alone.

Mabey more emphasis should be put on avoiding those situations in the first place than just building a tank that can handle anything.

Well put. As Bob K, Robert DeHate, and others have said on this thread and others, we often put ourselves at jeopardy by trying to avoid it--the more hardware and the heavier it becomes, the greater the loads, hence the need to make it stronger/heavier still. What could have been a 15# rocket now weighs 35. And if all does fail as it may sometime do, the kinetic energy has likely quadrupled in the process and therefore the rocket will sustain/inflict much more damage.

Good electronics with scrupulous attention to proper installation and use with some form of redundancy is an excellent start. Most other failures I've seen are motor related, flutter failure or wind shear issues--latter I have seen mostly on the Black Rock desert.

Having had an N casing land 50 feet from me after being way up there has me seriously considering Bob K's suggestion of having separate chutes attached to these. Even a small x-form that could have been seen might have prevented a potential death under these circumstances. Granted its a needle and the haystack scenario, but small comfort to those involved if it had happened.

JS

 

[fumoffu]

Originally posted by bobkrech

delta v / delta t = a a in Gs is simply delta v /(delta a * 32)
Bob

Did you mean 1 G = 32 f/s^2 or dv/(dt*32)?

These are very rough physics of what is going on there. I certain may do some more studying/experiments to get a better understanding of what's involved.

When I first started looking up physics envolved, I thought that the deployment of the parachute would be more like an impulse. But that quicky lead the insainly high loads of because the time of the force was so small.

Then I considered that the shockcord would stretch over time. This still produces a very short time with large forces. But certainly enlightened me to why stretch in important.

I am personally not conviced that these averages are the best way to design your gear. For example if you say that deployment takes 0.1s and your calculation yeild 25G _average_ force during that period. You presume you have a 4 lb rocket and use a 200lb work load U-bolt. The forces are not going to be constant during deployment! I could easily see an impulse for a small fraction of time being 25 or more! I don't know, I am making this up

I guess what I am getting at is I want to see a force curve at deployment. Like time on one axis and G (or Newtons) on the other. Any done this, or know where this information is? Certainly would be an interesting experiment.

Just some thoughts.

 

[bobkrech]

Originally posted by fumoffu
Did you mean 1 G = 32 f/s^2 or dv/(dt*32)?

These are very rough physics of what is going on there. I certain may do some more studying/experiments to get a better understanding of what's involved.

When I first started looking up physics envolved, I thought that the deployment of the parachute would be more like an impulse. But that quicky lead the insainly high loads of because the time of the force was so small.

Then I considered that the shockcord would stretch over time. This still produces a very short time with large forces. But certainly enlightened me to why stretch in important.

I am personally not conviced that these averages are the best way to design your gear. For example if you say that deployment takes 0.1s and your calculation yeild 25G _average_ force during that period. You presume you have a 4 lb rocket and use a 200lb work load U-bolt. The forces are not going to be constant during deployment! I could easily see an impulse for a small fraction of time being 25 or more! I don't know, I am making this up

I guess what I am getting at is I want to see a force curve at deployment. Like time on one axis and G (or Newtons) on the other. Any done this, or know where this information is? Certainly would be an interesting experiment.

Just some thoughts.

By definition acceleration is delta V / delta t which is in units of distance divided per unit time squared.

The acceleration due to gravity is 32 ft/s/s so if you divide the acceleration by g or the (delta V / delta t) / 32 fpsps you obtain the acceleration in units of the nondimensional G.

For modeling purposes you could treat a shock cord as a spring which develops a force F = k x where k is the spring constant and x is the distance of extention.

https://en.wikipedia.org/wiki/Hooke's_law

I'll leave it to you to do the math which shows that the retardation forces induced by the shock cord are relatively constant throughout the decelleration process if the shock cord is sized and rated accordingly.

Bob

 

[denverdoc]

Couple of thoughts about what's has not come up--where I fly at least, a common practice is to z fold shock cord and wrap in masking tape (or stronger) in several segments, so that it looks like a chain of sausages. I'm not convinced that a lot of energy goes into tearing the tape, and no matter how long I make the primary tether, each packet breaks cleanly.

But along the lines of stretch, if one made the packets progressively more difficult to rupture, you might reach a point where you have one or more remaining bundled.

I like Bob's idea of adding stretch in the system, but given other constraints, I like Kevlar near any EC's, and have used both strap and tubular with success. I can see adding some stretchier material in loops, but in the name of simplicity, wondering if anyone has tried the GLR product that adds some stretch to Kevlar?

JS

 

[fumoffu]

Originally posted by bobkrech



For modeling purposes you could treat a shock cord as a spring which develops a force F = k x where k is the spring constant and x is the distance of extention.

https://en.wikipedia.org/wiki/Hooke's_law

I'll leave it to you to do the math which shows that the retardation forces induced by the shock cord are relatively constant throughout the decelleration process if the shock cord is sized and rated accordingly.

Bob

Sorry if I have derailed this thread
But I think the question of why is important?

So F(t) = k x(t), and F(t) = m a(t) = m x''(t) therefore 0 = k x(t) + m x''(t). This is a classic second-order differential equation with the solution in the form of x(t) = A cos (C t) + B sin (C t) (more or less) where A and B are constants from the intial values of x(t), x'(t) and C is based on k and m (mass). I have forgoten much of my DiffEq class but is what I was able to look up and recall.

In this simple example it can be shown that the average is only 60% of the maximal force (I think).

Isn't there an article or a web site which goes those these physics of recovery?

 

[denverdoc]

I'm confused by the equations: OK if force =-K(x), as in a Hookian spring, there is no time depenence inherent to the equation and arguably it is still dependent on a forcing function. In this case of the quasi-impulse variety.

So jerk a string, the effect will eventually be felt as a big displacement and then will settle down to a damped oscillation.

This is a classic impulse reponse. The pluck on guitar strings depend on it. You still need to assume the jerk in the first place and not the reverbs in response when designing recovery systems.
JS