Shock!

Something I keep seeing in the rocketry forums is advice along the lines of “use a shock cord x times the length of the body tube”. Now, granted, this is a rule of thumb, not intended to be rigorous, and as a rule of thumb it may have given satisfactory results in most cases. But from a basic physics and engineering standpoint, to me it makes no sense at all.

Think about it: Suppose you have one long, slender rocket and it has the “perfect” length shock cord. And now you go and build a second rocket, just like the first in all respects except its body tube is only half as long… all other factors including total mass and nose cone mass being kept equal. We’ll suppose the result still is stable, and being the same mass and diameter as the first rocket its flight characteristics are roughly the same. So from a recovery perspective, what’s changed? What does the body tube length have to do with anything relating to the recovery ejection? Basically two things: First, for the shorter rocket, the volume of the rocket is smaller, which means the initial ejection pressure (with the same ejection charge) is larger (though drops off faster as the nose cone moves), and so the nose cone and body fly apart faster. Second, the shorter rocket presents less area, so is less subject to drag forces that will slow the components down as they separate. And yet, by the rule of thumb, the shock cord should be half as long?

It seems to me the main reason for making a shock cord “long enough” is to give the nose cone and rocket body a chance to lose some of their velocity to air drag before the cord loses slack and has to absorb the rest of the rocket’s kinetic energy. If that’s the case then the shorter-but-otherwise-same rocket with its faster ejection and reduced drag should need a longer shock cord, shouldn’t it?

Again, I realize this is simply a rule of thumb that people have had success with. It just seems to me that a better rule of thumb for determining what “long enough” is, based on criteria other than the irrelevant body tube length, ought to be possible.

I raised this question over on TRF and got some good responses, notably one from jcato who wrote:

… Surely there’s something better.

There is — it’s called the ‘Modulus of Rupture’ (M(r)) (which gives one the energy of rupture of a material – expressed in Joules). Another item to look at is the ‘Modulus of Elasticity’ (a smaller number than the Rupture modulus, but gives an idea of how a material will react to loading (up to the Elastic Limit)) — and is expressed in ‘in/in’ — which gives you some insight into how ‘long’ a linear tensile member should be (like a shock cord). A longer cord can handle more energy than a shorter (in addition to it’s other benefits) – and this is because, in textile materials, that total energy (Modulus of Rupture) is *mass* related. Cord ‘Y’ that is twice as long as cord ‘X’ will weigh twice as much and can absorb twice the energy, as well. Being longer, it also will have unit elongations that are half (that in/in thing) – thus lowering the stress (and dynamic shock) the system must take.

The bottom line in a lot of that research resulted in a number — 76mN/TEX for the M(r) of nylon. Another way of expressing that unit is ‘J/gm’ – which simplifies the analysis greatly – just weigh your (nylon) shock cord and you can calculate how many Joules of energy it will take to rupture. It is equally simple to calculate the energy (in Joules) of a particular rocket traveling at a particular velocity. Compare the numbers and make sure the shock cord is the larger of the two.

Now, that really only gives us the ultimate strength of the shock cord – not how long it must be (well, assuming the mass of the cord is enough). I’ve always felt that the tensile rating of the cord should be between 50 and 100 times the dead load of the vehicle. Some may say that is extreme, but I’ve got some examples of dynamic (shock) loading multiples of 82x the dead load, so I think my numbers have merit. …

But I’m not so sure it’s “simple” to calculate the relevant energy. It seems to me what’s relevant is the kinetic energy (in the center of mass frame) of the rocket components at ejection. You want to be sure the cord can handle that energy and bring the nose cone and body to a stop relative to one another. For that you need to know the velocities of the nose cone and body, which depend on lots of nasty factors: Ejection pressure, drag forces, and so on.

Well, let’s throw some freshman physics at it, shall we? And crunch some numbers. Now, a lot of these numbers are ones I’ll be pulling out of my a… out of thin air, so don’t take them too seriously, but they may be a decent start at figuring out how to look at the problem.

Assume a spherical rocket. No, wait, assume a typical low power rocket. Let’s get specific. Assume an Estes Big Bertha launching on an Estes B6-4. Here’s a Rocksim file for the Big Bertha which I’ll be using in the following.

Define some variables. Masses and dimensions are just being read out of the sim file; I don’t guarantee they correspond to any real life Big Bertha, but they’re reasonable:

  • M = Total rocket mass (at launch)
    = 92.5 g = 0.0925 kg (Calculations will be done in kg, but I’ll show the numbers here in g.)
  • Mm = Initial motor mass
    = 18.2 g
  • Mme = Empty motor mass
    = 12.6 g
  • Mp = Propellant mass = Mm – Mme
    = 5.6 g
  • Mr = Recovery system mass
    = 7.74 g
  • Mnc = Nose cone mass
    = 11.1 g
  • Mbe = Body mass at ejection = M – Mp – Mr – Mnc
    = 68.1 g

Now, what happens at ejection? It’s complicated. The nose cone gets popped out. Yes, that’s complicated.

Actually that’s not even the whole truth. The nose cone gets pushed forward and the body gets pushed backward. Action, reaction, right?

If we know the forces on the components during ejection we can calculate their velocities at the end. The principal forces, I’d expect, are: The gas pressure on the nose cone, the friction between the nose cone and body tube, and the forces on the body tube due to gases escaping out the front and back. The latter I don’t know how to figure out, so I’m going to hope they’re small and ignore them. I’ll also ignore air drag during the ejection event.

I assume someone’s measured ejection pressure due to a B6 ejection charge in a Big Bertha. I haven’t found it, although I see numbers on the order of 15 psi kicking around for high power rocketry. So let’s just use that number. But think about this: For ideal gases in a volume V, the pressure P is nRT/V, where n is the number of moles of gas, R is the gas constant, and T is the temperature. We start off with air in the part of the body tube between the motor mount and the nose cone shoulder. (Wadding and chute take up some space, but not much once you squeeze all the air out, so I’ll neglect them.) The ejection adds gas, which raises the pressure. It also heats up the gas, which also raises the pressure. If the volume is larger, then the fractional increase in the amount of gas is smaller, and the temperature increase is smaller, so the pressure increase is smaller. Roughly, the pressure increase is inversely proportional to the volume. So until better information comes along we’re going to say 15 psi for our Big Bertha/B6, but for other size rockets (with the same motor) we’ll scale that number by the volume ratio.

The frictional force on the nose cone shoulder of course is highly variable, depending on how the rocket was built, temperature, how much soot has gotten deposited in the body tube, and so on. But as stealth6 pointed out a couple years ago, there’s at least one other rule of thumb based on irrelevant parameters knocking around the hobby:

The often repeated “rule” is that one should be able to lift the rocket by the cone. If the cone pulls out, then more friction/tighter fit is needed. But the fit should also be just tight enough to hold; more so is too much.

He goes on to say why this is a silly rule, but it’s probably the rule by which our Big Bertha was built. Anyway, assuming so will give us the frictional force to within a factor of 2 or so. The force is just the weight (in Newtons) of the rocket body. Which is negligible! That is, if the ejection gas pressure is 15 psi, and the nose cone end area is around 2 square inches, that’s about 30 pounds of force on the nose cone. The rocket body, on the other hand, weighs in the neighborhood of 3 ounces. So the gas pressure completely overwhelms the shoulder friction. Or it should! Yet nose cones do fail to pop off sometimes. That’s if they’re jammed somehow, so the frictional force is many times what it should be, and/or the ejection charge gas is leaking somewhere.

OK, so we have something like this:

rect2985

where the two boxes represent the body and nose cone, the spring represents the pressure force (we think of it as a force acting on the nose cone, but remember an equal and opposite force acts on the body), and the damping piston represents the friction — which is negligible. In the center of mass frame both the body and nose cone start out at zero velocity and have equal and opposite forces — call their magnitude Fg — acting on them. As they move apart the gas volume will increase and the pressure will decrease, but I’ll neglect that too. So the force is constant up until the time the shoulder leaves the body tube; then the pressure drops to zero in our simplified model. We know the force and the distance over which it acts:

  • Ls = nose cone shoulder length
    = 1.9 cm

from which we can calculate the final velocities; they are

  • vnc = nose cone velocity at ejection
    = 18.8 m/s
  • vbe = body velocity at ejection
    = 3.1 m/s

(Left as an exercise for the reader, or if you really want me to I’ll write it up in a separate post.) I think these velocity estimates make sense. Possibly a little large, but right order of magnitude, anyway.

So now what? Well, the nose cone and body move apart. And their relative velocity decreases, due mainly to air drag on the body and the parachute, I would expect. Supposing for some reason there were no shock cord connecting the nose cone to the body; how far apart would they get as a function of time?

OpenRocket says the Big Bertha’s drag coefficient Cdr rises from its minimum of 0.50 at motor burnout to about 0.74 just about at apogee, then falls back to about 0.62 during the ¾ second from apogee to deployment. After deployment, OpenRocket is silent on the subject.

I’m presuming, by the way, that the chute’s shrouds are connected directly to the nose cone, not to a point along the (nonexistent) shock cord. So the nose cone is being affected by the chute, which is starting out packed but opening, which is a mess. Which is why I propose to start by focusing on the body and save the nose cone for later.

The drag force on the rocket body is

  • Fdb = – 1/2 Cdr * ρ * vb^2 * A

where ρ is the air density, taken as 1.2 kg/m^3. vb starts at vbe but decreases. A is the projected frontal area… which varies depending on the orientation of the body. Initially it would be the area seen from the front, but as it tumbles the area will change. (Or if you like A is a constant reference area and Cdr changes. Either way. I’ll do it the first way.)

If ejection occurs in a horizontal direction then gravity doesn’t enter into the equation of motion for the horizontal component:

Mbe * dvb/dt = –k * vb^2

where k = 1/2 Cdr * ρ * Aref. The solution is vb(t) = 1/(1/v(0)+k*t/Mbe). It approaches zero as t goes to infinity.

It doesn’t drop very fast, though. In the worst case the body turns to a 90 degree angle of attack and stays there, presenting an area of roughly 300 cm^2. Then after 1 second it’s still going 2 m/s and has moved 2.4 m from the ejection event. After 5 seconds, 0.9 m/s velocity and 6.4 m distance. Given that the nose cone starts with a much higher velocity, even with the parachute, it probably goes a comparable distance, so the separation is of order 10 m within a few seconds.

And who the heck puts a 10 m shock cord, even if it’s Kevlar, on a Big Bertha? Remember, this is assuming 90 degree angle of attack, so in reality the velocity and separation should be larger.

My guess is the 15 PSI pressure is too large, and maybe some other assumptions are less than realistic. But lacking anything better, I’ll press on. Again, saving the nose cone for later, what happens to the body velocity and distance when we change parameters of the rocket?

Consider a Baby Bertha. The main differences between Big and Baby are that the body tube is shorter, and the fins are a little smaller, on the Baby. I figure it weighs about 30 g less — ⅔ as much — and has about ¼ the pressurized volume and  ⅔ the area at 90 degree angle of attack.

We can run the numbers for a Baby Bertha. We can also run them for two other hypothetical designs: A Big Bertha lightened (in the body) to match the mass of a Baby Bertha, and a Baby Bertha with mass added (again in the body) to match the mass of a Big Bertha. What I get for the body motion is:

Big Bertha Baby Bertha Big Bertha Light Baby Bertha Heavy
time (s) velocity (m/s) distance (m) velocity (m/s) distance (m) velocity (m/s) distance (m) velocity (m/s) distance (m)
0 3.06 0.00 10.32 0.00 5.09 0.00 6.21 0.00
1 1.95 -2.43 3.20 -5.43 1.92 -3.01 3.52 -4.61
2 1.44 -4.09 1.89 -7.86 1.19 -4.50 2.45 -7.53
3 1.13 -5.37 1.34 -9.45 0.86 -5.51 1.88 -9.67
4 0.94 -6.40 1.04 -10.63 0.67 -6.26 1.53 -11.36
5 0.80 -7.26 0.85 -11.57 0.55 -6.87 1.29 -12.76

Decreased mass has two counteracting effects: It makes the rocket components fly apart faster for a given ejection charge, but it also makes them slow down faster for a given drag force. So for instance Big Bertha Light starts with a higher body velocity, but within about a second it’s dropped to a slower rate and by about four seconds its distance has become lower than for Big Bertha.

Decreased length, though, leads to decreased volume, hence higher ejection pressure, and therefore higher velocity; it also leads to decreased projected area, hence lower drag force, and therefore less deceleration. Comparing Big Bertha with Baby Bertha Heavy, the latter has much larger velocity to start with and maintains that advantage; its distances are about twice as big. This suggests, then, that — all else being equal — a shorter rocket needs a longer shock cord, as I handwavingly argued at the start.

This has been long, and there’s more to say. There’s the nose cone’s motion to try to model somehow. There’s getting reasonably accurate ejection pressures. There’s looking at this from a kinetic energy standpoint — connecting up with that energy of rupture, remember? I plan on coming back to this in a future post or two.

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