I’m designing a very basic pneumatic shock absorber, single throw, single use (until reset). Just designed to cushion the fall of an object. It will have a plunger going into a cylinder, with holes in the cylinder to let air escape at a controlled rate as the plunger goes in.
I know the mass and velocity of the falling object.
What formula do I use to consider the relationship between the the size of the holes in the cylinder and the forces that it will take to push in the plunger?
I think I’m allowed one little bump, aren’t I? There must be someone out there who knows about this stuff?
Try a little…
Trial and error, Give it your best judgment, then work with it. Who knows, maybe the challenge of judgment will be rewarding.
It sounds like you are, or imposing to be an engineer. Me being a mechanic, will question the circumstances surrounding the birth of many engineers.
The circumstances in which the device will operate are difficult to re-create on the test bench, and the consequences of failure are time consuming and expensive.
How much does the thing weigh, and how fast is it hitting?
Are we talking about a shock absorber made out of a cardboard mailing tube, or something that’s gonna cushion the re-entry impact of the ISS?
40 m/s, about 300 gms
OK, testing would be difficult. You’d need to drop it from a height of 80 m (~25 storeys) to get it to that speed.
How much decelleration can the thing stand? It takes 80 m at 1 gravity to get it up to, or slow it down from, that speed.
If your shock absorber has a travel of, say, 10 cm, the decelleration will be about 800 gravities. That’ll kill pretty much anything sensitive.
What’s wrong with a parachute?
Well, It should be a function of viscosity, orifice size for sure. Maybe also fluid density, ratio of orifice area to total cross sectional area, length of stroke.
Since it’s air, you’d need total starting volume of fluid. That’s because air is compressible, while the oil in an oil-filled shock absorber is incompressible. The compressibility of air will complicate matters.
But then again, the above information is necessary for derivation of the equation. Not all of it will necessarily end up in the final product. I’m an aero engineer, which means I’m a step away from a mech, but not close enough to know diddly about shock absorber design.
Disclaimer: I don’t know much about fluid flow.
But depending on just how expensive this falling object is, I would use either Bernoulli’s equation (cheap) or the full Navier-Stokes equations (expensive) to derive a differential equation for the motion of the plunger.
For example, Bernoulli’s equation for incompressible fluids (air is not incompressible, of course; but this may be a reasonable approximation in your case) is
Delta P = P[sub]1[/sub] - P[sub]2[/sub] = rho v[sup]2[/sup] / 2
(where rho is the density of the fluid, P[sub]i[/sub] are the pressures, and v is the fluid speed). This will give you an estimate of the airspeed through your piston relief holes.
An example of using this: We have Delta P = F/A = m x’’/A (where x’’ is the acceleration and A is the piston area) and a(x) v = A x’ (conservation of fluid volume, where a(x) is the area of relief holes in the cylinder below height x), so
x’’ = ((rho A[sup]3[/sup]) / (2 a(x)[sup]2[/sup] m)) (x’)[sup]2[/sup].
Now by choosing various a(x) you can tailor the acceleration curves x’’. For example, if you choose a(x)[sup]2[/sup] = b x, you get an equation
x’’ x = c (x’)[sup]2[/sup] = ((rho A[sup]3[/sup]) / (2 b m)) (x’)[sup]2[/sup]
which has a solution x = a t[sup]r[/sup] with r = 1/(1-c).
I may not have done all of this math correctly (and I know I’ve used some approximations without checking them), but this strikes me as a reasonable approach to finding the approximate needed a(x).
I myself would take the approach of yelling over the wall to the fluids guy (actually, I’d walk over to his desk). He’d probably tell me something along the lines of what Omphaloskeptic said.
You could use CFD code too, but that would probably be overkill for this problem.
Many thanks to Omphaloskeptic in particular. A few clarifications if I may, Omphaloskeptic . Be gentle with me, I haven’t done any real math for 20 years.
You haven’t defined:
m though I assume its mass
x’ though I suspect that is the speed of the piston
c?
r?
t’?
I can follow you through to your derivation of x’’ x = ((rho A3) / (2 b m)) (x’)2
I don’t know how you get the c (x’)2 in x’’ x = c (x’)2 = ((rho A3) / (2 b m)) (x’)2 [emphasis added]
And I have to admit that I am totally lost on the last bit (“which has a solution x = a tr with r = 1/(1-c)”)
Can you baby me through a little more?
Desmo, thanks for your input. As you have probably figured out (given other questions at various times) the object in question is a water rocket. It pulls anything up to a few hundred G towards the end of the “burn” anyway (when pressure is high and mass minimal).
And the only sensitive part (usually) is the timing and release mechanism for the chute. So if I don’t have a chute…
Chutes are OK, but (a) been there, done that, want to try something else (b) chutes tend to allow downrange drift which can cause tree problems, and © chutes are fiddly to deploy cleanly.
And I should add that I can probably deccelerate over about 30cm, which keeps things down to a manageable (!) 260G or so.