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Ok. But I can't tell if he wants us to see damage to the ball valve itself, or the inside of the body of the ball valve, or what.
As for the earlier discussion about different piston weights...
Numbers be damned. IMO, purely from empirical experience, of course... A heavier piston may move at the same speed, and the performance out the end of the barrel may not change... But the heavier piston WILL do more damage.
You have more mass, (Mass not being the size, but quantity of matter, and weight.), moved by the same force. You ARE going to get more of a hammering. Even if the heaver piston is moving slower, as Ragnarok pointed out. There is still enough force there, to turn that piston into a hammer.
Just like Tech's steel hammer, vs plastic hammer.
Same guy exerts the same force on each hammer, and the steel hammer is going to drive the nail, but the plastic one will probably just impale, or at least deform itself on the nail.
If you can't fix it, you don't own it.
It's not the ball valve that got crushed, but the bushing that was there as a piston stop. The o-ring that got shoved into the crushed bushing was the piston stop. I replaced the bushing with steel, and there are no signs of damage. I'll upload new pictures soon so that it's clear that the piston does not impact the pilot valve.
I think we're missing an important point in this discussion about pistons being like a hammer. The mass doesn't matter. That same mass that figures into the piston accelerating slower from the sealed position also increases its impact force into the stop. The masses cancel, and the only thing that really factors into the impact force is how much of the impact the piston absorbs (which is probably slightly greater with UHMW, but because of the terrible tensile strength, I'm still iffy on using it).
Are you using any kind of bumper besides the bushing? Deceleration force is inversely proportional to deceleration distance.
For example stopping from 60 to the intersection over a couple hundred feet is no problem. Stopping by hitting a parked car at the intersection involves much higher forces due to the much shorter distance. The higher forces are well known to be able to test the limits on the strength of steel.
If you use UHMW, it will need a safe stopping distance to keep forces in check.
Here is a photo of a car that decelerated in the distance of about 1.5 feet.
By design a longer deceleration zone can reduce deceleration forces.
Ahhhh, that's like nails on a chalkboard to an engineer.
It's easy to prove that in a captured system with fixed net force and travel distance, mass does not effect the final kinetic energy.
F = m*a
a = F/m
v<sub>f</sub> = SQRT(2*a*d)
v<sub>f</sub> = SQRT(2*(F/m)*d)
E<sub>k</sub> = 1/2*m*v<sub>f</sub><sup>2</sup>
E<sub>k</sub> = 1/2*m*SQRT(2*(F/m)*d)<sup>2</sup>
E<sub>k</sub> = F*d
So no, the heavier piston will not hit harder. It will hit with the same kinetic energy, meaning the valve materials will be required to absorb the same quantity of energy in order to bring the piston to rest.
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Now I know who to turn to, if I ever want some numbers crunched.
I'm happy for you engineer types who can "prove" things with numbers.
But in real life, I'm telling you, if you swing a steel hammer, and a plastic hammer, exactly the same distance, with exactly the same force...
The steel hammer is likely to do some damage.
And the plastic one is as likely to BE damaged, as to do any damage.
Why ? Simply because of the material that the hammer is made of.
They may land with exactly the same force. The numbers may all crunch the same on acceleration/deceleration, etc...
But it is still a fact that steel will be more likely than plastic, to do damage. And plastic will also be more likely to absorb some of the blow, itself.
Now, if your plastic hammer weighed the same as your steel hammer... not only would you be rich from patent royalties, but the plastic hammer would be some bit more likely to cause some damage.
In the end, if I want to keep from cracking off the end of a piston valve, I would probably put the same buffer behind them both. But STILL more likely to go with the plastic, (or wood), valve than the steel one.
If you can't fix it, you don't own it.
Often overlooked in the simple example is the local SOS, chamber size, and the mass of the accelerating air pressure.
Either could have higher energy.
The two roll off factors for both are, local SOS on the lighter piston limiting the accelerating, and the heavier piston with a small chamber will be experiencing a loss of force due to the drop of chamber pressure in the shot with the longer dwell time.
I'll leave it to the engineers to fight out which has higher KE.
I have found a heavier piston requires a heavier bumper to be effective. A lighter piston can be safely arrested with lighter materials. A bumper to stop a slow moving locomotive simply needs a big spring. A bumper to stop a flying practice arrow can be a simple net. Here is one to 303 fps.
What you're thinking is if you swing a steel hammer and a plastic hammer with the same velocity (or angular velocity) the steel hammer will do more damage. Force is vastly different than velocity. If you apply the same force to a plastic hammer and a steel hammer during the swing, they will deal the same amount of damage to whatever they hit... but the plastic hammer will be travelling MUCH faster the instant before it impacts the target.
That's probably because the lighter piston is plastic and therefore absorbs the impact energy more like a spring than a steel piston. And unfortunately for your nice analogy (which is very true) the impact force of a car and a practice arrow are nowhere near equal.
And I am using a bumper. See the o-ring that was shoved into the dent in the bushing? That was my bumper. The o-ring has some small stress cracks on the outside, so don't tell me it's not an effective bumper.
Better valve picture: The replacement for the crushed brass bushing is circled.
Not quite saefroch... Remember, that for a start the plastic hammer will deform. You have to factor that in when calculating what sort of damage to expect. Secondly, a high velocity, low weight projectile may apply an identical force to a slow moving, heavy one, but they will not cause the same damage. Due to the rate at which target materials can deform before failing, among other factors... They will behave differently.
You can figure it all out very accurately with numbers, but you have to factor EVERY variable in if you want to get a truly accurate result.
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How far is the piston moving? If it's more than 0.15" then you're just allowing the piston to accelerate and cause more damage to itself and the launcher for no performance benefit.
I was going to mention that they wouldn't deal the same amount of damage, and that the impact force of the plastic hammer would be slightly less because of deformation, but I thought that'd be molecule-fondling.
How did you get .15"? That's something to do with cross-sectional area or something flow path equaling the area of the valve port? The piston currently retracts ~.75".
In my current launchers, to limit the acceleration distance and thus the total piston kinetic energy. I limit the full travel to no more than 2/3 the valve seat. For example my 2 inch valve opens fully in about 1 inch (port size is one inch tall) and the bumper is cushioning the piston starting at about 3/4 inch to stop it before 1.2 inches of total travel.
A piston that retracts .75 inch with a valve seat of roughly .4 inches is moving way too far and thus picks up too much KE. That valve would have much more reasonable forces if it moved no more than .25 inches. It will take the same time to open the first 1/4 inch. It simply won't keep moving past full open to pick up extra KE.
That piston travel distance is about the same as I use on my 1 inch Marshmallow Cannon.
You may be able to reduce the piston travel by adding more bushings in the pilot area. I have a real long stopper assy in my cannons to limit the piston travel.
Oh, and don't overestimate the ability of a soft hammer to deform to limit impact force. I dented a car door with a marshmallow. It was not dried out, frozen, or anything. It was a fresh marshmallow.
Last edited by Technician1002 on Fri Nov 19, 2010 7:43 pm, edited 2 times in total.
I may be able to limit the displacement of the piston on triggering, but I'd like to see a calculation that determines the minimum distance it has to travel before it is no longer a limiting factor in flow... When I built it I constructed the piston an associated housing so that it retracts fully out of the flow path, but I guess that's overkill.
I hope that dent wasn't on a working car's door...
The math I used for my cannons is to make the area of the open port into the orifice about twice the area as the port. For example in my 2 inch cannon, the ID of the pipe using Pi X Radius Squared for the area gives me Pi X 1 inch squared. 1 squared is 1. Times Pi = 3.14 square inches.
I have 3 ports 1 inch tall (thus full open at one inch) by 2 inches wide for a total of 6 square inches. To have a full all around port open with the same area of the orifice the general measurement is 1/4 the Diameter for full open. Like you, I open further than that as I want the barrel to be the only restriction to flow. For you anything over 1/4 of the .364 is considered full open or about 0.1 inches of travel. For yours I recommend double that at about 0.2 to 1/4 inch because your port is not open to the chamber a full 360 degrees, but only on two sides.
The dent is on a working car door.. Oops.. I was chrono'ing some marshmallows and it was parked about 15 feet away. I didn't think it would bother it. It is an old (ancient) van, so no major harm. It already is crap.
The math for the surface area of a cylinder is here. Do not add the area of the ends of the cylinder as you want the area of the wall of the cylinder only. At 1/4 Diameter, it will equal your valve seat orifice.
Last edited by Technician1002 on Fri Nov 19, 2010 7:36 pm, edited 2 times in total.
Soooo how do I calculate how far my piston should retract before it's no longer the limiting factor if the piston is just retracting off the port into an open valve housing? I understand how you did your calculation, but yours seems far simpler, as your piston just uncovers part of a circular port...
The area of the wall of an imaginary cylinder extending between the piston and the opening to the barrel is the same size as the opening into the barrel at 1/4 the diameter.
This would mean the barrel and the imaginary cylinder has the same restriction to flow. One is the gap between the piston and valve seat and the other is the opening in the valve seat. Opening it further removes the cylinder from the flow and the opening into the barrel is the only significant restriction to flow. This is why I use 1/2 Diameter instead of 1/4 diameter.
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