SpudFiles

so I've got a 1.5" Bore cylinder with a 2" stroke, and I charge it to 100 psi. After its shot it goes up to 200 psi. what sort of energy would my projectile have?

bore area = 5.546025

final pressure 200psi

200psi dvided by 5.546025 =36.061

the projectile impacted with about 36 lbs of force

This sound right? i think it does

So the stroke length is irrelevant? if the cylinder gets pushed all the way back, which would not give accurate results, I would just up the pressure and try again?

I think so, the only significance of stroke length would be compressible volume of air.

I am no engineer just a 15 year old high school freshman with an inquisitive mind.

metalmeltr wrote:bore area = 5.546025

final pressure 200psi

200psi dvided by 5.546025 =36.061

the projectile impacted with about 36 lbs of force

This sound right? i think it does

I don't think so, because that calculation includes 100 psi that was not imparted by the projectile... wouldn't you just take the difference and use that?
Final pressure 200 psi
Starting pressure 100 psi
Difference, 100 psi
100 psi divided by whatever blah blah then get the answer.

And I don't think the stroke length has anything to do with it either, as long as there is enough room for it not to hit the back.

I see it as energy used to build up to 200 psig. If you start with a cylinder twice as long as the on preprsurized to 100psig with tha same bore and start it at 0 psig and in the end get 200 psig halfway down the stroke you will have 100psig and at the end you have 200psig. So by starting with 100psi just reduces the travel of the cylinder, the final pressure is what you want. Not sure, but this seems to make sense to me.

Well, both sides make some sense to me. I'm out
Someone with a bigger brain will come by and help. I'm just a 14 year old high school freshman

Nice to know were in the same boat

I'm with metalmeltr on this one. To use some physics speak, I'd look at it as an equilibrium system. If it is at equilibrium (no acceleration) the force of the air pushing back equals the force of the whatever pushing against the compressed air.

Of course there's some intermediate steps in there for the force transfer, but they can be ignored if the piston has been displaced.

lockednloaded wrote:So the stroke length is irrelevant? if the cylinder gets pushed all the way back, which would not give accurate results, I would just up the pressure and try again?

Correct.

metalmeltr wrote:I see it as energy used to build up to 200 psig. If you start with a cylinder twice as long as the on preprsurized to 100psig with tha same bore and start it at 0 psig and in the end get 200 psig halfway down the stroke you will have 100psig and at the end you have 200psig. So by starting with 100psi just reduces the travel of the cylinder, the final pressure is what you want. Not sure, but this seems to make sense to me.

i to agree with you, just as an example if you have 1square in. with 100 psi of air behind it then plance 100punds of wight on it it wont move, place another 100punds on it and the air pressure will be 200punds, so if you work out the area x that by your final pressure and you got your answer....... i think...

Ok yeah, the more I think about it, the more I think you're right... my bad!

I'm really considering building one of these, It would cost me fifty dollars to build it from a 1-1/16" bore 2" stroke cylinder. Do you think it would be a worth while way to calculate FPS of a projectile? I'd just do the energy calculations, then divide it by projectile mass

Lockednloaded wrote:It would cost me fifty dollars to build it from a 1-1/16" bore 2" stroke cylinder. Do you think it would be a worth while way to calculate FPS of a projectile?

No.

thats a good deal, but its nearly twice the price and lacks the DIY spirit. Its kind of like buying a QEV vs building your own piston valve

A chronograph would have the advantage of accuracy, some energy will be lost to heat in the foce plate cylinder aparatus.