Math Modification to Increase Air Efficiency in a Paintball Gun

AI Thread Summary
A modification to paintball guns aims to increase air efficiency by reducing the firing chamber's volume with orings, which increases pressure while using the same amount of air. This adjustment allows for decreased air input to maintain ball velocity, enhancing overall efficiency. Some participants in the discussion misunderstand this concept, arguing that it doesn't truly increase efficiency, despite the mathematical basis for the modification. The complexity of air gun systems means that various factors, including design and environmental conditions, influence efficiency outcomes. The original poster seeks a mathematical formula to demonstrate that a specific pressure can be achieved with less air, aiming to clarify misconceptions in the community.
Highschoolkid00
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hey guys, I am new

i didnt exactly which section to put this in...but i know it involves math so whatever

ok i was on a paintball forum and this guy came up with a modification to do to a gun to increase air effiency (which is a very important aspect in a gun)

In the firing chamber of the gun he placed lots of orings to create less volume in the chamber to create more pressure using the same amount of air. Then all he has to do is decrease the amount of air going in the chamber to get ball velocity back to normal (this is through a setting called dwell...its how long the valve is opened for in milliseconds).by using less air at the same pressure he saves air and DUR gets better effiency...

now with that said, some people...just...dont...understand how simple that is...they kept saying he wasnt increaseing effiency at all...he just somehow used more air in one place and less in another...ppl tryed equations but of course there incorrect if all the variables equal the same.

anywho i really want to know the correct equation or formula on how to solve this...PSI=pounds per inch^2...so would it be something like...150psi=(x amount of volume in inches squared) x (y amount of mass in pounds (which would very small since air weighs little)

thanks and sorry for long post and if i posted in wrong section

-luke
 
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An air gun is likely an extremely complicated system when discussing efficiency. The effects of what was done would probably depend greatly on the design and specifics of that model of air gun. In fact, the dimensions, materials, elasticity of the balls, humidity of the air, etc, etc, will all probably play a role in a discussion of efficiency.

In short, you would probably be confounded in a brief attempts to create a model. Bets to try and test instead.
 
or wait?

would u have to divide instead...sorry I am ususally a lil above average in this crap...cuz it doesn't make sense this way...the way i thought it was would meen that 150lbs of air fir inside something that's 1 square inch approx...

ok so it is division DUR:rolleyes:
 
no i tihnk i have it...

i wanted this done in...whatever its called when all conditions are perfect...

i just wanted to know the mathematical formula to show that u can get X amount of psi with less air


its a really simple thing that smart ppl would say DUHHHH to lol...i just wanted to prove the dummies with a formula...and of course figure it out for myslef haha


thx guys
 
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