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Old 04-04-2013, 10:08 AM #1
Ryn_Droma
 
 
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math question... WHAT?!??

I know right, you didn't expect a math question on a PB forum, well it's PB related and you guys are smart, so help if you can cause the people on math forums are viewing my thread, but not responding to it .

I am working on my own little dream f a paintball marker and am stuck on some math. I need to figure out the pressure required for operation given the volume of air I have and the velocity we all want. The formulas (yes I stole these from somewhere on here) I am trying to work with are;

E = 1/2(M)(V^2)

M in kg (.0033kg for paintball)
V in meters per sec (300 feet/sec, converted to 91.44 meters/sec)
E solved to be 27.5922kg m^2/sec^2

Also;

E = (V)(P)
V is meters cubed (2.2472in^3 converted to .00003683m^3)
P solved to be 749,788.043kg/m(sec^2)

I converted that to PSI using an online calculator and got 108.7PSI and that would mean 2.2472in^3 @ 110 PSI would give me a velocity just over 300 FPS.

I don't think that number is right. I was expecting to get in the range of 150-180 PSI

Can anyone see a mistake, am I using the wrong formulas? It's been over 10 years since high school and I never was that great at math. Any help will be greatly appreciated.
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Old 04-04-2013, 09:41 PM #2
ElPanda
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did you square your velocity?

I literally did exactly what you are doing about a year ago and my mistake was not squaring my velocity
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Old 04-04-2013, 11:22 PM #3
AndrewTheWookie
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Looks like you did mv^2 only, and forgot to multiply by .5 in the first equation (which works its way through to give us about 54psi). Referencing ElPanda's example, IIRC his equation was how much volume he needed for 300 fps at 180psi, and the correct equation worked out to about .678in^3, so given over 2in^3 in your dump chamber, even as low as 54psi doesn't sound too outlandish.
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Last edited by AndrewTheWookie : 04-05-2013 at 03:05 AM.
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Old 04-05-2013, 01:28 PM #4
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You're dead on on my mistake, thank you. I suppose you are also right on the pressure, I was flipped backwards on that . You are correct on what the original equation was for. I wrote it all down a long while back, but forgot to bookmark the thread.

ElPanda, thank you also
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