# Reply To: Please help me do the math…

Forums › Air Tanks, Pumps, Compressors, & Filters › Please help me do the math… › Reply To: Please help me do the math…

“kmd1984”No problem…

You still did not explain the math behind this though… 48 cubic inches and 3000 psi equals somehow 5.7 cubic feet? How?!

Wait, I have an idea!

48 CI is 0.0277778 CF and 3000 psi is 206.8427 bar, thus 0.0277778 x 206.8427 = 5.745! Correct?!

Again,

215 CC is 0.00759265 CF and 3000 psi is 206.8427 bar, so 0.00759265 x 206.8427 = 1.5704 CF!

I believe I get it now!

Thanks,

Kmd

OK – so this still doesn’t tell us how to come about the final #. I think I have that now, see below

Take the amount of air in source tank (Size in CC x BAR) Now if the Air is 4500PSI and you want to fill your tank to 3000psi then your “available air” would be 1500PSI or 103.4214Bar. (SInce once the source tank gets below the pressure you want to fill to it will no longer fill it will be equalized)

A tank rated at 88CF has a physical capacity (or fluid volume) of 7.9L so the “available air would be 7.9 (1000) = 7900CC x103.4214 (1500psi of available pressure) or 817029.06

Now take the size of your tank to be filled (in cc) and multiply by the amount of air needed in BAR. In this case it would be 215 times 68.9476 (1000psi- to take the tank from 2000 to 3000). or 14823.734.

Now divide the available air (817029.060) by the amount needed (14823.734) and you get 55.11 fills.

If you wanted to till to 3000 and refill at 1000 then you would need 137.8951bar (2Kpsi) times 215 which = 29647.4465 which when you divide your source air of (817029.06) by gives you 27.55 fills.

Now if you only want to go from 1000 up to 2000 each time there would be 1361714.31 available (7900cc x 172.3689bar (2500psi)) and it would take 215cc x 68.9476 (1Kpsi) 14823.734 to fill each time so if we divide 1361714.31 by 14823.734 we would get 91.86 fills.

Of course we would then have to figure out the volume of the fill hose and reduce the available air by that # times the # of fills as that will be “wasted” each time you De-pressure the line to remove the tank.

Quite honestly if I had to do all this figuring each time I used my tanks I would just shoot my R9 all the time.

But this was a good brain work out. Thanks for explaining the question better and not just letting it go.

"kmd1984"No problem... You still did not explain the math behind this though... 48 cubic inches and 3000 psi equals somehow 5.7 cubic feet? How?! Wait, I have an idea! 48 CI is 0.0277778 CF and 3000 psi is 206.8427 bar, thus 0.0277778 x 206.8427 = 5.745! Correct?! Again, 215 CC is 0.00759265 CF and 3000 psi is 206.8427 bar, so 0.00759265 x 206.8427 = 1.5704 CF! I believe I get it now! Thanks, KmdOK - so this still doesn't tell us how to come about the final #. I think I have that now, see below Take the amount of air in source tank (Size in CC x BAR) Now if the Air is 4500PSI and you want to fill your tank to 3000psi then your "available air" would be 1500PSI or 103.4214Bar. (SInce once the source tank gets below the pressure you want to fill to it will no longer fill it will be equalized) A tank rated at 88CF has a physical capacity (or fluid volume) of 7.9L so the "available air would be 7.9 (1000) = 7900CC x103.4214 (1500psi of available pressure) or 817029.06 Now take the size of your tank to be filled (in cc) and multiply by the amount of air needed in BAR. In this case it would be 215 times 68.9476 (1000psi- to take the tank from 2000 to 3000). or 14823.734. Now divide the available air (817029.060) by the amount needed (14823.734) and you get 55.11 fills. If you wanted to till to 3000 and refill at 1000 then you would need 137.8951bar (2Kpsi) times 215 which = 29647.4465 which when you divide your source air of (817029.06) by gives you 27.55 fills. Now if you only want to go from 1000 up to 2000 each time there would be 1361714.31 available (7900cc x 172.3689bar (2500psi)) and it would take 215cc x 68.9476 (1Kpsi) 14823.734 to fill each time so if we divide 1361714.31 by 14823.734 we would get 91.86 fills. Of course we would then have to figure out the volume of the fill hose and reduce the available air by that # times the # of fills as that will be "wasted" each time you De-pressure the line to remove the tank. Quite honestly if I had to do all this figuring each time I used my tanks I would just shoot my R9 all the time. But this was a good brain work out. Thanks for explaining the question better and not just letting it go.