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Posted: Fri May 16, 2008 10:23 pm
by daxspudder
well, sojo was just reiterating what i said in your quote there, but your using the exact kind of compressor i figured you were using, my math will do the trick for you. again to simplify it into a easy to use formula given the variables you gave:

LPM=liters per minute
C=chamber(vol)
Ts=time in seconds to add one volume
PSI=desired pressure
T=time to charge to desired PSI in seconds
X=number of atm required for given PSI

60/(LPM/C)=Ts;
PSI/14.7=X;
Ts*X=T;

so to use your example: C=1.8L, LPM=18, PSI=130:
60/(18/1.8)=6;
Ts=10;
130/14.7=8.84;
X=8.84;
6*8.84=53.04
T=53.04

so like i said in my first post, just under a minute
[/satisfied rant]
and like noname said, thermal expansion will be a small factor here, so it will probably drop a few, not more than 10psi when the air cools.

depending on the kind of pump you using, it might slow the LPM flow rate when pressure is on the other side, so it might take longer. if your compressor is used to fill car tires, it might take as long as 10 minutes, because design pressure while might max 100+psi, is designed to operate in 45-65psi range... just a thought worth adding in case you trying to gauge psi by the amount of time your pump runs, id recommend a gauge if that is the case....