Karl,
I think we all knew what you meant. Thanks for the correction factor.
However I've got two additional commnents.
1) You still got your slide rules? The last one I owned fell out of
my Triumph 650 motorcycle as I was accelerating back in '74. I then bought
an HP 45 which I still have and use.
2) Using my HP 45 and the ideal gas law, the atmospheric pressure
is 14.71 PSI at standard conditions, as follows:
P=nRT/V (ideal gas law)
P= pressure
n= moles of gas
R= Universal gas constant
T = temperature
P= 1 (lb-mole of air) * (10.73psia/(ft3)/lb-mole/(deg R) * (492 degR)
---------------------------------------------------------------
359 (ft3/lb-mole)
= 14.71 PSI
This is the well accepted number that most engineers use when dealing
with gaseous systems. Anyway, whether you use 14.8 or 14.7, the difference is
less than 1%, and both results give you correct "ballbark" answers. In fact,
maybe
14.8 is truly better. Since 14.7 is based on standard conditions (32deg F
and 1 atm),
after correcting it for a higher temperature, it may be 14.8 PSI.
Well, enough with boring details.
Jeff "Picky-Picky" Nathanson
At 11:05 AM 10/13/96 -0400, you wrote:
>In a message dated 96-10-12 23:29:56 EDT, you write:
>
><< A little correction, see below=
>
> >A 10:1 compression ratio thus should yield (14.8 X 10)-14.8 or about
> >13.3
> >PSI
>
> You mean 133 psi not 13.3 psi :)
> >>
>
>Oh, this is just swell. I act like Mr. Smarty Pants, sending a nit-picking
>post correcting Barry's formula, and I can't get the decimal point right.
> You can see that my work improved greatly in the mid-1970's with the advent
>of the cheap calculator. I still have all of my slide rules though, ready
>for those occasions when I don't care if I get the order of magnitude
>right...
>
>-Karl
>
>
Jeff C. Nathanson
Director of Product Development
Manufacturing Systems & Technologies, Inc.
|