If I recall, you really need to apply Boyle's Law with an
adiabatic cross component and then iterate using Lipshitz Theorum.
Of course you must hold all the lower order variables constant to
simulate steady state thermal dissipation.
BTW this will not tell you anything at all about your flux core
status or when the dilithium crystals are about to blow.
Mike L.
60A,67E,59Bug
----- Original Message -----
From: Bullwinkle <yd3@nvc.net>
To: Andrew B. Lundgren <Lundgren@iname.com>; mgs
<mgs@autox.team.net>
Sent: July 6, 2000 1:07 AM
Subject: Re: Octane and altitude ratios???
> Andrew:
>
> My brother and I spent several hours searching for a formula
that would convert
> sea level compression ratio, to that at a different altitude. We
couldn't really
> find anything. We found a few formulas that would give pressure
for a given CR
> but it would take several formulas, and calculations to get any
results.
> However the following might provide some insight.
>
> Air pressure at sea level is 14.7 lbs/sq. in.
> Air pressure at 10,000 ft is 10.2 lbs/sq. in.
>
> Air density in % at 10,000 ft is (14.7-10.2)/14.7 or 31% less.
If an engine is
> using a 10:1 CR at sea level, then at 10,000 feet the compessed
air's density
> would be 31% less. I would think a compression ratio of 6.9:1
would compress
> the air 31% less than the 10:1 ratio.
>
> So my guess is that a 10:1 CR at sea level equates to 6.9:1 at
10,000 feet.
>
> The decrease in air pressure as the altitude increases is not a
linear
> function. The correction at 5000 ft wouldn't be half of the
above. As a rough
> guess, I would say that decrease in air pressure from sea level
to 5,000 ft
> would be approximately 2 lbs./sq. in. So the equivalent CR of a
10:1 might be
> about 8.7:1.
>
> This is just a guess.
>
> Blake
>
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