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Re: Octane and altitude ratios???

To: "Andrew B. Lundgren" <Lundgren@iname.com>, mgs <mgs@autox.team.net>
Subject: Re: Octane and altitude ratios???
From: Bullwinkle <yd3@nvc.net>
Date: Wed, 05 Jul 2000 22:07:26 -0700
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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