I don't think it works quite like this. I was going to do some similar
calculations but then decided it was too much work to worry about. And I
try to avoid getting too scientific on this list although I've probably
over stepped my bounds a time or two. I only have to mention the infamous
discussion of batteries on cement floor discussion a few years ago as an
example that some will shudder to remember. Well, in retrospect, the list
has been doing really good this year after we survived that discussion,
cats, drivelines, and other forgotten topics.
But to make a long story short, there is the pressure decrease from the
altitude, lower temperatures, the need to compensate the fuel mixture for
high altitude. Probably too complex to worry about. I was hoping it would
be simplified in a book kind of like the "ideal" gas law (engineering
students learn that its not so ideal after all). Might make a good research
project for an engineering student though if it hasn't been done before.
David
67 BGT
71 BGT
At 10:07 PM 7/5/00 -0700, Bullwinkle wrote:
>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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