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Re: Octane and altitude ratios???

To: mgs@autox.team.net
Subject: Re: Octane and altitude ratios???
From: James Nazarian Jr <James.Nazarian@Colorado.EDU>
Date: Thu, 6 Jul 2000 10:23:16 -0600 (MDT)
If you were to look up compression ratio in a car dictionary it would be
defined by the change in volume during the stroke of the engine, ie. the
volume of the camber when piston is at TDC is 1/10 the volume at BDC for
10:1. That said, compression ratio would not change, only density of the
charge, this is why turbo/supercahrge cars are less effected by altitude
differences.  You are on the right track about the charge being 31% less,
but that does not affect the compression ratio.  It has to do with the
density of air being lower before the compression started, and therefore
it is still less dense when the compression is completed

James Nazarian
'71 B roadster
'71 BGT rust free and burnt orange
'63 Buick 215 

"Aerodynamics are for people who cannot build engines"
Enzo Ferrari

> 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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