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Re: compression readings above sea level

To: Barney Gaylord <barneymg@ntsource.com>
Subject: Re: compression readings above sea level
From: ENRIQUE CLAURE <span@datacom-bo.net>
Date: Wed, 13 May 1998 09:53:05 -0400
I think Barney deserves a round of applaus on the excelente information he
supplied on compression readings above sea level. Thank you Barney you have
made my day. Enrique 

At 11:45 PM 5/12/98, you wrote:
>At 05:26 PM 5/12/98 -0400, ENRIQUE CLAURE wrote:
>
>>I was wondering, if measuring car compression at sea level and at high
>altitud say like Pikes Peak in Colorado will give same values? I was told
>that measuring at higher than sea level will give lower readings? any
>comments?
>
>Atmospheric pressure decreases with altitude.  And in the range of altitude
>of interest (driveable roads) the pressure drop is nearly linear, so you
>can easily interpolate between the numbers below.
>
>The easiest number to work with is the ratio between pressure at altutude
>compared with pressure at sea level.  If you use this ratio, all the rest
>of the factors cancel out.  These ratios are as follows.
>
>   Atmospheric Pressure Ratio (APR)
>At sea level  1.0000
>At  5,000 Ft  0.8320
>At 10,000 Ft  0.6877
>At 15,000 Ft  0.5643
>
>Starting with the compression _gauge_ reading you get at sea level, just
>multiply by the APR for the alitude of interest to get the _gauge_ reading
>at that altitude.  For example, suppose you get a pressure reading of 140
>psig at sea level.  For other altitudes you would get the following results:
>
>   Compression Reading With Altitude
>At sea level  1.0000 x 140 = 140 psig
>At  5,000 Ft  0.8320 x 140 = 116 psig
>At 10,000 Ft  0.6877 x 140 =  96 psig
>At 15,000 Ft  0.5643 x 140 =  79 psig
>
>Because of temperature drop at higher altitudes, the air density is not
>reduced as much as the pressure.  The density ratio is also nearly linear
>within the range of interest:
>
>   Atmospheric Density Ratio (ADR)
>At sea level  1.0000
>At  5,000 Ft  0.8617
>At 10,000 Ft  0.7385
>At 15,000 Ft  0.6292
>
>These numbers indicate the reduction of the mass flow of air that your
>engine can take in at various altitudes, and in turn are roughly
>proportional to the power that your engine can make in the thinner air.
>Suppose your engine can make 80 HP at sea level.  You can multiply the sea
>level power output by the ADR to get the expect power output at other
>altitudes:
>
>   Power Output With Altitude
>At sea level  1.0000 x 80 = 80 HP
>At  5,000 Ft  0.8617 x 80 = 69 HP
>At 10,000 Ft  0.7385 x 80 = 59 HP
>At 15,000 Ft  0.6292 x 80 = 50 HP
>
>As I recall, the access road to Pikes Peak Park starts at about 8,500 feet
>and ends up at a little over 14,000 ft.  The result of the thin air at
>these altitudes is that you enter the gate with about 62 HP and reach the
>top with about 52 HP.  Now you can see why people who drive a lot at higher
>elevations can be such great fans of turbo chargers.
>
>Barney Gaylord
>1958 MGA with an attitude (and 72 HP an a good day)
>
>


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