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4PW's
8th Dec 2005, 19:39
Hello there Reader,

I'm a bit fuzzy on something from the dark ages, back when one studied Flight Planning and Meteorological subjects.

Hopefully someone closer to the subject might be able to help.

You land at a sea level aerodrome where the barometric pressure is 975Hpa, and the ambient temperature a bracing 4 degrees C.

Not a biggie, but what's the density altitude correction for the temperature variation?

Thirty feet per Hectopascal for pressure variations to 1013, but what of temperature variation from ISA?

Sorry to bore you all, and yes I have sought assistance from the considerable efforts of Mr John Tullamarine's most useful references, all to no avail.

The Standard Atmosphere link was interesting, John.

Not sure about what all those results mean though.... :hmm:

Thanks in advance.

PS Was it something like 320'/degree?

bookworm
8th Dec 2005, 19:48
1 degC at the same pressure changes the density by ~ 1/288 in fractional terms, so equivalent to about 4 hPa. So that's about a hundred feet, give or take.

4PW's
8th Dec 2005, 23:30
I thought I recalled something else, however thank you for that, Bookworm.

Rainboe
8th Dec 2005, 23:51
ISA is SL temp 15 deg C reducing 2 deg C /1000' PA down to -56 deg C

Old Smokey
9th Dec 2005, 02:24
First you must find the Pressure Height, and find the standard temperature there, for comparison with the actual temperature.

Sea Level at 975 hPa = 1061 ft Pressure Height

Standard Temperature at 1061 feet = 12.9°C

ISA Deviation = +4.0 - 12.9 = ISA - 8.9°C

Density Height Correction = - 8.9 X 120.9 = - 1076 feet

Density Height = 1061 -1076 = - 15 feet

In your example the low temperature has very nearly compensated for the low pressure, in giving a Density Height very near to the 'original' altitude.

(Note that the effect of QNH variation in calculating Pressure Height is logarithmic, not linear, and in this case was 27.74 feet per hPa).

Regards,

Old Smokey

4PW's
9th Dec 2005, 04:27
Thank you, Rainboe.

Quite right too, but it's Smokey's answer I was looking for.

Thanks for that Old Smokey.