altimeter settings

[bigelz1215]

can anyone point me to a specific caa or metoffice diocument where it states that in general near the surface a reduction in pressure of 1mb is equivalent to 27.5 ft ? most student PPL's are taught 1mb is equivalent to 30 ft but i am positive the actual correct value is 27.5 .

[Spitoon]

It's an approximation - I learned 27 ft but only as a rule of thumb. If you want the definitive answer it's part of the ICAO ISA which you can find in Doc 7488 Manual of the ICAO Standard Atmosphere (extended to 80 kilometres). It gives an equation that will give you an exact value for any level you want.

[bigelz1215]

thanks spitoon

[Mark 1]

The pressure lapse rate with altitude is equal to the density*g, and the density is proportional to pressure divided by (absolute) temperature.

At ISA sea level the density of air is 1.225 kg/m^3
1MBar = 100 N/M^2
g=9.80665M/s^2

Doing the calc thus gives change in height for 1MB = 8.324m or 27.3 ft

[bigelz1215]

superb - thanks a bunch Mark

[Old Smokey]

Doing the calculation for a 1 hPa change isn't quite enough. For a 1 hPa change from standard, 27.3 feet is indeed correct, but further change is NOT linear, but exponential.

The lower is the QNH, the higher is the correction, e.g. for 942.13 hPa (the equivalent of a 2000 ft correction, the mean correction becomes 28.1 ft per hPa.

The higher is the QNH, the lower is the correction, e.g. for 1050.41 hPa (the equivalent of a -1000 ft correction, the mean correction becomes 26.9 ft per hPa.

If we are to generalise and wish for 1 standard linear correction, it would be wise to opt for the higher correction for lower than standard QNH (i.e. 28.1 ft per hPa), and for the lower correction for higher than standard QNH (i.e. 26.9 ft per hPa).

Such a choice would always yield exact or slightly higher than correct Pressure Height for the QNH, at which performance would equal or be slightly less than the figures extracted from the performance data, i.e. on the "safe" conservative side.

If you find a formula desirous to be very close to the truth, here is a slightly simplified algorythm optimised between zero and 1000 feet -

If you use Natural Logarithms (base e -

PH Correction = 27603.8 X Ln (1013.25 / QNH)

If you use Logarithms to base 10 -

PH Correction = 63560.05 X Log (1013.25 / QNH)

If you happen to use Inches of Mercury in lieu of hPa on your altimeter, substitute 29.92 for 1013.25 in the formula without other change.

The calculation depends upon the Mean temperature of the air column between zero and the Pressure Height (Tm), introducing difficulty into this type of calculation as you don't know the answer in advance to calculate Tm, thus the optimisation for zero to the +1000 ft correction. It leads to slight conservatism in the range -1000 ft to zero, and +1000 ft to +2000 ft.

This formula for the -1000 to +2000 ft correction corresponds to a QNH range of 942.13 hPa to 1050.41 hPa, which should cover 99.99% of cases. Considering the change of geo-potential of the atmosphere (the change in 'G' with change of distance from Sea Level) is minor and not worth considering at these small corrections, certainly worth considering at high altitudes (although Airbus don't address it in their otherwise good tome "Getting to grips with Aircraft Performance").

The formula is exact at and between zero (1013.25 hPa) to +1000 ft (977.17 hPa), +5 ft in error at -1000 ft (1050.41 hPa), and +10 ft in error at +2000 ft (942.13 hPa), all on the CONSERVATIVE side. Note that the very small errors do not exceed 0.07 ft per hPa in the extreme cases.

Regards,

Old Smokey