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Jul 21, 2022 at 15:22 comment added Sacha T aerodrome = 273K, T isa = 288K. so dT = 15K. Applying the correct formula (see Namsters' answer below) gives 15/0.001981 * ln (1+ 1.981/288) = 51.90ft (since dH aerodrome = 0). The table from PANS OPS Doc 8168 uses an approximation which works for temps above -15C (4 times the alt above the aerodrome in thousands of feet, times dT which equals 60).
Jan 25, 2021 at 4:55 answer added Namster timeline score: 1
Jan 25, 2021 at 1:55 comment added Juan Jimenez I agree, you are mixing temperature scales (K and C) and hence your incorrect result.
Jan 24, 2021 at 16:26 history edited Namster CC BY-SA 4.0
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Jan 24, 2021 at 4:34 comment added skipper44 Infact 288degK at 0 ft altitude is = 15degC = ISA, so intended dev from ISA seems to be zero for purpose of the example question.
Jan 24, 2021 at 3:09 comment added Ralph J 259 degree temp dev from ISA is unreasonable by an order of magnitude. You may be mixing C and K units there. Please fix that & edit your question accordingly.
Jan 24, 2021 at 2:27 review First posts
Jan 24, 2021 at 3:10
Jan 24, 2021 at 2:23 history asked Namster CC BY-SA 4.0