Fortlaufende Fragennummer: Frage Nr. 100
Mahlzeit!
Irgendwo häng es bei mir und überseh irgendwas...
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The take-off distance of an aircraft is 600m in standard atmosphere, no
wind at 0 ft pressure-altitude.
Using the following corrections:
"± 20 m /1 000 ft field elevation"
"-5m/ kt headwind"
"+10m/kt tailwind"
"± 15 m / % runway slope"
"± 5 m / °C deviation from standard temperature"
The take-off distance from an airport at 1 000 ft elevation, temperature 17°C, QNH 1013,25 hPa, 1% up-slope, 10 kt tail wind is:
715 m
555 m
755 m
685 m
solution: 755m
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zu meiner denkweise:
600m (take-off distance) + 20m (± 20 m /1 000 ft field elevation) + 10m (± 5 m / °C deviation from standard temperature) + 15m (± 15 m / % runway slope) + 100m (+10m/kt tailwind) = 745m
wieso fehlen mir 10m?
muss nochmal 2° zur berechnung dazu wegen (2°/1000ft)? nur so würde es ein sinn ergeben.
Mahlzeit!
Irgendwo häng es bei mir und überseh irgendwas...
----------------------------------------------------
The take-off distance of an aircraft is 600m in standard atmosphere, no
wind at 0 ft pressure-altitude.
Using the following corrections:
"± 20 m /1 000 ft field elevation"
"-5m/ kt headwind"
"+10m/kt tailwind"
"± 15 m / % runway slope"
"± 5 m / °C deviation from standard temperature"
The take-off distance from an airport at 1 000 ft elevation, temperature 17°C, QNH 1013,25 hPa, 1% up-slope, 10 kt tail wind is:
715 m
555 m
755 m
685 m
solution: 755m
----------------------------------------------------
zu meiner denkweise:
600m (take-off distance) + 20m (± 20 m /1 000 ft field elevation) + 10m (± 5 m / °C deviation from standard temperature) + 15m (± 15 m / % runway slope) + 100m (+10m/kt tailwind) = 745m
wieso fehlen mir 10m?
muss nochmal 2° zur berechnung dazu wegen (2°/1000ft)? nur so würde es ein sinn ergeben.
Gruß
Chris
Chris
