Solche Aufgaben kommen neuerdings beim lba in gnav des öfteren vor...


The
departure is from an airfield at 2000 ft elevation. Temperature at the
field is +20°C, QNH = 1013 hPa. The plan is to climb to FL 290, outside
air temperature is -40°C. The CAS in the climb is 180 kt. Compressibility negligible.
The average TAS in the climb is ____


(Die Lösung dieser ganz neuen Frage ist beim LBA in ein Eingabefeld
einzutragen, bei der Austro-Control und anderen Behörden ist es eine
Multiple-Choice Frage).

Erklärung und Lösung:

Mean Wind Velocities:

Factors such as the Rate of Climb (ROC) or Rate of Descent (ROD) and
how the wind is changing with altitude affect the determination of the
mean wind velocity.
If the wind velocity was changing at a constant
rate (in terms of both direction and speed) and the ROC/D of the
aircraft was also constant, then the wind at the halfway altitude could
be used. In a situation however, where ROC decreases with increase in
altitude, or the wind velocity is more constant at the higher altitudes,
it may be better to use the wind velocity at an altitude of 2/3 of the
climb.

Mean TAS:
Obviously, the TAS will vary through the
climb/descent as the air density decreases/increases. Again the time
spent at any given altitude(s) will affect how much bias is given to any
particular TAS although the procedure is the same as for wind velocity.

Solution:
Average TAS (assuming decreasing rate of climb as climbing to FL290) = TAS at 2/3 level
2/3 level is at 2,000ft + 18,000ft = FL200
Temperature at 2/3 level = +20° - 40° = -20°C
Using these inputs for the mechanical computer, the average TAS = 248kts
Quelle: CAT