Calculate the headwind component for helicopter performance purposes, if the take-off direction is 360°M and the reported wind is 315°/20 kt?
The wind velocity is made of two components, one acting along the take-off direction called the headwind component, and the other acting across the take-off direction called the crosswind component.
It is possible to calculate the crosswind component taking the sine and the cosine of the angle difference between the take-off direction and the wind direction:
- Head/Tailwind component = Wind Speed x cos(Take off - Wind direction)
- Crosswind component = Wind Speed x sin(Take off - Wind direction)
Headwind component = 20 kt x cos (360° - 315°) = 14 kt.
According to CAT.POL.H.105 - General:
When showing compliance with the requirements, among others, account shall be taken of the environmental conditions, concerning the wind conditions:
(A) except as provided in (C), for take-off, take-off flight path and landing requirements, accountability for wind shall be no more than 50% of any reported steady headwind component of 5 kt or more;
Since, the reported headwind component is 14 kt, then 7 kt should be taken into account for performance purposes.
NOTE:
Reported wind is always magnetic.
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