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Drift of an Airplane

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z overview

An airplane travels with speed $ 800$ km/h due east while the wind velocity is $ 50$ km/h from WSW.

\includegraphics[width=.8\linewidth]{b_flugzeug_drift}

In the picture east is the direction of the horizontal axis and north is the direction of the vertical one. The velocity vector of the air plane itself is therefore

$\displaystyle \vec{x}=\left(\begin{array}{c}800 \\ 0 \end{array}\right) $

and the vector of the wind velocity is given by

$\displaystyle \vec{y}=\left(\begin{array}{c} 50\cos(\pi/8) \\ 50\sin(\pi/8) \end{array}\right) .$

Starting at $ O$ the coordinates of the air plane are after one hour (superposition of the motions)

$\displaystyle \vec{x}+\vec{y}=\left(\begin{array}{c} 800+ 50\cos(\pi/8) \\ 50\s... ...ray}\right) \approx \left(\begin{array}{c} 846.19 \\ 19.13 \end{array}\right) .$

If $ \vec{p}$ denotes the orthogonal projection of $ \vec{y}$ on the line with direction $ \vec{x}$ , then the velocity of the air plane relative to the earth is the sum of the component due east

$\displaystyle \vec{x}+\vec{p}=\left(\begin{array}{c} 800+ 50\cos(\pi/8) \\ 0 \end{array}\right) \approx \left(\begin{array}{c} 846.19 \\ 0\end{array}\right) $

and the drift

$\displaystyle \vec{d}=\vec{x}+\vec{y}-(\vec{x}+\vec{p})=\left(\begin{array}{c} ... ...end{array}\right) \approx \left(\begin{array}{c} 0 \\ 19.13 \end{array}\right) $

towards north.

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see also:

  automatisch erstellt am 28.  3. 2008