In other words, the velocity of the plane with respect to the ground is the vector sum of the plane's velocity relative to the air and the air's velocity relative to the ground. See Fig. 18. Note that, in general, is parallel to neither nor . Let us now consider how we might implement Eq. (81) in practice.

As always, our first task is to set up a suitable Cartesian coordinate system.
A convenient system for dealing with 2-dimensional motion parallel to the Earth's surface
is illustrated in Fig. 19. The -axis points northward, whereas the -axis points
eastward. In this coordinate system, it is conventional to specify a vector in
term of its magnitude, , and its *compass bearing*, . As illustrated in Fig. 20,
a compass bearing is the angle subtended between the direction of a vector and the direction to
the North pole: *i.e.*, the -direction. By convention, compass bearings
run from to . Furthermore, the compass bearings of North, East, South, and West
are , , , and , respectively.

According to Fig. 20, the components of
a general vector , whose magnitude is and whose compass bearing is , are simply

As an illustration, suppose that the plane's velocity relative to the air
is
, at a compass bearing of , and
the air's velocity relative to the ground is
, at a compass
bearing of . It follows that the components of and
(measured in units of km/h) are

(83) | |||

(84) |

According to Eq. (81), the components of the plane's velocity relative to the ground are simply the algebraic sums of the corresponding components of and . Hence,

(85) |

Our final task is to reconstruct the magnitude and compass bearing of vector ,
given its components
. The magnitude of follows
from Pythagoras' theorem [see Eq. (35)]:

(86) |

In principle, the compass bearing of is given by the following formula:

(87) |

(88) |

if . These expressions can be derived from simple trigonometry. For the case in hand, Eq. (89) is the relevant expression, hence

(90) |