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Obtaining Bearing from a Velocity Vector

Date: 08/01/2001 at 14:55:00
From: Michael McGuire
Subject: Obtaining bearing from a velocity vector

I have the x, y, and z components of a velocity vector of an airplane. 
(The center of the earth is the origin with the x- and y-axis passing 
through the prime meridian and 90 degrees east, respectively, and both 
pass through the equator.  The z-axis is Earth's axis.) I must use 
this vector to calculate the bearing of the plane. Bearing being zero 
degrees for north, 45 degrees for northeast, etc.

Date: 08/02/2001 at 14:03:53
From: Doctor Rick
Subject: Re: Obtaining bearing from a velocity vector

Hi, Michael.

If I understand you correctly, you also need to know the position of 
the airplane in this coordinate system. Let's say the position vector 
is P and the velocity vector is V. Then the great circle in whose 
plane the airplane is moving has normal PxV (the cross-product or 
vector product of the position and velocity vectors). The bearing is 
the angle between this plane and the plane of the line of longitude at 
the airplane's position. The normal to the latter plane is PxN, where 
N is a vector along the positive Z axis (pointing toward the north 
pole). The angle between the two planes (the bearing) is the angle 
between the normal vectors. We can find this angle using the fact that 
the dot product (scalar product) of two vectors is the product of 
their magnitudes times the cosine of the angle between them. Thus,

  bearing = arccos -----------

We have a little problem here: as it stands, we'll only get bearings 
between 0 and 180 degrees east of north. We've lost a sign (east or 
west). However, the sign of the z component of PxV tells you whether 
the bearing is east or west: if the sign is negative, then the bearing 
is west; if the sign is positive, the bearing is east.

- Doctor Rick, The Math Forum   
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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