Obtaining Bearing from a Velocity VectorDate: 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, (PxV).(PxN) bearing = arccos ----------- |PxV||PxN| 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 http://mathforum.org/dr.math/ |
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