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Azimuth Angle

Date: 09/08/98 at 01:24:30
From: Rachel Harrington
Subject: Calculating the azimuth angle

I am trying to find out more about the azimuth angle. I think that it 
is used to find the error between true north and magnetic north. I 
know that the formula is:

   AZ = (cos D*sin t)/[(cos L*sin D)-(sin L*cos D*cos t)]

I need to know what D, t and L stand for. I think that they are 
latitude, declination and meridian angle. I do not know what 
declination and meridian angle mean, however.

I am a high school math teacher working on a navigation project with 
my students. Any answers you could give me would be very helpful. 


Date: 09/08/98 at 13:16:45
From: Doctor Rick
Subject: Re: Calculating the azimuth angle

Hi, Rachel. I have done some celestial navigation, so I can explain 
the terms. Your formula for azimuth, however, does not match mine; 
there is a missing element, and I have not been able to complete it so 
as to give a formula equivalent to mine.

The formula gives the azimuth angle, which is the horizontal direction 
from the observer to the celestial object being observed - that is, 
the angle from North to the point on the horizon directly below the 
object. Another way to look at it is that it is the angle between the 
observer's line of longitude and the line (great circle) connecting 
the observer's position to the point on the earth directly below the 
celestial object. 

The azimuth angle has nothing directly to do with magnetic north, but 
by comparing the azimuth angle with the bearing on a magnetic compass, 
you can find the error between true north and magnetic north. (This 
error is called the magnetic declination, but it has nothing to do 
with the declination described below.)

L is the observer's latitude.

D is the declination of the celestial object being observed. 
Declination is the equivalent of latitude on sky maps - it is the 
latitude of the point on the earth where the object is straight 

t is the meridian angle, also called the local hour angle or LHA. It 
is the difference between the observer's longitude and the longitude 
of the celestial object (the latter is usually called the "Greenwich 
hour angle" or GHA).

Here is the formula I have. First you compute the altitude, called H -
this is the angle of the celestial object above the horizon:

  H = arcsin[sin(L)sin(D) + cos(L)cos(D)cos(t)]

Then you can compute the azimuth angle, Z:

  Z = arcsin[cos(D)sin(t)/cos(H)]

The same formulas can be used to find the distance and direction 
between two points on the earth given their latitudes and longitudes. 
The only difference is that H is 90 degrees minus the angular distance 
between the points, so the distance is R * (90 - H) where R is the 
radius of the earth.

I hope this information helps in your class project. I find the topic
fascinating, along with the history of navigation - the recent book,
_Longitude: The True Story of a Lone Genius Who Solved the Greatest 
Scientific Problem of His Time_, by Dava Sobel, has some good stories.

- Doctor Rick, The Math Forum   
Associated Topics:
High School Definitions
High School Euclidean/Plane Geometry
High School Geometry
High School Practical Geometry

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