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Radius of the Earth as an Ellipsoid


Date: 06/26/2000 at 12:36:01
From: Kathy St. Amand
Subject: Earth radius based on ellipsoid

I need to determine the radius of the earth for a given latitude.  
I've been given two equations based on ellipsoid model WGS84. I get 
different answers for each.

1) The equations appear totally different. Are they?

2) Which equation will answer my question?

     x (latitude) = 40 (example number only)
     a (semimajor axis) = 6378137
     b (semiminor axis) = 6356752.314245

   equation 1:
     radius = a(1-e^2)/(1-e^2sin^2x)^(3/2)
     e= (a^2-b^2)/a^2

   equation 2:
     radius = b*(1+tanlat2)^0.5 / ( (b^2/a^2)+tanlat2)^.5
     tanlat2 = (tanx)^2

Thank you.


Date: 06/27/2000 at 12:13:25
From: Doctor Rick
Subject: Re: Earth radius based on ellipsoid

Hi, Kathy. I'm not an expert in this subject, but I'm interested and 
I'd like to learn more.

I found a Web site that helps to clarify some of the issues here.

  Geodesy (The Hong Kong Polytechnic University)
  http://www.lsgi.polyu.edu.hk/cyber-class/geodesy/syllabus.htm   

When you speak of "the radius of the earth for a given latitude," 
there are two big questions:

(1) What do you mean by the radius of an ellipsoid? 
(2) What do you mean by latitude on an ellipsoid?

Each question could have several reasonable answers. 

(1) By "radius," you could mean (a) the distance from the center of 
the ellipsoid to the point, or (b) the radius of curvature of the 
ellipsoid at the point. In the latter case, the radius of curvature 
depends on the direction you are going. The Web site calls the radius 
of curvature in the north-south direction the "radius of curvature in 
the meridian," and in the east-west direction, the "radius of 
curvature in the prime vertical."

(2) By "latitude," you could mean (a) the angle between a line from 
the center of the earth and the equatorial plane (called the 
"geocentric latitude"), or (b) the angle with the equatorial plane 
that you'd get if the ellipsoid were stretched in a north-south 
direction into a sphere (called the "reduced latitude"), or (c) the 
angle between the equatorial plane and a line that is perpendicular to 
the surface of the ellipsoid at the point (called the "geodetic 
latitude").

According to the Web site, your equation 1 gives the radius of 
curvature in the meridian as a function of geodetic latitude x. 

Now for equation 2. The site doesn't discuss anything like it, but 
I transformed the equation into this form:

     r = 1/sqrt(cos^2(x)/a^2 + sin^2(x)/b^2)

If we let x be the geocentric latitude and let r be the distance from 
the center of the ellipsoid, then r*cos(x) is the distance out from 
the Z axis, and r*sin(x) is the Z distance from the equatorial plane. 
Using the equation of the ellipse,

     x^2/a^2 + y^2/b^2 = 1

you can show that the formula is correct. Thus, your second equation 
gives the distance from the center as a function of the geocentric 
latitude x.

Now it's up to you. Each equation is correct, in that it gives the 
radius of the earth as a function of latitude. But which kind of 
radius do you want, and which kind of latitude are you given? I can't 
tell - you have to decide.

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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