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### Radius of the Earth as an Ellipsoid

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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

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:
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.
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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 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/
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Associated Topics:
College Higher-Dimensional Geometry
High School Higher-Dimensional Geometry

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