Radius of the Earth as an EllipsoidDate: 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/ |
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