Distance Between Points on the EarthDate: 12/11/97 at 21:21:31 From: Cameron Hollingshead Subject: Proofs for Non-Euclidian Geometry The problem is to solve for the distance between two latitude/ longitude points with no parallels, say... 24N 70E and 65N and 30W. The actual points are irrelevant as long as they do not share a meridian or parallel. I began trying to solve this problem by trying to solve for the angle between the points and the center of the earth. Along that line, I planned to use that angle "X" in the formula: (X/360) (2pi Radius), like finding the length of a chord on the circle made when the sphere is intersected by the plane containing both points and the center of the earth. After realizing that there are x,y,and z- axis measurements for the points on the surface of the sphere, I realized it was a bit more complicated. I planned to break this line down into its three axis components by doing sort of a reverse-vector addition type maneuver, which miraculously worked. This involved making a cube with the center of the earth at one corner and inscribing 90 degree arcs in each of the three two-dimensional planes: XY, XZ, and YZ. Accordingly I assumed that since the radius of the earth is about 6400kM, this value could be placed as the three axes radiating from the earth's center point. Then I inserted the north latitude measurements (on separate graphs), solved using the side-angle-side method for the third side, used this for the height of the XY component, and then solved for the XZ and YZ components. This left me with two different Cartesian coordinates; one for each point (IN 3-D). Here is the real problem: I know how to find the angle between two points on a 2-D graph, relative to the origin; just find each individual angle and take the absolute value of one minus the other. But HOW ON DARWIN'S GREEN-AND-EVOLVING EARTH DO YOU DO THE SAME FOR THREE DIMENSIONS? If the Good Doctor, or anyone else knows any of the following... 1) Is my method reasonably accurate? 1.b) If not, why not? 2) Is there an easier formula than this? 2.b) If so, how much do I pay you for it? and 2.c) If terms 2 and 2b are respectively affirmative and reasonable, then can any type of geometric proof be applied in this situation to make sure that people don't give me looks when I whip out this 10+ page math problem? ....then I would greatly appreciate sharing the solutions. I have spent in excess of three hours of intense concentration deeply pondering this problem, and all of the above procedures are of my own creation. I don't want an answer to the question so much as a formula to solve it. Thank you, Doctor Math Date: 12/31/97 at 11:53:26 From: Doctor Rob Subject: Re: Proofs for Non-Euclidian Geometry 1. Your method can work. 1.b. Not applicable. 2. Yes, see below. 2.b. It is free. 2.c. The proof is embodied in the formulas connecting rectangular and spherical coordinates, together with the fact that the dot product of two vectors is the product of their lengths times the cosine of the angle between them. The formulas are: x = r*sin(theta)*sin(phi), y = r*cos(theta)*sin(phi), z = r*cos(phi), r = Sqrt[x^2+y^2+z^2], phi = arctan(Sqrt[x^2+y^2]/z), theta = arctan(x/y), where phi is the angle the line containing the origin and (x,y,z) makes with the z-axis, theta is the angle the plane containing (x,y,z) and the z-axis makes with the xz-plane, and r is the distance of (x,y,z) from the origin. The dot product of (x1,y1,z1) and (x1,y2,z2) is x1*x2 + y1*y2 + z1*z2 on the one hand, and on the other, it is Sqrt[x1^2+y1^2+z1^2]*Sqrt[x2^2+y2^2+z2^2]*cos(t), where t is the central angle. Assume the Earth is a perfect sphere. Let all angles be measured in degrees. If the latitude is North, let phi = 90 - latitude. If the latitude is South, let phi = 90 + latitude. The North Pole has phi = 0, the South Pole has phi = 180, and 0 <= phi <= 180. If the longitude is East, let theta = longitude. If the longitude is West, let theta = -longitude. Greenwich, England, has theta = 0, and -180 <= theta <= 180. Let the angles for the two points be (phi1, theta1) and (phi2, theta2). Then compute c = sin(phi1)*sin(phi2)*cos(theta1-theta2) + cos(phi1)*cos(phi2). c is the angle at the center of the sphere made by radii from the center to the two points. Then the shortest great circle distance between the two points is d = R*Arccos(c)*Pi/180, where R is the radius of the earth in miles (or kilometers), and the arccosine is taken between 0 and 180 degrees, inclusive. -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/