Longitude and Latitude to Determine DistanceDate: 5/21/96 at 21:9:50 From: Peter Atlas Subject:Longitude and Latitude to Determine Distance On the College Geometry page, a user writes in asking: I've been looking for the equation that finds the distance between two cities, given the latitude and longitude of both cities. I am a high school teacher, and quite interested in this question. Is there a way to solve it using rectangular coordinates (my students don't study spherical geometry)? I have the formula: distance (along a great circle) = R * ACos[sin(lat1)*sin(lat2) + cos(lat1)*cos(lat2)*cos(long1-long2)] but can't seem to derive it using only rectangular coordinates. Thank you for your help, Peter Atlas Date: 6/17/96 at 18:31:38 From: Doctor Anthony Subject: Re: Longitude and Latitude to Determine Distance I have used my own notation, but you should be able to see how it fits in with the equation you quote. This problem can be done by using the scalar product of two vectors to find the angle between those vectors. If the vectors are OA and OB where A and B are the two points on the surface of the earth and O is the centre of the earth. The scalar product gives: OA*OB*cos(AOB) = R^2*cos(AOB) where R = radius of the earth. Having found angle AOB the distance between the points is R*(AOB) with AOB in radians. To find the scalar product we need the coordinates of the two points. Set up a three dimensional coordinate system with the x-axis in the longitudinal plane of OA and the xy plane containing the equator, the z-axis along the earth's axis. With this system, the coordinates of A will be Rcos(latA), 0, Rsin(latA) and the coordinates of B will be Rcos(latB)cos(lonB-lonA),Rcos(latB)sin(lonB-lonA),Rsin(latB) The scalar product is given by: xA*xB + yA*yB + zA*zB = R^2cos(latA)cos(latB)cos(lonB-lonA) + R^2sin(latA)sin(latB) Dividing out R^2 will give cos(AOB): cos(AOB) = cos(latA)cos(latB)cos(lonB-lonA) + sin(latA)sin(latB) This gives AOB, and the great circle distance between A and B will be: R*(AOB) with AOB in radians. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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