Using Longitude and Latitude to Determine DistanceDate: 17 Apr 1995 17:47:06 -0400 From: Clarence Warren Subject: (none) I've been looking for the equation for finding the distance between two cities, given the latitude and longitude of both cities. For example: what is the formula one would use to find the distance between San Francisco (N37 37' 00" latitude, W122 22' 00" longitude) and Paris (N48 44' 00" latitude, E02 23' 00" longitude.)? I'm trying to write a program in Visual Basic that will include this distance calculation in it. Any help that members of "The Swat Team" can offer would be greatly appreciated. Date: 17 Apr 1995 21:38:43 -0400 From: Dr. Ken Subject: Re: distance on a sphere Hello there! This problem can be most easily solved by using spherical coordinates on the earth. Have you dealt with those before? Here's the transformation from spherical coordinates to normal rectangular coordinates, where a=latitude and b=longitude, and r is the radius of the earth: x = r Cos[a] Cos[b] y = r Cos[a] Sin[b] z = r Sin[a] Then we'll use the following property of the dot product (notated [p,q]): [p,q] = Length[p] * Length[q] * Cos[angle between p & q] Now, any vector that points to a point on the surface of the earth will have length r. So the right side we have r^2 * Cos[angle between p & q]. On the left side, we can pull the r's out of the dot product, and cancel them with the r's on the right side. Let t represent the angle between p and q. Then if the latitude and longitude of our two cities, p and q, are (a1,b1) and (a2,b2), we have Cos[a1] Cos[b1] Cos[a2] Cos[b2] + Cos[a1] Sin[b1] Cos[a2] Sin[b2] + Sin[a1] Sin[a2] = Cos[t] So you can compute the angle t as a function of a1, b1, a2, b2, which are the latitudes and longitudes of our cities p and q. Then visualize what you've got: draw a great circle through the points p and q. This is just a plain old Joe-Schmoe circle of radius r, and the angle t is the angle of the arc that subtends p and q. The problem from here on out is just figuring out what the arc length between p and q is. The relevant formula is Arc length = t/360 * 2Pi* r. So that's your formula. By substitution, we have Arccos[Cos[a1] Cos[b1] Cos[a2] Cos[b2] + Cos[a1] Sin[b1] Cos[a2] Sin[b2] + Sin[a1] Sin[a2]]/360 * 2Pi * r Oh, by the way, West longitude means negative values of b, and South latitude means negative values of a. Enjoy, and let us know if I've made mistakes, or something's not clear! -Ken "Dr." Math From: Dr. Ken Date: October 30, 1997 I thought I'd add a couple of remarks that may help some people use the above formula. First, of all, you need to make sure your calculator or computer is using degrees, not radians, to figure out the Sine, Cosine, and ArcCosine functions. If you're using a calculator or computer that uses radians, then use a different version of the formula: Arccos[Cos[a1] Cos[b1] Cos[a2] Cos[b2] + Cos[a1] Sin[b1] Cos[a2] Sin[b2] + Sin[a1] Sin[a2]] * r Also, keep in mind that these formulae don't take into account the squashed nature of the earth. As you may know, the earth is kind of fat around the equator, as a result of the centrifugal force it gets from spinning on its axis. So that will throw a little error into these calculations. I've never tried to come up with a formula that takes the squashing into account, and I suspect it might be hard. -Dr. Ken The Math Forum Date: 11/21/2001 at 09:28:12 From: Nat Keller Subject: Angular earth distance I saw Chris Michels' Latitude/Longitude Distance Calculation page at http://jan.ucc.nau.edu/~cvm/latlongdist.html which he says is based on the above answer. I recently needed to program a latitude, longitude, distance calculation and found a formula in the SAS tech page (as well as a few other places). I think that your formula can be reduced to the following (and they give the same results): A = LAT1, B = LONG1 C = LAT2, D = LONG2 (all converted to radians: degree/57.29577951) IF A = C AND B = D THEN DISTANCE = 0; ELSE IF [SIN(A)SIN(C)+COS(A)COS(C)COS(B-D)] > 1 THEN DISTANCE = 3963.1*ARCOS[1]; /* solved a prob I ran into. I haven't fully analyzed it yet */ ELSE DISTANCE=3963.1*ARCOS[SIN(A)SIN(C)+COS(A)COS(C)COS(B-D)]; With more time I would solidify and show that it equals your formula. NK Date: 11/21/2001 at 10:38:54 From: Doctor Rick Subject: Re: Angular earth distance Hi, Nat. If you go to our Dr. Math Search page at http://mathforum.org/mathgrepform.html and look for items with the words latitude, longitude, and distance, you will find several formulas, including yours. Chris Michels must have found the formula this way himself; it was not written in response to him. He did not find the simplest formula in our archives. He did not find the most reliable formula, either; another, the Haversine formula, is supposed to be less prone to rounding errors under certain conditions. You're correct that the formula he found can be simplified to yours. It's easy to start with yours and obtain his, by applying the angle-difference identity cos(b-d) = cos(b)*cos(d) + sin(b)*sin(d) You can also follow through the derivation of the more complicated formula and modify it by rotating the cartesian coordinate system so that point 1 is in the x-z plane. Then the y coordinate of point 1 is 0, which causes terms to vanish. The x coordinate of point 2 is r*cos(a1)*cos(b2-b1), and the y coordinate of point 2 is r*cos(a1)*sin(b2-b1). - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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