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 April 19th, 2014, 05:47 PM #1 Newbie   Joined: Apr 2014 From: orlando,fl Posts: 1 Thanks: 0 unifying latitude and longitude into a single digit Hello guys so I basically have a program that allows people to do a 30 mile radius search around a latitude and longitude (Lat,Long) point. I want to do an experiment and join both the latitude and longitude (LatLong) together. Let me illustrate how my program works right now the latitude,longitude of new york city is (40.752050,-73.994517) a radius obviously has 4 points southern,northern,eastern and western point and for a 30 mile radius for that lat,long is (40.317268,-74.568456) and (41.186832,-73.42057 as you can see the original Lat/Long fit within those figures . That translates into a Latitude change of .434782 (40.752050-40.31726 and longitude change of .573939 (-73.994517-(-)74.568456).My question is what formula or method can I use to combine the latitude and longitude into a single digit while keeping my overall change of proportion; I have tried to do the midpoint formula -73.994517+40.752050 / 2= -16.62 and then combined the change .573939+ .434782 /2= .504 so that now a 30 mile radius is between -16.116 through -17.124 . This however is not working for example asbury park,NJ is like 40 miles away and for some reason its inside my midpoint range (40.235268,-74.035669) = -16.90 , however compare that with the original NYC 30-mile radius and you see it doesn't fit in (40.317268,-74.568456) and (41.186832,-73.42057 . This must mean that my midpoint formula is not working correctly, what can i do to get this right? and I apologize for making this long but I wanted to give as much information as possible. I know this is possible because that Lat,Long changes are static for that exact point .434782,.573939 any help would be greatly appreciated. Last edited by romeo407; April 19th, 2014 at 05:55 PM.
 April 19th, 2014, 08:16 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms I think what you want is the distance formula d = sqrt(x^2 + y^2), so -73.994517 * -73.994517 + 40.752050 * 40.752050 = 7135.9181 and its square root is 84.47 or so.
 June 24th, 2014, 08:24 PM #3 Banned Camp   Joined: Feb 2013 Posts: 224 Thanks: 6 I'm only a high school graduate. I don't know the math, and don't know what you are talking about. However here is what I know. The Earth is 3D, it is spherical. To compute position, you need 3 dimensional geometry. The simplest is using radians and only orthogonal movements (Rook movements). Diagonal movements are not easily computed because of spherical geometry. The hypotenuse is different for different sizes of "right" triangles. You will need spherical geometry calculations.
 June 25th, 2014, 05:21 AM #4 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms It's true -- the Earth is better approximated by a sphere than a plane (and better still as an oblate spheroid, and better still as the EGM96 geoid, etc.). But the difference for a distance of only 30 miles will be small.

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