How to get accurate latitude and longitude in Android?

Latitude/ Longitude on map graphic

• I would like to know the latitude/ longitude values for this map, based on the X/Y parameters: http://www.findforward.com/image/map.gif Any idea how I can get those? E.g. a great answer would consist of a grid of say 100*100 sections (or less -- the more the better( defined as x1,y1,x2,y2 rectangles with the latidude, longitude values, like so: 10,20,30,10,20.000,30.000 [only sample values]

• Answer:

  Hi, j_philipp-ga: I analyzed the map.gif image which you posted by combining the few "country borders" that happen to coincide with lines of longitude or parallels of latitude with some other reasonably easily identifiable features, like the mouth of the Amazon or Cape Horn. In addition to searching out Web pages that give geograhic coordinates for such things, I had the use of a compact world atlas published in 1949 in Great Britain: The Handy Reference Atlas, edited by John Bartholomew (15th ed., Edinburgh) Before delving into the arcana of geographic coordinates, however, it's probably a good idea to begin with a refresher on the pixel coordinates of a GIF image. After saving the cited map.gif to my local hard disk, I used Microsoft's HTML Help Image Editor (v. 4.73) to examine its details. Of course many other programs would serve this purpose equally well. The image is 896 by 576 (width times height), and as explained in this tutorial: [Manual Clickable Image Maps] http://staff.washington.edu/larryg/Classes/R561/zz-click-man.html "Note that pixel values always start at 0,0 in the upper left corner and get larger as you move down and to the right." Thus our pixel coordinates will range from 0 to 895 in the horizontal "X" direction (left to right) and from 0 to 575 in the vertical "Y" direction (top to bottom). I'll assume a basic familiarity with the conventions of longitude and latitude; at bottom are a couple of links to earlier Google Answers threads where I've dealt with these to some extent. Per our earlier exchanges of Comments and Clarifications, the longitude and latitude angles are going to be provided in degrees. This would pass without mention except that mathematicians and numerical programmers prefer to work with angles measure in radians, for more or less the same reason that we'd pick "natural" logarithms over the common ones. However in using a pidgeon-Excel formulation of the algorithms, I think the context of degrees versus radians will be clear. After the horizontal (longitude) coordinates were modelled rather easily (linearly in pixel's X, as expected), the vertical (latitude) coordinates proved somewhat challenging. Fortunately the Google Search Engine is our friend, and the following page tells us pretty much everything we need to know: [Mercator Conformal Projection] http://www.ualberta.ca/~norris/navigation/Mercator.html "The Mercator Projection is named after its inventor: Gerhard Kremer, a Flemish cartographer who lived from 1512 to 1594. (Gerhardus Mercator was the latinized form of his name)." "He published the first map using this projection in 1569, but the projection did not become popular until 30 years later (1599), when Edward Wright published an explanation of it." Here are the formulas given there for a perfect sphere, in a slightly modified form convenient to our needs. For points in the northern hemisphere: L = angle of longitude in radians (east = positive) A = angle of latitude in radians (north = positive) E = distance east N = distance north we have: E = r L N = r ln( tan( (A + pi/2)/2 ) where r represents a distance (namely the radius of the sphere "to scale"). Note that angles here are measured by radians, which are "pure" numbers, lacking any identifying units of measurement. The angle pi/2 in radians is equal to 90 degrees. In the southern hemisphere we use essentially the same formulas, but A is replaced by |A| and the sign of N (distance north) switches to negative: N = -r ln( tan( (|A| + pi/2)/2 ) Finally the distance coordinates E and N are "affinely" related to pixel coordinates: X = horizontal gif coordinate in pixels (right = positive) Y = vertical gif coordinate in pixels (down = positive) This is a fancy way of saying that X,Y are first degree polynomials in E,N. In fact for some constant coefficients a,b,c,d: X = a E + b Y = c N + d where c < 0 in order to obtain the reversed direction of the vertical coordinate (latitude increases from bottom to top, but the Y coordinate decreases in that direction). By a least squares fitting of one dozen "measured" points of latitude and ten points of longitude, I arrived at these rounded approximations: DEGREES( L ) = (X - 430.25)/2.4 A = SIGN(395.5 - Y)*[2*ATAN(EXP(ABS(395.5 - Y)/137)) - PI/2] Here L and A are in radians, but Excel provides a function DEGREES( ) that converts to the familiar units (as well as a function RADIANS( ) to go the other way). In fact the formulas above are pretty much the ones I worked with in Excel, except for the necessity of cell references and the fact that PI must become the constant (nullary) function PI(). * * * * * * * * * * * * * * * * * * * * * * * * * * A Sample Grid ============= The mapping above puts: the Prime Meridian along X = 430.25, and the Equator at Y = 395.5. For the sake of illustration, let's begin with a single rectangle of size 89 x 55 that is roughly centered on that point. Bearing in mind that pixels are "fence posts", if we have N pixels in a line (horizontal or vertical), then the difference in the coordinates of the endpoints is N-1. Two adjacent pixels would have a difference of 1, for example. So if we took the left and right extremes of this first rectangle to be: X0 = 430 - 44 = 386 X1 = 430 + 44 = 474 and if we took the top and bottom extremes to be: Y0 = 396 - 27 = 369 Y1 = 396 + 27 = 423 then the "center" of the rectangle would be X=430,Y=396. [If one prefers to hit Y = 395.5, then a rectangle of even height would be required.] It is a nice feature that longitude L depends only on X, and latitude A only on Y. In the present circumstance: DEGREES( L ) = (430 - 430.25)/2.4 = -0.25/2.4 ~ -0.1 A = SIGN(395.5 - 396)*[2*ATAN(EXP(ABS(395.5 - 396)/137)) - PI/2] = - [ 2*ATAN( EXP( 0.5/137 ) ) - PI/2 ] ~ -0.00365 in radians DEGREES( A ) ~ -0.2 Here the negativeness of L means we're slightly West of the Prime Meridian, and the negativeness of A that we are slightly South of the Equator. The corresponding tabulation you requested would be: X0, Y0, X1, Y1, A , L ------------------------- 386,369,474,423,-0.2,-0.1 where I've rounded the angles to the nearest tenth of a degree. In fact we can readily work out the horizontal and vertical intervals of our grid separately, adding or subtracting 89 to the pairs of X coordinates until we reach the map's edges, and likewise adding or subtracting 55 to the pairs of Y coordinates: X0, X1,DEG(L) Y0, Y1,DEG(A) -------------- -------------- 0, 29,-173.2 0, 38, 82.7 30,118,-148.4 39, 93, 79.7 119,207,-111.4 94,148, 74.6 208,296, -74.3 149,203, 67.2 297,385, -37.2 204,258, 56.5 386,474, -0.1 259,313, 41.6 475,563, 37.0 314,368, 22.2 564,652, 74.1 369,423, -0.2 653,741, 111.1 424,478, -22.6 742,830, 148.2 479,533, -41.9 831,895, 180.3 534,575, -55.2 Note that the final longitudinal interval manages to squeak its center just beyond the International Dateline (or more precisely the meridian antipodal to the Prime Meridian, since the International Dateline is allowed to bend around somewhat). A coarse rectangular grid would be obtained by pairing any X interval with any Y interval, thus covering your map.gif. Our first rectangle, for example, is the combination of the sixth interval from the X column and the eight interval from the Y column. * * * * * * * * * * * * * * * * * * * * * * * * * * Additional Links ================ [Q: Identifying zip codes between two points] http://answers.google.com/answers/threadview?id=247783 [Q: Distance and angles "through" the globe.] http://answers.google.com/answers/threadview?id=233527 regards, mathtalk-ga