Transformation between (x,y) and (longitude, latitude)Date: 01/02/2002 at 03:32:53 From: Hing C So Subject: Transformation between (x,y) and (longitude, latitude) Dear Dr. Math, I have two questions on the transformation between (x,y) and (longitude, latitude). 1. Given 2 positions in the terms of longitude and latitude, say, (a1, b1) and (a2,b2) with the distance between them very small, say, within 10 meters, I would like to know if there are any methods that can compute their separation with the best accuracy. I converted the (a,b) to (x,y) using the following formulae: x = R*cos(a)*cos(b) y = R*cos(a)*sin(b) where R is the earth radius. The distance is calculated as the square root of (x1-x2)^2+(y1-y2)^2. Please comment on my method. 2. I have a digital map giving the four corners in terms of longitudes and latitudes, say, (a1,b1), (a1,b2), (a2,b1), (a2,b2). I would like to know how to compute the (x,y) of a particular point in the map. Thank you very much. H.C.So Date: 01/02/2002 at 11:19:16 From: Doctor Rick Subject: Re: Transformation between (x,y) and (longitude, latitude) Hi, Hing. I'll respond to each of your questions after the question: (1) A distance of 10 meters is *very* small compared to the radius of the earth, so we can definitely use a flat-earth approximation. Your formulas give the coordinates of the projection of a point onto the plane on the Greenwich meridian (longitude = 0). This is not what you want; you really want to project the point onto a plane parallel to the surface of the earth in the vicinity of the points of interest. I find it easiest to think a little differently. For the y coordinate, we can use the north-south distance between two lines of latitude: y = R*(b2-b1)*pi/180 Here, I have converted the latitude difference, (b2-b1), from degrees to radians, by multiplying it by (pi radians)/(180 degrees). The product of the angle in radians and the radius is the arc length in the same units as the radius. For the x coordinate, we can use the distance along a line of latitude from one line of longitude to the other: x = R*(a2-a1)*(pi/180)*cos(b1) Here we have an additional factor, the cosine of the latitude along which we are measuring. The line of latitude is a circle with a smaller radius than that of the equator; it is reduced by the factor cos(b1). Thus, you see that I have set up a coordinate system (x,y) that puts one of the points of interest at the origin. The distance from the origin to any other point (x,y) is the square root of (x^2 + y^2). (2) We'd have to know the particular projection used in the map in order to be completely accurate: is it a Mercator projection, for instance? If the scale is small enough (as I assume based on the context of the first question), I would assume it's fairly linear over the region of the map. Then, if you want x to vary from 0 to w and y to vary from 0 to h (with the origin at the bottom left), the coordinates of a point (a,b) are x = w*(a - a1)/(a2 - a1) y = h*(b - b1)/(b2 - b1) That's just a simple linear transformation. Again, it assumes that the map covers a small area. If you're asking something more complicated than this, please try again to explain it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 08/10/2004 at 05:51:36 From: Bryan Subject: Using Longitude and Latitude to Determine Distance I'm developing a digital map that encounters the same problem but the range is bigger. The area of the map is around 50 km^2. Could the formulas: x = w*(a - a1)/(a2 - a1) y = h*(b - b1)/(b2 - b1) be applied here? What is the accuracy factor in this case? Date: 08/10/2004 at 08:23:25 From: Doctor Rick Subject: Re: Using Longitude and Latitude to Determine Distance Hi, Bryan. I think a linear scaling will be fine for a map with maximum distances less than 10 miles, as your area suggests. It is hard to give a formula for the maximum error, but the following page from the Dr. Math Archives has a table that will help: Planar Approximation: Latitude and Longitude http://mathforum.org/library/drmath/view/62720.html There is one assumption I didn't mention with the linear scaling formula: it assumes that the latitudes and longitudes at the corners of the map have been chosen so that the map will have the same scale horizontally and vertically. The Planar Approximation formula in the page I just mentioned will help you do this. x = (lon2-lon1)*cos(lat1)*pi/180 y = (lat2-lat1)*pi/180 Here, x and y are in radians; if we want them in miles, we should use x = (lon2-lon1)*cos(lat1)*pi*R/180 y = (lat2-lat1)*pi*R/180 where R is the radius of the earth, R = 6367 km = 3956 mi. Solving for lon2,lat2 in terms of x,y: lat2 = lat1 + y*180/(pi*R) lon2 = lon1 + x*180/(pi*R*cos(lat1)) If (lon1,lat1) is the center of the map, w pixels across and h pixels high, and you want a scale of n pixels on the map = one mile on earth, the latitudes and longitudes of the corners of the map should be lat = lat1 +or- (w/2)*n*180/(pi*R) lon = lon1 +or- (h/2)*n*180/(pi*R*cos(lat1)) - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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