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    Re: Circle of position
    From: Frank Reed
    Date: 2011 Jan 23, 23:52 -0800

    Hello, Djamel.

    You wrote:
    "I am looking for algorithms concerning Circle of position plot on Mercator
    projection."

    There are various ways to proceed. Here's one...

    A circle of position is a circle with some radius R centered on some latitude, Lat0, and longitude, Lon0. The points of that circle would be very easy to define if Lat0 happened to be 90 degrees. So let's do that first. If the center is the North (or South) pole, then the circle of position is simply a circle of constant latitude, and obviously if the radius is R degrees then the latitude would be 90-R. So decide how many points you want to use to set up your circle, and suppose for this example that R is 10 degrees. If you pick 180 points (just for a convenient number --your actual choice would be based on display issues), then your circle of position is defined by pairs of lat, lon values as follows: (80,0), (80,2), (80,4), etc. Connect the dots and you have a nice approximation to the circle of points with latitude 80 degrees. Very simple and very easy to code up. Next, we can move our circle of points to any arbitrary point on the globe by doing a coordinate rotation. This is exactly the same thing you would do to rotate a set of points from equatorial to ecliptic coordinates. You can dig this up yourself probably (see for example: http://en.wikipedia.org/wiki/Ecliptic_coordinate_system --as a starting point), but if you have any questions, just ask. All we're doing is moving the "pole" of our coordinate to the point Lat0, Lon0. Apply the coordinate transformation to every pair of points that define the circle. Now you have the lat, lon values for the circle centered on that new pole. These are the actual lat, lon values for a circle of position centered at Lat0, Lon0. Finally, for plotting on a Mercator projection, you do the standard transformation to Mercator x, y for all the points in the shifted circle (see for example: http://en.wikipedia.org/wiki/Mercator_projection). Incidentally, unless you have a specific reason for applying a Mercator projection, there's really no reason to prefer it, and you might want to consider other projections.

    -FER


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