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What is the surface area?

What is the surface area of a region formed by N points on the earth's surface connected by the great circle?

  • How do I obtain the surface area of a region formed by N points on the earth's surface connected by the great circle? The points are such that only two of them have the same latitude or longitude. The remaining points are arbitrarily located. Example: Determine the area on the earth's surface formed by connecting New York to Seattle, Seattle to Honolulu, Honolulu to Houston, and Houston to New York with great circles. Here N = 4.

  • Answer:

    Let us assume the Earth to be a perfect sphere. Let the radius of Earth be ‘R’. Point O denotes the centre of the sphere. Let us calculate the spherical area of the spherical quadrilateral formed by joining the points A, B, C, and D on the Earth surface. Let us further assume that the positions are A: theta deg N latitude, phi deg E longitude B: theta deg S latitude, phi deg E longitude C: theta deg S latitude, phi deg W longitude D: theta deg N latitude, phi deg W longitude The area of spherical quadrilateral ABCD is given by A = [pi*R^2*(phi)*sin {pi*(theta)/180}]/45 [N.B. 'phi' and 'theta' are in degrees] For the detailed procedure give me your email id so that I can send my workings as email attachment.

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It will depend on how you join the points. If you join the points by tout string over the surface of the earth, area will be zero.

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Method 1. Triangulate the region. The area of a spherical triangle equals (A+B+C-pi)*R^2, where A, B and C are the angles in radians, and R is the radius of earth. This will work poor for areas which are small compared to the whole earth (i.e. are almost flat). A+B+C-pi will be VERY close to zero, and errors will be huge. If you want to use this method, calculate the angles to EXTREME precision. Method 2. Use grid on the surface. E.g. you can use parallels and meridians, if you're not too close to the poles. Intersect the region with each cell of the grid, and calculate the area of this part as if it was flat. Add them up. The result will be approximate. The smaller are the grid cells, the higher is the precision. Also, the earth is not really spherical. If you want precision, maybe you have to take this into account.

ringm

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