数学方程转换

Question

I'm attempting to write a method that determines the area of a polygon (complex or simple) on a sphere. I have a paper that was written by a few guys at the JPL that more or less give you the equations for these calculations.

The pdf file can be found here:

http://trs-new.jpl.nasa.gov/dspace/handle/2014/40409

The equation can be found on page 7, under "The Spherical Case - Approximation":

I also typed the equation in Word:

Spherical_Case_Equation

I need assistance with converting this equation into the standard form (I think that's the right terminology). I've already done something similar for the Planer Case:

private double calcArea(Point2D[] shape) {
    int n = shape.length;
    double sum = 0.0;

    if (n < 3) return 0.0;

    for (int i = 0; i < n-1 ; i++) {
        sum += (shape[i].getX() * shape[i+1].getY()) - (shape[i+1].getX() * shape[i].getY());
    }

    System.out.println(0.5 * Math.abs(sum));
    return 0.5 * Math.abs(sum);
}

I just need help with doing something similar for the spherical case. Any assistance will be greatly appreciated.

Answer 1

I haven't read the paper you referenced. The area of a spherical polygon is proportional to the angle excess .

Area = r²(Σ Aᵢ - (n - 2)π)

To compute the corner angles, you may start with the 3D coordinates of your points. So at corner i you have vertex p[i] = (x[i],y[i],z[i]) and adjacent vertices p[i-1] and p[i+1] (resp p[(i+n-1)%n] and p[(i+1)%n] to get this cyclically correct). Then the cross products

v₁ = p[i] × p[i-1]
v₂ = p[i] × p[i+1]

will be orthogonal to the planes spanned by the incident edges and the origin which is the center of the sphere. Noe the angle between two vectors in space is given by

Aᵢ = arccos(⟨v₁,v₂⟩ / (‖v₁‖ * ‖v₂‖))

where ⟨v₁,v₂⟩ denotes the dot product between these two vectors which is proportional to the cosine of the angle, and ‖v₁‖ denotes the length of the first vector, likewise ‖v₂‖ for the second.