google map 计算地图面积方法

花了几个小时把js的google计算地图面积的算法改成了c# 的。

  class Program
    {
        static void Main(string[] args)
        {
            // a = new qq.maps.LatLng(39.93, 116.44004334);
            //b = new qq.maps.LatLng(39.93, 116.35421264);
            //c = new qq.maps.LatLng(39.896775, 116.35421264);
            //d = new qq.maps.LatLng(39.89, 116.44004334);
            //29869510.924133233平方米

            List<LatLng> a = new List<LatLng>();
            a.Add(new LatLng() { Latitude = 39.93, Longitude = 116.44004334 });
            a.Add(new LatLng() { Latitude = 39.93, Longitude = 116.35421264 });
            a.Add(new LatLng() { Latitude = 39.896775, Longitude = 116.35421264 });
            a.Add(new LatLng() { Latitude = 39.89, Longitude = 116.44004334 });

            double area = computeArea(a, 0);

        }

        static double computeArea(List<LatLng> x, double y)
        {
            return Math.Abs(computeSignedArea(x, y));
        }

        static double computeSignedArea(List<LatLng> a, double b)
        {
            double c = 6378137;
            if (b > 0)
            {
                c = b;
            }
            double g = a.Count - 1;
            double e = 0;
            LatLng d = a[0];
            for (int i = 1; i < g; i++)
            {
                e += Zm(d, a[i], a[i + 1]);
            }
            return e * c * c;
        }

        static double Zm(LatLng a, LatLng b, LatLng c)
        {
            return Tm(a, b, c) * Um(a, b, c);
        }

        static double Tm(LatLng a, LatLng b, LatLng c)
        {
            List<LatLng> d = new List<LatLng>() { a, b, c, a };
            List<double> e = new List<double>();
            double f = 0;

            for (int i = 0; i < d.Count - 1; i++)
            {
                e.Add(Uf(d[i], d[i + 1]));
                f += e[i];
            }
            f /= 2;
            double g = Math.Tan(f / 2);

            for (int j = 0; j < d.Count - 1; j++)
            {
                g *= Math.Tan((f - e[j]) / 2);
            }

            return 4 * Math.Atan(Math.Sqrt(Math.Abs(g)));
        }
        static double Um(LatLng a, LatLng b, LatLng c)
        {
            List<LatLng> d = new List<LatLng>() { a, b, c };

            double[,] e = new double[3, 3];

            for (int i = 0; i < 3; i++)
            {
                LatLng f = d[i];
                double f1 = re(f);
                double f2 = se(f);
                e[i, 0] = Math.Cos(f1) * Math.Cos(f2);
                e[i, 1] = Math.Cos(f1) * Math.Sin(f2);
                e[i, 2] = Math.Sin(f1);
            }
            return 0 < e[0, 0] * e[1, 1] * e[2, 2] + e[1, 0] * e[2, 1] * e[0, 2] + e[2, 0] * e[0, 1] * e[1, 2] - e[0, 0] * e[2, 1] * e[1, 2] - e[1, 0] * e[0, 1] * e[2, 2] - e[2, 0] * e[1, 1] * e[0, 2] ? 1 : -1;
        }

        static double Uf(LatLng a, LatLng b)
        {
            var c = re(a);
            var d = re(b);
            return 2 * Math.Asin(Math.Sqrt(Math.Pow(Math.Sin((c - d) / 2), 2) + Math.Cos(c) * Math.Cos(d) * Math.Pow(Math.Sin((se(a) - se(b)) / 2), 2)));
        }

        static double re(LatLng a)
        {
            return Rd(a.Latitude);
        }
        static double se(LatLng a)
        {
            return Rd(a.Longitude);
        }

        static double Rd(double a)
        {
            return Math.PI / 180 * a;
        }

    }

    partial struct LatLng
    {
        private double _Longitude;
        public double Longitude
        {
            get
            {
                return _Longitude;
            }
            set
            {
                _Longitude = value;
            }
        }

        private double _Latitude;
        public double Latitude
        {
            get
            {
                return _Latitude;
            }
            set
            {
                _Latitude = value;
            }
        }

    }

 double computeDistanceBetween(LngLat a, LngLat b, double c)
        {
            return Uf(a, b) * c;
        }
        public double ComputeLength(List<LngLat> a, double b = 0)
        {
            if (a.Count < 2)
            {
                return 0;
            }
            double c = 6378137;
            if (b > 0)
            {
                c = b;
            }
            double g = a.Count - 1;
            double d = 0;

            for (var e = 0; e < g; ++e)
            {
                d += computeDistanceBetween(a[e], a[e + 1], c);
            }
            return d;
        }