java - J2ME - 计算2个纬度和经度之间的距离

Question

这个问题和这个很相似one .思路是在J2ME中计算2个经纬度点之间的距离。

我见过很多使用 Math.atan2 的示例，它在 Java 中很好，但在 JavaME 中不存在。

关于具体解决方案或 Math.atan2 替代方案的任何想法？

Answer 1

atan2 似乎有一个解决方案 here .

我重构了 atan2 的代码:

class TrigMath {

    static final double sq2p1 = 2.414213562373095048802e0;
    static final double sq2m1 = .414213562373095048802e0;
    static final double p4 = .161536412982230228262e2;
    static final double p3 = .26842548195503973794141e3;
    static final double p2 = .11530293515404850115428136e4;
    static final double p1 = .178040631643319697105464587e4;
    static final double p0 = .89678597403663861959987488e3;
    static final double q4 = .5895697050844462222791e2;
    static final double q3 = .536265374031215315104235e3;
    static final double q2 = .16667838148816337184521798e4;
    static final double q1 = .207933497444540981287275926e4;
    static final double q0 = .89678597403663861962481162e3;
    static final double PIO2 = 1.5707963267948966135E0;

    private static double mxatan(double arg) {
        double argsq = arg * arg, value;
        value = ((((p4 * argsq + p3) * argsq + p2) * argsq + p1) * argsq + p0);
        value = value / (((((argsq+q4)*argsq+q3)*argsq+q2)*argsq+q1)*argsq+q0);
        return value * arg;
    }

    private static double msatan(double arg) {
        return arg < sq2m1 ? mxatan(arg)
             : arg > sq2p1 ? PIO2 - mxatan(1 / arg)
             : PIO2 / 2 + mxatan((arg - 1) / (arg + 1));
    }

    public static double atan(double arg) {
        return arg > 0 ? msatan(arg) : -msatan(-arg);
    }

    public static double atan2(double arg1, double arg2) {
        if (arg1 + arg2 == arg1)
            return arg1 >= 0 ? PIO2 : -PIO2;
        arg1 = atan(arg1 / arg2);
        return arg2 < 0 ? arg1 <= 0 ? arg1 + Math.PI : arg1 - Math.PI : arg1;
    }
}