大圆距离问题

Question

我熟悉计算两点之间大圆距离的公式。

即

<?php
$theta = $lon1 - $lon2; 
$dist = sin(deg2rad($lat1)) * sin(deg2rad($lat2)) +  cos(deg2rad($lat1)) * cos(deg2rad($lat2)) * cos(deg2rad($theta)); 
$dist = acos($dist); 
$dist = rad2deg($dist); 
//convert degrees to distance depending on units desired
?>

我需要的是相反的。 给定起点、距离和简单的基本 NSEW 方向，计算目的地点的位置。 我已经很久没有上数学课了。 ;)

Answer 1

为了回答我自己的问题，为任何好奇的人提供一个从 Chad Birch 提供的 C 函数转换而来的 PHP 类：

class GreatCircle 
{

    /*
     * Find a point a certain distance and vector away from an initial point
     * converted from c function found at: http://sam.ucsd.edu/sio210/propseawater/ppsw_c/gcdist.c
     * 
     * @param int distance in meters
     * @param double direction in degrees i.e. 0 = North, 90 = East, etc.
     * @param double lon starting longitude
     * @param double lat starting latitude
     * @return array ('lon' => $lon, 'lat' => $lat)
     */
    public static function getPositionByDistance($distance, $direction, $lon, $lat)
    {
        $metersPerDegree = 111120.00071117;
        $degreesPerMeter = 1.0 / $metersPerDegree;
        $radiansPerDegree = pi() / 180.0;
        $degreesPerRadian = 180.0 / pi();

        if ($distance > $metersPerDegree*180)
        {
            $direction -= 180.0;
            if ($direction < 0.0)
            {
                $direction += 360.0;
            }
            $distance = $metersPerDegree * 360.0 - $distance;
        }

        if ($direction > 180.0)
        {
            $direction -= 360.0;
        }

        $c = $direction * $radiansPerDegree;
        $d = $distance * $degreesPerMeter * $radiansPerDegree;
        $L1 = $lat * $radiansPerDegree;
        $lon *= $radiansPerDegree;
        $coL1 = (90.0 - $lat) * $radiansPerDegree;
        $coL2 = self::ahav(self::hav($c) / (self::sec($L1) * self::csc($d)) + self::hav($d - $coL1));
        $L2   = (pi() / 2) - $coL2;
        $l    = $L2 - $L1;

        $dLo = (cos($L1) * cos($L2));
        if ($dLo != 0.0)
        {
            $dLo  = self::ahav((self::hav($d) - self::hav($l)) / $dLo);
        }

        if ($c < 0.0) 
        {
            $dLo = -$dLo;
        }

        $lon += $dLo;
        if ($lon < -pi())
        {
            $lon += 2 * pi();
        }
        elseif ($lon > pi())
        {
            $lon -= 2 * pi();
        }

        $xlat = $L2 * $degreesPerRadian;
        $xlon = $lon * $degreesPerRadian;

        return array('lon' => $xlon, 'lat' => $xlat);
    }


    /*
     * copy the sign
     */
    private static function copysign($x, $y)
    {
        return ((($y) < 0.0) ? - abs($x) : abs($x));
    }   

    /*
     * not greater than 1
     */
    private static function ngt1($x)
    {
        return (abs($x) > 1.0 ? self::copysign(1.0 , $x) : ($x));
    }   

    /*
     * haversine
     */
    private static function hav($x)
    {
        return ((1.0 - cos($x)) * 0.5);
    }

    /*
     * arc haversine
     */
    private static function ahav($x)
    {
        return acos(self::ngt1(1.0 - ($x * 2.0)));
    }

    /*
     * secant
     */
    private static function sec($x)
    {
        return (1.0 / cos($x));
    }

    /*
     * cosecant
     */
    private static function csc($x)
    {
        return (1.0 / sin($x));
    }

}

Answer 2

这是我发现的一个 C 实现，应该相当简单地转换为 PHP：

#define KmPerDegree         111.12000071117
#define DegreesPerKm        (1.0/KmPerDegree)
#define PI                  M_PI
#define TwoPI               (M_PI+M_PI)
#define HalfPI              M_PI_2
#define RadiansPerDegree    (PI/180.0)
#define DegreesPerRadian    (180.0/PI)
#define copysign(x,y)       (((y)<0.0)?-fabs(x):fabs(x))
#define NGT1(x)             (fabs(x)>1.0?copysign(1.0,x):(x))
#define ArcCos(x)           (fabs(x)>1?quiet_nan():acos(x))
#define hav(x)              ((1.0-cos(x))*0.5)              /* haversine */
#define ahav(x)             (ArcCos(NGT1(1.0-((x)*2.0))))   /* arc haversine */
#define sec(x)              (1.0/cos(x))                    /* secant */
#define csc(x)              (1.0/sin(x))                    /* cosecant */

/*
**  GreatCirclePos() --
**
**  Compute ending position from course and great-circle distance.
**
**  Given a starting latitude (decimal), the initial great-circle
**  course and a distance along the course track, compute the ending
**  position (decimal latitude and longitude).
**  This is the inverse function to GreatCircleDist).
*/
void
GreatCirclePos(dist, course, slt, slg, xlt, xlg)
    double  dist;   /* -> great-circle distance (km) */
    double  course; /* -> initial great-circle course (degrees) */
    double  slt;    /* -> starting decimal latitude (-S) */
    double  slg;    /* -> starting decimal longitude(-W) */
    double  *xlt;   /* <- ending decimal latitude (-S) */
    double  *xlg;   /* <- ending decimal longitude(-W) */
{
    double  c, d, dLo, L1, L2, coL1, coL2, l;

    if (dist > KmPerDegree*180.0) {
        course -= 180.0;
        if (course < 0.0) course += 360.0;
        dist    = KmPerDegree*360.0-dist;
    }
    if (course > 180.0) course -= 360.0;
    c    = course*RadiansPerDegree;
    d    = dist*DegreesPerKm*RadiansPerDegree;
    L1   = slt*RadiansPerDegree;
    slg *= RadiansPerDegree;
    coL1 = (90.0-slt)*RadiansPerDegree;
    coL2 = ahav(hav(c)/(sec(L1)*csc(d))+hav(d-coL1));
    L2   = HalfPI-coL2;
    l    = L2-L1;
    if ((dLo=(cos(L1)*cos(L2))) != 0.0)
        dLo  = ahav((hav(d)-hav(l))/dLo);
    if (c < 0.0) dLo = -dLo;
    slg += dLo;
    if (slg < -PI)
        slg += TwoPI;
    else if (slg > PI)
        slg -= TwoPI;

    *xlt = L2*DegreesPerRadian;
    *xlg = slg*DegreesPerRadian;

} /* GreatCirclePos() */

来源：http://sam.ucsd.edu/sio210/propseawater/ ppsw_c/gcdist.c

Answer 3

我认为，撤销半正矢公式然后生成自己的公式会更困难。

首先，通过在地表上行进“直线”，从地核产生的角度（你认为它是直的，但它是弯曲的）。

角度（弧度）= 弧长/半径。
角度 = ArcLen/6371 公里

纬度应该很简单，只需角度的“垂直”（北/南）分量即可。

Lat1 + Cos（方位角） * 角度

经度划分因纬度而异。 所以这就变得更难了。 您可以使用：

Sin(方位角) * 角度（东定义为负）来查找经度方向的角度，但转换回该纬度的实际经度会更加困难。

Answer 4

请参阅此网站上的给定距离和方位的目的地点部分：http://www.movable-type.co.uk/scripts/latlong.html