Question

我正在将具有半径的中心点（纬度，经度）传递给此函数，以获取圆周点（纬度，经度），以便我从中创建几何并将其存储在MYSQL中，以便稍后执行ST_Contains函数上。

在Google地图上绘制时，这使我得到的是椭圆形而不是圆形。

public static function convert($center, $radius, $numberOfSegments = 360)
{
    $n = $numberOfSegments;
    $flatCoordinates = [];
    $coordinates = [];
    for ($i = 0; $i < $n; $i++) {
        $flatCoordinates = array_merge($flatCoordinates, static::offset($center, $radius, 2 * pi() * $i / $n));
    }
    $flatCoordinates[] = $flatCoordinates[0];
    $flatCoordinates[] = $flatCoordinates[1];
    for ($i = 0, $j = 0; $j < count($flatCoordinates); $j += 2) {
        $coordinates[$i++] = array_slice($flatCoordinates, $j, 2);
    }

    return [
        'type' => 'Polygon',
        'coordinates' => [array_reverse($coordinates)]
    ];
}

public static function toRadians($angleInDegrees = null)
{
    return $angleInDegrees * pi() / 180;
}

public static function toDegrees($angleInRadians = null)
{
    return $angleInRadians * 180 / pi();
}

public static function offset($c1, $distance, $bearing)
{
    $lat1 = static::toRadians($c1[1]);
    $lon1 = static::toRadians($c1[0]);
    $dByR = $distance / 6378137; // distance divided by 6378137 (radius of the earth) wgs84
    $lat = asin(
        sin($lat1) * cos($dByR) +
            cos($lat1) * sin($dByR) * cos($bearing)
    );
    $lon = $lon1 + atan2(
        sin($bearing) * sin($dByR) * cos($lat1),
        cos($dByR) - sin($lat1) * sin($lat)
    );
    return [static::toDegrees($lon), static::toDegrees($lat)];
}

考虑到WGS84，我真的不需要像地球曲率那样精确，因为半径很小（25m-150m）。

Answer 1

我通过变量名猜测您是从this question获得函数的，但是您忘记了代码的一行，特别是引起该人问题的那一行。

此外，第一个函数中的数组代码比所需的要冗长得多。我在这里对此进行了概括，并使用内置的PHP函数进行弧度/度转换：

<?php
function convert($center, $radius, $numberOfSegments = 360)
{
    $n = $numberOfSegments;
    $flatCoordinates = [];
    $coordinates = [];
    for ($i = 0; $i < $n; $i++) {
        $bearing = 2 * M_PI * $i / $n;
        $flatCoordinates[] = offset($center, $radius, $bearing);
    }
    $flatCoordinates[] = $flatCoordinates[0];

    return [
        'type' => 'Polygon',
        'coordinates' => [$flatCoordinates]
    ];
}

function offset($c1, $distance, $bearing) {
    $lat1 = deg2rad($c1[0]);
    $lon1 = deg2rad($c1[1]);
    $dByR = $distance/6378137; // convert dist to angular distance in radians

    $lat = asin(
        sin($lat1) * cos($dByR) + 
            cos($lat1) * sin($dByR) * cos($bearing)
    );
    $lon = $lon1 + atan2(
        sin($bearing) * sin($dByR) * cos($lat1),
        cos($dByR) - sin($lat1) * sin($lat)
    );
    $lon = fmod(
        $lon + 3 * M_PI,
        2 * M_PI
    ) - M_PI;
    return [rad2deg($lon), rad2deg($lat)];
}

$result = convert([49, -122], 5000, 20);
foreach ($result['coordinates'][0] as $v) echo "$v[1]\t$v[0]\n";

输出：

49.044915764206 -122
49.042715494198 -121.97882560793
49.036330613043 -121.9597290805
49.026387559463 -121.94458290688
49.01386140937  -121.93486968973
48.999979747353 -121.93153702633
48.986101953242 -121.93490597976
48.973585929464 -121.94464162541
48.963655392709 -121.95978779908
48.957280637886 -121.97886189803
48.955084235794 -122
48.957280637886 -122.02113810197
48.963655392709 -122.04021220092
48.973585929464 -122.05535837459
48.986101953242 -122.06509402024
48.999979747353 -122.06846297367
49.01386140937  -122.06513031027
49.026387559463 -122.05541709312
49.036330613043 -122.0402709195
49.042715494198 -122.02117439207
49.044915764206 -122

我使用了this site for testing，它按预期绘制了一个圆圈。

获取圆多边形的圆周点（经度和纬度）

1 个答案: