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我正在寻找一个 PHP 库/PHP 脚本，它允许我计算给定中心点（纬度/经度）的准确边界框。

使用椭球公式（例如 WGS84）会很棒。我知道，必须有一个图书馆，但我找不到。

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2 回答 2

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//轴承位后函数错误，应该如下：

function getBoundingBox($lat_degrees,$lon_degrees,$distance_in_miles) {

    $radius = 3963.1; // of earth in miles

    // bearings - FIX   
    $due_north = deg2rad(0);
    $due_south = deg2rad(180);
    $due_east = deg2rad(90);
    $due_west = deg2rad(270);

    // convert latitude and longitude into radians 
    $lat_r = deg2rad($lat_degrees);
    $lon_r = deg2rad($lon_degrees);

    // find the northmost, southmost, eastmost and westmost corners $distance_in_miles away
    // original formula from
    // http://www.movable-type.co.uk/scripts/latlong.html

    $northmost  = asin(sin($lat_r) * cos($distance_in_miles/$radius) + cos($lat_r) * sin ($distance_in_miles/$radius) * cos($due_north));
    $southmost  = asin(sin($lat_r) * cos($distance_in_miles/$radius) + cos($lat_r) * sin ($distance_in_miles/$radius) * cos($due_south));

    $eastmost = $lon_r + atan2(sin($due_east)*sin($distance_in_miles/$radius)*cos($lat_r),cos($distance_in_miles/$radius)-sin($lat_r)*sin($lat_r));
    $westmost = $lon_r + atan2(sin($due_west)*sin($distance_in_miles/$radius)*cos($lat_r),cos($distance_in_miles/$radius)-sin($lat_r)*sin($lat_r));


    $northmost = rad2deg($northmost);
    $southmost = rad2deg($southmost);
    $eastmost = rad2deg($eastmost);
    $westmost = rad2deg($westmost);

    // sort the lat and long so that we can use them for a between query        
    if ($northmost > $southmost) { 
        $lat1 = $southmost;
        $lat2 = $northmost;

    } else {
        $lat1 = $northmost;
        $lat2 = $southmost;
    }


    if ($eastmost > $westmost) { 
        $lon1 = $westmost;
        $lon2 = $eastmost;

    } else {
        $lon1 = $eastmost;
        $lon2 = $westmost;
    }

    return array($lat1,$lat2,$lon1,$lon2);
}

我得到了这个功能感谢：http: //xoxco.com/clickable/php-getboundingbox

于 2010-06-30T11:56:59.497 回答
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如果您假设边界框在一个方向上是东西走向，而在另一个方向上是南北走向，那么这是一个相对容易解决的问题。你可以独立做纬度和经度。

对于纬度，将点从西向东排序。此时，您必须将列表视为循环缓冲区。您需要测试每个点并找到下一个点最远的点。所以假设十个点 a 0到 a 9，如果 a 4和 a 5最远，则纬度边界框是从5到4。称他们为w和 a e

对于经度，您只需要找到最北端和最南端，称它们为 a n和 a s。

a w和 a e的经度以及 a n和 a s 的纬度定义了边界框。

于 2010-04-13T08:34:35.587 回答