在PHP中测量两个坐标之间的距离
在PHP中测量两个坐标之间的距离
提问于 2012-04-07 17:35:26
嗨，我需要计算两个点之间的距离，具有经度和经度。
我希望避免任何对外部API的调用。
我尝试用PHP实现Haversine公式：
代码如下：
class CoordDistance
{
public $lat_a = 0;
public $lon_a = 0;
public $lat_b = 0;
public $lon_b = 0;
public $measure_unit = 'kilometers';
public $measure_state = false;
public $measure = 0;
public $error = '';
public function DistAB()
{
$delta_lat = $this->lat_b - $this->lat_a ;
$delta_lon = $this->lon_b - $this->lon_a ;
$earth_radius = 6372.795477598;
$alpha = $delta_lat/2;
$beta = $delta_lon/2;
$a = sin(deg2rad($alpha)) * sin(deg2rad($alpha)) + cos(deg2rad($this->lat_a)) * cos(deg2rad($this->lat_b)) * sin(deg2rad($beta)) * sin(deg2rad($beta)) ;
$c = asin(min(1, sqrt($a)));
$distance = 2*$earth_radius * $c;
$distance = round($distance, 4);
$this->measure = $distance;
}
}用一些有公共距离的给定点进行测试，我得不到可靠的结果。
我不知道是在原始公式中还是在我的实现中有错误
回答 11
Stack Overflow用户
发布于 2015-05-31 19:28:12
我找到了this code，它给了我可靠的结果。
function distance($lat1, $lon1, $lat2, $lon2, $unit) {
$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);
$miles = $dist * 60 * 1.1515;
$unit = strtoupper($unit);
if ($unit == "K") {
return ($miles * 1.609344);
} else if ($unit == "N") {
return ($miles * 0.8684);
} else {
return $miles;
}
}结果：
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "M") . " Miles<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "K") . " Kilometers<br>";
echo distance(32.9697, -96.80322, 29.46786, -98.53506, "N") . " Nautical Miles<br>";Stack Overflow用户
发布于 2016-12-02 17:55:54
它只是对@martinstoeckli和@Janith Chinthana answers的补充。对于那些好奇哪种算法是最快的人，我写了performance test。最佳性能结果显示来自codexworld.com的优化功能
/**
* Optimized algorithm from http://www.codexworld.com
*
* @param float $latitudeFrom
* @param float $longitudeFrom
* @param float $latitudeTo
* @param float $longitudeTo
*
* @return float [km]
*/
function codexworldGetDistanceOpt($latitudeFrom, $longitudeFrom, $latitudeTo, $longitudeTo)
{
$rad = M_PI / 180;
//Calculate distance from latitude and longitude
$theta = $longitudeFrom - $longitudeTo;
$dist = sin($latitudeFrom * $rad)
* sin($latitudeTo * $rad) + cos($latitudeFrom * $rad)
* cos($latitudeTo * $rad) * cos($theta * $rad);
return acos($dist) / $rad * 60 * 1.853;
}测试结果如下：
Test name Repeats Result Performance
codexworld-opt 10000 0.084952 sec +0.00%
codexworld 10000 0.104127 sec -22.57%
custom 10000 0.107419 sec -26.45%
custom2 10000 0.111576 sec -31.34%
custom1 10000 0.136691 sec -60.90%
vincenty 10000 0.165881 sec -95.26%Stack Overflow用户
发布于 2016-04-19 16:45:10
这里是计算两个纬度和经度之间距离的简单而完美的代码。可从此处找到以下代码- http://www.codexworld.com/distance-between-two-addresses-google-maps-api-php/
$latitudeFrom = '22.574864';
$longitudeFrom = '88.437915';
$latitudeTo = '22.568662';
$longitudeTo = '88.431918';
//Calculate distance from latitude and longitude
$theta = $longitudeFrom - $longitudeTo;
$dist = sin(deg2rad($latitudeFrom)) * sin(deg2rad($latitudeTo)) + cos(deg2rad($latitudeFrom)) * cos(deg2rad($latitudeTo)) * cos(deg2rad($theta));
$dist = acos($dist);
$dist = rad2deg($dist);
$miles = $dist * 60 * 1.1515;
$distance = ($miles * 1.609344).' km';页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/10053358
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