Php 如何得到N个围绕中心坐标和半径的坐标

我得到了一个中心坐标和一个距离/半径，我需要使用半径从中心得到N个坐标，例如如何在下图中得到12个红点坐标

您可以使用计算来获取点

我首先从所需的中心点计算轴承，然后foreach循环轴承并将其传递给函数

function destinationPoint($lat, $lng, $brng, $dist) {
    $rad = 6371; // earths mean radius
    $dist = $dist/$rad;  // convert dist to angular distance in radians
    $brng = deg2rad($brng);  // conver to radians 
    $lat1 = deg2rad($lat); 
    $lon1 = deg2rad($lng);

    $lat2 = asin(sin($lat1)*cos($dist) + cos($lat1)*sin($dist)*cos($brng) );
    $lon2 = $lon1 + atan2(sin($brng)*sin($dist)*cos($lat1),cos($dist)-sin($lat1)*sin($lat2));
    $lon2 = fmod($lon2 + 3*M_PI, 2*M_PI) - M_PI;  // normalise to -180..+180º
    $lat2 = rad2deg($lat2);
    $lon2 = rad2deg($lon2);


    echo "lat = ".$lat2."\n";
    echo "lon = ".$lon2."\n\n";
}

$lat = 0;
$lng = 0;
$dist = 1; // km
$n = 12;

$bearings = range(0, 360-(360/$n)  , 360/$n); // create array of all bearings needed from $lat/$lng

foreach($bearings as $brng){
    echo $brng ."\n";
    destinationPoint($lat, $lng, $brng, $dist);
}

在理论上，根据您需要的精确度，您只需要使用函数计算一半的值，然后您就可以使用基本计算来计算另一半值。
但是如果距离很大（不知道大到底意味着什么），可能会有不同

您可以使用计算来获取点

我首先从所需的中心点计算轴承，然后foreach循环轴承并将其传递给函数

function destinationPoint($lat, $lng, $brng, $dist) {
    $rad = 6371; // earths mean radius
    $dist = $dist/$rad;  // convert dist to angular distance in radians
    $brng = deg2rad($brng);  // conver to radians 
    $lat1 = deg2rad($lat); 
    $lon1 = deg2rad($lng);

    $lat2 = asin(sin($lat1)*cos($dist) + cos($lat1)*sin($dist)*cos($brng) );
    $lon2 = $lon1 + atan2(sin($brng)*sin($dist)*cos($lat1),cos($dist)-sin($lat1)*sin($lat2));
    $lon2 = fmod($lon2 + 3*M_PI, 2*M_PI) - M_PI;  // normalise to -180..+180º
    $lat2 = rad2deg($lat2);
    $lon2 = rad2deg($lon2);


    echo "lat = ".$lat2."\n";
    echo "lon = ".$lon2."\n\n";
}

$lat = 0;
$lng = 0;
$dist = 1; // km
$n = 12;

$bearings = range(0, 360-(360/$n)  , 360/$n); // create array of all bearings needed from $lat/$lng

foreach($bearings as $brng){
    echo $brng ."\n";
    destinationPoint($lat, $lng, $brng, $dist);
}

在理论上，根据您需要的精确度，您只需要使用函数计算一半的值，然后您就可以使用基本计算来计算另一半值。
但是如果距离很大（不知道大到底意味着什么），可能会有不同

---

斯威夫特相当于@Andreas的回答：

func destinationPoint(latitude: Double, longitude: Double, bearing: Double, dist: Double) -> CLLocationCoordinate2D {
    let lat2 = asin(sin(latitude) * cos(dist) + cos(latitude) * sin(dist) * cos(bearing))
    var lon2 = longitude + atan2(sin(bearing) * sin(dist) * cos(latitude),cos(dist) - sin(latitude) * sin(lat2))
        lon2 = fmod(lon2 + 3 * .pi, 2 * .pi) - .pi  // normalise to -180..+180º
    return CLLocationCoordinate2D(latitude: lat2 * (180.0 / .pi), longitude: lon2 * (180.0 / .pi))
}

let latRadian = coordinate.latitude * .pi / 180
let lngRadian = coordinate.longitude * .pi / 180
let distance = (radius / 1000) / 6371 // km
let n = 24

let coordinates = stride(from: 0.0, to: 360.0, by: Double(360 / n)).map {
    destinationPoint(latitude: latRadian, longitude: lngRadian, bearing: $0 * .pi / 180, dist: distance)
}

---

斯威夫特相当于@Andreas的回答：

func destinationPoint(latitude: Double, longitude: Double, bearing: Double, dist: Double) -> CLLocationCoordinate2D {
    let lat2 = asin(sin(latitude) * cos(dist) + cos(latitude) * sin(dist) * cos(bearing))
    var lon2 = longitude + atan2(sin(bearing) * sin(dist) * cos(latitude),cos(dist) - sin(latitude) * sin(lat2))
        lon2 = fmod(lon2 + 3 * .pi, 2 * .pi) - .pi  // normalise to -180..+180º
    return CLLocationCoordinate2D(latitude: lat2 * (180.0 / .pi), longitude: lon2 * (180.0 / .pi))
}

let latRadian = coordinate.latitude * .pi / 180
let lngRadian = coordinate.longitude * .pi / 180
let distance = (radius / 1000) / 6371 // km
let n = 24

let coordinates = stride(from: 0.0, to: 360.0, by: Double(360 / n)).map {
    destinationPoint(latitude: latRadian, longitude: lngRadian, bearing: $0 * .pi / 180, dist: distance)
}

---

问题出在哪里？你一定是带着这个来了什么地方？我的答案有什么遗漏吗？现在检查…，谢谢问题出在哪里？你一定是带着这个来了什么地方？我的答案有什么遗漏吗？现在检查…，谢谢