Question

我给出了一个由纬度和经度定义的位置。现在我想计算一个边界框，例如那个点10公里。

边界框应定义为latmin，lngmin和latmax，lngmax。

我需要这些东西才能使用panoramio API。

有人知道如何获得积分的公式吗？

编辑：我正在寻找一个公式/功能，需要使用lat＆amp; lng作为输入并返回一个边界框作为latmin＆amp; lngmin和latmax＆amp; latmin。 Mysql，php，c＃，javascript很好，但伪代码也应该没问题。

编辑：我不是在寻找能够显示2点距离的解决方案

Answer 1

我建议将地球表面局部近似为一个球体，其半径由给定纬度的WGS84椭球给出。我怀疑latMin和latMax的精确计算需要椭圆函数，并且不会产生明显的精度增加（WGS84本身就是近似值）。

我的实现如下（它是用Python编写的;我还没有测试过它）：

# degrees to radians
def deg2rad(degrees):
    return math.pi*degrees/180.0
# radians to degrees
def rad2deg(radians):
    return 180.0*radians/math.pi

# Semi-axes of WGS-84 geoidal reference
WGS84_a = 6378137.0  # Major semiaxis [m]
WGS84_b = 6356752.3  # Minor semiaxis [m]

# Earth radius at a given latitude, according to the WGS-84 ellipsoid [m]
def WGS84EarthRadius(lat):
    # http://en.wikipedia.org/wiki/Earth_radius
    An = WGS84_a*WGS84_a * math.cos(lat)
    Bn = WGS84_b*WGS84_b * math.sin(lat)
    Ad = WGS84_a * math.cos(lat)
    Bd = WGS84_b * math.sin(lat)
    return math.sqrt( (An*An + Bn*Bn)/(Ad*Ad + Bd*Bd) )

# Bounding box surrounding the point at given coordinates,
# assuming local approximation of Earth surface as a sphere
# of radius given by WGS84
def boundingBox(latitudeInDegrees, longitudeInDegrees, halfSideInKm):
    lat = deg2rad(latitudeInDegrees)
    lon = deg2rad(longitudeInDegrees)
    halfSide = 1000*halfSideInKm

    # Radius of Earth at given latitude
    radius = WGS84EarthRadius(lat)
    # Radius of the parallel at given latitude
    pradius = radius*math.cos(lat)

    latMin = lat - halfSide/radius
    latMax = lat + halfSide/radius
    lonMin = lon - halfSide/pradius
    lonMax = lon + halfSide/pradius

    return (rad2deg(latMin), rad2deg(lonMin), rad2deg(latMax), rad2deg(lonMax))

编辑：以下代码将（度，素数，秒）转换为度数+度数的分数，反之亦然（未经测试）：

def dps2deg(degrees, primes, seconds):
    return degrees + primes/60.0 + seconds/3600.0

def deg2dps(degrees):
    intdeg = math.floor(degrees)
    primes = (degrees - intdeg)*60.0
    intpri = math.floor(primes)
    seconds = (primes - intpri)*60.0
    intsec = round(seconds)
    return (int(intdeg), int(intpri), int(intsec))

Answer 2

我写了一篇关于找到边界坐标的文章：

http://JanMatuschek.de/LatitudeLongitudeBoundingCoordinates

本文解释了这些公式，并提供了Java实现。（这也说明了为什么费德里科的最小/最大经度公式是不准确的。）

Answer 3

我已将Federico A. Ramponi对C＃的回答转换为任何感兴趣的人：

public class MapPoint
{
    public double Longitude { get; set; } // In Degrees
    public double Latitude { get; set; } // In Degrees
}

public class BoundingBox
{
    public MapPoint MinPoint { get; set; }
    public MapPoint MaxPoint { get; set; }
}        

// Semi-axes of WGS-84 geoidal reference
private const double WGS84_a = 6378137.0; // Major semiaxis [m]
private const double WGS84_b = 6356752.3; // Minor semiaxis [m]

// 'halfSideInKm' is the half length of the bounding box you want in kilometers.
public static BoundingBox GetBoundingBox(MapPoint point, double halfSideInKm)
{            
    // Bounding box surrounding the point at given coordinates,
    // assuming local approximation of Earth surface as a sphere
    // of radius given by WGS84
    var lat = Deg2rad(point.Latitude);
    var lon = Deg2rad(point.Longitude);
    var halfSide = 1000 * halfSideInKm;

    // Radius of Earth at given latitude
    var radius = WGS84EarthRadius(lat);
    // Radius of the parallel at given latitude
    var pradius = radius * Math.Cos(lat);

    var latMin = lat - halfSide / radius;
    var latMax = lat + halfSide / radius;
    var lonMin = lon - halfSide / pradius;
    var lonMax = lon + halfSide / pradius;

    return new BoundingBox { 
        MinPoint = new MapPoint { Latitude = Rad2deg(latMin), Longitude = Rad2deg(lonMin) },
        MaxPoint = new MapPoint { Latitude = Rad2deg(latMax), Longitude = Rad2deg(lonMax) }
    };            
}

// degrees to radians
private static double Deg2rad(double degrees)
{
    return Math.PI * degrees / 180.0;
}

// radians to degrees
private static double Rad2deg(double radians)
{
    return 180.0 * radians / Math.PI;
}

// Earth radius at a given latitude, according to the WGS-84 ellipsoid [m]
private static double WGS84EarthRadius(double lat)
{
    // http://en.wikipedia.org/wiki/Earth_radius
    var An = WGS84_a * WGS84_a * Math.Cos(lat);
    var Bn = WGS84_b * WGS84_b * Math.Sin(lat);
    var Ad = WGS84_a * Math.Cos(lat);
    var Bd = WGS84_b * Math.Sin(lat);
    return Math.Sqrt((An*An + Bn*Bn) / (Ad*Ad + Bd*Bd));
}

Answer 4

我编写了一个JavaScript函数，它返回一个方形边界框的四个坐标，给定一个距离和一对坐标：

'use strict';

/**
 * @param {number} distance - distance (km) from the point represented by centerPoint
 * @param {array} centerPoint - two-dimensional array containing center coords [latitude, longitude]
 * @description
 *   Computes the bounding coordinates of all points on the surface of a sphere
 *   that has a great circle distance to the point represented by the centerPoint
 *   argument that is less or equal to the distance argument.
 *   Technique from: Jan Matuschek <http://JanMatuschek.de/LatitudeLongitudeBoundingCoordinates>
 * @author Alex Salisbury
*/

getBoundingBox = function (centerPoint, distance) {
  var MIN_LAT, MAX_LAT, MIN_LON, MAX_LON, R, radDist, degLat, degLon, radLat, radLon, minLat, maxLat, minLon, maxLon, deltaLon;
  if (distance < 0) {
    return 'Illegal arguments';
  }
  // helper functions (degrees<–>radians)
  Number.prototype.degToRad = function () {
    return this * (Math.PI / 180);
  };
  Number.prototype.radToDeg = function () {
    return (180 * this) / Math.PI;
  };
  // coordinate limits
  MIN_LAT = (-90).degToRad();
  MAX_LAT = (90).degToRad();
  MIN_LON = (-180).degToRad();
  MAX_LON = (180).degToRad();
  // Earth's radius (km)
  R = 6378.1;
  // angular distance in radians on a great circle
  radDist = distance / R;
  // center point coordinates (deg)
  degLat = centerPoint[0];
  degLon = centerPoint[1];
  // center point coordinates (rad)
  radLat = degLat.degToRad();
  radLon = degLon.degToRad();
  // minimum and maximum latitudes for given distance
  minLat = radLat - radDist;
  maxLat = radLat + radDist;
  // minimum and maximum longitudes for given distance
  minLon = void 0;
  maxLon = void 0;
  // define deltaLon to help determine min and max longitudes
  deltaLon = Math.asin(Math.sin(radDist) / Math.cos(radLat));
  if (minLat > MIN_LAT && maxLat < MAX_LAT) {
    minLon = radLon - deltaLon;
    maxLon = radLon + deltaLon;
    if (minLon < MIN_LON) {
      minLon = minLon + 2 * Math.PI;
    }
    if (maxLon > MAX_LON) {
      maxLon = maxLon - 2 * Math.PI;
    }
  }
  // a pole is within the given distance
  else {
    minLat = Math.max(minLat, MIN_LAT);
    maxLat = Math.min(maxLat, MAX_LAT);
    minLon = MIN_LON;
    maxLon = MAX_LON;
  }
  return [
    minLon.radToDeg(),
    minLat.radToDeg(),
    maxLon.radToDeg(),
    maxLat.radToDeg()
  ];
};

Answer 5

您正在寻找椭圆体公式。

我发现开始编码的最佳位置是基于CPAN的Geo :: Ellipsoid库。它为您提供了创建测试的基线，并将结果与结果进行比较。我在之前的雇主那里用它作为PHP类似库的基础。

Geo::Ellipsoid

查看location方法。叫它两次，你就得到了你的bbox。

您没有发布您使用的语言。可能已经有一个地理编码库可供您使用。

哦，如果你现在还没弄明白，Google地图会使用WGS84椭圆体。

Answer 6

由于我需要一个非常粗略的估计，所以为了在弹性搜索查询中过滤掉一些不必要的文档，我采用了以下公式：

Min.lat = Given.Lat - (0.009 x N)
Max.lat = Given.Lat + (0.009 x N)
Min.lon = Given.lon - (0.009 x N)
Max.lon = Given.lon + (0.009 x N)

在给定位置需要N = kms。对于你的情况N = 10

不准确但方便。

Answer 7

我改编了一个我发现这样做的PHP脚本。你可以用它来找到一个点周围的一个角落（例如，20公里外）。我的具体示例适用于Google Maps API：

http://www.richardpeacock.com/blog/2011/11/draw-box-around-coordinate-google-maps-based-miles-or-kilometers

Answer 8

这是一个使用javascript的简单实现，它基于纬度到kms的转换，1 degree latitude ~ 111.2 km。

我正在计算从给定纬度和经度宽度为10km的地图边界。

function getBoundsFromLatLng(lat, lng){
     var lat_change = 10/111.2;
     var lon_change = Math.abs(Math.cos(lat*(Math.PI/180)));
     var bounds = { 
         lat_min : lat - lat_change,
         lon_min : lng - lon_change,
         lat_max : lat + lat_change,
         lon_max : lng + lon_change
     };
     return bounds;
}

Answer 9

@Jan Philip Matuschek的插图出色的解释。（请将他的回答投票，而不是这个;我加上这个，因为我花了一点时间理解原来的答案）

优化寻找最近邻居的边界框技术需要得到距离为d的点P的最小和最大纬度，经度对。所有落在这些点之外的点肯定距离点大于d。这里要注意的一件事是Jan Philip Matuschek解释中突出显示的交叉纬度的计算。交叉纬度不在点P的纬度处，而是略微偏离它。这是确定距离d的点P的正确最小和最大边界经度时经常遗漏但重要的部分。这在验证中也是有用的。

P的（纬度，经度）与（纬度，经度）之间的半径距离等于距离d。

Python gist here https://gist.github.com/alexcpn/f95ae83a7ee0293a5225

Answer 10

我正在研究边界框问题，作为查找静态LAT，LONG点的SrcRad半径内的所有点的副问题。有很多使用

的计算

maxLon = $lon + rad2deg($rad/$R/cos(deg2rad($lat)));
minLon = $lon - rad2deg($rad/$R/cos(deg2rad($lat)));

计算经度界限，但我发现这不能给出所需的所有答案。因为你真正想做的是

(SrcRad/RadEarth)/cos(deg2rad(lat))

我知道，我知道答案应该是一样的，但我发现事实并非如此。似乎没有确定我正在做（SRCrad / RadEarth）首先然后除去Cos部分我留下了一些位置点。

获得所有边界框点后，如果你有一个计算给定纬度的点到点距离的函数，那么很容易只获得距固定点一定距离半径的那些点。这就是我做的。我知道它需要一些额外的步骤，但它帮助了我

-- GLOBAL Constants
gc_pi CONSTANT REAL := 3.14159265359;  -- Pi

-- Conversion Factor Constants
gc_rad_to_degs          CONSTANT NUMBER := 180/gc_pi; -- Conversion for Radians to Degrees 180/pi
gc_deg_to_rads          CONSTANT NUMBER := gc_pi/180; --Conversion of Degrees to Radians

lv_stat_lat    -- The static latitude point that I am searching from 
lv_stat_long   -- The static longitude point that I am searching from 

-- Angular radius ratio in radians
lv_ang_radius := lv_search_radius / lv_earth_radius;
lv_bb_maxlat := lv_stat_lat + (gc_rad_to_deg * lv_ang_radius);
lv_bb_minlat := lv_stat_lat - (gc_rad_to_deg * lv_ang_radius);

--Here's the tricky part, accounting for the Longitude getting smaller as we move up the latitiude scale
-- I seperated the parts of the equation to make it easier to debug and understand
-- I may not be a smart man but I know what the right answer is... :-)

lv_int_calc := gc_deg_to_rads * lv_stat_lat;
lv_int_calc := COS(lv_int_calc);
lv_int_calc := lv_ang_radius/lv_int_calc;
lv_int_calc := gc_rad_to_degs*lv_int_calc;

lv_bb_maxlong := lv_stat_long + lv_int_calc;
lv_bb_minlong := lv_stat_long - lv_int_calc;

-- Now select the values from your location datatable 
SELECT *  FROM (
SELECT cityaliasname, city, state, zipcode, latitude, longitude, 
-- The actual distance in miles
spherecos_pnttopntdist(lv_stat_lat, lv_stat_long, latitude, longitude, 'M') as miles_dist    
FROM Location_Table 
WHERE latitude between lv_bb_minlat AND lv_bb_maxlat
AND   longitude between lv_bb_minlong and lv_bb_maxlong)
WHERE miles_dist <= lv_limit_distance_miles
order by miles_dist
;

Answer 11

这是用于根据纬度/经度和距离获取边界框坐标的 javascript 代码。经测试，工作正常。

Number.prototype.degreeToRadius = function () {
    return this * (Math.PI / 180);
};

Number.prototype.radiusToDegree = function () {
    return (180 * this) / Math.PI;
};

function getBoundingBox(fsLatitude, fsLongitude, fiDistanceInKM) {

    if (fiDistanceInKM == null || fiDistanceInKM == undefined || fiDistanceInKM == 0)
        fiDistanceInKM = 1;
    
    var MIN_LAT, MAX_LAT, MIN_LON, MAX_LON, ldEarthRadius, ldDistanceInRadius, lsLatitudeInDegree, lsLongitudeInDegree,
        lsLatitudeInRadius, lsLongitudeInRadius, lsMinLatitude, lsMaxLatitude, lsMinLongitude, lsMaxLongitude, deltaLon;
    
    // coordinate limits
    MIN_LAT = (-90).degreeToRadius();
    MAX_LAT = (90).degreeToRadius();
    MIN_LON = (-180).degreeToRadius();
    MAX_LON = (180).degreeToRadius();

    // Earth's radius (km)
    ldEarthRadius = 6378.1;

    // angular distance in radians on a great circle
    ldDistanceInRadius = fiDistanceInKM / ldEarthRadius;

    // center point coordinates (deg)
    lsLatitudeInDegree = fsLatitude;
    lsLongitudeInDegree = fsLongitude;

    // center point coordinates (rad)
    lsLatitudeInRadius = lsLatitudeInDegree.degreeToRadius();
    lsLongitudeInRadius = lsLongitudeInDegree.degreeToRadius();

    // minimum and maximum latitudes for given distance
    lsMinLatitude = lsLatitudeInRadius - ldDistanceInRadius;
    lsMaxLatitude = lsLatitudeInRadius + ldDistanceInRadius;

    // minimum and maximum longitudes for given distance
    lsMinLongitude = void 0;
    lsMaxLongitude = void 0;

    // define deltaLon to help determine min and max longitudes
    deltaLon = Math.asin(Math.sin(ldDistanceInRadius) / Math.cos(lsLatitudeInRadius));

    if (lsMinLatitude > MIN_LAT && lsMaxLatitude < MAX_LAT) {
        lsMinLongitude = lsLongitudeInRadius - deltaLon;
        lsMaxLongitude = lsLongitudeInRadius + deltaLon;
        if (lsMinLongitude < MIN_LON) {
            lsMinLongitude = lsMinLongitude + 2 * Math.PI;
        }
        if (lsMaxLongitude > MAX_LON) {
            lsMaxLongitude = lsMaxLongitude - 2 * Math.PI;
        }
    }

    // a pole is within the given distance
    else {
        lsMinLatitude = Math.max(lsMinLatitude, MIN_LAT);
        lsMaxLatitude = Math.min(lsMaxLatitude, MAX_LAT);
        lsMinLongitude = MIN_LON;
        lsMaxLongitude = MAX_LON;
    }

    return [
        lsMinLatitude.radiusToDegree(),
        lsMinLongitude.radiusToDegree(),
        lsMaxLatitude.radiusToDegree(),
        lsMaxLongitude.radiusToDegree()
    ];
};

使用如下所示的 getBoundingBox 函数绘制边界框。

var lsRectangleLatLong = getBoundingBox(parseFloat(latitude), parseFloat(longitude), lsDistance);
            if (lsRectangleLatLong != null && lsRectangleLatLong != undefined) {
                latLngArr.push({ lat: lsRectangleLatLong[0], lng: lsRectangleLatLong[1] });
                latLngArr.push({ lat: lsRectangleLatLong[0], lng: lsRectangleLatLong[3] });
                latLngArr.push({ lat: lsRectangleLatLong[2], lng: lsRectangleLatLong[3] });
                latLngArr.push({ lat: lsRectangleLatLong[2], lng: lsRectangleLatLong[1] });
            }

Answer 12

只需访问panoramio网站，然后从panoramio网站打开世界地图，即可。只需前往指定所需纬度和经度的地点。

然后您在地址栏中找到了纬度和经度，例如在此地址中。

http://www.panoramio.com/map#lt=32.739485&ln=70.491211&z=9&k=1&a=1&tab=1&pl=all

lt = 32.739485 =＆gt;纬度 ln = 70.491211 =＆gt;经度

此Panoramio JavaScript API小部件围绕一对/长对创建一个边界框，然后返回所有这些边界内的照片。

另一种类型的Panoramio JavaScript API小部件，您还可以使用example and code is here更改背景颜色。

它没有表现出作曲情绪。它在出版后显示。

<div dir="ltr" style="text-align: center;" trbidi="on">
<script src="https://ssl.panoramio.com/wapi/wapi.js?v=1&amp;hl=en"></script>
<div id="wapiblock" style="float: right; margin: 10px 15px"></div>
<script type="text/javascript">
var myRequest = {
  'tag': 'kahna',
  'rect': {'sw': {'lat': -30, 'lng': 10.5}, 'ne': {'lat': 50.5, 'lng': 30}}
};
  var myOptions = {
  'width': 300,
  'height': 200
};
var wapiblock = document.getElementById('wapiblock');
var photo_widget = new panoramio.PhotoWidget('wapiblock', myRequest, myOptions);
photo_widget.setPosition(0);
</script>
</div>

Answer 13

如果感兴趣的话，我已经转换了Federico A. Ramponi给PHP的答案：

<?php
# deg2rad and rad2deg are already within PHP

# Semi-axes of WGS-84 geoidal reference
$WGS84_a = 6378137.0;  # Major semiaxis [m]
$WGS84_b = 6356752.3;  # Minor semiaxis [m]

# Earth radius at a given latitude, according to the WGS-84 ellipsoid [m]
function WGS84EarthRadius($lat)
{
    global $WGS84_a, $WGS84_b;

    $an = $WGS84_a * $WGS84_a * cos($lat);
    $bn = $WGS84_b * $WGS84_b * sin($lat);
    $ad = $WGS84_a * cos($lat);
    $bd = $WGS84_b * sin($lat);

    return sqrt(($an*$an + $bn*$bn)/($ad*$ad + $bd*$bd));
}

# Bounding box surrounding the point at given coordinates,
# assuming local approximation of Earth surface as a sphere
# of radius given by WGS84
function boundingBox($latitudeInDegrees, $longitudeInDegrees, $halfSideInKm)
{
    $lat = deg2rad($latitudeInDegrees);
    $lon = deg2rad($longitudeInDegrees);
    $halfSide = 1000 * $halfSideInKm;

    # Radius of Earth at given latitude
    $radius = WGS84EarthRadius($lat);
    # Radius of the parallel at given latitude
    $pradius = $radius*cos($lat);

    $latMin = $lat - $halfSide / $radius;
    $latMax = $lat + $halfSide / $radius;
    $lonMin = $lon - $halfSide / $pradius;
    $lonMax = $lon + $halfSide / $pradius;

    return array(rad2deg($latMin), rad2deg($lonMin), rad2deg($latMax), rad2deg($lonMax));
}
?>

Answer 14

感谢@Fedrico A.为Phyton实现，我已将其移植到Objective C类别类中。这是：

#import "LocationService+Bounds.h"

//Semi-axes of WGS-84 geoidal reference
const double WGS84_a = 6378137.0; //Major semiaxis [m]
const double WGS84_b = 6356752.3; //Minor semiaxis [m]

@implementation LocationService (Bounds)

struct BoundsLocation {
    double maxLatitude;
    double minLatitude;
    double maxLongitude;
    double minLongitude;
};

+ (struct BoundsLocation)locationBoundsWithLatitude:(double)aLatitude longitude:(double)aLongitude maxDistanceKm:(NSInteger)aMaxKmDistance {
    return [self boundingBoxWithLatitude:aLatitude longitude:aLongitude halfDistanceKm:aMaxKmDistance/2];
}

#pragma mark - Algorithm 

+ (struct BoundsLocation)boundingBoxWithLatitude:(double)aLatitude longitude:(double)aLongitude halfDistanceKm:(double)aDistanceKm {
    double radianLatitude = [self degreesToRadians:aLatitude];
    double radianLongitude = [self degreesToRadians:aLongitude];
    double halfDistanceMeters = aDistanceKm*1000;


    double earthRadius = [self earthRadiusAtLatitude:radianLatitude];
    double parallelRadius = earthRadius*cosl(radianLatitude);

    double radianMinLatitude = radianLatitude - halfDistanceMeters/earthRadius;
    double radianMaxLatitude = radianLatitude + halfDistanceMeters/earthRadius;
    double radianMinLongitude = radianLongitude - halfDistanceMeters/parallelRadius;
    double radianMaxLongitude = radianLongitude + halfDistanceMeters/parallelRadius;

    struct BoundsLocation bounds;
    bounds.minLatitude = [self radiansToDegrees:radianMinLatitude];
    bounds.maxLatitude = [self radiansToDegrees:radianMaxLatitude];
    bounds.minLongitude = [self radiansToDegrees:radianMinLongitude];
    bounds.maxLongitude = [self radiansToDegrees:radianMaxLongitude];

    return bounds;
}

+ (double)earthRadiusAtLatitude:(double)aRadianLatitude {
    double An = WGS84_a * WGS84_a * cosl(aRadianLatitude);
    double Bn = WGS84_b * WGS84_b * sinl(aRadianLatitude);
    double Ad = WGS84_a * cosl(aRadianLatitude);
    double Bd = WGS84_b * sinl(aRadianLatitude);
    return sqrtl( ((An * An) + (Bn * Bn))/((Ad * Ad) + (Bd * Bd)) );
}

+ (double)degreesToRadians:(double)aDegrees {
    return M_PI*aDegrees/180.0;
}

+ (double)radiansToDegrees:(double)aRadians {
    return 180.0*aRadians/M_PI;
}



@end

我测试了它，似乎工作得很好。 Struct BoundsLocation应该被一个类替换，我只是在这里用它来共享它。

Answer 15

上述所有答案都只是部分正确。特别是在像澳大利亚这样的地区，他们总是包括极点并计算一个非常大的矩形，甚至10kms。

特别是Jan Philip Matuschek在http://janmatuschek.de/LatitudeLongitudeBoundingCoordinates#UsingIndex的算法包括了一个非常大的矩形（-37，-90，-180,180），几乎在澳大利亚的每一个点。这会影响数据库中的大量用户，并且必须为全国近一半的所有用户计算距离。

我发现罗彻斯特理工学院的 Drupal API地球算法在极点和其他地方都能更好地工作，并且更容易实现。

https://www.rit.edu/drupal/api/drupal/sites%21all%21modules%21location%21earth.inc/7.54

使用上述算法中的earth_latitude_range和earth_longitude_range来计算边界矩形

并使用谷歌地图记录的距离计算公式来计算距离

https://developers.google.com/maps/solutions/store-locator/clothing-store-locator#outputting-data-as-xml-using-php

要以公里而不是里程搜索，请将3959替换为6371。 对于（Lat，Lng）=（37，-122）和带有列lat和lng 的Markers表，公式为：

SELECT id, ( 3959 * acos( cos( radians(37) ) * cos( radians( lat ) ) * cos( radians( lng ) - radians(-122) ) + sin( radians(37) ) * sin( radians( lat ) ) ) ) AS distance FROM markers HAVING distance < 25 ORDER BY distance LIMIT 0 , 20;

在https://stackoverflow.com/a/45950426/5076414

上阅读我的详细答案

Answer 16

这是Federico Ramponi在Go中的回答。注意：没有错误检查：（

import (
    "math"
)

// Semi-axes of WGS-84 geoidal reference
const (
    // Major semiaxis (meters)
    WGS84A = 6378137.0
    // Minor semiaxis (meters)
    WGS84B = 6356752.3
)

// BoundingBox represents the geo-polygon that encompasses the given point and radius
type BoundingBox struct {
    LatMin float64
    LatMax float64
    LonMin float64
    LonMax float64
}

// Convert a degree value to radians
func deg2Rad(deg float64) float64 {
    return math.Pi * deg / 180.0
}

// Convert a radian value to degrees
func rad2Deg(rad float64) float64 {
    return 180.0 * rad / math.Pi
}

// Get the Earth's radius in meters at a given latitude based on the WGS84 ellipsoid
func getWgs84EarthRadius(lat float64) float64 {
    an := WGS84A * WGS84A * math.Cos(lat)
    bn := WGS84B * WGS84B * math.Sin(lat)

    ad := WGS84A * math.Cos(lat)
    bd := WGS84B * math.Sin(lat)

    return math.Sqrt((an*an + bn*bn) / (ad*ad + bd*bd))
}

// GetBoundingBox returns a BoundingBox encompassing the given lat/long point and radius
func GetBoundingBox(latDeg float64, longDeg float64, radiusKm float64) BoundingBox {
    lat := deg2Rad(latDeg)
    lon := deg2Rad(longDeg)
    halfSide := 1000 * radiusKm

    // Radius of Earth at given latitude
    radius := getWgs84EarthRadius(lat)

    pradius := radius * math.Cos(lat)

    latMin := lat - halfSide/radius
    latMax := lat + halfSide/radius
    lonMin := lon - halfSide/pradius
    lonMax := lon + halfSide/pradius

    return BoundingBox{
        LatMin: rad2Deg(latMin),
        LatMax: rad2Deg(latMax),
        LonMin: rad2Deg(lonMin),
        LonMax: rad2Deg(lonMax),
    }
}

如何计算给定lat / lng位置的边界框？

16 个答案: