我是gis的新手。我在使用JavaScript。
假设有中心latlong。另一点是目标(latlong)远离中心的n米。如何计算这个位置差异(中心 - latlong)从中心和n在斜切。
伙计们请帮助我。让我们认为地球是正确的球形。
答案 0 :(得分:1)
function distance(lat1, lon1, lat2, lon2) {
var radius = 6371e3; // meters
var dLon = gis.toRad(lon2 - lon1),
lat1 = gis.toRad(lat1),
lat2 = gis.toRad(lat2),
distance = Math.acos(Math.sin(lat1) * Math.sin(lat2) +
Math.cos(lat1) * Math.cos(lat2) * Math.cos(dLon)) * radius;
return distance;
}
以上javascript函数用于计算距离,
function(lat, long, distance){
//can i calculate lat_difference, long difference
}
答案 1 :(得分:1)
这是你的代码: -
<script>
Math.radians = function(degrees) {
return degrees * Math.PI / 180;
};
function calculateDistance(lat,lon,lat_center,lon_center){
var distance = ( 6371 * Math.acos( Math.cos( Math.radians(lat) ) * Math.cos( Math.radians( lat_center ) )
* Math.cos( Math.radians( lon_center ) - Math.radians(lon) ) + Math.sin( Math.radians(lat) ) * Math.sin(Math.radians(lat_center)) ) )*1000;
console.log(distance+" meter");
return distance;
}
var lat = '38.898556';
var lon = '-77.037852';
var lat_center = '38.897147';
var lon_center = '-77.043934';
calculateDistance(lat,lon,lat_center,lon_center);//will return 549 meter
//for getting lat and lon from a distance from a given point
//lat1 = latitude of start point in degrees
//long1 = longitude of start point in degrees
//d = distance in KM
//angle = bearing in degrees
function get_gps_distance(lat1,long1,d,angle)
{
//# Earth Radious in KM
var R = 6378.14;
//# Degree to Radian
var latitude1 = lat1 * (Math.PI/180);
var longitude1 = long1 * (Math.PI/180);
brng = angle * (Math.PI/180);
latitude2 = Math.asin(Math.sin(latitude1)*Math.cos(d/R) + Math.cos(latitude1)*Math.sin(d/R)*Math.cos(brng));
longitude2 = longitude1 + Math.atan2(Math.sin(brng)*Math.sin(d/R)*Math.cos(latitude1),Math.cos(d/R)-Math.sin(latitude1)*Math.sin(latitude2));
// # back to degrees
latitude2 = latitude2 * (180/Math.PI);
longitude2 = longitude2 * (180/Math.PI);
// # 6 decimal for Leaflet and other system compatibility
lat2 = latitude2;
long2 =longitude2;
var tab = {};
// Push in array and get back
tab[0] = lat2;
tab[1] = long2;
return tab;
}
get_gps_distance('38.898556','-77.037852',.549,90);
</script>