如何计算与另一个点相距一定距离的点的纬度? [英] How to calculate the latlng of a point a certain distance away from another?
问题描述
为了在地图上画一个圆，我有一个中心 GLatLng (A) 和一个以米为单位的半径 (r).
这是一个图表:
-------------/---/-///r |*-------------*A/B//-/---/-------------如何计算位置 B 的 GLatLng?假设r平行于赤道.
使用 GLatLng.distanceFrom() 方法在给定 A 和 B 时获取半径是微不足道的 - 但反过来则不然.似乎我需要做一些更重的数学运算.
我们需要一个方法，当给定方位角和从源点行进的距离时返回目标点.幸运的是，Chris Veness 在 评论时，这是当圆圈环绕其中一个极点时发生的情况.
在北极附近绘制pointA，半径为1000公里:
var pointA = new google.maps.LatLng(85, 0);//靠近北极无功半径 = 1000;//1000公里pointA.destinationPoint(90, radius) 的截图:
To draw a circle on map I have a center GLatLng (A) and a radius (r) in meters.
Here's a diagram:
-----------
--/ --
-/ -
/
/
/ r
| *-------------*
A / B
/
/
- /-
-- /--
-----------
How to calculate the GLatLng at position B? Assuming that r is parallel to the equator.
Getting the radius when A and B is given is trivial using the GLatLng.distanceFrom() method - but doing it the other way around not so. Seems that I need to do some heavier math.
We will need a method that returns the destination point when given a bearing and the distance travelled from a source point. Luckily, there is a very good JavaScript implementation by Chris Veness at Calculate distance, bearing and more between Latitude/Longitude points.
The following has been adapted to work with the google.maps.LatLng class:
Number.prototype.toRad = function() {
return this * Math.PI / 180;
}
Number.prototype.toDeg = function() {
return this * 180 / Math.PI;
}
google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
dist = dist / 6371;
brng = brng.toRad();
var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();
var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) +
Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));
var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
Math.cos(lat1),
Math.cos(dist) - Math.sin(lat1) *
Math.sin(lat2));
if (isNaN(lat2) || isNaN(lon2)) return null;
return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
}
You would simply use it as follows:
var pointA = new google.maps.LatLng(25.48, -71.26);
var radiusInKm = 10;
var pointB = pointA.destinationPoint(90, radiusInKm);
Here is a complete example using Google Maps API v3:
<!DOCTYPE html>
<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=UTF-8"/>
<title>Google Maps Geometry</title>
<script src="http://maps.google.com/maps/api/js?sensor=false"
type="text/javascript"></script>
</head>
<body>
<div id="map" style="width: 400px; height: 300px"></div>
<script type="text/javascript">
Number.prototype.toRad = function() {
return this * Math.PI / 180;
}
Number.prototype.toDeg = function() {
return this * 180 / Math.PI;
}
google.maps.LatLng.prototype.destinationPoint = function(brng, dist) {
dist = dist / 6371;
brng = brng.toRad();
var lat1 = this.lat().toRad(), lon1 = this.lng().toRad();
var lat2 = Math.asin(Math.sin(lat1) * Math.cos(dist) +
Math.cos(lat1) * Math.sin(dist) * Math.cos(brng));
var lon2 = lon1 + Math.atan2(Math.sin(brng) * Math.sin(dist) *
Math.cos(lat1),
Math.cos(dist) - Math.sin(lat1) *
Math.sin(lat2));
if (isNaN(lat2) || isNaN(lon2)) return null;
return new google.maps.LatLng(lat2.toDeg(), lon2.toDeg());
}
var pointA = new google.maps.LatLng(40.70, -74.00); // Circle center
var radius = 10; // 10km
var mapOpt = {
mapTypeId: google.maps.MapTypeId.TERRAIN,
center: pointA,
zoom: 10
};
var map = new google.maps.Map(document.getElementById("map"), mapOpt);
// Draw the circle
new google.maps.Circle({
center: pointA,
radius: radius * 1000, // Convert to meters
fillColor: '#FF0000',
fillOpacity: 0.2,
map: map
});
// Show marker at circle center
new google.maps.Marker({
position: pointA,
map: map
});
// Show marker at destination point
new google.maps.Marker({
position: pointA.destinationPoint(90, radius),
map: map
});
</script>
</body>
</html>
Screenshot:
UPDATE:
In reply to Paul's comment below, this is what happens when the circle wraps around one of the poles.
Plotting pointA near the north pole, with a radius of 1,000km:
var pointA = new google.maps.LatLng(85, 0); // Close to north pole
var radius = 1000; // 1000km
Screenshot for pointA.destinationPoint(90, radius):
这篇关于如何计算与另一个点相距一定距离的点的纬度?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
