[英]Geographic Midpoint between two coordinates
我一直在使用Moveable-Type网站来帮助我进行一些Geocoordinate计算并且它非常有用,但是,我在计算两个坐标之间的中点时遇到了一个错误。 我的结果接近预期,但不够接近:
posA = {47.64570362, -122.14073746}
posB = {47.64316917, -122.14032175}
预期结果(取自可移动型计算器)= 47°38'40“N,122°08'26”W = {47.644444, -122.140556}
我的结果: {49.6054801645915, -122.14052959995759}
这是我的代码:
private Geocoordinate MidPoint(Geocoordinate posA, Geocoordinate posB)
{
Geocoordinate midPoint = new Geocoordinate();
double dLon = DegreesToRadians(posB.Longitude - posA.Longitude);
double Bx = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Cos(dLon);
double By = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Sin(dLon);
midPoint.Latitude = RadiansToDegrees(Math.Atan2(Math.Sin(DegreesToRadians(posA.Latitude)) + Math.Sin(DegreesToRadians(posB.Latitude)),
Math.Sqrt((Math.Cos(DegreesToRadians(posA.Latitude)) + Bx) * (Math.Cos(DegreesToRadians(posA.Latitude))) + Bx) + By * By));
midPoint.Longitude = posA.Longitude + RadiansToDegrees(Math.Atan2(By, Math.Cos(DegreesToRadians(posA.Latitude)) + Bx));
return midPoint;
}
我有几种私有方法可以在Degrees和Radians之间进行转换。 例如
private double DegreeToRadian(double angle)
{
return Math.PI * angle / 180.0;
}
我无法弄清楚为什么我的结果在Lat值上偏离了几度。 有任何想法吗?
谢谢
你把一些括号弄错了。 我在代码中标出了这个位置。
private Geocoordinate MidPoint(Geocoordinate posA, Geocoordinate posB)
{
Geocoordinate midPoint = new Geocoordinate();
double dLon = DegreesToRadians(posB.Longitude - posA.Longitude);
double Bx = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Cos(dLon);
double By = Math.Cos(DegreesToRadians(posB.Latitude)) * Math.Sin(dLon);
midPoint.Latitude = RadiansToDegrees(Math.Atan2(
Math.Sin(DegreesToRadians(posA.Latitude)) + Math.Sin(DegreesToRadians(posB.Latitude)),
Math.Sqrt(
(Math.Cos(DegreesToRadians(posA.Latitude)) + Bx) *
(Math.Cos(DegreesToRadians(posA.Latitude)) + Bx) + By * By)));
// (Math.Cos(DegreesToRadians(posA.Latitude))) + Bx) + By * By)); // Your Code
midPoint.Longitude = posA.Longitude + RadiansToDegrees(Math.Atan2(By, Math.Cos(DegreesToRadians(posA.Latitude)) + Bx));
return midPoint;
}
一些方法(例如WGS84惯例)包括地球的扁率。 ( 至少这就是它在这里所说的 )
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