我有两个地理坐标（起点和终点），我可以在它们之间创建一条线。假设这两个坐标距离为30米，那么我需要以3米的间隔找到每个点的地理位置。所以有10个这样的要点是必需的。
我可以通过一些公式找到点，但这些点与起点和终点形成的线的方向不同。
到目前为止我所做的是......
using System;
namespace Test
{
public class AzimuthCalculator
{
public const double range = 0.00186411F; // in Miles
public const double rangeInMeter = 3F;
public const double factor = 0.003F;
public static void Main(String[] args)
{
double sLatitude = 28.6187763214111F;
double sLongitude = 77.2093048095703F;
double eLatitude = 28.6191763153134F;
double eLongitude = 77.2097146511078F;
Console.WriteLine($"Start Point : {sLatitude}, {sLongitude}");
Console.WriteLine($"End Point : {eLatitude},{eLongitude}");
double distance = CalculateDistance(sLatitude, sLongitude, eLatitude, eLongitude);
Console.WriteLine($"Distance : {distance} miles, {MilesToMeter(distance)} meter, {(distance * 1.60934)} kilometer");
distance = distance * 1.60934;
double numberOfIDS = distance/factor;
double azimuthAngle = DegreeBearing(sLatitude, sLongitude, eLatitude, eLongitude);
Console.WriteLine($"Azimuth : {azimuthAngle}");
Console.WriteLine($"IDS : {numberOfIDS}");
double constantAzimuth = (azimuthAngle/numberOfIDS);
azimuthAngle = constantAzimuth;
Console.WriteLine($"Original Azimuth : {azimuthAngle}");
double[] getAnotherPoint = pointRadialDistance(sLatitude, sLongitude, azimuthAngle, distance);
Console.WriteLine($"End Point : {getAnotherPoint[0]},{getAnotherPoint[1]}");
distance = 0.003; // 3 meter
getAnotherPoint = pointRadialDistance(sLatitude, sLongitude, azimuthAngle, distance);
int totalIDS = (Int32)(numberOfIDS);
for(int i=0;i<totalIDS;i++)
{
sLatitude = getAnotherPoint[0];
sLongitude = getAnotherPoint[1];
distance = 0.003;
Console.WriteLine($"new PointLatLng({getAnotherPoint[0]},{getAnotherPoint[1]}),");
getAnotherPoint = pointRadialDistance(sLatitude, sLongitude, azimuthAngle, distance);
}
Console.ReadLine();
}
static double rEarth = 6371.01F; // Earth radius in km
static double epsilon = 0.000001F;
public static double[] pointRadialDistance(double lat1, double lon1, double bearing, double distance)
{
double rlat1 = ToRad(lat1);
double rlon1 = ToRad(lon1);
double rbearing = ToRad(bearing);
double rdistance = distance / rEarth; // normalize linear distance to radian angle
double rlat = Math.Asin( Math.Sin(rlat1) * Math.Cos(rdistance) + Math.Cos(rlat1) * Math.Sin(rdistance) * Math.Cos(rbearing));
double rlon = 0.0F;
if ( Math.Cos(rlat) == 0 || Math.Abs(Math.Cos(rlat)) < epsilon) // Endpoint a pole
rlon=rlon1;
else
rlon = ((rlon1 - Math.Asin( Math.Sin(rbearing) * Math.Sin(rdistance) / Math.Cos(rlat) ) + Math.PI ) % (2*Math.PI) ) - Math.PI;
double lat = ToDegrees(rlat);
double lon = ToDegrees(rlon);
double[] listNew = new double[2];
listNew[0] = lat;
listNew[1] = lon;
return (listNew);
}
public static GeoLocation FindPointAtDistanceFrom(GeoLocation startPoint, double initialBearingRadians, double distanceKilometres)
{
const double radiusEarthKilometres = 6371.01;
var distRatio = distanceKilometres / radiusEarthKilometres;
var distRatioSine = Math.Sin(distRatio);
var distRatioCosine = Math.Cos(distRatio);
var startLatRad = DegreesToRadians(startPoint.Latitude);
var startLonRad = DegreesToRadians(startPoint.Longitude);
var startLatCos = Math.Cos(startLatRad);
var startLatSin = Math.Sin(startLatRad);
var endLatRads = Math.Asin((startLatSin * distRatioCosine) + (startLatCos * distRatioSine * Math.Cos(initialBearingRadians)));
var endLonRads = startLonRad
+ Math.Atan2(
Math.Sin(initialBearingRadians) * distRatioSine * startLatCos,
distRatioCosine - startLatSin * Math.Sin(endLatRads));
return new GeoLocation
{
Latitude = RadiansToDegrees(endLatRads),
Longitude = RadiansToDegrees(endLonRads)
};
}
public struct GeoLocation
{
public double Latitude { get; set; }
public double Longitude { get; set; }
}
public static double DegreesToRadians(double degrees)
{
const double degToRadFactor = Math.PI / 180;
return degrees * degToRadFactor;
}
public static double RadiansToDegrees(double radians)
{
const double radToDegFactor = 180 / Math.PI;
return radians * radToDegFactor;
}
public static double CalculateDistance(double sLatitude, double sLongitude, double eLatitude, double eLongitude)
{
var radiansOverDegrees = (Math.PI / 180.0);
var sLatitudeRadians = sLatitude * radiansOverDegrees;
var sLongitudeRadians = sLongitude * radiansOverDegrees;
var eLatitudeRadians = eLatitude * radiansOverDegrees;
var eLongitudeRadians = eLongitude * radiansOverDegrees;
var dLongitude = eLongitudeRadians - sLongitudeRadians;
var dLatitude = eLatitudeRadians - sLatitudeRadians;
var result1 = Math.Pow(Math.Sin(dLatitude / 2.0), 2.0) + Math.Cos(sLatitudeRadians) * Math.Cos(eLatitudeRadians) * Math.Pow(Math.Sin(dLongitude / 2.0), 2.0);
// Using 3956 as the number of miles around the earth
var result2 = 3956.0 * 2.0 * Math.Atan2(Math.Sqrt(result1), Math.Sqrt(1.0 - result1));
return result2;
}
static double DegreeBearing(double lat1, double lon1,double lat2, double lon2)
{
var dLon = ToRad(lon2 - lon1);
var dPhi = Math.Log(Math.Tan(ToRad(lat2) / 2 + Math.PI / 4) / Math.Tan(ToRad(lat1) / 2 + Math.PI / 4));
if (Math.Abs(dLon) > Math.PI)
dLon = dLon > 0 ? - (2 * Math.PI - dLon) : (2 * Math.PI + dLon);
return ToBearing(Math.Atan2(dLon, dPhi));
}
public static double ToRad(double degrees)
{
return degrees * (Math.PI / 180);
}
public static double ToDegrees(double radians)
{
return radians * 180 / Math.PI;
}
public static double ToBearing(double radians)
{
// convert radians to degrees (as bearing: 0...360)
return (ToDegrees(radians) + 360) % 360;
}
public static double MeterToMiles(double meter)
{
return (meter / 1609.344);
}
public static double MilesToMeter(double miles)
{
return (miles * 1609.344);
}
}
}
为什么你从这里constantAzimuth = (azimuthAngle/numberOfIDS);计算错误的方位并稍后使用它？
您可以使用方法described here计算大圆路径上的中间点（基本上它是SLERP - 球面线性插值）
Formula:
a = sin((1−f)⋅δ) / sin δ
b = sin(f⋅δ) / sin δ
x = a ⋅ cos φ1 ⋅ cos λ1 + b ⋅ cos φ2 ⋅ cos λ2
y = a ⋅ cos φ1 ⋅ sin λ1 + b ⋅ cos φ2 ⋅ sin λ2
z = a ⋅ sin φ1 + b ⋅ sin φ2
φi = atan2(z, √x² + y²)
λi = atan2(y, x)
where
f is fraction along great circle route (f=0 is point 1, f=1 is point 2),
δ is the angular distance d/R between the two points.