سلام
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void DistanceFromLine(double cx, double cy, double ax, double ay ,
double bx, double by, double &distanceSegment,
double &distanceLine)
{
double r_numerator = (cx-ax)*(bx-ax) + (cy-ay)*(by-ay);
double r_denomenator = (bx-ax)*(bx-ax) + (by-ay)*(by-ay);
double r = r_numerator / r_denomenator;
//
double px = ax + r*(bx-ax);
double py = ay + r*(by-ay);
//
double s = ((ay-cy)*(bx-ax)-(ax-cx)*(by-ay) ) / r_denomenator;
distanceLine = fabs(s)*sqrt(r_denomenator);
//
// (xx,yy) is the point on the lineSegment closest to (cx,cy)
//
double xx = px;
double yy = py;
if ( (r >= 0) && (r <= 1) )
{
distanceSegment = distanceLine;
}
else
{
double dist1 = (cx-ax)*(cx-ax) + (cy-ay)*(cy-ay);
double dist2 = (cx-bx)*(cx-bx) + (cy-by)*(cy-by);
if (dist1 < dist2)
{
xx = ax;
yy = ay;
distanceSegment = sqrt(dist1);
}
else
{
xx = bx;
yy = by;
distanceSegment = sqrt(dist2);
}
}
return;
}
این هم یه برنامه که تحت کنسول هست
الان ویژوال سی شارپ نصب شده ندارم وگرنه برات تبدیلش میکردم
using System;
using System.Text;
public class CDistanceBetweenLocations
{
public static double Calc(double Lat1,
double Long1, double Lat2, double Long2)
{
/*
dlon = lon2 - lon1
dlat = lat2 - lat1
a = (sin(dlat/2))^2 + cos(lat1) * cos(lat2) * (sin(dlon/2))^2
c = 2 * atan2(sqrt(a), sqrt(1-a))
d = R * c
Where
* dlon is the change in longitude
* dlat is the change in latitude
* c is the great circle distance in Radians.
* R is the radius of a spherical Earth.
* The locations of the two points in
spherical coordinates (longitude and
latitude) are lon1,lat1 and lon2, lat2.
*/
double dDistance = Double.MinValue;
double dLat1InRad = Lat1 * (Math.PI / 180.0);
double dLong1InRad = Long1 * (Math.PI / 180.0);
double dLat2InRad = Lat2 * (Math.PI / 180.0);
double dLong2InRad = Long2 * (Math.PI / 180.0);
double dLongitude = dLong2InRad - dLong1InRad;
double dLatitude = dLat2InRad - dLat1InRad;
// Intermediate result a.
double a = Math.Pow(Math.Sin(dLatitude / 2.0), 2.0) +
Math.Cos(dLat1InRad) * Math.Cos(dLat2InRad) *
Math.Pow(Math.Sin(dLongitude / 2.0), 2.0);
// Intermediate result c (great circle distance in Radians).
double c = 2.0 * Math.Atan2(Math.Sqrt(a), Math.Sqrt(1.0 - a));
// Distance.
// const Double kEarthRadiusMiles = 3956.0;
const Double kEarthRadiusKms = 6376.5;
dDistance = kEarthRadiusKms * c;
return dDistance;
}
public static double Calc(string NS1, double Lat1, double Lat1Min,
string EW1, double Long1, double Long1Min, string NS2,
double Lat2, double Lat2Min, string EW2,
double Long2, double Long2Min)
{
double NS1Sign = NS1.ToUpper() == "N" ? 1.0 : -1.0;
double EW1Sign = NS1.ToUpper() == "E" ? 1.0 : -1.0;
double NS2Sign = NS2.ToUpper() == "N" ? 1.0 : -1.0;
double EW2Sign = EW2.ToUpper() == "E" ? 1.0 : -1.0;
return (Calc(
(Lat1 + (Lat1Min / 60)) * NS1Sign,
(Long1 + (Long1Min / 60)) * EW1Sign,
(Lat2 + (Lat2Min / 60)) * NS2Sign,
(Long2 + (Long2Min / 60)) * EW2Sign
));
}
public static void Main(string[] args)
{
if (args.Length < 12)
{
System.Console.WriteLine("usage: DistanceBetweenLocations" +
" N 43 35.500 W 80 27.800 N 43 35.925 W 80 28.318");
return;
}
System.Console.WriteLine(Calc(
args[0],
System.Double.Parse(args[1]),
System.Double.Parse(args[2]),
args[3],
System.Double.Parse(args[4]),
System.Double.Parse(args[5]),
args[6],
System.Double.Parse(args[7]),
System.Double.Parse(args[8]),
args[9],
System.Double.Parse(args[10]),
System.Double.Parse(args[11])));
}
}
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نوشته شده توسط jeson_park
