如何计算两个GPS坐标之间的距离(使用经纬度)?
当前回答
private double deg2rad(double deg)
{
return (deg * Math.PI / 180.0);
}
private double rad2deg(double rad)
{
return (rad / Math.PI * 180.0);
}
private double GetDistance(double lat1, double lon1, double lat2, double lon2)
{
//code for Distance in Kilo Meter
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
return (dist);
}
private double GetDirection(double lat1, double lon1, double lat2, double lon2)
{
//code for Direction in Degrees
double dlat = deg2rad(lat1) - deg2rad(lat2);
double dlon = deg2rad(lon1) - deg2rad(lon2);
double y = Math.Sin(dlon) * Math.Cos(lat2);
double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
if (direct < 0)
direct = direct + 360;
return (direct);
}
private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
return (Speed);
}
private double GetDuration(DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
return (TimeDifferenceInSeconds);
}
其他回答
private double deg2rad(double deg)
{
return (deg * Math.PI / 180.0);
}
private double rad2deg(double rad)
{
return (rad / Math.PI * 180.0);
}
private double GetDistance(double lat1, double lon1, double lat2, double lon2)
{
//code for Distance in Kilo Meter
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = Math.Abs(Math.Round(rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344 * 1000, 0));
return (dist);
}
private double GetDirection(double lat1, double lon1, double lat2, double lon2)
{
//code for Direction in Degrees
double dlat = deg2rad(lat1) - deg2rad(lat2);
double dlon = deg2rad(lon1) - deg2rad(lon2);
double y = Math.Sin(dlon) * Math.Cos(lat2);
double x = Math.Cos(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) - Math.Sin(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(dlon);
double direct = Math.Round(rad2deg(Math.Atan2(y, x)), 0);
if (direct < 0)
direct = direct + 360;
return (direct);
}
private double GetSpeed(double lat1, double lon1, double lat2, double lon2, DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Round(TimeDifference.TotalSeconds, 0);
double theta = lon1 - lon2;
double dist = Math.Sin(deg2rad(lat1)) * Math.Sin(deg2rad(lat2)) + Math.Cos(deg2rad(lat1)) * Math.Cos(deg2rad(lat2)) * Math.Cos(deg2rad(theta));
dist = rad2deg(Math.Acos(dist)) * 60 * 1.1515 * 1.609344;
double Speed = Math.Abs(Math.Round((dist / Math.Abs(TimeDifferenceInSeconds)) * 60 * 60, 0));
return (Speed);
}
private double GetDuration(DateTime CurTime, DateTime PrevTime)
{
//code for speed in Kilo Meter/Hour
TimeSpan TimeDifference = CurTime.Subtract(PrevTime);
double TimeDifferenceInSeconds = Math.Abs(Math.Round(TimeDifference.TotalSeconds, 0));
return (TimeDifferenceInSeconds);
}
一个T-SQL函数,我用来根据中心的距离选择记录
Create Function [dbo].[DistanceInMiles]
( @fromLatitude float ,
@fromLongitude float ,
@toLatitude float,
@toLongitude float
)
returns float
AS
BEGIN
declare @distance float
select @distance = cast((3963 * ACOS(round(COS(RADIANS(90-@fromLatitude))*COS(RADIANS(90-@toLatitude))+
SIN(RADIANS(90-@fromLatitude))*SIN(RADIANS(90-@toLatitude))*COS(RADIANS(@fromLongitude-@toLongitude)),15))
)as float)
return round(@distance,1)
END
打印稿版本
export const degreeToRadian = (degree: number) => {
return degree * Math.PI / 180;
}
export const distanceBetweenEarthCoordinatesInKm = (lat1: number, lon1: number, lat2: number, lon2: number) => {
const earthRadiusInKm = 6371;
const dLat = degreeToRadian(lat2 - lat1);
const dLon = degreeToRadian(lon2 - lon1);
lat1 = degreeToRadian(lat1);
lat2 = degreeToRadian(lat2);
const a = Math.sin(dLat / 2) * Math.sin(dLat / 2) + Math.sin(dLon / 2) * Math.sin(dLon / 2) * Math.cos(lat1) * Math.cos(lat2);
const c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1 - a));
return earthRadiusInKm * c;
}
你可以在f#的fssnip中找到这个实现(有一些很好的解释)
以下是重要的部分:
let GreatCircleDistance<[<Measure>] 'u> (R : float<'u>) (p1 : Location) (p2 : Location) =
let degToRad (x : float<deg>) = System.Math.PI * x / 180.0<deg/rad>
let sq x = x * x
// take the sin of the half and square the result
let sinSqHf (a : float<rad>) = (System.Math.Sin >> sq) (a / 2.0<rad>)
let cos (a : float<deg>) = System.Math.Cos (degToRad a / 1.0<rad>)
let dLat = (p2.Latitude - p1.Latitude) |> degToRad
let dLon = (p2.Longitude - p1.Longitude) |> degToRad
let a = sinSqHf dLat + cos p1.Latitude * cos p2.Latitude * sinSqHf dLon
let c = 2.0 * System.Math.Atan2(System.Math.Sqrt(a), System.Math.Sqrt(1.0-a))
R * c
我把上面的答案用在Scala程序中
import java.lang.Math.{atan2, cos, sin, sqrt}
def latLonDistance(lat1: Double, lon1: Double)(lat2: Double, lon2: Double): Double = {
val earthRadiusKm = 6371
val dLat = (lat2 - lat1).toRadians
val dLon = (lon2 - lon1).toRadians
val latRad1 = lat1.toRadians
val latRad2 = lat2.toRadians
val a = sin(dLat / 2) * sin(dLat / 2) + sin(dLon / 2) * sin(dLon / 2) * cos(latRad1) * cos(latRad2)
val c = 2 * atan2(sqrt(a), sqrt(1 - a))
earthRadiusKm * c
}
我对函数进行了压缩,以便能够轻松地生成具有两个固定位置之一的函数,并且只需要一对lat/lon来生成距离。
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