如何從兩個位置的經度和緯度找到距離？

R = earth’s radius (mean radius = 6,371km) 
Δlat = lat2− lat1 
Δlong = long2− long1 
a = sin²(Δlat/2) + cos(lat1).cos(lat2).sin²(Δlong/2) 
c = 2.atan2(√a, √(1−a)) 
d = R.c

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2009-09-14 07:00:15 mauris

5

應用Haversine公式來查找距離。請參閱下面的C＃代碼來查找2個座標之間的距離。更好的是，如果你想說在一定範圍內找到商店列表，你可以在SQL中應用WHERE子句或者在C＃中應用LINQ過濾器。

這裏的公式是以公里爲單位，你將不得不改變相關的數字，它將以英里爲單位。

例如：將6371.392896轉換爲里程。

DECLARE @radiusInKm AS FLOAT 
    DECLARE @lat2Compare AS FLOAT 
    DECLARE @long2Compare AS FLOAT 
    SET @radiusInKm = 5.000 
    SET @lat2Compare = insert_your_lat_to_compare_here 
    SET @long2Compare = insert_you_long_to_compare_here 

    SELECT * FROM insert_your_table_here WITH(NOLOCK) 
    WHERE (6371.392896*2*ATN2(SQRT((sin((radians(GeoLatitude - @lat2Compare))/2) * sin((radians(GeoLatitude - @lat2Compare))/2)) + (cos(radians(GeoLatitude)) * cos(radians(@lat2Compare)) * sin(radians(GeoLongitude - @long2Compare)/2) * sin(radians(GeoLongitude - @long2Compare)/2))) 
    , SQRT(1-((sin((radians(GeoLatitude - @lat2Compare))/2) * sin((radians(GeoLatitude - @lat2Compare))/2)) + (cos(radians(GeoLatitude)) * cos(radians(@lat2Compare)) * sin(radians(GeoLongitude - @long2Compare)/2) * sin(radians(GeoLongitude - @long2Compare)/2))) 
    ))) <= @radiusInKm

如果你想在C＃中執行haversine公式，

double resultDistance = 0.0; 
    double avgRadiusOfEarth = 6371.392896; //Radius of the earth differ, I'm taking the average. 

    //Haversine formula 
    //distance = R * 2 * aTan2 (square root of A, square root of 1 - A) 
    //     where A = sinus squared (difference in latitude/2) + (cosine of latitude 1 * cosine of latitude 2 * sinus squared (difference in longitude/2)) 
    //     and R = the circumference of the earth 

    double differenceInLat = DegreeToRadian(currentLatitude - latitudeToCompare); 
    double differenceInLong = DegreeToRadian(currentLongitude - longtitudeToCompare); 
    double aInnerFormula = Math.Cos(DegreeToRadian(currentLatitude)) * Math.Cos(DegreeToRadian(latitudeToCompare)) * Math.Sin(differenceInLong/2) * Math.Sin(differenceInLong/2); 
    double aFormula = (Math.Sin((differenceInLat)/2) * Math.Sin((differenceInLat)/2)) + (aInnerFormula); 
    resultDistance = avgRadiusOfEarth * 2 * Math.Atan2(Math.Sqrt(aFormula), Math.Sqrt(1 - aFormula));

DegreesToRadian是我自定義創建一個功能，它是一個簡單的1個班輪"Math.PI * angle/180.0

My blog entry - SQL Haversine

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2010-06-11 07:36:50 Zac

-1

只使用距離公式Sqrt((x2-x1)^2 + (y2-y1)^2)

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2015-01-08 10:57:49 RNR123

1

這裏是由給定的IP查找位置/近地點，經度/緯度小提琴：

http://jsfiddle.net/bassta/zrgd9qc3/2/

，這裏是我使用來計算直線距離的函數：

function distance(lat1, lng1, lat2, lng2) { 
     var radlat1 = Math.PI * lat1/180; 
     var radlat2 = Math.PI * lat2/180; 
     var radlon1 = Math.PI * lng1/180; 
     var radlon2 = Math.PI * lng2/180; 
     var theta = lng1 - lng2; 
     var radtheta = Math.PI * theta/180; 
     var dist = Math.sin(radlat1) * Math.sin(radlat2) + Math.cos(radlat1) * Math.cos(radlat2) * Math.cos(radtheta); 
     dist = Math.acos(dist); 
     dist = dist * 180/Math.PI; 
     dist = dist * 60 * 1.1515; 

     //Get in in kilometers 
     dist = dist * 1.609344; 

     return dist; 
    }

它返回以公里爲單位的距離

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2015-03-31 14:36:37

0

在此頁面您可以看到整個代碼和公式在Android位置中計算位置的距離類

http://grepcode.com/file/repository.grepcode.com/java/ext/com.google.android/android/4.4.4_r1/android/location/Location.java#Location.computeDistanceAndBearing%28double%2Cdouble%2Cdouble%2Cdouble%2Cfloat[]%29

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2015-07-03 07:25:31 Radon8472

0

以下是包含在以前的答案討論三個公式的模塊（F90編碼的）。你可以把這個模塊放在程序的頂部（PROGRAM MAIN之前），也可以單獨編譯並在編譯期間包含模塊目錄。

module spherical_dists 
contains 
subroutine haversine_formula(lon1,lat1,lon2,lat2,dist) 
implicit none 
real,intent(in)::lon1,lon2,lat1,lat2 
real,intent(out)::dist 
real,parameter::pi=3.141592,mean_earth_radius=6371.0088 
real::lonr1,lonr2,latr1,latr2 
real::delangl,dellon,dellat,a 
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.) 
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.) 
dellon=lonr2-lonr1 
dellat=latr2-latr1 
a=(sin(dellat/2))**2+cos(latr1)*cos(latr2)*(sin(dellon/2))**2 
delangl=2*asin(sqrt(a)) !2*asin(sqrt(a)) 
dist=delangl*mean_earth_radius 
end subroutine 
subroutine great_circle_distance(lon1,lat1,lon2,lat2,dist) 
implicit none 
real,intent(in)::lon1,lon2,lat1,lat2 
real,intent(out)::dist 
real,parameter::pi=3.141592,mean_earth_radius=6371.0088 
real::lonr1,lonr2,latr1,latr2 
real::delangl,dellon 
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.) 
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.) 
dellon=lonr2-lonr1 
delangl=acos(sin(latr1)*sin(latr2)+cos(latr1)*cos(latr2)*cos(dellon)) 
dist=delangl*mean_earth_radius 
end subroutine 
subroutine vincenty_formula(lon1,lat1,lon2,lat2,dist) 
implicit none 
real,intent(in)::lon1,lon2,lat1,lat2 
real,intent(out)::dist 
real,parameter::pi=3.141592,mean_earth_radius=6371.0088 
real::lonr1,lonr2,latr1,latr2 
real::delangl,dellon,nom,denom 
lonr1=lon1*(pi/180.);lonr2=lon2*(pi/180.) 
latr1=lat1*(pi/180.);latr2=lat2*(pi/180.) 
dellon=lonr2-lonr1 
nom=sqrt((cos(latr2)*sin(dellon))**2. + (cos(latr1)*sin(latr2)-sin(latr1)*cos(latr2)*cos(dellon))**2.) 
denom=sin(latr1)*sin(latr2)+cos(latr1)*cos(latr2)*cos(dellon) 
delangl=atan2(nom,denom) 
dist=delangl*mean_earth_radius 
end subroutine 
end module

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2017-06-07 09:26:43

0

我使用SQL查詢

選擇，（ACOS（罪（做input_lat 0.01745329）*罪（lattitude * 0.01745329）+ COS（input_lat * 0.01745329）* COS（lattitude * 0.01745329） * cos（（input_long-longitude）0.01745329）） 57.29577951）* 69.16由表D中的D

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2017-09-14 11:10:24

如何從兩個位置的經度和緯度找到距離？

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