calculate distance between two gps coordinates

* * @param LNG * @param lat * @return */public int[] Getxy (double LNG, double lat) {///compare with reference points longitude calculate coordinates if (LNG > D34[0]) {return getXY1 (LNG, LAT);} else {return getXY2 (LNG, LAT);}} /** * The fixed point is the baseline xy coordinate */private int[] getXY1 (double LNG, double lat) {//WillLatitude converted to Gaussian projection

: This article describes how to calculate the distance between two GPS points in PHP. For more information about PHP tutorials, see. The following is a function for PHP to calculate the distance between two GPS points: PHP calcula

1062. Calculate the Manhattan distance, 1062 calculate the Manhattan distance The question description shows the coordinates (x1, y1), (x2, y2) of the two points on the plane, and calculates the Manhattan distance between the

PHP calculates the distance between two coordinates (longitude, latitude), coordinates longitude This example describes how PHP calculates the distance between two coordinates (longitude, latitude). Share to everyone for your reference. Specific as follows: Here we use PHP

Calculate the distance between two points based on the latitude and longitude, and the distance between two points based on the latitude and longitudeCalculate the Distance Based on the longitude and latitude of two points How are these cable cables determined? The earth is constantly rotating around the Earth's axis (

Define (' Earth_radius ', 6378.137);//Earth radius Define (' PI ', 3.1415926); /** * Calculate the distance between the two sets of latitude coordinates * PARAMS:LAT1 latitude 1; lng1 longitude 1; lat2 latitude 2; lng2 longitude 2; Len_type (1:m or 2:km); * Return m or km */ function Getdistance ($lat 1, $lng 1, $lat 2, $lng 2, $len _type = 1, $deci

PHP searches nearby based on gps coordinates Define ('Earth _ RADIUS ', 6370.6935 ); Function vicinity ($ lng, $ lat, $ distance = 0.5 ){ // $ Distance = 0.5; // unit: 10 KM $ Radius = EARTH_RADIUS; $ Dlng = rad2deg (2 * asin (sin ($

Here is the function of PHP to calculate the distance of two GPS points: PHP calculates the distance between two GPS pointsMySQL calculates the distance between two GPS coordinatesJavaS

Example One function distance ($lat 1, $lng 1, $lat 2, $lng 2, $miles = True){$pi = m_pi/180;$lat 1 *= $PI 80;$LNG 1 *= $PI 80;$lat 2 *= $PI 80;$LNG 2 *= $PI 80;$r = 6372.797;$dlat = $lat 2-$lat 1;$DLNG = $lng 2-$LNG 1;$a = sin ($dlat/2) *sin ($dlat/2) +cos ($lat 1) *cos ($lat 2) *sin ($DLNG/2) *sin ($DLNG/2);$c = 2 * atan2 (sqrt ($a), sqrt (1-$a));$km = $r * $c;Return ($miles? ($KM * 0.621371192): $km);} Example Two /***

How to coordinate coordinates of GPS longitude and latitude Original article:Http://hi.baidu.com/myonlylovegg/blog/item/dacc24d4db648acc51da4b28.html It is easy to convert longitude and latitude to decimal.As follows:Decimal degrees = degrees + minutes/60 + seconds/3600Example: 57 ° 55 '56. 6 "= 57 + 55/60 + 56.6/3600 = 57.9323888888888114 ° 65 '24. 6 "= 114 + 65/60 + 24.6/3600 =

/** calculate the distance between two coordinates (m) * @param float $fP 1Lat starting point (latitude) * @param float $fP 1Lon Start (longitude) * @param float $fP 2Lat end point (latitude) * @param Float $fP 2Lon end point (longitude) * @return int*/functionDistancebetween ($fP 1Lat,$fP 1Lon,$fP 2Lat,$fP 2Lon){ $fEARTH _radius= 6378137; //Angle converted to

Calculate the distance between geographical locations and geographical distance . Net calculates the maximum and minimum latitude and longitude in the specified range based on coordinates, and calculates the distance between two points based on the latitude and longitude

several solutions to these problems, and finally recommended mongodb, which is also one of the common solutions of LBS. These two days I have been doing this kind of service, because I have a small amount of data, so I am using the mysql partial index method. set @x1 = 31.292491624635;set @y1 = 121.50865016331;set @r0 = 10;select *, AsText(pos), X(pos), Y(pos), 6378 * 2 *ASIN(SQRT( POWER(SIN((@orig_lat - abs(X(pos))) * pi()/180 / 2),2) + COS(@orig_lat * pi()/180 ) * COS(abs(X(pos)) * pi()/180)

A few days ago in the development of mobile phone site encountered such a demand, with a certain degree of map positioning function to see where they are now, at the same time to show themselves as the center, around 1000 meters within the business. function is very simple, get their own latitude and longitude, calculate the latitude and longitude coordinates of four points, from the database to take out th

getjwdb (double Glat, double Glon, double distance, double angle, out double BJD, out do Uble BWD) { Double dx = distance * + math.sin (angle * math.pi/180.0); Double dy = distance * Math.Cos (angle * math.pi/180.0); Double EC = 6356725 + 21412 * (90.0-glat)/90.0; 21412 is the difference between the equatorial

+$DLNG), ' left-bottom ' =Array(' lat ' =$lat-$dlat, ' LNG ' =$LNG-$DLNG), ' right-bottom ' =Array(' lat ' =$lat-$dlat, ' LNG ' =$LNG+$DLNG) ); //Use this function to calculate the resulting result, and bring in the SQL query. $squares = Returnsquarepoint ($LNG, $lat);//$info _sql = "Select Id,locateinfo,lat,lng from ' Lbs_info ' where LATL T;>0 and lat>{$squares [' Right-bottom '] [' lat ']} and lat}/** * Calc