计算2个GPS坐标之间的距离

如何计算两个GPS坐标之间的距离(使用经纬度)?

当前回答

一、关于“面包屑”方法

地球半径在不同的纬度上是不同的。在Haversine算法中必须考虑到这一点。考虑轴承的变化，它将直线变成拱门(更长的) 考虑到速度变化将把拱门变成螺旋(比拱门更长或更短) 高度变化将使平面螺旋变成3D螺旋(再次变长)。这对丘陵地区非常重要。

下面是考虑#1和#2的C语言函数:

double   calcDistanceByHaversine(double rLat1, double rLon1, double rHeading1,
       double rLat2, double rLon2, double rHeading2){
  double rDLatRad = 0.0;
  double rDLonRad = 0.0;
  double rLat1Rad = 0.0;
  double rLat2Rad = 0.0;
  double a = 0.0;
  double c = 0.0;
  double rResult = 0.0;
  double rEarthRadius = 0.0;
  double rDHeading = 0.0;
  double rDHeadingRad = 0.0;

  if ((rLat1 < -90.0) || (rLat1 > 90.0) || (rLat2 < -90.0) || (rLat2 > 90.0)
              || (rLon1 < -180.0) || (rLon1 > 180.0) || (rLon2 < -180.0)
              || (rLon2 > 180.0)) {
        return -1;
  };

  rDLatRad = (rLat2 - rLat1) * DEGREE_TO_RADIANS;
  rDLonRad = (rLon2 - rLon1) * DEGREE_TO_RADIANS;
  rLat1Rad = rLat1 * DEGREE_TO_RADIANS;
  rLat2Rad = rLat2 * DEGREE_TO_RADIANS;

  a = sin(rDLatRad / 2) * sin(rDLatRad / 2) + sin(rDLonRad / 2) * sin(
              rDLonRad / 2) * cos(rLat1Rad) * cos(rLat2Rad);

  if (a == 0.0) {
        return 0.0;
  }

  c = 2 * atan2(sqrt(a), sqrt(1 - a));
  rEarthRadius = 6378.1370 - (21.3847 * 90.0 / ((fabs(rLat1) + fabs(rLat2))
              / 2.0));
  rResult = rEarthRadius * c;

  // Chord to Arc Correction based on Heading changes. Important for routes with many turns and U-turns

  if ((rHeading1 >= 0.0) && (rHeading1 < 360.0) && (rHeading2 >= 0.0)
              && (rHeading2 < 360.0)) {
        rDHeading = fabs(rHeading1 - rHeading2);
        if (rDHeading > 180.0) {
              rDHeading -= 180.0;
        }
        rDHeadingRad = rDHeading * DEGREE_TO_RADIANS;
        if (rDHeading > 5.0) {
              rResult = rResult * (rDHeadingRad / (2.0 * sin(rDHeadingRad / 2)));
        } else {
              rResult = rResult / cos(rDHeadingRad);
        }
  }
  return rResult;
}

2有一种更简单的方法，效果很好。

按平均速度。

Trip_distance = Trip_average_speed * Trip_time

由于GPS速度是由多普勒效应检测的，与[Lon,Lat]没有直接关系，如果不是主要的距离计算方法，至少可以考虑作为次要的(备份或校正)。

2014-01-21 15:36:20

其他回答

这段Lua代码改编自维基百科和Robert Lipe的GPSbabel工具:

local EARTH_RAD = 6378137.0 
  -- earth's radius in meters (official geoid datum, not 20,000km / pi)

local radmiles = EARTH_RAD*100.0/2.54/12.0/5280.0;
  -- earth's radius in miles

local multipliers = {
  radians = 1, miles = radmiles, mi = radmiles, feet = radmiles * 5280,
  meters = EARTH_RAD, m = EARTH_RAD, km = EARTH_RAD / 1000, 
  degrees = 360 / (2 * math.pi), min = 60 * 360 / (2 * math.pi)
}

function gcdist(pt1, pt2, units) -- return distance in radians or given units
  --- this formula works best for points close together or antipodal
  --- rounding error strikes when distance is one-quarter Earth's circumference
  --- (ref: wikipedia Great-circle distance)
  if not pt1.radians then pt1 = rad(pt1) end
  if not pt2.radians then pt2 = rad(pt2) end
  local sdlat = sin((pt1.lat - pt2.lat) / 2.0);
  local sdlon = sin((pt1.lon - pt2.lon) / 2.0);
  local res = sqrt(sdlat * sdlat + cos(pt1.lat) * cos(pt2.lat) * sdlon * sdlon);
  res = res > 1 and 1 or res < -1 and -1 or res
  res = 2 * asin(res);
  if units then return res * assert(multipliers[units])
  else return res
  end
end

2008-12-13 22:26:04

这取决于你需要它有多准确。如果你需要精确到毫米的精度，最好看看使用椭球的算法，而不是球体，比如Vincenty的算法。

2008-12-13 22:28:50

我认为R中的一个算法版本仍然缺失:

gpsdistance<-function(lat1,lon1,lat2,lon2){

# internal function to change deg to rad

degreesToRadians<- function (degrees) {
return (degrees * pi / 180)
}

R<-6371e3  #radius of Earth in meters

phi1<-degreesToRadians(lat1) # latitude 1
phi2<-degreesToRadians(lat2) # latitude 2
lambda1<-degreesToRadians(lon1) # longitude 1
lambda2<-degreesToRadians(lon2) # longitude 2

delta_phi<-phi1-phi2 # latitude-distance
delta_lambda<-lambda1-lambda2 # longitude-distance

a<-sin(delta_phi/2)*sin(delta_phi/2)+
cos(phi1)*cos(phi2)*sin(delta_lambda/2)*
sin(delta_lambda/2)

cc<-2*atan2(sqrt(a),sqrt(1-a))

distance<- R * cc

return(distance)  # in meters
}

2020-04-06 11:32:17

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<[&ltMeasure>] 'u> (R : float<'u>) (p1 : Location) (p2 : Location) =
    let degToRad (x : float&ltdeg>) = System.Math.PI * x / 180.0&ltdeg/rad>

    let sq x = x * x
    // take the sin of the half and square the result
    let sinSqHf (a : float&ltrad>) = (System.Math.Sin >> sq) (a / 2.0&ltrad>)
    let cos (a : float&ltdeg>) = System.Math.Cos (degToRad a / 1.0&ltrad>)

    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

2011-08-22 15:40:03