使用haversine公式的纬度/经度点之间的距离

Question

我正在尝试查找两个经度和纬度点之间的距离。我正在尝试使用great circle distance。这是公式：

我不知道为什么，但我的程序不工作。这是我得到的结果：

Change Angle: 0.00016244370761414
Earth Radius: 6371

RESULTS: 
Correct  Distance: 24.883 km
Computed Distance: 1.0349288612097

来源：

$latStart = 44.638;
$longStart = -63.587;

$latFinish = 44.644;
$longFinish = -63.597;


# Convert Input to Radians
$latStart = deg2Rad($latStart);
$longStart = deg2Rad($longStart);

$latFinish = deg2Rad($latFinish);
$longFinish = deg2Rad($longFinish);

# Because the Earth is not perfectly spherical, no single value serves as its 
# natural radius. Distances from points on the surface to the center range from 
# 6,353 km to 6,384 km (≈3,947–3,968 mi). Several different ways of modeling the 
# Earth as a sphere each yield a convenient mean radius of 6371 km (≈3,959 mi).
# http://en.wikipedia.org/wiki/Earth_radius
$earthRadius = 6371;

# difference in Long/Lat
$latChange = $latFinish - $latStart;
$longChange = $longFinish - $longStart;



# haversine formula 
# numerically stable for small distances
# http://en.wikipedia.org/wiki/Great-circle_distance
$changeAngle = 2 * asin(
                sqrt(
                        pow(sin($latChange/2),2) +
                        cos($latStart) * cos($latFinish) * pow(sin($longChange/2),2)
                )
        );



echo "Change Angle: $changeAngle\n";
echo "Earth Radius: $earthRadius\n";

Answer 1

让我们使用平面近似进行粗略检查。纬度差为0.006°，经度差为0.01°，但乘以纬度的余弦得到0.0075°。应用毕达哥拉斯：

>>> sqrt(0.006 ** 2 + 0.0075 ** 2)
0.0096046863561492727

大约是 0.000167 弧度，非常接近您的计算。 （更粗略的检查：一个度数大约是 69 英里，比 100 公里多一点，所以 0.01° 应该比 1 公里多一点。）

所以我认为是你所谓的“正确距离”是错误的，而不是你的计算。

Answer 2

您的方法大致基于毕达哥拉斯定理——我总是以艰难的方式完成它，例如（实际上，我预先计算了轴的值并将它们与数据一起存储在数据库中） ：

$startXAxis   = cos(deg2Rad($latStart)) * cos(deg2Rad($longStart));
$startYAxis   = cos(deg2Rad($latStart)) * sin(deg2Rad($longStart));
$startZAxis   = sin(deg2Rad($latStart));
$finishXAxis   = cos(deg2Rad($latFinish)) * cos(deg2Rad($longFinish));
$finishYAxis   = cos(deg2Rad($latFinish)) * sin(deg2Rad($longFinish));
$finishZAxis   = sin(deg2Rad($latFinish));

$changeAngle = acos($startXAxis * $finishXAxis + $startYAxis * $finishYAxis + $startZAxis * $finishZAxis);

Answer 3

您的公式看起来与我的实现不同。但是我的在 .NET 中，但我已经对其进行了单元测试并且效果很好。

这是一个稍微改写的版本：http://megocode3.wordpress.com/2008/02/05/haversine-formula-in-c/

/// <summary>
/// Implementation of the Haversine formula
/// For calculating the distance between 2 points on a sphere
/// http://en.wikipedia.org/wiki/Haversine_formula
/// </summary>
public class Haversine
{
    /// <summary>
    /// Calculate the distance between 2 points in miles or kilometers
    /// http://megocode3.wordpress.com/2008/02/05/haversine-formula-in-c/
    /// 
    /// This assumes sea level
    /// </summary>
    public double Distance(LatLon pos1, LatLon pos2, DistanceType type)
    {
        const double RADIUS_OF_EARTH_IN_MILES = 3963.1676;
        const double RADIUS_OF_EARTH_IN_KILOMETERS = 6378.1;

        //radius of the earth
        double R = (type == DistanceType.Miles) ? RADIUS_OF_EARTH_IN_MILES : RADIUS_OF_EARTH_IN_KILOMETERS;

        //Deltas
        double dLat = ToRadian(pos2.Lat - pos1.Lat);
        double dLon = ToRadian(pos2.Lon - pos1.Lon);

        double a = Math.Sin(dLat/2)*Math.Sin(dLat/2) + Math.Cos(ToRadian(pos1.Lat))*Math.Cos(ToRadian(pos2.Lat)) * Math.Sin(dLon / 2) * Math.Sin(dLon / 2);
        double c = 2 * Math.Asin(Math.Min(1, Math.Sqrt(a)));

        double d = R*c;
        return d;
    }

    /// <summary>
    /// Convert to Radians.
    /// </summary>
    private double ToRadian(double val)
    {
        return (Math.PI / 180) * val;
    }
}