目前我正在开展一个项目，我需要计算两个维度为2d的位置之间的公里距离。
我很想看到一个javascript示例
我用一个例子制作了http://jsfiddle.net/ym2sn70u/5/ d f k f。我喜欢在KM中将响应作为字符串值。例如50.9公里。
我很感激
答案 0 :(得分：0)
你可以使用下面描述的3种算法中的任何一种（用C＃编码，可以用类似的语法转换成Java / Javascript）来找到地球上2个地理点之间的距离，以英里或公里为单位（re：{{3} }）：
/*****************************************************************************************
Module : GIS.cs |Class Lib
Description : Calculate distance between two geo-points on surface
*****************************************************************************************
Author : Alexander Bell
Copyright : 2011-2015 Infosoft International Inc
*****************************************************************************************
using System;
namespace BusNY
{
internal enum UnitSystem { SI = 0, US = 1 }
internal static class GIS
{
#region internal: properties (read-only)
internal static double EarthRadiusKm { get {return _radiusEarthKM;} }
internal static double EarthRadiusMiles { get { return _radiusEarthMiles; } }
internal static double m2km { get { return _m2km; } }
internal static double Deg2rad { get { return _toRad; } }
#endregion
#region private: const
private const double _radiusEarthMiles = 3959;
private const double _radiusEarthKM = 6371;
private const double _m2km = 1.60934;
private const double _toRad = Math.PI / 180;
#endregion
#region Method 1: Haversine algo
/// <summary>
/// Distance between two geographic points on surface, km/miles
/// Haversine formula to calculate
/// great-circle (orthodromic) distance on Earth
/// High Accuracy, Medium speed
/// re: http://en.wikipedia.org/wiki/Haversine_formula
/// </summary>
/// <param name="Lat1">double: 1st point Latitude</param>
/// <param name="Lon1">double: 1st point Longitude</param>
/// <param name="Lat2">double: 2nd point Latitude</param>
/// <param name="Lon2">double: 2nd point Longitude</param>
/// <returns>double: distance, km/miles</returns>
internal static double DistanceHaversine(double Lat1,
double Lon1,
double Lat2,
double Lon2,
UnitSystem UnitSys ){
try {
double _radLat1 = Lat1 * _toRad;
double _radLat2 = Lat2 * _toRad;
double _dLatHalf = (_radLat2 - _radLat1) / 2;
double _dLonHalf = Math.PI * (Lon2 - Lon1) / 360;
// intermediate result
double _a = Math.Sin(_dLatHalf);
_a *= _a;
// intermediate result
double _b = Math.Sin(_dLonHalf);
_b *= _b * Math.Cos(_radLat1) * Math.Cos(_radLat2);
// central angle, aka arc segment angular distance
double _centralAngle = 2 * Math.Atan2(Math.Sqrt(_a + _b), Math.Sqrt(1 - _a - _b));
// great-circle (orthodromic) distance on Earth between 2 points
if (UnitSys == UnitSystem.SI) { return _radiusEarthKM * _centralAngle; }
else { return _radiusEarthMiles * _centralAngle; }
}
catch { throw; }
}
#endregion
#region Method 2: Spherical Law of Cosines
/// <summary>
/// Distance between two geographic points on surface, km/miles
/// Spherical Law of Cosines formula to calculate
/// great-circle (orthodromic) distance on Earth;
/// High Accuracy, Medium speed
/// re: http://en.wikipedia.org/wiki/Spherical_law_of_cosines
/// </summary>
/// <param name="Lat1">double: 1st point Latitude</param>
/// <param name="Lon1">double: 1st point Longitude</param>
/// <param name="Lat2">double: 2nd point Latitude</param>
/// <param name="Lon2">double: 2nd point Longitude</param>
/// <returns>double: distance, km/miles</returns>
internal static double DistanceSLC(double Lat1,
double Lon1,
double Lat2,
double Lon2,
UnitSystem UnitSys ){
try {
double _radLat1 = Lat1 * _toRad;
double _radLat2 = Lat2 * _toRad;
double _radLon1 = Lon1 * _toRad;
double _radLon2 = Lon2 * _toRad;
// central angle, aka arc segment angular distance
double _centralAngle = Math.Acos(Math.Sin(_radLat1) * Math.Sin(_radLat2) +
Math.Cos(_radLat1) * Math.Cos(_radLat2) * Math.Cos(_radLon2 - _radLon1));
// great-circle (orthodromic) distance on Earth between 2 points
if (UnitSys == UnitSystem.SI) { return _radiusEarthKM * _centralAngle; }
else { return _radiusEarthMiles * _centralAngle; }
}
catch { throw; }
}
#endregion
#region Method 3: Spherical Earth projection
/// <summary>
/// Distance between two geographic points on surface, km/miles
/// Spherical Earth projection to a plane formula (using Pythagorean Theorem)
/// to calculate great-circle (orthodromic) distance on Earth.
/// central angle =
/// Sqrt((_radLat2 - _radLat1)^2 + (Cos((_radLat1 + _radLat2)/2) * (Lon2 - Lon1))^2)
/// Medium Accuracy, Fast,
/// relative error less than 0.1% in search area smaller than 250 miles
/// re: http://en.wikipedia.org/wiki/Geographical_distance
/// </summary>
/// <param name="Lat1">double: 1st point Latitude</param>
/// <param name="Lon1">double: 1st point Longitude</param>
/// <param name="Lat2">double: 2nd point Latitude</param>
/// <param name="Lon2">double: 2nd point Longitude</param>
/// <returns>double: distance, km/miles</returns>
public static double DistanceSEP(double Lat1,
double Lon1,
double Lat2,
double Lon2,
UnitSystem UnitSys ){
try
{
double _radLat1 = Lat1 * _toRad;
double _radLat2 = Lat2 * _toRad;
double _dLat = (_radLat2 - _radLat1);
double _dLon = (Lon2 - Lon1) * _toRad;
double _a = (_dLon) * Math.Cos((_radLat1 + _radLat2) / 2);
// central angle, aka arc segment angular distance
double _centralAngle = Math.Sqrt(_a * _a + _dLat * _dLat);
// great-circle (orthodromic) distance on Earth between 2 points
if (UnitSys == UnitSystem.SI) { return _radiusEarthKM * _centralAngle; }
else { return _radiusEarthMiles * _centralAngle; }
}
catch { throw; }
}
#endregion
}
}
第三种算法（球形地球投影）是最快/最简单的算法，提供相当好的准确度。
关于您编辑过的问题：虽然纬度/经度格式与原始3算法中使用的坐标的地理坐标系（GCS）十进制度单位不同，但Javascript代码段似乎没问题。以下是欧洲/美国一些知名地方的小数度坐标样本（使用此网络应用程序：http://www.codeproject.com/Articles/884183/enRoute-Real-time-NY-City-Bus-Tracking-Web-App）：
大都会博物馆，Lat：40.779437 / Lon：-73.963244
ΑκρόποληΑθηνών，Lat：37.971532 / Lon：23.725749
Brandenburger Tor，Lat：52.516275 / Lon：13.377704
Champs-Elysées，Lat：48.869576 / Lon：2.30825
Museo del Prado，Lat：40.413782 / Lon：-3.692127
Cappella Sistina，Lat：41.902947 /经度：12.454484
C＃代码段中的所有3个函数均已使用此Lat / Lon十进制度格式进行测试并正常工作，其结果以km或英里计算。
关于你的其他问题（数字到字符串转换）：你可以使用num .toPrecision(N)或num .toFixed(N) javascript函数（加上，Math.round()在数学计算中也很有用）。
希望这会有所帮助。最好的问候，