Question

我正在尝试使用半正公式计算两个地理坐标之间的距离。

代码：

Dim dbl_dLat As Double
Dim dbl_dLon As Double
Dim dbl_a As Double

dbl_P = WorksheetFunction.Pi / 180
dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1)
dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1)

dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)

dbl_Distance_KM = 6371 * 2 * WorksheetFunction.Atan2(Sqr(dbl_a), Sqr(1 - dbl_a))

我正在测试这些坐标：

dbl_Longitude1 = 55.629178
dbl_Longitude2 = 29.846686
dbl_Latitude1 = 37.659466
dbl_Latitude2 = 30.24441

代码返回20015.09，这显然是错误的。根据Yandex地图，它应该是642公里。

我哪里错了？经度和纬度是否格式错误？

Answer 1

据我所知，问题是atan2（）的参数顺序因语言而异。以下作品*对我来说：

Option Explicit

Public Sub Distance()
    Dim dbl_Longitude1 As Double, dbl_Longitude2 As Double, dbl_Latitude1 As Double, dbl_Latitude2 As Double

    dbl_Longitude1 = 55.629178
    dbl_Longitude2 = 29.846686
    dbl_Latitude1 = 37.659466
    dbl_Latitude2 = 30.24441

    Dim dbl_dLat As Double
    Dim dbl_dLon As Double
    Dim dbl_a As Double
    Dim dbl_P As Double

    dbl_P = WorksheetFunction.Pi / 180
    dbl_dLat = dbl_P * (dbl_Latitude2 - dbl_Latitude1)      'to radians
    dbl_dLon = dbl_P * (dbl_Longitude2 - dbl_Longitude1)    'to radians

    dbl_a = Sin(dbl_dLat / 2) * Sin(dbl_dLat / 2) + _
            Cos(dbl_Latitude1 * dbl_P) * Cos(dbl_Latitude2 * dbl_P) * Sin(dbl_dLon / 2) * Sin(dbl_dLon / 2)

    Dim c As Double
    Dim dbl_Distance_KM As Double
    c = 2 * WorksheetFunction.Atan2(Sqr(1 - dbl_a), Sqr(dbl_a))  ' *** swapped arguments to Atan2
    dbl_Distance_KM = 6371 * c

    Debug.Print dbl_Distance_KM
End Sub

*输出：2507.26205401321，虽然gcmap.com表示答案是2512公里。这可能是一个精确的问题---我认为这很接近于算作工作。（编辑也可能是gcmap使用的是本地地球半径而不是平均半径;我不确定。）

解释

我找到了大圆距离的半正式公式的this description，这是你正在实施的。该页面上的JavaScript实现为c提供了此计算：

var c = 2 * Math.atan2(Math.sqrt(a), Math.sqrt(1-a));

在JavaScript中，atan2()采用参数y，x。但是，在Excel VBA中，WorksheetFunction.Atan2采用参数 x ， y 。您的原始代码作为第一个参数传递Sqr(dbl_a)，就像在JavaScript中一样。但是，Sqr(dbl_a)必须是Excel VBA中的第二个参数。

关于命名的评论

基于@ JohnColeman的观点，有很多方法可以命名变量。在这种情况下，我建议使用单位前缀而不是类型：例如，deg_Latitude1，RadPerDeg = Pi/180和rad_dLat = RadPerDeg * (deg_Latitude2 - deg_Latitude1)。我个人认为这有助于避免unit-conversion mishaps。

Answer 2

我的VBA代码，以英尺为单位返回答案；但是，“ d”是千米的答案。

Imports System.Math
Module Haversine
Public Function GlobalAddressDistance(sLat1 As String, sLon1 As String, sLat2 As String, sLon2 As String) As String
    Const R As Integer = 6371
    Const cMetersToFeet As Single = 3.2808399
    Const cKiloMetersToMeters As Integer = 1000
    Dim a As Double = 0, c As Double = 0, d As Double = 0

    'Convert strings to numberic double values
    Dim dLat1 As Double = Val(sLat1)
    Dim dLat2 As Double = Val(sLat2)
    Dim dLatDiff As Double = DegreesToRadians(CDbl(sLat2) - CDbl(sLat1))
    Dim dLonDiff As Double = DegreesToRadians(CDbl(sLon2) - CDbl(sLon1))

    a = Pow(Sin(dLatDiff / 2), 2) + Cos(DegreesToRadians(dLat1)) * Cos(DegreesToRadians(dLat2)) * Pow(Sin(dLonDiff / 2), 2)
    c = 2 * Atan2(Sqrt(a), Sqrt(1 - a))
    d = R * c

    'Convert kilometers to feet
    Return Format((d * cKiloMetersToMeters * cMetersToFeet), "0.##").ToString
End Function

Private Function DegreesToRadians(ByVal dDegrees As Double) As Double
    Return (dDegrees * PI) / 180
End Function

最终模块

计算坐标之间的公里距离

2 个答案:

解释

关于命名的评论