请问，已知两点坐标如何求得两点距离，具体问题请进，感谢先 [问题点数：100分，结帖人zhptj]
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本版专家分：2结帖率 98.7% -
已知两个点的经纬度，请问如何得到两点的距离
地球弧度计算在内时公式如何
不计算在内时公式如何
望各位朋友指教问题点数：100分
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本版专家分：1059 -
呵呵，公式建议在网上google或者找本测量学的书看看，摄影测量或控制测量肯定都有哈！
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本版专家分：218 -
首先你需要将大地坐标转换为高斯平面坐标，然后通过平面坐标的x,y值，计算两点之间的直线距离
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本版专家分：1908
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黄花
2007年3月 VC/MFC大版内专家分月排行榜第二
2007年2月 VC/MFC大版内专家分月排行榜第二
2007年1月 VC/MFC大版内专家分月排行榜第二
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经纬度换算成大地坐标，然后计算距离。
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本版专家分：40 -
程序如下:
Public Function Distance(lon1 As Double, lat1 As Double, lon2 As Double, lat2 As Double) As Double
Dim a_2d As Double
Dim e_2d As Double
Dim h_2d As Integer
Dim DEG_2_RAD As Double
Dim RAD_2_DEG As Double
Dim x_rads As Double
Dim y_rads As Double
Dim n_2ds As Double
Dim x_2d As Double
Dim y_2d As Double
Dim z_2d As Double
Dim x_radm As Double
Dim y_radm As Double
Dim n_2dm As Double
Dim x_2d_mark As Double
Dim y_2d_mark As Double
Dim z_2d_mark As Double
Dim curdistance As Double
a_2d = 6378137
e_2d = 0.00669438
h_2d = 15
DEG_2_RAD = 0.01745329252
RAD_2_DEG = 57.2957795129
x_rads = Abs(lon1) * DEG_2_RAD
y_rads = Abs(lat1) * DEG_2_RAD
n_2ds = a_2d / Sqr(1 - e_2d * Sin(y_rads) * Sin(y_rads))
x_2d = (n_2ds + h_2d) * Cos(y_rads) * Cos(x_rads)
y_2d = (n_2ds + h_2d) * Cos(y_rads) * Sin(x_rads)
z_2d = (n_2ds * (1 - e_2d) + h_2d) * Sin(y_rads)
x_radm = Abs(lon2) * DEG_2_RAD
y_radm = Abs(lat2) * DEG_2_RAD
n_2dm = a_2d / Sqr(1 - e_2d * Sin(y_radm) * Sin(y_radm))
x_2d_mark = (n_2dm + h_2d) * Cos(y_radm) * Cos(x_radm)
y_2d_mark = (n_2dm + h_2d) * Cos(y_radm) * Sin(x_radm)
z_2d_mark = (n_2dm * (1 - e_2d) + h_2d) * Sin(y_radm)
curdistance = (x_2d_mark - x_2d) * (x_2d_mark - x_2d) + (y_2d_mark - y_2d) * (y_2d_mark - y_2d) + (z_2d_mark - z_2d) * (z_2d_mark - z_2d)
curdistance = Sqr(curdistance)
Distance = curdistance
End Function
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本版专家分：2
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d＝111.12cos{1/[sinΦAsinΦB十 cosΦAcosΦBcos(λB—λA)]}
其中A点经度，纬度分别为λA和ΦA，B点的经度、纬度分别为λB和ΦB，d为距离。