使用Python实现了一下我们同事的C++高斯投影正反算，实际跑通，可用。

#!/ usr/bin/python# -*- coding:utf-8 -*-import mathdef LatLon2XY(latitude, longitude): a = 6378137.0 # b = 6356752.3142 # c = 6399593.6258 # alpha = 1 / 298.257223563 e2 = 0.0066943799013 # epep = 0.00673949674227 #将经纬度转换为弧度 latitude2Rad = (math.pi / 180.0) * latitude beltNo = int((longitude + 1.5) / 3.0) #计算3度带投影度带号 L = beltNo * 3 #计算中央经线 l0 = longitude - L #经差 tsin = math.sin(latitude2Rad) tcos = math.cos(latitude2Rad) t = math.tan(latitude2Rad) m = (math.pi / 180.0) * l0 * tcos et2 = e2 * pow(tcos, 2) et3 = e2 * pow(tsin, 2) X = 111132.9558 * latitude - 16038.6496 * math.sin(2 * latitude2Rad) + 16.8607 * math.sin( 4 * latitude2Rad) - 0.0220 * math.sin(6 * latitude2Rad) N = a / math.sqrt(1 - et3) x = X + N * t * (0.5 * pow(m, 2) + (5.0 - pow(t, 2) + 9.0 * et2 + 4 * pow(et2, 2)) * pow(m, 4) / 24.0 + ( 61.0 - 58.0 * pow(t, 2) + pow(t, 4)) * pow(m, 6) / 720.0) y = 500000 + N * (m + (1.0 - pow(t, 2) + et2) * pow(m, 3) / 6.0 + ( 5.0 - 18.0 * pow(t, 2) + pow(t, 4) + 14.0 * et2 - 58.0 * et2 * pow(t, 2)) * pow(m, 5) / 120.0) return x, ydef XY2LatLon(X, Y, L0): iPI = 0.0174532925199433 a = 6378137.0 f= 0.00335281006247 ZoneWide = 3 #按3度带进行投影 ProjNo = int(X / 1000000) L0 = L0 * iPI X0 = ProjNo * 1000000 + 500000 Y0 = 0 xval = X - X0 yval = Y - Y0 e2 = 2 * f - f * f #第一偏心率平方 e1 = (1.0 - math.sqrt(1 - e2)) / (1.0 + math.sqrt(1 - e2)) ee = e2 / (1 - e2) #第二偏心率平方 M = yval u = M / (a * (1 - e2 / 4 - 3 * e2 * e2 / 64 - 5 * e2 * e2 * e2 / 256)) fai = u \ + (3 * e1 / 2 - 27 * e1 * e1 * e1 / 32) * math.sin(2 * u) \ + (21 * e1 * e1 / 16 - 55 * e1 * e1 * e1 * e1 / 32) * math.sin(4 * u) \ + (151 * e1 * e1 * e1 / 96) * math.sin(6 * u)\ + (1097 * e1 * e1 * e1 * e1 / 512) * math.sin(8 * u) C = ee * math.cos(fai) * math.cos(fai) T = math.tan(fai) * math.tan(fai) NN = a / math.sqrt(1.0 - e2 * math.sin(fai) * math.sin(fai)) R = a * (1 - e2) / math.sqrt( (1 - e2 * math.sin(fai) * math.sin(fai)) * (1 - e2 * math.sin(fai) * math.sin(fai)) * (1 - e2 * math.sin(fai) * math.sin(fai))) D = xval / NN #计算经纬度（弧度单位的经纬度） longitude1 = L0 + (D - (1 + 2 * T + C) * D * D * D / 6 + ( 5 - 2 * C + 28 * T - 3 * C * C + 8 * ee + 24 * T * T) * D * D * D * D * D / 120) / math.cos(fai) latitude1 = fai - (NN * math.tan(fai) / R) * ( D * D / 2 - (5 + 3 * T + 10 * C - 4 * C * C - 9 * ee) * D * D * D * D / 24 + ( 61 + 90 * T + 298 * C + 45 * T * T - 256 * ee - 3 * C * C) * D * D * D * D * D * D / 720) #换换为deg longitude = longitude1 / iPI latitude = latitude1 / iPI return latitude, longitude# # print LatLon2XY(40.07837722329, 116.23514827596)# print XY2LatLon(434760.7611718801, 4438512.040474475, 117.0)

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