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  • Python 高斯坐标转经纬度算法


    # 高斯坐标转经纬度算法
    # B=大地坐标X
    # C=大地坐标Y
    # IsSix=6度带或3度带
    def GetLatLon2(B, C,IsSix):
    #带号
    D = math.trunc(C / 1000000)
    #中央经线(单位:弧度)
    K = 0
    if IsSix:
    K = D * 6 - 3 #6度带计算
    else:
    K = D * 3 #3度带计算
    L = B/(6378245*(1-0.006693421623)*1.0050517739)
    M = L +(0.00506237764 * math.sin(2*L)/2-0.00001062451*math.sin(4*L)/4+0.0000002081*math.sin(6*L)/6)/1.0050517739
    N = L +(0.00506237764 * math.sin(2*M)/2-0.00001062451*math.sin(4*M)/4+0.0000002081*math.sin(6*M)/6)/1.0050517739
    O = L +(0.00506237764 * math.sin(2*N)/2-0.00001062451*math.sin(4*N)/4+0.0000002081*math.sin(6*N)/6)/1.0050517739
    P = L +(0.00506237764 * math.sin(2*O)/2-0.00001062451*math.sin(4*O)/4+0.0000002081*math.sin(6*O)/6)/1.0050517739
    Q = L +(0.00506237764 * math.sin(2*P)/2-0.00001062451*math.sin(4*P)/4+0.0000002081*math.sin(6*P)/6)/1.0050517739
    R = L +(0.00506237764 * math.sin(2*Q)/2-0.00001062451*math.sin(4*Q)/4+0.0000002081*math.sin(6*Q)/6)/1.0050517739
    S = math.tan(R)
    T = 0.006738525415*(math.cos(R))**2
    U = 6378245/math.sqrt(1-0.006693421623*(math.sin(R))**2)
    V = 6378245*(1-0.006693421623)/(math.sqrt((1-0.006693421623*(math.sin(R))**2)))**3
    W = 5+3*S**2+T-9*T*S**2
    X = 61+90*S**2+45*S**4
    Y = 1+2*S**2+T**2
    Z = 5+28*S**2+24*S**4+6*T+8*T*S**2
    Lat= (180/pi)*(R-(C-D*1000000-500000)**2*S/(2*V*U)+(C-D*1000000-500000)**4*W/(24*U**3*V)-(C-D*1000000-500000)**6*X/(7200*U**5*V))
    Lon= (180/pi)*(C-D*1000000-500000)*(1-(C-D*1000000-500000)**2*Y/(6*U**2)+(C-D*1000000-500000)**4*Z/(120*U**4))/(U*math.cos(P))
    Lat = Lat
    Lon = K + Lon
    return (Lon, Lat)
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  • 原文地址:https://www.cnblogs.com/defineconst/p/6026639.html
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