如何在python或Javascript中从UTM转换为LatLng

如何在python或Javascript中从UTM转换为LatLng

我最终从IBM找到了解决该问题的Java代码：http
:
//www.ibm.com/developerworks/java/library/j-coordconvert/index.html

仅供参考，这是我所需方法的python实现：

import mathdef utmToLatLng(zone, easting, northing, northernHemisphere=True):    if not northernHemisphere:        northing = 10000000 - northing    a = 6378137    e = 0.081819191    e1sq = 0.006739497    k0 = 0.9996    arc = northing / k0    mu = arc / (a * (1 - math.pow(e, 2) / 4.0 - 3 * math.pow(e, 4) / 64.0 - 5 * math.pow(e, 6) / 256.0))    ei = (1 - math.pow((1 - e * e), (1 / 2.0))) / (1 + math.pow((1 - e * e), (1 / 2.0)))    ca = 3 * ei / 2 - 27 * math.pow(ei, 3) / 32.0    cb = 21 * math.pow(ei, 2) / 16 - 55 * math.pow(ei, 4) / 32    cc = 151 * math.pow(ei, 3) / 96    cd = 1097 * math.pow(ei, 4) / 512    phi1 = mu + ca * math.sin(2 * mu) + cb * math.sin(4 * mu) + cc * math.sin(6 * mu) + cd * math.sin(8 * mu)    n0 = a / math.pow((1 - math.pow((e * math.sin(phi1)), 2)), (1 / 2.0))    r0 = a * (1 - e * e) / math.pow((1 - math.pow((e * math.sin(phi1)), 2)), (3 / 2.0))    fact1 = n0 * math.tan(phi1) / r0    _a1 = 500000 - easting    dd0 = _a1 / (n0 * k0)    fact2 = dd0 * dd0 / 2    t0 = math.pow(math.tan(phi1), 2)    Q0 = e1sq * math.pow(math.cos(phi1), 2)    fact3 = (5 + 3 * t0 + 10 * Q0 - 4 * Q0 * Q0 - 9 * e1sq) * math.pow(dd0, 4) / 24    fact4 = (61 + 90 * t0 + 298 * Q0 + 45 * t0 * t0 - 252 * e1sq - 3 * Q0 * Q0) * math.pow(dd0, 6) / 720    lof1 = _a1 / (n0 * k0)    lof2 = (1 + 2 * t0 + Q0) * math.pow(dd0, 3) / 6.0    lof3 = (5 - 2 * Q0 + 28 * t0 - 3 * math.pow(Q0, 2) + 8 * e1sq + 24 * math.pow(t0, 2)) * math.pow(dd0, 5) / 120    _a2 = (lof1 - lof2 + lof3) / math.cos(phi1)    _a3 = _a2 * 180 / math.pi    latitude = 180 * (phi1 - fact1 * (fact2 + fact3 + fact4)) / math.pi    if not northernHemisphere:        latitude = -latitude    longitude = ((zone > 0) and (6 * zone - 183.0) or 3.0) - _a3    return (latitude, longitude)

在这里，我认为这很简单

easting*x+zone*y

。