如何在python或Javascript中从UTM转换为LatLng [英] How to convert from UTM to LatLng in python or Javascript

问题描述

我有一堆坐标为UTM格式的文​​件。对于每个坐标，我有东，北和区。我需要将其转换为LatLng，以便与Google Map API一起使用，以便在地图中显示信息。

I have a bunch of files with coordinates in UTM form. For each coordinate I have easting, northing and zone. I need to convert this to LatLng for use with Google Map API to show the information in a map.

我找到了一些执行此操作的在线计算器，但没有实际代码或图书馆。 http://trac.osgeo.org/proj4js/ 是Javascript的投影库，但正在寻找在演示中，它不包括UTM投影。

I have found some online calculators that does this, but no actual code or libraries. http://trac.osgeo.org/proj4js/ is a projection library for Javascript, but looking at the demo it doesn't include UTM projection.

我对整个GIS域仍然很新鲜，所以我想要的是ala：

I am still pretty fresh to the entire GIS domain, so what I want is something ala:

(lat,lng) = transform(easting, northing, zone)

推荐答案

我最终找到了解决它的IBM代码： http://www.ibm.com/developerworks/java/library/j-coordconvert/index.html

I ended up finding java code from IBM that solved it: http://www.ibm.com/developerworks/java/library/j-coordconvert/index.html

仅供参考，这是我需要的方法的python实现：

Just for reference, here is my python implementation of the method I needed:

import math

def 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，我认为它很简单，比如 easting * x + zone * y 或其他东西。

And here I thought it was something simple like easting*x+zone*y or something.

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