如何计算给定经纬度位置的边界框? [英] How to calculate the bounding box for a given lat/lng location?

问题描述

我给了一个由纬度和经度定义的位置. 现在我想计算例如内的边界框到那点十公里.

I have given a location defined by latitude and longitude. Now i want to calculate a bounding box within e.g. 10 kilometers of that point.

边界框应定义为latmin，lngmin和latmax，lngmax.

The bounding box should be defined as latmin, lngmin and latmax, lngmax.

我需要这些东西才能使用 panoramio API .

I need this stuff in order to use the panoramio API.

有人知道如何获得积分吗?

Does someone know the formula of how to get thos points?

编辑:伙计们，我正在寻找一个公式/函数，该函数需要lat& lng作为输入，并返回边界框latmin& lngmin和latmax& latmin. mysql，php，c#，javascript很好，但伪代码也可以.

Guys i am looking for a formula/function which takes lat & lng as input and returns a bounding box as latmin & lngmin and latmax & latmin. Mysql, php, c#, javascript is fine but also pseudocode should be okay.

编辑:我不是在寻找可以显示2点距离的解决方案

I am not looking for a solution which shows me the distance of 2 points

推荐答案

我建议将地球表面局部近似为一个球体，其半径由WGS84椭球体在给定的纬度下给出.我怀疑latMin和latMax的精确计算将需要椭圆函数，并且不会产生明显的准确性提高(WGS84本身就是一个近似值).

I suggest to approximate locally the Earth surface as a sphere with radius given by the WGS84 ellipsoid at the given latitude. I suspect that the exact computation of latMin and latMax would require elliptic functions and would not yield an appreciable increase in accuracy (WGS84 is itself an approximation).

我的实现如下(它是用Python编写的；我尚未对其进行测试):

My implementation follows (It's written in Python; I have not tested it):

# degrees to radians
def deg2rad(degrees):
    return math.pi*degrees/180.0
# radians to degrees
def rad2deg(radians):
    return 180.0*radians/math.pi

# Semi-axes of WGS-84 geoidal reference
WGS84_a = 6378137.0  # Major semiaxis [m]
WGS84_b = 6356752.3  # Minor semiaxis [m]

# Earth radius at a given latitude, according to the WGS-84 ellipsoid [m]
def WGS84EarthRadius(lat):
    # http://en.wikipedia.org/wiki/Earth_radius
    An = WGS84_a*WGS84_a * math.cos(lat)
    Bn = WGS84_b*WGS84_b * math.sin(lat)
    Ad = WGS84_a * math.cos(lat)
    Bd = WGS84_b * math.sin(lat)
    return math.sqrt( (An*An + Bn*Bn)/(Ad*Ad + Bd*Bd) )

# Bounding box surrounding the point at given coordinates,
# assuming local approximation of Earth surface as a sphere
# of radius given by WGS84
def boundingBox(latitudeInDegrees, longitudeInDegrees, halfSideInKm):
    lat = deg2rad(latitudeInDegrees)
    lon = deg2rad(longitudeInDegrees)
    halfSide = 1000*halfSideInKm

    # Radius of Earth at given latitude
    radius = WGS84EarthRadius(lat)
    # Radius of the parallel at given latitude
    pradius = radius*math.cos(lat)

    latMin = lat - halfSide/radius
    latMax = lat + halfSide/radius
    lonMin = lon - halfSide/pradius
    lonMax = lon + halfSide/pradius

    return (rad2deg(latMin), rad2deg(lonMin), rad2deg(latMax), rad2deg(lonMax))

以下代码将(度，素数，秒)转换为度+分数的一部分，反之亦然(未测试):

The following code converts (degrees, primes, seconds) to degrees + fractions of a degree, and vice versa (not tested):

def dps2deg(degrees, primes, seconds):
    return degrees + primes/60.0 + seconds/3600.0

def deg2dps(degrees):
    intdeg = math.floor(degrees)
    primes = (degrees - intdeg)*60.0
    intpri = math.floor(primes)
    seconds = (primes - intpri)*60.0
    intsec = round(seconds)
    return (int(intdeg), int(intpri), int(intsec))

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