问题描述
我在python方面经验不足，但由于这个社区的原因，我对其进行了改进！我非常需要一个函数，该函数接受输入并给出以下输出：
输入：
1-圆1中心的纬度/经度坐标（例如（50.851295，5.667969））
2-圆1的半径，以米为单位（例如200）
3-圆2中心的纬度/经度坐标（例如（50.844101，5.725889））
4-以米为单位的圆2的半径（例如300）
- 交点是（50.848295，5.707969）和（50.849295，5.717969）
- 圆圈重叠
- 圆是切线的，交点是（50.847295，5.705969）
- 圆圈不相交
我研究了该平台，其他平台，库中的相似主题，试图结合不同的解决方案，但未能成功。任何帮助深表感谢！
编辑：
这个问题的解决非常感谢TurePålsson，他在下面进行了评论，并在此链接https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles中指导我进行了whuber的出色工作，我基于该工作编写了以下代码，并对其进行了测试。我想在这里分享它，以防有人觉得有用。感谢您提供任何反馈意见。
'''
FINDING THE INTERSECTION COORDINATES (LAT/LON) OF TWO CIRCLES (GIVEN THE COORDINATES OF THE CENTER AND THE RADII)
Many thanks to Ture Pålsson who directed me to the right source,the code below is based on whuber's brilliant logic and
explanation here https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles
The idea is that;
1. The points in question are the mutual intersections of three spheres: a sphere centered beneath location x1 (on the
earth's surface) of a given radius,a sphere centered beneath location x2 (on the earth's surface) of a given radius,and
the earth itself,which is a sphere centered at O = (0,0) of a given radius.
2. The intersection of each of the first two spheres with the earth's surface is a circle,which defines two planes.
The mutual intersections of all three spheres therefore lies on the intersection of those two planes: a line.
Consequently,the problem is reduced to intersecting a line with a sphere.
Note that "Decimal" is used to have higher precision which is important if the distance between two points are a few
meters.
'''
from decimal import Decimal
from math import cos,sin,sqrt
import math
import numpy as np
def intersection(p1,r1_meter,p2,r2_meter):
# p1 = Coordinates of Point 1: latitude,longitude. This serves as the center of circle 1. Ex: (36.110174,-90.953524)
# r1_meter = Radius of circle 1 in meters
# p2 = Coordinates of Point 2: latitude,-90.953524)
# r2_meter = Radius of circle 2 in meters
'''
1. Convert (lat,lon) to (x,y,z) geocentric coordinates.
As usual,because we may choose units of measurement in which the earth has a unit radius
'''
x_p1 = Decimal(cos(math.radians(p1[1]))*cos(math.radians(p1[0]))) # x = cos(lon)*cos(lat)
y_p1 = Decimal(sin(math.radians(p1[1]))*cos(math.radians(p1[0]))) # y = sin(lon)*cos(lat)
z_p1 = Decimal(sin(math.radians(p1[0]))) # z = sin(lat)
x1 = (x_p1,y_p1,z_p1)
x_p2 = Decimal(cos(math.radians(p2[1]))*cos(math.radians(p2[0]))) # x = cos(lon)*cos(lat)
y_p2 = Decimal(sin(math.radians(p2[1]))*cos(math.radians(p2[0]))) # y = sin(lon)*cos(lat)
z_p2 = Decimal(sin(math.radians(p2[0]))) # z = sin(lat)
x2 = (x_p2,y_p2,z_p2)
'''
2. Convert the radii r1 and r2 (which are measured along the sphere) to angles along the sphere.
By deFinition,one nautical mile (NM) is 1/60 degree of arc (which is pi/180 * 1/60 = 0.0002908888 radians).
'''
r1 = Decimal(math.radians((r1_meter/1852) / 60)) # r1_meter/1852 converts meter to Nautical mile.
r2 = Decimal(math.radians((r2_meter/1852) / 60))
'''
3. The geodesic circle of radius r1 around x1 is the intersection of the earth's surface with an Euclidean sphere
of radius sin(r1) centered at cos(r1)*x1.
4. The plane determined by the intersection of the sphere of radius sin(r1) around cos(r1)*x1 and the earth's surface
is perpendicular to x1 and passes through the point cos(r1)x1,whence its equation is x.x1 = cos(r1)
(the "." represents the usual dot product); likewise for the other plane. There will be a unique point x0 on the
intersection of those two planes that is a linear combination of x1 and x2. Writing x0 = ax1 + b*x2 the two planar
equations are;
cos(r1) = x.x1 = (a*x1 + b*x2).x1 = a + b*(x2.x1)
cos(r2) = x.x2 = (a*x1 + b*x2).x2 = a*(x1.x2) + b
Using the fact that x2.x1 = x1.x2,which I shall write as q,the solution (if it exists) is given by
a = (cos(r1) - cos(r2)*q) / (1 - q^2),b = (cos(r2) - cos(r1)*q) / (1 - q^2).
'''
q = Decimal(np.dot(x1,x2))
if q**2 != 1 :
a = (Decimal(cos(r1)) - Decimal(cos(r2))*q) / (1 - q**2)
b = (Decimal(cos(r2)) - Decimal(cos(r1))*q) / (1 - q**2)
'''
5. Now all other points on the line of intersection of the two planes differ from x0 by some multiple of a vector
n which is mutually perpendicular to both planes. The cross product n = x1~Cross~x2 does the job provided n is
nonzero: once again,this means that x1 and x2 are neither coincident nor diametrically opposite. (We need to
take care to compute the cross product with high precision,because it involves subtractions with a lot of
cancellation when x1 and x2 are close to each other.)
'''
n = np.cross(x1,x2)
'''
6. Therefore,we seek up to two points of the form x0 + t*n which lie on the earth's surface: that is,their length
equals 1. Equivalently,their squared length is 1:
1 = squared length = (x0 + t*n).(x0 + t*n) = x0.x0 + 2t*x0.n + t^2*n.n = x0.x0 + t^2*n.n
'''
x0_1 = [a*f for f in x1]
x0_2 = [b*f for f in x2]
x0 = [sum(f) for f in zip(x0_1,x0_2)]
'''
The term with x0.n disappears because x0 (being a linear combination of x1 and x2) is perpendicular to n.
The two solutions easily are t = sqrt((1 - x0.x0)/n.n) and its negative. Once again high precision
is called for,because when x1 and x2 are close,x0.x0 is very close to 1,leading to some loss of
floating point precision.
'''
if (np.dot(x0,x0) <= 1) & (np.dot(n,n) != 0): # This is to secure that (1 - np.dot(x0,x0)) / np.dot(n,n) > 0
t = Decimal(sqrt((1 - np.dot(x0,n)))
t1 = t
t2 = -t
i1 = x0 + t1*n
i2 = x0 + t2*n
'''
7. Finally,we may convert these solutions back to (lat,lon) by converting geocentric (x,z) to geographic
coordinates. For the longitude,use the generalized arctangent returning values in the range -180 to 180
degrees (in computing applications,this function takes both x and y as arguments rather than just the
ratio y/x; it is sometimes called "atan2").
'''
i1_lat = math.degrees( math.asin(i1[2]))
i1_lon = math.degrees( math.atan2(i1[1],i1[0] ) )
ip1 = (i1_lat,i1_lon)
i2_lat = math.degrees( math.asin(i2[2]))
i2_lon = math.degrees( math.atan2(i2[1],i2[0] ) )
ip2 = (i2_lat,i2_lon)
return [ip1,ip2]
elif (np.dot(n,n) == 0):
return("The centers of the circles can be neither the same point nor antipodal points.")
else:
return("The circles do not intersect")
else:
return("The centers of the circles can be neither the same point nor antipodal points.")
'''
Example: the output of below is [(36.989311051533505,-88.15142628069133),(38.2383796094578,-92.39048549120287)]
intersection_points = intersection((37.673442,-90.234036),107.5*1852,(36.109997,-90.953669),145*1852)
print(intersection_points)
'''
解决方法
根据您需要的精度，您可能会或可能不会将地球视为一个球体。在第二种情况下，计算变得更加复杂。
精确测量的最佳选择当半径较小时（例如您的示例）是使用投影（例如UTM），然后应用常见的平面欧几里得计算。
让我们首先从https://stackoverflow.com/a/55817881/2148416复制平面交点函数：
def circle_intersection(x0,y0,r0,x1,y1,r1):
d = math.sqrt((x1 - x0) ** 2 + (y1 - y0) ** 2)
if d > r0 + r1: # non intersecting
return None
if d < abs(r0 - r1): # one circle within other
return None
if d == 0 and r0 == r1: # coincident circles
return None
a = (r0 ** 2 - r1 ** 2 + d ** 2) / (2 * d)
h = math.sqrt(r0 ** 2 - a ** 2)
x2 = x0 + a * (x1 - x0) / d
y2 = y0 + a * (y1 - y0) / d
x3 = x2 + h * (y1 - y0) / d
y3 = y2 - h * (x1 - x0) / d
x4 = x2 - h * (y1 - y0) / d
y4 = y2 + h * (x1 - x0) / d
return (x3,y3),(x4,y4)
借助utm library，可以在UTM坐标中完成小半径（长达几公里）的精确计算。它处理了有关地球比球体更像椭球体的所有复杂问题：
import utm
def geo_circle_intersection(latlon0,radius0,latlon1,radius1):
# Convert lat/lon to UTM
x0,zone,letter = utm.from_latlon(latlon0[0],latlon0[1])
x1,_,_ = utm.from_latlon(latlon1[0],latlon1 [1],force_zone_number=zone)
# Calculate intersections in UTM coordinates
a_utm,b_utm = circle_intersection(x0,r1)
# Convert intersections from UTM back to lat/lon
a = utm.to_latlon(a_utm[0],a_utm[1],letter)
b = utm.to_latlon(b_utm[0],b_utm[1],letter)
return a,b
使用示例（半径稍大）：
>>> p0 = 50.851295,5.667969
>>> r0 = 2000
>>> p1 = 50.844101,5.725889
>>> r1 = 3000
>>> a,b = geo_circle_intersection(p0,p1,r1)
>>> print(a)
(50.836848562566004,5.684869539768468)
>>> print(b)
(50.860635308778285,5.692236858407678)