【发布时间】：2020-11-12 08:26:53
【问题描述】：
我在 python 方面没有那么丰富的经验，但感谢这个社区，我改进了它！我迫切需要一个接受输入并给出以下输出的函数：
输入：
1- 圆心 1 的经纬度坐标（例如 (50.851295, 5.667969) ）
2- 以米为单位的圆 1 的半径（例如 200）
3- 圆心 2 的经纬度坐标（例如 (50.844101, 5.725889) ）
4- 以米为单位的圆 2 的半径（例如 300）
输出：可能的输出示例可以是；
- 交点为 (50.848295, 5.707969) 和 (50.849295, 5.717969)
- 圆圈重叠
- 圆是相切的，交点是 (50.847295, 5.705969)
- 圆圈不相交
我检查了这个平台、其他平台、库中的类似主题，尝试组合不同的解决方案但未能成功。非常感谢任何帮助！
编辑：
非常感谢 Ture Pålsson 解决了这个问题，他在下面发表了评论，并在此链接 https://gis.stackexchange.com/questions/48937/calculating-intersection-of-two-circles 中指导我了解 whuber 的出色工作。基于该工作，我编写了下面的代码，并且据我测试它可以工作。我想在这里分享它，以防有人发现它有帮助。感谢您提供任何反馈。
'''
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 = (0,0,0) 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, longitude. This serves as the center of circle 1. Ex: (36.110174, -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, x0)) / np.dot(n,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,y,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)
'''
【问题讨论】：
-
向我们展示您到目前为止尝试了什么？本站不是代码补全服务
-
在 GIS 堆栈交换中有一个很好的答案：gis.stackexchange.com/questions/48937/…。如果你在 Python 中实现它时遇到问题，你可以回到这里并提出一个更有针对性的问题。
-
非常感谢@TurePålsson！基于 wuber 的出色工作，我解决了问题并添加了上面的代码。
标签： python geometry coordinates intersection