如何解决如何检查位置是否在地图上的圆圈内
我正在尝试使用Matlab检查纬度/经度是否在地图上的圆内。我用了这个公式:
d=R*acos(sin(S_lat(t)*d2r)*sin(N_lat*d2r)+cos(S_lat(t)*d2r)*cos(N_lat*d2r)*cos((S_long(t)-N_long)*d2r))
和以下公式:
a = sin((S_lat(t)-N_lat)*d2r / 2)^2 + cos(N_lat*d2r) * cos(S_lat(t)*d2r) * sin((S_long(t)-N_long)*d2r / 2)^2;
c = 2 * asin(sqrt(a));
d = R * c;
但是它没有给我正确的答案。
解决方法
与this page相比,您的第二个距离公式看起来几乎是正确的
a = sin²(Δφ/2) + cos φ1 ⋅ cos φ2 ⋅ sin²(Δλ/2)
c = 2 ⋅ atan2( √a,√(1−a) )
d = R ⋅ c
据我所知,对于范围为0..1的正参数,asin
的结果应与第二行中的atan2
公式相同。
如果a
的值可能略微超出此间隔,则反正弦将给出错误的结果,所以
如果atan2不可用,则可以从2⋅asin(min(1, √a))(包括防止舍入错误)。
您能举例说明有错误的数据点吗?
Python代码(ideone to play with)
import math
def eadist(lat1,lon1,lat2,lon2):
print(lat1,lon2)
d2r = math.pi / 180
R = 6378
dlat = (lat2 -lat1) * d2r
dlon = (lon2 -lon1) * d2r
a = (math.sin(dlat/2))**2 + math.cos(lat1*d2r)*math.cos(lat2*d2r) * (math.sin(dlon/2))**2
print("a=",a)
c = 2*math.asin(math.sqrt(a))
print("c=",c)
dist = c * R
return dist
print(eadist(45,90,45,91))
print(eadist(45,46,90))
print(eadist(45,-45,-90))
给出正确的结果
45 90 45 91
a= 3.8076210902190215e-05
c= 0.012341263173265506
78.71257651908739 //1 degree by parallel
45 90 46 90
a= 7.615242180438042e-05
c= 0.017453292519943295
111.31709969219834 //1 degree by meridian
45 90 46 91
a= 0.0001135583120069672
c= 0.021313151879913415
135.93528269008777 //diagonal step
45 90 -45 -90
a= 1.0
c= 3.141592653589793
20037.0779445957 //antipodal point,a half of meridian length
,
R = 6378; %地球 d2r = pi / 180; %度到弧度
N_lat = 45; N_long = 90;
rov = 70;
[latc,longc] = scircle1(N_lat,N_long,rov);%以(N_lat,N_long)为中心对一个圆进行半径补偿,并rov为其半径
D = zeros(1,length(latc)); %检查从圆心到圆的任何点的距离 使用此公式,%perimeter等于rov
对于我= 1:length(latc)
% Formula1
%D(i)=acos(sin(latc(i)*d2r)*sin(N_lat*d2r)+cos(latc(i)*d2r)*cos(N_lat*d2r)*cos((longc(i)-N_long)*d2r));
%Formula2
%D(i)= R * sqrt((latc(i)-N_lat)^ 2 +(longc(i)-N_long)^ 2);
%Formula3 %a = sin((latc(i)-N_lat) d2r / 2)^ 2 + cos(N_lat d2r)* cos(latc(i)* d2r)* sin((longc(i) -N_long) d2r / 2)^ 2; %c = 2 * asin(sqrt(a)); %D(i)= R * c; %Formula4 %D(i)= R 距离(latc(i),longc(i),N_lat,N_long);
结束
,通过矢量几何和球形坐标:
您可以通过以下方式将经度/纬度角转换为单位球面上的直角坐标:
X = cos(Θ) sin(φ),Y = sin(Θ) sin(φ),Z = cos(φ)
您可以为测试点和圆心计算这些值。然后,点积的反余弦值将为您提供圆心角,您可以将其与地球半径上的圆弧*进行比较。
*圆弧长度,而不是圆平面中的半径。
, R=6378; %Earth raius
d2r = pi/180; %degrees to radians
N_lat=45;
N_long=45;
rov=50;
%World map
worldmap world
load coastlines
[latcells,loncells] = polysplit(coastlat,coastlon);
plotm(coastlat,coastlon,'green')
hold on;
plotm(N_lat,N_long,'b*');
hold on;
[latc,longc] = scircle1(N_lat,rov);%compute a circle with (N_lat,N_long)as a center and rov its radius
plotm(latc,longc,'r-');
D=zeros(1,length(latc));
%Check if the distance from the center of the circle to any point of its
%perimeter is equal to rov using this formulas
for i =1:length(latc)
%
% % Formula1
% %D(i)=acos(sin(latc(i)*d2r)*sin(N_lat*d2r)+cos(latc(i)*d2r)*cos(N_lat*d2r)*cos((longc(i)-N_long)*d2r));
%
% % Formula2
% % D(i)=R*sqrt((latc(i)-N_lat)^2+(longc(i)-N_long)^2);
%
% %Formula3
% % a = sin((latc(i)-N_lat)*d2r / 2)^2 + cos(N_lat*d2r) * cos(latc(i)*d2r) * sin((longc(i)-N_long)*d2r / 2)^2;
% % c = 2 * asin(sqrt(a));
% % D(i) = R * c;
%
% %Formula4
%
% D(i)=distance('gc',latc(i),longc(i),N_lat,N_long);
% %Formula5
% x = (longc(i)-N_long)*d2r*cos((latc(i)+N_lat)*d2r/2);
% y = (latc(i)-N_lat)*d2r;
% D(i)=R*sqrt(x^2+y^2);
%Formula6
x=cos(latc(i)*d2r)*cos(longc(i)*d2r)-cos(N_lat*d2r)*cos(N_long*d2r);
y=cos(latc(i)*d2r)*sin(longc(i)*d2r)-cos(N_lat*d2r)*sin(N_long*d2r);
z=sin(latc(i)*d2r)-sin(N_lat*d2r);
c=sqrt(x^2+y^2+z^2);
angle=asin(c/2);
D(i)=R*angle;
end
enter code here
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