如何计算三角形（由三个（X，Y）对指定）和圆形（X，Y，R）之间的交集面积？我已经搜索了一些但没有结果。这是为工作而非学校。:)
在C#中，它看起来会像这样：
struct { PointF vert[3]; } Triangle;
struct { PointF center; float radius; } Circle;
// returns the area of intersection, e.g.:
// if the circle contains the triangle, return area of triangle
// if the triangle contains the circle, return area of circle
// if partial intersection, figure that out
// if no intersection, return 0
double AreaOfIntersection(Triangle t, Circle c)
{
...
}
你需要多精确？如果你可以用简单的形状来近似圆，那么问题就可以简化。例如，将圆建模为一组非常窄的三角形，在中心汇聚。
ϑ = 2 sin⁻¹(0.5 c / r)
A = 0.5 r² (ϑ - sin(ϑ))
其中c是弦长，r是半径，ϑ是通过圆心的角度，A是面积。请注意，如果超过半个圆被切断，则此解决方案将失效。
如果您只需要一个近似值，那么这可能不值得努力，因为它对实际交点的外观做出了几个假设。
供日后参考，这是我编写的一些Matlab代码来解决这个问题：
https://uk.mathworks.com/matlabcentral/fileexchange/126645-intersection-of-polygon-and-circle
它或多或少地实现了Brian Moths的答案。
以下是代码：
function [A, pointList, isArc, AP] = polygonCircleIntersection(R, P)
% polygonCircleIntersection - calculates area of intersection of circle
% of radius R centered in origin with a 2D polygon with vertices in P, and
% provides shape of intersection
%
% Inputs:
% R : radius of circle [scalar]
% P : 2 X nV matrix of polygon vertices. Each column is a vertex
% Vertices must be in counter-clockwise order or things will go wrong
%
% Outputs:
% A : area of intersection
% pointList : list of points on the perimiter of the intersection
% isArc : array of logicals. isArc(i) is true if the segment between
% i-1 and i is an arc.
% AP : the area of the polygon
%
% Used for FOV analysis
%
% See also: intersect, polyshape, triangulation
%
% Author: Simão da Graça Marto
% e-mail: simao.marto@gmail.com
% Date: 22/03/2023
nV = size(P, 2);
AP = 0;
for i = 1:nV
AP = AP + shoelace(P(:,i), P(:,mod(i,nV)+1));
end
% SANITY CHECK
% order of points
if(AP<0)
error("Polygon must be in counter-clockwise order");
%If this just means the polygon is not visible, uncomment the
%following and remove the error:
% A = 0;
% pointList = [];
% isArc = [];
% return
end
nT2 = sum(P.^2);
isOutside = nT2>R^2;
%1st edge case: all points inside, so polygon fully inside:
if(~any(isOutside))
A = 0;
for i = 1:nV
A = A + shoelace(P(:,i), P(:,mod(i,nV)+1));
end
pointList = P;
isArc = false(1,nV);
return;
end
% We must start from an outside vertex, so cycle vertices to be in a correct order
shiftK = 1-find(isOutside,1,"first");
P = circshift(P,shiftK, 2);
isOutside = circshift(isOutside,shiftK, 2);
% compute intersection
pointList = []; %list of points forming shape
isArc = [true]; % indicates if i-1 to i is an arc. if not, it's a line segment
outsideCircle = true;
iP = 1;
for i = 1:nV
x0 = P(:,i);
d = P(:,mod(i,nV)+1)-x0;
%segment circle intersection
[xI, onCircle] = segmentCircleIntersection(x0, d, R);
pointList = [pointList xI];
if(~isempty(xI))
%if point iP is outside circle, then segment from previous point to
%this one must be an arc
isArc(iP) = outsideCircle;
%point iP to iP+1 is always a line segment, by construction
isArc(iP+1) = false;
outsideCircle = onCircle(2); %if segment goes out of circle, we are now outside circle, otherwise we are inside
iP = iP + 2; %update point index
end
end
%edge cases: triangle perimeter fully outside, but is the circle
%inside or outside triangle?
if(isempty(pointList) && all(isOutside))
polygonContainsOrigin = inpolygon(0,0,P(1,:), P(2,:));
if(polygonContainsOrigin) %2nd edge case: circle fully inside triangle
A = pi*R^2;
pointList = [1;0]; %to allow drawing
else %3rd edge case: triangle fully outside circle
A = 0;
pointList = [];
isArc = [];
end
return;
end
%SANITY CHECKS
if(isempty(pointList) && any(isOutside) && ~all(isOutside))
error("there are no intersections, but some points are inside and others outside")
end
if(any(isArc & circshift(isArc, 1)))
error("there can't be two arcs in a row")
end
nI = size(pointList, 2);
%compute area as a sum of triangles (shoelace) and arcs
A = 0;
for i = 1:nI
iPrev = mod(i-2,nI)+1;
xPrev = pointList(:, iPrev);
xi = pointList(:, i);
if(isArc(i))
thPrev = atan2(xPrev(2),xPrev(1));
thi = atan2(xi(2),xi(1));
dth = wrapTo2Pi(thi-thPrev);
if(dth == 2*pi), dth = 0; end
A = A + dth*R^2/2;
else
A = A + shoelace(xPrev, xi); %can be negative. That is correct.
end
end
% SANITY CHECK
if(A<0 || A>pi*R^2 || A > AP)
error("invalid area")
end
end
%Always returns 2 or 0 points. If tangent, none are returned because it
%would have no effect on the area. If segment starts or ends inside circle,
%start or end point are returned. By circle understand it to include its
%interior, so intersection is the segment between the two points returned
function [xI, onCircle] = segmentCircleIntersection(x0, d, R)
a = sum(d.^2);
b = 2*x0'*d;
c = sum(x0.^2) - R^2; %could use precomputed nT2 instead
D = b^2 - 4*a*c;
if(D<=0) % line never goes in the circle
xI = [];
onCircle = [];
return
end
%intersection points (along segment coordinate)
ll = (-b + [-1 1]*sqrt(D) )/(2*a);
% SANITY CHECK
if( ll(1) >= ll(2) )
error("A mathematical impossibility. Did you give this thing complex numbers or something?")
end
if( ll(2)<0 || ll(1)>1) %intersection fully outside segment
xI = [];
onCircle = [];
return
end
%compute intersection
ll = max(0, ll);
ll = min(1, ll);
xI = x0 + ll.*d;
onCircle = [ll(1)>0 ll(2)<1];
end
%Shoelace formula for area of a straight segment with edges connected to
%origin, forming a triangle.
%can be negative if area is meant to be subtracted
function A = shoelace(p1, p2)
A = (p1(1)*p2(2) - p1(2)*p2(1))/2;
end