检查图像点是不是与圆相交

Posted 2023-02-25

技术标签:

【中文标题】检查图像点是不是与圆相交【英文标题】：Check if image points intersects with a circle检查图像点是否与圆相交 【发布时间】：2015-08-24 10:26:48 【问题描述】：

我正在尝试检测肩膀。在彩色图片中您可以看到 2 个圆圈，大圆圈与边缘图片相交 4 个点。我使边缘变薄并将所有边缘点（x，y）放在一个数组中。现在我正在尝试使用圆方程找到交点：(x1 - x) + (y1 -y) - r^2 = 0。问题是，方程只有一次为零（有时永远不会）。我将圆的中心点和半径四舍五入，这样我就可以得到一个特定的像素......没有帮助！

彩色图片：

边缘图片：

代码：

得到圈子：

%% im - colored image.
%  x1,y1,x2,y2,x3,y3 - are the right eye left eye and chin points.
%  The function calculates a new circule and sends its center points and 
%  radius back.



function [cx,cy,newRadius] = calcFaceCircle(im,x1,y1,x2,y2,x3,y3) 


m1 = (y1 - y3)/(x1-x3);
m2 = (y2 - y3)/(x2-x3);

d = sqrt((x2-x1)^2 + (y2-y1)^2);

centerX = ((m1*m2*(y2-y1) + m2*(x2 + x3) - m1*(x1 + x3))/(2*(m2-m1)));
centerY = (-(1/m1)*(centerX - (x3 + x2)/2) + (y3 + y2)/2);

radius = (sqrt((x1 - centerX)^2 + (y1 - centerY)^2));
%% new circle radius
newRadius = round(radius * 1.65);


newCircleX = (x1+x2)/2; 
newCircleY = (y1+y2)/2;
%% new circle center points
cx = round(newCircleX + sqrt(newRadius^2-(d/2)^2)*(y1-y2)/d);
cy = round(newCircleY + sqrt(newRadius^2-(d/2)^2)*(x2-x1)/d);

%% display the image with circles and points
figure(01);
imshow(im);
hold on; 

ang=0:0.01:2*pi; 
xp=radius*cos(ang);
yp=radius*sin(ang);
plot(x1,y1,'.');
plot(x2,y2,'.');
plot(x3,y3,'.');

plot(centerX+xp,centerY+yp);

xp=newRadius*cos(ang);
yp=newRadius*sin(ang);
plot(cx+xp,cy+yp);

hold off;

end

肩膀：

%% 
%edgeFun - array of edge image points (x,y).
%  x - circle's center X.
%  y - circle's center Y.
%  r - circle's radios
%
% The function calculates the interection points between edgeFunc and the
% circle.
% The funciton returns the first intersection and last intersection.

function [xL,xR] = getCropInfo(edgeFunc,x,y,r)

j=1;
flag = 0;
xL = -1;
xR = -1;
count = 0;

if(x > 0 && y > 0 && r > 0)


    X = edgeFunc(:,2);     

    [edgeX,edgeY] = size(X);


    for i = 1:edgeX
        x1 = edgeFunc(i,2);
        y1 = edgeFunc(i,1); 

        % check if the points intersect with the circle
        if((floor((x-x1)^2 + (-y+y1)^2 - r^2) == 0))
            if(flag == 0)  % check if first intersection point found
                disp('found');
                flag = 1;
                xL = i;  % first point
            end

            j = i;  % last point
            count = count + 1;
        end

        xR = j;

    end


else

    return;

end
if(count == 4)
    disp('ok');

end

【问题讨论】：

所以基本上，你有img的边缘，和同一个坐标系上的一个圆，你想找到所有的交点？正确！我有一种感觉，方程永远不会返回零......这不是代码问题由于坐标系中的分辨率，可能永远不会返回零。有一些方法可以通过 matlab 绘图和手动估计相交（为什么我什至建议这样做？）或者您可以查看 polyxpoly，将边缘变成折线 - uk.mathworks.com/help/map/ref/polyxpoly.html 好的，所以 polyxpoly(x1,y1,x2,y2) as x1,y1 是圆坐标？我怎么得到那个？在链接上：[latc, lonc] = scircle1(lat0, lon0, km2deg(rad)); ，但 km2deg 以 KM 为单位接收无线电，但我的无线电以像素为单位。你的圆也需要是一条折线，如果你看第一个例子，你基本上需要以x和y的格式给出你的圆的完整坐标。 【参考方案1】：

为了让您的生活更轻松，我尝试了代码：

x1 = [-4,-3,-2,-1,0,1,2,3]; %//My edge, which is no where as handsome as yours
y1 = [-5,3,6,5,5,6,3,-5];

ang = 1:0.1:2*pi;
xp=4*cos(ang);  %//My circle, which is no where as good looking as yours
yp=4*sin(ang);

[xi,yi] = polyxpoly(x1,y1,xp,yp);  //Evaluate the intersects by interpolation

figure 
plot(x1,y1,'r')
hold on
plot(xp,yp,'b')

mapshow(xi,yi,'DisplayType','point','Marker','o');