[英]How to get line rectangle intersection segment?
我想找到代数重建方法的权重矩阵。 为此,我必须找到与网格的线交点。 我可以找到直线与线的交点,但我必须存储相交的线段网格数。 因此,假设在网格第一个正方形中不与网格相交,则将零置于权重矩阵的第一个元素上。
这是我为线路交叉尝试的代码:
ak = 3:6
aka = 3:6
x = zeros(size(aka))
y = zeros(size(ak))
for k = 1:length(ak)
line([ak(1) ak(end)], [aka(k) aka(k)],'color','r')
end
% Vertical grid
for k = 1:length(aka)
line([ak(k) ak(k)], [aka(1) aka(end)],'color','r')
end
hold on;
X =[0 15.5]
Y = [2.5 8.5]
m = (Y(2)-Y(1))/(X(2)-X(1)) ;
c = 2.5 ;
plot(X,Y)
axis([0 10 0 10])
axis square
% plotting y intercept
for i = 1:4
y(i) = m * ak(i) + c
if y(i)<2 || y(i)>6
y(i) = 0
end
end
% plotting x intercept
for i = 1:4
x(i) = (y(i) - c)/m
if x(i)<2 || x(i)>6
x(i) = 0
end
end
z = [x' y']
我有一条线,由参数m, h
定义,其中y = m*x + h
这条线穿过网格(即像素)。
对于网格的每个方格(a, b)
(即方形[a, a+1]x[b, b+1]
),我想确定给定的线是否穿过这个方格 ,如果是, 广场中段的长度是多少,这样我就可以构造权重矩阵,这对于代数重建方法是必不可少的。
这是一个很好的方法,使一条线与矩形网格相交并得到每个交叉段的长度:我使用了来自此链接的第三个答案中的伪代码的线条交点
% create some line form the equation y=mx+h
m = 0.5; h = 0.2;
x = -2:0.01:2;
y = m*x+h;
% create a grid on the range [-1,1]
[X,Y] = meshgrid(linspace(-1,1,10),linspace(-1,1,10));
% create a quad mesh on this range
fvc = surf2patch(X,Y,zeros(size(X)));
% extract topology
v = fvc.vertices(:,[1,2]);
f = fvc.faces;
% plot the grid and the line
patch(fvc,'EdgeColor','g','FaceColor','w'); hold on;
plot(x,y);
% use line line intersection from the link
DC = [f(:,[1,2]);f(:,[2,3]);f(:,[3,4]);f(:,[4,1])];
D = v(DC(:,1),:);
C = v(DC(:,2),:);
A = repmat([x(1),y(1)],size(DC,1),1);
B = repmat([x(end),y(end)],size(DC,1),1);
E = A-B;
F = D-C;
P = [-E(:,2),E(:,1)];
h = dot(A-C,P,2)./dot(F,P,2);
% calc intersections
idx = (0<=h & h<=1);
intersections = C(idx,:)+F(idx,:).*repmat(h(idx),1,2);
intersections = uniquetol(intersections,1e-8,'ByRows',true);
% sort by x axis values
[~,ii] = sort(intersections(:,1));
intersections = intersections(ii,:);
scatter(intersections(:,1),intersections(:,2));
% get segments lengths
directions = diff(intersections);
lengths = sqrt(sum(directions.^2,2));
directions = directions./repmat(sqrt(sum(directions.^2,2)),1,2);
directions = directions.*repmat(lengths,1,2);
quiver(intersections(1:end-1,1),intersections(1:end-1,2),directions(:,1),directions(:,2),'AutoScale','off','Color','k');
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