Cross section plot for 2D plot

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Tan Has
Tan Has 2012-10-3
I have a code to plot the quiver data.
I want to plot a 1D line plot across the 2D plot to see how the value of quiver changes from left to right alone a line.
Here is the code:
function[d]=DOMAINGENERATION_assocwithDNAPL(alpha, beta, kappa,Yo,CTD,vo)
fid = fopen('domain¶meter.dat'); % open file that will contain grid coordinate system and associated parameters
% for i = 1:5:45;
i=1;
% fh=figure;
m=4+(i-1)*11844;
n=11843+(i-1)*11844;
M=dlmread('3d2pct.plt',',',m,0, [m 0 n 9]); %3d2pct.plt data file
xarray=M(:,1);
yarray=M(:,2);
DV_XX=M(:,8);
DV_YY=M(:,9);
FRONT=M(:,3);
SF=reshape(FRONT,80,148);
X=reshape(xarray,80,148);
Y=reshape(yarray,80,148);
% figure;
v = [1 0.95:1.00];
contour(X(1,:),Y(:,1),SF,1);
colormap([0 0 0]);
hold on
quiver(xarray(1:4:end),yarray(1:4:end),DV_XX(1:4:end),DV_YY(1:4:end),1, 'k','filled');
hold on
xlabel('(m)');
ylabel('(m)');
set(gcf,'color',[1 1 1]);
set(gca,'color','none');
axis equal;
%*************** Plot area ***********************************************
axis([0 0.37 0 0.20])
%*************************************************************************
end

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回答（1 个）

Babak
Babak 2012-10-3
Do you need to find the intersection of a 1D line and a curve that has been plotted in a 2D plane?
Did I understand your problem correctly?
The intersection will be a whole bunch of points if I understood you correctly.
  2 个评论
Tan Has
Tan Has 2012-10-3
编辑：Tan Has 2012-10-3
yes for question. you got it. Could you help me
Babak
Babak 2012-10-3
OK so here's my answer:
You need to have a boundry [xmin xmax] and [ymin ymax] that you are interested in finding the intersection of a line and your curve. Now, assume that the curve is given by its points as X=[x1 x2 x3...] and Y=[y1 y2 y3 ...] and the line you want to intersect it with is given by an equation y=a*x+b, the intersection includes points like (xj,yj) that satisfies yj=a*xj+b and if you find xj in X, it may be located somewhere between x(i) and x(i+1). The corresponding Y for x(i) and x(i+1) are y(i) and y(i+1). If y(i) and y(i+1) are close enough to yj and closer than y(i-1) and y(i+2) to yj then you can conclude that (xj,yj) is an intersection point.

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