intersection points between two curves

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Prakash Chettri
Prakash Chettri 2021-11-29
编辑： Prakash Chettri 2021-12-13
Any suggestions would be greatly accepted. Thank you in advance.
  1 个评论
Chunru
Chunru 2021-11-29
Can you make your code executable by including values for some variables?

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回答（2 个）

KSSV
KSSV 2021-11-29
You can use this file exchange to get intersection points: https://in.mathworks.com/matlabcentral/answers/621553-finding-the-intersec-point-using-interx
To get area, pick the required points, arrange them in a order and use polyarea. Also have a look on trapz.
Chunru
Chunru 2021-11-29
Ran in:
% These are the Input parameters used
w = 60;
h = 350;
T0 = 60;
rho = 600;
T= 3.175;
N= 8;
W = T*N;
X0 = w/2;
Y0 = 0;
We=50.9381;
T0R=62.2211;
T0L=58.0029;
% First curve
X01R= (X0+rho*cos(T0*pi/180)-(rho-W/2)*cos(T0R*pi/180))-We;
x01r=X01R; y01r=Y0; k=1;
while (x01r>=0) && (y01r<=h)
YR1(k)=Y0+(k-1)*h/500; y01r=YR1(k);
phy(k)= asind(y01r/(rho)+sin(T0R*pi/180));
XR1(k)=X01R-rho*(cos(phy(k)*pi/180)-cos(T0R*pi/180));
x01r=XR1(k); k=k+1;
end
plot(XR1,YR1, 'r')
Warning: Imaginary parts of complex X and/or Y arguments ignored.
%Second curve
a=1;
T1L= asind(rho*sin(T0*pi/180)/(rho+W/2-(a*T)));
X1L=X0+(rho*cos(T0*pi/180)-(rho+W/2-(a*T))*cos(T1L*pi/180));
x1l=X1L; y1l=Y0; m=1;
while (x1l>=0) && (y1l<=h)
Y1L(m)=Y0+(m-1)*h/500; y1l=Y1L(m);
phy(m)= asind(y1l/(rho)+sin(T1L*pi/180)); %% Fiber angle
XL1(m)=X1L-rho*(cos(phy(m)*pi/180)-cos(T1L*pi/180));
x1l=XL1(m); m=m+1;
end
hold on
plot(XL1,Y1L, '-y')
Warning: Imaginary parts of complex X and/or Y arguments ignored.
xlim([0 60])
% Check your program: The curve coordinates have complex number
% Use the fileexchange function to find the interaction points
[x0,y0] = intersections(real(XR1), real(YR1), real(XL1), real(Y1L));
plot(x0, y0, 'ro', 'MarkerFaceColor', 'r')
function [x0,y0,iout,jout] = intersections(x1,y1,x2,y2,robust)
%INTERSECTIONS Intersections of curves.
% Computes the (x,y) locations where two curves intersect. The curves
% can be broken with NaNs or have vertical segments.
%
% Example:
% [X0,Y0] = intersections(X1,Y1,X2,Y2,ROBUST);
%
% where X1 and Y1 are equal-length vectors of at least two points and
% represent curve 1. Similarly, X2 and Y2 represent curve 2.
% X0 and Y0 are column vectors containing the points at which the two
% curves intersect.
%
% ROBUST (optional) set to 1 or true means to use a slight variation of the
% algorithm that might return duplicates of some intersection points, and
% then remove those duplicates. The default is true, but since the
% algorithm is slightly slower you can set it to false if you know that
% your curves don't intersect at any segment boundaries. Also, the robust
% version properly handles parallel and overlapping segments.
%
% The algorithm can return two additional vectors that indicate which
% segment pairs contain intersections and where they are:
%
% [X0,Y0,I,J] = intersections(X1,Y1,X2,Y2,ROBUST);
%
% For each element of the vector I, I(k) = (segment number of (X1,Y1)) +
% (how far along this segment the intersection is). For example, if I(k) =
% 45.25 then the intersection lies a quarter of the way between the line
% segment connecting (X1(45),Y1(45)) and (X1(46),Y1(46)). Similarly for
% the vector J and the segments in (X2,Y2).
%
% You can also get intersections of a curve with itself. Simply pass in
% only one curve, i.e.,
%
% [X0,Y0] = intersections(X1,Y1,ROBUST);
%
% where, as before, ROBUST is optional.
% Version: 2.0, 25 May 2017
% Author: Douglas M. Schwarz
% Email: dmschwarz=ieee*org, dmschwarz=urgrad*rochester*edu
% Real_email = regexprep(Email,{'=','*'},{'@','.'})
% Theory of operation:
%
% Given two line segments, L1 and L2,
%
% L1 endpoints: (x1(1),y1(1)) and (x1(2),y1(2))
% L2 endpoints: (x2(1),y2(1)) and (x2(2),y2(2))
%
% we can write four equations with four unknowns and then solve them. The
% four unknowns are t1, t2, x0 and y0, where (x0,y0) is the intersection of
% L1 and L2, t1 is the distance from the starting point of L1 to the
% intersection relative to the length of L1 and t2 is the distance from the
% starting point of L2 to the intersection relative to the length of L2.
%
% So, the four equations are
%
% (x1(2) - x1(1))*t1 = x0 - x1(1)
% (x2(2) - x2(1))*t2 = x0 - x2(1)
% (y1(2) - y1(1))*t1 = y0 - y1(1)
% (y2(2) - y2(1))*t2 = y0 - y2(1)
%
% Rearranging and writing in matrix form,
%
% [x1(2)-x1(1) 0 -1 0; [t1; [-x1(1);
% 0 x2(2)-x2(1) -1 0; * t2; = -x2(1);
% y1(2)-y1(1) 0 0 -1; x0; -y1(1);
% 0 y2(2)-y2(1) 0 -1] y0] -y2(1)]
%
% Let's call that A*T = B. We can solve for T with T = A\B.
%
% Once we have our solution we just have to look at t1 and t2 to determine
% whether L1 and L2 intersect. If 0 <= t1 < 1 and 0 <= t2 < 1 then the two
% line segments cross and we can include (x0,y0) in the output.
%
% In principle, we have to perform this computation on every pair of line
% segments in the input data. This can be quite a large number of pairs so
% we will reduce it by doing a simple preliminary check to eliminate line
% segment pairs that could not possibly cross. The check is to look at the
% smallest enclosing rectangles (with sides parallel to the axes) for each
% line segment pair and see if they overlap. If they do then we have to
% compute t1 and t2 (via the A\B computation) to see if the line segments
% cross, but if they don't then the line segments cannot cross. In a
% typical application, this technique will eliminate most of the potential
% line segment pairs.
% Input checks.
if verLessThan('matlab','7.13')
error(nargchk(2,5,nargin)) %#ok<NCHKN>
else
narginchk(2,5)
end
% Adjustments based on number of arguments.
switch nargin
case 2
robust = true;
x2 = x1;
y2 = y1;
self_intersect = true;
case 3
robust = x2;
x2 = x1;
y2 = y1;
self_intersect = true;
case 4
robust = true;
self_intersect = false;
case 5
self_intersect = false;
end
% x1 and y1 must be vectors with same number of points (at least 2).
if sum(size(x1) > 1) ~= 1 || sum(size(y1) > 1) ~= 1 || ...
length(x1) ~= length(y1)
error('X1 and Y1 must be equal-length vectors of at least 2 points.')
end
% x2 and y2 must be vectors with same number of points (at least 2).
if sum(size(x2) > 1) ~= 1 || sum(size(y2) > 1) ~= 1 || ...
length(x2) ~= length(y2)
error('X2 and Y2 must be equal-length vectors of at least 2 points.')
end
% Force all inputs to be column vectors.
x1 = x1(:);
y1 = y1(:);
x2 = x2(:);
y2 = y2(:);
% Compute number of line segments in each curve and some differences we'll
% need later.
n1 = length(x1) - 1;
n2 = length(x2) - 1;
xy1 = [x1 y1];
xy2 = [x2 y2];
dxy1 = diff(xy1);
dxy2 = diff(xy2);
% Determine the combinations of i and j where the rectangle enclosing the
% i'th line segment of curve 1 overlaps with the rectangle enclosing the
% j'th line segment of curve 2.
% Original method that works in old MATLAB versions, but is slower than
% using binary singleton expansion (explicit or implicit).
% [i,j] = find( ...
% repmat(mvmin(x1),1,n2) <= repmat(mvmax(x2).',n1,1) & ...
% repmat(mvmax(x1),1,n2) >= repmat(mvmin(x2).',n1,1) & ...
% repmat(mvmin(y1),1,n2) <= repmat(mvmax(y2).',n1,1) & ...
% repmat(mvmax(y1),1,n2) >= repmat(mvmin(y2).',n1,1));
% Select an algorithm based on MATLAB version and number of line
% segments in each curve. We want to avoid forming large matrices for
% large numbers of line segments. If the matrices are not too large,
% choose the best method available for the MATLAB version.
if n1 > 1000 || n2 > 1000 || verLessThan('matlab','7.4')
% Determine which curve has the most line segments.
if n1 >= n2
% Curve 1 has more segments, loop over segments of curve 2.
ijc = cell(1,n2);
min_x1 = mvmin(x1);
max_x1 = mvmax(x1);
min_y1 = mvmin(y1);
max_y1 = mvmax(y1);
for k = 1:n2
k1 = k + 1;
ijc{k} = find( ...
min_x1 <= max(x2(k),x2(k1)) & max_x1 >= min(x2(k),x2(k1)) & ...
min_y1 <= max(y2(k),y2(k1)) & max_y1 >= min(y2(k),y2(k1)));
ijc{k}(:,2) = k;
end
ij = vertcat(ijc{:});
i = ij(:,1);
j = ij(:,2);
else
% Curve 2 has more segments, loop over segments of curve 1.
ijc = cell(1,n1);
min_x2 = mvmin(x2);
max_x2 = mvmax(x2);
min_y2 = mvmin(y2);
max_y2 = mvmax(y2);
for k = 1:n1
k1 = k + 1;
ijc{k}(:,2) = find( ...
min_x2 <= max(x1(k),x1(k1)) & max_x2 >= min(x1(k),x1(k1)) & ...
min_y2 <= max(y1(k),y1(k1)) & max_y2 >= min(y1(k),y1(k1)));
ijc{k}(:,1) = k;
end
ij = vertcat(ijc{:});
i = ij(:,1);
j = ij(:,2);
end
elseif verLessThan('matlab','9.1')
% Use bsxfun.
[i,j] = find( ...
bsxfun(@le,mvmin(x1),mvmax(x2).') & ...
bsxfun(@ge,mvmax(x1),mvmin(x2).') & ...
bsxfun(@le,mvmin(y1),mvmax(y2).') & ...
bsxfun(@ge,mvmax(y1),mvmin(y2).'));
else
% Use implicit expansion.
[i,j] = find( ...
mvmin(x1) <= mvmax(x2).' & mvmax(x1) >= mvmin(x2).' & ...
mvmin(y1) <= mvmax(y2).' & mvmax(y1) >= mvmin(y2).');
end
% Find segments pairs which have at least one vertex = NaN and remove them.
% This line is a fast way of finding such segment pairs. We take
% advantage of the fact that NaNs propagate through calculations, in
% particular subtraction (in the calculation of dxy1 and dxy2, which we
% need anyway) and addition.
% At the same time we can remove redundant combinations of i and j in the
% case of finding intersections of a line with itself.
if self_intersect
remove = isnan(sum(dxy1(i,:) + dxy2(j,:),2)) | j <= i + 1;
else
remove = isnan(sum(dxy1(i,:) + dxy2(j,:),2));
end
i(remove) = [];
j(remove) = [];
% Initialize matrices. We'll put the T's and B's in matrices and use them
% one column at a time. AA is a 3-D extension of A where we'll use one
% plane at a time.
n = length(i);
T = zeros(4,n);
AA = zeros(4,4,n);
AA([1 2],3,:) = -1;
AA([3 4],4,:) = -1;
AA([1 3],1,:) = dxy1(i,:).';
AA([2 4],2,:) = dxy2(j,:).';
B = -[x1(i) x2(j) y1(i) y2(j)].';
% Loop through possibilities. Trap singularity warning and then use
% lastwarn to see if that plane of AA is near singular. Process any such
% segment pairs to determine if they are colinear (overlap) or merely
% parallel. That test consists of checking to see if one of the endpoints
% of the curve 2 segment lies on the curve 1 segment. This is done by
% checking the cross product
%
% (x1(2),y1(2)) - (x1(1),y1(1)) x (x2(2),y2(2)) - (x1(1),y1(1)).
%
% If this is close to zero then the segments overlap.
% If the robust option is false then we assume no two segment pairs are
% parallel and just go ahead and do the computation. If A is ever singular
% a warning will appear. This is faster and obviously you should use it
% only when you know you will never have overlapping or parallel segment
% pairs.
if robust
overlap = false(n,1);
warning_state = warning('off','MATLAB:singularMatrix');
% Use try-catch to guarantee original warning state is restored.
try
lastwarn('')
for k = 1:n
T(:,k) = AA(:,:,k)\B(:,k);
[unused,last_warn] = lastwarn; %#ok<ASGLU>
lastwarn('')
if strcmp(last_warn,'MATLAB:singularMatrix')
% Force in_range(k) to be false.
T(1,k) = NaN;
% Determine if these segments overlap or are just parallel.
overlap(k) = rcond([dxy1(i(k),:);xy2(j(k),:) - xy1(i(k),:)]) < eps;
end
end
warning(warning_state)
catch err
warning(warning_state)
rethrow(err)
end
% Find where t1 and t2 are between 0 and 1 and return the corresponding
% x0 and y0 values.
in_range = (T(1,:) >= 0 & T(2,:) >= 0 & T(1,:) <= 1 & T(2,:) <= 1).';
% For overlapping segment pairs the algorithm will return an
% intersection point that is at the center of the overlapping region.
if any(overlap)
ia = i(overlap);
ja = j(overlap);
% set x0 and y0 to middle of overlapping region.
T(3,overlap) = (max(min(x1(ia),x1(ia+1)),min(x2(ja),x2(ja+1))) + ...
min(max(x1(ia),x1(ia+1)),max(x2(ja),x2(ja+1)))).'/2;
T(4,overlap) = (max(min(y1(ia),y1(ia+1)),min(y2(ja),y2(ja+1))) + ...
min(max(y1(ia),y1(ia+1)),max(y2(ja),y2(ja+1)))).'/2;
selected = in_range | overlap;
else
selected = in_range;
end
xy0 = T(3:4,selected).';
% Remove duplicate intersection points.
[xy0,index] = unique(xy0,'rows');
x0 = xy0(:,1);
y0 = xy0(:,2);
% Compute how far along each line segment the intersections are.
if nargout > 2
sel_index = find(selected);
sel = sel_index(index);
iout = i(sel) + T(1,sel).';
jout = j(sel) + T(2,sel).';
end
else % non-robust option
for k = 1:n
[L,U] = lu(AA(:,:,k));
T(:,k) = U\(L\B(:,k));
end
% Find where t1 and t2 are between 0 and 1 and return the corresponding
% x0 and y0 values.
in_range = (T(1,:) >= 0 & T(2,:) >= 0 & T(1,:) < 1 & T(2,:) < 1).';
x0 = T(3,in_range).';
y0 = T(4,in_range).';
% Compute how far along each line segment the intersections are.
if nargout > 2
iout = i(in_range) + T(1,in_range).';
jout = j(in_range) + T(2,in_range).';
end
end
% Plot the results (useful for debugging).
% plot(x1,y1,x2,y2,x0,y0,'ok');
end
function y = mvmin(x)
% Faster implementation of movmin(x,k) when k = 1.
y = min(x(1:end-1),x(2:end));
end
function y = mvmax(x)
% Faster implementation of movmax(x,k) when k = 1.
y = max(x(1:end-1),x(2:end));
end

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