Find intersection point between two plotted lines

Hi all,

I have been trying to find an intersection point between two plotted lines however, I used several functions but it doesnot work.. usually the output is 2*0 empty double matrix ... Actually i need the number of intersection ... ?

here is the code and attached are the data

clearvars
clc; 
close all 

Data = readmatrix('data_tangent.xlsx')
x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);
Binit = x(1:5) \ y(1:5)
Bymax = x(idx) \ y(idx)
Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;
figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
hold off
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x)  0 max(y)])
x1=x(Line_ymaxv)
y1=Line_ymax(Line_ymaxv)
P=InterX([x;y],[x1;y1])

The function i used is from "https://www.mathworks.com/matlabcentral/fileexchange/22441-curve-intersections"

Answer 1

[Star Strider]

The code I wrote for your previous post, Plot tangent line on part of the curve is designed as requested to have only one intersection, that being at the origin, since both tangent lines were requested to go through the origin, at least as I understood it.

How else would you want to define the two tangent lines?

[Mark Sc]

@Star Strider

Thank you for your feedback,

just one intersect point thats' what I wanna ...

I just wanna a code to let me know the intersection point value

Thank you once again

[Star Strider]

My pleasure!

The two lines I plotted (only one is shown here) both intersect at the origin, since that is how they were requested and designed. The intersection with the curve is designed to be at ‘ymax’ with the x-coordinate of that intersection being ‘x(idx)’, so nothing further needs to be computed.

% clearvars
% clc; 
% close all

format long

Data = readmatrix('data_tangent.xlsx')

Data = 13042×2

1.980321668300000 7.370000000000000 1.982328936500000 7.350000000000000 1.985318034200000 7.380000000000000 1.987310766000000 7.400000000000000 1.990343472900000 7.310000000000000 1.993270790900000 7.510000000000000 1.995263522700000 7.530000000000000 1.997299863700000 7.430000000000000 2.000270790900000 7.510000000000000 2.002187206600000 7.740000000000000

x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);

Binit = x(1:fix(idx/4)) \ y(1:fix(idx/4))

Binit = 
   3.644815761434447

Bymax = x(idx) \ y(idx)

Bymax = 
   3.216013064246002

Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;

Intersection_x = interp1(Line_init-Line_ymax, x, 0, 'linear','extrap')

Intersection_x = 
     2.273736754432321e-13

Intersection_y = interp1(x, Line_init, Intersection_x, 'linear','extrap')

Intersection_y = 
     9.094947017729282e-13

Intersection = [Intersection_x, Intersection_y]         % Desired Result

Intersection = 1×2

1.0e-12 * 0.227373675443232 0.909494701772928

figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
plot([0;x(Line_initv)], [0;Line_init(Line_initv)], '-r')
hold off
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x)  0 max(y)])

% x1=x(Line_ymaxv)
% y1=Line_ymax(Line_ymaxv)
% P=InterX([x;y],[x1;y1])

EDIT — (18 Apr 2023 at 20:03)

Added ‘Intersection’ calculation and result using interp1 to calculate the coordinates.

.

Answer 2

[Cameron]

Looks like a stress-strain curve. Depending on the material, you could adjust the variable I named cutoff to 0.3*ymax or whatever you want. This code takes the first intersection of the stress-strain curve and interpolates the value for yield.

clearvars
clc; 
close all 
Data = readmatrix('data_tangent.xlsx')
x = Data(:,1);
y = Data(:,2);
[ymax,idx] = max(y);
Binit = x(1:5) \ y(1:5);
Bymax = x(idx) \ y(idx);
Line_init = x*Binit;
Line_initv = Line_init <= ymax;
Line_ymax = x*Bymax;
Line_ymaxv = Line_ymax <= ymax;
Line_y = Line_ymax(Line_ymaxv);
trunc_x = x(1:find(Line_ymaxv,1,'last'));
trunc_y = y(1:find(Line_ymaxv,1,'last'));
ii = length(trunc_y);
cutoff = 0.5*ymax; %you can adjust this
SaveMe = [];
while trunc_y(ii) > cutoff
    if (Line_y(ii) > trunc_y(ii) && Line_y(ii-1) <= trunc_y(ii-1)) | ...
        (Line_y(ii) < trunc_y(ii) && Line_y(ii-1) >= trunc_y(ii-1))
        SaveMe = [SaveMe;ii];
    end
    ii = ii - 1;
end
m1 = polyfit(x(SaveMe(end)-1:SaveMe(end)),y(SaveMe(end)-1:SaveMe(end)),1);
m2 = polyfit(x(SaveMe(end)-1:SaveMe(end)),Line_y(SaveMe(end)-1:SaveMe(end)),1);
x_cross = (m2(2) - m1(2))/(m1(1) - m2(1));
y_cross = m1(1)*x_cross + m1(2);
figure
plot(x, y)
hold on
plot([0;x(Line_ymaxv)], [0;Line_ymax(Line_ymaxv)], '-r')
scatter(x_cross,y_cross,'filled')
grid
axis('equal')
xlabel('X')
ylabel('Y')
axis([0 max(x)  0 max(y)])
hold off

[Mark Sc]

@Cameron

Thank you for your answer,

I meant if I already have a two lines plotted so I just wanna know the point of intersection ..

So would you able to help me ?

Thank you very much

[Cameron]

The answer is in the code I provided as x_cross and y_cross.

[Mark Sc]

@Cameron

Thank you Cameron, but what If I already have my own x and y

x1_y1 as arrays.....

I would appreciate if I can just enter these arrays and got the point of intersection..

Thanks,