How to find intersection between a straight line and a curve?

Question

Sebastian Dahl Sandbu 2022-2-4

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https://ww2.mathworks.cn/matlabcentral/answers/1643405-how-to-find-intersection-between-a-straight-line-and-a-curve

评论： Star Strider 2023-8-18

I want to find the coordinates for the two intersection points between the horisontal line and the graph. I think the problem is that there is no matching value between the line and the elements in the vector, so when I try to use intersect() I get an empty set. Any idea how this can be done? Thx .

On the y-axis I have a vector with intensity levels, and on the x-axis position. The data is from a file.

az = load('profile_AZ_7cm_Full.mat');
el = load('profile_EL_7cm_Full.mat');
x = az.Position(1,1:end); 
intensity_az = 10*log10(sum(az.RF.^2));
normx_intensity = intensity_az-max(intensity_az);
y = el.Position(2,1:end);
intensity_el = 10*log10(sum(el.RF.^2));
normy_intensity = intensity_el-max(intensity_el);
l = -6;
isx = interp1(normx_intensity,l);
figure;
subplot(2,1,1)
plot(x*1e3,normx_intensity)
hold on
yline(l)
title('Azimuth scan of beam profile')
xlabel('x-axis (mm)')
ylabel('Intensity (dB)')

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Answer 1

Star Strider 2022-2-5

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https://ww2.mathworks.cn/matlabcentral/answers/1643405-how-to-find-intersection-between-a-straight-line-and-a-curve#answer_889315

在 MATLAB Online 中打开

Another approach —

x = linspace(-15, 15, 150);

y = sinc(0.5*x);

l = 0.6;

[~,idx] = max(y);

xq(1) = interp1(y(1:idx),x(1:idx), l);

xq(2) = interp1(y(idx+1:end), x(idx+1:end), l);

figure

plot(x, y)

hold on

plot(x, 0.6*ones(size(x)), ':k')

plot(xq, [1 1]*l, 'xr', 'MarkerSize',10)

hold off

text(xq, [1 1]*l, compose('\\leftarrow x = %+.3f',xq), 'Horiz','left', 'Vert','middle', 'Rotation',-90)

.

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Star Strider 2022-2-5

在 MATLAB Online 中打开

Thank you!

My solution is intended to work specifically for this problem, so yes.

To get multiple intersections in different problems, I always begin with:

zxi = find(diff(sign(y - l)));

This creates ‘zero-crossings’ where ‘y’ is approximately equal to ‘l’ (in the context of the current problem), and returns their indices. This then guides the rest of the code:

for k = 1:numel(zxi)

idxrng = max([1, zxi(k)-2]) : min([numel(y), zxi(k)+2]); % Index Range For Interpolation

xi(k) = interp1(y(idxrng), x(idxrng), l); % High-Precision Estimate Of Intersection Coordinate

end

Testing it:

x = linspace(-15, 15);

y = sinc(0.5*x);

l = 0.05;

zxi = find(diff(sign(y - l)));

for k = 1:numel(zxi)

idxrng = max([1, zxi(k)-2]) : min([numel(y), zxi(k)+2]); % Index Range For Interpolation

xi(k) = interp1(y(idxrng), x(idxrng), l); % High-Precision Estimate Of Intersection Coordinate

end

figure

plot(x, y)

hold on

plot(x, ones(size(x))*l, ':k')

plot(xi, ones(size(xi))*l, 'xr', 'MarkerSize',7.5)

hold off

This gets around the problem with all interpolation methods (even hand-coded original versions) in that they need to be given a specific region over which to interpolate. After that, the interpolation proceeds straight away! (The loop is obviously necessary.) The index range ‘idxrng’ calculation traps situations where a zero-crossing is at the beginning or end of the sequence so it will not reference points that do not exist. The accuracy of the interpolation to calculate the intersections is in part a function of the resolution of the data, so initially interpolating to a finer resolution (more data points) is occasionally required.

.

Priyam Mishra 2023-8-18

Sir, I have same kind of problem in which in which blue color line is the cone tip data which is plotted along the depth. A linear fitted line with the same data is also shown. I want to find the values of depth (y-axis) where they are intersecting to each other. I want to find out all the points of intersection. Please help me Sir.

Thanks for your kind response.

Priyam Mishra

Star Strider 2023-8-18

在 MATLAB Online 中打开

Change the variables —

y = linspace(-15, 15);

x = sinc(0.5*y);

l = 0.05;

zxi = find(diff(sign(x - l)));

for k = 1:numel(zxi)

idxrng = max([1, zxi(k)-2]) : min([numel(y), zxi(k)+2]); % Index Range For Interpolation

yi(k) = interp1(x(idxrng), y(idxrng), l); % High-Precision Estimate Of Intersection Coordinate

end

figure

plot(x, y)

hold on

plot(ones(size(x))*l, y, ':k')

plot(ones(size(yi))*l, yi, 'xr', 'MarkerSize',7.5)

hold off

.

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Answer 2

Davide Masiello 2022-2-4

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1643405-how-to-find-intersection-between-a-straight-line-and-a-curve#answer_889200

在 MATLAB Online 中打开

I am a bit ovewhelmed by your code, so I will give you a qualitative example instead of modifying it.

x = [x1,x2,x3,___,xn];      % Your array of x data
y = [y1,y2,y3,___,yn];      % Your array of y data
c = c0;                     % The y value of your straight horizontal line
f = @(z) interp1(x,y-c,z);  % Creates anonymous function representing the difference between curve and straight line
xi = fzero(f,x0);           % Finds zero of the function (i.e., the intersection point)

The value of xi depends on the the initial condition x0, so you might want to loop the fzero over several x0 values to find multiple intersection points. It could be like this:

x0 = linspace(x1,xn,5);
for i = 1:5
    xi(i) = fzero(f,x0(i));           % Finds zero of the function (i.e., the intersection point)
end

Then you can do

xi = unique(xi);

to remove repeated values.

Hope it makes sense.

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