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I fitted two planes to two datasets, and the planes are outputted as 1x1 sfits (see figure).

How do I get the intersection between these two planes? I want to be able to extract the xyz info along this line.

I'm not used to working with symbolic equations in MATLAB, so I tried this, without much success:

%

%Extract surface formula

cliff=formula(fit_cliff);

platform=formula(fit_platform);

%Extract formula Coefficients

coef_cliff=coeffvalues(fit_cliff);

coef_platform=coeffvalues(fit_platform);

c1=coef_cliff(1);

c2=coef_cliff(2);

c3=coef_cliff(3);

c4=coef_cliff(4);

c5=coef_cliff(5);

p1=coef_platform(1);

p2=coef_platform(2);

p3=coef_platform(3);

p4=coef_platform(4);

p5=coef_platform(5);

%Write equation and solve for intersection

syms x y

z1 = c1+(c2*x)+(c3*y)+(c4*(x^2))+(c5*(x*y));

z2 = p1+(p2*x)+(p3*y)+(p4*(x^2))+(p5*(x*y));

intersect = solve(z1==z2)

Star Strider
on 30 Aug 2016

I would subtract one from the other and then use the contour function on the result to determine the zero value. See the contour documentation for Display Single Contour Line but use:

v = [0, 0];

instead to find the intersection of the result of the subtraction. See the documentation on ContourMatrix to return these values to your workspace so you can use them in the rest of your code.

John D'Errico
on 30 Aug 2016

Edited: John D'Errico
on 30 Aug 2016

So what was the problem? :)

Let me guess. This problem will come down to solving what is effectively a 4th degree polynomial. The results will look nasty. Lots of roots. Lots of things that say root of. So the solution you got looked like a mess. (Because it is, and it must be so.) The fact is, there is no simple formula for the solution. Certainly not anything that will be easy to look at.

However, there is a simple solution. You know the range in x and y that you care about. So pick a sequence of values for x (or y.) Just loop over this set of values. So for any value of x, now use vpasolve to solve for the solution to your problem, with x substituted into that expression. Or you can use fzero.

For any x, you get the value of y, to a reasonably high degree of precision. You will actually get a set of probably 4 values for y for any x. Pick the REAL root that falls in the region of interest. This gives you a closely spaced set of solutions to the problem. Easy to do. It gives you what you want. And the nice thing is, all of these points on the curve are essentially exact, to within the ability of the solver to give you high accuracy.

You can use a spline to approximate the curve if you want something smooth. No matter what, you WON'T get a simple function, as I said before. So just use a spline in the end.

Teja Muppirala
on 30 Aug 2016

This works for me:

%%1. Making some test coefficients

c1=1; c2=2; c3=3; c4=4; c5=5;

p1=-1.1; p2=2.2; p3=-3.3; p4=4.3; p5=-5.5;

syms x y

z1 = c1+(c2*x)+(c3*y)+(c4*(x^2))+(c5*(x*y))

z2 = p1+(p2*x)+(p3*y)+(p4*(x^2))+(p5*(x*y))

%%2. Do The solving

y_of_x = solve(z1==z2,y)

z_of_x = simplify(subs(z1,y,y_of_x))

%%3. Plot the result

figure

h1 = fsurf(z1,[-2 2]);

set(h1,'facecolor','r','facealpha',0.5);

hold all

h2 = fsurf(z2,[-2 2]);

set(h2,'facecolor','b','facealpha',0.5);

hp = fplot3(x,y_of_x,z_of_x,[-2 2]);

set(hp,'LineWidth',2,'Color','g');

ylim([-2 2]);

This gives in the command window:

z1 =

2*x + 3*y + 5*x*y + 4*x^2 + 1

z2 =

(11*x)/5 - (33*y)/10 - (11*x*y)/2 + (43*x^2)/10 - 11/10

y_of_x =

(3*x^2 + 2*x - 21)/(105*x + 63)

z_of_x =

(x*(87*x + 44))/21

Replace "fsurf" with "ezsurf" and "fplot3" with "ezplot3" if your MATLAB is old and doesn't have these functions.

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