Finding the intersecting point of three surfaces.

[Topher925]

I'm faced with a problem where I need to determine the intersecting point of three different surfaces. Normally I would do this by using the incredibly painful method of Lagrange multipliers. However, this computation needs to be done relatively quickly and I can afford some error, in the range of about +/- 3%. I really don't want the computation requirement of the LM method and would prefer something more simple, does anyone know of something else? Preferably something that can be easily coded into matlab?

I was told about a method that's used for CAD systems, surface-surface-intersection which apparently is a numerical search method for defining this point. However, I just can't seem to find any info on it.

The surfaces I am working with are define by second order multi variable polynomials of the form: x²+ x + xy + y + y²+C. Can anyone help me out?

 

[Topher925]

Any input from anyone? At least one of you math people have to know something?

 

[mathman]

The surfaces I am working with are define by second order multi variable polynomials of the form: x²+ x + xy + y + y²+C. Can anyone help me out?

These don't look like they will intersect. Try to be more precise.

 

[HallsofIvy]

Yes, the surfaces you refer to are all "parallel". They have NO point in common since for any given x, y, the z values differ by C₁- C₂.

 

[Topher925]

I don't understand, how is it that two polynomials of that form can not intersect? On my plotted graphs they do and when they intersect they create a line.

Anyways, I should have been more clear about what I am actually doing. Only one of the surfaces is defined by a second order polynomial of the form I stated above. The other two are regular flat planes, one parallel to the X-Y axis and the other is somewhat perpendicular to X-Y plane. They all intersect at a single point which I need to find. LM is too computation intensive so I am looking for a simpler way.

 

[HallsofIvy]

I don't understand, how is it that two polynomials of that form can not intersect? On my plotted graphs they do and when they intersect they create a line.

Anyways, I should have been more clear about what I am actually doing. Only one of the surfaces is defined by a second order polynomial of the form I stated above. The other two are regular flat planes, one parallel to the X-Y axis and the other is somewhat perpendicular to X-Y plane. They all intersect at a single point which I need to find. LM is too computation intensive so I am looking for a simpler way.

You said:

The surfaces I am working with are define by second order multi variable polynomials of the form: x₂+ x + xy + y + y₂+C.

which I took to mean you had 3 paraboloids of the form z= x₂+ x + xy + y + y₂+C. Those will have no point of intersection. In general three surfaces may not intersect or may intersect in any number of points (or even along an entire curve).

Now you are saying you have one surface of that kind and two planes which is quite a different situation.

How you would find the point of intersection depends, of course, on the specific surfaces involved. Three surfaces means you have 3 equations in x, y, and z: solve those three equations simultaneously.

If two of the surfaces are planes, then you can solve those two linear equations to get two of the coordinates in terms of the other two. You can then put those into the equation of the paraboloid to get a single quadratic equation. Solve that using, perhaps, the quadratic formula.

If, for example, your surfaces are z= x₂+ x + xy + y + y₂+1, a plane parallel to the xy-plane is given by z= constant, for example z= 5. You say the other plane is "somewhat perpendicular to X-Y plane" so let's the plane x+ y= 0 which IS perpendicular to the xy-plane. x+y= 0 gives y= -x so if we put y= -x and z= 5 into the equation of the paraboloid, we get 5= x²+ x- x²- x+ x²+ 1 which reduces to the quadratic equation x²= 4 and so x= 2 or x= -2. If x= 2, y= -2 and one point where all three surfaces intersect is (2, -2, 5). If x= -2, then y= 2 and the other point where they intersect is (-2, 2, 5).

 

[Topher925]

Thanks for the reply HallsOfIvy. First, I should clarify that the second order polynomials I am referring to are described by C1x^2 + C2x + C3xy + C4y + c5Y^2 + C5 = z where the coefficients "Cx" changes in order to describe the polynomial. Actually what is happening is these polynomials are fitted to randomly generated data points.

The other two planes are derivatives (tangent planes) of other curves taken from that type of polynomial. All described using all three variables, so they are not perfectly parallel or perpendicular but are close. Since one of the polynomials is second order I obviously can't solve it like a system of linear equations.

It's apparent that I am not accurately describing the problem. I'll try to post an example tomorrow if/when I get my internet connection working.

 

[HallsofIvy]

Well, again, you can solve the two linear equations for two of the coordinates as a function of the third. Putting those into the non-linear equation will give you a quadratic equation for that third coordinate. Since a quadratic equation can have 0, 1, or 2 real solutions, the intersection can be 0, 1, or 2 points.

 

[mathman]

I'll describe a silar problem that I have worked on (a long time ago). The intersection of two planes will be a straight line. Thereforefore your problem is then the intersection of a line with a quadratic surface. This is straightforward enough. In any case the system can be reduced to one quadratic equation, which can be readily solved.

 

[Topher925]

Well, again, you can solve the two linear equations for two of the coordinates as a function of the third. Putting those into the non-linear equation will give you a quadratic equation for that third coordinate. Since a quadratic equation can have 0, 1, or 2 real solutions, the intersection can be 0, 1, or 2 points.

I was trying to avoid using quadratic equations and having multiple solutions but it seems like the best way to go about this problem. It is another few lines of code I will have to add but I guess it will work. Its certainly better than Lagrange multipliers now that I think about it.

Thanks for all the help.