Implicit surface

[Richard Baker]

How do you parametrize an implicit surface? The surface in question is quadratic in terms of x, y, and z. Ie., 0 = ax squared + bx + cy squared + dy + ez + fz + gxy + hxz + uyz +j. I want general parametrization for all arbitrary constants from a-j. Answer should probably involve some combination of trigonometric functions of uv.

I am asking this question for my son who does not have internet access. I will copy, paste, print, and send the answer to him. My mathematics background is far short of being able to understand this. Thank you.

[studiot]

Whomsoever it is for, is this homework or coursework ?

Before trying to generate a parametrisation you need to confirm some things.

1) Why are there two coefficients of z (e and f) ?

2) Why is the coefficient of yz (u) and what is its relationship to uv?

3) What is v ?

[Richard Baker]

Neither, independent study, preparing for college.  He is interested in ray-tracing.  He apologizes-- he wrote the formula wrong and I have a hearing loss.  This is by telephone. ax² + bx + cy² + dy + ez² + fz + gxy + hxz + iyz + j =0

  x,y,z are all functions of u and v  

He thanks you.  So do I.

 

 

 

[Sensei]

1 hour ago, Richard Baker said:

He is interested in ray-tracing.

Typical way ray-tracing is implemented is divide geometry to triangles (at startup) and find intersection between each triangle and ray (origin and direction vectors).

[Richard Baker]

I don't know how relevant the current area of exploration is to this stated interest.  But I will relay this.  Difficult, but no-one where he is has can answer these questions.  And neither can I.  So this is the tenth-best solution.

[studiot]

12 hours ago, Richard Baker said:

I don't know how relevant the current area of exploration is to this stated interest.  But I will relay this.  Difficult, but no-one where he is has can answer these questions.  And neither can I.  So this is the tenth-best solution.

It's very complicated because of the cross products (xy, xz, yz) in the general equation.

Surfaces have a general equation f(x,y,z) = 0

The equation you refer to is called a quadric and the surfaces are called quadric surfaces.

Obtaining a parameterisation means obtaining three equations

x = f₁ (u,v)

y = f₂(u,v)

z = f₃(u,v)

So that each RHS only contains the parameters u and v and each LHS only contains one of the coordinates.

For 3 dimensions there are many more possibilities than for 2 so the surfaces are often split into further sub categories, each with its own parametrisation.
As with 2D to eliminate the efects of the cross products you would normally perform a coordinate transformation first.

These are purely analytical methods,

I was preparing a detailed explanation, but this often takes me some time so here are five references to be going on with.

Sensei mentioned computer methods which work on a different plan and the last reference is about that.

https://opentextbc.ca/calculusv3openstax/chapter/quadric-surfaces/

https://www.win.tue.nl/~sterk/Bouwkunde/hoofdstuk3.pdf

https://tutorial.math.lamar.edu/classes/calciii/parametricsurfaces.aspx

https://www.researchgate.net/publication/241165348_A_Simple_Method_for_the_Parameterization_of_Quadric_Surfaces

http://staff.ustc.edu.cn/~dengjs/files/papers/47 mu23.pdf

 

What sort of college course has this stuff as a pre requisite ?

 

 

Edited July 15, 2021 by studiot

[Richard Baker]

Lol-- check your inbox.  He has a mini-computer in the form of a TI calculator.  Thank you.  Will copy and paste this.  We may return.  

Edited July 15, 2021 by Richard Baker

[Richard Baker]

I will copy and mail what is in those links.  Thank you.  

While he is waiting, he has another question: I have nine points in 3d space and would like to find with the same formula previously discussed the coefficients a-j.  I applied Kramer's rule but determinant equals 0.  Am I missing something? 

[studiot]

3 hours ago, Richard Baker said:

I will copy and mail what is in those links.  Thank you.  

While he is waiting, he has another question: I have nine points in 3d space and would like to find with the same formula previously discussed the coefficients a-j.  I applied Kramer's rule but determinant equals 0.  Am I missing something? 

In order that a quadric through 9 given points may be determinate:-

1)  No four of the nine points may lie on one straight line

2) No more than six may lie in one plane

3) No more than 5 may lie on one conic.

[Richard Baker]

If I added cubic terms and made it a cubic surface, would there be the same limitations?

If I added cubic terms and try to match it up with sixteen points, would there be the same limitations?

 

Edited July 16, 2021 by Richard Baker

[studiot]

12 minutes ago, Richard Baker said:

If I added cubic terms and made it a cubic surface, would there be the same limitations?

If I added cubic terms and try to match it up with sixteen points, would there be the same limitations?

 

Have you a particular example in mind ?

[Richard Baker]

Suppose I had a triangular mesh that formed a Tyrannosaurus Rex, for example.  Would it be possible to divide that model into quadric surfaces or cubic surfaces?  I am concerned about the problem of the double hyperboloid or analogous surface of the third degree.  That would split up the model and make the t. rex disconnected. 

[studiot]

18 minutes ago, Richard Baker said:

Suppose I had a triangular mesh that formed a Tyrannosaurus Rex, for example.  Would it be possible to divide that model into quadric surfaces or cubic surfaces?  I am concerned about the problem of the double hyperboloid or analogous surface of the third degree.  That would split up the model and make the t. rex disconnected. 

Start here for mesh modelling of real world solid objects. (if a T Rex is real world 🙂)

https://en.wikipedia.org/wiki/Coons_patch

[studiot]

Since we are leaning towards (computer) applications, rather the the pure Math of Analytical or Algebraic Geometry i would recommend the any/all of the following books.

The first three are particularly applicable to CAD, meshes and so on,

The last (Cundy) is a fun book about the practical side of making physical models of Mathematical objects such as polyhedra, the cone that runs uphill and many more.

[Richard Baker]

Those look good.  I will check out prices and send him a late birthday present.  Thank you for the suggestions.

[Richard Baker]

Thank you all.  One last question.  Relating to the case of the hyperbolic paraboloid.  I have a paraboloid in the form of a*x²-b*y²= c*z. I have a very messy parametrization for the hyperbolic paraboloid that has many if...then conditions.  Is there a simpler way to do parameterizations in terms of u and v?  

[joigus]

On 7/24/2021 at 1:14 AM, Richard Baker said:

Thank you all.  One last question.  Relating to the case of the hyperbolic paraboloid.  I have a paraboloid in the form of a*x²-b*y²= c*z. I have a very messy parametrization for the hyperbolic paraboloid that has many if...then conditions.  Is there a simpler way to do parameterizations in terms of u and v?  

Hyperbolic functions. It's done in relativity all the time. Whistle if you need more help.  

Clues: Rescale x and y; then assign hyperbolic cosine (cosh) and sine (sinh). z goes along for the ride. 

[Richard Baker]

Suppose I have that hyperbolic paraboloid rotated in space so the saddle is pointed to the right or left or any other direction.  I would have to parameterize that in terms of hyperbolic sine and hyperbolic cosine, correct?  How would I go about doing that?  I worked out the implicit expression by rotating the coordinate system but whenever I use hyperbolic sine or cosine in my parametrization it only captures the upper half of the saddle.  (he had to leave the phone, says you can figure out the rest of the question.)