CGAL users discussion list

[yanyajie <>]

Hi, Marc,

 

Thank you for replying! I’ll check your suggestion. Hope it solves

my problem.

 

\nabla is what you type in latex to print out the operator

for computing gradient, which is an upside-down triangle.

 

 

Yajie.

 

 

 

From:

Sent: Monday, September 12, 2011 5:50 PM

To:

Subject: Re: [cgal-discuss] eigen-decompose the gradient of normal field

 

Hi,

 

I dont understand your formula what is \nabla{n(x)}, ?

anyway, I dont see how you can get rid of the 1st and 2nd fundamental forms here and hence the weingarten operator.  A way to get around it and obtain "closed" formula is given there:

www.geometrictools.com/Documentation/PrincipalCurvature.pdf

(you may want to try Spivak or Porteous for proofs)

Best

Marc

On Sep 12, 2011, at 10:00 AM, yanyajie wrote:

Hi,

 

I try to locate ridge lines on point set surface. Now,

I have fitted an implicit surface F(x)=0, x=[x,y,z]’, to a point model (or, a point set).

According to Ohtake’s paper, “Ridge-valley lines on meshes via implicit surface fitting”,

one can obtain the normal n=n_x, n_y, n_z]’of a point on F using equation:

                                        n(x)=\nabla{F} / |\nabla{F(x)}|,

i.e., the normalized gradient vector. Then, the two non-zero

eigenvalues of 3x3 matrix \nabla{n(x)}, and the corresponding eigenvectors,

are the two principal curvatures and corresponding principal directions of x.

 

The results of normal computation are good. However, the eigen-decomposition of

the gradient of normal field gives me one real and two conjugate complex eigen-values

at many points. How should I deal with this situation? Does that mean these points

are umbilical points whose principal directions are undefined?

 

Has anyone read this paper and implemented the method described therein? Is my

understanding in terms of finding the principal curvatures correct? I have also looked

at the “approximating differential quantities” chapter in CGAL’s mannual, which estimates

the Weingarten Map. I think Weingarten Map is somehow equal to the gradient of normal

here. So, the method I am using here must make some sense.

 

Thanks in advance!

 

Yajie.