Comments on: Noe’s tutorial on deforming 3D geometry using RBFs

By: D SURESH BABU

D SURESH BABU — Wed, 15 Mar 2017 09:46:53 +0000

Hi,

It is excellent work. RBF interpolator is giving the same displacement vector value. So all depenedent points are moving to same magnitude. Am I doing nay mistake.

By: Nina

Nina — Tue, 11 Jun 2013 23:52:35 +0000

Thank you!

Also can you please explain what are “t” and “k” in the shifted Log Function?

By: admin

admin — Mon, 05 Nov 2012 11:53:53 +0000

In reply to Nina. To Nina The work we used it for was this: Surface membrane based bladder registration for evaluation of accumulated dose during brachytherapy in cervical cancer. K. Ø. Noe, K. Tanderup, T.S. Sørensen. IEEE International Symposium on Biomedical Imaging (ISBI) 2011. We adopted the RBF method from here: G. K. Matsopoulos, N. A. Mouravliansky, P. A. Asvestas, K. K. Delibasis, and V. Kouloulias. Thoracic non-rigid registration combining selforganizing maps and radial basis functions. Medical Image Analysis, 9(3):237 – 254, 2005.

By: Nina

Nina — Thu, 25 Oct 2012 01:00:33 +0000

Thanks for the great work!

I was wondering if you can post the references of the work too.

thanks

By: Damien

Damien — Thu, 09 Jun 2011 02:17:05 +0000

Thanks for the tutorial. I would’ve had trouble applying it if it wasn’t this detailed! I’ve been having similar problems and didn’t know how to resolve it until I came across your post. Kudos!

By: David

David — Mon, 23 May 2011 05:33:04 +0000

Thanks for this tutorial. This really helped a lot!

By: admin

admin — Fri, 29 Oct 2010 11:32:55 +0000

In reply to Minh Nguyen. Minh Nguyen: The size of the matrix depends on the number of control points and not the number of vertices to displace. So if you think of the control points you are right, but not if you are thinking about the vertices in the model to deform.

By: Minh Nguyen

Minh Nguyen — Tue, 14 Sep 2010 15:40:22 +0000

The code is very interesting. However, I doubt that this is practical for relatively large models (wouldn’t say too big) with 100Ks of vertices. Because your matrix G is a dense matrix, most computers won’t even have enough memory to allocate 100k x 100k x sizeof(float) just for the matrix G itself.

By: Belayouni

Belayouni — Fri, 02 Jul 2010 15:22:19 +0000

Thanks for this simple and well explained tutorial.
Actually I’m trying to apply this algorithm to deform a small Cartesian grid on MatLab. But since the gii terms of the radial function are zeros and the G matrix is not invertible (det(G)=0) because sqrt(log(0+k^2))=0 k=1).
I wanted to know how did you deal with this? 🙂

By: yaoyansi

yaoyansi — Mon, 08 Mar 2010 08:02:57 +0000

Thank you very very much for this source code.