ray tracing, indices and path length registration

3 visualizaciones (últimos 30 días)
Albert
Albert el 2 de Oct. de 2017
Comentada: Albert el 27 de Jun. de 2018
Dear all
Thank you for reading my following question and I greatly appreciate any of your help.
Problem: Given a grid (N*N) and a non-straight ray with known discrete coordinates, how to register the indices and path length in the pixels crossed by the ray? This is on a plane, what if it is in 3d space? Are there any existing MATLAB script to carry out this function?
Thank you very much.
Sincerely
Albert
  6 comentarios
Alexander Jones
Alexander Jones el 26 de Jun. de 2018
Hello, I am trying to do some microwave tomography, and am in need of a very similar thing - I was wondering if you found a solution to the problem of finding path length, and if so if you could please point me in the right direction.
Albert
Albert el 27 de Jun. de 2018
Hi Jones,
To be clear, are you trying to trace a straight or curve? If it is for straight, please see https://www.mathworks.com/matlabcentral/fileexchange/28496-tomobox?focused=5159748&tab=function
Anyhow you may want to have a look at this document for the theory behind the code, see here (much efficient than calculating the interaction) http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.42.3443&rep=rep1&type=pdf
If you are looking at curve rather than straight ray, as Anton Semechko mentioned, the curve ray may be represented by segments, for each segment you can use the above code to calculate index and path length, but you need truncate the result before you concatenate them together for one ray. Because the above code calculates the index and path length in each voxel for straight line across the whole image reconstruction volume.
The above code can handle 2D and 3D.
Good luck.

Iniciar sesión para comentar.

Respuesta aceptada

Anton Semechko
Anton Semechko el 26 de Jun. de 2018
Editada: Anton Semechko el 26 de Jun. de 2018
For 2D reconstruction, the problem can be approached follows:
1) Let's say you have a set of K rays {Rk}, where Rk is k-th ray parameterized by the equation Rk(t)=Xk+t*Dk, where Xk is the point of origin and Dk is the direction of travel.
2) Given a M-by-M grid (G) sampled on a plane passing through point P with unit normal N, find points of intersection between G and all of the rays {Rk}.
3) Find transformation that aligns N with the z-axis, and horizonal and vertical grid lines of G with the x- and y-axes, respectively.
4) Apply the transformation from (3) to the points of intersection in (2).
5) Bin data from (4) to get the intensity image corresponding to the number of hits in a given bin (i.e., grid cell/pixel).
This process can be easily generalized to get a volumetric image using a 3D grid. This is because 3D grid is just a stack of 2D grids.
  2 comentarios
Albert
Albert el 27 de Jun. de 2018
Thank you very much, Anton Semechko. You are right. Actually I did something similar to what you described in your step 2, but it was relatively slow, especially when the number of rays is very large. Hopefully, we are on the same page, but looks like we are not. I was trying to trace a curve through an image reconstruction volume, not a straight line. For a straight ray, it will be much easier to carry out. It is already done, please see here: https://www.mathworks.com/matlabcentral/fileexchange/28496-tomobox?focused=5159748&tab=function
Anton Semechko
Anton Semechko el 27 de Jun. de 2018
Yeah, I missed the part where you said that the rays aren't straight. In that case, rays can be (i) approximated as piecewise linear curves composed of several straight line segments, (ii) points intersection can be approximated by finding line segments that intersect with the plane containing the grid, (iii) if the rays have analytical representation, points of intersection can be refined using a root-finding algorithm.

Iniciar sesión para comentar.

Más respuestas (0)

Categorías

Más información sobre 3-D Volumetric Image Processing en Help Center y File Exchange.

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by