Thanks for the reply.
I dont have that tool box.
please guide / suggest some code example (s ).
regards,
from
Image Vision
"Huibao_Lin" <hlin@mathworks.com> wrote in message <jumbk1$sj4$1@newscl01ah.mathworks.com>...
> Hi,
>
> You can use the Computer Vision System Toolbox to solve these problems
> nicely.
>
> 1. You can use the "epipolarLine" function to compute the epipolar line. For
> example, the epipolar line in image 2 corresponding to a point, pt1, in
> image 1 is
> line2 = epipolarLine(F, pt1)
>
> 2. I guess your goal is to find the a point in image 2, pt2, which
> corresponds to another point, pt1, in image 1. We knew that pt2 will lie on
> the eipolar line, line2, for sure. We could test every location on line2 to
> find the best match for pt1. The matching score of a location and pt1 can be
> computed by comparing some features extracted from their surrounding
> regions. Because the epipolar line can have arbitrary orientation and the
> two images may have different scales, reliable features are difficult to
> find.
>
> A better solution is to rectify the images firstly. Then the corresponding
> points will have the same row position and the two images will have the same
> vertical scale. You can use the "estimateUncalibratedRectification" function
> to compute the rectification transformations. Please see the example of this
> function for the use of it. Then you can use simple block matching along the
> row direction to search the corresponding point. You can also use the
> "disparity" function to compute the disparity, which returns the location of
> all corresponding points in the image.
>
> Thanks,
>
> Huibao
>
> "imagevision " wrote in message
> news:julvf7$a5o$1@newscl01ah.mathworks.com...
>
> Hi,
>
> can any body suggest me some matlab code for the following problem:
>
> I have matrix F (transformation matrix), and two image views.
> (i)
> I want to find out epipolar line on image 2 corresponding to a point in
> image 1.
> (ii)
> by using a point in image 1 and matrix F, how i can find coordinate of a
> point in image 2
>
> thanks in advance for your help.
> regards,
> from
> Image Vision
