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Re: Mapping image into a polar-square coordinate [message #61182] Wed, 09 July 2008 13:00 Go to previous message
Camilo Mejia is currently offline  Camilo Mejia
Messages: 4
Registered: July 2008
Junior Member
On Jul 9, 12:43 pm, "jsch...@gmail.com" <jsch...@gmail.com> wrote:
>> yeah, but i dont know how to extract a rectangular matrix which rows
>> are radius and columns are angles
>
> Paolo's suggestion of bilinear is a good one.
>
> The best thing to do is construct a polar coordinate system and then
> transform that into a rectangular system that is equivalent to your
> pixel indices.
>
> Suppose there is a rectangular coordinate system, centered on the
> middle pixel of your 981 x 981 data. Then if we want to extract the
> annulus which is between 100 and 200 pixels from the center, we could
> do something like this.
>
> -----
>
> image: 981 x 981 (same as your ``data'' array)
> new_image: 4096 x 10
>
> ;; first construct the equivalent polar coordinates
>
> min_r = 100.0
> max_r = 200.0
>
> ;; this is theta = [0, 2*pi)
> new_th = rebin(dindgen(4096) / 4096d * (2d * !dpi), 4096, 10)
>
> ;; this is r = [r_min, r_max]
> new_r = rebin(transpose((max_r - min_r) * dindgen(10) / 9d + min_r),
> 4096, 10)
>
> ;; now convert to rectangular coordinates
> ;; and shift such that the origin lies not at the center
> ;; but at image[0,0]
>
> new_x = new_r * cos(new_th) + 490.0
> new_y = new_r * sin(new_th) + 490.0
>
> ;; new_x and new_y are fractional pixel coordinates
> ;; use bilinear to extract the values
>
> new_img = bilinear(image, new_x ,new_y)
>
> -----
>
> Hope that helps,
> Josiah

Thanks a lot Josiah and Paolo, it works awesome

Camilo
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