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I'm looking to create image warping effect that looks like this animation, which resembles a misshapen magnifying glass being moved over a background:

enter image description here

Not sure where to start, but here are some related posts:

J. M.'s missing motivation
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M.R.
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    Maybe also related https://mathematica.stackexchange.com/a/234660/72682. The hard part is working out what ImageTransformation you need to apply. – flinty Nov 15 '20 at 15:44
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    Another with distortion: https://mathematica.stackexchange.com/q/15065/363 – Chris Degnen Nov 18 '20 at 13:27
  • This is solely ImageTransformation. The transformation is a simplified form of blackhole effect with two or three moving blackholes. It is hard to reconstruct the trajectories exact. – Steffen Jaeschke Nov 20 '20 at 19:20
  • Try machine learning. Train a neural net with many such images and then ask your question (easier said than done). – Quasar Supernova Nov 21 '20 at 12:28
  • If you have access to the tool that generated this effect, then it may be possible to track randomly coloured pixels via frame-to-frame optical flow. I'd start off with: ImageResize[RandomImage[1, {32, 32}, ColorSpace -> "RGB"], {200, 200}, Resampling -> "Nearest"]. The up-sampling is to mitigate colour shifts when the filter resamples after distortion. It would be very difficult but it's probably the best way to reproduce this exact distortion. – flinty Nov 23 '20 at 17:03

2 Answers2

12

We can use cylindrical functions with ImageTransformation[] to show waves on the water, for example

img=

Figure 0

lst = Table[
   ImageTransformation[
    img, {#[[
        1]] (1 + 
         Sin[Pi x/40] BesselJ[1, 
           20 Sin[Pi x/
               40] Sqrt[(#[[1]] - .5)^2 + (#[[2]] - .5)^2]]), #[[
        2]] (1 + 
         Sin[Pi x/40] BesselJ[1, 
           15 Sin[Pi x/
               40] Sqrt[(#[[1]] - .5)^2 + (#[[2]] - .5)^2]])} &,  
    Padding -> "Periodic"], {x, 0, 40, 1/2}];

Figure 1

And with some azimuthal waves

lst = Table[
   ImageTransformation[
    img, {#[[
         1]] (1 + 
          Sin[Pi x/40] Cos[4 Pi ArcTan[#[[1]], 1 - #[[2]]]] BesselJ[1,
             20 Sin[
              Pi x/40] Sqrt[(#[[1]] - .5)^2 + (1 - #[[2]])^2]]), #[[
         2]] (1 + 
          Sin[Pi x/40] Sin[4 Pi ArcTan[#[[1]], 1 - #[[2]]]] BesselJ[1,
             20 Sin[
              Pi x/40] Sqrt[(#[[1]] - .5)^2 + (1 - #[[2]])^2]])} (1 + 
        x/20 Sin[Pi x/40]) &, Padding -> "Periodic"], {x, 0, 40, 
    1/2}];

Figure 2

Alex Trounev
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Here is the beginning of the transformation in slow motion. We see that the top right and bottom left quadrants are getting heavily distorted in a symetrical way, while the other two quadrants are tilting.

(The gif below was obtained by saving the original gif, then use Import to import the file as a list of 181 images, then Export the first 40 or so of these images as a new gif with the option DiplayDurations->0.6).

enter image description here

Does this help figure out how to do it? It is likely a clue, but I don't know what it means.

The following transformation shows some potential.

list = Table[
  ImageTransformation[img, Sin[x Pi #] Sin[ Pi Reverse[#]] &], {x, 
   0.3, 0.70, 0.01}]

Here is the list as a gif.

enter image description here

Of course there are endless possible variations:

list = Table[
  ImageTransformation[img, 
   0.70 # + 0.1  Sin[x Pi #] Sin[Pi Reverse[#]] + 0.1 &], {x, 0.3, 
   2.70, 0.01}]

enter image description here

Jean-Pierre
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  • Could you add some code to your answer? – Alex Trounev Nov 20 '20 at 10:31
  • This is really a question, not an answer. Does the fact that we have quadrants affected differently provide a clue as to the kind of function used in the transformation? The second image is just an example of a failed, but entertaining attempt. I did not keep the code. – Jean-Pierre Nov 20 '20 at 12:31
  • Maybe do background subtraction first? Also please post code for this we can ping pong ideas in that way... – M.R. Nov 20 '20 at 19:22