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I'm cropping a rotated image, but I want to keep the ascect ratio fixed and not have any "extra" black pixels. Here's an example:

image = ExampleData[{"TestImage", "House"}];
rot = ImageRotate[image, 5 Degree]
ImageCrop[rot, {250, 200}]

enter image description here

But this isn't a good solution because it specifies the exact crop and I'm not sure what that will be generally. So really, what I want is the take a big of a crop as possible, preserving some given aspect ratio (which by default would be the same as the original image's asepect).

I would manually do it like this:

enter image description here

Details

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  • $\begingroup$ How do you feel about computing it from the morphological components? How fast does this need to be? One other option is to first pad with a very specific random color then find the bounding pixels with that color from the image data. $\endgroup$ – b3m2a1 Sep 18 '17 at 20:50
  • $\begingroup$ @b3m2a1 We should be able to compute it with basic geometry no? $\endgroup$ – M.R. Sep 18 '17 at 20:50
  • $\begingroup$ Oh if you know the rotation angle, sure. Take the corner points of a rectangle, rotate those, take the second from smallest left, second from largest right, or something. I was assuming you had randomly rotated images. $\endgroup$ – b3m2a1 Sep 18 '17 at 20:51
  • $\begingroup$ @b3m2a1 I don't think its that easy. See stackoverflow.com/questions/16702966/… $\endgroup$ – M.R. Sep 18 '17 at 20:55
  • $\begingroup$ Oh I missed the AspectRatio restriction. I see the complication now. $\endgroup$ – b3m2a1 Sep 18 '17 at 21:09
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Simply use the option "SameRatioCropping" in ImageRotate : 

image = ExampleData[{"TestImage", "House"}];
ImageRotate[image, 5 Degree,"SameRatioCropping"]

enter image description here

"SameRatioCropping" is documented. There is also a good explanation in this Almsick's presentation, at time 10 minutes

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