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How can i blend multiple images that has same scene except one object on different positions in every image?

asked 2013-03-28 17:30:59 -0500

Dirhem gravatar image

I want to blend multiple photo shots of same scene but only one object is in different position on every shot. I want to know what kind of algorithm would give desired results. Here is an example

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answered 2013-03-29 05:17:47 -0500

The images seem already aligned (the camera didn't move?). I will do the following:

  1. subtract all images => you get a mask of nothing is moving
  2. get the bounding box of all moving part (where user is)
  3. for each bounding box: there is one image with the user and two images without. Copy the bounding box of the user (it's the image where subtraction is not "mostly" zero.
  4. Invert your mask and copy any of the three image to the final image => aka copy the "background".

I have no prior of the quality attended... but try if there is no illumination changes on your scene, it should works.

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Indeed, aligning would not be necessary if the actual position of the camera is constant :)

StevenPuttemans gravatar imageStevenPuttemans ( 2013-03-29 06:01:54 -0500 )edit

Actually this code will eventually be used on android phone so photos won't be perfectly aligned and i beleive there will be some exposure differences also.

Dirhem gravatar imageDirhem ( 2013-03-29 16:46:52 -0500 )edit
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Then you need matching for aligning the images, and eventually using a basic histogram normalization to get a more same level of lighting exposure.

StevenPuttemans gravatar imageStevenPuttemans ( 2013-03-30 08:01:43 -0500 )edit
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answered 2013-03-29 04:32:49 -0500

What you should do is trying to align the pictures based on feature points. What I would do is

  1. Use a manual step to select a bounding box of the tree in all images
  2. Calculate feature points in each bounding box (SURF, SIFT, ORB, ...)
  3. Align the picture by finding matching feature points
  4. Combine images based on this points --> merging them with an opacity of 50% or so.

However, I think that in order to get a great result, you will need to find more robust implementations. But this would give you a start.

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Asked: 2013-03-28 17:30:59 -0500

Seen: 1,073 times

Last updated: Mar 29 '13