Projection of an Image
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If I know the four corners of a rectangle that has been projected, how can I obtain a top-down view of that image using image projection. I tried to use fitgeotrans and imwarp but did not obtain the correct results. Below is the code I used:
U = double([topLeft; topRight; bottomRight; bottomLeft]);
topLeftNew = [1 1];
topRightNew = [1 width];
bottomLeftNew = [height 1];
bottomRightNew = [height width];
X = double([topLeftNew; topRightNew; bottomRightNew; bottomLeftNew]);
tform = fitgeotrans(U, X, 'projective');
B = imwarp(orig, tform);
Does anyone know what I could do to fix this or a different method? The image I'm trying to transform is:

The four corners I've obtained are shown as well. The result of the transformation is.

I'd like for the result to only contain the notecard and orient it correctly upright as well. Is this possible and how can I go about achieving this?
回答 (2 件)
Ben Drebing
2017 年 12 月 21 日
I think that the orderings of your fixedpoints and movingpoints do not match up. Try
topLeft = [360, 546];
topRight = [609, 109];
botRight = [873, 240];
botLeft = [645, 703];
U = [topLeft; topRight; botRight; botLeft];
width = size(I,2);
height = size(I,1);
topLeftNew = [1 1];
topRightNew = [1 width];
bottomLeftNew = [height 1];
bottomRightNew = [height width];
X = double([topLeftNew;bottomLeftNew; bottomRightNew; topRightNew ]);
tform = fitgeotrans(U, X, 'projective');
B = imwarp(I, tform);
imshow(B,[]);
1 件のコメント
Masoud Heidari
2021 年 3 月 5 日
Thank you very much Ben! it was a big help for me too.
Image Analyst
2017 年 12 月 21 日
0 投票
See attached demo.

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