Cylinder unwrapping with imtransform
7 views (last 30 days)
Show older comments
Hello everybody, I am trying to make the unwrapping of a photo of a cylindrical object (only one side), I applied this code I wrote:
if true
% code
I = imread('2258.jpg');
axis on
box on
imshow(I)
ndims_in = 2;
ndims_out = 2;
forward_mapping = []
f = @(x, unused) arcsin(1/x);
inverse_mapping = f;
tdata = [];
tform = maketform('custom', ndims_in, ndims_out, ...
forward_mapping, inverse_mapping, tdata);
udata = [-1 1];
vdata = [-1 1];
xdata = [-1.57 1.57];
ydata = [-1 1];
I2 = imtransform(I, tform, 'UData', udata, 'VData', vdata, ...
'XData', xdata, 'YData', ydata);
subplot(1,2,1)
imshow(I)
subplot(1,2,2)
imshow(I2)
end
but I received lots of errors: ??? Error using ==> mldivide Matrix dimensions must agree.
Error in ==> @(x,unused)arcsin(1/x)
Error in ==> maketform>inv_composite at 592 U = feval(t.tdata(i).inverse_fcn, U, t.tdata(i));
Error in ==> tform at 56 X = feval( t.(f.fwd_fcn), U, t );
Error in ==> tforminv at 68 varargout = tform('inv', nargout, varargin{:});
Error in ==> tformarray at 241 M = tforminv(G,T);
Error in ==> imtransform at 275 B = tformarray(args.A, args.tform, args.resampler, tdims_a, tdims_b, ...
Where is the error for you? How could I solve it? Thank you a lot, Phalaen
1 Comment
Accepted Answer
David Young
on 18 Sep 2014
Edited: David Young
on 18 Sep 2014
I haven't tested this, but here's a quick thought: try replacing
f = @(x, unused) arcsin(1/x);
with
f = @(x, unused) asin(1./x);
as you probably want the element-by-element division rather than the matrix inverse here.
[Edit: arcsin replaced with asin to take note of Youssef Khmou's helpful comment]
More Answers (2)
Elisa
on 19 Sep 2014
1 Comment
David Young
on 19 Sep 2014
Hard to know - maybe if you attach the image as suggested by Image Analyst it might be possible to figure it out.
See Also
Products
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!