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convert rose: -matte -virtual-pixel transparent -verbose \
-distort Perspective "0,0,0,0 0,46,0,46 70,0,60,10 70,46,60,30" \
+verbose null:Code: Select all
Perspective Projection:
-distort PerspectiveProjection \
'1.97143,-2.14114e-17,3.50196e-15,0.328571,
1,2.27881e-14,0.0185714,2.71608e-18'
Perspective Distort, FX Equivelent:
-fx 'ii=i+page.x+0.5; jj=j+page.y+0.5;
rr=-0.009420*ii -0.000000*jj + 1;
xx=(+0.507246*ii +0.000000*jj -0.000000)/rr;
yy=(-0.166667*ii +1.000000*jj -0.000000)/rr;
rr>0 ? p{ xx-page.x-.5, yy-page.x-.5 } : blue'
The numbers of the form ####e-19 may as well be equivalent to 0.0000 for all that it matters.i = ( 1.97143*x +0.000000*y -0.000000 ) / ( 0.0185714*x + 0.000000*y + 1 )
j = ( 0.328571*x +1.000000*y +0.000000 ) / ( 0.0185714*x + 0.000000*y + 1 )
For which I get:Anthony wrote:convert rose: -matte -virtual-pixel transparent -verbose \
-distort Perspective "0,0,0,0 0,46,0,46 70,0,60,10 70,46,60,30" \
+verbose null:
Except that most of those original control points are almost in a line. The error on the original set is enormous. They are not good control points! But then you noted that it should generate a unity matrix, (except for the error rate).fmw42 wrote:It would seem to me that the -fx inverse coefficients should be equivalent to what GIMP produces.
Already changed the output into three rows of 3,3,2 numbers, which you can see on the IM Examples page (when it appears)fmw42 wrote:Also as Anthony is only reporting 8 coefficients in the -distort PerspectiveProjection, I presume the 9th is 1. In order to make it easier to interpret, it might be nice to have it formatted so that one has 3 rows of 3 coefficients: numX, numY, den)
Already changed the output into three rows of 3,3,2 numbers, which you can see on the IM Examples page (when it appears)
Mapping Perspective Distortion
http://www.imagemagick.org/Usage/distor ... ve_mapping
The difference appears to be too big to be accounted for by a 0 to 0.5 change. But I could be wrong. Anthony, do you have a way to regenerate the -verbose information with the old code that did not account for the 0.5 shift?If I want to move and position the 0,0 pixel I would use 0.5,0.5 If I want to move the actual 'corner' of the image (not the corner pixel) I would then use 0,0. As this transform may eventually be used for 'panorama' matching, I wanted to be able to specify coodinates with fractions of a pixel.
This may be the source of the differences between IM and GIMP.
Sorry if I did that then their are too many terms for the FX PerspectiveProjection argument, and the 'Equivelent Command' will not work.fmw42 wrote:If you don't mind, please make it 3,3,3 (including the constant 1 term). Many people may not understand the default of that term is 1 and may assume it is zero, if left off.Already changed the output into three rows of 3,3,2 numbers, which you can see on the IM Examples page (when it appears)
Mapping Perspective Distortion
http://www.imagemagick.org/Usage/distor ... ve_mapping
The old code did it the same way! The difference was I did not document it!The difference appears to be too big to be accounted for by a 0 to 0.5 change. But I could be wrong. Anthony, do you have a way to regenerate the -verbose information with the old code that did not account for the 0.5 shift?If I want to move and position the 0,0 pixel I would use 0.5,0.5 If I want to move the actual 'corner' of the image (not the corner pixel) I would then use 0,0. As this transform may eventually be used for 'panorama' matching, I wanted to be able to specify coodinates with fractions of a pixel.
This may be the source of the differences between IM and GIMP.
- The numbers in red have nothing todo with the numbers from gimp, it was just an example.convert pic1.png -matte -virtual-pixel transparent -distort Perspective "0,0,0,0 0,50,0,50 0,90,0,90 90,90,90,90" pic1_pers.png
Yes, I see that now in your distort page. You assume the constant in the denominator is 1 and you only excect 8 coeffs. Unfortunately, it is a little inconsistent to have one set of coeffs from fx show 9 values and the other show 8, but I understand your issue.Anthony wrote:Sorry if I did that then their are too many terms for the FX PerspectiveProjection argument, and the 'Equivelent Command' will not work.
Unfortunately GIMP does not allow one to supply fractional pixel coordinates. It works only by GUI by dragging the corners, but does give a read out (in the lower left corner of the window) to show you the coordinates.Anthony wrote:If you want to do what GIMP is doing try adjusting the input/output coordinates by 0.5. I am just not exactly certian what GIMP does, as in IM the control points are used ASIS for the mathematical calculations.
Best I can suggest is to try adjusting the corners of the image in GIMP and write down the old and new coordinate values. The old ones are the original image corners and the new ones are the values shown in the bottom left corner of the window as you drag the corner. Too bad GIMP does not list out the control points, but only their version of the coefficients.rick0 wrote:Note: I'm just trying to find a visual and easy way of find the coordinates, and then implement them in IM.
how did you get the numbers ? 1.97,0,0,0.329,1,0,0.0185,0convert rose: -channel rgba -matte -virtual-pixel transparent -mattecolor none -distort perspectiveprojection "1.97,0,0,0.329,1,0,0.0185,0" rose_proj_coeffs.png
Image
THIS WORKS. So I am puzzled why Anthony says it is forward projecting?
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convert rose.png -distort perspectiveprojection ".507,0,0,-.167,1,0,-.00943,0" rose_fx_coeffs.png
The perspective coefficient matrix is not a listing of control points. It is the transformation coefficients for the perspective function which is a rational linear polynomial equation (first order in the numerator and first order in the denominator). See http://www.imagemagick.org/Usage/distor ... projection It will be all zeros with 1s along the diagonal for any original image as it shows that there is no change. The ones means that x stays at x and y stays at y for every point.This is before i drag the corners, (don't know why it show 1, but i guess every colum is a corner + middle)
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how did you get the numbers ? 1.97,0,0,0.329,1,0,0.0185,0 All I did was open the image in GIMP and drag the corners, but I recorded the coordinates that show in the lower left corner before moving the cursor away from the corner. See below:But if you could provide us a working example and how you got there, like you did above that would safe our lifes.
rick0 wrote:convert pic1.png -matte -virtual-pixel transparent -distort Perspective "0,0,0,0 0,50,0,50 0,90,0,90 90,90,90,90" pic1_pers.png
The numbers in red have nothing todo with the numbers from gimp, it was just an example.
Anthony, unfortunately, my example shows that they do not appear to be close enough. But it may be this particular example in which the right side control points are rather close together. Thus adding or not adding 0.5 to each coordinate may make a significant difference in this example. A larger image with less distortion may work out better.In any case the coefficients should be close enough to produce very very simular results. Just take the first 8 values and ignore the final 9th value which is always 1.0. I am not certain however if the vaules from gimp should be read row by row, or column by column without further input.