new coordinates of a point after image rotation

[manit]

Consider an image of breadth 'b' pixel and length 'l' pixel.
If this image is rotated clockwise by angle theta then coordinates of point (x,y) become (x cos theta - y sin theta , y cos theta + x sin theta).
For an image where each x,y is an integer such that x belongs to {0,1,...,b-1} and y belongs to {0,1,...,l-1}
So pixel x,y becomes ( (l-1) sin theta + x cos theta - y sin theta , y cos theta + x sin theta )

Is above statement true ?

Thanks.

[fmw42]

I think that is only true when rotating about the origin at x=y=0. But in Imagemagick, the original is at the top left corner. So you have to change the x,y values to offset them to the center before rotation, then shift them back to the new origin (top left corner) after the rotation.

[snibgo]

IM has multiple methods of rotating an image. The "most pure" method is "-distort SRT 0,0,n" where n is the degrees of rotation. This rotates about (0,0), which is top-left, eg:

Code: Select all

magick -size 200x200 xc:#8f8 -fill Red -draw "point 50,20" -virtual-pixel Black -distort SRT 0,0,20 +repage pure_rot.png

The red dot moves to the predicted location, (x * cos (theta) - y * sin (theta) , y * cos (theta) + x * sin (theta)).

Another rotation operation is "-rotate n", which rotates around the centre. The same prediction can be used, but first subtract (width/2,height/2) than add it at the end.

[manit]

(Question 1)Can i say that first pixel on top left of image is regarded as 0,0 in image magick . and horizontal axis is x while vertical is y , so bottom right pixel of a 512x512 image will be (511,511)?

(Question 2)Now when I run the command
convert input_image.png -rotate n rotated_image.png
where n is a number denoting degree of rotation in clockwise direction.
Then , rotation takes place about the center of image.
Here is my logic about effect of rotation on a particular point in image - https://ibb.co/dqZr2z .
So , I should calculate x and y after subtracting width/2 and height/2 respectively from pixel coordinates of imagemagick.
Now new_x should be calculated using formula mentioned in image .
The pixel coordinates of point after rotation can be found out by adding width/2 and height/2 to new_x and new_y respectively.

Is that right ?


Thanks.

[snibgo]

Yes, and yes.

But ... although that is true of pixel coordinates (which are always discrete integers), image coordinates are continuous floating-point, eg pixel coordinates are 0 to 511 but image coordinates are between 0.0 and 512.0. "-distort SRT" works in image coordinates, so we should use "-distort SRT 0.5,0.5,20" if we want the top-left pixel to remain in the same position.

[manit]

I get that a pixel in original image defined by an integer won't necessarily map to another integer coordinate .
I think if image is square shape and rotation angle is 90 degrees then i will get integer coordinates .
In other cases i am okay with slight error .
I don't want to skew the image just simple rotation so 'distort' option may not be useful in my case .

[manit]

I saw that rotating by multiple 90 degrees is most simple case which I have verified.
But when image is rotated by any other agnle then rotated image has higher dimension than original one .
In that case , what should be formula for new dimension of image ?
Also , is it that I have to subtract width/2 and height/2 of original image to get x and y respectively then add new_width/2 and new_height/2 to new_x and new_y respectively to get final coordinates ?

Thanks.

[snibgo]

But when image is rotated by any other agnle then rotated image has higher dimension than original one .

For the command I showed, the dimensions don't change. If you are using some other command, please show it.

Also , is it that I have to ...

It depends on your command. You also need to understand the canvas size and offset, eg:

Code: Select all

magick rose: -rotate 30 info:

rose: PNM 86x76 86x76-8-15 8-bit sRGB 0.016u 0:00.046

"rose:" is 70x46 pixels, and the canvas is larger than this, with offsets that are half the increase and negative.

[manit]

I am running following command

convert fp.png -rotate 15 fp-rotated-15degree.png

In this case fp.png is 512x512 (https://ibb.co/f8YvZe) while fp-rotated-15degree.png turns out to be 630x630 (https://ibb.co/dST7fK) .

[snibgo]

The basic formulae are:

new_width = w*cos(ang) + h*sin(ang)
new_height = h*cos(ang) + w*sin(ang)

I think IM rounds these up to the nearest integers ("ceiling"), and adds 2.

[manit]

The basic formulae are:

new_width = w*cos(ang) + h*sin(ang)
new_height = h*cos(ang) + w*sin(ang)

I think IM rounds these up to the nearest integers ("ceiling"), and adds 2.

512 * (cos15 + sin 15) = 627.06

So , I will try to find correspondence for a point in 512x512 image to its 15 degree rotated (clockwise) variant by following steps
Consider example of point 13,366 in 512x512 image.
So its x=13-(512/2)=-243 , y=366-(512/2)=110
new_x (according to https://ibb.co/dqZr2z) = -243 cos 15 + 110 sin 15 = −206.249880827 (note theta in our case is -15 degree)
new_y = 110 cos 15 - (-243) sin 15 = 169.144868852
-206+315=109
169+315=484

so new coordinates of 13,366 is (109,484)
Distance of 13,366 from 256,256 is 266.737698873
Distance of 109,484 from 315,315 is 266.45262243
Makes me think . I am doing right .

Is that okay ?

Is there a quick way to check like making a red dot on white 512x512 canvas then searching for red dot in rotated image ?

Thanks.

[snibgo]

Code: Select all

magick -size 512x512 xc:#8f8 -fill Red -draw "point 13,366" -virtual-pixel Black -rotate -15 +write info: +repage p1.png

Then use a GUI image editor such as Gimp to find the coordinates of the centre of the red dot.

Of course, you could do this with IM, but it's worth looking at the image to check it is as you intended.

[manit]

The basic formulae are:

new_width = w*cos(ang) + h*sin(ang)
new_height = h*cos(ang) + w*sin(ang)

I think IM rounds these up to the nearest integers ("ceiling"), and adds 2.

512 * (cos15 + sin 15) = 627.06

So , I will try to find correspondence for a point in 512x512 image to its 15 degree rotated (clockwise) variant by following steps
Consider example of point 13,366 in 512x512 image.
So its x=13-(512/2)=-243 , y=366-(512/2)=110
new_x (according to https://ibb.co/dqZr2z) = -243 cos 15 + 110 sin 15 = −206.249880827 (note theta in our case is -15 degree)
new_y = 110 cos 15 - (-243) sin 15 = 169.144868852
-206+315=109
169+315=484

so new coordinates of 13,366 is (109,484)
Distance of 13,366 from 256,256 is 266.737698873
Distance of 109,484 from 315,315 is 266.45262243
Makes me think . I am doing right .

Is that okay ?

Is there a quick way to check like making a red dot on white 512x512 canvas then searching for red dot in rotated image ?

Thanks.

The above calculation I posted is wrong.
I figured out the right one from fmw42 advise -

I think that is only true when rotating about the origin at x=y=0. But in Imagemagick, the original is at the top left corner. So you have to change the x,y values to offset them to the center before rotation, then shift them back to the new origin (top left corner) after the rotation.

It should be as follows
Suppose pixel location is (a,b) in original image
then shifting origin from top-left to center makes pixel coordinate (a-h_res/2,v_res/2-b)
Now use formula x_new=x cos theta + y sin theta , y_new= y cos theta - x sin theta
(NOTE - theta is angle of rotation (in degrees) in clockwise direction which is default for rotate option)
Now move origin back to top left . So final coordinates become - (x_new+new_h_res/2 , new_v_res/2 - y_new)

Here h_res = horizontal resolution , v_res = vertical resolution.

Thanks.