Photoshop USM sharpening with ImageMagick

[Carlos82]

What resize settings can I apply to get the Photoshop USM sharpening shown in the last picture of this example?
http://www.xs4all.nl/~bvdwolf/main/foto ... ample1.htm

Is antialias a default resize setting when downsizing?

Thanks.

[fmw42]

read these two pages from Anthony's notes about resizing, artifacts and using the resize expert setting to make things sharper

http://www.imagemagick.org/Usage/resize/#artifacts
http://www.imagemagick.org/Usage/resize/#filter

and especially the expert options.

http://www.imagemagick.org/Usage/resize/#filter_options

Also see Anthony's comments in the following post about Photoshop adding an unsharp masking to sharpen the pictures when it resizes.

viewtopic.php?f=1&t=13726&hilit=+photoshop

See also IM unsharp at http://www.imagemagick.org/script/comma ... hp#unsharp

[anthony]

Photo shop applied the resize operation, and then performs an unsharp operation to try to make the edges a little sharper.

I have not looked at, or experimented with unsharp so their is little in IM Examples. I have not found any other useful guides or tutorials on unsharp either, but haven't looked hard. If you find something, let me know.

See the raw notes Fred sent me in
http://www.imagemagick.org/Usage/convolve/#unsharp
(will appear in a couple of days)

[Carlos82]

There is a lot of info. But it would be very useful to know exactly what amount of Photoshop radius, amount and thresold applied the guy who wrote that article, and if he tried to reproduce with ImageMagick. Simply to avoid testing.

I've read that 0.2 or 0.3 is recommended, but in that image it seems more 0.6 - 1.0 radius.

Well, in the comments it is said that sigma equals radius:
http://redskiesatnight.com/2005/04/06/s ... ge-magick/

Regards.

[anthony]

Well, in the comments it is said that sigma equals radius:
http://redskiesatnight.com/2005/04/06/s ... ge-magick/

Ideally radius is about 2 times sigma. It controls how big the 'convolution' is and thus the
overall speed. If your set radius to 0x{sigma} IM makes a internal choice for the radius based on the sigma.
Actually it does not matter much for a small sigma. Regardless the radius will then be about '1' (rounded up integer) as it is only used to determine the 'neighbourhood'.

See IM Examples on blur.
http://www.imagemagick.org/Usage/convolve/#blur
Unsharp is actually subtraction of the 'blur' of the image, so blur info is VERY relevant for its parameters.

[fmw42]

I believe that if you use radiusxsigma=0xsigma, then the effective radius will be about 3xsigma. See http://www.imagemagick.org/script/comma ... s.php#blur

"As a guideline, set r to approximately 3σ. If a radius of 0 is specified, ImageMagick selects a suitable radius for you."

Also see http://en.wikipedia.org/wiki/Gaussian_blur

"In theory, the Gaussian function at every point on the image will be non-zero, meaning that the entire image would need to be included in the calculations for each pixel. In practice, when computing a discrete approximation of the Gaussian function, pixels at a distance of more than 3σ are small enough to be considered effectively zero."

If you use 0xsigma, you get a Gaussian roll-off or profile for the filter (and a circular region of effect)

However, if you use radiusx65000 (or some very large number compared to radius), you get a linear roll-off or profile (and a square region of effect) And the size of the filter is just radius.

[anthony]

I believe that if you use radiusxsigma=0xsigma, then the effective radius will be about 3xsigma. See http://www.imagemagick.org/script/comma ... s.php#blur

"As a guideline, set r to approximately 3σ. If a radius of 0 is specified, ImageMagick selects a suitable radius for you."

As a FYI, if radius is defined the kernel width is set to 2*ceil(radius)+1
If sigma very very small, width is set to 1 (usually no effect on result)

Otherwise the width calculated to be at least 5 (radius 3), which is then increased until overall sum of the bell curve (its weighted average) falls below numerical accuracy from the previous one. That is the weighted_average_sum*QuantumRange is less than 1.

In other words a Q8 ImageMagick (QuantumRange=255) will have a smaller blur kernel size than a Q16 version (QuantumRange=65355), as it does not have the precision needed for a Q16 kernel size. Of course Q16 kernel is also much smaller than a Q32 or Q64. HDRI does not effect this result as it is the Q level, used for output limits, that define this precision.

Not this is NOT linear, and could be larger than sigma*3. It is not only different for different IM, Q levels, but also different for a 1D kernel (as in -blur) verses a 2D kernel (as in -gaussian).

As an experiment I recompiled my IM with an added print statement for the final kernel 'width' that Im decided to use, in my Q16 IM
-blur 0x100 needed a 1D kernel size of 667 (radius = 333)
-gaussian 0x100 needed a 2D kernel size 667 as well!!! Even though they are different functions.

Remember while while doubling sigma effectivally doubles the width of the gaussian bell curve, the precise cut of may not be quite so linear. But in all my experiments the width always seems to be the same for 1D or 2D curves even though the functions are slightly different, Hmmm their seems to be a bug in the 2D version

Blur Sigma => radius IM decided on (minimum = radius 3)
1 => 4
2 => 8
3 => 12
5 => 20
10 => 34
20 => 75
50 => 176
100 => 333
1000 => 2558
10000 => 14885

For IM Q16 that is roughly ceil(sigma*3.33) for reasonable sigma's (less than 100)

[anthony]

Hmmm their seems to be a bug in the 2D version

After fixing the normalization in the 2D determination (IM version 6.5.3-3) I get...

-blur 0x100 => radius = 333
-gaussian 0x100 => radius 470

So my bug fix while making IM more correct and accurate also makes it slower.
At least for 2d Gaussian kernals (which is not commonly needed).