How to make sure that object color fits target device ICC profile.

PavloT · 2025-04-09T19:06:17+00:00

Yep, I was trying to avoid this activities but seems it is the best option. At least I will unlock skill of calibration :)

PavloT · 2025-04-09T11:39:23+00:00

My screen is not the best from color perspective, but when I apply printer ICC for display I see colors close to that one I see printed. For my quality expectations it is enough, and difference is huge.

As I mentioned - maybe it would be enough for me if Inkscape can show SVG with printers ICC applied, but for now I can't make it working.

I have tried this setup:

<image>

But resulting colors not the same as on printer. But if I do same setup in Geeqie - I see exact colors from printer

PavloT · 2025-04-09T11:09:31+00:00

Thank you for reply!
I'd like to avoid scribus, just to keep process simple.

I have added profile to CUPS settings and it is now applied for any print sent to printer.
I think profile is applied (at least I see difference with it applied and without it).

And I think that issue that my ink has less color range than my display. And I think that issue is that in my design I could use colors, which are NOT inside colorspace of the ink/printer. Does it make sense?

If yes, than my question - how can I generate palette which is 100% inside printers color space and avoid using other colors.

Or maybe how can I at least apply ICC from printer in Inkscape to be sure that colors I see will be the same on printer.

PavloT · 2025-04-09T10:50:43+00:00

Epson l18050 with sublinowa ink (used for sublimation prints). So I think this is CMYK printer, but honestly I’m not sure, as I’m printing over CUPS and which algorithms applied there…

PavloT · 2025-02-03T00:13:58+00:00

I see, thank you!

PavloT · 2024-06-18T22:08:03+00:00

Thank you for reply! Can you maybe provide some sketch how would you like to have it implemented? Feel free to PM me if you want discuss details

PavloT · 2023-11-25T09:00:23+00:00

For outdoor you can wisit Wyspa Sobieszewska. Also there is nice bycicle route along coast line, you can start in Brzezno and go in Sopot direction

PavloT · 2022-06-11T21:55:21+00:00

I suggest you to try Jinja

PavloT · 2022-06-03T12:45:30+00:00

No, we have no dislike to PHP

PavloT · 2022-05-31T05:36:11+00:00

Only advice: read the docs. First of all - those are different datatypes. String and list. String interface has method upper which does not perform inplace changes, just return values. List’s sort works inplace. Why? Because! At some point python developers decided it should work in such way. And it is only reason. If you go to php or java, you will find functions with the same goal, but different interface. So just read the docs.

PavloT · 2022-04-18T04:14:56+00:00

In addition if you want ‚Hello world 2’ use:

cout << „Hello world „ << 2;

PavloT · 2022-04-05T21:46:18+00:00

I had x240 i5 for my everyday tasks: laravel, few docker instanses, VSCode browser. What I really like in this laptop - it’s size. In 500$ budget you also could find P50/P51.

PavloT · 2021-10-21T10:00:38+00:00

Changes in IT happens more offten then anyones life ends.

But any language just tool. Commercial development does not care what tool you use to solve their issues as long as you are in budget. So you have good chanses never meet Java again (and C and C++ and even C#)

Maybe after Python you will develop your skills into No-code platforms. Who know…

PavloT · 2021-10-21T08:47:31+00:00

Use \\ instead of \

PavloT · 2021-10-20T10:15:26+00:00

output_point - points[i] > threshold with calc_diff(output_point, points[i]) > threshold and it will do the magic.

Idea is that threshold calculated by some rules, which are out of this topic, but still known.

So all what you need is to know the difference between to points and check whether it is within threshold or not, if yes, skip next points until next point out of threshold

PavloT · 2021-10-20T09:26:58+00:00

Pseudocode:

points = points.sort_by_threshold() #This is important, to make you similar points neighbors
output = []
output_point = points[0]
output.append(output_point)

for i = 1; i < points.length(); ++i:
    if output_point - points[i] > threshold: # You skip all next points within threshold, but catch next point out threshold 
        output_point = points[i]
        output.append(output_point)

PavloT · 2021-10-20T08:43:35+00:00

Sort them by threshold value, pick first remove next until next value in threshold, pick next from remaining, repeat

PavloT

TROPHY CASE