Hi
This is an offspring of samueljohnchia's thread on Camera Profiling,
http://www.luminous-landscape.com/forum/index.php?topic=75480.0.
I run into some questions which are too basic to fit in that thread. I want to use, at least explore ColorPerfect. The gray scale that the author David Dunthorn offers for calibrating images, contains only values between 13 and 242 RGB. Also the calibration process he advises is rather cumbersome.
http://www.c-f-systems.com/DunthornCalibration.htmlI think a usual ICC profile would be easier. But before that - the most basic (silliest) questions first:
1- When camera output is considered linear, and the calibration can, according to David even should be done on a gray scale alone - why do we need a scale as a target? Isn't it enough with a black point = Lab 0/0/0, a white point = Lab 100/0/0 and a straight line between them (gamma=1) ?
2- The best current sensors are said to cover 14 Exposure Values of dynamic range. My Canon 5D2 may have 10 or so. My wide gamut screen is calibrated to 100 candelas. If I create a synthetic image containing a black and a white patch, and point my spot meter to the screen image, it shows EV 4 1/3 and 9 respectively, that's 4 2/3 EVs of DR. How can I profile 14 EVs of DR with just 4 2/3 of them?
I have run into these questions trying to figure out a procedure to create a gray chart for camera profiling.
Here is the idea:
a-Create a synthetic copy of the CC24 and fill it with gray patches.
b-Calibrate the wide gamut screen to gamma=1, 5000 K
c-Display the chart on that screen, take a screen shot and a camera photo
d-Read the gray values off the screen shot, fill them into the Argyll .cie file for the CC24 and create the profile using the camera photo as the input.
Questions:
3- Would this be a valid procedure?
4- Which luminance (in candelas) should I choose for the screen? The max, or the 100 which is my standard?
5- I would like the gray patches to represent Exposure Values. Middle Gray is Lab 50/0/0 in linear space - how would I calculate the rest?
Thank you for your answers.