0

Is this saying correct ?

Raw camera images store colors in a logarithmic scale. The blacks in these images are nearly 0 and the whites are supposed to be infinity.

mattdm
  • 143,140
  • 52
  • 417
  • 741
max_user
  • 67
  • 1
  • 2
  • Possible duplicate of What is RAW, technically? – Romeo Ninov Dec 28 '15 at 13:50
  • http://photo.stackexchange.com/questions/1455/what-is-raw-technically http://photo.stackexchange.com/questions/2627/good-examples-of-raws-advantages-over-jpeg http://photo.stackexchange.com/questions/5410/why-can-i-adjust-the-white-balance-of-a-raw-file-but-not-a-jpeg-file http://photo.stackexchange.com/questions/15/what-are-the-pros-and-cons-when-shooting-in-raw-vs-jpeg/28#28 – Romeo Ninov Dec 28 '15 at 13:50
  • 1
    Your question makes no sense, RAW files are not images. One can create an image from a RAW file but in itself, it's not an image, it does not store colors but intensities for a specific color-component. – Itai Dec 28 '15 at 16:33
  • The simple way to think of it is that the sensor counts photons, and the RAW file stores these counts (or rather a number proportional to them) directly. – Szabolcs Dec 28 '15 at 16:37
  • 1
    Where did this statement come from? – mattdm Dec 28 '15 at 17:57
  • @Itai: you could make that claim about virtually any image file format. "One can create an image from a JPEG file, but in itself it's not an image, it does not store colors but discrete cosine transforms", etc. – whatsisname Dec 28 '15 at 19:55
  • The statement is from http://heroinewarrior.com/cinelerra/cinelerra.html#LINEARIZE – max_user Dec 28 '15 at 20:00
  • fascinating! they assumed that "normal" images are linear so, logically, raws must be logarithmic? :-O – szulat Dec 28 '15 at 20:31
  • Yeah dude. There are so many strange. – max_user Dec 28 '15 at 20:38
  • @whatsisname - RAW data has to be interpolated to be turned into an image, it does not represent an image until the missing color components are computed and, unlike compression, there is no one way to get an image out of it. That's why we need a RAW converter and the OS previews these files with an embedded JPEG thumbnail. – Itai Dec 28 '15 at 21:32
  • @Itai believe me, JPEG files need so much more processing than interpolation, and on the other hand, storing just single component for a pixel in a typical bayer RAW file might be considered a form of data compression. so... there is no significant difference, just words. – szulat Dec 29 '15 at 00:00
  • 1
    @szulat I think the people want to emphasize a terminology difference because while JPEG files require processing to render, there is only one correct result. For RAW files, external information is required to know which result is wanted (even if hints are included). – mattdm Dec 29 '15 at 19:20
  • @mattdm interesting take! but the raw file still has only one interpretation (derived from the fact it is a direct record of the physical measurement) even if we routinely use various algorithms and parameters to display its contents. what about high dynamic range 16 bit tiff or exr files? are they "images"? there is no single correct way to render them, assuming the usual 8-bit display. we modify raws (but also 16 bit tiffs!) more eagerly than jpegs only because of their higher information density (--> less processing artifacts), not because they are fundamentally different "non-images". – szulat Dec 30 '15 at 00:08

3 Answers3

3

No. The data that comes off a digital camera sensor is linear. Each sensel, or pixel well as they are often called, collects light energy in the form of photons and converts that energy to a voltage which is then measured when the sensor chip is read. If enough light falls on a particular sensel to create the maximum charge that the sensel can produce, additional light will not increase the signal produced by the sensel. With a CCD type image sensor, some of the excess energy may bleed into adjoining sensels and produce what is referred to as blooming. If adjoining sensels that are filtered by the Bayer mask to be sensitive to either red, green, or blue light all receive enough light to saturate them then the properly processed data will produce white for that area on the resulting photo. Increasing the amount of light will not make the white any brighter, since it is already as bright as the system is capable of producing.

Michael C
  • 175,039
  • 10
  • 209
  • 561
  • Thank you so much, Michael ! English is not my native language. Is it possible to interpret this "Raw camera images" as JPG/TIFFs from raw-camera ? JPG/TIFFs from raw-camera store colors in a logarithmic scale (for all I know). But even in this case, this assertion "The blacks in these images are nearly 0 and the whites are supposed to be infinity" is wrong. What do you think about this? – max_user Dec 28 '15 at 12:59
1

No. All raw data I have seen is in linear intensity. A value of 4028 indicates twice as much light as 2014, for example.

Olin Lathrop
  • 17,402
  • 1
  • 32
  • 68
  • Thank you very much ! English is not my native language. Is it possible to interpret this "Raw camera images" as JPG/TIFFs from raw-camera ? JPG/TIFFs from raw-camera store colors in a logarithmic scale (for all I know). – max_user Dec 28 '15 at 13:08
  • @max: JPG are usually post-processes from the raw data by the camera using heuristics and defaults. They usually are somewhat logarithmic, but they aren't raw. Raw data is linear. That's the way the sensors work. – Olin Lathrop Dec 28 '15 at 14:32
0

Well, here is an aditional No.

Raw camera images

Probably "files" is more suitable.

store colors

Colors is a complex term. Colors inslcusive has a phisiological component, so the idea could be "sensor data".

in a logarithmic scale.

The data is linear, that is why it is raw. The sensor indicates one value, the raw file stores that value. Does not process it.

The blacks in these images are nearly 0

I have some thoughts here.

a) It is very difficult to have a pure black. In theory a pure black could be a deep space photo, but we all know that is inclusive relative: http://hubblesite.org/gallery/album/entire/pr2012037a/

So, it is phisically difficult to have a pure black on terrestrial photo (where you have actually light to take a picture). You could try for example photographing a very dark hole on a building, and make a real long exposure and have some more info you did not expected to have before.

What I mean with this is that black is relative.

b) Inclusive you could have a photo taken with the lens cap on on a dark room and the sensor will send some noise.

and the whites are supposed to be infinity.

a) If you have heat on a body you have electromagnethic waves, you have radiant energy, photons, to photograph. (That is the principle on a black body temperature, for us photographers, color temperature).

If you have more heat you have more photons to photograph.

In theory you could have a temperature of about 141,000,000,000,000,000,000,000,000,000,000 °K https://www.youtube.com/watch?v=4fuHzC9aTik

So, you can always have more "white" than you could photograph... before you are converted to Quark–gluon plasma.

b) A 14 bit image is limited to 16384 values. Lets think the next-next generation of cameras have 24bit files, that would have 16 million per channel. You could have a 1000 bit raw file, and that is far(x10∞) from infinity. So you are limited.

So, the answer is No.

Rafael
  • 24,760
  • 1
  • 43
  • 81
  • On the last point, if the data were logarithmic, the highest value could indeed be very high and effectively "infinity". – mattdm Dec 28 '15 at 18:37
  • interestingly, some "raw" files use quasi-logarithmic encoding to compress data (e.g. sony), which also means they are not truly raw - but on the other hand, you could have a nonlinear ADC in your camera, producing logarithmic raw data - and some cctv cameras really do that! – szulat Dec 29 '15 at 00:21