I have heard about 18% gray tone — what is it really, and why 18% (and not 20% or some other value), and how can I make it in Photoshop?

  • \$\begingroup\$ possible duplicate of DIY Gray card for exposure/color correction. \$\endgroup\$
    – chills42
    Commented Jul 22, 2010 at 17:46
  • \$\begingroup\$ I think it could be worth having on its own as the "18% grey is how light meters are calibrated" is a very persistent myth, and it'd be great to get a proper answer up. \$\endgroup\$
    – ex-ms
    Commented Jul 24, 2010 at 17:45
  • \$\begingroup\$ @matt - where did your comment with bythom link go? \$\endgroup\$
    – Karel
    Commented Jul 26, 2010 at 18:00
  • \$\begingroup\$ @Karel - I made an error that made the comment quite misleading & didn't notice in time to edit it. Thought it best to delete it & add a longer answer (thanks for including it in yours, too). \$\endgroup\$
    – ex-ms
    Commented Jul 26, 2010 at 19:11
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    \$\begingroup\$ @ex-ms: Minolta, Sekonic, Pentax and Gossen work to 18% grey while Canon and Nikon work to 12%. I have been unable to determine the figure for Olympus. So to call 18% grey a myth is more than a little far fetched. \$\endgroup\$
    – labnut
    Commented Mar 26, 2011 at 17:52

10 Answers 10


Warning: this is a long, somewhat technical post that includes some math (but when you get past the superscripts and such, it's ultimately pretty simple math).

First of all, I should start with a simple idea of how I believe 18% was selected in the first place. I can't remember which one any more, but one of Ansel Adams' books mentions what I think is probably the origin.

About the most reflective naturally occurring substance on earth is fresh, clean snow, which reflects somewhere around 95% of the light falling on it (depending a little on exactly how fresh, how clean, how cold and/or moist it was when the snow formed, etc.)

At the opposite extreme, a surface covered in fresh, clean soot reflects about the least light of any naturally occurring substance. The range here is from about 3 to 4%. Let's again take the middle of that range, and call it 3.5%.

To can an overall average, we can then average those two. However, given such a wide range, the statisticians tell use an arithmetic mean produces a poor result (the larger number dominates almost completely, and the smaller one is nearly ignored). For numbers like this, a geometric mean is the "correct" way to do things.

The geometric mean of these works out as the square root of .95 * .035. Running that through the calculator, we get 0.1823458... Rounded to two places that's 18%.

Since the Thom Hogan article has been cited, I'll talk a bit about it. Some time ago, Thom Hogan published an article:


...that claims meters in Nikon digital cameras are calibrated for a mid-level grey that corresponds to 12% reflectance rather than the 18% grey of most standard grey cards.

Unfortunately, while the title and opening paragraph of the article are quite emphatic about 18% being a “myth”, the remainder of the article fails to provide much factual basis for this claim. Here’s what Thom gives as the basis for his statements:

ANSI standards (which, unfortunately, are not publically published--you have to pay big bucks to have access to them), calibrate meters using luminance, not reflection. For an ANSI calibrated meter, the most commonly published information I've seen is that the luminance value used translates into a reflectance of 12%. I've also seen 12.5% and 13% (so where the heck does Sekonic's 14% come from?), but 12% seems to be correct--one half stop lighter than 18%, by the way. I haven't seen anyone claim that ANSI calibration translates into a reflectance of 18%.

In the end, he seems to have no real basis for his claims, merely a statement that “12% seems to be correct,” with no real evidence, or even information about why he considers this correct. Despite this, however, this article is now widely cited on various photographically oriented web sites (among other places) as if it were absolute and indisputable fact.

Since this issue seems to be of interest to a fair number of photographers, I decided to see if I could find some real facts with evidence to support them. The first step in this journey was to find the standard in question. Doing some searching, I found what I believe is the relevant standard. Contrary to Thom’s implication above, this is really published by the ISO rather than ANSI. This may be trivial to most, but when I was looking for the standard it was somewhat important – I put in a fair amount of work trying to find an ANSI standard that apparently does not exist. In the end, however, I found the relevant ISO standard: ISO 2720-1974, “Photography - General purpose photographic exposure meters (photoelectric type) - Guide to product specification (First edition - 1974-08-15)”.

I also found that Thom was (at least from my viewpoint) quite mistaken about prices as well – a copy of this standard costs only $65 US. This didn't strike me as "big bucks" -- in fact, it seemed like a fair price to pay for some real enlightenment (pun noted by not really intended) on the subject.

The standard confirmed part of what Thom had to say, such as calibrating meters directly from sources that emit light rather than from reflected light. Unfortunately, other parts of what Thom had to say are not quite so closely aligned with the content of the standard. For example, at the conclusion of his article, he includes a comment from “lance” that mentioned a "'K' factor", without specifying its exact meaning or purpose. Thom replied by saying: “No manufacturer I've talked to knows anything about a K factor, though, and they all speak specifically about the ANSI standard as their criteria for building and testing meters.”

As stated, this may not be exactly wrong – but it’s certainly misleading at best. In reality, a large part of the ISO standard is devoted to the K factor. Much of the rest is devoted to the C factor, which corresponds to the K factor, but is used for incident light meters instead (the K factor applies only to reflected light meters). It would be utterly impossible to follow the standard (at least with respect to a reflected light meter) without knowing (quite a lot) about the K factor.

The standard specifies that: “The constants K and C shall be chosen by statistical analysis of the results of a large number of tests carried out to determine the acceptability to a number of observers, of a number of Photographs, for which the exposure was known, obtained under various conditions of subject matter and over a range of luminances.”

The standard also specifies a range within which the K factor must fall. The numbers for the range depend on the method used for measuring/rating film speed (or its equivalent with a digital sensor). For the moment, I’m going to ignore the DIN-style speeds, and look only at the ASA-style speed ratings. For this system, the allowable range for the K factor is 10.6 to 13.4. These numbers do not correspond directly to reflectance values (e.g. 10.6 doesn't imply a 10.6% grey card as mid-level grey), but they do correspond to different levels of illumination that will be metered as mid-level grey. In other words, there is not one specific level of reflectance that is required to be metered as mid-level grey – rather, any value within the specified range is allowable.

The K factor is related to a measured exposure by the following formula:

K = LtS / A2


K = K factor
L = Luminance in cd/m2
A = f-number
t = effective shutter speed
S = film speed

Using this formula and a calibrated monitor, we can find the K factor for a specific camera. For example, I have a Sony Alpha 700 camera and a monitor that’s calibrated for a brightness of 100 cd/m2. Doing a quick check, my camera meters the screen (displaying its idea of pure white) with no other visible light sources, at an exposure of 1/200th of a second at f/2. Running this through the formula, gives a K factor of 12.5 – just above the middle of the range allowed by the standard.

The next step is to figure up what level of “grey” on a card that corresponds to. Let’s do that based on the sunny f/16 rule, which says a proper exposure under bright sunlight is f/16 with a shutter speed that’s the reciprocal of the film speed. We can mathematically transform the formula above to:

L = A2K/tS

Let’s work things out for ISO 100 film:

L = 16x16xK/.01x100

The .01 and 100 cancel (and they will always cancel since the rule is that the exposure time is the reciprocal of the film speed), so this simplifies to: L = 256K.

Working the numbers for the lowest and highest allowable values for the K factor gives 2714 and 3430 respectively.

Now, we run into the reason the ISO standard specifies light levels rather than reflectance of a surface – even though we’ve all seen and heard the sunny f/16 rule, the reality is that clear sunlight varies over a considerable range, depending on season, latitude, etc. Clear sunlight has brightness anywhere from about 32000 to 100000 lux. The average of that range is about 66000 lux, so we’ll work the numbers on that basis. This has to be multiplied by the reflectance to give a luminance – but the result from that comes out in units of “apostilbs” rather than cd/m2. To convert from apostilbs to cd/m2, we multiply by 0.318:

L = I x R x 0.318.


R = reflectance
I = Illuminance (in Lux)
L = luminance (in cd/m2)

We already have the values for L that we care about, so we’ll rearrange this to give the values of R:

R = L / 0.318 I

Plugging in our minimum and maximum values for I, we get:

R1 = L / 10176
R2 = L / 31800

Then we plug in the two values for L to define our allowable range for R:

R1,1 = 2714 / 10176
R1,2 = 2714 / 31800
R2,1 = 3430 / 10176
R2,2 = 3430 / 31800

R1,1 = .27
R1,2 = .085
R2,1 = .34
R2,2 = .11

In other words, between the range of brightness of the sun and the range of K factors allowed by the ISO standard, a reflectance anywhere from about 8.5% to about 34% can fall within the requirements of the standard. This is obviously a very wide range of values – and one that clearly includes both the 12% Thom advocates and the 18% of a typical grey card.

To narrow the range a bit, let’s consider just the arithmetic and geometric mean of the range of brightness from the sun: 66000 and 56569 lux respectively. Plugging these into the formula for the range of possible reflectance values gives:

R1,1 = 2714 / 20988
R1,2 = 2714 / 17989
R2,1 = 3430 / 20988
R2,2 = 3430 / 17989

The results from those are:

R1,1 = .13
R1,2 = .15
R2,1 = .16
R2,2 = .19

An 18% grey card is close to one end of this range, but still falls within the range. A 12% grey card falls outside the range; we have to assume an above-average light level for it to work out. If we average the four numbers above together, we get a value of about 16% grey as being the "ideal" – one that should work out reasonably well under almost any condition.

To summarize:

  1. The ISO standard allows a range of calibrations, not just one level
  2. Normal daylight brightness covers a fairly wide range as well
  3. 18% grey is justifiable based on average light levels
  4. 12% grey is not justifiable based on average light levels
  5. Based on average light levels, the ideal value for a grey card would be about 16%
  6. You meter might be calibrated to 18%, but probably isn't (and shouldn't be) calibrated to 12%.
  • \$\begingroup\$ This is a great explanation, but I think the argument is circular in that you assume sunny/16, but vary the illuminance to 'normal daylight'. Seems like it shouldn't be necessary to redefine that term. What about assuming any fixed exposure: EV = lg(IS/C) (incident) = lg(LS/K) (reflected) => L/I = K/C? R should be derivable from L/I & constant factor, hence the implied reflectance of the standard directly from the ranges for K/C in the standard? \$\endgroup\$
    – ex-ms
    Commented Jul 28, 2010 at 1:25
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    \$\begingroup\$ Tracked it down: the ANSI standard he's likely referring to was ANSI PH3.49-1971, superseded by (but not radically different to) ISO 2720-1974. \$\endgroup\$
    – ex-ms
    Commented Jul 28, 2010 at 1:44
  • \$\begingroup\$ @matt: you could be right about the standard, but frankly, I kind of doubt he had any specific standard in mind, especially one that had been obsolete for nearly 30 years when he wrote the article. \$\endgroup\$ Commented Jul 28, 2010 at 3:42
  • \$\begingroup\$ @matt: I hadn't considered comparing the results for incident and reflected readings to determine the reflectance range. It sounds reasonable, but I'll have to reread the standard to figure out what it comes to. \$\endgroup\$ Commented Jul 28, 2010 at 4:05
  • \$\begingroup\$ I think ex-ms brings up a good point in his first comment -- you've got a stop-and-a-half range for sunlight. If you want your argument to be more convincing, you should find more exact values, rather than synthesizing your own by averaging. If the sunny 16 rule were calibrated for one end of the range instead of the middle, as you arbitrarily chose, then 12% would be more correct than 18%. \$\endgroup\$
    – Evan Krall
    Commented Mar 25, 2011 at 20:39

what it real is, why 18% ?

This is the amount of light used by most cameras for determining exposure. This was chosen (instead of 20%, etc) because, on average, most "photographs" used by average photographers tend to work out to roughly the same amount of light exposure as solid, 18% gray.

If, however, you're shooting something that has a lot of white, or a lot of dark, you're exposure will be off. For example, if you take a photo of a large white building, you'll probably want to adjust your exposure to compensate, since the default will target 18% gray, see all of the white, and lower your exposure (to make the entire picture average the same light content as 18% gray). You'll want to have a higher-than-default exposure to compensate.

You can use a solid card that's colored the appropriate gray tone to help compensate for this in your camera. Many cameras have exposure compensation functions that will let you configure your camera by pointing it at something with the right amount of color saturation.

If you want to make your own card, you'll want to saturate the card with 18% grey. This will work out to a solid fill (in RGB) of about 46 for R, G, and B. Just be aware, however, that most printers will distort your color somewhat -so when you print, you may want to check the results against your original.

  • 7
    \$\begingroup\$ You probably want to just buy a grey card - they're cheap, under $5. \$\endgroup\$
    – Reid
    Commented Jul 22, 2010 at 19:00
  • 3
    \$\begingroup\$ I agree completely - but the OP did ask how to make it work ;) \$\endgroup\$ Commented Jul 22, 2010 at 19:04
  • \$\begingroup\$ Because of average ? the 50% is the average of black and white or not ? \$\endgroup\$
    – Aristos
    Commented Jul 23, 2010 at 5:27
  • \$\begingroup\$ @Aristos: I don't quite understand your question. Basically, if you make your photos B&W, and took the average amount of light across all of them, you'd end up with 18% gray. (More white than black - since we usually photograph lit subjects) \$\endgroup\$ Commented Jul 26, 2010 at 21:48
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    \$\begingroup\$ see the Hogan article mentioned in the accepted answer re: 18% and meters. I think you also misunderstand the nature of the '18%,' it's the amount of light reflected from the card, not the brightness of the image. rgb(46,46,46) is much too dark. \$\endgroup\$
    – ex-ms
    Commented Jul 27, 2010 at 4:39

The cards are designed to reflect about 18% of the incoming light, which to a human appears half way between max white and darkest black and happens to be a fairly good guess at the average reflectance of typical natural scenes - L*50 as has been mentioned correctly above.

The next question, and where the 12% comes from, is this: If we meter off the 18% gray card, which to a human appears as mid-grey but really has about 18% the intensity of maximum diffused white, what value should this information be recorded at in our raw files. Remember that film has a gentle rolloff in the highlights, while digital has an absolute cutoff. So they decided to give a half stop of extra headroom to safeguard the highlights (possibly specular) and, if desired, provide a half stop rolloff. It was decided that luminance coming from a gray card reflecting 18% of incoming light, aka L*50, aka mid-gray should be actually recorded at 18%/sqrt(2) = about 12.8% of max diffused white - in the linear raw file.

As to what happens to the data after that, it becomes very messy and the standards have really made a mess of it, imho.


Think of the tonal scale from black to white. Instead of an even gradient, break it into 11 parts (called zones). Zone 0 is solid black with no detail. Zone 10 is solid white with no detail. Zone 5 in the middle is 18% gray. Google "zone system" for more info.

It's very likely that the gray tone you actually care about is 12% as this is what camera meters are most probably calibrated to measure. See Thom Hogan's article about gray cards.

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    \$\begingroup\$ Technically, this is incorrect. The zone system has 11 zones, from 0 to 10 (not 9) with 5 in the middle. \$\endgroup\$
    – user456
    Commented Jul 26, 2010 at 15:00
  • \$\begingroup\$ I updated my answer according to matt smillie's comment regarding light meters' calibration. Thanks matt. I don't know where the comment disappeared. \$\endgroup\$
    – Karel
    Commented Jul 26, 2010 at 18:07
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    \$\begingroup\$ is 18% because the eye see logarithm and the 18% is for the eye the middle gray, the middle point between white and black \$\endgroup\$
    – Aristos
    Commented Jul 26, 2010 at 19:01
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    \$\begingroup\$ Minolta, Sekonic, Pentax and Gossen work to 18% grey while Canon and Nikon work to 12%. I have been unable to determine the figure for Olympus. Hogan's article is singularly inept. \$\endgroup\$
    – labnut
    Commented Mar 26, 2011 at 17:48

The OP asked: why does the standard grey card have a reflectance of 18%?

The short answer is that a significant number of manufacturers calibrated their light meters in the belief that a standard scene averages 18% reflectance.

The following manufacturers work to 18% reflectance:

This information was taken from their light meter manuals. Follow the links to see my references.

The following are believed to work to 12% reflectance, though I was unable [to confirm this. Information derived from the light meter Wikipedia article. See also this article from photo.net

While I have no information for Olympus.

So the next question becomes: why do some manufacturers choose 18% and others choose 12%?

The answer can be found in ISO 2720, which states that:

The constants K and C shall be chosen by statistical analysis of the results of a large number of tests carried out to determine the acceptability to a large number of observers, of a number of photographs, for which the exposure was known, obtained under various conditions of subject manner and over a range of luminances.

This means that each manufacturer is free to determine by measurement what the average grey level of the standard scene is. Given that they have used independent measurements of the calibration constants (K and C) it is surprising (and gratifying) that there is so much agreement.

K and C are the calibration constants for reflected light and incident light meters.
K has recommended values of 10.6 to 13.4
C has recommended values of 320 to 540

Now it turns out that the two groups of manufacturers, though their own testing, have arrived at different values of K and C. And these values, through the simple application of the laws of physics, result in either 18% or 12% reflectance for the standard scene.

For the interested, the formulas can be found in the Light Meter Wikipedia article, so I won't repeat them here.

So what is the 'correct' value? 18% or 12%?

a) you don't have much choice but to work with the value your manufacturer has chosen.
b) the difference is small enough to have little practical effect.
c) it seems no one has noticed the difference anyway.

The bottom line is that the 18% or 12% values for average reflectance were arrived at by measuring the average reflectance (photographically) of a large number of scenes. So these are numbers that were arrived at experimentally and it is not surprising that there are some differences.

Is there any way to arrive at the number theoretically?

In the Lab colour space L* (brightness) can range from 0 (black) to 100 (diffuse white). I choose Lab colour space because it is designed to approximate human vision. If one makes the assumption that average brightness falls mid-way between these two extremes then one has a starting point of L* = 50.

Now, using Bruce Lindbloom's excellent CIE Color Calculator, we can calculate the corresponding luminance and the sRGB pixel values. This gives values of 18.4% luminance (Y on the CIE XYZ scale) and 118.9 pixels for sRGB.
enter image description here

Of course, to say that the average scene's average brightness is mid-way between white and black is a big assumption and over simplifies the real world. One really needs some kind of experimental basis for this assumption. But it is certainly interesting that this calculation arrives at a result close to that of many manufacturers.

  • 1
    \$\begingroup\$ I opened my camera's manual - a canon 550D - and searched for 12% and 18% (having a pdf version of the manual is fun), 12% doesn't appear anywhere, 18% appears 3 times, once in the context of white balance and twice in the description for a setting where "the dynamic range is expanded from the standard 18% gray to bright highlights", so, it looks like canon uses 18% and not 12% (more likely than the other possibilities: that they use a different standard for the 550D than for the rest of their line or that the manual is incorrect) \$\endgroup\$
    – Nir
    Commented Mar 27, 2011 at 11:38

18% grey is the shade that through the lens (TTL) metering base their exposure figures from -- you can also use it to check white balance if you want to calibrate for a shoot.

When you haven't got one to hand, you can usually substitute with an area of concrete, if it's in the scene or at least under similar lighting circumstances.

  • \$\begingroup\$ And the downvote was why? \$\endgroup\$ Commented Jul 26, 2010 at 18:08
  • \$\begingroup\$ See comments at the top of the question re: 18% and meters. \$\endgroup\$
    – ex-ms
    Commented Jul 26, 2010 at 19:35
  • \$\begingroup\$ @matt My answer refers to TTL metering though (after all, if you where measuring the incident light, you wouldn't need an 18% grey card) \$\endgroup\$ Commented Jul 26, 2010 at 20:15
  • \$\begingroup\$ All I can suggest is reading Hogan's article & references more carefully; he's not talking about incident meters either. \$\endgroup\$
    – ex-ms
    Commented Jul 26, 2010 at 23:24
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    \$\begingroup\$ @ex-ms, Minolta, Sekonic, Pentax and Gossen work to 18% grey while Canon and Nikon work to 12%. I have been unable to determine the figure for Olympus. Hogan's article is singularly inept. \$\endgroup\$
    – labnut
    Commented Mar 26, 2011 at 17:44

For the most part, standards aren't designed to explain theory. Their purpose is to describe how to do something, determine film speed, calibrate an exposure meter, etc, and they are based on research which can be found in scientific papers in scientific journals. Three papers that describe the theory of meter calibration are:

Stimson, Allen, An Interpretation of Current Exposure Meter Technology, Photographic Science and Engineering, vol 6, No 1, Jan-Feb 1962.

Scudder, Nelson, Stimson, Re-evaluation of Factors Affecting Manual or Automatic Control of Camera Exposure, Journal of the SMPTE, vol 77, Jan 1968.

Connelly, D, Calibration Levels of Films and Exposure Devices, The Journal of Photographic Science, vol 16, 1968..


On how to print an 18% card, with no theory or rationale as to why 18 not some other number...

Following any theory or advice on setting RGB values in a graphics program isn't reliable. Monitors and printers are designed to make graphics look good and fall short of scientific accuracy. Even if your whole system is calibrated - well, I never quite trust such things to be accurate, especially not for physical optical properties of printouts.

Ultimately, you'll have to make a big gray rectangle of some chosen RGB value and print it out. How to know what RGB value?

First use your graphics program to print a fine grid of black squares on an empty white background. Make the squares cover 18% of the area. The gap between squares should be 1.59 times the width or height of the squares. Make this grid small but large enough to have good control over accurate geometry, and make it cover a whole page.

With good dark ink in the printer, the white will be almost 100% reflective and the black almost 0% (but nothing is perfect) so the overall reflectance averages out to 18%. Photograph this black & white printout out of focus, letting the camera do the averaging.

Guess an RGB value, make the whole page that value of gray and print it out. Photograph it, out of focus, next to your black and white grid. Based on whether it's lighter or darker than the grid, refine your RGB guess. Repeat until they match.

Be careful to have uniform lighting and avoid vignetting effects in the optics.


To summarize an answer.

From white to black, the eyes see a range of greys. Because the eyes see logarithmic (and the ears hear logarithmic), what the eyes looks like the middle - this is actually not contains 50% black + 50% white, but 18%.

The middle point for the eye contain 18% Black On White.

To make this on Photoshop you fill a white background with 18% black pattern. So in photoshop if you fill with half black a white background, you do not get the middle gray that eye can see.

Many years ago I was made a page base on that 18% rule to calibrate the monitor. The different with other calibrations was that, I use 18% to fill the background with black and not 50%

I still have this gamma calibration page online. Blur your eyes and try to make the circle inside to disappear.


"18%" is the amount of light reflected by the midpoint of black and white (Adams' Zone V). The precise convention of 18% (not 17%, 19%, etc) comes from the graphics industry (probably - see links).

In photography, it has two main uses:

  • An object known to be a neutral colour can help when correcting white balance. This doesn't have to be middle grey, but any grey.
  • An object of known reflectance can help with metering. The palm of your hand is a good substitute for this function of a grey card: it's about 1 stop brighter.

However, in the largest typeface possible here:

18% grey is not what your meter is calibrated to.

Really and truly.

This is a persistent myth, but is not actually the case. Your in-camera meters are calibrated closer to 12% grey, which is a difference of approximately half a stop. This is the ANSI standard.

There's a practical explanation by Thom Hogan here: http://www.bythom.com/graycards.htm

And for those who prefer to deal in foot-candles and foot-lamberts, a mathier version here: http://www.richardhess.com/photo/18no.htm

That's a shocking revelation, what should I do?

Probably nothing. Most grey cards sit unused in the closet. Even in use, a half stop is pretty negligible in most situations. So if you're happy how things are, continue!

If this difference matters to you, the Hogan article has practical advice on a sidebar, excerpted here:

If you shoot digitally, shoot a gray card under even lighting at the metered value, and at third-stop increments (use only spot or center-weighted metering, and make sure the card is angled slightly towards the light [to ensure you're seeing reflected light]). Look at the histograms for each exposure (on the camera, not in Photoshop, which uses a different method of generating histograms). If you're using a 18% gray card, pick the exposure setting that generates a centered value and set that in your exposure compensation control.

  • 3
    \$\begingroup\$ If there's a myth that deserves to be proclaimed in "the largest typeface possible", it's that Thom Hogan's article has any basis in fact. The claim that: "This is the ANSI standard" falls right on the border between misleading and outright false. \$\endgroup\$ Commented Jul 28, 2010 at 0:40
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    \$\begingroup\$ At key points Thom clearly states "I believe" and does not declare it as fact. In the end the issue boils down to what is written 1) in ANSI standard (some might have the access and willingness to read it) and 2) in camera firmware (to which the access is very limited). So the facts can only come from camera engineers and not internet discussion :) \$\endgroup\$
    – Karel
    Commented Jul 28, 2010 at 15:30

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