winterrose Loc: Kyneton, Victoria, Australia

ETTR is a fallacy.

The thinking is that because each successive f/stop records double the number of photons recorded during an exposure, by overexposing the image, (OK then, subjecting the sensor to a greater number of photons than is normally necessary to accurately register any given light value in order to record an image), we gain the advantage of doing our image processing using a vastly greater volume of data.

Utter nonsense.

The facts………

The camera sensor converts the quantity of photons striking it during the exposure and stores the result as a voltage.

That voltage is read and converted to say, a 12bit binary number.

A binary number is to the base 2 which means that there are only two possible values for each successive entry.

Each entry is a multiple of 2, or double the value of the previous one in the line.

A 12 bit number is equal to 4096 to the base 10.

Consider a 12 f/stop sensor.

Each successive f/stop represents a doubling of the light that “it” will pass.

F/stops are not values. An f/stop in isolation is not quantifiable; it has relevance only when in reference to a known value.

Each successive f/stop represents a doubling, (or halving, all else being equal), of the volume of photons that it will pass reference its neighbor.

Voila! That’s the same deal as that binary number….Isn’t it?

Well, no, it isn’t.

The numbers of photons passed by different f/stops increases logarithmically whereas the binary description of the equivalent tonal values the progression is linear.

By the time we can gain access to the data recorded by the sensor it has been irreversibly divided into 4096 steps of tonality ranging from black to white.

There is now the same number of steps for each of the f/stops regardless of where they “fall” and the actual number of photons involved takes no further part or has any further influence.

In RAW PP all you are doing is changing the value of the binary number which describes the tonal value of each pixel in the image.

The notion that by ETTR we gain the advantage of doing our image processing using a vastly greater volume of data is therefore invalid.

Rob.

tainkc Loc: Kansas City

I understand your logic completely, but are we not adding or subtracting photons by adjusting the focal length at any given aperture, thereby creating a trap door? I really don't care since I shoot in manual anyway

winterrose Loc: Kyneton, Victoria, Australia

I don't think you understand at all. What has focal length or shooting manual to do with the discussion?

KM6VV Loc: Central Coast, CA

I fully agree with you. The last statement still escapes me a little.

Can't we say that a sensor can handle a certain number of F-stops at any given length of exposure (similar to film?) and that the goal is to get the range of intensities that it can capture to fall within that range? So, shifting right (ETTR) covers more highlights at the expense of detail in the shadows? Or is there more to it then that?

I never cared for the "twice as much data" at a higher f-stop, explanation; a binary value is just a binary value, "mapped" to an intensity (or small range of intensities).

winterrose Loc: Kyneton, Victoria, Australia

The thinking or claim or contention by the proponents of exposing to the right or ETTR, is that as each f/stop "contains" twice the data of that of the previous one, much is to be gained by using as much of the right hand end of the histogram as possible thus capturing a great deal, twice or three times more I have seen claimed, more data from which to extract detail.

That is simply not true as I have shown in my thread.

Rongnongno Loc: FL

Ok, it is official W/R has a bug up his neither region over ETTR.

Second thread on the same topic.

winterrose Loc: Kyneton, Victoria, Australia

Others have had no trouble so please either try to offer something intelligent or if you choose not to I'm sure that there are lots of other threads you can join.

Cheers, Rob.

winterrose Loc: Kyneton, Victoria, Australia

[quote=KM6VV]

"I fully agree with you. The last statement still escapes me a little.

Can't we say that a sensor can handle a certain number of F-stops at any given length of exposure (similar to film?) and that the goal is to get the range of intensities that it can capture to fall within that range?"

Absolutely....

"So, shifting right (ETTR) covers more highlights at the expense of detail in the shadows?"

NO. All that exposing to the right achieves is.....an overexposed image. It does not gather any extra data for post editing.

TucsonCoyote Loc: Tucson AZ

I noticed that....wonder if he's under the weather....or something!?

winterrose Loc: Kyneton, Victoria, Australia

Others have had no trouble so please either try to offer something intelligent or if you choose not to I'm sure that there are lots of other threads you can join.

Cheers, Rob.

blackest Loc: Ireland

The way light levels (and sound) works is logarithmically.
It's probably simplest to think in terms of lens apertures 1 stop change equals a doubling or halving of the area of the opening of the aperture.

e.g f2.8 is twice the area of f4 and f5.6 is half the area of f4

the handy thing with binary is each time you add a digit you double the range of values
e.g
0
1 gives you 2 values
00
01
10
11 gives 4 values

000-111 gives 8 values 0000 1111 gives 16 and so on

if you think of light as rain and you have a container with a funnel with an open area of 1 square foot and you collect 10cc's of water with an identical set up you would catch another 10cc's if instead of two set ups
you changed the funnel for one with a 2 foot area that would collect twice as much as a 1 foot area funnel or 20cc's assuming rain falls steadily and evenly distributed. which it pretty much will do.

if you take 8 bits and set them up as decimal columns then you can see that each bit doubles the range so 8 bit could represent 8 stops and 12 bits 12 stops and 14 bits 14 stops. that would be that if it wasn't for noise. the tonal bands are different widths e.g with our 8 bit representation any sensor site that collects enough light to fill the left most column is our brightest tone
ie from 1 000 0000 to 1 111 1111 our second brightest tone covers values 01 00 0000 to 01 11 1111 and so on.
essentially the brightest tone can collect twice as much light as the next one down. It also means that noise values make a greater contribution to the count the lower the signal value is.

winterrose Loc: Kyneton, Victoria, Australia

Yes?

Darkroom317 Loc: Mishawaka, IN

Coming from using the Zone System, overexposing to get more detail makes sense to me. However, given the way the curve works with digital it makes less sense. Unlike film digital has neither a toe nor shoulder therefore the highlights eventually just blowout rather than having the slow buildup to absolute white that can't be retrieved. A digital curve can only be adjusted in post processing but even it cannot change the fact that is no toe or shoulder when it comes to digital but rather just a straight line curve that suddenly cuts off. I feel that you may be correct.

winterrose Loc: Kyneton, Victoria, Australia

Thank you 317.

steve_stoneblossom Loc: Rhode Island, USA

Am I alone, or do others find technical analysis such as this even more difficult to comprehend without proper punctuation?

Additionally, you may as well have written your final paragraph in Greek. The correlation between noise, tonal bands, bits, stops, sensor sites, and "decimal columns" seems not very well explained. Maybe this thread is intended for those more technically inclined than myself.