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Expose to the right

Van Kamper

2 posts

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"RE: Expose to the right"

Van Kamper Registered since 15th Apr 2012
Sun 15-Apr-12 03:07 AM | edited Sun 15-Apr-12 03:52 AM by Van Kamper

>I don't get how ETTR takes exposure out of the hands of the
>photographer. It's not auto, it's a specific exposure choice
>that has to be tightly controlled in order to prevent blown
>highlights. It IS a choice to see the camera as a data
>collector, not something that makes pictures. With ETTR you
>are making the conscious choice to collect data at capture,
>and make the picture in post.
>ETTR is a method that uses your camera to optimize data
>capture. With optimum data, your pictures don't necessarily
>look as intended out of camera, but you have the most data
>which gives you the most flexibility in post processing.
>Whether you shoot this way, or shoot for results out of
>camera, both are conscious choices and neither is auto. I'm
>not sure what look you'd be going for in post that you
>couldn't achieve with ETTR, unless you blew the exposure and
>lost highlights.
>As to why, there's a reason behind the prevailing wisdom.
>There are in fact a couple reasons.
>Light is a geometric scale, mapped by our cameras into a
>linear scale. 1 stop of light is a doubling of the amount of
>light. Converted to a numeric value by analog to digital
>conversion, that means the difference between 16 and 32 is one
>stop, 32 to 64 is one more stop, and 64 to 128 is one more
>stop, etc. In a 12-bit conversion, the numeric value can range
>from 0-(2^12-1) or 0-4095. The brightest stop of light that
>can be recorded therefore contains a numeric range of
>2048-4095. That's a full half of the numeric scale! The next
>stop has a range of 1024-2047. Your brightest 2 stops contain
>3/4 of your numeric range! Now consider the numeric range of
>all 12 stops:
>2048-4095 or 2048 counts
>1024-2047 or 1024 counts
>512-1023 or 512 counts
>256-511 or 256 counts
>128-255 or 128 counts
>64-127 or 64 counts
>32-63 or 32 counts
>16-31 or 16 counts
>8-15 or 8 counts
>4-7 or 4 counts
>2-3 or 2 counts
>0-1 or 1 count
>So, on a 12-bit capture, your brightest stop has the ability
>to contain 2048 gradations, and your darkest stop only 1. This
>is why you expose to the right, because let's say your data
>spans 8 stops. You really want it to occupy the top 2/3 of
>that table, not the middle, and certainly not the bottom! 8
>stops of data in your file could contain a numeric range of
>16-4095 (ETTR, 4079 potential gradations), or 4-1023 (centered
>histogram, 1019 potential gradations) or 0-255 (exposed to the
>left, 255 potential gradations, and equivalent to jpeg bit

Michael Reichman doesn't know his math, and a whole bad set of wrong ideas developed.

1. Mathematically a "doubling" of something (in this case LIGHT INTENSITY, and not tonal values as Michael thinks)is shown as 2,4,8,16,32, 64....2048,etc. Each interval is the SAME (a doubling effect). There is NO increase in qty of light between interals, and it DOES NOT refer to tonal values. It only says that it is twice as bright at 2048 compared to 1024 (1 stop diff). The same holds true that it is twice as bright at 1024 compared to 512 (1 stop diff). It does not mean 512 tones or 1024 tones....that scale has nothing to do with qty.

2. Expose a black card next to a white card, each if IDENTICAL size. Now look at it on the histogram. You will see that the peaks are just as high for both the white and black patc. The peaks on a histogram represent number of pixels exposed for a particular tonal value between o-255. You will find that the QTY of pixels exposed for the black patchy is the same as for the white patch (if they were identical size). The peaks for both are equal.

3. Shadow noise is due to sensor performance, not because less information is recorded in the darkest zones compared to brightest zones.

4. A sensor is like film in recording data. The light falls where it must on the sensor to create the image, and each image will show a different histogram. Try plotting Reichmans scale 1,2,4,8,16...2048 which he considers as a QTY of something on a histogram, and you always get the same result (straight line ascending to the right which is nonsense). Light falls where it wants to on the sensor, not in any order as he suggests.

I recommend people read Ansels books on sensitometry, or pick up some math books. Michael is a photographer, and his buddy Thomas Knoll apparently were on a trip together when they came up with this nonsense. They should have got a piece of paper, and wrote down what the x and y axis mean on a histogram (one is for qty, the other for tonal value). Two cards photographed at zone 3 and 7 will have identical peaks (same quantities of pixels exposed for that tonal value). There isno descendinig order as the table above shows.

The histogram shows it all....read what is there.... Qty vs tone value (0-255). Do you see on the histogram anywhere the scale 1,2,4,,8,16,32....2048? For further reading check out math books on statistics "interval data".

I rushed through this, and hope it makes sense.I would stick to the cameras exposure (Nikon knows a few things), but i agree if you got time it won't hurt to use ETTR, but I prefer to see the mood or affects as is on the cameras lcd and worry less about blocking highlights when rushing.

A general, generic topic Expose to the right [View all] , nwcs Awarded for his in-depth knowledge in various areas, including Landscape and Wildlife Photography , Thu 29-Mar-12 12:12 PM
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