Why shoot to the right

An article explaining why to shoot to the right with digital cameras

 

http://www.adidap.com/2006/09/13/why-shoot-to-the-right/

 

hope it helps someone

The LL site has been recommending the same thing for some time now. Of course, the people over there use dslr mostly.

 

In the real world, some P&S digicams overexpose. Hence the need for -1/3 to -2/3 exposure compensation for these camera.

It's true about exposing to the left for P&S. My Fuji F10 captures

incredible amount of detail in shadows you'ld think by the looks

of the LCD preview would clip to absolute black.

 

The shot linked below was taken at forced ISO 80 exposing for

the sky which gave a very dark LCD preview resembling a RAW

capture off a scanner. The bottom portion is a 100% crop of an

area lightened to show the shadow detail captured.

I'm not sure why any member of the online photographic community would visit the referenced site, which appears to rip off without attribution or payment the intellectual property of existing sites. Moreover, it does a half-assed job of its dirty work, garbling or mangling the material -- a sign its creators don't know what they are talking about.

Not necessarily good advice!

 

Many times there is information to the right that is totally unimportant and can (and needs) to be clipped. Example: Light streaming through a window of an interior would show as badly clipping the highlights. If you followed their recommendation, the rest of your picture would be way too dark! Or perhaps light from a back lit subject blowing out. Expose for that light and you have a picture that could be under exposed by 3 stops. Yikes.

Mr. Reichmans concept of shooting to the right is ridiculous. Where does he come up with this stuff? He starts with 12 bits allowing for 4096 data points, then makes this huge leap of science believing it is okay to apply a geometric progression scale to describe his science that the brightest first stop contains half the data points (so expose to the right). Some weird stuff can develop when you think this way. The geometric scaling 1,2,4,8,16,32,64,128,256,512....2048 is a simple concept. The value 512 represents twice the exposure compared to the value 256, and the same comparing 32 to 16. That is all it means. It does not measure the TOTAL AMOUNT OF TONES left available with each 1 stop change. Instead it is a PROPORTIONAL SCALE showing a doubling progression, nothing more, each stop is "twice as bright" as the first, and shown on a numeric scale! In Ansels Book, The Negative, Pg 10, he shows how film reacts to light (and we can apply this to digital also). The XY graph has the X axis shown as a log scale in base 10, with each DOUBLING of exposure representing a .3 log shift to the right. He then below that scale shows the same effect using the EXPOSURE UNIT scale (0,1,2,4,8..2048), and then again below that another Zone System scale representing the same. All 3 scales along the x axis refer to the same ....a doubling of exposure as we move to the right, and the corresponding increase in density in the y-axis referring to brightness (same for digital and film). None of these scales refer to the measure of the 'quantity' of tonal values available at each stop, but rather only the tonal brightness value at a specific point on the scale when exposure is doubled (shown on y-scale)! He also implies digital is linear and film is not. I don't believe this, because digital must emulate film as close as possible. If we had this curve moving instead of from the normal bottom left to top right and suddenly the curve was reversed.... well were going to get some really weird photos. Film and digital both have to be pretty close to linear, both recording proportionally with each f/stop increase/decrease. How close they follow the same curve is clear when we realize our exposure meters function with film and digital cameras alike. Slight differences exist between digital and film, no more then differences between different film types. The only major difference is digital falls off sharply in the highlights (straight line).However, both record in a linear fashion along the straight line portion of the characteristic curve. They have to!

 

If his theory were true, where 1/2 the tones are recorded in the first brightest stop, then this is what I expect:

 

1. The histogram would be "always" heavily skewed to the right due the geometric level of decay after just 5 stops, but it isn't as I will show in the example below where most of the dark tones will record on the left side of the histogram.

 

2. The quality would severely deteriorate geometrically as we approach the shadows, approaching pixelation very quickly. Within 5 stops the qty of dark tones recorded would be very low if you follow his belief.

 

Example. Shoot a white vase against a large dark background, and you will find all the tones to the left on the histogram, and only a small portion to the right representing the white vase. In this case you cannot expose to the right without loosing highlight detail in the white vase. The opposite is also true, pick a dark vase and large light background and all the tones on the histogram will record to the right. Taking this extreme contrast situation, there is no chance of exposing to the right. Does that mean my quality will suffer because I cannot expose to the right? Or do my white tones record with high quality (because it received the first brightest stop and therefore 2048 tones) while darker tones suffer due to few tones remaining in the last few stops? I don't think so.

 

Exposing to the right just overexposes the scene. The only advantage would be digital shows less noise once you avoid deep shadows. Have a quality camera and most of this is gone. The general shape of both histograms is identical in before and after. The remaining data points for dark tones remains the same in both before and after shots

 

 

Van Camper:

 

Interesting and spirited argument, but I will leave its refutation to those more proficient than I.

 

However, I note that your "Example" is irrelevant because it concerns a case where the dynamic range is maximal by construction and therefore Reichman's rule wouldn't apply. The rule only applies to cases where less than the available dynamic range is used for an image. His point, which I should note is shared by the chief software architect of Photoshop, is that for those images the dynamic range can be maximized to improve the quality of the image, and then the image can be manipulated in your favorite editor to lessen the apparent overexposure. In other words, exposing to the right is a technique to increase the amount of information that an image captures.

 

Of course, if you don't shoot RAW or don't like to post-process, then just ignore the rule and use the default values provided by the camera.

 

Interestingly enough, the site that cribbed luminous-landscape.com's original material doesn't have a clue about your argument, so perhaps you should bring it to their attention.

"However, I note that your "Example" is irrelevant because it concerns a case where the dynamic range is maximal by construction and therefore Reichman's rule wouldn't apply."

 

In my example, I intentionally used a high contrast scene to make the argument that if we can't expose to the right, do we then suffer a quality loss by his theory? By his argument, zone 7 would be high quality, but by zone 4 we have 1/16 (1/2x1/2x1/2) of the tones left, imagine a 5 stop range. I have never seen such a picture (many different quality levels in one photo as you move from lightest to darkest). Even if we used a 1:1 light ratio, we still need to record all the zones from 0-8 with equal quality.

 

Your point that his view is shared by the chief architect for photoshop is appreciated, two heads are better then one. But programmers and camera designers/builders are distant worlds apart. His views are no better then mine, and of course I am not an expert either, but we can all read a histogram. I agree in that exposing to the right helps get better shadow detail, and less noise, and then you print it down (and remember noise is a problem with scanners too). This is like using shadow masking (Howard Bond) to raise the shadow level and then use a higher contrast paper to print it down again, thereby increasing apparent sharpness?. with snappier shadow areas to boot! But exposing to the right for better shadow detail is not the same as arguing you must do so because of loss of tonal values because of his half rule principle (each lower f/stop gets ? the remaining tones).

 

The answer is in the histogram, it shows all the information as recorded by the camera. The height of the bars indicates how prevalent (total number) each shade (0-255) is compared to the others in the image (How to Wow, Photoshop for Photography, pg 54).

Just look at the histogram, at the valleys and peaks. Any high peaks found on the left side of the histogram will tell you Mr. Reichmans theory is WRONG (it just PROVED to you the brightest first stop DID NOT receive the most data)! Look at some of your own images, if the brightest stop records ? the data, then the darker left side of the histogram could not record the highest peaks. Yet I have MANY images where the peaks are at the highest (and widest) on the left side of the histogram in the darkest tones, totally contradicting his theory the brightest tones capture 1/2 the data. Study the histogram, decide for yourself.

 

I find it hard to believe that so many of us accept his theory simply on his comment "BECAUSE IT IS LINEAR". He failed to even define it so some of us might accept it. It appears if he says so, then it must be! This is too short an explanation in my book. Yet I see his misinterpretation spreading. It is found in the book DIGITAL PHOTOGRAPHY, Industrial Strength Digital Photography Techniques, 2nd edition, pg 298-299. It is also mentioned in the website at www.adidap.com (exposing to the right) which is shown below. Again, this site accepted his argument based on the comment "BECAUSE of the linearity". Sorry, film and digital are both linear (having proportional intervals) along the straight line portion of the curve, only the tail and shoulders differ, but then again they also do with different films as well. Meters are also linear in order to follow the same interval scale used in film/digital media so everything matches.

 

TITLE: Result of the Digital camera?s sensor linearity (www.adidap.com)- SUMMARY:

 

So the sensor of the Digital Camera is linear, what does it mean?

Well it simply means that if our digital camera?s sensor can capture 12 bits (that?s 2^12 = 4098) of data in RAW and assuming a Dynamic Range of 5 stops.

* The 1st stop will hold 1/2 of that i.e. 2048 - yes 1/2 the data recorded is in the highest stop

* The 2nd stop will hold 1024

* The 3rd stop will hold 512

* The 4th stop will hold 256

* The 5th stop will hold 128

To keep things simple, every time we push our histogram to the left (toward the dark area) we are giving our digital camera?s less data to record (BECAUSE of the linearity explained above) and, since a stop is still a stop, this can only be compensated by equally spreading this available data over the stop range which explains why, in digital photography, blacks are always noisier