What does 'exposing to the right' mean?

Hi, I was wondering what it means to 'expose to the right' and how do you do

it? And will it help decrease the noise in the shadows?

 

Thank you

Ester

there is an excellent explanation at

 

http://www.luminous-landscape.com/tutorials/expose-right.shtml

 

yes, it will decrease the noise in the shadows.

 

this article talks about using the in camera histogram to choose the most "right shifted" curve that still retains pleasing highlight detail. I find it easier to bracket and work out which exposure to keep in adobe camera raw.

THank you Brent, just read the article, just one more question. Does this mean, that I could use matrix (canon 30D) and then use plus ec compensation until my histogram is as far right as it will go without blowing the highlights? And then fix in pscs2. I shoot raw. Thank you again Brent.

Ester,

 

"Does this mean, that I could use matrix (canon 30D) and then use plus ec compensation until my histogram is as far right as it will go without blowing the highlights?"

 

The answer is maybe. It is possible that the histogram will you you that your overall exposure is "safe" when in reality, one of the channels (RGB) is actually blown out. Set your 30D to give you an RGB histogram to give you a better idea what is going on. In my experience, the camera histogram doesn't always agree with the histogram in Adobe Camera Raw and blown out channels are still possible. (I've read that the RGB histogram on the 30D is a translation of a jpeg conversion rather than based on actual RAW data.) Once you have biased your exposure to the right and taken a photo, bracket your exposure by backing off a half stop and more just to be on the safe side.

 

Check the histogram illustrations here:

 

http://www.photo.net/equipment/canon/30D/review_rsa_page3.html

 

and here:

 

http://www.dpreview.com/reviews/canoneos30d/page8.asp

 

Personally, I don't like to get my histogram curves WAY over to the right unless the subject happens to have a lot of really light tonalities (like a white sand beach).

 

 

Jim

Jim Doty, "Personally, I don't like to get my histogram curves WAY over to the right unless the subject happens to have a lot of really light tonalities."<br>

<br>

Why not? It would seem that the perfect digital exposure is where one pixel in one channel just maxes out the analog-digital converter (in other words, the histogram just touches the right edge). That would give you the absolute maximum dynamic range and the lowest noise.<br>

<br>

Of course, the histogram shown on the back of all digi-cams that I know of has already been run through some kind of gamma-curve, so the histogram doesn't really represent the linear sensor data. Unfortunately, this just makes the histogram less sensitive in the all-important highlight region.<br>

<br>

A couple of articles I like about digital exposure:

<ul>

<li><a href="http://www.adobe.com/products/photoshop/pdfs/understanding_digitalrawcapture.pdf">Understanding Digital Raw Capture</a>

<li><a href="http://www.adobe.com/products/photoshop/pdfs/linear_gamma.pdf">Raw Capture, Linear Gamma, and Exposure</a>

</ul>

Cheers,<br>

<br>

Geoff S.

THank you Jim for your very useful advice, especially on using RGB histograms. Ive been using brightness and it hasnt been informative enough and rgb might be.

 

Geoff, thank you as well. I am going to give it a try because I shoot mainly people and detail in the highlights is very important to me and as little noise in the shadows is equally important. I like what you said 'That would give you the absolute

maximum dynamic range and the lowest noise'.

 

I also agree with Jim that the cameras histogram is different to the raw so I am probably going to push to the right a bit more.

 

Thank you both so much for the good advice. Ester

Hi Ester, I'm in no way disagreeing with anything anyone's said about "exposing to the right". I understand the rationale behind it completely. However, I find that, for me, *some* types of shots I want to take do not lend themselves to that technique - for example, low-key, dark, moody portraits. Now I know I can still "expose to the right" and crank everything back in Photoshop but for me, it just doesn't look the same. Here's an example where I looked at the histogram I was getting before this shot, and while there were still technically no blown highlights, I didn't like the look, so I cranked it back, thus "exposing to the left" a little more: http://www.photo.net/photo/5053172 Again, I'm not disagreeing with what anyone's said or the rationale behind "exposing to the right". I'm just saying that, for me, it doesn't always work in every situation that I personally am shooting for. Some rules can occasionally be bent, if not broken, with interesting results. Best wishes . . .

Hi Beau, thank you for your very useful comment. I would only do it if I knew shadow noise was going to be an issue, like in the forest, especially when I have to use a high iso. I actually did give it a try today and it worked pretty well actually. I did have some noise but it was from my iso and not from shadows. I would probably do it again but maybe just use center metering and not matrix. Thank you again for your advice.

Geoff,

 

In answer to your question regarding my statement: "I don't like to get my histogram curves WAY over to the right"

 

I guess the operative word in my statement is "WAY". If the RGB histograms are bunched way over the the right, even with no clipping, some of the images just don't look "right" to my eye when I've finished with the post processing.

 

Once I get my base exposure with curves to the right, I bracket back down (less light) in my exposures. Sometimes one of the bracketed down versions works better than one with curves that are more to the right. It is a matter of personal preference and it varies from image to image.

 

Thanks for the links to the two fine Adobe articles.

 

Jim

Jim Doty:

<ul>

<i>In answer to your question regarding my statement: "I don't like to get my histogram curves WAY over to the right"<br>

<br>

I guess the operative word in my statement is "WAY". If the RGB histograms are bunched way over the the right, even with no clipping, some of the images just don't look "right" to my eye when I've finished with the post processing.</i>

</ul>

Jim,

 

Thanks for the reply. From what I understand about the way ACR handles exposure it <i>should</i> be the same as changing the exposure in the camera, but I can imagine any number of reasons why that wouldn't be the case. I'll have to do some more reading and experiments with different exposures to see if I can discover anything.<br>

<br>

It seems to me that a key part of making good photos with digital centers around being able to understand and adapt to the (dare I say, "unique") characteristics of the medium. I spent a long time learning about Tri-X and Kodachrome, and I don't expect the transition to digital to be much different.<br>

<br>

Cheers,<br>

<br>

Geoff S.

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 (our meters are linear also). 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. The histogram would be ALWAYS heavily skewed to the right due the geometric level of decay he proposes, but it isn't as I will show in the example below.

 

Example. Shoot a white vase against a dark background, and you will find all the tones to the left on the histogram (with the highest peaks), and only a small portion to the right representing the white vase (obviously ? the tones were not recorded in the 1st brightest stop, because the histogram doesn't show this). 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 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 the decay process as he describes. In both examples, the amount of total information recorded on the y axis of the histogram is the same, regardless if it is a white or black card. What does change is how the histogram must show this data. The histograms job is to duplicate what it saw, and if the background is black, then the histogram will be heavily skewed to the left, and if white skewed to the right. The histogram is not meant to draw a geometric progression showing a decay, which he implies. This decay process is not visible in my histograms. I often get 90% of my tones in the darkest areas (not the 1st brightest stop), simply because most of the scene is dark, and the histogram shows that, and nothing else.

 

To summarize, tudy 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). Look 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 (but you see it all the time in darker subject matter). 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 reproduced 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

 

______________________________________________

 

 

 

Sorry for confusion. In the above article, where you see the "?" symbol, please substitute for "1/2".

Shees, I need some sleep, lets try the correction again.

 

"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 (but you see it all the time in darker subject matter). "

 

Please switch the "?" symbol for "1/2" when you read this section in the above article. Sorry for the confusion.