Histograms

I will probably get a more technical answer than I can handle at this time but

I'll try anyway. Histograms. How do I read one and what valuble information

can I obtain from them. Would an advanced knowledge of a histogram improve my

photography. Thanks in advance.

A simple google search: histogram +photoshop

Histograms are fairly simple; horizontally they represent the tonal range of an image, and vertically they represent the quantity of a particular tone. Photographers are mainly concerned with the horizontal aspect. I'm not sure how much "advanced knowledge" is available, but a basic, working familiarity with histograms is invaluable. They are wonderfully useful in determining if parts of a photo are over or under exposed. I find the histogram to be much more accurate than just eyeing the photo on the small LCD screen on the back of the camera. It's the light meter for the 21st century.

Check out <a href="http://www.luminous-landscape.com/tutorials/understanding-series/understanding-histograms.shtml" >this</a> link from Luminous Landscape for a good description of histograms.

I will assume that you are thinking about the in-camera histogram? The best exposure - at least when it comes to flexibility in post-processing, is one that touches the right hand side of the chart (assuming you don't want any blown highlights - for some effects, they too might be desirable to avoid underexposure of other parts of the image). <p>The "problem" with digital camera histograms is of course that they don't show the actual tones captured by the RAW file if you shoot RAW, but of a standard conversion jpeg, meaning that even if the histogram and overexposure blinking indicates that you have blown highlights, they might still be salvageable. For critical work, bracketing is in any case a good idea, but if you can read the histogram, you can usually tell instantly if your exposure is spot on.

In my mind this is what I believe histograms show. The darks are on the left side and go

to the brightest ending on the right. I don't like to see the graph hit either the left or right

wall. The intensity is represented by the height of the curves.

 

It depends on the image as to what shows on the histogram. Take a pic of something

black and what does the histogram look like? Take a pic of a bright sunny sky and what

does the histogram show?

 

In my way of thinking the histogram is a report card of the image showing the various

levels of light intensity recorded for the given photo.

 

Here is a good article to read:

 

http://www.luminous-landscape.com/tutorials/understanding-series/understanding-histograms.shtml

 

 

Hope this helps you!

A histogram is a bar chart with 256 bars (0 to 255 ) the distribution on the horizontal scale tells you the relative exposure levels and the verical scale lets you know , how many pixels are exposed at that level.

 

There is no suckh thing as a perfect shape t oa histogram --that's dependent on the content of the individual photograph -- beyond these general rules: it's better to have more data on the right side of the histogram than on the left side -- Without getting really technical, if you shot film this corresponds to the idea that with negative film you should expose t orecord good detail in the shadows --which you can always print down to a darker tone -- and to watch out for even moderate peaks at either end of the histogram. If you are having peaks there you are losing shadow detail (on the left end of the h'gram) or highlight detail (on the right end of the h'gram).

Bill wrote: "Take a pic of something black and what does the histogram look like?"

 

Bill, as a side note, don't forget to take metering into account - you might think that if you photograph something black you'll end up with a spike at the left hand end of the histogram - but in reality if it's ALL black the camera will meter (and record it) it as a neutral gray and put the spike smack bang in the middle (same for an all white object).

Gosh Colin you're correct.

 

Run the camera in manual mode!

 

Best!

"it's better to have more data on the right side of the histogram than on the left side "

 

That is according to luminous landscape. Alhough it is okay to expose to the right so you gain more shadow detail, and reduce noise, you do not lose 1/2 your tones with each stop decrease as he proposed. He is reading a mathematical progression incorrectly, then comes up with some weird views. If you look at a progression such as 1,2,4....1024,2048,4096...the difference between 4096 and 2048 is not 2048 tones, but rather 1 stop. The difference between 2048 and 1024 is not 1024 tones remaining, but rather 1 stop. That is all that those numbers can imply. He for his own convenience came up with a theory based on an inappropriate use of those numbers. His argument was that's the way it is, no explanation, no proof. Saying you lose 1/2 your tones in 1 stop, and again lose 1/2 in the next stop is incorrect (the histogram would be skewed if it were true all the time). Sorry, this has been discussed over and over, everyone with their own views. It would have been better if it was discussed in the form of 100%, 200%,400%,etc, or Zone 1,2,3, or gamma .3,.6,.9 to represent 1 stop changes. Then it would have ben less confusing.

 

The exposing to the right is partly correct, you get better shadow detail so long as you protect the highlights. You get less noise. But you do not lose your tonal values with each 1 stop decreas as he describes. So he is half right.

The main thing to consider with histograms is that you should <b>not</b> see pixels piled up at either the left edge of the graph (indicating pixels gone to pure black) or at the right edge of the graph (indicating pixels gone to pure white.)

 

If you're talking about in-camera histograms there is another issue to be aware of, and that is that the histogram you're seeing is built from a weighted <b>average</b> of the three color channels. The implication here is that even a nice-looking histogram may well have pixels within certain color channels that have been pegged at zero ("black") or full-scale ("white.") Bottom line, for in-camera histograms, you want to be conservative -- try to get a "shape" that tails off comfortably above zero and comfortably below full scale. Easier said than done...

Ellis Vener wrote: <i>There is no suckh thing as a perfect shape t oa histogram --that's dependent on the content of the individual photograph</i><p>

Worth repeating.<p>

I've got a simple histogram '<a href="http://www.kevinconnery.com/classes/histogram/index.htm">quiz</a>' I use in class.

If you're up for it Benjamin, there was a <a href="http://www.photo.net/bboard/q-and-a-fetch-msg?msg_id=00MjK8">discussion</a> a while ago about the histogram and the "Expose to the right" technique of digital capture. Van was arguing what he is saying here, and a few others were arguing the opposite side.<br><br>

 

This was just to let everyone know that there are opposing points of view to Van's argument, but I just couldn't be bothered writing them all down again.

Bernie, the math is wrong. You cannot derive a formula/theory from a mathematical representation that only represents a doubling of something.If you do not understand this, that is not my problem. I tried explaining it as simple as possible. I suggest you pick up a math book. But don't imply I have no right to comment in here about it or any other place. We hear enough of the same thing about Epson scanners, why are you not complaining then? If something is misleading, I will continue to correct them. Mr. Reichman and Thomas Knoll have a few things to learn about math. Although Knoll knows programming, it doesn't make him a mathematician. As for Reichman, he is a photographer, that doesn't make him an expert. Neither is qualified. As before, the differenced between two points is 1 stop, 1 zone, .3 gamma, not the amount of tones.

 

Regardless, it doesn't really matter. By the time you check your histogram, your R/G/b channels, I will have captured the sunset (bracketed) and be gone.

I just started shooting RAW and was told by several folks in numerous forums I visit not to

trust the incamera histogram on my Pentax K100D because it's based on incamera

processing and not the true RAW data captured by the sensors.

 

Not true from my experience. Shot a brightly lit limestone brick wall and adjusted

exposure so the histogram just barely showed highlites blown. In fact there were tiny

areas flashing on the LCD telling me some highlites were gone.

 

Opened it in RAW Developer and turned off all settings except the color matrix and the

default tone curve. The image was very dark with RD's histogram data slammed to the left

side. Applied a tone curve to see if the highlite areas that were flashing on the LCD were

truly posterized which means they were blown and they were.

 

I even adjusted exposure slider and also tried isolating the posterized flat highlites

applying fine curve tweaks to get extra levels to darken and provide a tonal roll off.

Couldn't do it.

Van, I wasn't having a go at you, I was just letting the OP know there is another side to the story.

 

Tim, I'm not sure how RAW Developer works, but it sounds like it is giving you the linear raw output, in which case it will be very dark and stacked to the left. Is there a way for you to uncheck the linear option? If you do that, it will apply the correct gamma correction to the linear data before white balancing and rendering to a colour space. It should look a lot better then.

Bernie,

 

I intentionally applied the linear response to rule out any questions blown highlites still

reside in the image without any additional RAW processing. My point being I've been told

so many things about digital capture and histograms that I'm not sure if anyone really

knows what they're talking about. Or maybe they just don't fill in the details as to what

they mean.

 

Should I trust the incamera histograms when shooting RAW? It's the only thing anyone has

to go by at the time of capture. I demonstrated that you CAN trust the histogram in

regards to blown highlites. Yes, expose to the right making sure you can see within the

incamera histogram data tapering off or not spiking to absolute white without touching

the right side.

 

Most aren't concerned about the miniscule amount of highlite detail indicated in my

previous post. I just brought it up to nail down what has been constantly

miscommunicated about what to trust from the sensor with each shutter release. Everyone

resorts to math in these discussions rather than filling in the details on what's important-

how much real image detail did the photographer capture, how can he/she tell at the time

of capture and is it enough to work with in post.

 

Histograms are very useful in this regard.

[[if something is misleading, I will continue to correct them]]

 

You have not responded to Colin's post of Sep 29, 2007; 08:11 a.m. Why is that? I'm not saying this to be argumentative. It seems to me that he describes the whole process quite succinctly for that thread. I take it to mean that he is correct in his response.

Here's a RAW Developer screencapture showing blown highlites in RAW capture linear

preview.<div></div>

Tim, as far as the in-camera histogram goes, this is what i know: It is based on the in-camera jpg obtained from the raw->jpg conversion. The jpg will be converted according to whatever settings you have set in camera. In general, you will see clipping earlier in a jpg than in the raw data. So, the implication is that if you are doing Expose to the Right, then you might not get all the way to the right edge of the histogram by judging off the in-camera jpg. This is why some people suggest setting the in-camera jpg to be of the lowest contrast possible, so that the highlight levels will be as close as possible to what is available in the raw data. But in practice, does the tiny difference between the in-camera jpg and the post-capture raw data really matter that much?

Bernie,

 

It does when you want to edit what you can't see on tiny LCD. Once in jpeg it's rendered

and the data is gone if there ever was any captured. Just would like to know for sure what

the camera was capable of capturing in any given scene. The incamera histograms on my

Pentax for both jpeg and RAW are identical when taking two shots of the same scene at

the same exposure. With RAW you have far more to work with afterward.

 

Another issue shooting jpegs are compression artifacts at even the highest quality level

that show up in the form of blocky patterns in textures. However, these artifacts would

only show up in poster sized enlargements.

I wasn't advocating jpg over raw. I was talking about the jpg <i>histogram</i> vs. the raw <i>histogram</i>. My point was that the difference between the two will be so little as to not be concerned with. ie. if the in-camera jpg histogram is just beginning to clip, then the raw histogram will likely be so close to the right edge, as to not need anymore +ve EC to nail "Expose to the right". Infact, going by the jpg histogram is probably the safest thing to do (ie. just beginning to clip, or just short of this point), as it should ensure you don't clip highlights in your raw data (coz there is less clipping in the raw).

Bernie, sorry for the misunderstanding.

Bernie,

 

I get your point. Another tip I was given in the past was to address a concern constantly

adjusting exposure to nail "Expose to the right" and I was advised to just set the camera to

Program mode and just start shooting in RAW and fix in post. That doesn't work either

because that's when I get the MOST blown highlites depending on the dynamic range of

the scene.

 

I'm still shooting manual and adjusting exposure according to the histogram. This is what

works for me even though it slows me down when shooting outdoors on top of wearing

the batteries out.

 

Thanks for the exchange and nice talking with you.

"[[if something is misleading, I will continue to correct them]]

 

"You have not responded to Colin's post of Sep 29, 2007; 08:11 a.m. Why is that? I'm not saying this to be argumentative. It seems to me that he describes the whole process quite succinctly for that thread. I take it to mean that he is correct in his response."

 

Hi Rob, I was preparing for the last few days for a trip to Utah back then. When I no longer responded, it was because I was gone for a 3 week trip.

 

However, being absent doesn't make me wrong. I am also not the only one that disagrees. As mentioned before I believe the mathematical progression can not be used for anything other then what it was meant to explain. The progression 1,2,4....1024,2048,4096 can only represent one thing. It represent interval data, the difference between two points is always 100%, and you cannot subtract one from the other to get a different value(eg- 2048-1024 does not equal 1024 tones remaining, it only means twice/or half as bright). This scale does not allow us to subtract the values between 2 points and give it some meaning.

 

Although exposing to the right is beneficial when possible (due to highlight restrictions), the advantage of noise reduction by exposing to the right I do not feel can be explained due to reasons cited (only 64 levels remaining at the lower end). I believe it is for different reasons....electronic equipment has a natural tendency towards noise (scanners, electronic hum, radio telescopes, car radios exhibit plenty noise when the alternator brushes are failing) when approaching max sensitivity.

 

Well, I believe exposing to the right helps reduce noise when you can use it, I just do not believe in Riechmans levels theory.In my opinion when he says the first stop records 1/2 the information, the next stop down records half again....it should mean the first stop is twice as bright as the other. When the light gets low expect noise due to the nature of electronic gear.

 

Please note, I am not a camera mfr, I don't know how to build a camera, and how many in here do? So I really believe we all have an opinion, but I doubt we can explain it correctly with absolute certainty. Knoll and Reichman I respect for their contributions, b ut at the same time they would have limited knowledge like the rest of us about the mfrng and science of a printer, lenses, scanners, cameras, etc. They are firstly a programmer and a photographer. I'm outta here. Cheers.