Best quality - Overexpose or underexpose?

[arumdevil]

I've just recently got a 300D and haven't done a lot of photography before, so I'm finding many usefull hints here for which I would like to thank everyone.

One thing I found confusing is that I saw one post that said to shoot in favor of underexposure so as to keep detail rather than washing out the whites. Then I read another article that suggests the opposite as this will reduce noise in the darker areas.

I'm guessing that the best way probably depends on weather you shoot JPG or RAW. At the moment I shoot in JPG as I have only 1x 128 CF card and the photos I'm taking are not worthy of RAW yet (most of them anyway)

Anyway, to get to the point - which is right? and if the format (JPG/RAW) does make a difference then what should I do for each?

Coming from a Digital Audio background my instinct is to get levels as hot as possible without overshooting (ie. as bright as possible without overexposing). this is exactly what I would do when recording audio into my computer and it seems reasonable to assume it's the same with light, but of course I could be totally wrong - what do I know ;)

Those of you who have an opinion please tell me what you think, and WHY so If there are fairly mixed results I can see for myself arguments either way.

thanks in advance for any input, and thanks to the forum admins and members for the forum!

[PacAce]

Coming from a Digital Audio background my instinct is to get levels as hot as possible without overshooting (ie. as bright as possible without overexposing). this is exactly what I would do when recording audio into my computer and it seems reasonable to assume it's the same with light, but of course I could be totally wrong - what do I know ;)


Yup, that's exactly how it should be done..."going as close to the red as possible without actually going into the red". However, that's the key. How do you know when the needle is going to go into the red until it's already gone there? Fortunately you have the histogram to look at but it won't always tell you if you've blown a highlight, especially when there aren't that many highlight pixels to begin with relative to the other shades of gray.

Another thing I noticed is that even if the histogram shows that the highlights have NOT been blown, it is possible to blow details in one or more of the color channels, especially the red channel. And this isn't evident until you process the image in an editor and discovered the washed out color details.

[RichardtheSane]

I always try to over expose where possible, but not to the point where the whites will be burnt out. This does retain most detail in the image. I know the theory, but I am not to sure about how to explain it - have a read about the thoeru from the same place I learnt it

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

I would also read up on Histograms, invaluable info in the digital photography world :)

[CyberDyneSystems]

There is no "right" answer here.. as the best results would be an exposure where there are no blown out highlights,.. and no over dark areas...

But that is not allways possible.

"Blown highlights" are impossible to recover from. Once the area is "white" it will never regain any detail.

Overly dark areas however may have recoverable detail hidden within...

...but underexposing does mean loosing color info AND increasing noise.

This is why many recomend "exposing to the right"... (not necesarily overexposed... but that certainly can happen)

The best balance I guess is to try to expose as "bright" as you can without blowing out an highlights. (don't let it go "white" anywhere)

Good luck :wink:

[arumdevil]

thanks for those replies.

I just had a quick look at some of those links. As far as I can see my suspicions were correct - high exposure without clipping, reason being the top "Stops" contain the most dynamic range. This is exactly the same as digital audio recording so I'm not so surprised really ;)

thanks :)

[arumdevil]

I've got another question - I'm not sure if I am supposed to start another thread but it seems related to me so I'll put it in here.

What is meant by "high key" / "Low key" ?

I first heard this in "Better Digital Photography" that I bought at the airport the other day (I won't be buying that rat turd again). they mention "High Key" photos but give no explanation whatsoever. I got the gist of it form the histogram tutorial but could someone just clarify for me?

and finally, what is a good digital photography magazine in the UK?

cheers

[DocFrankenstein]

I don't get the tip from the luminous landscape.

The histogram on the camera is the same as in photoshop (looks the same) How does exposing it to the right will make better noise/image ratios?

[cmM]

Well, most of the time you'll see noise in the shaddows (or any darker parts) of your picture. Exposing it a little more will have more light reflected from those darker spots, therefore being in the "top stops". More light = more detail, because as you said, the top stops have the most dynamic range.

Am I making sense :roll:

[Volatile]

The scale of the histogram is logrithmic, so there is more data in the right hand side. Later, when you adjust levels, you will come out with more overall detail if you exposure was "to the right".

My problem always seems to be scenes that have a dynamic range that exceeds the sensor's capability, so you are forced to choose between "blown highlights" and "lost in shadow". The solution is exposure bracketing, ie shooting the scene twice, once to capture highlight details, second to capture shadow details. Tedious PSing ensues...

[Tom W]

thanks for those replies.

I just had a quick look at some of those links. As far as I can see my suspicions were correct - high exposure without clipping, reason being the top "Stops" contain the most dynamic range. This is exactly the same as digital audio recording so I'm not so surprised really ;)

thanks :)

Not only is it that similar to digital recording, but it also parallels analog recording where you recorded as high as possible without clipping to keep the softer sound passages above the noise floor. Same idea though - using as much of your dynamic range as possible.

The histogram is digital photography's answer to the VU meter. :)

[Jesper]

I don't get the tip from the luminous landscape.

The histogram on the camera is the same as in photoshop (looks the same) How does exposing it to the right will make better noise/image ratios?

Here is another explanation: Tonal quality and dynamic range in digital cameras (http://www.normankoren.com/digital_tonality.html)

If you look at the number of levels per exposure zone (scroll down a bit on the page mentioned above), you'll see that in the darker zones, you'll have much less levels than in the brighter zones. So in very dark parts of your photos, you only have 4 or 8 shades of black or gray, while in the brightest parts you have 2048 or 1024 shades of white or gray.

If you overexpose (but not blow it out), you'll have more shades of color available. When you correct the exposure on the computer, you'll have more image data available to work with.

You'll benefit most of this when you shoot RAW and convert your RAW image to a 16-bit image and use software that supports working with 16-bit images (such as Photoshop CS).

[arumdevil]

Not only is it that similar to digital recording, but it also parallels analog recording where you recorded as high as possible without clipping to keep the softer sound passages above the noise floor. Same idea though - using as much of your dynamic range as possible.

The histogram is digital photography's answer to the VU meter. :)

that's not quite true, it's definitely closer to digital audio. Most Analog media (ie. tape) doesn't have the nasty clipping effects you would get in a digital recording. In analog people often purposefully "go into the red" in order to oversaturate the sound, giving it a "warm" or "compressed" quality but without the harsh "clipping" you get in digital audio if you go over the red line.

Also in digital audio the lower end of the dynamic range has fewer bits to describe the levels and so anything recorded below the top 48 db (in 16 bit) will sound terrible if amplified.

I seem to be straying OT so I'll stop :)

[Tom W]

Not only is it that similar to digital recording, but it also parallels analog recording where you recorded as high as possible without clipping to keep the softer sound passages above the noise floor. Same idea though - using as much of your dynamic range as possible.

The histogram is digital photography's answer to the VU meter. :)

that's not quite true, it's definitely closer to digital audio. Most Analog media (ie. tape) doesn't have the nasty clipping effects you would get in a digital recording. In analog people often purposefully "go into the red" in order to oversaturate the sound, giving it a "warm" or "compressed" quality but without the harsh "clipping" you get in digital audio if you go over the red line.

Also in digital audio the lower end of the dynamic range has fewer bits to describe the levels and so anything recorded below the top 48 db (in 16 bit) will sound terrible if amplified.

I seem to be straying OT so I'll stop :)

I wasn't really talking about deliberately distorting sound like you might do with an electric guitar, but rather retaining the original fidelity as one might do when recording a concert or copying music from a CD or vinyl record (yes, I used to do that) with a plain, ordinary reel-to-reel or decent casette recorder. One could also use highlight clipping on a camera to get an artistic, if not accurate result.

Anyway, listen to the sound of a sine wave vs. a square wave of the same fundamental frequency and you'll see just how nasty analog clipping can be.

[DocFrankenstein]

The scale of the histogram is logrithmic, so there is more data in the right hand side.
That's what the article says and this is what I thought at the time I read it.

Basically the article says that the histogram is logarithmic in the camera...

Then I have a problem. I took a picture and I compared photoshop's histogram to the histogram displayed in the camera. They are identical. Absollutely the same. No shifts whatsoever.

But I know that photoshop's histogram is linear :shock:

[arumdevil]

I wasn't really talking about deliberately distorting sound like you might do with an electric guitar, but rather retaining the original fidelity as one might do when recording a concert or copying music from a CD or vinyl record (yes, I used to do that) with a plain, ordinary reel-to-reel or decent casette recorder. One could also use highlight clipping on a camera to get an artistic, if not accurate result.

Anyway, listen to the sound of a sine wave vs. a square wave of the same fundamental frequency and you'll see just how nasty analog clipping can be.

OK, fair point :)

[Tom W]

I wasn't really talking about deliberately distorting sound like you might do with an electric guitar, but rather retaining the original fidelity as one might do when recording a concert or copying music from a CD or vinyl record (yes, I used to do that) with a plain, ordinary reel-to-reel or decent casette recorder. One could also use highlight clipping on a camera to get an artistic, if not accurate result.

Anyway, listen to the sound of a sine wave vs. a square wave of the same fundamental frequency and you'll see just how nasty analog clipping can be.

OK, fair point :)

After I reread your post, I realize that while there are similarities, the digital and analog situations really aren't exactly the same. I guess that my original post kind-of implied that, but I was only trying to point out similar-ness rather than same-ness. And, I was trying to put it in perspective that us old analog folks can understand. :)

Now, when is Canon going to put 3 big VU meters on the back of the camera? :)

[Tom W]

The scale of the histogram is logrithmic, so there is more data in the right hand side.
That's what the article says and this is what I thought at the time I read it.

Basically the article says that the histogram is logarithmic in the camera...

Then I have a problem. I took a picture and I compared photoshop's histogram to the histogram displayed in the camera. They are identical. Absollutely the same. No shifts whatsoever.

But I know that photoshop's histogram is linear :shock:

Of course, someone has to remain on topic! :D

So, by linear, do you mean that the histogram is divided into several equal steps of "brightness" along the X axis? That would imply, of course, a linear scale. But how about the data? Does each equal increase in brightness correspond to a linear or logarithmic increase in exposure "intensity"? Or to put it another way, does each subsequent doubling in brightness show up on the scale as a linear doubling of distance from "0" or does it show up as an increasingly larger gain (3 db gain) as you get farther from "0" even though the scale is marked linearly?

I'm just trying to put this into a perspective that I can relate to. I haven't visited the log scale in quite a while.

[PaulN]

Tom,

Here's my understanding: The cmos chip (or ccd for that matter) is a linear device. Hence if you double the incident light intensity, you double the output voltage. Each pixel's output voltage is converted to a digital value (either 8 or 12 bits). Hence, for a 12 bit 10D photo, the range of intensity values spans 2^12 = 4096 possible values. Now, consider the highest possible intensity value (let's say it is 4096). One stop in intensity below this is at 2048, and one stop below this is 1024, another stop below this, 512, etc. Therefore we have the following:

a) Brightest 1 stop (from 4096 to 2048): 2048 different possible intensity levels

b) Next Brightest stop (from 2048 to 1024): 1024 different values
c) Next brightest stop (from 1024 to 512): 512 possible values
d) Next Brightest stop (from 512 to 256): 256 possible values
e) Next brightest stop (from 256 to 128) 128 possible values
etc, etc.

As you can see, the brightest stop has the largest range of different intensity levels, twice as many as the next stop, etc. Hence the origin of
"expose to the right" (without saturating the brightest channel.)

-paul

[Tom W]

Thanks,, Paul. When you put it like that, its pretty simple.

Now, are those lines on the camera's histogram (the verticle ones equally spaced along the X axis) representative of one stop each or are they just lines with no numeric reference? I hope that they are stops as that would be quite useful in determining how far an exposure might be improved, at least in some cases.

[Saturn]

Hello:

You could write 500 pages on this one! If you over expose, your highlights will not have any pixels. Going by the numbers in Photoshop, anything less than a 3% dot means paper white.

If you underexpose, lightening in Photoshop will greatly increase noticeable noise in your lower mid-tones and shadows. Using a program like Dfine will help reduce noise, without softening the edges in an image. I've also found that PhaseOne's raw processing abilities far exceed any other program to bring out shadow detail, without flattening, or increasing shadow noise in an image.

I learned to use the histogram function on my Canon digicam. Meter your scene and if you get an overexposure warning for the highlights, just stop down 1/2 stop. That usually takes care of the problem.

I've also gotten heavily into split exposure shooting. Camera on tripod, and shoot sometimes 2, 3 or 4 exposures with varying exposures. I then combine the files in Photoshop. The range of tones achievable with this technique easily exceeds film.

Thanks.

[PaulN]

Tom,

The 10D manual does not say whether these horizontally spaced (but vertically oriented) lines are in stops. There are 4 lines on the 10D's LCD histogram, thus dividing the brighness space into 5 regions. If these lines are arranged in 1 stop divisions from zero to maximum intensity, that would say that there are only 5 stops of dynamic range available, and this is inconsistent with producing a 12 bit image. Hence, I would say that these lines are not arranged in 1 stop intervals (i could be wrong since Canon may have decided not to start the histogram at zero intensity).
We could test this ourselves with any manual exposure; just take a picture (on a tripod) so that a nice peaked histogram results. Then expose again with -1 stops of compensation. Check the histogram. Does it move to the left by one line? If so, I am dead wrong in my previous assertion. If I had my tripod here next to me I could test this right now, alas...

side note: One would think that if a camera produces a 12 bit image that one should be able to get at MOST 12 stops of dynamic range. Due to noise, I would expect that this limit is of course reduced. Some people have claimed that the 10D actually produces 8.5 stops of dynamic range, but I'd be interested to see the data and the the testing setup that produced this result.

-paul

[PacAce]

I have a feeling people are talking about apples and oranges in the same sentence and then trying to makes comparisons between the two. But you can't. The histogram is nothing more than a representation of how many pixels in an image is of a certain shade. The different shades of gray are represented in the histogram along the horizontal (X) axis. The number of pixels in a particular shade of gray are presented along the vertical (Y) axis.

This data in the histogram has nothing to do with the linear behavior of a sensor to light. The amount of light that strikes the sensor has to, in one way or another, get conveted to data that represents a certain shade of gray. And that is what is displayed on the histogram, not the intensity of the light that struck the sensor at a certain point.

[PaulN]

Leo,

Assuming you are right about the histogram a shade of gray, isn't it still true that a shade of gray is proportional to an intensity? The only way I can see this not to be the case is if the camera is converting the raw digitized values to a 8 bit histogram with a gamma correction applied)

I'll test the histogram tomorrow to see if the horizonal spacing on the 10D camera is in 1 stop or 2 stop (if the histogram is 8 bit, the lines would be 2 stop spacings) increments.

-paul

[PacAce]

Leo,

Assuming you are right about the histogram a shade of gray, isn't it still true that a shade of gray is proportional to an intensity? The only way I can see this not to be the case is if the camera is converting the raw digitized values to a 8 bit histogram with a gamma correction applied)

I'll test the histogram tomorrow to see if the horizonal spacing on the 10D camera is in 1 stop or 2 stop (if the histogram is 8 bit, the lines would be 2 stop spacings) increments.

-paul

I just tried it and it moves one line for every stop I adjusted my flash. In other words, I started with -3 FEC and the histogram hump was on the left side of the screen. -2 fec was on the 2nd line. I did this for -1, 0, +1, +2 and +3 and each time it moved up one line except on both ends where the hump was actually off the screen with only a small section of the hump visible.

[PaulN]

Leo,

was this on a 10D or a MKII? I ask because the 10D has only 4 lines on the camera's histogram.

-paul

[PacAce]

Leo,

Assuming you are right about the histogram a shade of gray, isn't it still true that a shade of gray is proportional to an intensity? The only way I can see this not to be the case is if the camera is converting the raw digitized values to a 8 bit histogram with a gamma correction applied)
-paul

Paul, yes, the shade of gray is proportional to the instensity of the light. However, the shades of gray are resented linearly. However, the voltage levels that are actually registered to represent these shades are not. In other words, pure white is represented by a voltage level between 2048 and 4096 where as pure black is represented by levels of 0 to 3 or something like that (sorry, I don't have that reference chart handy in front of me but you get the idea, right?). So, you've got 2048 levels of pure white range squished into that one segment of the histogram on the right and a handful of black levels spread out in the left segment of the histogram. And the other shades make up the 3 in-between.

[PacAce]

Leo,

was this on a 10D or a MKII? I ask because the 10D has only 4 lines on the camera's histogram.

-paul

Yes, you're right. I have the 10D and when I said 2nd, I meant 1st. I mistakenly said 2nd because I took the left edges of the histogram as the 1st and the right edge as the 6th.