As I learn to use my histogram, I am trying out a lazy version of "expose to the right", not wanting to mess with exposure compensation in post-processing unless this is needed to hold important shadow detail. I could call it keep away from the left.

In this laziness, the only time I feel a need to increase exposure is when I have both significant stuff near the left end of the histogram AND spare room at the right of the histogram; an "empty" part at the right.

How close one can safely go to the left is something to work out for the histogram of a particular camera (and maybe for different settings on the same camera, like RAW vs JPEG output). Maybe a useful threshold is the point on the histogram corresponding to about two or three stops below the mid-tones, or wherever your shadows start to crumble.

Calibrated like that, it does not matter if the scale is linear, logarithmic, or whatever.


That is, if the histogram is almost emply at the left, then it does not matter if the right end of the histogram is also empty, it just means that you have an easy, moderate contrast subject and so plenty of exposure latitude. Sometimes, this latitude might be best used to increase sharpness, through increased DOF or increased shutter speed at the lowest safe exposure level for the subject. (Similar to what Ansel Adams suggests with low contrast subjects, despite the very different technology!)

Quote (Jonathan Wienke @ April 14 2004,19:57)

The reason "expose to the right" is so important is that every digital image sensor in existence records linear values for light levels, so the data straight off the sensor always looks like the first image.

(actually, data straight out of processing with gamma of 2.2, which is what histograms are almost always dealing with)

Thanks Jonathan for a compelling way to visualise the hazards of leaving important details stranded in the low levels.

Oops, keyboard cursors have strange meaning in this post window... one more try:

Hi,

I still have to say that all (though quite amateurish I have to admit) tools I use have histograms with linear intensity scale:

Camera: Minolta Dimage A2
SW: Dimage Viewer, PaintShop Pro, Zoner Media Explorer

Jonathan showed us a picture with spread spectrum, that shows clearly the potential of picture processing, but can't show us what I have in mind.

Also, it reminds me of my most favourite painting - Kuindzhi's
"Night over Dnepr" - the most mysterious play with the light a man did with a brush. See:

http://www.center-gallery.ru/works/middle/kuindzhi8-middle.jpg
http://www.center-gallery.ru/book/kuindzhi-eng.shtml

and don't miss it when you visit Moscow (original in Tretiakovskaya gallery) or Sankt-Peterburg (copy in Ermitage).

Back to the subject:
Please do the following test with your cameras to see what do you have:

1.
Take a picture of a white (gray) sheet, lit with a colour neutral light (to eliminate colour corrections and channel differences), with automatic exposure. Most probably you will see the histogram of such a picture to have a very narrow peak near - the camera most probably decides to take the exposure in the middle of its dynamic range.

2.
Inspect the picture in your processing software. Most probably you will see the narrow peak at the same position as it is in the camera.

3.
If your SW allows this, check the average intensity value, or median value of the histogram. It should confirm that it is close to the 1/2 of the maximum intensity value.

There are 2 possible results of this test:

A.
- Peak in the camera in 1/2 of the histogram
- Peak in the SW at the same place
- Average histogram value close to the 1/2 of the camera linear dynamic range

B.
- Peak at the border or 4th and 5th segment of the histogram
- Native SW histogram at the same place, general SW like I use shows linear intensity values, so it will be in the middle.
- Average histogram value again close to the 1/2 of the camera linear dynamic range

In my case, it is A. - average histogram intensity value is 126 (8-bit, range 0-255), shown in the middle of the histogram both in-camera and in all 3 SW I use. I tested both JPG and RAW camera output.

If your camera displays middle value on the border of 4th and 5th segment, it indeed uses logarithmic intensity scale.

There is of course the possibility that cameras take a different exposure - it would be then good to compenstate so that the peak is somewhere at well-defined position of the histogram to make later readout/comparison easier.

By no means I question the "expose to the right" idea. I just want to understand things better, and feel that there is some myth to be explained.

Please let us know here what your results are.

Thanks, Jan

Quote (Jan @ April 15 2004,16:16)

I still have to say that all (though quite amateurish I have to admit) tools I use have histograms with linear intensity scale: Camera: Minolta Dimage A2 SW: Dimage Viewer, PaintShop Pro, Zoner Media Explorer 1. Take a picture of a white (gray) sheet, lit with a colour neutral light (to eliminate colour corrections and channel differences), with automatic exposure. Most probably you will see the histogram of such a picture to have a very narrow peak near - the camera most probably decides to take the exposure in the middle of its dynamic range. 2. Inspect the picture in your processing software. Most probably you will see the narrow peak at the same position as it is in the camera. 3. If your SW allows this, check the average intensity value, or median value of the histogram. It should confirm that it is close to the 1/2 of the maximum intensity value.

A  peak near the middle of the histogram in this case means the histogram is NOT linear. A linear histogram would have the peak around the 18% mark - middle gray is 18% reflective.

A value of 128 out of 255 in an imqge editor is NOT 1/2 the linear sensor value, but close to 18% gray since standard monitors have a gamma value near 2.2 for PC monitors or 1.8 for Macintosh monitors.

If a camera or image editing program had a linear histogram, the histogram  would be bunched up on the left side for every picture, making it useless. (the right side would only contain the very brightest highlights of the photo, the midtones would be 1/5 of the way from the left, so 1/2 of the visible data would be in the leftmost 1/5 of the histogram)

If you want to see a linear histogram, take a RAW picture, convert it with a RAW converter to a TIFF with 1.0 gamma (which will look very dark), load that TIFF into an image editing program and look at the histogram - you will see the histogram bunched up in the left 1/5 of the display.

Hi,

One more late-Sunday night comment:

The Night over Dnepr painting is irreproducible, to see its strength you have to see the original.

It's all made from tiny horizontal yellow stripes on a black background, some of them cut deep into the black paint.

Maybe soemone who lives nearby could post here a detail how it's done - it would be a very interesting old-fashioned "pixel-peeping" this time.

Jan

Hello,

I have one comment to the above tutorial:

The tutorial suggests that 1/2 of the signal intensity values are grouped in the rightmost fifth of a histogram. This is correct if histograms have logarithmic scale.

So far, in processing software and few cameras I had in hands, I found histograms that are linear - in other words (imagine 8-bit scale 0-255), 0 value on the left, 128 in the middle, 255 on the right.

In these histograms, spreading of signal to the left is not that dramatic as if the histograms were logarithmic (the leftmost fifht of histogram contains equal number of values as the rightmost fifth - approx 51, who counts.)

But the main message of the tutorial of course holds also for these histograms - the signal/noise ratio is indeed better for higher intensity values, so it is always advisable to shoot ()right, if one can manage the post-processing.

Have a good light,

Jan

No camera in the world has a linear histogram display. If they did, the data would alway be very crowded on the left side and there would be practically nothing in the rightmost 4/5 of the histogram unless the image was massively overexposed. This is a linear conversion of a properly exposed RAW image:



If you look at the histogram of this image, it's all bunched on the left. Here's what it looks like after a tone curve has been applied to bring the pixel values in line with the 2.2 gamma of the sRGB color space:



If you look at the histogram of this image, it starts making sense; the distribution of values is much more even across the spectrum from minimum to maximum.

The reason "expose to the right" is so important is that every digital image sensor in existence records linear values for light levels, so the data straight off the sensor always looks like the first image. This means that all of the shadow tones are crammed into relatively few discrete values in the raw sensor data, and any underexposure makes this situation even worse.

As long as you don't crowd the histogram up against the right edge and lose color saturation in the highlights (the more saturated the highlights, the more room you need to leave on the right) or clip, the better the final image will be. The optimum tradeoff between highlight desaturation and clipping and shadow noise will vary somewhat from camera to camera, but biasing the histogram to the right is almost always a good idea.

Hi,

I still have to say that all (though quite amateurish I have to admit) tools I use have histograms with linear intensity scale:

Camera:Minolta Dimage A2
Dimage Viewer
PaintShop Pro

I don't think the histograms on the LCD of a DSLR are sufficiently detailed and accurate for this kind of analysis; at least not on my D60.

For example, I recently took a few shots of water skiers on a nearby recreational lake, with Canon 100-400 IS zoom. Using the 'evaluative' meter setting, the histogram looked like a narrow, tall mountain peak slap bang in the middle of the range. Hmm! Plenty of scope to increase exposure, thinks I. So I use half the shutter speed and the narrow peak shifts somewhat to the right, but it's still not all the way there.

Unfortunately, the blown highlight indicator starts flashing and I assume that's the white turbulence in the waves which represents a very small proportion of the total picture area and is therefore not noticeable on the camera's histogram.

However, Photoshop CS raw converter's histogram does show clearly what's happened and to retrieve detail in the frothy turbulence I have to pull back exposure by one stop to exactly the exposure the evaluative meter gave me in the first instance. Basically, the shot has been saved from ruin.

However, as Jonathan has mentioned, if this overexposure had been on a saturated color, say the richly tanned skin of a model, I would probably not have been able to retrieve the detail and/or color so successfully.

Quote (Jan @ April 15 2004,15:43)

Hi, I still have to say that all (though quite amateurish I have to admit) tools I use have histograms with linear intensity scale: Camera:Minolta Dimage A2 Dimage Viewer PaintShop Pro

You have no clue what you are talking about. NO digital camera has a linear intensity scale. Human vision operates on a logarithmic basis, not linear. The most common color spaces have a gamma of 2.2; this means that if you double the numeric color value (say from 50 to 100), the number of photons emitted from a properly  calibrated moinitor will increase by a factor of 2^2.2, or approximately 4.59. All color spaces and histograms are designed with a gamma factor built in, but all digital sensors have a linear response; the number of photons required to change the sensor output value from 50 to 100 only has to increase by a factor of 2, not 4.59. Converting from linear to logarithmic is where "expose to the right increases image quality; light intensity values are always disproportionately distributed toward the bright end of the scale. So keeping exposure at the bright end of the linear scale exponentially increases the accuracy of linear to logarithmic conversion process.

Quote (Jonathan Wienke @ April 16 2004,02:46)

All color spaces and histograms are designed with a gamma factor built in

I apparently missed this fact.
When all displaying devices take into account gamma factor (for image data that require it), and all histograms show values AFTER expansion, that's it. Histograms in relation to the expanded data are linear, but in relation with raw data taken from sensors (which matters here), logarithmic indeed.

I was fooled by the idea that histograms show always real values of related data.

Thanks for teaching me something!