What exactly does 800 ISO rating mean?

[Tim Nolte]

Apologies for what may seem like a basic question.

I've read countless times that the MX sensor is rated at 800 ISO, or the sweet spot is 800, or one should light for 800, or some such variation. My question is, since it's shooting RAW and I can use tools like FLUT and curves to push or pull the image six ways to Sunday, what exactly does it mean that the MX sensor is an 800 ISO sensor? Is 800 ISO the exposure at which exactly half of the dynamic range falls above medium gray and exactly half falls below? I.e. if I light a gray card for 800, will it show up as a spike smack dab in the middle of the RAW histogram? Or is it more subjective than that, and just that 800 feels like the right balance between noise and highlight protection?

 

[jimhare]

My understanding is that if the histogram is compressed at the top or bottom, you are limiting your dynamic range plus losing highlight protection.

Makes sense if you consider the dynamic range is finite so it's not so much about achieving balance, it's maximizing the range.

Feed the sensor what it wants and it will treat you extremely well!

 

[Ben Manning]

Have a read of this (Exposing the RED MX) then at the bottom there is a link to an interview with Ted from RED....Well worth a listen.

 

[Tim Nolte]

My understanding is that if the histogram is compressed at the top or bottom, you are limiting your dynamic range plus losing highlight protection.

Makes sense if you consider the dynamic range is finite so it's not so much about achieving balance, it's maximizing the range.

Feed the sensor what it wants and it will treat you extremely well!

This makes sense to me and is what I'm inclined to do, however it also means I'm more or less ignoring the whole 800 ISO thing and any traditional metering, and simply lighting using the exposure tools on the EVF. Since getting things spread across the RAW histogram is the best way to expose, I would hope that lighting to 800 ISO does just that, because then a light meter becomes a useful tool again. Is this true?

 

[Richard Lowther]

I have found when I set the light meters to 800 the raw histogram always looks under exposed. This doesn't seem to effect the MX sensor as its so clean, but I always wonder if having the histogram so back end weighted will this effect the DR of the camera? Is it better to have the Raw histogram so it is spread out from just under white clip to just above black clip and does this give you more to work with in post?
Thanks

 

[Pietro Impagliazzo]

Tim, found an article that briefly mentions that M-X is more in tune with light meters:

http://www.reduser.net/forum/showthread.php?t=42233

 

[Tim Nolte]

Tim, found an article that briefly mentions that M-X is more in tune with light meters:

http://www.reduser.net/forum/showthread.php?t=42233

Thanks, Pietro, this looks fantastic. I'm headed to work, but I can't wait to come back and read it.

 

[Tim Nolte]

I have found when I set the light meters to 800 the raw histogram always looks under exposed. This doesn't seem to effect the MX sensor as its so clean, but I always wonder if having the histogram so back end weighted will this effect the DR of the camera? Is it better to have the Raw histogram so it is spread out from just under white clip to just above black clip and does this give you more to work with in post?
Thanks

Great question, Richard. I've noticed that myself. You got me thinking about why exactly that is, and I have a guess now.

To be clear, I do not know if this is how the histogram works! That is exactly what I am trying to figure out. The following is to help illustrate that question. So.

Is this how it works?

The histogram displays the exposure data linearly. That is, since the brightest stop is twice as bright as the next brightest stop, it takes twice as much space on the histogram. If this is true, it explains the scrunched-to-the-left histogram. As you can see in my image, 50% brightness falls way to the left.

So, if this is true, does it still mean one should go for an evenly spread histogram, or the somewhat bunched to the left?

 

[Clint Lealos]

In the early days of Red it was sort of a general rule to expose to the right. Meaning bunch your histogram as much to the right as possible. The reason is that unlike film, RAW images are linear. Red has 4096 brightness values available. The brightest stop has 2048 brightness values. The next stop has 1024 brightness values and so on. Your diagram illustrates the point quite nicely You can see that the vast majority of brightness values are in the two boxes to the right. You want to fill up those boxes with as much information as you can without clipping the highlights. Essentially, making a "fat" negative to work with in post. Of course, certain scenes will never be bunched to the right. For example, night exteriors generally will be bunched to the left unless you are in Las Vegas Also, the MX sensor has changed the playing field a bit, but the math is still valid. Expose to the right without clipping and you should always be good to go.

 

[Alexander Ibrahim]

Great question, Richard. I've noticed that myself. You got me thinking about why exactly that is, and I have a guess now.

To be clear, I do not know if this is how the histogram works! That is exactly what I am trying to figure out. The following is to help illustrate that question. So.

Is this how it works?

<image excised>

The histogram displays the exposure data linearly. That is, since the brightest stop is twice as bright as the next brightest stop, it takes twice as much space on the histogram. If this is true, it explains the scrunched-to-the-left histogram. As you can see in my image, 50% brightness falls way to the left.

So, if this is true, does it still mean one should go for an evenly spread histogram, or the somewhat bunched to the left?

No, that is NOT how it works.

You've described a non-linear scaling of a histogram.

It's non-linear because you have weighted it to represent photographic stops.

What you have actually created is a geometric scale.

As Clint described above it does sort of represent a reality of how photography works... but your description does NOT represent how a histogram works.

So ... as usual with digital cameras you must expose to the right (ETTR)

That is to say, you want your image as far to the right as practical without clipping anything in the image.

As a secondary goal, you need to add enough light in so that the blacks don't clip either.

You don't generally want your histogram to "pile up" on either side ... you want it nicely and evenly balanced. When I am hired to color a picture usually the first thing I do- and the one that takes the most technical judgement- is fix exposure.

Speaking of which... I am supposed to be doing that- not fiddling around on RedUser.

 

[Tim Nolte]

No, that is NOT how it works.

You've described a non-linear scaling of a histogram.

It's non-linear because you have weighted it to represent photographic stops.

What you have actually created is a geometric scale.

Doesn't it depend on your units of measurement? Yes, if your unit is stops, each stop should take equal space on the scale, but AFAIK stops themselves are a logarithmic measurement of light energy. We use them because they reflect the way we perceive light, i.e. logarithmically. If you measure in energy (photons hitting the sensor), then for every stop of increase, there are twice as many photons hitting the sensor. Displayed linearly, that stop is twice as long as the previous one on the scale.

As Clint described above it does sort of represent a reality of how photography works... but your description does NOT represent how a histogram works.

I agree, the histogram on my DSLR definitely does not work that way. It devotes equal space to each stop of light, which is convenient, because it relates to how I perceive light. So the RAW histogram also works this way? I could see it working either way, especially since it is raw data. I was just speculating that it works the way I described, because that would explain the common left-biased-with-long-tail-to-the-right look of footage exposed for 800 ISO. I don't really care which way it works, it's just helpful to know! :laugh:

So ... as usual with digital cameras you must expose to the right (ETTR)

That is to say, you want your image as far to the right as practical without clipping anything in the image.

As a secondary goal, you need to add enough light in so that the blacks don't clip either.

You don't generally want your histogram to "pile up" on either side ... you want it nicely and evenly balanced. When I am hired to color a picture usually the first thing I do- and the one that takes the most technical judgement- is fix exposure.

This all makes perfect sense, thank you. Especially if the histogram is scaled in terms of light energy, not stops, it seems to me you'd want it spread across the whole range without bunching, like you said, because that means you're making the most use of your available bits.

Speaking of which... I am supposed to be doing that- not fiddling around on RedUser.

Addicting, ain't it? I just fiddled away my whole lunch hour. :biggrinjester:

 

[paulherrin]

the histogram is just a graph of your luminance data, based on a build up of per-pixel values. you want as much a range of values as you can get. the far left is pure black, the far right pure white. the middle represents .5 where black = 0 and white =1. with 8-bit RGB, when black is 0 and white is 255, the middle will be at 128. our vision is more logarithmic, so it doesn't necessarily match up to perception of light and dark.

what you generally want is to capture and spread out as much data as you can across the histogram, usually trying to avoid clipping blacks or whites. that's why the expose to the right (ETTR) usually works well as a catchall method. what you probably want is to use the available range to capture as many nuances between values as possible. open up photoshop, open up the histogram view, and manipulate an image to see how the histogram responds to different adjustments. you can also look at the individual color channel histograms, not just the combined luminance values.

sometimes you get an image that looks really good in frame, but then the histogram is maybe mostly in the middle and down towards the blacks, and nothing is clipping in the whites. what i like to do is go brighter than i know i want it, and go ahead and fill it up across there. if you adjust your exposure slowly brighter you will see it spread out across there. it's just picking up more data variation between values, which is good for color correction, among other things.

 

[Michael Kromschroeder]

Thank you Tim for the Question, this is something I was thinking about as well and didn't figure it out yet.

So what does "sweet spot" ISO 800 mean?

Michael

 

[Dan Hudgins]

midtone

midtone

the histogram is just a graph of your luminance data, based on a build up of per-pixel values. you want as much a range of values as you can get. the far left is pure black, the far right pure white. the middle represents .5 where black = 0 and white =1. with 8-bit RGB, when black is 0 and white is 255, the middle will be at 128. our vision is more logarithmic, so it doesn't necessarily match up to perception of light and dark.

Midtone (Kodak 18% gray card) is only 0.5 or 127 for 0 to 255 if your output is corrected for a gamma 2.5 display.

Since TVs and newer LCD monitors are corrected for gamma 2.2, midtone is about 0.458 not 0.5, that gives a 8bit value between 116 and 118 depending on how you round the numbers, I am using 116 as midtone reference gray in my programs.

Any time you grade the image and you have an 18% gray card in the shot, it should read code 116 in the 8bit monitor data or about 0.458 of the full signal for other bit depths.

Kodak used a 16% gray standard for the FILM LAD, and on their digital D-LAD they used code 445 but state that 18% gray should be code 470 (on 10 bit data gamma about 1.7 clipped between 95 and 685).

Anyway, if the 18% gray card does not read midtone for the output gamma it is over or under exposed for the ISO used.

If you grade the 18% gray card lighter or darker you are in effect changing the ISO of the exposure in post, so if the gray card is dark at exposure of 800 then when you make it brighter you have increased the ISO to maybe 1600 or as much as you lighten it.

Any grading has the effect of changing the ISO at the time of exposure if it makes the midtones lighter or darker.

 

[Dan Hudgins]

"sweet spot"

"sweet spot"

Thank you Tim for the Question, this is something I was thinking about as well and didn't figure it out yet.

So what does "sweet spot" ISO 800 mean?

Michael

The "sweet spot" is the exposure between over and under exposure.

In the digital camera it is where you are below the white clip enough to hold two to three stops of highlight detail, and still have enough bits under 2% black subject to get shadow detail and not too much noise.

In a 12bit sensor like M the "sweet spot" is about ISO 320.

M-X used a 14bit ADC on the sensor board (?) and it seems log encoding before being recorded 12bit, so you get some extra shadow detail and less compression effected noise in the dark areas, so you can "underexpose" the sensor more going down about 1.5 bits (~=to stops also) so you get 800.

You need to be 10-11 bits above the bottom to have a good 8 bits after gamma 2.2 adjustment since you don't want histogram gaps in the dark areas of the image (10bits linear = about 8 bits at gamma 2.2 before histogram gaps show up).

If max esposure is 160 before green pixels clip (on 90% white subject), then 320 is one stop down giving about 10 bits from the bottom in a 12bit ADC output.

14bit ADC has 2 more bits, so you can get maybe 2 more stops, but other sources of noise may not drop enough to get the same fixed and moving noise at 1280, so 800 is between 640 and 1280 in the "sweet spot" where you avoid shadow noise and histogram gaps and have enough underexposure of the sensor to avoid blown out highlights two to three stops above a 90% white subject.