is this right?

[george013]

In another thread, which is closed by a friendly mod, some calculations where used. I had my doubts about those. It's a way of calculating one see by time on the internet. Also in that article about ETTR of Reichmann which can't be found anymore.

Let's show the post.

As my raw histograms fade into obscurity, I recall that the max values were about 3700 presumably out of 4096 levels (12-bit ADC). Half a stop more is a factor of x1.414; therefore the max sensor raw exposure values would have been 3700*1.414 = 5233 meaning clipped highlights because the ADC can not render anything > 4095. In other words, half a stop up from Brian's original setting would have blown some highlights at the sensor.

What's going on here.
1) an assumption that by multiplying the digital values one can simulate another exposure.
2) that simulating a half stop means multiplying those digital values with a factor 1.414

Is there anybody with some knowledge who can help me out?

George

[xpatUSA]

In another thread, which is closed by a friendly mod, some calculations where used. I had my doubts about those. It's a way of calculating [that] one [sees from time to time] on the internet. Also in that article about ETTR of Reichmann which can't be found anymore.

Let's show the post.
<see OP for text>

What's going on here:
1) an assumption that by multiplying the digital values one can simulate another exposure.

It was not an assumption! It was a statement taken from general knowledge.

Sensor:

"Digital sensors are very linear for the vast majority of the response curve, with deviations at the very bottom (due to noise, depending on where the black point is set) and at the very top near saturation. Digital sensors are particularly linear when compared to film, which has a pronounced 'S' shape response curve."

https://photo.stackexchange.com/ques...e-dslr-sensors

Prove to us that digital camera sensors are not essentially linear.

ADC output:

The ADC used in digital cameras is linear:

http://kronometric.org/phot/post/CiC...rge/AD9235.pdf

Prove to us that a general-purpose ADC is non-linear.

Anticipating that you can offer no such proof: it should be obvious, even to your good self, that the ADC output is linear with respect to sensor exposure (unless the sensor is saturated). It should also be obvious that if an exposure of x causes an ADC output of y, then an exposure of 2x will cause an ADC output of 2y.

2) that simulating a half stop means multiplying those digital values with a factor 1.414

Since I have proved above that the ADC output is linear with respect to sensor exposure, it follows that if an exposure of x causes an ADC output of y, then an exposure of 1.414x will cause an ADC output of 1.414y.

Is there anybody with some knowledge who can help me out?

George

[Manfred M]

Also in that article about ETTR of Reichmann which can't be found anymore.

Luminous Landscape turned into a pay site a few years ago; cost is $1US per month, so if is not exactly expensive and it remains as one of the most informative sites out there.

As a paid up member, I can still access the article, but unfortunately it is no longer available for non-members.

[Polar01]

Yes it is well worth the $12.00 US per year. A lot of great articles and features.

Cheers: Allan

[george013]

Enough views for a thread without responses.

Let me make it a bit simpler. Does an increase of exposure of 0.5 result in 1.414 times more light on the sensor or 1.5 times.
Dan is invited too for this "discussion".

George

[revi]

Increasing EV by 1 doubles the light intensity reaching the film/sensor.
Thus, light intensity vs EV is an exponential function, of the form I = 2^(EV) (leaving out some constants to simplify)
Increasing the EV by .5 gives us then I = 2^(EV+0.5) = 2^(EV)*2^(0.5),
so increasing the EV by 0.5 multiplies the intensity by 2^(0.5)
2^(0.5)=sqrt(2) ≈ 1.414.

And, to a good approximation, digital sensors have a linear response to the incoming light.

[george013]

Increasing EV by 1 doubles the light intensity reaching the film/sensor.
Thus, light intensity vs EV is an exponential function, of the form I = 2^(EV) (leaving out some constants to simplify)
Increasing the EV by .5 gives us then I = 2^(EV+0.5) = 2^(EV)*2^(0.5),
so increasing the EV by 0.5 multiplies the intensity by 2^(0.5)
2^(0.5)=sqrt(2) ≈ 1.414.

And, to a good approximation, digital sensors have a linear response to the incoming light.

Thanks for the response. I've been calculating too.
First the examples of Ted.

Let's shoot the computer monitor, say 300 cd/m^2

0.65*300*1/50 / (2.8^2) = 0.497 lux-sec where time = 1/50 sec and f-number = 2.8

Let's up the exposure half a stop:

0.65*300*1/50 / (2.4^2) = 0.677 lux-sec where time = 1/50 sec and f-number = 2.4, from:

https://en.wikipedia.org/wiki/F-numb...f-number_scale

According to you, George, that second number should be 0.746 lux-sec which it obviously is not.

Here he's saying that according my figures the second number would be 0.746. He doesn't say that according his numbers it would be 0.703.
I made a spreadsheet where I calculated the f-number my self, out of the aperture value, exposure value without exposure time. Getting than f-numbers of 2.378414.. and 2.8284...gave me a result of 1.414 too. I assumed that was the factor f-number are changed with. Just by coincidence.
Back to that formule 0.65*300*1/50 / (2.8^2) = Hm, the focal plane exposure, in lx*sec. If I want to enlarge that with half a stop, so 50%, I must create a condition where Hm1=1.5Hm2. Is this right?
In this example the only variable is the f-number. And within that f-number the aperture diameter.
When I do some calculations based on that aperture area I got D1^2=1.5D2^2. That would give 1.22D2=D1.

The sensor can have a lineair response to the incoming light. Even the digitized analogue values. But I doubt one can calculate with them.
A simple A/D converter takes an analogue range and a division number. That range can be between 10 and 14 stops, just to mention something.
Going further with that calculation stuff would mean that when shooting a greycard it would result in a line at 127. I only have 1 stop left. I can do much more on my camera.

George

[xpatUSA]

Back to that formule 0.65*300*1/50 / (2.8^2) = Hm, the focal plane exposure, in lx*sec. If I want to enlarge that with half a stop, so 50%, I must create a condition where Hm1=1.5Hm2. Is this right?

NO. Half a stop is not 50% more. You have already been told this by Revi.

https://en.wikipedia.org/wiki/Exposure_value

Please provide a link to where it says 1/2 stop = times 1.5

Even the digitized analogue values. But I doubt one can calculate with them.

LOL

Going further with that calculation stuff would mean that when shooting a grey card it would result in a line at 127.

Please show us your further calculation in full ... not holding my breath, though.

Meanwhile, with the modern SOS method of ISO determination, that number is 118, not 127:

http://www.cipa.jp/std/documents/e/DC-004_EN.pdf

With the older saturation-based method it is even less, about 100.

I only have 1 stop left.

No, you do not. "Stops" only apply to exposure of the sensor, not to the resulting RGB image pixel values.

Staying with your number of 127 and expressing exposure Hmean as a fraction of Hsaturated:

Hm/Hsat = (127/255)^2.2 = an exposure of 21.57%, where 2.2 is a simple value of gamma.

Now for the maximum value:

Hm/Hsat = (255/255)^2.2 = an exposure of 100.00%

So the number of "stops left" is:

EV = log to the base 2 of (100/21.57) = 2.2 stops.

That proves that you will have 2.2 stops of exposure "left", not 1.

It's a pity that you can not accept what more knowledgeable people keep telling you, George.

[DanK]

Let me make it a bit simpler. Does an increase of exposure of 0.5 result in 1.414 times more light on the sensor or 1.5 times.
Dan is invited too for this "discussion".

Revi answered this more completely in this thread, but I gave you a sufficient answer in the last thread. So you have now had the reason why it is ~ 1.4 explained to you in at least two different forms, and I also gave you in the last thread a concrete example showing that it cannot be 1.5. I really don't know what more to say. Unless you can show us that 1.5 x 1.5 = 2, it CAN'T be 1.5.

[xpatUSA]

Revi answered this more completely in this thread, but I gave you a sufficient answer in the last thread. So you have now had the reason why it is ~ 1.4 explained to you in at least two different forms, and I also gave you in the last thread a concrete example showing that it cannot be 1.5. I really don't know what more to say. Unless you can show us that 1.5 x 1.5 = 2, it CAN'T be 1.5.

It's a great pity that George has cast doubt on some of my work which is quoted in the OP of his thread. That forces me to respond again and again to George's continuing recalcitrance. Otherwise, I'd just shut up and laugh at his outrageous claims, LOL.

[george013]

Ok. I'm confinced. About the fractional stops. My point of view was that if stops are optical values, one stop is double or half the exposure, the fractional stops would be something alike. It's more a mathematical approach. I didn't try it yet, I wonder how to calculate the according shutter speed.. I don't need to know, my light meter knows.

Next one.
If it is ok multiplying a pixel value wit 1.414, half a stop, why is it not ok to multiply with 2, a full stop.?

George

[xpatUSA]

If it is ok multiplying a pixel value wit 1.414, half a stop, why is it not ok to multiply with 2, a full stop.?

It is not OK to multiply an image pixel value by 1.414 and call it half a stop and never has been; therefore, it is not OK to multiply a pixel value by 2 and call that a full stop.

Stops only apply to sensor or film exposure. Between the sensor and an RGB image, gamma is applied; the value of gamma is often different for the common color spaces. Let's think in terms of exposure as a fraction and let's use an 8-bit image (0-255 levels); let's use Adobe RGB, for which the gamma is a simple 1/2.2.

Let's use an exposure of 1/8 and then an exposure of +1 stop which is two times 1/8 = 1/4.

We convert from fractional exposure to pixel value as follows:

pixel value = (exposure fraction raised to the power of 1/2.2) times 255. Trust me.

pixel value = (1/8)^(1/2.2) * 255 = 99

For 1 stop more, pixel value = (1/4)^(1/2.2) * 255 = 136

You, George, might have expected the pixel value to double but it does not because of the gamma. In other words, a pixel value of 136 compared to 99 represents a one stop higher sensor exposure, but the ratio 136/99 is not 1 EV.

[george013]

It is not OK to multiply an image pixel value by 1.414 and call it half a stop and never has been; therefore, it is not OK to multiply a pixel value by 2 and call that a full stop.

Stops only apply to sensor or film exposure. Between the sensor and an RGB image, gamma is applied; the value of gamma is often different for the common color spaces. Let's think in terms of exposure as a fraction and let's use an 8-bit image (0-255 levels); let's use Adobe RGB, for which the gamma is a simple 1/2.2.

Let's use an exposure of 1/8 and then an exposure of +1 stop which is two times 1/8 = 1/4.

We convert from fractional exposure to pixel value as follows:

pixel value = (exposure fraction raised to the power of 1/2.2) times 255. Trust me.

pixel value = (1/8)^(1/2.2) * 255 = 99

For 1 stop more, pixel value = (1/4)^(1/2.2) * 255 = 136

You, George, might have expected the pixel value to double but it does not because of the gamma. In other words, a pixel value of 136 compared to 99 represents a one stop higher sensor exposure, but the ratio 136/99 is not 1 EV.

I didn't expect anything. I doubted that what you where doing was right, multiplying the pixel value to simulate another exposure. That means the assumptions in that thread where all wrong.
And I'm still not convinced. As said before in an A/D conversion you have a range and a divider. That divider can be anything, 8 bit, 12 bit etc. But that range is specific for the sensor. It covers the dynamic range of it be it 10 or 15 stops just to say something.
This use of the gamma correction is still something I don't understand. But let that be another discussion/lesson.
I don't believe you can use pixel values to calculate with as you did and do.
When I remember well it was Reichmanns article of ETTR where he used this. The last stop covers the pixel values 127-255, the second last stop 64-127 etc. Strange, mostly nothing disappears on the internet, but this does.

George

[dje]

I didn't expect anything. I doubted that what you where doing was right, multiplying the pixel value to simulate another exposure. That means the assumptions in that thread where all wrong.
And I'm still not convinced. As said before in an A/D conversion you have a range and a divider. That divider can be anything, 8 bit, 12 bit etc. But that range is specific for the sensor. It covers the dynamic range of it be it 10 or 15 stops just to say something.
This use of the gamma correction is still something I don't understand. But let that be another discussion/lesson.
I don't believe you can use pixel values to calculate with as you did and do.
When I remember well it was Reichmanns article of ETTR where he used this. The last stop covers the pixel values 127-255, the second last stop 64-127 etc. Strange, mostly nothing disappears on the internet, but this does.

George

George have a look at the following four screen shots from RawDigger. They are raw histograms of images of a grey card. The first two were captured with 14 bits and and the difference in exposure is 1 stop. The third and fourth shots were captured with 12 bits and once again the exposure difference is 1 stop.









Take the green channels

Shot 1 : Avg 538
Shot 2 : Avg 1074
Shot 3 : Avg 132
Sgot 4 : Avg 265

You can see from this that pixel digital values for the higher exposure are double those of the lower exposure, for both 14 bit and 12 bit.

The LL article you refer to may be available to paid up members only, I don't know. Michael Reichmann passed away a couple of years ago.

Dave

[george013]

George have a look at the following four screen shots from RawDigger. They are raw histograms of images of a grey card. The first two were captured with 14 bits and and the difference in exposure is 1 stop. The third and fourth shots were captured with 12 bits and once again the exposure difference is 1 stop.









Take the green channels

Shot 1 : Avg 538
Shot 2 : Avg 1074
Shot 3 : Avg 132
Sgot 4 : Avg 265

You can see from this that pixel digital values for the higher exposure are double those of the lower exposure, for both 14 bit and 12 bit.

The LL article you refer to may be available to paid up members only, I don't know. Michael Reichmann passed away a couple of years ago.

Dave

I don't know where I'm looking at.
If EV means exposure value it's the situation of the camera at the moment of the shot: the used f-number and the used shutter speed.
If that's a greycard why have the different channels have such a different value?
If the horizontal ax gives the pixel values in 14 bit, then the value of 350 would become 5 in 8 bit. Very dark.
If the horizontal ax has two representations, pixel value and EV, and every EV is a doubling that value the question still is is that right?
I did have a fast look in the RawDigger manual but can't find an explanation of it. I might be there, but I didn't find it yet.

George

[dje]

Yes the exposure is quite low. I'll take some more shots in the morning with higher values. The x axes are in 14 or 12 bit numbers.

Are you saying the RawDigger people don't know what they are doing?

The channel values are different from each other due to the spectrum of the illumninant and the differing CFA filter responses (and the sensor silicon). This is where White Balance comes in. The red and blue channels are multiplied by factors to bring all channels equal.

[xpatUSA]

This use of the gamma correction is still something I don't understand.

I take it that you do know what gamma correction is. If not, our discussion is pointless.

https://www.cambridgeincolour.com/tu...correction.htm

Please study and understand this diagram:



Note that the output says "recorded value in a file" - that is the pixel value you keep referring to. Now, in my discussion, instead of 'original scene luminance' I have chosen to use fractional exposure - this is called "normalization" and is common in engineering and allows one diagram to be used for all cases. This diagram is actually normalized because units of luminance are not shown on the x-axis and units of pixel level are not shown on the y-axis.

DO YOU AGREE THAT THE ABOVE DIAGRAM IS VALID FOR OUR DISCUSSION?

CAN YOU SEE AND UNDERSTAND THE RELATIONSHIP BETWEEN IMAGE PIXELS AND EXPOSURE?

Please answer these questions and from there we can proceed ...

[george013]

Yes the exposure is quite low. I'll take some more shots in the morning with higher values. The x axes are in 14 or 12 bit numbers.

Are you saying the RawDigger people don't know what they are doing?

The channel values are different from each other due to the spectrum of the illumninant and the differing CFA filter responses (and the sensor silicon). This is where White Balance comes in. The red and blue channels are multiplied by factors to bring all channels equal.

No, I don't say that. I just can't find the explanation.
Take a shot with that card somewhere in the middle of the diagram, something as 127 in a 8 bits value.

George

[xpatUSA]

If the horizontal axis has two representations, pixel value and EV, and every EV is a doubling that value the question still is is that right?

In RawDigger, the histogram's horizontal axis does not represent pixels. it represents raw values captured by the sensor; you can not call these values "pixels" - that term only applies to an image and, as you yourself have told us many times, a set of raw values is not an image.

[george013]

In RawDigger, the histogram's horizontal axis does not represent pixels. it represents raw values captured by the sensor; you can not call these values "pixels" - that term only applies to an image and, as you yourself have told us many times, a set of raw values is not an image.

Sorry. Digitized analogue sensel values, in Dave's example 12 or 14 bits. So good?

George