Update on the vanishing blue.

[xpatUSA]

Not that it matters, but I compensate in steps of 1/3

As "most of us" do these days; but it might matter a tiny bit.

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.

Whether or not that matters with that particular scene is moot. Had it been a motor-cycle in full sunlight ... probably not.

I make the simple assumption that 4095 out of the ADC means a saturated sensor and gives 255/255 after conversion to 8-bit RGB.

At the same time, any shadowy part of the scene that was, say, 8/4096 at Brian's exposure would have increased to a whopping 11/4096 (8*1.414/4096) ... hardly enough to affect any conversion in terms of gamut clipping, IMNSHO.

[DanK]

As "most of us" do these days; but it might matter a tiny bit.

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.

Whether or not that matters with that particular scene is moot. Had it been a motor-cycle in full sunlight ... probably not.

I make the simple assumption that 4095 out of the ADC means a saturated sensor and gives 255/255 after conversion to 8-bit RGB.

At the same time, any shadowy part of the scene that was, say, 8/4096 at Brian's exposure would have increased to a whopping 11/4096 (8*1.414/4096) ... hardly enough to affect any conversion in terms of gamut clipping, IMNSHO.

Wandering off topic, but this got me thinking because I stick with increments of half a stop, with the rationale that 1/6 stop shouldn't matter and half stops are what I was used to from the old days. So that got me wondering: what if one uses 1/3 stop in this case? The relevant factor is the cube root of 2, which is 1.26. So, by the same logic, an increase of 1/3 stop would be a max exposure of 3700 x 1.26 = 4662: still clipping, but less severe by a bit.

[xpatUSA]

Wandering off topic, but this got me thinking because I stick with increments of half a stop, with the rationale that 1/6 stop shouldn't matter and half stops are what I was used to from the old days. So that got me wondering: what if one uses 1/3 stop in this case? The relevant factor is the cube root of 2, which is 1.26. So, by the same logic, an increase of 1/3 stop would be a max exposure of 3700 x 1.26 = 4662: still clipping, but less severe by a bit.

Quite so, Dan.

Ah, the old days ... my M42 lenses, marked in full stops with intermediate half-stop detents, if you're lucky.

My first Sigma DSLR only had 1/2-stop increments and raw-only files . . . quite nostalgic, really.

[george013]

As "most of us" do these days; but it might matter a tiny bit.

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.

Whether or not that matters with that particular scene is moot. Had it been a motor-cycle in full sunlight ... probably not.

I make the simple assumption that 4095 out of the ADC means a saturated sensor and gives 255/255 after conversion to 8-bit RGB.

At the same time, any shadowy part of the scene that was, say, 8/4096 at Brian's exposure would have increased to a whopping 11/4096 (8*1.414/4096) ... hardly enough to affect any conversion in terms of gamut clipping, IMNSHO.

I'm not sure. I wonder if this is correct.

George

[xpatUSA]

I'm not sure. I wonder if this is correct.

George

Classic George. Please tell us what "this" refers to and prove that it is not correct.

[george013]

Classic George. Please tell us what "this" refers to and prove that it is not correct.

A stop more means doubling the light, 100% more light. Half a stop more would mean 50% more light, a factor of 1.5. Which also can be gained by multiplying the exposure time with 1.5.
As you write: classic George. I'm not sure.

George

PS.
Beside this I'm not sure you can calculate with those digital values as you do.

George

[DanK]

George,

I'll show you concretely first. If you were right, what would happen if you opened up half a stop twice? Let L be the original amount of light.

First half stop: 1.5 times as much light, (1.5 x L)
Second half stop: 1.5 times as much light as the previous, 1.5 x (1.5 x L) = 2.25 x L [which of course should be 2 x L)

now try this same thing with the square root of 2, which is roughly 1.4, as Ted indicated.

You are thinking in additive terms, not multiplicative. The the relationship between f'/stops and light is logarithmic: one stop is always the same multiple of the amount of light, not the same additive increment

Dan

[george013]

George,

I'll show you concretely first. If you were right, what would happen if you opened up half a stop twice? Let L be the original amount of light.

First half stop: 1.5 times as much light, (1.5 x L)
Second half stop: 1.5 times as much light as the previous, 1.5 x (1.5 x L) = 2.25 x L [which of course should be 2 x L)

now try this same thing with the square root of 2, which is roughly 1.4, as Ted indicated.

You are thinking in additive terms, not multiplicative. The the relationship between f'/stops and light is logarithmic: one stop is always the same multiple of the amount of light, not the same additive increment

Dan

I think you're mixing up with f-numbers. There you've a factor 1.41 being V2.
The amount of light is determined by a) the f-number and b) the exposure time. A stop is half or double the light. Due to the formule of the area of the aperture, double or halve is done with a factor V2 or 1.41... But this is only for the f-number, not for the light.

And yes, 1.5x1.5=2.25.

George

[xpatUSA]

Beside this I'm not sure you can calculate with those digital values as you do.

George

Please prove that I can not "calculate with those digital values".

[george013]

Please prove that I can not "calculate with those digital values".

Classic Ted.

First thing first.

George

[xpatUSA]

I think you're mixing up with f-numbers. There you've a factor 1.41 being V2.
The amount of light is determined by a) the f-number ** and b) the exposure time. A stop is half or double the light. Due to the formule of the area of the aperture, double or halve is done with a factor V2 or 1.41... But this is only for the f-number, not for the light.

George

** Nope. The amount of light is determined by the reciprocal of the f-number squared . .

Exposure H = 0.65*scene luminance*exposure time / (f-number squared).

Proof:

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

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.

Now, if we change the monitor brightness instead of the aperture, it is equally obvious that the exposure in lux-sec will change in direct proportion but that can NOT be equated to changes measured in EV; therein lies your error.

[george013]

** Nope. The amount of light is determined by the reciprocal of the f-number squared . .

Exposure H = 0.65*scene luminance*exposure time / (f-number squared).

Proof:

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

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.

Now, if we change the monitor brightness instead of the aperture, it is equally obvious that the exposure in lux-sec will change in direct proportion but that can NOT be equated to changes measured in EV; therein lies your error.

You're right that the f-number and shutter speed don't determine the amount of light. They deal with the variation of amount of light.


Try this same formula with a full stop and it nearly is how I would expect. Don't trust the 2.4 and 2.8. They're numbers with 1 decimal.
The first part is a constant. The second part are the f-numbers in quadrat. With whole numbers no problem.

If you want double the light on your sensor you have to multiply the diameter of the aperture with 1,4..... or halve the shutter speed. It's in your formula too, except to less decimals.

It took some time for I had to see the blood moon.

George

[xpatUSA]

Originally Posted by xpatUSA Please prove that I can not "calculate with those digital values".

Classic Ted.

First thing first

Classic! See post #111 for my response to "first thing".

Now please prove that I can not "calculate with those digital values".

[xpatUSA]

You're right that the f-number and shutter speed don't determine the amount of light. They deal with the variation of amount of light.

Correct - that is obvious in the formula I quoted above. The "variation of amount" (gain) from outside to the image plane is a ratio, normally expressed in EV where EV = log base two of (t/N^2) where t = shutter time and N = f-number.

Try this same formula with a full stop and it nearly is how I would expect.

What on earth does that mean? I don't understand it.

Don't trust the 2.4 and 2.8. They're numbers with 1 decimal.

Please explain in more detail than just "They're numbers with 1 decimal". One decimal point is close enough for the purposes of illustration. In your camera, do select an f-number to more than one decimal place? ... no! Do you trust your selection? ... yes!

Have you heard of an 'Aunt Sally' or a 'Straw Man'? Perhaps there is a Dutch equivalent to raising an issue that is seemingly relevant but actually is not.

The first part is a constant. The second part are the f-numbers in quadrature. With whole numbers no problem.

I'm sick and tired of your "no problems", stated without explanation, examples or proof.

If you want double the light on your sensor you have to multiply the diameter of the aperture with 1,4..... or halve the shutter speed. It's in your formula too, except to less decimals.

The aperture diameter is NOT explicitly in the formula and that formula is not mine. If you disagree with it, go and argue with the Japanese Camera & Imaging Products Association (CIPA).

I fold, George. (poker term).

[DanK]

Personally, I’m placing my bets with John Napier and Leonard Euler.

Time to move on.


Sent from my iPad using Tapatalk

[george013]

Classic! See post #111 for my response to "first thing".

Now please prove that I can not "calculate with those digital values".

Post 101.

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.

Light is coming through the aperture. By changing the F-number one is changing the aperture diameter.
F-numbers, the so called "whole" F-numbers, represent a series starting from 1 that archives a double exposure from the one before. Typical 1,1.4,2,2.8,4,5.6. Based on the formula of the surface of a circle the constant factor is 1.414 or V2.
One stop means doubling the exposure. So half a stop would mean increasing the exposure with a factor 1.5. This is archived by changing the aperture diameter. I'm not telling what correcting factor that will be, I get dizzy of those numbers.

Post 111.

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

In your calculations the difference is 0.677/0.497=1.362. I don't know how to deal with that.

My second point.
Often seen calculation: doing multiplying with the digital numbers. I don't think that can be done. I''m always looking for an answer on next situation.
I shoot a greycard. Under controled conditions this would give me a histogram with 1 line in the middle at 128. When I double that one I'm on te maximum of 255. Is this 1 stop or 2 stops? Or just rubbish? And the used camera has no influence on it?

George

[xpatUSA]

Post 101.

Light is coming through the aperture. By changing the F-number one is changing the aperture diameter.
F-numbers, the so called "whole" F-numbers, represent a series starting from 1 that archives a double exposure from the one before. Typical 1,1.4,2,2.8,4,5.6. Based on the formula of the surface of a circle the constant factor is 1.414 or V2.
One stop means doubling the exposure. So half a stop would mean increasing the exposure with a factor 1.5.

Wrong. This has already been explained by Dan; why are you repeating this error?

This is archived by changing the aperture diameter. I'm not telling what correcting factor that will be, I get dizzy of those numbers.

If numbers make you dizzy, please avoid reading my technical posts.

Post 111.

In your calculations the difference is 0.677/0.497=1.362. I don't know how to deal with that.

My second point.
Often seen calculation: doing multiplying with the digital numbers. I don't think that can be done. I''m always looking for an answer on next situation.

I shoot a greycard. Under controlled conditions, this would give me a histogram with 1 line in the middle at 128.

Wrong. The output value for "middle gray" is dependent on the ISO method used by the camera manufacture:

For Standard Output Saturation it is 118/255, sRGB.

For ISO Sensor Saturation-based, it is 100/255.

For ISO Recommended Exposure Index, it is anything Chuck Norris says it is.

When I double that one I'm on the maximum of 255. Is this 1 stop or 2 stops?

"Stops" do not apply to output images - only sensor exposure. You are forgetting the gamma factor which, for sRGB is approximately 1/2.2.

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

Try calculating the change in sRGB for an exposure change of 1/3 EV; show us your formula, the numbers, and the result. Testing, testing, this is test ...

[JBW]

you two are making me dizzy with all your numbers. The fellow that designed the Lancaster had a simple theory: If it looks right it will work right.

There must be a way to transpose that theory to photography

[DanK]

With all due respect, I suggest shutting this thread down. It is going in circles--unless 1.5 x 1.5 = 2--and no longer has anything to do with the questions posed in the opening post.

[xpatUSA]

With all due respect, I suggest shutting this thread down. It is going in circles--unless 1.5 x 1.5 = 2--and no longer has anything to do with the questions posed in the opening post.

Indeed. Let's let George have the coveted Last Word and be done with it. I'm out.