Need a Photographer Equivalent English Professor

What makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor?
Because photography is the craft of exposing a piece of photosensitive material to an image projected by a lens/pinhole.

It is not the craft of measuring DOF (which, let's not forget, is a bit subjective), sharpness, or noise.
And what matters more, in terms of the visual properties of the photo, the total amount of light falling on the sensor or the density of the light falling on the sensor?
I would never describe the visual properties of an image by such crude measurement but yes, the density of the light falling on the sensor is more readily apparent than the total amount of light.
So, you are saying, for example, that a photo of a scene at 50mm f/2.8 1/100 ISO 400 on mFT will look more similar to a photo of the same scene from the same position using 100mm f/2.8 1/100 ISO 400 on FF than 100mm f/5.6 1/100 ISO 1600 on FF?
Obviously that's not what I'm saying. Your question was "what makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor" - and my answer is that a difference in exposure is much more visible than a difference in DOF (which, again, is a bit subjective), diffraction (which might be or might not be a problem at the set aperture) or noise (not an issue except in low light, for large sensor cameras)
Really? The exposures for 50mm f/2 1/100 ISO 400 and mFT and 100mm f/4 1/100 ISO 1600 are two stops apart (but the same total light). What makes this difference in exposure more visible than the differences that we would see if the FF photo were instead taken at 100mm f/2 1/100 ISO 400, which has the same exposure?
I see... I never said the ISO should be considered constant, thinking it would be obvious - and that was a mistake from my part.
Why would you set the ISO to anything other than what would give the desired brightness in the photo (aside from ISOless shooting with a camera that has an ISOless sensor)?
Why would you change the ISO in a scenario where this wasn't mentioned?
I would change the ISO as needed (or let the camera change the ISO for me in Auto ISO) to get the desired output brightness. I mean, isn't that painfully obvious?
By differences in exposure I actually mean deviations from the "correct" (intended) exposure, not numeric differences between some EVs.
If f/2 1/100 gave the "correct exposure" on mFT, why wouldn't f/4 1/100 give the "correct exposure" on FF?
"Correct" (intended) exposure depends on the effective ISO;
And there we have it -- why many, if not most, of the anti-Equivalence crowd can't come to terms with Equivalence -- a complete and total misunderstanding of what exposure is and what role the ISO setting plays.
what you say could, or could not be true.
What I said is true, and what I said is that the ISO setting merely sets the output brightness of the photo.
However, the format is not included in the equation; only scene luminance, aperture (f-stop), shutter speed and ISO are.
What is correct about 50mm f/2 1/100 ISO 400 on mFT that is not "correct" about 100mm f/4 1/100 ISO 1600 on FF?
In a strict sense, exposure is indeed no more than light per area, that's not very meaningful by itself. But then, total light is exposure multiplied by area, not very meaningful either.
The total amount of light that falls on the sensor is indeed rather meaningful since, combined with sensor efficiency, it results in the same noise. In any case, the total amount of light falling on the sensor has far more to do with the visual properties of the photo than does the exposure, with regards to cross-format comparisons (for comparisons within a format, either measure works just as well).
But the exposure value, combined with sensitivity/ISO, is giving the image "brightness", variations of which are usually more apparent than noise differences.

If you only knew the total light you would not be able to take a properly exposed image. You might be able to approximate noise, though ;-)
What is the difference in brightness between 50mm f/2 1/100 ISO 400 on mFT and 100mm f/4 1/100 ISO 1600 on FF?
However, it's components are important. Same exposure, or same total light can be obtained with very different settings - one can vary aperture, shutter speed and even scene luminosity.
You're quite skilled at pushing the conversation on a convenient track.
What you mean to say is that I'm quite skilled at pointing out that you do not understand how exposure and total light relate to the visual properties of the photo.
I wonder...
While you're wondering, if you could let us know how the exposures compare with total light as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I find the way you're trying to "teach" me things instead of discussing them quite arrogant. You already have your answers, try not to ignore them.
More likely is that you find questions that reveal fundamental misconceptions you have about exposure, total light, ISO, and brightness to be uncomfortable.
The answer may just clear up some rather important fundamental misunderstandings.
Why, are you misunderstanding something?
It's pretty obvious who is misunderstanding here. So, I ask again:

Could let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
 
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What makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor?
Because photography is the craft of exposing a piece of photosensitive material to an image projected by a lens/pinhole.

It is not the craft of measuring DOF (which, let's not forget, is a bit subjective), sharpness, or noise.
And what matters more, in terms of the visual properties of the photo, the total amount of light falling on the sensor or the density of the light falling on the sensor?
I would never describe the visual properties of an image by such crude measurement but yes, the density of the light falling on the sensor is more readily apparent than the total amount of light.
So, you are saying, for example, that a photo of a scene at 50mm f/2.8 1/100 ISO 400 on mFT will look more similar to a photo of the same scene from the same position using 100mm f/2.8 1/100 ISO 400 on FF than 100mm f/5.6 1/100 ISO 1600 on FF?
Obviously that's not what I'm saying. Your question was "what makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor" - and my answer is that a difference in exposure is much more visible than a difference in DOF (which, again, is a bit subjective), diffraction (which might be or might not be a problem at the set aperture) or noise (not an issue except in low light, for large sensor cameras)
Really? The exposures for 50mm f/2 1/100 ISO 400 and mFT and 100mm f/4 1/100 ISO 1600 are two stops apart (but the same total light). What makes this difference in exposure more visible than the differences that we would see if the FF photo were instead taken at 100mm f/2 1/100 ISO 400, which has the same exposure?
I see... I never said the ISO should be considered constant, thinking it would be obvious - and that was a mistake from my part.
Why would you set the ISO to anything other than what would give the desired brightness in the photo (aside from ISOless shooting with a camera that has an ISOless sensor)?
Why would you change the ISO in a scenario where this wasn't mentioned?
I would change the ISO as needed (or let the camera change the ISO for me in Auto ISO) to get the desired output brightness. I mean, isn't that painfully obvious?
Not in (virtual) tests which should be controlled. If we said that one factor (e.g. exposure) would be changed, the others are usually assumed constant.

But I already assumed responsibility for not explicitly mentioning this, and make the necessary clarifications. Which you don't seems to be willing to accept, instead going on with your convenient scenarios.
By differences in exposure I actually mean deviations from the "correct" (intended) exposure, not numeric differences between some EVs.
If f/2 1/100 gave the "correct exposure" on mFT, why wouldn't f/4 1/100 give the "correct exposure" on FF?
"Correct" (intended) exposure depends on the effective ISO;
And there we have it -- why many, if not most, of the anti-Equivalence crowd can't come to terms with Equivalence -- a complete and total misunderstanding of what exposure is and what role the ISO setting plays.
I wonder, why Equivalence? Is that some sort of religion? :-)

The misunderstanding is yours, what I say is correct.
what you say could, or could not be true.
What I said is true, and what I said is that the ISO setting merely sets the output brightness of the photo.
If constant exposure. Or you could adjust the exposure to match the ISO.
However, the format is not included in the equation; only scene luminance, aperture (f-stop), shutter speed and ISO are.
What is correct about 50mm f/2 1/100 ISO 400 on mFT that is not "correct" about 100mm f/4 1/100 ISO 1600 on FF?
As I said, the exposure triangle does not depend on the format; it's designed to be independent of the format. Thus, "on mFT" and "on FF" is just useless information.

The focal length is useless information.
In a strict sense, exposure is indeed no more than light per area, that's not very meaningful by itself. But then, total light is exposure multiplied by area, not very meaningful either.
The total amount of light that falls on the sensor is indeed rather meaningful since, combined with sensor efficiency, it results in the same noise. In any case, the total amount of light falling on the sensor has far more to do with the visual properties of the photo than does the exposure, with regards to cross-format comparisons (for comparisons within a format, either measure works just as well).
But the exposure value, combined with sensitivity/ISO, is giving the image "brightness", variations of which are usually more apparent than noise differences.

If you only knew the total light you would not be able to take a properly exposed image. You might be able to approximate noise, though ;-)
What is the difference in brightness between 50mm f/2 1/100 ISO 400 on mFT and 100mm f/4 1/100 ISO 1600 on FF?
The question is unrelated to what I said.
However, it's components are important. Same exposure, or same total light can be obtained with very different settings - one can vary aperture, shutter speed and even scene luminosity.
You're quite skilled at pushing the conversation on a convenient track.
What you mean to say is that I'm quite skilled at pointing out that you do not understand how exposure and total light relate to the visual properties of the photo.
I wonder...
While you're wondering, if you could let us know how the exposures compare with total light as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I find the way you're trying to "teach" me things instead of discussing them quite arrogant. You already have your answers, try not to ignore them.
More likely is that you find questions that reveal fundamental misconceptions you have about exposure, total light, ISO, and brightness to be uncomfortable.
You have nothing to teach me, except perhaps arrogance - thanks, but I'll pass.

Unfortunately, I am unable to teach you how to read.
The answer may just clear up some rather important fundamental misunderstandings.
Why, are you misunderstanding something?
It's pretty obvious who is misunderstanding here. So, I ask again:

Could let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.

Alex
 
As explained to you before, the f-number is, indeed, directly connected to DOF:

http://en.wikipedia.org/wiki/Depth_of_field#DOF_formulas

In case you missed it, the "N" in the formulas is the f-number.
As explained to you before:

The hyperfocal distance formula used in the DOF-formula is a simplification of the formula Louis Derr derived in 1906. It is considered to be the strictly correct one in modern times.
Watch what's going to happen.
It goes like this:

p=((D+d)*f)/d

p= hyperfocal distance

D= aperture diameter

d= diameter of CoC

f= focal lenght

As you see, there is no f-number in the formula, but as it contains both focal lenght and aperture diameter it can be rewritten with the use of f-number. As f-number is easily available in the field, it makes it easier to use the formula, along with easier integration with the DOF formula that relies on the hyperfocal formula.

You can easily rewrite the formula for calculating hot water in the same way using electric current from the solar panel, still it doesnt make the electric current a part of what defines the heating of the water.
Here's Wikipedia's formula for the Hyperfocal Distance to compare and contrast with the one you gave above:

http://en.wikipedia.org/wiki/Hyperfocal_distance

In optics and photography, hyperfocal distance is a distance beyond which all objects can be brought into an "acceptable" focus. There are two commonly used definitions of hyperfocal distance, leading to values that differ only slightly:

Definition 1:
The hyperfocal distance is the closest distance at which a lens can be focused while keeping objects at infinity acceptably sharp. When the lens is focused at this distance, all objects at distances from half of the hyperfocal distance out to infinity will be acceptably sharp.

Definition 2:
The hyperfocal distance is the distance beyond which all objects are acceptably sharp, for a lens focused at infinity.

The distinction between the two meanings is rarely made, since they have almost identical values. The value computed according to the first definition exceeds that from the second by just one
focal length.

As the hyperfocal distance is the focus distance giving the maximum
depth of field, it is the most desirable distance to set the focus of a fixed-focus camera.

For the first definition,

H = f²/(Nc) + f

where

H is hyperfocal distance

f is focal length

N is f-number (f/D for aperture diameter D)

c is the circle of confusion limit

For any practical f-number, the added focal length is insignificant in comparison with the first term, so that

H ≈ f²/(Nc)

This formula is exact for the second definition, if H is measured from a thin lens, or from the front principal plane of a complex lens; it is also exact for the first definition if H is measured from a point that is one focal length in front of the front principal plane. For practical purposes, there is little difference between the first and second definitions.


Let's fist work out and example computing the hyperfocal distance for 50mm f/2 on mFT (aperture diameter = 50mm / 2 = 25mm, CoC = 0.015mm) using both your formula and Wikipedia's.

OK, here we go. First, your formula:

D=25mm, d=0.015mm, f=50mm --> p=((25mm + 0.015mm)*50mm) / 0.015mm = 83383mm = 83.4m.

Wikipedia's formula:

f=50mm, N=2, c=0.015mm --> H = (50mm)² / (2 x 0.015mm) + 50mm = 83383mm = 83.4m.

Huh -- seems like they give the same result. I wonder why that might be? Let's take a look, shall we?
Of course its the same result, the last one is a rewriting based on the first one like I said.
Looks like it's the same formula as the Wikipedia formula but the Wikipedia formula is written in terms of the f-number. How about that?
Of course its the same formula, the wikipedia formula is a rewriting of the strictly correct one derived by Louis Derr.
So what was the point of you presenting your formula that is "considered to be the strictly correct one in modern times" when it's the same as the Wikipedia formula I gave?
The original one does not contain f-number.
That a formula using the f-number can be rewritten in terms of the aperture diameter and focal length? Well, duh -- the f-number is the quotient of the focal length and aperture diameter, as has been explained to you multiple times.

In other words, your claim that DOF is not a function of the f-number is complete, and total, bunk.
As I have explained: Changing aperture diameter changed both the resulting f-number and resulting DOF. The resulting f-number is not what changes the resulting DOF, the aperture diameter does that. This is the key to understanding why "f/2 is equivalent to f/4" is clumsy and wrong. What follows what is important if you want to get your detail right.
Your formula, p=((D+d)*f)/d, would read H = (f/N + c) · f / c using Wikipedia's variables. You're good with algebra 1, right? First, we distribute the f in the numerator to get:

H = (f²/N + c·f) / c

Next, we split up the numerator:

H = (f²/N) / c + (c·f) / c

Now we simplify:

H = f²/Nc + f
This is already done in your wikipedia link. Scroll down to "History - Derr 1906"
So it is!
Another rewriting of formula:

Definition 1: Weight (DOF) = mass * acceleration

Definition 2: Momentum (f-number) = mass * velocity

This can lead to the formula:

Weight = momentum / velocity * acceleration

Still, momentum is not what defines weight, they just have mass i common.

My suggestion: "this momentum is mass-equivalent to this momentum". (same mass with different velocity) -> "f/2 is size-equivalent to f/4" Mass is after all what defines weight(DOF).

The clumsy one:

"this momentum is equivalent to this momentum" in terms of weight... -> f/2 is equivalent to f/4 in terms of DOF...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
 
...


As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.

Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
 
What makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor?
Because photography is the craft of exposing a piece of photosensitive material to an image projected by a lens/pinhole.

It is not the craft of measuring DOF (which, let's not forget, is a bit subjective), sharpness, or noise.
And what matters more, in terms of the visual properties of the photo, the total amount of light falling on the sensor or the density of the light falling on the sensor?
I would never describe the visual properties of an image by such crude measurement but yes, the density of the light falling on the sensor is more readily apparent than the total amount of light.
So, you are saying, for example, that a photo of a scene at 50mm f/2.8 1/100 ISO 400 on mFT will look more similar to a photo of the same scene from the same position using 100mm f/2.8 1/100 ISO 400 on FF than 100mm f/5.6 1/100 ISO 1600 on FF?
Obviously that's not what I'm saying. Your question was "what makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor" - and my answer is that a difference in exposure is much more visible than a difference in DOF (which, again, is a bit subjective), diffraction (which might be or might not be a problem at the set aperture) or noise (not an issue except in low light, for large sensor cameras)
Really? The exposures for 50mm f/2 1/100 ISO 400 and mFT and 100mm f/4 1/100 ISO 1600 are two stops apart (but the same total light). What makes this difference in exposure more visible than the differences that we would see if the FF photo were instead taken at 100mm f/2 1/100 ISO 400, which has the same exposure?
I see... I never said the ISO should be considered constant, thinking it would be obvious - and that was a mistake from my part.
Why would you set the ISO to anything other than what would give the desired brightness in the photo (aside from ISOless shooting with a camera that has an ISOless sensor)?
Why would you change the ISO in a scenario where this wasn't mentioned?
I would change the ISO as needed (or let the camera change the ISO for me in Auto ISO) to get the desired output brightness. I mean, isn't that painfully obvious?
Not in (virtual) tests which should be controlled. If we said that one factor (e.g. exposure) would be changed, the others are usually assumed constant.

But I already assumed responsibility for not explicitly mentioning this, and make the necessary clarifications. Which you don't seems to be willing to accept, instead going on with your convenient scenarios.
By differences in exposure I actually mean deviations from the "correct" (intended) exposure, not numeric differences between some EVs.
If f/2 1/100 gave the "correct exposure" on mFT, why wouldn't f/4 1/100 give the "correct exposure" on FF?
"Correct" (intended) exposure depends on the effective ISO;
And there we have it -- why many, if not most, of the anti-Equivalence crowd can't come to terms with Equivalence -- a complete and total misunderstanding of what exposure is and what role the ISO setting plays.
I wonder, why Equivalence? Is that some sort of religion? :-)

The misunderstanding is yours, what I say is correct.
what you say could, or could not be true.
What I said is true, and what I said is that the ISO setting merely sets the output brightness of the photo.
If constant exposure. Or you could adjust the exposure to match the ISO.
However, the format is not included in the equation; only scene luminance, aperture (f-stop), shutter speed and ISO are.
What is correct about 50mm f/2 1/100 ISO 400 on mFT that is not "correct" about 100mm f/4 1/100 ISO 1600 on FF?
As I said, the exposure triangle does not depend on the format; it's designed to be independent of the format. Thus, "on mFT" and "on FF" is just useless information.

The focal length is useless information.
In a strict sense, exposure is indeed no more than light per area, that's not very meaningful by itself. But then, total light is exposure multiplied by area, not very meaningful either.
The total amount of light that falls on the sensor is indeed rather meaningful since, combined with sensor efficiency, it results in the same noise. In any case, the total amount of light falling on the sensor has far more to do with the visual properties of the photo than does the exposure, with regards to cross-format comparisons (for comparisons within a format, either measure works just as well).
But the exposure value, combined with sensitivity/ISO, is giving the image "brightness", variations of which are usually more apparent than noise differences.

If you only knew the total light you would not be able to take a properly exposed image. You might be able to approximate noise, though ;-)
What is the difference in brightness between 50mm f/2 1/100 ISO 400 on mFT and 100mm f/4 1/100 ISO 1600 on FF?
The question is unrelated to what I said.
However, it's components are important. Same exposure, or same total light can be obtained with very different settings - one can vary aperture, shutter speed and even scene luminosity.
You're quite skilled at pushing the conversation on a convenient track.
What you mean to say is that I'm quite skilled at pointing out that you do not understand how exposure and total light relate to the visual properties of the photo.
I wonder...
While you're wondering, if you could let us know how the exposures compare with total light as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I find the way you're trying to "teach" me things instead of discussing them quite arrogant. You already have your answers, try not to ignore them.
More likely is that you find questions that reveal fundamental misconceptions you have about exposure, total light, ISO, and brightness to be uncomfortable.
You have nothing to teach me, except perhaps arrogance - thanks, but I'll pass.

Unfortunately, I am unable to teach you how to read.
The answer may just clear up some rather important fundamental misunderstandings.
Why, are you misunderstanding something?
It's pretty obvious who is misunderstanding here. So, I ask again:

Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.

I'll now head over to Off Topic to see if I can't find a thread or two on Evolution vs Creationism. :-D
 
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As explained to you before, the f-number is, indeed, directly connected to DOF:

http://en.wikipedia.org/wiki/Depth_of_field#DOF_formulas

In case you missed it, the "N" in the formulas is the f-number.
As explained to you before:

The hyperfocal distance formula used in the DOF-formula is a simplification of the formula Louis Derr derived in 1906. It is considered to be the strictly correct one in modern times.
Watch what's going to happen.
It goes like this:

p=((D+d)*f)/d

p= hyperfocal distance

D= aperture diameter

d= diameter of CoC

f= focal lenght

As you see, there is no f-number in the formula, but as it contains both focal lenght and aperture diameter it can be rewritten with the use of f-number. As f-number is easily available in the field, it makes it easier to use the formula, along with easier integration with the DOF formula that relies on the hyperfocal formula.

You can easily rewrite the formula for calculating hot water in the same way using electric current from the solar panel, still it doesnt make the electric current a part of what defines the heating of the water.
Here's Wikipedia's formula for the Hyperfocal Distance to compare and contrast with the one you gave above:

http://en.wikipedia.org/wiki/Hyperfocal_distance

In optics and photography, hyperfocal distance is a distance beyond which all objects can be brought into an "acceptable" focus. There are two commonly used definitions of hyperfocal distance, leading to values that differ only slightly:

Definition 1:
The hyperfocal distance is the closest distance at which a lens can be focused while keeping objects at infinity acceptably sharp. When the lens is focused at this distance, all objects at distances from half of the hyperfocal distance out to infinity will be acceptably sharp.

Definition 2:
The hyperfocal distance is the distance beyond which all objects are acceptably sharp, for a lens focused at infinity.

The distinction between the two meanings is rarely made, since they have almost identical values. The value computed according to the first definition exceeds that from the second by just one
focal length.

As the hyperfocal distance is the focus distance giving the maximum
depth of field, it is the most desirable distance to set the focus of a fixed-focus camera.

For the first definition,

H = f²/(Nc) + f

where

H is hyperfocal distance

f is focal length

N is f-number (f/D for aperture diameter D)

c is the circle of confusion limit

For any practical f-number, the added focal length is insignificant in comparison with the first term, so that

H ≈ f²/(Nc)

This formula is exact for the second definition, if H is measured from a thin lens, or from the front principal plane of a complex lens; it is also exact for the first definition if H is measured from a point that is one focal length in front of the front principal plane. For practical purposes, there is little difference between the first and second definitions.


Let's fist work out and example computing the hyperfocal distance for 50mm f/2 on mFT (aperture diameter = 50mm / 2 = 25mm, CoC = 0.015mm) using both your formula and Wikipedia's.

OK, here we go. First, your formula:

D=25mm, d=0.015mm, f=50mm --> p=((25mm + 0.015mm)*50mm) / 0.015mm = 83383mm = 83.4m.

Wikipedia's formula:

f=50mm, N=2, c=0.015mm --> H = (50mm)² / (2 x 0.015mm) + 50mm = 83383mm = 83.4m.

Huh -- seems like they give the same result. I wonder why that might be? Let's take a look, shall we?
Of course its the same result, the last one is a rewriting based on the first one like I said.
Looks like it's the same formula as the Wikipedia formula but the Wikipedia formula is written in terms of the f-number. How about that?
Of course its the same formula, the wikipedia formula is a rewriting of the strictly correct one derived by Louis Derr.
So what was the point of you presenting your formula that is "considered to be the strictly correct one in modern times" when it's the same as the Wikipedia formula I gave?
The original one does not contain f-number.
Oh my. That one can make such a self-harming statement when the equivalence of the formulas, both by example and mathematics, was demonstrated, is, well, disappointing.
That a formula using the f-number can be rewritten in terms of the aperture diameter and focal length? Well, duh -- the f-number is the quotient of the focal length and aperture diameter, as has been explained to you multiple times.

In other words, your claim that DOF is not a function of the f-number is complete, and total, bunk.
As I have explained: Changing aperture diameter changed both the resulting f-number and resulting DOF.
As I explained, changing the f-ratio on the camera (or lens) results in the diaphragm on the lens opening or closing, which changes the effective aperture diameter.
The resulting f-number is not what changes the resulting DOF, the aperture diameter does that. This is the key to understanding why "f/2 is equivalent to f/4" is clumsy and wrong. What follows what is important if you want to get your detail right.
Yes or know: Does 50mm f/2 on mFT have the same [diagonal] angle of view and same aperture diameter as 100mm f/4 on FF? My goodness!
Your formula, p=((D+d)*f)/d, would read H = (f/N + c) · f / c using Wikipedia's variables. You're good with algebra 1, right? First, we distribute the f in the numerator to get:

H = (f²/N + c·f) / c

Next, we split up the numerator:

H = (f²/N) / c + (c·f) / c

Now we simplify:

H = f²/Nc + f
This is already done in your wikipedia link. Scroll down to "History - Derr 1906"
So it is!
Another rewriting of formula:

Definition 1: Weight (DOF) = mass * acceleration

Definition 2: Momentum (f-number) = mass * velocity

This can lead to the formula:

Weight = momentum / velocity * acceleration

Still, momentum is not what defines weight, they just have mass i common.

My suggestion: "this momentum is mass-equivalent to this momentum". (same mass with different velocity) -> "f/2 is size-equivalent to f/4" Mass is after all what defines weight(DOF).

The clumsy one:

"this momentum is equivalent to this momentum" in terms of weight... -> f/2 is equivalent to f/4 in terms of DOF...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
As has been explained, and demonstrated, to you countless times:


The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).
 
...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
...I'll spell it out to you, once again, the same as I did for Fygaren:

As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).

And yet, some will fight to the death against the obvious. Indeed, upthread, one of the opponents of Equivalence even refused to answer the following simple question:

Please tell us how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I can only wonder why. ;-)
 
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As explained to you before, the f-number is, indeed, directly connected to DOF:

http://en.wikipedia.org/wiki/Depth_of_field#DOF_formulas

In case you missed it, the "N" in the formulas is the f-number.
As explained to you before:

The hyperfocal distance formula used in the DOF-formula is a simplification of the formula Louis Derr derived in 1906. It is considered to be the strictly correct one in modern times.
Watch what's going to happen.
It goes like this:

p=((D+d)*f)/d

p= hyperfocal distance

D= aperture diameter

d= diameter of CoC

f= focal lenght

As you see, there is no f-number in the formula, but as it contains both focal lenght and aperture diameter it can be rewritten with the use of f-number. As f-number is easily available in the field, it makes it easier to use the formula, along with easier integration with the DOF formula that relies on the hyperfocal formula.

You can easily rewrite the formula for calculating hot water in the same way using electric current from the solar panel, still it doesnt make the electric current a part of what defines the heating of the water.
Here's Wikipedia's formula for the Hyperfocal Distance to compare and contrast with the one you gave above:

http://en.wikipedia.org/wiki/Hyperfocal_distance

In optics and photography, hyperfocal distance is a distance beyond which all objects can be brought into an "acceptable" focus. There are two commonly used definitions of hyperfocal distance, leading to values that differ only slightly:

Definition 1:
The hyperfocal distance is the closest distance at which a lens can be focused while keeping objects at infinity acceptably sharp. When the lens is focused at this distance, all objects at distances from half of the hyperfocal distance out to infinity will be acceptably sharp.

Definition 2:
The hyperfocal distance is the distance beyond which all objects are acceptably sharp, for a lens focused at infinity.

The distinction between the two meanings is rarely made, since they have almost identical values. The value computed according to the first definition exceeds that from the second by just one
focal length.

As the hyperfocal distance is the focus distance giving the maximum
depth of field, it is the most desirable distance to set the focus of a fixed-focus camera.

For the first definition,

H = f²/(Nc) + f

where

H is hyperfocal distance

f is focal length

N is f-number (f/D for aperture diameter D)

c is the circle of confusion limit

For any practical f-number, the added focal length is insignificant in comparison with the first term, so that

H ≈ f²/(Nc)

This formula is exact for the second definition, if H is measured from a thin lens, or from the front principal plane of a complex lens; it is also exact for the first definition if H is measured from a point that is one focal length in front of the front principal plane. For practical purposes, there is little difference between the first and second definitions.


Let's fist work out and example computing the hyperfocal distance for 50mm f/2 on mFT (aperture diameter = 50mm / 2 = 25mm, CoC = 0.015mm) using both your formula and Wikipedia's.

OK, here we go. First, your formula:

D=25mm, d=0.015mm, f=50mm --> p=((25mm + 0.015mm)*50mm) / 0.015mm = 83383mm = 83.4m.

Wikipedia's formula:

f=50mm, N=2, c=0.015mm --> H = (50mm)² / (2 x 0.015mm) + 50mm = 83383mm = 83.4m.

Huh -- seems like they give the same result. I wonder why that might be? Let's take a look, shall we?
Of course its the same result, the last one is a rewriting based on the first one like I said.
Looks like it's the same formula as the Wikipedia formula but the Wikipedia formula is written in terms of the f-number. How about that?
Of course its the same formula, the wikipedia formula is a rewriting of the strictly correct one derived by Louis Derr.
So what was the point of you presenting your formula that is "considered to be the strictly correct one in modern times" when it's the same as the Wikipedia formula I gave?
The original one does not contain f-number.
Oh my. That one can make such a self-harming statement when the equivalence of the formulas, both by example and mathematics, was demonstrated, is, well, disappointing.
Was that a poor attempt of using domination techniques? Its not working.
That a formula using the f-number can be rewritten in terms of the aperture diameter and focal length? Well, duh -- the f-number is the quotient of the focal length and aperture diameter, as has been explained to you multiple times.

In other words, your claim that DOF is not a function of the f-number is complete, and total, bunk.
As I have explained: Changing aperture diameter changed both the resulting f-number and resulting DOF.
As I explained, changing the f-ratio on the camera (or lens) results in the diaphragm on the lens opening or closing, which changes the effective aperture diameter.
You may "change the f-number on your camera" but whats happening on a more tecnical level is this:

f-number can only be a result of aperture diameter and not the other way around. When you twist the aperture ring on your lens or turn the control dial on your camera, this changes the aperture diameter and the resulting lens speed is shown to you either on the lens or in your VF. You keep twisting and turning until the wanted result shows.
The resulting f-number is not what changes the resulting DOF, the aperture diameter does that. This is the key to understanding why "f/2 is equivalent to f/4" is clumsy and wrong. What follows what is important if you want to get your detail right.
Yes or know: Does 50mm f/2 on mFT have the same [diagonal] angle of view and same aperture diameter as 100mm f/4 on FF? My goodness!
Yes, so is "f/2 is equivalent to f/4" in terms of FOV not clumsy then?
Your formula, p=((D+d)*f)/d, would read H = (f/N + c) · f / c using Wikipedia's variables. You're good with algebra 1, right? First, we distribute the f in the numerator to get:

H = (f²/N + c·f) / c

Next, we split up the numerator:

H = (f²/N) / c + (c·f) / c

Now we simplify:

H = f²/Nc + f
This is already done in your wikipedia link. Scroll down to "History - Derr 1906"
So it is!
Another rewriting of formula:

Definition 1: Weight (DOF) = mass * acceleration

Definition 2: Momentum (f-number) = mass * velocity

This can lead to the formula:

Weight = momentum / velocity * acceleration

Still, momentum is not what defines weight, they just have mass i common.

My suggestion: "this momentum is mass-equivalent to this momentum". (same mass with different velocity) -> "f/2 is size-equivalent to f/4" Mass is after all what defines weight(DOF).

The clumsy one:

"this momentum is equivalent to this momentum" in terms of weight... -> f/2 is equivalent to f/4 in terms of DOF...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).
As has been explained, and demonstrated, to you countless times:

You mix f-number and aperture diameter in a very clumsy way.

Saying "f/2 is equivalent to f/4" and then meaning other results of aperture diameter (DOF, diffraction etc), is the same as saying "f/2 is equivalent to f/4" and then meaning other results of focal lenght (FOV)
 
...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
...I'll spell it out to you, once again, the same as I did for Fygaren:

As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).

And yet, some will fight to the death against the obvious. Indeed, upthread, one of the opponents of Equivalence even refused to answer the following simple question:

Please tell us how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I can only wonder why. ;-)
The brightness will be the same on all 3 examples, but the 100mm f/2 1/100 ISO 400 on FF has an aperture diameter that is twice that of the other two resulting in 4 times as mutch light hitting the sensor. It will collect 4 times as mutch light because the aperture area is 4 times that of the others resulting in equivalent lens speed as the 50mm and the need for bumping ISO 2 stops on the other FF to get the same brightness. This is very obvious and should not require a reply.

But that does not change the clumsy, inaccurate and wrong way you will fight to death for, explaining why "f/2 is equivalent to f/4"

Saying "f/2 is equivalent to f/4" and then meaning other results of aperture diameter (DOF, diffraction etc), is the same as saying "f/2 is equivalent to f/4" and then meaning other results of focal lenght (FOV)

f-number is after all the result of focal lengsh and diameter and a quantitative measure of lens speed.
 
What makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor?
Because photography is the craft of exposing a piece of photosensitive material to an image projected by a lens/pinhole.

It is not the craft of measuring DOF (which, let's not forget, is a bit subjective), sharpness, or noise.
And what matters more, in terms of the visual properties of the photo, the total amount of light falling on the sensor or the density of the light falling on the sensor?
I would never describe the visual properties of an image by such crude measurement but yes, the density of the light falling on the sensor is more readily apparent than the total amount of light.
So, you are saying, for example, that a photo of a scene at 50mm f/2.8 1/100 ISO 400 on mFT will look more similar to a photo of the same scene from the same position using 100mm f/2.8 1/100 ISO 400 on FF than 100mm f/5.6 1/100 ISO 1600 on FF?
Obviously that's not what I'm saying. Your question was "what makes exposure more important than DOF, diffraction, and the total amount of light projected on the sensor" - and my answer is that a difference in exposure is much more visible than a difference in DOF (which, again, is a bit subjective), diffraction (which might be or might not be a problem at the set aperture) or noise (not an issue except in low light, for large sensor cameras)
Really? The exposures for 50mm f/2 1/100 ISO 400 and mFT and 100mm f/4 1/100 ISO 1600 are two stops apart (but the same total light). What makes this difference in exposure more visible than the differences that we would see if the FF photo were instead taken at 100mm f/2 1/100 ISO 400, which has the same exposure?
I see... I never said the ISO should be considered constant, thinking it would be obvious - and that was a mistake from my part.
Why would you set the ISO to anything other than what would give the desired brightness in the photo (aside from ISOless shooting with a camera that has an ISOless sensor)?
Why would you change the ISO in a scenario where this wasn't mentioned?
I would change the ISO as needed (or let the camera change the ISO for me in Auto ISO) to get the desired output brightness. I mean, isn't that painfully obvious?
Not in (virtual) tests which should be controlled. If we said that one factor (e.g. exposure) would be changed, the others are usually assumed constant.

But I already assumed responsibility for not explicitly mentioning this, and make the necessary clarifications. Which you don't seems to be willing to accept, instead going on with your convenient scenarios.
By differences in exposure I actually mean deviations from the "correct" (intended) exposure, not numeric differences between some EVs.
If f/2 1/100 gave the "correct exposure" on mFT, why wouldn't f/4 1/100 give the "correct exposure" on FF?
"Correct" (intended) exposure depends on the effective ISO;
And there we have it -- why many, if not most, of the anti-Equivalence crowd can't come to terms with Equivalence -- a complete and total misunderstanding of what exposure is and what role the ISO setting plays.
I wonder, why Equivalence? Is that some sort of religion? :-)

The misunderstanding is yours, what I say is correct.
what you say could, or could not be true.
What I said is true, and what I said is that the ISO setting merely sets the output brightness of the photo.
If constant exposure. Or you could adjust the exposure to match the ISO.
However, the format is not included in the equation; only scene luminance, aperture (f-stop), shutter speed and ISO are.
What is correct about 50mm f/2 1/100 ISO 400 on mFT that is not "correct" about 100mm f/4 1/100 ISO 1600 on FF?
As I said, the exposure triangle does not depend on the format; it's designed to be independent of the format. Thus, "on mFT" and "on FF" is just useless information.

The focal length is useless information.
In a strict sense, exposure is indeed no more than light per area, that's not very meaningful by itself. But then, total light is exposure multiplied by area, not very meaningful either.
The total amount of light that falls on the sensor is indeed rather meaningful since, combined with sensor efficiency, it results in the same noise. In any case, the total amount of light falling on the sensor has far more to do with the visual properties of the photo than does the exposure, with regards to cross-format comparisons (for comparisons within a format, either measure works just as well).
But the exposure value, combined with sensitivity/ISO, is giving the image "brightness", variations of which are usually more apparent than noise differences.

If you only knew the total light you would not be able to take a properly exposed image. You might be able to approximate noise, though ;-)
What is the difference in brightness between 50mm f/2 1/100 ISO 400 on mFT and 100mm f/4 1/100 ISO 1600 on FF?
The question is unrelated to what I said.
However, it's components are important. Same exposure, or same total light can be obtained with very different settings - one can vary aperture, shutter speed and even scene luminosity.
You're quite skilled at pushing the conversation on a convenient track.
What you mean to say is that I'm quite skilled at pointing out that you do not understand how exposure and total light relate to the visual properties of the photo.
I wonder...
While you're wondering, if you could let us know how the exposures compare with total light as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I find the way you're trying to "teach" me things instead of discussing them quite arrogant. You already have your answers, try not to ignore them.
More likely is that you find questions that reveal fundamental misconceptions you have about exposure, total light, ISO, and brightness to be uncomfortable.
You have nothing to teach me, except perhaps arrogance - thanks, but I'll pass.

Unfortunately, I am unable to teach you how to read.
The answer may just clear up some rather important fundamental misunderstandings.
Why, are you misunderstanding something?
It's pretty obvious who is misunderstanding here. So, I ask again:

Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.
It's sad you'd rather leave pretending to yourself you "won" some internet dispute, instead of trying to understand where you got it wrong. Such a simple thing, just a small assumption you're always making because your thinking is constrained by equivalence...
I'll now head over to Off Topic to see if I can't find a thread or two on Evolution vs Creationism. :-D
What, are you going to preach Creationism too? :-p

Alex
 
Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.
It's sad you'd rather leave pretending to yourself you "won" some internet dispute, instead of trying to understand where you got it wrong. Such a simple thing, just a small assumption you're always making because your thinking is constrained by equivalence...
See the question? Information or "entertainment" -- the ball is in your court.
 
Yes or no: Does 50mm f/2 on mFT have the same [diagonal] angle of view and same aperture diameter as 100mm f/4 on FF? My goodness!
Yes...
Thank you for the simple and direct answer to the simple and direct question.
...so is "f/2 is equivalent to f/4" in terms of FOV not clumsy then?
Very. However, in terms of DOF, diffraction, and total light projected on the sensor, it is quite elegant.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).
As has been explained, and demonstrated, to you countless times:

You mix f-number and aperture diameter in a very clumsy way.
Really? What's clumsy about the f-ratio being the quotient of the focal length and aperture diameter? What's clumsy about the equivalent f-ratio scaling in the exact same manner as the equivalent focal length?
Saying "f/2 is equivalent to f/4" and then meaning other results of aperture diameter (DOF, diffraction etc), is the same as saying "f/2 is equivalent to f/4" and then meaning other results of focal lenght (FOV)
Perhaps my gifts in physics, mathematics, and photography make statements like:

50mm f/2 1/100 ISO 400 on mFT is equivalent to 100mm f/4 1/100 ISO 1600 on FF

more obvious and simple to me than to others who lack my gifts and find it "clumsy". They find statements like:

50mm f/2 1/100 ISO 400 on mFT is equivalent to 100mm f/2 1/100 ISO 400 on FF

to be much more simple, even if the photos look rather more different, which would cause someone like me to wonder why someone would use a term like "equivalent" to describe such photos.

Ah, well -- it happens.
 
...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
...I'll spell it out to you, once again, the same as I did for Fygaren:

As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).

And yet, some will fight to the death against the obvious. Indeed, upthread, one of the opponents of Equivalence even refused to answer the following simple question:

Please tell us how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I can only wonder why. ;-)
The brightness will be the same on all 3 examples, but the 100mm f/2 1/100 ISO 400 on FF has an aperture diameter that is twice that of the other two resulting in 4 times as mutch light hitting the sensor. It will collect 4 times as mutch light because the aperture area is 4 times that of the others resulting in equivalent lens speed as the 50mm and the need for bumping ISO 2 stops on the other FF to get the same brightness. This is very obvious and should not require a reply.
Thank you for the simple and direct answer to the simple and direct question. All correct. However, you did not say anything about DOF. Would you like to elaborate?
But that does not change the clumsy, inaccurate and wrong way you will fight to death for, explaining why "f/2 is equivalent to f/4"
I should think that the fact that options 1 and 3 had the same perspective, framing (more or less -- a little different due to the differences in aspect ratio), DOF, diffraction, shutter speed, brightness, and more or less the same noise (if the sensors had more or less the same efficiency) would be the explanation that you might be looking for.
Saying "f/2 is equivalent to f/4" and then meaning other results of aperture diameter (DOF, diffraction etc), is the same as saying "f/2 is equivalent to f/4" and then meaning other results of focal lenght (FOV)
Really? You think it's unnatural to compare different focal lengths on different formats? How very curious!
f-number is after all the result of focal lengsh and diameter and a quantitative measure of lens speed.
Aside from DOF and noise, why do we care about lens speed? Saying, for example, that f/2 on mFT is "faster than" f/4 on FF is like saying a winner in a high school 100m race is "faster than" a silver medalist in the Olympics.
 
A 25mm f/1.4 lens for four thirds has the same lens speed as a 50mm f/1.4 for full frame.
Only for exposure. Not for DOF, total light captured, or diffraction. If those are to be equated (and I'd argue that the image is much more important than the exposure information in the EXIF data) then you have to also multiply the f-stop by the crop factor just like you did for focal length.
There is a difference between the aperture setting you'd choose to achieve a particular effect and the speed of the lens, which is simply the widest f-stop available. As in the example I gave Erik, I might need f/16 on a 50mm lens on a full-frame camera to get the same depth of field as f/8 on a 25mm lens on four thirds, but that has nothing to do with the speed of the two lenses, which might both be f/1.4 lenses.
Okay, I'll say it differently.

Why do we call it "speed"? That's right - because of the shutter speed it enables - a fast lens enables a fast shutter speed.

But here's the thing. In digital, what limits our shutter speed? It's the noise we're willing to tolerate in the final image.
An oversimplification.

I could shoot at that limit in some circumstances. I could also have enough light to take much shorter exposures.
That would put you over the limit.
So that's it, there is a shutter "limit" determined by noise? Nonsense.
Well, it is for me, usually.
Even if that's true, it's not generally true.
I usually have to keep a shutter speed fast enough to avoid motion/shake blur, not noise.
And so you do that irrespective of noise? Must be nice to not need fast cameras or fast lenses. Heck you just need a cell phone.
Depends on the light. It always depends on the light.

But the cell phone comment... are you even trying to understand what I'm saying? No, really - such things are described in pretty much every beginner's photography book.
And I usually have to open the aperture enough to avoid too much noise.
And if it doesn't open enough to get you below "too much noise"?
I could want to use certain shutter speeds, e.g. in order to give a plane's propeller the exact amount of motion blur I want. I could be forced to surpass the "noise we're willing to tolerate" limit, because otherwise motion blur would become an even bigger problem.
Then you're just changing the limit.
What limit? In that case the limit is the one given by motion blur.
You might say that f/1.4 is the same speed on both formats, but I'd say you are wrong.
Things are much simpler if f/1.4 is the same speed.
Siimpler, yes, but also unfair and wrong.
What's wrong is "equivalence". The f-number meaning what it always meant is right.
Equivalence works, in real life. You've been shown that. To say that a simple mathematical relationship is wrong after having been shown that it's right is just foolishness.
I never saw an example of equivalence being useful, except for writing articles, fighting about it on the Internet and "proving" that "FF" is cheaper/better/whatever.

And I explained countless times why equivalence is wrong. It's a combination of forcing a set of parameters, and pretending they give you "equivalent" images.
You'd take the simple basic notions and apply them according to the situation, instead of inventing a complex system based on many assumptions.
How is multiplying by crop factor complex?
The focal length being mixed with angle of view is added complexity. F-stops given the responsibility of measuring DOF is added complexity. Writing countless articles - not all agreeing with each other - about "equivalence" is added complexity.
It requires countless articles because of hard-headed people that won't listen to a very simple topic.
Or perhaps because some people are constantly attempting to push this pseudo-science on us.
The exposure equation, though, is simple and elegant.
No, it's actually not. It's more complex, and it's logarithmic. You're just used to it and have accepted it for a long time.
It actually is (but indeed I understand it well). You just love equivalence too much to see differently.
So is the notion that focal length is just a measure of how strong an optical lens would converge light rays. So is the notion that "APS-C crop" is no different than cutting a print with a pair of scissors...
Question: Does cutting with scissors change DOF?

Answer: Yes it does.
Irrelevant question. Irrelevant answer. DOF is subjective.
I can get four times the shutter speed (1/4th the shutter period) with f/1.4 on full frame as I can on 4/3 given the same amount of noise in the final image. This is why my f/4 lens on full-frame is effectively way, way faster than the f/2 lens on my pocket compact - I can use the f/4 lens at an exposure index of 12,800, whereas I'll only tolerate the f/2 lens at an EI of perhaps 400, giving the f/4 lens a 3-stop speed advantage.
I would never buy a "full frame" camera to shoot it one ISO step above what I'm using on my APS-C camera.
Then you're missing a big part of the reason people buy full-frame cameras. I bought one specifically for this reason.
And you never shoot at ISO 100?
Almost never. Once in a while.
All right...
Let me ask you a very simple question. If I mount a 1.4x teleconverter to my camera, and then mount a 200mm/2.8 lens to that, would you consider the combination to be a 280mm/4 equivalent system?
If you put a 1.4x teleconverter and a 200mm f/2.8 lens on it, then you'd have 280mm f/4. Not equivalent.
No, you are wrong.

The teleconverter goes behind the lens. It has no ability to change what the lens does at all since it comes after the lens in the light path. The lens is still a 200/2.8 and it has no idea what's behind it.

The teleconverter is making that 200/2.8 behave as a 280/4 would behave, but it isn't changing the lens at all.
You are the one who's wrong.

The "lens" attached to the camera is no longer the 200mm f/2.8, but the 200mm f/2.8 plus the teleconverter. Optically, it's no different than having a lens comprising the 200mm's elements, plus the TC's elements added to the rear.

You cannot think of it as a part of the camera, because it's an optical component. It would be no different than thinking that half of the lens is "part of the camera".
In fact, the teleconverter is between the lens and the camera. It can be thought of as part of the lens, or as part of the camera. If you think of it as part of the lens, it makes the focal length of the combined optical system longer without changing the aperture (which is why the f-stop changes - f-stop=focal length/aperture). If you think of it as part of the camera, it makes the sensor appear smaller from the lens' perspective. These two interpretations are equivalent, in every way - multiplying focal length and f-stop by (crop/teleconverter) factor, or shrinking the sensor by the same factor.
Except for one small detail: the focal length is actually changing (and f-stop has to be adjusted, because physical aperture remains the same), while the sensor is not shrinking.
Have you ever looked through a teleconverter from the lens' perspective? It just shrinks the sensor from that point of view.

Teleconverter%20optical%20reduction.jpg


This is why the simplest explanation for equivalence is that crop factor = teleconverter.
Nonsense. Crop factor is nothing else than using a smaller portion of the image projected by the optical lens (teleconverter included).
And no one argues that a teleconverter changes BOTH focal length and f-stop. Oddly, the same people will argue that shrinking the sensor only changes equivalent focal length, but not equivalent f-stop. From my point of view, that's irrational, especially in the face of photographic and mathematical evidence to the contrary.
Nonsense. Shrinking sensors, teleconverters being part of the camera but then changing lens' properties... that's kettle logic.
--
Lee Jay
Alex
 
Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.
It's sad you'd rather leave pretending to yourself you "won" some internet dispute, instead of trying to understand where you got it wrong. Such a simple thing, just a small assumption you're always making because your thinking is constrained by equivalence...
See the question? Information or "entertainment" -- the ball is in your court.
See the answer?

Sigh... I'll point it out to you again, not that it will matter. By 'exposure' you mean keeping exposure constant, because you're thinking in equivalence and in your mind things must be "equal". By 'total light' you mean keeping the total light constant, because you're thinking in equivalence.

But, exposure being more important than total light simply means knowing (measuring) what the exposure is is much more important than knowing what the total light is (sure, you can derive one from another if you know the sensor's area). It's the act of finding the "correct" (intended) exposure vs. computing noise based on total light - that's what makes your question irrelevant, and a red herring.

The fact that people were using exposure meters since ages and not shot noise meters are illustrating my point.

Alex
 
Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.
It's sad you'd rather leave pretending to yourself you "won" some internet dispute, instead of trying to understand where you got it wrong. Such a simple thing, just a small assumption you're always making because your thinking is constrained by equivalence...
See the question? Information or "entertainment" -- the ball is in your court.
See the answer?
I see the dodge. Are dodges "equivalent to" answers?
Sigh... I'll point it out to you again, not that it will matter. By 'exposure' you mean keeping exposure constant, because you're thinking in equivalence and in your mind things must be "equal".
Um, I'm not thinking to keep exposure constant at all. Where'd you get that idea from?
By 'total light' you mean keeping the total light constant, because you're thinking in equivalence.
Yes.
But, exposure being more important than total light...
In what way? You see, if you were to answer the questions above, rather than dodge them, this could go from "entertainment" to "information".
...simply means knowing (measuring) what the exposure is is much more important than knowing what the total light is (sure, you can derive one from another if you know the sensor's area). It's the act of finding the "correct" (intended) exposure vs. computing noise based on total light - that's what makes your question irrelevant, and a red herring.
Which exposure is "correct"?
  • f/2 1/100 ISO 400
  • f/4 1/100 ISO 1600
After all, the exposures in the above example are a full two stops apart. That said, I should think that you would need to know something about the desired intent with regards to the visual properties of the resulting photo to answer the question, would you not? This brings us full circle to the original question I've asked that you've been dodging.
The fact that people were using exposure meters since ages and not shot noise meters are illustrating my point.
Neither here nor there, and the fact that you think it's here and there illustrates my point.
 
Could you let us know how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
Of course, as always, answers are voluntary, not compulsory.
I'd rather not discuss red herrings. You already have the answers and more than enough clarifications; you have no excuse for refusing to understand.
Like I said, answers are voluntary, not compulsory. This freedom is so essential, the US Constitution even has an Amendment guaranteeing it.
It's sad you'd rather leave pretending to yourself you "won" some internet dispute, instead of trying to understand where you got it wrong. Such a simple thing, just a small assumption you're always making because your thinking is constrained by equivalence...
See the question? Information or "entertainment" -- the ball is in your court.
See the answer?
I see the dodge. Are dodges "equivalent to" answers?
Sigh... I'll point it out to you again, not that it will matter. By 'exposure' you mean keeping exposure constant, because you're thinking in equivalence and in your mind things must be "equal".
Um, I'm not thinking to keep exposure constant at all. Where'd you get that idea from?
For sure you are. You're assuming that's my point, exposure value constant - instead of "correct" (intended) exposure. I've explained that it's not, countless times.
By 'total light' you mean keeping the total light constant, because you're thinking in equivalence.
Yes.
But, exposure being more important than total light...
In what way? You see, if you were to answer the questions above, rather than dodge them, this could go from "entertainment" to "information".
I explained this countless times, even in the paragraph you're citing below.
...simply means knowing (measuring) what the exposure is is much more important than knowing what the total light is (sure, you can derive one from another if you know the sensor's area). It's the act of finding the "correct" (intended) exposure vs. computing noise based on total light - that's what makes your question irrelevant, and a red herring.
Which exposure is "correct"?
  • f/2 1/100 ISO 400
  • f/4 1/100 ISO 1600
I'm saying "correct" (intended) because how you expose an image depends on the photographer's intention; there is no one correct exposure. Obviously, both of your examples would generate the same image "brightness".
After all, the exposures in the above example are a full two stops apart. That said, I should think that you would need to know something about the desired intent with regards to the visual properties of the resulting photo to answer the question, would you not? This brings us full circle to the original question I've asked that you've been dodging.
You don't even read what I'm writing, are you?
The fact that people were using exposure meters since ages and not shot noise meters are illustrating my point.
Neither here nor there, and the fact that you think it's here and there illustrates my point.
You decided from the beginning that you "won" and I'm "wrong". So you're now just ignoring what I'm saying, and go on with your predefined path.

Arguing with you, unfortunately, is like playing chess with a pigeon.

Alex
 
...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
...I'll spell it out to you, once again, the same as I did for Fygaren:

As has been explained, and demonstrated, to you countless times:

http://www.josephjamesphotography.com/equivalence/#dof

The DOF (depth of field) is the distance between the near and far points from the focal plane that appear to be in critical focus and is a central player in the amount of detail rendered in an image. It is also important not to confuse DOF with background blur (which is discussed further down). Photos with:
will have the same DOF (and diffraction). Alternatively, photos with:
will also have the same DOF (and diffraction).

And yet, some will fight to the death against the obvious. Indeed, upthread, one of the opponents of Equivalence even refused to answer the following simple question:

Please tell us how the exposures compare with total light, and how the brightness compares, as a measure for the visual properties of the following three photos of the same scene from the same position displayed at the same size:
  • 50mm f/2 1/100 ISO 400 on mFT
  • 100mm f/2 1/100 ISO 400 on FF
  • 100mm f/4 1/100 ISO 1600 on FF
I can only wonder why. ;-)
The brightness will be the same on all 3 examples, but the 100mm f/2 1/100 ISO 400 on FF has an aperture diameter that is twice that of the other two resulting in 4 times as mutch light hitting the sensor. It will collect 4 times as mutch light because the aperture area is 4 times that of the others resulting in equivalent lens speed as the 50mm and the need for bumping ISO 2 stops on the other FF to get the same brightness. This is very obvious and should not require a reply.
Thank you for the simple and direct answer to the simple and direct question. All correct. However, you did not say anything about DOF. Would you like to elaborate?
The question only said brightness and total light, but Id be happy to elaborate. Maby this time Ill get through :)

Some place in there I could have said: ...so because the mFT and the FF have the same size physical aperture, it will lead to the same DOF, total light etc. However, f/2 on mFT is not equivalent to f/4 on FF, so you have to raise the ISO by two stops to get an equivalent brightness on the final image...
But that does not change the clumsy, inaccurate and wrong way you will fight to death for, explaining why "f/2 is equivalent to f/4"
I should think that the fact that options 1 and 3 had the same perspective, framing (more or less -- a little different due to the differences in aspect ratio), DOF, diffraction, shutter speed, brightness, and more or less the same noise (if the sensors had more or less the same efficiency) would be the explanation that you might be looking for.
Saying "f/2 is equivalent to f/4" and then meaning other results of aperture diameter (DOF, diffraction etc), is the same as saying "f/2 is equivalent to f/4" and then meaning other results of focal lenght (FOV)
Really? You think it's unnatural to compare different focal lengths on different formats? How very curious!
No, I said "this lens speed is equivalent to this lens speed" and then meaning it had the same FOV...how is it you can twist that in to me thinking its unatural to compare different focal lenghts?
f-number is after all the result of focal lengsh and diameter and a quantitative measure of lens speed.
Aside from DOF and noise, why do we care about lens speed? Saying, for example, that f/2 on mFT is "faster than" f/4 on FF is like saying a winner in a high school 100m race is "faster than" a silver medalist in the Olympics.
Ok
 
...
As told to you countless times, and explicitly spelled out in the article, f/2 on mFT is not merely equivalent to f/4 on FF in terms of DOF, but also in terms of diffraction and the total amount of light projected on the sensor -- equivalences that the anti-equivalence crowd are always keen to go out of their way to not mention, for reasons I've already explained.
As I told you countless times, its the aperture diameter that give the equivalent DOF, diffraction and total light. Its not the f-number, that is a quantitative measure of lens speed.
Thanks for pointing out again and again that aperture diameter is the physical parameter that affects the physical phenomenon DoF, etc. F-ratio is a number that has no causal effect. Mass of the earth will affect your weight; density of the earth does not, even though you can calculate the mass of earth back if you know the volume.

"anti-equivalence crowd" This differentiation of pro and anti-equivalences finally convinced me that the motivation of wanting to call F/2 on mFT is equivalent to F/4 on FF so that mFT is to be considered equivalent to FF like Fiat and Ferrari both going at 60mph makes Fiat and Ferrari equivalent.

Further evidence is the refusal of pro-equivalence folks to add either shutter speed change or ISO (could be both, I suppose). Whenever I asked just that, it's typically if I knew anything about F-stop, I would know F/2 is not F/2 anymore.

Sigh.
mostly,

My original reply was with a suggestion, what do you think?

"So how about f/2 is size-equivalent to f/4?

The size of the aperture is after all what determins total light, diffraction, DOF and f-number."

I think it could work,

A: - f/2 on mFT is equivalent to f/2 on FF

B: - True, but f/2 on mFT is size-equivalent to f/4 on FF

A: - True

End of discussion.

I think size works well as its a bit vague (?), it can mean both diameter and area(?)

Most imortant is that its not confusing, and aperture size and f-number is directly connected as focal lenght and FOV

Cheers!
 
Fygaren wrote:However, f/2 on mFT is not equivalent to f/4 on FF, so you have to raise the ISO by two stops to get an equivalent brightness on the final image...
so, what you're saying at such great length is that it isn't equivalent with respect to exposure, which no one denies. However, it is equivalent in every other respect. Assume perfect lens and camera (constrained only by diffraction and the quantum nature of light) and if given an image, to same aspect ratio, from an FF camera set to f/4, 400 ISO from one taken with mFT, f/2, 100 ISO, assuming the same ( perfectly calibrated) shutter speed, wou would be unable to tell them apart. That sounds to me to be equivalent in everything but exposure.
No, I said "this lens speed is equivalent to this lens speed" and then meaning it had the same FOV...how is it you can twist that in to me thinking its unatural to compare different focal lenghts?
The difference here is that some think that the limit to shutter speed is acceptability of noise, while others think it is achieving some arbitrary exposure. For me, the idea that all f/2 lenses have the same speed, i.e., let you use the same shutter speed, without reference to frame size seems absurd.
 

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