define DOF
define DOF
define DOF
DOF = what is acceptably sharp infront of and behind the focal plane given that only what is at the focal plane is actually in focus.
Now, how do you define DOF?
@DonaldB has written:the understanding of most photographers about equivalence is DoF, the problem is no one really wants to discuss equivalence practically.
If there's something you believe about equivalence that others here don't believe, state exactly what that is.
If there's something you don't believe about equivalence that others here believe, state exactly what that is.
Either way, I'm sure I can stage a practical demonstration that shows the reality.
Look at the equations can be quite instructive. damienfournier.co/dof-the-simplified-formula-to-understand-dof/
Take ASPC and FF. For the same FOV at a fixed distance d, then the "equivalent" focal length of the focal length on a FF for a focal length of f on an APSC is
1.5 f. So at the same f-number, N the DOF =2 (d/f)^2 N c, c the circle of confusion. So on the FF,
DOF_ff=2(d/1.5 f)^2 N_ff c_ff. DOF_apsc=2 (d/f)^2 N_apsc c_apsc
Hence DOF_ff/DOF_apsc=(1/1.5)^2 (N_ff/N_apsc) (c_ff/c_apsc).
Case 1 same sensor MP count, the c_ff/c_apsc)=1.5 for pixel peaking.
In this case DOF_ff/DOF_apsc=(1/1.5) (N_ff/N_apsc).
For printing to the same print size, there is a 1.5 enlargement ratio penalty for apsc which enlarges the c_apsc and hence
DOF_ff/DOF_apsc=(1/1.5)^2 (N_ff/N_apsc).
There are other conditions to consider but in general equivalence does not really hold up when considering DOF.
@DonaldB has written:define DOF
DOF = what is acceptably sharp infront of and behind the focal plane given that only what is at the focal plane is actually in focus.
Now, how do you define DOF?
same
There are other conditions to consider but in general equivalence does not really hold up when considering DOF.
Can you elaborate? For low magnification (say, the distance to the subject is at least 10x the focal length used), Equivalence really does hold up with regards to DOF:
@sybersitizen has written: @DonaldB has written:the understanding of most photographers about equivalence is DoF, the problem is no one really wants to discuss equivalence practically.
If there's something you believe about equivalence that others here don't believe, state exactly what that is.
If there's something you don't believe about equivalence that others here believe, state exactly what that is.
Either way, I'm sure I can stage a practical demonstration that shows the reality.
Look at the equations can be quite instructive. damienfournier.co/dof-the-simplified-formula-to-understand-dof/
It's a simplified formula, meaning it's not always applicable to reality. What does the unsimplified formula say?
There are other conditions to consider but in general equivalence does not really hold up when considering DOF.
The shorthand way in which most of us are accustomed to describing the principles of equivalence is indeed incomplete (because it's simplified). A question in another recent thread here prompted me to stage a practical test, and the results initially surprised me. It demonstrated that some of what we routinely say about equivalence is probably only strictly true at infinity focus. In order to avoid being surprised, we just have to be attentive to the terms we use.
exif data is finally showing 😊
edit, no its not ,its intermittent 🤔
What do you see in those results?
What do you conclude from them?
Do you believe they disprove things that others here believe about equivalence? If so, what things?
@PeteW has written:Paging @GreatBustard.
What?! TC misrepresented Equivalence (again) and locked a thread when this misrepresentation was challenged?! Say it isn't so! 😁😁😁
Anyway, let's keep this simple for the PM exchange. People rarely argue the validity of perspective, framing, DOF, and exposure time with regards to Equivalence -- it's always the noise end, so let's just concentrate on that point. To that end, the main disconnect for the anti-Equivalence crowd is two-fold:
1) What is the difference between and relevance of Exposure vs Total Light.
2) What the sources of noise in a photo are.The problem is, no matter how well explained, "they" simply refuse to change their preconceptions. That is, exposure is the be-all and end-all and noise is entirely a result of the sensor. You can create an extended syllogism (a=b, b=c, c=d, d=e, e=f) and they may or may not accept each link in the chain, but at the end, when you conclude that this means a=f, they go right back to square one.
So, let's create the chain. For most all photography, there are two primary sources of noise: the noise that is inherent in the light itself (a law of physics thing, not a technology thing), called "photon [shot] noise", and the additional electronic noise, often called "read noise", added by the sensor and supporting hardware. The former is typically the dominant of the two sources, and the only role the sensor plays in this source of noise is what proportion of light projected on the sensor is recorded (QE -- "Quantum Efficiency") which is pretty much the same for sensors of the same generation and has been relatively unchanged in the past decade.
The photon noise for a particular portion of the photo is entirely a function of the total amount of light that made up that portion of the photo. For example, if 4x as much light makes up a portion of the photo, that portion will be half as noisy (not including the electronic noise).
The electronic noise matters more in the portions of the photo made with very little light (e.g. the deep shadows at base ISO and progressively more of the photo as we use higher ISO settings to "compensate" for the lesser light). This has a huge effect on the DR of a photo, and the differences in electronic noise between sensors varies considerably more than the differences in QE, thus the greater variability of DR, which we will see when pushing shadows heavily and progressively more and more of the photo as we use higher ISO settings to "compensate" for lower and lower light (typically "significant" past, say, light levels where you'd be using ISO 6400 and higher).
The exposure is a function of the scene luminance, f-number, and exposure time whereas the total light is a function of the scene luminance, aperture diameter (yes, measured in millimeters), and exposure time. It is important to realize that the aperture diameter is also intimately involved with DOF, so, for a given scene and exposure time, DOF and noise are directly related. For example, for a given scene, perspective, framing, and exposure time, f/2.8 on mFT will have twice the DOF and be twice as noisy as f/2.8 on FF because the aperture diameter for FF is 2x larger, whereas the DOF and noisiness will be the same for f/2.8 on mFT and f/5.6 on FF, as the aperture diameters are the same.
If these points are not clearly understood, then further [productive] discussion is not possible, so that must be tackled first in any honest discussion.
Thanks GB, that is a great summary!
I had already responded to the other forum member (EZGritz) with this (which I hope I got right):
You have so much wrong that I honestly do not know where to start a response!
You need to go back and re-read the Joseph James web page on equivalence, slowly. I know some of it is quite technical but you will find that what I have said is right.
Your comment:
"F-stop is not the physical size of the aperture. It's the ratio between the front element and the back element in the hole made by the aperture blades"
...is only half right. Yes, the f-stop is a ratio but it has nothing to do with the front and back elements. The f-stop number is calculated from the focal length in mm and the physical measurement of the aperture opening in mm. Google this and you will find I am correct.
Example:
50mm focal length ÷ 18mm aperture = f/2.8Turn it the other way around:
50mm focal length ÷ f/2.8 = 18mm apertureAs previously mentioned, focal lengths are not equivalent across camera formats. A 50mm focal length on full frame is equivalent to a 25mm focal length on m4/3.
So let's do the same maths for m4/3 as I did above for FF:
25mm focal length ÷ f/2.8 = 9mm apertureYou should now be able to see that if both cameras are used at f/2.8, the FF camera will be getting more total light because the physical aperture is larger.
For these cameras to take an equivalent photograph using the same shutter speed, the settings would have to be as follows:
FF: 50mm at f/5.6 (50mm ÷ f/5.6 = 9mm)
m4/3: 25mm at f/2.8 (25mm ÷ f/2.8 = 9mm)At these settings, the photographer can use the same shutter speed on both the FF and m4/3 cameras. But note that the FF camera will necessarily need to use a higher ISO. If ISO on the m4/3 is set at 200, the FF camera will need ISO 800.
And this is his response:
Just take a small sensor and a large sensor camera with the same lens speed, set the same ISO, and let the camera choose the shutter speed for the correct exposure you can check with the histogram, and you will see both cameras choose the same shutter speed. Lock the shutter speed and let the ISO float and you will see they choose the same ISO. The rest is measurebating, math and theories to prove bad science. The physically bigger kit will collect more total light but it has to be spread over a larger surface. Each pixel may be hit with the same amount of light. A finite amount of light falls on any point on the surface of the sensor. The end result is what matters if you are trying to figure out which camera is going to work in the lowest light before both of them fail. It can be the M43 camera using CDAF because it will find focus in lower light as long as there is contrast in the area you are focusing on, virtually in the dark. It may be a noisy image though. If it's stacking images to make the photo it doesn't have to be. That's a noise canceling process. You can bring out details from an area that looks entirely black without bringing out the noise. The quality of the glass plays a role in this. Better glass passes more light.
😵 😵 😵
And this is his response:
EZGritz has written:Just take a small sensor and a large sensor camera with the same lens speed, set the same ISO, and let the camera choose the shutter speed for the correct exposure you can check with the histogram, and you will see both cameras choose the same shutter speed. Lock the shutter speed and let the ISO float and you will see they choose the same ISO. The rest is measurebating, math and theories to prove bad science. The physically bigger kit will collect more total light but it has to be spread over a larger surface. Each pixel may be hit with the same amount of light. A finite amount of light falls on any point on the surface of the sensor. The end result is what matters if you are trying to figure out which camera is going to work in the lowest light before both of them fail. It can be the M43 camera using CDAF because it will find focus in lower light as long as there is contrast in the area you are focusing on, virtually in the dark. It may be a noisy image though. If it's stacking images to make the photo it doesn't have to be. That's a noise canceling process. You can bring out details from an area that looks entirely black without bringing out the noise. The quality of the glass plays a role in this. Better glass passes more light.
😵 😵 😵
Remember when I said, "The problem is, no matter how well explained, 'they' simply refuse to change their preconceptions. That is, exposure is the be-all and end-all and noise is entirely a result of the sensor."? EZGritz response is a textbook example of that. Note how he says that the 4x greater amount of light is spread over 4x as much area (which is true), so it's all the same in the end (which is false).
He, and the rest of the anti-equivalence crowd, simply cannot let go of exposure as being central, when, in fact, it is completely irrelevant in cross format comparisons (except inasmuch as exposure is a component of total light: Total Light = Exposure x Sensor Area). The reason exposure is used, rather than Total Light, is because for a given format, exposure and Total Light are essentially interchangeable, like mass and weight are when in the same acceleration field. But when comparing different formats, the distinction is central, just as is the distinction between mass and weight if, say, you are comparing a person on the Moon vs a person on Earth.
So, EZGritz is uninterested in your "mathematical wizardry" -- he's only thinking about exposure. Thus, the question to pose to him is the following:
Consider the following three scenarios:
1) mFT at 25mm f/2.8 1/400 ISO 1600
2) FF at 50mm f/2.8 1/400 ISO 1600
3) FF at 50mm f/5.6 1/400 ISO 6400
How do the exposure, lightness, noise, perspective, [diagonal] framing, DOF, and motion blur compare for all three? I'll give you my answers first. The perspective, [diagonal] framing, lightness, and motion blur are the same for all three. The DOF and noisiness of (1) and (3) are the same, and the DOF and noisiness of (2) is half of that. The exposure of (1) and (2) are the same, but the photo in (2) is made with 4x as much light as the photo in (1), which is why it's half as noisy. The exposure of (3) is two stops lower than the exposure of (1), but both are made with the same total amount of light, and that's why they are equally noisy.
Quick prediction: he won't care. TC might step in, but not to tell EZGritz to stop arguing against the facts and/or just drop it, but to once again whine about how everyone already knows all this and is tired of hearing it over and over. What do I win? 😁
Remember when I said, "The problem is, no matter how well explained, 'they' simply refuse to change their preconceptions. That is, exposure is the be-all and end-all and noise is entirely a result of the sensor."? EZGritz response is a textbook example of that. Note how he says that the 4x greater amount of light is spread over 4x as much area (which is true), so it's all the same in the end (which is false).
He, and the rest of the anti-equivalence crowd, simply cannot let go of exposure as being central, when, in fact, it is completely irrelevant in cross format comparisons (except inasmuch as exposure is a component of total light: Total Light = Exposure x Sensor Area). The reason exposure is used, rather than Total Light, is because for a given format, exposure and Total Light are essentially interchangeable, like mass and weight are when in the same acceleration field. But when comparing different formats, the distinction is central, just as is the distinction between mass and weight if, say, you are comparing a person on the Moon vs a person on Earth.
So, EZGritz is uninterested in your "mathematical wizardry" -- he's only thinking about exposure. Thus, the question to pose to him is the following:
Consider the following three scenarios:
1) mFT at 25mm f/2.8 1/400 ISO 1600
2) FF at 50mm f/2.8 1/400 ISO 1600
3) FF at 50mm f/5.6 1/400 ISO 6400How do the exposure, lightness, noise, perspective, [diagonal] framing, DOF, and motion blur compare for all three? I'll give you my answers first. The perspective, [diagonal] framing, lightness, and motion blur are the same for all three. The DOF and noisiness of (1) and (3) are the same, and the DOF and noisiness of (2) is half of that. The exposure of (1) and (2) are the same, but the photo in (2) is made with 4x as much light as the photo in (1), which is why it's half as noisy. The exposure of (3) is two stops lower than the exposure of (1), but both are made with the same total amount of light, and that's why they are equally noisy.
Quick prediction: he won't care. TC might step in, but not to tell EZGritz to stop arguing against the facts and/or just drop it, but to once again whine about how everyone already knows all this and is tired of hearing it over and over.
You are right, he won't care and will just shift his goalposts around to confuse the situation more than it already is.
Thankfully this debate isn't happening in the open forum. After TC locked the thread, EZGritz wanted to keep going so he sent me a PM to tell me I was wrong, didn't understand etc. So this has all been playing out in PMs. If it had been in the open forum, I think I'd have been put into the sandbox with you by now!
What do I win? 😁
Ha ha, I am pretty sure you will win. If it was within my power, I would make you DPR m4/3 moderator for a week. That would be fun to watch.
The end result is what matters if you are trying to figure out which camera is going to work in the lowest light before both of them fail. It can be the M43 camera using CDAF because it will find focus in lower light as long as there is contrast in the area you are focusing on, virtually in the dark.
my a7iv blows any camera FF or M43 away ive owned by a huge amount in low light focusing.
What do you see in those results?
What do you conclude from them?
Do you believe they disprove things that others here believe about equivalence? If so, what things?
shooting close ups is a lot less noticeable from the first set of images i posted.
same focal length moved camera for same fov. is my exif data showing ? because its not on my computer. 🤨but everyone elses is.
Those photos mean zero and show nothing really.
If you want to compare DOF's you should string out a tape measure on the ground for at least a couple of metres and photograph it under different conditions to compare the DOF's along the tape measure in the various shots.
@DanHasLeftForum has written:No, DPReview moderators cannot see members' PMs under any circumstances.
PMs is where members can exchange personal information if they choose to and moderators have no right to potentially see that.
Yes, some members break the rules in PMs and in those cases I have always referred the PM to DPR Admin who do have access to view PMs.
DPR Admin have then dealt with the offending members.
As you seem to be a moderator on DPR, maybe you can tell me why I had a thread deleted on the Open Forum where I made a one line question. "Why do some photographers use medium format". I was accused of "Trolling". This has earnt me what seems to be permanent "sandboxing".
Many people cannot seem to understand the whys and wherefores, of DPR moderation. I made a complaint to Admin. but got no reply.
the mods are pulling all the tricks to make MF look better compared to FF. i just reprocessed 2 images that were posted as a comparision . the WB was deliberately set off to show more natural colours and more detail favering MF. as soon as i took them both into ACR and corrected WB you could barely tell the 2 images apart. so i posted my findings and my post was deleted instantly 😁 i knew it would, which makes DPR site as dishonest as a dodgy car sales yard.
@NCV has written: @DanHasLeftForum has written:No, DPReview moderators cannot see members' PMs under any circumstances.
PMs is where members can exchange personal information if they choose to and moderators have no right to potentially see that.
Yes, some members break the rules in PMs and in those cases I have always referred the PM to DPR Admin who do have access to view PMs.
DPR Admin have then dealt with the offending members.
As you seem to be a moderator on DPR, maybe you can tell me why I had a thread deleted on the Open Forum where I made a one line question. "Why do some photographers use medium format". I was accused of "Trolling". This has earnt me what seems to be permanent "sandboxing".
Many people cannot seem to understand the whys and wherefores, of DPR moderation. I made a complaint to Admin. but got no reply.
the mods are pulling all the tricks to make MF look better compared to FF. i just reprocessed 2 images that were posted as a comparision . the WB was deliberately set off to show more natural colours and more detail favering MF. as soon as i took them both into ACR and corrected WB you could barely tell the 2 images apart. so i posted my findings and my post was deleted instantly 😁 i knew it would, which makes DPR site as dishonest as a dodgy car sales yard.
You are asking people to take your word for it regarding the 2 images you posted were not dodgy.
Maybe they were, maybe they weren't.
You can't expect everyone to just take your word for it. Some will and some won't.