[altopiet]

It would maybe help to answer this question: Why would you buy an aps-c DSLR (or mirrorless camera) over a superzoom bridge camera with a new, very-efficient BSI sensor that offers a 28-300 f/2.8 "equivalent" fixed lens?

That "28-300 f/2.8" is pretty powerful, no? It would allow the same exposure over the same FOV as a 28-300 lens on aps-c, and that sensor is brand new and very efficient. If you didn't like primes, didn't care about VF and AF speed and were just concerned with IQ, why would you ever consider a larger sensor DSLR or MILC? Wouldn't it be a waste of money?

Now here's the thing - any answer you give that involves IQ, even if you're not not using the term equivalence, has the concept baked in. As does the 'low light' score in DXOmark. The Total Light is crucial to the reason why.

What this means is that you can just say, "because the aps-c DSLR has a larger sensor, and it's going to have the capability to give me better images than that P&S". Equivalence/Total LIght is implicit in that answer, just like horsepower is (usually) going to be baked into "Car A is faster than Car B."

Now, where the background knowledge of why may help you is if some neighbor or salesman says "why bother with a DSLR, this brand new P&S with this 28-300 2/8 lens is just as good, gets you the same exposure, etc." Might be good to have some facts and methodologies at your disposal to be able to call BS. It also would help you if the sensor sizes are much closer than P&S and aps-c - like aps-c & FF, because depending on the lens + body combo you're considering the larger sensor combo may not be better.

Of course it's completely up to you. IMO, it's unseemly to go through life with half-understandings when the truth is so readily available.

.

I do not say the discussion per se is worthless, but as hobbyist, I personally started with a K1000, and in later years, when I had some money, I bought my first DSLR and Pentax again because I had some old lenses that I could use. None of anything discussed here even had a remote influence over subsequent purchases, in my case it was affordability that counted in purchasing decisions for me.

I use my camera mainly as when I started years ago, with Shutter speed, f stop and ISO in mind. The rest like instantaneous review of the image taken, the fact that you could change the ISO after every photo taken, and other technical advances, was, and still is just an added bonus to me.

Hence my question if anybody have an idea what, if any, of the discussion in this thread, have any influence in purchasing decisions of other hobbyist and or professionals?

If it really have a major influence, I'll have to take some of my limited time to try and understand all sides of the argument, before I make my next purchase , but if not, I might just as well take that time and go out and shoot, because as it is, I don't get enough time to do that

[ElJamoquio]

My dear Jshermann999, FTR this forum is not your bar and you don't decide that everyone should embrace your "equivalence", or leave.

I understand that you "must" support a fellow "equivalentionist" no matter what, and since you cannot attack my arguments:

you choose to attack me/shut me up instead.

Well, have fun playing that game - by yourself. You still won't be able to predict resolution from the aperture opening and sensor size alone (like a certain poster wants us to believe).

Sweet jeebus no one believes you can predict resolution from aperture opening and sensor size alone.

Please stop repeating your 'only photon noise' falsehood.

---------- Post added 10-20-14 at 08:37 AM ----------

Hence my question if anybody have an idea what, if any, of the discussion in this thread, have any influence in purchasing decisions of other hobbyist and or professionals?

Absolutely critical in my purchases.

Comes up all the time in my day-to-day use, particularly when I'm running two cameras but also when I'm determining whether to use a shorter faster lens or a longer slower lens in low light.

I doubt I've convinced anyone else of anything ever. This is a religion to people.

[jsherman999]

Hence my question if anybody have an idea what, if any, of the discussion in this thread, have any influence in purchasing decisions of other hobbyist and or professionals?

Didn't I answer that with my example? Would you buy a 28-300 f/2.8 superzoom over an aps-c DSLR if your primary concern was IQ? Why not?

If you had the money available and came apon a used 645D + 25 f/4 kit, would that 645D + 25 f/4 get you better high-ISO noise than a D800 + 70-200 f/2.8? Or D800 paired with a 35 f/1.8? Or f/1.4? Would knowing those things influence your decision if you were buying 'larger' to try to get better low-light performance?

[BrianR]

Now, where the background knowledge of why may help you is if some neighbor or salesman says "why bother with a DSLR, this brand new P&S with this 28-300 2/8 lens is just as good, gets you the same exposure, etc." Might be good to have some facts and methodologies at your disposal to be able to call BS.

Facts won't help, you're probably better off just smiling politely and walking away quietly. If you're feeling friendly you could offer to share a wine cooler and talk about a less controversial topic like politics or religion.

[Kunzite]

Sweet jeebus no one believes you can predict resolution from aperture opening and sensor size alone.

What else should be added? We're prolonging this discussion because you can't be bothered to explain what you mean.
And if you can't add anything else, well...

Please stop repeating your 'only photon noise' falsehood.

Just stating it without any kind of explanation won't make it a falsehood.
Only the photon noise depends on "total light" in the manner described by "equivalence"; it is equal to the square root of the signal, thus guaranteed to be the same between any two different sensors - provided the signal is the same.
If there's something wrong with the paragraph above, provide a reasonably well explained correction and I'll be glad to accept it. After all, I'm a programmer, not a sensor engineer.

[jsherman999]

Facts won't help, you're probably better off just smiling politely and walking away quietly. If you're feeling friendly you could offer to share a wine cooler and talk about a less controversial topic like politics or religion.

Works at the sales counter, doesn't work as well in an online forum (I've tried)

I do wish we still had the political subforum though.

[ElJamoquio]

Only the photon noise depends on "total light" in the manner described by "equivalence"; it is equal to the square root of the signal, thus guaranteed to be the same between any two different sensors - provided the signal is the same.
If there's something wrong with the paragraph above, provide a reasonably well explained correction and I'll be glad to accept it. After all, I'm a programmer, not a sensor engineer.

'same technology'. Like it or lump it.

[Kunzite]

Hiding behind some vague words when asked to provide explanations won't do.
Let me help you explain; a conveniently found article will assist us:

The CCD signal-to-noise ratio calculation in the tutorial uses the following equation:
SNR = PQet / [ PQet + Dt + Nr2 ]1/2
where P is the incident photon flux (photons/pixel/second), Q(e) represents the CCD quantum efficiency, t is the integration time (seconds), D is the dark current value (electrons/pixel/second), and N(r) represents read noise (electrons rms/pixel).
Examination indicates that the equation above is simply structured as a ratio of total signal generated during the exposure time divided by the combined noise attributable to the three noise components described previously. The three noise sources are not correlated, and the denominator incorporates appropriate values for each noise component: the square-root of the signal accounts for the photon noise, dark noise is equivalent to the square-root of the product of dark current and integration time, and the square-root of N(r)-squared corresponds to the read noise component.

We can simplify the formula by renaming PQet as S(ignal).
Which means:
SNR = S / [ S + Dt + Nr2 ]1/2
or:
N = [ S + Dt + Nr2 ]1/2

"Same technology" have to be interpreted as "same pixels", and not some kind of magic which allows two imaginary sensors to nicely match "equivalence"; so we're only allowed to vary P (or S, in the simpler form). (Note: "equivalence" works with a constant t and varies the photon flux P instead - by changing the f-stop)

Let's consider the case with two sensors with the "same technology" i.e. "same pixels"; f is the crop factor between the two. Reminder: that's per-pixel SNR.

On the other sensor, let's get the same "total light" over the entire sensor area:
SNRcrop = (S * f^2) / [ (S * f^2) + Dt + Nr2 ]1/2

The pixel count will necessarily be:
Pixels_crop = Pixels / f^2

So far, I'm pretty confident with what I wrote

Now, could you (or anyone else) explain how "non-cropped sensor's" total SNR might be equal to "crop factor sensor's" total SNR, in conditions where Dt and Nr are significant?

I'm genuinely curious. Really. Enough not to care who's right and who's wrong. And because of this, please show me the courtesy of responding only if you have a proper explanation.

[ElJamoquio]

Hiding behind some vague words when asked to provide explanations won't do.
Let me help you explain; a conveniently found article will assist us:

We can simplify the formula by renaming PQet as S(ignal).
Which means:
SNR = S / [ S + Dt + Nr2 ]1/2
or:
N = [ S + Dt + Nr2 ]1/2

"Same technology" have to be interpreted as "same pixels", and not some kind of magic which allows two imaginary sensors to nicely match "equivalence"; so we're only allowed to vary P (or S, in the simpler form).

That's false. "Same Technology" refers to the read noise, so we are not constrained to only vary S. Dark current affects both images as well but in things that aren't, say, astrophotography, the dark current is not a large factor so in that sense I reject your premise.

If equivalence was false for the reasons you claim then these two lines would be offset from one another. Of course they are not.

[jsherman999]

Hiding behind some vague words when asked to provide explanations won't do.
Let me help you explain; a conveniently found article will assist us:

We can simplify the formula by renaming PQet as S(ignal).
Which means:
SNR = S / [ S + Dt + Nr2 ]1/2
or:
N = [ S + Dt + Nr2 ]1/2

"Same technology" have to be interpreted as "same pixels", and not some kind of magic which allows two imaginary sensors to nicely match "equivalence"; so we're only allowed to vary P (or S, in the simpler form). (Note: "equivalence" works with a constant t and varies the photon flux P instead - by changing the f-stop)

Let's consider the case with two sensors with the "same technology" i.e. "same pixels"; f is the crop factor between the two. Reminder: that's per-pixel SNR.

On the other sensor, let's get the same "total light" over the entire sensor area:
SNRcrop = (S * f^2) / [ (S * f^2) + Dt + Nr2 ]1/2

The pixel count will necessarily be:
Pixels_crop = Pixels / f^2

So far, I'm pretty confident with what I wrote

Now, could you (or anyone else) explain how "non-cropped sensor's" total SNR might be equal to "crop factor sensor's" total SNR, in conditions where Dt and Nr are significant?

What point are you trying to make, first? It's muddled, and I don't know what that CCD link has to do with anything you previously wrote that you're "confident" in. So, summarize, tie-in, before I dive in answering the wrong thing.

(If you think you've discovered something that disproves Total Light as a determinant of shot noise, you're gravely mistaken and contradicted by that very article you linked... but I won't assume you meant that yet.)

[Kunzite]

That's false. "Same Technology" refers to the read noise, so we are not constrained to only vary S. Dark current affects both images as well but in things that aren't, say, astrophotography, the dark current is not a large factor so in that sense I reject your premise.

If equivalence was false for the reasons you claim then these two lines would be offset from one another. Of course they are not.

"Same Technology" refers to read noise... how exactly? If it doesn't mean the same per-pixel read noise. And how and why do you want to vary read noise? Please explain.

"Equivalence" doesn't work for astrophotography, noted.

I see where you're getting those ideas, hmm... noise as a "function of DOF" and so on. But, you're basically saying "here's a chart, accept it with no explanation" - which is exactly what I asked you not to do.

jsherman999, please read my posts:

the photon noise depends on "total light" in the manner described by "equivalence"; it is equal to the square root of the signal, thus guaranteed to be the same between any two different sensors - provided the signal is the same.

[jsherman999]

I see where you're getting those ideas, hmm... noise as a "function of DOF" and so on.

Pardon the interjection in your convo with El J but it's not "noise as a function of DOF" but (shot) noise as a function of total light, and when total light is varied by varying physical aperture, DOF is affected simultaneously and correlates. Don't assume the chart shows a cart leading a horse.

jsherman999, please read my posts:

the photon noise depends on "total light" in the manner described by "equivalence"; it is equal to the square root of the signal, thus guaranteed to be the same between any two different sensors - provided the signal is the same.

Sorry, still muddled. Photon shot noise is guaranteed to be the same if total light is the same. Total light can easily be the same between any two sensors of varying sizes or efficiency by varying a number of parameters, like exposure time or physical aperture or distance to the light source. Please un-muddle your question/point further so we're not playing a guessing game about what you're really trying to say

EDIT: I actually just went back and read everything you've written in the last few pages for a third time and I still don't see what your point is. Looks like you started out trying to "catch" El J saying reslution is a factor hat can be determined by equivalence alone, something he never said, and then you segued into photon shot noise, seemingly to 'prove' something that no-one as far as I can see disputes. Please link your concepts more precisely, I'm interested in hearing what you're driving at.
.

[Kunzite]

You will actually find things like "Sensor Noise as a function of Depth of Field" on that blog, which is a silly thing to say. We agree, photon noise is a function of light flux which can be varied by varying physical aperture. We agree, DoF is affected as well.

If you give up on the assumption that it's all about photon noise, it should be clearer.
The question was: if Dt and Nr are significant (or, to make ElJamoquio happy, if Nr is significant) can we still say that both sensors' total SNRs are equal?

And that's the smallest problem, a curiosity of mine - wanting to know the limits of "equivalence" when it's supposed to work. After that - and one of the two things which actually started this discussion - is that technology is not necessarily equal. Same format, significant differences.

[ElJamoquio]

But, you're basically saying "here's a chart, accept it with no explanation" - which is exactly what I asked you not to do.

Your premise is again incorrect. I explained where your conclusions were wrong in theory and gave you empirical data that disproves your hypothesis.

[clackers]

the effect of read noise he's not conflating it with shot noise

They are both just components, summed as ... what to call it?

I know ... Total Noise!

The dude abides.

Whatever your beliefs, to me you're a Pentaxian, and with coolness points.