I finally 'get' equivalence

Started May 22, 2012 | Discussions thread
Forum ProPosts: 10,046
so so wrong
In reply to Mr Lumière, May 23, 2012

We started with a great post. Simple and accurate. And you've shown you don't understand it. Indeed, you tried to make it more complicated, no doubt because you don't like the truth.

How more simple can it be? It's 4 times the total light. What's not to get?

People have a really hard time understanding this concept and it seems you didn't graps it well either, please let me explain:

Please don't

FF sensor capture more light in absolute terms, 4x more light.
FF sensors capture THE SAME AMOUNT OF LIGHT in rellative term.

I have no ideal what "relative terms" means. What is that? Should have stopped with the first statement, which is all that counts. Four times the light.

Let me see. In relative terms, a 4 liter V8 has the same power as a 2 liter 4 cylinder. But they don't have equal performance . .

In fact per pixel noise is directly rellated to pixel pitch not to sensor size, a 36mp FF sensor will have more noise per pixel than a 8mp m43 sensor of the same built for example.

No one cares about noise per pixel. We don't print pixels. It's the total image noise, for a given image size. Which is total light.

The reason more light gets to the sensor is because the sensor is bigger, it doesn't get more light per pixel, at all.

No. The reason the sensor gets more light is because there is a bigger lens gathering the light, which typically goes with a bigger sensor You could make a small sensor and a big lens and theoretically it could work just fine (assuming you could keep the sensor from saturating from the light density.)

Sensors don't gather light. Lenses do, and concentrate the light on a sensor. Take a tiny lens. Put a huge sensor behind it. Won't help at all, no matter how much you redesign the lenses.

You do not have more DR or less noise because of sensor size, you get that out of pixel pitch.

No. You get noise out of the total light used. The Nikon D800 has lower noise than the Canon Mk III, but over 1.5X the pixels. How do you think they did that?

However if a FF sensor has 12mp and a m43 sensor has 12mp it is safe to assume that the FF sensor will have better dynamics, but only because the pixels will be bigger and let more light in.

Funny. Pixels let light in? Please.

In a more simple way, if you crop a FF picture by a factor of 4 do you get more noise? Less DR? No.

Yes you do. Because if you crop it by 4 it's 4 times smaller. Meaning you have to magnify it by four to get back to the same image size. Meaning the noise becomes more visible by a factor of 4.

If a scene requires a FF sensor equiped camera to use an f2 aperture at ISO 400 it will also require a m43 sensor equipped camera to use an f2 aperture at ISO 400.

No. It will require a m43 sensor to have F1.0 at ISO 200 for the same noise. That's equivalence 101.

Noise and DR are the result of pixel pitch not sensor size, and to some extent processing, and that is as long as sensor built is identical.

No. For the same size image, noise is independent of pixel pitch. Noise only appears to go down with larger pixels because you print the image smaller. Of course you see less noise! It's not very fair to compare a large image to a small one.

Equivalence in term of distance from subject and DoF are affected by a 2x factor, a 25mm lens will require to be as far as a 50mm lens on a FF camera, an f2 aperture on m43 will give you the same DoF as an f4 aperture on FF.

Equivalence is a total theory that states equivalence for noise, resolution, and DOF. It doesn't cover dynamic range, but it could.

F4 on M43 is F8 on FF. No quibbling, no ifs, no buts. That's for DOF, resolution (assuming resolution is theoretically limited by lens performance), and noise.

Equivalence is really normalizing something that should never have been normalized. The front lens objective. Once you understand the the system pe5formance is tied to the diameter of the front objective you understand everything. F-stop obscures this because you have normalized the lens diameter to the system size.

Saying a lens is F2.8 is like saying an engine has 100 HP per liter. But how much horsepower does that lens have? F2.8 means the diameter is 2.8 times the focal length. A normalized number that does not tell you performance.

Do you really think that a pinkie sized F2.8 lens in a cell phone is equal to a battery size F2.8 4/3rd lens which is in turn equal to a flashlight size FF lens?

Did you know the Hubble telescope is an F24 lens? Do you think your F2.8 lens is better? Wanna bet?

This is all rather simple really.

Evidently not for you. Was this some type of contest to see how many things you could get wrong in a single post?

Lets get really simple. You know what determines the performance of a telescope? The front objective diameter. Scopes are advertized as 200" or 20" or 2" or whatever. You ever see a telescope vendor advertizing the CCD sensor size? Pixel pitch? Nope. Because it doesn't matters. Astronomers are smart dudes.

Larger lens gather more light. A properly design sensor uses all that light. No more, no less (unless you put a FF lens on a crop sensor, then you throw away some of the light.)

Yes, it really is that simple.

Now, if your camera is good enough for you, fine. That's all that's counts. A cell phone, given good light, can take great photos. But it isn't equivalent to a FF, even w/a F2.8 lens.

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