Is Full Frame sharper than APS-C? My answer here

[rs]

"As I already explained, it is a question of how much enlargement is necessary for a print."

this is about pixels, no film there you are enlarging grain etc , as long you have 1,6 times better lenses and use f8  and the signal/noise is equal from the  APS 24 mp  and the  24 mp 24x36 there is no enlargements advantages with 24x36
It is a pure optical question= we need 1,6 better lenses on the APS

Whoa !! Come on guys, Mikael is right here. I'm amazed how many people still confuse format with digital image size. It must all us old gits who come from film.

To use 18 MP as example, 18MP of digital info is 18MP - it doesn't matter what format it's in ! A 5D mk1 file has to be "enlarged" more than that from a 7D. Talking of enlarging an APS file "40" times is rubbish - who ever said that does not understand what is happening to enable them to view this digital data as a picture. Enlarge an APS Film negative 40 times - yes.

I'm surprised Neuro hasn't jumped in and put that one straight !

If we're talking about two cameras, both with 18MP of digital perfection, then presuming each has a lens up to the demands of the sensor, there will be no difference. But we are far from perfection, and as lots of this process is analogue, we never will reach digital perfection. Any two sensors of the same tech generation of different sizes will have very different characteristics. The smaller sensor, even if it has the same number of MP (such as a 7D vs a 1D X) will inevitably have smaller individual pixels, each able to gather less light than their FF equivalent and require more amplification to get the same rated sensitivities. This extra amplification of a weaker signal leads to more noise.

Just as adding MP to any one sensor doesn't harm as the final image is the sum total of all the pixels (the entire sensor), the larger sensor will always have an overall advantage - the larger sensor as a whole is able to capture more light. In that way, the analogy of two different size cuts from the same film (read same tech generation) is spot on. The larger one will always win.

Since sensors reached about 10+ MP, magnification of the sensor became much more important than how big each individual pixel is printed. If not, a Nokia Pureview 808 would be better than all but the latest medium format digital backs.

What you say is correct, but the enlargement required to produce a given picture size from a 7D and a 1DX is the same, irrespective of whether the resulting picture from the 1DX is better or not.

It depends upon whether you're talking about enlargement of each digital pixel in the file, or enlargement of the sensor/film. Enlargement of the sensor (with any moderately large MP sensor) is much more closely linked to image quality than enlargement of the digital file.

[rs]

If we're talking about two cameras, both with 18MP of digital perfection, then presuming each has a lens up to the demands of the sensor, there will be no difference.

Eh, no.

Not even close.

Camera A has a square sensor, 1" x 1". Camera B has a square sensor, 2" x 2". Both cameras have a 1024-by-1024 pixel array, for a grand total of one megapickle. Both sensors are digital nirvana, counting photons at the very limit of what quantum mechanics allows.

Each of camera B's pixels can gather four times as many photons as camera A's pixels. Camera B has two stops more dynamic range and the gradients it captures are two stops smoother, with two stops less pixelation. Make a 2" x 2" print from both, and the one from camera B will show two stops more detail and appear that much sharper.

Still not convinced?

Imagine that the sensors are so small and so densely packed that a single photon is enough to fill one of camera a's pixel wells. It can only record black and white, but it does so at amazingly high resolution. Camera B's pixels, on the other hand, are twice the linear dimension, so each photosite can record the presence of 0, 1, 2, 3, or 4 photons. The end result for camera A is a 1 megapickle black and white image; the end result for camera B is a 1 megapickle four-bit grayscale image. Obviously, camera B's image contains a great deal more information -- and it looks it, too.

Cheers,

b&

I fully agree with you. That's why I followed that first sentence with 'But we are far from perfection, and as lots of this process is analogue, we never will reach digital perfection', and then went on to say 'the larger sensor as a whole is able to capture more light'

[TrumpetPower!]

Mikael, I've explained enlargement multiple times.

But I'll give it yet another whack. Why not?

Let's say your client has hired you to photograph his sailboat tied to the dock at sunset, and he wants you to make a 48" x 72" print.

If you use a 135 format camera, such as a 5DIII, the image recorded on the sensor will have to be enlarged roughly 50 times to make said print. If you use an 8x10 view camera, the image will be enlarged about 6 times.

Let's say your view camera has a low-resolution sensor that is all of 500 ppi, making it an effective 20 megapickles -- slightly less than the 5DIII's 22 megapickles.

Also assume you're using a 50mm lens on the 5DIII and a 400mm lens, so both have the same field of view. And assume you're shooting the 5DIII at f/8 and the view camera at f/64, so they both have the same depth of field. And even go ahead and assume that you're shooting the 5DIII at ISO 100 and the view camera at ISO 6400, so the shutter speed is the same. For the sake of this discussion, also assume that the technology is comparable for both -- the sensors are the same generation, the lenses are both superb, and so on.

So, two setups almost identical. On the one hand, we've got a 22 megapickle 5DIII at ISO 100; on the other, a 20 megapickle view camera at ISO 6400. Same depth of field, same shutter speed, same framing, same everything else. And, at the end of the day, the same 48" x 72" print.

What, if any, difference do you think there will be between the two prints, and why?

Cheers,

b&

[tortilla]

If we're talking about two cameras, both with 18MP of digital perfection, then presuming each has a lens up to the demands of the sensor, there will be no difference.

Eh, no.

Not even close.

Camera A has a square sensor, 1" x 1". Camera B has a square sensor, 2" x 2". Both cameras have a 1024-by-1024 pixel array, for a grand total of one megapickle. Both sensors are digital nirvana, counting photons at the very limit of what quantum mechanics allows.

Each of camera B's pixels can gather four times as many photons as camera A's pixels. Camera B has two stops more dynamic range and the gradients it captures are two stops smoother, with two stops less pixelation. Make a 2" x 2" print from both, and the one from camera B will show two stops more detail and appear that much sharper.

Sure, but you are not always dynamic range limited, for instance in a studio or in a foggy landscape situation. The only difference will be that camera B's sensor needs four times less exposure (and even that doesn't matter when using lighting or a tripod).

[Sporgon]

Well I know Im not the Fiery Finn, but here goes:

The enlargement required to produce the prints would be similar, the 10x8 would require slightly more "enlargement" because it's less megapixel.

You're shooting the view camera at f64, not sure how the very large pixels on this hypothetical 10x8 sensor would negate the diffraction of this aperture. Similarly in this theoretical scenario the light capturing ability of this sensor will be amazing.

However you've wacked the ISO up to 6400 on the view camera. Again will the very large pixels cope with this - we don't know.

Actually just thinking about this as I go along, you actually cannot say what the difference will be, apart from the fact the enlargement is about the same, because of the very small aperture and very high ISO being used on the theoretical view camera.

IF the very large pixels resulted in no more diffraction than f8 on the 5D, and IF iso 6400 resulted in no more noise etc, then the image from the view camera would exhibit smoother tones, smoother graduation and more perceived resolution due to the lens projecting light in a much larger image onto the 10 x 8 chip.

[AlanF]

Let me have a stab at this, as I don’t mind making a mistake of I can learn from it. The 50 mm lens at f/8 has exactly the same open aperture as a 400 mm at f/64. Therefore, both let in the same number of photons at the same shutter speed. These are spread over the same number of pixels for each sensor. Therefore the signal to noise is the same for both sensors as both follow the square root of the number of photons. Each final enlargement has the same number of pixels. So each final enlargement has the same number of pixels with same signal to noise. So, they are of identical quality, the lenses being of equal quality..

[Meh]

Some of you are confusing enlargement and sensor resolution.  Do you know who you are?

[Sporgon]

Some of you are confusing enlargement and sensor resolution.  Do you know who you are?

As digital sensor potential resolution is physically defined by pixels, enlargement is indubitably related to sensor resolution.

This is totally different to film, where the image recorded on the film emulsion was physically enlarged to whatever size you were printing. ( or projecting ).

[Meh]

Some of you are confusing enlargement and sensor resolution.  Do you know who you are?

well it is a optical question and it is hard to get a APS lens that are 1,6 times better than a 24x36mm lens regarding resolution, contrast etc. If  there was one there will  not be any differences results wise between a 24MP APS sensor and a 24Mp 24x36mm sensor and if the signal/noise ratio was the same from the two sensors

What exactly do you mean by an "APS lens that is 1.6 times better than 24x36mm lens"?  Are you talking about two different lenses (e.g. an EF-S lens and an EF lens) or the same lens on different bodies?

Under what circumstances do you suppose the SNR would be the same for an APS-C sensor and a FF sensor that each had the same number of pixels?

[Meh]

Some of you are confusing enlargement and sensor resolution.  Do you know who you are?

As digital sensor potential resolution is physically defined by pixels, enlargement is indubitably related to sensor resolution.

This is totally different to film, where the image recorded on the film emulsion was physically enlarged to whatever size you were printing. ( or projecting ).

What makes it so different?

[Sporgon]

You are not physically enlarging each pixel from your sensor. The pixels are the electronic capture devices. This information is then interpreted by your computer program. Enlarging them to see your picture would be about as useful as enlarging the processor in your computer and having a look at that.

[Meh]

You are not physically enlarging each pixel from your sensor. The pixels are the electronic capture devices. This information is then interpreted by your computer program. Enlarging them to see your picture would be about as useful as enlarging the processor in your computer and having a look at that.

Are you talking about the sensor or the image? 

[Sporgon]

Well if I can answer with a question. When you look at your picture at 100% on your computer do you think you are looking at the pixels from your sensor ?

[AprilForever]

Mikael, I've explained enlargement multiple times.

But I'll give it yet another whack. Why not?

Let's say your client has hired you to photograph his sailboat tied to the dock at sunset, and he wants you to make a 48" x 72" print.

If you use a 135 format camera, such as a 5DIII, the image recorded on the sensor will have to be enlarged roughly 50 times to make said print. If you use an 8x10 view camera, the image will be enlarged about 6 times.

Let's say your view camera has a low-resolution sensor that is all of 500 ppi, making it an effective 20 megapickles -- slightly less than the 5DIII's 22 megapickles.

Also assume you're using a 50mm lens on the 5DIII and a 400mm lens, so both have the same field of view. And assume you're shooting the 5DIII at f/8 and the view camera at f/64, so they both have the same depth of field. And even go ahead and assume that you're shooting the 5DIII at ISO 100 and the view camera at ISO 6400, so the shutter speed is the same. For the sake of this discussion, also assume that the technology is comparable for both -- the sensors are the same generation, the lenses are both superb, and so on.

So, two setups almost identical. On the one hand, we've got a 22 megapickle 5DIII at ISO 100; on the other, a 20 megapickle view camera at ISO 6400. Same depth of field, same shutter speed, same framing, same everything else. And, at the end of the day, the same 48" x 72" print.

What, if any, difference do you think there will be between the two prints, and why?

Cheers,

b&

Why are you writing about 50mm lens , 500 PPI  etc etc etc. it has nothing to do with the question
I have answered , and it started here http://www.canonrumors.com/forum/index.php?topic=12448.15
it is all about the optical resolution, a  optical issue with the given parameters I presented - and that I have wrote more than once here.

Explain now the enlargement factor  with the given parameters I have presented since http://www.canonrumors.com/forum/index.php?topic=12448.15 .
and please, keep your self to the subject.

and cheers/skål

From my viewpoint, APS-C delivers me from both teleconverters and high fstops on many occasions.

[Meh]

Well if I can answer with a question. When you look at your picture at 100% on your computer do you think you are looking at the pixels from your sensor ?

Why would you answer with a question?