Canon Testing a 75+ Megapixel EOS-1 Body? [CR1]

[jrista]

right answer will always be, smaller pixels, lower noise
NOISE
simple as that

Is it that simple?  What about the signal?

now it is motor cycles again
please learn to discuss the subject, noise
small pixel always gives lower noise than bigger, the signal has not been up to discussion
You and Jrista has some problem to know what we discuss or not
Must it be like this?
and are you going to make you funny about me again?
head room, signal/noise , fluorite glass ,  etc etc  what next?

Ok, lets get to the root of this here.


1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?


This is a test. Your answer does matter. You will be judged upon it. Go!

[neuroanatomist]

please learn to discuss the subject

Sage advice....try following it, instead of DRagging every thread down into the dolDRums.

[jrista]

right answer will always be, smaller pixels, lower noise
NOISE
simple as that

Is it that simple?  What about the signal?

now it is motor cycles again
please learn to discuss the subject, noise
small pixel always gives lower noise than bigger, the signal has not been up to discussion
You and Jrista has some problem to know what we discuss or not
Must it be like this?
and are you going to make you funny about me again?
head room, signal/noise , fluorite glass ,  etc etc  what next?

Ok, lets get to the root of this here.


1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?


This is a test. Your answer does matter. You will be judged upon it. Go!

test? what kind of tests?

I got guests here in my summer house, what kind of answer do you want?
A  pragmatic answer will be-YOU don't se any difference from middle grey up to high light, (I can call it infinity)
But you are going to se better results in the shadows - then there are a huge among of other answers
BUT first my guests who are staying here  2  days more.

and PS You will be judged upon it. Go!
what in h... is that?

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

[neuroanatomist]

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

I think you're reading too much into the response.  The only takeaway I got was that we'll have to wait two more days for new pictures of QPcards on barbecues.

[ragmanjin]

right answer will always be, smaller pixels, lower noise
NOISE
simple as that

Is it that simple?  What about the signal?

now it is motor cycles again
please learn to discuss the subject, noise
small pixel always gives lower noise than bigger, the signal has not been up to discussion
You and Jrista has some problem to know what we discuss or not
Must it be like this?
and are you going to make you funny about me again?
head room, signal/noise , fluorite glass ,  etc etc  what next?

Ok, lets get to the root of this here.


1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?


This is a test. Your answer does matter. You will be judged upon it. Go!

test? what kind of tests?

I got guests here in my summer house, what kind of answer do you want?
A  pragmatic answer will be-YOU don't se any difference from middle grey up to high light, (I can call it infinity)
But you are going to se better results in the shadows - then there are a huge among of other answers
BUT first my guests who are staying here  2  days more.

and PS You will be judged upon it. Go!
what in h... is that?

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

Boys who do you think you are.
well Neuro and Jrista if  you don't understand my answer about signal noise and results below middle grey and  above middle grey here will come a quick answer
As I said it is one answer of many.

but here is one 
translate it can you do by your self
swedish to english  use google




Det enda som egentligen händer är ju att det då är helt och hållet [effektiv QE] som bestämmer bruset...
Brusvärdet är då alltid roten av ljusmängden (+tillskott normalt sett, men inte i detta fallet då). Bruset vid 18% grå eller ljusare kommer knappt påverkas alls av att det inte finns något läsbrus, det är en så liten del av bruset totalt sett här - även på ISO1600.

För t.ex 6D är vitpunkten i råfilen på ISO1600 ca 4400e- (FWC/16, eftersom ISO1600 är 16ggr mer än bas-ISO).
Mellangrått blir då 18% av 4400 = 792e-, och fotonbruset av detta är sqrt(792) = 28,1e- för 18% grå

Eftersom läsbruset i en normal kamera är ca 3e- vid ISO1600 hade detta knappt påverkat signalen. Skulle man lagt till 3e- till 28.1e- fotonbrus hade det blivit:
sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e- inte en märkbar skillnad, knappt mätbar)

Under 18% grå kommer bruset växa saktare när man går neråt mot det mörkare om man INTE har något läsbrus. Detta är skillnaden jmf med läsbrus.
Vid 1% exponering (-6.6Ev) är den infångade ljusmängden bara 44e-, så bruset blir:
sqrt(44) = 6.6e-
Här är läsbruset märkbart, då
sqrt(6.6^2 + 3^2) = 7.3e-.

Skillnaden är då 6.6e- utan läsbrus vs 7.3e- med normalt läsbrus - ingen jättestor skillnad här heller, men antagligen märkbar
6.6e- av 44e- signal = 20*log(44/6.6) = 16.4dB brusavstånd
7.3e- av 44e- signal = 20*log(44/7.3) = 15.6dB med läsbrus.

Den viktiga skillnaden kommer ju när läsbruset hade börjat dominerar fotonbruset, nere i skuggorna. Samma sak som DR-punkten alltså.
På -10Ev har den "läsbrusfria" kameran fortfarande ca 6dB SNR. Kameran med 3e- läsbrus kommer då ha nästan noll i SNR, dvs lika mycket brus som signal. Här syns det mycket skillnad.

Den läsbrusfria pixeln kan helt utan förluster delas upp i hur små pixlar som helst utan att bruset över bildytan (per detalj) ökar, har man läsbrus kommer detta vid en viss pixelmängd börja bli "för många brustillskott" för att det ska finas någon vinst med det.

right answer will always be, smaller pixels, lower noise
NOISE
simple as that

Is it that simple?  What about the signal?

now it is motor cycles again
please learn to discuss the subject, noise
small pixel always gives lower noise than bigger, the signal has not been up to discussion
You and Jrista has some problem to know what we discuss or not
Must it be like this?
and are you going to make you funny about me again?
head room, signal/noise , fluorite glass ,  etc etc  what next?

Ok, lets get to the root of this here.


1. Assume we have a hypothetical sensor that introduces ZERO read noise whatsoever...no dark current noise, no high frequency noise...no read noise of any kind from any electronic source in the camera, on the sensor die or anywhere else...just for discussions sake.

2. That sensor captures an image projected by a lens in dim light, at an ISO setting of 1600.

3. Is that image noise free, or is there an intrinsic component of noise that is a very part of the image itself?


This is a test. Your answer does matter. You will be judged upon it. Go!

test? what kind of tests?

I got guests here in my summer house, what kind of answer do you want?
A  pragmatic answer will be-YOU don't se any difference from middle grey up to high light, (I can call it infinity)
But you are going to se better results in the shadows - then there are a huge among of other answers
BUT first my guests who are staying here  2  days more.

and PS You will be judged upon it. Go!
what in h... is that?

And there you go, peoples! No answer, more obfuscation, and some beating around the bush about "guests".

I think we can safely come to the conclusion that Mikael does not understand the concept of intrinsic noise in an image signal.

Boys who do you think you are.
well Neuro and Jrista if  you don't understand my answer about signal noise and results below middle grey and  above middle grey here will come a quick answer
As I said it is one answer of many.

but here is one 
translate it can you do by your self
swedish to english  use google




Det enda som egentligen händer är ju att det då är helt och hållet [effektiv QE] som bestämmer bruset...
Brusvärdet är då alltid roten av ljusmängden (+tillskott normalt sett, men inte i detta fallet då). Bruset vid 18% grå eller ljusare kommer knappt påverkas alls av att det inte finns något läsbrus, det är en så liten del av bruset totalt sett här - även på ISO1600.

För t.ex 6D är vitpunkten i råfilen på ISO1600 ca 4400e- (FWC/16, eftersom ISO1600 är 16ggr mer än bas-ISO).
Mellangrått blir då 18% av 4400 = 792e-, och fotonbruset av detta är sqrt(792) = 28,1e- för 18% grå

Eftersom läsbruset i en normal kamera är ca 3e- vid ISO1600 hade detta knappt påverkat signalen. Skulle man lagt till 3e- till 28.1e- fotonbrus hade det blivit:
sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e- inte en märkbar skillnad, knappt mätbar)

Under 18% grå kommer bruset växa saktare när man går neråt mot det mörkare om man INTE har något läsbrus. Detta är skillnaden jmf med läsbrus.
Vid 1% exponering (-6.6Ev) är den infångade ljusmängden bara 44e-, så bruset blir:
sqrt(44) = 6.6e-
Här är läsbruset märkbart, då
sqrt(6.6^2 + 3^2) = 7.3e-.

Skillnaden är då 6.6e- utan läsbrus vs 7.3e- med normalt läsbrus - ingen jättestor skillnad här heller, men antagligen märkbar
6.6e- av 44e- signal = 20*log(44/6.6) = 16.4dB brusavstånd
7.3e- av 44e- signal = 20*log(44/7.3) = 15.6dB med läsbrus.

Den viktiga skillnaden kommer ju när läsbruset hade börjat dominerar fotonbruset, nere i skuggorna. Samma sak som DR-punkten alltså.
På -10Ev har den "läsbrusfria" kameran fortfarande ca 6dB SNR. Kameran med 3e- läsbrus kommer då ha nästan noll i SNR, dvs lika mycket brus som signal. Här syns det mycket skillnad.

Den läsbrusfria pixeln kan helt utan förluster delas upp i hur små pixlar som helst utan att bruset över bildytan (per detalj) ökar, har man läsbrus kommer detta vid en viss pixelmängd börja bli "för många brustillskott" för att det ska finas någon vinst med det.

What, nobody willing to hit up FreeTranslation?
The only thing that really happens is, of course, the fact that it is completely [effective QE] which determines noise ...
Brusvärdet is always the root of ljusmängden ( addition normally, but not in this case).  Noise at 18% gray or lighter will hardly be affected at all of the fact that there is no läsbrus, it is such a small proportion of overall noise here - even at ISO1600.

For example 6D is white spot in råfilen on ISO1600 approximately 4400 e- (attractive FWC/16, since ISO1600 is 16x more than base-ISO).
Mellangrått will be 18% of 4400 = 792e-, and fotonbruset of this is sqrt(792) = 28.1e- for 18% gray

as läsbruset in a normal camera is approximately 3e- at ISO1600 had barely had an impact on the signal. It would have added to the 3e- to 28.1e- fotonbrus had it been:
sqrt(28.1 ^2 3 ^ 2) = 28.3e- ( 0.2e- not a noticeable difference, barely measurable)

under 18% gray noise will grow more slowly when it goes down to the darker if you do not have a läsbrus. This is the difference comparison with läsbrus.
At 1% exposure (-6.6Ev) is the captured ljusmängden only 44e-, so noise becomes:
sqrt(44) = 6.6e-
this is läsbruset noticeably, then
sqrt(6.6) ^2 3 ^ 2) = 7.3e-.

The difference is then 6.6e- without läsbrus vs 7.3e- with normal läsbrus - no enormous difference here either, but probably noticeable
6.6e- of 44e- signal = 20 * log(44/ 6.6) = 16.4dB brusavstånd
7.3e of 44e- signal = 20 * log(44/ 7.3) = 15.6dB with läsbrus.

The important difference, of course, will when läsbruset had begun to dominate fotonbruset, down in the shadows. Same as DR-point then.
-10Ev has the "läsbrusfria" the camera is still approximately 6dB SNR. The camera with 3e- läsbrus will have almost zero in SNR, i.e. , as much noise as a signal. This can be seen very difference.

The läsbrusfria pixel can completely without losses are divided up in how small pixels at any time without the noise of image area (per detail) increases, it has läsbrus this will at a certain pixelmängd start to become "too many brustillskott" for it to appear any profit with it.


Not all of those words translated right, but the just of it looks like the maths. Nothing hurts my brain more than maths. All I know is that I have cameras with sensor sizes ranging from the size of my pinky nail to the size of your average card reader, and every time I print a print outside of that camera's native resolution, the outcome is exponentially better with bigger sensors and bigger pixels. As an example: http://motionblurdaily.com/2013/06/18/why-we-use-what-we-use-medium-format-vs-full-frame-vs-aps-c/

Or, as another example, check out the photos below. One is from a Canon SX40, with a sensor the size of my pinky nail and a pixel size of 1.5 µm, at base ISO. One is from my Canon T2i with an APS-C sensor and a 4.2µm pixel size, and the third is from a medium-format Phase One camera with a huge sensor (37mm x 49mm) and a 6.8µm pixel pitch. Can you guess which is which?

[eha]

Det enda som egentligen händer är ju att det då är helt och hållet [effektiv QE] som bestämmer bruset...
Brusvärdet är då alltid roten av ljusmängden (+tillskott normalt sett, men inte i detta fallet då). Bruset vid 18% grå eller ljusare kommer knappt påverkas alls av att det inte finns något läsbrus, det är en så liten del av bruset totalt sett här - även på ISO1600.

För t.ex 6D är vitpunkten i råfilen på ISO1600 ca 4400e- (FWC/16, eftersom ISO1600 är 16ggr mer än bas-ISO).
Mellangrått blir då 18% av 4400 = 792e-, och fotonbruset av detta är sqrt(792) = 28,1e- för 18% grå

Eftersom läsbruset i en normal kamera är ca 3e- vid ISO1600 hade detta knappt påverkat signalen. Skulle man lagt till 3e- till 28.1e- fotonbrus hade det blivit:
sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e- inte en märkbar skillnad, knappt mätbar)

Under 18% grå kommer bruset växa saktare när man går neråt mot det mörkare om man INTE har något läsbrus. Detta är skillnaden jmf med läsbrus.
Vid 1% exponering (-6.6Ev) är den infångade ljusmängden bara 44e-, så bruset blir:
sqrt(44) = 6.6e-
Här är läsbruset märkbart, då
sqrt(6.6^2 + 3^2) = 7.3e-.

Skillnaden är då 6.6e- utan läsbrus vs 7.3e- med normalt läsbrus - ingen jättestor skillnad här heller, men antagligen märkbar
6.6e- av 44e- signal = 20*log(44/6.6) = 16.4dB brusavstånd
7.3e- av 44e- signal = 20*log(44/7.3) = 15.6dB med läsbrus.

Den viktiga skillnaden kommer ju när läsbruset hade börjat dominerar fotonbruset, nere i skuggorna. Samma sak som DR-punkten alltså.
På -10Ev har den "läsbrusfria" kameran fortfarande ca 6dB SNR. Kameran med 3e- läsbrus kommer då ha nästan noll i SNR, dvs lika mycket brus som signal. Här syns det mycket skillnad.

Den läsbrusfria pixeln kan helt utan förluster delas upp i hur små pixlar som helst utan att bruset över bildytan (per detalj) ökar, har man läsbrus kommer detta vid en viss pixelmängd börja bli "för många brustillskott" för att det ska finas någon vinst med det.

My translation:

"
What really happens then, is that it's the effective QE that determines the noise. The amount of noise is always the square root of the amount of light (+ some additions normally, but not in this case). The noise at 18% gray or brighter won't be affected at all by the fact that there's no read noise, because it's such a small portion of the total noise in this case - even at ISO 1600.

For 6D, for example, the white point of the raw file at ISO 1600 is approx. 4400e- (FWX/16 since ISO 1600 is 16x more than base ISO). Middle gray is then 18% of 4400e- = 792e-, and the photon noise of this is the square root of 792e- = 28.1e- for 18% gray.

Since the read noise of a normal camera is approx. 3e- at ISO 1600, this would barely affect the signal. If you were to add the 3e- onto the 28.1e- of photon noise it would be:
Sqrt(28.1^2 + 3^2) = 28.3e- (+0.2e-, i.e. no noticable difference, barely measurable)

Below 18% gray the noise will grow more slowly as one approaches the darker if there is NO read noise. This is the difference compared to read noise.
At 1% exposure (-6.6Ev) the captured amount of light is only 44e-, so the noise will be:
Sqrt(44) = 6.6e-
In this case the read noise is noticable, because
Sqrt(6.6^2 + 3^2) = 7.3e-.

The difference is then 6.6e- without read noise, vs. 7.3e- with normal read noise -- no great difference here either, but probably noticable
6.6e- of 44e- signal = 20*log(44/6.6) = 16.4dB noise difference
7.3e- of 44e- signal = 20*log(44/7.3) = 15.6dB with read noise.

The important difference thus appears when the read noise starts to dominate the photon noise, down in the shadow area. The same business as with the DR point, in other words.
At -10Ev the "read noise free" camera still has approx. 6dB SNR. The camera with 3e- read noise will then have almost no SNR, i.e. the same amount of noise as signal. Here, there difference will be apparent.

The noise free pixel can losslessly be divided into as small pixels as possible withouth increasing the noise relative to the IQ (per detail), if there's read noise however, at a certain pixel number there'll bee "too many noise additions" for there to be any benefit of it.
"

Sorry for the poor translation. I'm Norwegian, not Swedish :p

[Don Haines]

While you were arguing about sensor technologies, quantum efficiency, and dark current noise, I was outside taking pictures.

Note that this picture was taken with a (gasp) crop camera, at ISO3200, and with a 20 second exposure. The northern lights way off in the distance came as a surprise when I looked at the picture on the computer, the naked eye could not see them.

The point I am trying to make is that even four year old sensor technology is amazing..... Go out and use it. Save the rancorous debate on new products until after it goes on sale and you get to use it.  You are arguing the fine points of vaporware.

[sharka23]

thanx Don.
now everything is fine .... again

[jrista]

While you were arguing about sensor technologies, quantum efficiency, and dark current noise, I was outside taking pictures.

Note that this picture was taken with a (gasp) crop camera, at ISO3200, and with a 20 second exposure. The northern lights way off in the distance came as a surprise when I looked at the picture on the computer, the naked eye could not see them.

The point I am trying to make is that even four year old sensor technology is amazing..... Go out and use it. Save the rancorous debate on new products until after it goes on sale and you get to use it.  You are arguing the fine points of vaporware.

Congrats on catching the northern lights! In Colorado, I'm too far south to even get a glimpse of them like that, even during big solar storms.

People do indeed complain too much about their cameras. Relative to each other there may be some differences, but relative to the past, every camera on the market these days is amazingly good. People should just get out and take pictures...

[Pi]

While you were arguing about sensor technologies, quantum efficiency, and dark current noise, I was outside taking pictures.

Note that this picture was taken with a (gasp) crop camera, at ISO3200, and with a 20 second exposure. The northern lights way off in the distance came as a surprise when I looked at the picture on the computer, the naked eye could not see them.

How do you know that this is not dark current noise? 

BTW, is there a town below the lights?

[Don Haines]

While you were arguing about sensor technologies, quantum efficiency, and dark current noise, I was outside taking pictures.

Note that this picture was taken with a (gasp) crop camera, at ISO3200, and with a 20 second exposure. The northern lights way off in the distance came as a surprise when I looked at the picture on the computer, the naked eye could not see them.

How do you know that this is not dark current noise? 

BTW, is there a town below the lights?

The auroral oval was fairly large and extended south of James Bay.. ( I am a bit north-west of Ottawa, Canada) and we sometimes see the lights here. There are no towns in that direction for a couple of hundred kilometers....  and they are very small towns. The glow on the horizon was distant northern lights.

[stewy]

Just got my 5D3 and tried it today; had a 40D since 2008. Its a nice improvement, but it doesn't blow me away. With that, I feel that the 5D3 is a great, all rounded camera. I would love to add a medium format Canon camera to the list. Since I don't do sports, the 1Dx isn't worth it for me. If Canon released a medium format camera then that would be awesome, but they would have to release at the same time some nice prime lens for landscape and portrait photographers.

Anyway, I'm off to go use my 5D3 some more.

[neuroanatomist]

I like your way to challenge statements

no comparison with any other camera

and when is good enough, with a other camera you might se the motive  twice better or worse


this is taken with a iPhone, some people like it

WOOOSSHHHHHHHHHH...

That was the sound of the point of those posts flying over your head.  You missed it, completely.

BTW, you should really try to pull out the detail in those shadows of your iPhone pic. I'm disappointed...   

[bdunbar79]

I ducked

[Don Haines]

I ducked

and here is some lower levels !!

no banding- nice

I like the first picture better.... pulling the details out of the shadow makes the sky less dramatic....