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How to measure ISO



 
 
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  #751  
Old November 29th 15, 11:23 AM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Eric Stevens wrote:

Savageduck:
Aaaaagh!!!!


Enough already!!


Yes. If he doesn't get it this time I wil give.


Please just leave the thread, you probably don't want to learn, even though you
at times make it seem like you do.

Remember, I am the one that have supported everything I've said. With examples,
math, web links to other people saying the exact same thing etc etc. You and
nospam are the ones that have provided nothing but "nuh-uh" in response. No
support, nothing but hot air.

So, no, you have nothing more to provide in this thread and you are doing the
right thing to leave.

--
Sandman
  #752  
Old November 29th 15, 08:33 PM posted to rec.photo.digital
Eric Stevens
external usenet poster
 
Posts: 12,469
Default How to measure ISO

On 29 Nov 2015 10:23:22 GMT, Sandman wrote:

In article , Eric Stevens wrote:

Savageduck:
Aaaaagh!!!!


Enough already!!


Yes. If he doesn't get it this time I wil give.


Please just leave the thread, you probably don't want to learn, even though you
at times make it seem like you do.

Remember, I am the one that have supported everything I've said. With examples,
math, web links to other people saying the exact same thing etc etc. You and
nospam are the ones that have provided nothing but "nuh-uh" in response. No
support, nothing but hot air.

So, no, you have nothing more to provide in this thread and you are doing the
right thing to leave.


I give up.

You are a believer, not an understander.
--

Regards,

Eric Stevens
  #753  
Old November 29th 15, 10:01 PM posted to rec.photo.digital
Eric Stevens
external usenet poster
 
Posts: 12,469
Default How to measure ISO

On 29 Nov 2015 09:09:55 GMT, Sandman wrote:

This is a hell of a long thread in distinct chunks. About half way
along a realised what has really been going on and chopped things off
there. I have placed my conclusions at the end for anyone who has the
stamina to geth there.


In article , Eric Stevens wrote:

Eric Stevens:
"You have to apply the crop factor (squared) to get the same
amount of light into a smaller sensor."

The above statement is quite true if you really do want to get
the same amount of light into the smaller sensor but I don't
know why on earth you would want to do this.

Sandman:
To create an equal image. With not enough light, the smaller
sensor will have to amplify the signal more and this creates more
noise. It's just something that most people don't think about.


The amount of light entering the camera depends on the area of the
entrance pupil. If it's all focussed on the sensor the intensity of
the light on the sensor depends on the area of the sensor. If you
scale the camera up or down by a factor k the amount of light
entering the camera varies as k^2. The area of the sensor also
varies as k^2. In other words the area of the sensor varies as the
entering light varies.


Yes, but as you know, exposure is amount of light per unit area. If the
exposure is the same, the amount of light collected by the sensors will be
different. The smaller sensor needs more light.


The smaller sensor needs more light to do what?

http://www.josephjamesphotography.com/equivalence/

"The only factors in the exposure are the scene luminance, f-ratio,
shutter speed, and transmissivity of the lens (note that neither
sensor size nor ISO are factors in exposure)."

Which means that in order to create an equal *image*, you have to adjust the
*exposure* to fit the size of the sensor.


How on earth can you say that when you have just quoted "note that
neither sensor size nor ISO are factors in exposure." What do you
mean by "equal image"? Equal in what way?

Eric Stevens:
The truth of the matter is that if you take a camera and scale
it either up or down, the lens f/value stays the same and the
level of illumination on the sensor stays the same. If the
sensitivity of the sensor has not changed then the invariant
level of illumination of the sensor means that exposure time
always remains the same.

Sandman:
But with the same f-stop, the smaller lens gets less light. So you
either need a larger f-stop or longer exposure time to match the
amount of light. This means you adjust the ISO down to the crop
factor square.


Who cares about the sensor getting less light?


Those that care about noise, since less light means more amplification and more
noise.


Why should there be more noise? Providing everything has been kept in
proportion as the camera has been scaled up or down, the light
intensity on the sensor remains the same irrespective of the size of
the sensor. Lighting of the sensor remains te same. Sensor remains the
same ...

The point is that the amount of light falling on each square
millimetre (or each square inch) remains exactly the same. As far as
the sensor is concerned there has been no change.


Not sure what supposed "change" you are talking about here? Changing the size
of the sensor means that comparing "exposure" falls apart.


The behaviour of a small patch of sensor is not affected by how much
more sensor there is around it. Certainly the sensitivity is not
affected. As I almost said "the point is that the amount of light
falling on each square millimetre remains exactly the same. As far as
the sensor is concerned there has been no change." Certainly it needs
neither more nor less amplification than it did before".

http://www.josephjamesphotography.com/equivalence/#8

"For a given scene, perspective, and framing, the total light
depends only on the aperture diameter and shutter speed (as opposed
to the f-ratio and shutter speed for exposure).


That's fine, if you are talking about a porthole letting light into a
room. But we are talking about a lense which takes light from a series
of point sources in limited external area and focusses that light onto
a series of point images on the sensor.

Fully equivalent
images on different formats will have the same brightness and be
created with the same total amount of light.


Not so, in the case of a camera.

Thus, the same total
amount of light on sensors with different areas will necessarily
result in different exposures on different formats, and it is for
this reason that exposure is a meaningless measure in cross-format
comparisons."


Having started with an incorrect premise he has finished with an
incorrect conclusion.

Eric Stevens:
Alternatively, if you have an invariant f/ value, an invariant
exposure time and an ivariant level of illumination of the
sensor then it follows that the ISO has remained the same. That
is, ISO value is not a function of sensor size.

Sandman:
Of course it isn't. But with the same exposure, sensor
amplification is a function of the sensor size, so while the ISO
number are similar, the sensor amplification is not.


Not so. Go back to my statement above and think about it. You will
conclude that amplification has nothing to do with sensor size.


http://www.josephjamesphotography.com/equivalence

"The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100
is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance,
regardless of the focal length or the sensor size. However, the
brightness for the two photos will be the same since the 4x lower
exposure is brightened 4x as much by the higher ISO setting. If the
sensor that the f/5.6 photo was recorded on has 4x the area as the
sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total
amount of light will fall on both sensors, which will result in the
same noise for equally efficient sensors"

Same amount of total light = same amount of amplification = same amount of
noise.


I assume you have been using the terms in the same way as James,
although you have not previously made that clear.

His important definitions a

"Exposu The total light per area (photons / mm˛) that falls on
the sensor while the shutter is open, which is usually expressed
as the product of the illuminance of the sensor and the time the
shutter is open (lux · seconds). The only factors in the exposure
are the scene luminance, t-stop (where the f-ratio is often a good
approximation for the t-stop), and the shutter speed (note that
neither sensor size nor ISO are factors in exposure)."

Here is a trap:

"Brightness: The brightness of an image (what people usually think
of as "exposure" -- same units as exposure): Brightness =
Exposure x Amplification."

Most people would assume that brightness = "The total light per area
(photons / mm˛) that falls on the sensor while the shutter is open..."
In fact James has been using the term as above, where 'magnification'
is a factor of arbitrary value. "Brightness" in this sense is the
value in a data array external to the sensor. (it might be part of the
sensor but it it is part of the electronics processing the output of
the sensor.)

"Total Light: The total number of photons that falls on the sensor
(lumen·seconds, or, equivalently, photons):
Total Light = Exposure x Effective Sensor Area."

--- long tail snipped (Savageduck heaves a sigh of relief) ---

I've finally realised where the problem lies in all this.

A. The terms as defined above are not entirely the same as those used
in the wider world. That this discussion was based on certain exact
definitions was not made clear from the outset. As James wrote:

"Many of the misunderstandings come from people using different
definitions for the same words. In particular, "f-ratio" is often
confused with "aperture", and "exposure" is confused with
"brightness" and "total light". The importance of these
distinctions is often overlooked or simply not understood, ... "

B. It has been assumed that noise is inversely proportional to photons
received: the more photons the less noise. To describe this as a hairy
approximation is a gross understatement, but no matter.

C. All this pseudo-mathematics is centred about an attempt to set out
the rules relating the properties of cameras of different sizes which
produce not only the the equivalent optical geometry but also equal
image noise.

D. If sensors on smaller cameras are required to accept (say) four
times as much light as a sensor in the larger camera, they will have
to have larger 'buckets' in the individual photosites. In the real
world their buckets will not be four times as large so will be
overflowed, with resulting highlight burnout.

E. Nobody operates a camera so as to obtain constant noise so the
conclusions, while interesting, are irrelevant to the real world.


I've had more than enough of this. Now that I've finally worked out
what Sandman has been talking about I'm happy to leave him to go his
own way and achieve his own state.
--

Regards,

Eric Stevens
  #754  
Old November 29th 15, 11:16 PM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Eric Stevens wrote:

Savageduck:
Aaaaagh!!!!

Enough already!!

Eric Stevens:
Yes. If he doesn't get it this time I wil give.


Sandman:
Please just leave the thread, you probably don't want to learn,
even though you at times make it seem like you do.


Remember, I am the one that have supported everything I've said.
With examples, math, web links to other people saying the exact
same thing etc etc. You and nospam are the ones that have
provided nothing but "nuh-uh" in response. No support, nothing
but hot air.


So, no, you have nothing more to provide in this thread and you
are doing the right thing to leave.


I give up.


Finally!

--
Sandman
  #755  
Old November 29th 15, 11:42 PM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Eric Stevens wrote:

Eric Stevens:
"You have to apply the crop factor (squared)
to get the same amount of light into a smaller sensor."

The above statement is quite true if you really do want to
get the same amount of light into the smaller sensor but I
don't know why on earth you would want to do this.

Sandman:
To create an equal image. With not enough light, the
smaller sensor will have to amplify the signal more and this
creates more noise. It's just something that most people
don't think about.

Eric Stevens:
The amount of light entering the camera depends on the area of
the entrance pupil. If it's all focussed on the sensor the
intensity of the light on the sensor depends on the area of the
sensor. If you scale the camera up or down by a factor k the
amount of light entering the camera varies as k^2. The area of
the sensor also varies as k^2. In other words the area of the
sensor varies as the entering light varies.


Sandman:
Yes, but as you know, exposure is amount of light per unit area.
If the exposure is the same, the amount of light collected by the
sensors will be different. The smaller sensor needs more light.


The smaller sensor needs more light to do what?


To create an equivalent exposure.

Sandman:
http://www.josephjamesphotography.com/equivalence


"The only factors in the exposure are the scene luminance,
f-ratio, shutter speed, and transmissivity of the lens (note that
neither sensor size nor ISO are factors in exposure)."


Which means that in order to create an equal *image*, you have to
adjust the *exposure* to fit the size of the sensor.


How on earth can you say that when you have just quoted "note that
neither sensor size nor ISO are factors in exposure." What do you
mean by "equal image"? Equal in what way?


There's a difference between "exposure" and "equal exposure". An exposure is a
given amount of light in a given unit area. With a smaller sensor, you create
an equal image by making an equal exposure, meaning more light on that smaller
area.

Eric Stevens:
The truth of the matter is that if you take a
camera and scale it either up or down, the lens f/value
stays the same and the level of illumination on the sensor
stays the same. If the sensitivity of the sensor has not
changed then the invariant level of illumination of the
sensor means that exposure time always remains the same.

Sandman:
But with the same f-stop, the smaller lens gets less
light. So you either need a larger f-stop or longer exposure
time to match the amount of light. This means you adjust the
ISO down to the crop factor square.

Eric Stevens:
Who cares about the sensor getting less light?


Sandman:
Those that care about noise, since less light means more
amplification and more noise.


Why should there be more noise?


Amplification leads no noise. Always.

Providing everything has been kept in proportion as the camera has
been scaled up or down, the light intensity on the sensor remains
the same irrespective of the size of the sensor. Lighting of the
sensor remains te same. Sensor remains the same ...


If the amount of total light is the same, the exposure is equal, but not the
*same* exposure (see above).

Using the same amount of light, you adjust the ISO down by the crop factor
squared, which means the signal is amplified the same amount (give or take,
sensor tech).

Eric Stevens:
The point is that the amount of light falling on each square
millimetre (or each square inch) remains exactly the same. As
far as the sensor is concerned there has been no change.


Sandman:
Not sure what supposed "change" you are talking about here?
Changing the size of the sensor means that comparing "exposure"
falls apart.


The behaviour of a small patch of sensor is not affected by how much
more sensor there is around it.


Of course not, but on a smaller sensor, that "small patch" is a part of the
whole image that is rendered by the camera. On a larger sensor, the same region
of the resulting image is made out of a larger "patch" of the sensor, which
receives more light. So there is more signal going in to that part of the
sensor that makes out that part of the image.

Certainly the sensitivity is not affected.


No, but to create an equally bright image, the smaller sensor needs to amplify
the signal more, since it has received less light to being with. I'm not sure
how many times I've said this in various forms.

As I almost said "the point is that the amount of light
falling on each square millimetre remains exactly the same. As far
as the sensor is concerned there has been no change." Certainly it
needs neither more nor less amplification than it did before".


Take it to the extreme, then. You have two sensors. One is 100x100mm and one is
10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor
will receive one hundredth of the "signal" as the larger sensor.

Both generate a 10MP image out of this signal, but the smaller sensor can't
possibly make it as bright as the larger sensor without amplification. And that
amplification creates noise.

Sandman:
http://www.josephjamesphotography.com/equivalence/#8


"For a given scene, perspective, and framing, the total light
depends only on the aperture diameter and shutter speed (as
opposed to the f-ratio and shutter speed for exposure).


That's fine, if you are talking about a porthole letting light into
a room. But we are talking about a lense which takes light from a
series of point sources in limited external area and focusses that
light onto a series of point images on the sensor.


Yeah? Do you have a point?

Sandman:
Fully equivalent images on different formats will have the same
brightness and be created with the same total amount of light.


Not so, in the case of a camera.


How so? Be specific.

Sandman:
Thus, the same total amount of light on sensors with different
areas will necessarily result in different exposures on different
formats, and it is for this reason that exposure is a meaningless
measure in cross-format comparisons."


Having started with an incorrect premise he has finished with an
incorrect conclusion.


Haha.

Eric Stevens:
Alternatively, if you have an invariant f/
value, an invariant exposure time and an ivariant level of
illumination of the sensor then it follows that the ISO has
remained the same. That is, ISO value is not a function of
sensor size.

Sandman:
Of course it isn't. But with the same exposure,
sensor amplification is a function of the sensor size, so
while the ISO number are similar, the sensor amplification is
not.

Eric Stevens:
Not so. Go back to my statement above and think about it. You
will conclude that amplification has nothing to do with sensor
size.


Sandman:
http://www.josephjamesphotography.com/equivalence


"The exposure (light per area on the sensor) at f/2.8 1/100 ISO
100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene
luminance, regardless of the focal length or the sensor size.
However, the brightness for the two photos will be the same since
the 4x lower exposure is brightened 4x as much by the higher ISO
setting. If the sensor that the f/5.6 photo was recorded on has 4x
the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then
the same total amount of light will fall on both sensors, which
will result in the same noise for equally efficient sensors"


Same amount of total light = same amount of amplification = same
amount of noise.


I assume you have been using the terms in the same way as James,
although you have not previously made that clear.


I have said so many many many many many times.

His important definitions a


"Exposu The total light per area (photons / mm˛) that falls on
the sensor while the shutter is open, which is usually expressed
as the product of the illuminance of the sensor and the time the
shutter is open (lux · seconds). The only factors in the exposure
are the scene luminance, t-stop (where the f-ratio is often a
good approximation for the t-stop), and the shutter speed (note
that neither sensor size nor ISO are factors in exposure)."


Indeed.

Here is a trap:


"Brightness: The brightness of an image (what people usually think
of as "exposure" -- same units as exposure): Brightness
Exposure x Amplification."


Exactly like I've said.

Most people would assume that brightness = "The total light per area
(photons / mm˛) that falls on the sensor while the shutter is
open..." In fact James has been using the term as above, where
'magnification' is a factor of arbitrary value. "Brightness" in this
sense is the value in a data array external to the sensor. (it might
be part of the sensor but it it is part of the electronics
processing the output of the sensor.)


"Brightness" is in the resulting image, like he said. Brightness = (Exposure +
amplification). 100% what I've said this entire time.

"Total Light: The total number of photons that falls on the sensor
(lumen·seconds, or, equivalently, photons): Total Light =
Exposure x Effective Sensor Area."


I.e. exactly what I've said.

I've finally realised where the problem lies in all this.


I won't hold my breath.

A. The terms as defined above are not entirely the same as those
used in the wider world. That this discussion was based on certain
exact


"Many of the misunderstandings come from people using different
definitions for the same words. In particular, "f-ratio" is often
confused with "aperture", and "exposure" is confused with
"brightness" and "total light". The importance of these
distinctions is often overlooked or simply not understood, ... "


Yes, these are some of the terms you and nospam have repeatedly misunderstood,
I agree.

B. It has been assumed that noise is inversely proportional to
photons received: the more photons the less noise. To describe this
as a hairy approximation is a gross understatement, but no matter.


Noise has nothing to do with amount of photons. It has everything to do with
amplification.

C. All this pseudo-mathematics is centred about an attempt to set
out the rules relating the properties of cameras of different sizes
which produce not only the the equivalent optical geometry but also
equal image noise.


I.e. creating equivalent images using different sensor sizes.

D. If sensors on smaller cameras are required to accept (say) four
times as much light as a sensor in the larger camera, they will have
to have larger 'buckets' in the individual photo sites.


This is false unless you're playing with highly overexposed images. A normally
exposed image, using the settings I have repeatedly mentioned, will not
saturate the photo sites of a MFT camera.

In the real world their buckets will not be four times as large so
will be overflowed, with resulting highlight burnout.


This is a false assumption. We are talking about a MFT sensor that has
amplified its signal four times as much as a FF sensor to create a "normal"
photo. Lowering the ISO and opening up the aperture will result in the exact
same photo, only with the exact same noise as the FF camera.

E. Nobody operates a camera so as to obtain constant noise so the
conclusions, while interesting, are irrelevant to the real world.


It's very relevant to a number of things:

1. The myth that smaller sensors are noisier, they aren't.
2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not.
3. The problem with using "35mm equivalent" terms for the focal length but not
for aperture to mislead users.
4. The myth that MFT sensors have larger DOF (a part of #3)

I've had more than enough of this. Now that I've finally worked out
what Sandman has been talking about I'm happy to leave him to go his
own way and achieve his own state.


No one is happier than me, believe me. And even if you didn't really understand
all of it, you came a long way, and have restored some of your maturity in the
process. Well done.

--
Sandman
  #756  
Old November 30th 15, 04:44 AM posted to rec.photo.digital
Eric Stevens
external usenet poster
 
Posts: 12,469
Default How to measure ISO

On 29 Nov 2015 22:42:49 GMT, Sandman wrote:

In article , Eric Stevens wrote:

Eric Stevens:
"You have to apply the crop factor (squared)
to get the same amount of light into a smaller sensor."

The above statement is quite true if you really do want to
get the same amount of light into the smaller sensor but I
don't know why on earth you would want to do this.

Sandman:
To create an equal image. With not enough light, the
smaller sensor will have to amplify the signal more and this
creates more noise. It's just something that most people
don't think about.

Eric Stevens:
The amount of light entering the camera depends on the area of
the entrance pupil. If it's all focussed on the sensor the
intensity of the light on the sensor depends on the area of the
sensor. If you scale the camera up or down by a factor k the
amount of light entering the camera varies as k^2. The area of
the sensor also varies as k^2. In other words the area of the
sensor varies as the entering light varies.

Sandman:
Yes, but as you know, exposure is amount of light per unit area.
If the exposure is the same, the amount of light collected by the
sensors will be different. The smaller sensor needs more light.


The smaller sensor needs more light to do what?


To create an equivalent exposure.

Sandman:
http://www.josephjamesphotography.com/equivalence


"The only factors in the exposure are the scene luminance,
f-ratio, shutter speed, and transmissivity of the lens (note that
neither sensor size nor ISO are factors in exposure)."


Which means that in order to create an equal *image*, you have to
adjust the *exposure* to fit the size of the sensor.


How on earth can you say that when you have just quoted "note that
neither sensor size nor ISO are factors in exposure." What do you
mean by "equal image"? Equal in what way?


There's a difference between "exposure" and "equal exposure". An exposure is a
given amount of light in a given unit area. With a smaller sensor, you create
an equal image by making an equal exposure, meaning more light on that smaller
area.


I'm breaking my vows but I sense progress. You may be saying that the
amount of information carried to the camera depends upon the number of
photons which reach it. From this it follows that the small sensor has
to handle the same amount of light as the larger sensor. Have I got it
right?

Eric Stevens:
The truth of the matter is that if you take a
camera and scale it either up or down, the lens f/value
stays the same and the level of illumination on the sensor
stays the same. If the sensitivity of the sensor has not
changed then the invariant level of illumination of the
sensor means that exposure time always remains the same.

Sandman:
But with the same f-stop, the smaller lens gets less
light. So you either need a larger f-stop or longer exposure
time to match the amount of light. This means you adjust the
ISO down to the crop factor square.

Eric Stevens:
Who cares about the sensor getting less light?

Sandman:
Those that care about noise, since less light means more
amplification and more noise.


Why should there be more noise?


Amplification leads no noise. Always.


Noise is always generated in a camera. The camera deals with this, in
part, by setting a noise floor. It sets a minimum level below which it
will ignore whatever is emerging from the sensor and it's
amplification. Of course this can never be entirely successful.

Providing everything has been kept in proportion as the camera has
been scaled up or down, the light intensity on the sensor remains
the same irrespective of the size of the sensor. Lighting of the
sensor remains te same. Sensor remains the same ...


If the amount of total light is the same, the exposure is equal, but not the
*same* exposure (see above).


It's this sort of statement which makes it hard to follow exactly what
you are saying. James defines exposure as (see below) "The total light
per area (photons / mm˛) that falls on the sensor while the shutter is
open". In the scaling up/down situation, if the total light is the
same, the total light per area (photons / mm˛) cannot be the same.
There is something wrong with your terminology.

Using the same amount of light, you adjust the ISO down by the crop factor
squared, which means the signal is amplified the same amount (give or take,
sensor tech).


I understand you better now that I know that by 'brightness' you don't
mean the density of the light which falls on the sensor, but instead
the amplified output of the sensor. At least James does. Do you mean
that also?

Eric Stevens:
The point is that the amount of light falling on each square
millimetre (or each square inch) remains exactly the same. As
far as the sensor is concerned there has been no change.

Sandman:
Not sure what supposed "change" you are talking about here?
Changing the size of the sensor means that comparing "exposure"
falls apart.


The behaviour of a small patch of sensor is not affected by how much
more sensor there is around it.


Of course not, but on a smaller sensor, that "small patch" is a part of the
whole image that is rendered by the camera. On a larger sensor, the same region
of the resulting image is made out of a larger "patch" of the sensor, which
receives more light. So there is more signal going in to that part of the
sensor that makes out that part of the image.


There comes a point where the message has got through, no matter how
many extra photons you have poured through.

Certainly the sensitivity is not affected.


No, but to create an equally bright image, the smaller sensor needs to amplify
the signal more, since it has received less light to being with. I'm not sure
how many times I've said this in various forms.


And the statement seemed meaningless, mad or daft before I realised
what you meant by brightness.

As I almost said "the point is that the amount of light
falling on each square millimetre remains exactly the same. As far
as the sensor is concerned there has been no change." Certainly it
needs neither more nor less amplification than it did before".


Take it to the extreme, then. You have two sensors. One is 100x100mm and one is
10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor
will receive one hundredth of the "signal" as the larger sensor.

Both generate a 10MP image out of this signal, but the smaller sensor can't
possibly make it as bright as the larger sensor without amplification. And that
amplification creates noise.


It is likely that even at their native ISO both sensors will require
amplificaton. It is also a big jump to assume that noise is
proportional to amplification, but never mind. Now that the
terminology is being defined I am beginning to understand you.


Sandman:
http://www.josephjamesphotography.com/equivalence/#8


"For a given scene, perspective, and framing, the total light
depends only on the aperture diameter and shutter speed (as
opposed to the f-ratio and shutter speed for exposure).


That's fine, if you are talking about a porthole letting light into
a room. But we are talking about a lense which takes light from a
series of point sources in limited external area and focusses that
light onto a series of point images on the sensor.


Yeah? Do you have a point?


Providing everything has been scaled correctly, the intensity of that
point of light on the sensor remains the same. The image is
constructed by the array of those points of light on the sensor.
Despite the fact that there are fewer photons finding the way to the
smaller sensor the intensity of the light (what I thought you meant by
brightness) remains the same across all sizes of cameras. If that is
the case, and I believe it is, all sizes of sensors will require the
same amount of amplification to reach the same brightness as defined
by James.

Sandman:
Fully equivalent images on different formats will have the same
brightness and be created with the same total amount of light.


Not so, in the case of a camera.


How so? Be specific.


I hope I just did. However I expect that my explanation will not
satisfy James' definition of equivalence.

Sandman:
Thus, the same total amount of light on sensors with different
areas will necessarily result in different exposures on different
formats, and it is for this reason that exposure is a meaningless
measure in cross-format comparisons."


Having started with an incorrect premise he has finished with an
incorrect conclusion.


Haha.


I think the chain of logic will have to be reworked with some care
given to definitions if this is going to be resolved.

Eric Stevens:
Alternatively, if you have an invariant f/
value, an invariant exposure time and an ivariant level of
illumination of the sensor then it follows that the ISO has
remained the same. That is, ISO value is not a function of
sensor size.

Sandman:
Of course it isn't. But with the same exposure,
sensor amplification is a function of the sensor size, so
while the ISO number are similar, the sensor amplification is
not.

Eric Stevens:
Not so. Go back to my statement above and think about it. You
will conclude that amplification has nothing to do with sensor
size.

Sandman:
http://www.josephjamesphotography.com/equivalence


"The exposure (light per area on the sensor) at f/2.8 1/100 ISO
100 is 4x as great as f/5.6 1/100 ISO 400 for a given scene
luminance, regardless of the focal length or the sensor size.
However, the brightness for the two photos will be the same since
the 4x lower exposure is brightened 4x as much by the higher ISO
setting. If the sensor that the f/5.6 photo was recorded on has 4x
the area as the sensor as the f/2.8 photo (e.g. FF vs mFT), then
the same total amount of light will fall on both sensors, which
will result in the same noise for equally efficient sensors"


Same amount of total light = same amount of amplification = same
amount of noise.


I assume you have been using the terms in the same way as James,
although you have not previously made that clear.


I have said so many many many many many times.


Yes you have said that about total light many many times but you have
never explained it. At several points in this thread I variously
enquired after or taken you to task over the terms you have been using
and you have responded by accusing me of semantic trolling and
similar. It wasn't semantic trolling: it was me trying to find out
*exactly* what you were trying to tell me.

His important definitions a


"Exposu The total light per area (photons / mm˛) that falls on
the sensor while the shutter is open, which is usually expressed
as the product of the illuminance of the sensor and the time the
shutter is open (lux · seconds). The only factors in the exposure
are the scene luminance, t-stop (where the f-ratio is often a
good approximation for the t-stop), and the shutter speed (note
that neither sensor size nor ISO are factors in exposure)."


Indeed.

Here is a trap:


"Brightness: The brightness of an image (what people usually think
of as "exposure" -- same units as exposure): Brightness
Exposure x Amplification."


Exactly like I've said.


But you have never explained that by 'brightness' you meant the output
of the amplifier.

Most people would assume that brightness = "The total light per area
(photons / mm˛) that falls on the sensor while the shutter is
open..." In fact James has been using the term as above, where
'magnification' is a factor of arbitrary value. "Brightness" in this
sense is the value in a data array external to the sensor. (it might
be part of the sensor but it it is part of the electronics
processing the output of the sensor.)


"Brightness" is in the resulting image, like he said. Brightness = (Exposure +
amplification). 100% what I've said this entire time.


The only place in the camera where there is an image is on the sensor.
After the sensor is scanned the image is converted into a stream of
data which takes whatever form the camera designer has thought of.

"Total Light: The total number of photons that falls on the sensor
(lumen·seconds, or, equivalently, photons): Total Light =
Exposure x Effective Sensor Area."


I.e. exactly what I've said.


I think you will find you have talked about 'total light' only once,
im message

I've finally realised where the problem lies in all this.


I won't hold my breath.

A. The terms as defined above are not entirely the same as those
used in the wider world. That this discussion was based on certain
exact


You have missed the end of the sentence which originally read

"That this discussion was based on certain exact definitions was
not made clear from the outset. As James wrote:"

"Many of the misunderstandings come from people using different
definitions for the same words. In particular, "f-ratio" is often
confused with "aperture", and "exposure" is confused with
"brightness" and "total light". The importance of these
distinctions is often overlooked or simply not understood, ... "


Yes, these are some of the terms you and nospam have repeatedly misunderstood,
I agree.


And you consistently failed to explain *exactly* what you meant to
many more people than me and nospam.

B. It has been assumed that noise is inversely proportional to
photons received: the more photons the less noise. To describe this
as a hairy approximation is a gross understatement, but no matter.


Noise has nothing to do with amount of photons. It has everything to do with
amplification.


Which depends on the number of photons received.

C. All this pseudo-mathematics is centred about an attempt to set
out the rules relating the properties of cameras of different sizes
which produce not only the the equivalent optical geometry but also
equal image noise.


I.e. creating equivalent images using different sensor sizes.


And lens sizes also.

D. If sensors on smaller cameras are required to accept (say) four
times as much light as a sensor in the larger camera, they will have
to have larger 'buckets' in the individual photo sites.


This is false unless you're playing with highly overexposed images. A normally
exposed image, using the settings I have repeatedly mentioned, will not
saturate the photo sites of a MFT camera.


I haven't got a MFT camera but one of these days I will try comparing
the Nikon D750 with my wife's Canon G12 (7.6mm x 5.7mm).

In the real world their buckets will not be four times as large so
will be overflowed, with resulting highlight burnout.


This is a false assumption. We are talking about a MFT sensor that has
amplified its signal four times as much as a FF sensor to create a "normal"
photo. Lowering the ISO and opening up the aperture will result in the exact
same photo, only with the exact same noise as the FF camera.


Here is something you can't prove. It is probably an assumption.

E. Nobody operates a camera so as to obtain constant noise so the
conclusions, while interesting, are irrelevant to the real world.


It's very relevant to a number of things:

1. The myth that smaller sensors are noisier, they aren't.
2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not.


The concept that ISO is a certain amplification is a myth (I hope).

3. The problem with using "35mm equivalent" terms for the focal length but not
for aperture to mislead users.


Look up the definition of f/ number (and think about it).

4. The myth that MFT sensors have larger DOF (a part of #3)

I've had more than enough of this. Now that I've finally worked out
what Sandman has been talking about I'm happy to leave him to go his
own way and achieve his own state.


No one is happier than me, believe me. And even if you didn't really understand
all of it, you came a long way, and have restored some of your maturity in the
process. Well done.


Thank you for those kind words. Now, please go and peddle your
nostrums somewhere else. I'm not buying.
--

Regards,

Eric Stevens
  #757  
Old November 30th 15, 08:48 AM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Eric Stevens wrote:

Sandman:
There's a difference between "exposure" and "equal exposure". An
exposure is a given amount of light in a given unit area. With a
smaller sensor, you create an equal image by making an equal
exposure, meaning more light on that smaller area.


I'm breaking my vows but I sense progress. You may be saying that
the amount of information carried to the camera depends upon the
number of photons which reach it. From this it follows that the
small sensor has to handle the same amount of light as the larger
sensor. Have I got it right?


Pretty much. In order to create an equal exposure, the smaller sensor needs the
same amount of total light (equal exposure), not the same amount of light per
unit area (same exposure).

Sandman:
Those that care about noise, since less light means
more amplification and more noise.

Eric Stevens:
Why should there be more noise?


Sandman:
Amplification leads no noise. Always.


Noise is always generated in a camera.


As a product of the signal amplification, i.e. the ISO setting.

Eric Stevens:
Providing everything has been kept in proportion as the camera
has been scaled up or down, the light intensity on the sensor
remains the same irrespective of the size of the sensor.
Lighting of the sensor remains te same. Sensor remains the same
...


Sandman:
If the amount of total light is the same, the exposure is equal,
but not the *same* exposure (see above).


It's this sort of statement which makes it hard to follow exactly
what you are saying. James defines exposure as (see below) "The
total light per area (photons / mm˛) that falls on the sensor while
the shutter is open". In the scaling up/down situation, if the total
light is the same, the total light per area (photons / mm˛) cannot
be the same.


Exactly.

"Same exposure" = "Same amount of light pr unit area"
"Equal exposure" = "Same amount of light on whole sensor"

Making an equal exposure means you're not using the same light per unit area.
This is what I have been saying the entire time. You need to give the smaller
sensor the same amount of *total light*, not the same exposure.

This is the same exposu

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f5.6, ISO 800

That will create two images that are equally bright. But they are ONLY equally
bright because the MFT sensor has amplified its signal MORE than the FF sensor
has, since it has received less light.

This is an equal exposu

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f2.8, ISO 200

Now we have given the MFT sensor the same amount of *total* light, not light
per unit area. This means that during the exposure, it has received as many
photons totally as the FF sensor (using an equivalent aperture). To balance
this, we have to adjust the ISO by the crop factor squared, to amplify the
signal LESS. This means that at ISO 200, the MFT sensor is amplifying the
signal equally to the FF sensor at ISO 800.

This tells us a couple of things:

1. "ISO" says nothing about sensor amplification (called "sensitivity" to most
people) when comparing between sensor formats

2. Smaller sensors aren't noisier, unlike what most people think. They just
need the same amount of light to create an equal image

Sandman:
Using the same amount of light, you adjust the ISO down by the
crop factor squared, which means the signal is amplified the same
amount (give or take, sensor tech).


I understand you better now that I know that by 'brightness' you
don't mean the density of the light which falls on the sensor, but
instead the amplified output of the sensor. At least James does. Do
you mean that also?


I have only ever talked about brightness in relation to the generated image.
I.e. I have mentioned several times that the ISO setting is not so much a way
to know how sensitive (amplified) the sensor is, but rather what brightness
level to expect.

So yes, "Brightness" has nothing to do with exposure (other than being a result
of it). It is how bright the image created by the camera is. Brightness is
exposure + amplification.

Eric Stevens:
The point is that the amount of light falling
on each square millimetre (or each square inch) remains
exactly the same. As far as the sensor is concerned there
has been no change.

Sandman:
Not sure what supposed "change" you are talking about
here? Changing the size of the sensor means that comparing
"exposure" falls apart.

Eric Stevens:
The behaviour of a small patch of sensor is not affected by how
much more sensor there is around it.


Sandman:
Of course not, but on a smaller sensor, that "small patch" is a
part of the whole image that is rendered by the camera. On a
larger sensor, the same region of the resulting image is made out
of a larger "patch" of the sensor, which receives more light. So
there is more signal going in to that part of the sensor that
makes out that part of the image.


There comes a point where the message has got through, no matter how
many extra photons you have poured through.


Yes, but the fewer the photons, the "weaker" the message, and without more
amplification, it will not be as bright.

Eric Stevens:
Certainly the sensitivity is not affected.


Sandman:
No, but to create an equally bright image, the smaller sensor
needs to amplify the signal more, since it has received less
light to being with. I'm not sure how many times I've said this
in various forms.


And the statement seemed meaningless, mad or daft before I realised
what you meant by brightness.


Well, good we got that sorted out, then.

Eric Stevens:
As I almost said "the point is that the amount of light falling
on each square millimetre remains exactly the same. As far as
the sensor is concerned there has been no change." Certainly it
needs neither more nor less amplification than it did before".


Sandman:
Take it to the extreme, then. You have two sensors. One is
100x100mm and one is 10x10mm, both are 10MP sensors. Using the
same exposure, the smaller sensor will receive one hundredth of
the "signal" as the larger sensor.


Both generate a 10MP image out of this signal, but the smaller
sensor can't possibly make it as bright as the larger sensor
without amplification. And that amplification creates noise.


It is likely that even at their native ISO both sensors will require
amplification.


No, that's not "likely" at all. That would only be "likely" if you knew exactly
how many photons that were transmitted, which wasn't a factor in the input.

It is also a big jump to assume that noise is
proportional to amplification,


You have different kind of noise, of course. Usually grouped together by
"photon noise" (different kind of noise from how the light hits the sensor,
travels through the lens etc) and "read noise" (noise created by the sensor and
hardware). But even though photon noise affect the end result, the read noise
is usually a lot higher and affects the resulting image much more.

Sandman:
Sandman:
http://www.josephjamesphotography.com/equivalence/#8

"For a given scene, perspective, and framing, the total light
depends only on the aperture diameter and shutter speed (as
opposed to the f-ratio and shutter speed for exposure).

Eric Stevens:
That's fine, if you are talking about a porthole letting light
into a room. But we are talking about a lense which takes light
from a series of point sources in limited external area and
focusses that light onto a series of point images on the sensor.


Sandman:
Yeah? Do you have a point?


Providing everything has been scaled correctly, the intensity of
that point of light on the sensor remains the same.


But that's the thing, using the same exposure *doesn't* scale everything
correctly.

Using a 25mm/f1.4 lens on MFT is usually called "equivalent" to a 50mm on FF,
but it's actually equivalent to a 50mm/f2.8 if scaled "correctly".

So:

FF: 50mm, f2.8, 1/250, ISO 800
MFT: 25mm, f2.8, 1/250, ISO 800

Will result in the *same* exposure, i.e the same amount of light per unit area.
But:

FF: 50mm, f2.8, 1/250, ISO 800
MFT: 25mm, f1.4, 1/250, ISO 200

Will result in an *equal* exposure, meaning you get the same amount of total
light through the lens on the sensor, you get the same depth of field, you get
the same signal amplification and the same field of view (i.e. collecting the
same scene).

This is what I have been saying for the last two weeks.

The image is constructed by the array of those points of light on
the sensor. Despite the fact that there are fewer photons finding
the way to the smaller sensor the intensity of the light (what I
thought you meant by brightness) remains the same across all sizes
of cameras.


*only* if you adjust the exposure as outlined above.

If that is the case, and I believe it is, all sizes of
sensors will require the same amount of amplification to reach the
same brightness as defined by James.


Absolutely, which is what I've said this entire time.

Eric Stevens:
I assume you have been using the terms in the same way as James,
although you have not previously made that clear.


Sandman:
I have said so many many many many many times.


Yes you have said that about total light many many times but you
have never explained it. At several points in this thread I
variously enquired after or taken you to task over the terms you
have been using and you have responded by accusing me of semantic
trolling and similar. It wasn't semantic trolling: it was me trying
to find out *exactly* what you were trying to tell me.


I "accused" you of semantic trolling when you cut out the context and claimed
that you can't change the sensitivity of the sensor. Which, while technically
true, was beside the point. A lot of people call the ISO setting on the camera
as a way to change the sensor sensitivity. If you hadn't cut out the context,
and just made a remark but signaled that you understood what I meant even so,
then you wouldn't have been trolling.

Eric Stevens:
His important definitions a


"Exposu The total light per area (photons / mm˛) that falls
on the sensor while the shutter is open, which is usually
expressed as the product of the illuminance of the sensor and
the time the shutter is open (lux · seconds). The only factors
in the exposure are the scene luminance, t-stop (where the
f-ratio is often a good approximation for the t-stop), and
the shutter speed (note that neither sensor size nor ISO are
factors in exposure)."


Sandman:
Indeed.


Eric Stevens:
Here is a trap:


"Brightness: The brightness of an image (what people usually
think of as "exposure" -- same units as exposure): Brightness
Exposure x Amplification."


Sandman:
Exactly like I've said.


But you have never explained that by 'brightness' you meant the
output of the amplifier.


I was not aware that it was unclear. This is the first time I used it:

Sandman
How to measure ISO
11/11/2015

"ISO was created decades ago measures amount of light
gathered per square inch on film. On a smaller sensor, ISO
breaks down because there are less such "inches". Using the
same ISO setting may result in similar brightness, but much
higher signal to noise ratio.

To get the same picture with two different size sensors, you
need to adjust the total amount of light, which ISO can not
do, it can only handle the amount of light given to it.

Smaller sensors needs a lower ISO to gather the same amount
of light as a larger sensor."

Note the date, that was 19 days ago.

This is the very first time I wrote it in a reply to you:

"A smaller sensor needs a higher sensitivity to produce the
same level of brightness as a larger sensor, meaning that
ISO 400 on MFT is more sensitive (more amplified) than ISO
400 on FF."
/ Sandman- 11/14/2015

Note the "to *produce* the same level of brightness".

To my knowledge, I have only ever used it in relation to the result, i.e. the
generated photo.

Eric Stevens:
"Total Light: The total number of photons that falls on the
sensor (lumen·seconds, or, equivalently, photons): Total
Light Exposure x Effective Sensor Area."


Sandman:
I.e. exactly what I've said.


I think you will find you have talked about 'total light' only once,
im message


Of course not, I've talked about this many many times:

Sandman
11/11/2015

"With less total amount of light, the signal to noise ratio
differs between sensor sizes, meaning that ISO 200 on MFT
has the same s/n ratio as ISO 800 on FF."

Sandman
11/14/2015

"Now, adjust the ISO by the crop factor squared, AND give
each sensor the exact same *total* amount of light:"

Sandman
11/14/2015

"So suddenly, higher pixel density isn't noisier at all.
because if you give the sensors the same amount of *total*
light, the noise level is more or less the same. So with
the same total amount of light, the end result is roughly
the same."

Sandman
11/15/2015

"Each square cm is hit by 10 photons in a given exposure. The
smaller sensor is hit by fewer *total* photos, regardless
of how many photo sites it has. It could be a 10MP MFT or a
24MP MFT, it matters not. The total amount of light that
hits the sensor is what is different."

And so on, and so on. A quick search found more then thirty occupancies of me
talking about the total amount of light. I didn't want to list them all here.

Eric Stevens:
"Many of the misunderstandings come from people using different
definitions for the same words. In particular, "f-ratio" is
often confused with "aperture", and "exposure" is confused
with "brightness" and "total light". The importance of these
distinctions is often overlooked or simply not understood,
... "


Sandman:
Yes, these are some of the terms you and nospam have repeatedly
misunderstood, I agree.


And you consistently failed to explain *exactly* what you meant to
many more people than me and nospam.


Maybe, and that's on me, for sure. But there's a fair level of stubbornness in
play here as well, especially when it comes to nospam. I.e. there is nothing
really "new" brought to the table in these last posts than haven't already been
said by me many many times. You have ignored the support, snipped the support
or just flat out denied the support without offering up a reason
(substantiation) for why the support isn't valid.

Eric Stevens:
B. It has been assumed that noise is inversely proportional to
photons received: the more photons the less noise. To describe
this as a hairy approximation is a gross understatement, but no
matter.


Sandman:
Noise has nothing to do with amount of photons. It has everything
to do with amplification.


Which depends on the number of photons received.


Only if the number of photons require amplification.

Eric Stevens:
C. All this pseudo-mathematics is centred about an attempt to
set out the rules relating the properties of cameras of
different sizes which produce not only the the equivalent
optical geometry but also equal image noise.


Sandman:
I.e. creating equivalent images using different sensor sizes.


And lens sizes also.


Indeed. I talked about that above as well.

Eric Stevens:
In the real world their buckets will not be four times as large
so will be overflowed, with resulting highlight burnout.


Sandman:
This is a false assumption. We are talking about a MFT sensor that
has amplified its signal four times as much as a FF sensor to
create a "normal" photo. Lowering the ISO and opening up the
aperture will result in the exact same photo, only with the exact
same noise as the FF camera.


Here is something you can't prove. It is probably an assumption.


I have proven it, remember:

http://jonaseklundh.se/files/iso_adjusted.png

Eric Stevens:
E. Nobody operates a camera so as to obtain constant noise so
the conclusions, while interesting, are irrelevant to the real
world.


Sandman:
It's very relevant to a number of things:


1. The myth that smaller sensors are noisier, they aren't.
2. The myth the "ISO" is a certain "sensitivity" (amplification),
it is not.


The concept that ISO is a certain amplification is a myth (I hope).


Yes, that's what I just said

Sandman:
3. The problem with using "35mm equivalent" terms for the focal
length but not for aperture to mislead users.


Look up the definition of f/ number (and think about it).


No need, the f-number is a physical function of the lens, but the f-ratio is
what is important when making equivalent comparison between cameras. See above.

--
Sandman
  #758  
Old November 30th 15, 01:32 PM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Whisky-dave
wrote:

Sandman:
To create an equal image. With not enough light, the smaller
sensor will have to amplify the signal more and this creates more
noise. It's just something that most people don't think about.


Those that started taking pictures using film know this


No, because if you shot with a 35 mm camera, all "sensors" were the same size. It
wasn't an issue back then. It's only an issue now.

ISO 400 was ISO 400 regardless of camera used.

Eric Stevens:
The truth of the matter is that if you take a camera and scale
it either up or down, the lens f/value stays the same and the
level of illumination on the sensor stays the same. If the
sensitivity of the sensor has not changed then the invariant
level of illumination of the sensor means that exposure time
always remains the same.


Sandman:
But with the same f-stop, the smaller lens gets less light. So you
either need a larger f-stop or longer exposure time to match the
amount of light. This means you adjust the ISO down to the crop
factor square.


Crop factor is irrelivent. Not sure why you're getting so confused
by this.


Whoosh.

Eric Stevens:
Alternatively, if you have an invariant f/ value, an invariant
exposure time and an ivariant level of illumination of the
sensor then it follows that the ISO has remained the same. That
is, ISO value is not a function of sensor size.


Sandman:
Of course it isn't. But with the same exposure, sensor
amplification is a function of the sensor size, so while the ISO
number are similar, the sensor amplification is not. So, to
repeat: FF: 1/250, f5.6, ISO 800 MFT: 1/250, f5.6, ISO 800


so no differnce then.


Great way to snip out the other example that illustrates the difference, troll.

What you snipped:

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f5.6, ISO 800

The above is the exact same *exposure*, but with different signal
amplification, the smaller sensor needs to amplify the signal more than the FF
sensor to present an equally bright image.

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f2.8, ISO 200

The above is the same *amount of light* on the sensor, which creates as
identical image as possible using different sensor technologies. Also, you
adjust the ISO by the crop factor squared to match the signal amplification of
the larger sensor, so you will get very equivalent noise.

--
Sandman
  #759  
Old November 30th 15, 04:14 PM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Whisky-dave
wrote:

To create an equal image. With not enough light, the
smaller sensor will have to amplify the signal more and this
creates more noise. It's just something that most people
don't think about.

Whisky-dave:
Those that started taking pictures using film know this


Sandman:
No, because if you shot with a 35 mm camera, all "sensors" were
the same size.


They didn;t have sensors, although some had inbuilt meters.


Whoosh

Sandman:
It wasn't an issue back then. It's only an issue now.


It's not an issue now, if anything it was more of an issue in the
past. That's why ISO came about.


Whoosh

Sandman:
ISO 400 was ISO 400 regardless of camera used.


Still the same today. ASA400 is ISO 400 which is 27 DIN.


Whoosh

--
Sandman
  #760  
Old November 30th 15, 04:17 PM posted to rec.photo.digital
Sandman
external usenet poster
 
Posts: 5,429
Default How to measure ISO

In article , Whisky-dave
wrote:


Great way to snip out the other example that illustrates the
difference, troll. What you snipped:

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f5.6, ISO 800

The above is the exact same
*exposure*, but with different signal amplification,


How do you know what signal amplification is being applied, you
don't.


Simple math and observed result.

Sandman:
the smaller sensor needs to amplify the signal more than the FF
sensor to present an equally bright image.


No it does NOT.


Incorrect.

Sandman:
FF: 1/250, f5.6, ISO 800
MFT: 1/250, f2.8, ISO 200


wrong.


Incorrect.

Sandman:
The above is the same *amount of light* on the sensor, which
creates as identical image as possible using different sensor
technologies. Also, you adjust the ISO by the crop factor squared
to match the signal amplification of the larger sensor, so you
will get very equivalent noise.


You're talking bull****.


Incorrect.

Take a sensor full frame cut it in two. They don;t halve in
sensitivity because you've cut it in two !


Since the photosites remain the same size, but will generate a photo with half
the fidelity (resolution) instead of more noise.

MFT sensors aren't one fourth the resolution of a FF sensor. Simple physics,
simple math.

Sensor size has NOTHING to do with the ISO rating.


Whoosh.

--
Sandman
 




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