How to measure ISO

In article , Eric Stevens wrote:

Sandman:
It's also important here to note that light isn't
projected as a "cone" in a camera. This misconception comes
from images like these:

https://c2.staticflickr.com/6/5442/17491457679_4003be7f39_o.png

Which just describes how light is focused by the lens. If you
looked at that, you might be led to think that the image you
are taking is being focused in a small point in the center of
the focal plane. Or that you're using a magnifier to burn
some ants.

Fact is, all light comes in from different angles, and is
projected over the entire sensor, not just the middle.

Eric Stevens:
But it is projected in cones, from different directions.

Sandman:
Not really. A photon always travels in a straight line (well,
almost always) and the lens in the camera changes its direction.
Or rather, the *lenses* in the camera changes its direction many
times.

The only cone shape you could talk about is the inverse one:

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

That's field of view.

The large one, yes. The smaller one is that field of view as focused on the focal
plane.

Previously you were describing the paths of ray bundles.
http://www.edmundoptics.com/images/a.../fig-9a-rs.gif shows
bundles of rays coming from different parts of the object and
constructing the image by being focussed on different parts of the
sensor.

That's also quite a misleading image, which make it appear that photons from
different places (say, one from your hair, and one from your chin) is focused
*together* at one specific point. They aren't.

--
Sandman

On 2015-11-18 22:28:36 +0000, Sandman said:

In article 2015111810114199157-savageduck1@REMOVESPAMmecom, Savageduck wrote:

nospam:
nope. you might *think* you're correct, but you are not.

what you stated is *wrong* and demonstrates a serious
misunderstanding of what is actually happening.

the brightness is the same regardless of sensor size. this is
easily demonstrated but you refuse to listen.

The problem is Jonas is talking about noise, and how it is
introduced due to sensor size differences and signal amplification
to match light sensitivity.

That is indeed what I am talking about, and I don't see it as much of a
problem.


ISO in the digital world is different to ASA/DIN/ISO for film.

Very correct.

However, as a guideline digital ISO is close enough to provide
equivalent sensitivity in all sensor sizes that making a full manual
digital camera exposure setting using the data from a handheld
incident light meter is going to give very close to equally exposed
images from a M4/3, an APS-C, or a FF regardless of MP density.

If you exchange "equivalent sensitivity" to "equivalent brightness",
then I'm all
with you. As I've shown, a smaller sensor needs to be more sensitive to
light in
order to produce an equally bright image at the same "ISO".

Noise is a completely different issue, and each of those sensors is
going to perform differently in that regard, in accord with the
particular sensor characteristics, and that is a result of pixel
density and CPU amplification.

Pixel density doesn't affect noise as much as people think it does, and is
marginal when compared to the ISO-related noise issue I have been
talking about.

A photographer shooting at a specific ISO on any size sensor is
going to be able to expose correctly according to that sensitivity,

Correct, but at that brightness level, a smaller sensor will be noisier
since it
receives less signal. Ergo - ISO is not a level of sensitivity of the
sensor, but
an expected brightness of the resulting image.

and obtain images which appear equally exposed. One might well be
noisier than another, by that has nothing to do with exposure.

A smaller sensor in the same exposure and same "ISO" will always be noisier,
since it is receiving less signal and its reported "ISO" is amplifying
the signal
more than a larger sensor, meaning you get more noise. It's a simple result of
physics.

....and yet for the photographer the ISO and the exposure settings
remain the same, and that is due to sensor signal amplification by the
camera CPU, and that negates the other scene issues. So it doesn't
matter if a smaller sensor is exposed to less total light, or the
larger more. The photographer can set the exposure on both cameras
manually with the same settings given by that incident lightmeter for a
particular ISO and aperture, knowing the results, with regard to
exposure are going to be similar. This whole question is about
exposure not noise. The noise is going to be different, but is
irrelevant when it comes to exposure and methods of minimizing or
fixing noise should be understood as a seperate issue.

BTW: I have always thought that attempting to cover up sensor noise by
going to a B&W conversion just gives you a noisy, bad looking B&W
image. Sensor noise looks nothing like film emulsion grain. I am a guy
who used to love the grain in Tri-X, but these days I find a
less-noisy, smoother, finer grain B&W image much more pleasing.

--
Regards,

Savageduck

In article , Eric Stevens wrote:

Sandman:
Yes, I am in reference to signal amplification when saving the RAW
file to the memory card. You have a charge in the sensor, which
when read is amplified to match the camera-set ISO. Unless it's
an ISO-less camera, but that's another topic.

When you say 'sensor' I take it that you mean that rectangular
object behind the lens. If you do, this is where we can start homing
in on the error. You say "You have a charge in the sensor, which
when read is amplified to match the camera-set ISO". This is not
correct. It is the charge in each individual photosite which is
amplified by a predetermined amount corresponding to the camera-set
ISO. If the photosite is empty the charge is zero and the amplified
voltage is also zero. That photosite is black. If another photosite
is full, the charge is amplified to the voltage level corresponding
with white. The size of the sensor has nothing to do with the amount
of amplification required by the individual photosites.

I agree with all of that, but it is also irrelevant to my point. This is also
why all my examples have been with an evenly distributed exposure, not an image
of a cat, but of a flat white wall for instance. Or even mathematically equal.

Sandman:
And here's where you go wrong. But you're on the verge of getting
it.

Even talking about photo sites is a bit misleading, really, given
the multitude of different resolutions we have out there. But for
the sake of simplicity, let's assume we have a 10MP FF sensor and
a 10MP MFT sensor (again, resolution/pixel density isn't
important, but lets keep it simple).

So, a given exposure is amount of light over a given unit area,
right? So a 1/250 f2.8 exposure gives 20,000,000 photons over a
square cm as an example.

Now, a FF sensor is 3.5 x 2.4 cm large, so it has a total of 8.4
square cm's. This means that the FF sensor will receive a total
of 168,000,000 photons.

The MFT sensor is 1.8 x 1.35 cm large, so it has a total of 2.43
square cm's, which is four times smaller than the FF sensor (not
exactly, it's a different aspect ratio, but close enough for our
example) and will thus receive a total of 48,600,000 photons,
right?

So, again - both are 10MP, so we divide those received photons by
the resolution:

FF: 168,000,000 / 10,000,000 = 16.8 photons per photo site
MFT: 48,600,000 / 10,0000,000 = 4.86 photons per photo site.

Again, four times less photons per photo site

But you are not comparing apples with apples. You think you are
comparing large sensors with small sensors but in fact you are
really comparing large photosites with small photo sites.

Again, the size of the photo sites doesn't matter. The math is the same even if
the photo sites were identical in size, but just fewer of them in the smaller
sensor.

FF: 8.4 cm^2/10Mp = 8.4E^-5mm^2 =.0092mm per side square.
MFT: 2.43cm^2/10Mp = 2.43E^-6mm^2 =.00156mm per side square

The diameter of your FF photosite is 5.88 times that of the MFT.

If the photosites remain of constant size then we have:

FF: 8.4cm^2 10Mp =16.8 photons per site
MFT: 2.43cm^2 2.9Mp =16.8 photons per site

Correct, and if MFT sensors always were one fourth the resolution of FF
sensors, we wouldn't have this discussion.

Unfortunately, they aren't, they are about the same resolution, which is why I
used that in my math above, and then we have the observable problem.

The amplification required is the same in both cases.

Your general idea is correct but you have expressed it entirely
incorrectly.

Which is odd of you to say, you seem to agree with it 100%. You were forced to
cut the resolution of the MFT sensor to a fourth to get comparable data.

It's not the change in size of sensor that requires an
adjustment. It's the change in size of photosite which requires an
adjustment.

Which in the real world is what happens when you change the size of the sensor.

If you think about it, you have highlighted the problem
that manufatcturers have encountered in maintaining sensitivity
while reducing the size of photosites (i.e. higher Mp).

Which, incidentally, is what I've been talking about this entire time



--
Sandman

In article 2015111815154678698-savageduck1@REMOVESPAMmecom, Savageduck wrote:

Sandman:
A smaller sensor in the same exposure and same "ISO" will always
be noisier, since it is receiving less signal and its reported
"ISO" is amplifying the signal more than a larger sensor, meaning
you get more noise. It's a simple result of physics.

...and yet for the photographer the ISO and the exposure settings
remain the same

Correct, and that's more important to the photographer. But it has led to the
idea that smaller sensors are noisier. But they aren't, if you give them the same
amount of light.

and that is due to sensor signal amplification by
the camera CPU, and that negates the other scene issues. So it
doesn't matter if a smaller sensor is exposed to less total light,
or the larger more. The photographer can set the exposure on both
cameras manually with the same settings given by that incident
lightmeter for a particular ISO and aperture, knowing the results,
with regard to exposure are going to be similar. This whole
question is about exposure not noise. The noise is going to be
different, but is irrelevant when it comes to exposure and methods
of minimizing or fixing noise should be understood as a seperate
issue.

It is an issue that is related to ISO, which is related to sensor size. The OP
posted a question about getting different result from different cameras. Others
noted that ISO is a bit arbitrary (which it is), and I also added that with
different sensor sizes, the same ISO setting will give you different results due
to the different levels of amplifications.

BTW: I have always thought that attempting to cover up sensor noise
by going to a B&W conversion just gives you a noisy, bad looking
B&W image. Sensor noise looks nothing like film emulsion grain. I
am a guy who used to love the grain in Tri-X, but these days I find
a less-noisy, smoother, finer grain B&W image much more pleasing.

Totally agree.

In fact, Exposure (from Alien Skin) has a nice grain function that can cover up
the bad noise quite nicely

--
Sandman

On Thu, 19 Nov 2015 08:08:48 +1300, Eric Stevens
wrote:

On 18 Nov 2015 08:01:18 GMT, Sandman wrote:

In article , Eric Stevens wrote:

Sandman:
It's also important here to note that light isn't projected as a
"cone" in a camera. This misconception comes from images like
these:

https://c2.staticflickr.com/6/5442/17491457679_4003be7f39_o.png

Which just describes how light is focused by the lens. If you
looked at that, you might be led to think that the image you are
taking is being focused in a small point in the center of the
focal plane. Or that you're using a magnifier to burn some ants.

Fact is, all light comes in from different angles, and is
projected over the entire sensor, not just the middle.

But it is projected in cones, from different directions.

Not really. A photon always travels in a straight line (well, almost always) and
the lens in the camera changes its direction. Or rather, the *lenses* in the
camera changes its direction many times.

The only cone shape you could talk about is the inverse one:

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

That's field of view. Previously you were describing the paths of ray
bundles. http://www.edmundoptics.com/images/a.../fig-9a-rs.gif
shows bundles of rays coming from different parts of the object and
constructing the image by being focussed on different parts of the
sensor. The fact that it's a multi-element sensor doesn't change what
happens.

Oops - read 'multi-element lens'.

Where the larger "cone" is the angle of view of the lens, and the smaller one on
the other side of the lens is the image focused.
--

Regards,

Eric Stevens

On 18 Nov 2015 22:56:36 GMT, Sandman wrote:

In article , Eric Stevens wrote:

Sandman:
It's also important here to note that light isn't
projected as a "cone" in a camera. This misconception comes
from images like these:

https://c2.staticflickr.com/6/5442/17491457679_4003be7f39_o.png

Which just describes how light is focused by the lens. If you
looked at that, you might be led to think that the image you
are taking is being focused in a small point in the center of
the focal plane. Or that you're using a magnifier to burn
some ants.

Fact is, all light comes in from different angles, and is
projected over the entire sensor, not just the middle.

Eric Stevens:
But it is projected in cones, from different directions.

Sandman:
Not really. A photon always travels in a straight line (well,
almost always) and the lens in the camera changes its direction.
Or rather, the *lenses* in the camera changes its direction many
times.

The only cone shape you could talk about is the inverse one:

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

That's field of view.

The large one, yes. The smaller one is that field of view as focused on the focal
plane.

Previously you were describing the paths of ray bundles.
http://www.edmundoptics.com/images/a.../fig-9a-rs.gif shows
bundles of rays coming from different parts of the object and
constructing the image by being focussed on different parts of the
sensor.

That's also quite a misleading image, which make it appear that photons from
different places (say, one from your hair, and one from your chin) is focused
*together* at one specific point. They aren't.

I agree that photons from different places are not all focussed at the
one point.

In the diagram, those bundles of rays are not each emanating from the
object as a whole: the blue are coming from the eyes, the green is
from the teeth and the red is from the chin. The rays in each entry
bundle are not parallel but come together at points on the object, at
the eyes, teeth and chin respectively.
--

Regards,

Eric Stevens

On 2015-11-18 23:25:30 +0000, Sandman said:

In article 2015111815154678698-savageduck1@REMOVESPAMmecom, Savageduck wrote:

Sandman:
A smaller sensor in the same exposure and same "ISO" will always
be noisier, since it is receiving less signal and its reported
"ISO" is amplifying the signal more than a larger sensor, meaning
you get more noise. It's a simple result of physics.

...and yet for the photographer the ISO and the exposure settings
remain the same

Correct, and that's more important to the photographer. But it has led to the
idea that smaller sensors are noisier. But they aren't, if you give
them the same
amount of light.

The only trouble with that is if you are going to use the same exposure
settings for an APS-C and a FF camera. (ISO/apperture/Shutter speed)
for the same scene with the same light, the equalization is going to
made by the individual camera CPUs, and the only evidence of different
sensors is going to be resolution and possible noise. The "brightness
of the image for both should be the same.

and that is due to sensor signal amplification by
the camera CPU, and that negates the other scene issues. So it
doesn't matter if a smaller sensor is exposed to less total light,
or the larger more. The photographer can set the exposure on both
cameras manually with the same settings given by that incident
lightmeter for a particular ISO and aperture, knowing the results,
with regard to exposure are going to be similar. This whole
question is about exposure not noise. The noise is going to be
different, but is irrelevant when it comes to exposure and methods
of minimizing or fixing noise should be understood as a seperate
issue.

It is an issue that is related to ISO, which is related to sensor size. The OP
posted a question about getting different result from different cameras. Others
noted that ISO is a bit arbitrary (which it is), and I also added that with
different sensor sizes, the same ISO setting will give you different
results due
to the different levels of amplifications.

BTW: I have always thought that attempting to cover up sensor noise
by going to a B&W conversion just gives you a noisy, bad looking
B&W image. Sensor noise looks nothing like film emulsion grain. I
am a guy who used to love the grain in Tri-X, but these days I find
a less-noisy, smoother, finer grain B&W image much more pleasing.

Totally agree.

In fact, Exposure (from Alien Skin) has a nice grain function that can cover up
the bad noise quite nicely

....and Exposure 7 is a well used part of my plugin tool kit. As a Mac
user I have also been playing with Tonality Pro which also does a very
good job, just a little bit differently.

This was done with Exposure 7:
https://db.tt/rYHrYPzQ

--
Regards,

Savageduck

On 18 Nov 2015 22:42:11 GMT, Sandman wrote:

In article , nospam wrote:

put simply: you are wrong.

So "simply" that nospam can't even conjure up one single substantiation for his
claim that my support is incorrect.

Here are the facts:

FACT #1: Exposure is the amount of light per unit area

Not exactly ...

FACT #2: A smaller sensor has less "unit area" than a larger sensor

FACT #3: Thus, a smaller sensor, exposed identically, receives less light

FACT #4: Brightness in an image is the direct result of the about of light
received by the sensor

No. It's the direct result of the amount of light received by the
photosites.

FACT #5: In order to create an equally bright image as a larger sensor, the
smaller sensor needs to amplify its signal more

Not if it is more sensitive i.e. has a higher quantum efficiency.

FACT #6: Thus, the "ISO" chosen is NOT the sensors sensitivity, it's the expected
brightness level.

This conclusion fails as a result of the errors in your prior
assumptions.

Support for these facts:

SUPPORT #1: http://jonaseklundh.se/files/same_iso.png When equally exposed at
the same ISO level, the smaller the sensor is, the more noise we observe. This is
the result of the aforementioned facts.

Not true.

SUPPORT #2: http://jonaseklundh.se/files/iso_adjusted.png When ISO is adjusted
by the crop factor squared, and the exposure is changed to give the smaller
sensors the same amount of total light, the end result is as identical of an
image you could get with different sensor technologies.

Now - I know you will probably snip this away and ignore it, but if you don't,
you have to falsify most or all of the aforementioned facts, and also falsify the
support for them to have anything but hot air to provide.

Most of them have already been falsified.

I'm waiting (not really).
--

Regards,

Eric Stevens

On 11/18/2015 05:42 PM, Sandman wrote:
In article , nospam wrote:

put simply: you are wrong.

So "simply" that nospam can't even conjure up one single substantiation for his
claim that my support is incorrect.

Here are the facts:

FACT #1: Exposure is the amount of light per unit area

FACT #2: A smaller sensor has less "unit area" than a larger sensor

FACT #3: Thus, a smaller sensor, exposed identically, receives less light

Exposure is the amount of light that strikes each individual grain on
the film (or photosite of the sensor). If the grains or photosites are
more densely packed, the quality of the resulting image will be better
(less grainy or noisy), but the exposure will be the same.

If I put ISO 200 film in my 35mm camera and in my 6x7cm camera and shoot
the same scene with both cameras at the same shutter speed and f/stop,
the exposure characteristics will be the same. The 6x7cm neg will yield
a better quality (less grainy) image, but the exposure will be the same.


FACT #4: Brightness in an image is the direct result of the about of light
received by the sensor

Brightness in an image is the direct result of the amount of light
received by each photosite of the sensor or grain of the film.


FACT #5: In order to create an equally bright image as a larger sensor, the
smaller sensor needs to amplify its signal more

Assuming the small sensor and the large sensor are the same technology:
that is, the individual photosites on each sensor have the same
sensitivity to light, no. However, the image resolution from the smaller
sensor will likely be worse than the larger sensor.

FACT #6: Thus, the "ISO" chosen is NOT the sensors sensitivity, it's the expected
brightness level.

There are several ISO standards that can be used for digital cameras.
One of them is basically how does the digital image compare to a film
image with a given ISO.

Support for these facts:

SUPPORT #1: http://jonaseklundh.se/files/same_iso.png When equally exposed at
the same ISO level, the smaller the sensor is, the more noise we observe. This is
the result of the aforementioned facts.

SUPPORT #2: http://jonaseklundh.se/files/iso_adjusted.png When ISO is adjusted
by the crop factor squared, and the exposure is changed to give the smaller
sensors the same amount of total light, the end result is as identical of an
image you could get with different sensor technologies.

Now - I know you will probably snip this away and ignore it, but if you don't,
you have to falsify most or all of the aforementioned facts, and also falsify the
support for them to have anything but hot air to provide.

I'm waiting (not really).



--
Ken Hart

On 18 Nov 2015 23:21:29 GMT, Sandman wrote:

In article , Eric Stevens wrote:

Sandman:
Yes, I am in reference to signal amplification when saving the RAW
file to the memory card. You have a charge in the sensor, which
when read is amplified to match the camera-set ISO. Unless it's
an ISO-less camera, but that's another topic.

When you say 'sensor' I take it that you mean that rectangular
object behind the lens. If you do, this is where we can start homing
in on the error. You say "You have a charge in the sensor, which
when read is amplified to match the camera-set ISO". This is not
correct. It is the charge in each individual photosite which is
amplified by a predetermined amount corresponding to the camera-set
ISO. If the photosite is empty the charge is zero and the amplified
voltage is also zero. That photosite is black. If another photosite
is full, the charge is amplified to the voltage level corresponding
with white. The size of the sensor has nothing to do with the amount
of amplification required by the individual photosites.

I agree with all of that, but it is also irrelevant to my point. This is also
why all my examples have been with an evenly distributed exposure, not an image
of a cat, but of a flat white wall for instance. Or even mathematically equal.

Sandman:
And here's where you go wrong. But you're on the verge of getting
it.

Even talking about photo sites is a bit misleading, really, given
the multitude of different resolutions we have out there. But for
the sake of simplicity, let's assume we have a 10MP FF sensor and
a 10MP MFT sensor (again, resolution/pixel density isn't
important, but lets keep it simple).

So, a given exposure is amount of light over a given unit area,
right? So a 1/250 f2.8 exposure gives 20,000,000 photons over a
square cm as an example.

Now, a FF sensor is 3.5 x 2.4 cm large, so it has a total of 8.4
square cm's. This means that the FF sensor will receive a total
of 168,000,000 photons.

The MFT sensor is 1.8 x 1.35 cm large, so it has a total of 2.43
square cm's, which is four times smaller than the FF sensor (not
exactly, it's a different aspect ratio, but close enough for our
example) and will thus receive a total of 48,600,000 photons,
right?

So, again - both are 10MP, so we divide those received photons by
the resolution:

FF: 168,000,000 / 10,000,000 = 16.8 photons per photo site
MFT: 48,600,000 / 10,0000,000 = 4.86 photons per photo site.

Again, four times less photons per photo site

But you are not comparing apples with apples. You think you are
comparing large sensors with small sensors but in fact you are
really comparing large photosites with small photo sites.

Again, the size of the photo sites doesn't matter. The math is the same even if
the photo sites were identical in size, but just fewer of them in the smaller
sensor.

Forced?

The only forcing came from the arithmetic.


FF: 8.4 cm^2/10Mp = 8.4E^-5mm^2 =.0092mm per side square.
MFT: 2.43cm^2/10Mp = 2.43E^-6mm^2 =.00156mm per side square

The diameter of your FF photosite is 5.88 times that of the MFT.

If the photosites remain of constant size then we have:

FF: 8.4cm^2 10Mp =16.8 photons per site
MFT: 2.43cm^2 2.9Mp =16.8 photons per site

Correct, and if MFT sensors always were one fourth the resolution of FF
sensors, we wouldn't have this discussion.

Unfortunately, they aren't, they are about the same resolution, which is why I
used that in my math above, and then we have the observable problem.

The amplification required is the same in both cases.

Your general idea is correct but you have expressed it entirely
incorrectly.

Which is odd of you to say, you seem to agree with it 100%. You were forced to
cut the resolution of the MFT sensor to a fourth to get comparable data.

It's not the change in size of sensor that requires an
adjustment. It's the change in size of photosite which requires an
adjustment.

Which in the real world is what happens when you change the size of the sensor.

Half the problem is that you do not write clearly. When you say "you
change the size of the sensor" you don't actually mean that. You mean
"change the size of the sensor AND the photosites". What you really
mean is changing the size of the photosites which matters. Changing
the size of the sensor is irrelevant.

If you think about it, you have highlighted the problem
that manufatcturers have encountered in maintaining sensitivity
while reducing the size of photosites (i.e. higher Mp).

Which, incidentally, is what I've been talking about this entire time

Nonsense.

When on 10 Nov 2015 you wrote in Message-ID:
you wrote:

"What are the cameras? If the sensor size is different, then ISO as
a value is out of the window and can't be compared. You have to
apply the crop factor (squared) to get the same amount of light
into a smaller sensor."

You said absolutely nothing about photosites or which could be
inferred as referring to the dimensions of photosites. As far as you
were concerned it was only sensor dimensions which mattered. What is
more, at the early stages of the discussion, you were entirely obtuse
to the efforts of the many others to point out to you that you were
wrong in both your claims and your arguments. You may have had an
insight since but I expect that few people will believe you when you
now claim that the change in the dimensions of photosites is what you
have been talking about this entire time.
--

Regards,

Eric Stevens