Finally got to the point where no new camera holds my interest(waiting for specific offering)

In article , Eric Stevens
wrote:

yes there is. sensors are linear devices, therefore a 14 bit a/d has a
theoretical maximum of 14 stops dynamic range, with real world less
than that.

Sensors are linear devices as you say, and it is not necessary that
their output can be scaled to any range or form of digtal output.

since it's linear, 14 stops is the theoretical maximum. period.

Why?

physics and math.

Oh, I see.

no, you really don't.

You are assuming 1 stop per bit. It doesn't have to work
that way.

actually, it does.

Nikon cameras commonly use 12 bits but their top-end cameras can also
use 14. Changing from 12 to 14 does not in any way alter the dynamic
range of which the sensor is capable.

the adc doesn't change the physical properties of the sensor, but it
*does* change the dynamic range of the output.

for a 12 bit adc, that is 12 stops, which is why the better cameras use
a 14 bit adc or even 16 bit, to remove that limitation.

All it does is subdivide the
range into smaller intervals. Exactly the same thing is done with
monitors, most of which use 8 bit data with the better monitors using
10 bits.

false all around. ****ty displays are 8 bpc (some 6 even bpc), but
anything that would be used for photography is almost certainly 10 bpc
and possibly 12 bpc.

https://www.eizo.com/library/basics/maximum_display_colors/
An inexpensive LCD monitor will employ an LUT table with eight bits
per RGB color; an LCD monitor designed for color reproduction
applications will incorporate an LUT with more than eight bits (i.e.,
10 or 12 bits) per RGB color and employ internal calculations at 10
or more bits to map input signals to output signals.

So what camera employs an LCD monitor?

almost every camera has an lcd monitor.

some of the cheapos don't due to cost, and there's a couple that
pretend that not having one, how it used to be in the dark ages of
film, is somehow 'better', so they leave it off and try to claim it's a
feature.

In any case, if an LCD monitor can map out input signals to suit its
gamut why cannot a camera sensor map out its ouput data to suit its
dynamic range?

the sensor is a *linear* *device*. full stop.

it's possible that a camera can modify the signal from the sensor
*before* the adc, but there's no valid reason to do so and the nikon
cameras in question definitely do not do that.

sigma, however, does do that, but that's because they're intentionally
trying to deceive users into thinking their cameras are better than
they actually are. sigma cameras have a 12 bit adc, yet the 'raw' files
contain values well in excess of 4095, the maximum value a 12 bit adc
can output. that ain't raw. it's actually the output of a lookup table
due to the very complex transform from the overlapping spectra to rgb.

os x has always supported 16 bpc (64 bpp), with mac and iphone/ipad
displays 10 bpc or 30bpp and supporting a wide gamut.

note that both the internal display (retina) *and* external display
(non-retina) supports 30 bpc:
https://forums.macrumors.com/attachments/screen-shot-4-png.593804/

In any case the 14 bits of image data is further transformed by the
raw convertor and there is no reason why this cannot restore the full
dynamic range inherent in the sensor.

of course there is. you can't put back what isn't there.

But why have you lost it?

because of the adc.

if you have an image with 12 stops of dynamic range, you can't expand
it to 20:1 without artifacts.

You can map out a dynamic range of 20:1 with 14 bits if you want to.
You can do just as easily with 12 bits. It's merely that the image
brightness/density intervals are different in each case.

not with a linear device.

since you agree the sensor is linear, what you describe cannot happen.

any claim that exceeds the theoretical maximum, as dxo has done, is
simply *not* *possible*.

Have you considered that your understanding of what is being done is
not correct?

since it's exactly correct, there is no reason to consider that.

A slightly circular argument wouldn't you say?

nope.

you are wrong. nothing circular about it.

meanwhile, *your* understanding is completely wrong, so it's *you* who
should not only consider it, but take action and learn something.

Even if it is wrong (which I deny) you haven't explained why.

yes i have, and more than once.

here it is once again, in case you missed it the last several times:
since it's linear, 14 stops is the theoretical maximum. period.

and you agree that they're linear:
Sensors are linear devices as you say,

therefore 14 stops is the theoretical maximum for the nikon slrs under
discussion (plus several others that i didn't look up). the real world
maximum is less than that.

dxo claims several nikon cameras exceed 14 stops, something which is
*not* *possible*.

--- Sigma cameras snipped ---

of course you snipped it, because it further proves you wrong.

I don't wish to go from the general to the particular.

no surprise there, because as i said, it further proves you wrong.

You have made a valid comment but all it does is raise questions which
cannot be answered without knowing more about both Nikon's and DxO's
methodology. In other wortds you are currently going off half-cocked.

nope.

i've clearly shown that dxo is fully cocked, as are you.


There is nothing wrong with manufacturers to ace DxO's synthetic
tests. This can only make for a better lens, unless of course you
know
of a better system suitable for testing a wide range of lenses.

you can't be serious.

Are you trying to say that they could make a better lens but they
don't because that would give them a lower test score?

not even remotely close.

stop making up ****.

Well explain more clearly what you think is wrong with lens
manufacturers optimizing their lens so as to perform better when
evaluated by DxO. Does this result in an inferior lens?

possibly.

if the tests can be gamed, then the results are meaningless.

they will not represent how a lens or camera or anything else will
perform in real world situations.

Thats twice now you have said that. Please explain.

i did already, more than once (and more than twice for that matter).

samsung detected when a benchmark app was run and disabled cpu
throttling, resulting in misleadingly high scores.

That's cheating. Its not a consequence of the nature of the tests.

according to you, doing that would be 'optimizing' to 'perform better'
on a test.

--- Samsung and Trump snipped ---

if only that could be done without issue.

That's cheating. I remember when S3 did something like that to enable
them ace a popular graphics card test. But merely optimising something
for a particular test is not cheating.

yes it is.

I understand you to have university qualifications. Did you, when
confronted with an examination, study all and every topic known to man
or did you focus in very clearly only on those which were relevant to
the examination? If, for the purpose of the examination, you focused
only on topics relevant to the examination, were you cheating?

straw man.

A valid analogy.

no.

the question to ask is if a student obtained a *copy* of the test to be
given and then *memorizing* the answers, versus studying the actual
material and taking the text without knowing what will be asked.

ASs far as DxO is concerned the memorizing of tests amounts to
designing a lens which optimizes its DxO test results. I ask yet agai:
what's wrong with that? Does it result in an inferior lens?

it could.

what matters is real world use, not detecting and optimizing for a
specific benchmark or regulatory test to mislead people.

^^this^^

Please translate this for me. Are you saying that the DxO tests do not
properly reflect the real-world requirements of a lens?

very little of what dxo does has any bearing in reality.

Yet again; please explain. I know you won't so .... :-(

i already did. more than once.

On Fri, 28 Dec 2018 00:19:03 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

yes there is. sensors are linear devices, therefore a 14 bit a/d has a
theoretical maximum of 14 stops dynamic range, with real world less
than that.

Sensors are linear devices as you say, and it is not necessary that
their output can be scaled to any range or form of digtal output.

since it's linear, 14 stops is the theoretical maximum. period.

Why?

physics and math.

Oh, I see.

no, you really don't.

You are assuming 1 stop per bit. It doesn't have to work
that way.

actually, it does.

An assertion, not an explanation.

Nikon cameras commonly use 12 bits but their top-end cameras can also
use 14. Changing from 12 to 14 does not in any way alter the dynamic
range of which the sensor is capable.

the adc doesn't change the physical properties of the sensor, but it
*does* change the dynamic range of the output.

for a 12 bit adc, that is 12 stops, which is why the better cameras use
a 14 bit adc or even 16 bit, to remove that limitation.

All it does is subdivide the
range into smaller intervals. Exactly the same thing is done with
monitors, most of which use 8 bit data with the better monitors using
10 bits.

false all around. ****ty displays are 8 bpc (some 6 even bpc), but
anything that would be used for photography is almost certainly 10 bpc
and possibly 12 bpc.

https://www.eizo.com/library/basics/maximum_display_colors/
An inexpensive LCD monitor will employ an LUT table with eight bits
per RGB color; an LCD monitor designed for color reproduction
applications will incorporate an LUT with more than eight bits (i.e.,
10 or 12 bits) per RGB color and employ internal calculations at 10
or more bits to map input signals to output signals.

So what camera employs an LCD monitor?

almost every camera has an lcd monitor.

Made by Eizo or it's competitors?

some of the cheapos don't due to cost, and there's a couple that
pretend that not having one, how it used to be in the dark ages of
film, is somehow 'better', so they leave it off and try to claim it's a
feature.

In any case, if an LCD monitor can map out input signals to suit its
gamut why cannot a camera sensor map out its ouput data to suit its
dynamic range?

the sensor is a *linear* *device*. full stop.

Which can be linearly mapped to another scale as required.

it's possible that a camera can modify the signal from the sensor
*before* the adc, but there's no valid reason to do so and the nikon
cameras in question definitely do not do that.

How do you know?

sigma, however, does do that, but that's because they're intentionally
trying to deceive users into thinking their cameras are better than
they actually are. sigma cameras have a 12 bit adc, yet the 'raw' files
contain values well in excess of 4095, the maximum value a 12 bit adc
can output. that ain't raw. it's actually the output of a lookup table
due to the very complex transform from the overlapping spectra to rgb.

os x has always supported 16 bpc (64 bpp), with mac and iphone/ipad
displays 10 bpc or 30bpp and supporting a wide gamut.

note that both the internal display (retina) *and* external display
(non-retina) supports 30 bpc:
https://forums.macrumors.com/attachments/screen-shot-4-png.593804/

In any case the 14 bits of image data is further transformed by the
raw convertor and there is no reason why this cannot restore the full
dynamic range inherent in the sensor.

of course there is. you can't put back what isn't there.

But why have you lost it?

because of the adc.

if you have an image with 12 stops of dynamic range, you can't expand
it to 20:1 without artifacts.

Why not? Its a linear image.

You can map out a dynamic range of 20:1 with 14 bits if you want to.
You can do just as easily with 12 bits. It's merely that the image
brightness/density intervals are different in each case.

not with a linear device.

If you are going from linear analog to digital you end up with a
series of brightness steps.

since you agree the sensor is linear, what you describe cannot happen.

We need a blackboard.


--- vaster snip ---


There is nothing wrong with manufacturers to ace DxO's synthetic
tests. This can only make for a better lens, unless of course you
know
of a better system suitable for testing a wide range of lenses.

you can't be serious.

Are you trying to say that they could make a better lens but they
don't because that would give them a lower test score?

not even remotely close.

stop making up ****.

Well explain more clearly what you think is wrong with lens
manufacturers optimizing their lens so as to perform better when
evaluated by DxO. Does this result in an inferior lens?

possibly.

if the tests can be gamed, then the results are meaningless.

they will not represent how a lens or camera or anything else will
perform in real world situations.

Thats twice now you have said that. Please explain.

i did already, more than once (and more than twice for that matter).

You have not explained. You have merely stated.

samsung detected when a benchmark app was run and disabled cpu
throttling, resulting in misleadingly high scores.

That's cheating. Its not a consequence of the nature of the tests.

according to you, doing that would be 'optimizing' to 'perform better'
on a test.

Not so. Its not optimizing if it merely falls back to its usual
performance when not being tested.

--- Samsung and Trump snipped ---

if only that could be done without issue.

That's cheating. I remember when S3 did something like that to enable
them ace a popular graphics card test. But merely optimising something
for a particular test is not cheating.

yes it is.

I understand you to have university qualifications. Did you, when
confronted with an examination, study all and every topic known to man
or did you focus in very clearly only on those which were relevant to
the examination? If, for the purpose of the examination, you focused
only on topics relevant to the examination, were you cheating?

straw man.

A valid analogy.

no.

Designing a course of study to optimize your chances of getting good
exam results is not the in any way like optimizing a lens design to
get better test results? Come on.

the question to ask is if a student obtained a *copy* of the test to be
given and then *memorizing* the answers, versus studying the actual
material and taking the text without knowing what will be asked.

ASs far as DxO is concerned the memorizing of tests amounts to
designing a lens which optimizes its DxO test results. I ask yet agai:
what's wrong with that? Does it result in an inferior lens?

it could.

How? This is at the heart of the argument.

what matters is real world use, not detecting and optimizing for a
specific benchmark or regulatory test to mislead people.

^^this^^

Please translate this for me. Are you saying that the DxO tests do not
properly reflect the real-world requirements of a lens?

very little of what dxo does has any bearing in reality.

Yet again; please explain. I know you won't so .... :-(

i already did. more than once.

You have merely made the same claim every time. That is not an
explanation.
--

Regards,

Eric Stevens

On Sun, 30 Dec 2018 20:09:16 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

You are assuming 1 stop per bit. It doesn't have to work
that way.

actually, it does.

An assertion, not an explanation.

it's a statement of fact.

if you don't understand the basics, then the rest is a waste of time.


See
https://digital-photography-school.c...right-for-you/
"12-bit RAW lossy compressed – This format stores 4,096 tonal values
for each color"

.... and "12-bit uncompressed – Also stores 4,096 tonal values for each
color, but does not throw out any data to shrink the file size."

i.e 2^12 = 4096.

Also:

"14-bit uncompressed – The best option most cameras offer ... stores
16,384 tonal values for each color per pixel and does not throw any
away, giving you the highest possible amount of information, to
work with in post-production."

.... and 2^14 16,384.

Note that the article discusses 'tonal values'. The range of tonal
values are whatever the camera is capable of producing. ie. the
camera's dynamic range. There is absolutely no reason why the dynamic
range should be any particular range of light values. It's whatever
the camera is capable of producing. In the D850 (or whatever it was
you were citing) the dynamic range was measured as being 14.3 stops.
Again, there is absolutely no reason why a dynamic range of 14.3 stops
should not be mapped by a 14 bit data stream. It can also be done with
12 bits with the only difference being that that the steps in
gradation are 4 times the size of the data stream coded with 14 bits.

--- everything else snipped ---
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

You are assuming 1 stop per bit. It doesn't have to work
that way.

actually, it does.

An assertion, not an explanation.

it's a statement of fact.

if you don't understand the basics, then the rest is a waste of time.


See

https://digital-photography-school.c...hich-is-right-
for-you/
"12-bit RAW lossy compressed * This format stores 4,096 tonal values
for each color"

... and "12-bit uncompressed * Also stores 4,096 tonal values for each
color, but does not throw out any data to shrink the file size."

i.e 2^12 = 4096.

Also:

"14-bit uncompressed * The best option most cameras offer ... stores
16,384 tonal values for each color per pixel and does not throw any
away, giving you the highest possible amount of information, to
work with in post-production."

... and 2^14 16,384.

Note that the article discusses 'tonal values'. The range of tonal
values are whatever the camera is capable of producing. ie. the
camera's dynamic range.

in other words, 14 bit can resolve a wider dynamic range than 12 bit,
just like i said.

too bad you don't understand any of it, since you seem to think it
contradicts me, when it actually confirms it.

There is absolutely no reason why the dynamic
range should be any particular range of light values. It's whatever
the camera is capable of producing. In the D850 (or whatever it was
you were citing) the dynamic range was measured as being 14.3 stops.
Again, there is absolutely no reason why a dynamic range of 14.3 stops
should not be mapped by a 14 bit data stream. It can also be done with
12 bits with the only difference being that that the steps in
gradation are 4 times the size of the data stream coded with 14 bits.

no.

--- everything else snipped ---

only because you don't understand it.

On Mon, 31 Dec 2018 19:17:27 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

You are assuming 1 stop per bit. It doesn't have to work
that way.

actually, it does.

An assertion, not an explanation.

it's a statement of fact.

if you don't understand the basics, then the rest is a waste of time.


See

https://digital-photography-school.c...hich-is-right-
for-you/
"12-bit RAW lossy compressed Â* This format stores 4,096 tonal values
for each color"

... and "12-bit uncompressed Â* Also stores 4,096 tonal values for each
color, but does not throw out any data to shrink the file size."

i.e 2^12 = 4096.

Also:

"14-bit uncompressed Â* The best option most cameras offer ... stores
16,384 tonal values for each color per pixel and does not throw any
away, giving you the highest possible amount of information, to
work with in post-production."

... and 2^14 16,384.

Note that the article discusses 'tonal values'. The range of tonal
values are whatever the camera is capable of producing. ie. the
camera's dynamic range.

in other words, 14 bit can resolve a wider dynamic range than 12 bit,
just like i said.

Not at all. You are continuing to not understand.

14 bits can be used to resolve a dynamic range of 5 stops. Or it can
be used to resolve a range of 37 stops. The difference between the two
cases is that the size of the steps will be smaller in the case of the
5 stop example.

too bad you don't understand any of it, since you seem to think it
contradicts me, when it actually confirms it.

There is absolutely no reason why the dynamic
range should be any particular range of light values. It's whatever
the camera is capable of producing. In the D850 (or whatever it was
you were citing) the dynamic range was measured as being 14.3 stops.
Again, there is absolutely no reason why a dynamic range of 14.3 stops
should not be mapped by a 14 bit data stream. It can also be done with
12 bits with the only difference being that that the steps in
gradation are 4 times the size of the data stream coded with 14 bits.

no.

--- everything else snipped ---

only because you don't understand it.

One of us doesn't.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

https://digital-photography-school.c...aw-which-is-ri
ght-
for-you/
"12-bit RAW lossy compressed - This format stores 4,096 tonal values
for each color"

... and "12-bit uncompressed - Also stores 4,096 tonal values for each
color, but does not throw out any data to shrink the file size."

i.e 2^12 = 4096.

Also:

"14-bit uncompressed - The best option most cameras offer ... stores
16,384 tonal values for each color per pixel and does not throw any
away, giving you the highest possible amount of information, to
work with in post-production."

... and 2^14 16,384.

Note that the article discusses 'tonal values'. The range of tonal
values are whatever the camera is capable of producing. ie. the
camera's dynamic range.

in other words, 14 bit can resolve a wider dynamic range than 12 bit,
just like i said.

Not at all. You are continuing to not understand.

nope. i understand it *very* well. you do not.

14 bits can be used to resolve a dynamic range of 5 stops. Or it can
be used to resolve a range of 37 stops. The difference between the two
cases is that the size of the steps will be smaller in the case of the
5 stop example.

nope. it's *not* *possible* to resolve 37 stops with a 14 bit adc.

5 stops is obviously possible, since 5 stops is much less than the
theoretical maximum of 14 stops. you'd also have 9 stops of latitude,
so you could be super-sloppy in exposure.

however, if your scene had 37 stops of dynamic range, then you'd either
clip the highlights and/or block up the shadows, likely both.




--- everything else snipped ---

only because you don't understand it.

One of us doesn't.

that statement is true.

nospam wrote:
Eric wrote:
14 bits can be used to resolve a dynamic range of 5 stops. Or it can
be used to resolve a range of 37 stops. The difference between the two
cases is that the size of the steps will be smaller in the case of the
5 stop example.

nope. it's *not* *possible* to resolve 37 stops with a 14 bit adc.

How about an IRL wager to resolve this disagreement? I can offer Welch-proof terms.

Given all the history, nospam will be required to deposit first.

-hh

On Jan 1, 2019, hh wrote
(in ):

nospam wrote:
Eric wrote:
14 bits can be used to resolve a dynamic range of 5 stops. Or it can
be used to resolve a range of 37 stops. The difference between the two
cases is that the size of the steps will be smaller in the case of the
5 stop example.

nope. it's *not* *possible* to resolve 37 stops with a 14 bit adc.

How about an IRL wager to resolve this disagreement? I can offer Welch-proof
terms.

Given all the history, nospam will be required to deposit first.

-hh

Good luck with that.

--
Regards,
Savageduck

On Tue, 01 Jan 2019 18:40:36 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

https://digital-photography-school.c...aw-which-is-ri
ght-
for-you/
"12-bit RAW lossy compressed - This format stores 4,096 tonal values
for each color"

... and "12-bit uncompressed - Also stores 4,096 tonal values for each
color, but does not throw out any data to shrink the file size."

i.e 2^12 = 4096.

Also:

"14-bit uncompressed - The best option most cameras offer ... stores
16,384 tonal values for each color per pixel and does not throw any
away, giving you the highest possible amount of information, to
work with in post-production."

... and 2^14 16,384.

Note that the article discusses 'tonal values'. The range of tonal
values are whatever the camera is capable of producing. ie. the
camera's dynamic range.

in other words, 14 bit can resolve a wider dynamic range than 12 bit,
just like i said.

Not at all. You are continuing to not understand.

nope. i understand it *very* well. you do not.

14 bits can be used to resolve a dynamic range of 5 stops. Or it can
be used to resolve a range of 37 stops. The difference between the two
cases is that the size of the steps will be smaller in the case of the
5 stop example.

nope. it's *not* *possible* to resolve 37 stops with a 14 bit adc.

5 stops is obviously possible, since 5 stops is much less than the
theoretical maximum of 14 stops. you'd also have 9 stops of latitude,
so you could be super-sloppy in exposure.

however, if your scene had 37 stops of dynamic range, then you'd either
clip the highlights and/or block up the shadows, likely both.

You are obviously wedded to 1 stop per bit. Why is that?

Why for example can you not have 2 stops per bit, or pi stops per bit?
As long as you scale the entire brightness range with the available 14
stops.




--- everything else snipped ---

only because you don't understand it.

One of us doesn't.

that statement is true.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

You are obviously wedded to 1 stop per bit. Why is that?

math.

Why for example can you not have 2 stops per bit, or pi stops per bit?
As long as you scale the entire brightness range with the available 14
stops.

because it doesn't work that way.

think about what a stop means.