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

In article , Eric Stevens
wrote:


I will give it one more try.

stop digging yourself a deeper hole.

All that the ADC is doing is digitizing the level of charge which has
accumulated in the sensor cells. It does not need to know, nor would
it care, how much exposure to light that represents. Hook it up to a
sensor and a 14-bit ADC will give you a 14-bit datastream which will
tell you all about the charge and nothing about the anount of light
which has fallen on the individual cells. As far as the ADC is
concerned, the important thing is whether or not the maximum voltage
of the charges to which it is exposed is within the ADC's input range.
If it is the ADC will digitize it. How much light it required does not
matter. The ADC will as happily digitize 2 levels of charge as it will
20,000 levels of charge. No matter what it digitizes it will output a
14-bit data stream. THat's why it won't care whether it is digitizing
a collection of charges with a dynamic range of 5 or 50. All will be
OK as long as the peak is within the range of the ADC's input.

still not getting it.

And you never explain why.

false.

In article , Eric Stevens
wrote:

You are still missing the point: lens aperture, shutter speeds or ISOs
are not identical to stops.

At this point my old physics professor would ask for dimensional analysis.
What's the dimensional analysis of "EV" vs "stop" ?

First define a stop.

use the standard definition.

Evasive and unhelpful response.

it's not in any way evasive nor is it unhelpful.

it's *you* who is evasive, since it's clear you don't know what the
definition even is.

If you mean f-number, a stop is dimensionless.

stop and f/stop are not the same.

Evasive and unhelpful response.

again, it's not in any way evasive nor is it unhelpful.

they are not the same. very simple.

If you mean sqrt(2) a stop is dimensionless.

If you mean 1/sqrt(2) a stop is still dimensionless.

Basically EV = Constant, or

(Shutter Speed) x f-number = another constant.
The dimensions of (Shutter Speed) are 1/T, f-number is dimensionless,
so the dimensions of EV should be 1/T.

nope.

Evasive and unhelpful response.

your repetition is what's evasive an unhelpful.

In article , Eric Stevens
wrote:

The fstop is [Image distance]/lens aperture.

also wrong. f/stop = focal length/aperture.

Aha! Your correction of me is an approximation.

it's not in any way an approximation.

See
https://en.wikipedia.org/wiki/F-number

from that link,
The f-number of an optical system (such as a camera lens) is the
ratio of the system's focal length to the diameter of the entrance
pupil.

exactly what i said it is.

"... as one focuses closer, the lens' effective aperture becomes
smaller, making the exposure darker. The working f-number is often
described in photography as the f-number corrected for lens
extensions by a bellows factor. This is of particular importance
in
macro photography".

selective snipping. that's *not* cool.

Do you think I should not have snipped the several thousand irrelevant
words, tables and images between the two parts I quoted? At least I
marked that I snipped.

your snipping altered its meaning.

The several thousand words, tables and images between the two parts I
quoted contained a great deal of irrelevant meaning. That's why I
snipped them.

nope.

*one* paragraph prior to what you quoted, in the *same* section, states
that the limitation is *ignored*.

that's not even 100 words, let alone 'several thousand', without any
tables or images in between either.

https://en.wikipedia.org/wiki/F-number#Working_f-number
The f-number accurately describes the light-gathering ability of a
lens only for objects an infinite distance away.[16] This limitation
is typically ignored in photography, where f-number is often used
regardless of the distance to the object.

since you somehow missed the first few times, note this part:
This limitation is typically ignored in photography, where f-number
is often used regardless of the distance to the object.

you also mistakenly claimed that f/stop is an approximation. it is not,
which *your* link confirms, earlier in the link:
The f-number of an optical system (such as a camera lens) is the
ratio of the system's focal length to the diameter of the entrance
pupil.

it's no wonder you snipped so much. it all proves you wrong.

The focal length of a lens is the image distance when focussed at
infinity. This is the conventional basis for specifying the broad
properties of a lens.

correct.

However one of the original purposes for an
f-number was to assist with the calculation of exposure. Large cameras
taking closeup photographs need the f-number to be determined for
exposure and that was the (usually fixed) stop diameter divided by the
focal length. So there is the nominal f-number with which we are
familiar and the effective (i.e. working) f-number which depends on
the setup.

large cameras aren't the issue.

As it says above the difference is typically ignored in
photography

in other words, you agree with what i originally said.

(with today's short focal length lenses)

not just short focal length, but all lenses at normal working distances
because the difference is too small to matter, below the tolerance of
the various components.

but you do it at
your peril in macro photography.

there is no peril. there's this thing called an exposure meter...

In article , Eric Stevens
wrote:

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

a) f/3
b) 1 stop
c) 1 EV
d) a and b.
e) b and c.

C'mon _E_ric!
I'm rooting for you!

Try (8*f - 11*f)/88 or (-3*f)/88

The correct reply was e).

But thanks for playing.

Seriously this time: b) is always correct. c) is also correct if the
lens is mounted in a camera with (presumably) a shutter speed and ISO
set. But, without a shutter speed and ISO there can be no EV.

false.

What then do you require to establish an EV?

knowledge.

In article , Eric Stevens
wrote:

Allright then. Please explain to your readers how you set a
to an EV of 20.

For what ISO and speed?

No, no. No ISO or speed. The lens calibration is equivalent to stop
settings according to nospam so it must be possible to set a lens
to a particular EV. I picked 20 as an example.

by picking 20 (or any number), you demonstrate you don't understand
it.

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

You are fortunate. I happen to have a lens on my desk. I've just
measured the difference is about 4mm.

good work!

since we now know that 1ev = 4mm, it's a very simple calculation to set
a lens to your desired ev 20:
simply zoom the lens to 80mm, or alternately, choose a fixed focal
length 80mm lens.

Oh no. You have it wrong. It's 4mm netween the f/8 mark and the f/11
mark on the lens ring. The diameter of the lens ring is 60mm which
gives it a circumference of 188mm. using your calculatios, 8omm is 152
degrees of rotation of the ring. I've just tried that but the ring
won't rotate that much so I have concluded that my 105mm Micro Nikkor
will not do an EV of 20.

so you're saying your initial measurements are wrong.

Boy! You can't follow a logical argument, even when it is crazy.

there's nothing to follow nor was it in any way logical.

In article , Eric Stevens
wrote:


Well DxO have explained what they are doing (quite clearly I would
have thought) and they are measuring it at the sensor. You should
write to them and explain they have got it wrong. No ordinary
photographer is interested in the DR of the sensor.

exactly. photographers are interested in the dynamic range of the
camera, which is one reason why what dxo is doing is bogus.

Well, go on. Write and explain it to them.

you're the one who said it, so it's on you.

The sensor will always do do
whatever it can do and it's DR can be scaled up or down to fit the
output bandwidth of the ADC. It is because of the scaling that you can
have the output of a sensor with a 14.8 stop DR scaled down to to
suit a 14 bit ADC. It's not a big deal.

except that it's *not* scaled.

Of course it is. In any sensible design the DR of the input to the ADC
will be designed to accept the DR of the sensor's output.

actually, the adc will be designed to match that of the sensor's
capabilities.

I'm glad we have agreed on that. So now we go from transforming the
number of electrons in each light well into a binary representation
using 14 bit. I have called this process scaling. I would like to know
what you call it.

what any ee would call it, quantization.

Are you an ee?

yes

The scaling occurs within the quantization.

no

it's definitely not scaling.

How do they decide the number of photons each additional bit in the
ADC?

majority vote

if this was not the case, the 'fiddling' would be well known since the
camera would perform differently than previous cameras, ...

But they do, they do

no they don't. they perform as expected.

Better and better with each new model in a series. Are you claiming
this is not behaving differently?

except that it's consistent with the normal progress of technology,
nothing out of the ordinary.

on the other hand, if a new camera doubled the dynamic range of its
predecessor, that would be well outside the norm, suggesting there's
some fiddling going on.

It could be an improvement, especially it was Apple.

apple already did it with smart hdr, which takes multiple images in
rapid succession at varying exposures, then combined via their isp and
nsp.

which is of course, entirely different than fiddling with the data
between the sensor and adc.

there are no breakthroughs other than computationally, such as google
night sight, which has nothing to do with the sensor technology or adc.

So, the technology is settled. Make way for a bunch of unemployed
physicists and engineers.

and taxi cab drivers

... and it would
likely be marketed as a benefit (e.g., 'new hdr sensor'), and hotly
argued because the camera is 'not pure' or some such.

It is merely your assumption that every individual design advance
would be trumpeted by the marketing department.

of course they will. it's a competitive advantage over the other
products.

There are probably hundreds of improvements in each new design. You
can't expect them to extoll them all.

nobody said all of them, but they will very definitely hype the major
ones to set it apart from the competition.

The hype the resulting benefit "The new improved Gizmo4 processor",
but they rarely hype what it is that makes iy improved. They would be
silly to do so.

gizmo4 is so last year.

a dramatic increase in dynamic range due to some sort of preprocessing
would be that.

But it's not dramatic. It's been slowly creeping up.

that's the point.

a breakthrough gets noticed. slow creeping doesn't since it's expected.


No glass (or in this case the material of the AA filter) offers 100%
transmissability to the electromagnetic spectrum. You continuing to
argue otherwise is plain silly.

you're really grasping at straws now.

nobody said 100%.

You implied 100% when you denied that the AA filter would have any
affect on the light falling on the sensor.

you're seriously grasping at straws.

You are seriously evading the point.

not at all.

look out your window, both with the window closed (through the glass)
and with it open (no glass).

do you notice a difference in dynamic range, colour balance, amount of
detail, etc? no.

You mean you don't?

nope, but then again, my windows aren't filthy.

But 16 months later, do they both have the same sensor?

yes, as did many other cameras at the time.

also, the d50 and d70s came out at the same time, so any supposed
improvement would be in *both* cameras, yet there's a difference.

https://www.dxomark.com/Cameras/Nikon/D50
https://www.dxomark.com/Cameras/Nikon/D70s

The dates of release are listed in
https://en.wikipedia.org/wiki/Nikon#...ompact_cameras

¤ Nikon D50, April 20, 2005 - Discontinued
¤ Nikon D70S, April 20, 2005 - Discontinued

That's rather useless information.

nope. it confirms what i said, thereby making it very useful,
especially when it's provided by someone *else*.

If you follow the links from that

i did

article you will find:

lots of stuff

"The D50 is a now-discontinued 6.1-megapixel entry-level digital
single-lens reflex camera, sold from June 2005 until November 2006
by Nikon.

read a bit further in your link and you will find:
https://en.wikipedia.org/wiki/Nikon_D50#Market
The D50 was announced on April 20, 2005 and went on sale in June
2005

which matches what i said:
€ Nikon D50, April 20, 2005 - Discontinued

"The Nikon D70 is a digital single-lens reflex camera, introduced at
the 2004 PMA Annual Convention and Trade Show, as Nikon's first
consumer-level digital SLR" (The 2004 show was held from Feb 12
to Feb 15.)

to quote you:
That's rather useless information.


Feb 2004 to June 2005 is 16 months.

you *really* are confused.

read it again...

On Tue, 15 Jan 2019 01:17:21 -0500, nospam
wrote:

In article , Eric Stevens
wrote:


Well DxO have explained what they are doing (quite clearly I would
have thought) and they are measuring it at the sensor. You should
write to them and explain they have got it wrong. No ordinary
photographer is interested in the DR of the sensor.

exactly. photographers are interested in the dynamic range of the
camera, which is one reason why what dxo is doing is bogus.

Well, go on. Write and explain it to them.

you're the one who said it, so it's on you.

I'm not equipped to do that. It's contrary to my beliefs.

The sensor will always do do
whatever it can do and it's DR can be scaled up or down to fit the
output bandwidth of the ADC. It is because of the scaling that you can
have the output of a sensor with a 14.8 stop DR scaled down to to
suit a 14 bit ADC. It's not a big deal.

except that it's *not* scaled.

Of course it is. In any sensible design the DR of the input to the ADC
will be designed to accept the DR of the sensor's output.

actually, the adc will be designed to match that of the sensor's
capabilities.

I'm glad we have agreed on that. So now we go from transforming the
number of electrons in each light well into a binary representation
using 14 bit. I have called this process scaling. I would like to know
what you call it.

what any ee would call it, quantization.

Are you an ee?

yes

The scaling occurs within the quantization.

no

it's definitely not scaling.

How do they decide the number of photons each additional bit in the
ADC?

majority vote

Evasion

if this was not the case, the 'fiddling' would be well known since the
camera would perform differently than previous cameras, ...

But they do, they do

no they don't. they perform as expected.

Better and better with each new model in a series. Are you claiming
this is not behaving differently?

except that it's consistent with the normal progress of technology,
nothing out of the ordinary.

on the other hand, if a new camera doubled the dynamic range of its
predecessor, that would be well outside the norm, suggesting there's
some fiddling going on.

It could be an improvement, especially it was Apple.

apple already did it with smart hdr, which takes multiple images in
rapid succession at varying exposures, then combined via their isp and
nsp.

So what's new about that?

which is of course, entirely different than fiddling with the data
between the sensor and adc.

You have slipped! A mistake! You have really understood what I have
been saying.

--- crap snipped ---
--

Regards,

Eric Stevens

On Mon, 14 Jan 2019 12:34:42 -0500, Alan Browne
wrote:

On 2019-01-13 21:07, Eric Stevens wrote:
On Sun, 13 Jan 2019 10:06:13 -0500, Alan Browne
wrote:

On 2019-01-12 21:15, Eric Stevens wrote:
On Sat, 12 Jan 2019 09:25:23 -0500, Alan Browne
wrote:

On 2019-01-11 18:18, Eric Stevens wrote:

The recorded output of the ADC is limited by the capabilities of the
ADC. But these have no effect on the capabilities of the sensor. If

If there is no way to encode the information, then that is the mootest
of moot points.

That may well be but, as I have several times said, it is possible to
scale the dynamic range of the sensor to fit the narrower dynamic
range of the ADC.

To which I've replied numberous times. In a nutshell, you're trading
one form of noise for another.

The 'scaling' is done during the conversion of analog to digital in
the ADC and involves no more noise than is inherent in any analog to
digital conversion.

I've pointed out quantization noise to you several times. Ignore it.
You ignore everything else.

You always get quantization noise when you digitize. That's what I
meant by 'inherent'.
--

Regards,

Eric Stevens

On Tue, 15 Jan 2019 01:17:20 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

a) f/3
b) 1 stop
c) 1 EV
d) a and b.
e) b and c.

C'mon _E_ric!
I'm rooting for you!

Try (8*f - 11*f)/88 or (-3*f)/88

The correct reply was e).

But thanks for playing.

Seriously this time: b) is always correct. c) is also correct if the
lens is mounted in a camera with (presumably) a shutter speed and ISO
set. But, without a shutter speed and ISO there can be no EV.

false.

What then do you require to establish an EV?

knowledge.

Evasion.
--

Regards,

Eric Stevens

On Tue, 15 Jan 2019 01:17:20 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

Allright then. Please explain to your readers how you set a
to an EV of 20.

For what ISO and speed?

No, no. No ISO or speed. The lens calibration is equivalent to stop
settings according to nospam so it must be possible to set a lens
to a particular EV. I picked 20 as an example.

by picking 20 (or any number), you demonstrate you don't understand
it.

Now if I said, for example f/11 you would understand it. But that is
not an EV.

as i said, you don't understand it.

here's a hint:
what's the difference between f/8 and f/11 ?

You are fortunate. I happen to have a lens on my desk. I've just
measured the difference is about 4mm.

good work!

since we now know that 1ev = 4mm, it's a very simple calculation to set
a lens to your desired ev 20:
simply zoom the lens to 80mm, or alternately, choose a fixed focal
length 80mm lens.

Oh no. You have it wrong. It's 4mm netween the f/8 mark and the f/11
mark on the lens ring. The diameter of the lens ring is 60mm which
gives it a circumference of 188mm. using your calculatios, 8omm is 152
degrees of rotation of the ring. I've just tried that but the ring
won't rotate that much so I have concluded that my 105mm Micro Nikkor
will not do an EV of 20.

so you're saying your initial measurements are wrong.

Boy! You can't follow a logical argument, even when it is crazy.

there's nothing to follow nor was it in any way logical.

You can't even recognise a logical argument!
--

Regards,

Eric Stevens