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

On Sun, 13 Jan 2019 07:48:35 -0800, Savageduck
wrote:

On Jan 13, 2019, Alan Browne wrote
(in ):

On 2019-01-12 22:39, Eric Stevens wrote:

After 313 posts to this mind numbing thread without any end in sight, can you
guys just pick up your cameras, and shoot something, anything?

I would love to but my camera is with the Nikon Repair Centre in
Australia being modified to deal with the two recalls to which it is
subject.

I may yet have to get around to settling this matter by producing some
diagrams. nospam is fundamentally wrong and so too are his conclusions
about DxO's DR tests. Unfortunately others are following his lead.
--

Regards,

Eric Stevens

In article , Eric Stevens
wrote:

After 313 posts to this mind numbing thread without any end in sight, can
you
guys just pick up your cameras, and shoot something, anything?

I would love to but my camera is with the Nikon Repair Centre in
Australia being modified to deal with the two recalls to which it is
subject.

only one camera?

use the camera in your phone.

I may yet have to get around to settling this matter by producing some
diagrams. nospam is fundamentally wrong and so too are his conclusions
about DxO's DR tests.

if anyone is fundamentally wrong, it's you. just about everything
you've said so far has been wrong.

Unfortunately others are following his lead.

now ask yourself why that is, when they usually bash me to hell.

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.


--
"2/3 of Donald Trump's wives were immigrants. Proof that we
need immigrants to do jobs that most Americans wouldn't do."
- unknown protester

On Sun, 13 Jan 2019 17:04:35 -0500, nospam
wrote:

In article , Eric Stevens
wrote:


A meat grinder will alter a sheep. But before the grinder the sheep
will remain a sheep and after the grinder the sheep will be whatever
the grinder produces. If you are going to determine the height of a
sheep would you prefer to do it before or after the grinder?

it's entirely irrelevant, in every possible way.

it's also not correct. the sheep cease to be sheep well before it
reaches the meat grinder.

Oh no. In my example the sheep are thrown straight into the meat
grinder.

that's going to make for some very unpalatable meat, one reason of many
why your example is both irrelevant and absurd.

In any case: it's not irellevant.

it is, and absurd too.

You will argue that this is irellevant, but it's not.

it is. very much so.

It's an analogy.

a bad one. horrible and nonsensical.

The place to
measure the DR of a sensor is at the sensor and not at the output of
an ADC. DxO's method measures the DR at the sensor.

only for those designing cameras and sensors, and it doesn't.
for *everyone* else (including you, since you don't do either of
those), the place to measure dynamic range are from *the* *images*,
which means after the adc + isp + anything else in the image path.

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.

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? The scaling occurs within the quantization.

it's definitely not scaling.

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

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.

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 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.

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.

Besides, there are some they may
not want to bring to the attention of the opposition.

oh, they know, and they'll reverse engineer it if they have to.

and if it really *is* the sensor they're measuring, then it should
be
the *same* for the *same* sensor, and it is not.

Not when you shove another piece of glass in front of one of the
sensors.

no effect on dynamic range.

So you keep saying. That doesn't make it true.

it's true because it is true.

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.

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?




But,
nevertheless, I would be surprised if 16 months of development did not
result in the slight improvement of the sensor DR range shown by the
D50.

except that your claim is that dxo measures the sensor, not the camera,
so if both have the same sensor, the dynamic range would be the same.

at least try to be consistent.

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. If you follow the links from that
article you will find:

"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.

"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.)

Feb 2004 to June 2005 is 16 months.

perhaps you use a different calendar than the rest of the world, but
where i come from, that's what is known as 'at the same time'.

In any case, arguing with you is leaving a foul taste in my mouth.
This is more than enough.

only because you've been eating meat ground from whole sheep. nobody
wants to eat wooly meat. blech.

In this case it's the taste of wooly arguments.
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 17:04:31 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

The physical truth of the matter is that deep down at the shreds lies
noise. Usually a lot more noise than signal.

Yep. An interesting point: according to DxO, when the test uses a
paper target, some of the noise may actually be the texture of the
paper of the target.

that would be a flaw in their testing.

Yep. Which is one of the reasons that DxO do it the way they do.

no.

But that say that it is! Do you still want to contradict them?
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 21:19:20 -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?
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 17:04:32 -0500, nospam
wrote:

In article , Eric Stevens
wrote:

Allright then. Please explain to your readers how you set a lens
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.

math is fun.

Especially when it is disconnected from reality.

then you should enjoy it more than most.


--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 17:04:30 -0500, nospam
wrote:

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. 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. As it says above the difference is typically ignored in
photography (with today's short focal length lenses) but you do it at
your peril in macro photography.
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 17:04:33 -0500, nospam
wrote:

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.

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

stop and f/stop are not the same.

Evasive and unhelpful response.

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.

In other words we must take language seriously.
Imprecise language discloses the lack of precision of thought."

yep.
--

Regards,

Eric Stevens

On Sun, 13 Jan 2019 17:04:36 -0500, nospam
wrote:

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.
--

Regards,

Eric Stevens