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

On Tue, 15 Jan 2019 01:17:19 -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.

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.

Your statement is too sweeping. It matters with a 10" lens normally
used for used for portraits and it matters more for a 105mm macro
lense normally used for dental photography.

but you do it at
your peril in macro photography.

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

TTL avoids the problem but if you use an exposure meter you have to
know the effective f-number. The nominal f-number is of no use to you.
--

Regards,

Eric Stevens

On Tue, 15 Jan 2019 01:17:18 -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.

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.

You don't mean i'm evasive. You mean I'm ignorant.

In fact I asked you as hopefully you won't argue about your own
definition.

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.

It's unhelpful in that it is meaningless in the context of the
question.

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.

Mere denial is not helpful.
--

Regards,

Eric Stevens

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!

there is nothing logical about the above exchange, which was the point.

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.

you are indeed equipped, since all it takes is a computer, or a
typewriter, or a pen.



it's definitely not scaling.

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

majority vote

Evasion

whoosh.




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?

that it hasn't been done before.

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.

oh, i understand what you've been saying, which is why i've
consistently said it's wrong.

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.

no.

In article , Eric Stevens
wrote:


(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.

Your statement is too sweeping. It matters with a 10" lens normally
used for used for portraits and it matters more for a 105mm macro
lense normally used for dental photography.

nope, and dentists don't use 105mm macro lenses anymore.

but you do it at
your peril in macro photography.

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

TTL avoids the problem

there's no need for ttl metering.

all that matters is that the meter knows what the lens is set to.

but if you use an exposure meter you have to
know the effective f-number. The nominal f-number is of no use to you.

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.

You don't mean i'm evasive. You mean I'm ignorant.

if you prefer, but avoiding answering the question is normally called
evasion.

In fact I asked you as hopefully you won't argue about your own
definition.

it's not my definition.

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.

It's unhelpful in that it is meaningless in the context of the
question.

it's not at all meaningless.

they are not the same. very simple.

it doesn't get any clearer than that.

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.

Mere denial is not helpful.

then why do you do it?

On 2019-01-15 18:29, Eric Stevens wrote:
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'.

No. A sample is a sample. It contains signal and _sampling_ source
noise (the noise of the thing being sampled and the noise of the ADC),
but not quantization noise.

Quantization noise is an artifact of data manipulation (such as scaling)
and is completely unavoidable when doing such operations.

--
"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 2019-01-15 18:46, Eric Stevens wrote:
On Tue, 15 Jan 2019 01:17:18 -0500, nospam

your repetition is what's evasive an unhelpful.

Mere denial is not helpful.

Yet that's all you do by repeating the same erroneous facts and conclusions.

--
"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 Tue, 25 Dec 2018 02:47:01 -0500, nospam
wrote:

--- snip ---

also, their 'tests' claim what is physically impossible, making them
untrustworthy and their scale is whatever they want it to be, with
newer cameras scoring higher and higher.

Suspicion and innuendo. That's not evidence.

the evidence is quite clear that they cannot be trusted *at* *all*.

The evidence appears to be that you can/will not produce any evidence
to support that statement.

false, there is extensive evidence that dxo is a sham, but as usual,
you refuse to acknowledge that and just want to argue, now having
resorted to ad hominem attacks, as usual.

their 'tests' claim what is physically not possible. that alone makes
them a scam

Example? - assuming of course that you are able to cite one.

of course i'm able. i do not make false claims. period.

--- Claim by nospam: ----

dxo measured 14.8 stops of dynamic range on the nikon d810 and d850,
which is *higher* than the theoretical maximum of 14 stops (14 bit a/d)
and in the real world, it won't actually get 14 stops.

I challenged this comment which lead to an enormous thread which
fanned out in all directions. My contention was (and is) that the
number of bits that are used to code an image have nothing to do with
the dynamic range of the sensor. You can code it with as many bits as
you like with factors other than dynamic range determining the choice.
In particular there is no reason why a sensor should not have a
dynamic range wider than implied by the number of bits with which it's
output is encoded.

At one stage Ron C suggested I should explain my views with diagrams.
With some reluctance I have been getting round to doing this.
Preparing the diagrams so as to be able to deal with nospams of tghis
world is no mean task. In the course of my background research I found
the following thread from dpreview dated Mar 25, 2012.
https://www.dpreview.com/forums/thread/3170233

If you read that you will see that the maximum number of bits with
which it is worthwhile coding an image is determined by (a) read noise
and (b) pixel well size. DR vs number of bits is touched upon but no
one has suggested that sensor DR is limited by bit number.

I also came across http://www.onmyphd.com/?p=analog.digital.converter
which is a good summary of the basics of analog to digital convertors
(ADC). Under the heading "What is an analog-digital convertor" the
author has written:

"An Analog-Digital Converter (ADC) is a widely used electronic
component that converts an analog electric signal (usually a
voltage) into a digital representation. The ADCs are at the
front-end of any digital circuit that needs to process signals
coming from the exterior world.

..... ---- complex text snipped ----

Therefore, more bits leads to more precision in the digital
representation. Here we simplify the range to be between 0 and
Vref, although the range may be between any two values."

That last has always been my point. What is being encoded is a signal
between gthe upper and lower limits to the sensitivity of the sensor.
Their mathematical relationship (as in their ratio = dynamic range)
does not come into the question. it is only their difference which
matters.

I'm not going to write any more about the original argument. nospam
can go and find another playmate. I may be tempted to come back in if
a sensible and rational discussion of a new aspect gets under way.
Other than tghat I've finished with this thread.


--

Regards,

Eric Stevens