On 29 Nov 2015 09:09:55 GMT, Sandman wrote:

This is a hell of a long thread in distinct chunks. About half way
along a realised what has really been going on and chopped things off
there. I have placed my conclusions at the end for anyone who has the
stamina to geth there.


In article , Eric Stevens wrote:

Eric Stevens:
"You have to apply the crop factor (squared) to get the same
amount of light into a smaller sensor."

The above statement is quite true if you really do want to get
the same amount of light into the smaller sensor but I don't
know why on earth you would want to do this.

Sandman:
To create an equal image. With not enough light, the smaller
sensor will have to amplify the signal more and this creates more
noise. It's just something that most people don't think about.

The amount of light entering the camera depends on the area of the
entrance pupil. If it's all focussed on the sensor the intensity of
the light on the sensor depends on the area of the sensor. If you
scale the camera up or down by a factor k the amount of light
entering the camera varies as k^2. The area of the sensor also
varies as k^2. In other words the area of the sensor varies as the
entering light varies.

Yes, but as you know, exposure is amount of light per unit area. If the
exposure is the same, the amount of light collected by the sensors will be
different. The smaller sensor needs more light.

The smaller sensor needs more light to do what?

http://www.josephjamesphotography.com/equivalence/

"The only factors in the exposure are the scene luminance, f-ratio,
shutter speed, and transmissivity of the lens (note that neither
sensor size nor ISO are factors in exposure)."

Which means that in order to create an equal *image*, you have to adjust the
*exposure* to fit the size of the sensor.

How on earth can you say that when you have just quoted "note that
neither sensor size nor ISO are factors in exposure." What do you
mean by "equal image"? Equal in what way?

Eric Stevens:
The truth of the matter is that if you take a camera and scale
it either up or down, the lens f/value stays the same and the
level of illumination on the sensor stays the same. If the
sensitivity of the sensor has not changed then the invariant
level of illumination of the sensor means that exposure time
always remains the same.

Sandman:
But with the same f-stop, the smaller lens gets less light. So you
either need a larger f-stop or longer exposure time to match the
amount of light. This means you adjust the ISO down to the crop
factor square.

Who cares about the sensor getting less light?

Those that care about noise, since less light means more amplification and more
noise.

Why should there be more noise? Providing everything has been kept in
proportion as the camera has been scaled up or down, the light
intensity on the sensor remains the same irrespective of the size of
the sensor. Lighting of the sensor remains te same. Sensor remains the
same ...

The point is that the amount of light falling on each square
millimetre (or each square inch) remains exactly the same. As far as
the sensor is concerned there has been no change.

Not sure what supposed "change" you are talking about here? Changing the size
of the sensor means that comparing "exposure" falls apart.

The behaviour of a small patch of sensor is not affected by how much
more sensor there is around it. Certainly the sensitivity is not
affected. As I almost said "the point is that the amount of light
falling on each square millimetre remains exactly the same. As far as
the sensor is concerned there has been no change." Certainly it needs
neither more nor less amplification than it did before".

http://www.josephjamesphotography.com/equivalence/#8

"For a given scene, perspective, and framing, the total light
depends only on the aperture diameter and shutter speed (as opposed
to the f-ratio and shutter speed for exposure).

That's fine, if you are talking about a porthole letting light into a
room. But we are talking about a lense which takes light from a series
of point sources in limited external area and focusses that light onto
a series of point images on the sensor.

Fully equivalent
images on different formats will have the same brightness and be
created with the same total amount of light.

Not so, in the case of a camera.

Thus, the same total
amount of light on sensors with different areas will necessarily
result in different exposures on different formats, and it is for
this reason that exposure is a meaningless measure in cross-format
comparisons."

Having started with an incorrect premise he has finished with an
incorrect conclusion.

Eric Stevens:
Alternatively, if you have an invariant f/ value, an invariant
exposure time and an ivariant level of illumination of the
sensor then it follows that the ISO has remained the same. That
is, ISO value is not a function of sensor size.

Sandman:
Of course it isn't. But with the same exposure, sensor
amplification is a function of the sensor size, so while the ISO
number are similar, the sensor amplification is not.

Not so. Go back to my statement above and think about it. You will
conclude that amplification has nothing to do with sensor size.

http://www.josephjamesphotography.com/equivalence

"The exposure (light per area on the sensor) at f/2.8 1/100 ISO 100
is 4x as great as f/5.6 1/100 ISO 400 for a given scene luminance,
regardless of the focal length or the sensor size. However, the
brightness for the two photos will be the same since the 4x lower
exposure is brightened 4x as much by the higher ISO setting. If the
sensor that the f/5.6 photo was recorded on has 4x the area as the
sensor as the f/2.8 photo (e.g. FF vs mFT), then the same total
amount of light will fall on both sensors, which will result in the
same noise for equally efficient sensors"

Same amount of total light = same amount of amplification = same amount of
noise.

I assume you have been using the terms in the same way as James,
although you have not previously made that clear.

His important definitions a

"Exposu The total light per area (photons / mm²) that falls on
the sensor while the shutter is open, which is usually expressed
as the product of the illuminance of the sensor and the time the
shutter is open (lux · seconds). The only factors in the exposure
are the scene luminance, t-stop (where the f-ratio is often a good
approximation for the t-stop), and the shutter speed (note that
neither sensor size nor ISO are factors in exposure)."

Here is a trap:

"Brightness: The brightness of an image (what people usually think
of as "exposure" -- same units as exposure): Brightness =
Exposure x Amplification."

Most people would assume that brightness = "The total light per area
(photons / mm²) that falls on the sensor while the shutter is open..."
In fact James has been using the term as above, where 'magnification'
is a factor of arbitrary value. "Brightness" in this sense is the
value in a data array external to the sensor. (it might be part of the
sensor but it it is part of the electronics processing the output of
the sensor.)

"Total Light: The total number of photons that falls on the sensor
(lumen·seconds, or, equivalently, photons):
Total Light = Exposure x Effective Sensor Area."

--- long tail snipped (Savageduck heaves a sigh of relief) ---

I've finally realised where the problem lies in all this.

A. The terms as defined above are not entirely the same as those used
in the wider world. That this discussion was based on certain exact
definitions was not made clear from the outset. As James wrote:

"Many of the misunderstandings come from people using different
definitions for the same words. In particular, "f-ratio" is often
confused with "aperture", and "exposure" is confused with
"brightness" and "total light". The importance of these
distinctions is often overlooked or simply not understood, ... "

B. It has been assumed that noise is inversely proportional to photons
received: the more photons the less noise. To describe this as a hairy
approximation is a gross understatement, but no matter.

C. All this pseudo-mathematics is centred about an attempt to set out
the rules relating the properties of cameras of different sizes which
produce not only the the equivalent optical geometry but also equal
image noise.

D. If sensors on smaller cameras are required to accept (say) four
times as much light as a sensor in the larger camera, they will have
to have larger 'buckets' in the individual photosites. In the real
world their buckets will not be four times as large so will be
overflowed, with resulting highlight burnout.

E. Nobody operates a camera so as to obtain constant noise so the
conclusions, while interesting, are irrelevant to the real world.


I've had more than enough of this. Now that I've finally worked out
what Sandman has been talking about I'm happy to leave him to go his
own way and achieve his own state.
--

Regards,

Eric Stevens