On 29 Nov 2015 22:42:49 GMT, Sandman wrote:
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.
Eric Stevens:
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.
Sandman:
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?
To create an equivalent exposure.
Sandman:
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?
There's a difference between "exposure" and "equal exposure". An exposure is a
given amount of light in a given unit area. With a smaller sensor, you create
an equal image by making an equal exposure, meaning more light on that smaller
area.
I'm breaking my vows but I sense progress. You may be saying that the
amount of information carried to the camera depends upon the number of
photons which reach it. From this it follows that the small sensor has
to handle the same amount of light as the larger sensor. Have I got it
right?
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.
Eric Stevens:
Who cares about the sensor getting less light?
Sandman:
Those that care about noise, since less light means more
amplification and more noise.
Why should there be more noise?
Amplification leads no noise. Always.
Noise is always generated in a camera. The camera deals with this, in
part, by setting a noise floor. It sets a minimum level below which it
will ignore whatever is emerging from the sensor and it's
amplification. Of course this can never be entirely successful.
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 ...
If the amount of total light is the same, the exposure is equal, but not the
*same* exposure (see above).
It's this sort of statement which makes it hard to follow exactly what
you are saying. James defines exposure as (see below) "The total light
per area (photons / mm˛) that falls on the sensor while the shutter is
open". In the scaling up/down situation, if the total light is the
same, the total light per area (photons / mm˛) cannot be the same.
There is something wrong with your terminology.
Using the same amount of light, you adjust the ISO down by the crop factor
squared, which means the signal is amplified the same amount (give or take,
sensor tech).
I understand you better now that I know that by 'brightness' you don't
mean the density of the light which falls on the sensor, but instead
the amplified output of the sensor. At least James does. Do you mean
that also?
Eric Stevens:
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.
Sandman:
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.
Of course not, but on a smaller sensor, that "small patch" is a part of the
whole image that is rendered by the camera. On a larger sensor, the same region
of the resulting image is made out of a larger "patch" of the sensor, which
receives more light. So there is more signal going in to that part of the
sensor that makes out that part of the image.
There comes a point where the message has got through, no matter how
many extra photons you have poured through.
Certainly the sensitivity is not affected.
No, but to create an equally bright image, the smaller sensor needs to amplify
the signal more, since it has received less light to being with. I'm not sure
how many times I've said this in various forms.
And the statement seemed meaningless, mad or daft before I realised
what you meant by brightness.
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".
Take it to the extreme, then. You have two sensors. One is 100x100mm and one is
10x10mm, both are 10MP sensors. Using the same exposure, the smaller sensor
will receive one hundredth of the "signal" as the larger sensor.
Both generate a 10MP image out of this signal, but the smaller sensor can't
possibly make it as bright as the larger sensor without amplification. And that
amplification creates noise.
It is likely that even at their native ISO both sensors will require
amplificaton. It is also a big jump to assume that noise is
proportional to amplification, but never mind. Now that the
terminology is being defined I am beginning to understand you.
Sandman:
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.
Yeah? Do you have a point?
Providing everything has been scaled correctly, the intensity of that
point of light on the sensor remains the same. The image is
constructed by the array of those points of light on the sensor.
Despite the fact that there are fewer photons finding the way to the
smaller sensor the intensity of the light (what I thought you meant by
brightness) remains the same across all sizes of cameras. If that is
the case, and I believe it is, all sizes of sensors will require the
same amount of amplification to reach the same brightness as defined
by James.
Sandman:
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.
How so? Be specific.
I hope I just did. However I expect that my explanation will not
satisfy James' definition of equivalence.
Sandman:
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.
Haha.
I think the chain of logic will have to be reworked with some care
given to definitions if this is going to be resolved.
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.
Eric Stevens:
Not so. Go back to my statement above and think about it. You
will conclude that amplification has nothing to do with sensor
size.
Sandman:
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.
I have said so many many many many many times.
Yes you have said that about total light many many times but you have
never explained it. At several points in this thread I variously
enquired after or taken you to task over the terms you have been using
and you have responded by accusing me of semantic trolling and
similar. It wasn't semantic trolling: it was me trying to find out
*exactly* what you were trying to tell me.
His important definitions a
"Exposu The total light per area (photons / mm˛) 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)."
Indeed.
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."
Exactly like I've said.
But you have never explained that by 'brightness' you meant the output
of the amplifier.
Most people would assume that brightness = "The total light per area
(photons / mm˛) 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.)
"Brightness" is in the resulting image, like he said. Brightness = (Exposure +
amplification). 100% what I've said this entire time.
The only place in the camera where there is an image is on the sensor.
After the sensor is scanned the image is converted into a stream of
data which takes whatever form the camera designer has thought of.
"Total Light: The total number of photons that falls on the sensor
(lumen·seconds, or, equivalently, photons): Total Light =
Exposure x Effective Sensor Area."
I.e. exactly what I've said.
I think you will find you have talked about 'total light' only once,
im message
I've finally realised where the problem lies in all this.
I won't hold my breath. 
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
You have missed the end of the sentence which originally read
"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, ... "
Yes, these are some of the terms you and nospam have repeatedly misunderstood,
I agree.
And you consistently failed to explain *exactly* what you meant to
many more people than me and nospam.
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.
Noise has nothing to do with amount of photons. It has everything to do with
amplification.
Which depends on the number of photons received.
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.
I.e. creating equivalent images using different sensor sizes.
And lens sizes also.
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 photo sites.
This is false unless you're playing with highly overexposed images. A normally
exposed image, using the settings I have repeatedly mentioned, will not
saturate the photo sites of a MFT camera.
I haven't got a MFT camera but one of these days I will try comparing
the Nikon D750 with my wife's Canon G12 (7.6mm x 5.7mm).
In the real world their buckets will not be four times as large so
will be overflowed, with resulting highlight burnout.
This is a false assumption. We are talking about a MFT sensor that has
amplified its signal four times as much as a FF sensor to create a "normal"
photo. Lowering the ISO and opening up the aperture will result in the exact
same photo, only with the exact same noise as the FF camera.
Here is something you can't prove. It is probably an assumption.
E. Nobody operates a camera so as to obtain constant noise so the
conclusions, while interesting, are irrelevant to the real world.
It's very relevant to a number of things:
1. The myth that smaller sensors are noisier, they aren't.
2. The myth the "ISO" is a certain "sensitivity" (amplification), it is not.
The concept that ISO is a certain amplification is a myth (I hope).
3. The problem with using "35mm equivalent" terms for the focal length but not
for aperture to mislead users.
Look up the definition of f/ number (and think about it).
4. The myth that MFT sensors have larger DOF (a part of #3)
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.
No one is happier than me, believe me. And even if you didn't really understand
all of it, you came a long way, and have restored some of your maturity in the
process. Well done.
Thank you for those kind words. Now, please go and peddle your
nostrums somewhere else. I'm not buying.
--
Regards,
Eric Stevens