How to measure ISO

On 2015-11-25 02:42:19 +0000, Eric Stevens said:


Based on past experience, the only way to put this to rest is to
kill-file the arrogant pig.

....or kill the thread.


--
Regards,

Savageduck

In article , Eric Stevens
wrote:


Based on past experience, the only way to put this to rest is to
kill-file the arrogant pig.

the problem is that his bogus statements can mislead others.

On Tue, 24 Nov 2015 19:03:31 -0800, Savageduck
wrote:

On 2015-11-25 02:42:19 +0000, Eric Stevens said:


Based on past experience, the only way to put this to rest is to
kill-file the arrogant pig.

...or kill the thread.

It's not just the thread. I prefer to kill-file Sandman. It worked
last time.
--

Regards,

Eric Stevens

On Tue, 24 Nov 2015 22:05:11 -0500, nospam
wrote:

In article , Eric Stevens
wrote:


Based on past experience, the only way to put this to rest is to
kill-file the arrogant pig.

the problem is that his bogus statements can mislead others.

That's what tends to keep me going.
--

Regards,

Eric Stevens

In article , Eric Stevens
says...
It's not just the thread. I prefer to kill-file Sandman. It worked
last time.

I've already done that. It's impossible to discuss with Sandman on a
rational basis.
--
Alfred Molon

Olympus E-series DSLRs and micro 4/3 forum at
http://tech.groups.yahoo.com/group/MyOlympus/
http://myolympus.org/ photo sharing site

On Wed, 25 Nov 2015 08:09:48 +0100, Alfred Molon
wrote:

In article , Eric Stevens
says...
It's not just the thread. I prefer to kill-file Sandman. It worked
last time.

I've already done that. It's impossible to discuss with Sandman on a
rational basis.

His ego alwys takes command.
--

Regards,

Eric Stevens

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.

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.

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.

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.

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

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.

Same exposure = different amount of amplification = different amount of noise.

Sandman:
So, to repeat:

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f5.6, ISO 800

The above is the exact same *exposure*, but with different signal
amplification, the smaller sensor needs to amplify the signal more
than the FF sensor to present an equally bright image.

Nonsense.

Based on?

http://admiringlight.com/blog/full-f...why-it-doesnt-
matter/

"One final nitpick that people like to point out on aperture
equivalence is that it also shows you the settings that not only
yield a similar image, but also allow for the same total amount of
light used to make an image. For instance, a full frame sensor is
four times larger in area than a Micro 4/3 sensor. Therefore, if
the f-stops are the same, and thus the intensity of the light is
the same (and the exposure is the same), then the full frame camera
will be using four times the total amount of light to make the
image because it?s got four times the total area. For the smaller
sensor to have the same total amount of light, they need two stops
faster aperture or two stops lower ISO with a longer shutter speed.
This is why, often, it?s said that full frame sensors will have two
stops better ISO performance over a Micro 4/3 sensor."


Sandman:
FF: 1/250, f5.6, ISO 800
MFT: 1/250, f2.8, ISO 200

The above is the same *amount of light* on the sensor, ...

But you don't want the same amount of light on the sensor.

To create an equal image, this is exactly what you want.

You want the same 'intensity' of light on the sensor and that's a
different thing entirely.

Not sure what you think is different.

Some more info about achieving the same amount of total light on differently
sized sensors:

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

Sandman:
... which creates as identical image as possible using different
sensor technologies. Also, you adjust the ISO by the crop factor
squared to match the signal amplification of the larger sensor,
so you will get very equivalent noise.

As a sidenote, while the aperture is different, when you take the
sensor format into account, that's actually the same. There's
this huge confusion about aperture sizes and what is
"equivalent". It's easy to see a MFT lens advertised as "12-35mm
f2.8 (24-70mm equivalent)" which leads to people thinking that
it's equivalent to a 24-70 mm/f2.8, but it's not. The crop factor
applies to the aperture as well.

f/ is aperture divided by focal length. If you change the focal
length and the f/ remains the same then you have also changed the
diameter of the lens.

Yes? What's your point?

Sandman:
The aperture of the 12-35mm lens *IS* 2.8, but if you're going to
state it's focal length equivalence, you should also state its
aperture equivalence, and write it as such:

"12-35mm f2.8 (24-70mm f5.6 equivalent)"

But that wouldn't sell as many lenses, of course.

Of course not: it's wrong.

It's not. Maybe this is the source of your confusion.

f2.8 on a MFT lens is equivalent to f5.6 on a FF lens. That means that with
those two settings you get the same amount of total light AND the same depth of
field through the lens.

Aperture and focal length is a function of the physical lens. When advertised,
lens manufacturer often tells the customer what the "equivalent" focal length
is, but "forgets" to tell the customer what the equivalent aperture is as well.

Here's some more information:

http://petapixel.com/2014/03/28/conc...affects-focal-
length-aperture/

And here's an example of misleading information:

http://www.slrlounge.com/wp-
content/uploads/2014/05/Vollbild_31_05_14_01_39.jpg

I.e. the advertiser state the 35mm equivalent focal length, but not the 35mm
equivalent aperture.

This is, of course, also the reason why most smaller sensor cameras is said to
have a larger depth of field, it's because the 2.8 aperture on a MFT lens
doesn't yield the same DOF as a 2.8 aperture on a FF sensor for obvious
reasons.

So when they advertise it as a "25mm f1.4 (50mm equivalent)" it is *not*
equivalent to a 50mm/f1.4, it is equivalent to a 50mm/f2.8.

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

"Equivalent lenses are lenses that produce Equivalent photos on the
format they are used on which means they will have the same AOV and
the same aperture diameter. Let's consider, for example,

25mm f/1.4 1/100 ISO 100 on mFT (4/3),
31mm f/1.8 1/100 ISO 150 on 1.6x (Canon APS-C),
33mm f/1.9 1/100 ISO 180 on 1.5x (all other APS-C), and
50mm f/2.8 1/100 ISO 400 on FF.

The focal lengths all have the same AOV, so we say the focal
lengths are equivalent.

The apertures (entrance pupils) all have the same diameters (25mm /
1.4 = 31mm / 1.6 = 33mm / 1.9 = 50mm / 2.8 = 18mm), so we say the
f-ratios are equivalent.

The photos have the same brightness, so we say the exposures are
equivalent."

This is 100% what I've been saying this entire time.

Sandman:
So, the above "total light" example, is an example where the
"exposure" is different (since that's per unit area, regardless
of sensor size) but the aperture and total light is *equivalent*
between the cameras, producing an almost identical image (sensor
tech permitting).

Well. I tried.

With what?

--
Sandman

In article , nospam wrote:


as for the stuff he said, some of it is true and some isn't. he is
as confused as you are.

Haha, more hot air from nospam.

--
Sandman

In article , nospam wrote:

Sandman:
It's like when you said this:

nospam Apple is purchasing Twitter analytics firm 12/09/2013


"film slrs aren't made anymore."

Which was technically incorrect as a statement on its own, but for
the context, it was clear what you meant.

that was made in good faith.

As was my claim. See how this works, yet? No? Didn't think so.

--
Sandman

In article , nospam wrote:

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.

they don't think about it because noise is so low that it doesn't
make any difference.

Who are "they"?

as for amplification, a nikon d800 proves you wrong. the
amplification is the same in crop mode as it is in full frame mode,
yet the sensor area is smaller, the exposure is the same, the
amplification is the same, the brightness of the image is the same
as is the noise.

Because it has the same size photo sites, which contemporary FF and MFT doesn't
have.

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.

that's equivalency, not exposure.

you *still* confuse the two concepts.

Yes you are, I have only ever talked about equivalency. To create equal images,
you can't use the same exposure per unit area, you need to expose the sensors
equally.

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.

previously, you said it is.

Quote? No? Ok, more hot air then.

Sandman:
But with the same exposure, sensor amplification is a function of
the sensor size,

no it isn't.

Source? No? Ok, more hot air.

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

"Equivalent lenses are lenses that produce Equivalent photos on the
format they are used on which means they will have the same AOV and
the same aperture diameter. Let's consider, for example,

25mm f/1.4 1/100 ISO 100 on mFT (4/3),
31mm f/1.8 1/100 ISO 150 on 1.6x (Canon APS-C),
33mm f/1.9 1/100 ISO 180 on 1.5x (all other APS-C), and
50mm f/2.8 1/100 ISO 400 on FF.

The focal lengths all have the same AOV, so we say the focal
lengths are equivalent.

The apertures (entrance pupils) all have the same diameters (25mm /
1.4 = 31mm / 1.6 = 33mm / 1.9 = 50mm / 2.8 = 18mm), so we say the
f-ratios are equivalent.

The photos have the same brightness, so we say the exposures are
equivalent."

They are equivalent since the signal has been equivalently amplified using
different ISO settings for each.

Sandman:
so while the ISO number are similar, the sensor amplification is
not.

you have *no* way to know what the amplification is outside of lab
instruments and it doesn't matter anyway.

It is easily shown when you notice the noise. Noise is a direct result of
signal amplification, and more noise means more amplification. And using the
same exposure per unit area, you get the same brightness but different amounts
of noise:

http://jonaseklundh.se/files/same_iso.png

But if you expose the sensors equally, adjusting the aperture to let the same
amount of total light in:

http://jonaseklundh.se/files/iso_adjusted.png

You get the same sensor amplification.

Sandman:
So, to repeat:

FF: 1/250, f5.6, ISO 800
MFT: 1/250, f5.6, ISO 800

The above is the exact same *exposure*, but with different signal
amplification, the smaller sensor needs to amplify the signal more
than the FF sensor to present an equally bright image.

nope.

the same exposure will produce an equally bright image. period.

Yes, since the smaller sensor amplifies its signal more. Simple physics. With
the same exposure, it receives less signal than a bigger sensor.

again, nikon d800 in crop and full frame mode. different size
sensors, same exposure, same iso, same brightness.

Funny how you are on and on about the D800 but you won't use an actual
different sized sensor in your supposed "rebuttals".

A D800 in crop mode has the same photo site size as a D800 in normal mode. A
MFT sensor hasn't got the same photo site size as a FF camera

i don't know why you still get this wrong, or even why you got it
wrong at all. it's very simple.

Indeed it is. And I have tons of support from all around the web to back my
claims. As opposed to you.

Sandman:
FF: 1/250, f5.6, ISO 800
MFT: 1/250, f2.8, ISO 200

The above is the same *amount of light* on the sensor, which
creates as identical image as possible using different sensor
technologies. Also, you adjust the ISO by the crop factor squared
to match the signal amplification of the larger sensor, so you
will get very equivalent noise.

that's equivalency.

No ****, Sherlock?

Sandman:
The aperture of the 12-35mm lens *IS* 2.8, but if you're going to
state it's focal length equivalence, you should also state its
aperture equivalence, and write it as such:

"12-35mm f2.8 (24-70mm f5.6 equivalent)"

But that wouldn't sell as many lenses, of course.

that's because it's misleading.

Yes it is, the 2.8 is for the smaller lens, and they tell you the equivalent
focal length but not the equivalent aperture.

*if* someone wants to match noise and dof, *then* the equivalent
f/stop would be used. otherwise no.

And, coincidentally, that's what I've been talking about this entire time.
Maybe there's a chance that nospam sees the light at the end of the tunnel?
Probably not.

--
Sandman