SD800IS - Why oh why the smaller sensor???

I was thinking the noise floor is not randomness for example between 0 and 5 at iso100 but rather varies for example between 15-20 as it would in my example with the averaged 32 pixels of the much higher resolution sensor. So i didnt mean you should see detail below the noisefloor but rather the noisefloor should not be shown and that 15-20 should be shifted to -2.5 +2.5 with the dynamic range determined by this randomness. Can the 5D show absolute 0 per pixel or does it indeed vary between higher levels?
 
Not overexposing, it just looks bad.
Oh. Remember, this is a crop and the whole image is well-balanced. Also this is a very high DR image. The white flowers are in full sun while the leaves in the background are dark and in the shade.

Yes, you can easily preserve the highlights if you want to, even without manual mode, just by using -EC.

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Lee Jay
(see profile for equipment)
 
I was thinking the noise floor is not randomness for example
between 0 and 5 at iso100 but rather varies for example between
15-20 as it would in my example with the averaged 32 pixels of the
much higher resolution sensor. So i didnt mean you should see
detail below the noisefloor but rather the noisefloor should not be
shown and that 15-20 should be shifted to -2.5 +2.5 with the
dynamic range determined by this randomness. Can the 5D show
absolute 0 per pixel or does it indeed vary between higher levels?
There's always noise in the RAW data, but when you convert you can choose to set the black point above the noise floor so you get pure blacks with no noise at all.



--
Lee Jay
(see profile for equipment)
 
However, in two dimensions, it's 1.55x (going by the size values in
the dpr glossary). Since the sensor is a 2d surface, not a line :)
that's the number that matters, isn't it? The bigger sensor
receives 55% more light.
It's still 2/3 of a stop (log base 2 (1.25^2)).
I guess that comes down to another way of saying "2/3 of a stop is actually a lot of difference in the amount of light admitted". :)

(Um, for the bored, it's actually a bit less than 2/3 of a stop, even -- 0.63 or so.)
 
However, in two dimensions, it's 1.55x (going by the size values in
the dpr glossary). Since the sensor is a 2d surface, not a line :)
that's the number that matters, isn't it? The bigger sensor
receives 55% more light.
It's still 2/3 of a stop (log base 2 (1.25^2)).
I guess that comes down to another way of saying "2/3 of a stop is
actually a lot of difference in the amount of light admitted". :)
Okay.
(Um, for the bored, it's actually a bit less than 2/3 of a stop,
even -- 0.63 or so.)
Yeah...since the calbration is + - 1/3 stop at least, I usually round!

--
Lee Jay
(see profile for equipment)
 
In the future with new technology even the low iso advantage of
larger sensors will disapear so i wonder if there will be any
reason for them.
The above-mentioned advantage will remain.
I was thinking about sensors with unlimited exposure. One difference between sensor sizes might still be the limit of intensity that the sensor can accept, the small ones might not be able to interpret the light fast anough when you point the camera to something bright anough, note that this would be a speed specification of the sensors and not a capacity specification as iso levels are.
 
I was thinking about sensors with unlimited exposure. One
difference between sensor sizes might still be the limit of
intensity that the sensor can accept, the small ones might not be
able to interpret the light fast anough when you point the camera
to something bright anough, note that this would be a speed
specification of the sensors and not a capacity specification as
iso levels are.
Don't think so. The physics of the photoelectric effect will allow an almost arbitrary charge rate production. It's a matter of being able to store it until measure or, measure it as it is produced. This second option holds promise IMHO as a method of arbitrarily increasing DR - continuously reseting the cell allows an arbritrary amount of charge to be measured.

--
Lee Jay
(see profile for equipment)
 
Well i thought there might be a speed limit of reading and resetting, do you think it could be done fast anough when pointing the camera at the sun with a 600mm f/4 equivalent lens and focused on a tiny sensor?
 
Our retina has a speed limit, thats why we shouldnt look at the sun.
Yes, but Si doesn't (exactly). It does have a thermal limit and that's why you shouldn't point a 600/4 at the sun, though an 85/1.2 would be worse.

--
Lee Jay
(see profile for equipment)
 
Why do you think higher mp sensors are worse? On the same process
more pixels will result in less chip coverage of the light
sensitive area but i think we are probably 2 generations further
now wich means the circuitry is 1/4th the size so even if you
double the mp you still have higher efficiency.
I'd have no problem agreeing with you if the pictures weren't proof that "2 generations" isn't enough to stuff twice the megapixels into the same sensor and still have the same results.
 
I guess that comes down to another way of saying "2/3 of a stop is
actually a lot of difference in the amount of light admitted". :)
Okay.
Okay, so, here's what I still don't understand. Does this mean that "f/2.8" on the SD900 is a better f/2.8 than that on the SD800is? (Equivalent to being f/2.2 or so?) Or same question another way: does this mean that the f/3.6 on SD900 lets in as much light as the f/2.8 on the SD800is, and the SD900's (slowest-end) f/4.9 as much as approx. f/4 on the SD800is?

Or, is this already taken into account by something?

Image stabilization aside for all of this, of course.
 
Okay, so, here's what I still don't understand. Does this mean that
"f/2.8" on the SD900 is a better f/2.8 than that on the SD800is?
Probably, yes. You can probably shoot the SD900 at one stop higher ISO (doesn't have third stop increments) and get more noise and more detail. You can then apply noise-reduction is software to reduce the detail and the noise, then resample the 10MP image to 7MP and end up about as good as the SD800is at one stop lower ISO.
(Equivalent to being f/2.2 or so?) Or same question another way:
does this mean that the f/3.6 on SD900 lets in as much light as the
f/2.8 on the SD800is,
Total light, yes. Light intensity is a function of f-stop but more area means more light (more photons). More photons means more detail or less noise depending on pixel count.
and the SD900's (slowest-end) f/4.9 as much
as approx. f/4 on the SD800is?
About.
Or, is this already taken into account by something?
Not sure what you mean, but metering is dependent on f-stop so the camera should meter the same.

This is why f2.8 on the 5D is so much more powerful than f2.8 on the S3IS - the larger sensor means much more total light collected. So ISO 3200 on the 5D is sort-of like ISO 100 on the S3, if you equalize detail.

--
Lee Jay
(see profile for equipment)
 

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