Re: he histogram as the basis of automatic exposure.
Eric Stevens <firstname.lastname@example.org> wrote:
> That this is, or should be, done has been recently discussed. I though
> about this while this afternoon unpacking and sorting by year a large
> box of old Science Fiction magazines on a large table.
> I removed the first magazine and immediately wondered where to put it.
> 1962: this was old but was it the oldest? How far from the left-hand
> end of the table should I place it to leave room for (how many?) older
> The next one was 1975. Well, that gave some indication of the minimum
> length of the line of books but, how much longer would the line grow?
> It wasn't long before 1954 required that I shift everything along the
> table. Soon came 1989. And so I went.
OK, so you decided that shifting on the table was a low-cost
> Only when I had finished did I know how many of which year that I had.
If you had thought that shifting was a high-cost operation,
you'd have sampled the magazines to get an idea. If you had
assumed that shifting was not possible at all, you'd have
sampled the magazines to a higher confidence and left spill
over areas at the borders of the table. Or first catalogued
your magazines without putting them on the table, then decided
by the cataloguing where each magazine had to go.
> Arriving at a histogram in a camera is even worse.
> The range of light
> values which may be detected is enormous
At 14 bit it's only 16k values. (And that's assuming it's not
doing a histogram for the resulting JPEG, where there are only
256 values --- which are way more relevant for out-of-the-camera
JPEGs.) That doesn't strike me as enormous, given that you pay
much more for a new car and manage to handle single cents, too.
Additionally, it's a *computer* doing the counting.
Computers are *good* with numbers.
> and the histogram engine has
> no way of knowing in advance of where the histogram will end up being
It doesn't care. If every last pixel is completely full,
it'll just mark "0:0,1:0,2:0,...,16383:0,16384:0,16385:36.000. 000"
assuming the sensor in question has 36M pixels.
> The only way to determine the histogram for exposure purposes is by
> taking a trial image first and determining it's histogram.
However that trial image neither needs the same sensor nor
the same resolution.
> But that
> trial image requires that an initial exposure be determined (probably
> by some form of matrix metering) followed by the taking of the trial
You can also guess an initial exposure.
> A histogram is taken from the trial image and after evaluation
> it is used to adjust the initial exposure. This is then used to take
> the final image.
> That's an awful lot of huffing and puffing for the camera,
No, it's not.
> not to
> mention the shoveling of electrons and I expect only cameras of the
> highest capabilities might undertake such a procedure.
How did you think cameras with an EVF or lifeview worked?
May I remind you that practically all the P&S and mobile
phone cameras do that every time they even display an image?
Otherwise they'd not be able to do autoexposure ...
It seems you have no idea of the cost of operations for the
different things a camera's processor performs.
> I suspect that it is more likely that modern high end cameras with
> more than a thousand sensing points may use these to arrive at a crude
> histogram upon which the final exposure will be based.
Unless these cameras are completely stupid, they'll use much
more information from the sensors, like *where* the values
are recorded and how they relate to each other.
For your homework, pick a couple of your typical photos, run
a histogram and then see how many pixels you can average ---
starting at 2x2, 3x3 and so on --- until you get relevant
differences in the histogram.
> It won't be as
> accurate as using a histogram from a complete image but it's going to
> be a lot easier than doing it the fancy way.
Think again how cameras which only have one sensor --- i.e.
most low end cameras --- display their images on the screen
for framing ...
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