Austin wrote:
> Color isn't relevant. The sensor doesn't have any color
> information, only intensity information. The color is
> deterministic...ie, a particular sensor has a particular
> color filter over it.
Colour could be relevent if the sensor has poor sensitivity
to a particular frequency range, or produces more noise in
that range (eg. blue, which I often hear contains more noise
than other channels :).
(I'm only noting this for humorous purposes, no flames please :)
>Basically, true. Typically, that 'number of steps' is called the
>resolution, ie, if you have an 8 volt range, and you have three bits
>(therefore can represent 8 different values), your resolution is 1V.
So why do you have difficulty with my suggestion that more bits
gives greater resolution? Decreasing the size of the steps
makes the conversion closer to the original analog curve.
> No, that's wrong (and you were doing so well ;-). Dynamic
> range IS resolution over ANY range, and 8 bits won't give
> you a DMax of 4. In fact, 8 bits is 48db, or a DMax of
> log10(2^8) or 2.4. Perhaps you are confusing the meaning
> of DMax?
Maybe I am. I thought that the optical density range was
a logarithmic conversion of light intensity. If it is,
I don't see any reason why an 8 bit system can't represent
an optical density of 4 using a value of 255. Or zero,
and represent an OD of 0 with 255. They're arbitrary numbers
representing something in the analogue realm. All the number
of bits determines is the number of steps between the
maximum and minimum values converted. I fail to see why
the bit depth is necessarily related directly to the
analogue realm - it depends entirely on how one is mapped
to the other. I *thought* that was the whole point of
Julian's argument.
>higher resolution. That is the resolution I am talking about in this
>discussion, not the X by Y resolution of the image.
Yup, I understood that.
Rob
Rob Geraghty [EMAIL PROTECTED]
http://wordweb.com
