[Leica] Retry and missing link (3)

[imx <imxputs@knoware.nl>]

The physical dimension is attached to the matrix when we print or display on
a screen. Easiest to understand is the monitor sceen. This has a fixed size
of screen dots of 0.26mm (some more some less!!). Normally we say 96 dots
per inch for this output medium. And now we add dimensions to our
dimensionless pixes in the matrix.The dot size is by the way almost
identical to the diameter of the CoC.
Because we now make a one to one correspondence between the N x M matrix and
the screen dots. If we have a 100 by 100 matrix (again for the sake of
argument), we get on the screen a 100 x 0.26 by 100 x 0.26mm image, composed
of 10000 pixels. The size is 26 by 26mm and the resolution is 1/0.26 = 3.8
lines per mm. It is of no importance if the screen itself is a 14 inch or a
21 inch monitor. 
If we need to fill a whole screen with a picture of 20x25cm, with this
resolution and dot size, we need  a matrix of 200mm/0.26mm by 250mm/0.26mm
that is 769 x 961 or rougly 800 x 1000.
Now we can combine matrix size, print size and resolution. If we want a
print (in this case screen) of dimension 20x25cm, that can show image
details of size 0.26mm (the value of the CoC or threshold of sharpness
impression) at a normal viewing distance, we need an image file of 800 x
1000 rows and columns (or dimensionless pixels).
Now we have to add the size of the sensor elements that we used to digitize
the image in the first place. If these sensors had a diameter of 0.26mm too,
we may assume that the sensors could only registrate quite course detail,
like outlines of buildings or cars or people at a distance.
If the sensors had a diameter of 0.026mm, we can registrate much finer
detail. but to cover the same image area, we need ten times as much pixels
or matrix elements. Now we need a matrix of 8000 by 10000 rows and columns!
With film and a resolution of 40 lines/mm on the negative (that is 1.3
million image pounts) I can easily enlarge 8 times to get a 20x25cm print
and then hold 5 lines per mm in the enlargement.
To get that level of detail in a digital print I need a matrix of 200mm
times 5 lines by 250mm times 5 lines equals 1000 by 1250 true matrix cells.
With current CCD chip technology that is a CCD count of 2000 by 2500 pixels.
This number is close to the current 3 Megabyte standard of digital cameras.
If I would improve my image quality on the negative I can with the same film
and lens get a resolution of 10 lines in the print over a 30 by 40cm print.
That would amount to a matrix of 3000 by 4000 or 12 million times two or a
CCD with 24 Mb of pixels.
I hope this clarifies.
My original intent was to show that the normal quality of handheld 35mm
photography can be quantified as having a resolution of 40 lines/mm or 1.3
million image points.
To transfer this requirement to the digital world needs a CCD with at least
a 3 MB pixel count, when you put the dimensions in the equation and not only
the number of sensors in rows and columns. The whole argument based on the
equivalence of my 1.3 image points to a ccd with 960 by 1440 sensors is
invalid because of its lack of accounting for the physical size/dimensions.
My original intent has not been to say that digital technology will surpass
in the near future the detail defintion and clarity of a good silverbased
negative. On the conrary. Above you can see that a 24 Mb chip would be
needed to come close to film performance.

Erwin