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Re: paper on CRT technology
- To: "telecine internet group" <telecine at alegria.com>
- Subject: Re: paper on CRT technology
- From: "Martin Euredjian" <martin at hollydig.com>
- Date: Thu, 29 Oct 1998 03:08:03 -0700
- Resent-Date: Thu, 29 Oct 1998 05:07:53 -0600
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>I've been sent an illustrated paper on technology of the CRT...
I submit the following review and opinion:
" There is strictly only one way to analyse this, and that is by Mathematics
and Physics. These tools are the backbone of the professional engineer, and
from this he/she will be able to distinguish between what is achivable and
what is not achivable in a practical and operational sense. "
Ever try to use that one on a client?
" The maths is simple within the CRT, as is the physics. The tube heater is
best analysed using quantum physics "
" It is clear that the smaller "theta" the better geometry and linearity one
can achieve, not forgetting focus. "
This comes after a long (and mostly pointles, IMO) mathematical derivation.
No part of that derivation makes it "clear" that better geometry can be
achieved by a smaller "theta". We all know that this is the case...I'm just
saying that the formulas have very little to say about this.
With reasonable deflection, the geometry and linearity achieved are mostly a
result of how refined a scan control system you have on the tube (assuming
decent coils, etc., etc.).
"Consider the "line pair" in a 2K resolution system scanning a 35m/m film.
In an academy aperture of 20.955m/m (horizontal) the "line pair" space would
be 10microns (approx.). The sampling would have to be locked to spot
position. This all is assuming that there is no afterglow. If we take
nyquist into consideration then we need a spot size of less than or equal to
5 microns, and a corresponding sampling rate dependent on the frame rate."
First of all, we must not confuse the term "line pair" with the number of
horizontal or vertical pixels carried by the digitized signal. To represent
two distinctly visible lines digitally at any resolution you need three
pixels, not two. Film is good for more than 800 lp/mm.
Secondly. Henry Nyquist, in his 1928 "Topics in Telegraph Transmission
Theory", postulated a theorem which proposed that a sample of twice the
highest signal frequency rate captures the signal perfectly. The "signal"
in our case is FILM not any given output data stream's orthogonal image
representation. You cannot say that a 100x100 scan of the film requires a
sampling of 200x200 !!!! By definition a 100 x 100 scanning of an image
requires exactly that many picture elements to be looked at. Again,
sampling theory applies to the relationship between the signal (film) to be
sampled and the required number of samples to represent it accurately. To
capture film "perfectly" you need in excess of 4000 samples horizontally.
Assuming I am not out of my mind, then the above example --2K scan of an
academy aperture of 20.955-- would require a 10.23 micron spot to be
projected onto the film. This is just to get 2048 pixels out of each line.
A "line pair" would be 20.46 microns apart (three pixels, center to center).
"Consider a spot size of say 40 microns which gives the best compromise of
light output for given phosphor as used in a 2K horizontal resolution
platform. Then using the 5 micron ref. the scan size required to cover the
20.955m/m would be 167.64m/m (6.6inches). "Theta" is now significant, as is
the dynamic focus and scanning linearity. The tangent at the point of impact
in the corners is now eliptical."
The new math, I guess. This is what I get with my HP calculator:
40 microns = 0.000040 meters
20.955 mm = 0.20955 meters (this is the projector aperture, BTW)
In his reference to the five micron scan size requirement the author implies
that 4096 samples are required. I calculate that, with a 40 micron spot,
this produces a horizontal scan patch of 163.84 mm (0.16384 meters), not
167.64 mm as the paper reads.
I argue that to scan 2048 horizontal pixels out of a piece of film you need
2048 spots on the scanning CRT. This means that a CRT using a 40 micron
spot would need to produce a horizontal scan patch of 81.92 mm of length.
Funny I should say that because, from memory, that's about the size of the
scan patch on a C-Reality!
With an 81.92 mm patch "Theta" is now less significant (although its effects
are always with us for good measure).
" Theta, for an active length of 254m/m ... "
How did we get to 254 mm ?
"If these problems exist at 2K, what problems exist at 6K? Yes, in theory 6K
is possible, but practically is it sensible."
Those of you desperately wanting to scan film at 6K please raise your
Yeah, that's what I thought.
" Can it compete in a commercial market against CCD? We can not allow
sentiment to come into this! "
Having just looked at the C-Reality at Cintel I'd have to say YES.
" However, no CRT to date can produce the Zenon intensity and spectral
What are we comparing here? "However, no LED to date can produce the
Incandescent-lamp intensity and spectral response ..."
They ARE different. We know that. The end result is what matters.
In my (never humble) opinion I have to say that this paper is
pseudo-scientific garbage and does not belong on the TIG.
My opinions are only my own and not necessarily those of my employer,
parents, wife, etc.
Not paid by any manufacturer, etc., etc.
[ Martin Euredjian ]
ToddAO - Hollywood Digital, CA, USA
Tel: (213) 465-0101
martin at hollydig.com
martinfx at msn.com
Thanks to Howard Lukk for support in 1998.
No product marketing allowed on the main TIG. Contact rob at alegria.com
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