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Re: filmscanners: What causes this and is there any easy solution ?
Oops! Sorry about the confusing nomenclature. What I meant to express was
that at sufficiently small angles, the 'side opposite' leg is effectively
the far leg of a 90deg triangle (its an assumption but its pretty darn
close). That means that you can apply the basic geometric rule of
Side Opposite
Tan(theta) = 
Side Adjacent.
So for the average human eye (something like 90% of the population  with
folks like Chuck Yeager being the radical exceptions, which largely explains
his survival and success in Korea), the limit of resolution occurs when
Theta is 1/60 of a degree of arc  which as you correctly pointed out is 1
minute of arc.
At 500dpi, the interline spacing is 0.002inches. So this means that we can
solve for Side Adjacent
Side Opposite 0.002inches
Side Adjacent =  =  = 6.875 Inches.
Tan(1/60deg) 0.000291
This is a very well researched formulae  the Society for Information
Display has lots more info on the mathematical models of how the eye works 
fairly precise models too.
So if we go from this equation, we know that if we look at the original film
at a big enough enlargement (say 8x10), then we will be able to see any
artifacts in the film of the kind being described, if we look more closely
than 6.5". Since folks have been doing this in the darkroom for decades 
using grain focussers and the like, and these artifacts have not been
observered, it is reasonable to conclude that they don't exist except in the
optoelectronics of the scanner process
 Original Message 
From: "Arthur Entlich" <artistic@ampsc.com>
To: <filmscanners@halftone.co.uk>
Sent: Wednesday, May 16, 2001 1:41 PM
Subject: Re: filmscanners: What causes this and is there any easy solution ?
>
>
> Karl Schulmeisters wrote:
>
> > I don't think this is the case. Otherwise you would have seen this
> > phenomenon from enlargements made from transparencies long ago.
Consider
> > this, the human eye can resolve about 1 minute of 1 degree of arc (1/60
of a
> > degree) in the horizontal plane (most sensitive  less in the vertical)
So
> > take a 35mm slide (which is about 1" tall) and enlarge it full frame to
> > 8x10" that's an enlargement factor of about 8. So a 4000dpi scan of a
35mm
> > slide is about the same as a 500dpi scan of an 8x10.
> >
> > So plugging 1/500th of an inch into the formula X = TanTheta Y where X
is
> > the lines/inch and Y is the eye's distance from the 8x10 enlargement, we
get
> > ..002 = Tan(1/60deg) Y or Y (max eye resolution) = .002/.000291 =
6.875".
> >
>
> If human eyes actually functioned based upon a mathematical formula,
> you'd have it all solved! ;)
>
> Just so some people unfamiliar with the nomenclature won't get too
> confused, the symbol ["] can be used to both indicate inches as a linear
> measurement and minutes (1/60th of one degree) of an arc. Your first
> reference (1") is one inch, your second reference (6.875") is 6.875
> minutes of a degree of an arc.
>
> The big problem we all face in analyzing the artifacts or other
> information we see in a scan is that we are looking at the scan in a
> translated format, either via a monitor (via a software package) or a
> print (usually inkjet for most people, via a print driver program) each
> of which add other confounding factors to what is being provided by the
> scanner. SInce none of these are purely optical in nature, we're in
> "uncharted waters", with no sextant to get us ashore, or is that asure?
;)
>
> Art
>
>
> > IOW, anyone who has looked at a full frame 8x10 enlargement of a 35mm
image,
> > closer than 7" is in essence 'scanning' the 35mm slide at greater than
> > 4000dpi. And since we don't have reports of folks seeing this sort of
> > difference in enlargements at this level (remmember folks use grain
> > focussers to get even higher resolution during focussing of an
> > enlargement)  I don't think there is any 'real information' there.
>
>
