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Re: [photo-3d] Digest Number 102 to Boris Starosta
- From: Olivier Cahen <o_cahen@xxxxxxxxxxxxxxxx>
- Subject: Re: [photo-3d] Digest Number 102 to Boris Starosta
- Date: Fri, 02 Jun 2000 11:53:03 +0200
Hi Boris. The question was not obvious in your messages, and the answer
is not easy. The maximum acceptable deviation much depends upon many
conditions.
First it changes versus viewing conditions. It is not the same in a
viewer and in projection, and even less in public projections. It( is
not the same for printed pictures, and how the reader woould watch them.
A typical figure, given by many authors, is that the deviation should
never exceed the thirtieth of the viewing distance. If you want to
publish anaglyphs or other printed images, you can accept this value as
a maximum. For a public projection, you can keep this ratio versus the
first row of viewers.
But I have already seen excellent images, in polarised projection, with
figures of maximum deviation up to twice that limit: there was a
continuous variation of the deviation from one end of the image to the
other end, i.e. as if you take a picture of the lawn in your garden,
with a foreground at one meter only and the background very far away,
but without any element in the foreground close to one in the
background.
Add to this that some people accept larger deviations than some other
people.
> Message: 17
> Date: Wed, 31 May 2000 22:24:48 -0400 (EDT)
> From: boris@xxxxxxxxxxxx
> Subject: ofd calculator
>
> > Date: Tue, 30 May 2000 18:06:24 +0200
> > From: Olivier Cahen <o_cahen@xxxxxxxxxxxxxxxx>
> ..
> > The value of the maximum deviation is easy to calculate: the stereo
> >base, multiplied by the lens-to-film distance, multiplied by the
> >difference between the reciprocals of the foreground and background
>
> Thanks, Olivier, I will try this calculation. You must mean "the value of
> the deviation is easy to calculate..." Given all of the variables, your
> formula would give THE deviation, not a maximum deviation. But that is
> what I am looking for, essentially.
>
> (Actually, I want to work backwards. I know what deviation I want, and I
> know the near and far distances. Given these I would select a focal length
> and a stereobase.)
>
> Boris
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