>Hope this is worth your time. If the image works for you as is then, heck
>it works! But try this as an experiment :-)


Looks good too. I can't really tell the difference though. I had to increase
the Z because the cross wouldn't all fit in -99. Here it is for your viewing
pleasure, Harold: your own personal Jesus. :)

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Attachments:
Download 'haroldjesus.jpg' (27 KB)

Preview of image 'haroldjesus.jpg'

Tony,
Thank you for my own Jesus! The image you posted is in parallel
format, but when I downloaded it and rearranged it for cross-eye
viewing it was good. Much more depth in this one. Thanks for the
effort.

Harold

TonyB <ben### [at] panamaphoenixnet> wrote in message
news:387fd349@news.povray.org...
> >Hope this is worth your time. If the image works for you as is then, heck
> >it works! But try this as an experiment :-)
>
>
> Looks good too. I can't really tell the difference though. I had to
increase
> the Z because the cross wouldn't all fit in -99. Here it is for your
viewing
> pleasure, Harold: your own personal Jesus. :)
>
>
>

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Much better!

--
Homepage: http://www.faricy.net/~davidf/
___     ______________________________
 | \     |_       <dav### [at] faricynet>
 |_/avid |ontaine      <ICQ 55354965>

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This explaination was in regard to red/cyan 3D pair but it seems to
work for cross-eyed stuff too.
From my post back on 5/14/99 (I was making sure I had Grimdude's
explaination right).
//---------------------------------------------------------
I think I did it right, it seemed to work. But in case I 
just got lucky;
let me make sure I have your directions straight. So it's:

1. Find the distance between the camera location and look_at.
   ie. if the look_at is <0,0,0> and my camera location is
   <0,.5,-4> then the result I want is along z which = 4 .

2. Divide the result of step 1. by 11.8(12.0 respectively).
   I get 0.33893 

3. Divide step 2. by 2(the number 2, Not step2).
   which = 0.1694915 

4. The result of step 3. is added to the x value in the camera
   location for the right.tga. And subtracted for the left.tga.
   left.tga = camera{ location<-0.1694915,.5,-4>
   right.tga= camera{ location<0.1694915,.5,-4>


-- 
Phil
...coffee?...yes please! extra sugar,extra cream...Thank you.

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>Thank you for my own Jesus! The image you posted is in parallel
>format, but when I downloaded it and rearranged it for cross-eye
>viewing it was good. Much more depth in this one. Thanks for the
>effort.

No problem. :) But, would you run that by me again? The original image I
posted was in _what_ format? How does one correctly view the other format?

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This will only work if your objects are close to the look_at point. If
there are objects much closer the viewer will not be able to fuse
the image. Also, is there a reason for the numbers you use? It
produces a separation more than twice the 1:30 ratio, which has
been widely recognized for decades as the normal separation for
humans. The results of that formula may often work but will result
in "stretch" and more people will have difficulty viewing the stereo
pair.

Harold

Phil Clute <pcl### [at] tiacnet> wrote in message
news:387FFA97.8C48E76B@tiac.net...
> This explaination was in regard to red/cyan 3D pair but it seems to
> work for cross-eyed stuff too.
> From my post back on 5/14/99 (I was making sure I had Grimdude's
> explaination right).
> file://---------------------------------------------------------
> I think I did it right, it seemed to work. But in case I
> just got lucky;
> let me make sure I have your directions straight. So it's:
>
> 1. Find the distance between the camera location and look_at.
>    ie. if the look_at is <0,0,0> and my camera location is
>    <0,.5,-4> then the result I want is along z which = 4 .
>
> 2. Divide the result of step 1. by 11.8(12.0 respectively).
>    I get 0.33893
>
> 3. Divide step 2. by 2(the number 2, Not step2).
>    which = 0.1694915
>
> 4. The result of step 3. is added to the x value in the camera
>    location for the right.tga. And subtracted for the left.tga.
>    left.tga = camera{ location<-0.1694915,.5,-4>
>    right.tga= camera{ location<0.1694915,.5,-4>
>
>
> --
> Phil
> ...coffee?...yes please! extra sugar,extra cream...Thank you.

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This is really for 3D red/blue glasses but it has worked for me in 
stereo too so I thought I would mention it anyhow...
-- 
Phil
...coffee?...yes please! extra sugar,extra cream...Thank you.

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Ok, I mean it will work often. The same psychophysics apply to anaglyphic
stereo (red/blue) as to polarized, cross-eye, HMD, LCS, etc., it is all
stereoscopic imaging and involves presenting each eye with an image
rendered from an angle within the range we can fuse into a single
stereoscopic perception. I just think you'll get better results in the long
run if you use the 1:30 ratio.

Harold

Phil Clute <pcl### [at] tiacnet> wrote in message
news:38800691.1991CFEF@tiac.net...
> This is really for 3D red/blue glasses but it has worked for me in
> stereo too so I thought I would mention it anyhow...
> --
> Phil
> ...coffee?...yes please! extra sugar,extra cream...Thank you.

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Tony,

Oh, well stereo images can be presented in many formats, there are
two "side-by-side" formats: parallel- where the left eye image is on
the left and right on the right; cross-eyed where the left eye image
is place on the right. Some people (like me) can more easily view a
cross-eye presentation than a parallel presentation, especially if the
width of the image is more than 2.5 inches (~65mm). Parallel
presentation is pretty much limited to images about that size, where
as cross-eyed presentation doesn't have that limitation. On the other
hand many people can't cross their eyes very easily. To view a
parallel stereo pair you have to relax your eyes. Technically, you
uncouple your convergence and accommodation, meaning that you
focus close but your eyes are parallel as if looking at infinity, this
is not normal and it takes practice for most people- remember the
"hidden image" random dot stereogram fad? If those were easy for
you then you will be able to free-view parallel stereo pairs.
To view cross-eye pairs you, well, cross your eyes. If you cross your
eyes for a parallel pair then you will see "pseudo-stereo" which is
the term for reversed pairs where depth is reversed. This is difficult
for the brain to interpret, so it looks odd or just flat.

Many stereo photography web sites present the images as a trio,
Left-Right-Left. That way the viewer can use parallel free-viewing
or cross-eyed. The version of the Jesus that you made for me has the
left eye image on the left, so it is a parallel rather than cross-eye pair.
I know that because when I cross-eye view it I see it in pseudo-stereo.

Harold

TonyB <ben### [at] panamaphoenixnet> wrote in message
news:387ffde9@news.povray.org...
> >Thank you for my own Jesus! The image you posted is in parallel
> >format, but when I downloaded it and rearranged it for cross-eye
> >viewing it was good. Much more depth in this one. Thanks for the
> >effort.
>
> No problem. :) But, would you run that by me again? The original image I
> posted was in _what_ format? How does one correctly view the other format?
>
>

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>remember the
>"hidden image" random dot stereogram fad? If those were easy for
>you then you will be able to free-view parallel stereo pairs.

I was never able to see them and ended up throwing away the only book I had.

>Many stereo photography web sites present the images as a trio,
>Left-Right-Left. That way the viewer can use parallel free-viewing
>or cross-eyed. The version of the Jesus that you made for me has the
>left eye image on the left, so it is a parallel rather than cross-eye pair.
>I know that because when I cross-eye view it I see it in pseudo-stereo.

Oh! Now I get it! I put the left on the left and the right on the right for
both of these images. Silly me! I thought that when you said left and right
they had to go on their respective sides, but now I understand. <Hurries to
try to view them correctly. "Wow!" he exclaims!>


Thanks for the explanation.

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