Couldn't figure out how to get the checkered plane to show up.

-- 
Tim Cook
http://empyrean.scifi-fantasy.com
mirror: http://personal.lig.bellsouth.net/lig/z/9/z993126

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

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Attachments:
Download 'mirror.jpg' (92 KB)

Preview of image 'mirror.jpg'

What exactly am I looking at? :)

Rune
--
3D images and anims, include files, tutorials and more:
Rune's World:  http://rsj.mobilixnet.dk (updated May 20)
POV-Ray Users: http://rsj.mobilixnet.dk/povrayusers/
POV-Ray Ring:  http://webring.povray.co.uk

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Rune wrote:
> What exactly am I looking at? :)

The inside of a mirrored cube with spheres at the corners and
centre, with a camera FOV of 170 degrees.  Recursion limit 64.

-- 
Tim Cook
http://empyrean.scifi-fantasy.com
mirror: http://personal.lig.bellsouth.net/lig/z/9/z993126

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

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Oh yeah, and a moderate focal blur (1.05 apeture).

There's also a checkered plane below the cube, but it causes
rather unpleasent feedback unless no_reflection is set, in
which case it doesn't show up at all...

-- 
Tim Cook
http://empyrean.scifi-fantasy.com
mirror: http://personal.lig.bellsouth.net/lig/z/9/z993126

-----BEGIN GEEK CODE BLOCK-----
Version: 3.12
GFA dpu- s: a?-- C++(++++) U P? L E--- W++(+++)>$
N++ o? K- w(+) O? M-(--) V? PS+(+++) PE(--) Y(--)
PGP-(--) t* 5++>+++++ X+ R* tv+ b++(+++) DI
D++(---) G(++) e*>++ h+ !r--- !y--
------END GEEK CODE BLOCK------

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I once did something like this myself, and came up with this interesting
thought:

let's say you placed a point (sphere with radius zero) on every integral
point in 3D space. (like, at the origin, and at <1,0,0>, and at <-2,1,0>,
and at <4052,153,398>, you get the idea), so you have an infinite 3D grid of
points. Now, you remove the point at the origin and place yourself there.

Is it possible to look in any direction without seeing a point? (meaning,
can you shoot a ray out from the origin that doesn't intersect a point on
the grid?)

 - Slime
[ http://www.slimeland.com/ ]

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For irrational solutions, yes.  For Diophantine ones, no.

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip


"Slime" <slm### [at] slimelandcom> wrote in message news:3cf40019@news.povray.org...
> I once did something like this myself, and came up with this interesting
> thought:
>
> let's say you placed a point (sphere with radius zero) on every integral
> point in 3D space. (like, at the origin, and at <1,0,0>, and at <-2,1,0>,
> and at <4052,153,398>, you get the idea), so you have an infinite 3D grid of
> points. Now, you remove the point at the origin and place yourself there.
>
> Is it possible to look in any direction without seeing a point? (meaning,
> can you shoot a ray out from the origin that doesn't intersect a point on
> the grid?)
>
>  - Slime
> [ http://www.slimeland.com/ ]
>
>

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> Is it possible to look in any direction without seeing a point?

In fact, if you looked in "most" directions you wouldn't see a point (that
is, with 100% probability). There are only countably many points, but
uncountably many directions to look in.

Anders

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"Slime" <slm### [at] slimelandcom> wrote in message
news:3cf40019@news.povray.org...
[snip]
> and at <4052,153,398>, you get the idea), so you have an infinite 3D
grid of
> points. Now, you remove the point at the origin and place yourself
there.
>
> Is it possible to look in any direction without seeing a point?
(meaning,
> can you shoot a ray out from the origin that doesn't intersect a
point on
> the grid?)

I would say no. To guarantee missing any point you would have to look
along a path parallel to the rows of the grids.

Alf

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Perhaps I should clarify my answer from before.  For Diophantine equations,
there is an infinite number of solutions (these are all-integer equations) with
the most notable ones being 3-4-5 triangles and their siblings.  Let's take a
two dimensional case to see how it works.
    Imagine a flat grid of points that are all spaced at one unit.  These points
will be in the intersections of all 1 unit vertical and horizontal grid lines.
Now, place yourself at the origin (0,0) and draw an imaginary line in some
arbitrary direction.  Will you hit a point on our grid?  Remember that all
points have coordinates that are only integer X and Y values, no fractional
values allowed!
    The answer is that only angles that are generated by integer ratios will
intersect a point. Any angle with an X and Y that are integer values will be
guaranteed to hit a point, and in fact an infinite number of points in that
direction.  Let's pick a test case- x = 71, y = -54.  Clearly that angle will
hit that point, but it will also hit every copy that is a multiple of that
triangle.  You will hit the point that is 2x those values (142,-108) and three
times those values (213,-162) and so on forever.
    So any integer sized triangle will have an infinite set of points that form
a line with respect to the origin.  In fact, any line in this scenario will
either intersect an infinite set of points that fits this criterion or no points
at all.  How can it hit no points at all?
    Simple- pick an angle that is generated by an irrational value.  By that, I
mean a value that cannot be calculated with a single division.  It doesn't even
have to be an integer division, as you can see with a little thought.  Let's
pick e as our value.
    Since e cannot be generated by a pair of integers (one dividing the other)
there cannot be a point that is on our grid that will generate an angle
corresponding to e.  We can come arbitrarily close, but that is another matter.
Since any irrational number cannot be represented in a Diophantine equation (muc
h less a division of two numbers at all!) it therefore represents an angle for
which there can be NO solution that results in intersecting a point on our grid.
Furthermore, since there is an infinite number of irrational values, there is an
infinite number of lines from our origin that will not intersect any point on
our grid ever.
QED.

Cheers!

Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip


"Alf Peake" <alf### [at] peake42freeservecouk> wrote in message
news:3cf4ed0c@news.povray.org...
> "Slime" <slm### [at] slimelandcom> wrote in message
> news:3cf40019@news.povray.org...
> [snip]
> > and at <4052,153,398>, you get the idea), so you have an infinite 3D
> grid of
> > points. Now, you remove the point at the origin and place yourself
> there.
> >
> > Is it possible to look in any direction without seeing a point?
> (meaning,
> > can you shoot a ray out from the origin that doesn't intersect a
> point on
> > the grid?)
>
> I would say no. To guarantee missing any point you would have to look
> along a path parallel to the rows of the grids.
>
> Alf
>
>
>

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> blah blah blah blah blah

Nice analysis! I had always considered the answer to this self-thought-up
question to be 'no', but never considered the posibility of lines with
irrational slopes. Very interesting.

 - Slime
[ http://www.slimeland.com/ ]

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