Well, you're talking to an ignoramus here anyway, so that doesn't help at
all.
The text spacing suffered in the translation but I guess I get you're point
anyhow.  What's perplexing is why the "x" ever changes if I'm understanding
this whatsoever.  Idea is that it gets seen together with whatever equation
then?  x=6 is also (2*x)=3 just simply sounds like division, (x/2)=3, in the
usual sense.  Maybe that's where other people are going wrong in the
thinking.
Oh!  Do you mean that "x" is the only thing checked in some way?  So despite
the resulting math it looks back at "x" alone??
Maybe I'm catching on, but you need not try and explain further unless
someone else is trying figuring this stuff out.

Bob

"Chris Huff" <chr### [at] yahoocom> wrote in message
news:chrishuff_99-D08175.16055207012000@news.povray.org...
> In article <38764dcf@news.povray.org>, "omniVERSE" <inv### [at] aolcom>
> wrote:
>
> > Whoa, did I not catch on to this reasoning at all!
> > If x=3 it will be x=3 when 2*x??
> > Okay, I'm lost.  Unless it's a average of the noise function....
>
> No, if the function returns a certain value at 6 units along the x
> axis(x=6), than if it is given 2*x instead of x, it will think x=6 when
> it is really x=3. So the value that showed up at 6 units now shows up at
> 3 units.
>
> N=What the function "sees"
> V=The real value
>
> With N=V:
>
>   0   1   2   3   4   5   6
> N |   |   |   |   |   |   |
>
>   0   1   2   3   4   5   6
> V |   |   |   |   |   |   |
>
>
> With N=2*V
>
>   0 1 2 3 4 5 6
> N | | | | | | |
>
>   0   1   2   3   4   5   6
> V |   |   |   |   |   |   |
>
> See what I mean now? If the function returned A at x=6 when given x, it
> will give A at x=3 when given 2*x.
>
> Sorry, I am not good at explaining this kind of thing.
>
> --
> Chris Huff
> e-mail: chr### [at] yahoocom
> Web page: http://chrishuff.dhs.org/

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