It's best understood by using it. Figuring out what bounded_by does, what
threshold changes and so on, can be learned through simple experimentation.

I guess the explanation of it would be 'math that forms a surface.' There
are points involved, but instead of say, x making a point at a certain
place, it makes a plane instead. Bill, I don't think I understand your
question fully. Any theory I have come up with regarding the isosurface I
learned through using it.

If you make an isosurface sphere, it is a sphere bounded by the bounded_by
object. If you make a y plane that has a sin(x) function on it, it is just
that, still bounded by an object. All isosurfaces must be bounded_by
something. If they weren't, they might take forever to render.

Chris Huff wrote:

> It is "show all points where the function is equal to the threshold."
> Bounding and clipping shapes will alter it if the surface is not totally
> contained in the bounding shape.

--
Samuel Benge

E-Mail: STB### [at] aolcom
Website: http://members.aol.com/stbenge

"Ni!"

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SamuelT. <STB### [at] aolcom> wrote :
> Bill, I don't think I understand your
> question fully.

    Actually you probably do, maybe even better than I do myself 8-) I am
notorious for running out questions before I know enough to ask questions
which could receive informative answers.

    I think Chris's answer while certainly true and exactly what I asked
for, is maybe one step too basic for me to understand all at once.

    But I'm getting there. The following is a little long and may be
tedious, but if someone could check me on it I would be grateful...

    Using the example "function(x+1.0)", going in the x direction, starting
at x=(-1.0), all points get the value 1.0 added to them. So at x=(-1.0), the
point equals 0.0 (less than the threshold of 1.0), then it moves right to
the middle, where x usually equals 0.0 but now equals 1.0, at that point,
all points are greater than the threshold and so aren't shown. (if x > 0.0
then Point > 1.0)

    Using that way of talking about it, the example
function{ (sin(x*3)+y)+y }" (parenthesis added for emphasis) and starting
from the inmost set of parenthesis(x*3), all x values are tripled. So you
get a similar effect as above, with one third the bounding box filled with
points whose values are greater than 1.0 and so are not shown. (if x > 0.333
then Point > 1.0)

    Then finding the sine of (x*3) you get values which start at -1.0 and
smoothly go to +1.0 then back to -1.0. This gives all points within the
bounding box a value of 1.0 or less and so all points are shown ( If you
drop the threshold below 1.0, you can see where the value of sin(x*3) is
greater than the threshold but never gets above 1.0 ).

    Adding y to the above begins to show the sine curve because at the
bottom of the bounding box where y is -1.0, adding y to it makes it -2.0, in
the middle height, where it is y=0, nothing is added to nothing. But for all
values of y which are greater than 0, y now equals twice as much as before.

    So the area where previously, x -almost- made it above 1.0, is now goes
just over 1.0 (where y>0) and is excluded by the threshold. Further up, as Y
gets exponentially greater, it becomes easier for x to be greater than 1.0
so the area that is excluded becomes wider, giving the look of a sine curve.

    Finally, adding y again flattens the curve even more, bringing all of
it's height into the bounding box. Adding more and more y makes flatter and
flatter waves.

    And I finally figured out that the doubling effect of abs and sqr are
because a negative number times a positive number equals a positive
number...

    All this makes sense to me. Which is either a sign that I am finally
getting a grasp on the whole thing or that I am way wrong. I don't know if
my math skills will go much further since I almost broke a blood vessel
trying to get this far, but if someone can confirm that I am right about
this, I'll consider this a working theory...

> Chris Huff wrote:
>
> > It is "show all points where the function is equal to the threshold."
> > Bounding and clipping shapes will alter it if the surface is not totally
> > contained in the bounding shape.
>
> --
> Samuel Benge
>
> E-Mail: STB### [at] aolcom
> Website: http://members.aol.com/stbenge
>
> "Ni!"
>
>

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Wow. You lost me quickly with all that math. Good luck building your theory.

Bill DeWitt wrote:

> SamuelT. <STB### [at] aolcom> wrote :
> > Bill, I don't think I understand your
> > question fully.
>
>     Actually you probably do, maybe even better than I do myself 8-) I am
> notorious for running out questions before I know enough to ask questions
> which could receive informative answers.
>
>     I think Chris's answer while certainly true and exactly what I asked
> for, is maybe one step too basic for me to understand all at once.
>
>     But I'm getting there. The following is a little long and may be
> tedious, but if someone could check me on it I would be grateful...
>
>     Using the example "function(x+1.0)", going in the x direction, starting
> at x=(-1.0), all points get the value 1.0 added to them. So at x=(-1.0), the
> point equals 0.0 (less than the threshold of 1.0), then it moves right to
> the middle, where x usually equals 0.0 but now equals 1.0, at that point,
> all points are greater than the threshold and so aren't shown. (if x > 0.0
> then Point > 1.0)
>
>     Using that way of talking about it, the example
> function{ (sin(x*3)+y)+y }" (parenthesis added for emphasis) and starting
> from the inmost set of parenthesis(x*3), all x values are tripled. So you
> get a similar effect as above, with one third the bounding box filled with
> points whose values are greater than 1.0 and so are not shown. (if x > 0.333
> then Point > 1.0)
>
>     Then finding the sine of (x*3) you get values which start at -1.0 and
> smoothly go to +1.0 then back to -1.0. This gives all points within the
> bounding box a value of 1.0 or less and so all points are shown ( If you
> drop the threshold below 1.0, you can see where the value of sin(x*3) is
> greater than the threshold but never gets above 1.0 ).
>
>     Adding y to the above begins to show the sine curve because at the
> bottom of the bounding box where y is -1.0, adding y to it makes it -2.0, in
> the middle height, where it is y=0, nothing is added to nothing. But for all
> values of y which are greater than 0, y now equals twice as much as before.
>
>     So the area where previously, x -almost- made it above 1.0, is now goes
> just over 1.0 (where y>0) and is excluded by the threshold. Further up, as Y
> gets exponentially greater, it becomes easier for x to be greater than 1.0
> so the area that is excluded becomes wider, giving the look of a sine curve.
>
>     Finally, adding y again flattens the curve even more, bringing all of
> it's height into the bounding box. Adding more and more y makes flatter and
> flatter waves.
>
>     And I finally figured out that the doubling effect of abs and sqr are
> because a negative number times a positive number equals a positive
> number...
>
>     All this makes sense to me. Which is either a sign that I am finally
> getting a grasp on the whole thing or that I am way wrong. I don't know if
> my math skills will go much further since I almost broke a blood vessel
> trying to get this far, but if someone can confirm that I am right about
> this, I'll consider this a working theory...
>
> > Chris Huff wrote:
> >
> > > It is "show all points where the function is equal to the threshold."
> > > Bounding and clipping shapes will alter it if the surface is not totally
> > > contained in the bounding shape.
> >
> > --
> > Samuel Benge
> >
> > E-Mail: STB### [at] aolcom
> > Website: http://members.aol.com/stbenge
> >
> > "Ni!"
> >
> >

--
Samuel Benge

E-Mail: STB### [at] aolcom
Website: http://members.aol.com/stbenge

"Ni!"

Post a reply to this message

SamuelT. <STB### [at] aolcom> wrote in message
news:37C5FF63.4B74E991@aol.com...
> Wow. You lost me quickly with all that math. Good luck building your
theory.
>

    It boils down to adding 1 to a fraction of 1 . It's figuring out what
fraction of 1 is where that causes all the mental sweat.

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Thanks Samuel this is just what I need, I'll have a look at it as
soon as this render is done, it's been parsing for a day, and
started rendering now at 0 pps.  So in a few days I'll be able to
have a look. 

"SamuelT." wrote:
> 
> Hello all. I'm in the process of making a website.
> 
> I have one gallery and a complete tutorial ready. The tutorial is for
> learning how to make isosurfaces. I encourage all who want to learn
> about isosurfaces to visit it. Enjoy!
> 
> The url is:
> 
> http://members.aol.com/stbenge/
> 
> --
> Samuel Benge
> 
> STB### [at] aolcom
> 
> "While you were sleeping
> he went on keeping the final line in his mind."
>  -Tortoise and the Hare, The Moody Blues

-- 
Cheers       email: mailto:sjl### [at] ndirectcouk
Steve           or: mailto:zer### [at] startat

%HAV-A-NICEDAY Error not enough coffee.   0 PPS

web: http://www.ndirect.co.uk/~sjlen/   
 or: http://start.at/zero-pps

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