POV-Ray : Newsgroups : povray.binaries.images : 6 volumes : Re: 6 volumes Server Time
1 Aug 2024 08:20:23 EDT (-0400)
  Re: 6 volumes  
From: milco2006
Date: 18 Jan 2009 17:55:01
Message: <web.4973b2be1026618d3b1a12740@news.povray.org>
> Can you explain a bit how, I fail to understand it's construction.
> Is there some ratio from green to blue/red ?

Firstly let me apologise as the term superellipsoid is slightly incorrect as it
is really more of a rounded box in the strictest sense. The reason I left it
split into blue and red and green is that this splits the shape down into 8
parts of spheres (green), 12 parts of cylinders (red), 6 small boxes (blue) and
one big box (which is inside). This identifies the different parts of which the
shape is composed.

> I have nothing against a superellipsoid, as long as you provide its
> parameters (and hopefully they are some kind of 'nice', even a cubic
> root of a fraction might match), and by the same way demonstrate that
> the volume is exactly and mathematicaly 7pi/3.

As I mentioned the different parts are varying shapes. To calculate the volume
one merely requires to add up the volumes of the whole or part shapes. To get
an exact formulation I used 'l' to be the length between the centres of the
corner spheres and 'r' to be their radius. We also know that l+2*r=2*R which
allows us to construct a cubic equation solely in l. This can be solved to find
required values and thus the shape can be constructed. The solution is not just
a little nasty, but rather very nasty but does exist and is expressible using
cube roots and fractions of polynomials in pi.

The nastiness of the solution somewhat detracts from the problem at hand
somewhat. So although I have presented a shape that has the correct volume it
is not a very satisfying solution.


> (and not an approximate value close enough)

The value can be calculated exactly due to the ability to split the shape into
standard geometrical sections. And as the solution to the cubic can be given in
a series of constants and mathematical operators it can technically be
calculated without approximation.

I am still keen to find a more natural shape that fulfils your conditions
however I hope this explanation is helpful.

Malcolm


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