Hi.

Well is more than good for me :)

Thanks for both the reply and the help.

Salu2.
Ricardo.


news:cjameshuff-D095A7.22565126052004@news.povray.org...
> In article <40b55d56@news.povray.org>,
>  "Ricardo Miguel Pereyra" <no### [at] spamcom> wrote:
>
> > Somebody can tell me the maths for tile a plane with regulars hexagons
> > registered in a circle of x radius leavin a space of x/10.
>
> My Spanish is nowhere good enough to explain this...hopefully this is
> clear enough:
>
> Take regular hexagons inscribed in circles of radius R, and put them in
> rows along one of their major axii...point to point. Each row of
> hexagons will consist of hexagons separated by a gap equal to the length
> of one side of the hexagons, which is equal to the radius. (Divide the
> hexagon into 6 equilateral triangles to prove this.)
>
> This gives you 3 radii between centers along this direction: you have a
> hexagon at < 0, 0>, one at < 3*R, 0>, one at <6*R, 0>, etc...
>
> The next row is offset "vertically" by half the height of a hexagon,
> which is sin(pi/3)*R, and "horizontally" by 1.5 radii.
>
> union {
>     #local R = 0.1;
>     #local XN = 10;
>     #local YN = 20;
>     #local J = 0;
>     #while(J < YN)
>         #local K = 0;
>         #while(K < XN)
>             sphere {< K*3*R + mod(J, 2)*1.5*R, sin(pi/3)*R*J, 0>, R}
> //          sphere {< K*3*R + 1.5*J*R, sin(pi/3)*R*J, 0>, R}
>             #local K = K + 1;
>         #end
>         #local J = J + 1;
>     #end
>     texture {
>         pigment {color rgb 0.98}
>     }
> }
>
> -- 
> Christopher James Huff <cja### [at] earthlinknet>
> http://home.earthlink.net/~cjameshuff/
> POV-Ray TAG: <chr### [at] tagpovrayorg>
> http://tag.povray.org/

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