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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