How to specify the basises of triangular lattice?

>
> Dear Johnson,
>     Recently I focus on the triangular lattice and its reciprocal lattice.
> By using the lattice properties from "Tutorial", just like:
>
>      (set! geometry-lattice (make lattice (size 3 3 no-size)
>                               (basis1 (/ (sqrt 3) 2) 0.5)
>                               (basis2 (/ (sqrt 3) 2) -0.5)
>
> and I fill the geometries manually to see its position in cartesian
> coordinate:
>      (set! geometry (list
>                      (make cylinder
>                        (center 0 0 0) (radius 0.2) (height infinity)
>                        (material (make dielectric (epsilon 12))))
>                      (make cylinder
>                        (center 1 0 0) (radius 0.1) (height infinity)
>                        (material (make dielectric (epsilon 12))))
>                      (make cylinder
>                        (center 0 1 0) (radius 0.3) (height infinity)
>                        (material (make dielectric (epsilon 12))))
>                      (make cylinder
>                        (center 1 1 0) (radius 0.4) (height infinity)
>                        (material (make dielectric (epsilon 12))))
>
> then I obtain the epsilon.h5 file and convert it to png. I found it
> inexplicable!
> where the cylinder centered at (1 0 0) with radius of 0.1 is actually
> located at (1, 0) in cartesian coordinate;
> the cylinder centered at (0 1 0) with radius of 0.3 is actually located at
> (0.5, sqrt(3)/2);
> and the cylinder centered at (1 1 0) with radius of 0.4 is actually located
> at (1.5, sqrt(3)/2).
>
 Even the basis vector coordinate is changed from right-hand type to
> left-hand type, why does this happen?
>
> I also alternate the basises to
>
> (set! geometry-lattice (make lattice (size 3 3 no-size)
>                          (basis1 1 0)
>                          (basis2 0.5 (/ (sqrt 3) 2))
> ))
>
> and obtain the same picture.
>
> Besides, I use the (set! ensure-periodicity false), but it come no effect.
>
>
> Thank you for reply!
> Regards,
> 陈浩.
>

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