🚀 go-pugleaf

Paul Nuber wrote:
> 
> uncle Al wants it to work in windows, he may not have a fortran compiler.
> It is more easy with python.  You can download it from www.python.org.
> Read the tutorial and here you go, it allows you to do other cool things too.
> In the python shell you can type (translate from fortran below)
> 
>           from math import *
>           ff = open("te.xyz","w")
>           a = 4. ; c=5. ; r=150.
>           iamin = int(r/a) + 1 ; icmin = int(r/c) + 1
>           for i in range(iamin):
>               for j in range(iamin):
>                   for k in range(icmin):
>                       x = a*i + a*j*cos(2.*pi/3.)
>                       y =       b*j*sin(2.*pi/3.)
>                       z = c*k
>                       if x**2 + y**2 + z**2 < r**2 :
>                             ff.write("%8s%12.6f%12.6f%12.6f\n" % ("Te",x,y,z))
>           ff.close()
> 
> (ctrl-Z to exist shell)
> It takes about 30 seconds on my pentium 4 machine to generate
> the required file (about 10 Mbytes).
> best
> paul

I was looking for an elegant general solution given *.CIF data in
existing code.  Silly me.  I am entirely happy with a brute force
solution for the specific case.  The "obvious" solution for a regular
lattice is to take an existing large Cartesian rendering, look at the
constant x,y,z increments at its far fringes (significant figures),
and run a triple arithmetic increment from the origin to to any radius
desired.

The program will roar to its conclusion whether compiled from Fortran,
C++, Python, or (hold your nose) Basic.  No trig at all.

"8^>)  Thanks folks.  To the keyboard!

-- 
Uncle Al
http://www.mazepath.com/uncleal/
 (Toxic URL! Unsafe for children and most mammals)
"Quis custodiet ipsos custodes?"  The Net!