Akismet spam checking has broken down, so comments are now restricted to registered users.
This curve-fitting thingy is one of several projects on which I’ve made progress in rare fits over several years. It ran into two big snags. I haven’t found how to determine which gridpoints are within the pen-width of a blending arc; two methods that ought to work don’t. (What would help: tutoring in drawing pictures in a MacOS display, so that I might have a better idea where they go wrong. And a pony.)
The other snag is this: For each pair of arcs, there is an infinite family of blending arcs; how to choose the osculation points to minimize rapid changes in curvature, while meeting the gridpoint constraints? ( . . more . . )
In Watch on the Rhine (1943; screenplay by Dashiell Hammett from a play by Lillian Hellman) the penniless Count remarks,
Blecher, we do not like each other.
The Nazi to whom he hopes to sell information replies,
But that will not stand in the way of our doing business.
To link such a sentiment to fascism implies a remarkable kind of snobbery.
In a dream, while strolling among bookshops and the like, I chance to meet someone who hates me but is constrained to be polite. With a venomous smile that person asks, the better to avoid them, on what streets one is most likely to meet me. I name two streets in Oakland, one in Berkeley and one in San Francisco.
On waking, I remember that none of those streets exists.
I noticed that Shapeways had 13 models of the roundest of the fullerenes (one of the 1812 forms of C60), but none of the less regular forms; so I made some.
Each of the white pieces has mirror symmetry; the red pieces are chiral. Not shown (because it hasn’t been printed yet): the blue set, which is a reflection of the red set. The idea is that you buy both red and blue if and only if you count reflected chiral forms separately.
These figures have 12 pentagons and up to 8 hexagons. They include the two smallest forms with no nontrivial symmetries, and the two smallest with no ‘peaks’ where three pentagons meet.
I could have used this a week ago.
def fccstack(): newlimit = 0 while True: oldlimit = newlimit newlimit += 1 # extend z for x in xrange(oldlimit): for y in xrange((x+newlimit+1)&1, oldlimit, 2): yield (x,y,oldlimit) # extend y for x in xrange(oldlimit): for z in xrange((x+newlimit+1)&1, newlimit, 2): yield (x,oldlimit,z) # extend x for y in xrange(newlimit): for z in xrange((y+newlimit+1)&1, newlimit, 2): yield (oldlimit,y,z) g = fccstack() for dummy in xrange(512): p,q,r = g.next() print "%d %d %d\t%d" % (p,q,r, p+q+r)
This lists coordinates of sites in a face-centred cubic lattice, filling the smallest cube that contains the number of sites required.
My bundle of 19 fullerenes is arranged in a face-centred cubic lattice, each ball occupying one of the 24 even-numbered cells of a 4×4×3 array (and parts of the adjacent cells). The spacing of the grid planes is determined by the sums of the radii of the circumspheres of nearest-neighbor balls.
There appears to be much more ‘daylight’ between balls than there needs to be. That’s not a problem here, as the size of the bounding box is not near the limits of the process; but still it got me thinking about how to tighten the packing. ( . . more . . )