Hi Wei, It occurs to me your question about how to make the hkcage hexagons more uniform in size when projected to a sphere is closely related to a simulation of unstretchable cloth that I tried last week. I've attached a picture where I draped a cloth made up of a uniform hexagonal mesh of triangles over a sphere (a 100 x 100 grid of triangles). All the equilateral triangles of the cloth surface keep exactly their same size. Of course the cloth billows and ripples to cover the sphere. This is just another way of seeing that the hkcage hexagons that are exactly of uniform size on the icosahedral surface with its 20 flat triangular faces cannot be mapped onto a sphere while keeping all the hexagons the same size, since the sphere has curvature and the triangle faces of the icosahedron are flat. But this also suggests a way to get a more uniform size of the hkcage hexagons on a sphere. The cloth doesn't conform to the sphere because the circle of radius r from the center of the flat cloth has too large a circumference to hug the sphere tightly, it has to ripple. So how about I just uniformly shrink the circle of radius r in the cloth so it matches the circumference of the sphere? My cloth triangles won't be equilateral anymore but will be as close as I can get. We could do the same thing with the flat mesh made by hkcage on an icosahedron, about each 5-fold vertex. At a 5-fold vertex there are 5 equilateral triangles of an icosahedron, each with 60 degree angles, so 300 degrees along the cone surface about the 5-fold axis. Flattened on the sphere that needs to become 360 degrees. The idea is to map polar coordinates on the flat icosahedron centered on the 5-fold vertex to polar coordinates on the sphere. The 300 degree angle will be expanded uniformly to 360 degrees. But unlike the draping cloth we also need to map the radius from the icosahedral 5-face cone to the sphere in a special way that depends on the angle if we want the sphere mesh to have exact icosahedral symmetry. The radius mapping along a ray towards the 3-fold axes and the 2-fold axes nearest the 5-fold will be different. One idea is to take the flat triangle formed by neighbor 5-fold, 2-fold and 3-fold axes (there are 120 such triangles) and map it to the spherical triangle where those 3 axes intersect the sphere. I already described how to map the polar angle (30 degrees -> 36 degrees), and for the radius we could map the line segment from 5-fold triangle vertex to a point in the triangle edge between 2-fold and 3-fold vertices so it is uniformly stretched to the longer arc on the spherical triangle, ie the arc where the sphere intersects the plane through the sphere radius, 5-fold vertex, and point on the 2-fold to 3-fold triangle edge. That description was probably a bit hard to follow, but basically it maps the flat icosahedron triangle bounded by 5-fold, 2-fold and 3-fold vertices to an equivalent spherical triangle trying to preserve uniformity of angles about the 5-fold axis, and also uniform stretch along rays. The math to do that mapping is probably a page or two of Python code. It would be interesting to see what the result would look like. I would expect the hkcage hexagons would have more uniform overall size but they will be skewed somewhat. Tom
On Dec 9, 2020, at 8:28 PM, Wei Zhang <zhangwei@umn.edu> wrote:
Hi Tom,
Thanks so much. I have found the Python file of Chimera on linux. I currently do not have ChimeraX yet. I think if one version can be worked out, it is not too hard to make the other version working. I do not know Python. I would need to ask my collaborator to look into the code.
I am thinking for the cases k=0, or h=k, It is not too hard to implement a spherical icosahedral cage. This is the diagram for a planner icosahedron definition: <Screen Shot 2020-12-09 at 9.02.59 PM.png> We just need to think the H axis is an arc between two neighboring 5 fold vertices, and K is another arc connecting the origin and the other nearby 5 fold vertice. So the positions of 1, 2, 3, 4 ... would be defined by angles or length on the arc, rather than the length on the line connecting the two 5-fold points.
A quick test can be made by modifying the code of this routine (Linux Chimera): Tools => Higher-order Structure => Icosahedron Surfaces. When the sphere factor = 1. The subdivision can be considered to be a subdivision of the arc that connects two 5 folds. I am not sure what is the best method to draw the lattice lines when the sphere factor is >0 and less than 1, some kind of interpolation function has to be developed.
I do not know how to handle the cases of skewed icosahedron either, ie. h and k are different. It seems very difficult to define the references points and determine how to draw those parallel lines on the spherical surface. I am thinking we have to start with a big spherical triangle. <Screen Shot 2020-12-09 at 9.56.03 PM.png> For example, if we want to define h=3, k=2 lattice (red triangle), we need to draw arcs on a sphere according to the purple triangle. The question is if we think the red triangle is the outline of a spherical triangle of three 5-fold vertices, where the vertices of the purple triangle are. The current version of Chimera can draw any icosahedral lattice beautifully. I think this must have been worked out for the planner cases. The python code must have some clues. I have to think about it more.
Best regards,
Wei
On Wed, Dec 9, 2020 at 5:48 PM Tom Goddard <goddard@sonic.net <mailto:goddard@sonic.net>> wrote: Hi Wei,
You are right the Chimera and ChimeraX hkcage command "sphere" option simply moves the cage vertices outward to lie on a sphere. The sphere passes through the 12 vertices of a plain icosahedron so those 12 vertices don't get moved. But the midpoint of one of the triangular faces of the icosahedron is at radius 0.79 (= sqrt( (1+2/sqrt(5))/3 )) if the vertices are at radius 1 (ie the ratio of inner icosahedron radius to outer radius is 0.79) so projecting a shape near the middle of a triangular icosahedron face will expand its size by 21% relative to if it is near an icosahedron vertex.
It is an interesting problem to try to reduce that distortion. It looks impossible to remove it. For that you would need an isometric map from an icosahedron triangle to a spherical triangle, but no such isometry exists since the triangle is flat and the sphere has curvature 1/R. Let me know what you figure out.
The code for computing the cages in Chimera is all Python and is in your Chimera distribution, on Linux it is here
chimera/share/IcosahedralCage/cages.py
You would be better off working on the successor of Chimera, called ChimeraX. If you improve hkcage we won't update Chimera as we no longer add new features to it since all development is on ChimeraX. The ChimeraX hkcage command is in the ChimeraX distribution, on Linux it is here
chimerax/lib/python3.8/site-packages/chimerax/hkcage/cage.py
Tom
On Dec 8, 2020, at 6:58 PM, Wei Zhang <zhangwei@umn.edu <mailto:zhangwei@umn.edu>> wrote:
Hello,
I use the command hkcage to generate an icosahedral lattice on a sphere:
hkcage 5 7 sphere 1
However, this only expands the strict icosahedral lattice (ie when using sphere=0) radially to match a spherical surface. As a result, the sizes of the hexagons and pentagons are not the same. The effect would be very dramatic if the parameter sphere is larger than 1.
I am wondering if the code can be modified to generate a truly icosahedral lattice on a spherical surface, or if that piece of code is open to access?
Thank you and best regards,
Wei
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