3
$\begingroup$

I would realize the papercraft of a geodesic sphere like this:

Illustration from Wikimedia Commons

It is the dual of the one discussed in THIS OTHER QUESTION .

Where can I find the printable nets, or the online resources to create them?

In the other discussion I learned that there are 3 classes of possible tessellations. One of these classes needs less pentagons to tessellate the sphere?

I don't exactly know how much faces I need, indicatively between 100 and 200. Which chord factors I should consider?

Thanks to all

$\endgroup$
2
$\begingroup$

Where can I find the printable nets, or the online resources to create them?

Sorry, don't know that. But once you know how to compute the relevant 3D coordinates, it should be possible to construct them yourself.

In the other discussion I learned that there are 3 classes of possible tessellations. One of these classes needs less pentagons to tessellate the sphere?

No: all of them are based on the symmetry of the icosahedron, so they all require 12 pentagons, matching the 12 vertices of the icosahedron.

I don't exactly know how much faces I need, indicatively between 100 and 200. Which chord factors I should consider?

For 100 to 200 faces in total, you'd want as many vertices in the dual. This means 196 to 396 triangles in that dual, computed using $\frac{6a-12}3$. The idea is that you count vertex-edge combinations, which are the same in both primal and dual. You can count these by multiplying the number of faces by the number of corners they each have, taking care of the 12 pentagons, and then dividing by the number of corners in a triangle. The easiest nets are Class I, I think, even though the Image you included in your question is Class III. For Class I, the number of triangles per icosahedron face has to be a square number. The only square number in the desired range is $4^2=16$ triangles per icosahedron face, leading to 320 triangles in total, which translates into 162 faces for your surface: 150 hexagons and 12 pentagons.

So here is what I'd do:

  1. Start with the 3D corners of the icosahedron. Place these corners on the unit sphere.
  2. Divide each icosahedron edge into 4 segments of equal length.
  3. Compute the inner points by joining matching points along the edges, and intersecting such lines inside the icosahedral face.
  4. Project all these inner points out onto the sphere.
  5. Compute the barycenter of each small triangle.
  6. Project these again out onto the sphere.
  7. Join the resulting points to form your hexagons and pentagons.
  8. Compute edge lengths from 3D coordinates.
  9. Place them one next to the other in the plane to get a printable grid.

Notice that 3. and 4. can be done together using homogenous coordinates. If you want to compute the intersection between some line $AB$ and another line $CD$, and you have 3D coordinates for both of them, simply compute $(A\times B)\times(C\times D)$ and normalize the result to length $1$. The first cross products $A\times B$ will give you a vector orthogonal to both $A$ and $B$. It is the normal vector of the plane spanned by these two points and the origin. The connection has to lie within that plane. The outer cross product gives you a vector orthogonal to two such normal vectors, i.e. a vector along the line where two planes intersect. By normalizing the result, you choose the point on that line which also lies on the sphere. Some care might be neccessary to get the sign right; you don't want the point on the opposite side of the sphere.

Steps 5. and 6. can be combined into summing three vectors and normalizing the result. 5., 7. and 9. are the steps where you have to get your combinatorics right, which will probably take the most time, even though there is nothing deep going on there.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.