Wondering if anyone can explain/confirm my understanding of dihedral angles relating to woodworking. as an example, if we are talking about a truncated icosahedron, the published dihedral angles are:
hexagon-hexagon face: 138.1896851
hexagon-pentagon face: 142.6226319
Now these are the face to face angles for each of the two required polygons making up the faces of the truncated icosahedron. If I want to construct a wooden model with 20 hexagons and 12 pentagons made of square sticks, how do I find the required bevels on the hexagons and pentagons? (see attached figures) After a LOT of confusion I think this is one way to do so.
First, take the hexagon-hexagon face and divide in half which gives 69.09.... then subtract this from 90 to get the required bevel angle on the hex-hex sides of each hexagon. This will be 20.9.....
Doing the same for the hexagon-pentagon faces gives a bevel of 18.68.....
Is this the correct method to find these angles? PLEASE HELP!!!
Wondering if anybody could show me how to calculate the face angles of the dodecahedron. I recently got a copy of Mathematical Models and it has a table with some of the face angles calculated. However it does not show how they got the angles. Also does anybody know what a re-entrant angle is?
bobby hello again,
yes i believe that it is the core of the elevated icosidodecahedron. from another site - "It is a nonconvex construction of 120 equilateral triangles arranged with icosahedral symmetry. Three-sided and five-sided pyramids have been erected on the faces of the underlying icosidodecahedron, which consists of twenty triangles and twelve pentagons."
I need to start from the very beginning. Hopefully after someone can walk me through the procedure for this model I will then be able to tackle any polyhedra on my own. So I guess to start with I need to figure out the angles need for the regular icosidodecahedron.
after seeing george harts website and finding da vinci's polyhedra drawings, I have become obsessed with trying to figure out how to build one for myself. I figured out how to construct an intersecting 5 tetrahedra model with help on a timber framers forum, however I still cannot make sense of how they derive the angles needed. (I do have a BS in Computer Science, but even with this background all of the abbreviations and the way they find the necessary angles make no sense to me?) I would like begin by making a model of the elevated icosidodecahedron as shown in the drawing by da vinci. sorry about giving you the wrong name bob.
Can anyone help me or point me in the right direction to find the angles needed in order to build a wood model of an icosidodecahedron, truncated icosahedron, and/or the 3V form of an icosahedron. I found Soapy Joes posting on the subject and could not gather how to figure out the angles from his discussions. I would like to be able to derive the angles so that I can later construct the rest of da vincis polyhedra.