DIVIDE SPHERE WITH EQUAL TRIANGLES 1

[Sarnath]

Dear friends,
i have a typical problem. the crisp of the problem is as follows:
Am trying to construct a sphere with sets of equilataral triangles. the sphere is of dia

11000 and the triangle size to be used should have a max height of 600mm. I tried to fix

triangles on to sphere but it does not get divided properly. Am every time forced to use a

of different sizes in each layers. Please note that its not meshing, its embedding the

triangle on to the sphere surface. well you see its like our football only difference is

that its a hexagon. there is a calculation for embedding hexagons on to spheres but i dont

know that, i feel that could lead me to the solution.
is there any body who can help me?
thanks.

Sarnath

sarnath@sarengg.com

 

[chicopee]

Is the triangle with a max height of 600mm a spherical or plane triangle? Is the max height measured along the surface of the sphere or is it a projected length? Also are dealing with right triangles?

 

[IRstuff]

A football, if I remember correctly, actually has pentagons embedded between the hexagons.

In any case, have you looked at geodesic dome geometry?

http://www.grunch.net/synergetics/domes/domegeo.html

TTFN

 

[Kapitan]

There are numerous ways of triangulating a sphere but using only plane equilateral triangles can never achieve it. If any number of plane equilateral triangles are assembled the resultant set will always be planar, it needs some non-equilateral types to be introduced into the group, to induce some curvature. The truncated icosahedron (football) case is made up of twelve pentagons and twenty hexagons. Dividing each of these polygons up into triangles will naturally give six equilaterals in the hexagon and five isosceles in the pentagon. However, if the centre point of each polygon is to be on the spherical surface the point has to be pulled out of the polygon plane creating a new set of only isosceles triangles in each case. So what all this means is that a facetted spherical form can be made of of just two types of isosceles triangles.
If the polygons have unit sides then the isosceles triangles will have a unit base length. This will give give a radius of 2.478 for the sphere that all the points at the corners of the triangles lay on, or if the sphere is of unit radius then the triangle base will be 0.4035 long.
The apex angles of the isosceles triangles are 58.583 and 70.731 degrees.
Other versions can be made up of characteristic triangles (left and right hand) of one size.

 

[TheTick]

I once read an article about a program that was developed to evenly distribute a set number of points around a sphere. The crux of the problem is that there is no formula for this process.

Gravity is a harsh mistress.

 

[IFRs]

Check ou this program. It is wonderful for designing triangulated spherical frames.

http://www.cadreanalytic.com/cadregeo.htm

 

[Sarnath]

Mr.Chicopee
the triangle is spherical and its the length of triangle when developed. the triangles are not flat. i know that may be the triangle when developed may not have straight edges but never mind.
thanx for your reply and sorry for delayed response.

 

[Sarnath]

Thanks Mr.Kapitan, that was quiet useful piece of info. well i actually did the same, i took the football model only, where i did as you said. as you rightly said we can make a sphere with sets of hexagon and pentagon, but as you said one set of the triangles (for pentagon) will have 70.731 degree angle on the isosceles triangle, i wonder if the problem can be achieved only with only one set of isoceles triangles.
any way thanks for you suggestion and sorry for late response.

 

[Kapitan]

The sphere can be triangulated with equilateral triangles but only in a primitive way. Technically, a Tetrahedron (4 equilateral triangular faces) is a geodesic dome, so is a Octahedron (8 faces) and the nearest to a sphere using these triangles is the Icosahedron (20 faces).
A good place to start if using only isosceles triangles is one of the Archimedean Polyhedra and sub-divide its faces into triangles and then pull the vertex of each set out to the sphere. Inside the 'Football' or Truncated Icosahedron is the Dodecahedron (12 pentagons). From this framework a reasonably smooth dome of 60 isosceles triangles can be constructed, this would require 2 types of hub (12 of 5 elements) and (20 of 6). This particular example is called a Pentakis Dodecahedron and a GOOGLE search will give plenty of geometric data and an idea of what it will look like.
This triangulation can be taken further but the thing will become too expensive and time consuming to build so an optimum number of triangles has to be set.
Another method of constructing a model is to visit this website :

http://www.netlib.org/polyhedra/

It is very informative, for numerous polyhedra there are files containing all the vertices coordinates. They can be pasted into a file and imported into a 3D modeller. From the website file extract only the last block of numbers up to the :EOF statement, these are X,Y and Z values. Near the beginning of the file is the line ( :svertices ), the first number on the line below it is the total of points defined in the block of data to use.
In my case it was CATIA V5, the wireframe built on the points and then the triangular faces on that. Any scaling function can be incorporated at the input data file stage or later on when the 3D model is complete.

 

[GregSmith]

If you want to build up a sphere with equilateral triangles you can only use the tetrahedron. All the other regular solids use different face geometries. Clearly a 600mm sided triangle is not going to give a tetrahedron inscribed on a sphere of 11000mm diameter.

If approximations are allowed, however, the situation is much better since the distortion of plane triangles reduces when the radius of the sphere becomes much larger than the triangle's side. I would start with the pentagons of the icosahedron, dividing them up with the best approximations to equilateral triangles, then project thos out on to the sphere.

Good luck.

GregSmith

 

[Kapitan]

Greg,
As stated earlier the Tetrahedron is just one of the Five Regular Platonic Polyhedra. The vertices of each of them inscribe the sphere, the faces of the tetrahedron, octahedron and icosahedron are Equilateral Triangles. It is the dodecahedron that has pentagon faces, the fourth is the cube.
There is no need for any approximations or distortions in this exercise, it is Pure geometry.

 

[TheAnswerGuy]

Sarnath,

If the above answers to your question does not give you the answer. Perhaps you could call Walt Disney World. Spaceship Earth in EPCOT Center is designed and built under the same premis as you are looking for. WDW has a technical office located at the base of the Spaceship Earth exhibit that answers questions just like you are asking. They should be able to give you the formula.

Yes! I live in Orlando, FL

 

[IRstuff]

EPCOT Center's display is the only complete public Geodesic dome building

TTFN