Low Prim Hexagon

I want to make bucky balls (buckminster -fullerine)- i need a perfect hexagon to do this.

A single prim Hexagon is impossible but....

A two prim hexagon may be possible.

I've tried squashing, sizing but still cant get opposite sides parallel.

Any ideas?

Thanks in anticipation of your help

I havent tried it, but wouldnt you just use two cubes and taper the outer edge of each by the same amount then link them by the longer edge?

If it is buckyballs you seek, you may have an easier time if you form the lattice rather than a solid. There are scripts out there which rez geodesics automatically out of cylinders and spheres.

Two-prim hexagons are relatively simple. Top-size two boxes to around 0.6-0.65, place bottom to bottom, then stretch as needed to ensure the sides match. They may not ever *perfectly* match, but for most purposes, they will do just fine.

ehm, Hexagons are 2D shapes. Maybe an hexahedron?

Selkit's answer is the way to go to get a 2-prim hexagon, but let me interject to offer some numbers to help you make it equalateral (or as close to equalateral as SL math will allow). Geometric law tells us that the height of a hexagon is roughly .866 times it's width, and that the length of each side is 1/2 the width. Therefore, the height of each of your two peices must be .433 times the width, and the top-size must be .50. So, to make the largest possible 2-prim hexagon, you'd use 2 cubes, each 10M wide by 4.33M tall, with their tops shrunk to .50. Place them bottom to bottom, and you've got a perfect hexagon, 8.66M tall.



As for taking a bunch of these and making them into a buckyball, that's gonna be considerably more complicated. Correct me if I'm wrong, but I don't think I've ever seen one made of just hexagons. I've always seen them with pentagons mixed in to fill in the gaps (or with all pentagons, like a soccer ball) since hexagons don't possess the proper angles to cover a sphere completely by themselves.

This one, for example, is made of 20 hexagons and 12 pentagons:



I don't feel like taking all the time to test whether it's possible with some amount of hexagons, so let me ask you if you know something I don't. Is there a magic number of hexagons that works out to make a hexagon-only buckyball?


EDIT: If you do need pentagons as well as hexagons, here are the measurements you'll want to use. Assuming you're going with the large hexagon size I just outlined, the dimensions of this 2-prim pentagon should work to fill the gaps. If you're going smaller, you can do the reduction math yourself, or else just start big and then select it all and shrink everything at once in SL. 12 of these pentagons, combined with 20 of the hexagons above will give you a 64-prim buckyball.



Keep in mind though that SL is problably going to have rounding issues with this. I'm not inworld to test right now, but I doubt that a top size of .618 will be acceptable. It may round to .62, or maybe even .60 or .65. You'll probably have some not so tight seams as a result, but it will be close.

I'm assuming you can figure out how to jigsaw puzzle them together on your own. If you need help with that though, ask. I can give you some pointers on how to do it by hand, as I'm sure can some other people, and there are plenty of people capable of explaining how to write a script to place all the pieces automatically.

No, it isn't possible. Tessellating hexagons alone will always give you a flat plane (there is an argument involving Euler characteristics if anyone is interested).

Ah, I should have realized that. Makes perfect sense.

Hey, since your math skills probably beat mine by at least tenfold, care to look over those measurement numbers I posted, and double check that they're correct?

I get the same measurements for the hexagon. For the pentagon with side length 5, I get 2.939 instead of 3, 8.090 instead of 8 and 4.755 instead of 4.77. This gives a topsize on the bottom prim of 0.618 (I think that's really the golden ratio there), which you certainly won't be able to get perfect. 0.62 will be pretty good though.

Ah believe that it isna possible wi`out exactly twelve pentagons.

If there are 3 faces around each vertex then yes, if we allow distorted faces and use more than 3 faces around a vertex then it can be done with more than 12 pentagons. For example three hexagons around a vertex together have 12 external edges, as do 4 pentagons around a vertex, so one could swap out the three hexagons for four pentagons.

Thanks for looking it over. Sounds like I made a rounding error. I remember I did originally have .618 for that top size, but somehow I must have rounded 8.09 to 8 when I double checked it myself, and then pythagorus gave me .625. The rest of the errors followed suit after that. Thanks again for checking it. I'll fix the picture.