Kepler's Conjecture

[archermoo]

As a note to the people posting about this on the Rules forum (of you are looking here that is): Only Steve Long can reply to messages there.

That and the original topic isn't really a rules question, but more of a "how does this work" question, so isn't really appropriate there, so I'm sure Steve will get around to moving it here eventually anyway.

To answer the question (how many Hero hexes will fit in a 10 meter diameter 65 meter long tube): 608. Which leaves 1 meter at the end of the tube without hexes in it. And some amount of space around the inside edge of the tube unoccupied as well.

:)

[Cancer]

What precisely do you mean by "fit in" there? My reading of the question left me confused.

Is this taking a stack of circular cross-sections spaced equally along the tube's length, and taking the number of hexes that can be inscribed entirely within the 10-meter diameter circle, and multiplying it by the number of layers in the stack?

An alternate interpretation is to draw hexes on the curved surface of the tube, and do a coaxial "stack" of such hexified tubes with each successive tube being one hex-height different in radius.

[archermoo]

What precisely do you mean by "fit in" there? My reading of the question left me confused.

Is this taking a stack of circular cross-sections spaced equally along the tube's length, and taking the number of hexes that can be inscribed entirely within the 10-meter diameter circle, and multiplying it by the number of layers in the stack?

An alternate interpretation is to draw hexes on the curved surface of the tube, and do a coaxial "stack" of such hexified tubes with each successive tube being one hex-height different in radius.

You just hurt my brain. :)

I was assuming the question was asking about how many three dimensional hexes (2 meter across hex 2 meters high) fit entirely within the cylinder. I don't think I can wrap my brain around your question. :doi:

[Cancer]

Ah, OK, you took the fastest way out, which is simply dividing the volume of the cylinder by the volume of a game-hex prism.

I interpreted it as being the literal task of dividing the tube into segments of hexsheet (perhaps curved hexsheet) which, appropriately stacked, did the best job of letting you map the volume of the tube. That is a brain-injuring task, and one with multiple ways to go about it. Which way is best probably depends on what's going to be done inside that tube.

[archermoo]

Ah, OK, you took the fastest way out, which is simply dividing the volume of the cylinder by the volume of a game-hex prism.

I interpreted it as being the literal task of dividing the tube into segments of hexsheet (perhaps curved hexsheet) which, appropriately stacked, did the best job of letting you map the volume of the tube. That is a brain-injuring task, and one with multiple ways to go about it. Which way is best probably depends on what's going to be done inside that tube.

I will certainly admit that at my heart, I am lazy, and therefor will generally take the quick way out if possible. :)

[Sean Waters]

I'm assuming it is a Hero question and so, how much AoE will we need to fill up that cylinder, assuming we bought he AoE with 'any area'.

737.

Mind you, what do I know, I'm a lawyer.

[Sean Waters]

Kepler was probably on about whole hexes, but we don't have to. We are just interested in filling the space :D

Do you know I'm not even sure Kepler ever played Hero. Not even first edition. B+%*£&d.

Probably played DnD in which case it was probably a percentile roll x 10.

[archermoo]

Kepler was probably on about whole hexes, but we don't have to. We are just interested in filling the space :D

Do you know I'm not even sure Kepler ever played Hero. Not even first edition. B+%*£&d.

Probably played DnD in which case it was probably a percentile roll x 10.

Actually, he was on about spheres rather than hexes. :)

[Sean Waters]

Actually, he was on about spheres rather than hexes. :)

So what game system uses spheres?

[archermoo]

So what game system uses spheres?

Reality. :)

The rules take a while to learn. And they're a bit obscure...

:D

[Cancer]

... but you can get university credit for learning the rules.

[archermoo]

... but you can get university credit for learning the rules.

But not necessarily. :)

[Sean Waters]

The rules are easy. You take a cylinder and pack in as many spheres of the right size as you can.

The problem is not doing it, the problem is proving you can do it.

Fortunately reality does not require your workings.

[Cancer]

But that omits the possibility of flaws in the packing / crystal structure. That "possibility" is a certainty under most physically realizeable situations....

[Sean Waters]

But that omits the possibility of flaws in the packing / crystal structure. That "possibility" is a certainty under most physically realizeable situations....

Hero.

[Checkmate]

Oh my bad, I thought this was the free pie thread.

[jguerin]

You mean free Π I think? :)

[Gummibear]

I'm assuming it is a Hero question and so, how much AoE will we need to fill up that cylinder, assuming we bought he AoE with 'any area'.

737.

Mind you, what do I know, I'm a lawyer.

Heya Sean, are you filling up all the little spaces that won't hold a whole hex around the circumference of the cylinder? What calculation/formula did you use?

The following was my own post reposted here since I obviously can't read a rules forum faq list. :hush:

"Well unfortunately I don't have permissions or some such thing so here's one possible response.

To find the volume of a cylinder of hexes 10m in dia and 65 m tall:

Volume = π • r² • height. And Volume = π • r² for the circle

Take each "slice to be one hex high and you get a cylinder that is 32.5 hexes high. I am doing the hexes as cylinders not as spherical objects. Then calculate to volume of one "slice" which is pi(2.5^2) or 19.64 which you then multiply by 32.5 and you get 638.3 hexes.

Is this what you were looking for jgeurin? Or was this already answered and my measly permissions didn't allow me to notice the response? "


So my math may be off but now I'm just curious where the difference is. :nonp: