Trias
Donkey with three behinds
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- Oct 1, 2008
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- 594
I have, and it is not much of a problem. Ever heard of a map?
Hexagon Tiles: Like them, love them, or RAGGGEEEE!!!
- Thread starter Jakerg23
- Start date
I have, and it is not much of a problem. Ever heard of a map?
I have my left hand sit on the W key (and thumb on the spacebar), because I turn off unit cycling in options. I think I am fairly unique in this regard though.
I order units with the mouse, sometimes tell units to wait with W and sometimes tell them to skip turn with space. I think I sometimes press F for sleep/wakeup too.
Anyway, I'm going off on a tangent... Point is I don't use the numpad either.
However, I did once try for a brief time issuing orders via voice commands. Having the numpad keys to assign to directions "North", "North West" etc. was handy. Too much work to set up though, and too awkward (and confusing to anyone within earshot) compared to simple mouse movements.
(I shouldn't laugh though, some people might have a disability and they use voice commands out of necessity)
SammyKhalifa
Deity
- Joined
- Sep 18, 2003
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- 6,308
Not sure. Do people have one of these "map" things on them at all times? Do they feel comfortable using it driving alone at high speeds in bumper-to-bumper traffic?
Trias
Donkey with three behinds
- Joined
- Oct 1, 2008
- Messages
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No, but most people do come equipped with a nifty little feature called memory. It also helps that an organic road layout has more unique features which you can recognize from the map and help you navigate. Road signs also are an amazing invention.
Or how about hexes vs. Squares?
The evidence is clear. Firaxis should have gone with pentagons. That way they could have tiled the sphere using tiles all the same size and shape!*
* Conditions apply.
skyline5k
Warlord
If you can drive at high speeds in bumper-to-bumper traffic and not crack up your car, then you're a far better driver than I am.
Sonnybonds
Prince
Ahriman
Tyrant
Ahovking
Cyber Nations
love it
Some people have claimed that if you put the pentagons near the poles you end up with a cigar shaped world. Is anyone able to do a 3d model of that? I can't imagine anything beyond the basic dodecahedron where each pentagon is successively broken up into larger and larger numbers of hexagons with a smaller pentagon in the middle.
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Please read the forum rules: http://forums.civfanatics.com/showthread.php?t=422889
Please read the forum rules: http://forums.civfanatics.com/showthread.php?t=422889
sturmhauke
Chieftain
Euler has been mentioned a few times in this thread regarding the use of hexagons to cover a sphere, so let me go into more detail. All polyhedra have an Euler characteristic, which is a constant that relates the number of vertices, edges, and faces. The Euler characteristic is represented with the Greek letter chi, but it looks like an x so I'll just use that. This is the Euler formula:
x = V - E + F
It has been proven that all convex polyhedra (that is, polyhedra without any "dents" or "spikes", and therefore suitable for representing a sphere) have an Euler characteristic of 2. Here is a list of proofs.
Now, let's assume for the moment that you can tessellate, or tile, a sphere with only hexagons. Each edge is shared between two hexes, and each vertex is shared between three. Therefore we have this set of equations:
E = 6F/2 = 3F
V = 6F/3 = 2F
Substituting back into the Euler formula gives us this:
x = 2F - 3F + F
x = 0
Therefore the Euler characteristic of a hexagon tiling is 0. However, it has been proven that all convex polyhedra have an Euler characteristic of 2. Therefore you cannot tile a sphere with only hexagons. This site has a proof that if you use pentagons and hexagons, there will always be 12 pentagons.
Now, that doesn't mean you have to use pentagons. You could use some really stupid shape instead. If you take the standard Civ map and put hexes on it, you can think of it as a polyhedron with however many hexes and two whatever-gons at the poles. Pentagons make for a more uniform grid, but then you're back to the problem of representing the grid on a plane, which I discussed earlier.
x = V - E + F
It has been proven that all convex polyhedra (that is, polyhedra without any "dents" or "spikes", and therefore suitable for representing a sphere) have an Euler characteristic of 2. Here is a list of proofs.
Now, let's assume for the moment that you can tessellate, or tile, a sphere with only hexagons. Each edge is shared between two hexes, and each vertex is shared between three. Therefore we have this set of equations:
E = 6F/2 = 3F
V = 6F/3 = 2F
Substituting back into the Euler formula gives us this:
x = 2F - 3F + F
x = 0
Therefore the Euler characteristic of a hexagon tiling is 0. However, it has been proven that all convex polyhedra have an Euler characteristic of 2. Therefore you cannot tile a sphere with only hexagons. This site has a proof that if you use pentagons and hexagons, there will always be 12 pentagons.
Now, that doesn't mean you have to use pentagons. You could use some really stupid shape instead. If you take the standard Civ map and put hexes on it, you can think of it as a polyhedron with however many hexes and two whatever-gons at the poles. Pentagons make for a more uniform grid, but then you're back to the problem of representing the grid on a plane, which I discussed earlier.
so you could have a hexagonal sphere of infinite size and it still has 12 pentagons?
You can imagine a cylinder made of hexagons, right? Like a sheet of hex graph paper where the right edge is bent around to meet the left edge. That's your basis. Presumably you can make caps for this tube out of pentagons, though i can't build that in my mind's eye.
sturmhauke
Chieftain
One way to do it is to take one of the hexagon-pentagon constructions that have already been shown, split it in half along the equator, and use those for the caps. If you flatten that construction out it would look somewhat like this map from upthread, except that the equatorial areas would be taller than the polar areas. Although at that point you might as well just ignore the complication of the cut-up polar area and its pentagons and just stick with a cylinder, or just forget about trying to represent a flattened sphere at all and just go for an actual sphere.
Thanks both of you. That explained it really well, and I'm surprised I didn't think of it myself.
By the way, i should have taken the image of "cigar" more literally (essentially a cylinder with spherical caps on each end). I just assumed people meant a more eccentric ellipsoid.
sturmhauke
Chieftain
Yes indeed.
You're welcome. This thread has given me a chance to work out my neglected math muscles.
It is actually quite easy to show that you need exactly 12 pentagons.
Let N=no. of hexagons
n= no. of pentagons.
E.g: In a soccer ball N=20 and n=12.
F=N+n
E=6N+5n/2
V=6N+5n/3
F-E+V = 2 and if you write this out you will get n=12. Also if you let N go to infinity then n still remains the same.
About a cigar shaped world, it also has an Euler Characteristic =2. Euler char remains unchanged under "nice" transformations and you can easily imagine squashing a cigar into a sphere.
A torus on the other hand has an E.C = 0 and in fact can be tiled by hexagons, but of course that is quite easy.
BB
Let N=no. of hexagons
n= no. of pentagons.
E.g: In a soccer ball N=20 and n=12.
F=N+n
E=6N+5n/2
V=6N+5n/3
F-E+V = 2 and if you write this out you will get n=12. Also if you let N go to infinity then n still remains the same.
About a cigar shaped world, it also has an Euler Characteristic =2. Euler char remains unchanged under "nice" transformations and you can easily imagine squashing a cigar into a sphere.
A torus on the other hand has an E.C = 0 and in fact can be tiled by hexagons, but of course that is quite easy.
BB
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