change hexagons to octagons

expuddle

Chieftain
Joined
Sep 30, 2010
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9
Don't know if it's been suggested before, but why not use octagons instead of hexagons? It seems like the balance between 1UPT and the number of units is just not right, more tiles = better balance no?
 
And how are you going to divide the map into octagons? (Maybe you can post a picture ;))
 
It has been posted before, (not by me), that it isnt possible to make a coherant map out of octogons.
 
It has been posted before, (not by me), that it isnt possible to make a coherant map out of octogons.

Of course it applies to regular octagons, you can divide a surface into octagons of some other shapes, but it won't look any good.
 
The only ways you can tile a plane with a single regular shape is with triangles, squares or hexagons. That's it. Octogons don't work.
 
This is a pretty simple result in mathematics, namely that there are exactly three tessellations of the plane by regular polygons. If I'm not mistaken, I can show this is true using some fairly simple geometry and integer factorization.

Any point in a plane tessellated must have equal vertex angles at the point. We can divide 360 in such a way that we choose the possible angles to be those less than 180 (because this is the only case which really makes sense, lest we divide it simply into a half-plane), and note the following:

First, any non-trivial n-gon must have an interior angle of at least 60 degrees, that is a triangle. So we note that the only values we care about amongst the following list of possible angle values are those greater than or equal to 60 and less than 180. By factorizing 360, and using simple combinatorics you have this:

1,2,4,5,8,9,10,12,15,18,20,24,30,36,40,60,90,120

as the possible interior angles of your tessellating polygons, note by the previous we can reduce this list now to 60, 90 and 120.

And we now refer to the fact that a regular n-gon has interior angle equal to (n-2)*180 *(1/n) and this must give:

(n1 - 2)*3 = n1
(n2 - 2)*2 = n2
(n3 - 2)*(3/2) = n3

For which we get the solutions n1 = 3, n2 = 4, n3 = 6.

This establishes that the only possible regular tessellations are by equilateral triangles, squares and regular hexagons.

EDIT: I apologize if I forgot some possible angle values when writing them down, I was trying to compute them fairly quickly, and might have missed some particular combination of factors, either way it wouldn't affect the outcome.
 
why stop at octagons? I say change the tiles to cubes.
 
I vote for phallus-shaped tiles!
 
The movement implied by these (impossible) octagonal tiles would basically be the same as if you had squares and could move both diagonally and orthogonally. It would never catch on...
 
Lets have circles for tiles!
 
Triangles, cities can use only 4 tiles!!!

LT-300.300.jpg
 
Seriously speaking, I was thinking about "truly spherical" maps divided into hexagons and pentagons, like a soccer ball.

This was probably discussed at this forum before, can anyone who is better at geometry than me tell me if it's possible to "scale" such maps, i.e. create maps of different sizes (number of tiles)? I guess they would look like different types of fullerenes (buckyballs), but can they scale indefinitely?
 
Seriously speaking, I was thinking about "truly spherical" maps divided into hexagons and pentagons, like a soccer ball.

This was probably discussed at this forum before, can anyone who is better at geometry than me tell me if it's possible to "scale" such maps, i.e. create maps of different sizes (number of tiles)? I guess they would look like different types of fullerenes (buckyballs), but can they scale indefinitely?
Yes, it is possible to create a spherical shape with regular polygons, but the largest is a 20-faced one using equilateral triangles. Larger than that, you have to mix polygons, which begs the question of how useful can it be when the game rules have to take into account facets that don't have equal numbers of sides?
 
PawelS, if I understand your question correctly, the answer is no. A buckyball has precisely the same amount of sides regardless of the sphere it envelops, the size of tiles is a function of the radius of this sphere. So "scaling", i.e., adding more tiles of the same pattern will not produce the same shape. As a matter of fact, it will more than likely close at very ugly boundaries.
 
Strangely enough... here's a recommendation i've made in August'07 to Stardock devs for an upcoming GC3;

planethexasaa0.png


Gamers do think alike in many ways!

Brad hates hexagons, so he'll probably really hate triangles. :)
 
Haven't you heard? The frog is trying to re-think his most elemental design choices to start a magical war with conventions when the GC3's dev team is ready for action.
Learned his lesson, i guess.

Seriously, Wardell can do whatever he pleases - i don't mind. Let's just hope the bad aura around EWM launch will dissipate soon enough for the next project StarDock has on the budget predictions table.
 
Haven't you heard? The frog is trying to re-think his most elemental design choices to start a magical war with conventions when the GC3's dev team is ready for action.
Learned his lesson, i guess.

Seriously, Wardell can do whatever he pleases - i don't mind. Let's just hope the bad aura around EWM launch will dissipate soon enough for the next project StarDock has on the budget predictions table.

GC3 better not end up as "Elemental in Space"...although I love EWoM.
 
Um, civ4 and previous civ versions sort of were octagons, because of the permitted diagonal moves.

Octagons may not tessellate a 2d plane, but you can pretend they do.
 
Yes, it is possible to create a spherical shape with regular polygons, but the largest is a 20-faced one using equilateral triangles. Larger than that, you have to mix polygons, which begs the question of how useful can it be when the game rules have to take into account facets that don't have equal numbers of sides?
Not true at all: you can have an infinite number of triangles forming a "sferical" (closed soccer ball) shape. The only thing you have to take into account is that you have 12 pentagons formed by triangles, in the structure - that's all! The rest can be triangles forming hexagons, and they can be INFINITE and still balling up.
 
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