Old School Monsters

[hcobb]

The gelatinous cube has a volume of 10x10x10 = 1000 ft^3

A ten foot high megahex has a volume of 16x7x10 = 1120 ft^3

Ergo the gelatinous cube is a megahex sized monster.

[Skarg]

It has a 3-hex counter though. I expect part of the reason may be that in a corridor of megahexes, how would you move a megahex creature less than three hexes forward?

[phiwum]

A hex is four foot across. Now, it's not clear to me what "across" means, since a line segment from border to border through the center of the hex varies in length according to where it intersects (how near a vertex).

But you concluded a hex is 16 sq ft.

If the side of a hexagon is 2' long (which is what I assume and which would make the vertex-to-vertex measure 4'), its area is

3 sqrt(3)/2 * 2^2

which is about 10.4 sq. ft, not 16 sq. ft.

I mean, nitpicking, sure, but if we're going to do math...

Of course, that makes it considerably larger than a 10' megahex volume.

[Skarg]

4 feet is side to side or center point to center point.

[phiwum]

Okay, then that makes vertex to vertex 8/sqrt(3) or about 4' 7 1/2".

And the length of a side is 4/sqrt(3), making the area

3 sqrt(3)/2 * (4/sqrt(3))^2 = 8 sqrt(3) sq ft

or about 13.9 sq. ft. Still less than 16, but more than I was estimating. 1000 ft^3 is more than 10 * 7 * 13.9 by a small amount (the latter is about 970 sq ft). A megahex is pretty darn close to 10 feet cubed.

So, roughly, the volume is right for a megahex creature. For game play, it has to be reasonably mobile in a megahex corridor. A four hex critter would do fine, flipping it every hex. Or a three hex critter does fine for a game that fudges conversion rates every time we change the scale of the map (from room to labyrinth to village, say).

That fudging of scales brought me no end of misery when I first started mapping things for this game. In the end, I surrendered and promised myself not to think too hard about it. I'm sure that sometimes, I call a distance between two villages six miles and other times five miles, but the players don't seem to notice. Nonetheless, I feel like a lesser person for doing so.

ETA: And now, thanks to Skarg, I realize that all of the furnishings I've put in the maps have been about two inches off per foot in each dimension. I thought that table was 4' across, only to learn it's 4'8". Who builds a table that's 4' 8"? It's crazy. My chairs are all too big. My chests are 5" too long. Dammit.

[Terquem]

A four foot hex (edge to edge not vertex to vertex) has an area of 13.8564 feet

This was calculated in AutoCad drawing a hexagram circumscribed around a circle with a radius of 2 feet.

the area of a 1000 cu foot creature ten feet high fit into hex shaped areas would be 7.2 hexes (approximately) the actual dimensions of a megahex across its widest point would be 12 feet and the the 10x10x10 creature would fit easily inside that dimension but would then encroach upon the four hexes attached to the inner sections of the megahex.

I even drew the mega hex and imposed a 10 x 10 square onto it. That was fun

[phiwum]

Originally Posted by Terquem

A four foot hex (edge to edge not vertex to vertex) has an area of 13.8564 feet

This was calculated in AutoCad drawing a hexagram circumscribed around a circle with a radius of 2 feet.

the area of a 1000 cu foot creature ten feet high fit into hex shaped areas would be 7.2 hexes (approximately) the actual dimensions of a megahex across its widest point would be 12 feet and the the 10x10x10 creature would fit easily inside that dimension but would then encroach upon the four hexes attached to the inner sections of the megahex.

I even drew the mega hex and imposed a 10 x 10 square onto it. That was fun

It's a lot easier than to use AutoCad to calculate the area. It's six equilateral triangles. It's trivial to calculate the area that way.

Now, to be fair, I didn't do the calculation because Google has made me less than a real man and I looked up the formula. So it's still a triviality, but I wimped out.

[Terquem]

a lot of things are easier than using AutoCAd. I use AutoCad a lot because it helps my brain work out problems in peculiar ways that are the way my brain wants to work.

[phiwum]

Eh, different strokes. Obviously, it gave you quite a good estimate for the area. For all I know, it actually represented the exact area internally, involving sqrt(3) and everything.

I have a background in mathematics, so I tend to go for that kind of reasoning. In fact, I feel kinda bad that I looked up the formula for area when it's so damned obvious that you take six times the area of an equilateral triangle.

[Terquem]

The difficult part of a hand calculation is determining the congruent sides measurements of a hex with a four foot across dimension