[Space] Mapping Large Flat Areas

[Brandy]

Here's something that might be of interest:

Average distance between two randomly-chosen points in a unit square.

[Brandy]

Originally Posted by RustedKitsune

Or should I just presume that the 2d ftl space has an effective height of 1 and then use the equations for a 3D space?

If you have hyperspace coordinates in the form (x,y) for two planets, the calculation for distance between them is d=√((x2-x1)^2+(y2-y1)^2).

[RustedKitsune]

Originally Posted by Brandy

If you have hyperspace coordinates in the form (x,y) for two planets, the calculation for distance between them is d=√((x2-x1)^2+(y2-y1)^2).

Yes, I know that, but what is the generalisation for when I don’t have coordinates? Since I’m looking at random distribution of a very large number of points, and I don’t care about creating their coordinates yet, is there a 2d equivalent to the 3D formula given in GURPS Space?

And stack exchange has been useless to me, since I’m trying to write my material as if somebody other than me was going to be running the game, and I would rather not force calculus down my throat or somebody else’s.

[Rupert]

I don't know the answer to this problem, but here's what I understand it to be:

We have a two-dimensional space (so x,y co-ordinates) upon which are scattered a large number of points, at random. How do we determine the average distance between closest neighbours?

[RustedKitsune]

Originally Posted by Rupert

I don't know the answer to this problem, but here's what I understand it to be:

We have a two-dimensional space (so x,y co-ordinates) upon which are scattered a large number of points, at random. How do we determine the average distance between closest neighbours?

That is exactly what I’m asking. However, asking it in Stack Exchange just got me a link to a calculus lesson, and I’m hoping for something like the equation for this problem in 3D space (you know, the side par on page 72 of Space that takes up half the page?).

Edit: also, hey Rupert. Nice to see someone from TML.

[thrash]

Originally Posted by RustedKitsune

Yes, I know that, but what is the generalisation for when I don’t have coordinates? Since I’m looking at random distribution of a very large number of points, and I don’t care about creating their coordinates yet, is there a 2d equivalent to the 3D formula given in GURPS Space?

Ah, okay. Given an area A and number of stars N, the mean area per star is A/N. The tightest tiling that will cover an area is hexagonal, so you just need the distance between the centers of a regular array of hexagons of area A/N: 1.07*sqrt(A/N).

[Rupert]

Reading that box, and seeing where the formula came from, I think that the 2D equivalent would be based off the area of a circle, etc. rather than volume, and would thus come out at about: 0.8 x (A/N)^0.5

I could be utterly wrong, though.

[RustedKitsune]

Originally Posted by thrash

Ah, okay. Given an area A and number of stars N, the mean area per star is A/N. The tightest tiling that will cover an area is hexagonal, so you just need the distance between the centers of a regular array of hexagons of area A/N: 1.07*sqrt(A/N).

Originally Posted by Rupert

Reading that box, and seeing where the formula came from, I think that the 2D equivalent would be based off the area of a circle, etc. rather than volume, and would thus come out at about: 0.8 x (A/N)^0.5

I could be utterly wrong, though.

Thank you. Thrash, your formula gives me about 1.5 hexes between stars at 50% density. Rupert, yours gives about (CORRECTED) 1.13 hexes in the same situation. I think Thrash has provided the answer here.
Off to finish my work now, thank you both.

[Agemegos]

Originally Posted by Rupert

Reading that box, and seeing where the formula came from, I think that the 2D equivalent would be based off the area of a circle, etc. rather than volume, and would thus come out at about: 0.8 x (A/N)^0.5

I could be utterly wrong, though.

The p.d.f. for nearest neighbour at distance x, for a uniform random distribution (i.e a Poisson process) is given here as 1-e^-(density * volume). For the case of a Poisson process on a 2-D plane that ends up implying that the average nearest-neighbour distance is 0.5/sqrt(number/area).

[RustedKitsune]

Originally Posted by Agemegos

The p.d.f. for nearest neighbour at distance x, for a uniform random distribution (i.e a Poisson process) is given here as 1-e^-(density * volume). For the case of a Poisson process on a 2-D plane that ends up implying that the average nearest-neighbour distance is 0.5/sqrt(number/area).

Average distance at 50% density, 1.4 hexes.
Huh, close enough to Thrash.