05-28-2019, 05:58 PM | #21 |
Join Date: May 2005
Location: Oz
|
Re: [Space] Mapping Large Flat Areas
It seems unlikely. With a 50% chance of a star in each adjacent hex there is 98.4375% chance of a nearest-neighbour at distance one, and if there isn't one at distance 1, then 99.9755859% chance of one at distance 2 (i.e. an overall probability of 1.56% chance that the nearest neighbour is at 2). The chance that the nearest neighbour is further than that is negligible (0.00038%), so the expected nearest-neighbour distance is 1.016 hexes (for stars constrained to be in the centres of hexes).
__________________
Decay is inherent in all composite things. Nod head. Get treat. |
05-28-2019, 06:08 PM | #22 | |
Join Date: Oct 2018
|
Re: [Space] Mapping Large Flat Areas
Quote:
Edit: I like all three formulas, but I’m using Thrash’s to make it obvious that a neighbouring star could be 1 or 2 hexes away. Close enough for vaguely mapping an area. |
|
05-28-2019, 06:22 PM | #23 |
Join Date: Oct 2018
|
Re: [Space] Mapping Large Flat Areas
As an example, consider a map of 36x36 hexes, for an area of 1,296 hexes. Average stellar density is 50%, for 648 systems. We’re interested in the habitable ones, which average 14.8% of all systems (these worlds are far more common in Traveller than GURPS), for 95.9 worlds. Eh, 96 worlds.
For all three formula, average distance between them is: Thrash: 3.93 hexes. (4 hexes) Rupert: 2.94 hexes (3 hexes) Agemegos: 0.55 hexes Umm... Age? I think this needs to be checked. It is 0.5/(sqrt(N/A)), right? |
05-28-2019, 06:45 PM | #24 | |
Join Date: May 2005
Location: Oz
|
Re: [Space] Mapping Large Flat Areas
Quote:
Three things to note. 1. The formula is for the case of a uniform distribution where any point can have a star at it, not for the queer case in which stars are constrained to be at the centres of hexes. 2. You are using "hex" as both a unit of distance and a unit of area, where your hex-area is not the square of your hex-distance. That is an error. The formula I gave you is for distance and area in consistent units: used the way you used it it gives distances in terms of the square root of the area of a hex. 3. The formula I gave is for actual Euclidean distance, not for the measure produced by counting hexes.
__________________
Decay is inherent in all composite things. Nod head. Get treat. Last edited by Agemegos; 05-28-2019 at 06:57 PM. |
|
05-28-2019, 06:55 PM | #25 |
Join Date: Oct 2018
|
Re: [Space] Mapping Large Flat Areas
You’re right, I was inputting it wrong. Now I just need to figure out how to convert it’s value into hexes if each hex face to opposing face =1
|
05-28-2019, 07:13 PM | #26 | |
Join Date: May 2005
Location: Oz
|
Re: [Space] Mapping Large Flat Areas
Quote:
There are eighteen other hexes within two hexes of a hex, and if each has 50% of a star with 14.8% of a habitable planet that means 0.926 chance that each is uninhabitable and 25% chance that they are all uninhabitable. There is 75% of a neighbour at distance one hex or two hexes, which makes it seem to me that an average nearest-neighbour distance of 2.94 or 3.93 hexes seems implausibly large.
__________________
Decay is inherent in all composite things. Nod head. Get treat. |
|
05-28-2019, 07:46 PM | #27 | |
Join Date: Oct 2018
|
Re: [Space] Mapping Large Flat Areas
Quote:
Although, if you have an circle of hexes, r2.5 hexes, that’s 19 hexes, for 9.5 systems, and 1.406 inhabitable planets. Going out to 3.5, that’s 31 hexes, 15 systems, and 2.22 inhabitable planets. Although that distance is just between easily inhabitable worlds, not all worlds. |
|
05-28-2019, 10:06 PM | #28 |
Join Date: May 2005
Location: Oz
|
Re: [Space] Mapping Large Flat Areas
To have a nearest neighbour at distance x you have to have a neighbour at distance x and no neighbours at any distance less than x.
A hex has 6x other hexes at distance x for integer x, so the probability of a neighbour there is 1-(1-p)^(6x). A hex has 3x(x-1) other hexes closer than x, and the probability that they are all empty is (1-p)^(3x(x-1)). So the probability of a nearest neighbour at distance x is (1-(1-p)^(6x))((1-p)^(3x(x-1))). So the expected nearest-neighbour distance is the limit of an infinite sequence with the nth term n(1-(1-p)^(6n))((1-p)^(3n(n-1))). That's a pig of a thing, but it converges pretty sharply for moderate p, so we can solve it numerically. I summed the first hundred terms for each of a selection of probability values up to 95%, which will put the errors in the 12th decimal place even for the lowest P tabulated below. I hope the results will be useful. Probability of a feature in each hex — E(hex count to nearest feature) 0.1% — 16.19
__________________
Decay is inherent in all composite things. Nod head. Get treat. Last edited by Agemegos; 05-29-2019 at 06:57 PM. Reason: extra precision |
05-29-2019, 10:17 AM | #29 |
Join Date: Feb 2016
|
Re: [Space] Mapping Large Flat Areas
One problem that I see is that you can have locations overlapping in hyperspace that are hundreds of ly apart in normal space. For example, star A and star B are 141 ly from Earth and are 200 ly apart from each other, but star A is directly above star B. On a two dimensional system, they overlap, meaning that it should be easier to travel from Star A to Star B than from Earth to either (just going by real world distance is still a 3D system).
|
06-01-2019, 01:39 AM | #30 | |
Join Date: May 2005
Location: Oz
|
Re: [Space] Mapping Large Flat Areas
Quote:
__________________
Decay is inherent in all composite things. Nod head. Get treat. |
|
Tags |
mapping, space |
Thread Tools | |
Display Modes | |
|
|