Glyph-space - calling all maths geniuses

[PTeppic]

Okay, I admit, my maths is somewhat rusty... but I know out there amongst you there are all sorts of wunderkind with meths coming out of your ears.

Here is a little project to keep you busy one evening, if you don't mind.

The earth is a sphere (roughly). How would you go about positioning say 40 points around the surface, so that they are equally separated in longitude and latitude. They would be 9 degrees apart in longitude, but then they would have to be spread in lattitude too, so as to be approximately equidistant from each other...

a) how far apart would they be from each other when so located?


Now, once that is done, model lines linking THROUGH the earth from each point to each other point.

b) how many of the lines intersect?

c) how far apart is the closest group of three from intersecting?


[This all goes to the intersecting glyhps ideas in one of the other threads about how the addresses thing works]

 

[Jedispara]

so basically you want to know how you could create a wormhole system on earth. based on locations in lat and long???

 

[PTeppic]

Not exactly - I am using the earth as a physical sphere that people can relate to, and we know the geometry of. With the figures that hopefully someone can provide we can then scale up to something a few hundred light years across to see how feasible the glyph address system actually is...

I have a personal feeling that across the immensity of even our local corner of the galaxy, there will not be many, if ANY, intersections of three pairs the 3D points, let alone the hundreds suggested by the show.

 

[Jedispara]

ok, well think of it this way, set a gate up in every city over 100,000 people, that woudl be a good estimate of how many plaents are out there... so then see how many cross becasue there could be millions of planets out there.

 

[PTeppic]

Yes, but we know that there are 38 "address" glyphs on the Earth gate, and a point of origin. Just from this earth gate we are told there hundreds of permutations (which we see on the dialling computer)... I reckon there isn't even one!

It is just basic geometry - 38 equidistant points simply doesn't create 3 pairs which intersect except through the centre, as far as I can see.

 

[Jedispara]

ok, what about centroids on a sphere between 3 points. only the gate system has it built in that it has to be on the surface of teh sphere?

 

[PTeppic]

Sorry - that went about a foot over my head!

 

[Jedispara]

oh, sorry, well basically the centroid is the gravitational center of a triangle so if you take the centroid (a centroid goes through the vertex and the middle point of the line oposite from it) of 3 points you get a location... there for if the 3 points were spread over a sphere than the point would be inside the sphere, but the gate system could have another one of those rules built into it so that it would create a wormhole location at that point in the middle of the triangle on teh surface of teh sphere. ok?

 

[PTeppic]

Okay - I think I got that...

Of course we "know" that the gates actually use the intersection of 6 points... which is total overkill, since the intersection of 3 or possibly 4 points gives an absolute and irrevocable point in 3D space.

 

[Jedispara]

true, but even with 6 points... the theory still works, also one thing our galexy is a sphere its more a flat plane so.. you wouldn't need the surface rule in there.. but it still holds true and it would create the number of permiations that they have found even to now.

 

[PTeppic]

That's my big point - would you get hundreds of permutations from intersections of only 38 or so fairly evenly spaced points around a nominal circle, on either a plane or a sphere.

 

[Jedispara]

consevably yes... because you would have from just the 1st chevron you would have well over 100 places, just based on that one base chevron... then look at the fact that there is 38 chevrons... so the answer is yes you would get that many periations.

 

[PTeppic]

I'm not just talking about permutations or combinations, I am talking about intersections.

For example, with four points, there is only single intersection between any two pairs. For five points, there are five intersections, for six points there are twelve (or 13 or totally equidistant). So, going bigger will get more intersections, but what about for three lines intersecting... hence my request for a bored mathematician.

 

[Jedispara]

good point, ok when i get back from choir tonight i will work on teh math of it... ok?

 

[PTeppic]

Oooh goody!

I suppose I probably could work it out, eventually... on the following basis:

a) assume a circle, with the origin at 0,0
b) plot say 40 points equi-distant around the circumference, with calculatable positions ON the circle
c) create lines from every point to every other one, retaining the equation for each of these lines
d) compare all combinations of three of these lines

 

[Jedispara]

ok, do know how many equations your talking about... with a 24 point polygon i thyink there are over 100 lines....

 

[PTeppic]

OH YES!!

I started fiddling around with some software today - I think a 40 point polygon has 2.7E9 lines, but I am having some trouble with maths underflow in the intersection routines... :blush:

Is there a formula for number of vertices for a N faced polygon?

 

[shu_hunter]

well, there is if you know how many edges there are;
V + F = E + 2
v=verticies
f=faces
e=edges
But, you probably don't, so this is worthless.

~Shu Hunter
:upto: stargate fan{atic}

 

[Jedispara]

what is the program??? mabye i can find it some where....

 

[PTeppic]

Originally posted by Jedispara
what is the program??? mabye i can find it some where....

I wrote it! :blush:

It works, insofar as it proves conclusively that for a flat plane with a circle with 40 equally spaced points, joining three pairs of separate points, there ARE combinations which intersect. In fact, there were about 5000 going through the centre, and 2500 which went through other points, and all to within an accuracy of 14 decimal places! However, there are some repeats in that list, and I can't be bothered to process the results to remove them - I would estimate about 1/3 of the combinations can be removed.

Now I just have to work it out for a 3d model, ie. a sphere.