Math puzzle

[9082365]

Hang on a minute. Doesn't this all rather depend on just how far this field is projected? A force field which is literally flush with the surface of the projector is, after all, kinda useless, certainly for impact damage as it will simply transmit the force as if it were part of the hull. And, of course, for every centimetre of projection not only does the surface area covered grow but the overlap and relative angles become increasingly important (far from the negligible suggested). Overall I'd have to say that anyone who turned up to a space science convention suggesting this as the way to project a seamless spherical force field might well cause the deaths of several eminent professors from either conniptive fits or laughing themselves stupid! And that's even before you consider the additional mass added by thousands of projectors before you even add engines, a crew, a life support system, and the other million or so things necessary to sustained existence in space. You may be able to get the thing moving (assuming it's built in space and stays there) but with all that momentum it'll be an absolute swine to stop at anything above a crawl!! What would Sheldon say?

[Railbot]

Oh no, he's trying to build a Dyson Sphere around a small star. Then put a shield up around it.

[ssa ed]

Andrew - very close. The Lower Bound: If you imagine a geo grid, and place intersecting circles on its surface (i.e the force field coincides with the sphere's surface), it will form a tessellation of near perfect hexagons where each side is 2m. The area of each "hexagon" = 10.3923 m^2. So least number of sensors is protection at the sphere's surface, and exactly 244,863 +/- 1 (rounding) sensors are needed. Not only do we have a lower bound on the answer, but the upper bound would be an imaginary sphere projected out to whatever distance desired. For example, if you want to project the force field 450 meters above the sphere, the surface area of the projected sphere would be 8 times as large, hence 8 times as many sensors would be needed.

[MrChug]

Simulate the problem following Ken Perlin's lead. Keep adding projectors until the surface is covered.

They will never have seen anything like us them there. - M. Spirito

[ramachandraniyers]

Area of sphere = 4PI r**2 projector area is 4PI Hence Number of projectors = r**2

[BotReject]

This is a well-known maths problem that is used in the architecture of geodesic domes and in biological virus construction. Like a virus you can use an icosahedron of 20 equilateral triangular sides (although some other shapes work too, for example some viruses are dodecahedrons, but the icosahedron is probably easier to work with) that encloses the sphere. The individual shield areas will be hexagons, but with 12 pentagons, one at each vertex. Then it's simply a case of fitting the hexagons to each triangular face, with those along the edge centered on the edge. For example, you could have two hexagons along each edge, one in the center and a pentagon centered on each corner, such that the edge of each equilateral triangle would be three times the hexagon diameter. Of course you can use more hexagons for each face of a larger icosahedron. There is, no doubt, a mathematical relationship, but I once used this bottom-up approach to code a graphics algorithm that generates icosahedrons from spheres and or/ hexagonal/pentagonal prisms.

Bot

[jschell]

It isn't that simple. First the sphere is curved. That means that the intersection with the surface is not exactly 4PI. It would be 4PI if the sphere was plane. Second that calculation presumes that the intersection is mallable in that it fills the space completely with the adjacent projectors. That would be true if the projections (and intersection) was a rectagle/square and was on a plane. Visually one can see the problem by drawing two circles that just touch side by side. Then attempt to draw another circle such that the 'gap' at the closest edge is also covered. The overlap, which must happen, is area that is lost (per your calculation) and will need to be made up by adding more projectors.

[Stefan_Lang]

An approximation would be easy enough to calculate (using r rather than the actual radius of the sphere to arrive at a formula): 1. Since the surface of the sphere is effectivey flat compared to the force fields, the most efficient pattern for placing them would be a hexagonal grid. 2. Each hexagon would have a side length of 2m, and an area of 6*sqrt(3) m2. 3. For covering a surface area of 4πr2, you need a minimum of ceil(4πr2 / (6*sqrt(3) m2)) hexes. With r=450 you get 244863 hexes. 4. Of course you cannot cover the surface of a sphere with hexes without overlapping the hexes themselves, so you need additional hexes due to the losses from that overlap. The amount of overlap varies depending on the strategy used for covering this surface. In any case, the more you minimize that overlap, the harder it gets to calculate the real number of hexes that you need. For your purposes, I'd just add some more or less reasonable amount of about 5-10% to the number calculated above. That should be close enough.

[Stefan_Lang]

In either variant you're going to have multiple projectors at the corners of each 'cell': in a square grid, if you want one projector on each corner and one in each center that results in an average of 2 projectors per square, not 5. See illustration:

1---2---3---4---5---6
| | | | | |
| 7 | 8 | 9 | 10| 11|
| | | | | |
12--13--14--15--16--17
| | | | | |
| 18| 19| 20| 21| 22|
| | | | | |
23--24--25--26--27--28

In this grid, projectors 1 and 7 belong to the first cell, 2 and 8 to the second, 12 and 18 to the 7th, 13 and 19 to the 8th. This covers all 5 projectors around the first (top left) cell. So you'd end up with 2/5 as many projectors, or about 319K. For a triangular grid, each corner is part of 6 adjacent triangles, so you only need to count 1 projector per two triangles, rather than three for one. Additionally, you forgot to add the central projector. So instead of 3 projectors per triangle you'd need an average of 1.5, or about 550K in total. For the hexagonal setup, each corner is part of three adjacent hexagons, so you only need to count 2 projectors per hexagon for the corners, plus one for the center. However, a 5% adjustment is not sufficient to close the gaps: the projectors on adjacent corners barely touch at the center of the sides, and, similarly, the central projectors circle barely touches each of the other circles. This will cover only about 90.69% of the total surface, not 95% (see http://en.wikipedia.org/wiki/Circle_packing[[^](http://en.wikipedia.org /wiki/Circle_packing "New Window")] ). Also, reucing the side length by 9.31% does not suffice, as parts of the circles will overlap, and only some amount serves for closing the gaps. You need to reduce the length to only 2*sqrt(3)m instead of 4m, and the area of such a hexagon would be 18*sqrt(3) or 31.177. The required amount of projectors would thus be 2544961*3/31.177 = 244889.

[TheGreatAndPowerfulOz]

Yeah, I was trying to figure out how to take that into account with the squares. As for the triangles, I don't believe there's a need for a central projector since the center of the triangle will be less than 2m from any one projector. Thanks for your input, I learned something today! LOL.

If your actions inspire others to dream more, learn more, do more and become more, you are a leader.-John Q. Adams
You must accept one of two basic premises: Either we are alone in the universe, or we are not alone in the universe. And either way, the implications are staggering.-Wernher von Braun
Only two things are infinite, the universe and human stupidity, and I'm not sure about the former.-Albert Einstein

[Stefan_Lang]

You assumed triangles with 4m sides, so the projected circles from the corners would just touch on the center of each side. You'd get exactly the same gaps as you would for the hexagons *with* the central projector. To avoid the gaps, the solution would be the same as for the hexagons: reduce side length to 2*sqrt(3)m and you're good. All of this gets easier if you use hexagonal projections instead of circular projections: hexagons are the highest order regular polygons usable for a periodic partitioning of the plane. There are tiles or sets of tiles for other forms of partitioning, but for the purpose of minimizing the overlap of circles, there's no point looking at any but regular polygons. Starting from these hex projections, you simply have to tile the surface into a grid of hexes and calculate the number you need for covering the entire surface: just divide the total surface by the are of a single hex. Each hex has a side length of 2m, so the area is 6*sqrt(3). You'll find that going by that number you get exactly the same number of projectors as you get from your hex grid distribution. Of course, this number is just a lower limit, as it is impossible to tile a sphere perfectly with hexes: you'll have those hexes overlap more or less over the entire area of the surface. How much extra hexes (i. e. projectors) you need due to those overlaps is hard to tell; you could make a better estimate based on a given strategy. E. g. you could place a row of hexes around the equator and then more rows at lower and higher latitudes. However, you'd inevitably run into latitudes where you could reduce the number of hexes - but doing so would disturb the hex grid you've used, and the only way to solve this is placing the new row closer to the last one. This gets even trickier when you're closing in on the poles... Another strategy is to project a dodecahedron or other regular polyhedron on the surface of the sphere and partition each side of the dodecahedron with hexes. The projected hexes of course will be deformed, and grow smaller towards the edges, but it will provide you with a basic plan to partition the surface of the sphere with a means to actzually calculate the number of hexes (projectors) that will indeed suffice.

[Gregory Gadow]

Wow, I am rather surprised that this is still being discussed :wtf: Anyhow, after a lot of searching, I've found an algorithm that seems to suit my needs. For a large sphere with radius R and circles of radius r, apply these steps: 1. Calculate the circumference of the sphere, C = 2πR 2. Calculate the number of circles necessary to cover any great circle on the sphere. In this formula, the circles are separated by r, meaning that the circumference of any given circle will pass through the midpoint of the two adjacent circles: N = C/r 3. Plug N into this formula to count the number of circles needed to provide complete overlapping coverage of the sphere: S = 2 + 2N*SUM(i=1 to N-1)[sin(π * i/N)] The VB code looks like this:

Dim Circ As Double = 2 * Math.PI * sphereR 'circumference
Dim GC As Double = Circ / circleR ' number of circles on a Great Circle
Dim N As Double = Math.Round(GC, 0, MidpointRounding.AwayFromZero) 'the N value
Dim Sum as Double = 0

For i As Integer = 1 To N - 1
Sum += Math.Sin(Math.PI * (i / N))
Next

Dim Result As Double = 2 + (2 * N) * Sum

With the original parameters (a sphere with a radius of 450m and circles with a radius of 4m) this gives 636,429 projectors needed. I don't need to worry how they are arranged, just that there is enough for the failure of any non-adjacent projectors to not cause breaches in the bubble. It is the bubble that moves at superluminal speed, the contents is just along for the ride; so breaks, with the result of about 2 km3 very quickly decelerating to the speed of light is.... well, quite dramatic. And the last thing you want in space travel is drama. Thanks for all the help. :thumbsup:

[TheGreatAndPowerfulOz]

You are correct that the center of the triangle would be too far from the corners with 4m sides (3.46m or 73% too far), but even with 2*sqrt(3) sides the center is too far from the corner (2.29m or 14.5% too far). Making the sides 3m gives a center of 1.984m from each corner. However, if you place the projectors at the centers of the sides of 4m equilateral triangle, then that covers the center as well as the corners. The center of such a triangle would be 1.73m (sqrt(3)) from the center of each side. Yeah, I've been reading up on geodesic spheres and using binary division of the dodecahedron to get the "optimal" projection into the sphere. It gets difficult when you have a minimum size constraint. I think to solve some of the issues you pointed out you'd have to use differently sized "hexes" here and there to make everything fit.

If your actions inspire others to dream more, learn more, do more and become more, you are a leader.-John Q. Adams
You must accept one of two basic premises: Either we are alone in the universe, or we are not alone in the universe. And either way, the implications are staggering.-Wernher von Braun
Only two things are infinite, the universe and human stupidity, and I'm not sure about the former.-Albert Einstein

[Stefan_Lang]

The triangle center circle doesn't need to be 2m from the corner, it needs to be 2m from the outermost parts of the gap. These are the midpoints of the sides, which are only 2/sqrt(3)m removed from the center:

|<------2m------>|<------2m------>|
| | |
O----------------*----------------O----------
\ | / ^ ^
\ | / | |
\ | / 2/sqrt(3)m |
\ | / v |
\ _O_----------/------- |
\ __-- --__ / |
\ _-- --_ / |
* * 2*sqrt(3)m
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
\ / |
\ / v
O---------------------------

Placing the projectors at the centers of the sides results in less overlap, but then you get an irregular distribution of projectors:

__a_____b_____c___
/\ /\ /\ /\
d e f g h i j k
/_l__\/_m__\/_n__\/_o__\
\ /\ /\ /\ /
p q r s t u v w
\/_x__\/_y__\/_z__\/

Now look at the distribution without the triangles to get a better view:

 a     b     c

d e f g h i j k

l m n o

p q r s t u v w

 x     y     z

You'll note how at the locations of the triangle corners there are now obvious gaps. Each trangle corner is surrounded by 6 projectors at a distance of 2m each, just close enough to not require another projector at the corner itself (just as you intended). E. g. look at the triangle corner between projectors l and m. It's surrounded by projectors e, f, l, m, q, and r. There's some overlap within that hexagon, but it isn't too bad. But then, look at projectors d, e, and l: they all cover most of the triangular area between them, creating a triple overlap for more than 80% of that area! Basically what you created with this strategy is a mixed grid made of hexagons and triangles.