Math puzzle
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.42, obviously or 75000ish.
"If you think it's expensive to hire a professional to do the job, wait until you hire an amateur." Red Adair. nils illegitimus carborundum me, me, me
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.You could easily figure out the minimum needed, assuming you ignored overlap. I'm assuming that the projectors are producing a circular force field. It would just be:
Projectors * 2 * 2 * Pi = 450 * 450 * 4 * Pi
Projectors = 450 * 450
Projectors = 202,500The hard part would be determining how you arrange the circles so that you minimize the overlap. Using the overlap depicted in the last picture of this[^], it means you'd only really use 81.83% of each circle with the rest being wasted.
Projectors * 2 * 2 * Pi * .8183 = 450 * 450 * 4 * Pi
Projectors = (450 * 450) / .8183
Projectors = 247,461.265Thus, you would need 247,462 (rounded up) projectors for your spaceship. :)
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That will be useful, but the shortest vertex ends up being 81.859m. I need to get the formulas for something on the order of V8 or V9, I think. Shouldn't be too difficult after that.
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That will be useful, but the shortest vertex ends up being 81.859m. I need to get the formulas for something on the order of V8 or V9, I think. Shouldn't be too difficult after that.
Shoudnt each length of the triangle be equal to the radius of you force field?
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.How far does the projector project the shield off the surface? Surely the surface projected surface will be greater than the space ship surface. Or do you mean to say that
Gregory.Gadow wrote:
Embedded in the surface are projectors that each create a force field 4m in diameter.
means a spherical shield?
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That will be useful, but the shortest vertex ends up being 81.859m. I need to get the formulas for something on the order of V8 or V9, I think. Shouldn't be too difficult after that.
This seems even easier: http://en.wikipedia.org/wiki/Geodesic_grid[^]
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.Would not each force field deform each neighbouring force field where they touched?
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
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This seems even easier: http://en.wikipedia.org/wiki/Geodesic_grid[^]
Brilliant, thanks.
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.If the force field is created by projectors, how far out from the surface would each field be projected, as this will increase the area needed to be covered by adding d (distance from the surface where the projection originates to where the field is instantiated) to r making the formula A = 4π(r+d)^2 I am thinking that if you were not really concerned with overlapping, you would divide the area needed to be covered by the area of the largest hexagon that would fit into a circle with a 4m diameter. Hexagons come to mind as being the most relevant shape to use for this, because of their obvious relationship to 360 degree coordinate systems.
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Brilliant, thanks.
Well, I just started thinking, and I think that you would have to resort to some kind of numerical approximation. The only condition that you will have is that the length of the triangle must be equal or lower to 3. You could set it equal to 3 but than the fit to the sphere wouldnt hit excactly, so you woul dhave spots on the circle that wouldnt be covered. However I think that you could use 6 triangles that would have to fit excactly to the ecvator length. Infact, I think this would be the only way you could find a fit.
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Shoudnt each length of the triangle be equal to the radius of you force field?
Kenneth Haugland wrote:
Shoudnt each length of the triangle be equal to the radius of you force field?
No, diameter. Radius would give >100% overlap.
Did you ever see history portrayed as an old man with a wise brow and pulseless heart, waging all things in the balance of reason? Is not rather the genius of history like an eternal, imploring maiden, full of fire, with a burning heart and flaming soul, humanly warm and humanly beautiful? --Zachris Topelius Training a telescope on one’s own belly button will only reveal lint. You like that? You go right on staring at it. I prefer looking at galaxies. -- Sarah Hoyt
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Kenneth Haugland wrote:
Shoudnt each length of the triangle be equal to the radius of you force field?
No, diameter. Radius would give >100% overlap.
Did you ever see history portrayed as an old man with a wise brow and pulseless heart, waging all things in the balance of reason? Is not rather the genius of history like an eternal, imploring maiden, full of fire, with a burning heart and flaming soul, humanly warm and humanly beautiful? --Zachris Topelius Training a telescope on one’s own belly button will only reveal lint. You like that? You go right on staring at it. I prefer looking at galaxies. -- Sarah Hoyt
I think that you either way would have more than a 100 % overlap. If you take 6 triangles and form a Hexagon, this would generate the minimum overlap and still fit on a sphere.
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Would not each force field deform each neighbouring force field where they touched?
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
GuyThiebaut wrote:
Would not each force field deform each neighbouring force field where they touched?
Yes. The force fields are "tuned" with their neighbors, so overlap causes them to merge. The result is that the entire ship is encased in an almost perfectly spherical bubble of... well, the short explanation is that the warp and weft of the universe have been slightly realigned at a very local level. The bubble is at the 450m mark; the projectors themselves are a bit within. Propulsion occurs by giving a few thousand force fields a little "twist", in effect creating a gravity well that the bubble falls into. To continue with the fabric of space analogy, the bubble is able to slide along the outer surface rather than have to dodge the threads inside, so it is able to go significantly faster than light. Relativity, it turns out, is a feature of electromagnetism; since electromagnetism cannot cross the bubble except as quantum fluctuations, time dilation is minimal. Gravity can cross, weakly, which allows the ship to get its bearings and travel a known route in reasonable safety. It cannot see what is ahead (below?) because of the artificial gravity well, so blazing new trails can be dangerous. The volume inside the bubble is just along for the ride. Hey, it's no hokier than warp drive or hyperspace ;P My plot requires that ships be huge and very expensive, and that space travel be a risky venture. This works for what I need.
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GuyThiebaut wrote:
Would not each force field deform each neighbouring force field where they touched?
Yes. The force fields are "tuned" with their neighbors, so overlap causes them to merge. The result is that the entire ship is encased in an almost perfectly spherical bubble of... well, the short explanation is that the warp and weft of the universe have been slightly realigned at a very local level. The bubble is at the 450m mark; the projectors themselves are a bit within. Propulsion occurs by giving a few thousand force fields a little "twist", in effect creating a gravity well that the bubble falls into. To continue with the fabric of space analogy, the bubble is able to slide along the outer surface rather than have to dodge the threads inside, so it is able to go significantly faster than light. Relativity, it turns out, is a feature of electromagnetism; since electromagnetism cannot cross the bubble except as quantum fluctuations, time dilation is minimal. Gravity can cross, weakly, which allows the ship to get its bearings and travel a known route in reasonable safety. It cannot see what is ahead (below?) because of the artificial gravity well, so blazing new trails can be dangerous. The volume inside the bubble is just along for the ride. Hey, it's no hokier than warp drive or hyperspace ;P My plot requires that ships be huge and very expensive, and that space travel be a risky venture. This works for what I need.
When the story is completed - post a link here as it sounds fascinating. I have read some space opera science fiction, which I really enjoyed, and what you are describing fits into that category very well. Falling into a gravity well is a brilliant idea!:thumbsup:
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
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GuyThiebaut wrote:
Would not each force field deform each neighbouring force field where they touched?
Yes. The force fields are "tuned" with their neighbors, so overlap causes them to merge. The result is that the entire ship is encased in an almost perfectly spherical bubble of... well, the short explanation is that the warp and weft of the universe have been slightly realigned at a very local level. The bubble is at the 450m mark; the projectors themselves are a bit within. Propulsion occurs by giving a few thousand force fields a little "twist", in effect creating a gravity well that the bubble falls into. To continue with the fabric of space analogy, the bubble is able to slide along the outer surface rather than have to dodge the threads inside, so it is able to go significantly faster than light. Relativity, it turns out, is a feature of electromagnetism; since electromagnetism cannot cross the bubble except as quantum fluctuations, time dilation is minimal. Gravity can cross, weakly, which allows the ship to get its bearings and travel a known route in reasonable safety. It cannot see what is ahead (below?) because of the artificial gravity well, so blazing new trails can be dangerous. The volume inside the bubble is just along for the ride. Hey, it's no hokier than warp drive or hyperspace ;P My plot requires that ships be huge and very expensive, and that space travel be a risky venture. This works for what I need.
Yeah, You'd want a kind of Voronoi algorithm runninng to get that bubble optimized. Or else all the ornaments would be clinking against each other and possibly breaking into shards and falling off the boughs onto the electric grate under the Christmas tree. Don't want slivers between the shocked toes too, right? Ouch.
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Yeah, You'd want a kind of Voronoi algorithm runninng to get that bubble optimized. Or else all the ornaments would be clinking against each other and possibly breaking into shards and falling off the boughs onto the electric grate under the Christmas tree. Don't want slivers between the shocked toes too, right? Ouch.
Quote:
You'd want a kind of Voronoi algorithm runninng to get that bubble optimized.
You actually want the exact opposite of a Voronoi diagram.
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When the story is completed - post a link here as it sounds fascinating. I have read some space opera science fiction, which I really enjoyed, and what you are describing fits into that category very well. Falling into a gravity well is a brilliant idea!:thumbsup:
“That which can be asserted without evidence, can be dismissed without evidence.”
― Christopher Hitchens
Thanks. I have a website up at http://hyperpedia.gregory-gadow.net/[^], which is going to be my selling point / source of meta info once the novel is written. I'm taking a break from the book to improve the articles (and they desperately need improvement) which is why I want to know about the math. I'd like to have the articles good enough to show off to publishers by the end of this summer. With luck, the book itself will be ready to start shopping around by the end of the year.
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.None of the below takes into account the curvature of the sphere. If you make a square with 4m sides with a projector at each corner then you'd have a small area in the center of the square that would not be covered. So, place a projector there. Yes there would be some overlap, and the "edges" of the fields where they meet would probably be weak spots That gives 5 projectors for each 16m2. So 2,544,961m2 * 5 projectors / 16m2 = 795,301 projectors, in this configuration. This is probably the miminal and most efficient configuration resource-wise. Formula: P = 5A/16, where A = surface area of square, P = projectors. If instead you have a square where the line between two opposite corners is 4m, then you wouldn't need a center projector. Such a square would have sides equal to the square root of 8m (a2 + b2 = (4m)2, but a == b, so 2a2=16m2, so then a = sqrt(16/2)m = sqrt(8)m). That would give you a square that is 8m2. Placing a projector at each corner gives you 2,544,961m2 * 4 projectors / 8m2 = 1,272,481 projectors. This is probably the configuration that gives the best overall coverage or protection. Formula: P = 0.5A /edit/ On the otherhand, you could use a triangle configuration. An equilateral triangle with 4m sides would have an area of approximately 6.928m2. So, you'd need 2,544,961m2 * 3 / 6.928m2 = 1,102,001 projectors. Formula: P = 3A/6.928 /edit 2/ With the hexagonal configuration, with radius 4m you would have six small gaps about 2m in from the center of each side. If you made the radius 3.8m, I'm guessing that would overcome the small gaps. With that 5% adjustment, that would be 2,544,961m2 * 7 / 37.5m2 = 474,854 projectors Formula: P = 14A/75 (14/75 == 7/37.5), or if you want to round-up, P = 15A/75 = A/5
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.Jeeze, it'll be a really interesting story if you go into that level of technical detail. Given that characters and character interactions count ten times more than technical details, try having one of them say "Yup, must be a couple o' gozillion projectors, so you'd better start polishin'." Writing Sci-Fi isn't about explaining every detail of technology; it's about not describing the bits that don't work: "Captain! The dilithium crystals are cracking under the strain! The warp engines are failing!" Um, what exactly are dilithium crystals and warp engines, again? And light sabres, for that matter?
I wanna be a eunuchs developer! Pass me a bread knife!
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I'm working on something for a science fiction piece, and I'm hoping the biggest gathering of geeks I'm a part of can help. The specific problem: I have a large sphere (a space ship, thanks for asking) with a radius of 450 meters. This gives it a surface area of 2,544,961 m2 (
A = 4πr2) Embedded in the surface are projectors that each create a force field 4m in diameter. I want to know how many projectors I would need to completely encase the sphere with absolutely no breaks. Overlap is fine, and variables such as the relative angle of neighboring forcefields or the distance of the force fields from the surface of the sphere can be considered negligible and thus irrelevant. The general problem: If the answer could come with a generalized algorithm that I can apply to spheres of other sizes, I would be most appreciative.A minimum bound can be found by the area ratio (each projector protects πp², p = 4m, so 16π, and the total area is 4.(450²).π, so the minimum number would be (4×450²)/16=50,625), and maximal circle packing uses 90.69% of the space[^], so that sets a lower bound of 55822.03. (That ignores the non-planar nature of the surface, but with such a large number that isn't going to be important.) As for actually building such a ship, I'd use an icosahedral basis for the ship's hull. Each triangular face can then be tiled to an appropriate density, and then filled by putting a projector at each vertex. The minimum separation between projectors in this configuration is such that the half-bisection of a triangle (point to centre) has a length of 4m; this length on a triangle of side length 2 is ½√5, i.e. l = ¼s√5 or s = 4l/√5. If l = 4 then s = 16/√5 = 7.16m. The radius of a sphere touching an icosahedral shell of edge length a is 0.951a[^], so for r = 450m, a = 450/0.951 = 473m. We actually want the arc length if we're building a spherical ship, not an icosahedral one, as the tiling will be done on arc not icosahedron edge. That's an angle of 2 arcsin (a/2r), and a/2r = 1 / 2*0.951, so the angle is 1.11rad and the arc length 498.2m. We're ignoring the small non-planar nature of our tiled triangles. So we need to tile at a level of 498.2/7.16 = 69.6, or 70 as it must be an integer. (I.e. each face will be split into triangles, with 70 per side.) The number of vertices on a face, including those on shared edges, is the 71st triangular number, which is ½.70.71 = 2485; subtracting the half share of the 207 vertices which are shared between two faces (on an edge) and the 4/5 of 3 vertices shared among 5 faces (the corners) there are 2379.1 projectors per face, or a total of 47582. Hm, I went wrong somewhere, as that's less than the minimum bound. Can someone spot the mistake? :~