Math puzzle
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Hell, I'm almost convinced that I can build a matter transmitter. There's just one little detail that I need to figure out, and then it's sorted. (I don't bother telling readers 1/50th of the details that I have worked out, though -- working it all out for me helps me to keep it consistent)
I wanna be a eunuchs developer! Pass me a bread knife!
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Exactly this. Consider a car with automatic transmission. Can you tell me how a planetary gearset works? More importantly, will I still be awake when you're done? So work out your technology, drop a couple of throwaway hints how it works, or at least what happens when it goes wrong, but for dogs sake don't explain it to me. A second pet peeve of mine is authors who think future generations will be fixated on what we are currently doing. With all of history to choose from, the odds of anyone choosing to visit 2013 is so implausible that the 4th wall is shattered to smithereens. Take two iconic authors, Frank Herbert, and Iain Banks. Dune. This has to be one of the best Sci-Fi books ever written. Does he explain how the space ships fold space? No, he explains that they need navigators who are permanently high to pilot them. Does he care what their ancestors were doing in the 20th century? Bahahahaha, you probably couldn't pinpoint the second millenium with his calendar. Excession. I would vote this the best Sci-Fi book ever written. Does he explain how they built those defense drones so small? No, he throws it into a fight, and as it dies, he mentions that the entire battle took half a second. Hm, sorry, spoilers :-O
ThePotty1 wrote:
the best Sci-Fi book ever written
Not unless Gateway is unwritten :P
I wanna be a eunuchs developer! Pass me a bread knife!
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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!
A good science fiction author knows exactly how things work, even when the details do not make it into the books.
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Even better; cue Stargate: Universe
Software Zen:
delete this; -
Hell, I'm almost convinced that I can build a matter transmitter. There's just one little detail that I need to figure out, and then it's sorted. (I don't bother telling readers 1/50th of the details that I have worked out, though -- working it all out for me helps me to keep it consistent)
I wanna be a eunuchs developer! Pass me a bread knife!
Mark_Wallace wrote:
working it all out for me helps me to keep it consistent
I imagine you've read Larry Niven's accounts of the critiques he's received for his Ringworld design. A veritable pile of well-known scientists from a variety of disciplines have written lengthy discussions on how to build one. He's incorporated some of their ideas in the later novels in the series.
Software Zen:
delete this; -
Mark_Wallace wrote:
working it all out for me helps me to keep it consistent
I imagine you've read Larry Niven's accounts of the critiques he's received for his Ringworld design. A veritable pile of well-known scientists from a variety of disciplines have written lengthy discussions on how to build one. He's incorporated some of their ideas in the later novels in the series.
Software Zen:
delete this;That kind of feedback I could live with, but I imagine that a lot of it comes from giving too much information in the story, which allows people to see that you've got it wrong :D The most I've ever given about matter transmitters is: "Everybody was so happy that I had got him to explain the Booths' functions so well. I remember it very clearly. He went to great pains to tell me how the statis effect stopped all movement in solid objects, and how that made the objects massless, but of infinite mass. Or was it indefinite mass? Anyway, it was something to do with making things to not exist any more, so that the universe stopped caring about where the things were. I had not understood a word, so I kept asking the idiot questions that it turned out everybody wished would be answered." The last line pushes you to believe that they work, because everyone else had an epiphany about their workings, and you associate yourself with them, even though almost no information whatsoever is provided in the story (it's similar to a vid "reaction shot", like where you don't show a great basketball pass, you just show the reaction of some kid in the spectator seats -- everyone leaves the theatre believing that they saw a great basketball pass). That kind of technique is usually referred to as "Getting away with murder".
I wanna be a eunuchs developer! Pass me a bread knife!
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That kind of feedback I could live with, but I imagine that a lot of it comes from giving too much information in the story, which allows people to see that you've got it wrong :D The most I've ever given about matter transmitters is: "Everybody was so happy that I had got him to explain the Booths' functions so well. I remember it very clearly. He went to great pains to tell me how the statis effect stopped all movement in solid objects, and how that made the objects massless, but of infinite mass. Or was it indefinite mass? Anyway, it was something to do with making things to not exist any more, so that the universe stopped caring about where the things were. I had not understood a word, so I kept asking the idiot questions that it turned out everybody wished would be answered." The last line pushes you to believe that they work, because everyone else had an epiphany about their workings, and you associate yourself with them, even though almost no information whatsoever is provided in the story (it's similar to a vid "reaction shot", like where you don't show a great basketball pass, you just show the reaction of some kid in the spectator seats -- everyone leaves the theatre believing that they saw a great basketball pass). That kind of technique is usually referred to as "Getting away with murder".
I wanna be a eunuchs developer! Pass me a bread knife!
Mark_Wallace wrote:
That kind of technique is usually referred to as "Getting away with murder"
I imagine it's all in knowing how much to bribe someone at the Artistic License Bureau :-D.
Software Zen:
delete this; -
I calculated out that you would need 318,120.125 to cover the entire area. So that would be 318,121 to cover it totally. I used my Amusement park job experience to help me... ou all know the game where you have the large red dot that you have to cover with 5 silver circles. I can email you the math, later. No time to type it in here right now.
Steve Maier
Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet.
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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.This is it. That moment they told us in high school where one day, algebra would save our lives.
CEO at: - Rafaga Systems - Para Facturas - Modern Components for the moment...
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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.Gregory.Gadow wrote:
I have a large sphere
Why does it need to be a sphere?
Gregory.Gadow wrote:
each create a force field 4m in diameter.
Is the force field itself a sphere? Or is it flat?
Gregory.Gadow wrote:
Overlap is fine
Overlap is required to insure complete coverage.
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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.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?
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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. :)
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.
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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.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
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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.Area of sphere = 4PI r**2 projector area is 4PI Hence Number of projectors = r**2
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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.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
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Area of sphere = 4PI r**2 projector area is 4PI Hence Number of projectors = r**2
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.
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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.An approximation would be easy enough to calculate (using
rrather 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 of6*sqrt(3) m2. 3. For covering a surface area of4πr2, you need a minimum ofceil(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. -
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
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 areIn 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
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| 7 | 8 | 9 | 10| 11|
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12--13--14--15--16--17
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| 18| 19| 20| 21| 22|
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23--24--25--26--27--28In 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.
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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--28In 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.
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