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Math puzzle

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  • G Gregory Gadow

    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.

    B Offline
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    Bassam Abdul Baki
    wrote on last edited by
    #4

    Have you looked into this[^]?

    Web - BM - RSS - Math - LinkedIn

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    • G Gregory Gadow

      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.

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      Kenneth Haugland
      wrote on last edited by
      #5

      This kind of problem arises in the design of omnidirectional loudspeakers[^], the nomal way is to use a dodekaedron[^], so I suppose that you just need to find the next way of partition the space.

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      • G Gregory Gadow

        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.

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        Steve Maier
        wrote on last edited by
        #6

        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

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        • G Gregory Gadow

          I majored in applied math in college *mumble* years ago, and I spent almost an hour trying to figure this one out. No luck. It is kind of a specialty problem :|

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          AspDotNetDev
          wrote on last edited by
          #7

          It would be pretty easy to approximate an answer, but an exact answer may be very difficult. For one, I'm not sure there's an exact spacing between projectors that is easily quantifiable (e.g., may depend on size of sphere and size of projector). You could probably decide on some spacing strategy, such as a honeycomb-like pattern, but I doubt it would be perfectly optimal (i.e., there'd be more overlap than necessary). How exact of an answer are you looking for?

          Thou mewling ill-breeding pignut!

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          • G Gregory Gadow

            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.

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            RedDk
            wrote on last edited by
            #8

            Seems to me you would adjust the radius in each of the psychircular entities one see's in this illustration ... frinstance: http://en.wikipedia.org/wiki/File:Apollonian_spheres.jpg[^]

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            • B Bassam Abdul Baki

              Have you looked into this[^]?

              Web - BM - RSS - Math - LinkedIn

              G Offline
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              Gregory Gadow
              wrote on last edited by
              #9

              That just might. If I think of the ship as a geodesic sphere having a radius of 450 meters, made up of polygons having edges of 4 meters or less, I can put a projector at each vertex and get what I need. That may be too many projectors but redundancy is definitely good. Thanks.

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              • G Gregory Gadow

                That just might. If I think of the ship as a geodesic sphere having a radius of 450 meters, made up of polygons having edges of 4 meters or less, I can put a projector at each vertex and get what I need. That may be too many projectors but redundancy is definitely good. Thanks.

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                Kenneth Haugland
                wrote on last edited by
                #10

                http://www.domerama.com/calculators/6v-geodesic-dome-calculator/[^]?

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                • G Gregory Gadow

                  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.

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                  R Giskard Reventlov
                  wrote on last edited by
                  #11

                  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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                  • G Gregory Gadow

                    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.

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                    Andrew Rissing
                    wrote on last edited by
                    #12

                    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 = 202,500

                    The 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 = 247,461.265

                    Thus, you would need 247,462 (rounded up) projectors for your spaceship. :)

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                    • K Kenneth Haugland

                      http://www.domerama.com/calculators/6v-geodesic-dome-calculator/[^]?

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                      Gregory Gadow
                      wrote on last edited by
                      #13

                      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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                      • G Gregory Gadow

                        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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                        Kenneth Haugland
                        wrote on last edited by
                        #14

                        Shoudnt each length of the triangle be equal to the radius of you force field?

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                        • G Gregory Gadow

                          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.

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                          W Offline
                          wizardzz
                          wrote on last edited by
                          #15

                          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?

                          Twitter[^]

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                          • G Gregory Gadow

                            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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                            Kenneth Haugland
                            wrote on last edited by
                            #16

                            This seems even easier: http://en.wikipedia.org/wiki/Geodesic_grid[^]

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                            • G Gregory Gadow

                              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.

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                              G Offline
                              GuyThiebaut
                              wrote on last edited by
                              #17

                              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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                              • K Kenneth Haugland

                                This seems even easier: http://en.wikipedia.org/wiki/Geodesic_grid[^]

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                                Gregory Gadow
                                wrote on last edited by
                                #18

                                Brilliant, thanks.

                                K 1 Reply Last reply
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                                • G Gregory Gadow

                                  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.

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                                  Duraplex
                                  wrote on last edited by
                                  #19

                                  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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                                  • G Gregory Gadow

                                    Brilliant, thanks.

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                                    K Offline
                                    Kenneth Haugland
                                    wrote on last edited by
                                    #20

                                    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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                                    • K Kenneth Haugland

                                      Shoudnt each length of the triangle be equal to the radius of you force field?

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                                      Dan Neely
                                      wrote on last edited by
                                      #21

                                      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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                                      • D Dan Neely

                                        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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                                        K Offline
                                        Kenneth Haugland
                                        wrote on last edited by
                                        #22

                                        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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                                        • G GuyThiebaut

                                          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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                                          G Offline
                                          Gregory Gadow
                                          wrote on last edited by
                                          #23

                                          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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