Math puzzle
-
Now that's the kind of practical application of math that I wish they would teach in school! ;) Marc
Latest Article: C# and Ruby Classes: A Deep Dive
My BlogI 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 :|
-
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. -
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 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.
-
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.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
-
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 :|
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?
-
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. -
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.
-
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.
-
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
-
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. :)
-
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.
-
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?
-
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?
-
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[^]
-
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
-
This seems even easier: http://en.wikipedia.org/wiki/Geodesic_grid[^]
Brilliant, thanks.
-
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.
-
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.
-
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
-
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.
