https://community.wolfram.com RSS Feed for Wolfram Community showing questions tagged with Geometry with no replies sorted by most viewed. https://community.wolfram.com/groups/-/m/t/392411 I need to use the same modal in grasshopper and manipulate it. I can only export the modal as a mesh but I need surface in rhino. I found this plug-in which helps mathematica to work with grasshopper: Mantis V 0.5 . But I don't think it is the right thing to use. Basically the question is: **Is it possible to write the code and manipulate it in grasshopper?** Here is the code: Manipulate[ Module[{\[CurlyEpsilon] = 10^-6, c1 = Tan[a1], c2 = Tan[a2], c3 = Tan[a3], c4 = Tan[a4], c5 = Tan[a5], c6 = Tan[a6]}, ContourPlot3D[ Evaluate[ c6 Sin[3 x] Sin[2 y] Sin[z] + c4 Sin[2 x] Sin[3 y] Sin[z] + c5 Sin[3 x] Sin[y] Sin[2 z] + c2 Sin[x] Sin[3 y] Sin[2 z] + c3 Sin[2 x] Sin[y] Sin[3 z] + c1 Sin[x] Sin[2 y] Sin[3 z] == 0], {x, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {y, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, {z, \[CurlyEpsilon], Pi - \[CurlyEpsilon]}, Mesh -> False, ImageSize -> {400, 400}, Boxed -> False, Axes -> False, NormalsFunction -> "Average", PlotPoints -> ControlActive[10, 30], PerformanceGoal -> "Speed"]], {{a1, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(1\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a2, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(2\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a3, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(3\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a4, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(4\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a5, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(5\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, {{a6, 1, "\!\(\*SubscriptBox[\(\[Alpha]\), \(6\)]\)"}, -Pi/2 - 0.01, Pi/2 + 0.01, ImageSize -> Tiny}, AutorunSequencing -> {1, 3, 5}, ControlPlacement -> Left] ![enter image description here][1] [1]: /c/portal/getImageAttachment?filename=ScreenShot2014-11-18at1.19.24PM.png&userId=11733 Aysu Aysoy 2014-11-18T18:06:44Z https://community.wolfram.com/groups/-/m/t/1140551 # Predicting fractal Dimension using Machine Learning ---------- ## Motivation Fractal Dimension is a way to measure the complexity of a fractal. This project's purpose it too estimate it with machine learning. Traditional methods such as box counting have shortcomings, such as a lack of versatility with many different types of fractals, and inefficient compute times. Machine learning was used to give good approximations of the fractal dimension of any fractal with reasonable speed. ## Data Collection I decided to use an image of a fractal as the input, and the dimension as the output. This allowed the algorithm to classify many types of fractals. ### Wikipedia Wikipedia has a set of fractal images, as well the corresponding fractal dimensions. I looked through them to pick the ones with high quality images for my training set, however it was not enough. ![enter image description here][1] ### Randomly Generated Julia Sets I chose to generate random Julia Sets, as there is a built in function for it in Wolfram Language. Julia Sets have a single parameter called c, and changing it generates unique fractals. I generated Julia Sets with random c values, both real and complex, to give the sets some variety. CreateJuliaFractals[size_] := Module[{jFunction = {JuliaSetPlot[#], #} &, jFractals}, jFractals = jFunction /@ Table[RandomComplex[], Floor[size/2]] \[Union] jFunction /@ Table[RandomReal[{-1, 1}], Ceiling[size/2]]; Association[#[[1]] -> JuliaFractalDimension[#[[2]]] & /@ jFractals] ] Calculating the fractal dimension from a given c value of a Julia Set is relatively simple. JuliaFractalDimension[c_] := 1 + Abs[c]^2/4 Log[2] ## Training I split the data into train and test and used the Predict function in Wolfram Language, as it has the capability of returning a result that differs from the training set. FractalDimensionPredictor[training_] := Predict[List @@ training] ## Evaluation I returned the real and predicted fractal dimension as well as the margin of error for every fractal in the test set. FractalMapEval[pred_, test_] := Keys[KeyMap[# -> {test[#], pred[#]} &, test]] ## Deployment I created a micro site in which people can put in pictures of any fractal and get an approximation of it's dimension. CloudDeploy[FormFunction[{"Fractal" -> "Image"}, Column[{#image, pred[#image]}] &, "PNG", AppearanceRules -> <|"Title" -> "Predict Fractal Dimension"|>], Permissions -> "Public"] ## Conclusion Before doing this project, I had no idea what fractal dimension was, and I learned many things including data generation and continuous classification. In the future I hope to use iterated function system to generate more fractals. Visit the micro site [here](https://www.wolframcloud.com/objects/7102eabb-7b80-4bb1-9652-eebcd75e90cd). Link to source code [here](https://github.com/LordDarkula/FractalNet). [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=fractalData.png&userId=515558 Aubhro Sengupta 2017-07-07T13:47:06Z https://community.wolfram.com/groups/-/m/t/383178 **Hello, I'm trying to discretize a very fine ring with "ImplicitRegion", but Mathematica does not let me.** in: ?x=ImplicitRegion[?0.12?^2?x^2+y^2??0.125?^2,{{x,-0.125,0.125},{y,-0.125,0.125},{z,0,1.}}] out: ImplicitRegion[0.0144?x^2+y^2?0.015625&&-0.125?x?0.125&&-0.125?y?0.125&&0?z?1.,{x,y,z}] **Answer:** DiscretizeRegion::drf: DiscretizeRegion was unable to discretize the region ImplicitRegion[<<2>>]. >> **BUT is a 3D Región, Why was it unable?.** **Equal if i use:** in: ?x=RegionDifference[Cylinder[{{0,0,0},{0,0,1}},0.125],Cylinder[{{0,0,0},{0,0,1}},0.12]] out: ImplicitRegion[0.0144?x^2+y^2?0.015625&&-0.125?x?0.125&&-0.125?y?0.125&&0?z?1.,{x,y,z}] **Answer:** DiscretizeRegion::drf: DiscretizeRegion was unable to discretize the region ImplicitRegion[<<2>>]. >> How can i do?, Thanks. Jesús. Jesús Manuel Cillero Ares 2014-11-03T19:36:16Z https://community.wolfram.com/groups/-/m/t/550930 Hi, I have drawn these two tiling the plane pattens (A & B), are these just considered variations of some of the current, or are these new? Geometrically they are appear to be different (but with similarities) to all the ones I have seen. Type A is similar to type 4, but my type A has three equal sides, and the other two make the same length as one of the equal. Type 4 has four equal sides. Type B is similar to type 8, but my type B has two 90 deg angles and the primary unit can overlap its self at each quadrant and maintain the patten. I would be interested in any opinion about these two Plane Tile Pattens. Thanks! Reece Hill 2015-08-21T11:20:42Z https://community.wolfram.com/groups/-/m/t/175990 Hi all, we are in need of some help (contract or otherwise) in recreating a series of classical mathematical models. we've done some work in rapid prototyping per the link below [url=http://formpig.com/blog/category/mathematical-models/][b]Formpig mathematical models[/b][/url] and with some assistance we would like to finish a couple of projects. if there is a forum or list of consultants that would be able to do this on a fee basis we would be happy to get a URL link in order to post there as well. please put "Mathematica" in your header and email us at "info" with the suffix formpig.com or log onto universaljoint.nu and contacts us via the website(s). thanks in advance for any help! - jon [img]/c/portal/getImageAttachment?filename=artifactsofresearch_flickr.png&userId=11733[/img] Jonathan Chertok 2013-12-28T19:05:07Z https://community.wolfram.com/groups/-/m/t/1184139 Why this question affects my choice of finite element solver ------------------------------------------------------------ In my project, I am importing a complex geometry from an STL file into Mathematica as a `MeshRegion`. I would like to edit this mesh significantly: For instance, drastically reduce or increase the number of elements (for finite element analysis) or make boolean operations. But I am not sure if Mathematica can re-mesh a `MeshRegion` that is obtained from discrete data. This question is of personal importance for me because it will help decide if I can go ahead with using `NDSolve` for the finite element analysis of my project: If re-meshing is very problematic in Mathematica, it may be wiser to import volume meshes from external software, or do all of the finite element analysis in a dedicated solver like Comsol. Re-meshing an already discretized `MeshRegion` -------------------------------------------- Let me illustrate. When I work with an `ImplicitRegion`, I seem to have full control, I can make a mesh with very many or few elements, and apply boolean operations too: IR = ImplicitRegion[ x^6 - 5 x^4 y + 3 x^4 y^2 + 10 x^2 y^3 + 3 x^2 y^4 - y^5 + y^6 + z^2 <= 1, {x, y, z}]; << NDSolve`FEM` ToElementMesh[IR, MaxCellMeasure -> Infinity, AccuracyGoal -> 0]["Wireframe"] ToElementMesh[RegionDifference[IR, Cuboid[]], MaxCellMeasure -> 0.001]["Wireframe"] ![enter image description here][1] Next, I create a `MeshRegion` by applying `DiscretizeRegion`. This leaves me in basically the same situation as importing some STL mesh into Mathematica as `MeshRegion`: MR = DiscretizeRegion[IR]; (* same as MR = Import["filename.stl", "MeshRegion"] *) Now I can no longer downsample: ToElementMesh[MR, MaxCellMeasure -> Infinity, AccuracyGoal -> 0] ToElementMesh[MR] ![enter image description here][2] Same thing! No downsampling appears to have happened. Possibly no re-meshing whatsoever. Mathematica returns basically the same number of elements (`TetrahedronElement["<" 16793 ">"]` vs `TetrahedronElement["<" 16841 ">"]`, respectively). Also I can no longer apply boolean operations: RegionDifference[MR, Cuboid[]] // DiscretizeRegion returns an error ![enter image description here][3] Is there some way I could gain control of the `DiscretizeRegion` in the same way as `ImplicitRegion`? In practice, I am presented with discrete data (STL file) and I would like to be able to re-mesh it, do boolean operations, and run FEM in full control of my mesh. Is that possible? Bad brute force solution ----- You can brute force Mathematica to re-mesh using this hack: MR2 = DiscretizeGraphics[ RegionPlot3D[ RegionMember[MR, {x, y, z}] == True && RegionMember[Cuboid[], {x, y, z}] == False, {x, -2, 2}, {y, -2, 2}, {z, -2, 2}, PlotPoints -> 20]] ToElementMesh[MR2, MaxCellMeasure -> Infinity, AccuracyGoal -> 0] ![enter image description here][4] However, converting the `MeshRegion` (picture above, left) with `ToElementMesh` hopelessly overmeshes things (picture above, right). I get about 500k tetrahedral elements. Once again I don't know how to control my mesh for finite element analysis. Disclaimer ----- I have posted this question to Mathematica Stackexchange [here][5], where it has gone unanswered for several days. Because of the importance of this question for my modeling work (I need to decide if I can go ahead with Mathematica for the finite element analysis part of my project), I decided to post it to the Wolfram Community to hopefully get some developer feedback. I am happy to summarise the insights of this this discussion for the Stackexchange community once we have (hopefully) reached some conclusions. [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=1.png&userId=1184102 [2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=2.png&userId=1184102 [3]: http://community.wolfram.com//c/portal/getImageAttachment?filename=3.png&userId=1184102 [4]: http://community.wolfram.com//c/portal/getImageAttachment?filename=4.png&userId=1184102 [5]: https://mathematica.stackexchange.com/questions/155484/is-it-possible-to-re-mesh-downsample-upsample-a-discretizeregion-object Alexander Erlich 2017-09-13T17:30:32Z https://community.wolfram.com/groups/-/m/t/1379392 #Summary Detailed digital information about water is of great important not only for urban design and development, but also for [geopolitical strategic plans][1]. A large portion of world population is now living in [water stress regions][2]. There are alarming patterns between the water crisis and political/civil unrest in parts of the world (e.g., Yemen, [Syria][3], even [Iran][4]). Therefore, having digital model of water sources is essential. My main goal here is to extract the hydrographic basin (also called, the drainage basin) out of digital elevation models (DEM). Detailed watershed segmentation of a geographical region is obtained using built-in morphological functions of Wolfram language that I applied on the DEM of the region. The watershed lines dividing neighboring basins, together with their elevation values, are extracted. We treated the centroid of sub-basins as vertexes of a graph where connections between neighboring vertexes depend on the land surface characteristics and boundary lines. We assumed that a valid connection between two vertexes (say, two sub-basins) can be made only if the two vertexes have minimum elevation at the boundary compared to all neighboring vertexes. Our results show a great match with the the watershed segmentation obtained manually by appropriate choice of markers (say, using a seasoned geographer knowledge). # DEM of Lake Urmia (Orumiyeh, Persian: ??????) The region that I choose for my analysis is [lake Urmia][5] on the north west of Iran. It is a UNESCO biosphere reserve which is now dangerously shrinking due to [water mismanagement and also drought][6]. We set the region a box between two points with altitudes and longitudes {35.5,44} and {39,48}. We extract the DEM using GeoElevationData function. region = {{35.5, 44}, {39, 48}}; data = N[QuantityMagnitude@ GeoElevationData[region, UnitSystem -> "Metric"]]; reliefmap = ReliefPlot[data, DataReversed -> True, PlotLegends -> BarLegend[Automatic, LegendLabel -> "elevation(m)"]] ![enter image description here][7] # Watershed segmentation We covert the DEM data into an image. mimg = ImageAdjust[Image[data]]; From now on, we shall treat our DEM as an image with colors corresponds to elevations. We find watershed segmentation (i.e., a detailed sub-basin structure) using Immersion method, mainly because the labeling of segments correspond to the relative value for a sub-basin's minimum elevation. wsc = WatershedComponents[mimg, Method -> "Immersion"]; bnds = ColorNegate@Image[wsc, "Bit"]; ![enter image description here][8] Next, we label the boundary between two adjacent sub-basins by the indexes of corresponding sub-basins. To do so, we use black-and-white image of watersheds where the boundaries (say, lines) have value one (white lines). Then, we assign two indexes to the line separating two sub-basins. These indexes are the indexes of adjacent sub-basins separated by the line. We shall do index assignment of lines in two steps: 1. Shift the sub-basin with higher index into the lines (using Erosion). bindex = Erosion[Replace[wsc, 0 -> Max[wsc] + 1, {2}], 1]; 2. Shift the sub-basin with lower index into the lines (using Dilation) tindex = Dilation[wsc, 1]; The first step has one subtlety: to shift the higher index into the lines, first we need to set the boundary value higher than any value in the image; otherwise the Erosion function will extend the boundaries only. doubleIndexArray = Replace[Transpose[{bindex, tindex}, {3, 1, 2}], {n_, n_} -> {0, 0}, {2}]; doubleIndexPairs = Prepend[DeleteCases[ DeleteDuplicates[Flatten[doubleIndexArray, 1]], {0, 0}], {0, 0}] After labeling the boundaries by two indexes, we find the minimum elevation along the boundaries separating two sub-basins. Now we have the information of adjacent sub-basins and the minimum elevation along the boundaries. bndSegs = Replace[doubleIndexArray, Dispatch@MapIndexed[#1 -> First[#2] - 1 &, doubleIndexPairs], {2}]; MinATborders = Thread[Rest[doubleIndexPairs] -> ComponentMeasurements[{mimg, bndSegs}, "Min"][[All, 2]]] # Graph made out of centroids of sub-basins From now on, we shall treat a sub-basin as a vertex. For the sake of representation, the location of the vertex is set as the centroids of sub-basins. The most important question is how the vertexes are connected together and what are the corresponding clusters made out of connected vertexes. In other word, the clusters represent larger basins that form a unique hydro-graphical ecosystem with a unique water graph. g0=Graph[Apply[UndirectedEdge,Rest@doubleIndexPairs,{1}]]; g1=WeaklyConnectedGraphComponents[g0][[1]]; basinConnect=Sort[Map[# -> AdjacencyList[g1,#]&,VertexList[g1]]] sortBasin[x_]:=Keys[x]->SortBy[Values[x],Association[MinATborders][Sort[{#,Keys[x]}]]&]; sortedBasins=sortBasin[#]&/@basinConnect ##Algorithm to make connections between sub-basins (how/where to connect vertexes and form edges) 1. Built a set for each sub-basin whose elements are the adjacent sub-basins. 2. Sort them by the minimum value of the boundary separating them. 3. Make a connection between two adjacent sub-basin by choosing the first element of the sorted list in step#2 4. Check for possible cycles in the graph (i.e., loops). If any, replace the original connection by the making the connection with the second element in the list of sorted sub-basins obtained in step#2. 5. Repeat step#5 until there is no more change in the graph Here is the graph that we obtain following above steps: ![enter image description here][9] # Comparison of our result with a basin structure obtained by pre-determined markers Let us assume that we are a seasoned geographer with a good knowledge of large basins (e.g., where the rivers are running and what branches of a river are independent and where they merge) and how they are connected on a map. Then, one shall obtain the following basins for our selected region: ![enter image description here][10] Now, let us overlay our graph on the top of map and compare different results ![enter image description here][11] # Concluding remarks Detailed watershed segmentation of a geographical region is obtained using built-in morphological functions of Wolfram language that we applied on the raw Digital Elevation Model (DEM) of the region. The watershed lines dividing neighboring basins, together with their elevation values, are extracted from DEM. We treated the centroid of sub-basins as vertexes of a graph where connections between neighboring vertexes depend on the land surface characteristics and boundary lines. We assumed that a valid connection between two vertexes (say, two sub-basins) can be made only if the two vertexes have minimum elevation at the boundary compared to all neighboring vertexes. Our results show a great match with the the watershed segmentation obtained manually by appropriate choice of markers (say, using a seasoned geographer knowledge). #Future works With the detailed dynamic relationships of hydrographic basins in force, we will be able to develop models describing the water flow (or water level) among basins using the precipitation data. Next, the water usage for irrigation and urban uses can be added into the model. Using these information, one will be able to develop an elaborate model for water management of a large hydrographic basin. Our approach can be improved further by treating a few loops (cycles) that appear in the graph. In this regard, one can remove the connection between two vertexes where the initial one has higher elevation and replace it with the second vertex in the list of neighboring basins sorted by the minimum boundary elevation. [1]: https://geopoliticalfutures.com/water-geopolitical-imperatives/ [2]: https://www.theguardian.com/cities/2016/jul/29/where-world-most-water-stressed-cities-drought [3]: http://www.pnas.org/content/112/11/3241 [4]: https://www.nytimes.com/2018/01/18/climate/water-iran.html [5]: https://www.theguardian.com/world/gallery/2016/sep/02/iran-lake-urmia-in-pictures [6]: https://link.springer.com/article/10.1007/s11269-011-9909-8 [7]: http://community.wolfram.com//c/portal/getImageAttachment?filename=1665ReliefMap.png&userId=1351917 [8]: http://community.wolfram.com//c/portal/getImageAttachment?filename=watershed.png&userId=1351917 [9]: http://community.wolfram.com//c/portal/getImageAttachment?filename=graph.png&userId=1351917 [10]: http://community.wolfram.com//c/portal/getImageAttachment?filename=geo-basin.png&userId=1351917 [11]: http://community.wolfram.com//c/portal/getImageAttachment?filename=overlay.png&userId=1351917 Mads Bahrami 2018-07-11T19:47:32Z https://community.wolfram.com/groups/-/m/t/55094 In the process of developing [url=http://demonstrations.wolfram.com/author.html?author=Jaime%20Rangel-Mondragon]some demonstrations I published recently[/url] about bisecting the perimeter of triangles I came across the following result. Consider the arc AB of a circle, the segment AB not necesarily being a diameter of the circle. If M is the apex in the middle of the arc and we join it to the center N of AB, then clearly AN = BN. In other words, if we join the middle M of arc AB with the middle N of segment AB, segments MN and AB are perpendicular and AN = BN. Perhaps the following manipulate would clarify the idea.[mcode] Manipulate[  Block[{r}, r = Sqrt[1 + h^2];   {c1, c2} = If[h < 0, {6, 5}, {5, 6}];   Graphics[{{EdgeForm[Thick], ColorData[2, c1], Disk[{0, -h}, r],      ColorData[2, c2],      Disk[{0, -h}, r, {ArcTan @@ {1, h}, ArcTan @@ {-1, h}}]},     ColorData[2, c1], Rectangle[{-1, -h}, {1, 0}], Black,     Line[{{-1, 0}, {1, 0}}], Line[{{0, -h + r}, {0, 0}}],     Style[{Text["A", {-1, -.1}], Text["B", {1, -.1}],       Text["M", {0, -h + r + .1}], Text["N", {0, -.1}]}, 20, Bold]},    PlotRange -> 1.2]],    {{h, 0, "circle"}, -1, 2, Slider}] [/mcode] [img=width: 411px; height: 455px;]/c/portal/getImageAttachment?filename=pic1.png&userId=48389[/img] Now for the interesting part. Suppose that there is a point C on the arc AB and instead of defining N as the perpendicular from M to AB, we take N as the perpendicular from M to AC and now relabel as D the middle point of segment AB (if point C is at the left of M we take N to be the perpendicular from M to BC). The following manipulate illustrates this. [mcode] Manipulate[  r = Sqrt[1 + h^2];  al = If[h == 0, Pi/2, ArcTan[1/h]];  a = If[h > 0, {al - Pi/2,     3 Pi/2 - al}, {Pi/2 - al, Pi/2 + al}];  ac = cc Last[a] + (1 - cc) First[a];  c = {0, h} + r {Cos[ac], Sin[ac]};  n = If[First[c] > 0, {-1, 0} +     Projection[{1, h + r}, c + {1, 0}], {1, 0} +     Projection[{-1, h + r}, c - {1, 0}]];  Graphics[{EdgeForm[Thick],    Circle[{0, h}, r, a], Black, Line[{{-1, 0}, c, {1, 0}, {-1, 0}}],    Line[{{0, h + r}, n}],    Style[{Text["A", {-1, -.1}], Text["B", {1, -.1}],      Text["M", {0, h + r + .1}], Text["D", {0, -.1}], Text["C", c],      Text["N", n]}, 20, Bold]},   PlotRange -> {{-1.5, 1.5}, {-.2, 2.5}}],    {{h, 0, "circle"}, -3, .9, Slider},  {{cc, .3, "point C"}, 0, 1, Slider}] [/mcode] [img=width: 411px; height: 447px;]/c/portal/getImageAttachment?filename=pic2.png&userId=48389[/img] Surprisingly, considering lengths, AN = NC + CB (or BN=NC+AC if C is to the left of M). Perhaps some reader can come up with a proof of this result?. What has this to do with bisecting perimeters of triangles and what is the importance of point D to even having received a label? Well, it turns out that the segment DN bisects the perimeter of triangle ABC. In seeking for an algebraic proof of this result, we set a = (1, 0), b = (-1, 0), d=(0,0) and the center of the circle at (0, h) with variable h. The radius of the circle is then r = Sqrt[1 + h^2]. Thus m = (0, h + r) and c = (0, h) + r (cos (t), sin (t)) for some value of t not exceeding pi/2 (the other bound will be mentioned later). The point n is then obtained as[mcode]{a, b} = {{-1, 0}, {1, 0}}; r = Sqrt[1 + h^2]; m = {0, h + r}; c = {0, h} + r {Cos[t], Sin[t]}; n = FullSimplify[a + (m - a).(c - a) (c - a)/((c - a).(c - a))] [/mcode][mcode] {1/2 (-1 + (h + Sqrt[1 + h^2]) Cos[t] + Sin[t]), 1/2 (-Cos[t] + (h + Sqrt[1 + h^2]) (1 + Sin[t]))} [/mcode] With these values, the expression we want to verify being equal to zero is tested as:[mcode] FullSimplify[Norm[n - a] - Norm[n - c] - Norm[c - b] /. Abs[xx_]^2 -> xx^2, t < Pi/2] [/mcode][mcode] -Sqrt[(-1 + h (-h + Sqrt[1 + h^2])) (-1 + Sin[t])] + Sqrt[(1 +     h (h + Sqrt[1 + h^2])) (1 + Sin[t])] -  Sqrt[2] Sqrt[1 + h^2 + Sqrt[1 + h^2] (-Cos[t] + h Sin[t])] [/mcode]So, we are left with the following (probably) true theorem [img=width: 692px; height: 96px;]/c/portal/getImageAttachment?filename=pic3.png&userId=48389[/img] but this takes us further away from the proof we want. Jaime Rangel-Mondragon 2013-07-02T22:05:08Z https://community.wolfram.com/groups/-/m/t/440476 I need the Transverse Cylindrical Equal-Area project for a map. Mathematica supplies the Cylindrical Equal-Area projection, but not its transverse. Does anyone know how to manipulate the Cylindrical Equal-Area projection to produce its transverse? Charles Elliott 2015-02-11T22:47:35Z https://community.wolfram.com/groups/-/m/t/756942 During WTC 2015, some of us had our heads 3D scanned, which was quite fun and educational. But I (we?) never received the 3D file back. Is this still planned? I was hoping to do some computations on my head, like: Plot[DragCoeficient[GeometricTransformation[myHead, RotationTransform[alpha]]],{alpha, 0, 2Pi}] (probably not yet on version 10.4, but hopefully, on a not so distance future, where N-S + turbulence models will be easily solvable :-) Pedro Fonseca 2015-12-12T20:46:00Z https://community.wolfram.com/groups/-/m/t/933907 I make two Cuboids into a box. This 'RegionDifference' works ! tray = Cuboid[-1.0 {998.542, 205.604, 0}, 1.0 {998.542, 205.604, 233.433}]; trayCutout = Cuboid[-1.0 {990.969, 198.032, 7.572}, 1.0 {990.969, 198.032, 225.861}]; reg1 = Fold[RegionDifference, tray, {trayCutout}]; ![enter image description here][1] I then make a 'spoon' region to use as a cutout, ![enter image description here][2] but this 'RegionDifference' fails ... spoon = RegionPlot3D[0.0452 (x^2 + y^2) - z <= 0, {x, -30.231, 30.231}, {y, -30.231, 30.231}, {z, 0, 41.345}]; cutout = Translate[First[spoon], {525.466, 0, -41.345}]; reg2 = Fold[RegionDifference, reg1, cutout]; **What do I need to do to 'reg1' after the 1st RegionDifference to make it usable in the 2nd RegionDifference & how can I make the 'cutout' usable in a RegionDifference ? Any ideas how to make this work ? Thanks** [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2016-10-03at1.42.06PM.png&userId=929271 [2]: http://community.wolfram.com//c/portal/getImageAttachment?filename=ScreenShot2016-10-03at2.13.47PM.png&userId=929271 Rick Tschudin 2016-10-03T21:25:59Z https://community.wolfram.com/groups/-/m/t/602402 Awhile ago I wrote the Demonstration [Packing Squares with Side 1/n](http://demonstrations.wolfram.com/PackingSquaresWithSide1N/). It is based on the Paulhus* method, which starts with a rectangle slightly larger than all the fractional squares, then packs all the squares into it via a clever method. $$\sum\limits_{a=2}^\infty \frac{1}{a^2} = \frac{\pi^2}{6}-1 \approx 0.644934067 \approx 0.644934069 \approx \frac{545}{467} \times \frac{21}{38}$$ ![Method of Paulhus][1] Oblongs of size $ \frac{1}{1} \times \frac{1}{2}$, $ \frac{1}{2} \times \frac{1}{3}$, $ \frac{1}{3} \times \frac{1}{4}$, $ \frac{1}{4} \times \frac{1}{5}$, ... have a total area of 1. $\sum\limits_{a=1}^\infty \frac{1}{a (a+1)} =1$ I believe it's still an unsolved question whether the infinite rectangles can fit in the unit square. If it's been solved, I'd love to see the paper. Anyways, I haven't figured out how to modify my code to intelligently handle oblongs instead of squares. Can anyone figure that out and use the code to pack the first 2000 oblongs in the unit square? (*) M. M. Paulhus, "An Algorithm for Packing Squares," Journal of Combinatorial Theory, Series A, 82(2), 1998 pp. 147–157. [1]: http://community.wolfram.com//c/portal/getImageAttachment?filename=PaulhusPacking.gif&userId=21530 Ed Pegg 2015-10-30T15:32:06Z https://community.wolfram.com/groups/-/m/t/295266 Is there any setting that could be used to improve the results of TransformedRegion? For some transformations it appears to perform very poorly by default. Below is one example taken from [MMA.SE][1] (the transformation function specifically is written by [Michael E2][2]). Transformation function: flow = Module[{x, y}, Function[{t, a, b}, Evaluate[ { x[t, a, b], y[t, a, b]} /. First@NDSolve[{ D[x[t, a, b], t] == y[t, a, b], D[y[t, a, b], t] == -Sin[x[t, a, b]], x[0, a, b] == a, y[0, a, b] == b }, {x, y}, {t, 0, 20}, {a, -1/2, 1/2}, {b, 1/2, 3/2} ] ], Listable]]; Region: object = Disk[{0, 1}, 0.5]; Background: splot = StreamPlot[{y, -Sin[x]}, {x, -4, 4}, {y, -3, 3}, StreamColorFunction -> "Rainbow"] Somewhat correct transformations given by `ParametricPlot` for $t=0$ and $t=10$: Show[ splot, ParametricPlot[ flow[t, r Cos[theta], 1 + r Sin[theta], t], {r, 0, 0.5}, {theta, 0, 2 Pi} ] ] ![t=0][3] ![t=10][4] Incomplete transformations given by `TransformedRegion`: Show[ splot, RegionPlot[ TransformedRegion[ object, flow[t, #, #2] & ] ] ] ![t=0][5] ![t=10][6] [1]: http://mathematica.stackexchange.com/questions/34837/how-to-draw-the-image-of-a-circle-under-the-action-of-a-transformation-of-the-ph [2]: http://mathematica.stackexchange.com/users/4999/michael-e2 [3]: /c/portal/getImageAttachment?filename=IUzHN.png&userId=295249 [4]: /c/portal/getImageAttachment?filename=hKVQu.png&userId=295249 [5]: /c/portal/getImageAttachment?filename=JTHBm.png&userId=295249 [6]: /c/portal/getImageAttachment?filename=rg7Dj.png&userId=295249 Calle E 2014-07-15T21:24:05Z https://community.wolfram.com/groups/-/m/t/994225 I have plotted `ChromaticityPlot["sRGB"]` but the color space boundary does not show up..though the legend is there. What shall I do? anyone please help. Thanks in advance. Timun Mas 2017-01-12T22:09:24Z https://community.wolfram.com/groups/-/m/t/2298518 Mathematica already packs some nice and strong functionality to handle meshes and it is probably an area that will be expanding in the upcoming versions. I am quite curious to know from the computational geometry development team whether there are any plans to incorporate the tools/functionality for handling discrete differential geometry in the future version 13. There are extensive libraries which can be integrated with Mathematica to expand its computational geometry based functionality: https://www.cgal.org/ and https://www.cs.cmu.edu/~kmcrane/index.html#code and https://www.cs.cmu.edu/~kmcrane/Projects/DDG/ Some features that will be indispensable to have would be: 1. Computing discrete gradients of volume and surfaces at the mesh vertices 2. determining Gaussian and Mean curvatures of a mesh 3. implementing Half-edge structures 4. Curvature Flow 5. represent vector fields on meshes It would be great if someone from the development team can engage here or provide some kind of a roadmap for the functionality in sight for discrete differential geometry. Ali Hashmi 2021-06-24T13:11:14Z https://community.wolfram.com/groups/-/m/t/983595 I seem to only be able to show the end points of a meshed line --how can I make those middle points (should be 2 through 4 in blue) appear ? See attached. Regards and Thanks, John ;-) Needs["NDSolve`FEM`"]; ?=ImplicitRegion [True,{{x,0,1}}]; mesh1=ToElementMesh[?,MaxCellMeasure->0.25,"MeshOrder"->1]; Show[ mesh1["Wireframe"["MeshElementIDStyle"->Red]], mesh1["Wireframe"["MeshElementIDStyle"->Blue,"MeshElement"->"PointElements","MeshElementIDStyle"->Blue]]] john massucci 2016-12-20T11:40:08Z https://community.wolfram.com/groups/-/m/t/501086 I have data which are the longitudes (l) and latitudes (b) for the vertices of the edges of a shape as it appears to an observer at a given location (i.e., as it appears on the observer's sky plane). Question: If I have some general idea what the original 3D object is (e.g., a surface of revolution formed from a lemniscate), how do I find the parameters of the 3D shape that best reproduce the 2D image the observer sees? The data are (in degrees) data = {{0, 0}, {-5.5, 5.}, {-10.7, 10.}, {-12.9, 15.}, {-15., 20.}, {-16.3, 25.}, {-16., 30.}, {-15.5, 35.}, {-15., 40.}, {-15., 45.}, {-10.6, 50.}, {-3.7, 52.5}, {6.3, 53.5}, {13.8, 50.}, {21.8, 45.}, {25.3, 40.}, {26.7, 35.}, {26.3, 30.}, {25.6, 25.}, {23., 20.}, {18.8, 15.}, {13.8, 10.}, {0, 0}} In 3-space the observer is located at {xobs,yobs,zobs} = {-8,0,0} and I can assume the 3D object emanates from {0,0,0} The transformation from general {x,y,z} to the angular position, {l,b}, on the observer's sky plane is given by long[{x_, y_, z_}, {xobs_, yobs_, zobs_}] := ArcTan[(y - yobs)/(x - xobs)] lat[{x_, y_, z_}, {xobs_, yobs_, zobs_}] := ArcTan[(z - zobs)/Sqrt[(x - xobs)^2 + (y - yobs)^2]] As an example, take it that I know the 3D object is a surface of revolution formed from a lemniscate lemSurface[x_, y_, z_, xa_, ya_, za_, c_] := (x^2 + y^2 + z^2)^2-2c^2(-(xa y - x ya)^2-(xa z - x za)^2-(ya z - y za)^2+(x xa + y ya + z za)^2) The surface is given implicitly by lemSurface[x, y, z, xa, ya, za, c]==0 where {xa, ya, za} is the vector around which the lemniscate is rotated and c scales its size. So my question reduces to: how can I find the values for xa, ya, za, c such that the projected image best fits data ? One way to find the {x,y,z} coordinates of the edge of the object as it appears to the observer is that these are given implicitly by the requirement that these are i) on the surface and ii) the observer is in the tangent plane for that coordinate on the surface. The surface normal is given by D[lemSurface[x, y, z, xa, ya, za, c], {{x, y, z}}] The tangent plane at any point on the surface {xs,ys,zs} is then D[lemSurface[xs, ys, zs, xa, ya, za, c], {{xs, ys, zs}}]).{x-xs,y-ys,z-zs} so that the {x,y,z} coordinates of the edge of the object seen by the observer are given implicitly by \[ScriptCapitalR] = ImplicitRegion[lemSurface[x, y, z, xobs, yobs, zobs, xa, ya, za, c] == 0 && D[lemSurface[x, y, z, xa, ya, za, c], {{x, y, z}}]).{xobs-x,yobs-y,zobs-z}== 0, {x, y, z}] I can plot the [ScriptCapitalR] is 3 space via DiscretizeRegion[\[ScriptCapitalR]] but this still does not give me the {l,b} coordinates of the projected image or allow me to fit the image of the implicit region to data. I guess ideally what I want is a function for the appearance of the silhouette from the observer position that can be fitted to the observed data, i.e., silhouette[l_ ,b_, xa_, ya_, za_, c_, xobs_, yobs_, zobs_] == 0 Roland C 2015-05-21T05:20:48Z https://community.wolfram.com/groups/-/m/t/455706 This question is posted in [mathematica.stackexchange][1] 1. For $1\le i<j<\infty$ let $X_{i,j}$ be independant and indentically distributed (i.i.d) real random variables with mean $0$ and variance $1$ and set $X_{j,i}=X_{i,j}$. Let $X_{i,i}$ be i.i.d. real random variables (with possibly a different distribution) with mean $0$ and variance $1$. Then $M_n=\left[ X_{i,j}\right]_{i,j=1}^n$ will be a random $n\times n$ symmetric matrix. ---------- 2. There are $n$ random eigenvalues which we will denote by $$ \lambda_1\le \lambda_2\le \dots \lambda_n. $$ The empirical spectral measure is $$ \nu^*_n=\frac1n \sum_{i=1}^n \delta_{\lambda_i}. $$ This is a random discrete probability measure which puts $n^{-1}$ mass to each (random) eigenvalue. ---------- 3. **theorem:(Wigner's semicircle law)** Let $$ \nu_n=\frac1n \sum_{i=1}^n \delta_{\frac{\lambda_i}{\sqrt{n}}}. $$ be the normalized empirical spectral measure. Then as $n\to \infty$ we have $$ \nu_n \Rightarrow \nu \qquad \text{a.s.} $$ where $\nu$ has density $$ \frac{d\nu}{dx}=\frac{1}{2\pi} \sqrt{4-x^2}\,\chi_{\{|x|\le 2\}}. $$ ($\chi_A$ denote the indicatrice function of $A$) ---------- **Question** Using simulation to understand this theorem in other word: Do the same thing that was done for the central limit theorem as it is shown in this page [CLT][2] I have already posted [here][3] about the same subject. Any help is welcome. [1]: http://mathematica.stackexchange.com/questions/76806/simulation-of-the-semicircle-wigner-theorem [2]: http://demonstrations.wolfram.com/TheCentralLimitTheorem/ [3]: http://mathematica.stackexchange.com/questions/75711/how-can-i-make-an-animated-demonstration-of-the-wigner-semicircle-distribution Labair Abdelkader 2015-03-09T14:39:31Z https://community.wolfram.com/groups/-/m/t/1968936 This post is to spark some discussion: To anyone who has used Mathematica for computational geometry and has had some experience with CGAL: https://www.cgal.org/ How much of the functionality from CGAL has been implemented in Mathematica till date? And is it possible to link CGAL with Mathematica to increase the strength of Computational Geometry in Mathematica? Ali Hashmi 2020-05-09T12:51:29Z https://community.wolfram.com/groups/-/m/t/829750 I've created a model, on a flat coordinate plane, the movement of the sun and moon using their known positions throughout the year. Both the sun and moon are orbiting the north pole with their respective frequencies. What I was surprised to find was the moon orbiting an apparent second center. Even though it is revolving around the north center, it also appears to orbit around the orbiting sun. I was curious if this type of movement could be expressed in a formula. It seems this double orbit effect would only be possible if their respective speeds are as they are. If I were to change the orbital period of one or the other, this effect would not happen and I'm curious to see how a formula would show this but don't know if something like this has been described before with a formula. Hoping one of the math wizards here will have a better clue than I :) https://www.youtube.com/watch?v=jTB6w6CHhIk TIA! Steve Torrence 2016-03-25T11:41:27Z