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12

Rotation about the origin MapIndexed[N@Nest[r, #1, First[#2-1]] &, points] {{0., 0., 0.}, {0.984808, 0.173648, 0.}, {1.87939, 0.68404, 0.}, {2.59808, 1.5, 0.}, {3.06418, 2.57115, 0.}, {3.21394, 3.83022, 0.}} ListPlot[%[[All, {1, 2}]]] the norm of the vectors is conserved. Rotation about the last point Ok, with the new request, rotating ...


12

Using Composition I can apply RotationTransform, TranslationTransform , ShearingTransform one after the other. Graphics3D[{ Opacity[1] , Red , Arrow[{{0, 0, 0}, {1, 0, 0}}] , Green , Arrow[{{0, 0, 0}, {0, 1, 0}}] , Blue , Arrow[{{0, 0, 0}, {0, 0, 1}}] , Opacity[0.2] , GeometricTransformation[Cuboid[-{1, 1, 1}/4, {1, 1, 1}/4], ...


10

Here is the general answer for any shaped object of surface genus-0, though it can have holes as long as it's an outer boundary (maybe its more general and someone can correct me). I will first describe the general UV mapping. This is usually done for a surface with a pre-chosen boundary, you need to choose which points are part of the boundary and give ...


7

im1 = Import["http://i.stack.imgur.com/78jWB.png"]; {cx, cy} = {50, 50}; cen = ComponentMeasurements[im1, "Centroid"][[All, 2]][[1]]; im2 = ImageForwardTransformation[im1, (# + {cx, cy} - cen) &, DataRange -> Full]; (* or ImageForwardTransformation[im1, TranslationTransform[{cx, cy}-cen], DataRange -> Full] *) im3 = ImageTransformation[im1, (# - ...


7

The easiest way to think about it is to iteratively bend the "line" of points as the program "moves" down the list. "Moves" is accomplished with Rest. The result regrettably has extra data that needs to be discarded, which is done with the ...[[All, 1]] bit of code. rot = N@NestList[ Rest@RotationTransform[10 Degree, {0, 0, 1}, #1[[1]]][#1] &, ...


7

Just use MeshFunction. Manipulate[ParametricPlot3D[{Sin[\[Theta]] Cos[\[Phi]], Sin[\[Theta]] Sin[\[Phi]], Cos[\[Theta]]}, {\[Theta], 0, \[Pi]}, {\[Phi], 0, 2 \[Pi]}, PlotStyle -> Opacity[0.5], Mesh -> {{0.}}, MeshStyle -> {Red, Thick}, MeshFunctions -> {Sin[a] Cos[b] #1 + Sin[a] Sin[b] #2 + Cos[a] #3 &}], {a, 0, \[Pi]}, {b, 0, 2 \[Pi]}] ...


6

Here's an unwrapping of the side of the cylinder, if that is what you are after. With the OP's polys and pts (CoordinateTransform taken from @Kuba's answer), we begin by selecting only the polygons on the side of the cylinder (sidepolys). (There's no noise in this data, at least as far as the end caps go, so the tolerance tol could be set to 0.) pts2 = ...


5

Looks to me to be a bug in RegionPlot. I say fhis beccause DiscretizeRegion @ TransformedRegion[Rectangle[{0, 0}, {1, 1}], {#1^(1/3) + #2, 1 + #2} &] gives as expected.


4

Some days ago,I look at a python's module called turtle.And I think weather we can simulate the behavier of some functions in turtle. First,I define some pre-function: initial[position_, th_] := Module[{θ = th, pos = position}, left := Function[theta, θ = θ + theta]; right := Function[theta, θ = θ - theta]; forward := Function[r, pos = pos + r*{Cos[θ], ...


4

I reported the issue to WRI on November 21. On December 16 they confirmed that it is a known bug. (I forgot to update the question till now.)


3

if you want to maintain the original distances between points I would also suggest the following: points1 = points = Most /@ {{0, 0, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}, {4, 0, 0}, {5, 0, 0}}; l = Length[points]; n = 1; While[n++; n <= l , points[[n ;;]] = RotationTransform[10 Degree, points[[n - 1]]][#] & /@ points[[n ;;]]]; ...


3

If (a,b) is a point in Affine space, then its projective coordinates are [a:b:1]. To select the representative on the unit sphere we can choose (a/r,b/r,1/r) where r=sqrt(a^2+b^2+1). Revering the transformation, given a point on the unit sphere (x,y,z) we can recover the affine point [a:b:1] as [x/z:y/z:1]. Using this idea try: ContourPlot3D[x^2 + y^2 + z^2 ...


3

Here's how I arrived at this (admittedly, quite ugly) mapping function: The "right" function to map points from cylinder coordinates to image coordinates in a pinhole camera model is the one MarcoB linked to in his comment. In a nutshell: {Sin[u],Cos[u],v,1} converts from cylinder coordinates u,v to (homogeneous) 3d coordinates which you then multiply by ...


2

Here is another approach (to simplify i give a 2D solution - your problem is in 2D - but it can be easily extended in 3D) : (* the coordinates of the first point *) ori = {0, 0}; (* the initial angles between the succesive points *) angles = {0, 0, 0, 0} (* the function to build the coordinates from the angles *) mkpts[ang_] := FoldList[(#1 + {Cos@#2, ...


2

You have to use the TransformationMatrix instead of the TransformationFunction: TransformationMatrix@result[[2]] {{0., -1., 0., 4.}, {1., 0., 0., 1.}, {0., 0., 1., 0.}, {0., 0., 0., 1.}} TransformationMatrix@result[[2]].m {{0., -1., 0., 2.}, {1., 0., 0., 2.}, {0., 0., 1., 0.}, {0., 0., 0., 1.}}


1

I guess there is some floating-point-related issue here... This works: f[t_] := RegionPlot[ TransformedRegion[ Rectangle[{-1, -1}, {1, 1}], { Indexed[#1, {1}] (1 + t (Indexed[#1, {2}]^2 - 1)) + 2 t, Indexed[#1, {2}] (1 + t (Indexed[#1, {1}]^2 - 1)) + 2 t } & ], PlotRange -> {{-1 + 2 t, 1 + 2 t}, {-1 + 2 t, 1 + 2 t}} ] ...


1

You can extract the matrix from result using Part: result[[2, 1]].m {{0., -1., 0., 2.}, {1., 0., 0., 2.}, {0., 0., 1., 0.}, {0., 0., 0., 1.}}



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