New answers tagged

0

This should work: g = {g0, g1, g2, g3, g4}; s = {S0, S1, S2, S3, S4}; Animate[Labeled[Graphics3D[g[[i]], ViewPoint -> Front], s[[i]],Top], {i, 1, 5, 1}, AnimationRunning -> False,PreserveImageOptions -> False]


0

I think what you are asking is a default feature. For example ListAnimate[Table[Graphics3D[{Hue[x], Cuboid[]}], {x, 0, 1, .01}]] If you pause it, rotate the Cuboid and then play again, you will see that it will resume its initial position.


4

You may do this by Plotting multiple figures and show them one by one via using Manipulate, Animate or Dynamic. Animate[ ListPlot[coord[[;; t]], PlotMarkers -> {Automatic, 20}, PlotRange -> {{0, 10}, {0, 3000}}], {t, 1, Length@coord, 1}] or Dynamic[ With[{t = Clock[{1, Length[coord], 1}]}, ListPlot[coord[[;; t]], ...


4

See PlotRange hh = 3; ll = 10; Animate[Graphics[{Thick, Blue, Line[{{0, 0}, {0, hh}}] , Thick, Green , Line[{{xr, 0} , {xr - (ll*xr)/Sqrt[xr^2 + hh^2], (ll*hh)/Sqrt[xr^2 + hh^2]}}]} , Axes -> True , PlotRange -> {{-4, 10}, {0, 4}}] , {xr, 3, Sqrt[ll^2 - hh^2]}]


3

I came up with a solution using ReflectionTransform[]. Here's my code: TabView[{ "Square" -> Manipulate[Graphics[ {{Opacity[1], Red, Rectangle[{a,b},{c,d}]}, Line[{{5, 5}, {5, -5}}], GeometricTransformation[{Opacity[1], Blue,Large,Rectangle[{a,b},{c,d}]}, ReflectionTransform[{5, 0}, {5,0}]]}],{a,0,2},{b,0,2},{c,1,4},{d,1,4} ], "Triangle" -&...


2

There was no problem exporting at the correct frame rate when I initially tried it with a small number of test frames, given that the code in the question doesn't produce any actual frames. However, I now tried again with a large table of frames to replicate the specific number of 979 frames in the question. Indeed, in this case the resulting movie has the ...


3

I admit that I am uncertain what the aim is. I post this with the hope that it may prompt correction/clarification with respect to the goal. In the following the mesh shading reflects mesh functions y=Sin[x] (the 1 was used to avoid difficulties with 0). Either one certainly would 'cut the sphere' in half (but any great circle could be chosen without the ...


1

In addition I added a point animation and observed some oddly behaviour of Dynamic and ContourPlot3D. Functions and Conditions. yPath has to be defined immidiately (without ":") cause the later \ recursive defintion wouldn' t work otherwise yPath[x_] = -((3 Sqrt[25 - x^2])/5); zPath[x_] := Sin[x] Sin[yPath[x]] comparison[x_] := yPath[x] == (-yPath[x]) ...


4

I am not entirely sure what the aim is here. I am choosing a different ellipse. However, this could be easily changed. p3 = Plot3D[Sin[x] Sin[y], {x, -5, 5}, {y, -5, 5}, PlotStyle -> {Green, Opacity[0.5]}, Mesh -> False]; cp = ContourPlot3D[ x^2/25 + y^2/9 == 1, {x, -5, 5}, {y, -5, 5}, {z, -1, 1}, Mesh -> None, ContourStyle -> Opacity[...


3

For changing the position of the cylinder you could add a dynamic Slider. Regard that "Show" won't work within "Dynamic", so thats why I wrote the trigonometric function in the contour argument. dx = 0; dz = 0; Row[{Slider2D[Dynamic[{dx, dz}], {{-10, -10}, {10, 10}}], "dx =" Dynamic[dx], ", dz =" Dynamic[dz]}] Dynamic[ContourPlot3D[{((x + dx)/5)^2 + ((z ...


2

You might look into ContourPlot3D: p1 = ContourPlot3D[(x/5)^2 + (y/3)^2 == 1, {x, -10, 10}, {y, -10, 10}, {z, -10, 10}, ContourStyle -> LightBlue]; p2 = Plot3D[Sin[x] Sin[y], {x, -10, 10}, {y, -10, 10}]; Show[p1, p2] Of course, if you want to do a little more with this, it helps to put both in a contourplot. So, how do we get the intersection of ...


0

Change 2 Pi in the PlotRange for the line with t0 Animate[Show[pl1, ParametricPlot3D[Block[{u = 3 t, v = 4 t}, {Cos[v] (r + Cos[u]), Sin[v] (r + Cos[u]), Sin[u]}], {t, 0, t0}, PlotStyle -> Thick], Graphics3D[{Green, Sphere[Block[{u = 3 t0, v = 4 t0}, {Cos[v] (r + Cos[u]), Sin[v] (r + Cos[u]), Sin[u]}], 0.1]}]], {t0, 0, 2 Pi}]


0

Animate[Show[pl1, ParametricPlot3D[Block[{u = 3 t, v = 4 t}, {Cos[v] (r + Cos[u]), Sin[v] (r + Cos[u]), Sin[u]}], {t, 0, t0}, PlotStyle -> Thick], Graphics3D[{Green, Sphere[Block[{u = 3 t0, v = 4 t0}, {Cos[v] (r + Cos[u]), Sin[v] (r + Cos[u]), Sin[u]}], 0.1]}]], {t0, 0, 2 Pi}]



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