geometry - Projection of a 3d curve to 2d

This question is related to:

How to obtain the coordinates of the intersection line of two surfaces?

Read the link first.

First, I have tried MaxCellMeasure->0.5, MaxCellMeasure->{"Length"->0.5} and MaxCellMeasure->{"Area"->0.5}, and change the number 0.5. It seems that the numbers of point doesn't change.

Secondly, it's very fast to run your code, but it takes about minutes to run my code with my f[x,y,z], g[x,y,z]. And this is my code:

Clear[f,g];
f[x_,y_,z_,t_]:=x^2+y^2+(z-1)^2-t^2/4;
g[x_,y_,z_]:=Piecewise[{{(x-1/2)^2+y^2+z^2-1/16,(x-1/2)^2+y^2-1/16<=0&&z>=0}},z];
reg=DiscretizeRegion[ImplicitRegion[{f[x,y,z,2]==0, g[x, y, z] == 0}, {x, y, z}]];
MeshCoordinates[reg]

Is it because I used Piecewise in g function that it needs more time to run the code?

Third, what I want to do next is below. In fact, I want to plot the projection of the intersection line on xy plane. And this is point projection, as shown below

Connect (0,0,1) and one point on the intersection line, extend it to intersect with xy plane. The yellow pint is want I am looking for. Repeat this for all points on the intersection line, then I can get a closed curve on xy plane. And I need to plot the curve in a 2-dimensional figure, instead of three-dimensional figure here.(I don't want to use Plot3D+Viewpoint to show a 2d curve in 3d coordinate, because the exported pdf file will be large.) I can calculate the yellow points' coordinate (x',y') based on (x,y,z) of red points on the intersection line. x'=x/(1-z),y'=y(1-z). But how to achieve this?

Or maybe I should use another way: use ParametricPlot

ParametricPlot[{x/(1-z),y/(1-z)},{x,y,z}] and x,y,z satisfy f[x,y,z]==g[x,y,z]

Comments

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.

Blog

Search This Blog