stylesheet - How to use version 8 default.nb style sheet in place of version 9 default.nb

I do not like the new default.nb style sheet in version 9, and would like to use the older one that was used in version 8 for any new notebook and for any notebook created since I started using V9.

I saw in the StyleSheets folder that V8 default.nb is still there

enter image description here

So I went to preferences and changed the following to say Default_8.0.nb

enter image description here

And restarted Mathematica. Now when I open a new notebook, it seems to be Default_8.0.nb since Section headers are in black and text font is times new Roman. So I thought that is good. Problem fixed. (even thought the Style sheet Default does not have the check mark on next to the it, so I am worried I need to do something more)

enter image description here

However, when I opened an existing notebook that I made some changes in since V9 was installed, I noticed that it still using V9 default.nb. Even though the Style sheet says Default.

enter image description here

Question is What is the correct and recommended steps to use Version 8 default.nb in version 9, so that any notebook that was used version 9 default.nb will now automatically use Default.8.0.nb

(I really do not understand how can WRI make this change for users without asking them. I do not want someone changing the style sheet of my documents without at least asking me. This should have an option for the user to decide on)

thanks

Answer

I believe DefaultStyleDefinitions only selects which style sheet to use for new Notebooks. It does not actually change which sheet is represented by "Default" in the list. This is why you see that item unchecked.

I believe you will need to replace the Default.nb file with the v8 version if you wish to force that behavior on existing Notebooks, but I don't know what you might break doing so.

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