Skip to main content

graphics3d - How to improve visibility of surfaces with opacity?


Consider this example


Graphics3D[{EdgeForm[None], Opacity[0.5],
Blue, Cuboid[{0, 0, 0}, {1, 1, 1/3}],
Green, Cuboid[{0, 0, 1/3}, {1, 1, 2/3}],

Red, Cuboid[{0, 0, 2/3}, {1, 1, 1}],
Gray, Opacity[1], Sphere[{1/2, 1/2, #/3 - 1/6}, 0.1] & /@ {1, 2, 3}},
Boxed -> False]

enter image description here


For the green box, the edges and the surfaces in the back are also quite visible. For blue and red boxes the contrast is much smaller. How to improve it such a way that the edges would be visible from the difference in colour and also make the balls inside visible?


One possible way can be using an extra layer with slightly different opacity


Graphics3D[{EdgeForm[None], Opacity[0.5],
Blue, Cuboid[{0, 0, 0}, {1, 1, 1/3}],
Green, Cuboid[{0, 0, 1/3}, {1, 1, 2/3}],

Red, Cuboid[{0, 0, 2/3}, {1, 1, 1}],
Opacity[0.6],
Blue, Polygon[{{0, 0, 0}, {1, 0, 0}, {1, 1, 0}, {0, 1, 0}}],
Red, Polygon[{{0, 0, 2/3}, {0, 1, 2/3}, {0, 1, 1}, {0, 0, 1}}],
Gray, Opacity[1], Sphere[{1/2, 1/2, #/3 - 1/6}, 0.1] & /@ {1, 2, 3}},
Boxed -> False]

enter image description here


For a lot of boxes, it does not seem to be a good option.



Answer




One (tedious) way is to define a box with different Opacity for different surfaces.


opqbox[{x1_, y1_, z1_}, {x2_, y2_, z2_}, {d_, u_, l_, r_, b_, f_}] := {EdgeForm[None],
Opacity[b], Polygon[{{x1, y1, z1}, {x2, y1, z1}, {x2, y2, z1}, {x1, y2, z1}}],
Opacity[u], Polygon[{{x1, y1, z2}, {x2, y1, z2}, {x2, y2, z2}, {x1, y2, z2}}],
Opacity[r], Polygon[{{x1, y1, z1}, {x2, y1, z1}, {x2, y1, z2}, {x1, y1, z2}}],
Opacity[l], Polygon[{{x1, y2, z1}, {x2, y2, z1}, {x2, y2, z2}, {x1, y2, z2}}],
Opacity[o], Polygon[{{x1, y1, z1}, {x1, y2, z1}, {x1, y2, z2}, {x1, y1, z2}}],
Opacity[f], Polygon[{{x2, y1, z1}, {x2, y2, z1}, {x2, y2, z2}, {x2, y1, z2}}]}



opc={0.5,0.15,0.35,0.20,0.25,0.25};
Graphics3D[{EdgeForm[None],Opacity[0.5],
Blue,opqbox[{0,0,0},{1,1,1/3},opc],
Green,opqbox[{0,0,1/3},{1,1,2/3},opc],
Red,opqbox[{0,0,2/3},{1,1,1},opc],
Gray,Opacity[1],Sphere[{1/2,1/2,#/3-1/6},0.1]&/@{1,2,3}},
Boxed->False]

enter image description here


d, u, l, r, b, f are the Opacity for down, up, left, right, back and front face.



enter image description here


This cube is made by the method suggested by Michael E2 in this answer.


Comments

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...

calculus and analysis - Integral of the Sinc product

Let us consider the following integral $$ B_n = \int_0^\infty \prod_{k=1,3,5,\dots}^n\frac{\sin (x/k)}{x/k}dx $$ By definition, Sinc[x] == Sin[x]/x , therefore B[n_?OddQ] := Integrate[Product[Sinc[x/k], {k, 1, n, 2}], {x, 0, Infinity}] B /@ Range[1, 13, 2] {Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2, Pi/2} OK, everything is fine. But... B[15] 467807924713440738696537864469 Pi/935615849440640907310521750000 What's going on? Answer As Eckhard wrote in comments B[n] is the n-th Borwein integral . (The letter B was not accidental :) ) This funny properties of Borwein integrals is related to the Fourier transform of Sinc function FourierTransform[Sinc[x], x, k] 1/2 Sqrt[Pi/2] (Sign[1 - k] + Sign[1 + k]) Plot[%, {k, -2, 2}, Filling -> 0] which is the box function. The result is $\pi/2$ while the sum $$ 1/3+1/5+\dots+1/n $$ B_n = \frac{\pi}{2} - \pi \bigg(\sum_{k=3,5,\ldots}^n\frac{1}{k} -1\biggr)^\frac{n-1}{2}\prod_{k=3,5,\dots}^n\frac{k}{k-1}. $$ For $n=15$ it is equal to $$ \frac{46780...

list manipulation - Selecting multiple columns from a matrix?

Sample data: data = { {{2013, 1, 1}, 24.13, 167.67, 231.82}, {{2013, 1, 2}, 32.15, 170.92, 225.99}, {{2013, 1, 3}, 35.43, 172.68, 221.67}, {{2013, 1, 4}, 36.73, 173.05, 218.32}, {{2013, 1, 5}, 58.19, 165.96, 197.05}, {{2013, 1, 6}, 69.99, 163.50, 187.52}, {{2013, 1, 7}, 71.37, 154.21, 175.58}, {{2013, 1, 8}, 72.51, 149.66, 163.25}}; I want a DateListPlot with three graphs, so for a matrix formed by columns 1 and 2, one for columns 1 and 3, and 1 for columns 1 and 4. At the moment I'm using this code: data2 = Transpose[{data[[All, 1]], data[[All, 2]]}]; data3 = Transpose[{data[[All, 1]], data[[All, 3]]}]; data4 = Transpose[{data[[All, 1]], data[[All, 4]]}]; DateListPlot[{data2, data3, data4}, Joined -> True, Filling -> {3 -> {1}}] but I have a hunch that this can be done more efficiently. I don't like the Transpose s in particular. Any ideas? edit (for extra credit) What if I need to multiply the second column by 2, which in my solution is simp...