Isometric perspective for Graphics3D

Question

Is it possible to render Graphics3D in an isometric projection? I know that the ViewPoint option can be used for orthogonal projection by specifying, e.g. ViewPoint -> {0, Infinity, 0}. This doesn't take multiple infinities though, so I can't do ViewPoint -> {Infinity, -Infinity, Infinity}, for instance.

I realise that I could achieve this by rotating the entire scene about two axes and using an orthogonal projection:

Graphics3D[
  Rotate[
    Rotate[
      Cuboid[{-.5, -.5, -.5}], 
      Pi/4, 
      {0, 0, 1}
    ], 
    ArcTan[1/Sqrt[2]], 
    {0, 1, 0}
  ], 
  ViewPoint -> {-Infinity, 0, 0}
]

However, this is rather cumbersome and it's harder to figure out the correct rotations for the viewpoint I'm interested in. I'd rather just specify the octant from which to view scene isometrically. Is there actually a "proper" way to achieve this?

Answer 1

As of V11.2 we can use a combination of ViewProjection and ViewPoint:

Graphics3D[Cuboid[], ViewProjection -> "Orthographic", ViewPoint -> {1, 1, 1}]

Various vantages:

v = Tuples[{Tuples[{-1, 1}, 3], IdentityMatrix[3]}];

Graphics3D[Cuboid[{-.5, -.5, -.5}, {1., 2., 4}], ViewProjection -> "Orthographic", 
  ViewPoint -> #1, ViewVertical -> #2] & @@@ v

Answer 2

[Edit notice: Updated to allow the setting of the vertical direction of the plot and to fix an error.]

Here is a slight generalization of my answer to Isometric 3d Plot. To get an isometric view, we need to construct a ViewMatrix that will rotate a vector of the form {±1, ±1, ±1} to {0, 0, 1} and project orthogonally onto the first two coordinates.

ClearAll[isometricView];
isometricView[
    g_Graphics3D,                                 (* needed only for PlotRange *)
    v_ /; Equal @@ Abs[N@v] && 1. + v[[1]] != 1., (* view point {±1, ±1, ±1} *)
    vert_: {0, 0, 1}] :=                          (* like ViewVertical; default: z-axis *)
 {TransformationMatrix[
   RescalingTransform[
     EuclideanDistance @@ 
       Transpose[Charting`get3DPlotRange@ g] {{-1/2, 1/2}, {-1/2, 1/2}, {-1/2, 1/2}}].
    RotationTransform[{-v, {0, 0, 1}}].
    RotationTransform[{vert - Projection[vert, v], {0, 0, 1} - Projection[{0, 0, 1}, v]}].
    RotationTransform[Mod[ArcTan @@ Most[v], Pi], v].
    TranslationTransform[-Mean /@ (Charting`get3DPlotRange@ g)]],
  {{0, 1, 0, 0}, {1, 0, 0, 0}, {0, 0, 1, 0}, {0, 0, 0, 1}}};

foo = Graphics3D[Cuboid[{-.5, -.5, -.5}, {1., 2., 4}]];
Show[foo, ViewMatrix -> isometricView[foo, {1, 1, 1}, {0, 0, 1}], 
 ImagePadding -> 20, Axes -> True, AxesLabel -> {x, y, z}]
Show[foo, ViewMatrix -> isometricView[foo, {-1, 1, 1}, {1, 1, 0}], 
 ImagePadding -> 20, Axes -> True, AxesLabel -> {x, y, z}]

All combinations of viewpoints and vertical axes:

Notes:

Getting an accurate plot range that includes the padding is important to computing the correct view matrix. There are alternatives to the undocumented internal function Charting`get3DPlotRange. Alexey Popkov has a method here: How to get the real PlotRange using AbsoluteOptions? I used PlotRange /. AbsolutOptions[g, PlotRange] and multiplied by 1.02 (I don't recall why not something like 1.04) to approximate the padding in my answer to Isometric 3d Plot.

My go-to resource for understanding ViewMatrix has been especially Heike's answer to Extract values for ViewMatrix from a Graphics3D.

This update is in response to Yves' comment. Working with the axes made me realize that the coordinate system is flipped (from "right-handed" to "left-handed). Hence I changed the projection from IdentityMatrix[4] to one that flips the x & y coordinates.

It might be a good idea to Deploy the graphics to prevent rotation by the mouse. When the graphics are rotated, the front end resets the ViewMatrix in a rather ugly way.

Answer 3

You can use the following post-process function to apply a general parallel projection:

parallelProjection[g_Graphics3D, axes_, pad_: 0.15] := 
  Module[{pr3, pr2, ar, t},
   pr3 = {-pad, pad} (#2 - #) & @@@ # + # &@Charting`get3DPlotRange@g;
   pr2 = MinMax /@ Transpose[[email protected]];
   ar = Divide @@ Subtract @@@ pr2;
   t = AffineTransform@Append[Transpose@axes, {0, 0, -1}];
   t = RescalingTransform@Append[pr2, pr3[[3]]].t;
   Show[g, AspectRatio -> 1/ar, 
    ViewMatrix -> {TransformationMatrix[t], IdentityMatrix[4]}]];

Here axes defines the projection of x,y,z axes to 2d plane and pad makes a room to display axes labels.

Isometric projection:

g = Graphics3D[Cuboid[], Axes -> True, AxesLabel -> {X, Y, Z}];
parallelProjection[g, {{-Sqrt[3]/2, -1/2}, {Sqrt[3]/2, -1/2}, {0, 1}}]

Cabinet projection:

α = π/4;
parallelProjection[g, {{1, 0}, {0, 1}, -{Cos[α]/2, Sin[α]/2}}]

Answer 4

Just in case you aren't looking for a completely correct solution, but instead, just a cheap workaround.

I was looking for a ViewPoint->{Infinity,Infinity, Infinity} kind of solution. Replacing Infinity with a number big enough (in my case 500) I could get the results I was looking for.