Analytical Solution for Scattering Cross-Section of Randomly Oriented Cylinder (or Improved Brute Force)

Question

I'm trying to calculate the sum of the shadows that a cylinder can make when rotated through both theta and phi.

I have the following code:

dia = 1;
length = 2;
dstart = -length;
dmat = N@4/200;
SamplingMatrix = ConstantArray[0,{201,201}];
Do[
  cyl = Cylinder[
    {
      {-length/2 * Cos[theta], 0, -length/2 * Sin[theta]},
      {length/2 * Cos[theta], 0, length/2 * Sin[theta]}
    },
    dia/2
  ];
  g = Resolve[Exists[z,{x,y,z}\[Element]cyl],Reals];
  f = ImplicitRegion[g,{x,y}];
  Do[
    Do[
      If[
        RegionMember[f,{dstart+dmat*(i-1),dstart+dmat*(j-1)}],
        SamplingMatrix[[i,j]] = SamplingMatrix[[i,j]] + 1;,
        Unevaluated[Sequence[]]
      ],
      {j,1,201}
    ]
    ,
    {i,1,201}
  ],
  {theta,0,Pi,Pi/180}
]
MatrixPlot[SamplingMatrix]

Which is clunky but it works. I then use the following to rotate the results from this to obtain the full radial distribution (using the function from here):

Do[
SamplingMatrix = SamplingMatrix + SquareMatrixRotate[rotatematrix,phi];
,
{phi,pi/180,pi,pi/180}
]
MatrixPlot[SamplingMatrix]

This seems to give a satisfactory result, but it definitely feels like I'm using a club to solve a problem that should have a more elegant solution. Is there a better way to do this? To be clear, I'm not a mathematician, so this could very well be a trivial problem which has a known solution. If so, I would appreciate it if someone would point me in the right direction.

Thanks

EDIT 1 I included the following image to show the output that I'm looking for:

EDIT 2 Revised the title to better reflect the nature of my problem

Answer 1

The shadow of a cylinder is two half-ellipses and the shadow of a rectangle at an angle. Under the reasonable assumption that the light rays of the sun are perpendicular to the flat ground, the two half-ellipses are two halves of the same ellipse, and the shadow of the rectangle is another rectangle. In other words, you don't have to take "perspective" into account, because the shadow is an orthogonal projection of the cylinder.

This reduces your task to finding the ellipse and the rectangle corresponding to the shadow of the cylinder at an angle $\theta$ to the ground, which you can probably solve. Once you do that, use the grade school formulas for the areas and add.

I now prove that rotating a circle gives you an ellipse. You need three facts from geometry.

First, for a vector $\hat{v} = \begin{bmatrix}x\\ y\\ z \end{bmatrix}$, you can rotate the vector about, say, the $x$-axis by multiplying it by the rotation matrix

$$R_\theta = \begin{bmatrix}1 & 0 & 0\\ 0 & \cos \theta & -\sin \theta\\ 0 & \sin \theta & \cos \theta \end{bmatrix}.$$

Second, the set of all vectors of the form

$$\hat{v} = \begin{bmatrix}x\\ y\\ z \end{bmatrix}$$

having $x^2 + y^2 = r^2$ and $z=z_0$ (a constant) are exactly the set of points in a circle that is parallel to the $xy$-plane with radius $r$. (Notice that $z$ is a constant equal to the distance the circle is above the ground.)

Finally, conveniently, for any vector $\hat{v} = \begin{bmatrix}x\\ y\\ z \end{bmatrix}$, the projection of that vector onto the ground (the $xy$-plane) is just $\text{proj}(\hat{v})= \begin{bmatrix}x\\  y\\ 0 \end{bmatrix}$, i.e. set the $z$-coordinate to zero.

Let's use $R_\theta$ with our circle vector and then project onto the ground:

$$R_\theta\; \hat{v} = \begin{bmatrix}1 & 0 & 0\\ 0 & \cos \theta & -\sin \theta\\ 0 & \sin \theta & \cos \theta \end{bmatrix}\begin{bmatrix}x\\ y\\ z \end{bmatrix} = \begin{bmatrix}x\\ y \cos \theta - z \sin \theta \\ y \sin \theta + z \cos \theta \end{bmatrix},$$ giving $$\text{proj}\left( R_\theta\; \hat{v} \right) = \begin{bmatrix}x\\ y \cos \theta \\ y \sin \theta\end{bmatrix}.$$ (Remember, we set the $z$-coordindate to zero, not the $\tilde{z}$-coordinate.)

Let's name the rotated coordinates:

$$R_\theta\; \hat{v} = \begin{bmatrix}\tilde{x}\\ \tilde{y}\\ \tilde{z} \end{bmatrix} = \begin{bmatrix}x\\ y \cos \theta - z \sin \theta \\ y \sin \theta + z \cos \theta \end{bmatrix}$$.

Now the question is, if we trace out a circle in the rotated $\tilde{x}\tilde{y}$-coordinates, i.e. if $\tilde{x}^2 + \tilde{y}^2 = r^2$ for $\tilde{x}=x$ and $\tilde{y} = y \cos \theta - z \sin \theta$ , what do we have for the projection in $x$-$y$ coordinates? We compute:

$$r^2=\tilde{x}^2 + \tilde{y}^2 = x^2 + (y \cos \theta - z \sin \theta)^2$$.

Projecting by setting $z$ to zero gives $x^2 + (\cos^2 \theta)\,y^2 = r^2,$ an ellipse, or a line if $\cos \theta = 0$.

Answer 2

Here a faster solution without Do using RegionPlot3D

First define cylinder (rotation [CurlyPhi] around {1,0,0} )

zyl[\[CurlyPhi]_] :=Cylinder[{-{0, -Sin[\[CurlyPhi]],Cos[\[CurlyPhi]]}, {0, -Sin[\[CurlyPhi]], Cos[\[CurlyPhi]]}},1/2]

Second list of 3D-regions

zi = Table[DiscretizeRegion[ zyl[\[CurlyPhi]],MaxCellMeasure -> Pi/2/500], {\[CurlyPhi],Subdivide[-Pi, Pi, 10]}] ;

Third 3d-Plot projected into plane {1,0,0},{0,1,0}

Show[{Map[RegionPlot3D, zi ] ,ParametricPlot3D[Sqrt[5/4] {Cos[t], Sin[t], 0}, {t, -Pi, Pi}]},ViewVector -> {1000 e3, 0 e3} ]

Rotation of this region gives a Circle[{0,0},Sqrt[5/4]] as the final result for the shadow you are looking for!