[Math] the position of a unit cube so that its projection on the ground has maximal area

3dgeometryoptimizationprojective-geometrysolid-geometry

Consider a unit cube in $\Bbb R^3$. What is a position (up to translation, etc.) of the cube such that its projection on the $Oxy$-plane has a maximal area?

Here is a picture: if the sides of the cube are parallel to the three axis, then the projection is the unit square, with has area $1$. But it is possible to project it so that we get a regular hexagon (like the orange one, below).

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I believe that the regular hexagon could be a local maximum for the area of the projection, but I'm not sure. I don't even know how to compute the corresponding area of the orange hexagon (what should be the length of the orange side of one of the six equilateral triangles in the hexagon?).

I solved the problem for a unit disk in the space: the area of the projection is $\pi \sin(\theta)$, maximal when the angle $\theta$ is $\pi/2$ ($\theta$ is indicated on the right of the picture).

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I also tried for a unit square in $2D$, the maximal area being achieved at $\theta=\pi/4$.

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A similar question could be asked for a cylinder or a cone instead of a cube (the answer is trivial for a sphere, by the way).

Thank you for your comments!

Best Answer

For a convex body, its projected surface area $P$ in viewing direction $v$ is equal to the surface integral $$P = \oint (n\cdot v)^+\,\mathrm dA,$$ where $n$ and $\mathrm dA$ are the normal and area of the differential surface element, and $$x^+ = \max(x,0) = \begin{cases}x & \text{if $x>0$,} \\ 0 & \text{otherwise}.\end{cases}$$

For a unit cube, $n$ only takes six possible values $(\pm1,0,0)$, $(0,\pm1,0)$, $(0,0,\pm1)$, each over an area of $1$ square unit, so the integral reduces to $$P = |v_x| + |v_y| + |v_z|.$$ Subject to the constraint that $\|v\|=1$, this is maximized at $v = \frac1{\sqrt3}(\pm1,\pm1,\pm1)$, i.e. projection along a space diagonal, which indeed gives the regular hexagon as the projected shape.

Related Solutions

[Math] Maximal area covered by two triangles in unit circle

I whipped up a quick Matlab program that computes area stochastically (generate 20000 points; count how many are inside either triangle), and then ran it on 10,000 randomly generated triangle pairs (i.e., 6 random points on the unit circle; the first 3 define one triangle; the next three define the other. WLOG, I made the first angle be 0). Each time a new "largest area" was found, I printed out the triangles. It instantly found an area of ~1.25; then steadily improved this to 1.87 over the next couple of minutes. A previous instance of the program reached 1.97 or so...and this one just jumped to ~1.92, with a solution that's very nearly the two 45-90-45 triangles erected on a diameter.

That suggests to me that the correct answer is indeed this pair of triangles, with area 2 as the optimum.

>> testCircles
area = 1.2587; angles are [0 4.13265878886387 1.69578766384379 2.78030964099446 4.4988452659303 2.81878425776269]
area = 1.4833; angles are [0 1.03997694447012 4.97128628805005 5.53558413650189 4.31034051247746 2.03741390656017]
area = 1.5065; angles are [0 2.583714405482 3.95028117360417 3.41185810372066 4.81492678793123 1.47312279640955]
area = 1.8173; angles are [0 4.99055083947817 1.93427156493409 1.86260140259647 3.04043013930809 5.0276959720494]
area = 1.8517; angles are [0 3.01875721392096 1.30210531664363 4.12564514448203 2.38462146306761 5.8391459560257]
area = 1.87; angles are [0 1.78060248058588 4.66075033844476 2.08372315689924 5.4697587833186 3.76689571284346]
area = 1.9184; angles are [0 4.43108946197308 1.4092884007527 4.31113371053254 2.84429076671212 1.32017294630991]

Here's a picture of the most recent success: a near-optimal pair of triangles

The associated angles/area are

area = 1.9428; angles are [0 1.51926494165392 3.23093144636552 0.0773804130776986 3.34779844425289 4.92681116635565]

Geometry – Calculating the Area of the Limiting Polygon

We calculate explicitly that the area of the limiting polygon is $\dfrac{4}{7}A$ where $A$ is the area of the original triangle.

Note that on the $(n+1)$-th iteration we cut off twice as many triangles as we did in the $n$-th iteration. Consider the triangles we cut off in the $(n+1)$-th iteration. Each of these has $\dfrac{1}{3}$ the height and $\dfrac{1}{3}$ the base of a triangle cut off in the $n$-th iteration. Hence the ratio of the areas of these triangles is $\dfrac{1}{9}$. Since the total area of the first triangles we cut off is $\dfrac{1}{3}A$, we have that the area of the limiting polygon is

$$A-\sum\limits_{n=0}^\infty2^n\left(\frac{1}{3}A\right)\left(\frac{1}{9}\right)^n=A-\frac{1}{3}A\frac{1}{1-\frac{2}{9}}=\frac{4}{7}A$$

Whether this actually helps us determine the shape of the limiting figure I am not so sure.

This was motivated by Hagen von Eitzen's calculation and Michael's subsequent observation.

$\\$

EDIT: Explanation of $\dfrac{1}{3}$ base and $\dfrac{1}{3}$ height statement.

Triangles pic

Consider the polygon at a particular vertex $V$ just before making the $n$-th iteration cuts. Choose a particular edge bordering $V$ and let it have length $9x$.

Now take the $n$-th and $(n+1)$-th iteration cuts. In the figure above, the green triangle is one that is cut off in the $n$-th iteration cuts, and the red triangle is one cut off in the $(n+1)$-th iteration cuts.

Because we trisect each time, one side of the green triangle has length $3x$ and one side of the red triangle has length $x$. Call these sides the base of each triangle. Consider now the length of the altitudes corresponding to these bases, letting the altitude of the green triangle be of length $3y$. By similar triangles we find that the altitude of the red triangle has length $y$ (the similar triangles are outlined in black).

Hence the ratio of area of the red triangle to the area of the green triangle is $$\frac{\frac{1}{2}xy}{\frac{1}{2}(3x)(3y)}=\frac{1}{9}$$ as claimed.

Fun note: we did not use the fact that the triangles in question are isosceles.

Best Answer

Related Solutions

[Math] Maximal area covered by two triangles in unit circle

Geometry – Calculating the Area of the Limiting Polygon

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