My general question concerns what we can learn about an arbitrary, three-dimensional convex polytope (or convex hull of an arbitrary polytope) strictly from the surface areas of its two-dimensional projections on a plane as it 'tumbles' in 3-space (i.e. as it rotates along an arbitrary, shifting axis).

If it's helpful, please imagine the following physical set-up:

We take an arbitrary three-dimensional convex polytope, and fix the center of mass to a coordinate in 3-space, $(x_0, y_0, z_0)$, located at some distance, $D$, above a flat surface. While this prohibits translation of the center of mass, the polytope is still allowed to tumble freely (i.e. it is allowed rotation around an arbitrary axis centered at the fixed coordinate).

There is no 'gravity' or other force to stabilize the tumbling polytope in a particular orientation. Over time it will continue to tumble randomly. (The 'physical' set-up is only meant for descriptive reasons.)

We shine a beam of coherent light on the tumbling polytope, larger than the polytope's dimensions, and continually record the area of the resulting shadow, or two-dimensional projection on the surface. To be clear, the area of the two-dimensional projection is the only information we are allowed to observe or record, and we are allowed to do so over an arbitrary length of time.

My question is - From observing the area of the tumbling polytope's shadow, or two-dimensional projection over time, what can we learn about it's geometry? To what extent can we characterize and/or reconstruct the polytope from its changing shadow (extracting the surface area for example - hat tip to Nurdin Takenov)?

Do we gain anything by watching the evolution of the convex polytopes shadow as it tumbles (part of the point for the physical example), as opposed to an unordered collection of two-dimensional projections?

Update - Nurdin Takenov (and Sergei Ivanov in later comments) nicely points out that we can use the average surface area of the two-dimensional projection to find the surface area of the tumbling convex polygon. Might we be able to find it's volume?

(Addendum - I would be really neat if somebody could point me to any algorithms in the literature... or available software.... that let's me calculate and characterize two-dimensional surface projections of convex polytopes!)

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    $\begingroup$ Such polyhedra are called unistable or monostatic, and there are some references here: en.wikipedia.org/wiki/Monostatic_polytope $\endgroup$ May 14, 2010 at 14:30
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    $\begingroup$ Joseph, thanks for your commment! But may I ask you to clarify what you mean by 'such polyhedra'? I'm thinking about simple characterizations / what you can learn about arbitrary polytopes (or perhaps, arbitrary convex polytopes to simplify things). $\endgroup$
    – Rob Grey
    May 14, 2010 at 14:42
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    $\begingroup$ I think it's likely best to restrict to convex polytopes. If I'm thinking about this correctly, you could symmetrically indent two opposite sides of a cube (two so as not to move the center of mass) and be unable to distinguish the resulting object from the cube based only on its shadows. $\endgroup$ May 14, 2010 at 17:06
  • $\begingroup$ Cam, thanks - I agree with you, and will update the question accordingly. $\endgroup$
    – Rob Grey
    May 14, 2010 at 17:12
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    $\begingroup$ The "random tumbling" part is confusing. Do you know which rotations of the polytope gave you particular projection areas? If not, do you know some probability distribution or you just know the set of possible areas? (In the latter case, you will only learn what is the maximum and minimum projection area, and this too little information). $\endgroup$ May 14, 2010 at 20:27

4 Answers 4


You cannot recover a convex polytope from its projection areas, even if you know the whole function (unit vector) $\mapsto$ (area of projection along this vector). There exist two different polytopes $P_1$ and $P_2$ such that for every unit vector $v\in\mathbb R^3$, the areas of the two projections along $v$ are equal.

Let me begin with a two-dimensional example. In this case, the projections are one-dimensional, and the "projection area" is width. Consider a regular triangle $T$ with side 1 and regular hexagon $H$ with side 1/2, positioned so that their sides are parallel. Their widths are the same in every direction. One can prove this without computation: the hexagon is the Minkowski symmetrization of the triangle and the symmetrization preserves widths.

Now go to dimension 3. Let $P_1=T\times[0,1]$, the prism with height 1 based on the regular triangle with side 1. Let $P_2=(\sqrt{2/3}H)\times[0,\sqrt{3/2}]$, the prism with height $\sqrt{3/2}$ based on the regular hexagon with side $1/\sqrt6$ (I hope that I get the constants right). I claim that they have the same projection area in every direction.

More generally, consider a prism of height $h$ based on a convex figure $F$ of area $A$ such that the base is parallel to the $xy$-plane. Its area of projection along a vector $v=(\cos\alpha\cos\theta,\sin\alpha\cos\theta,\sin\theta)$, is given by $$ A\cdot \sin\theta+w(\alpha)\cdot h \cdot \cos\theta $$ where $w(\alpha)$ is the width of $F$ in the horizontal direction that forms oriented angle $\alpha$ with the $x$-axis. The constants in the above example are chosen so that the two prisms have the same $A$ and the difference in $w(\alpha)$ is compensated by the difference in $h$.

  • $\begingroup$ In the 3-dimensional case: the area of the base is the same, the height is different! So the volumes are different too! Although the projections have the same area! Also, in 2-dimensional case, the area of the triangle and the area of the hexagon are not the same. Just to conclude that the initial question about volumes is also anwered here. $\endgroup$
    – Olga
    Dec 6, 2013 at 15:20

Hej, there seem to be a number of variations of the original question out there. Let me tell you what is known (to me):

Let us assume that an (unknown) convex polytope $P$ in $R^3$ is given. I assume that we can observe orthogonal projections onto planes (think of shadows that we see on a screen that is orthogonal to the direction of the incoming parallel light). If the direction of the incoming light is a unit vector u, then the information we have is the data $ F(P,u)=area(proj_u(P)) $ where $proj_u(P)$ is the orthogonal projection of $P$ along the direction $u$. This function is called projection function or brightness function in convex geometry. What do we know about $P$, when we know the projection function for all directions u?

1) The surface area of $P$ is proportional to the averaged function $F(P,u)$, averaged over all directions (was remarked before, and is a consequence of Cauchy's projection formula, even true in all dimensions and when the polytope is some convex set).

2) If $P$ is translated then $F(u)$ does not change, so the best we can hope for is determination up to translations. In general $P$ is not uniquely determined up to translations by $F(P,u)$. Also this was remarked earlier. It is also clear. Take some polytope $P$ which does not have a centre of symmetry. Then its reflection at the origin $\hat P$ is not a translation of the original $P$, but $F(P,u)=F(\hat P,u)$ for all $u$.

3) Amazingly, the central symmetry is the crucial property: If $P$ is central symmetric the $F(P,u)$ (known for all u) determines $P$ uniqely up to translations. This was already shown by Alexandrov and is often called Alexandrov's projection theorem. It holds in all dimensions and for arbitrary central symmetric convex bodies. See Richard Gardner's book on geometric tomography.

4) Under the central symmetry assumption, there is an algorithm to determine $P$ from $F(P,u)$, see the article RECONSTRUCTION OF CONVEX BODIES FROM BRIGHTNESS FUNCTIONS by R. Gardner and Peyman Milanfar.

Best wishes, Markus Kiderlen


If the polytope is convex, we certainly could find the surface area of the polytope: Let $u_i$ - vectors, such that for every $i$ vector $u_i$ is perpendicular to face $\Gamma_i$ and $|u_i|=area(\Gamma_i)$. Then, if we project the polytope to the plane with normal vector $n$, the area of projection would be:

$Pr(n)=\frac{1}{2}\sum_i |(u_i,n)|$

Then consider integral above all the possible $n$(they are on the unit sphere U). It's not hard to show, that for every $a$:

$\int_{U} |(a,n)|dS=2\pi |a|$


$\int_{U} Pr(n)dS=2\pi\sum_i |u_i|$

I could't remember name of this formula.


This is a nice question, and I look forward to seeing a definitive answer. Meanwhile, let me point out that it is known that a convex polytope in $E^n$ is not always determined by a finite set of its projections, but it is determined by all of its 2D projections. Richard Gardner, in Geometric Tomography, puts the first point this way: "it is generally not possible to choose a finite set of subspaces in such a way that the corresponding projections distinguish $P$ from every other convex polytope" [p.93].

For some reason I cannot add a comment to reply to Rob Grey's latest. My apologies for not understanding the MathFlow conventions. Here is my answer:

Yes, I wonder also. That's why what I posted is not an answer to your interesting question. You have rather less info--just the area, not the actual projection--but you have the "evolution" as you put it, presumably an association of the direction of projection with the area. This could be viewed as an area associated with each point on the unit sphere (the point representing the projection direction). I guess here I am presuming a particular answer to Sergei Ivanov's question.

So I would rephrase your question: Does the map from $S^2$ to projection area uniquely determine the polytope, or more generally, a convex body?

There is a new paper on the arXiv by Gardner, Gronchi, and Theobald that addresses a very specific case of Rob's original question: "Determining a rotation of a tetrahedron from a projection" arXiv:1111.7100. They determine the conditions under which the (orthogonal projection) shadow of a known tetrahedron permits reconstructing its orientation in space. This is already a richly complex situation.

  • $\begingroup$ Sorry, I meant that to be a Comment, not an Answer! $\endgroup$ May 14, 2010 at 15:21
  • $\begingroup$ Thanks Joseph! However, I wonder what one gains by looking at the evolution through a finite set of projections (as you rotate along one axis, then another)... $\endgroup$
    – Rob Grey
    May 14, 2010 at 19:22
  • $\begingroup$ Joseph, I think you're having problems because you have a new account and therefore a sub-15 rep. score... Nevertheless, thank you for your persistence in trying to help answer my question! I like your rephrasing... but I feel like a positive answer is too optimistic. My hope is that one could definitively extrapolate something like the volume or surface area of the convex body, or perhaps narrow down the convex body to a subset of possibilities. $\endgroup$
    – Rob Grey
    May 14, 2010 at 20:55

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