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.




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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!)



Characterizing a tumbling convex polytope from the surface areas of its two-dimensional projections
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 $h$.


