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A square wheel rolling on a catenary road maintains the wheel center at a fixed height, a well-known construction previously discussed on MO (e.g., "Generalizing square wheels rolling on inverted catenaries"). I wondered if this fundamentally one-dimensional example could be generalized to two dimensions, in the following sense:

Is there a solid body $B$ and a non-flat surface $S$ which together have the property that, from some one, special fixed position of $B$ resting on $S$, $B$ can roll on $S$ in any horizontal direction $v$ so that some point in $B$ (its center) remains at a fixed height?

Of course if $B$ is a sphere and $S$ is a plane, then the constant-height property holds. Note I am asking that this only hold for some special initial position of $B$, but demand that rolling in any direction of the full $360^\circ$ retains constant height along that ray.

My guess is that the requirement that this hold for every position of $B$ on $S$ forces a sphere on a plane.

The following is meant to be suggestive only!
What brought this to mind is the traditional Easter Egg Roll. :-)

Addendum. Here is another suggestive image, not metrically accurate, of a revolved square diamond that can roll on a revolved catenary, as per Anton's answer.
           Square Egg

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up vote 7 down vote accepted

It happens if and only if in the initial position it is the surfaces or revolution around $z$-axis and the intersection of $B$ with say $(yz)$-plane is rollable plane figure.

Note that $B$ and its initial position overdetermines the surface; it gives the surface as the foot points of rolling in all directions plus it gives its gradients at this foot points. Both of these data are periodic along each ray. For close rays they the foot-point-trajectories spread apart, but the gradients stays perodic. Therefore the data can feet together only if the gradient is parallel to the ray. The later implies that we have a surface of revolution.

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Brilliant! So a square revolved around a diagonal should roll on a revolved catenary. (See Addendum to question.) Happy Easter! –  Joseph O'Rourke Apr 8 '12 at 13:12
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