Suppose you have a length $L$ of metal pipe at your disposal, and you would like to build a wireframe unit-radius sphere, by bending, cutting, and welding the pipe into a connected structure $F$. Your goal is to minimize the height difference between where the center of the unit sphere would be ($1$) and where the center of your structure $F$ is, in any position resting on flat ground.

To be more precise, let $S$ be a unit radius sphere with center $o$, and $F$ a connected structure, built from pieces that are topological arcs, inscribed in contained within $S$. Let $H$ be the convex hull of $F$, required to enclose $o$, and define $\delta$ to be $1$ minus the minimum distance from $o$ to any point on $H$. So $\delta$ is the difference between the spheres centered on $o$ circumscribed about and inscribed in $H$.

Q1. For a given $L$, what is $\delta_{\min}(L)$, the minimum value of $\delta$ for any connected structure $F$? Equivalently, given $\delta$, what is $L_{\min}(\delta)$, the minimum length of all pieces in $F$ together that achieve $\delta$?

For example, given $L=4 \pi$, one might construct two orthogonal hoops:
This achieves $\delta = 1 - \sqrt{2}/2 = 0.29$. But surely this is not optimal. For example, one could remove some pipe near the poles and add it to the equator to lower $\delta$. This suggests this question:

Q2. Is an optimal frame structure $F$ always composed of straight segments? I.e., does it ever help to bend the pipe?

I think not.

The question can be generalized to any dimension. Even in $\mathbb{R}^2$ it seems not uninteresting:

Q3. What are the optimal structures $F$ inscribed in a unit circle that minimize $\delta$ for a given $L$? [Added:] Could it be that a regular $n$-gon minus one edge is optimal for that $L$, as suggested by the hexagon example above?

This two-dimensional version especially feels like it should have been investigated previously, but I am not finding any literature on it. Ideas and/or pointers welcomed—Thanks!

Addendum. Thanks to Gerhard Paseman and Aaron Meyerowitz, it is now clear that for the 2D question Q3, the Steiner tree spanning the vertices of a regular $n$-gon is a strong candidate for optimality, and certainly improves upon the $n$-gon minus an edge (except for $n=6$, as per Aaron's remark).
    Steiner Tree

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    $\begingroup$ Perhaps Steiner trees answer the question in two dimensions. It would be interesting to compare Steiner trees in higher dimensions to other attempts. Gerhard "Ask Me About System Design" Paseman, 2012.04.12 $\endgroup$ – Gerhard Paseman Apr 12 '12 at 18:54
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    $\begingroup$ I think you also want to explicitly require $o$ to be contained in $H$, as otherwise we could cheat. $\endgroup$ – Michael Biro Apr 12 '12 at 19:03
  • $\begingroup$ @Gerhard: Yes, you must be right: The Steiner tree for the vertices of a regular $n$-gon, which must be known. $\endgroup$ – Joseph O'Rourke Apr 12 '12 at 19:32
  • $\begingroup$ @Michael: Thanks, you are correct: to make $\delta$ meaningful, $o$ should be inside $H$. Corrected; thanks! $\endgroup$ – Joseph O'Rourke Apr 12 '12 at 19:44

For Question 3, in the case that $L=3\sqrt{2} \approx 4.24$ one can do better than 3 sides of a square. The two diagonals have length 4 with the extra left for improving thigs a bit. Probably a Steiner tree would be even more efficient (I think that your 5/6 of a hexagon example with all the 120 degree angles lucks into being a Steiner tree itself.)


Here is a structure F with $L=5$ and $\delta=0.08897$ I think this construction gives the minimum $\delta$ at least for certain $L.$ It certainly is an improvement over 5/6 of a hexagon and works for any $L.$

alt text

The illustration shows the unit circle and a concentric circle whose radius happens to be $r=0.9110275245$ along with three tangents. The straight segments run from $[2r^2-1,\pm2r\sqrt{1-r^2}]$ to $[r,\pm\sqrt{1-r^2}]$.

For $L \le \frac{\pi+2}{\sqrt{2}} =3.635655$ We achieve $r=\frac{L}{\pi+2} $ using the (left) half of a circle of radius $r$ along with two horizontal segments. The Steiner tree competition would presumably achieve $r=\frac{L}{6}$ using three segments of length$\frac{L}3$ to get convex hull an equilateral triangle.

  • $\begingroup$ @Aaron: Brilliant! Why not mirror your right-half construction to the left? $\endgroup$ – Joseph O'Rourke Apr 16 '12 at 2:56
  • $\begingroup$ Mirrored to the right would not be connected which seems to be part of the rules. Notice that mentioned case of L=3.635 is slightly better than a Steiner tree for a square in getting $r=\frac1{\sqrt{2}}$. Curious that this and the Steiner tree construction seem close. $\endgroup$ – Aaron Meyerowitz Apr 16 '12 at 3:39
  • $\begingroup$ @Aaron: Oh, I see. Very clever construction! $\endgroup$ – Joseph O'Rourke Apr 16 '12 at 12:24
  • $\begingroup$ Upon reflection, I think we should call this the Meyerowitz Tree. :-) $\endgroup$ – Joseph O'Rourke Apr 16 '12 at 20:40

It is unlikely that one can solve that type of problem. But one can consider this as a competition; who get a better upper/lower bound. For lower bound there is a nice construction which I learn from Zalgaller's paper.

Take regular $n$-gon in $xy$-plane with center at the origin and a side parallel to $y$-axis. Consider the circles formed by revolution of its vertices around $x$-axis in $xyz$-space. Cut the circles along $xy$-plane and leave only the upper part. Repeat the construction for the part below $xy$-plane, but for the line obtained by rotation of $x$-axis by angle $\tfrac\pi{n}$. You get a spiral-like curve.

I am sure this gives a better estimate than your example. One can do bit better but not much better.

  • $\begingroup$ Can you tell which n gives the best result and what it is? $\endgroup$ – Aaron Meyerowitz Apr 17 '12 at 6:20
  • $\begingroup$ @Aaron, all $n$ are good. $L_n=\pi/\tan \frac\pi {2n}$ and $\delta_n=1-\cos\frac\pi {2n}$. I would bet that this construction gives the best values if you the wire stays in the sphere. $\endgroup$ – ε-δ Apr 18 '12 at 2:12

Guessing that this is the construction (p.90) to which ε-δ refers. (I cannot read Russian):
  Zalgaller Fig. 3

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    $\begingroup$ Can't you use Google translate!!! looks like it is Czech and starts out: The sequence kaqestve wego xaga indicators let em come to korotku it osmotra swap wu imenno NGOs on post high p Rassmotrim KUB with center O takogo razmera qto ego rib kasa ts sphere with my gran Otsekaemye togo Cuba xapoqki I will be geodeziqeskie krug sphere radius s IMET Zanumeru glad to help! $\endgroup$ – Aaron Meyerowitz Apr 12 '12 at 16:08
  • $\begingroup$ When you say inscribed in S do just mean contained in it? It seems that you want to have some points of the frame at distance 1 from the center but must all be such? The circle question does not have this feature. For your hexagon less a side example for L=5, could you use the outer half of each extreme side along with 2/3 of a circle of radius $1-\sqrt{3}/2$ using $1+2\pi/\sqrt{3} \approx 4.6276$ of length to obtain $\delta=1-\sqrt{3}/2$? Then you could modify to decrease $\delta$ $\endgroup$ – Aaron Meyerowitz Apr 12 '12 at 17:57
  • $\begingroup$ I meant "contained within." It will touch $S$ as well, so I thought "inscribed" meant the same. But I can see my language was not precise. Now corrected. Thanks! $\endgroup$ – Joseph O'Rourke Apr 12 '12 at 18:42

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