🚀 go-pugleaf

In article <C0Fo36.GyE@gabriel.keele.ac.uk> radar@grumpy (Reidar J. Husmo) writes:
> Robert Weinberg (robertw@informix.com) wrote:
> : Manhole covers are manufactured in circular shape, rather than oval or square,  
to  
> : satisfy one paramount condition.
> 
> It is therefore tempting to say that any (2n+1) polygon which edges are part
> of a circle, with centre being the opposite vertex, will do.

I can't prove it, but I came to the same supposition as Reidar.

In a (2n+1)-sided "rounded" polygon, every vertex is equidistant to its two  
"opposite" vertices.

When such a polygon (the manhole cover) is tipped up to the vertical, the minimum  
cross-section it can present to the horizontal plane is the distance between one  
vertex and one of its "opposites" - when you rotate the polygon, the distance  
between a vertex and an arc presents the same cross-section.

This distance is equal to the maximum cross-section of the same polygon when laid  
out (the manhole itself). So given a lip of the kind allowed for circular manholes,  
a (2n+1) "rounded" polygon would never fall through, it seems to me.
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