Actually, this construction works for a 2-gon (line segment) as well, so long as the little caps are semicircles of the same radius. This produces the traditional college/high school running track, which is two semicircular caps and 2 straightaways. So it appears that closed convex C1 curves can be constructed from n circular arcs, so long as n>=4. The proof a non-existence for n=3 needs to be tightened up a little. At 10:41 AM 3/14/2012, Michael Kleber wrote:

And since it's possible to make two segments that join without using this construction, we can make any number >=5 this way.

--Michael

On Wed, Mar 14, 2012 at 1:38 PM, Henry Baker <hbaker1@pipeline.com> wrote:

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Yes, that construction occurred to me as well. E.g., take any n-gon & replace each corner with a circular arc of the same angle, but small enough to not interfere with other corners.

Since line segments are "arcs of infinite radius", this produces an acceptable figure of 2n segments. So we can trivially construct any even number.

At 10:25 AM 3/14/2012, Michael Kleber wrote:

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Wait -- if you have any convex C0 curve made of circle segments, can't you always change it into a C1 of the type Henry wants, changing each corner into a circle arc, by pushing sufficiently small circles into each corner until they are tangent to both arcs that meet there?

If I'm understanding this discussion correctly, this makes a 5-segment loop easily: take any pair of different-curvature circle arcs A and B joined without a corner, connect their remote end-points with any circle arc C, then smooth out the AC and BC corners using small circles D and E.

--Michael

On Wed, Mar 14, 2012 at 12:46 PM, Henry Baker <hbaker1@pipeline.com> wrote:

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I'm not constructing Reuleaux triangles; they are only C0, not C1, curves.

At 09:31 AM 3/14/2012, Fred lunnon wrote:

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Why "divisible by 2 or 4", when a Reuleaux triangle has 6 arcs? WFL

On 3/14/12, Henry Baker <hbaker1@pipeline.com> wrote:

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I'll see what I can do. On a long bicycle ride yesterday, I began to have doubts about 5; perhaps the number needs to be divisible by 2 or 4. But the construction below doesn't require r1=r2, r3=r4 (the mechanical drawing approach), so this gives me some hope.

At 01:25 AM 3/14/2012, Bill Gosper wrote: >Henry, can you exhibit a smooth loop with five radius changes? >A special case of four is the "four point ellipse" from mechanical drawings of yore. >These can co-rotate in continuous tangential >contact<http://gosper.org/pump1.gif>. > >Similarly, "six point Reuleaux triangles <http://gosper.org/reuleaux.gif>". >(Rich's observation.) >--rwg > >hgb> >I now think that it is impossible to create a simple closed C1 curve from >only 3 circular arc segments. The following construction for 4 segments >shows why this is. 1. Draw a circle of radius r1. 2. Draw a circle of >radius r2 that intersects circle #1. 3. Draw a circle of radius r3 inside >the intersection that is tangent to the first 2 two circles. 4. Draw >another circle of radius r4 inside the intersection of #1 & #2 that is >tangent to #1 and #2. A circular arc segment is taken from each of the 4 >circles to produce a closed C1 curve. Basically, it is the boundary of the >intersection region, with both sharp ends cut off by circular arcs from >circles #1 & #2. The construction shows that r3<r1, r3<r2, r4<r1, r4<r2. >There are probably interesting relationships between the centers of these >circles, considered as complex numbers, and the various radii. There is a >paper by someone at Bell Labs that showed some similar relationships of >tangent circles & complex coordinates.

-- Forewarned is worth an octopus in the bush.

-- Forewarned is worth an octopus in the bush.