[math-fun] Closed C1 curve made of nonempty nonfull circular arc segments
Simple proof of the original impossibility claim: If you have a C1 closed curve made of 3 circular-arc segments (each strictly between 0 and 360 degrees), then it must consist of those 3 arcs in decreasing order of radii (starting from someplace). Wlog regard the largest-radius arc as centred at (0,0) hence it is r=constant in polar coordinates; the other two arcs must therefore decrease r. This prevents closure, contradicting our original assumption. Hence such a curve is impossible, QED. Possibility claim: If N>3 then a C1 closed curve made of N circular arcs, all different in radius, is possible. Let the Kth arc have radius RK>0 and traverse angle AK with 0<AK<360 degrees. Let B0=0, B1=A1, B2=A1+A2, B3=A1+A2+A3, ..., BN=A1+A2+...+AN. Wlog take R1=1. The C1 closure conditions are BN = 360 degrees, sum(K=1..N) RK * (sin(BK)-sin(B[K-1])) = 0, sum(K=1..N) RK * (cos(BK)-cos(B[K-1])) = 0. This is 4 equations in 2N unknowns. We can also equivalently regard this as follows. Choose N distinct points on the unit circle in the complex plane, call them z1,z2,...,zN and demand sum(K=1..N) RK * (zk-z[k-1]) = 0 and R1=1. This is N unknowns (if the z's are regarded as known) and 3 equations. Obviously these equations have a solution (all Rj=1) and the solution set must consist of a (>=N-3) dimensional space containing the all-1s point since each equation can reduce the dimensionality of the solution set by at most 1. A random point near enough to the all-1s point in the space, does the job. QED. On 3/14/12, math-fun-request@mailman.xmission.com <math-fun-request@mailman.xmission.com> wrote:
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Today's Topics:
1. Re: circular-arc splines, again (Henry Baker) 2. Re: circular-arc splines, again (Andy Latto) 3. Re: circular-arc splines, again (Michael Kleber) 4. Re: circular-arc splines, again (Henry Baker) 5. Re: circular-arc splines, again (Henry Baker)
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Message: 1 Date: Wed, 14 Mar 2012 09:46:07 -0700 From: Henry Baker <hbaker1@pipeline.com> To: Fred lunnon <fred.lunnon@gmail.com> Cc: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] circular-arc splines, again Message-ID: <E1S7rLm-0005WC-K8@elasmtp-spurfowl.atl.sa.earthlink.net> Content-Type: text/plain; charset="us-ascii"
I'm not constructing Reuleaux triangles; they are only C0, not C1, curves.
At 09:31 AM 3/14/2012, Fred lunnon wrote:
Why "divisible by 2 or 4", when a Reuleaux triangle has 6 arcs? WFL
On 3/14/12, Henry Baker <hbaker1@pipeline.com> wrote:
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.
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Message: 2 Date: Wed, 14 Mar 2012 13:15:38 -0400 From: Andy Latto <andy.latto@pobox.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] circular-arc splines, again Message-ID: <CAKqg3U2xNs19EpOE=h7k-UW=oFv2XKXhk9GM1CfNsYCBvGK7=w@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
On Wed, Mar 14, 2012 at 10:53 AM, Henry Baker <hbaker1@pipeline.com> wrote:
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.
I don't have a proof yet, but I think that it might be true that if an arc of radius r1 lies between arcs of radius r2 and r3, then either r1 is less than both of r2 and r3, or greater than both. This would imply that the total number of arcs must be even.
Andy
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Message: 3 Date: Wed, 14 Mar 2012 13:25:27 -0400 From: Michael Kleber <michael.kleber@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] circular-arc splines, again Message-ID: <CAPmvm8yxQSFcdqDjEWmK_ayoru-yLOEgsQz5hjnwk2ELOTTpHw@mail.gmail.com> Content-Type: text/plain; charset=ISO-8859-1
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:
I'm not constructing Reuleaux triangles; they are only C0, not C1, curves.
At 09:31 AM 3/14/2012, Fred lunnon wrote:
Why "divisible by 2 or 4", when a Reuleaux triangle has 6 arcs? WFL
On 3/14/12, Henry Baker <hbaker1@pipeline.com> wrote:
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.
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-- Forewarned is worth an octopus in the bush.
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Message: 4 Date: Wed, 14 Mar 2012 10:38:10 -0700 From: Henry Baker <hbaker1@pipeline.com> To: Michael Kleber <michael.kleber@gmail.com> Cc: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] circular-arc splines, again Message-ID: <E1S7sAL-0001PG-Sf@elasmtp-junco.atl.sa.earthlink.net> Content-Type: text/plain; charset="us-ascii"
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:
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:
I'm not constructing Reuleaux triangles; they are only C0, not C1, curves.
At 09:31 AM 3/14/2012, Fred lunnon wrote:
Why "divisible by 2 or 4", when a Reuleaux triangle has 6 arcs? WFL
On 3/14/12, Henry Baker <hbaker1@pipeline.com> wrote:
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.
------------------------------
Message: 5 Date: Wed, 14 Mar 2012 10:40:37 -0700 From: Henry Baker <hbaker1@pipeline.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] circular-arc splines, again Message-ID: <E1S7sCs-00081U-9w@elasmtp-curtail.atl.sa.earthlink.net> Content-Type: text/plain; charset="us-ascii"
"any n-gon" should have been "any convex n-gon"
At 10:38 AM 3/14/2012, Henry Baker wrote:
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:
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:
I'm not constructing Reuleaux triangles; they are only C0, not C1, curves.
At 09:31 AM 3/14/2012, Fred lunnon wrote:
Why "divisible by 2 or 4", when a Reuleaux triangle has 6 arcs? WFL
On 3/14/12, Henry Baker <hbaker1@pipeline.com> wrote:
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.
------------------------------
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