March 2012 - mathfun@mailman.xmission.com

Re: [math-fun] OT: Elsevier
by Dan Asimov 16 Mar '12

16 Mar '12

I don't think there are any real obstacles, just a matter of researchers getting organized -- as it sounds from Fred's post that particle physicists have already done -- and also for academics to arrange to receive credit from their employers for service to the profession for the effort they may put into the editing and reviewing that will be needed. --Dan P.S. My preference would be for the arXiv to offer *optional* peer-editing / reviewing instead of its being mandatory. Gene wrote: << There is no longer any need for a journal to have a printing press, and the only useful feature they provide is peer-editing and review. Can someone in the academic community answer this: What are the obstacles that prevent authors from bypassing these price-gouging journals and publishing electronically, with peer review? >> ________________________________________________________________________________________ It goes without saying that .

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[math-fun] doublet strings
by Fred lunnon 15 Mar '12

15 Mar '12

From the formula for A(n, k) given in http://oeis.org/A183135 it can easily be deduced that the number of "doublet strings" of length 2 n on k = 2 symbols equals the number of binary strings with n zeros and n ones. But if you were not told this, what constructive bijection between these sets might establish it directly? For n = 3 the sets are (in lexicographical order!) : AAAAAA OOOIII AAAABB OOIOII AAABBA OOIIOI AABAAB OOIIIO AABBAA OIOOII AABBBB OIOIOI ABAABA OIOIIO ABBAAA OIIOOI ABBABB OIIOIO ABBBBA OIIIOO BAAAAB IOOOII BAABAA IOOIOI BAABBB IOOIIO BABBAB IOIOOI BBAAAA IOIOIO BBAABB IOIIOO BBABBA IIOOOI BBBAAB IIOOIO BBBBAA IIOIOO BBBBBB IIIOOO Fred Lunnon

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[math-fun] Huge software upgrades, again
by Henry Baker 15 Mar '12

15 Mar '12

I'm in the process of downloading iPad 5.1 software: 729 Mbytes! I think that this must be a record for a casual software update. Microsoft will certainly want to beat them for the first Gigabyte software update. I guess that pretty much kills the data plan for the month. Has Apple never heard of rsync ? (Hint: it's already part of the Unix SW that Apple uses.)

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[math-fun] pump [was circular arc splines]
by Bill Gosper 15 Mar '12

15 Mar '12

Fred Lunnon> At last I have understood what Bill's "pump" graphic was about! Not so sure about how/whether it actually pumps stuff or not, mind ... WFL Julian> http://gosper.org/pumper.mp4 (193K) --rwg

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[math-fun] Fwd: circular-arc splines, again
by Bill Gosper 15 Mar '12

15 Mar '12

Henry> The proof a non-existence for n=3 needs to be tightened up a little. ----------------------------------------------------------------------------------------- ES>A C1 configuration of circular arcs remains so under inversion about a point, other than possibly extending to infinity. Invert about one of the arc junctions. The two arcs meeting at that junction become parallel rays, one from a finite point P and extending infinitely far to the left, the other from a finite point Q and extending infinitely far to the right. The rays cannot lie on the same line, as then the two arcs would be part of the same circle. If a 3-arc configuration exists, then P and Q are joined by a single circular arc A. The directed tangents to A at P and Q are parallel, so A must undergo an angle change of 2πn. If n=0, A is a line segment, and this is impossible because the rays do not lie on the same line. If n is not zero, A is (one or more) complete circles, and P and Q coincide, again impossible for the same reason. -- Gene This morning Julian sent me: ---------- Forwarded message ---------- From: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Date: Wed, Mar 14, 2012 at 10:38 AM Subject: Re: [math-fun] circular-arc splines, again To: billgosper(a)gmail.com Without the "simple" condition, making a C1 curve out of k circular arcs reduces to finding k circles, each tangent to the next (and the last to the first), with an even number of external tangencies (zero if you want to avoid inflections). This turns into the equations |z_i-z_(i+1)|=±r_i±r_(i+1) (same sign for external tangent, different for internal). I can't figure out how to solve this with Mma, and it's too complicated to do by hand except in simple cases. Can Macsyma do better? You can't do three arcs, even if you use two external tangencies. You need at least one internal tangency. Then you need another circle either internally or externally tangent to both of them, which would need to be tangent at the same point (you can get a different one that's internal to one and external to the other, but that gives you a cusp in the resulting curve), leading to a non-simple curve. Julian

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[math-fun] arc complexity of a knot
by Warren Smith 14 Mar '12

14 Mar '12

> This discussion leads me to wonder that if a finite number > of circular arcs in R^3 are fitted together to make a C^1 > closed curve that's knotted, what is the smallest number > of distinct arcs that can do this? > > --Dan --that might actually be an interesting new measure of knot complexity: the min number of arcs needed to make that knot. I can prove (but I'll instead leave it as a puzzle) ArcComplexity <= 2*PolygonalComplexity. Can a bound of the same nature be found (with some appropriate constant replacing "2") in the opposite direction?

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[math-fun] Circular arc splines, knotted case
by Dan Asimov 14 Mar '12

14 Mar '12

This discussion leads me to wonder that if a finite number of circular arcs in R^3 are fitted together to make a C^1 closed curve that's knotted, what is the smallest number of distinct arcs that can do this? --Dan ________________________________________________________________________________________ It goes without saying that .

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[math-fun] circular arc splines, 4-segment case
by Henry Baker 14 Mar '12

14 Mar '12

As before, if you start with 2 intersecting circles (circles I & II, of radii r1 & r2), and then draw 2 more circles (III & IV, with radii r3 & r4) inside the I-II intersection which are tangent to both I & II, then you can construct a closed convex C1 curve of 4 segments. By construction, the line of the _centers_ of I & II is perpendicular to the line of _centers_ of III & IV. Now if you simply start with 4 points (P1, P2, P3, P4), such that the segment P1P2 is perpendicular to the segment P3P4, and the two line segments intersect, then we can construct the configuration of circles as above by choosing appropriate radii. In fact, once we choose 1 of the radii, the others can be calculated. In fact, I think that all of these configurations are equivalent to only 2 configurations, modulo Moebius transforms. The 2 different configurations include one in which circles III & IV intersect, and one in which they do not intersect. --- It would be interesting to find something like the "handles"/"control points" of quadratic & cubic splines to design such curves. If I have a line L1 and a point P1 on that line, and a line L2 and a point P2 on that line, then consider the point of intersection P3 (assume that they intersect). We can construct a "hairpin" P1P3P2, which consists of the 2 line segments P1P3 and P3P2. Now replace P3 with a tiny circular arc, and grow that arc until it touches either P1 or P2 (or both). "Growing the arc" consists of increasing the radius -- but not the angle -- of the arc, so that the ends remain C1 continuous with the P1 and P2 line segments, which also shrink. Our "hairpin" now consists of a line segment and a circular arc, and the join point is C1; the line segment may be only a point. This shows how to construct a spline between 2 given points which must meet those points at a particular angle. We note that if we do this inside of a triangle & minimize the number of segments, we produce the incircle.

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[math-fun] Closed C1 curve made of nonempty nonfull circular arc segments
by Warren Smith 14 Mar '12

14 Mar '12

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(a)mailman.xmission.com <math-fun-request(a)mailman.xmission.com> wrote: > Send math-fun mailing list submissions to > math-fun(a)mailman.xmission.com > > To subscribe or unsubscribe via the World Wide Web, visit > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > or, via email, send a message with subject or body 'help' to > math-fun-request(a)mailman.xmission.com > > You can reach the person managing the list at > math-fun-owner(a)mailman.xmission.com > > When replying, please edit your Subject line so it is more specific > than "Re: Contents of math-fun digest..." > > > 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) > > > ---------------------------------------------------------------------- > > Message: 1 > Date: Wed, 14 Mar 2012 09:46:07 -0700 > From: Henry Baker <hbaker1(a)pipeline.com> > To: Fred lunnon <fred.lunnon(a)gmail.com> > Cc: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] circular-arc splines, again > Message-ID: <E1S7rLm-0005WC-K8(a)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(a)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. > > > > > ------------------------------ > > Message: 2 > Date: Wed, 14 Mar 2012 13:15:38 -0400 > From: Andy Latto <andy.latto(a)pobox.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] circular-arc splines, again > Message-ID: > <CAKqg3U2xNs19EpOE=h7k-UW=oFv2XKXhk9GM1CfNsYCBvGK7=w(a)mail.gmail.com> > Content-Type: text/plain; charset=ISO-8859-1 > > On Wed, Mar 14, 2012 at 10:53 AM, Henry Baker <hbaker1(a)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 > > > > ------------------------------ > > Message: 3 > Date: Wed, 14 Mar 2012 13:25:27 -0400 > From: Michael Kleber <michael.kleber(a)gmail.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] circular-arc splines, again > Message-ID: > <CAPmvm8yxQSFcdqDjEWmK_ayoru-yLOEgsQz5hjnwk2ELOTTpHw(a)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(a)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(a)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. >> >> >> _______________________________________________ >> math-fun mailing list >> math-fun(a)mailman.xmission.com >> http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun >> > > > > -- > Forewarned is worth an octopus in the bush. > > > ------------------------------ > > Message: 4 > Date: Wed, 14 Mar 2012 10:38:10 -0700 > From: Henry Baker <hbaker1(a)pipeline.com> > To: Michael Kleber <michael.kleber(a)gmail.com> > Cc: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] circular-arc splines, again > Message-ID: <E1S7sAL-0001PG-Sf(a)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(a)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(a)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(a)pipeline.com> > To: math-fun <math-fun(a)mailman.xmission.com> > Subject: Re: [math-fun] circular-arc splines, again > Message-ID: <E1S7sCs-00081U-9w(a)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(a)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(a)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. > > > > > ------------------------------ > > _______________________________________________ > math-fun mailing list > math-fun(a)mailman.xmission.com > http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > End of math-fun Digest, Vol 109, Issue 17 > ***************************************** > -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step) and math.temple.edu/~wds/homepage/works.html

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[math-fun] Fwd: circular-arc splines, again
by Fred lunnon 14 Mar '12

14 Mar '12

Warning --- Henry Baker's mailer is still somehow foxing the return address! ---------- Forwarded message ---------- From: Fred lunnon <fred.lunnon(a)gmail.com> Date: Wed, 14 Mar 2012 20:23:25 +0000 Subject: Re: [math-fun] circular-arc splines, again To: Henry Baker <hbaker1(a)pipeline.com> What I (and Bill Gosper, I think) have in mind here is illustrated by the second, static diagram at http://www.cut-the-knot.org/do_you_know/cwidth.shtml These curves are C^1, with 4k+2 segments and constant width. WFL On 3/14/12, Henry Baker <hbaker1(a)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(a)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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