July 2015 - mathfun@mailman.xmission.com

[math-fun] General planar gear configurations (Somsky refutes Smith's proof)
by Warren D Smith 12 Jul '15

12 Jul '15

On 7/12/15, WRSomsky <wrsomsky(a)speakeasy.net> wrote: > Wait... In your Theorem 3, in regards to the gear angle/phases, you seem > to be making the > claim that any set of abutted gears will automatically mesh properly? > That if, say, I have four > gears of radius 1 with, say 8 teeth each, that as long as I place their > axles such that the A-B, > B-C, C-D, and D-A distances are all two, that they will always mesh > properly? I don't believe > that is true. --that would indeed be what I was claiming. You do not believe? > Ah, I see where your reasoning is faulty. > You're measuring the exterior angles. > However, for measuring the phase-angle of the gears, > you need to "follow the flow" > and you need to measure the *interior* angle > of every other gear. --Oh dear. So it is kind of an alternating sum, and I omitted the minus signs every other term? Shit. The exterior "turn" angle and interior polygon corner angles are exact negatives of each other (mod 180) so it does not matter in my argument which one you use. (I prefer the former.) However, omitting minus signs in every other term, is something that really does matter! You have refuted my proof. However, G is a bipartite graph, i.e. has two classes of vertices ("pink" and "blue") and as we travel round any cycle in G, we alternately hit pink and blue vertices. So for each cycle, there really are two sums we need to worry about: the pink-angles sum p and the blue-angles sum b. My old argument was that p+b=0 (mod 360) which was automatic for Euclidean polygons. What we really want is that p-b=rational (mod 360). That sadly is not automatic. But it is implied by, for example, p=rational. However, the pink vertices happen to be the sun and outer-ring gear (which I'll call the "antisun") in planetary systems; the planets then are the blue vertices. So long as the planets viewed from the sun are spaced at rational angles apart (measured in degrees) and ditto for the antisun, we are ok. Indeed it suffices if each sun-viewed angle plus its corresponding antisun-viewed angle, is a rational. These demands are certainly true in Somsky's all-collinear case where p=0, or any distorted version of it where, e.g, his angle "Omega" is rational.

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[math-fun] so-called "Somsky gear controversy"; moving on to more gear problems
by Warren D Smith 12 Jul '15

12 Jul '15

Yes, I had already decided myself Rokicki was right, etc, in correspondence to him, Somsky, Deventer; which I'd also mentioned in a post to math-fun. But Julian Ziegler-Hunts may also be confused, in the sense there are TWO kinds of motion possible in Somsky's configuration, namely (a) turning the gears, (b) moving the sun-gear's position. My error was only considering motions of type (b); JZH may be making the opposite error. Type-b motion inherently has bounded travel. Type-a allows unlimited turning. The animations at https://www.youtube.com/watch?v=ApD2pSTWVjk and http://gosper.org/blockysomsky2.gif make the latter especially clear, also showing other people were also working on this. This also confirms my general-N planets claim in two particular cases. The main results I am aware of at present at (1) Somsky theorem about movable sun (2) the analogous (?) less interesting theorem that if you make a quadrilateral ABCD out of 4 meshed spur gears, then it cannot flex -- except in the degenerate case where A & C are racks, then the motion is arguably a bit more interesting than usual; and it perhaps is interesting that both this & Somsky's case involved a two-angles-are-zero degenerate quadrilateral; (3) my general theorem that any planar bipartite graph is attainable as a gear-meshing graph for gears in the plane that do not intersect, and can be turned in sync thru infinite angle. And this attainability in a countably infinite number of inequivalent ways for each such graph. Not yet answered: (4) I presume/hope, but have not investigated/checked, that Somsky further showed his theorem was best possible, i.e. the sun can never move unless his condition is obeyed? I.e. in a quadrilateral of 4 meshed gears, 3 being spur, the 4th being an "inner ring" complemented-spur, the only way the quadrilateral can flex, is if Somsky condition obeyed? So it is "if and only if"? (5) One could also consider bevel gears, which naturally live not on a plane, but on the surface of a sphere; their axes all pass thru the sphere center. Then re-pose all the same questions in that scenario. [True masochists could also consider the hyperbolic plane as the remaining nonEuclidean geometry.] In particular, re my old proof ("theorem 2") about unsymmetric planetary gear systems... I think it should be possible to construct functioning unsymmetric planetary bevel gears on a sphere, I now outline how to try. I am not sure if this now will be possible for any number N of planets; on the sphere there might be a finite maximum possible N. The "stereographic projection" maps sets of circles on the plane to sets of circles drawn on the sphere, and it maps rational coordinates to rational coordinates. https://en.wikipedia.org/wiki/Stereographic_projection http://www.isallaboutmath.com/proof.aspx (annoying long video of ster-proj proof, couldn't bear to watch it all the way thru; but lovely graphics & audio) However, it usually maps rational circle-radii to irrational circle radii (radii measured angularly). That seems like it is usually going to kill you for gear purposes. However, then you realize that what matters for gears is not the angular radius R of the circle drawn on the unit sphere, but rather its CIRCUMFERENCE, which equals 2*pi*sin(R). So what matters is the sine of the angular radius. Since sin(2*arctan(x)) = 2*x/(1+x*x) is a rational function, it is quite possible to map circles with rational radii in the plane, to circles on the unit sphere whose angular-radii have rational sine. In fact this ALWAYS happens if the plane-circle is centered at the origin which is the south pole of the sphere. Then sin((a-b)/2) = sqrt( (1-cos(a-b))/2 ) = sqrt( (1+sin(a)*sin(b)-cos(a)*cos(b))/2 ) indicates that in order to make it happen for off-center circles, you need to be somewhat lucky, the luck is similar to the luck needed in choosing a random right triangle with integer legs and hoping it has integer hypotenuse. So anyhow, I suspect this amount of luck should be attainable after computer search, at least for some small number of planets. (6) My (3) totally settles the question of which "topologies" (or "meshing combinatorial structures," whatever you want to call it) are attainable for functioning coplanar gears. However if we allow gears of different thicknesses and in offset parallel planes, then nonplanar graphs also become possible and I do not have a characterization of which graphs then are attainable. This question likely is too difficult to solve completely, but it is probably feasible to say some interesting stuff that partially solves it. In particular, Kuratowski's theorem tells us that in some (admittedly not very relevant) sense the "only" nonplanar graphs are K5 and K33, and only the latter is bipartite. K23 is attainable as a planetary gear system with 3 planets, 1 sun, and 1 outer ring. Indeed we can even achieve "K33 with a single edge missing." Is it possible to achieve K33 as circle tangency relations among 6 distinct circles in the plane? Yes... but the trouble is that one of them is going to have to be tangent on BOTH its inside AND outside, which is a pretty strange kind of gear, so I do not think you can do this mechanically. If we ask for K33 where each edge is allowed to be a gear CHAIN not just two gears directly meshed, then it is no problem to add the missing edge of K33 in the form of a chain with 2 intermediate gears. It similarly is no problem to do K5 where each K5 edge is done as a chain with 1 or 3 intermediate gears. So I think therefore that any graph can be instantiated using gears provided each graph edge is implemented as a long-enough gear chain, not a direct meshing.

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Re: [math-fun] math-fun Somsky gear controversy
by Bill Gosper 12 Jul '15

12 Jul '15

On Sat, Jul 11, 2015 at 11:59 PM, Julian Ziegler Hunts <julianj.zh(a)gmail.com > wrote: > Rokicki's right; Smith's misreading the proof (more specifically, I think > he's noticing that two angles are the same, but thinking that those two > angles change when the mechanism is rotating, which they don't—they're > related to the relative positions of the gears). > > The hard parts in animating gears are (i) the fact that gears don't > actually move as if they're non-slipping adjacent circles and (ii) deciding > what the teeth should look like, paying attention to the fact that, because > (i) is most easily solved by ignoring it, you have to put effort into > making it look realistic (no intersections, no gaps). > > I mostly ignored these problems in creating the attached gif, which shows > a complete cycle. It's not very smooth, but it gets the point across and > it's already 1.6MB. > > Julian > gosper.org/blockysomsky2.gif Superb. Thank you! Considering the spectrum of possible working configurations, it looks like the revolution (not rotation) period of the sun gear goes to infinity as its radius of gyration goes to zero. Another pathway to van Deventer's 11 million to 1 reducer? --Bill

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[math-fun] Interesting tessellation patterns on Pluto
by Warren D Smith 11 Jul '15

11 Jul '15

http://ichef.bbci.co.uk/news/800/cpsprodpb/7D8E/production/_84224123_untitl…

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Re: [math-fun] curly loops
by Warren D Smith 11 Jul '15

11 Jul '15

> From: Dan Asimov <asimov(a)msri.org> > Let a C^oo closed curve in R^3 be called a "curly loop" if its curvature is > nowhere vanishing. > Two curly loops are for instance A) the unit circle in the plane and B) the > unit circle in the plane traversed twice around. > PUZZLE: Can A and B be continuously deformed one into the other ? through > curly loops ? so that at all stages of the deformation the tangent > directions vary continuously? --Call a "noncurly loop" a C^oo closed curve in R^3 whose curvature is allowed to vanish. If you can do a curly-->curly deformation using noncurly loops, doesn't it automatically follow that you can do it with curly loops? I mean, just replace any tiny segment that looks locally like the graph y=x^3 or y=x^4 i.e has a zero-curvature point at x=0, with a helix-like vine wound round the old curve near there. In particular for your problem, your curve B could be "unfolded" 180 into a figure-8, then "untwisted" 180 into "an 8 with the middle sawn thru," then "inflated" into a circle A, where actually I'll assume slight distortions at all stages so that the curve never self-intersects and stays C^oo. That process may involve a small number of zero-curvature points on the curve, arising near the self intersection of the 8. But if in that vicinity we replace with a very highly wound helical patch during all times when there was any worry, then aren't we obviously safe?

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[math-fun] Mr. Watson come here I want you.
by Bill Gosper 11 Jul '15

11 Jul '15

(The contiguous Watson's Theorem) Thank you! The following message has been sent to the *MathWorld* team: *Source URL:* http://mathworld.wolfram.com/WatsonsTheorem.html *E-mail:* billgosper(a)gmail.com *Name:* --Bill Gosper *Comment type:* ADD *Organization:* dis- *Country:* United States *Comments:* Simply including one contiguous identity, e.g., HypergeometricPFQ[{a, b, c}, {a/2 + b/2, 2 c}, 1] == Sqrt[\[Pi]] Gamma[(a + b)/2] Gamma[1/2 + c] Gamma[ 1/2 (-a - b) + c] (1/( Gamma[a/2] Gamma[(1 + b)/ 2] Gamma[-(a/2) + c] Gamma[(1 - b)/2 + c]) + 1/( Gamma[(1 + a)/2] Gamma[b/ 2] Gamma[(1 - a)/2 + c] Gamma[-(b/2) + c])) lets you add integers to all five parameter slots, via the three-term recurrences. This covers a mere factor of ∞^5 more cases. I can't understand why nobody does this! You should also do it for Dixon, Pfaff, Whipple, Dougall, ... . *MathWorld usage:* weekly *Mathematica usage:* daily

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[math-fun] Somsky gears rotating forever
by Warren D Smith 09 Jul '15

09 Jul '15

Wait a minute, thinking a bit more re Rokicki's argument and my counter-argument: Seems to me TR's "constant rim speed" argument works, and my counter-argument is inapplicable, PROVIDED we regard all axes as fixed locations, and the outer ring rotates about its fixed center as the "input" or "drive." In that case Somsky's angle "Omega" will be constant 180 degrees.

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Re: [math-fun] math-fun Digest, Vol 149, Issue 26
by Warren D Smith 09 Jul '15

09 Jul '15

> However, if the radii b and b' are unequal then Somsky's proof > indicates the two gears with > centers B and B' will rotate at constant angular velocity hence > necessarily eventually collide. TR: I just don't see this at all; why must they rotate at constant angular velocity? For any flat 2D system of gears not slipping, the speed of movement of the contact points on each gear is fixed across all gears, WDS: Only if all axes have fixed location are the rim speeds constant! And if you look at the Deventer video https://www.youtube.com/watch?v=M_BUn4TDns8 about 2:37-3:10 you can see with own eyes that the sun gear is about to hit the outer rim when he pulls it over... in the pictorial proof by Somsky he calls the angle Omega... -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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[math-fun] Do black holes exist?
by Fred Lunnon 09 Jul '15

09 Jul '15

Very thoughtful and approachable recent explanatory survey at Shuang-Nan Zhang Astrophysical Black Holes in the Physical Universe http://arxiv.org/pdf/1003.0291v3.pdf I particularly appreciated the punchy discussion on page 7 (#169) of practical criteria for deciding whether to accept a theoretical model of a physical observation. Fred Lunnon

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[math-fun] Backwards W?
by James Propp 09 Jul '15

09 Jul '15

Typography is a bit removed from math, but I know some of you are font geeks, and in any case, the ability to attend to minutiae of notation is a part of the mathematician's toolkit. So without further apology, let me ask: Is the "W" in http://jamespropp.org/backwards-w.jpg truly backwards, as I think it is? Or does this sort of font-asymmetry admit variations? What's the relevant nomenclature describing the thickness of ascending and descending strokes? (If, ignoring my wife's counsel, I end up entering the Hollingworth 5 and 10 to tell them that their W is backwards, I want to sound like I know what I'm talking about, even if I actually don't.) Jim Propp

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