July 2015 - mathfun@mailman.xmission.com

Re: [math-fun] Draft of "The Lessons of a Square-Wheeled Trike"
by Henry Baker 14 Jul '15

14 Jul '15

The Egyptians experimented with catenaries for St Louis-type arches perhaps 3000-4000 years ago. I've seen a picture in a book somewhere showing a quite good approximation to a catenary arch from ancient Egypt. Of course, they didn't know it as a "catenary", and obviously knew nothing about cosines -- hyperbolic or otherwise. It is conceivable that the ancient Egyptians discovered the Gaudi's principle, wherein hanging catenaries become catenary arches when flipped over and compression & tension exchange places. If some clever Egyptian had discovered this principle, it was quickly lost, as the ancient world dnd't seem to be filled with catenary curves (excepting ship rigging, which the ancient Egyptians excelled at). It is extremely interesting to me that once the Greeks & Babylonians discovered semicircular arches & parabolic arches, no more experimentation seems to have been done with other shapes. In particular, the ancient Romans (to my knowledge) used ONLY circular arches, even though the amount of material required was substantially greater than what would be required for other types of arches. The Romans were outstanding engineers, and while waiting for their slave labor to carry the excess material required to build a circular arch, these smart people had plenty of time to ponder whether there was a better way. There are several possible explanations why they stuck with circular arches: 1. The Greeks became enamored of the elegance of circles, and -- whether or not another shape would have been more "efficient" -- they preferred circles for aethetic/moral/religious reasons. Analogous problems limited astronomers until Galileo risked death/excommunication to argue for ellipses. 2. The ancient Romans may have depended upon a method of building arches that depended critically upon their being arcs with a consistent radius. I have conjectured that the ancient Romans devised a system of "centering" that did NOT require a huge amount of scaffolding. This is because the Romans built so many arches, in so many locations -- including locations without any lumber for scaffolding -- that they required a modular method using a minimal amount of lumber that had to be carried along with the troops for other purposes -- including building siege towers and catapults. In this case, the Romans weren't worrying about minimizing material within the arch itself, but minimizing the material required for the "centering" of the arch. https://en.wikipedia.org/wiki/Centering 3. Strict liability. The engineers who designed & built structures in the ancient world were required to stand under them when load was first applied. This strict liability principle guaranteed significant overbuilding -- which is why some Roman roads & arches are still in use today after 2000 years -- but so severely penalized experimentation that no one was brave enough to consider alternatives. http://www.commonlaw.com/Hammurabi.html 229. If a builder has built a house for a man, and has not made his work sound, and the house he built has fallen, and caused the death of its owner, that builder shall be put to death. 230. If it is the owner's son that is killed, the builder's son shall be put to death. 231. If it is the slave of the owner that is killed, the builder shall give slave for slave to the owner of the house. At 12:35 PM 6/30/2015, James Propp wrote: >I've finished a draft of installment #1 (as opposed to #0) of "Mathematical Enchantments", and I'd welcome comments.

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Re: [math-fun] New challenge: Offset Sonsky Gears
by rcs＠xmission.com 14 Jul '15

14 Jul '15

<Another mystery reject. Including a diagram seems to be a factor, but not the whole story. I've excised the diagram. --rich> ----------- Date: Mon, 13 Jul 2015 23:06:45 -0700 Subject: Re: New challenge: Offset Sonsky Gears From: Tom Rokicki <rokicki(a)gmail.com> To: William Somsky <wrsomsky(a)gmail.com> Can you share any support for your belief, or is this a puzzle for us to resolve? -tom

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Re: [math-fun] New challenge: Offset Sonsky Gears
by M. Oskar van Deventer 13 Jul '15

13 Jul '15

Hi Tom, Thank you for accepting the challenge, and for responding so quickly. > do you mean, they won't mesh with the same centers as in the original (36-18-10-8-6) ... That is indeed what I mean. I want to 3D print a Somsky-style contraption, where the sun gear is coaxial with another sun gear, and the same for each other planetary gears and the annulus. I can follow your reasoning for the individual (35-17-9-9) meshes as does (35-17-11-7). However, I doubt whether your reasoning is correct for both together. How can we check that? > Maybe what happens is the two 11's clash? I am not worrying about clashes. Mathematically, gears can overlap. Moreover, if your reasoning is correct, then there should exist plenty “Rokicki Gears” that don’t overlap. Or am I misunderstanding you? Best regards, Oskar From: Tom Rokicki Sent: Monday, July 13, 2015 9:18 PM To: M. Oskar van Deventer Cc: Warren Smith ; William Somsky ; Julian Ziegler Hunts ; Bill Gosper ; math-fun Subject: Re: New challenge: Offset Sonsky Gears Oskar, Thanks for the challenge! When you say the (35-17-11-9-7) gears don't mesh---do you mean, they won't mesh with the same centers as in the original (36-18-10-8-6), or they don't mesh in *general* (that is, permitting the centers of the gears to move)? Clearly by Somsky (35-17-9-9) meshes as does (35-17-11-7), and I would think therefore so would (35-17-11-9-7) with only one 11 and one 7, and then by symmetry so would (35-17-11-9-7) with two 11s and two 7s. That is, I would *expect* that the (35-17-9-9) has enough freedom of the sun gear to permit an 11-7 pair to also mesh. Maybe what happens is the two 11's clash? I'm not doing this mathematically, but strictly intuitively, which of course is always dangerous . . . but I'm curious where my intuition is breaking down. -tom On Mon, Jul 13, 2015 at 12:00 PM, M. Oskar van Deventer <m.o.vandeventer(a)planet.nl> wrote: Gentlemen, While you are still discussing new theorems about the Somsky Gears (which I am unable to parse as non-mathematician), I would like to take the liberty and pose a new challenge: Offset Somsky Gears. What Bill Somksy has proven, is that there are plenty of exact solutions for planetary gears where the sun is offset from the annulus gear, with exactly meshing gears. Bill sent me the below 34-18-10-8-6 example mid 2012. So how about offsetting the generating circle of each gear as shown in the image below? In this example, I offset Bill’s 34-18-10-8-6 geometry into a 35-17-11-9-7 geometry. So the circles fit in this geometry. However, when drawing the corresponding gears, you will discover that they won’t mesh. So offsetting these Somsky Gears does not yield more Somsky Gears. For regular planetary gears, the classic threefold symmetrical (120-degrees) concentric geometry has many solutions with different gearing ratios that all mesh. Now, the challenge is to find asymmetric concentric planetary-gear geometry and/or a Somsky geometry that meshes, AND where the above-described offset yields another exactly meshing configuration. I hope that the challenge is a bit clear. Probably, a professional mathematician can provide a proper definition of the challenge. Enjoy! Oskar -----Original Message----- From: Warren D Smith Sent: Monday, July 13, 2015 3:31 PM To: M. Oskar van Deventer ; William Somsky ; math-fun Subject: Gear topologies main theorem, revised & corrected I got tired of posting wrong theorem+proofs that get refuted by return mail, so I wrote it more carefully, put it in a file and stuck it on the web via dropbox. You can read the theorem & proof here: https://dl.dropboxusercontent.com/u/3507527/GearTopologies.txt -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step) -- -- http://cube20.org/ -- [Golly link suppressed; ask me why] --

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[math-fun] Gear topologies main theorem, revised & corrected
by Warren D Smith 13 Jul '15

13 Jul '15

I got tired of posting wrong theorem+proofs that get refuted by return mail, so I wrote it more carefully, put it in a file and stuck it on the web via dropbox. You can read the theorem & proof here: https://dl.dropboxusercontent.com/u/3507527/GearTopologies.txt -- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)

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Re: [math-fun] New challenge: Offset Sonsky Gears
by William R Somsky 13 Jul '15

13 Jul '15

Let me take a look and see if I still have my program which would tabulate the 4-gear exact matches for small tooth-count gears (I think I did up to an annulus of 50 teeth). Scanning that should see if there are any that match your criterion. On 07/13/15 12:00, M. Oskar van Deventer wrote: > Gentlemen, > While you are still discussing new theorems about the Somsky Gears > (which I am unable to parse as non-mathematician), I would like to > take the liberty and pose a new challenge: Offset Somsky Gears. > What Bill Somksy has proven, is that there are plenty of exact > solutions for planetary gears where the sun is offset from the annulus > gear, with exactly meshing gears. Bill sent me the below 34-18-10-8-6 > example mid 2012. > G34-18-10-8-6 > So how about offsetting the generating circle of each gear as shown in > the image below? In this example, I offset Bill’s 34-18-10-8-6 > geometry into a 35-17-11-9-7 geometry. So the circles fit in this > geometry. However, when drawing the corresponding gears, you will > discover that they won’t mesh. So offsetting these Somsky Gears does > not yield more Somsky Gears. > Offset Somsky Gears - view 1 > For regular planetary gears, the classic threefold symmetrical > (120-degrees) concentric geometry has many solutions with different > gearing ratios that all mesh. > Now, the challenge is to find asymmetric concentric planetary-gear > geometry and/or a Somsky geometry that meshes, AND where the > above-described offset yields another exactly meshing configuration. > I hope that the challenge is a bit clear. Probably, a professional > mathematician can provide a proper definition of the challenge. > Enjoy! > Oskar

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[math-fun] superconductivity at 203K alleged in rotten eggs
by Warren D Smith 13 Jul '15

13 Jul '15

http://arxiv.org/abs/1506.08190 hydrogen sulfide allegedly is superconductive at 203K, by far a new record high temperature, but the catch is you need to pressurize it over 90 GPa in a diamond anvil.

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[math-fun] Fwd: Re: math-fun Somsky gear controversy
by rwg 13 Jul '15

13 Jul '15

[Delayed in vain anticipation of Julian's blessing.] Fred, stop polishing those damned specs! The bug isn't the sizes, it's the physically impossible square teeth, as Julian admitted. Another reason to hate orthodontists. It's ironic that mathematically correct teeth sell the mathematically incorrect Looney gears. --rwg JZH animation: gosper.org/blockysomsky2.gif Somsky's 5 planets(!): gosper.org/somsky5.gif -------- Original Message -------- Subject: Re: [math-fun] math-fun Somsky gear controversy Date: 2015-07-12 04:02 From: Fred Lunnon <fred.lunnon(a)gmail.com> To: math-fun <math-fun(a)mailman.xmission.com> Reply-To: math-fun <math-fun(a)mailman.xmission.com> There's a bug in the JZH animation: the sun gear is slightly up-scaled. An animation of Deventer's 3-planet "Looney gears" (sic) lacking the extra radius-13 planet is at https://www.youtube.com/watch?v=ApD2pSTWVjk The Somsky proof at 3:35 in Deventer's video https://www.youtube.com/watch?v=M_BUn4TDns8 establishes that the gear teeth actually mesh together correctly. In an elementary 3-planet mechanism, this meshing imposes two additional (continuous) constraints on the mechanism, leaving only a _single_ degree of freedom modulo isometry for given planet radii. It follows that if a frame of reference is chosen in which (say) centres of outer ring and one planet are fixed in the plane, then all centres are necessarily fixed; the remaining freedom is expressed by the rotation angle of the ring. Fred Lunnon

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[math-fun] huge black hole which formed quite quickly
by Warren D Smith 13 Jul '15

13 Jul '15

This hole is claimed to have attained about 10% of its host galaxy's mass, when universe was only 2 Gyr old, as opposed to usual story which is attaining 0.2% to 0.5% of galaxy mass by now (universe age 13.8 Gyr). How? Benny Trakhtenbrot, C. Megan Urry, 7 others: An over-massive black hole in a typical star-forming galaxy, 2 billion years after the Big Bang Science 349, 6244 (10 July 2015) 168-171 ABSTRACT Supermassive black holes (SMBHs) and their host galaxies are generally thought to coevolve, so that the SMBH achieves up to about 0.2 to 0.5% of the host galaxy mass in the present day. The radiation emitted from the growing SMBH is expected to affect star formation throughout the host galaxy. The relevance of this scenario at early cosmic epochs is not yet established. We present spectroscopic observations of a galaxy at redshift z=3.328, which hosts an actively accreting, extremely massive BH, in its final stages of growth. The SMBH mass is roughly one-tenth the mass of the entire host galaxy, suggesting that it has grown much more efficiently than the host, contrary to models of synchronized coevolution. The host galaxy is forming stars at an intense rate, despite the presence of a SMBH-driven gas outflow.

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Re: [math-fun] math-fun Somsky gear controversy
by rcs＠xmission.com 12 Jul '15

12 Jul '15

<This note is from van Deventer. I approved the non-member post, but then it was apprehended by the Math-Fun Magical Mystery Spam Filter. I had to remove the cute diagram. It's four gears of different sizes, surrounding a fifth sun gear that also rotates and has a moving center, all inside a sixth concave-circular gear. --rich> Hi Bill (Gosper), Thank you for sharing this with me. I am including Bill (Somsky) and Warren Smith at the cc, as they may be interested in this too. Oskar -------- Original Message -------- Subject: Re: [math-fun] math-fun Somsky gear controversy Date: 2015-07-12 02:30 From: Bill Gosper <billgosper(a)gmail.com> To: Julian Ziegler Hunts <julianj.zh(a)gmail.com> Cc: math-fun(a)mailman.xmission.com Reply-To: math-fun <math-fun(a)mailman.xmission.com> 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 > [diagram removed --rich] 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] General planar gear configurations
by Warren D Smith 12 Jul '15

12 Jul '15

I. If we are not interested in movable gear-axes, then it seems to me that Somsky's sum-of-radii condition is not needed, which makes my theorem 2 work without needing to choose any "mates" obeying Somsky's condition, and then for any N>1, including odd N, number of planets, in Deventer's "unsymmetric planetary gears." II. Also as I remarked before, after unconfusion supplied with aid by Rokicki, my theorem 1 was false, or more precisely: it is true that Somsky's type of Omega-altering motion has bounded travel, but the other kind of motion with all gear-axis locations fixed, permits infinite turning. ----- III. I now believe I can prove this highly general THEOREM 3: Let G be _any_ planar bipartite graph. ( https://en.wikipedia.org/wiki/Planar_graph , https://en.wikipedia.org/wiki/Bipartite_graph .) * With each vertex of G we can associate a circular gear in the plane. * With each edge (a,b) of G we can demand the gears associated with vertices a and b be perfectly meshed. * With each non-edge (a,b) of G we can demand the gears a,b do not mesh and do not conflict by overlapping. * Any one of the vertices of G can be demanded to be an "outer ring" type gear with all the others being conventional filled-circle type gears. * All gear radii can be chosen to be integers in a countably infinite number of scaling-inequivalent ways. * The gears will all rotate in sync with their centers (axes) fixed at the specified vertex locations, with infinite angle of turning permitted. And suitable initial angles exist such that all gear teeth mesh. Note, for non-bipartite G, there would be an odd-cycle in the graph, and then gear rotation would be impossible because of a sign problem. E.g. 3 gears meshed in a triangle are jammed and cannot rotate, but 4 gears meshed according to a quadrilateral can rotate. Also for non-planar G, no such configuration is possible in the sense that some gears would need to overlap. (However, with offset parallel planes some such overlaps could be permissible, in which case nonplanar G could be permissible, and I have not attempted to determine which nonplanar graphs G then would be permissible. It is easy to see that some such nonplanar G do exist.) Thus in these senses THEOREM 3 is "best possible." In particular my theorem 2 about the unsymmetrical planetary gears is only a very special case of it. HOW TO PROVE IT: Many of the underlying ideas were described in a paper I wrote many years ago, which you can read here: paper #6 here: http://rangevoting.org/WarrenSmithPages/homepage/works.html the new ideas beyond that paper involve showing it can always be done with integer radii, in a countably infinite number of ways, and that suitable angles exist. Wlog let G be "polyhedral" having V vertices, E edges, F faces, where V-E+F=2 by Euler theorem. (Basically because if G's edge-set is a subset of such a polyhedral graph the proof will only get easier.) Each face is a 2k-gon for integers k>1. This enables seeing that F<=V-2, and then that E<=2V-4. Now consider supplementing G's edge-set with extra "red" edges to convert it to a maximal (fully triangulated polyhedral) graph H. The number of extra red edges will be at least F (if all faces are quadrilaterals) and indeed at least 2E-3F=2(V-2)-F=2V-F-4>=V-2. Embed H using the Andreev-Thurston theorem (or my paper's polynomial time algorithmic version of it). By an appropriate inversive transformation and real-rescaling, any particular G-vertex can be demanded to be "outer" i.e. an outer-ring type gear, of specified radius, let us say radius=1, with all the other G-vertex circles lying inside it, and two of these interior circles also have specified sizes which I can and will demand to be rational numbers, by using the 3 continuous degrees of freedom of the 3 parameters specifying the inversive transformation and rescaling. Now, there are V-3 gears (G vertices) whose radii have not yet been specified. These are variable because the red edges of H are not present in G, and for each one we remove from H there is a new degree of freedom of motion of the G-vertices while still satisfying the conditions that all the Andreev-Thurston circles associated with the G-vertices are tangent if and only if there is a G-edge between those two vertices, otherwise they do not touch. These motions are permissible within some perhaps small, but nonzero range. We keep removing the red edges of H one by one, using the resulting freedom to specify one more gear-radius associated with 1 more G-vertex, to be (again each time) a rational number. This can be done because the rationals are everywhere dense in the reals, so no matter how small the range of motion we get by removing a red H-edge, there always is a rational radius inside that range. We can do this for at least the 3 initially specified rationals, plus at least V-2 more. And since V-2+3>V, we clearly have enough degrees of embedding freedom to make all gear radii rational in this way, indeed with at least one degree of freedom left over. Now multiply everything by LCM(denominators) to make all gear radii be integers. Now at this point we need to associate an "initial angle of rotation" with each gear. If you have a gear, at a specified rotation angle A to the x-axis-direction, then any gear that is meshed with it has a rotation angle that then is determined (up to rotations by multiples of 360/T degrees if T is its number of teeth) by A and by the relative locations of the two gears. If we go round a polygonal cycle in the graph G, fortunately the total sum of exterior "bend" angles in this polygon must sum to 360 degrees, which means we never get a conflict at the end of the cycle when we get back to the original gear; the final demand is always the same modulo 360 as the initial demand. So we can color "chains" of edges of G "blue" one chain at a time, each chain linking two previously-blued vertices as its endpoints. Each chain we add will not generate a conflict. Eventually the entirety of G is colored blue, i.e. we have chosen acceptable initial angles of rotation for every gear, without ever suffering a conflict. In that argument, you might worry that there are E demands about angles, but only V angles which is too many equations for too few unknowns. However, the "sum of bends in polygon=360" Euclidean theorem means that really, many of these E demands are redundant, i.e. are consequences of others, not independent. The number of independent demands is F, due to the well known theorem that G's faces form a "cycle basis"; and since F<=V-2 this worry is obliterated. At this point we have specified the integer radii, the center (axis) locations, and the initial rotation angles, for every gear. Finally, we can turn one of the gears, and the others will all turn too with the invariant that the rim-speed is equal for every gear. (And due to bipartiteness, the speeds all have the correct signs.) This invariant shows that there is no "jam" and it turns freely forever. QED. 3+F>2+F<=V

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