March 2012 - mathfun@mailman.xmission.com

Re: [math-fun] circular-arc splines, again
by Henry Baker 14 Mar '12

14 Mar '12

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(a)pipeline.com> 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. > >-- >Forewarned is worth an octopus in the bush.

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Re: [math-fun] circular-arc splines, again
by Henry Baker 14 Mar '12

14 Mar '12

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.

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Re: [math-fun] circular-arc splines, again
by Henry Baker 14 Mar '12

14 Mar '12

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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Re: [math-fun] circular-arc splines, again
by Henry Baker 14 Mar '12

14 Mar '12

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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[math-fun] circular-arc splines, again
by Bill Gosper 14 Mar '12

14 Mar '12

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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[math-fun] circular-arc splines, again
by Henry Baker 13 Mar '12

13 Mar '12

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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[math-fun] Re library of Babel
by Stuart Errol Anderson 13 Mar '12

13 Mar '12

... Winners? Losers? I'd rather a conversation than a competition anyday.

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[math-fun] Shape of the Colosseum, part 2 (circular splines)
by Henry Baker 13 Mar '12

13 Mar '12

If a simple closed curve is composed of a finite number of circular arcs that are joined with continuous derivatives (C1 continuity), then it has an interesting property. Consider the face and hour-hand of a circular clock. Each point on the circular border of the clock face is uniquely related to a 12-hour time, and a normal vector pointing towards the center of the clock is in 1-1 correspondence with a 12-hour time. Now consider a "clock" whose border consists of a finite number of circular arcs joined in with C1 continuity and such that the center of the radius of curvature is always "to the right" as we traverse the curve in "clockwise" manner. Note that the radii of curvature of the different arc segments may be different. Yet the normal to any point on any circular arc which points towards the center of curvature is still related uniquely to the angle of the normal, and hence to a clock-time. So a point moving along such a border can always be given a unique clock time which is determined by its absolute angle relative to the center of curvature of its arc segment. Note that this is well-defined even at the join points, since both arcs agree on what time it is at their join points. Now it remains to be seen if we can construct such curves. We can obviously construct a circular arc with no join points. We can't construct one with one join point, unless it is still a single circle. We also can't construct one with two join points, unless the two radii of curvature match, because we can't achieve C1 continuity. I conjecture that such curves can always be constructed with 3 or more join points. The Colosseum has 8 such join points. Here is a way to get a "cheaper" Colosseum from 4 join points. We can construct two of the arcs to go from 2:00 o'clock to 4:00 o'clock and 8:00 o'clock to 10:00 o'clock. The 3rd and 4th arcs go from 4:00 o'clock to 8:00 o'clock and from 10:00 o'clock to 2:00 o'clock. We choose radius of curvature r1 for the first two arcs and radius of curvature for the second two arcs. If r1=r2, we get a circular clock face, but if r1<r2 or r1>r2, then we get an approximation to an elliptical clock face. Note that C0 continuity is assured by symmetry of the structure, while C1 continuity is assured by the continuity of the angles at the join points. Note that all the join points are determined by symmetry, once we pick on join "time" -- e.g., 2:00 o'clock. However, we are free to pick this particular join point anywhere between 12:00 o'clock and 3:00 o'clock. Using the same ideas, we can interpolate 4 additional join points & choose one additional radius of curvature r3. Once again, we can choose this third radius of curvature without further constraint. When we allow infinite radii of curvature, things get more interesting; also, if we allow negative radii of curvature, we can start "counting the clock backward" for those arc segments.

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[math-fun] The Library of Babel
by Stuart Anderson 13 Mar '12

13 Mar '12

Is the Library of Babel a sphere whose consummate center is any hexagon, and whose circumference is inaccessible.? >From Wikipedia; "Borges's narrator describes how his universe consists of an enormous expanse of interlocking hexagonal rooms, each of which contains the bare necessities for human survival—and four walls of bookshelves. Though the order and content of the books is random and apparently completely meaningless, the inhabitants believe that the books contain every possible ordering of just a few basic characters (letters, spaces and punctuation marks). Though the majority of the books in this universe are pure gibberish, the library also must contain, somewhere, every coherent book ever written, or that might ever be written, and every possible permutation or slightly erroneous version of every one of those books. The narrator notes that the library must contain all useful information, including predictions of the future, biographies of any person, and translations of every book in all languages. Conversely, for many of the texts some language could be devised that would make it readable with any of a vast number of different contents. Despite — indeed, because of — this glut of information, all books are totally useless to the reader, leaving the librarians in a state of suicidal despair. This leads some librarians to superstitions and cult-like behaviour, such as the "Purifiers", who arbitrarily destroy books they deem nonsense as they scour through the library seeking the "Crimson Hexagon" and its illustrated, magical books. Another is the belief that since all books exist in the library, somewhere one of the books must be a perfect index of the library's contents; some even believe that a messianic figure known as the "Man of the Book" has read it, and they travel through the library seeking him." A fulltext copy of the story is here;http://www.daimi.au.dk/~doina/literature/Jorge_Luis_Borges_-_The_Library_of_Babel.html >From the third paragraph of the story: "Each book contains 410 pages; each page, 40 lines; each line, about 80 black letters." That makes 410 x 40 x 80 = 1,312,000 characters. The fifth paragraph tells us that "there are 25 orthographic symbols" including spaces and punctuation. So the library contains 25^1,312,000 books. "Five shelves correspond to each one of the walls of each hexagon; each shelf contains thirty-two books of a uniform format;" "Twenty shelves - five long shelves per side - cover all sides except two;" So there are 640 books in a hexagon. Borges states what he calls "the classic dictum: The Library is a sphere whose consummate center is any hexagon, and whose circumference is inaccessible." However the total number of books and the hexagons are not commensurate, the classic dictum is not strictly speaking true. To show this, it suffices to divide the number of books in the library by the number of books in a hexagon, ignore the large quotient and focus on the remainder. Using mathematica (and continued fractions - could have done it without them) I got a remainder of 385, that is there is at least one hexagon with only 385 books - 12 shelves of 32 books and one book left over. Put that book on its own shelf and there are seven empty shelves left. Nice numbers The phrase "The library is a sphere whose exact center is any hexagon and whose circumference is unattainable" seems to rule out the existence of a uniquely special hexagon. But such a hexagon must exist, and it could be designated as the spacial origin ie the center. The remainder can be calculated with continued fractions. In mathematica ' ContinuedFraction[(5^2623999)/128] The result was a 182165 digit number followed by 5 terms of the continued fraction expansion of the remainder ie {305631237.... ..... 339715957641, 1, 1, 1, 1, 25} then convert the continued fraction terms into a fraction; FromContinuedFraction[{0,1, 1, 1, 1, 25}] = 77/128 the result; 77/128 (of a hexagon) 640 * 77/128 = 385 books I wrote to Bloch, William Goldbloom author of The Unimaginable Mathematics of Borges’ Library of Babel (2008) and he did the same calculation using mod arithmetic in Maple; irem(25^1312000,640); or 25&^1312000 mod 640; Both of them quickly return 385. The mod calc was straighforward, I still find it strange doing arithmetic on such large numbers, I wonder what the practical size limits are for large numbers in software like maple and mathematica? Some thoughts; It seems to me that the 385 book hexagon complicates Borges story in some interesting ways. On one hand you can say its nothing special - plenty of hexagons are missing books - from the book destruction by Purifier librarians, so what makes this hexagon unique? How do we even know that the 'missing' 255 book slots are not spread across the whole library? Just an empty slot on a shelf in 255 different disparate locations (or any possible combination) that are to all intents and purposes inaccessible. One can say Borges was not a mathematician so he did not embed the remainder hexagon in the story. He picked the dimensions of the books and hexagon shelving from the municipal library where he worked when writing the story. On the other hand the stupendous scale of the library focusses attention on the design of its essential elements ie the books, hexagons and librarians. The numerical relationship between numbers of books and numbers of hexagons is a central design element of the library, how can it not be deemed significant? Why not explore how this new information sits with the existing ideas, themes and concepts found in the text? Readers bring new interpretations and the meaning of a text shifts with time, details of a texts creation and the authors intent are not the whole story. And Borges may have known his choice of numbers would produce a remainder, without knowing what it was (though I doubt this). The Crimson Hexagon is very interesting when considered in relation to the 385 book hexagon, ".. the Crimson Hexagon --- books smaller than natural books, books omnipotent, illustrated and magical" the 385 book hexagon could segue into the Crimson Hexagon, if the books are illustrated they are not part of the 80 char 40 line 410 page book collection, and need separate shelving. If they are all magical then perhaps they could act as a catalog, providing an alternative to considering the library as its own catalogue. The notion of a first book and a last book is also interesting. The first book if done lexigraphically might be all aaaaaaaa... and the last book all spaces (or whatever the 25th symbol is). The 385 hexagon could accentuate its uniqueness by containing the first book or last book or both. Perhaps not, books seem randomly distributed around the library. There is also The Book (of the Book-Man) ""On some shelf in some hexagon, it was argued, there must exist a book that is the cipher and perfect compendium of all other books" If there are 385 books in a hexagon and they are stacked together, the 385th book has to go on its own shelf. This particular book is a good candidate for The Book, given the uniqueness. Then there is the circular book, though on reflection bringing that in to this 385 hexagon is pushing it, as even the narrator discounts the idea of a circular book. The empty space in the 385 book hexagon is also another angle. There is space for 255 more books. Books that are invisible, books that use different symbols, books that are yet to be written ... 255 is a Mersenne number, 255 is a repeating digit in base 2 .... of course there are many ways of looking at the library, Wikipedia mentions some of the philosphical implications; "There are numerous philosophical implications within the idea of the infinite library. One of these is that the librarians may have been horribly mistaken about the nature of some of the books of nonsense. Some of these are assuredly copies of other books, some written in a substitution cipher, others phonetically, some in made-up languages, etc. Every book in the library is intelligible, if one decodes it right. Any book in the library can be decoded into any other book in the library, using a third book in the library as a One-time pad. This lends itself to the philosophical idea proposed by Immanuel Kant, that by defining rules for the universe, we create rules of the universe. Because the librarians assumed that the books of nonsense were exactly that, they may have tossed away several copies of Directions to the crimson hexagon from where you are now standing, simply because it was written in a cipher. Additionally, because there are by definition all books, there are certainly also books of lies and falsehoods. For each copy of the catalog to the library, there will be many copies of false codices, claiming some false books to be true and some true books to be false. In short, any room in the library could be the crimson hexagon. Hidden in the gibberish of the library, there are works beyond human capacity to write, simply by definition that it contains all possible books, of which these are a possibility. The library cannot be damaged by the destruction of any of its books because even though a single book is unique, there are also similar books differing by a single letter. The library is a temptation, because it offers these gems of enlightenment, and buries them in deception. One can consider any text as being pulled from the library by the act of the author defining the search letter by letter until they reach a text close enough to the one they intended to write. The text already existed theoretically, but had to be found by the act of the author's imagination.[7] Another implication is an argument against certain proofs of the existence of God, as it is carried out by David Hume using the thought experiment of a similar library of books generated not by human mind, but by nature. [8] Stuart Anderson www.squaring.net

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[math-fun] the shape of the Colosseum
by Henry Baker 12 Mar '12

12 Mar '12

In case anyone ever asks, the shape of the Colosseum in Rome is a pretty decent approximation to an ellipse, given by a sequence of 8 circular segments which are continuous & have continuous first derivatives. It is well known that the evolute of the ellipse is an astroid (though not from outer space); the evolute traces the centers of curvature, hence the centers of the circular arcs that form the circumference of the Colosseum. As an approximation, the Romans chose 8 points on the astroid & constructed 8 arcs that were continuous. Thus, at each join point, the _radius_ of the arc changes discretely to the next point on the astroid, while the arc itself remains continuous with continuous derivative to the next arc. The 8 points were the 4 "corners" of the astroid, plus 4 intermediate points; the points are symmetric in both the x & y directions. The major axis is approx. 187.2 meters; the minor axis is approx. 157.3 meters. Since the seating has approximately the same seat-seat distance all around the "ellipse", the internal oval is _not_ an ellipse, as it does not scale linearly with the exterior elliptical walls. It's highly unlikely that the Romans knew anything about evolutes or astroids, but they did seem to like the ellipse. At some point in the future, I'm going to attempt to calculate the maximum deviation from "elliptical" of this Roman construction. I suspect that the error is of the order of single digit feet. I don't know if the foci of the Colosseum have any particular role in the events of the Colosseum. Due to the elliptical shape of the Colosseum, and the fact that the highest (wooden) rank of seats was very steep, the Colosseum would have some aspects of a "whispering gallery", whereby sound at one focus would be easily heard at the other focus. This effect would be even more pronounced when the canvas covering was in use. One can only wonder if the Romans ever performed Arthur C Clarke's "sunstroke" in the Colosseum by having all the spectators use mirrors to burn up someone on the field. Archimedes had already shown them how it could be done. Perhaps this is why ancient Romans didn't take cellphones to the Colosseum -- to avoid the heated displeasure of the crowd ? http://variety-sf.blogspot.com/2007/09/arthur-clarkes-slight-case-of-sunstr… Taylor, Rabun. Roman Builders: A Study in Architectural Process. Cambridge U. Press, 2003, p. 139.

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