March 2012 - mathfun@mailman.xmission.com

[math-fun] Magic square expert?
by rcs＠xmission.com 21 Mar '12

21 Mar '12

I have a note from a Korean high school student, asking if the results in his paper are new. The paper is about panmagic squares. Can anyone recommend an expert to consult? Rich

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Re: [math-fun] Amazing pi, how sweet the sound
by Dan Asimov 21 Mar '12

21 Mar '12

Ha! If only that same judge had been around to tell Matthew Cook he had every right to publish his universality proof of rule 110, we wouldn't have had to wait two years. --Dan Hilarie wrote: << Its digits will be free, You can sing without a bound, But does this hold for e? < http://lightyears.blogs.cnn.com/2012/03/21/can-you-copyright-music-of-pi-ju… > >> ________________________________________________________________________________________ It goes without saying that .

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Re: [math-fun] Julian pimped my pump
by Neil Bickford 21 Mar '12

21 Mar '12

BG>Perhaps NeilB can convert this to a wmv for us. Video: https://vimeo.com/38945542 (See Download>Original for the original .wmv) --Neil On Wed, Mar 21, 2012 at 9:37 AM, Bill Gosper <billgosper(a)gmail.com> wrote: > On Tue, Mar 20, 2012 at 12:28 AM, Bill Gosper <billgosper(a)gmail.com> > wrote: > > I got Now Live msgs for both .mp4 and .mov. Which got posted? Did you > > squish > > it 2x vertically on purpose? > > Since the axis of the pump was not parallel to the vertical axis of > the video, the > squish caused a distortion which is nonrigid over time, else I'd alert > math-fun > that you beat Neil to the punch. > --Bill > > > Can it be looped? > > --Bill > > > > On Mon, Mar 19, 2012 at 8:41 AM, Stephen Jones <sjones(a)sevcom.com> > wrote: > >> > >> I'm going to upload the 15mb AVI to toobnix and see how it does through > >> the > >> conversion to 'web friendly player' format. I'll give you the link. > >> toobnix > >> is still in alpha, but it could prove to be a lot of 'advertiser free > >> fun.' > >> > > >

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[math-fun] Amazing pi, how sweet the sound
by Hilarie Orman 21 Mar '12

21 Mar '12

Its digits will be free, You can sing without a bound, But does this hold for e? http://lightyears.blogs.cnn.com/2012/03/21/can-you-copyright-music-of-pi-ju…

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Re: [math-fun] Optimum shapes -- some myths debunked & clarity gained
by Henry Baker 19 Mar '12

19 Mar '12

You are correct about the catenary not having constant stress, only constant weight/mass per unit length. I think that long anchor chains for boats/ships may thin out towards the anchor itself, because the lower portions don't have to carry as much load. I think that the reason for the non-constant cross-section of the St Louis arch has more to do with wind loads than static loads. Ditto for the Eiffel Tower, which was designed more for wind loads than static loads. Note that the "inversion" principle doesn't depend upon constant stress or constant cross-section, but is more general. I would be interested in your solution for the constant stress case, because that is more like a real arch made of constant (?) bulk modulus material. I'll have to study it more carefully. Check out this guy's efforts on "fractal" structures of real materials: http://www.london-institute.org/people/farr/fractals.shtml At 06:18 PM 3/19/2012, Warren Smith wrote: >> H.Baker: >> There is a principle of "inversion" for structures in gravity: under the >> transformation y <-> -y, compression becomes tension & vice versa. I think >> that Hooke remarked on this principle. >> >> So, the catenary (y=cosh(x)) curve assumed by a hanging chain becomes the >> St. Louis arch or the Gaudi cathedral (Gaudi actually designed his cathedral >> upside down with weights & strings tied together; these models are on view >> in the basement of his cathedral in Barcelona). >> >> This principle must have been known to the Greeks & Romans; they need only >> look at the necklaces on their spouses, or the chains holding their slaves >> together. > >--well... no. > >Perhaps the inverted catenary is indeed an optimum arch, but >definitely NOT for a constant-cross-section thin beam... instead the >beam would have to become thicker near its base, with a certain >roughly exponential dependence of thickness on height. >This should be obvious by considering infinitely large arch, obviously there >is a finite upper limit to size at given strength, which the catenary >fails to tell you >about hence it is wrong. >Put another way, obviously the catenary does not have constant tension, it grows >along the chain; hence cannot be optimum in the sense of having constant stress. > >Wikipedia says: >"a catenary is the ideal shape for a freestanding arch of constant thickness" >which my analysis would claim was a flat-out lie. >http://en.wikipedia.org/wiki/Catenary#The_inverted_catenary_arch > >To try to get constant stress, cross-sectional-area >instead would have to vary proportional to total mass of portion >of arch lying above that point, times the |y'| / sqrt(1+y'^2) >geometry factor. >The St Louis arch indeed does have such thickening, but the trouble is that >when you put in such thickening, that destroys the validity of the >"inversion principle" since you'd then instead want a hanging chain >with lighter >links near the middle... Wikipedia says the St Louis arch actually is >"close to a more general curve called a flattened catenary." > >So... what is going on here? How can we reconcile conflicting claims? >I think answer is this. The cycloid is the optimal freestanding arch >of constant thickness IF we are only interested in uniformizing >*compressive* (along beam) stress. But it suffers bending-type >stresses which are not uniform. >The inverted catenary is optimum IF we are only interested in keeping >the bending stress zero everywhere but do not care about having highly >nonuniform along-arch >compressive (or along-chain tensional) stress. > >I think either goal could be correct. I.e. it could easily be that >bending is not a concern, >for example because arch is thick enough, in which case only >compression is a concern. > >If you want to design a thin-beam arch which has uniform bending-type >stress everywhere AND also uniform compressive-along-beam stress, >that'd >require a coupled system of 2 differential equations to tell you both >the beam thickness >(which varies along the arch) and also the curve shape. > >Oho! Wikipedia > http://en.wikipedia.org/wiki/Catenary#Catenary_of_equal_strength >seems to realize that (rather deeply buried) claiming that for hanging chain of >small variable thickness where strength is proportional locally to >mass per unit length of chain, the uniform-tensional-stress chain >shape, with zero bending-type-stress, >has shape > y = C * ln(sec(x/C)) where C=constant. >See >http://books.google.com/books?id=3N5JAAAAMAAJ&pg=PA329#v=onepage&q&f=false >where this is derived and attributed to Davies Gilbert, Philosophical >Transactions part iii (1826) page 202. > >And I compute that this would have > CrossSectionalArea = A*sec(x/C)*|tan(x/C)|^(-B) where A,B constants >and > ArcLength(From 0 to x) = C*ln(sec(x/C) + tan(x/C)). > >Note |x|<pi*C/2 otherwise we get singularities, i.e. the span cannot >be too long. >Note that these are quite unusual curves... > >You may also enjoy the exercise of proving that there is no optimum dome shape >for a constant-thickness thin-shell dome (if "optimum" means "constant stress"). >I.e, thickness variation with height is necessary for an optimal dome. >Hence describing >it requires a coupled system of two differential equations one for >thickness and one for location... and there are really two kinds of >stress, compressive along a meridian, and tensional along a line of >latitude, of a dome, and by "uniform" we need to mean some real-valued >function of these two kinds of stress numbers, is constant everywhere >on >the dome surface. > >Wikipedia > http://en.wikipedia.org/wiki/Catenary#Suspension_bridge_curve >also claims the best shape for a suspension bridge is a parabolic cable >for constant thickness cable in heavy-deck light-cable limit. They cite > Routh, Edward John (1891). "Chapter X: On Strings". A Treatise on >Analytical Statics. University Press. >But this parabola claim is also false, as you again can see by >considering the infinite-x limit which clearly cannot exist, the cable >would break beyond a certain critical bridge length. > >Huh? So again it appears that the parabola claim is true if you do >not care about nonuniform tensional stress in the cable (or, if >inverted, compressional stress along >arch is nonuniform); but as I derived in earlier post, if you do want >that uniform, >then optimal shape for thin-beam arch of constant thickness is circular arc, >BUT this will have bending stresses in the arch which will not be uniform! > >(The fact is, the Greeks and Romans did not build catenary arches, so >you seem wrong about them appreciating that.) > >The Eiffel tower is an example of an exponential pillar, roughly >speaking -- that >is the optimum shape for a thin vertical beam in a constant gravity >field (constant stress) under vertical gravity forces only. However, >in reality, tall towers' main problem usually is not collapsing >vertically like the world trade center, it instead is wind loads and >buckling instabilities. Hence the "optimality" of the exponential >pillar is largely irrelevant. >Optimal pillar shapes with respect to buckling (i.e. maximizing a >buckling eigenvalue) >have in fact been found in closed form and the cycloid shape arises in >that problem too. >I consider solving that to have been quite an impressive feat...

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[math-fun] Optimum shapes -- some myths debunked & clarity gained
by Warren Smith 19 Mar '12

19 Mar '12

> H.Baker: > There is a principle of "inversion" for structures in gravity: under the > transformation y <-> -y, compression becomes tension & vice versa. I think > that Hooke remarked on this principle. > > So, the catenary (y=cosh(x)) curve assumed by a hanging chain becomes the > St. Louis arch or the Gaudi cathedral (Gaudi actually designed his cathedral > upside down with weights & strings tied together; these models are on view > in the basement of his cathedral in Barcelona). > > This principle must have been known to the Greeks & Romans; they need only > look at the necklaces on their spouses, or the chains holding their slaves > together. --well... no. Perhaps the inverted catenary is indeed an optimum arch, but definitely NOT for a constant-cross-section thin beam... instead the beam would have to become thicker near its base, with a certain roughly exponential dependence of thickness on height. This should be obvious by considering infinitely large arch, obviously there is a finite upper limit to size at given strength, which the catenary fails to tell you about hence it is wrong. Put another way, obviously the catenary does not have constant tension, it grows along the chain; hence cannot be optimum in the sense of having constant stress. Wikipedia says: "a catenary is the ideal shape for a freestanding arch of constant thickness" which my analysis would claim was a flat-out lie. http://en.wikipedia.org/wiki/Catenary#The_inverted_catenary_arch To try to get constant stress, cross-sectional-area instead would have to vary proportional to total mass of portion of arch lying above that point, times the |y'| / sqrt(1+y'^2) geometry factor. The St Louis arch indeed does have such thickening, but the trouble is that when you put in such thickening, that destroys the validity of the "inversion principle" since you'd then instead want a hanging chain with lighter links near the middle... Wikipedia says the St Louis arch actually is "close to a more general curve called a flattened catenary." So... what is going on here? How can we reconcile conflicting claims? I think answer is this. The cycloid is the optimal freestanding arch of constant thickness IF we are only interested in uniformizing *compressive* (along beam) stress. But it suffers bending-type stresses which are not uniform. The inverted catenary is optimum IF we are only interested in keeping the bending stress zero everywhere but do not care about having highly nonuniform along-arch compressive (or along-chain tensional) stress. I think either goal could be correct. I.e. it could easily be that bending is not a concern, for example because arch is thick enough, in which case only compression is a concern. If you want to design a thin-beam arch which has uniform bending-type stress everywhere AND also uniform compressive-along-beam stress, that'd require a coupled system of 2 differential equations to tell you both the beam thickness (which varies along the arch) and also the curve shape. Oho! Wikipedia http://en.wikipedia.org/wiki/Catenary#Catenary_of_equal_strength seems to realize that (rather deeply buried) claiming that for hanging chain of small variable thickness where strength is proportional locally to mass per unit length of chain, the uniform-tensional-stress chain shape, with zero bending-type-stress, has shape y = C * ln(sec(x/C)) where C=constant. See http://books.google.com/books?id=3N5JAAAAMAAJ&pg=PA329#v=onepage&q&f=false where this is derived and attributed to Davies Gilbert, Philosophical Transactions part iii (1826) page 202. And I compute that this would have CrossSectionalArea = A*sec(x/C)*|tan(x/C)|^(-B) where A,B constants and ArcLength(From 0 to x) = C*ln(sec(x/C) + tan(x/C)). Note |x|<pi*C/2 otherwise we get singularities, i.e. the span cannot be too long. Note that these are quite unusual curves... You may also enjoy the exercise of proving that there is no optimum dome shape for a constant-thickness thin-shell dome (if "optimum" means "constant stress"). I.e, thickness variation with height is necessary for an optimal dome. Hence describing it requires a coupled system of two differential equations one for thickness and one for location... and there are really two kinds of stress, compressive along a meridian, and tensional along a line of latitude, of a dome, and by "uniform" we need to mean some real-valued function of these two kinds of stress numbers, is constant everywhere on the dome surface. Wikipedia http://en.wikipedia.org/wiki/Catenary#Suspension_bridge_curve also claims the best shape for a suspension bridge is a parabolic cable for constant thickness cable in heavy-deck light-cable limit. They cite Routh, Edward John (1891). "Chapter X: On Strings". A Treatise on Analytical Statics. University Press. But this parabola claim is also false, as you again can see by considering the infinite-x limit which clearly cannot exist, the cable would break beyond a certain critical bridge length. Huh? So again it appears that the parabola claim is true if you do not care about nonuniform tensional stress in the cable (or, if inverted, compressional stress along arch is nonuniform); but as I derived in earlier post, if you do want that uniform, then optimal shape for thin-beam arch of constant thickness is circular arc, BUT this will have bending stresses in the arch which will not be uniform! (The fact is, the Greeks and Romans did not build catenary arches, so you seem wrong about them appreciating that.) The Eiffel tower is an example of an exponential pillar, roughly speaking -- that is the optimum shape for a thin vertical beam in a constant gravity field (constant stress) under vertical gravity forces only. However, in reality, tall towers' main problem usually is not collapsing vertically like the world trade center, it instead is wind loads and buckling instabilities. Hence the "optimality" of the exponential pillar is largely irrelevant. Optimal pillar shapes with respect to buckling (i.e. maximizing a buckling eigenvalue) have in fact been found in closed form and the cycloid shape arises in that problem too. I consider solving that to have been quite an impressive feat...

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Re: [math-fun] Optimum arch shapes -- Romans not so stupid
by Henry Baker 19 Mar '12

19 Mar '12

There is a principle of "inversion" for structures in gravity: under the transformation y <-> -y, compression becomes tension & vice versa. I think that Hooke remarked on this principle. So, the catenary (y=cosh(x)) curve assumed by a hanging chain becomes the St. Louis arch or the Gaudi cathedral (Gaudi actually designed his cathedral upside down with weights & strings tied together; these models are on view in the basement of his cathedral in Barcelona). This principle must have been known to the Greeks & Romans; they need only look at the necklaces on their spouses, or the chains holding their slaves together. At 12:35 PM 3/19/2012, Warren Smith wrote: >This is the same as my previous post but with errors corrected and >some other stuff added. Please discard earlier erroneous post! > >OPTIMUM ARCH SHAPES >===Warren D. Smith March 2012======= > >The optimum shape for an arch under assumptions 1-4 below is a cycloid. >[Apparently Galileo knew this...] > >1. "thin beam of constant cross section, constant density, and >constant material properties" >2. supports only its own weight, not other stuff ("freestanding"); >constant downward >gravity field. >3. "optimum" means constant stress regardless of position on the arch-curve. >4. No other (i.e. nongravitational) forces to worry about. > >Another important use of arches is for arch bridges, where there is >also a flat bridge deck exerting (to oversimplify it) a constant >downward load per unit of horizontal length. > >The equation of the curve of an optimum arch satisfying (1) would be >that the total weight of the part of the arch (plus the external >load) above some height y >(which is evaluable via an obvious integral) must equal a constant times >1/sqrt(1 + x'^2) where x=horizontal location is regarded as a function >of y=vertical height >to describe the arch curve. > >Equivalently, the total weight of the part of the arch (plus the external >load) between x and -x (which is evaluable via an obvious integral) >must equal a constant times >|y'|/sqrt(1 + y'^2) >where y=vertical location is regarded as a function >of x=horizontal location to describe the arch curve centered at x=0. > >So for example, with both the flat bridge deck and arch itself >contributing to the load > >integral( A*sqrt(1+y'(u)^2) + B , u=-x..x) = 2*C*|y'(x)|/sqrt(1+y'(x)^2) > >would define the function y(x) defining the curve >for an arch centered at x=0. Here A=weight per unit arclength of arch, >B=weight per unit horizontal length of deck, and C=stress (constant compressive >force along the arch curve). Use fundamental theorem of calculus to rewrite >as a differential equation > >A*sqrt(1 + diff(y(x),x)^2) + B + C * diff( diff(y(x),x) / sqrt( >1+diff(y(x),x)^2 ), x ) = 0 > >valid for x>0, and solve: > >deq := A*sqrt(1 + diff(y(x),x)^2) + B + C*diff( diff(y(x),x) / >sqrt(1+diff(y(x),x)^2), x ) = 0; >dsolve(deq, y(x)); > >MAPLE9's dsolve facility succeeds in finding a closed form >in the case A=0 using only sqrt: >dsolve(subs(A=0,deq), y(x)); >result: > > y(x) = sqrt([C/B]^2 - x^2) - C/B + H > >where H is the height of the arch at its center x=0. >THIS IS A SEMICIRCLE. > >In the case B=0 we solve with >dsolve(subs(B=0,deq), y(x)); >and MAPLE only manages to express y(x) via an integeral of a "RootOf." >This seems to me to be something of a failure of dsolve >since I happen to know the solution is a cycloid and thus is >expressible in closed form, >albeit that is best done by writing both x and y as a function of a >third variable >(you can do this explicitly and plug into differential eq to verify is >solution). If I rewrite the equation with B=0 in terms of x(y) not y(x) >[inverse function] then > >integral( A*sqrt(1+x'(u)^2) , u=y..H) = C/sqrt(1+x'(y)^2) > >whence > >A*sqrt(1+x'(y)^2) + diff( C/sqrt(1+x'(y)^2), y) = 0 > >and MAPLE9 solves again now via > deq2 := A*sqrt(1+diff(x(y),y)^2) + diff( C/sqrt(1+diff(x(y),y)^2), y) = 0; > dsolve(deq2, x(y)); >whereupon it now succeeds in finding a rather messy closed form which >I presume is >equivalent to a cycloid, though have not verified that. > >In the general-A,B case re-expressed using x(y) I find >A*sqrt(1+x'(y)^2) + B*x'(y) + diff( C/sqrt(1+x'(y)^2), y) = 0 > > deq2 := A*sqrt(1+diff(x(y),y)^2) + B*diff(x(y),y) + diff( >C/sqrt(1+diff(x(y),y)^2), y) = 0; > dsolve(deq2, x(y)); > >In the general case (both A and B nonzero) dsolve pretty much fails to >solve the differential >equation in either x(y) or y(x) form. >But perhaps you will be better-able to solve differential equations >than MAPLE9's >dsolve, I await your reports. > >Re Baker on the Romans whom, he claims, preferred circles, >we see above semicircle is optimum arch in the heavy-deck light-arch >limit (B=0). >The circle also is the optimum shape for an arch under the following >(different) assumptions > >1'.same >3'.same >2'.replace by assumption that the weight of the arch itself is tiny compared to >the weight of everything else it supports, and that other stuff exerts >a constant hydrostatic pressure. In other words, circle is the optimum >shape for the wall of a "submarine" at high depth or for a deep >underground tunnel >through mud. > >Many Roman arches are small doorways inside massive walls, and for >them the assumption 2' might seems pretty well satisfied -- and this >shape is still used today in applications of that nature. > >So the circle is optimum in two different senses as an arch shape >(although not optimum in every >sense) and it also is the easiest shape to fabricate. >I conclude the Romans were not so stupid. > >If you want to know whether the Romans had a more sophisticated and >more general theory, I would suggest >(a) try to read Vitruvius (ancient Roman engineering textbook) > http://www.gutenberg.org/ebooks/20239 >(b) consider the Pantheon, the world's largest dome for a very long time, >constructed by Hadrian and still in use today. > >I think the Romans did not really understand what they were doing >compared to what I understand, but they had a lot of knowledge nonetheless...

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Re: [math-fun] Wikiproofia
by Henry Baker 19 Mar '12

19 Mar '12

I've been told that Java (not Javascript) routinely "proves" that new modules can't overrun their boundaries when the modules are loaded. I've also been told that Microsoft is routinely checking many/most of its device drivers to have certain properties. Microsoft is apparently a fan of "model checking", which is a certain restricted type of theorem proving. I've also been told that there are several types of operating systems that have been proven "secure" -- whatever that means. I've been told that Intel has routinely been checking their floating point implementations ever since the infamous Pentium bug incident. The only thing holding "Wikiproofia" back right now is a small, dedicated "open source" project to focus on this for a good 5-10 years to develop the basic core system and a good library of "lemmas" to cover many of the more obvious things that need to be proven. "Artificial Intelligence" isn't needed right this second; that would only be required if you didn't want to have to explain every last detail to the proof checker; but many of the details would already be handled by a large enough library of already proven lemmas. At 07:47 PM 3/17/2012, Fred lunnon wrote: >On 3/18/12, Marc LeBrun <mlb(a)well.com> wrote: >>>="Henry Baker" <hbaker1(a)pipeline.com> >>> mechanically _check_ submitted proofs >> >> I agree this is a good idea, but have wondered how far off the dream is. >> ... >> Is Wikiproofia's first customer release date inevitably post-singularity? > > There are certainly plenty of projects aimed in this direction --- just search >on "automated theorem proving". But I think they're unrealistic, for much >the same reasons Marc advanced. > > If you can't solve the problem, look for an easier one instead. In this case, >the easier problem is automatic verification that a computer program implements >its specification. Here the universe is extremely circumscribed, and well >documented. When we've got that question satisfactorily answered, then it >might be time to look seriously at automated theorem proving. > > I have a vague memory of some bigwig in the UK MoD --- or maybe it was >the USA DoD --- announcing grandiosely in public that in future all defence >contracts would include a clause demanding that any software be proved correct. >That was some time in the 1970's. When we'd all stopped laughing, we said --- >quietly, of course --- don't hold your breath! > > But after all the ballyhoo has been deflated, and the journalists >and politicians >have departed to look for other mirages to chase, things do eventually come >quietly to pass in a lab somewhere out of the spotlight. > > In the end, Deep Blue defeated Kasparov ... > >Fred Lunnon

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[math-fun] Optimum arch shapes -- Romans not so stupid
by Warren Smith 19 Mar '12

19 Mar '12

This is the same as my previous post but with errors corrected and some other stuff added. Please discard earlier erroneous post! OPTIMUM ARCH SHAPES ===Warren D. Smith March 2012======= The optimum shape for an arch under assumptions 1-4 below is a cycloid. [Apparently Galileo knew this...] 1. "thin beam of constant cross section, constant density, and constant material properties" 2. supports only its own weight, not other stuff ("freestanding"); constant downward gravity field. 3. "optimum" means constant stress regardless of position on the arch-curve. 4. No other (i.e. nongravitational) forces to worry about. Another important use of arches is for arch bridges, where there is also a flat bridge deck exerting (to oversimplify it) a constant downward load per unit of horizontal length. The equation of the curve of an optimum arch satisfying (1) would be that the total weight of the part of the arch (plus the external load) above some height y (which is evaluable via an obvious integral) must equal a constant times 1/sqrt(1 + x'^2) where x=horizontal location is regarded as a function of y=vertical height to describe the arch curve. Equivalently, the total weight of the part of the arch (plus the external load) between x and -x (which is evaluable via an obvious integral) must equal a constant times |y'|/sqrt(1 + y'^2) where y=vertical location is regarded as a function of x=horizontal location to describe the arch curve centered at x=0. So for example, with both the flat bridge deck and arch itself contributing to the load integral( A*sqrt(1+y'(u)^2) + B , u=-x..x) = 2*C*|y'(x)|/sqrt(1+y'(x)^2) would define the function y(x) defining the curve for an arch centered at x=0. Here A=weight per unit arclength of arch, B=weight per unit horizontal length of deck, and C=stress (constant compressive force along the arch curve). Use fundamental theorem of calculus to rewrite as a differential equation A*sqrt(1 + diff(y(x),x)^2) + B + C * diff( diff(y(x),x) / sqrt( 1+diff(y(x),x)^2 ), x ) = 0 valid for x>0, and solve: deq := A*sqrt(1 + diff(y(x),x)^2) + B + C*diff( diff(y(x),x) / sqrt(1+diff(y(x),x)^2), x ) = 0; dsolve(deq, y(x)); MAPLE9's dsolve facility succeeds in finding a closed form in the case A=0 using only sqrt: dsolve(subs(A=0,deq), y(x)); result: y(x) = sqrt([C/B]^2 - x^2) - C/B + H where H is the height of the arch at its center x=0. THIS IS A SEMICIRCLE. In the case B=0 we solve with dsolve(subs(B=0,deq), y(x)); and MAPLE only manages to express y(x) via an integeral of a "RootOf." This seems to me to be something of a failure of dsolve since I happen to know the solution is a cycloid and thus is expressible in closed form, albeit that is best done by writing both x and y as a function of a third variable (you can do this explicitly and plug into differential eq to verify is solution). If I rewrite the equation with B=0 in terms of x(y) not y(x) [inverse function] then integral( A*sqrt(1+x'(u)^2) , u=y..H) = C/sqrt(1+x'(y)^2) whence A*sqrt(1+x'(y)^2) + diff( C/sqrt(1+x'(y)^2), y) = 0 and MAPLE9 solves again now via deq2 := A*sqrt(1+diff(x(y),y)^2) + diff( C/sqrt(1+diff(x(y),y)^2), y) = 0; dsolve(deq2, x(y)); whereupon it now succeeds in finding a rather messy closed form which I presume is equivalent to a cycloid, though have not verified that. In the general-A,B case re-expressed using x(y) I find A*sqrt(1+x'(y)^2) + B*x'(y) + diff( C/sqrt(1+x'(y)^2), y) = 0 deq2 := A*sqrt(1+diff(x(y),y)^2) + B*diff(x(y),y) + diff( C/sqrt(1+diff(x(y),y)^2), y) = 0; dsolve(deq2, x(y)); In the general case (both A and B nonzero) dsolve pretty much fails to solve the differential equation in either x(y) or y(x) form. But perhaps you will be better-able to solve differential equations than MAPLE9's dsolve, I await your reports. Re Baker on the Romans whom, he claims, preferred circles, we see above semicircle is optimum arch in the heavy-deck light-arch limit (B=0). The circle also is the optimum shape for an arch under the following (different) assumptions 1'.same 3'.same 2'.replace by assumption that the weight of the arch itself is tiny compared to the weight of everything else it supports, and that other stuff exerts a constant hydrostatic pressure. In other words, circle is the optimum shape for the wall of a "submarine" at high depth or for a deep underground tunnel through mud. Many Roman arches are small doorways inside massive walls, and for them the assumption 2' might seems pretty well satisfied -- and this shape is still used today in applications of that nature. So the circle is optimum in two different senses as an arch shape (although not optimum in every sense) and it also is the easiest shape to fabricate. I conclude the Romans were not so stupid. If you want to know whether the Romans had a more sophisticated and more general theory, I would suggest (a) try to read Vitruvius (ancient Roman engineering textbook) http://www.gutenberg.org/ebooks/20239 (b) consider the Pantheon, the world's largest dome for a very long time, constructed by Hadrian and still in use today. I think the Romans did not really understand what they were doing compared to what I understand, but they had a lot of knowledge nonetheless...

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[math-fun] optimum arch shapes
by Warren Smith 19 Mar '12

19 Mar '12

sorry, my earlier post contained errors. Will re-post corrected version shortly.

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