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...