*This site has been archived for historical purposes. These pages are no longer being updated.*

## Two half twists in a cylinder

Ruled surfaces (called "scrolls") have been studied by geometers such as Eisenhart, Cayly, Emch and Monge. In 1850 Gaspar Monge {in his book "Application de l'Analysis à la Géometrie" 5 ed. (1850), pp. 83-89} wrote of a particular type of ruled surface useful for sculptors and architects who could design their curved surfaces while using a straight edge anchored along a fixed line, such as the z-axis. He proved two theorems concerning a surface whose generator passes through the z-axis.

Theorem I The Surface equation satisfies the partial differential equation

x(1)^{2}z_{xx}+ 2xyz_{xy}+ y^{2}z_{yy}= 0

Theorem II The solution to this equation has the algebraic form

z = xF(y/x) + G(y/x)(2)

In 1920 Arnold Emch (American Journal of Mathematics42pp. 189-210), described such a surface using the z-axis asC, the first generator and_{1}C, a circle of radius a lying in the xy-plane as the second generator. A ruling passing through the z-axis at an angle_{2}ameets C_{2}at pointm. A constant angle a would form a cone, but if the angle a varied as the point m moved around the circumference C_{2}so thata = p/q q, a p/q twist surface results. The equation for such a surface is given in cylindrical coordinates.

r = a + z tan p/q thetaorz = (r-a) cot p/q thetaThe moebius (half twist) surface

## One half twist in a box

If p/q = 1/2 a moebius band would result. This surface of one half twist has fascinated artists such as Escher because of its peculiar unifacial property, having only one surface and only one edge: the inside is not really distinct from the outside. So Escher formed a moebius endless belt around which a column of ants could march forever. The surface is usually formed by giving a half twist to a narrow ribbon of paper and then pasting the ends together. We find that this process will not work for a square sheet of paper as it refuses to take shape unless it is cut because the moebius surface must self-intersect or it will self-destruct. In fact this half twist case of a p/q twist surface self-intersects in a straight line. A convenient way to obviate this problem is to construct a ruled surface allowing it to intersect itself as the rulings pass unhindered between each other.## A quarter twist in a cylinder

Using tedious but cunning manipulation of algebra and trigonometry formulas Emch derives formulas needed to prove his theorems and determine the order of the surface (the number of times a straight line intersects the surface) and the degree of the surfaces' equations. He also studies the curves of self-intersection made by all the p/q twist surfaces. To verify the theorems and illustrate the interesting properties he describes, however, the algebra and calculus needed is daunting and any heuristic geometrical models are not available.

Computer programs to study the p/q twist surfaces

Mathematica has the power not only to display these elusive properties, thereby making these interesting surfaces accessible, but also to display the surface and the curves of self- intersection on a screen. Mathematica can also illustrate any number of rulings for the cylindrical and Cartesian coordinates, thereby facilitating the actual construction of these models enclosed in a box or in a cylinder. By use of its calculus and algebra packages Mathematica can also verify Monge's first theorem concerning the partial differential equation and second theorem concerning the form of the surface equation.

Monge offers a proof of his first theorem mentioned already:

Theorem 1 The Ruled Surface equation satisfies the partial differential equation

x(1)^{2}z_{xx}+ 2xyz_{xy}+ y^{2}_{yy}= 0

Here is presented a proof of the second Monge theorem.

Theorem 2 Any solution to this equation (1) has the algebraic form

z = xF(y/x) + G(y/x)(2)

Proof that (1) -> (2) using the substitutions; u = ln x and v = ln y

(and proceeding in a similar manner to a Cauchy-Euler ODE)

soz and in a similar way_{x}= z_{u}u_{x}= z_{u}/xz _{y}= z_{v}v_{y}= z_{y}/y

z and in a similar way_{xx}= z_{uu}/x^{2}- z_{u}/x^{2}z _{yy}= z_{vv}/y^{2}- z_{v}/y^{2}

z _{xy}= z_{uv}

so the equation

x(1)^{2}z_{xx}+ 2xyz_{xy}+ y^{2}z_{yy}= 0

## Three half twists in a cylinder

z(3)_{uu}+ 2z_{uv}+ z_{vv}- z_{u}- z_{v}= 0

or [Dwhich breaks into two linear factors^{2}_{u}+ 2D_{u}D_{v}+ D^{2}_{v}D - D_{u}- D_{v}]z=0

L(z) = L1(z) L2(z)

=[(D_{u}+ D_{v})(D_{u}+ D_{v}-1)z] = 0

Now for each of these linear operators L1 and L2 use the Linear PDE I {Constant Coefficient} method of solution for Az_{x}+ Bz_{y}+ Cz = 0 which is:

z= e so z^{(-Cx/A)f(Bx-Ay) }Z = exp(-Cx/A)f(Bx-Ay)_{1}=e^{0 g(u-v) }= g(lnx - lny) = g(ln[y/x]) = G[y/x]

and z_{2}= e^{(-u)f(u-v)}= [exp(lnx)](lnx - lny) = xF[y/x]

then the solution

z = z _{1}+z_{2}= xF(y/x) + G(y/x)(2)

Now for these surfaces Z = r cot (p theta/q) - a cot (p theta/q) we can replace cot (p theta/q) by functions of y/x in the following manner.

Take the case of p/q = 1/1

z = r cot (theta) - a cot (theta)

Since tan t = y/x cot t = x/y and r = x sec t = x (1 + tan^{2}t)

= x (1+[y/x]^{2})

so z = x (1+[y/x]^{2}){1/(y/x) } + (-a){1/(y/x) }

and soz = xF(y/x) + G(y/x)

In some of the old trigonometry books one finds a meticulous (and boring) examination of "sub-multiple angles such as

tan A/2 = ( [1+ tan^{2}A] - 1)/tan A

and tan A = {(3tan [A/3] - tan^{3}[A/3])/(1-3tan^{2}[A/3])}

so all the p/q are not unreachable for the stout hearted, but there is no need to present the details here for all fractions p/q theta

Then one could use Emch's equation (8) on p192 concerning tan rw which is an expansion of binomial functions.

Using w to represent theta and theta/2

and using r to represent an assortment of values of p.

In this way we could show that Mathematica demonstrates Monge's second theorem that our ruled surfaces

z = r cot (p theta/q) - a cot (p theta/q) also satisfy the predicted formz =for all cases such as 1/1, 2/1, 3/1, . . 1/2, 3/2, 5/2. . .xF(y/x) + G(y/x)

Return to Polyhedra Page