aa r X i v : . [ phy s i c s . c l a ss - ph ] M a r From Elastica to Floating Bodies of Equilibrium
Franz WegnerInstitute for Theoretical PhysicsRuprecht-Karls-University Heidelberg, GermanyMarch 4, 2020
Abstract
A short historical account of the curves related to the two-dimensio-nal floating bodies of equilibrium and the bicycle problem is given. Bor,Levi, Perline and Tabachnikov found, quite a number had already beendescribed as
Elastica by Bernoulli and Euler and as
Elastica under Pres-sure or Buckled Rings by Levy and Halphen. Auerbach already realizedthat Zindler had described curves for the floating bodies problem. Aneven larger class of curves solves the bicycle problem.The subsequent sections deal with some supplemental details: Severalderivations of the equations for the elastica and elastica under pressureare given. Properties of Zindler curves and some work on the problemof floating bodies of equilibrium by other mathematicians are considered.Special cases of elastica under pressure reduce to algebraic curves, asshown by Greenhill. Since most of the curves considered here are bicyclecurves, a few remarks concerning them are added.
Contents ρ = 1 / ρ = 1/2 . . . . . . . . . . . . . . . . . . . . . . . 122.5 The Bicycle Problem . . . . . . . . . . . . . . . . . . . . . . . . . 17 ρ = 1 / Mikl´os R´edei introduces in his paper ”On the tension between mathematicsand physics”[42] the ”supermarket picture” of the relation of mathematics andphysics: that mathematics is like a supermarket and physics its customer.When in 1938 Auerbach solved the problem of floating bodies of equilibrium[2]for the density ρ = 1 / ρ = 1 / Elastica and
Elastica under Pressure or Buckled Rings . These2urves, known since the seventeenth and nineteenth century, first as solutionsof elastic problems, have shown up as solutions of quite a number of otherproblems, in particular as boundaries of two-dimensional bodies which can floatin equilibrium in all orientations; this later problem is also solved by
Zindlercurves .These classes of curves yield solutions to the bicycle problem. One asksin this problem for front and rear traces of a bicycle, which do not allow toconclude, in which direction the bicycle went, that is the traces are identicalin both directions. The curves that give the traces of the front wheel and thetraces of the rear wheels are the boundary and the envelope of the water lines,resp., of the floating bodies of equilibrium.The subsequent sections deal with some supplemental details: In section 3several derivations of the equations for the elastica and elastica under pressureare given. Section 4.1 deals with the Zindler curves and work on the problemof floating bodies of equilibrium by other mathematicians, including criticism.Section 5 is devoted to work by Greenhill, who found that special cases of theelastica under pressure can be represented by algebraic curves. Since most ofthe curves considered here are special cases of bicycle curves, section 6 bringssome remarks on these curves.
In this paper we consider a class of planar curves, which surprisingly showup in quite a number of different physical and mathematical problems. Thesecurves are not generally known, since they are represented by elliptic functions,although special cases can be represented by more elementary functions.
These curves appear in two flavors; they obey in the linear form the equation1 p y ′ = ay + b (1)in Cartesian coordinates ( x, y = y ( x )), y ′ = d y/ d x . The circular form is de-scribed by 1 √ r + r ′ = ar + b + cr , (2)in polar coordinates ( r, φ ) with r = r ( φ ) and r ′ = d r/ d φ . Eq. (1) yields thecurvature κ , 1 y ′ dd x p y ′ = − y ′′ (1 + y ′ ) / = κ = 2 ay. (3)From (2) we obtain the curvature1 rr ′ dd φ r √ r + r ′ = r + 2 r ′ − rr ′′ ( r + r ′ ) / = κ = 4 ar + 2 b. (4)3e relate the polar coordinates to Cartesian coordinates and shift y by r , r cos φ = r + y, r sin φ = x. (5)Then (4) reads κ = 4 ar + 8 ar y + 4 a ( x + y ) + b. (6)If we now choose b = − ar , a = ˜ a/ (4 r ) , (7)and perform the limit r → ∞ , then eq. (6) reads κ = 2˜ ay, (8)which is the linear form (3). Thus the linear form is a limit of the circular form,where the radius r goes to infinity.The equation of the curves can be formulated coordinate-independent,2 κ ′′ + κ − µκ − σ = 0 , (9)where the prime now indicates the derivative with respect to the arc length.Multiplication by κ ′ allows integration, κ ′ + κ − µ κ − σκ = 2 ˆ E. (10)The coefficient σ vanishes in the linear case. The derivation will become appar-ent, when we formulate the physical and/or mathematical problems solved bythe curves. But the relation between eqs. (1, 2) and eqs. (9, 10) can also begiven directly.[7] The linear curves show up in
Elastica . The question is: How does an elasticbeam (or wire or rod) of given length bend? It may be fixed at two ends andthe directions of the wire at both ends are given, or it is fixed at one end andloaded at the other end. Bending under load was already considered in the13th century by Jordanus de Nemore, around 1493 by Leonardo da Vinci[3], in1638 by Galileo Galilei, and in 1673 by Ignace-Gaston Pardies.[37, 53], althoughthey could not give correct results.[37, 52] James Bernoulli following Hooke’sideas obtained a correct differential equation assuming that the curvature isproportional to the bending moment, and partially solved it in 1691-1692. [4],d y d x = x √ a − x . (11)Huygens[31] argued in 1694 that the problem had a larger variety of solutionsand sketched several of them. The more general differential equation was givenby Bernoulli[5] in 1694-1695,d y d x = x ± ab p a − ( x ± ab ) , (12)4his equation yields 1 q d x d y ) = x ± aba , (13)which is equation (1) with x and y exchanged and different notation for theconstants. James Bernoulli also realized ”... among all curves of a given lengthdrawn over a straight line the elastic curve is the one such that the center ofgravity of the included area is the furthest distant from the line, just as the cate-nary is the one such that the center of gravity of the curve is the furthest distant...”[5]. Thus he found that the cross section of a volume of water contained ina cloth sheet is bounded (below the water line) by the elastica curve.In 1742 Daniel Bernoulli (nephew of James) proposed in a letter to LeonhardEuler that the potential energy of a bent beam is proportional to the integral ofthe square of the curvature κ integrated over the arclength s of the beam R d sκ .In a 1744 paper Euler[14] used variation techniques to solve the problem. Hefound eq. (12), classified the solutions, discussed the stability, and realized thatthe curvature is proportional to x , κ = y ′′ (1 + y ′ ) / = 2 xa . (14)This property plays a role in at least two other physical phenomena:In 1807 Pierre Simon Laplace[36] investigated the shape of the capillary. Thesurface of a fluid trapped between two parallel vertical plates obeys also (14),since the difference of pressure inside and outside the fluid is proportional tothe curvature of the surface.Charges in a linearly increasing magnetic field move due to the Lorentz forceon trajectories with curvature proportional to the magnetic field and thus againon trajectories given by the curves of elastica. Without being aware of thisequivalence Evers, Mirlin, Polyakov, and W¨olfle considered them in their paperon the semiclassical theory of transport in a random magnetic field.[15]In 1859 Kirchhoff[33] introduced a kinetic analogue by showing that theproblem of elastica is related to the movement of a pendulum. Set˙ x = cos θ, ˙ y = sin θ, (15)where the dot indicates the derivative with respect to the arc s . Then oneobtains ˙ θ = κ, y ′ = tan θ. (16)Thus using eqs. (3) and (1) one obtains˙ θ = 4 a y = 4 a (cos θ − b ) . (17)Multiplication by mr / m for the mass and r for the length of thependulum) and a corresponding choice of the constants a and b , yields mr θ − mgr cos θ = E. (18)5igure 1: Examples of linear Elastica (i)Figure 2: Examples of linear Elastica (ii)Figure 3: Examples of linear Elastica (iii)6his is the energy of a pendulum, if we substitute time t for the arc s in thederivative. Thus θ ( t ) is the time dependence of the angle of the pendulumagainst its lowest position. The periodic movement is obvious. Suppose a ispositive. Then for − < b < − θ ≤ θ ≤ + θ . These are the inflectional solutions. If b < − θ = π . These are the non-inflectional solutions.The limit case b = − b = −
1. It iscalled syntractrix of Poloni (1729). The last row of figure 1 and figures 2 and 3show inflectional cases corresponding to periodic oscillations without reachingthe highest point. This yields a large variety of shapes including the
Eight in themiddle of figure 2 called lemnoid . The second drawing in figure 3 correspondsto the case where the pendulum moves up to a horizontal position. It is the rectangular elastica or right lintearia . The pendulum swings below the horizontalposition in the last two rows of figure 3. In all cases the curves show equilibriumpositions of the elastic beam. However, only sufficiently short pieces of thecurves correspond to a stable equilibrium or even the absolute minimum of thepotential energy. x The hyperbolas are shown in magenta,their asymptotes in black. The verticesof the two branches are connected by amagenta line.Figure 4: Rectangular elastica as roulette of hyperbola
Rectangular elastica as roulette of hyperbola
The rectangular elasticcurve is the locus of the center of a rectangular hyperbola rolling without slippingon a straight line. The upper branch of the hyperbola is shown in figure 4 intwo different orientations. The midpoint between the branches lies on the bluerectangular linteria. (Sturm 1841, see [16], Greenhill 1892 [23]).An excellent survey with many figures on the history of the elastica has beengiven by Raph Levien.[37] Also Todhunter[52] and Truesdell[53, 54] give reviewsof the history of elasticity. Many details are found in the treatise by Love[39]on the mathematical theory of elasticity. In his PhD thesis[8] (1906) Max Borninvestigated elastic wires theoretically and experimentally in the plane and alsoin three dimensional space. The solution of eq. (1) or equivalently eqs. (11, 12)7as given by Saalsch¨utz[44] in 1880 by means of elliptic functions. The ellipticfunctions were developed by Abel and Jacobi mainly in the years 1826 – 1829in several articles in Crelles Journal.[12]. Abel died in 1829, Jacobi publishedhis fundamental work in the same year.[32] The explicit solutions are not givenhere. They can be found, e.g., in sect. 263 of [39], in sect. 13 of [37] and in ref.[60, 61]. Engineers often call ’Bernoulli-Euler beam theory’ approximations, inwhich the beam is only slightly bent.[62]
By now we considered elastica to which only forces acted at the ends. A moregeneral problem considers elastic wires, on which forces act along the arc. Mau-rice Levy realized in 1884[38] that the case, where a constant force P per arclength acts perpendicularly on the wire, yields the differential equation (2). Heshowed that this problem could be solved by elliptic functions and found twotypes of solutions. They are called buckled rings, if the wire is closed. Theconstraints are equivalent if instead the perimeter and the area inside the ringare given. Halphen worked out the results in some detail in the same year[27]and included it in his ’Trait´e des fonctions elliptic et de leurs applications’.[28]Some elastica under pressure are shown in figure 5. Their symmetry is givenby the dieder groups D p with p = 2 , , p = 4 , , , p = 5 , , r = a cos(3 φ ) (19)with the curvature κ = 4 r /a (20)in polar coordinates ( r, φ ), which may be written( x + y ) = a x ( x − y ) (21)in Cartesian coordinates ( x, y ). This Kiepert curve (W. Roberts and L. Kiepert1870, see [16]) looking like a cloverleaf is shown in fig. 6. Area-constrained planar elastica in biophysics
Cells in biology have usually nearly constant volume and constant surface area.Their shape is to a large extend determined by the minimum of the mem-brane bending energy, see e.g. Helfrich[29] and Svetina and Zeks[49]. Thetwo-dimensional analogue was considered by Arreaga, Capovilla et al.[1, 11]and by Goldin et al.[22]. Since pressure and area are conjugate quantities, the8igure 5: Examples of Elastica under Pressure (buckled rings)Figure 6: Kiepert curve r = a cos(3 φ )shapes are also given by those of elastica under pressure. Now of course, con-stant area and constant length of the bounding loop are given. The authorsrefer for the determination of the shape to the Lagrange equations (9, 10) asreported by Langer and Singer for elastica[35] and for buckled rings [34], see alsothe references [7, 47]. The equations for the elastica under pressure can also beobtained by considering the forces and momenta in the rods.[51] The solutionof the equation (2) for buckled rings in terms of elliptic functions can be foundin [38, 27, 28, 24] and in [60, 61]. Reference [61] contains further figures.9 .4 Floating Bodies of Equilibrium The curves mentioned in the previous subsections appear in two other problems:the problem of
Floating Bodies of Equilibrium and the
Bicycle Problem . Thefirst of these problems is related to the problem 19 in the Scottish Book byUlam[55]: ”Is a solid of uniform density which will float in water in every positiona sphere?” The two-dimensional version of the problem concerns a cylinder ofuniform density ρ which floats in water in equilibrium in every position withits axis parallel to the water surface. Sought is the curve different from a circleconfining the cross section of the cylinder perpendicular to its axis.The density of the log be ρ (more precisely ρ is the ratio of the density ofthe log over that of the liquid). The area of the cross section be A , the partabove and below the water-line are denoted by A and A . Then Archimedes’law requires A = (1 − ρ ) A, A = ρA. (22)The distance of the center of gravity of the cross section above the water-linebe h , that below the water-line h , the length of the log L . Then the potentialenergy is ρ (1 − ρ ) ALg ( h + h ) . (23)Thus h + h has to be constant. The line connecting the two centers of gravityhas to be perpendicular to the water-line. Rotation by an infinitesimal angleyields that the length 2 l of the water-line obeys23 l (cid:18) A + 1 A (cid:19) = h + h . (24)Thus the length of the water-line does not depend on the orientation of the log.The conditions that the area below the water-line and the length of the water-line are constant implies that the part of the perimeter of the cross-section belowthe water-line is constant. It also implies that the envelope of the water-lines isgiven by the midpoints of the water-lines.This two-dimensional problem has attracted many mathematicians. ρ = 1 / ρ = 1 /
2. The solutions are not related tothe elastica, but it seems worthwhile to mention them. Basically the solutionswere found by Zindler[63], although he did not consider this physical problem,but found convex curves which have the property that chords between twopoints on the boundary which bisect the perimeter have constant length 2 l andsimultaneously cut the enclosed area in two halves. They can be parameterizedby x ( α ) = l cos( α ) + ξ ( α ) , y ( α ) = l sin( α ) + η ( α ) , (25) ξ ( α ) = Z α d β cos( β )ˆ ρ ( β ) , η ( α ) = Z α d β sin( β )ˆ ρ ( β ) , (26)10 P QQN N
Figure 7: Lower part of the boundary (black) of the floating bodyTwo waterlines P Q and P Q are in blue and cyan. The midpoints N and N lie on the envelope (red) of the water-lines.with parameter α , where ξ and η obey ξ ( α + π ) = ξ ( α ) , η ( α + π ) = η ( α ) . (27)( ξ, η ) are the coordinates of the envelope of the water lines, and | ˆ ρ ( α ) | is theradius of curvature of the envelope. Typically the envelope has an odd numberof cusps.Figure 8: Heart-shaped Zindler curve,see Auerbach[2] Figure 9: Zindler curve, see Auer-bach[2]Condition (27) implies ˆ ρ ( β + π ) = − ˆ ρ ( β ). The chords run from ( x ( α ) , y ( α ))to ( x ( α + π ) , y ( α + π )). Zindler did not consider the centers of gravity of bothhalfs of the area. Otherwise he would have realized that their distance does notdepend on the angle α and the line between them is always perpendicular tothe chord. This class of curves was also found by Auerbach[2] in 1938 and byGeppert[19] in 1940. Special cases were given by Salkowski[46] in 1934 and bySalgaller and Kostelianetz[45] in 1939. Examples of Zindler curves are shownin figures 8 to 11. They are due to Auerbach[2], Zindler[63], and Salgaller and11igure 10: Zindler curve, five-fold sym-metry Figure 11: Zindler curve, see Salgal-ler and Kostelianetz[45], similarly byZindler[63]Kostelianetz[45]. The envelopes of the water lines are shown in red, the waterlines in blue and cyan. ρ = For a long time it was not clear, whether solutions for ρ = 1 / ’Different heart-shapedcross sections work for other densities (he means densities different from / )and there are other solutions that are not heart-shaped.’ Indeed, there are cross-sections that are not heart-shaped for density 1/2 and densities different from1/2. But I do not know a heart-shaped solution for density different from 1/2and I doubt that at the time he wrote the paper a solution for densities differentfrom 1/2 was known. At least he does not give reference to such a solution.Attempts to find solutions for ρ = 1 / /
2. They consider a carousel, which isa dynamical equilateral polygon in which the midpoint of each edge travelsparallel to it. The trace of the vertices describe the boundary and the midpointsoutline the envelope of the waterline. In this way they found solutions wherethe chords form an equilateral pentagon. However, their solutions were notsufficiently convex, since the water line cuts the cross section in some positionsseveral times.For special densities ρ one can deform the circular cross-section into one with12 -fold symmetry axis and mirror symmetry. r ( φ ) = r (1 + 2 ǫ cos( pφ ) + 2 ∞ X n =2 c n cos( pnφ )) , (28)where the coefficients c n are functions of ǫ and p with c n = O ( ǫ n ). The corre-sponding p − ǫ . Surprisingly, the perturbation expansionin ǫ yielded (up to order ǫ ) one and the same solution for all p − ρ and 1 − ρ . Thepresent author reported this result in [58]. This result was unexpected. It wasprobably true to all orders in ǫ and thus deserved further investigation. (In eq.(83) of [58]v3 c δ should be replaced by s δ ).In a first step the limit p → ∞ was considered with r ∝ p and ǫ ∝ /p . Thiscorresponds to the transition from the circular case to the linear case mentionedin subsection 2.1. In this limit only terms P n,k c n,k p n +2 k − ǫ n +2 k cos( pnφ ) inthe expansion (28) contribute (with odd n ). Property of constant distance
These curves have a remarkable property,which I call the property of constant distance:Consider two copies of the curves. Choose an arbitrary point on each curve.Then in the linear case there exists always a length δu by which the curves canbe shifted against each other, and in the circular case there exists an angle δφ by which the two curves can be rotated against each other, so that the distance2 l between the two points stays constant, if they move on both curves by thesame arc distance s .Considering this procedure the other way round, we may shift curves inthe linear case continuously against each other and watch how the distance 2 l increases with δu , or we may rotate the curves in the circular case continuouslyagainst each other and watch how 2 l varies with δφ . When δφ is increasedby 2 π/p , then both curves fall unto themselves, and a solution for the floatingbodies is found, provided the curve is sufficiently convex, so that the chordbetween the two points does not intersect the curve at another point.Figure 12: Property of constant distanceThe first figure of 5 and figure 6 are shown in two copies rotated against eachother by 45 , 60 and 30 . 13igure 13: Floating bodies of equilibrium, p = 3 and p = 4In figure 12 three examples for the property of constant distance are shown.Two copies of the first of the figures 5 are shown in black and blue. The lines ofconstant distance 2 l are drawn in cyan and green switching color in the middle,where they touch the red envelope. Similarly these lines are shown for copiesof the Kiepert curve, figure 6, rotated against each other by δφ = 60 and δφ = 30 , resp. The length 2 l of the chord for the Kiepert curve is given by2 l = a | sin(3 δφ/ | / .The distance 2 l shrinks for the buckled ring (first figure of figure 12) afterrotation by 2 π/p = 180 to zero. Therefore it cannot serve as floating body ofequilibrium. The buckled rings in figure 13 of symmetry D and D and thosein figure 14 of symmetry D are boundaries of floating bodies of equilibrium.Rotation by 2 π/p yields a non-zero distance l . The waterlines are shown ingreen and cyan, the envelope of the waterlines in red. The figures with odd p are also solutions for ρ = 1 /
2, thus special Zindler curves. The figures with p = 5 are besides solutions for ρ = 1 / ρ > / ρ < / p = 5We turn to the linear case with examples in figure 15. In the first to thirdrow examples of figures from the second to fourth row of figure 1 are shown.They are shifted by a distance δu and in one case one curve is reflected. Thisreflected curve is solution of eq. 1 with the same constants a and b . In the fourthand fifth row two examples are shown, where the figures were shifted so far thatthey fall on each other, together with lines of length 2 l and the envelopes.The derivation of the differential equations (2,1) for the curves are containedin [57] based on [58, 59]. First the linear case was dealt with, where first largedistances, then infinitesimal distances, and finally arbitrary distances were con-sidered (section 2 of [59]). It yields eq. (1) (Eq. (17) of [59] and Eq. (27) of [57]).14igure 15: Property of constant distance for the linear caseThe circular case is considered in section 3 of [59] and in section 3.2 of [57]. Inderiving this equation the author assumed that also for non-integer periodicity p such chords (of infinitesimal length) exist between the curves rotated againsteach other by nearly 2 π . This assumption yields a differential equation of order3, (43) of [59] and (33) of [57]. It can be integrated easily to eq. (2) (Eq. (47)of [59] and Eq. (37) of [57]). Explicit solution of these equations showed thatthe property of constant distance holds.[60, 61]The problem is originally non-local, since it connects end-points of the chordsgenerally without the necessecity of closing them to a polygon of chords. Theeqs. (1, 2), however, reduce it to a local problem: The equations connect onlylocus and direction of the curve at the same point.It came to my big surprise, when Bor, Levine, Perline, and Tabachnikov[7]pointed out, that the problem of elastica under pressure and the problem offloating bodies of equilibrium in two dimensions are governed by the same dif-15erential equation (2). Charges in magnetic fields
The curvature of the boundary curves isquadratic in the radius r , κ = 4 ar + 2 b according to eq. (4). Thus chargesmoving in a perpendicular magnetic field of such an r-dependence, will movealong these curves.Figure 16: Magnetic billiard for perpendicular incidenceThe red boundary is given by the envelope of the chords.Figure 17: Magnetic billiard for perpendicular incidenceThe red boundary is given by the envelope of the chords.A different system is a dynamical billiard. There a particle alternates be-tween free motion and specular reflections at a boundary (angle of incidenceequals angle of reflection). Circles are boundaries with the property that theparticle leaves and arrives at the boundary at the same angle δ . Gutkin[25]found boundaries of billiards with this property, but different from circles.In a magnetic billiard the particle is charged and subject to a constantperpendicular magnetic field. Thus it does not move on straight lines, but onLarmor circles of radius R given by its charge and mass, and by the strengthof the magnetic field. If the angle δ with the boundary of the billiard is a rightangle, then billiards bounded by the (red) envelopes of the chords have thisGutkin δ property, if the radius equals half the length of the chords, R = l .16Bialy, Mironov, and Shalom[6]). These Larmor arcs can be inside (shown indark blue), but also outside the boundary, (shown in dark green) in figure 16.Such Larmor arcs are also possible at the envelopes of the chords of linearelastica, see figure 17. P’ P M Q Q’ Figure 18: Magnetic billiards for δ = 90 The dark red parallel curves are the boundaries of the billiards. Right figure:Cut-out.If the particle does not meet the boundary at right angle, then the boundaryis described by a parallel curve to the envelope. These parallel curves are shownin figure 18 in dark red. The chord
P P ′ touches the red envelope at the midpoint M . The parallel curves and the Larmor circles go through Q and Q ′ . Thetangents at these curves at Q and Q ′ are indicated by red lines. The anglebetween the tangents at Q be δ , that at Q ′ is δ ′ . Both angles add up to180 . The angles in the rhomb P QP ′ Q ′ with midpoint M obey Q ′ P M = QP M = Q ′ P ′ M = QP ′ M = 90 − δ . It is now obvious that the radius ofthe Larmor arcs obey l = R sin δ and the distance between the red envelope andthe boundaries of the billiards is R cos δ .This construction is restricted to angles δ sufficiently close to 90 and en-velopes sufficiently convex.[6] Obviously the boundaries are not allowed to havedouble points. The bicycle problem is closely related to the problem of finding floating bodiesof equilibrium. It was addressed by Finn[17, 18, 50]. The problem goes backto a criticism of the discussion between Sherlock Holmes and Watson in
TheAdventure of the Priory School [13] on which way a bicycle went whose tires’traces are observed. Let the distance between the front and the rear wheel ofthe bicycle be l . The end points of the tangent lines of length l to the trace ofthe rear wheel in the direction the bicycle went yields the points of the traces17f the front wheel. Thus if the tangent lines in both directions end at the traceof the front wheel, it is open which way the bicycle went. Thus curves γ for therear wheel (in red in Fig. 7) and Γ for the front wheel (in black in fig. 7) aresolutions for such an ambiguous direction of the bicycle. The tire track problemconsists in finding such curves Γ and γ different from the trivial solutions ofcircles and straight lines.Obviously the solutions of the two-dimensional floating body problem solvethe bicycle problem, but also the linear elastica and the Zindler curves solve theproblem. There are more solutions to the bicycle curves. Finn argues that thevariety of bicycle curves is much larger: Draw from one point ( N ) of the reartire track the tangent to the front wheel in both directions and give an arbitrarysmooth tire track between these two points ( P ) and ( Q ) in figure 7. Then onecan continue tire track curves in both directions.[17, 18].We shortly explain why Zindler curves are bicycle curves. Let the bicyclistgo in one direction so that the front wheel is at ( x ( α ) , y ( α )) given in eq. (25)and with the rear wheels at ( ξ ( α ) .η ( α )). Then the bicyclist going in the oppositedirection is with its front wheels at x − ( α ) = − l cos( α ) + ξ ( α ) , y − ( α ) = − l sin( α ) + η ( α ) . (29)Since ( x − ( α ) , y − ( α )) = ( x ( α + π ) , y ( α + π )) and the traces of the rear wheelsagree due to (27), both bicyclists use the same traces for their wheels and onecannot determine, which way the bicyclist went.Zindler curves, but also a number of curves from elastica and buckled ringsyield envelopes, that is traces of the rear wheels with cusps. Then the rear wheelhas to go back and forth. For these curves the front wheel has to be turned bymore than the right angle. Driving along these traces requires artistic skills anda suitable bicycle. Apart from the Zindler curves (figs. 8 to 11) this is the casefor the first, second, and fourth buckled ring of figure 13, the inner rear trace offigure 14, and the elastica of the fourth row of fig. 15. However, the third tracesof fig. 13, the outer trace of fig. 14, and the traces of the fifth row of fig. 15can be easily traversed. They constitute good solutions of the bicycle problem. The equations governing the elastic beam (wire) have been given in differentways. They are developed in this section. These differential equations havebeen given in numerous papers apart from their original derivation in numerouspapers. I mention only [7, 34, 35, 37, 39, 47, 48, 51] and references therein.
James Bernoulli considered initially a beam AB loaded by a weight C at theend assuming the beam to be horizontally at the point of the load, see figure19. He assumed the curvature κ to be a function f of the moment. Hence,18 BC xy Figure 19: Beam AB with load C κ = d φ d s = dd x y ′ p y ′ = f ( x ) , (30)where φ is the angle of the tangent at the curve against the x-axis. Integrationyields y ′ = S ( x ) p − S ( x ) , S ( x ) = Z x d ξf ( ξ ) . (31)Assuming a linear relation between the curvature and the moment gives f ( x ) = 2 xa , S ( x ) = x ± aba (32)and thus equation (12). The rectangular elastica primarily considered by JamesBernoulli is obtained for b = 0. Here we consider the force and torque acting in the elastic wire similar to thatgiven by Levy[38] and derive eqs. (3, 4). Let us cut out a piece from r to r ′ (figure 20). At the ends act forces F and − F ′ . In addition a force per length P perpendicular to the wire exerts the force F P = P e × ( r ′ − r ) (33)on the piece of wire, where e is the unit vector perpendicular to the plane.The total force vanishes in the static case, F − F ′ + P e × ( r ′ − r ) = 0 . (34)Integration yields the force acting on the wire, F ( r ) = F + P e × r . (35)Next we consider the torque acting on the piece of wire. Due to the curvatureof the wires there are torques M and − M ′ at the ends of the wires. Moreover19 MF-M’-F’ rr’ nt
Figure 20: Wire, forces, torques, pressure, and tangent and normal vectors r × F and − r ′ × F ′ are exerted by the forces at the ends and the force on thepiece exerts M P = Z d M P = Z ( − r × d F P ) = − P Z r × ( e × d r )= − P Z e ( r d r ) = − P e ( r ′ − r ) . (36)The total torque vanishes, M − M ′ + r × F − r ′ × F ′ + M P = 0 . (37)This yields M ( r ) = − r × F − P e r + M . (38)Let us introduce M = e M, M = e M , e × F = F n . (39)Multiplication of (38) by e yields M ( r ) = r · F n − P r + M = − P ( r − F n P ) + F n P + M . (40)The torque is proportional to the curvature κM = − EIκ, (41)where E is the elastic modulus and I the second moment with respect to theaxis in e -direction through the center of gravity of the cross section of the wire.Love calls this the ordinary approximate theory and discusses it in sections 255– 258 of his book.[39]If there is no external force, P = 0, then the first equation (40) yields κ = − EI ( r · F n + M ) (42)20n agreement with eq. (3). Hence κ increases linearly with r parallel to F n .If P = 0, then we replace r − F n /P → r and obtain κ = 4 ar + 2 b, a = P/ (8 EI ) , b = − ( F n / (2 P ) + M ) / (2 EI ) . (43)Hence the curvature κ increases quadratically with r in accordance with eq. (4). We start with the Frenet-Serret formula for the tangential vector t and thenormal vector n of the wire d t d s = κ n , d n d s = − κ t . (44)We express the force as F = F t t + F n n . (45)Then going along the wire (beam) we obtaindd s ( F t t + F n n ) = P n . (46)Thus d F t d s − κF n = 0 , (47)d F n d s + κF t = P. (48)The torque obeys d M d s = − F n . (49)Finally the torque is assumed to be proportional to the curvature M = − EIκ. (50)This yields EI d κ d s = F n . (51)We insert this relation in eq. (47) and obtaind F t d s − EIκ d κ d s = 0 , (52)which can be integrated to F t − EI κ = c ′ . (53)Equations (51) and (48) yield EI d κ d s = d F n d s = − κF t + P = κ ( − c ′ − EI κ ) + P. (54)21ence d κ d s + 12 κ + c ′ EI κ − PEI = 0 . (55)If we multiply this equation by d κ/ d s and integrate, then we obtain12 (cid:18) d κ d s (cid:19) + 18 κ + c ′ EI κ − PEI κ = ˆ E. (56)These are the equations (9) and (10). In this subsection we will derive this equation requiring the extreme of theintegral over the square of the curvature with appropriate side conditions. Weperform a purely geometrical derivation. As a function of the arc parameter s we introduce the angle φ of the tangent against the x-axis and the Cartesiancoordinates. The origin is at s = 0. The curve starts with the angle φ . φ ( s ) = φ + Z s d s ′ κ ( s ′ ) , (57) x ( s ) = Z s d s ′ cos( φ ( s ′ )) , (58) y ( s ) = Z s d s ′ sin( φ ( s ′ )) . (59)The length of the arc be s . The area between the arc and the straight linefrom the origin to the endpoint of the arc ( x ( s ) , y ( s )) is given by A = 12 Z s d s ( x ( s ) sin( φ ( s )) − y ( s ) cos( φ ( s ))) . (60)We may have several side conditions on the curve: the angle at the end point φ ( s ), the coordinates of the end point and the area fixed. Thus the correspond-ing quantities have to be subtracted from the integral over κ / λ ...λ , I = 12 Z s d sκ ( s ) − λ φ ( s ) − λ x ( s ) − λ y ( s ) − λ A − λ x ( s ) + y ( s )) . (61)In total these may be too many conditions. But for those we do not take intoaccount, we set λ i = 0. The variation of I has to vanish, δI = Z s d sδκ ( s ) F ( s ) , (62) F ( s ) = κ ( s ) − λ + ( λ + λ x ( s )) Z s s d s ′ sin( φ ( s ′ )) − ( λ + λ y ( s )) Z s s d s ′ cos( φ ( s ′ ))22 λ Z s s d s ′ ( x ( s ′ ) cos( φ ( s ′ )) + y ( s ′ ) sin( φ ( s ′ )))+ λ (cid:18)Z s s d s ′ sin( φ ( s ′ )) (cid:19) + λ (cid:18)Z s s d s ′ cos( φ ( s ′ )) (cid:19) (63)= κ ( s ) − λ + λ ( y ( s ) − y ( s )) − λ ( x ( s ) − x ( s ))+ λ x ( s )( x ( s ) − x ( s )) + y ( s )( y ( s ) − y ( s ))]+ λ ( x ( s ) y ( s ) − y ( s ) x ( s )) . (64)The equation F ( s ) = 0 has to be solved.One immediately sees from eq. (64)that κ depends (for λ = 0) quadratically on the distance from some point. Thederivatives of F with respect to s , indicated by a dot vanish,˙ F ( s ) = ˙ κ ( s ) + K = 0 , (65) K = cos( φ ( s ))[ λ + λ x ( s ) − λ x ( s ) − λ y ( s )]+ sin( φ ( s ))[ − λ + λ y ( s ) − λ y ( s ) + λ x ( s )] , (66)¨ F ( s ) = ¨ κ ( s ) + K κ + λ = 0 , (67) K = − sin( φ ( s ))[ λ + λ x ( s ) − λ x ( s ) − λ y ( s )]+ cos( φ ( s ))[ − λ + λ y ( s ) − λ y ( s ) + λ x ( s )] , (68)... F ( s ) = ... κ ( s ) + K ˙ κ ( s ) − K κ ( s ) = 0 . (69)Elimination of K and K yields κ ˙ F ( s ) − κ ˙ κ ¨ F ( s ) + 1 κ ... F ( s ) = κ ˙ κ − κ ˙ κ ¨ κ − λ κ ˙ κ + 1 κ ... κ = 0 . (70)This expression is a complete derivative. Integration and multiplication by κ gives 12 κ + ¨ κ + λ + C κ = 0 . (71)This is the Lagrange equation for the relative extrema of the bending energy.Multiplication of this expression by ˙ κ and another integration yields12 ˙ κ + 18 κ + λ κ + C κ = C , (72)which agrees with eqs. (55) and (56) by renaming the constants. Apart fromeq. (70) also boundary conditions for F = 0, ˙ F = 0, and ¨ F (0) at s = 0 or s = s have to be fulfilled. Thus κ , ˙ κ , and ¨ κ at s = 0 or s = s are expressedby λ i and x ( s ), y ( s ), φ at s = 0 or s = s . Insertion in (71) and (72) yieldsthe corresponding constants C and C .23 .5 Water contained in a cloth sheet James Bernoulli showed that water contained in a long cloth sheet (in y-direction)is bound in the xz-direction by an elastic curve. (See Levien[37], footnotes 3 and5, and Truesdell[53], p. 201). This problem goes under the name of lintearia .The cloth ends should be fixed at ( ξ , ζ ) and ( ξ , ζ ). The height of thewater-line is h . The water surface ranges from x to x as in figures 21 and 22. z x(x ,h )(x ,h ) (ξ ,ζ )(ξ ,ζ ) Figure 21: Cross section of the cloth filled with water for x > ξ and x < ξ . x x min max z x(x ,h )(x ,h ) (ξ ,ζ )(ξ ,ζ ) Figure 22: Cross section of the cloth filled with water for x < ξ and x > ξ .The cyan curve indicates in both cases the continuation of the elastica curve.We give two derivations, a longer one using variational techniques, and ashorter one, considering forces and pressure as in subsection 3.3. We denote the area of the cross section by A , the breadth of the cloth by s , thepotential energy by V , the length of the cloth in y-direction L y , the gravitationalconstant g , and the density of the water ρ . Then the area of the cross section A , the potential energy V and the breadth s of the cloth are given by24 = Z x x d x ( h ( x ) − z ( x )) , (73) V = L y gρ Z x x d x ( h ( x ) − z ( x )) , (74) s = s + s + s , (75) s := p ( ξ − x ) + ( ζ − h ) , (76) s := p ( ξ − x ) + ( ζ − h ) , (77) s := Z x x d x p z ′ , (78) z ′ := d z d x , h ′ := d h d x . (79)Of course we know that h ( x ) does not depend on x . However, it is of advantageto have h ( x ) and h ( x ) as two variables, which can be varied independently.We look for the extreme of V − αA − σS . The variations are δA = Z x x d x ( δh ( x ) − δz ( x )) , (80) δV = L y gρ Z x x d x ( hδh − zδz ) , (81) δs = δx ( x − ξ ) + δh ( h − ζ ) s , (82) δs = δx ( x − ξ ) + δh ( h − ζ ) s , (83) δs = δx q z ′ − δx q z ′ + Z x x d x z ′ √ z ′ d δz d x , (84) z ′ i := z ′ ( x = x i ) , h ′ i := h ′ ( x = x i ) . (85)We have not included contributions proportional to δx i in δA and δV , since h i = z i at x i . Partial integration transforms the integral to Z x x d x z ′ √ z ′ d δz d x = δz z ′ p z ′ − δz z ′ p z ′ − Z x x d xδz dd x (cid:18) z ′ √ z ′ (cid:19) . (86)The variation of V − αA − σS contains contributions proportional to δx i , δh i ,and δz i and an integral over x , Z x x d x ( L y gρh ( x ) − α ) δh ( x ) (87) − Z x x d x (cid:18) L y gρz ( x ) − α − σ dd x (cid:18) z ′ √ z ′ (cid:19)(cid:19) δz ( x ) (88)25he variation of δh ( x ) yields constant h as expected, α = L y gρh. (89)The variation of δz ( x ) yields L y gρ ( z ( x ) − h ) = σ dd x (cid:18) z ′ √ z ′ (cid:19) = σ z ′′ (1 + z ′ ) / = σκ. (90)Thus the curvature increases proportional to the depth measured from thewater-surface.Since the variation of h ( x i ) = z ( x i ) yields δh i + h ′ i δx i = δz i + z ′ i δx i , δz i = δh i − z ′ i δx i , (91)we obtain the contributions δx x − ξ s ∓ p z ′ ! + δh h − ζ s ∓ z ′ p z ′ ! + δx x − ξ s ± z ′ p z ′ ! + δh h − ζ s ± z ′ p z ′ ! . (92)The factors of δx i and δh i have to vanish. They yield the direction of thelines from the lines of the suspension to the lines where the cloth touches thewaterline. If x > ξ and ξ > x as in fig. 21, then the upper signs in (92)apply, and the slope is continuous across the waterline as expected.If instead x < ξ and ξ < x as in fig. 22, then z ( x ) is double-valued withvalues z − ( x ) and z + ( x ). Then the x -integral of s reads s = Z x max x min d x q z ′ − ( x )+ Z x x min d x q z ′ ( x )+ Z x max x d x q z ′ − ( x ) . (93)Accordingly the contributions from s change sign and the lower signs in (92)apply. Again the slope is continuous across the waterline. The expression for thearea and similarly for the potential have different signs in front of the integrals, A = Z x max x min d x ( h ( x ) − z − ( x )) − Z x x min d x ( h ( x ) − z + ( x )) − Z x max x d x ( h ( x ) − z + ( x )) . (94) A simpler derivation can be given by considering the forces and the pressure asin subsection 3.3.Since the cloth can be bent without exerting any forces or torques, one has F n = 0, M = 0, c = 0. Thus eqs. (47) and (48) readd F t d s = 0 , κ F t L y = P, (95)26here F t is the total force over the length L y . The pressure acts from insideand depends on z , P = − ρg ( h − z ) . (96)This yields F t = const , κ = L y ρg ( z − h ) F t (97)in agreement with equation (3). We determine the differential equation for the roulette of the hyperbola. Weparametrize the coordinates ( x h , y h ) of the hyperbola by a parameter p , x h = a cosh( p ) , y h = b sinh( p ) , (98)d x h d p = a sinh( p ) , d y h d p = b cosh( p ) . (99) a and b are the semi axes of the hyperbola. The arc length s of the hyperbolareadsd s = −√ q d p, q = a sinh ( p ) + b cosh ( p ) = a + b p ) + b − a . (100)The coordinates ( x, y ) of the midpoint between the branches of the hyperbolarolling on the x-axis without slipping obey (cid:18) xy (cid:19) = (cid:18) s (cid:19) + 1 √ q (cid:18) a sinh( p ) b cosh( p ) − b cosh( p ) a sinh( p ) (cid:19) (cid:18) x h y h (cid:19) , (101)which yields x = s + a + b √ q sinh(2 p ) , y = − ab √ q . (102)The derivatives d x d p = a b q / , d y d p = ab ( a + b )2 q / sinh(2 p ) (103)yield d y d x = a + b ab sinh(2 p ) . (104)We express sinh(2 p ) by q and q by y , q = (cid:18) aby (cid:19) , (cid:18) q + a − b (cid:19) = (cid:18) a + b p ) (cid:19) + (cid:18) a + b (cid:19) . (105)27his yields the differential equation for the Sturm roulette for the hyperbolarolling on a straight lined y d x = 1 ab s(cid:18) q + a − b (cid:19) − (cid:18) a + b (cid:19) = 1 ab p ( q + a )( q − b )= 1 ab s(cid:18) a b y + a (cid:19) (cid:18) a b y − b (cid:19) = 1 y p ( b + y )( a − y ) . (106)For a rectangular hyperbola one has b = a and this equation reduces tod y d x = p a − y y , (107)which is eq. (11) for the rectangular elastica, only the coordinates x and y exchanged. The hyperbolas are de-picted in magenta, theasymptotes in black,and the elastica inblue.Figure 23: Full period of the rectangular elastica as roulette of hyperbolaThe roulette shown in figure 4 shows only half a period of the rectangularelastica. One obtains the full period, if one first rolls the upper branch of thehyperbola, as shown in figure 23 starting with hyperbola 1 over hyp. 2 to hyp.3, where for 1 and 3 one asymptote is along the x-axis. Now we roll the lowerbranch from hyp. 3 to 4 and 5. This yields the second half of the elastica. Thehyperbola 5 has the same orientation as number 1, only shifted by one periodalong the x-axis. Now one may continue for another period, and so on.28
A A CC Figure 24: Two halves of the areaThe area is cut in halves by the diameter D ( α ) = A AA . The coordinates of A are ξ ( α ) , η ( α ), and those of A , A , C , C are (x,y) at α, α + π, γ, γ + d γ , resp.The infinitesimal triangle of eq. (109) is AC C . ρ = 1 / The boundaries of the two-dimensional floating bodies of equilibrium with den-sity 1 / Zindler considers mainly in sects. 6, 7 and 10 of [63] convex plain areas withthe property that any chord between two points bisecting the perimeter hasconstant length and bisects the area.The envelope of the chords is defined by Equation (26) with the constraint(27) which implies ξ ( π ) = ξ (0), η ( π ) = η (0). The boundary can be parameter-ized by equation (25). Hence x ( α + π ) = ξ ( α ) − l cos( α ) , y ( α + π ) = η ( α ) − l sin( α ) . (108)The diameters D ( α ) ranges from ( x ( α ) , y ( α )) to ( x ( α + π ) , y ( α + π )). It bisectsthe perimeter and the area enclosed by the boundary, provided l is sufficientlylarge, so that the diameters cut the boundary only at the end points of thediameter. 29e consider now the two regions cut by the diameter D ( α ). They consist ofinfinitesimal triangles AC C , (fig. 24) A = ( ξ ( α ) , η ( α )) , C = ( x ( γ ) , y ( γ )) , C = ( x ( γ + d γ ) , y ( γ + d γ )) , (109)with γ = α = α...α + π for region 1 and γ = α = α − π...α for region 2.Then the perimeter is given by u , = Z α , + πα , d γ p x ′ ( γ ) + y ′ ( γ ) = Z α , + πα , d γ p ˆ ρ ( γ ) + l . (110)Since ˆ ρ ( γ − π ) = − ˆ ρ ( γ ), one obtains the same integral. Hence u = u . Thediameter bisects the perimeter.The area of the infinitesimal triangle (109) is given byd A = 12 d γ [( x ( γ ) − ξ ( α )) y ′ ( γ ) − ( y ( γ ) − η ( α )) x ′ ( γ )] . (111)With the abbreviations ξ γ,α := ξ ( γ ) − ξ ( α ) , η γ,α := η ( γ ) − η ( α ) (112)we obtain x ( γ ) − ξ ( α ) = l cos( γ ) + ξ γ,α ,y ( γ ) − η ( α ) = l sin( γ ) + η γ,α ,x ′ ( γ ) = − l sin( γ ) + ˆ ρ ( γ ) cos( γ ) ,y ′ ( γ ) = l cos( γ ) + ˆ ρ ( γ ) sin( γ ) (113)and d A = 12 d γ [ l + lξ γ,α cos( γ ) + lη γ,α sin( γ )+ ξ γ,α ˆ ρ ( γ ) sin( γ ) − η γ,α ˆ ρ ( γ ) cos( γ )] . (114)One finds that both areas A and A are equal A = A = π l − ∆ , ∆ = Z α + πα d γη ( γ )ˆ ρ ( γ ) cos( γ )= − Z α + πα d γξ ( γ )ˆ ρ ( γ ) sin( γ ) . (115)They are independent of α , since the integrand is a periodic function of γ withperiod π . The term proportional to l vanishes, since it is a total derivative of afunction, which vanishes at the limits, Z α , + πα , d γ [ ξ γ,α cos( γ ) + η γ,α sin( γ )]= Z α , + πα , d γ dd γ [ ξ γ,α sin( γ ) − η γ,α cos( γ )] . (116)30 .2 Centers of Gravity Zindler did not consider the centers of gravity of the two half areas. It isimportant for the floating body problem that their distance is constant andthat the straight line between the two centers is normal to the chord.The centers of gravity (¯ x ( γ ) , ¯ y ( γ )) of the infinitesimal triangles (109) aregiven by¯ x ( γ ) = 13 ξ ( α ) + 23 ( l cos( γ ) + ξ ( γ )) = ξ ( α ) + 23 ( l cos( γ ) + ξ γ,α ) , ¯ y ( γ ) = 13 η ( α ) + 23 ( l sin( γ ) + η ( γ )) = η ( α ) + 23 ( l sin( γ ) + η γ,α ) . (117)Thus the centers of gravity ( ¯ X, ¯ Y ) of the halves of the area are given by theintegrals A , ¯ X , ( α ) = Z d A ¯ x = Z d Aξ ( α ) + 13 Z d γ [ l cos( γ )+ l ξ γ,α (1 + cos ( γ )) + l η γ,α cos( γ ) sin( γ )+ lξ γ,α cos( γ ) + lξ γ,α η γ,α sin( γ )+ lξ γ,α ˆ ρ ( γ ) sin( γ ) cos( γ ) − lη γ,α ˆ ρ ( γ ) cos ( γ )+ ξ γ,α ˆ ρ ( γ ) sin( γ ) − ξ γ,α η γ,α ˆ ρ ( γ ) cos( γ )] , (118) A , ¯ Y , ( α ) = Z d A ¯ y = Z d Aη ( α ) + 13 Z d γ [ l sin( γ )+ l ξ γ,α cos( γ ) sin( γ ) + l η γ,α (1 + sin ( γ ))+ lξ γ,α η γ,α cos( γ ) + lη γ,α sin( γ )+ lξ γ,α ˆ ρ ( γ ) sin ( γ ) − lη γ,α ˆ ρ ( γ ) sin( γ ) cos( γ )+ ξ γ,α η γ,α ˆ ρ ( γ ) sin γ − η γ,α ˆ ρ ( γ ) cos( γ )] . (119)The result can be written A , ¯ X , ( α ) = ± ( l ˆ x + l ˆ x ) + l ˆ x + ˆ x ,A , ¯ Y , ( α ) = ± ( l ˆ y + l ˆ y ) + l ˆ y + ˆ y , (120)which yields ˆ x = −
23 sin( α ) , ˆ y = 23 cos( α ) . (121)The integral for ˆ x and ˆ y can be writtenˆ x = 23 Z d γ dd γ [ ξ γ,α ( ξ γ,α sin( γ ) − η γ,α cos( γ ))] , ˆ y = 23 Z d γ dd γ [ η γ,α ( ξ γ,α sin( γ ) − η γ,α cos( γ ))] . (122)They vanish at the limits. Thus ˆ x = ˆ y = 0. Thus the distance h betweenboth centers of gravity obeys A , h = l and the line between both centers isperpendicular to the chord between the two areas.31 P QQN N
Figure 25: Darboux butterflyUsually P , P , Q are given and Q is constructed. Compare this figure withfig. 7. At least seven papers[2, 19, 20, 30, 43, 45, 46] appeared from 1933 to 1940, whichdiscuss (i) characteristic properties of the circle, and (ii) which (convex) plainregions have the property that chords between two points of the boundary ofconstant length cut the bounded region in two pieces of constant area. I shortlyreport on them. The first one by Hirakawa[30] stated two theorems:Theorem I. A closed convex plane curve with the property that all chords offixed length span arcs of equal length, is a circle.Theorem II. A plane oval, in which the areas cut off by chords of equal lengthhave the same content, is a circle.Apparently it is meant that this should hold for all lengths of chords. Salkowskiemphasizes that considerable weakening of the conditions yields similar results.
Salkowski[46] started in 1934 with these two theorems and sharpened them.First he introduces what is now known as Darboux butterfly:Consider a polygonal line P , P , P , ...P n with constant side length P i P i +1 = s . Then one connects P with Q = P n by the line P Q = 2 d . Then a point P n +1 = Q is determined so that Q Q = s, P Q = P Q = 2 d . Q is thepoint which lies on the parallel to P P n through P . The arcs P P and Q Q in fig. 7 are equal. They are replaced by straight lines of equal length in theDarboux butterfly fig.25. In the limit considered here, where P P = s tends tozero, the ratio of arc and distance tends to one. He argues that then Theorem Iis equivalent to Theorem II and that this remains true when s tends to 0 (andcorrespondingly n to ∞ ). He restricts the corresponding curve c to a curvewithout turning point. He considers the isosceles trapezoid P i P i +1 Q i Q i +1 withcircumcircles with centers M i . This center is intersection point of the middlenormals on P i P i +1 , Q i Q i +1 , but also on P i Q i , P i +1 Q i +1 . Denote the midpointof P i Q i by N i and that of P i +1 Q i +1 by N i +1 . In the limit s → N i yield the curve ( N ). The point M i yield the evolute ( M ) of ( N ). (I think, here32s a misprint in the paper: Instead of ”die Punkte N liegen auf einer Evolute derKurve” it should read ”die Punkte M liegen auf einer Evolute der Kurve”. Alittle bit later seems to be another misprint: Instead of ”Mittelpunkt der Sehne P N ” one should read ”Mittelpunkte N der Sehne P Q ”.)Salkowski continues: It may happen that the trapezoid degenerates to arectangle. In this case the curve ( N ) has a cusp. The tangents to the oval at theend points P and Q are parallel and perpendicular to P Q . If the arc
P Q is lessthan half of the circumference of the total circumference, then there exists an arc P ′ Q ′ of the same length with P ′ Q ′ parallel P Q , but shorter chord P ′ Q ′ < P Q .Thus the cusp of ( N ) is only possible, if P Q bisects the circumference. Such anexample for ( N ) is Steiner’s hypocycloid with three cusps.He shows nowTheorem III. If a plane regular piece of curve has the property that three setsof chords of constant length 2 d , d , d cut off constant lengths of curve andform a triangle, then the curve is a circle.(Gericke[20] gave in 1936 another proof of this theorem).Theorem IV. If all chords over constant arcs of length 2 s of a curve have thesame length 2 d and the chords over arcs of length s have the same length 2 d ′ ,then the curve is a circle.Theorem IV’. If a set of quadrangles P QRS with constant side lengths 2 d , d ,2 d , d can be inscribed in an oval with corresponding constant arcs of curve,then the oval is a circle. In particular one findsTheorem V. If all chords of an oval, which cut off one fourth of the circumference,have the same length, then the curve is a circle.Finally Salkowski asks the general question: Are there ovals c , in which an n -gon with equal edges 2 d can be shifted so that the corner points divide theperimeter in equal parts? One realizes that the area of the n -gon has to haveconstant size, further at least one angle is larger than a right angle. Considerthree consecutive corners P, Q, R of the figure with an obtuse angle at Q . Shiftthe chord P Q continuously to QR , then the midpoint N describes a piece of anoval ( N ) and its evolute describes the curve ( M ) of the midpoints of the circles,which touch the oval in the end points of the chords. Denote the midpoint of P Q by N , the midpoint of QR by N . M , M are the intersections of the mid-normals on P Q and QR , resp. M is the intersection of the two mid-normals.Then the points M , M, M constitute a triangle with obtuse angle at M . Thecurve ( M ) touches the mid-normals at M and M .So far I agree with the construction. But now Salkowski continues: Thusthe piece of curve M M (to my opinion it should read curve M M , since M is generally not in the curve) is longer than the chord M M , thus larger thanthe distance M M . Since M is the midpoint of the circumcircle of the isoscelestriangle P QR , thus
M N = M N , such a configuration is not possible unless M, M , M coincide. Then the triangle P QR transforms into its neighboringposition by an infinitesimal rotation, thus it remains unchanged during the shiftalong the curve c , hence remaining a circle.I do not see a reason, why M, M , M coincide in general. They will coincideat points where the curvature of ( P ) at Q has an extreme. Then the neighboring33 N N P QR MM M N N P QR M N N P QR
Figure 26: The construction by SalkowskiThe curves ( P ) are in black, ( N ) in red, and ( M ) in blue for s Q = π/ π/ P R is notconstant.Salkowski has based his argument on an obtuse angle at P QR . Such obtuseangles appear in several cases considered in ref. [58]. The curves (28) are convexfor values ǫ up to approximately 1 / (2 p ). I choose the curve with p = 7 foldsymmetry and δ = 52 . , that is θ = 37 . . With ǫ = 1 /
98 the angle P QR varies between 102 . and 108 . and thus is always obtuse. I use anapproximation in linear order in ǫ for the curve (P), r ( φ ) = 1 + 2 ǫ cos( pφ ) , φ = s − ǫp sin( ps ) (123)and choose s P = s Q − θ , s R = s Q + 2 θ . (124)The corresponding curves are shown in figure 26 for s Q = π/ π/
14, and 0.The curves ( P ) are in black, ( N ) in red, and ( M ) in blue.For s Q = 0 and s Q = π/ M , M , and M coincide. Butin between, in particular for s Q = 1 /
14 the three points M , M , and M differ.Note that curve ( M ) has cusps. Thus the proof of his last theorem fails. Thisdoes not mean that his theorem is disproved. Since in terms of eqs. (123) therange for ǫ for convex boundaries becomes smaller with increasing p , it may bethat there are not such ovals. Zindler had derived curves whose chords, which bisect the perimeter, have con-stant length and bisect the area. Auerbach[2] 1938 rederived the solution byZindler, but he showed that these curves had also the property, that any chordacting as water-line yields the same potential energy, and thus all orientations34re in equilibrium. The first five sections are for general densities ρ . He obtainsfor the curvatures at P and Qκ Q − κ P = 2 d θ d s , κ Q + κ P = 4 d sin θ, (125)where d/ l and θ the angle between chord and tangent. From section 6on he considers the case ρ = 1 /
2. Auerbach derives the expressions for the coor-dinates x, y of the boundary, eqs. (25, 26). One has to replace ( d/ θ ) → ˆ ρ and d/ → l . Eugene Gutkin gives a few remarks of personal and socio-historical characterat the end of his paper[26]. He reports the cruel death of Herman Auerbachunder the Nazi regime. He also reports that the archimedean floating problemwas popular among older mathematics students around 1939 in Leningrad. Theresults of three of them, Ruban, Zalgaller and Kostelianetz were published inthe Proceedings (Doklady) of the Soviet Academy of Sciences in a Russian and ashorter French version.[43, 45] Only one of the authors, Zalgaller (often writtenSalgaller), could continue his career after the war. Kostelianetz did not returnfrom the war. Ruban returned as invalid from the war, no longer able to domathematics.Although Ruban[43] and Salgaller and Kostelianetz[45] published side byside, they obtained contradictory results. Salgaller and Kostelianetz obtainedsolutions for ρ = 1 / sinθ of the angle θ between chord and tangent are equal atboth ends of the tangent. Erroneously he concluded that both angles are equal,but they add up to π .In sect. 5 of his paper[43] Ruban introduces the angle θ between chord andtangent and curvature κ (Ruban uses k instead of κ ) of a circle of length S and claims without proof or explanation: If θ ctg θ = πnlS ctg πnlS (126)for n = 1 , , ... , then there exists a number ǫ > | κ ( s ) − κ | < ǫ yields the equality κ ( s ) = κ , where l is the length of the arc. Since θ = πl/S , the inequality may be rewritten C n = 0 with C n = cos( θ ) sin( nθ ) − n cos( nθ ) sin θ . (127)It is likely that Ruban, who has derived the relation b [ κ ( s + l ) + κ ( s )] = 4 sin θ ( s ) (128)in eq. (5) ( s is the arc parameter) and used a Fourier expansion for κ ( s ) con-sidered κ ( s ) = κ + a n cos 2 πnsS + b n sin 2 πnsS . (129)35ogether with 2 θ ( s ) = R s + ls d tκ ( t ) and restricting to a n and b n in linear orderyields condition C n = 0 for nontrivial solutions a n , b n . This is the starting pointfor non-circular perturbative solutions, as given in [58], where θ = π/ − δ .Thus Ruban was close to a solution, if he would have performed a perturbationexpansion. Obviously C n = 0 is always fulfilled for n = 1. It corresponds toa translation of the curve, compare sect. 4.2 of [58]. Thus Ruban’s statementshould not include the case n = 1.In 1940 Geppert[19] gave the solutions for ρ = 1 /
2, but erroneously arguedthat there are no solutions for ρ = 1 /
2. He simply overlooked that in this latercase the points on the boundary are end-points of two different chords, not one.A general obstacle to find solutions for ρ = 1 / n of equal parts. This was very good for n = 2, ρ = 1 /
2, but it is not at allnecessary for ρ = 1 / In his 1899 paper[24] Greenhill gives special solutions expressed by pseudo-elliptic functions, in which the cosine and sine of the angle nθ and nθ/
2, resp.are algebraic functions of the radius r . Thus the curves are algebraic.I do not attempt to go through the theory of the pseudo-elliptic functions,but refer only to the main results.Starting point is the expression for the polar angle θ as function of the radius r , θ = 12 Z d r Ar + Br + Cr √ R , R = r − ( Ar + Br + C ) . (130)The integral is divided into two contributions, θ = θ ′ + ( B − B ′ ) u, (131)with θ ′ = 12 Z d r Ar + B ′ r + Cr √ R , u = 12 Z d r √ R , (132)where u is the arc length.The shape of the curves depends on two independent parameters. Thesemay be the dimensionless AC and BC , or ǫ and µ in [61], or x and y or β and γ by Greenhill.[24]For a given periodicity in θ ′ , that is by an increase of θ ′ by 2 π/n (class I) orby 4 π/n , n odd (class II), a certain relation between these parameters is fixed.Since at Greenhill’s time the integral for the arc length u , expressed as ellipticfunction of the first kind, was tabulated, one could easily calculate θ for thesecases.If moreover one requires B = B ′ , in order to obtain an algebraic curve, onehas a second condition, which allows only for single solutions.36 .1 Class I For class I one finds solutions of the formsin( nθ ′ ) = H ( q ) √ P Qq n/ , (133)cos( nθ ′ ) = L ( q ) √ P Qq n/ , (134) H ( q ) = q n − + h q n − + ... + h n − , (135) L ( q ) = q n − + l q n − + ... + l n − , (136) P = − q + 2(2 γ + 1) q − , (137) P = q + 2(2 β − γ − q + (2 β − , (138) P = P P = − ( q + 2( β − γ − q − β + 1) + 16 β γq. (139)For γ > q = r and for γ < − q = − r . P is related to R by P = 16 β | γ | R = 16 β γr − h β p | γ | ( Aq ± Bq + C ) i . (140)Hence A, B, C are related to β and γ by4 β p | γ | = A, ± β p | γ | ( β − γ −
1) = B, β p | γ | (1 − β ) = C. (141)The zeroes q , of P are obtained as q , = 2 γ + 1 ± p γ ( γ + 1) . (142)This yields the extreme values r , of r , r , = (cid:26) √ γ + 1 ± √ γ γ > , √− γ ± √− γ − γ < − . (143)The scale of q is chosen so that q q = 1. The other extreme values q , are thezeroes of P , q , = − β + 2 γ + 1 ± p γ (2 β − γ − . (144)One may exchange P and P by simultaneously rescaling q . This yields P ( β, γ, q ) = − (1 − β ) P ( β ′ , γ ′ , q ′ ) , (145) P ( β, γ, q ) = − (1 − β ) P ( β ′ , γ ′ , q ′ ) , (146) P ( β, γ, q ) = (1 − β ) P ( β ′ , γ ′ , q ′ ) , (147) H ( β, γ, q ) = − (1 − β ) n − L ( β ′ , γ ′ , q ′ ) , (148) L ( β, γ, q ) = − (1 − β ) n − H ( β ′ , γ ′ , q ′ ) (149)with β ′ = − β − β , γ ′ = γ − β , q ′ = q − β . (150)37he derivative of θ ′ calculated from eqs. (133, 134) yields n d θ ′ d q = 2 q d H d q P + qH d P d q − nHP qL √ P P (151)= − q d L d q P + qL d P d q − nLP qH √ P P . (152)Thus one requires 2 q d H d q P + qH d P d q − nHP = nLP ′ , (153)2 q d L d q P + qL d P d q − nLP = − nHP ′ , (154) P ′ := q + 2( β − γ ′ − q − β + 1 , (155)so that d θ ′ d q = P ′ q √ P = R ′ q √ R , (156) P ′ = 4 β p | γ | R ′ = Ar + B ′ r + C, (157) B − B ′ = ± β p | γ | ( γ ′ − γ ) . (158)If one multiplies eq. (133) by H and eq. (134) by L , and adds both equations,then one obtains q n +1 dd q H P + L P q n = 0 , (159)which yields after integration H P + L P q n = const . (160)Denoting this constant by Q , one obtainscos ( nθ ′ ) + sin ( nθ ′ ) = 1 , (161)as required.From eqs. (133,134) one obtainscos(2 nθ ′ ) = L ( q ) P − H ( q ) P Q q n . (162)Since q n cos(2 nθ ′ ) = ℜ (( x + i y ) n ) (163)and q = x + y , these curves are algebraic curves.To obtain solutions for eqs. (133, 134) one has to solve eqs. (153-155). Thecoefficients β and γ are solutions of algebraic equations with integer coefficients.Several curves of class I are shown in figures 27 – 34. They are listed in table1. 38 β γ § and fig. in [24] fig. this paper2 -1.36602540 -1.57735027 23 8 273 -0.37948166 -1.14356483 24 9 284 -0.19053133 -1.06944356 26 10 304 4.19179270 1.34344652 - - 295 -0.11633101 -1.04168414 27 11 315 -4.26375725 -2.97763686 - - 326 -0.07884381 -1.02799337 - - 336 2.21204454 0.41789155 - - 34Table 1: Constants of curves class IFigure 27: Curve with n = 2 Figure 28: Curve with n = 3Figure 29: Curve with n = 4 Figure 30: Curve with n = 4Figure 31: Curve with n = 5 Figure 32: Curve with n = 539igure 33: Curve with n = 6 Figure 34: Curve with n = 6 For class II the angle θ ′ can be written for special values A, B, C and odd n sin( n θ ′ ) = H + p R + Qr n/ , (164)cos( n θ ′ ) = H − p R − Qr n/ , (165) H ± = r n − ± h r n − + h r n − ± ... ± h n − , (166) R ± = r ± ( Ar + Br + C ) , (167) R = R + R − . (168)Differentiating (164, 165) one obtains n d θ ′ r = ± r d H ± d r R ± + H ± r d R ± d r − nH ± R ± rH ∓ p R − R + (169)This expression should yieldd θ ′ d r = A ′ r + B ′ r + C ′ r √ R (170)This requires2 r d H ± d r R ± + H ± r d R ± d r − nH ± R ± = ± nH ∓ ( A ′ r + B ′ r + C ′ ) . (171)If we multiply the equation with the upper signs by H + and that with the lowersigns by H − , and add both then we obtain r n +1 dd r H R + + H − R − r n = 0 , (172)which after integration yields H R + + H − R − r n = const . (173)40hich we set to Q . If we set H = r n − + h r n − + ..., H = h r n − + h r n − + ..., P = Ar + Br + C, (174)so that H ± = H ± H , R ± = r ± P, (175)then H R + + H − R − = 2( H + H ) r + 4 H H P, (176)which is a polynomial of order 2 n − r . Thus eq. (173) can only be fulfilled for odd n .From eqs. (164, 165) one obtainscos( nθ ′ ) = H − R − − H R + Q r n . (177)Since r n cos( nθ ′ ) = ℜ (( x + i y ) n ) (178)and H − R − − H R + = − H + H ) P − H H r (179)is a polynomial even in r , these curves are algebraic, too.Eq. (171) is an equation for a polynomial of order n + 2 in r . Equating thecoefficients of the powers n + 2 and n + 1 yields nA = nA ′ and ( n − Ah = − nA ′ h . A = 0 would yield a constant curvature of the curve, thus only a circleor straight line, we require A = 0. Hence A ′ = A , h = 0. The coefficients ofthe zeroth power in r yield − nCh n − = − nC ′ h n − . Thus if h n − = 0, then C ′ = C .The simplest case is given by n = 3. For this case one obtains A ′ = A , B ′ = B/ C ′ = − C/
3. Here one does not obtain necessarily C ′ = C , since h = 0. Requiring B ′ = B , C ′ = C yieldssin(3 θ/
2) = r √ r + Ar √ r = r Ar , (180)cos(3 θ/
2) = r √ r − Ar √ r = r − Ar . (181)Hence we obtain cos(3 θ ) = − Ar (182)and with A = − /a r = a cos(3 θ ) . (183)This curve is shown in figure 6.For n = 5 we find B = 14 C − AC , q = √ C (184)41nd B ′ = − C − AC . (185)Thus B = B ′ is obtained for AC = 316 , BC = − . (186)A remark on n = 5 of Greenhill[24]. §
14: The ratio of minimal radius andmaximal radius is ( √ − / . §
17 containsnumerical errors: c = ( √ − / . c ) / (1 − c ) = 2 . §
14. The curves of §
14 and §
17 are notonly of the same character, but they are the identical.Figure 35: Curve with n=5 Figure 36: Curve with n=7Figure 37: Curve with n=7 Figure 38: Curve with n=9 n § and fig. in ref. [24] fig. this paper3 13 - 65 14 3 357 15 4 367 15 4 379 16 5 38Table 2: Constants of curves class II42f we follow the curve with a continuous change of the direction of its tangent,then we have to circle twice around the origin in the figures 35, 37, 38, until wereturn to the start of the curve, whereas in figure 36 we have to do this onlyonce. The curves are listed in table 2. Most of the elastica with and without pressure and the Zindler curves are bicyclecurves. Finn[17] pointed out, that the class of bicycle curves is probably muchlarger. We will first comment on his ideas, but also on an important observationby Varkonyi[56] before we give some generalizations due to Mampel[40] and someclosed curves with winding numbers different from one.
Tabachnikov[50] and also Salkowski[46] argue that one can give a segment ofthe front tire track. One draws the tangent at one point of the track of theback wheel in both directions and gives a smooth but arbitrary track of thefront wheel between these two points. Then one can continue the tracks in bothdirections as for example described by Salkowski. Finn[17] instead starts witha segment of the back tire track, which is so long that the corresponding piecesof the front tire tracks meet themselves. In both cases certain conditions at theends of the segments have to be met.
P P QQN CC
C C
P Q
Figure 39: Bicycle tire tracks: Front track in black, rear track inred. Centroids of the areas enclosed by the cyan-green chords andthe black track lie on the magenta circleIn section 2 and also section 6 of ref. [56] Varkonyi considered closed bicycletracks. In fig. 39 two nearby locations of the bicycle moving to the left is shownin cyan and those of the bicycle moving to the right in green. They representchords of length 2 l of the black front track curve. These chords and the blackline enclose areas A and A of constant size A . Their centroids are the points C and C . Going from P Q to P Q the area P P N is cut away and Q Q N is added. These areas are l α/ α between the twochords. Their centroids C P and C Q are 2 l/ of the two chords. Thus the centroid moves from C to C by the distance rα with r = 2 l / (3 A ) parallel to the chords. Since l and A are constant, thecentroids lie on a circle of radius r .Varkonyi now argues: If the bicycle curve is closed, then the same argumentsapply for the complementary area of size A = A − A , where A is the areaenclosed by the black front track. The centroids of the complementary area lieon a circle of radius r = 2 l / (3 A ). In general it is not clear, whether thesetwo circles are concentric. Thus it is not clear whether closed bicycle curvescorrespond to the boundaries of homogeneous floating bodies of equilibrium.The centroid of the whole area lies on the connecting line between the centersof the two circles. If one allows for an inhomogeneus body, then the volumecentroid and the mass centroid can differ. Then one obtains floating bodies ofequilibrium, if the mass centroid lies in one of the circle centers. For density ρ = 1 / ρ = 1 /
2. havedieder symmetry. Thus also the centers of the circles are concentric. Whetherthere are other solutions for closed bicycle tracks is not known.Figure 40: Nearly a triangle withrounded corners Figure 41: Nearly a square withrounded cornersWe give several examples of bicycle curves in the following. The segmentof the front tire track from which we start is shown in blue. The following tiretrack is determined by adding Darboux butterflies and shown in black. (ForDarboux butterflies see subsection 4.3.1 and figure 25). Chords are drawn afterevery fifth application of the butterfly. The centroids of the enclosed area areconnected and shown in magenta. The appearence resembles part of a circle.In figures 40 to 43 the initial segment consists of three pieces, first a straightline of length s/
3, then a circular arc of the same length s/ s/
3. The circular arc changes the direction by an angle2 π/n with n = 3 , , , n rounded corners.The butterfly procedure generates nearly circular arcs from the (nearly) straightlines and nearly straight lines from the arcs. These curves are very similar to44igure 42: Nearly a pentagonwith rounded corners Figure 43: Nearly a hexagonwith rounded cornersthe figures 9 and 10, the first and third of fig. 13, the first of fig. 14, and fig.51.Figure 44: Based on eq. (187) with c = 2 Figure 45: Based on eq. (187) with c = 4Figure 46: Based on eq. (187) with c = 4 . c = 5 . c = 5Figure 49: Based on eq. (187) with c = 6 Figure 50: Based on eq. (187) with c = 8A second class of front tire tracks are shown in figures 44 to 50. The initialfront track segment is given by y = b (1 − x )(1 − cx ) . (187)We have chosen b = 1 / c = 2 , , . , . , , , c = 5 (fig. 48) the ares on bothsides of the chord are equal. Thus a centroid is not defined in this case or liesat infinity. From fig. 50 we see that the tracks may become very wild. This isalso the case, if we use larger values of b in eq. (187). Of course the curves we found for the floating bodies of equilibrium had to beclosed after one revolution. This is not required for bicycle curves. Also theZindler curves can be generalized to a larger class of bicycle curves, which I callZindler multicurves.The definition of the Zindler curves is generalized by replacing eqs. (25, 26)to x ( α ) = l cos( mα ) + ξ ( α ) , y ( α ) = l sin( mα ) + η ( α ) , (188) ξ ( α ) = Z α d β cos( mβ )ˆ ρ ( β ) , η ( α ) = Z α d β sin( mβ )ˆ ρ ( β ) , (189)46igure 51: Zindler curve with m = 1 , n = 3 Figure 52: Zindler curve with m = 1 , n = 5Figure 53: Zindler multicurve with m = 3 , n = 5 Figure 54: Zindler multicurve with m = 3 , n = 1Figure 55: Zindler multicurve with m = 5 , n = 1 Figure 56: Zindler multicurve with m = 5 , n = 3where m is an odd integer. The radius of curvature is now | ˆ ρ/m | . As in (27) werequire ξ ( α + π ) = ξ ( α ) , η ( α + π ) = η ( α ) , (190)which again implies ˆ ρ ( β + π ) = − ˆ ρ ( β ). The curves for m = 1 are Zindlercurves. For larger m the curves are no longer double point free. Generally theyrepeat only after m revolutions. These curves are also bicycle curves, since theargument around eq. (29) applies again.Examples are ˆ ρ ( β ) = ( m − n ) sin( nβ ) (191)47ith odd n , n = m , and n, m coprime. One obtains for the envelopes (traces ofthe rear wheels) ξ ( α ) = n − m n + m ) α ) + n + m n − m ) α ) , (192) η ( α ) = n − m n + m ) α ) − n + m n − m ) α ) . (193)These envelopes are known as hypocycloids for n > m and as epicycloids for n < m . They wind m times around the origin and have n cusps. These cuspspoint outward for hypocycloids and inward for epicycloids. With α = u/ z = x + i y = n − m i( n + m ) u/ + n + m i( m − n ) u/ + l e i mu/ . (194)In particular for m = 1 , n = 3 one obtains z = e u + 2e − i u + l e i u/ . (195)Examples of such Zindler curves, m = 1, are shown in figures 51 and 52. Weshow four examples of Zindler multicurves with m = 3 and m = 5 in figures 53to 56. Figure 57: Envelopewith one cusp pointingoutside Figure 58: Curve ac-cording to (195) with l = 1 .
6. Figure 59: Three cuspspointing to the left.Mampel[40] considers generalized Zindler curves. He introduces envelopes(denoted as ’Kern’ K ) with an odd number of cusps. He attaches tangents withconstant length l in both directions. The endpoints of the tangents form hisgeneralized Zindler curves Z , irrespective of any convexity. Examples similarto Mampel’s figures 8a, 8b, and 9, are shown in figures 57 to 59. The curves 57and 59 consist of circular arcs. The envelope of fig. 58 is a hypocycloid.48 .4 Other bicycle curves Figure 60: Curves with ratio = 1.72 Figure 61: Curves with ratio = 1.582We show some buckled rings, which turn around the center several times.The ratio of the maximal radius and the minimal radius is given. The buckledring figure 60 is very similar to the Zindler multicurve figure 56.Figures 61 to 67 show buckled rings which turn around the center nine times.These buckled rings have two different envelopes. The smaller one has five cuspspointing inward, figure 61. The outer envelope, figure 64 has no cusps. If theratio of the largest distance to the smallest distance from the center is not toolarge, then the outer trace for the rear tire is without cusps. This is the casefor the figures 62 to 64. If the ratio becomes larger, then cusps appear as seenin figures 65 to 67.Figure 62: Curves with ratio = 1.1546 Figure 63: Curves with ratio = 1.399
A short historical account of the curves related to the two-dimensional floatingbodies of equilibrium and the bicycle problem is given in this paper. Bor, Levi,Perline and Tabachnikov found that quite a number of the boundary curves49igure 64: Curves with ratio = 1.582 Figure 65: Curves with ratio = 1.8637Figure 66: Curves with ratio = 2.0869 Figure 67: Curves with ratio = 2.4451had already been described as
Elastica and
Elastica under Pressure or BuckledRings . Auerbach already realized that curves described by Zindler are solutionsfor the floating bodies problem of density 1/2. An even larger class of curvessolves the bicycle problem.The subsequent sections deal with some supplemental details: Several deriva-tions of the equations for the elastica and elastica under pressure are given. Theproperties of Zindler curves and some work on the problem of floating bodiesof equilibrium by other mathematicians is discussed. Special cases of elasticaunder pressure lead to algebraic curves as shown by Greenhill. Since most ofthe curves considered here are bicycle curves, we added some remarks on them.
Acknowledgment
I am indebted to Sergei Tabachnikov for many usefuldiscussions. I am grateful to J.A. Hanna and M. Bialy for useful information.50 eferences [1] G. Arreaga, R. Capovilla, C. Chryssomalakos, J. Guven,
Area-constrainedplanar elastica , Phys. Rev. E 65 (2002) 031801[2] H. Auerbach,
Sur une probl‘eme de M. Ulam concernant l’‘equilibre descorps flottants.
Studia Math. 7 (1938) 121-142[3] R. Balarini,
The Da Vinci-Euler-Bernoulli Beam Theory?
Mechanical En-gineering Magazine Online (April 18, 2003)[4] J. Bernoulli,
Quadratura curvae, e cujus evolutione describtur inflexae lam-inae curvatura . In
Die Werke von Jakob Bernoulli , Birkh¨auser[5] J. Bernoulli,
Jacobi Bernoulli, Basiliensis, Opera vol. 1 (1744) Cramer &Philibert, Geneva[6] M. Bialy, A.E. Mironov, L. Shalom,
Magnetic billiards: Non-integrabilityfor strong magnetic field; Gutkin type examples , arXiv: 2001.02119v1[7] G. Bor, M. Levi, R. Perline, S. Tabachnikov,
Tire track geometry and in-tegrable curve evolution , arxiv: 1705.06314[8] M. Born,
Untersuchungen ¨uber die Stabilit¨at der elastis-chen Linie in Ebene und Raum, unter verschiedenen Grenzbe-dingungen , PhD Thesis. Universit¨at G¨ottingen (1906)https://archive.org/details/untersuchungenb00borngoog/page/n5[9] J. Bracho, L. Montejano, D. Oliveros,
A classification theorem for ZindlerCarousels , J. Dynam. Control Systems 7 (2001) 367[10] J. Bracho, L. Montejano, D. Oliveros,
Zindler curves and the floating bodyproblem , Period. Math. Hungar 49 (2004) 9[11] R. Capovilla, C. Chryssomalakos, J. Guven,
Elastica hypoarealis , Eur. Phys.J. B 29 (2002) 163-166[12] Journal f¨ur die reine und angewandte Mathematik (Crelles Journal) vol. 1(1826) – 4(1929)[13] A.C. Doyle,
The Adventure of the Priory School in The Return of SherlockHolmes , McClure, Philips & Co. (New York 1905), Georges Newnes, Ltd.(London 1905)[14] L. Euler,
Additamentum: De curvis elasticis in Methodus inve-niendi lineas curvas maximi minimive proprietate gaudentes, Lausanne(1744); Translated and annotated by W.A. Oldfather, C.A. Ellis, andD.M. Brown
Leonhard Euler’s elastic curves
Abhandlungen ¨uber das Gleichgewicht und die Schwingungen der ebenenelastischen Kurven , Ostwald’s Klassiker der exakten Wissenschaften 175(Leipzig 1910) 5115] F. Evers, A.D. Mirlin, D.G. Polyakov, P. W¨olfle,
Semiclassical theoryof transport in a random magnetic field , Phys. Rev. B60 (1999) 8951;cond-mat/9901070[16] R. Ferr´eol,
Encyclop´edie des formes math´ematiques remarquables
Which way did you say that bicycle went? ∼ finn/research/bicycle/tracks.html[18] D.L. Finn, Which way did you say that bicycle went?
Math. Mag. 77 (2004)357-367[19] H. Geppert, ¨Uber einige Kennzeichnungen des Kreises , Math. Z. 46 (1940)117-128[20] H. Gericke,
Einige kennzeichnende Eigenschaften des Kreises , Math. Z. 40(1936) 417[21] E.N. Gilbert,
How Things Float , Am, Math. Monthly 98 (1991) 201[22] I. Goldin, J-M. Delosme, A.M. Bruckstein,
Vesicles and Amoebae: OnGlobally Constrained Shape Deformation , J. Math. Imaging and Vision 37(2010) 112[23] A.G. Greenhill,
The applications of elliptic functions , MacMillan & Co,London, New York (1892)[24] A.G. Greenhill,
The elastic curve, under uniform normal pressure , Mathe-matische Annalen LII (1899) 465[25] E. Gutkin,
Capillary Floating and the Billiard Ball Problem , J. Math. FluidMech. 14 (2012) 363-382[26] E. Gutkin,
Addendum to: Capillary Floating and the Billiard Ball Problem ,J. Math. Fluid Mech. 15 (2013) 425[27] G.-H. Halphen,
Sur une courbe elastique , Journal de l’ecole polytechnique,54 e cahier (1884) p.183 (available via gallica.bnf.fr)[28] G.-H. Halphen, La courbe elastique plane sous pression uniforme , Chap. Vin
Traite des fonctions elliptiques et de leurs applications, Deuxieme partie.Applications a la mecanique, a la physique, a la geodesie, a la geometrie etau calcul integral . p. 192[29] W. Helfrich,
Elastic Properties of Lipid Bilayers: Theory and Possible Ex-periments , Z. Naturforsch. 28c (1973) 693-703[30] J. Hirakawa,
On a characteristic property of the circle , The Tˆohoku Math.Journal 37 (1933) 175 5231] C. Huygens,
Constructio universalis problematis ... , Acta EruditorumLeipzig 1694 p. 338[32] C.G.J. Jacobi,
Fundamenta nova theoriae functionum ellipticarum , K¨onigs-berg, Borntr¨ager 1829[33] G. Kirchhoff, ¨Uber das Gleichgewicht und die Bewegung eines unendlichd¨unnen elastischen Stabes , J. reine u. angewandte Math. 56 (1859) 285-313[34] J. Langer,
Recursion in Curve Geometry , New York J. of mathematics 5(1999) 25-51[35] J. Langer, D.A. Singer,
The total squared curvature of closed curves , J.Differential Geometry 20 (1984) 1-22[36] P. S. Laplace,
Oeuvres compl´etes de Laplace , vol. 4, Gauthiers-Villars (1880)[37] R. Levien,
The elastica: A mathematical history
Memoire sur un nouveau cas integrable du probleme de l’elastiqueet l’une des ses applications , Journal de Mathematiques pures et appliquees3 e serie, tome 10 (1884) p. 5-42[39] A.E.H. Love, A Treatise on the Mathematical Theory of Elasticity , Cam-bridge University Press (1927), p. 263[40] K.L. Mampel, ¨Uber Zindlerkurven
Jour. f¨ur die reine und angewandteMathematik 234 (1967) 12-44[41] D. Oliveros, L. Montejano,
De volantines, espir´ographos y la flotaci´on delos cuerpos , Revista Ciencias 55-56 (1999) 46-53[42] M. R´edei,
On the tension between mathematics and physics ,http://philsci-archive.pitt.edu/16071/ (2019)[43] A.N. Ruban,
Sur le probl`eme du cylindre flottant , Comptes Rendus (Dok-lady) de l’Acad´emie des Sciences de l’URSS, XXV (1939) 350[44] L. Saalsch¨utz,
Der belastete Stab unter Einwirkung einer seitlichen Kraft ,B.G. Teubner, Leipzig (1880)[45] V. Salgaller and P. Kostelianetz,
Sur le probl`eme du cylindre flottant ,Comptes Rendus (Doklady) de l’Acad´emie des Sciences de l’URSS, XXV(1939) 353[46] E. Salkowski,
Eine kennzeichnende Eigenschaft des Kreises , Heinrich Lieb-mann zum 60. Geburtstag, Sitz.berichte der Heidelberger Akademie derWissenschaften (1934) 57-62[47] D.A. Singer,
Lectures on elastic curves and rods , Curvature and variationalmodeling in physics and biophysics, AIP Conference Proc. 1002 (2008) 3-325348] H. Singh, J.A. Hanna,
On the planar Elastica, Stress, and Material Stress ,Jour. of Elasticity 136 (2019) 87-101[49] S. Svetina, B. ˇZekˇs,
Membrane bending energy and shape determinationof phospholipid vesicles and red blood cells , Eur. Biophysics J 17 (1989)101-111[50] S. Tabachnikov,
Tire track geometry: variations on a theme , Israel J. ofMath. 151 (2006) 1-28 archive math.DG/0405445[51] I. Tadjbakhsh, F. Odeh,
Equilibrium states of elastic rings , J. Math. Anal.Appl. 18 (1967) 59-74[52] I. Todhunter, (K. Pearson, ed.),
A History of the Theory of Elastic-ity and of the Strength of Materials , 2 volumes, Cambridge Univer-sity Press (1886, 1893). See also: en.wikiquote.org/wiki/A History ofthe Theory of Elasticity and of the Strength of Materials[53] C. Truesdell,
The rational mechanics of flexible or elastic bodies: 1638-1788, in: Leonhard Euler, Opera Omnia , Orell F¨ussli Turici, ser. 2, vol. XI,2 (1960)[54] C. Truesdell,
Der Briefwechsel von Jacob Bernoulli , chapter Mechanics,especially Elasticity, in the correspondence of Jacob Bernoulli with Leibniz.Birkh¨ausser (1987)[55] S. Ulam, Problem 19 in
The Scottish Book , ed. R.D. Mauldin, Birkh¨auser(2015)[56] P.L. Varkonyi,
Floating Body Problems in Two Dimensions , Studies inAppl. Mathemetics 122 (2009) 195-218[57] F. Wegner,
Floating Bodies of Equilibrium , Studies in Applied Mathematics111 (2003) 167-183[58] F. Wegner,
Floating bodies of equilibrium I , arxiv: physics/0203061[59] F. Wegner,
Floating bodies of equilibrium II , arxiv: physics/0205059[60] F. Wegner
Floating bodies of equilibrium. Explicit Solution , arxiv:physics/0603160[61] F. Wegner,
Floating bodies of equilibrium in 2D, the tire track problem andelectrons in a parabolic magnetic field , arxiv: physics/0701241v3[62] Wikipedia contributors,
Euler-Bernoulli beam theory , Wikipedia, The FreeEncyclopedia, https://en.wikipedia.org/wiki/Euler-Bernoulli beam theory[63] K. Zindler, ¨Uber konvexe Gebilde. II. Teil¨Uber konvexe Gebilde. II. Teil