MATH ESSAY WRITING_14 PAGES

ar X

iv :p

hy si

cs /0

50 71

05 v1

[ ph

ys ic

s. bi

o- ph

] 1

3 Ju

l 20

05

Lancret helices

Alexandre F. da Fonseca, C. P. Malta ∗

Instituto de F́ısica, Universidade de São Paulo, USP

Rua do Matão, Travessa R 187, Cidade Universitária, 05508-900, São Paulo,

Brazil

Abstract

Helical configurations of inhomogeneous symmetric rods with non-constant bend-

ing and twisting stiffness are studied within the framework of the Kirchhoff rod

model. From the static Kirchhoff equations, we obtain a set of differential equations

for the curvature and torsion of the centerline of the rod and the Lancret’s theorem

is used to find helical solutions. We obtain a free standing helical solution for an

inhomogeneous rod whose curvature and torsion depend on the form of variation of

the bending coefficient along the rod. These results are obtained for inhomogeneous

rods without intrinsic curvature, and for a particular case of intrinsic curvature.

Key words:

Kirchhoff rod model, inhomogeneous rod, Lancret’s theorem, tendrils of climbing

plants

PACS: 46.70.Hg, 87.15.La, 02.40.Hw

∗ C. P. Malta

Email address: [email protected] (C. P. Malta).

Preprint submitted to Elsevier Science 2 February 2008

1 Introduction

Helical filaments are tridimensional structures commonly found in Nature.

They can be seen in microscopic systems, as biomolecules [1], bacterial fibers [2]

and nanosprings [3], and in macroscopic ones, as ropes, strings and climbing

plants [4,5,6]. Usually, the axis of all these objects is modeled as a circular

helix, i. e. a 3D-space curve whose mathematical geometric properties, namely

the curvature, kF , and the torsion, τF , are constant [7,8,9]. This kind of helical

structure has been shown to be a static solution of the Kirchhoff rod model [7].

The Kirchhoff rod model [10,11] has been proved to be a good framework to

study the statics [7,12,13] and dynamics [29] of long, thin and inextensible

elastic rods. Applications of the Kirchhoff model range from Biology [1,14,5]

to Engineering [15] and, recently, to Nanoscience [16]. In most cases, the rod or

filament is considered as being homogeneous, but the case of nonhomogeneous

rods have also been considered in the literature. It has been shown that non-

homogeneous Kirchhoff rods may present spatial chaos [17,18]. In the case of

planar rods, Domokos and collaborators have provided some rigorous results

for non-uniform elasticae [19] and for constrained Euler buckling [20,21]. Devi-

ations of the helical structure of rods due to periodic variation of the Young’s

modulus were verified numerically by da Fonseca, Malta and de Aguiar [22].

Nonhomogeneous rods subject to given boundary conditions were studied by

da Fonseca and de Aguiar in [23]. The effects of a nonhomogeneous mass

distribution in the dynamics of unstable closed rods have been analyzed by

Fonseca and de Aguiar [24]. Goriely and McMillen [25] studied the dynamics

of cracking whips [26] and Kashimoto and Shiraishi [27] studied twisting waves

in inhomogeneous rods.

2

The stability analysis of helical structures is of great importance in the study

of the elastic behavior of filamentary systems and has been performed both

experimentally [28] and theoretically [29,30,31]. It has been also shown that

the type of instability in twisted rods strongly depends on the anisotropy of

the cross section [32,33,34].

Here, we consider a rod with nonhomogeneous bending and twisting coef-

ficients varying along its arclength s, B(s) and C(s), respectively. We are

concerned with the following question: is there any helical solution for the sta-

tionary Kirchhoff equations in the case of an inhomogeneous rod ? The answer

is ‘yes’ and it will be shown that the helical solution for an inhomogeneous

rod with varying bending coefficient cannot be the well known circular helix,

for which the curvature, kF , and torsion, τF , are constant. To this purpose,

we shall derive a set of differential equations for the curvature and the torsion

of the centerline of an inhomogeneous rod and then apply the condition that

a space curve must satisfy to be helical: the Lancret’s theorem. We shall ob-

tain the simplest helical solutions satisfying the Lancret’s theorem and show

that they are free standing helices, i.e., helices that are not subjected to axial

forces [30]. A resulting helical structure different from the circular helix, from

now on, will be called a Lancret helix.

According to the fundamental theorem for space curves [9], the curvature

kF (s), and the torsion, τF (s), completely determine a space curve, but for

its position in space. We shall show that the kF (s) and τF (s) of a Lancret he-

lix depend directly on the bending coefficient, B(s), an expected result since

the centerline of the rod does not depend on the twisting coefficient (see for

example, Neukirch and Henderson [13]).

3

Some motivations for this work are related to defects [35] and distortions [36]

in biological molecules. These defects and distortions could be modeled as

inhomogeneities along a continuous elastic rod.

In Sec. II we review the general definition of a space curve, the Frenet basis

and the so-called Lancret’s theorem. In Sec. III we present the static Kirchhoff

equations for an intrinsically straight rod with varying stiffness, and derive

the differential equations for the curvature and torsion of the rod. In Sec.

IV we use the Lancret’s theorem for obtaining helical solutions of the static

Kirchhoff equations and we show that they cannot be circular helices if the

bending coefficient is not constant. As illustration, we compare a homogeneous

rod with two simple cases of inhomogeneous rods: (i) linear and (ii) periodic

bending coefficient varying along the rod. The circular helix has a well known

relation of the curvature and torsion with the radius and pitch of the helix. In

Sec. V we define a function involving all these variables in such a way that for

the circular helix its value is identically null. We have verified, numerically,

that this function approaches zero for the inhomogeneous cases considered

here. In Sec. VI we analyse the cases of null torsion (straight and planar

rods). Since helical solutions of intrinsically straight rods are not dynamically

stable [30], in Sec. VII we consider a rod with a given helical intrinsic curvature

and we obtain, for this case, a helical solution of the static Kirchhoff equations

similar to that of an intrinsically straight inhomogeneous rod. In Sec. VIII we

summarize the main results.

4

2 Curves in space

A curve in space can be considered as a path of a particle in motion. The

rectangular coordinates (x, y, z) of the point on a curve can be expressed as

function of a parameter u inside a given interval:

x = x(u) , y = y(u) , z = z(u) , u1 ≤ u ≤ u2 . (1)

We define the vector x(u) ≡ (x(u), y(u), z(u)). If u is the time, x(u) represents

the trajectory of a particle.

2.1 The Frenet frame and the Frenet-Serret equations

The vector tangent to the space curve at a given point P is simply dx/du. It

is possible to show [9] that if the arclength s of the space curve is considered

as its parameter, the tangent vector at a given point P of the curve x(s) is

a unitary vector. So, using the arclength s to parametrize the curve, we shall

denote by t its tangent vector

t = dx

ds , (2)

‖t‖ = 1. The tangent vector t points in the direction of increasing s.

The plane defined by the points P1, P2 and P3 on the curve, with P2 and P3

approaching P1, is called the osculating plane of the curve at P1 [9]. Given a

point P on the curve, the principal normal at P is the line, in the osculating

plane at P , that is perpendicular to the tangent vector at P . The normal

vector n is the unit vector associated to the principal normal (its sense may

5

be chosen arbitrarily, provided it is continuous along the space curve).

From t.t = 1, differentiating with respect to s (indicated by a prime) it follows

that:

t.t′ = 0 , (3)

so that t and t′ are orthogonal. It is possible to show that t′ lies in the

osculating plane, consequently t′ is in the direction of n. This allows us to

write

t ′ = kF n , (4)

kF being called the curvature of the space curve at P .

The curvature measures the rate of change of the tangent vector when moving

along the curve. In order to measure the rate of change of the osculating plane,

we introduce the vector normal to this plane at P : the binormal unit vector

b. At a point P on the curve, b is defined in such a way that

b = t × n . (5)

The frame {n, b, t} can be taken as a new frame of reference and forms the

moving trihedron of the curve. It is commonly called the Frenet frame. The rate

of change of the osculating plane is expressed by the vector b′. It is possible

to show that b′ is anti-parallel to the unit vector n [9]. So we can write

b ′ = −τF n , (6)

τF being called the torsion of the space curve at P .

6

The rate of variation of n [9] can be obtained straightforwardly. It is given by

n ′ = −kF t + τF b . (7)

The set of differential equations for {t, n, b} is

t′ = kF n ,

n′ = −kF t + τF b ,

b′ = −τF n ,

(8)

and are known as the formulas of Frenet or the Serret-Frenet equations [9].

2.2 The Fundamental theorem of space curves

A space curve parametrized by its arclength s is defined by a vectorial func-

tion x(s). The form of x(s) depends on the choice of the coordinate system.

Nevertheless, there exists a form of characterization of a space curve given

by a relation that is independent of the coordinates. This relation gives the

natural equation for the curve.

kF (s) gives the natural equation in the case of planar curves. Indeed, if ϕ is

the angle between the tangent vector of the planar curve and the x-axis of

the coordinate system, it is possible to show that kF = dϕ/ds. Since cos(ϕ) =

dx/ds and sin(ϕ) = dy/ds, knowing kF (s), then ϕ(s), x(s), and y(s) of the

planar curve can be obtained immediatly:

ϕ(s) =

s ∫

s0

kF (s) ds, x(s) =

s ∫

s0

cos ϕ(s)ds, y(s) =

s ∫

s0

sin ϕ(s)ds. (9)

7

In the case of non-planar curves, if we have two single valued continuous func-

tions kF (s) and τF (s), s > 0, then there exists one and only one space curve,

determined but for its position in space, for which s is the arclength, kF (s)

the curvature, and τF (s) the torsion. It is the Fundamental theorem for space

curves [9]. The functions kF (s) and τF (s) provide the natural equations of the

space curve.

2.3 Curves of constant slope: the Lancret’s theorem

A space curve x(s) is a helix if the lines tangent to x make a constant angle

with a fixed direction in space (the helical axis) [8,9]. Denoting by a the unit

vector of this direction, a helix satisfies

t.a = cos α = constant . (10)

Differentiating Eq. (10) with respect to s gives a.n = 0. Therefore a lies in

the plane determined by the vectors t and b:

a = t cos α + b sin α . (11)

Differentiating Eq. (11) with respect to s, gives

0 = (kF cos α − τF sin α)n ,

or

kF τF

= tan α = constant. (12)

8

This result says that for curves of constant slope the ratio of curvature over

torsion is constant. Conversely, given a regular curve for which the equation

(12) is satisfied, it is possible to find [9] a constant angle α such that

n (kF cos α − τF sin α) = 0 ,

d

ds (t cos α + b sin α) = 0 ,

implying that the vector a = t cos α + b sin α is the unit vector along the axis.

Moreover, a.t = cos α=constant, so that the curve has constant slope. This

result can be expressed as:

A necessary and suficient condition for a space curve to be a curve of constant

slope (a helix) is that the ratio of curvature over torsion be constant. It is the

well known Lancret’s theorem, dated of 1802 and first proved by B. de Saint

Venant [9,37].

If a helical curve x(s) is projected onto the plane perpendicular to a, the

vector x1(s) representing this projection is given by

x1(s) = x − (x.a)a . (13)

It is possible to show [9] that the curvature k1 of the projected curve is given

by:

k1(s) = kF (s)

sin2 α . (14)

The shape of the planar curve obtained by projecting a helical curve onto the

plane perpendicular to its axis is used to characterize it. For example, the well

known circular helix projects a circle onto the plane perpendicular to its axis.

9

The spherical helix projects an arc of an epicycloid onto a plane perpendicular

to its axis [9]. The logarithmic spiral is the projection of a helical curve called

conical helix [9].

3 The static Kirchhoff equations

The statics and dynamics of long and thin elastic rods are governed by the

Kirchhoff rod model. In this model, the rod is divided in segments of infinites-

imal thickness to which the Newton’s second law for the linear and angular

momentum are applied. We derive a set of partial differential equations for

the averaged forces and torques on each cross section and for a triad of vectors

describing the shape of the rod. The set of PDE are completed with a linear

constitutive relation between torque and twist.

The central axis of the rod, hereafter called centerline, is represented by a

space curve x parametrized by the arclength s. A Frenet frame is defined for

this space curve as described in the previous section. For a physical filament

the use of a local basis, {d1, d2, d3}, to describe the rod has the advantage of

taking into account the twist deformation of the filament. This local basis is

defined such that d3 is the vector tangent to the centerline of the rod (d3 = t),

and d1 and d2 lie on the cross section plane. The local basis is related to the

Frenet frame {n, b, t} through

(d1 d2 d3) = (n b t)





























cos ξ − sin ξ 0

sin ξ cos ξ 0

0 0 1





























, (15)

10

where the angle ξ is the amount of twisting of the local basis with respect to

t.

In this paper, we are concerned with equilibrium solutions of the Kirchhoff

model, so our study departs from the static Kirchhoff equations [38]. In scaled

variables, for intrinsically straight isotropic rods, these equations are:

F ′ = 0 , (16)

M ′ = F × d3 , (17)

M = B(s) k1 d1 + B(s) k2 d2 + C(s) k3 d3 , (18)

the vectors F and M being the resultant force, and corresponding moment

with respect to the centerline of the rod, respectively, at a given cross section.

As in the previous section, s is the arclength of the rod and the prime ′ denotes

differentiation with respect to s. ki are the components of the twist vector,

k, that controls the variations of the director basis along the rod through the

relation

d ′

i = k × di , i = 1, 2, 3 . (19)

k1 and k2 are related to the curvature of the centerline of the rod (kF = √

k21 + k 2 2) and k3 is the twist density. B(s) and C(s) are the bending and

twisting coefficients of the rod, respectively. In the case of macroscopic fila-

ments the bending and twisting coefficients can be related to the cross section

radius and the Young’s and shear moduli of the rod. Writing the force F in

the director basis,

F = f1d1 + f2d2 + f3d3 , (20)

the equations (16–18) give the following differential equations for the compo-

11

nents of the force and twist vector:

f ′1 − f2 k3 + f3 k2 = 0 , (21)

f ′2 + f1 k3 − f3 k1 = 0 , (22)

f ′3 − f1 k2 + f2 k1 = 0 , (23)

(B(s) k1) ′ + (C(s) − B(s)) k2 k3 − f2 = 0 , (24)

(B(s) k2) ′ − (C(s) − B(s)) k1 k3 + f1 = 0 , (25)

(C(s) k3) ′ = 0 . (26)

The equation (26) shows that the component M3 = C(s) k3 of the moment

in the director basis (also called torsional moment), is constant along the

rod, consequently the twist density k3 is inversely proportional to the twisting

coefficient C(s)

k3(s) = M3 C(s)

. (27)

In order to look for helical solutions of the Eqs. (21–26) the components of

the twist vector k are expressed as follows:

k1 = kF (s) sin ξ , (28)

k2 = kF (s) cos ξ , (29)

k3 = ξ ′ + τF (s) , (30)

where kF (s) and τF (s) are the curvature and torsion, respectively, of the space

curve that defines the centerline of the rod and ξ is given by Eq. (15). If the

rod is homogeneous, the helical solution has constant kF and τF , and ξ ′ is

proved to be null [5].

Substituting Eqs. (28–30) in Eqs. (21–26), extracting f1 and f2 from Eqs. (25)

and (24), respectively, differentiating them with respect to s, and substituting

in Eqs. (21), (22) and (23), gives the following set of nonlinear differential

equations:

12

[M3 kF (s) − B(s) kF (s) τF (s)] ′ − (B(s) kF (s))

′ τF (s) = 0 , (31)

(B(s) kF (s)) ′′ + kF (s) τF (s)[M3 − B(s) τF (s)] − f3(s) kF (s) = 0 , (32)

(B(s) kF (s)) ′ kF (s) + f

′

3(s) = 0 . (33)

Appendix A presents the details of the derivation of Eqs. (31–33).

The Eqs. (31–33) for the curvature, kF , and torsion, τF do not depend on the

twisting coefficient, C(s). Therefore, the centerline of an inhomogeneous rod

does not depend on the twisting coefficient like in the case of homogeneous

rods (see, for example, Eqs. (13) and (14) of Ref. [13]).

Langer and Singer [39] have obtained a set of first-order ordinary differential

equations for the curvature and torsion of the centerline of a homogeneous

rod that contains terms proportional to k2 F

and τ 2 F . The Eqs. (31–33) have the

advantage of involving only terms linear in kF and τF .

4 Helical solutions of inhomogeneous rods

In order to find helical solutions for the static Kirchhoff equations, we apply

the Lancret’s theorem to the general equations (31–33). We first rewrite the

Lancret’s theorem in the form:

kF (s) = β τF (s) , (34)

with β 6= 0. From Eq. (12),

β ≡ tan α = Constant . (35)

13

Substituting Eq. (34) in Eq. (31) we obtain

τ ′ F

(M3 − B τF ) − 2 τF (B τF ) ′ = 0 . (36)

Substituting Eq. (34) in Eq. (32) and extracting f3, we obtain

f3 = (B τF )

′′

τF + τF (M3 − B τF ) . (37)

Differentiating f3 with respect to s and substituting in Eq. (33) we obtain the

following differential equation for τF :

(B τF ) ′′′

τF −

(B τF ) ′′τ ′

F

τ 2 F

+ (β2 + 1) τF (B τF ) ′ = 0 , (38)

where the Eq. (36) was used to simplify the above equation. One immediate

solution for this differential equation is

(B τF ) ′ = 0 , (39)

that substituted in Eq. (36) gives

τ ′ F

(M3 − BτF ) = 0 . (40)

For non-constant τF , the Eq. (40) gives the following solution for τF :

τF (s) = M3

B(s) . (41)

Substituting the Eqs. (39) and (41) in Eq. (37) we obtain that

f3(s) = 0 . (42)

14

Substituting Eq. (41) in (34) we obtain:

kF (s) = β M3

B(s) . (43)

Substituting Eq. (15) in Eq. (20), the force F becomes

F = (f1 cos ξ − f2 sin ξ) n + (f1 sin ξ + f2 cos ξ) b + f3 t , (44)

where {n, b, t} is the Frenet basis. Using the Eqs. (A.7) and (A.8) for f1 and

f2 (Appendix A), we obtain

F = −(B kF ) ′ n + kF [M3 − B τF ] b + f3 t , (45)

where f3, in the inhomogeneous case, must satisfy the Eq. (33).

Substituting the Eqs. (41–43) in the Eq. (45), and using Eq. (35), it follows

F = 0. Therefore, the helical solutions satisfying (39) are free standing.

Now, we prove that a circular helix cannot be a solution of the static Kirchhoff

equations for a rod with varying bending stiffness. If a helix is circular, k′ F

= 0

and τ ′ F

= 0, and from Eq. (31) we obtain:

2 kF τF B ′ = 0 . (46)

Since B′(s) 6= 0, Eq. (46) will be satisfied only if kF = 0 and/or τF = 0.

Therefore, it is not possible to have a circular helix as a solution for a rod

with varying bending coefficient.

The solutions for the curvature kF , Eq. (43), and the torsion τF , Eq. (41), can

be used to obtain the unit vectors of the Frenet frame (by integration of the

Eqs. (8)). From Eqs. (43), (35) and (11), we can obtain α and a once kF (0),

15

M3 and B(s) are given. By choosing the z-direction of the fixed cartesian basis

as the direction of the unit vector a, we can integrate t in order to obtain the

three-dimensional configuration of the centerline of the rod.

Figure 1 displays the helical solution of the static Kirchhoff equations for rods

with bending coefficients given by

Fig 1a: Ba(s) = 1 , (47)

Fig 1b: Bb(s) = 1 + 0.007 s , (48)

Fig 1c: Bc(s) = 1 + 0.1 sin(0.04s + 2) . (49)

The case of constant bending (47) produces the well known circular helix

displayed in Fig. 1a. Figs. 1b–1c show that non-constant bending coefficients

(Eqs. (48–49)) do not produce a circular helix.

The helical solutions displayed in Fig. 1 satisfy the Lancret’s theorem, Eq. (12).

The tridimensional helical configurations displayed in Fig. 1 were obtained by

integrating the Frenet-Serret equations (8) using the following initial condi-

tions for the Frenet frame: t(s = 0) = (0, sin α, cos α), n(s = 0) = (−1, 0, 0)

and b(s = 0) = (0, − cos α, sin α). This choice ensures that the z-axis is par-

allel to the direction of the helical axis, vector a. The centerline of the helical

rod is a space curve x(s) = (x(s), y(s), z(s)) that is obtained by integration

of the tangent vector t(s). We have taken the helical axis as the z-axis and

placed the initial position of the rod at x(0) = 1/k1(0), y(0) = 0 and z(0) = 0

(in scaled units), where k1(0) is the curvature of the planar curve at s = 0

obtained by projecting the space curve onto the plane perpendicular to the

helical axis (Eq. (14)). From Eq. (14) we have

k1(0) = kF (0)

sin2 α . (50)

16

Using the Eq. (35), it follows that

sin2 α = β2

1 + β2 . (51)

From Eq. (43), setting s = 0, we get

β = kF (0) B(0)

M3 . (52)

Substituting Eqs. (51) and (52) in Eq. (50), we obtain:

x(0) = 1

k1(0) =

kF (0) B 2(0)

M 23 + k 2 F (0)B2(0)

. (53)

kF (0) and M3 are free parameters that have been chosen so that the helical

solutions displayed in Fig. 1 have the same angle α. The parameters kF (0) =

0.24 and M3 = 0.05 give x(0) ≃ 4 for the helical solutions displayed in Figs.

1a and 1b, and the parameters kF (0) = 0.22 and M3 = 0.05 give x(0) ≃ 4.36

for the helical solution displayed in the Fig. 1c.

For short the projection of the space curve onto the plane perpendicular to the

helical axis will be called projected curve. As mentioned in Sec. II, the circle

is the projected curve of the most common type of helix, the circular helix.

Fig. 2 displays the projected curves related to the helical solutions displayed in

Fig. 1. Fig. 2a shows that the helical solution of the inhomogeneous rod with

constant bending coefficient projects a circle onto the plane perpendicular to

the helical axis.

If required, the natural equations for the projected curves displayed in Fig. 2

are easily obtained, for instance, by substitution of the solution (Eq. (43)) for

the curvature kF (s) of the helical rod into Eq. (14). The natural equation of

17

the projected curve is given by its curvature,

k1(s) = βM3 B(s)

sin−2 α , (54)

where β and sin−2 α can be obtained by Eqs. (51) and (52). Then, in the

Eq.(54), setting B(s) = Bi(s), i = a, b, c, as given in Eqs. (47–49), produces the

natural equation for the corresponding projected curve displayed in Fig. 2. The

helical rod displayed in Fig. 1b is a conical helix since the radius of curvature

of its projected curve (inverse of k1(s)) is a logarithmic spiral (1/k1(s) is a

linear function of s [9]).

From Eqs. (28–30), (27) and (41) we obtain the variation of the angle ξ between

the local basis, di, i = 1, 2, 3, and the Frenet frame, {n, b, t}:

ξ′ = k3(s) − τF (s) =

(

M3 C(s)

− M3

B(s)

)

= M3 B(s) − C(s)

B(s) C(s) . (55)

Eq. (55) shows that ξ′ 6= 0 for the general case of B(s) 6= C(s), i. e. helical

filaments corresponding to inhomogeneous rods are not twistless. The circular

helix is a helicoidal solution for the centerline of an inhomogeneous rod having

constant bending coefficient. We emphasize that the inhomogeneous rod is not

twistless in contrast with the homogeneous case where it has been proved that

ξ′ = 0 [5].

A homogeneous rod has B(s) and C(s) constant so that k3 = Constant (from

Eq. (26)). Since ξ′ has been proved to be null for a helical solution of a ho-

mogeneous rod (see reference [5]), Eq. (30) shows that the torsion τF must

be a constant. In order to satisfy the Lancret’s theorem (Eq. (12)) the curva-

ture kF of the helical solution must also be a constant. Therefore, the only

type of helical solution for a homogeneous rod is the circular helix, while an

18

inhomogeneous rod may present other types of helical structures.

5 Radius and Pitch of the helical solution

The radius R of a helix is defined as being the distance of the space curve to

its axis. The pitch P of a helix is defined as the height of one helical turn, i.e.,

the distance along the helical axis of the initial and final points of one helical

turn.

For a circular helix, R, P, kF and τF are constant, and it is easy to prove that

λ = (

√

R2 + P2/(4π2) )

−1

= √

k2 F

+ τ 2 F

. (56)

For other types of helix, it constitutes a very hard problem in differential

geometry to obtain the relation between the curvature kF and the torsion τF

with the radius R and the pitch P. We have seen in Sec. II that the definitions

of curvature and torsion involve the calculation of the modulus of the tangent

and normal vectors derivative with respect to the arclength of the rod. We also

saw that the Frenet-Serret differential equations for the Frenet frame depend

on the curvature and torsion. The difficulty of integration of the Frenet-Serret

equations for the general case where kF (s) and τF (s) are general functions of

s poses the problem of finding an analytical solution for the centerline of the

rod, thus the difficulty of relating non constant curvature and torsion with non

constant radius and pitch. Due to this difficulty we shall test the possibility

of generalizing the relation (56) to the present inhomogeneous case. In order

to do so, from the equation (56) we define:

gλ(s) ≡ (

√

R2(s) + P2(s)/(4π2) )

−1

− √

k2 F (s) + τ 2

F (s) , (57)

19

where R(s) and P(s) are the radius and the pitch of the helical structure as

function of s. In the case of a circular helix, from Eq. (56), gλ(s) = 0 for all s.

Since the z-axis is defined as being the axis of the helical solution we can

calculate the radius R(s) through: R(s) = √

x2(s) + y2(s), where x(s) and

y(s) are the x and y components of the vector position of the centerline of the

helical rod.

The pitch of the helix is the difference between the z-coordinate of the initial

and final positions of one helical turn. A helical turn can be defined such that

the projection of the vector position of the spatial curve along the xy-plane

(vector x1 of Eq. (13)), rotates of 2π around the z-axis.

Fig. 3 shows gλ(s) for the free standing helix of Fig. 1b. We see that gλ(s)

oscillates, its maximum amplitude being smaller than 0.006. For the helical

shape displayed in Fig. 1c we found that the maximum value of gλ(s) is smaller

than 0.008 (data not shown). While for a circular helix gλ = 0, for the free

standing helices displayed in Fig. 1b and Fig. 1c the function gλ oscillates

around zero with small amplitude.

The small amplitude of these oscillations suggests that the relations R(s) ≃

kF (s)[ √

k2 F (s) + τ 2

F (s)]−1 and P(s) ≃ 2πτF (s)[

√

k2 F (s) + τ 2

F (s)]−1, valid for cir-

cular helices, could be used to derive approximate functions for the radius and

the pitch of different types of helical structures, but the oscillatory behavior

indicates that these relations are not simple functions of the geometric features

of the helix.

20

6 Straight and planar inhomogeneous rods

Straight rods (kF = 0), and planar rods (kF 6= 0), have null torsion (τF = 0),

and constitute particular cases of helices. In both cases there is at least one

direction in space that makes a constant angle α = π/2 with the vector tangent

to the rod centerline.

The straight inhomogeneous rod is a solution of the static Kirchhoff equations

that has non-constant twist density (Eq. (27)), in contrast with the homoge-

neous case for which the twist density is constant.

The twisted planar ring (kF = Constant) is a solution of the static Kirchhoff

equations only if the bending coefficient can be written in the form:

B(s) = A0 cos(kF s) + B0 sin(kF s) + CI /k 2 F

, (58)

with A0, B0 and CI constant. If kF is function of s (instead of being a constant)

there exist no solutions for Eqs. (21–26). So, the existence of a planar solution

related to the general form of the components of the twist vector given by

equations (28–30) requires kF = Constant.

7 Helical structure with intrinsic curvature

The helical shape displayed in Fig. 1b resembles that exhibited by the tendrils

of some climbing plants. In these plants the younger parts have smaller cross

section diameter, giving rise to non-constant bending coefficient. The main

difference between the solution displayed in Fig. 1b and the tendrils of climbing

plants is that the solution in Fig. 1b was obtained for an intrinsically straight

21

rod while the tendrils have intrinsic curvature [5].

The tendrils of climbing plants are stable structures while the helical solution

displayed in Fig. 1b is not stable because the rod is intrinsically straight [30].

We shall show that a rod with intrinsic curvature may have a static solution

of the Kirchhoff equations similar to that displayed in Fig. 1b.

The intrinsic curvature of a rod is introduced in the Kirchhoff model through

the components of the twist vector, k(0), in the unstressed configuration of the

rod as

k (0) 1 = k

(0) F

(s) sin ξ , (59)

k (0) 2 = k

(0) F

(s) cos ξ , (60)

k (0) 3 = ξ

′ + τ (0) F

(s) , (61)

where k (0) F

(s) and τ (0) F

(s) are the curvature and torsion of the space curve that

represents the axis of the rod in its unstressed configuration, simply called

intrinsic curvature of the rod. We consider that the unstressed configuration

of the axis of the rod forms a helical space curve with the intrinsic curvature

satisfying

B(s) k (0) F

(s) = K0 , (62)

B(s) τ (0) F

(s) = T0 , (63)

where K0 and T0 are constant and B(s) is the bending coefficient of the rod.

The linear constitutive relation (Eq. (18)) becomes

M = B(s)(k1 − k (0) 1 )d1 + B(s)(k2 − k

(0) 2 )d2 + C(s)(k3 − k

(0) 3 )d3 , (64)

where C(s) is the twisting coefficient of the rod. The static Kirchhoff equations

for this case, Eqs. (16), (17) and Eq. (64), are given by

22

f ′1 − f2 k3 + f3 k2 = 0 , (65)

f ′2 + f1 k3 − f3 k1 = 0 , (66)

f ′3 − f1 k2 + f2 k1 = 0 , (67)

(B(s)(k1 − k (0) 1 ))

′ − B(s)(k2 − k (0) 2 )k3 + C(s)(k3 − k

(0) 3 ) − f2 = 0 , (68)

(B(s)(k2 − k (0) 2 ))

′ + B(s)(k1 − k (0) 1 )k3 − C(s)(k3 − k

(0) 3 ) + f1 = 0 , (69)

(C(s)(k3 − k (0) 3 ))

′ + B(s)(k (0) 1 k2 − k

(0) 2 k1) = 0 . (70)

The components of the twist vector are expressed as:

k1 = kF (s) sin χ , (71)

k2 = kF (s) cos χ , (72)

k3 = χ ′ + τF (s) . (73)

In order to obtain the simplest solution for the static Kirchhoff equations Eqs.

(65–70) with the intrinsic curvature given by Eqs. (59–61) and (62–63) we shall

look for a solution such that χ = ξ in Eqs. (71–73). This solution preserves

the intrinsic twist density of the helical structure. In this case, the Eq. (70)

becomes simply [C(s)(τF − τ (0) F

)]′ = 0 or

M3 = C(s)(τF − τ (0) F

) = Constant , (74)

and we obtain the following differential equations for the curvature kF (s), and

the torsion τF (s), of the rod:

[M3 kF − B τF (kF − k (0) F

)]′ − [B(kF − k (0) F

)]′ τF = 0 ,

[B(kF − k (0) F

)]′′ + τF [M3 kF − B τF (kF − k (0) F

)] − f3 kF = 0 ,

[B(kF − k (0) F

)]′ kF + f ′

3 = 0 ,

(75)

where we have omitted the dependence on s to simplify the notation. In order

to obtain a helical solution of these equations we apply the Lancret’s theorem,

23

Eq. (12), to the Eqs. (75). We obtain the following results:

f3(s) = 0 , (76)

[B(s)(kF (s) − k (0) F

(s))]′ = 0 ⇒ kF (s) − k (0) F

(s) = K

B(s) , (77)

[B(s)(τF (s) − τ (0) F

(s))]′ = 0 ⇒ τF (s) − τ (0) F

(s) = T

B(s) , (78)

where K and T are integration constants. From Eqs. (74) and Eq. (78) we

obtain

T = B(s)

C(s) M3 , (79)

so that the ratio B(s)/C(s) has to be constant.

From Eqs. (77), (78), (62), (63) and (12) we have

kF (s)

τF (s) =

K + K0 T + T0

= tan α . (80)

From Eqs. (68), (69), (77) and (78) it follows that

f1 = 0 ,

f2 = 0 .

(81)

Therefore, the helical solution given by Eqs. (76–78) (obtained imposing χ =

ξ) is a free standing helix (F = (f1, f2, f3) = 0).

It follows from Eqs. (76–78), (62) and (63) that the solutions for the curvature

kF (s), and the torsion τF (s), of the rod with helical intrinsic curvature are

similar to those of intrinsically straight rods, Eqs. (41–43). Therefore, rods

with intrinsic curvature and a non-constant bending coefficient given by Eq.

24

(48) (Eq. (49)) can have a three-dimensional configuration similar to that

displayed in Fig. 1b (Fig. 1c).

8 Conclusions

The existence of helical configurations for a rod with non-constant stiffness has

been investigated within the framework of the Kirchhoff rod model. Climbing

and spiralling solutions of planar rods have been studied by Holmes et. al. [40].

Here, we have shown that helical spiralling three-dimensional structures are

possible solutions of the static Kirchhoff equations for an inhomogeneous rod.

From the static Kirchhoff equations, we derived the set of differential equations

(31–33) for the curvature and the torsion of the centerline of a rod whose

bending coefficient is a function of the arclength s. We have shown that the

circular helix is the type of helical solution obtained when B(s) is constant,

independently of the rod being homogeneous or inhomogeneous.

Though the differential equations for the curvature and torsion are general,

we have obtained only the simplest helical solutions (Eqs. (39) and (41–43)),

obtained when the Lancret’s theorem is applied to the differential equations.

We show that these solutions are free standing and that the curvature and

torsion depend directly on the form of variation of the bending coefficient.

Figures 1b and 1c are examples of helical solutions of inhomogeneneous rods

whose bending coefficients are given by Eqs. (48) and (49). The helical struc-

ture displayed in Fig. 1b is a conical helix since the projected curve onto the

plane perpendicular to the helical axis is a logarithmic spiral, i. e., 1/k1(s) is

a linear function of s [9].

25

In the particular case of an inhomogeneous rod with the intrinsic curvature

defined by Eqs. (59–61) and (62–63), with B(s)/C(s) constant, we also ob-

tain the helical solutions displayed in Figs. 1b and 1c. The tendrils of some

climbing plants present a three-dimensional structure similar to that displayed

in Fig. 1b. In these plants, the cross-section diameter of the tendrils varies

along them, giving rise to non-constant bending coefficient, and the differen-

tial growth of the tendrils produces intrinsic curvature [5]. The bending and

twisting coefficients of a continuous filament with circular cross-section are

proportional to its moment of inertia I. It implies that B(s)/C(s) is constant

for an inhomogeneous rod. Therefore, the tendrils of climbing plants can be

well described by the Kirchhoff model for an inhomogeneous rod with a linear

variation of the bending stiffness.

Acknowledgements

This work was partially supported by the Brazilian agencies FAPESP, CNPq

and CAPES. The authors would like to thank Prof. Manfredo do Carmo for

valuable informations about the Lancret’s theorem.

A Appendix: The differential equations for the curvature and tor-

sion

Here, we shall derive the Eqs. (31–33). Substitution of Eqs. (28–30) into Eqs.

(21–26) gives:

f ′1 − f2 (ξ ′ + τF ) + f3 kF cos ξ = 0 , (A.1)

f ′2 + f1 (ξ ′ + τF ) − f3 kF sin ξ = 0 , (A.2)

26

f ′3 − f1 kF cos ξ + f2 kF sin ξ = 0 , (A.3)

(B(s) kF sin ξ) ′ + (C(s) − B(s)) kF cos ξ (ξ

′ + τF ) − f2 = 0 , (A.4)

(B(s) kF cos ξ) ′ − (C(s) − B(s)) kF sin ξ (ξ

′ + τF ) + f1 = 0 , (A.5)

(C(s) (ξ′ + τF )) ′ = 0 . (A.6)

First, we extract f1 and f2 from Eqs. (A.5) and (A.4), respectively:

f1 = −(B(s) kF ) ′ cos ξ + [M3 kF − B(s) kF τF ] sin ξ , (A.7)

f2 = (B(s) kF ) ′ sin ξ + [M3 kF − B(s) kF τF ] cos ξ , (A.8)

where M3 = C(s) (ξ ′ + τF ) is the torsional moment of the rod that is con-

stant by Eq. (A.6). Differentiating f1 and f2 with respect to s, substituting in

Eqs. (A.1) and (A.2), respectively, and using Eqs. (A.7) and (A.8), gives the

following equations:

{−(B(s) kF ) ′′ − τF [M3 kF − B(s) kF τF ] + f3 kF } cos ξ +

{[M3 kF − B(s) kF τF ] ′ − τF (B(s) kF )

′} sin ξ = 0 ,

(A.9)

{(B(s) kF ) ′′ + τF [M3 kF − B(s) kF τF ] − f3 kF } sin ξ +

{[M3 kF − B(s) kF τF ] ′ − τF (B(s) kF )

′} cos ξ = 0 .

(A.10)

Multiplying Eq. (A.9) (Eq. (A.10)) by sin ξ (cos ξ) and then adding the re-

sulting equations, we obtain the Eq. (31) for the curvature and torsion:

[M3 kF − B kF τF ] ′ − (B kF )

′τF = 0 . (A.11)

Multiplying Eq. (A.9) (Eq. (A.10)) by − cos ξ (+ sin ξ) and then adding the

resulting equations, we obtain the Eq. (32):

(B kF ) ′′ + kF τF (M3 − BτF ) − f3 kF = 0 . (A.12)

27

Finally, the Eq. (33) is obtained by substituting Eqs. (A.7) and (A.8) in Eq.

(A.3):

(B kF ) ′ + f ′3 = 0 . (A.13)

References

[1] T. Schlick, Curr. Opin. Struct. Biol. 5 (1995) 245. W. K. Olson, Curr. Opin.

Struct. Biol. 6 (1996) 242.

[2] C. W. Wolgemuth, T. R. Powers, R. E. Goldstein, Phys. Rev. Lett. 84 (2000)

1623.

[3] D. N. McIlroy, D. Zhang, Y. Kranov, M. Grant Norton, Appl. Phys. Lett. 79

(2001) 1540.

[4] A. Goriely, M. Tabor, Phys. Rev. Lett. 80 (1998) 1564; A. Goriely, M. Tabor,

Physicalia 20 (1998) 299.

[5] T. McMillen, A. Goriely, J. Nonlin. Sci. 12 (2002) 241.

[6] P. Pieranski, J. Baranska, A. Skjeltorp, Eur. J. Phys. 24 (2004) 613.

[7] M. Nizette, A. Goriely, J. Math. Phys. 40 (1999) 2830.

[8] M. P. do Carmo, Differential Geometry of Curves and Surfaces, Prentice Hall,

Inc. Englewood Cliffs, New Jersey, 1976.

[9] D. J. Struik, Lectures on Classical Differential Geometry, 2nd Edition, Addison-

Wesley, Cambridge, 1961.

[10] G. Kirchhoff, J. Reine Anglew. Math. 56 (1859) 285.

[11] E. H. Dill, Arch. Hist. Exact. Sci. 44 (1992) 2; B. D. Coleman, E. H. Dill, M.

Lembo, Z. Lu, I. Tobias, Arch. Rational Mech. Anal. 121 (1993) 339.

28

[12] G. H. M. van der Heijden, J. M. T. Thompson, Nonlinear Dynamics 21 (2000)

71.

[13] S. Neukirch, M. E. Henderson, J. Elasticity 68 (2002) 95.

[14] I. Tobias, D. Swigon, B.D. Coleman, Phys. Rev. E 61 (2000) 747; B.D. Coleman,

D. Swigon, I. Tobias, Phys. Rev. E 61 (2000) 759.

[15] Y. Sun, J. W. Leonard, Ocean. Eng. 25 (1997) 443; O. Gottlieb, N. C. Perkins,

ASME J. Appl. Mech. 66 (1999) 352.

[16] A. F. da Fonseca, D. S. Galvão, Phys. Rev. Lett. 92 (2004) art. no. 175502.

[17] A. Mielke, P. Holmes, Arch. Rational Mech. Anal. 101 (1988) 319.

[18] M. A. Davies, F. C. Moon, Chaos 3 (1993) 93.

[19] G. Domokos, P. Holmes, Int. J. Non-Linear Mech. 28 (1993) 677.

[20] G. Domokos, P. Holmes, B. Royce, J. Nonlinear Sci. 7 (1997) 281.

[21] P. Holmes, G. Domokos, J. Schmitt, I. Szeberényi, Comput. Methods Appl.

Mech. Engrg. 170 (1999) 175.

[22] A. F. da Fonseca, C. P. Malta, M. A. M. de Aguiar, Physica A 352 (2005) 547.

[23] A. F. da Fonseca, M. A. M. de Aguiar, Physica D 181 (2003) 53.

[24] A. F. Fonseca, M. A. M. de Aguiar, Phys. Rev. E 63 (2001) art. no. 016611.

[25] A. Goriely, T. McMillen, Phys. Rev. Lett. 88 (2002) art. no. 244301.

[26] A whip is a nonhomogenous thread with varying radius of cross section.

[27] H. Kashimoto, A. Shiraishi, J. Sound and Vibration 178 (1994) 395.

[28] J. M. T. Thompson, A. R. Champneys, Proc. R. Soc. London, Ser. A 452 (1996)

117.

29

[29] A. Goriely, M. Tabor, Nonlinear Dynamics 21 (2000) 101.

[30] A. Goriely, M. Tabor, Proc. R. Soc. London, Ser. A 453 (1997) 2583.

[31] A. Goriely, P. Shipman, Phys. Rev. E 61 (2000) 4508.

[32] G. H. M. van der Heijden, J. M. T. Thompson, Physica D 112 (1998) 201.

[33] A. Goriely, M. Nizette, M. Tabor, J. Nonlinear Sci. 11 (2001) 3.

[34] N. Chouäieb, PhD Thesis, École Polytechnique Fédérale de Lausanne, 2003.

[35] W. A. Kronert et al, J. Mol. Biol. 249 (1995) 111.

[36] V. Geetha, Int. J. Biol. Macromol. 19 (1996) 81.

[37] B. de Saint Venant, J. Ec. Polyt. 30 (1845) 26.

[38] The references in [11] can be seen for a complete derivation of the Kirchhoff

equations.

[39] J. Langer, D. A. Singer, SIAM Rev. 38 (1996) 605.

[40] P. Holmes, G. Domokos, G. Hek, J. Nonlinear Sci. 10 (2000) 477.

30

(a) (b) (c)

Fig. 1. Helical solutions of the Kirchhoff equations using the Lancret’s Theorem. (a)

circular helix solution for an inhomogeneous rod with constant bending coefficient

Ba = 1 (47); (b) and (c) Lancret helices for inhomogeneous rod with bending

coefficient given by Eqs. (48) and (49), respectively. The parameters, in scaled units,

are M3 = 0.05, Γ = 0.9, and the total length of the rod is L = 130. kF (0) = 0.24 for

the helical solutions displayed in panels (a) and (b), and kF (0) = 0.22 for panel (c).

(a) (b) (c)

Fig. 2. (a), (b) and (c) are projected curves of the helical solutions displayed in

Fig. 1a, Fig. 1b and Fig. 1c, respectively. We used Eq. (13) to obtain the projected

curves.

20 40 60 80 100 s

-0.006

-0.002

0.002

0.006

gλ

Fig. 3. gλ(s) for the free standing helix solution displayed in Fig. 1b.

31