盲鰻不是鰻魚

東港約從1998年開始捕盲鰻。盲鰻俗稱「龍筋」,不論是大火快炒韮黃,或是香酥鹽烤、蜜烤三杯、麻油快炒、薑片清湯,吃起來很有嚼勁,脆脆的,相當美味可口。盲鰻黏液清洗時不易清乾淨,有的人使用啤酒洗。
盲鰻（Myxinidae），名字裡有個“鰻”字，但並不是鰻魚。它最特別的技能，大概就是遭遇襲擊的時候，可以在0.4秒內釋放出大量黏液，充滿捕食者的嘴和鰓，讓對方難以呼吸。這樣一來，鯊魚就算再捨不得，也只能把獵物放生。
一隻盲鰻身上，大約有100個黏液腺（slime gland）負責釋放黏液。起初，它們分泌的黏液可能不到5毫升，連小小的一個茶匙都盛不滿。但在受到攻擊的0.4秒過後，幾毫升的黏液就可以變成滿滿一桶史萊姆，比原來的體積擴大了約10000倍。
當生存遇到壓力時，盲鰻會產生大量的黏液，人們對這個現象可能並不陌生。不過對盲鰻來說，“壓力”究竟是指怎樣一種狀況，又為什麼能讓黏液急劇擴張？
回答這個問題之前，我們先要知道盲鰻製造的是怎樣的黏液。這種可以迅速變大一萬倍來打擊敵人的黏液中，除了水和黏液素等常規成分之外，還有一種十分關鍵的要素——纖維
這些纖維是由盲鰻黏液腺裡的端絲細胞（gland thread cell）製造的，又細又長。平時，這些細絲成捆存在，就像一團一團精心纏好的紗線。而一旦盲鰻被捕食者攻擊，或者感覺到周遭的威脅，線團會在毫秒級的時間裡解散開來。當然，無事發生的時候，線團們也會悄悄地鬆動，只是速度要慢許多。 

Unravelling hagfish slime

Gaurav Chaudhary

, 

Randy H. Ewoldt

 and 

Jean-Luc Thiffeault

Published:16 January 2019https://doi.org/10.1098/rsif.2018.0710

Abstract

Hagfish slime is a unique predator defence material containing a network of long fibrous threads each ∼10 cm in length. Hagfish release the threads in a condensed coiled state known as skeins (∼100 µm), which must unravel within a fraction of a second to thwart a predator attack. Here we consider the hypothesis that viscous hydrodynamics can be responsible for this rapid unravelling, as opposed to chemical reaction kinetics alone. Our main conclusion is that, under reasonable physiological conditions, unravelling due to viscous drag can occur within a few hundred milliseconds, and is accelerated if the skein is pinned at a surface such as the mouth of a predator. We model a single skein unspooling as the fibre peels away due to viscous drag. We capture essential features by considering simplified cases of physiologically relevant flows and one-dimensional scenarios where the fibre is aligned with streamlines in either uniform or uniaxial extensional flow. The peeling resistance is modelled with a power-law dependence on peeling velocity. A dimensionless ratio of viscous drag to peeling resistance appears in the dynamical equations and determines the unraveling time scale. Our modelling approach is general and can be refined with future experimental measurements of peel strength for skein unravelling. It provides key insights into the unravelling process, offers potential answers to lingering questions about slime formation from threads and mucous vesicles, and will aid the growing interest in engineering similar bioinspired material systems.

1. Introduction

Marine organisms present numerous interesting examples of fluid–structure interactions that are necessary for their physiological functions such as feeding [1,2], motion [3], mechanosensing [4] and defence [5]. A rather remarkable and unusual example of fluid–structure interaction is the production of hagfish slime, also known as hagfish defence gel. The hagfish is an eel-shaped deep-sea creature that produces the slime when it is provoked [6]. Slime is formed from a small amount of biomaterial ejected from the hagfish’s slime glands into the surrounding water [7]. The biomaterial expands by a factor of 10 000 (by volume) into a mucus-like cohesive mass, which is hypothesized to choke predators and thus provide defence against attacks (figure 1a) [8]. Such defence mechanisms have been observed in several species of hagfish [8,9].

Figure 1. 

• Download figure
• Open in new tab
• Download PowerPoint

The secreted biomaterial has two main constituents—gland mucus cells and gland thread cells—responsible for the mucus and fibrous component of slime, respectively [6,10]. The plasma membranes of both kinds of cells shear off when secreted from the slime glands [10,11]. In the present study, we focus on the secreted thread cells, referred to as skeins from here on. Skeins possess a remarkable structure wherein a long filament (10–16 cm in length) is efficiently packed in canonical loops into a prolate spheroid (120–150 μm by 50–60 μm) [7,10] (figure 1b). When mixed with the surrounding water, the fibre (1–3 μm thread diameter) unravels from the skein (figure 1c) and forms a fibrous network with other threads and mucous vesicles. This process occurs on time scales of a predator attack (100–400 ms), as apparent from video evidence [8,12].

While several studies have revealed the mechanical and biochemical aspects [13–17] of slime, little is known about the mechanisms involved in its rapid deployment. More recently, efforts have been made to understand the mechano-chemical aspects of the mucus component, and its swelling and rupture [18,19], but such an approach is yet to be extended to the mechano-chemical processes in the unravelling of skeins. Newby [20] postulated that the fibre is coiled under a considerable pressure and the rupture of the cell membrane allows the fibre to uncoil. However, later studies [12,21,22] have shown that convective mixing is essential for the production of fibres and slime. More recently, Bernards et al. [11] experimentally demonstrated that Pacific hagfish skeins can unravel even in the absence of flow, potentially due to chemical release of the adhesives holding the fibre together, but the time scales observed in their work are orders of magnitude larger than physiological time scales during the attack. Therefore, the key question about the fast time scales involved in this process remains to be answered. Deeper insights into the remarkable process of slime formation will aid the development of bioinspired material systems with novel functionality, such as materials with fast autonomous expansion and deployment. Motivated by the aforementioned experimental studies, our objective in this paper is to investigate the role of viscous hydrodynamics in skein unravelling via a simple physical model, and thus supply a qualitative understanding of the unravelling process.

The key question we answer here is whether the viscous hydrodynamic unravelling alone can account for the fast unravelling time scales that are observed in physiological scenarios. We hypothesize that suction feeding in marine predators creates sufficient hydrodynamic stresses to aid in the unravelling of skeins and set up the slime network. We develop fundamental insight by considering only the simplest flow fields—uniform flow and extensional flow. Our modelling framework, however, generalizes to complex flow fields that occur in physiological conditions.

In §2, we present a simple qualitative experiment demonstrating the force-induced unravelling of a hagfish skein. This motivates the model paradigm that follows. Section 3 outlines the problem statement, and we derive the general governing equations. In §4, the equations are solved for skein unravelling in simple flows under different physically relevant scenarios. In §5, we discuss the results in more detail, including the influence of constitutive model parameters for the peel strength, and comment on the qualitative comparisons between the experimental studies and theoretical work.

2. Unravelling experiment

To motivate the mathematical modelling, we perform a simple experiment demonstrating the force-induced unravelling of thread from a skein (figure 2; see also electronic supplementary material, video). A skein, obtained from Atlantic hagfish, is held in place by weak interactions with the substrate, and a force is applied to the dangling end using a syringe tip that naturally sticks to the filament. Figure 2 shows the unravelling skein at different time frames. Frame 1 shows the unforced and stable configuration, with no unravelling. Unravelling occurs only when a force is applied from frame 2 onwards. There are events when the thread peels away in clumps, but the orderly unravelling recovers quickly. A minimum peeling force seems required to unravel the thread from the skein. A simple estimate of the minimum peeling force based on weak adhesion (van der Waals interaction) between unravelling fibre and skein gives an estimate of 0.1 μN (see electronic supplementary material, section II).

Figure 2. 

• Download figure
• Open in new tab
• Download PowerPoint

3. Problem formulation

To determine if viscous hydrodynamic forces can account for fast skein unravelling, we consider a model of an inextensible slender thread unravelling from a spherical skein. The thread unravels and separates from the skein in response to a local force due to a viscous fluid flow surrounding the connected thread and skein. A schematic is shown in figure 3. Here 𝑥(𝑠,𝑡)$\mathit{x}(s,t)$ is the Eulerian (laboratory) coordinate of the centreline of the filament as a function of the Lagrangian (material) thread arclength s, 0≤𝑠≤𝐿(𝑡)$0\le s\le L(t)$, with 𝐿(𝑡)$L(t)$ the time-dependent unravelled thread length. The thread is peeling from the skein at the Eulerian point 𝑥(𝐿(𝑡),𝑡)=𝑋(𝑡)$\mathit{x}(L(t),t)=\mathit{X}(t)$, which may depend on time if the skein is allowed to move.

Figure 3. 

• Download figure
• Open in new tab
• Download PowerPoint

3.1. Hydrodynamic force balance

We assume inertial effects, filament self-interactions, and external Brownian and gravitational forces to be negligible. The fluid dynamics in this situation are described by the Stokes equations. For the most general case of a thread in viscous flow, a local balance of filament forces and viscous forces (using local drag theory for a slender filament) is given by Tornberg & Shelley [23]

8𝜋𝜇𝛿(𝑥𝑡−𝑢(𝑥,𝑡))=−((1+2𝛿)𝕀+(1−2𝛿)𝑠̂ 𝑠̂ )⋅𝑓.$$8\pi \mu \hspace{0.17em}\delta \hspace{0.17em}({\mathit{x}}_t-\mathit{u}(\mathit{x},t))=-((1+2\delta )\hspace{0.17em}\mathbb{I}+(1-2\delta )\hspace{0.17em}\hat{\mathit{s}}\hat{\mathit{s}})\cdot \mathit{f}.$$

3.1

Here the tangent to the thread is 𝑠̂ $\hat{\mathit{s}}$, the dynamic viscosity is μ and

𝛿=−1log(𝜀2e)>0,with𝜀=𝑟𝐿,$$\delta =-\frac{1}{\log ({\epsilon }^2\text{e})}>0,\text{with}\hspace{0.17em}\epsilon =\frac{r}{L},$$

3.2

are the slenderness parameter and thread aspect ratio, respectively, with r the thread radius.

The internal net force per unit length, f, of an inextensible filament is expressed by the Euler–Bernoulli bending theory for an elastic beam, and has both tensile and bending components,

𝑓(𝑠)=−(𝑇𝑥𝑠)𝑠+𝐸𝑥𝑠𝑠𝑠𝑠,|𝑥𝑠|=1.$$\mathit{f}(s)=-{(T\hspace{0.17em}{\mathit{x}}_s)}_s+E\hspace{0.17em}{\mathit{x}}_{ssss},|{\mathit{x}}_s|=1.$$

3.3

Here E is the bending modulus of the thread, 𝑇(𝑠,𝑡)$T(s,t)$ is the tension in the filament, and each subscript s denotes one derivative, e.g. xs = ∂x/∂s. The inextensibility condition is |xs| = 1, so s and distance along the thread must always coincide.

In the spirit of rheology, we consider the response to simple flows to isolate key features of the complex behaviour, obtain analytical results and gain an understanding of the unravelling process. We only consider cases with zero curvature, xss = 0, immersed in one-dimensional flow fields, with the thread aligned with the flow streamlines. Equation (3.1) then reduces to a one-dimensional statement that the component of internal net filament force per unit length f(s) along the streamline (taken as the x direction) is equal to the local viscous drag per unit length,

4𝜋𝜇𝛿(𝑥𝑡−𝑢(𝑥,𝑡))=−𝑓.$$4\pi \mu \hspace{0.17em}\delta \hspace{0.17em}(x_t-u(x,t))=-f.$$

3.4

Then the one-dimensional form of equation (3.3) with xssss = 0 and the inextensibility condition xs = 1 gives f = −Ts, so that

𝑇𝑠=4𝜋𝜇𝛿(𝑥𝑡−𝑢(𝑥,𝑡)).$$T_s=4\pi \mu \hspace{0.17em}\delta \hspace{0.17em}(x_t-u(x,t)).$$

3.5

With xs = 1 and 𝑥(𝐿,𝑡)=𝑋(𝑡)$x(L,t)=X(t)$ we have 𝑥=𝑋−𝐿+𝑠$x=X-L+s$, where X is the skein position, and thus 𝑥𝑡=𝑋˙−𝐿˙$x_t=\dot{X}-\dot{L}$. We integrate (3.5) from s = 0 to 𝐿$L$ to find

𝑇|𝑠=𝐿−𝑇|𝑠=0=4𝜋𝜇𝐿𝛿(𝑋˙−𝐿˙−1𝐿∫𝐿0𝑢(𝑋−𝐿+𝑠,𝑡)d𝑠).$$T{|}_{s=L}-T{|}_{s=0}=4\pi \mu L\hspace{0.17em}\delta \hspace{0.17em}(\dot{X}-\dot{L}-\frac{1}{L}{\int }_0^{L}u(X-L+s,t)\hspace{0.17em}\text{d}s).$$

3.6

We then change the integration variable to 𝑥=𝑋−𝐿+𝑠$x=X-L+s$, and finally obtain

𝑇𝐿−𝑇0=−4𝜋𝜇𝐿𝛿(𝐿˙−𝑋˙+𝑢¯(𝐿,𝑋,𝑡)),$$T_L-T_0=-4\pi \mu L\hspace{0.17em}\delta \hspace{0.17em}(\dot{L}-\dot{X}+\overline{u}(L,X,t)),$$

3.7

where 𝑇𝐿=𝑇|𝑠=𝐿$T_L=T{|}_{s=L}$, T0 = T|s=0 and 𝑢¯(𝐿,𝑋,𝑡)$\overline{u}(L,X,t)$ is the average velocity on the filament,

𝑢¯(𝐿,𝑋,𝑡):=1𝐿∫𝑋𝑋−𝐿𝑢(𝑥,𝑡)d𝑥.$$\overline{u}(L,X,t):=\frac{1}{L}{\int }_{X-L}^{X}u(x,t)\hspace{0.17em}\text{d}x.$$

3.8

Equation (3.7) expresses the balance between the tension forces at the end of the thread and the drag force on the thread. We shall use this equation to derive a peeling formula for different thread–skein configurations in §4. But first, we need to examine how the thread will peel from the skein to unravel.

3.2. Unravelling from the skein

The relationship between R and 𝐿$L$, respectively, the radius of the spherical skein and the length of the unravelled thread, is described by volume conservation

dd𝑡(43𝜋𝜂𝑅3+𝜋𝑟2𝐿)=0⟹𝐿˙=−4𝜂𝑅2𝑅˙/𝑟2.$$\frac{\text{d}}{\text{d}t}(\frac{4}{3}\pi \eta R^3+\pi r^{2}L)=0\hspace{0.17em}⟹\hspace{0.17em}\dot{L}=-4\eta R^2\dot{R}/r^2.$$

3.9

Here r is the thread radius and 0 < η ≤ 1 is the packing fraction of thread into the spherical skein, assumed independent of R. (In this section, we keep the packing fraction as a variable, but in all later numerical simulations we take η = 1, since the skein is fairly tightly packed.) Explicitly, we have

𝑅3=𝑅30−34(𝐿−𝐿0)𝑟2/𝜂$$R^3=R_0^3-\frac{3}{4}(L-L_0)r^2/\eta$$

3.10

with R0 the initial skein radius and 𝐿0$L_0$ the initial unravelled length. A convenient way of relating R and 𝐿$L$ is

𝑅=𝑅0(𝐿max−𝐿𝐿max−𝐿0)1/3and𝐿max:=𝐿0+43𝜂𝑅30/𝑟2,$$R=R_0(\frac{L_{\max }-L}{L_{\max }-L_0}{)}^{1/3}\text{and}\hspace{0.17em}{L}_{\max }:=L_0+\frac{4}{3}\eta R_0^3/r^2,$$

3.11

where 𝐿max$L_{\max }$ is the total length of thread that can be extracted and 𝐿0$L_0$ is the initial unravelled length.

Next, we use a modified form of the work-energy theorem [24] to describe the unravelling dynamics,

𝐸˙total=(𝑇𝐿−𝐹P(𝑉))𝑉,𝑉=𝐿˙,$${\dot{E}}_{\mathrm{total}}=(T_L-F_{\mathrm{P}}(V))\hspace{0.17em}V,V=\dot{L},$$

3.12

where 𝐸˙total${\dot{E}}_{\mathrm{total}}$ is the rate of change in total energy of the system, 𝑇𝐿$T_L$ is the net force (given by equation (3.7)) drawing out the thread at a peeling velocity V, and FP(V) is a velocity-dependent peeling force acting at the peeling site. Neglecting the inertia and changes to the elastic energy of the peeling thread gives

𝑇𝐿=𝐹P(𝑉),𝑉=𝐿˙.$$T_L=F_{\mathrm{P}}(V),V=\dot{L}.$$

3.13

A natural dimensionless quantity that will determine the dynamics of the unravelling process is given by the ratio of the net viscous drag force on the thread and the resisting peel force, each of which depends on a characteristic velocity U,

℘:=𝐹D(𝑈)𝐹P(𝑈).$$\mathrm{\wp }:=\frac{F_{\mathrm{D}}(U)}{F_{\mathrm{P}}(U)}.$$

3.14

The functional form of the peeling force, FP(V), in general, is dependent on parameters such as the chemistry of peeling surfaces, velocity of peeling, etc. In the absence of a known functional form for hagfish thread peeling, we use a simple constitutive form of peeling force that includes a wide range of behaviour, given by

𝐹P(𝑉)=𝛼𝑉𝑚,0≤𝑚≤1,$$F_{\mathrm{P}}(V)=\alpha V^m,0\le m\le 1,$$

3.15

for constant α > 0 and m. Such a power-law form of peeling force has been observed in several engineered and biological systems [25–29]. Several other parametric forms of velocity-dependent peeling force exist that are functionally more complex [30,31]. However, to obtain simple and insightful solutions, we use the power-law form defined above. The form (3.15) allows for the limiting case m = 0, a constant peeling force, e.g. to simply counteract van der Waals attractions at the peel site.

For m > 0, we can rearrange equation (3.13) for the velocity, 𝑉=𝐿˙=(𝑇𝐿/𝛼)1/𝑚$V=\dot{L}={(T_L/\alpha )}^{1/m}$. Using (3.9), we can then obtain a solution for the case where the tension at the peeling point, 𝑇𝐿$T_L$, is constant,

43(𝑅30−𝑅3)=(𝑇𝐿𝛼)1/𝑚𝑟2𝑡𝜂,$$\frac{4}{3}(R_0^3-R^3)=(\frac{T_L}{\alpha }{)}^{1/m}\frac{r^{2}t}{\eta },$$

3.16

where R0 = R(0). From (3.16), we can easily extract the ‘depletion time’ or ‘full-unravelling time’ 𝑡dep$t_{\mathrm{dep}}$ by setting R = 0,

𝑡dep=4𝜂𝑅303𝑟2(𝑇𝐿𝛼)−1/𝑚.$$t_{\mathrm{dep}}=\frac{4\eta R_0^3}{3r^2}(\frac{T_L}{\alpha }{)}^{-1/m}.$$

3.17

In the next section, we compute this time scale when the skeins are subjected to different hydrodynamic flow scenarios, which cause different time histories of tension, 𝑇𝐿(𝑡)$T_L(t)$.

4. Skein in one-dimensional flow

Having described the unravelling dynamics in §3.2 for the case of constant tension, 𝑇𝐿$T_L$, we now consider a skein in a hydrodynamic flow where generally 𝑇𝐿$T_L$ varies in time as the thread–skein geometry changes during unravelling. In physiological scenarios the flow can arise from the hagfish–predator motion, or the suction feeding of the predator, or a combination of both. To simplify the problem we assume an incompressible flow of the form

𝑢(𝑥,𝑦,𝑡)=(𝑢(𝑥,𝑡),−𝑦𝑢𝑥(𝑥,𝑡)).$$\mathit{u}(x,y,t)=(u(x,t),-y{u}_x(x,t)).$$

4.1

The thread will be assumed to lie along the x-axis. We solve for the depletion time for four relevant cases: pinned thread in uniform flow (§4.1); pinned skein in uniform flow (§4.2); free skein and thread in extensional flow (§4.3); and free skein splitting into two smaller skeins in extensional flow (§4.4).

4.1. Pinned thread

The simplest case to consider is the thread pinned at s = 0 in figure 3, with a uniform flow to the right, 𝑢(𝑥,𝑡)=𝑈$u(x,t)=U$. This situation can arise in a controlled experiment if the thread is pinned down, or in the physiological unravelling process if the end of the thread is caught in the network of other threads, or stuck on the mouth of a predator.

The tension in the thread at 𝑠=𝐿$s=L$ balances the Stokes drag on the skein of radius R, 𝑇𝐿=𝑇(𝐿(𝑡),𝑡)=6𝜋𝜇𝑅(𝑢(𝐿,𝑡)−𝐿˙)$T_L=T(L(t),t)=6\pi \mu R\hspace{0.17em}(u(L,t)-\dot{L})$. Using (3.13) and (3.15), we obtain the governing equation for unravelling as

(𝐿˙)𝑚=6𝜋𝜇𝛼−1𝑅(𝐿)(𝑢(𝐿,𝑡)−𝐿˙).$${(\dot{L})}^m=6\pi \mu {\alpha }^{-1}R(L)\hspace{0.17em}(u(L,t)-\dot{L}).$$

4.2

From (3.9), since 𝐿˙>0$\dot{L}>0$ (the thread never ‘re-spools’), the unspooling speed satisfies 𝐿˙≤𝑢(𝐿,𝑡)$\dot{L}\le u(L,t)$, i.e. the thread cannot unspool faster than the ambient flow speed. The radius 𝑅(𝐿)$R(L)$ is given by (3.10).

We non-dimensionalize (4.2) using a characteristic length scale R0 and flow speed U, which gives

(𝐿˙∗)𝑚=℘𝑅∗(𝐿∗)(𝑢∗(𝐿∗,𝑡∗)−𝐿˙∗),$${({\dot{L}}^{\ast })}^m=\mathrm{\wp }\hspace{0.17em}{R}^{\ast }(L^{\ast })\hspace{0.17em}(u^{\ast }(L^{\ast },t^{\ast })-{\dot{L}}^{\ast }),$$

4.3

where 𝐿˙∗=𝐿˙/𝑈${\dot{L}}^{\ast }=\dot{L}/U$, 𝑅∗=𝑅/𝑅0$R^{\ast }=R/R_0$ and 𝑢∗(𝐿∗,𝑡∗)=𝑢(𝐿,𝑡)/𝑈$u^{\ast }(L^{\ast },t^{\ast })=u(L,t)/U$ are the non-dimensional unravelling rate, skein radius and flow rate, respectively. The non-dimensional time scale naturally results from these choices as 𝑡∗=𝑡/(𝑅0/𝑈)$t^{\ast }=t/(R_0/U)$. The dimensionless quantity ℘$\mathrm{\wp }$ on the right-hand side of (4.3) is given by

℘=6𝜋𝜇𝑅0𝑈𝛼𝑈𝑚=6𝜋𝜇𝑅0𝑈1−𝑚𝛼−1.$$\mathrm{\wp }=\frac{6\pi \mu R_0\hspace{0.17em}U}{\alpha U^m}=6\pi \mu R_0\hspace{0.17em}{U}^{1-m}{\alpha }^{-1}.$$

4.4

This is the ratio of characteristic drag to peeling force, as defined in (3.14). If ℘$\mathrm{\wp }$ is large (e.g. zero resistance to peeling), then (4.2) implies 𝐿˙≈𝑢(𝐿,𝑡)$\dot{L}\approx u(L,t)$, that is, in this drag-dominated limit, the drag force so easily unravels the skein that it advects with the local flow velocity. In the opposite limit of small ℘$\mathrm{\wp }$, we get 𝐿˙≈0$\dot{L}\approx 0$ and the skein cannot unravel. Hence, we require ℘ ≫ 1 for a fast unravel time.

To achieve the criterion ℘ ≫ 1, at a flow of speed U = 1 m s−1 and a skein of initial radius R0 = 50 μm, we require the peeling resistance at this velocity to satisfy FP(1 m s−1) ≪ 1.4 × 10−6 N. The estimated van der Waals peeling force is much lower than this threshold, FvdW ∼ 0.1 μN. At such a flow speed a skein containing 16.7 cm of thread (an upper bound physiological value) will unravel affinely (kinematically matching the flow speed) in roughly 167 ms. This lower bound estimate is commensurate with the rapidity with which hagfish slime is created (100–400 ms).

In figure 4, we show a numerical solution of (4.2) with a uniform flow for some typical physical parameter values, and assuming a moderately large force ratio ℘ = 10. (Equation (4.2) is an implicit relation for 𝐿˙$\dot{L}$ which must be solved numerically at every time step; it is a differential--algebraic equation rather than a simple ODE [30].) For these parameters, the kinematic lower bound on the depletion time is 𝐿max/𝑈≈167$L_{\max }/U\approx 167$ ms, and the numerical value is 𝑡dep≈194$t_{\mathrm{dep}}\approx 194$ ms.

Figure 4. 

• Download figure
• Open in new tab
• Download PowerPoint

There is a mathematical oddity where the skein might not get depleted in finite time, depending on the exponent m. To see this, consider a skein close to depletion, 𝐿=𝐿max−𝑈𝜏$L=L_{\max }-U\tau$, where τ > 0 is small. The equation for τ is

(−𝜏˙)𝑚=℘(𝑈𝜏𝐿max−𝐿0)1/3(1+𝜏˙),𝜏˙<0.$${(-\dot{\tau })}^m=\mathrm{\wp }(\frac{U\tau }{L_{\max }-L_0}{)}^{1/3}\hspace{0.17em}(1+\dot{\tau }),\dot{\tau }<0.$$

4.5

Since τ is small and we expect the thread to be drawn out slowly as it is almost exhausted, we take 1+𝜏˙≈1$1+\dot{\tau }\approx 1$. Hence, we have the approximate form

(−𝜏˙)𝑚≈C𝑚𝜏1/3,C𝑚:=℘(𝑈(𝐿max−𝐿0))1/3$${(-\dot{\tau })}^m\approx {\text{C}}^m\hspace{0.17em}{\tau }^{1/3},{\text{C}}^m:=\mathrm{\wp }(\frac{U}{(L_{\max }-L_0)}{)}^{1/3}$$

4.6

for some constant C > 0, with solution

𝜏(𝑡)≈[𝜏1−1/3𝑚0−(1−13𝑚)C𝑡]3𝑚/(3𝑚−1).$$\tau (t)\approx {[{\tau }_0^{1-1/3m}-(1-\frac{1}{3m})\text{C}\hspace{0.17em}t]}^{3m/(3m-1)}.$$

4.7

The behaviour of this solution as the skein is almost depleted depends on m. For m > 1/3, the exponent 3m/(3m − 1) in (4.7) is greater than 1, so 𝜏(𝑡)→0$\tau (t)\to 0$ as 𝑡$t$ approaches the depletion time, with 𝜏′(𝑡dep)=0${\tau }^{\prime }(t_{\mathrm{dep}})=0$ so that 𝐿(𝑡)$L(t)$ has slope zero when the skein is depleted (as can be seen at the very end in figure 4). We can thus rewrite (4.7) as

𝜏(𝑡)≈[(1−13𝑚)C(𝑡dep−𝑡)]3𝑚/(3𝑚−1),𝑚>13,𝑡↗𝑡dep.$$\begin{array}{r}\tau (t)\approx {[(1-\frac{1}{3m})\text{C}\hspace{0.17em}(t_{\mathrm{dep}}-t)]}^{3m/(3m-1)},m>\frac{1}{3},\hspace{0.17em}t\nearrow t_{\mathrm{dep}}.\end{array}$$

4.8

For m < 1/3, the exponent 3m/(3m − 1) is negative, but the factor 1 − 1/3m inside the brackets is also negative, so that 𝜏(𝑡)$\tau (t)$ asymptotes to zero as 𝑡→∞$t\to \mathrm{\infty }$ and the skein never gets fully depleted. In that case, we write (4.7) as

𝜏(𝑡)≈[(13𝑚−1)C𝑡]−3𝑚/(1−3𝑚),𝑚<13,𝑡→∞.$$\tau (t)\approx {\left[(\frac{1}{3m}-1)\text{C}t\right]}^{-3m/(1-3m)},m<\frac{1}{3},\hspace{0.17em}t\to \mathrm{\infty }.$$

4.9

Physically, for m < 1/3 the drag force (∼τ1/3) is decreasing faster than the peeling force (∼𝜏˙𝑚$\sim {\dot{\tau }}^m$).

In practice, it is difficult to see the difference between 𝑚≶1/3$m\lessgtr 1/3$ numerically. The thread appears to get depleted even for m < 1/3 because of limited numerical precision as 𝐿$L$ approaches 𝐿max$L_{\max }$. The symptom of a problem is that the depletion time starts depending on the numerical resolution for m < 1/3. Of course, the skeins in the hagfish slime do not need to get fully depleted to create the gel, so a power m < 1/3 is still applicable. When comparing the different flow scenarios we will explore a range of m and define an ‘effective deployment’ time 𝑡dep,50%$t_{\mathrm{dep},50\text{\%}}$, when 50% of the thread length is unravelled.

4.2. Pinned skein

When the skein is pinned and the thread is free at the other end, the tension arises from hydrodynamic drag on the thread. Such a scenario can arise if the skein is arrested in the network of other fibres or in a mucus network.

Consider a free thread ending at s = 0 and a pinned skein at 𝑠=𝐿(𝑡)$s=L(t)$ (figure 3), so that the Eulerian skein position X is fixed and is thus not a function of time. Unlike the pinned thread case in §4.1, where a shrinking skein led to a decreasing drag, here the tension increases with time as the extended thread provides more drag.

We formulate the problem by imposing boundary conditions at the free end, 𝑇0=𝑇(0,𝑡)=0$T_0=T(0,t)=0$, and pinned end, 𝑥(𝐿(𝑡),𝑡)=𝑋$x(L(t),t)=X$. From (3.7) with 𝑇0=𝑋˙=0$T_0=\dot{X}=0$ and 𝑇𝐿=𝛼(𝐿˙)𝑚$T_L=\alpha \hspace{0.17em}{(\dot{L})}^m$, the equation for the growth of the thread is

(𝐿˙)𝑚=−4𝜋𝜇𝛼−1𝐿𝛿(𝐿)(𝐿˙+𝑢¯(𝐿,𝑋,𝑡)).$${(\dot{L})}^m=-4\pi \mu \hspace{0.17em}{\alpha }^{-1}L\hspace{0.17em}\delta (L)\hspace{0.17em}(\dot{L}+\overline{u}(L,X,t)).$$

4.10

The slenderness parameter δ depends on 𝐿$L$ through its definition (3.2). Because the thread extends to the left in figure 3, we must have 𝑢¯(𝐿,𝑋,𝑡)<0$\overline{u}(L,X,t)<0$ to avoid unphysical respooling. The pinned thread equation (4.2) and the pinned skein equation (4.10) have a very similar form, though the drag in the former (∼𝑅(𝐿)$\sim R(L)$) decreases with 𝐿$L$ and that in the latter (∼𝐿𝛿(𝐿)$\sim L\delta (L)$) increases with 𝐿$L$.

Using a characteristic velocity U and a characteristic length scale 𝐿0$L_0$, we obtain the non-dimensional form of (4.10) as

(𝐿˙∗)𝑚=−℘𝐿∗𝛿(𝐿∗)(𝑢¯∗(𝐿∗,𝑋∗,𝑡∗)+𝐿˙∗),$${({\dot{L}}^{\ast })}^m=-\mathrm{\wp }{L}^{\ast }\delta (L^{\ast })({\overline{u}}^{\ast }(L^{\ast },X^{\ast },t^{\ast })+{\dot{L}}^{\ast }),$$

4.11

where 𝐿˙∗=𝐿˙/𝑈${\dot{L}}^{\ast }=\dot{L}/U$, 𝐿∗=𝐿/𝐿0$L^{\ast }=L/L_0$ and 𝑢∗(𝐿∗,𝑡∗)=𝑢(𝐿,𝑡)/𝑈$u^{\ast }(L^{\ast },t^{\ast })=u(L,t)/U$ are the non-dimensional unravelling rate, unravelled length and flow rate, respectively. The natural dimensionless force ratio (3.14) is

℘=4𝜋𝜇𝐿0𝑈1−𝑚𝛼−1.$$\mathrm{\wp }=4\pi \mu \hspace{0.17em}{L}_0\hspace{0.17em}{U}^{1-m}\hspace{0.17em}{\alpha }^{-1}.$$

4.12

This differs from ℘$\mathrm{\wp }$ in (4.4) by replacing R0 with 𝐿0$L_0$. It is sensible in this pinned skein case to use the initial thread length 𝐿0$L_0$, since drag on the thread controls the unravelling rate.

Figure 5 shows a numerical solution of (4.10) for our reference parameter values using a constant velocity field, 𝑢(𝐿,𝑋,𝑡)=−𝑈=−1$u(L,X,t)=-U=-1$ m s−1. As before, the lower bound on the depletion time is 𝐿max/𝑈≈167$L_{\max }/U\approx 167$ ms, and now the numerical value is 𝑡dep≈226$t_{\mathrm{dep}}\approx 226$ ms. This is slower than what we observed in the pinned thread case (𝑡dep≈194$t_{\mathrm{dep}}\approx 194$ ms), but here we are using the much smaller force ratio ℘ = 1/2. This shows that the pinned skein case can unravel almost as fast as the lower bound for a much smaller value of ℘$\mathrm{\wp }$, since the drag on the thread increases with 𝐿$L$, as reflected by the accelerating speed 𝐿˙$\dot{L}$ in figure 5. This is in contrast to the deceleration in figure 4 for the pinned thread, where drag decreases as the skein radius diminishes.

Figure 5. 

• Download figure
• Open in new tab
• Download PowerPoint

4.3. Free skein and thread

In the previous two cases, we took either the thread or skein to be pinned; here we consider the case where neither is pinned, and both are free to move with the flow. This scenario is possible if both the thread and the skein are not stuck to the existing fibrous network, or at the beginning of the slime formation when none of the skeins have unravelled to a significant extent.

The force at the peeling point 𝑥(𝐿(𝑡),𝑡)=𝑋(𝑡)$x(L(t),t)=X(t)$ is then determined by the balance of two forces: Stokes drag on the spherical skein, 𝐹1=6𝜋𝜇𝑅(𝑢(𝑋,𝑡)−𝑋˙)$F_1=6\pi \mu R\hspace{0.17em}(u(X,t)-\dot{X})$, and drag on the thread, 𝐹2=−4𝜋𝜇𝐿𝛿(𝐿˙−𝑋˙+𝑢¯(𝐿,𝑋,𝑡))$F_2=-4\pi \mu L\delta \hspace{0.17em}(\dot{L}-\dot{X}+\overline{u}(L,X,t))$. The latter is obtained from (3.7) with 𝑇𝐿=𝐹2$T_L=F_2$ and T0 = 0. Since both the skein and thread are free and we have neglected inertia, F1 + F2 = 0, which we can use to solve for 𝑋˙$\dot{X}$, the velocity of the peeling point in the Eulerian (laboratory) frame. Coupling this with the peel force constitutive model (3.13)–(3.15), the unspooling rate equation is then 𝛼(𝐿˙)𝑚=𝐹1=−𝐹2$\alpha {(\dot{L})}^m=F_1=-F_2$. The dynamics of this scenario are governed by the system

(𝐿˙)𝑚=−12𝜋𝜇𝛼−1𝑅𝐿𝛿2𝐿𝛿+3𝑅(𝐿˙+𝑢¯(𝐿,𝑋,𝑡)−𝑢(𝑋,𝑡))$${(\dot{L})}^m=-\frac{12\pi \mu {\alpha }^{-1}RL\delta }{2L\delta +3R}\hspace{0.17em}(\dot{L}+\overline{u}(L,X,t)-u(X,t))$$

4.13a

and𝑋˙=2𝐿𝛿2𝐿𝛿+3𝑅(𝐿˙+𝑢¯(𝐿,𝑋,𝑡))+3𝑅2𝐿𝛿+3𝑅𝑢(𝑋,𝑡),$$\text{and}\hspace{0.17em}\dot{X}=\frac{2L\delta }{2L\delta +3R}\hspace{0.17em}(\dot{L}+\overline{u}(L,X,t))+\frac{3R}{2L\delta +3R}\hspace{0.17em}u(X,t),$$

4.13b

where 𝑢¯(𝐿,𝑋,𝑡)$\overline{u}(L,X,t)$ is the thread-averaged velocity (3.8). The velocity (4.13b) for the thread–skein system is the average of a velocity 𝐿˙+𝑢¯(𝐿,𝑋,𝑡)$\dot{L}+\overline{u}(L,X,t)$ arising from drag on the thread and a velocity 𝑢(𝑋,𝑡)$u(X,t)$ arising from drag on the skein, weighed by the relative strength of the drags.

The difference 𝑢¯(𝐿,𝑋,𝑡)−𝑢(𝑋,𝑡)$\overline{u}(L,X,t)-u(X,t)$ that appears in (4.13a) implies that adding a constant to the velocity field does not change the unspooling dynamics, as expected since the thread–skein system is freely advected by the flow, and only relative velocities generate drag. Hence, unlike our previous two cases in §§4.1 and 4.2, a spatially varying flow field is required for unravelling. For a linear velocity field 𝑢(𝑥,𝑡)=𝜆𝑥$u(x,t)=\lambda x$, i.e. uniaxial extensional flow with extensional strain rate λ, we have 𝑢(𝑋,𝑡)−𝑢¯(𝐿,𝑋,𝑡)=𝜆𝐿/2$u(X,t)-\overline{u}(L,X,t)=\frac{\lambda }{L/2}$ independent of X, so that we can solve the 𝐿˙$\dot{L}$ equation (4.13a) by itself,

(𝐿˙)𝑚=6𝜋𝜇𝛼−1𝑅𝐿𝐿+(3𝑅/2𝛿)(12𝜆𝐿−𝐿˙).$${(\dot{L})}^m=\frac{6\pi \mu {\alpha }^{-1}RL}{L+(3R/2\delta )}(\frac{1}{2}\lambda \hspace{0.17em}L-\dot{L}).$$

4.14

The mass conservation equation (3.10) then relates R to 𝐿$L$, and the slenderness parameter (3.2) relates δ to 𝐿$L$.

To define a characteristic length scale for this problem, should we use R0 or 𝐿0$L_0$ as a length scale? Both are important for the unravelling process to start quickly, but typically 𝐿0$L_0$ is a bit larger than R0. A compromise is to use R0 as the viscous drag length scale and 𝑈=𝜆𝐿0$U=\lambda L_0$ as the velocity scale. The choice of R0 emphasizes the magnitude of the drag on the skein, and 𝜆𝐿0$\lambda L_0$ reflects the amplitude of velocity gradients over the longer length 𝐿0$L_0$. We thus obtain the dimensionless form of (4.14) as

(𝐿˙∗)𝑚=℘𝑅∗𝐿∗𝐿∗+(3𝑅∗/2𝛿)(12𝐿∗−𝐿˙∗),$${({\dot{L}}^{\ast })}^m=\mathrm{\wp }\hspace{0.17em}\frac{R^{\ast }{L}^{\ast }}{L^{\ast }+(3R^{\ast }/2\delta )}(\frac{1}{2}{L}^{\ast }-{\dot{L}}^{\ast }),$$

4.15

where 𝐿˙∗=𝐿˙/𝜆𝐿0${\dot{L}}^{\ast }=\dot{L}/\lambda L_0$, 𝑅∗=𝑅/𝑅0$R^{\ast }=R/R_0$ and 𝐿∗=𝐿/𝑅0$L^{\ast }=L/R_0$ are the non-dimensional unravelling rate, skein radius and unravelled length, respectively. The natural dimensionless number in this case is

℘=6𝜋𝜇𝑅0𝑈𝛼𝑈𝑚=6𝜋𝜇𝑅0(𝜆𝐿0)1−𝑚𝛼−1.$$\mathrm{\wp }=\frac{6\pi \mu \hspace{0.17em}{R}_0\hspace{0.17em}U}{\alpha \hspace{0.17em}{U}^m}=6\pi \mu \hspace{0.17em}{R}_0\hspace{0.17em}{(\lambda \hspace{0.17em}{L}_0)}^{1-m}\hspace{0.17em}{\alpha }^{-1}.$$

4.16

Assuming as before that 𝐿˙≥0$\dot{L}\ge 0$ (the thread does not ‘re-spool’), the right-hand side of (4.14) implies 𝐿˙≤𝜆𝐿/2$\dot{L}\le \frac{\lambda }{\hspace{0.17em}}L/2$, which gives the constraint that 𝐿(𝑡)≤𝐿0e(1/2)𝜆𝑡$L(t)\le L_0\hspace{0.17em}{\text{e}}^{(1/2)\lambda t}$. This constraint is the kinematic limit where the thread extends at a rate dictated by the strain rate in the flow. This implies that the depletion time satisfies

𝑡dep≥2𝜆−1log(𝐿max𝐿0).$$t_{\mathrm{dep}}\ge 2{\lambda }^{-1}\log (\frac{L_{\max }}{L_0}).$$

4.17

In the two pinned cases we considered before, the lower bound on the depletion time was of the form 𝑡dep≥𝐿max/𝑈$t_{\mathrm{dep}}\ge L_{\max }/U$, independent of 𝐿0$L_0$. The lower bound (4.17) depends explicitly on the ratio 𝐿max/𝐿0$L_{\max }/L_0$, so a very short initial thread length will take a long time to unravel, even if ℘$\mathrm{\wp }$ is large.

When the thread is almost depleted, the unspooling rate decreases due to the factor of R in (4.14). To see this explicitly, put 𝐿=𝐿max−𝑈𝜏$L=L_{\max }-U\tau$ in (4.14) and assume τ and 𝜏˙$\dot{\tau }$ are small,

(−𝜏˙)𝑚≈12℘(𝑈𝜏𝐿max−𝐿0)1/3𝐿max𝐿0,𝜏≪1.$${(-\dot{\tau })}^m\approx \frac{1}{2}\mathrm{\wp }(\frac{U\tau }{L_{\max }-L_0}{)}^{1/3}\frac{L_{\max }}{L_0},\tau \ll 1.$$

4.18

This is exactly the same form as (4.6), with a different constant C. We conclude that once again the criterion for finite-time complete unravelling is m > 1/3, as it was for the pinned thread case (§4.1). But as before this is not very physically consequential, as it only applies to the last phase of unspooling when the skein is almost completely unravelled.

Figure 6 shows a numerical solution of (4.14) for our reference parameter values and with a strain rate λ = 10 s−1 for ℘ = 10. (We choose λ such that 𝜆𝐿max$\lambda L_{\max }$ is of the same order of magnitude as U = 1 m s−1 in the pinned cases.) The lower bound (4.17) on the depletion time is 1.48 s, and the numerical value is 𝑡dep≈1.73$t_{\mathrm{dep}}\approx 1.73$ s. This is slower than what we observed in the two pinned cases (𝑡dep<1$t_{\mathrm{dep}}<1$ s), due to the factor 2log(𝐿max/𝐿0)≈14.8$2\log (L_{\max }/L_0)\approx 14.8$. The slowdown due to the short initial thread length is thus considerable in this case. A longer initial length or a higher strain rate would be needed to make the times comparable.

Figure 6. 

• Download figure
• Open in new tab
• Download PowerPoint

4.4. Two free skeins (skein splitting)

Another scenario of unravelling is when a skein splits into smaller connected fractions, which then unravel. This scenario is possible since the hagfish ejects the skein through its slime gland and the resulting shear forces (from the ejection process and the fluid’s viscous drag) can break the skein into two.

Here we consider the simple case of a skein breaking into two halves. The unravelling may be faster since the initial viscous drag is dominated by two skeins, rather than a skein and a small initial length of thread. A diagram of this configuration is show in figure 7: we model the broken skein as two spheres, of radius R1 and R2, respectively, connected by an unravelled length of thread, which can unspool at both ends. We fix a reference point s = 0 between the two skeins such that 𝑥(0,𝑡)=𝑋(𝑡)$x(0,t)=X(t)$. The thread then extends a length 𝐿1(𝑡)$L_1(t)$ towards the first skein (right) and 𝐿2(𝑡)$L_2(t)$ towards the second skein (left), with 𝐿=𝐿1+𝐿2$L=L_1+L_2$ the total unravelled length. Without loss of generality we take 𝐿1(0)=𝐿2(0)=𝐿0/2$L_1(0)=L_2(0)=L_0/2$. The peeling force at the first skein (𝑠=𝐿1$s=L_1$, 𝑥=𝑋+𝐿1$x=X+L_1$) is the sum of the drag forces due to the second skein (𝑠=−𝐿2$s=-L_2$, 𝑥=𝑋−𝐿2$x=X-L_2$) and drag on the thread,

𝑇|𝑠=𝐿1=−6𝜋𝜇𝑅2(𝑢(𝑋−𝐿2,𝑡)−(𝑋˙−𝐿˙2))−4𝜋𝜇𝐿𝛿(𝑢¯(𝐿,𝑋,𝑡)−𝑋˙),$$T{|}_{s=L_1}=-6\pi \mu R_2(u(X-L_2,t)-(\dot{X}-{\dot{L}}_2))-4\pi \mu L\delta \hspace{0.17em}(\overline{u}(L,X,t)-\dot{X}),$$

4.19

where now 𝑢¯(𝐿,𝑋,𝑡):=(1/(𝐿1+𝐿2))∫𝑋+𝐿1𝑋−𝐿2𝑢(𝑥,𝑡)d𝑥$\overline{u}(L,X,t):=(1/(L_1+L_2)){\int }_{X-L_2}^{X+L_1}u(x,t)\hspace{0.17em}\text{d}x$. Since the thread and skeins are free, the peeling force at s = L1 (4.19) must balance the viscous drag force on the first skein,

𝑇|𝑠=𝐿1=6𝜋𝜇𝑅1(𝑢(𝑋+𝐿1,𝑡)−(𝑋˙+𝐿˙1)).$$T{|}_{s=L_1}=6\pi \mu R_1(u(X+L_1,t)-(\dot{X}+{\dot{L}}_1)).$$

4.20

Equating (4.19) and (4.20), we can solve for 𝑋˙$\dot{X}$,

𝑋˙=3𝑅1(𝑢(𝑋+𝐿1,𝑡)−𝐿˙1)+3𝑅2(𝑢(𝑋−𝐿2,𝑡)+𝐿˙2)+2𝐿𝛿𝑢¯(𝐿,𝑋,𝑡)3(𝑅1+𝑅2)+2𝐿𝛿.$$\dot{X}=\frac{3R_1(u(X+L_1,t)-{\dot{L}}_1)+3R_2(u(X-L_2,t)+{\dot{L}}_2)+2L\delta \hspace{0.17em}\overline{u}(L,X,t)}{3(R_1+R_2)+2L\delta }.$$

4.21

We use this to eliminate 𝑋˙$\dot{X}$ from (4.20),

𝑇|𝑠=𝐿1=6𝜋𝜇𝑅13(𝑅1+𝑅2)+2𝐿𝛿(3𝑅2(𝑢(𝑋+𝐿1,𝑡)−𝑢(𝑋−𝐿2,𝑡)−𝐿˙)+2𝐿𝛿(𝑢(𝑋+𝐿1,𝑡)−𝑢¯(𝐿,𝑋,𝑡)−𝐿˙1)).$$\begin{array}{rl}T{|}_{s=L_1}& =\hspace{0.17em}\hspace{0.17em}\frac{6\pi \mu R_1}{3(R_1+R_2)+2L\delta }(3R_2(u(X+L_1,t)-u(X-L_2,t)-\dot{L})\\ & \hspace{0.17em}\hspace{0.17em}+2L\delta (u(X+L_1,t)-\overline{u}(L,X,t)-{\dot{L}}_1)).\end{array}$$

4.22

We can then also carry out the same calculation for the second skein, at 𝑠=−𝐿2$s=-L_2$, and find

𝑇|𝑠=−𝐿2=6𝜋𝜇𝑅23(𝑅1+𝑅2)+2𝐿𝛿(3𝑅1(𝑢(𝑋+𝐿1,𝑡)−𝑢(𝑋−𝐿2,𝑡)−𝐿˙)−2𝐿𝛿(𝑢(𝑋+𝐿2,𝑡)−𝑢¯(𝐿,𝑋,𝑡)+𝐿˙2)).$$\begin{array}{rl}T{|}_{s=-L_2}& =\hspace{0.17em}\hspace{0.17em}\frac{6\pi \mu R_2}{3(R_1+R_2)+2L\delta }(3R_1(u(X+L_1,t)-u(X-L_2,t)-\dot{L})\\ & -2L\delta (u(X+L_2,t)-\overline{u}(L,X,t)+{\dot{L}}_2)).\end{array}$$

4.23

Now it is a matter of solving the coupled peeling equation 𝛼(𝐿˙1)𝑚=𝑇|𝑠=𝐿1$\alpha {({\dot{L}}_1)}^m=T{|}_{s=L_1}$, 𝛼(𝐿˙2)𝑚=𝑇|𝑠=−𝐿2$\alpha {({\dot{L}}_2)}^m=T{|}_{s=-L_2}$. To keep things simple, let us take a symmetric configuration centred on x = X = 0 where the two skeins are initially of equal size. (Unequal splitting would result in a depletion time in between this case of even splitting and the free skein–thread of §4.3.) We take an antisymmetric velocity field 𝑢(𝑥,𝑡)=−𝑢(−𝑥,𝑡)$u(x,t)=-u(-x,t)$ that pulls apart the skeins, such as for an extensional flow 𝑢=𝜆𝑥$u=\lambda x$. Then R1 = R2 and 𝐿1=𝐿2$L_1=L_2$ for all time, and ū = 0. The tensions (4.22) and (4.23) are then equal and greatly simplify to 𝑇|𝑠=𝐿1=𝑇|𝑠=−𝐿2=6𝜋𝜇𝑅1(𝑢(𝐿1,𝑡)−𝐿˙1)$T{|}_{s=L_1}=T{|}_{s=-L_2}=6\pi \mu R_1\hspace{0.17em}(u(L_1,t)-{\dot{L}}_1)$. Thus, the dynamics for this case is governed by

𝛼(𝐿˙1)𝑚=6𝜋𝜇𝑅1(𝑢(𝐿1,𝑡)−𝐿˙1).$$\alpha {({\dot{L}}_1)}^m=6\pi \mu R_1\hspace{0.17em}(u(L_1,t)-{\dot{L}}_1).$$

4.24

The drag force on the thread has dropped out, since the antisymmetric velocity field leads to cancelling forces on the thread. Another way to think of (4.24) is to observe that in making a symmetric configuration, with the two skeins being pulled apart by a straining flow centred on the origin, we have effectively ‘pinned’ the thread at x = 0. We have thus recovered our pinned thread equation (4.2) from §4.1, with the notable difference that now we cannot use a constant velocity field U, but must resort to a straining flow 𝜆𝑥$\lambda x$ or some other non-uniform flow.

Figure 7. 

• Download figure
• Open in new tab
• Download PowerPoint

We non-dimensionalize (4.24) using a characteristic length scale R0 and obtain

(𝐿˙∗1)𝑚=℘𝑅∗1(𝑢∗(𝐿∗1,𝑡∗)−𝐿˙∗1),$${({\dot{L}}_1^{\ast })}^m=\mathrm{\wp }{R}_1^{\ast }\hspace{0.17em}(u^{\ast }(L_1^{\ast },t^{\ast })-{\dot{L}}_1^{\ast }),$$

4.25

where 𝐿˙∗=𝐿˙/𝜆𝑅0${\dot{L}}^{\ast }=\dot{L}/\lambda R_0$ and 𝑅∗=𝑅/𝑅0$R^{\ast }=R/R_0$ are the non-dimensional unravelling rate and skein radius, respectively. The natural dimensionless number in this case is

℘=6𝜋𝜇𝜆𝑅20𝛼(𝜆𝑅0)𝑚=6𝜋𝜇𝑅2−𝑚0𝜆1−𝑚𝛼−1.$$\mathrm{\wp }=\frac{6\pi \mu \hspace{0.17em}\lambda \hspace{0.17em}{R}_0^2}{\alpha \hspace{0.17em}{(\lambda \hspace{0.17em}{R}_0)}^m}=6\pi \mu \hspace{0.17em}{R}_0^{2-m}\hspace{0.17em}{\lambda }^{1-m}\hspace{0.17em}{\alpha }^{-1}.$$

4.26

Figure 8 shows the unravelling dynamics associated with a split skein using parameter values similar to the free configuration of §4.3 and figure 6. The free skein unravels faster when split, as expected (0.756 s versus 1.73 s), owing to a stronger effective drag force and a kinematic upper bound with a rate λ rather than 𝜆/2$\lambda /2$. There is an important difference between using the free thread–skein equation (4.13) and the free split-skein equation (4.24): the former has a drag slaved to a short initial thread length, whereas for the latter the drag depends on the initial radius of the split skein, which can easily be larger.

Figure 8. 

• Download figure
• Open in new tab
• Download PowerPoint

In addition to the four cases discussed in this section, we also analysed a slightly more realistic scenario of suction flow where velocity decays away from the mouth of the predator and we consider a pinned skein at different locations away from the mouth. We use an approximate flow profile from the experimental data available in the literature (electronic supplementary material, section III). In general, such a flow profile is both spatially and temporally varying, but we neglect time-dependent variations for our analysis. The peak velocity (at the mouth of the predator) was chosen to match the characteristic velocity (U = 1 m s−1). The velocity decays over a characteristic length scale of the order of the gape size (e.g. opening size of the mouth). We estimate gape size from the video evidence [8] of slime deployment, resulting in extensional strain rates between 0.28–2.2 s−1, with the rate being highest at the predator's mouth. These are smaller extension rates than considered in the earlier cases of this section. The choice of pinning location drastically affects the unravelling time. A depletion time of ∼0.4 s was obtained for the case where the skein is pinned at a distance equal to one-third of the gape size of the predator. This is longer than the unravel time we found for the pinned skein in a uniform flow (≈0.22 s), due to the decaying velocity away from the predator’s mouth, but the unravelling time still falls close to natural unravelling time scales. More complicated spatially and temporally varying flow fields can be treated in a similar way; we expect the time scales in such cases to be of the order of those we found, given that in real scenarios the thread–skein system can be very close to or in the mouth of the predator, and the non-dimensional quantity ℘$\mathrm{\wp }$ is likely to be sufficiently large.

5. Discussion

5.1. The role of the dimensionless parameters ℘$\mathrm{\wp }$ and m

The unravelling times for various cases discussed in §4 depend on the dimensionless quantity ℘$\mathrm{\wp }$, but also separately on the model parameter m in the peeling force law. This is clear from the dimensionless governing equations in §4 that depend on these two dimensionless parameters separately, although m also appears in the definition of ℘$\mathrm{\wp }$. The power-law exponent m determines the peeling force dependence on the unravelling rate. Such a rate dependence exists in peeling scenarios due to the viscoelastic nature of adhesion at the peeling site. In the case of hagfish thread peeling from the skein, the dependence can possibly arise from viscoelastic time scales involved in the deformation of mucous vesicles or the polymeric solution of mucus [15], or the protein adhesive between the loops of thread [11]. The peeling resistance also depends on the dimensional constant factor, α, but its influence on unravelling is built into the dimensionless factor ℘$\mathrm{\wp }$, for which ℘ ∼ α−1.

Figure 9 compares all four flow scenarios of §4 as a function of ℘$\mathrm{\wp }$ and m in terms of the effective deployment time, i.e. the time to unravel half of the thread length, 𝑡dep,50%$t_{\mathrm{dep},50\text{\%}}$. (This effective time is used because some flows cannot fully deplete the skein in finite time for m < 1/3, as discussed in §4. Moreover, in practice, the threads do not need to be fully unravelled to create slime.) The flow parameters are identical to those previously described. For all cases, the limit of high drag and low peel resistance, ℘ ≫ 1, converges to the kinematic limit of unravelling where unconstrained portions of the skein–thread system exactly advect with the local flow velocity. At the other extreme, viscous drag is weak compared with peel resistance and for some small value of ℘$\mathrm{\wp }$ unravelling is too slow to match physiological time scales.

Figure 9. 

• Download figure
• Open in new tab
• Download PowerPoint

The power-law exponent m is a secondary effect compared with ℘$\mathrm{\wp }$. In general, for ℘ > 10, m has negligible effect on unravelling times. For ℘ < 10, the dependence on m is case specific. For the cases of pinned thread (§4.1), pinned skein (§4.2) and free skein–thread (§4.3), a larger value of m leads to a smaller unravelling time (while keeping the same value of ℘$\mathrm{\wp }$). For the case of skein splitting in an extensional flow (§4.4), such a monotonic trend is not observed and above a critical value of ℘$\mathrm{\wp }$ the unravelling is faster for small values of m. This presumably arises from the nonlinearity in the peel force constitutive equation. For example, taking m = 1 as a reference case, making m < 1 increases the dimensionless peel resistance for 𝐿˙∗<1${\dot{L}}^{\ast }<1$, but decreases the peel resistance for 𝐿˙∗>1${\dot{L}}^{\ast }>1$. As such, whether the unravelling rate is 𝐿˙∗≷1${\dot{L}}^{\ast }\gtrless 1$, the exponent m can accelerate or decelerate the unravelling process.

The value of m affects the minimum required ℘min to achieve unravel times comparable to physiological time scales (i.e. 𝑡dep,50%$t_{\mathrm{dep},50\text{\%}}$ at or below the dotted lines in figure 9). For the uniform velocity field cases (U = 1 m s−1), ℘min is a weaker function of m for the pinned thread case, ℘min = 0.29–1.32, compared with the pinned skein case, ℘min = 0.03–3. For the cases of a free skein–thread in extensional flow, even with splitting, 𝑡dep,50%$t_{\mathrm{dep},50\text{\%}}$ never falls below 400 ms even at high ℘$\mathrm{\wp }$. That is, 𝑡dep,50%$t_{\mathrm{dep},50\text{\%}}$ is higher than the physiological unravel time scales by a factor of 2 or 3. However, as stated earlier, the time scales in such cases are determined by the specific choice of strain rate, λ, and the initial unravel length 𝐿0$L_0$. Such kinematic and geometric parameters are certainly variable in reality, and small changes could easily decrease the unravel time scales, as previously discussed. In any case, m becomes important only when depletion time scales are much larger than the kinematic limit, clearly showing that m is of secondary concern compared with ℘$\mathrm{\wp }$.

An important caveat to the m = 0 case of FP = α = constant is that peeling cannot occur (𝐿˙=0$\dot{L}=0$) if the viscous drag falls below a critical value. For example, if either the initial skein radius R0 or thread length 𝐿0$L_0$ is too small, the viscous drag force is less than FP and unravelling cannot occur. Thus, in figure 9 a minimum value of ℘$\mathrm{\wp }$ is needed for the m = 0 cases. The minimum values range from about 1 to 10, depending on the case and the corresponding definition of ℘$\mathrm{\wp }$ for the flow and geometry. In three cases, the viscous drag can potentially increase during unravelling as the thread elongates (figure 9b–d). In these cases, the minimum ℘$\mathrm{\wp }$ is associated with initiating the peeling process. For the other case of the pinned thread, figure 9a, the viscous drag decreases during unravelling, since it is slaved to the skein radius, which decreases in size during the process. Unravelling here will eventually stop at a critical value of R. This, therefore, feeds back to requiring a larger critical initial value of ℘$\mathrm{\wp }$ to unravel by 50%, and is used in figure 9a to determine the domain of ℘$\mathrm{\wp }$ for the m = 0 case.

5.2. Estimating the parameter ℘$\mathrm{\wp }$

A key question remains: what is ℘$\mathrm{\wp }$ in physiological scenarios? For this, we must know the peeling force parameters in the constitutive model, and no direct experimental measurements are yet available. Here we make estimates for the two extreme conditions of m = 1 and m = 0, i.e. a linear dependence on velocity (akin to a constant viscous damping coefficient) and a constant peel force, respectively.

For the m = 1 case, we consider viscous resistance acting at the peel site with stress 𝜎=𝜇𝑢𝜖˙$\sigma ={\mu }_u\dot{ϵ}$, where μu is the uniaxial extensional viscosity between the separating thread and the skein (related to shear viscosity as μu = 3μ), and 𝜖˙=𝐿˙/𝐿𝑐$\dot{ϵ}=\dot{L}/L_c$ is the local extensional strain rate that depends on the peel velocity 𝐿˙$\dot{L}$ and the characteristic velocity gradient length Lc. The stress acts over the characteristic thread–thread contact area, which we assume scales as A ≈ d2, i.e. contact across the diameter and the length of contact along the thread also scales with the diameter. The peel force is then

𝐹P≈𝜇𝑢(𝐿˙𝐿𝑐)𝑑2.$$F_{\mathrm{P}}\approx {\mu }_u\left(\frac{\dot{L}}{L_c}\right)d^2.$$

5.1

Comparing with the peeling law in (3.15), 𝐹P=𝛼𝐿˙𝑚$F_{\mathrm{P}}=\alpha {\dot{L}}^m$, we get

𝛼=𝜇𝑢𝑑2𝐿𝑐;𝑚=1.$$\alpha ={\mu }_u\frac{d^2}{L_c};m=1.$$

5.2

Substituting this into ℘ = FD/FP, and considering the majority of cases where drag is set by the skein radius R0, i.e. equations (4.4), (4.16), (4.26), we obtain

℘=6𝜋𝜇𝑅0𝑈𝜇𝑢(𝐿˙/𝐿𝑐)𝑑2=6𝜋𝜇𝜇𝑢𝑈𝐿˙𝑅0𝐿𝑐𝑑2,$$\mathrm{\wp }=\frac{6\pi \mu R_0\hspace{0.17em}U}{{\mu }_u(\dot{L}/L_c)d^2}=6\pi \frac{\mu }{{\mu }_u}\frac{U}{\dot{L}}\frac{R_0{L}_c}{d^2},$$

5.3

where important ratios have been grouped. The simplest case is peel viscosity arising from the surrounding viscous liquid at the peel site. In other words, the viscosity causing drag is also resisting peeling, and to cast in terms of extensional viscosity we take μu = 3μ, a result for a Newtonian fluid. The viscosity μ may be that of sea water, or a surrounding mucous vesicle solution with higher viscosity, but under these assumptions the ratio μ/μu is still the same. Furthermore, typically 𝑈/𝐿˙≳1$U/\dot{L}\hspace{0.17em}\gtrsim \hspace{0.17em}1$, and the velocity gradient length scale is likely set by the thread radius, Lc ≈ r. Then (5.3) dramatically simplifies to

℘=2𝜋𝑅0𝑑,$$\mathrm{\wp }=2\pi \frac{R_0}{d},$$

5.4

which clearly estimates ℘ ≫ 1, or more specifically for R0 = 50 μm and d = 2 μm, ℘ ≈ 160. If drag is instead dominated by the thread length 𝐿0$L_0$, e.g. for the pinned skein case of §4.2, then the numerator in (5.3) would be modified by replacing 6πR0 with 4πL0. We expect 𝐿0$L_0$ to be of the same order as R0, e.g. a single curl in the coil. But if 𝐿0$L_0$ is smaller, it would decrease ℘$\mathrm{\wp }$ accordingly.

Our specific assumptions can modify the details, but in general we estimate that physiological conditions for m = 1 would give ℘ > 1, if not ℘ ≫ 1. The velocity gradient length scale Lc could be smaller than the thread radius r. A decrease in Lc/r makes ℘$\mathrm{\wp }$ proportionally smaller, but it is difficult to imagine this being more dramatic than, say, a factor of 10. The velocity ratio 𝑈/𝐿˙$U/\dot{L}$, if anything, will be larger than 1, and this proportionally increases the estimate of ℘$\mathrm{\wp }$. The viscosity ratio μ/μu could be smaller, e.g. if the viscosity of proteins between thread wrappings is larger than the surrounding viscous liquid. However, we note that the surrounding viscous liquid can have a very large viscosity, e.g. the measured extensional viscosity of hagfish mucous vesicle solutions obtained by Böni et al. [15] is μu ≈ 10 Pa s. This is much higher than water, μu ≈ 3 mPa s. An additional mechanism of increasing drag, and ℘$\mathrm{\wp }$, is for mucous vesicles to bind on the thread during the unravelling process (figure 10), which would transmit additional forces to the drag term, as suggested by Winegard & Fudge [22]. Such a scenario is possible since mucous vesicles and thread cells are densely packed inside the slime glands and are released simultaneously. For all of these variations, ℘ > 1 seems very likely for physiological conditions in this constant viscosity estimate for m = 1.

Figure 10. 

• Download figure
• Open in new tab
• Download PowerPoint

To estimate physiological ℘$\mathrm{\wp }$ for m = 0, the other extreme of a constant force resisting peel, we consider peel strength interactions between the skein fibres solely due to van der Waals forces. We estimate FP = α ≈ 10−7 N (see electronic supplementary material, section II). Substituting into ℘ = FD/FP, and considering cases where drag is set by the skein radius, i.e. equations (4.4), (4.16), (4.26), with R0 = 50 μm, water viscosity μ = 1 mPa s, and U = 1 m s−1, gives ℘ ≈ 90. We see that ℘ ≫ 1 with these assumptions. Even if the force resisting peeling is larger by a factor of 10 or 100, still ℘≳1$\mathrm{\wp }\hspace{0.17em}\gtrsim \hspace{0.17em}1$ and viscous hydrodynamics can provide rapid unravelling that can be very close to the kinematically derived lower bounds on unravelling time.

6. Conclusion

Our analysis shows that, under reasonable physiological conditions, unravelling due to viscous drag can occur within a few hundred milliseconds and is accelerated if the skein is pinned at a surface, such as the mouth of a predator. A dimensionless ratio of viscous drag to peeling resistance, ℘ = FD/FP, appears in the dynamical equations and is the primary factor determining unravelling time scales. Large ℘$\mathrm{\wp }$ corresponds to fast unravelling that approaches a kinematic limit wherein free portions of the thread–skein system directly advect with the local flow velocity. For characteristic velocity U, the bound is 𝑡dep≥𝐿max/𝑈$t_{\mathrm{dep}}\ge L_{\max }/U$, whereas for extensional flows with strain rate λ, 𝑡dep≥𝜆−1log(𝐿max/𝐿0)$t_{\mathrm{dep}}\ge {\lambda }^{-1}\log (L_{\max }/L_0)$, where 𝐿0$L_0$ is the initial thread length.

The modelling approach captures essential features and insights by considering a single skein unravelling in idealized flow fields. Future modelling efforts could build on our work by expanding and detailing several aspects, primarily with new experimental measurements of peel strength for skein unravelling, but also details of physiological flow fields including characteristic velocities and strain rates. Real physiological scenarios are more complex due to chaotic flows and multi-body interactions (multiple skeins, mucous vesicles). Our model does not consider such interactions, or the important feature of unravelled threads interacting to create a network. At leading order, we expect such modelling to require more complex flow fields that create extension (to unravel fibres) but also bring different fibres together. Mixing flows would be excellent candidates for theoretical analysis, and any experimental characterization of physiological flow fields should keep this perspective in mind, e.g. simple suction flow with extension, but no mixing, may not be sufficient to create a network of unravelled threads.

Although the physiological flow fields may be different from the ones that were used in the analysis, our results underline the importance of viscous hydrodynamics and boundary conditions on the process. Recent work [11] found that Pacific hagfish skeins undergo spontaneous unravelling in salt solution. However, the observed unravelling time scales (∼min) are much larger than the physiological time scales (approx. 0.4 s) during the attack. It is possible that ion transport to the peeling site may help in peeling the adhesive contacts, which may be diffusion limited without flow. Although it is known that flow is required to accelerate unravelling to 𝑡dep,50%<1$t_{\mathrm{dep},50\text{\%}}<1$ s, it is not yet clear whether flow-enhanced ion transport may also contribute to a faster unravelling, in addition to the drag effects. The effects of various salt ions on the swelling and rupture of mucin vesciles have been studied in the past [18,19], but the influence of such ionic effects on the skeins and their deployment is not yet known. If ion transport and chemistry are important, this would modify the FP behaviour and require modelling of transport at the peel site due to flow. Our results do not rule out the possibility of ion-mediated unravelling but provide an alternative mechanism of unravelling which may be occurring alone or in conjunction with a multitude of other processes.