Preliminary calculations

Under classical stability analysis, an equilibrium is stable if and only if the potential energy is at a minimum. To analyze this system, we suppose that each of the muscles has a length-dependent potential energy function. Defining x₁ and x₂ as the lengths of muscle 1 and muscle 2, respectively, and letting the pendulum mass be m with length ℓ. Under these assumptions, we have the potential energy function: (1) Where m and ℓ are the mass and length of the pendulum, respectively, and g is the acceleration due to gravity (9.81 m/s²). The spring potential energies, V₁ and V₂, are functions of the spring lengths, x₁ and x₂, which are themselves functions of the pendulum’s vertical angle, θ. For convenience, we suppose that the origins of the muscles are at O_i = (±a, 0), and that the insertions are both b units vertical from their origins onto the inverted pendulum in the neutral configuration. This gives closed-form expressions for their lengths: (2a) (2b)

An important property derivable from these lengths are the springs’ moment arms, which are the derivatives of these functions with respect to the angle, θ: (3)

Since we are analyzing the system about θ = 0, it is also helpful to define the length (L) and moment arms (r) of the springs in that position: (4)

A final useful observation we can make about this system is that, in an upright position, the second derivative of the spring lengths vanishes.

(5)

Classical stability analysis

For this system to be stable in an upright position (θ = 0), requires that the gradient of Eq 1 vanishes, and that its second derivative is positive. Using the chain-rule, we can evaluate the first derivative: (6) Where we have used Eq 3 to rewrite the derivatives of x_i with respect to angle as the moment arms. We require that the upright position is in equilibrium, so that Eq 6 evaluated at θ = 0 vanishes. Eq 6 can be further simplified by recognizing that the muscles’ force, F_i, is the gradient of the potential energy, or that . Doing so results in the familiar moment-balance required for equilibrium: (7)

We next consider the second derivative of Eq 1, which needs to be positive to ensure stability. We evaluate the derivative of Eq 6 using the product and chain rules: (8)

Because of the geometry of the pendulum, notably Eq 5, the terms involving the derivatives of the moment arms with respect to angle vanish once we evaluate this expression at θ = 0. Further, we recognize the muscles’ stiffness, , which yields the stability condition: (9)

Together, Eqs 7 and 9 place restrictions on the muscle forces and stiffnesses required to maintain stability in the upright position.

Force-stiffness relationships and the Hill model

So far, this analysis has been very general with no assumptions made about the nature of the springs supporting the inverted pendulum. For skeletal muscles, Bergmark (1989) [6] used a relationship between the muscles’ active force (F), length (x), and stiffness (K) derived from cross-bridge theory, which we call the force-stiffness relationship [30,33,36]: (10) Where q is a dimensionless constant of proportionality, often chosen between 1.0 and 40 [6,14,15,37–39]. Using this relationship, Eqs 7 and 9 can be recast as constraints on muscle forces [40,41]. For our simple pendulum model, equilibrium (Eq 7) requires that both spring forces are equal, and stability (Eq 9) requires that: (11)

This muscle force solution was a triumph of this stability approach since Eq 11 predicts considerable antagonistic muscle co-contraction to stabilize an upright posture in the absence of a net joint moment. Supporting this prediction are countless in vitro and in silico studies which have demonstrated that the spine is inherently unstable when it is not supported by active musculature [2,5,8,41–43].

However, the relationship in Eq 11 is problematic for forward-dynamics modelling purposes because the basic Hill-type muscle model simply does not capture short-range stiffness [44]. Consider a Hill-type muscle model, omitting the passive structures for brevity. The muscle’s force is given by the product of its activation (α), together with the maximum isometric force (F₀), the force-length (f_L) and force-velocity (f_V) relationships [24,45]: (12)

One can show that, since stiffness is the derivative of this expression with respect to x_i, the Hill model is only consistent with the force-stiffness relationship if its force-length relationship is a power law (i.e. f_L(x) = cx^q, where c is a constant of integration and q is the dimensionless constant earlier). Unfortunately, this result may only be locally consistent with experimental evidence, and thus precludes a conventional Hill-type muscle model from exhibiting the short-range stiffness phenomenon that theoretically stabilizes the upright pendulum. Resolving this issue requires a departure from conventional Hill-type models, and exploring the mechanisms that give rise to Eq 10 in the first place.

The Huxley model

An alternative to the Hill-type muscle model is the Huxley muscle model that is the foundation of the sliding filament theory [31]. Although it is more widely used to investigate muscle function at the molecular level [46–48], it has been used in several investigations at the whole muscle level [32,49–52]. We briefly describe the expression for the Huxley model with two-states, analyze some important results, and demonstrate that the Huxley model satisfies the force-stiffness relation.

The Huxley model describes a half-sarcomere, where a large population of myosin-actin cross-bridges reside (Fig 3A). Each bound cross-bridge has its own displacement from an equilibrium position, s (Fig 3B), and can alternate between bound and unbound states through the attachment and detachment rate functions, f(s) and g(s), respectively (Fig 3C). The model considers only bound cross-bridges, which is used to construct a displacement distribution function over all the cross-bridges, n(s, t) (Fig 3D). This is so that n(s, t)Δs can be thought of as, roughly, the number of cross-bridges that have stretches between s and s + Δs. The time-evolution of n(s, t) is given by the partial differential equation [52]: (13) Where α and f_L(x) are, as before, the muscle’s activation and force-length curve, and v is the half-sarcomere’s shortening velocity. Myosin cross-bridges possess a ‘characteristic stretch,’ denoted h, typically on the order of 5–10 nm [53], beyond which the reaction kinetics favour the detached state (Fig 3C).

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Fig 3. An overview of the two-state Huxley model used in this investigation.

The myosin heads can exist in two states: (A) unbound, or (B) bound, to actin. Once bound, the model tracks the displacements among the myosin molecules, which have an associated stiffness of k_m ≈ 0.2 to 5.0 pN/nm. The rates between these two states are characterized by an attachment rate, parameterized by f(s), and a detachment rate, parameterized by g(s), both of which are graphically depicted in (C). A hypothetical displacement distribution function (D), where the area under this graph between displacements s₁ and s₂ is approximately the proportion of myosin heads that are bound with displacements between s₁ and s₂.

https://doi.org/10.1371/journal.pone.0307977.g003

Force and stiffness in the Huxley model

Zahalak (1981) [49] presented expressions that map between the bond distribution, n(s, t) and macroscopic variables like the muscle’s force and stiffness. Notably, in our Supplemental Material, we follow a modified approach to Blangé et al., (1972) [33], the Huxley model satisfies the force-stiffness relationship, and is thus a candidate for stabilizing the inverted pendulum. At any instant, the muscle’s active force (P) is given by the expression: (14) Where Γ is a constant that depends on the sarcomere length, myosin stiffness—generally on the order of 0.2 to 5 pN/nm [19,54,55]—myosin concentration, cross-sectional area, and spacing between actin-myosin binding sites [49]. The integral in Eq 14 can be thought of as adding up the contributions to the total force from all the currently bound cross-bridges with Γ playing the role of an effective cross-bridge stiffness. Similarly, the stiffness (K) of the half-sarcomere is given by another integral [32]: (15) Where N_s is the number of sarcomeres arranged in series along the muscle fibre. An important solution for the Huxley model is the steady state () isometric (v = 0) condition, which is a uniform distribution. One can plug this analytic solution into Eqs 14 and 15 to derive the force-stiffness equation (see Supplemental Material): (16) Which is similar to Eq 10, and identical in the case when the pendulum is upright. This property of the Huxley model gives rise to the short-range stiffness phenomenon. Of note, the theoretical dimensionless q is the ratio of the sarcomere length (~2.8 μm) [56] to the characteristic bond length (~5–10 nm) [57], which yields values of approximately 240–560 [16]: more than tenfold larger than previous investigations have assumed [6,30,36,58]. In generalizing this expression for other activations, we find that the short-range stiffness is tunable based on muscle activation, which is consistent with other investigations [44,59].

Simulations

Numerical simulations were carried out in ArtiSynth [60], where custom muscle elements were programmed that incorporated a Distribution Moment (DM) approximation to the Huxley model. In particular, we used Donovan’s uniform distribution approximation [61], as it has the advantage of having the same steady-state behaviour as the underlying Huxley model.

The inverted pendulum had a mass of 30.0 kg and a length of 0.30 m, supported, by two 0.35 m long muscles (fibre lengths of 14 mm) with 0.05 m long moment arms, on either side of the single degree-of-freedom pivot joint (Table 1). With the muscle parameters in Table 1, along with the classical stability analysis (Eqs 7, 9 and 11), gives that the pendulum is stable if the muscle activations exceed 0.7%. To test this threshold, we ran simulations where the muscle activations were 0.5%, 1.0%, 10%, 50% and 100%. The pendulum was initially upright and, after one second, a 5 Nm torque was applied to the pendulum for 50 ms to test the stability of the pendulum in the numerical simulation. In addition, we compared simulation results to an unsupported inverted pendulum and one with equivalent linear springs with stiffnesses that match the muscle at 100% activation to give an example of a stable solution, and another where with the same springs and an additional damper in parallel with damping coefficient arbitrarily set to 10 Ns/mm, to give context on an asymptotically stable solution.

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Table 1. Parameters that were used throughout the simulations.

The muscle properties were selected to be representative of average muscle properties for lumbar spine extensors (erector spinae).

https://doi.org/10.1371/journal.pone.0307977.t001

A related analytical stability analysis

We initially presented a stability analysis that classifies equilibria as stable or unstable based on minimizing the potential energy [4,6,15]. Yet our numeric simulations conflicted with the predicted stability. A partial explanation for this mismatch is that our muscle elements do not provide produce conservative forces, which precludes a stability analysis based on potential energy minima. Therefore, we conducted a broader analysis of the inverted pendulum as if it were supported by standard viscoelastic solids, which are a decent approximation to a constant-activation Huxley model (Fig 7). We found that this systems’ stability is independent of the short-range stiffness magnitude and now offer this as an explanation for why the Huxley muscles did not stabilize the inverted pendulum.

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Fig 7. Approximation of the inverted pendulum in this analysis (left) with one supported by standard viscoelastic solid models (right).

https://doi.org/10.1371/journal.pone.0307977.g007

We impose that the springs labelled S_i (Fig 7) can be approximated by standard-linear viscoelastic solid elements, whose springs are linear and have the steady-state muscle stiffness (Fig 1) [17,24,25]. On the other hand, the parallel Maxwell elements are linear with spring constant , and relaxation times . This arrangement ensures that if these elements are slowly lengthened, their force-elongation curve will follow the steady-state force-length relation; but, when lengthened quickly, will have a transient short-range stiffness . Here, k_i is the slope of the tangent to the force-length curve, which we called the static stiffness (c.f. Fig 1). Like before, we let x_i be the total muscle length, and ξ_i be the Maxwell-element springs deflections, so that the muscle tensile forces, F_i, can be calculated from the relationships: (17a) (17b) Where f_L(x) is the steady-state force-length relationship. Next, we apply Newton’s second law to arrive at the equation of motion: (18) Where r_i are the muscles’ moment arms, and the negatives ensure that positive forces in muscle 1 produces a negative moment. Like before, the muscle lengths, x_i, are functions of the joint angle, θ (Eqs 2a and 2b). Similarly, the moment arms r_i are the derivatives of x_i with respect to θ. At last, we introduce and define the state vector, q = (θ ω ξ₁ ξ₂)^T. We linearize this system about the upright steady-state, q₀ = (0 0 0 0)^T, which requires that the forces in S₁ and S₂ are equal, yielding the linear system: (19)

Like before, a represents the moment arm in the upright position, and k₁ and k₂ are the slopes of the usual force-length expression. We can determine whether the system, of the form , is asymptotically stable by analyzing its eigenvalues, the solutions to the equation 0 = det(λI − A). This polynomial is: (20)

Of course, solving this quartic equation for λ would be needlessly laborious, so we employ a useful analytic result. From the Routh-Hurwitz criterion [68], this system will be unstable if any characteristic polynomial coefficients (including the intercept) are negative. Assuming that the short-range stiffnesses (), time constants (τ_i), mass (m) and pendulum length (ℓ) are all positive, we find that the system is unstable if: (21)

Which is a result similar to the static analysis (Eq 9). We strengthen this claim in Appendix I, where we show that the converse is also true: if k₁ + k₂ > mgℓ/a², then the Routh-Hurwitz criteria are satisfied, and the pendulum is asymptotically stable. Interestingly, this result does not depend on the short-range stiffness and suggests that the pendulum is unstable if both muscles are on the plateau or descending limbs of the force-length relationship, whereas the ascending limb may be more stable. If this criterion in Eq 21 is met, the pendulum is asymptotically stable, a stronger notion of stability than minimum potential energy suggests, meaning that perturbations converge on the equilibrium rather than oscillate about it.

Alternative stabilizing mechanisms

Short-range stiffness offers an attractive stabilizing mechanism because it produces a force that resists a perturbation almost instantaneously and without any input from the central nervous system. This effect, along with the amplification of lengthening muscle force from the force-velocity relationship, seems like a plausible stabilizing mechanism for many biomechanical systems, and models that treat short-range stiffness have found that it helps in responding to a perturbation [14,16,30,44]. However, our simulations and analysis are strongly suggestive that short-range stiffness plays a critical supporting role in establishing stability by rapidly dissipating a perturbation, as it cannot stabilize the system on its own. Therefore, another mechanism must be responsible for providing mechanical stability to these systems. In this section, we speculate on a few mechanisms that seem most plausible.

The simplest hypothesis that imposes joint stability might be that stabilizing muscles exist on the ascending limb of the force-length relationship, so they already meet the stability criteria. This property is certainly true for some muscles, in the spine, for example, the multifidus, which Ward et al. [69] found is soundly stationed on the ascending limb in a neutral posture. Similarly, Burkholder and Lieber (2001) [70] found that many human skeletal muscles operate on the ascending limb near the plateau region. Unfortunately, morphometry studies have found that this finding is not always true. The sarcomere lengths in a neutral posture for several muscles, particularly in the cervical spine, are located near the plateau region or on the descending limb [70–73]. These outcomes raise an interesting paradox when combined with our analysis: the active component of muscles on the plateau or descending limbs cannot use force-length properties to stabilize a joint, and yet, many muscles operate in this region and our joints behave in a stable manner. This paradox strongly suggests that the inverted pendulum model may be missing an important feature of muscle mechanics, other viscoelastic soft tissues, joint configuration, posture, or require a controller to maintain stability.

Another hypothesis might be the residual force enhancement or depression phenomena of muscles. In this analysis, we assumed that the steady-state force produced by a lengthened active muscle would, after a transient, tend toward the force-length relationship. However, experimental evidence is abundant at the sarcomere, fibre, and whole muscle levels, which suggests that this is not the case [74–82]. Instead, actively lengthened muscles tend toward more force, called residual force enhancement, and actively shortened muscles tend toward less, called force depression. Interestingly, this change in steady-state force is proportional to the change in stretch, at least on the descending limb [83], and can exceed 50% of the maximum isometric force [81]. This proportionality suggests that its effects may be approximated with another spring, in parallel with the muscles and Maxwell elements, whose resting length is the initial length of the muscle [18,67]. This mechanism would be stabilizing since the proportionality constant would add to the static stiffnesses and, unlike short-range stiffness, not be transient. For our simple inverted pendulum, the stability condition with force enhancement would be: (22) Where k_FE is the proportionality constant relating stretch to force enhancement. This argument demonstrates the plausible stabilizing effect of residual force enhancement and depression but is not comprehensive. Experimentally verifying the potential role of this phenomenon, and its molecular basis, remains an avenue for future work.

Perhaps the most direct hypothesis is that the central nervous system manages joint stability through motor control and reflexes [43,84,85]. Using the spine as an example, the small muscles of the spine, like the rotatores and intertransversarii, have some of the densest clusters of muscle spindles in the human body, allowing them to quickly transmit postural signals to the spinal cord [86,87]. Further, some computer studies have obtained stable cervical spine behaviour using controls that respond to vestibular reflexes with Hill-type muscle models that typically omit short-range stiffness [88]. The most compelling evidence for this hypothesis, at least for the spine, comes from Moorhouse and Granata (2007) [43], who used a system identification technique with experimental data to determine that spinal reflexes accounted for 42% of the total stabilizing trunk stiffness. While short-range stiffness may still play a role in this paradigm, it may function as a damper that constrains and dissipates the perturbation and buys the central nervous system time to respond to a disturbance.