# Idealized walking and running gaits minimize work

Manoj Srinivasan, Andy Ruina

## Abstract

Even though human legs allow a wide repertoire of movements, when people travel by foot they mostly use one of two locomotor patterns, namely, walking and running. The selection of these two gaits from the plethora of options might be because walking and running require less metabolic energy than other more unusual gaits. We addressed this possibility previously using numerical optimization of a minimal mathematical model of a biped. We had found that, for a given step-length, the two classical descriptions of walking and running, ‘inverted pendulum walking’ and ‘impulsive running’, do indeed minimize the amount of positive work required at low and high speeds respectively. Here, for the case of small step-lengths, we establish the previous results analytically. First, we simplify the two-dimensional particle trajectory problem to a one-dimensional ‘elevator’ problem. Then we use elementary geometric arguments on the resulting phase plane to show optimality of the two gaits: walking at low speeds and running at high speeds.

Keywords:

## 1. Introduction

When people travel by foot from point A to point B, and are not rushed, they walk. When they are in a hurry, they run. Classically, human walking has been idealized as a gait in which the hip (or the body centre of mass) travels in a series of inverted pendulum circular arcs about the foot on the ground as shown in figure 1a (e.g. Alexander 1976). In one energy-consistent version of this idealization, only one foot is on the ground at a time, and the transition from one circular arc to the next is accomplished by the trailing leg pushing off just before the heel of the leading leg strikes the ground (Kuo 2001; Kuo et al. 2005; Ruina et al. 2005; Srinivasan & Ruina 2006). Analogously, as shown in figure 1b, running is simply idealized as a series of parabolic free-flights of the body, interrupted by brief bounces. In idealized running, the work absorbed in the downward part of a bounce is regenerated on the way back up (Rashevsky 1948; Ruina et al. 2005; Srinivasan & Ruina 2006). Among the essentially infinite variety of motions that our two legs are capable of, why do we choose to walk and run in ways that are somewhat close to these two idealizations?

Figure 1

Idealized motions for walking and running. (a) In inverted pendulum walking, the body vaults in circular arcs on a straight leg. One leg is on the ground at a time. At the change from one circular arc to the next the trailing leg pushes off and then the leading leg heelstrikes the ground. (b) In impulsive running, the body travels from one parabolic arc to the next with a bounce in between. Adapted from Srinivasan (2006) and Srinivasan & Ruina (2006).

A common and common-sense conjecture is that humans and other animals do things in a way that minimizes metabolic energy expenditure (e.g. Borelli 1680; Alexander 1989). To explore this conjecture, we previously (Srinivasan & Ruina 2006) formulated a simple mathematical model of a bipedal animal consisting of a point mass upper body and two massless legs (figure 2). The legs can do and absorb work by extending and contracting. The model was a simplification of earlier models (Alexander 1980, 1992; Minetti & Alexander 1997). Using numerical optimization, we identified the gaits that minimized the amount of positive work required to travel with a given speed, step-length and maximum-allowed leg extension. We found that, indeed, inverted pendulum walking is energetically optimal at low speeds and step-lengths and impulsive running is energetically optimal at higher speeds.

Figure 2

Minimal biped model. The upper body is modelled as a point mass. The two legs are modelled as massless telescoping actuators, capable of changing length, exerting force and performing work on the upper body. Figure adapted from Srinivasan (2006) and Srinivasan & Ruina (2006).

In this paper, we examine the same model, this time analytically, for the special case of small step-lengths. Our aims are to more firmly establish the previous numerical result and also to reveal more of the essential structure of the bipedal work-minimization problem.

In §2, we present problem A, the gait-optimization problem, and add some symmetry assumptions. In §3, we present problem B, the simplification to small step-lengths, further discussed in §4. In §5, we simplify the locomotion problem to problem C, that of riding a descending elevator given some constraints. Finally, in §6, we show that the energy-optimal ways to ride the elevator correspond, in problem A, to inverted pendulum walking and impulsive running.

## 2. Problem A: model and symmetry assumptions

### (a) Biped model and metabolic cost

The minimal biped model (figure 2) has a point mass upper body and two massless legs. Each leg is a telescoping actuator that can change its length. The legs can apply force and perform work on the upper body when they are in contact with the ground. When not in contact with the ground, the legs can be swung around arbitrarily. We assume that, at most, one of the two legs is in contact with the ground at a time. In our previous numerical work, we enforced a limit on the length of the leg lmax. Here, to simplify the mathematical argument, we replace the maximum leg-length constraint with a specification of the height of the centre of mass at mid-step l0. The point of contact of the foot with the ground is xc=0 throughout the duration of the step under consideration. A step is defined as half a stride. The flight phase, if it exists for a given gait, is the portion of the contact phase in which the leg force is zero.

The idealized gaits such as inverted pendulum walking and impulsive running, discussed in §1, require infinite forces as they involve instantaneous velocity changes corresponding to the kinks in the centre of mass trajectories. Because unbounded forces can be troublesome analytically, we impose simple bounds on the leg force FA(t) : FminFA(t)≤Fmax, whenever it is convenient to do so. However, allowing infinite forces frees the model of the above force-bound parameters, making the results simpler and more general. We assume that the forces are bounded until the end of §5 where we allow impulsive forces.

We model the metabolic cost as being separable into three parts.

1. Animals have a positive metabolic expenditure even at rest. Related to this cost, we posit a constant metabolic rate (per unit time) term .

2. Leg-swing cost. Legged locomotion may require swinging the leg with active muscle work. Here, we assume that this leg-swing metabolic cost, Cswing per step, is purely a function of the speed and step-length or, equivalently, of speed and frequency (Alexander 1976; Kuo 2001).

3. Stance cost. This is incurred by a leg when it is in stance phase, that is, in contact with the ground. We model this cost, Cstance per step, as being proportional to a weighted sum of the total positive work and the total negative work performed by the telescoping actuator leg.

If the average forward (horizontal) speed over a step is v, the step-length is dstep and the step period is tstep=dstep/v, the total metabolic cost C per step is(2.1)In the calculation here, we prescribe the forward speed v and the step-length dstep. Step-length is the horizontal distance covered during a step, which may include flight phases.1 Hence, it does not matter whether we minimize the energetic cost per step C, the energetic cost per unit distance C/dstep or the energetic cost per unit time Cv/dstep. All of those minimizations result in the same optimal solution. Further, the first two terms in the energy-cost equation (2.1) drop out of the optimization because they are constant for a given forward speed v and step-length dstep. Prescribing v and dstep is not a loss of generality because we will observe the roles of v and dstep by seeing how the optimal solutions depend on them.

The leg's instantaneous mechanical power output is , where FA is the leg force, with compression taken as positive, and lA is the leg-length (figure 2; the subscript ‘A’ denotes problem A). The stance cost over a step (with duration −tstep/2≤ttstep/2) is given by(2.2)where [P]+ registers positive leg power, equalling P when P≥0 and equalling zero when P<0. [P]≡[−P]+ registers negative leg power. Typically, muscles are assumed to have b1>0 (work costs) and b2>0 (absorbing mechanical work also has a metabolic cost) but here we only need that b1+b2>0.

### (b) Symmetry assumptions

We assume periodic gaits with each step similar to the next. To simplify the analytic argument here, we assume that the trajectory of the centre of mass within a step is symmetric about the ‘mid-step’ (figure 3a). Mid-step (at t=0) is when the body is directly above the foot-contact point. Hence, only half a step—from mid-step till the end of the step—contains all of the information about the gait. Although this symmetry assumption limits the generality of the analytic demonstration, we note that all the numerical optima we previously found had this symmetry about mid-step (Srinivasan & Ruina 2006).

Figure 3

Problem A. (a) One step of a gait that is symmetric about the mid-step. (b) Assuming periodic steps that are symmetric about mid-step implies that the vertical component of the velocity is zero at mid-step and at the end of the step.

### (c) Direct implications of the model assumptions

The periodicity assumption, for level-ground locomotion, implies that there is no energy change from one mid-step to the next. The boundedness of the leg force FA(t) implies that the acceleration is always bounded and that the velocity of the body is continuous. Because the velocity vector along the gait-trajectory is continuous, the symmetry conditions above require that the vertical component of the velocity vanishes both at mid-step and at the end of the step (figure 3b).

Because there are no passive dissipation mechanisms, the net work of the legs in one step is zero: and the net positive work equals the net negative work: . Therefore, the cost in equation (2.2) is and is proportional to the total positive work. In other words, minimizing a weighted sum of the positive and negative work is equivalent to minimizing either the total positive work or the total negative work and does not depend on the values of b1 and b2 (as long as b1+b2>0).

The symmetry assumption implies that the amount of positive work in the first half of the stride equals the amount of negative work in the second half of the stride and vice versa. Hence, the total stance cost from equation (2.2) can be rewritten as(2.3)Without loss of generality, we take b1+b2=1, so that the cost to be minimized is the sum of the positive and the negative work over half a step: .

### (d) Optimization problem A

The position of the centre of mass is given by (xA(t), yA(t)) with mid-step at xA(0)=0. We wish to determine those values of and FA(t) over t∈[0,tstep/2] that together minimize the total cost CA:(2.4)subject to the restriction that xA, yA, and FA all obey the differential equations(2.5)(2.6)and the boundary conditions that xA(0)=0, yA(0)=l0, xA(tstep/2)=dstep/2, and .

The specification of the initial conditions and the single control function leg force FA(t) determine (xA(t), yA(t)) through the differential equation. We proceed to solve this optimization problem, with the added ‘small angle’ assumption that dl0, by considering a sequence of equivalent replacement problems.

Trajectories can be completely specified by dstep, tstep and .

Using yA(0)=l0, , we can obtain the vertical velocity component and the vertical position yA(t) by integrating the vertical acceleration . Having determined yA(t), we can determine FA(t)/lA(t) from equation (2.6)(2.7)We can use this f(t) in equation (2.5) to obtain for xA,(2.8)Solving for xA(t) requires two initial conditions. One of these, the horizontal position at mid-step, is given: xA(0)=0. The other initial condition can uniquely be determined as a function of f(t), dstep and tstep as follows.

First, observe that equation (2.8) is linear in xA. This linearity implies that the final horizontal position is given by , where k1,2 depend on f(t). Since xA(0)=0, we have . This proportionality implies that any given dstep=2xA(tstep/2) is obtained by a unique choice of the initial speed . Determination of xA(t) completes the description of the trajectory.

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## 3. Problem B: riding on a circular track with a vertical extendible leg

Now we simplify problem A to obtain problem B, motivated by the following two observations. First, in problem A, motion in a circular arc about the foot contact point does not require mechanical work ( since ). We would like a similar property for problem B. Second, we have assumed that the leg makes small angles with the vertical.

We combine these two ideas to obtain problem B (figure 4a). Imagine a circular track of radius l0 centred at (0,0) in the sagittal plane and a point mass body riding the circular track with a vertical telescoping leg. The foot of the vertical telescoping leg rides on the circular track with constant horizontal speed. This constant horizontal velocity component is assumed to be enforced externally, without any energetic cost to the biped.

Figure 4

Problem B: riding a circular track. (a) The radial telescoping leg of problem A has been replaced by a vertical telescoping leg in problem B. The foot of the vertical telescoping leg moves on the circular track with constant horizontal speed, enforced externally. (b) The action of the vertical telescoping leg riding on a circular track is shown in detail. The vertical velocity components must vanish at mid-step and at the end of step.

As shown in figure 4a, yB is the vertical position of the point mass from the ground. yV(t) is the length of the vertical telescoping leg and yC(t)=yB(t)−yV(t) is the height of the circular track from the ground, corresponding to where the point mass is at time t. FB is the vertical force transmitted by the vertical telescoping actuator.

### (a) Optimization problem B

The governing equation for the vertical motion is(3.1)where the vertical force FB(t) is bounded FminFB(t)≤Fmax. The goal is to minimize the sum of the positive and negative work CB of the vertical telescoping actuator(3.2)subject to the same boundary conditions as problem A(3.3)The one extra constraint in problem A, namely xA(tstep/2)=dstep/2, is automatically satisfied here by the assumption of constant horizontal speed v in problem B. A formula for in equation (3.1) might be obtained by twice differentiating with respect to time, where xB(t)=vt.

## 4. Small step-lengths: problem B is similar to problem A

An optimization problem is characterized by the objective function to be minimized and the constraints defining the space of feasible solutions. Here, we will show only the similarity of the objective functions in problems A and B rigorously, and provide only a heuristic justification for the similarity of the set of feasible trajectories.

### (a) Definition: analogous trajectories

As established in lemma 2.1, every trajectory in problem A is completely described by the specification of the speed v, step-length dstep and the vertical acceleration . For every such trajectory in problem A, we define a unique analogous trajectory in problem B as that having the same speed v, step-length dstep and vertical motion yB(t)=yA(t). Thus, a trajectory in problem B can be uniquely described.

### (b) Relation between the space of feasible trajectories in problems A and B

The vertical forces are equal for analogous trajectories in problems A and B because the vertical accelerations are the same, by definition. Therefore, |FB(t)|=|FAy|≤|FA(t)|. As a consequence of this inequality and the boundedness of the forces, the feasible region of problem A is not in general identical to the feasible region of problem B. However, we imagine that the disparity between the two feasible sets vanishes in some sense in the limit of small step-lengths and as we let the force bounds go to their respective infinities. We shall neither state this claim more precisely nor prove it rigorously.

### (c) A one-parameter family of gaits: limit of small step-lengths

In order to meaningfully take the limit of small step-lengths, we need to specify what is kept constant as we take the limit. In problem A, given a particular trajectory with speed v, step-length d0, step-duration t0 and vertical acceleration , we can generate a one-parameter family of gaits, all with the same average speed v but with different step-lengths, by simply using a time-stretched version of the vertical acceleration g(t). In other words, the vertical acceleration for a gait with a step-length dstepd0 is given by(4.1)Clearly, this one-parameter family is uniquely defined for every v, d0 and g(t). Having defined a one-parameter family of gaits parametrized by the step-length, we can meaningfully ask how the energetic costs scale as dstep→0. In the following, we use dstep=ϵ, as ϵ→0.

For analogous trajectories of problems A and B, we have CA/CB=1+O(ϵ2) when dstep=ϵ. In other words, for analogous trajectories of problems A and B,(4.2)

Note, simply showing that CACB is not meaningful as both CA and CB approach zero as dstep→0.

Lemma 4.4 shows that and lemma 4.5 shows that FA=(1+O(ϵ2))FB for all analogous trajectories. Together, these two lemmas establish the equality of corresponding terms in equation (4.2) up to a factor of (1+O(ϵ2)), as required. We now show lemmas 4.2 and 4.3, which will be used in lemma 4.4.

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In problem A, for any given and v, the fluctuations in the leg-length are much smaller than the initial leg-length in the limit of small step-lengths, i.e. lA(t)=l0(1+O(ϵ2)).

Noting that from equation (4.1), we have(4.3)because t′≤tstep=O(ϵ) and g(t) is bounded and independent of dstep. Combining this with xA(t)≤dstep and , we have lA(t)=l0+O(ϵ2)=l0(1+O(ϵ2)).

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In problem A, for any given g(t) and v, the fluctuations of the horizontal velocity component become much smaller than the average horizontal speed in the limit of small step-lengths. In particular, .

g(t) determines yA(t), which in turn determines f(t) in the differential equation for : (equation (2.8)). This gives since both t and xA(t) are O(ϵ). In other words, the fluctuations of the horizontal speed about the initial speed are O(ϵ2). However, this implies . Because v does not depend on ϵ, we have .

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In problems A and B, for analogous trajectories. In other words, the leg's extension rate in problem A is asymptotically equal to the extension rate of the vertical telescoping leg in problem B.

1. . Differentiating this equation with respect to t, we have . Noting that lA=l0(1+O(ϵ2))=yA(1+O(ϵ2)) from lemma 4.2, we have . Equivalently, .

2. The equation for the circular track of radius l0 is . This implies . Because yC=l0(1+O(ϵ2)), we have .

3. yV=yByC and yB=yA for analogous trajectories. Therefore, . Substituting the results of items (i) and (ii) above into this equation, we haveHowever, from lemma 4.3 we have and, as a corollary, xA(t)=xB(t)(1+O(ϵ2)). Using these, we have

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FA=FB(1+O(ϵ2)).

If FAy is the component of FA in the vertical direction, . By definition, the vertical accelerations and, therefore, the vertical forces for analogous trajectories in problems A and B are identical. Thus FB=FAy=FA(1+O(ϵ2)).

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## 5. Problem C: riding an elevator that accelerates down

### (a) Replace the circular path by a parabolic path

In the discussion of the equivalence of problems A and B, only item (ii) of lemma 4.4 above uses anything about the foot of the vertical telescoping actuator travelling in a circular path, and the circular path yC(t) enters the proof of lemma 4.4 only as its first derivative . Differentiating , we have .

We replace the circular track by a parabolic track, yp(t)=l0v2t2/2l0. We see that yp(t)=l0v2t2/2l0 describes a parabolic track in the xy plane by substituting t=xB/v in the expression for yp(t). If the foot of the vertical telescoping actuator travels with constant horizontal speed on this parabolic track, the corresponding vertical velocity component of the foot will be(5.1)Hence, replacing the circle yC by the parabola yp does not falsify lemma 4.4 or proposition 4.1. We make this replacement because the vertical acceleration of the foot () is conveniently constant when travelling on the parabolic track with constant horizontal speed.

Summarizing, the original problem A has been reduced to finding the path of the body that entails the least cost while riding a parabolic track with a vertical telescoping leg and constant horizontal velocity. The constraints are that the initial and final vertical speeds must be zero.

### (b) Galilean relativity

Because the horizontal speed is a constant in problem B, the body will have no horizontal motion when observed from an inertial frame moving with horizontal speed v. As the foot of the vertical telescoping actuator is riding the parabolic arc, when the constant speed is subtracted out, the foot will be simply seen to be accelerating downwards with , say.(5.2)

Note further that the cost to be minimized (equation (3.2)) and the boundary conditions (equation (3.3)) also depend on only the vertical motion of the telescoping leg. Therefore, we can rewrite problem B with the parabolic track, just in terms of the vertical coordinates.

### (c) Problem C: elevator problem

The foot moving downwards with constant downward acceleration ae=v2/l0 is most conveniently represented as being attached to an elevator moving downwards with constant acceleration ae (figure 5). At time t=0, both the person and the elevator are at the same position (without loss of generality). This initial state corresponds to mid-step, the apex of the parabolic or circular arc. The person can push or pull on the elevator with his vertical telescoping legs without affecting the elevator's motion. The person must have zero vertical velocity at time t=tstep/2. The objective is to meet this zero vertical velocity constraint by reacting against the elevator in a manner that incurs the least cost.

Figure 5

Problem C: riding an accelerating elevator. Both the point mass and the elevator start at the same position (START) with zero vertical speeds. The elevator maintains a constant downward acceleration ae. The point mass can push or pull against the elevator using arbitrarily strong vertical telescoping legs. When the elevator reaches END, the vertical speed of the point mass should again be zero. The objective is to ensure this by doing the least amount of work with the vertical telescoping legs.

Figure 5 shows the key variables in the elevator problem. Downward displacements and velocities are considered positive. ye is the position of the elevator, ym is the position of the mass m, and yr=yeym is the relative position of the elevator with respect to the mass. The vertical motion of the point mass is governed by the equation , where F is the compressive leg force. The differential equation for the relative position yr is(5.3)

At mid-step t=0, the positions and the velocities of both the elevator and the mass are equal to zero: ye(0)=ym(0)=yr(0)=0 and . Thus, this starting configuration is represented by the origin O in the yr plane (figure 6ac). Because , we have . The constraint that the mass must have zero velocity at the end of the step, , can be restated as . That is, in the yr plane, the final state should lie somewhere on the line AB corresponding to (figure 6ac).

Figure 6

Solution to the elevator problem. (a) Case 6.2, ae<g. The thin parabolic contours are constant energy lines. The goal is to go from the origin O to the dotted line AB with the least positive work. Optimal strategy is the vertical line OM. Two alternate suboptimal strategies ON and OPQ are shown as thin dashed lines. (b) Case 6.3, ae>g. Optimal strategy is the path ORS. An alternate path OT is shown as a thin dotted line. (c) Case 6.4, ae=g. All trajectories (e.g. OF, OG, OH) that do not backtrack have the same cost and hence are optimal. (d) The set of accessible velocity directions—directions in which a trajectory can proceed. In the upper half-plane, the trajectories can never move to the left. In the lower half-plane, the trajectories can never move to the right. No magnitude information is intended by the equality of the length of the arrows.

Thus, graphically, the goal in the elevator problem is to take the state of the system from the origin O in the yr plane to any point on the line AB in a given amount of time tstep/2 and with minimum cost.

Summarizing, the elevator problem is to minimize the total cost(5.4)subject to the terminal constraint , no explicit constraint on the force F(t), and the relative position yr(t) being determined by the differential equation and the initial conditions yr(0)=0 and . Recall that [P]+=P when P≥0, [P]+=0 when P<0, and [P]=[−P]+.

Equation (5.4) tells us that work is performed only when there is change in the distance between the elevator and the person.

### (d) Limit of infinite force-bounds

As in Srinivasan & Ruina (2006) and as alluded to in §2a, we formally wish to determine the limit of the sequence of optimal solutions as the force-bounds in the elevator problem increase without bound (Fmax→∞ and Fmin→−∞) for every combination of speed v and step-length dstep. To show the similarity of problem A to problem B (§4) in the limit of small step-lengths, we assumed that the leg forces were bounded. However, in the following discussion, we find it convenient to allow infinite leg forces, in particular, impulses that change the vertical speed instantaneously.

## 6. Optimal ‘gaits’ in the elevator problem

### (a) Walking and running in the elevator problem

We now describe how the two idealized gaits, inverted pendulum walking and impulsive running, described earlier in §1 and illustrated in figure 1, can be most naturally described in the context of the elevator problem. Inverted pendulum walking can be most naturally described as riding the elevator till t=tstep/2 and then pushing off impulsively against the elevator at exactly t=tstep/2 so that the vertical velocity of the person is reset to zero. Impulsive running, on the other hand, is jumping impulsively off the elevator at t=0 giving the point mass an initial vertical velocity that ensures that the vertical speed at t=tstep/2 equals zero. Before we show that these strategies are optimal (proposition 6.1), we present some results, to be used later, about the structure of the optimization problem in the yr plane.

### (b) Constant energy contours

Rearranging equation (5.3), we get . The instantaneous mechanical power of this force (as used in equation (5.4)) is given by(6.1)

Equation (6.1) simply notes that the leg power P is the time derivative of the total energy . Contours of constant energy E in the yr plane play a key role in the proof of optimality of walking and running below. Generically, the constant energy contours are parabolas in the yr plane (figure 6a,b). When ae<g, the parabolas open leftward (figure 6a), when ae>g the parabolas open rightward (figure 6b), and when ae=g, the parabolas degenerate into straight lines (figure 6c). The constant energy contours coincide with the trajectories corresponding to gravitational free-fall, with F(t)=0 and P(t)=0. The total cost Ce is simply the absolute sum of all positive and negative increments of E along a trajectory. A trajectory that goes from contour E1 to contour E2 incurs a cost of at least |E1E2|.

### (c) Feasible directions of progress in the yr− plane

The trajectory of the point mass in the plane is determined by the following equations (the second equation among which is the same as equation (5.3)).(6.2)The set of feasible tangent vector directions is shown in figure 6d. Because in the upper half-plane, any trajectory can only move to the right. Similarly, a trajectory can move only to the left in the lower half-plane (). Because the force can be unbounded, all rightward tangent vector directions are accessible in the upper half-plane and all leftward directions are accessible in the lower half-plane. On the line , a trajectory necessarily has a vertical tangent.

No information about the feasible magnitudes of the tangent vectors to a trajectory is intended by the equality of arrow lengths in figure 6d. Not every tangent vector magnitude is possible in a given tangent vector direction (equation (6.2)).

### (d) Time duration of a given trajectory in the yr− plane

The time taken to go between two points P1 and P2 along a trajectory in the yr plane is given by the integral if everywhere along the trajectory.

By choosing F=mgmae and , the right-hand sides of the equation (6.2) become identically zero. Thus, for this choice of leg force, the line becomes a line of fixed points. In other words, arbitrary lengths of time can be spent at any point on the line. Hence, if a trajectory P1P2 intersects the axis, the total time duration for the trajectory needs to be formally calculated as(6.3)The specific instances of the improper integral in equation (6.3) that we will consider will either be convergent by construction or their convergence will not affect the ensuing arguments.

For the elevator problem, when ae<g, ‘inverted pendulum walking’ minimizes the cost Ce, when ae>g, ‘impulsive running’ minimizes the cost Ce, and when ae=g, there are infinitely many strategies that have the same minimum cost Ce.

ae<g. Recall that the goal is to move from the origin O to the line AB with the least cost in figure 6a. We claim that the optimal strategy is to ‘walk’: that is, ride the elevator with zero relative speed (remaining at the origin O) and then at the last moment t=tstep/2, push off impulsively to make the vertical velocity of the point mass zero (jumping instantaneously from O to M).

By construction, it is clear that OM is a feasible trajectory. We now need to establish that every other feasible trajectory (such as the dotted lines ON and OPQ in figure 6a) necessarily incurs a greater cost. We shall consider two types of feasible trajectories.

1. Trajectories entirely in the upper half-plane. Trajectories entirely in the upper half-plane can never move leftward, as discussed earlier. Starting from O, such a trajectory can reach any point on AB in the right half-plane (yr=0 axis included). The point on the right half of AB that is on the least-energy parabolic contour is M. Hence, for instance, reaching N requires higher net positive work. The path OM reaches M by performance of only positive work and so achieves the minimum possible cost |EMEO|. Further, the only way to reach M from O while remaining in the upper half-plane (therefore, never going left) is the straight line OM, thus establishing uniqueness of the minimum.

2. Trajectories not entirely in the upper half-plane. Consider any such path OPQ, where PQ is the part of the path that lies entirely in the upper half-plane and Q is on AB. Then,(6.4)The cost of going from P to AB when restricted to the upper half-plane is minimized when Q is directly above P, repeating arguments from item 1. This minimum cost will be the same as that of OM, COM=|EMEO| (because the energy contour parabolas are equally spaced). Thus, COPQCOP+COM. If OP does not degenerate to the point O, the trajectory OP requires some positive work or negative work or both, so that the cost associated with OP is positive. Therefore, the cost COPQ is greater than the cost COM, establishing optimality for OM among all possible trajectories. ▪

ae>g. We claim that the optimal strategy in this case is to ‘run’. In other words, jump impulsively at t=0+ (going from O to R in figure 6b) and then fly freely without doing any further work (constant-E parabola RS with F=0). Here, R has , so that at t=tstep/2, and on AB, as required. Hence, path RS requires a time duration of exactly tstep/2.

By construction, the trajectory ORS is feasible (satisfies all constraints). A necessary condition for an alternate path OT to have a lower positive work than ORS is that OT never touches RS or an E-contour of higher energy. Hence, path OT lies entirely below the path ORS (figure 6b). We will now show that any such path OT with a lower positive work, whether or not it lies entirely in the upper half-plane, will necessarily take longer than tstep/2 to go from O to AB.

We use the equation (6.3) for the time of a trajectory. By construction of trajectory ORS, the time spent at O or along OR is equal to zero. Hence, the time duration for the path ORS is simply given by . Because yr(S)<yr(T) and at a given yr, on OT (when defined) is less than that on ORS, we have(6.5)Therefore, time for ORS<time for O+T≤time for OT. Summarizing, no path OT that requires less positive work than ORS can also reach AB in time. Hence, any trajectory that reaches AB at the same time as ORS must require at least as much positive work.

Cost CORS=COR=|EREO| is all due to positive work. Because any other feasible trajectory (satisfying all constraints) requires at least as much positive work, any other feasible trajectory incurs at least as much cost. Thus, ORS minimizes the cost. Uniqueness of the minimum ORS is easily demonstrated: any path not identical to ORS but only touching ORS from below and requiring same positive work or cost will require a longer time, repeating earlier arguments.

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ae=g. For this case, we show below by construction that there are infinitely many globally optimal trajectories.

Consider ‘non-backtracking’ trajectories that go from O to AB, over which never decreases (e.g. OF, OG and OH in figure 6c). In other words, and . Named examples of such trajectories include inverted pendulum walking, impulsive running, pendular running and level walking (Srinivasan 2006; Srinivasan & Ruina 2006).

When ae=g, the equation of motion reduces to . For non-backtracking trajectories, first, because , we have F≥0, i.e. non-tensional leg forces. Next, for such a trajectory to reach AB at t=tstep/2, we need . Clearly, infinitely many functions F(t)>0 satisfy this equation. Further, when ae=g, the power P from equation (6.1) simplifies to , which for non-backtracking trajectories is greater than zero by definition. In other words, all these trajectories reach AB by performance of no negative work and only positive work. As line AB is an isoenergy contour, we see that the cost incurred by every non-backtracking trajectory equals the energy difference between O and AB, clearly the minimum possible cost. Summarizing, all feasible gaits with non-tensional leg forces (F>0) are optimal when ae=g. ▪

## 7. Discussion

Recall that the elevator acceleration ae is the centripetal acceleration of the point mass body while travelling on a circular arc: ae=v2/l0 from equation (5.2). Thus, the conditions for walking (ae<g), running (ae>g) and the transition ae=g reduce to conditions on the so-called Froude number v2/(gl0) (Alexander 1976) or the non-dimensional forward speed (Srinivasan & Ruina 2006). We have shown that the gait transition (at least for low step-lengths) occurs at V=1.

One classical theory of why animals switch from a walk to a run (Alexander 1976; Usherwood 2005; Srinivasan 2006) is that travelling in a circular arc requires physically unreasonable tensional forces at higher speeds. This classical theory does indeed show why pendular walking is undesirable at high speeds, but it does not explain why another walking gait, one without such a tightly curved trajectory (like level walking), might not be optimal at higher or lower speeds.

Here, despite allowing for tensional leg forces, we find that all the optimal gaits have only compressive leg forces and, purely due to energetic reasons, that there is still a gait transition. Interestingly, the speed above which walking requires tension (V=1) and when walking becomes energetically unfavourable coincide, at least for small step-lengths (see discussion of longer step-lengths in Srinivasan & Ruina (2006)).

At the transition speed V=1 or ae=g, the elevator problem shows that all gaits that use only non-tensional leg forces (F≥0) have the same cost. The cost landscape is perfectly flat at the optimum. The flatness of the cost landscape in this simple model suggests that the actual optimal solution may be very sensitive to small changes in the model, at least near V=1. Indeed, numerical optimizations indicate that at slightly larger step-lengths an apparently unique third gait, ‘a pendular run’ seems optimal at these intermediate speeds (Srinivasan & Ruina 2006). In the calculation here, the pendular run is one of an infinite number of optimal gaits at exactly V=1.

The analytic demonstration in this paper depends on the limit of small step-lengths. However, the essential structure of the optimal solutions seem to be preserved as we go from small step-lengths to large step-lengths (as seen in the numerical explorations of Srinivasan & Ruina (2006)), except for details near V=1.

The allowance of infinite (impulsive) forces was another simplification. Presumably, more realistic finite force bounds will ‘soften’ the impulses, resulting in a smooth change in velocities instead of discontinuous changes. The essential structure of the optima seems to remain the same (Srinivasan & Ruina 2006), again except for details near V=1.

Finally, the biped model itself is obviously a gross simplification of a real bipedal animal, not taking into account, for instance, the presence of elastic elements (tendons that can store and release mechanical energy reducing requirements for muscle work), leg dynamics (except for simple leg swing costs in §2), muscle properties (including a more realistic metabolic cost not exactly aligned with mechanical work) and other anatomical realism (Anderson & Pandy 1999). Obviously, it is of interest which generalizations of the simple model here change the conclusions and in what way.

## 8. Conclusions

Legs of real animals perform positive and negative work during locomotion at finite speed. One might hope for ever-improving coordination strategies that could reduce this mechanical work to arbitrarily small levels. The present calculation suggests that animal legs cannot get away with zero positive work, even the absence of dissipation, when travelling at finite speeds and step-lengths (see also Chatterjee & Garcia (2000) for a related discussion in the context of passive–dynamic robots). We started with a minimal model of a bipedal animal and asked what the energy-optimal ways are for it to travel on its legs. We reduced this locomotion problem to a simpler and somewhat entertaining problem of finding the work-minimizing way of riding a downward accelerating elevator while ensuring that the vertical velocity at the end of the ride is zero. We solved this simpler problem analytically with elementary geometric arguments. We showed that the optimal solutions for the elevator problem are analogous to the solutions obtained for the original locomotion optimization problem of Srinivasan & Ruina (2006). In other words, we discovered that the two classic idealized gaits, inverted pendulum walking and impulsive running, indeed minimize a work-based metabolic cost. That both of these two classic idealizations that have been around for a few decades in various forms should prove to be work-minimizing for the same biped model at different regimes is interesting and is perhaps a tribute to those earlier researchers (Rashevsky 1948; Alexander 1976; Margaria 1976; Cavagna et al. 1977).

## Acknowledgments

This research was supported by an NSF robotics grant to A.R. (NSF CISE 0413139). M.S. was supported by NSF FIBR 0425878 during revisions of this manuscript. An anonymous reviewer pointed out a key error in an earlier version of this paper.

## Footnotes

• In running, ‘step-length’ includes the horizontal distance covered while one foot is in contact with the ground in addition to the distance travelled in flight.