Running is a speedy form of locomotion in which the body moves in a periodic fashion. The period of steady-state running is the time duration in between consecutive landing of the same foot on the ground. The term gait cycle refers to running events (changes in the configuration of the body and various transfers of energy) that occur during that period of time. In running, all feet are in the air sometime during the gait cycle. During the rest of the cycle only one foot is on the ground at a time. In contrast, in walking there is always at least one foot on the ground. Figure 8.13 presents the configuration of a runner's body in four distinct instances during a gait cycle.

The body configuration shown in Fig. 8.13a belongs to a time toward the end of the stance phase. As shown in the diagram, frictional force acting on the runner is propulsive during this later phase of the stance

CD o

T3 C

CD o

T3 C

0.1 0.2 time (s)

cd 2W o

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Figure 8.13a-f. Instances during a gait cycle of a middistance runner (a-d). The vertical and horizontal components of the ground force acting on an athlete running at 4 m/s are illustrated in (e). The average ground force is shown to increase with the speed of the runner (f).

period. Swing phase begins once the foot leaves the ground. There are brief periods of time at the beginning and toward the end of the swing phase when both feet are off the ground. Body configurations in these phases are shown in Fig. 8.13b and 8.13c. At the initial stages of the stance phase, friction force is in the opposite direction of locomotion (Fig. 8.13d). The vertical and horizontal ground forces acting on the runner during the stance phase are shown in Fig. 8.13e as a function of time. As can be seen, the vertical ground force is much greater in magnitude than the horizontal ground force. The sum of horizontal ground force in a gait cycle during steady-state running on a horizontal surface adds up to zero. Average ground force in a gait cycle increases with the speed of running (Fig. 8.13d). Thus, forces resisted by the muscles, tendons, and bones of the lower limbs increase with increasing running speed. This is why overuse injuries become more apparent at high running speeds.

Alexander (1992) estimated that about one-third of the energy turnover during each stance of a 75-kg person running at a speed of 4.5 m/s is stored as elastic energy in the Achilles tendon. Using a free-body diagram of the foot in contact with the ground, one can compute the tension in the Achilles tendon in terms of the ground force measured during running. The elastic strain energy stored in the tendon is proportional to the square of the tensile force carried by this tendon. The Achilles tendon is not the only component of the lower limbs where strain energy is stored. The arch of the heel, the quad tendon, and the patellar ligament also have the capacity to store considerable elastic energy. As a result, when muscles act as brakes during the initial phase of stance, about half the energy used in muscle contraction is returned by the tendons during the final phase of the stance, making running an efficient activity.

Example 8.8. Minimum Time Running: An Optimal Approach. Maronski (1996) introduced an optimal control approach to address this question: How should a runner vary their speed with distance to minimize the time during which they cover a given distance? The author developed an optimal velocity profile by formulating and solving a problem in optimal control theory. In his model, the racer is regarded as a particle of mass m. The vertical displacements of the body associated with the cyclic nature of the stride pattern are neglected. The equation of motion in the direction of running is assumed to be in this form:

in which v denotes the speed of the runner, t is the time, and fo and t are constants. The parameter mfo-q is the propulsive force, and mv/t is the resistive force. The propulsive force is the product of the mass m, the maximum propulsive force per unit mass fo, and the propulsive force setting

This is a dimensionless parameter with values in the range

The parameter t appearing in Eqn. 8.29a is called the constant damping coefficient. The overall resistive force may include air resistance (assistance) on the runner as well as the steady slope of the track. For simplicity, we will focus on a horizontal track. Admittedly, the mechanical model of running adopted by this article is rather elementary. The model could not be used, for example, in investigating overuse injuries during running. However, the analysis presented here may provide insights into the decision making process on minimizing running time during a competition.

The energy transformations in the competitor's body are represented by a differential equation on the power balance:

in which e denotes the actual reserves of chemical energy per unit mass in excess of the nonrunning metabolism, b is the recovery rate of chemical energy per unit mass, v fov is the actual mechanical power per unit mass used by the runner, and f is the efficency of transforming the chemical energy into the mechanical one.

The following intitial conditions are assumed:

in which x denotes the distance the runner has covered at time t. The runner should adjust their speed v during a race over a given distance Do to minimize the time t* of the event. This is possible because of the variations of the propulsive force setting v, which may be adjusted arbitrarily in any point of the race. In mathematical terms, this is formulated as follows.

Find v = v(x), v = v(x), and e(x) so that t*, given by the equation t* = idt = idx/v (8.32)

is minimized. The details of the optimization procedure may be found in Maronski (1996). His results indicate that the distance of the race may be broken into three phases:

1. The early phase of running (acceleration phase) during which the competitor moves with maximal propulsive force (v = 1).

2. The middle phase of the race (cruise phase) when the runner moves with partial propulsive force and the velocity is constant.

3. The negative kick phase where reserves of chemical energy has been depleted and the recovery rate of chemical energy is used to propel the runner at the final stage of the competition.

Let us illlustrate this model with a simple case in which the runner completes the race by going through phase 1 and ends the race at phase 2. The solution of Eqn. 8.29a for v = 1 and the initial conditions specified in Eqn. 8.31 can be shown to be equal to v = for [1 - exp (—t/r)j

Integration of this equation with respect to time gives the distance covered by the runner at time t:

The energy balance during the acceleration phase (q = 1) is given by the relation de/dt = b - (fo/p) v = b - (fo2 r/p) [1 - exp (-t/r)]

Integrating this equation with respect to time, we find e = eo + bt - (fo2 rt/p) - (fo2 r2/p) exp (-t/r)

Let us denote by t1 the time duration of phase 1. Then the velocity of the runner at the end of phase 1, the distance covered by the runner during phase 1, and the remaining reserves of chemical energy at the end of phase 1 are given by the relations:

e1 = eo + bt1 - (fo2 rtx/p) - (fo2 r2/p) exp (-t1/r) (8.34c)

Next, we consider the constant velocity phase. The equations governing the speed v and the rate of change of chemical energy e are then as follows:

(de/dt) = b - v12/(pr) e = e (t) = e1 + [b - v12/(pr)](t - t1) (8.35b)

The second phase ends when e reduces to zero.

Let us consider a competitor in a 400-m race on a horizontal track. The parameters indicating the energy and force producing capacity of the athlete are fo = 12.0 m/s2, r = 0.9 s, eo = 2,400 m2/s2, b = 42 m2/s3, p = 0.3

The athlete would like to know with which velocity he should run so that he finishes the race in phase 2, with eo all used up at the end of the run.

We first look at the equations of phase 1 and see how long would it take for the runner to complete phase 1. In this phase, the speed of the runner is given by the equation v1 = 10.8[1 - exp (-t1/0.9)]

This equation shows that the racer reaches the optimum speed within a fraction of a second into phase 1. For example, 2 s into the race, if phase 1 were to continue, that competitor would have a speed of 9.63 m/s. Thus, we will neglect the time and energy spent in phase 1 and assume that the competitor runs the race with constant speed. We want to determine his speed. As the chemical energy at the end of the second phase must be equal to zero, we set Eqn. 8.35b equal to zero. In this equation we set the chemical energy at the end of phase 1 (e1) equal to the initial chemical energy eo. This assumption is justified because the first phase takes only a few seconds. Thus, we have eo + [b - v2/^)] t* = 0 (8.36a)

The distance D of the racing event must be equal to the constant velocity times t*:

Eliminating t* from Eqns. 8.36a and 8.36b, we obtain eov + bD - D v2/(^t) = 0 (8.37a)

using the parameter values supplied as input this equation reduces to the following algebraic equation:

Solution of this equation yields the result that the runner completes the race in 48 s with an average speed of 8.32 m/s.

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Getting Started With Dumbbells

Getting Started With Dumbbells

The use of dumbbells gives you a much more comprehensive strengthening effect because the workout engages your stabilizer muscles, in addition to the muscle you may be pin-pointing. Without all of the belts and artificial stabilizers of a machine, you also engage your core muscles, which are your body's natural stabilizers.

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