Motion

is abduction, decreasing it is adduction (as in Figs. 1.9 and 1.10). Increasing $\theta _{\mathrm {knee}}$ is flexion, decreasing it is extension. Increasing $\theta _{\mathrm {ankle}}$ is dorsiflexion (flexion), decreasing it is plantarflexion (extension).

Fig. 3.1

OneKnees setAnkles ofHip conventionsRunning for hip (thigh), knee (lower leg), and ankle (foot) angles. Other conventions are also used, such as with the thigh angle being relative to the vertical and the ankle angle defined as $90^{\circ }$ less than here (based on [98])

Table 3.1

MusclesHip that move the thigh (upper leg) ( $\theta _{\mathrm {thigh}}$ , femur relative to hip)

Location, muscle	Action	Approximate origin/insertion
Pelvic girdle
Psoas major $^\mathrm{a}$	Flexes thigh	Ilium/upper femur
Iliacus $^\mathrm{a}$	Flexes thigh	Ilium/upper femur
Anterior thigh
Pectineus	Flexes, adducts thigh	Pubis/upper femur
Adductor longus	Adducts, rotates, flexes thigh	Pubis/upper femur
Adductor brevis	Adducts thigh	Pubis/upper femur
Adductor magnus	Adducts, rotates thigh	Pubis, ischium/upper femur
Gracilis	Adducts thigh, flexes lower leg	Pubis/upper tibia
Lateral thigh
Tensor fasciae latae	Aids thigh flex, abduction, rotation	Ilium/upper tibia
Posterior thigh
Gluteus maximus	Extends, rotates thighStability; stabilizesKnees knee	Hip/top femur
Gluteus medius	Abducts, rotates thigh	Ilium/GT femur
Gluteus minimus	Abducts, rotates thigh	Ilium/GT femur
Piriformis	rotates, abducts flexed thigh	Sacrum, ilium/GT femur
Obturator internus	Rotates, abducts flexed thigh	Obdurator foramen/GT femur
Obturator externus	Rotates thigh	Obdurator foramen/upper femur
Quadratus femoris	Rotates thigh	Ischium/GT femur

GT is the greater trochanter of the femur

$^\mathrm{a}$ TheMuscles iliopsoasThigh consistsFemur ofHip thePoints of origin psoasPoints of insertion majorFlexor muscles, ilicacus, and psoas minor

See [102] and [133] for more details

TheKinematics hipDegrees of freedom hasHip three degrees of freedom of rotation, as does a ball-and-socket joint (Fig. 1.4). This sagittal plane flexion/extension motion about the mediolateral axis of the hip in Fig. 3.1 can be larger in amplitude than the other two rotations (adduction–abduction about an anteroposterior axis and internal–external rotation about the longitudinal axis of the femur) (Table 1.10). While these two rotations are not negligible, we can ignore them here. TheKnees knee sagittal plane flexion/extension motion about the mediolateral axis is the one angular degree of freedom of the knee described in Fig. 1.4 as a “hinge.” The knee joint actually has three degrees of freedom, but the other two (internal–external rotation and adduction–abduction (varus-valgus)) are much less important because of soft tissue and bony constraints. AnkleAnkles motion of interest here is restricted to extension (plantarflexion) and flexion (dorsiflexion). These three major sagittal plane rotations ofKnees hipAnkles, kneeHip, and ankle constitute a three-segment model of the leg, which is in contrast to the one segment models of levers examined in Chap. 2. TheWalking head, armsRunning, and trunk also move relative to the rest of the body during the walking and running cycles.

Table 3.2

MusclesMuscles that moveLegs theThigh lowerHip Knees legQuadriceps ( $\theta _{\mathrm {knee}}$ , lowerHamstrings legPoints of origin relativePoints of insertion toFlexor muscles thigh)

Location, muscle	Action	Approximate origin/insertion
Anterior
Rectus femoris $^\mathrm{a}$	Extends leg, flexes thigh	Hip/patellar tendon
Vastus lateralis $^\mathrm{a}$	Extends leg	Upper femur/patellar tendon
Vastus medialis $^\mathrm{a}$	Extends leg	Upper femur/patellar tendon
Vastus intermedius $^\mathrm{a}$	Extends leg	Upper femur/patellar tendon
Sartorius	flexes leg; flexes, rotates thigh	Ilium/upper tibia
Gracilis (Table 3.1)
Posterior
Biceps femoris $^\mathrm{b}$	Flexes leg, extends thigh	Ischium, upper femur/upper fibula, tibia
Semitendinosus $^\mathrm{b}$	Flexes leg, extends thigh	Ischium/upper tibia
Semimembranosus $^\mathrm{b}$	Flexes leg, extends thigh	Ischium/upper tibia

$^\mathrm{a}$ The quadriceps femoris consists of the rectus femoris, vastus lateralis, vastus medialis, and vastus intermedius

$^\mathrm{b}$ The hamstrings consist of the biceps femoris, semitendinosus, and semimembranosus (see [102] and [133] for more details)

Table 3.3

MusclesAnkles that move the foot ( $\theta _{\mathrm {ankle}}$ , foot relative to lower leg) and toe

Location, muscle	Action	Approximate origin/insertion
Anterior
Tibialis anterior	Dorsiflexes, inverts foot	Upper tibia/first metatarsal
Extensor hallucis longus	Extends great toe, dorsiflexes foot	Mid fibula/great toe
Extensor digitorum longus	Extends toes, dorsiflexes foot	Tibia, fibula/small toe phalanges
Peroneus tertius	Dorsiflexes, everts foot	Lower fibula/metatarsal V
Peroneus longus	Plantar flexes, everts foot	Fibula/metatarsal I
Peroneus brevus	Plantar flexes, everts foot	Fibula/metatarsal V
Posterior
Gastrocnemius $^\mathrm{a}$	Plantar flexes foot, flexes leg	Lower femur/calcaneal tendon
Soleus $^\mathrm{a}$	Plantar flexes foot	Fibula, tibia/calcaneal tendon
Plantaris	Plantar flexion of foot/flexion of leg	Femur/calcaneal tendon
Popliteus	Rotates leg	Lower femur/upper tibia
Tibialis posterior	Inverts foot, plantar flexion of foot	Tibia, fibula/foot
Flexor hallucis longus	Flexes great toe, plantar flexes foot $^\mathrm{b}$	Fibula/phalanx of great toe
Flexor digitorum longus	Flexes toes, plantar flexes foot	Tibia/small toe distal phalanges

TheFeet calcanealLegs (AchillesCalcaneus) tendonAchilles tendon attachesGastrocnemius toSoleus thePoints of origin calcaneusPoints of insertion (heelFlexor muscles)

$^\mathrm{a}$ TheExtensor muscles triceps surae consists of the lateral and medial heads of the gastrocnemius and the soleus

$^\mathrm{b}$ Active in takeoff and tip-toeing

See [102] and [133] for more detailsMuscles

KinematicsKinematics entailsMuscles trackingSkeletal muscles the motion of the center of mass of the body, whether or not it is rigid, and tracking these angles (described here only in the sagittal plane) versus time and versus each other during aWalking walking orRunning running cycle. These three angles are changed by forces controlled by the different sets of muscles outlined in Tables 3.1, 3.2 and 3.3. Several of these muscles are depicted in Figs. 1.8, 2.24, 3.2, 3.3 and 3.4 (see [133] for more details).

Fig. 3.2

a AnteriorQuadriceps Muscles andThigh medial muscles of the thigh, with b quadriceps femoris, with much of the rectus femoris removed, c deeper muscles of the medial thigh muscles, d the iliopsoas (psoas major and iliacus) and pectineus muscles, and e articular muscles of theKnees knee (from [101]. Used with permission)

Fig. 3.3

PosteriorMuscles thigh and gluteal region muscles, showing a superficial muscles, b deeper muscles, and c even deeper muscles (from [101]. Used with permission)

Fig. 3.4

LateralMuscles viewsLegs ofFeet theTendons right leg, showing a muscles of the lower leg and foot and b the tendons extending into the foot (from [101]. Used with permission)

ManyMuscles of these muscles cross only one joint. However, muscles that cross more than one joint can be very important in motion and in multisegment modelingMultisegment modeling of motion. For example, in Chap. 2, we modeled the forces on systems with the leg by focusing only on theHip hip (with the leg straight), kneeKnees, orAnkles ankle joint and ignored the multisegment nature of the leg and multi-joint muscles. A model of the leg involving the knee andHip hip is shown in Fig. 3.5 [43]. Muscles 1 and 3 are hip and knee extensors and muscles 2 and 4 are hip and knee flexors. Each of these muscles crosses one joint. MusclesBiarticulate muscles 5 and 6 cross two joints, and so they are biarticulate muscles. Muscle 5 is both a hip flexor and knee extensor, while muscle 6 is both a hip extensor and knee flexor. Each model muscle corresponds to one or more real muscles, and can be activated to different degrees at different times to achieve the desired torques and motions of the leg. Muscles 1 and 3 can be activated to extend the hip and the knee, either with or without muscle 5 activation. With muscle 5 activated, the torque at the hip is reduced and that at the knee is increased, so this two-joint muscle can be thought to be redistributing some muscle torque and joint power from the hip to the knee. If instead muscle 6 is activated along with muscles 1 and 3, torque could be thought to be redistributed from the knee to the hip. The torque about one joint can be increased without affecting the other joint by activating one of these two-joint muscles in conjunction with one-joint muscles, as in Problem 3.21. (TheCycling importanceBicycle of two-joint muscles in bicycling is addressed in Chap. 5 and Fig. 5.39.)

Fig. 3.5

Model ofMuscles humanFlexor muscles legExtensor muscles withHip sixKnees effectiveBiarticulate muscles muscles that connect the hip and knee to the upper and lower legs. Muscles 1–4 cross one joint and muscles 5 and 6 cross two joints (based on [43])

Multiple muscles are activated in a coordinated manner during motion. For example, the muscle and joints involved in the up phase of a pushupPushups are shown in Table 3.4. The concentric action of these muscles provide the force for the upward motion [64]. (As seen in Chap. 5, concentric and eccentric muscle action entail muscle shortening and lengthening, respectively.) In the dip phase the eccentric action of these same muscles provide the braking action acting to control the downward force of gravity. In addition, several muscles are kept static to keep the body straight from head to heels during the dip: the cervical extensors (to keep the head and neck fixed), rectus abdominus and obliques (lumbar spine), and hip flexors (hips).

Table 3.4

Joints and muscles involved in the up phase of aExercise push-upPushups

Joint	Joint action	Segment moved	Major muscles
Shoulder joint	Horizontal adduction	Trunk	Pectoralis major
			Anterior deltoid
			Coracobrachialis
Shoulder girdle	Abduction	Scapula	Pectoralis minor
Shoulder girdle	Abduction	Scapula	Serratus anterior
Elbows	Extension	Upper arm	Triceps
Elbows	Extension	Upper arm	Anconeus
Wrists	Hyperextension	Forearm	Extensor carpi radialis
			Longus and brevis
			Extensor carpi ulnaris

Using data from [64]

3.2 Standing

3.2.1 StabilityStability

StabilityStanding isStability essential during standing, as well as during any type of motion. We will examine the overall stability of the body and then local stability.

Fig. 3.6

StabilityStability during standing for a model person a standing upright (stable) and with b torque diagrams, c leaning to her left (unstable), and d leaning to her right (unstable)

Overall Stability

FirstStability consider the body as a rigid mass. The criterion for overall stability during standing is for theCenter of mass center of mass to be over the area spanned by the feet, which means that a vertical line passing through the center of mass (the line of gravity) passesLine of gravity in this area of the support base (Figs. 3.6 and 3.7). OtherwiseTorques there would be torques that would not be balanced, and a “rigid” human would be unstable and topple over. Let us examine the torques about a rotation axis emanating from the center of mass. We see in Fig. 3.6 that when the center of mass is above this area spanned by the feet, the right foot causes a negative torque and the left foot causes a positive torque, and they cancel. (The torque due to the weight of the center of mass about this axis is zero.) When the center of mass is to the left of the foot area both torques are negative, and when it is to the right both torques are positive. The torques cannot balance in either case, and there is instability. Try it! (Stand and try to lean over.) We are most stable when the line of gravity is near the center of the support base. When we lean (to one side, forward, and so on) the line of gravity passes can pass through the periphery of this base (a less stable position) or outside this base (an unstable position). (TheCenter of gravity term center of gravity is also often used in stability analysis; technically it refers to theCenter of mass center of mass in the direction of gravity, which is better expressed as the line of gravity.)

Fig. 3.7

AStability standing person is stable when her center of mass is over the cross-hatched region spanned by her feet, as shown here with her feet a together, b apart, c and apart, with stability also provided by a cane or crutch (based on [20])

WeStability routinely alter our positions to try to maintain stability. When weCarrying carry an object in one hand, we lean or bend our bodies to the other side to keep the line of gravity of our body and the load within this support base. When weWalking walk into the wind, we improve stability by leaning forward so our line of gravity shifts to the edge of the support base nearest the oncoming wind (or in general, nearest the force we are encountering). Then, if the forcePulling, pushing pushes us back a bit, we can still maintain stability [64].

WeStability generally try to improve stability by adopting these and related strategies by (1) lowering our center of mass, such as aJumping jumper preparing to land (which also increases the collision time, as is discussed later this chapter), aWrestling wrestler trying to remain stable, and aGymnastics gymnast on a balance beam who feels balance being lost; (2) widening the base of support in the direction of force, such as aBoxing boxer adopting a forward-backward stance when delivering or preparing to receive a punch and someonePulling, pushing pushing and pulling heavy objects orCatching catching a ball; (3) modifying our posture so the line of gravity is near the edge of the support base, such as aFootball football player bending toward the person about to hit him, a person in a tug-of-war leaning backward when expecting to be pulled forwardBasketball, and thoseRunning running and changing directions frequently, as inBasketball basketball; (4) purposely increasing our mass, such asFootball football players who frequently suffer collisions deciding to gain weight; (5) aligning all segmented body parts or other components along the line of gravity and the support base, such as for people standing on each other; and (6) increasingFriction friction between all components in contact with the ground or other contacting surfaceBaseball, such as runners choosing to wear cleats and rubber-soled shoes and baseball hitters deciding to wearBatting batting gloves [64].

MobilityStability is often achieved at the expense of some stabilityStability. For example, asRunning runners andSwimming swimmers prepare to initiate forward motion, they lean forward to a barely stable position. In fact, wrestlersWrestling and judo practitioners (judokas) try to throw their opponents so their centers of mass move outside the stability regions [16]. In contrast, when we try to stop we adopt positions to make us more stable, such as when we adopt the stability position with one leg forward and one backward to help stop our forward motion when we throw a ball forward.

Fig. 3.8

SimpleStability models of people facing to the right: a an upright person with mass fairly well balanced in front and back, with the person’s center of mass shown—which is normally 58% of the person’s height over the soles of his feet, and b one with unbalanced mass, such as a person with a beer belly or a woman who isPregnancy pregnant, with separate centers of mass shown for the main body and the additional mass (which are approximately, but not exactly, at the same height). Clearly, the overall center of mass of the person in (b) is displaced to the right relative to that in (a). To prevent this being in front of the balls of the feet, the person often contorts his position, which can lead toInjuries muscle strain and a badBack back (see [20] for more details)

FigureStability 3.8 suggests different stability conditions for an upright person with mass fairly well balanced in front and back and one with unbalanced mass, such as one with a “beer belly” or a woman who isPregnancy pregnant. With its five disks, the lumbar curve in our spines (Fig. 2.34) is long and curved enough to keep our centers of mass over ourHip hips. This curvature is further enabled by the dorsal wedge shape of the two lowest lumbar disks (L4 and L5) in men and the three lowest in women (L3, L4 and L5). (These wedges make them thinner in the dorsal, backBack region.) ThisPregnancy extra wedge-shaped disk enables standing pregnant women to change their spine curvatures to keep their centers of mass over their hips. Otherwise these centers of mass would move forward by $\sim$ 3.2 cm by the end of pregnancy. Women do this by extending their lower backs to increase the lumbar lordosis (which is the angle between the tops of the highest and lowest lumbar disks) from $\sim$ 32 $^{\circ }$ as pregnancy starts to $\sim$ 50 $^{\circ }$ at full-term pregnancy [87, 139].

Local Stability

SuchStability overall stability presumes a rigid body and consequently rigid joints. Such stable jointsKnees indicate local stability. TheLigaments design of the human knee is an example of good human design. Four ligaments provide much of the needed lateral and cross support.

TheStability kneeFemur isTibia theFibula junctionSynovial joints ofKnees the femur in the upper leg, and the tibia and fibula bones in the lower leg (Fig. 1.3, right knee). The tibial (or medial) collateral ligament connects the femur and tibia and the fibular (or lateral) collateral ligament connects the femur and the fibula. These collateral ligaments prevent left/right sliding and provide overall leg tautnessStanding. They are slightly posterior, so they are taut during extension, and can produce the straight leg needed during standing.

TheseStability twoKnees ligaments are not sufficient for stability. There are two cruciate ligaments that cross each other, and prevent twisting and forward/backward sliding of theKnees knee. They cross in between the lateral and medial menisci, which are two curved half sockets of cartilage on the tibia. The posterior cruciate ligament connects the medial side of the femur to a more lateral part of the tibia. The anterior cruciate ligament connects a lateral part of the femur to a more medial part of the tibia. The anterior cruciate ligament is so-named because it attaches to the tibia at a position anterior to that of the posterior cruciate ligament (which is attached more posteriorly).

Local stability of theStability kneeKnees is also provided by the oblique ligaments, as well as others. ThePatella kneecapKneecap (patella), attached below to the tibia by thePatellar ligament patellar ligament and above to theQuadriceps tendonTendons of the quadriceps femorisMuscles muscle, also provides stability. Other functions of the kneecap are to lessen the compressivestress on the femur by increasing the area of contact between theKnees patellar tendon and the femur and to aidKnees knee extension by lengthening the lever arm of theMuscles quadriceps muscle via an anterior displacement of the quadriceps tendon (Fig. 2.26).

TheStability directionsMuscles of muscles (Chap. 5) can also affect stability and joint motion. For example, the biceps brachii provide a force that has a component normal to the lower arm that causes rotary motion and a component parallel to the lower arm that tends to stabilize the lower arm for joint angles $$${>}90^{\circ }$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_3_Chapter_IEq30.gif”></SPAN>, as in Fig. <SPAN class=InternalRef><A href=$ 3.46c below, or destabilize it (with destabilization tending to cause dislocation) for joint angles ${<}90^{\circ }$ , as in Fig. 3.46a [48]. This is explored in Problem 3.9.

How do our bone and muscle structures help promote local stability during locomotion? One reasonStability Muscles Stability humans are stable when upright on one or two feet is because we have relatively narrowHip hips and a hip ilium that is short and faces sideways, so that the small gluteal muscles emanate from it at the side of theHip hip [87]. This stabilizes the upper body over each leg duringWalking walking when only leg is on the ground (the stance phase). (In chimpanzees, the ilium is longer and points backwards so these muscles are in the back and do not provide stability when they try toStanding stand and toWalking walk. To promote some stability they need to tilt the hip above the leg off the ground and this swaying motion costs energy during standing and walking.) Stability duringRunning running is aided by the gluteus maximus, the largest muscle in the body in terms of mass and strength, which contracts during running to prevent the trunk from falling forward (flexion) and to decelerate the swinging leg; it is fairly inactive during level walking and even uphill walking [84, 87]. Our heads and gaze are stabilized during running byEyes eye and neck muscles that counter head rotational accelerations sensed by the semicircular canals and, each time your foot hits the ground, by the nuchal (neck) ligaments to theShoulders shoulder and arm on that same side that pull back your head as they lower [15, 87].

3.2.2 Forces on the Feet

ConsiderForces Feet Standing a person weighing 700 N (160 lb). During standing, each foot of the person must support 350 N. The total cross-sectional area of the two feet is about 350 cm $^{2}$ , so the average force/area or pressure on the feet is about 2 N/cm $^{2}$ . However, not all of the foot touches the ground. Most of the contact is at the ball and heel of each foot. Because of this much smaller area, the pressure on these more limited points of contact is much higher, about 10 N/cm $^{2}$ during standing on both feet.

The peak-forces on the feet are much higher during walking. DuringWalking walking there is usually only one foot on the ground, so the force on that foot is twice that with both feet on the ground and consequently the pressure is twice that during standing. Also, the normal forces involved in braking and forward propulsion during walking can be twice those during standing, which are only those needed to balance gravity. Overall, this leads to a peak force of about 1,400 N per foot, which translates to a pressure of about 40 N/cm $^{2}$ . Because the whole foot is not flat on the ground during most phases of walking, the contact area is less than during standing and the peak-pressure duringWalking walking can be about 60 N/cm $^{2}$ . ClearlyRunning, these forces are larger during running. Force plots are routinely measured during motion by using pressure sensors. (One such force plot is shown later in Fig. 3.19.)

3.3 WalkingWalking Running

We now examine the physics of walking and then running. But before we start this, let us let us look at our physiological features that help us move on land. As bipeds with long legs, we humans are very well designed to walkStability stably and to do so over long distances with low energy consumption (see Chap. 6). The multiple bones in the human foot are suspended by a series of muscles and ligaments, like cables as in a suspension bridge, to form the foot arch that assists stability during standing. It also helps walking; when the foot hits the ground the arch flattens, and as the arch is restored the foot is propelled forward by the toes. Standing and walking on two feet are also enabled by our vertical necks, our knees being angled under the hips, large hip and knee joints, and by our large heel bones [87].

We are not optimized to run very fast or stably, but we can run long distances at moderate speeds very well, as inMarathon running Running marathons. Our long legs (with diameters large enough to minimize stressFractures fracture, Chap. 4), largeHip hip, kneeKnees and ankleAnkles joints (to minimize damage to cartilage, Chap. 4) and protruding noses (to control water vapor levels, Chap. 9) benefit both walking and running. In addition, our short toes, foot arch, long Achilles tendonAchilles tendon (for energy storage, Chap. 4), largeMuscles gluteus maximus muscles (for torso and leg control), narrow waists, low shoulders, stabilized heads during running, slow twitch muscles in the leg (that favor endurance, Chap. 5), and ability to lose heat quickly for long times (by sweating, Chap. 6) benefit our running. The downsides of spreading our weight on two feet instead of four include proclivity to lowerInjuries backBack pain, sprained ankles, and knee problems [87].

In contrastWalking, chimpanzees use ${\sim } 4\times$ as much energy as we do to move a given distance when they walk on two feet or four feet (knuckle walking). When on two feet, they expend extra energy to maintain stabilityStability by swaying back and forth sideways and to prevent their very bentAnkles Hip Knees hips, knees, and ankles from collapsing due to gravity [132]. Chimps are optimized for speed (when galloping on their knuckles), power, agility and stability in motion, andClimbing; ascending, descending stairs climbing. They and other quadrupeds canRunning run much faster and, at these very fast speeds, for longer times than humans [87].

We willWalking see that there are similarities and differences in the cycles, kinematics, forces, and energetics of walking and running. One similarity is that in both the footPulling, pushing pushes back to get a forward reaction force. It is difficult to walk in sand because of the small reaction force on the feet until the sand is compressed; this compression is seen by your footprints. Footprints in sand are deeper in the front that at the heel because the foot is pushing off (backward) in the sand. At theRunning start of a race, sprintersSprinting Running push off from starting blocks for forward propulsion of the reaction force.

Fig. 3.9

3.3.1 Kinematics

The photographsWalking in Fig. 3.9 show the stages of walking during a step by the right foot. The stages of walking are diagrammed in Fig. 3.10. The hipKnees Ankles Hip, knee, and ankle sagittal plane angles are usually referenced to the time the foot touches the ground (foot strike, FS—or heel strike/contact) and when the toe leaves the ground (toe off, TO or foot offRunning) (Fig. 3.10). From foot strike to toe off is the stance phase (foot flat or loading, midstance, terminal stance, and then preswing). From toe off to foot strike is the swing phase (initial swing, midswing, and then terminal swing). For walking, the stance phase occurs during the first 60–65% of the cycle for each leg and the swing phase occurs during the last 35–40% of this gait cycle. There are two feet on the ground for the first and last $\sim$ 10% of the stance phase for walking, just after foot strike and just before toe off. Both feet are never off the ground at the same time. The evolution of the three motion angles during a walking cycle is shown in the top row of Fig. 3.11.

Fig. 3.10

WalkingWalking and runningRunning gait sequences. For walking, from foot strike (or heel strike) to toe off (or foot off) is the stance phase, with subdivisions sometimes called: foot flat or loading, midstance, terminal stance, and then preswing. From toe off to foot strike is the swing phase, with subdivisions called: initial swing, midswing, and then terminal swing (from [125]. Used with permission)

Fig. 3.11

AverageWalking Knees Ankles Hip values for sagittal plane joint rotation (in $^{\circ }$ ), moment (torque) per mass (in N-m/kg body mass), and power per mass (in W/kg body mass) for the hip, knee, and ankle during a step in walking (based on [55]. Also see [11])

AtWalking Hip Knees Ankles contact the upper leg (hip) is flexed by about $30^{\circ }$ (relative to the vertical in this analysis), then extends to about $10^{\circ }$ extension, and during preswing and then most of swing it flexes to $35^{\circ }$ (Fig. 3.11). At contact the lower leg (knee) is nearly extended ( ${\sim } 5^{\circ }$ ), flexes to $20^{\circ }$ during midstance, extends again, flexes to $40 ^{\circ }$ during preswing, peaks at 60– $70^{\circ }$ flexion at midswing, and then extends again. At contact, the foot (ankle) is slightly plantarflexed (0– $5^{\circ }$ ), extends a bit more to $\sim$ $7^{\circ }$ , dorsiflexes as the leg rotates forward over the planted foot to $15^{\circ }$ , dorsiflexes to $15^{\circ }$ for propulsion near the end of stance; after toe off, it dorsiflexes to get toe clearance and plantarflexes to prepare for contact again.

3.3.2 Muscular Action

Each flexion/extensionWalking Muscles action described in the kinematics is caused by the several flexor/extensorFlexor muscles Extensor muscles muscles for that joint. The muscular activity at each phase of the walking cycle is shown for the three sagittal joints in Figs. 3.12 and 3.13. Tables 3.1, 3.2 and 3.3 describe these muscle groups [102, 122].

Fig. 3.12

JointReaction forces Walking reaction forces transmitted through tibial plateau during one gait normal walking (solid line), along with the muscles forces—averaged for 12 subjects (based on [103, 112]Walking)

Fig. 3.13

Electromyographic (EMG)Muscles Electromyogram (EMG) Walking activity of the muscles of the a hipHip, b kneeKnees, and c ankleAnkles and foot during walking in a healthy person. Dotted regions represent activation $$<$$

20% of maximum voluntary contractions, while black regions represent

$$>$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_3_Chapter_IEq53.gif”></SPAN>20% activation. <SPAN class=EmphasisTypeItalic>White regions</SPAN> in (<SPAN class=EmphasisTypeBold>c</SPAN>) for the intrinsic muscles of the foot show at least some level of activation. The four phases of stance are, in time sequence, loading, midstance, terminal stance, and preswing; the three phases of swing are initial swing, midswing, and terminal swing (Also see Fig. <SPAN class=InternalRef><A href=

3.10.) (based on [11, 116])

Let us discussWalking Muscles Flexor muscles Extensor muscles which muscle groups are important during the stance phase and then the swing phase during walking. TheAnkles ankle dorsiflexors help to lower the leg to the ground at heel strike (the initial contact, and the first part of stance) by eccentric muscle contraction (see Chap. 5). The hipHip extensors act after contact in early stance to continue this deceleration by reversing the forward swing of the leg, while the intrinsic muscles and long tendons of the foot preserve the shape of the arch of the foot. The kneeKnees extensors help the person accept the body weight during loading and then stabilizeStability the knee during midstance. The ankle plantarflexors decelerate mass and control dorsiflexion up to mid-stance, while the hipHip abductors stabilize the pelvis. In terminal stance (heel off) and then preswing (toe off), the body is accelerated by the ankle plantarflexors by concentric muscle contraction (see Chap. 5) and then by the long flexors of the toes, while the pelvis is still stabilized, and the thigh is decelerated by eccentric contraction in preswing.

In initialWalking or preswing to midswing, the hipHip flexors advance the thigh by concentric contraction, theAnkles ankle dorsiflexors enable foot clearance from initial swing to midswing, and in terminal swing theHip hip extensors decelerate the thigh by eccentric contraction, theKnees knee-flexors decelerate the lower leg by eccentric contraction, the ankle dorsiflexors help position the foot, and the knee extensors extend the knee to prepare for foot contact.

AlternativelyWalking Muscles, we could analyze this motion by describing when each muscle group is activated. TheHip hip extensors act after contact in early stance and the hip flexors then advance the limb during preswing to midswing; from stance to swing the hip adductors and hamstrings are activated. The quadriceps muscles are activated before contact, then resist extensiveKnees knee flexion (by eccentric muscle contractions) during stance to avoid buckling due to gravity, and then extend the knee (by concentric muscle contractions) through early midstance as the leg rotates over the planted foot; they also control hip flexion prior to stance. The dorsiflexors fire during the swing phase to achieve foot clearance and during contact to control the foot placement. The plantar flexing soleus and gastrocnemius muscles fire during stance to control the motion of the leg over the foot and then to propel the foot and body forward. Figure 3.11 shows the joint moment and power generated during the gait cycle.

3.3.3 FrictionFriction

Is friction good or bad for walking and other locomotion? ItWalking Friction is both.

FirstWalking Friction, what is friction? On a microscopic scale it is a complex set of interactions between two surfaces and any medium in between them and it is not easily modeled. On a macroscopic scale it is modeled using aCoefficient of friction coefficient of friction $\mu$ . An object of weight $W_{\mathrm {b}}=m_{\mathrm {b} }g$ feels a normal force $N=m_{\mathrm {b}}g$ (see Fig. 3.14). Initially the object is static and the friction interaction is characterized by a static coefficient of friction $\mu _{\mathrm {s}}$ (Table 3.5). Consider a lateral force on that object F. For low forces, the frictional force is equal in magnitude and opposite in direction to this force and the object does not move. The object will move only if this force exceeds $F_{\mathrm {f} }=\mu _{\mathrm {s}}N$ . With the object moving, the net lateral force on the object is $F-F_{\mathrm {f}}$ , where now $F_{\mathrm {f}}=$ $\mu _{\mathrm {k}}N$ . $\mu _{\mathrm {k}}$ is the kinetic, dynamic, or translational coefficient of friction, which can be different from $\mu _{\mathrm {s}}$ .

Fig. 3.14

a StaticWalking Friction equilibrium of an object on a table and b lateral force required to overcome static friction to enable motion

Table 3.5

Static and kinetic (when noted) coefficient of friction, $\mu$ Friction Coefficient of friction Cartilage Basketball Tennis Running Walking Skiing Skating Knees Hip Tennis

Common objects and the body
Rubber (tire) on dry (wet) concrete road, static	1.0
Brake material on cast ion, dry; static	0.4
Brake material on cast ion, with mineral oil; static	0.1
Graphite on steel, static	0.1
Steel on unlubricated steel	0.7
Steel on lubricated steel or ice	0.15
Teflon on teflon (or steel)	0.04
Ice on ice	0.1
Ice on ice, 4 m/s, 0 $^{\circ }$ C, kinetic	0.02
Wood on wood	0.25–0.50
Articular cartilage in the human knee, kinetic	0.005–0.02
Articular cartilage in the human hip, kinetic	0.01–0.04
Athletic equipment (kinetic)
Skates on ice	0.003–0.007
Skis on snow	0.05–0.20
Tennis balls on wood	0.25
Basketball shoes on clean (dusty) wooden floor	1.0–1.2 (0.3–0.6)
Cleated shoes on astroturf	1.2–1.7
Jogging shoes on felt carpet, clay, asphalt	0.9–1.1, 0.3–0.5, 0.6–0.8

Using data from [19, 20, 110]

Fig. 3.15

In synovialWalking Boundary layers Cartilage Friction joints a boundary layer lubrication of the articular cartilage surfaces for relatively flat and nearby surfaces and b mixed lubrication at articular cartilage, showing boundary lubrication when the separation is on the order of the surface roughness and fluid-film lubrication in areas of more widely separated surfaces (based on [105])

Internal friction is usually troublesome. FrictionWalking in the kneeKnees joints during walking is bad. In fact, it is terrible. The knee is a synovial joint where the cartilage on the femur and tibia, and the surrounding synovial fluid contained in a sac, provide a very lowFriction Coefficient of friction coefficient of kinetic friction $\mu _{\mathrm {k}}=0.003-0.01$ . (This is smaller than most—if not all—man-made systems. See Table 3.5.) SuchSynovial joints Knees Hip synovial joints are found in the knees and hipsHip, which have relatively thick $\ge$ 5 mm cartilage, and in the finger joints, which have relatively thin cartilage layers, ${\sim }1-2$ mm. LubricationSynovial fluid at such articulating surfaces can occur by fluid-film lubrication, in which the surfaces are relatively far apart and are separated by synovial fluid; boundaryBoundary layers lubrication, in which the surfaces are very close together (1–100 nm) and the lubrication is provided by the lubricating glycoprotein on the surfaces (Fig. 3.15a); or by a mixed lubrication, in which there is a mixture of both (Fig. 3.15b). SynovialFriction Coefficient of friction joints with fluid-film lubrication have an extremely low coefficient of friction of $\simeq$ 0.02, whereas boundary-lubricatedBoundary layers surfaces typically have coefficients of friction two orders of magnitude higher than those that are fluid-film lubricated. (Articular cartilage is a fairly complex and dynamic material, containing synovial fluid, etc. (Fig. 4.51).)

LossWalking Friction Injuries of the synovial fluid from traumatic injury, lossAging of cartilage from long-term wear or injury, and excessive bone growth in the joint region—as in osteoarthritis—produceDiseases and disorders major pain during attempted motion. (The synovialInjuries fluid about the author’sKnees knee left the containing sac during a pantomime in junior high school, as a result of a collision of his knee with a door. Until his leg was stabilizedStability by a cast, his severe pain was lessened during locomotion by his walking backward, with no rotation in the afflicted joint.)

External friction can be necessary. WithoutWalking Friction Walking Running friction we could not walk or run. We would merely slipSlipping. When the heel of the foot touches the ground during walking, friction from the ground must slow it and then stop the forward motion of the foot (Fig. 3.16). The fraction of a second or so later when the ball of the foot is about to leave the ground, friction from the ground helps propel the body forward. This initialDeceleration deceleration and subsequent accelerationAcceleration of the foot typically requires forces of about 0.15 mg and so a static coefficientFriction Coefficient of friction of friction $$$\mu _{\mathrm {s}}> 0.15$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_3_Chapter_IEq70.gif”></SPAN> is necessary to walk. Figure <SPAN class=InternalRef><A href=$ 3.17 shows how this required minimum coefficient of friction increases with walking speed; it is about 0.17 for a typical walking speed of 3 mph (miles per hour) (a 20 min mile). Figure 3.18 shows how it varies during the course of a step, and how it increases when pushingPulling, pushing different loads [25]. Table 3.5 lists the kinetic coefficients of friction for several cases. ObviouslySlipping, walking on ice is difficult. (For other reasons, walking on thin ice is also ill-advised.)

Fig. 3.16

a HeelWalking contact stage and decelerating the foot and b toe-off stage and accelerating the foot

Fig. 3.17

CoefficientCoefficient of friction Walking of friction required during walking at different speeds (from [24]. Reprinted with permission of Wiley)

More generallyWalking Friction Walking Running Slipping, in every type of forward locomotion, the reaction force to pushingPulling, pushing back on the ground is what propels you forward (Newton’s Third Law of Motion) [16], and maintaining sufficient friction between the foot and ground prevents slippingSlipping. Depending on the activity, this can be aided by wearing shoes with rubber soles, which are often contoured to maximize friction and surface contact, or, on softer surfaces (grass), by wearing cleats and spikes that dig into the ground. Minimizing surface friction is the goal when you want to move fast on ice (skatingSkating in ice skating and hockeyHockey) or snow (skiingSkiing), but creating friction is sometimes important. When ice skaters need to accelerate or push off, they use the edge or tip of the skate blade to dig into the ice and push off, and use opposite motions to decelerate. DownhillSkiing skiers accelerate due to gravity, and use the ski edges and ski poles to change direction and tweak speed. In swimmingSwimming, swimmers propel forward by pushing water backward by using their arms and legs.

Fig. 3.18

DynamicPulling, pushing Walking Friction Coefficient of friction coefficient of friction required to push a weighted cart with different loads (from [23]. Reprinted with permission of Wiley)

3.3.4 Energetics

OurEnergetics Walking motion during walking is surprisingly and deceptively simple when we analyze the kinetic and potential energy of the body during a walking cycle. Let us choose the horizontal direction of walking as the x direction and the vertical direction as the z direction. The kinetic energyKinetic energy KE of an object isWalking $m_{\mathrm {b}}v^{2}/2$ , where $m_{\mathrm {b}}$ is the body mass and v is the magnitude of the velocity, the speed, of the center of mass. This total KE can be separated into that due to motion in the horizontal and vertical directions, as

$\begin{aligned} \mathrm{KE}= & {} \mathrm{KE}_{\mathrm {H}}+\mathrm{KE}_{\mathrm {V}},\end{aligned}$

(3.1)

$\begin{aligned} \mathrm{KE}_{\mathrm {H}}= & {} \frac{1}{2}m_{\mathrm {b}}v_{x}^{2}=\frac{1}{2}m_{\mathrm {b}}\left( \frac{\mathrm{d}x}{\mathrm{d}t}\right) ^{2},\end{aligned}$

(3.2)

$\begin{aligned} \mathrm{KE}_{\mathrm {V}}= & {} \frac{1}{2}m_{\mathrm {b}}v_{z}^{2}=\frac{1}{2}m_{\mathrm {b}}\left( \frac{\mathrm{d}z}{\mathrm{d}t}\right) ^{2}. \end{aligned}$

(3.3)

ThePotential energy Walking potential energy is

$\begin{aligned} \mathrm{PE}=m_{\mathrm {b}}gz_{\mathrm {CM}}, \end{aligned}$

(3.4)

where $z_{\mathrm {CM}}$ is the vertical position of the center of mass of the body.

WhenWalking we describe a system where energy is not supplied, say by an external driving force, or lost, such as to frictionFriction, the total energy E of this system is constant, where

$\begin{aligned} E=\mathrm{KE}+\mathrm{PE}. \end{aligned}$

(3.5)

We would not expect $\mathrm{KE}\,{+}\,\mathrm{PE}$ to be constant during walking, but it is almost constant.

TheWalking photographs in Fig. 3.9 show the stages of walking during a step by the right foot. In stage (a), the right foot has just decelerated, much of the heel of the foot is on the ground, and the foot is in front of the body’s center of mass. In stage (b), the right foot is squarely on the ground and under the center of mass. In stage (c), the right foot is behind the center of mass, mostly the ball is in contact with the ground, and forward propulsion is beginning. In stage (d), the front of the right foot is propelling the body forward, while the heel of the left foot has made contact with the ground and has begun to decelerate. We will examine the horizontal and vertical forces ( $F_{\mathrm {H}}$ and $F_{\mathrm {V}}$ ) and energies at each of these stages (Fig. 3.19).

Fig. 3.19

Fig. 3.20

Changes in mechanical energy during walkingWalking from force plates, with heel strike and toe off shown, for (from top to bottom curve) forward kinetic energy, gravitational potential energy plus vertical kinetic energy, and total energy (based on [22, 93])

The horizontalWalking forces are due toFriction friction. During stage (a) in Fig. 3.19, $F_{\mathrm {H}}$ slows the right foot, and during stages (c) and (d) it accelerates it. Because the horizontal speed of the center of mass of the whole body decreases when the right foot decelerates and it increases when the foot accelerates, during stage (a) KE $_{\mathrm {H}}$ decreases and during stages (c) and (d) it increases.

The center of massWalking Center of mass (CM) of the body is changing during this step. The right leg is becoming more vertical and straighter from stage (a) to stage (b), so the center of mass ( $z_{\mathrm {CM}}$ ) is rising and PE is increasing. From stages (b) to (c) to (d), the right leg is becoming less vertical and is also bending, so the center of mass is falling and PE is decreasing. After stage (d) the left leg gets straighter and PE starts to increase again.

BecauseWalking changes in the height of the center of mass must be the result of vertical forces, we see that from stages (a) to (b), $z_{\mathrm {CM}}$ begins to rise, which must mean that there is a net positive vertical force and $$$F_{ \mathrm {V}}>m_{\mathrm {b}}g$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_3_Chapter_IEq81.gif”></SPAN>. From stages (b) to (c), <SPAN id=IEq82 class=InlineEquation><IMG alt=$ begins to fall, which must mean that there is a net negative vertical force and $F_{ \mathrm {V}}<m_{\mathrm {b}}g$ .

So we see that fromWalking stage (a) to (b), the center of mass is rising and PE is increasing, while the right foot is decelerating and KE $_{\mathrm {H}}$ is decreasing. During stage (c), the center of mass is falling and PE is decreasing, while the right foot is accelerating and KE $_{\mathrm {H}}$ is increasing. We see that PE and KE $_{\mathrm {H}}$ are out of phase, one is decreasing, while the other in increasing, and vice versa. This also occurs in harmonic motion, such as that for a mass on a spring or a pendulum.

Figure 3.20 shows that duringWalking a step $\mathrm{PE}\,+\,\mathrm{KE}_{\mathrm {V}}$ and KE $_{\mathrm {H}}$ are out of phase. (KE $_{\mathrm {V}}$ also changes during these stages, as a result of the motion of the center of mass $z_{\mathrm {CM}}$ ; however, it changes relatively little.) More surprisingly, their sum $\mathrm{PE} + \mathrm{KE}_{\mathrm {V} } + \mathrm{KE}_{\mathrm {H}}$ is fairly constant during the step. Again, these two observations are reminiscent of harmonic motion, during which the PE and KE are out of phase and their sum does not change during a cycle. This exchange of kinetic and potential energy also occurs in a pendulum and for a rolling egg, as in Fig. 3.21, and this observation has led some to compare the leg during walking to a pendulum, by what is known as the ballisticBallistic model of walking Pendulum model of walking or pendulum model of walking. Before pursuing this analogy further, we will review harmonic and pendulum motion.

Fig. 3.21

The cyclic exchange of kinetic and potential energy in a pendulum and a rolling egg, with constant total energy, is similar to that in walkingWalking (from [99]. Used with permission)

3.3.5 Review of Harmonic Motion, Pendulums, and Moments of Inertia

Simple Harmonic Oscillator

TheHarmonic oscillator Simple harmonic oscillator position x from equilibrium for a body of mass m attached by a spring with spring constant k (Fig. 3.22) is determined from Newton’s Second Law of Motion $$F=ma$$

, where F is the restoring force on the mass due to the spring and $a=\mathrm{d}^{2}x/\mathrm{d}t^{2}$ is the acceleration of the mass. So

$\begin{aligned} m\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}= & {} -kx \end{aligned}$

(3.6)

$\begin{aligned} \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}= & {} -\frac{k}{m}x=-\omega ^{2}x, \end{aligned}$

(3.7)

where $\omega =(k/m)^{1/2}$ is the resonant frequency of this harmonic oscillator. The solution can be written as

$\begin{aligned} x=A\cos (\omega t+\phi ), \end{aligned}$

(3.8)

where A is an arbitrary amplitude and $\phi$ is a phase (which is not very significant here) (see Appendix C for more information about the solution).

Fig. 3.22

Simple harmonic oscillator, with an object with mass m attached to a spring with spring constant k

The kinetic energyKinetic energy during this motion is

$\begin{aligned} \mathrm{KE}=\frac{1}{2}mv^{2}=\frac{1}{2}m\left( \frac{\mathrm{d}x}{\mathrm{d}t}\right) ^{2}=\frac{1}{2} mA^{2}\omega ^{2}\sin ^{2}(\omega t+\phi ), \end{aligned}$

(3.9)

because $\mathrm{d}x/\mathrm{d}t=-A\omega \sin (\omega t+\phi )$ . The potential energy isPotential energy

$\begin{aligned} \mathrm{PE}=\frac{1}{2}kx^{2}=\frac{1}{2}kA^{2}\cos ^{2}(\omega t+\phi )=\frac{1}{2} mA^{2}\omega ^{2}\cos ^{2}(\omega t+\phi ) \end{aligned}$

(3.10)

using $\omega =(k/m)^{1/2}$ . The PE and KE are clearly $90^{\circ }$ out of phase during this oscillatory motion. The total energy,

$\begin{aligned} E=\mathrm{KE}+\mathrm{PE}=\frac{1}{2}mA^{2}\omega ^{2}\sin ^{2}(\omega t+\phi )+\frac{1}{2} mA^{2}\omega ^{2}\cos ^{2}(\omega t+\phi )=\frac{1}{2}mA^{2}\omega ^{2}, \end{aligned}$

(3.11)

is constant during this motion.

The radial frequency $\omega$ has units of radians per second. The cycle frequency f is $\omega /2\pi$ , which has units of cycles per second (cps) or Hz (Hertz). This means that a complete cycle (period) occurs in a time $T_{ \mathrm {period}} = 1/f = 2 \pi /\omega$ , during which time $\omega t$ changes by $2\pi$ rad.

Mass on a Pendulum

WePendulum will use the pendulum to model walking. In a simple pendulum, a ball of mass m is at the end of a plumb (string) of length L (Fig. 3.23a). For now we will assume that the plumb is massless. Assume that the plumb is at an angle $\theta$ from the vertical. The position of the ball along the arc of the plumb is $L\theta$ . The speed of the ball normal to the plumb is $v=L\mathrm{d}\theta /\mathrm{d}t$ , and the acceleration of the ball normal to the plumb is $L\mathrm{d}^{2}\theta /\mathrm{d}t^{2}$ . (For small angles, $|\theta | \ll 1$ , the lateral coordinate is $x\simeq L\theta$ , and so $\mathrm{d}^{2}x/\mathrm{d}t^{2}\simeq L\mathrm{d}^{2}\theta /\mathrm{d}t^{2}$ .) From the diagram, the tension T in the string is balanced by $mg\cos \theta$ , and there is a net acceleration of the ball $mg\sin \theta$ that tends to decrease the magnitude of $\theta$ . Newton’s Second Law $$F=ma$$

, can be written as (with $$ma=F$$

)

$\begin{aligned} mL\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}=-mg\sin \theta . \end{aligned}$

(3.12)

Fig. 3.23

a Mass on a simple pendulum and b complex pendulum

For small angle motion $(|\theta | \ll 1)$ , so $\sin \theta \approx \theta$ and

$\begin{aligned} mL\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}= & {} -mg\theta \end{aligned}$

(3.13)

$\begin{aligned} \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}= & {} -\frac{g}{L}\theta =-\omega ^{2}\theta . \end{aligned}$

(3.14)

Physically and mathematically this is similar to the simple harmonic oscillator, now with a resonant frequency $\omega =(g/L)^{1/2}$ and solution

$\begin{aligned} \theta (t)=B\cos (\omega t+\beta ), \end{aligned}$

(3.15)

where B is an arbitrary amplitude and $\beta$ is a phase (which again is not very significant here) (see Appendix C for more information about the solution).

The kinetic energyKinetic energy during this motion is

$\begin{aligned} \mathrm{KE}=\frac{1}{2}mv^{2}=\frac{1}{2}m\left( L\frac{\mathrm{d}\theta }{\mathrm{d}t}\right) ^{2}= \frac{1}{2}mB^{2}L^{2}\omega ^{2}\sin ^{2}(\omega t+\beta ), \end{aligned}$

(3.16)

because $v=L \mathrm{d}\theta /\mathrm{d}t=-BL\omega \sin (\omega t+\beta )$ . The potential energy isPotential energy

$\begin{aligned} \mathrm{PE}=mgL(1-\cos \theta )\approx \frac{1}{2}mgL\theta ^{2}, \end{aligned}$

(3.17)

because for small $\theta \; (|\theta | \ll 1),1-\cos \theta \approx 1-\theta ^{2}/2$ , and so

$\begin{aligned} \mathrm{PE}\approx \frac{1}{2}mgLB\cos ^{2} (\omega t+\beta )=\frac{1}{2} mB^{2}L^{2}\omega ^{2}\cos ^{2}(\omega t+\beta ) \end{aligned}$

(3.18)

using $\omega =(g/L)^{1/2}$ . Again, the PE and KE are out of phase during this oscillatory motion. As with the harmonic oscillator, the total energy

$\begin{aligned} E= & {} \mathrm{KE}+\mathrm{PE} \nonumber \\= & {} \frac{1}{2}mB^{2}L^{2}\omega ^{2}\sin ^{2}(\omega t+\beta )+\frac{1}{2}mB^{2}L^{2}\omega ^{2}\cos ^{2}(\omega t+\beta ) \end{aligned}$

(3.19)

$\begin{aligned}= & {} \frac{1}{2}mB^{2}L^{2}\omega ^{2} \end{aligned}$

(3.20)

is constant during the motion of the pendulum.

During walking the leg can be modeled as a pendulum because of this interplay between the KE and PE. We will use this pendulum (ballistic) model to determine the oscillation frequency of the leg $\omega =(g/L)^{1/2}$ and from this we will obtain the walking speed. Before doing this, we need to modify this simple pendulum model because the leg cannot be approximated by a mass (foot) at the end of a massless string (upper and lower legs). (We will also use this refined model in our analysis of throwing a ball.)

More Complex Pendulum

If we multiply both sides of (3.12) by L, we get

$\begin{aligned} mL^{2}\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}=-mgL\sin \theta . \end{aligned}$

(3.21)

The right side is actually the net torque $\tau$ acting on the ball of the simple pendulum about an axis at the top of the string, where the magnitude of the distance vector $$r=L$$

, the force on the ball is mg, and the angle from this $\mathbf {r}$ to $\mathbf {F}$ is $360^{\circ }-\theta$ , according to our convention (Fig. 3.23b). With $\sin (360^{\circ }-\theta )=-\sin \theta$ ,

$\begin{aligned} \tau =mgL\sin (360^{\circ }-\theta )=-mgL\sin \theta =mL^{2}\frac{ \mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}. \end{aligned}$

(3.22)

The $mL^{2}$ term is theMoment of inertia moment of inertia I of the ball about the pivot axis, which is at the hipHip. More generally, the moment of inertia of a complex object can be written as the sum of contributions from distinct masses $m_{i}$ each a distance $R_{i}$ from the axis

$\begin{aligned} I=\sum _{i}m_{i}R_{i}^{2} \end{aligned}$

(3.23)

or as an integral of mass density $\rho (\mathbf {r})$ at positions $\mathbf {r }$ that are a distance R from the axis, integrated over the volume V

$\begin{aligned} I=\int \rho (\mathbf {r})R^{2}\mathrm{d}V. \end{aligned}$

(3.24)

For these complex objects with moment of inertia I, total mass m, and center of mass a distance d from the axis

$\begin{aligned} \tau =-mgd\sin \theta =I\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}} \end{aligned}$

(3.25)

and with the small angle approximation $\sin \theta \approx \theta$

$\begin{aligned} \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}=-\frac{mgd}{I}\sin \theta \approx -\frac{mgd}{I} \; \theta . \end{aligned}$

(3.26)

The resonant frequency is now $\omega =(mgd/I)^{1/2}$ (see Appendix C for more information about the solution).

Also, the angular momentum of the pendulum $L=I\varOmega$ with $\varOmega =\mathrm{d}\theta /\mathrm{d}t$ , and $\tau =\mathrm{d}L/\mathrm{d}t$ .

Fig. 3.24

a TheRadius of gyration parallel axis theorem is illustrated for an object of mass m for an axis about the center of mass normal to the page and about an arbitrary parallel axis, and b the radius of gyration is schematically illustrated for moments of inertia about the center of mass and an arbitrary axis

3.3.6 Ballistic (or Pendulum) Model of Walking

Figure 3.10 shows that during the stance phase the center of mass moves similar to an inverted pendulumWalking Ballistic model of walking Pendulum model of walking, while in the swing phase the swinging leg moves like a pendulum. A simple form of the ballistic model says that during the swing cycle your leg is like a pendulum with radial frequency $\omega$ . This means that a complete cycle (period) occurs in a time $T_{\mathrm {period}}=2\pi /\omega$ . One forward swing of the leg corresponds to a half cycle, which takes $T_{\mathrm {half \;period}}=\pi /\omega$ . For a leg of length $L_{\mathrm {leg}}$ swinging through an arc of $\Delta \theta$ , the step length is ${\sim } L_{\mathrm {leg}}(\Delta \theta )$ for small $\Delta \theta$ (Fig. 3.25a), so weWalking might say that the stepping or walking speed v is this step length divided by $T_{\mathrm {half \;period}}$ or

$\begin{aligned} v= \frac{L_{\mathrm {leg}}(\Delta \theta )}{T_{\mathrm {half \;period}}} = \frac{\omega }{\pi }L_{\mathrm {leg}}(\Delta \theta ). \end{aligned}$

(3.29)

(We will revisit this equation later.) For ball on a string $\omega =(g/L_{\mathrm {leg}})^{1/2}$ , so $T_{\mathrm {half \;period}}=\pi (L_{\mathrm {leg}}/g)^{1/2}$ . For a body height $$H = 1.8$$

m, Fig. 1.15 and Table 1.6 show that the leg length is $L_{\mathrm {leg}} = 0.53\,H = 0.95$ m. With $g = 9.8\,\mathrm{m/s^{2}}$ , $T_{\mathrm {half \;period}} = 0.98$ s.

How does this compare to real walking? TheWalking speed for a slow walk is about $0.5\,\mathrm{m/s}\,=\,1.1$ mph. (Test this!) With a step length of 0.3 m, the time for a step is $T_{\mathrm {half \;period}} \simeq 0.60$ s. For a fast walk the speed is about $2.0\,\mathrm{m/s}\,=\,4.5$ mph and the step length is larger, 0.7 m, and so the step time is about 0.35 s. The model predictions are fair for the slow walk with this ball on a string model (0.98 s model result compared to 0.6 s actual time). Fast walking cannot be modeled as a free pendulum; it is clearly a forced pendulum. You can feel this in your legs as you change from a slow, leisurely pace to a fast, vigorous pace.

Can we improve the model of slow walking by using a moreWalking refined model for the mass distribution of the leg? For a leg of length L with a uniform linear density (and the same thickness throughout, which is not exactly true) and mass $m_{\mathrm {leg}}$ , as suggested by Fig. 3.25a, we need to use the expression from the more general, refined model (3.26) with $\omega =(m_{\mathrm {leg} }gd/I)^{1/2}$ and $T_{\mathrm {half \;period}}=\pi (I/m_{\mathrm {leg}}gd)^{1/2}$ . The distance of the center of mass from the axis is $d=L_{\mathrm {leg}}/2$ . The moment of inertia $I=m_{\mathrm {leg}}L_{\mathrm {leg}}^{2}/3$ . This is obtained by integrating over a uniform linear mass density (mass per unit length) $\lambda =m_{\mathrm {leg}}/L_{\mathrm {leg}}$ from $$R = 0$$

to $L_{\mathrm { leg}}$ in (3.24):

$\begin{aligned} I=\int \rho (\mathbf {r})R^{2}\mathrm{d}V=\int _{0}^{L_{\mathrm {leg}}}\lambda R^{2}\mathrm{d}R= \frac{m_{\mathrm {leg}}}{L_{\mathrm {leg}}}\frac{L_{\mathrm {leg}}^{3}}{3}=\frac{1}{3} m_{\mathrm {leg}}L_{\mathrm {leg}}^{2}. \end{aligned}$

(3.30)

(WithWalking a uniform cross sectional area A, $\rho (\mathbf {r})=\lambda /A$ and $dV=A \; \mathrm{d}R$ , so $\rho (\mathbf {r})\mathrm{d}V=\lambda \; \mathrm{d}R$ .) Then $\omega =(m_{\mathrm { leg}}gd/I)^{1/2} =(m_{\mathrm {leg}}g(L_{\mathrm {leg}}/2)/(m_{\mathrm {leg}}L_{ \mathrm {leg}}^{2}/3))^{1/2} = (3g/2L_{\mathrm {leg}})^{1/2}$ and $T_{\mathrm {half \;period}}=\pi (2L_{\mathrm {leg}}/3g)^{1/2} = 0.80$ s, again for $L_{\mathrm {leg}} = 0.95$ m. This prediction is fairly close to the estimated 0.60 s step time, and agreement is much better than for the ball on string model. We should remember that this is just a model; energy recovery is not perfect, and $\sim$ 70%, for walking (Fig. 3.26).

What happens if we refine the model a bit more? The leg doesWalking not have a uniform linear mass density; the upper leg (thigh) is heavier per unit length than the lower leg (with the foot) (Fig. 3.25b). The entire leg has a mass $m_{ \mathrm {leg}} = 0.161\,m_{\mathrm {b}}$ and length $L_{\mathrm {leg}} = 0.530\,H$ . Tables 1.6 and 1.7 and Fig. 1.15 show that the upper leg has a mass $m_{ \mathrm {u,leg}} = 0.10\,m_{\mathrm {b}} = 0.621\,m_{\mathrm {leg}}$ and has a length $L_{\mathrm {u,leg}} = 0.245\,H = 0.462\,L_{\mathrm {leg}}$ ( $$H =$$

body height), while the lower leg and foot have a mass $m_{\mathrm {l,leg}} = 0.061\,m_{\mathrm {b}} = 0.379\,m_{\mathrm {leg}}$ and a length $L_{\mathrm {l,leg}} = 0.285\,H = 0.538\,L_{\mathrm {leg}}$ .

The centerWalking of mass distance d from the hipHip joint and the moment of inertia about the hip are needed. TheCenter of mass center of mass of two objects is in general given by

$\begin{aligned} r_{\mathrm {CM}}=\frac{m_{1}r_{1}+m_{2}r_{2}}{m_{1}+m_{2}}, \end{aligned}$

(3.31)

which here is

$\begin{aligned} d= & {} \frac{m_{\mathrm {u,leg}}(L_{\mathrm {u,leg}}/2)+m_{\mathrm {l,leg}}(L_{\mathrm { u,leg}}+L_{\mathrm {l,leg}}/2)}{m_{\mathrm {u,leg}}+m_{\mathrm {l,leg}}} \end{aligned}$

(3.32)

$\begin{aligned}= & {} \frac{(0.621\,m_{\mathrm {leg}})(0.231\,L_{\mathrm {leg}})+(0.379\,m_{\mathrm {leg}})(0.731\,L_{\mathrm {leg}})}{m_{\mathrm {leg}}}=0.421\,L_{\mathrm {leg}}.\qquad \end{aligned}$

(3.33)

Fig. 3.26

Gait parameters as a function of speed for humans (thin curves) and horses (thick curves), for different gaits: walkingWalking (W), runningRunning (R), trotting (T), galloping (G), and human skippingSkipping (open circles) (from [99]. Used with permission. Also see [59])

The momentWalking Moment of inertia of inertia is (Fig. 3.25b)

$\begin{aligned} I= & {} \int _{0}^{L_{\mathrm {u,leg}}}\lambda _{\mathrm {u,leg}} R^{2}\mathrm{d}R+\int _{L_{\mathrm {u,leg}}}^{L_{ \mathrm {leg}}}\lambda _{\mathrm {l,leg}} R^{2}\mathrm{d}R=\frac{1}{3}\lambda _{\mathrm {u,leg}}L_{\mathrm {u,leg}}^{3}\nonumber \\&+\,\frac{1}{3}\lambda _{\mathrm {l,leg}}(L_{\mathrm {leg}}^{3}-L_{\mathrm {u,leg}}^{3}), \end{aligned}$

(3.34)

where $\lambda _{\mathrm {u,leg}}=m_{\mathrm {u,leg}}/L_{\mathrm {u,leg}} = 0.62\,m_{ \mathrm {leg}}/0.46 L_{\mathrm {leg}} = 1.34\,m_{\mathrm {leg}}/L_{\mathrm {leg}}$ and $\lambda _{\mathrm {l,leg}}=m_{\mathrm {l,leg}}/L_{\mathrm {l,leg}} = 0.38\,m_{ \mathrm {leg}}/0.54 L_{\mathrm {leg}} = 0.70\,m_{\mathrm {leg}}/L_{\mathrm {leg}}$ . With $L_{\mathrm {u,leg}} = 0.46\,L_{\mathrm {leg}}$ , we see that $I = 0.256\,m_{ \mathrm {leg}}L_{\mathrm {leg}}^{2}$ . (Problem 3.27 compares these models of the moments on inertia with measured data, using the parallel axis theorem and the radii of gyration in Table 1.9.)

ThenWalking $\omega =(m_{\mathrm {leg}}gd/I)^{1/2}=(m_{\mathrm {leg}}g(0.421\,L_{ \mathrm {leg}})/(0.256\,m_{\mathrm {leg}}L_{\mathrm {leg}}^{2}))^{1/2}=(1.64 g/L_{ \mathrm {leg}})^{1/2}$ . We see that $T_{\mathrm {half \;period}}= \pi (0.61\,L_{ \mathrm {leg}}/g)^{1/2} = 0.76$ s, which is 4.5% faster than the uniform density model, and not a great improvement over it.

What is wrong with this model? We have made some dubious assumptions. The swing phase of each leg occurs only over 35% of the gait cycle, so the legs are swinging only during 70% of the cycle and not the whole cycle. Equation (3.29) actually describes how fast the swinging foot moves and not the center of mass of the body. Figure 3.10 suggests that during the swing phase the center of mass moves only about half of the step length, and so if the center of mass is assumed to move at a constant speed during the gait cycle the speed v would be about half that given in (3.29). More important, the assumptions that the leg just passively swings and that it remains straight are far too simple. We should not expect this model to be quantitatively accurate [100, 129].

Fig. 3.27

3.3.7 Inverted Pendulum Model

DuringWalking Inverted pendulum model stance the leg is like an inverted pendulum, with the center of mass traveling on a circular arc of radius R (Fig. 3.27). To maintain the circular motion, a centripetal force with magnitude $m_{\mathrm {b}}v^{2}/R$ is needed, where v is approximately the walking speed. (This is the same as an upward “centrifugal force” in a frame that is rotating with the leg.) When this exceeds $m_{\mathrm {b}}g$ , the center of mass will not come down. This sets a maximum walking speed of $v_{\mathrm {max}}=(gR)^{1/2} \sim 3.0$ m/s for $$R = 0.9$$

m; at faster speeds people have to move in a different way: jogging, Runningrunning, etc. On the moon ( $g_{\mathrm {Moon}}=0.17\,g_{\mathrm { Earth}}$ ) and Mars ( $g_{\mathrm {Mars}}=0.4\,g_{\mathrm {Earth}}$ ) this limit is smaller, 1.2 m/s and 1.9 m/s, respectively. This low upper limit for the walking speed on the moon explains why Astronautsastronauts had to resort to Hoppinghopping on the moon to move fast. (This is the real “moonwalking,” for those who also recall Michael Jackson’s “moonwalk.”) Children (on Earth) have smaller R and consequently have to start running at lower speeds than adults (and sometimes need to run to keep up with adults). Adults usually start running at speeds even slower than the 3.0 m/s estimate, about 2.0 m/s, possibly in an effort by the body to minimize the metabolic energy cost per unit distance traveled (seeWalking Fig. 3.26). Speeds near 4 m/s are attainable in Race walkingrace walking because the racers distort their Hiphips at midstance in a way that flattens the trajectory of the center of mass—thereby increasing R.

3.4 Running

RunningRunning is not just fast walking; it is qualitatively different. DuringWalking walking, each foot is on the ground for more than half the time, and sometimes both are on the ground at the same time. Both feet are never off the ground at the same time. During running, each foot is on the ground for less than half the time, and—unlike walking—sometimes neither foot is on the ground. Race walkingThis distinction can have serious implications. Jane Saville, an Australian race walker, was in first place 200 m from the finish line in the 20 km race walking competition in the 2000 Summer Olympics. As photographs later confirmed, both of her feet were off the ground at the same time and she was disqualified. (Following standard practice, she had received two prior cautions before her red card disqualification.) She later won the bronze medal in this event in the 2004 Summer Olympics.

Sprinting RunningIn running a sprint (up to 400 m), the runner starts in a four-point stance and Pulling, pushingpushes off the starting block with both legs to get forward acceleration, with the body fairly horizontal at first to keep pushing forward. The sprinter then accelerates as fast as possible with the body becoming vertical to increase stride length, to attain and maintain maximum speed over the short time to the finish line [16]. In running a long-distance race, fast acceleration at the beginning is less important. The goal instead is to maintain a maximum speed with the needed endurance. (For world record running speeds see Table 6.34 below.)

3.4.1 Kinematics

RunningFigure 3.28 shows snapshots of different stages of running. One differenceWalking between walking and running is that the leg becomes almost straight during walking at midstance, but it never becomes straight during running. Another difference is that during walking the stance is longer than the swing, while the reverse is true for running. As seen in Fig. 3.10, during running the stance occurs during the first 40% of the cycle after foot strike, and the foot is off the ground for the next 60% of this cycle, from toe off to the next foot strike. The leg floats for the first 15% of the cycle after toe off, swings forward during the next 30%, and then floats for 15% until foot strike.

Fig. 3.28

Fig. 3.29

MeanKnees Ankles Running locus of a thigh angle versus knee angle and b ankle angle versus knee angle during running at 3.57 m/s, using joint angles as defined in Fig. 3.1 (based on [98, 141])

Running Joints Knees Ankles HipThe changes of sagittal joint angles are qualitatively the same as inWalking walking (Fig. 3.11 top), but with differences, such as with the knee angle. Figure 3.29 shows a representative example of how the locus of these angles varies with respect to each other during a gait cycle. For a running speed of 3.4 m/s ( $\sim$ 8 min mile), the thigh angle varies from $39^{\circ }$ (flexion) to $-21^{\circ }$ (extension) for a range of $60^{\circ }$ . At footstrike the thigh is about $25^{\circ }$ from the vertical because the hip is flexed; any initial angle change after FS is relatively small. During the extension prior to footstrike the knee angle is $13^{\circ }$ and increases to $41^{\circ }$ during cushioning flexion (for a range of $27 ^{\circ }$ during cushioning flexion), and then decreases to $15^{\circ }$ during extension during the propulsive phase. At swing phase flexion the angle is $104^{\circ }$ , for a total knee range of $90^{\circ }$ . For the ankle, the dorsiflexion prior to footstrike is $91^{\circ }$ , that at the stance phase is $112^{\circ }$ , and that at plantar flexion is $64^{\circ }$ , for a total range of ankle motion of $48^{\circ }$ . These values change with running speed and with grade for uphill and downhill running. Table 3.6 shows typical loci of the thigh and ankle angles versus the knee angle during a stride.

Table 3.6

Knees Ankles Hip RunningMaximum thigh, knee, and ankles angles (in degrees) for running at different speeds and grades on a treadmill Treadmill

	Level			Ramp
Speed (m/s)	3.4	4.2	5.0	3.4	3.4	3.4
Grade	0%	0%	0%	20%	0%	20%
Thigh
Flexion	39.0	41.0	46.6	25.8	39.0	54.5
Extension	20.6	23.1	28.0	18.4	20.6	18.7
Knee
Extension before footstrike	13.4	16.7	15.5	4.0	13.4	37.7
Cushioning flexion	40.6	39.4	42.9	41.2	40.6	46.3
Propulsion phase extension	15.1	16.8	14.2	32.1	15.1	13.0
Swing phase flexion	103.9	108.1	117.3	99.6	103.9	113.2
Ankle
Dorsiflexion before footstrike	91.3	91.8	92.5	–	91.3	97.6
Stance phase dorsiflexion	112.4	108.7	110.8	103.5	112.4	116.4
Plantar flexion	64.2	58.7	57.1	76.4	64.2	59.4

Using data from [98]

Your legsWalking Running are not the only things that move during running and fast walking (along with your center of mass). Your arms swing back and forth and your torso rotates, both $180^{\circ }$ out of phase with your legs. Why? Consider what would happen if this did not happen. With every stride by your right leg you would need to gain much positive angular momentum about the vertical axis, while with every stride with your left leg you would need to acquire negative angular momentum, as is evident in Fig. 3.30. This is just like the rotor in a washing machine. The motion in your arms and torso cancels $\sim$ 90% of these changes, as seen in this figure. Your swinging arms serve as passive dampers and are not driven by shoulder muscles [119].

Fig. 3.30

NormalizedRunning mean vertical angular momentaAngular momentum of the arms, head-plus-trunk, legs, and whole body during a running cycle, at 4.5 m/s. The normalized vertical angular momentum is obtained by dividing the vertical angular momentum (in kg-m $^{2}$ /s) by body mass (in kg) and the square of the runner’s standing height (height in m) (based on [69, 70])

3.4.2 Muscular Action

MusclesEach flexion/extension action described by kinematics is caused by the Flexor muscles Extensor musclesflexor/extensor muscles of that joint. The muscular activity at each phase of the cycle is qualitatively similar to that shown for the three sagittal joints forWalking walking in Figs. 3.11, 3.12 and 3.13. Tables 3.1, 3.2 and 3.3 describe these muscle groups. Figure 3.31 shows muscle activity during Runningrunning.

AgingWith advancing age people generally walk and run slower because they take shorter strides. In walking, there is less activation of knee extensor and ankle plantar flexor muscles, and some increased hip extensor activation to partly compensate for this [38]. In running, stride length and running speed typically decrease by $\sim$ 0.08 m and $\sim$ 0.10 m/s per decade from 20 to 60 years of age, due to lessened ankle muscle activation, without changes in the activation of the hip and knee muscles [39]. (Also see the discussions in Chaps. 6 and 9 on aerobic capacity and oxygen in the blood with advancing age and Fig. 6.12 on how world-class marathon running times increase with age.)

Fig. 3.31

MuscleMuscles Electromyogram (EMG) Running Muscle length Soleus Gastrocnemius length and electromyographic (EMG) activity versus time during one running cycle of the right leg, for the gluteus maximus (GM), vastus lateralis (VL), biceps femoris (BF), vastus medialis (VM), combined VL and VM (V), rectus femoris (RF), semitendinosus (ST), combined semimembranosus (SM) and ST (SMT), gastrocnemius (GA), soleus (SO), and tibialis anterior (TA) (from [65]. Used with permission). Also see [91]

3.4.3 Energetics

Running EnergeticsFigure 3.28 shows that the body slows down during stage (a) when the right foot is planted on the ground, and it accelerates during stage (c) as the right foot propels the body forward, as inWalking walking. This means that the horizontal kinetic energyKinetic energy

$\begin{aligned} \mathrm{KE}_{\mathrm {H}}(t=\mathrm {stage \;b})<\mathrm{KE}_{\mathrm {H}}(t=\mathrm {stage \;d}), \end{aligned}$

(3.35)

because stage (b) is right after (a) and stage (d) is right after (c). During stage (d) both feet are off the ground, the body is in the air, and the body Center of masscenter of mass and Potential energypotential energy are relatively high. In contrast, during stage (b) the right foot is on the ground, the right leg is bent (while it is straight duringWalking walking) and so the center of mass and potential energy are relatively low. Consequently,

$\begin{aligned} \mathrm{PE}(t=\mathrm {stage \;b})<\mathrm{PE}(t=\mathrm {stage \;d}) \end{aligned}$

(3.36)

and soRunning

$\begin{aligned} \mathrm{KE}_{\mathrm {H}}(t=\mathrm {stage \;b})+\mathrm{PE}(t=\mathrm {stage \;b})\ll \mathrm{KE}_{\mathrm {H}}(t=\mathrm {stage \;d})+\mathrm{PE}(t=\mathrm {stage \;d}). \end{aligned}$

(3.37)

We see that during running KE $_{\mathrm {H}}$ and PE (or really $\mathrm{PE}+\mathrm{KE}_{\mathrm { V}}$ , with KE $_{\mathrm {V}}$ being relatively small) are not out of phase and their sum changes much during each stride (see Fig. 3.32). The pendulum model cannot be applied to running at all [5].

Fig. 3.32

Changes in mechanical energy during runningRunning at two different speeds from force plates, for (from top to bottom curve) horizontal kinetic energy; gravitational potential energy plus vertical kinetic energy and gravitational potential energy; and total energy and gravitational potential energy plus horizontal kinetic energy (based on [22, 93])

Fig. 3.33

Vertical forceReaction forces Running exerted on the ground during a running stride (solid line, with dashed line approximation). This was calculated from force plates and was averaged for five-runners running at 4.5 m/s (based on [5])

RunningNot all of the lessened KE and PE during stage (b) is lost; some is saved within the body as stretched tendons and ligaments, as in a pogo stick. We will examine this now and revisit it again later after we have learned a bit more about the elastic properties of parts of the body in Chap. 4.

RunningFigure 3.33 shows a force plot of the Reaction forcesvertical normal forces on the foot on the ground during a run at 4.5 m/s, which corresponds to a competitive speed during a Marathon running Runningmarathon (2 h 37 min over 42.2 km or 26 miles, 385 yd). The peak force in this example is $2.7\,m_{\mathrm {b}}g=2.7$ $W_{\mathrm {b}}$ . For a 70 kg person (700 N, 160 lb), this is $\sim$ 1,900 N. Sprinting RunningDuring sprinting, this force can increase to $3.6\,m_{\mathrm {b}}g$ . (These forces are estimated later in this chapter in the discussion of collisions.)

Fig. 3.34

RunningThe torque balance in the quasistatic condition when this normal force peaks at 1,900 N is depicted in Fig. 3.34, which is similar to the second class lever discussed in Chap. 2. In equilibrium, the Achilles tendonAchilles tendon exerts an upward force of 4,700 N and there is a 6,400 N reaction force down on the foot from the rest of the body, because 4,700 $\sim$ 1,900 N (120/47 mm).

RunningThe Achilles tendonAchilles tendon is also known as the calcaneal (kal-kane’-ee-ul) tendon. It attaches the triceps surae (sur-ee’), which includes the Soleussoleus and the medial and lateral heads of the Gastrocnemiusgastrocnemius (gas-trok-nee’-mee-us) Musclesmuscles (Figs. 1.8 and 3.5, Table 3.3), to the Calcaneuscalcaneus (kal-cane’-ee-us), also known as the heel.

During this Marathon running Runningmarathon-type run, there is a loss of about 100 J of kinetic energy each time the foot touches the ground and decelerates to a stop. Not all of it is lost. Some of it is stored in the body, and a portion of that is recoverable, mostly from body components that store the energy like springs. To maintain a steady average running speed, the portion that cannot be recovered must be supplied by the body in the acceleration phase of the stride. How much does the body slow down during this step? For given mass and speed, just calculate the change in speed when 100 J of kinetic energy is lost (see Problem 3.34).

RunningApproximately 35 J of this 100 J is stored in stretching the Achilles tendonAchilles tendon. We will see in the discussion about mechanical properties in Chap. 4 how this value is consistent with the forces exerted on the Achilles tendon. Of this 35 J, about 93% is recoverable during the acceleration phase, so this tendon acts almost like a friction-free spring. During the step, the foot arch flattens and it takes about 17 J to do this. About 80% of this is recoverable. Some energy is also lost to the quadriceps tendon over the kneecap.Muscles Knees Patella Kneecap Much of the remainder is lost to muscles acting as brakes, and this is not recoverable. Overall, about 50 J is recoverable from the body’s springs and about 50 J must be supplied by the muscles during the acceleration phase. The notion that 50% of the energy is recoverable energy is not universally accepted; Fig. 3.26 suggests that only about 5% is recoverable. The materials properties of the body are clearly significant—for analyzing the recoverable energy storage and the needed braking-action of muscles. We will examine both soon. The long bones in the leg can also store energy, but we will see that they store little energy. There are also questions about the efficiency and timing of the energy return, and potential confusions between the loss of kinetic energy during a step with the much larger ( $\sim$ $5\times$ ) metabolic energy needed to replace this lost kinetic energy.

RunningGood running shoes lessen the impact in each step. Can they also help by storing energy as springs, thereby lessening the amount of energy needed in each step? It is not clear that this occurs. Good running shoes are squeezed up to at most 10 mm in the shoe sole under the forefoot upon impact. They can store about 7 J, of which about 54–66% could be returned, so it would seem that they can help a bit. However, by cushioning the forces felt by the feet, significantly less recoverable energy is stored by the body. Even more cushioning would produce instability during stance and take off. Furthermore, even if much energy could be stored elastically in shoes, it is not clear if much of it would be recoverable. Stored energy has to be released at the right time, over the right duration (frequency), and at the right position [110]. The resonance frequencies of unloaded shoes are $\sim$ 100–200 Hz and loaded shoes likely 20–30 Hz. Because any stored energy in the heel needs to be released 100–200 ms after foot contact and it is really released 50–100 ms ( $\sim$ 1/20 Hz) after contact, the timing does not seem right. Also, if energy is stored in one part of the shoe because of impact (say the heel) it does little good if the extra “kick” is needed in the front of the foot for forward propulsion. Sprinting RunningSprinters use running shoes with essentially no Paddingpadding during races because their body returns more energy without the extra padding. Sports injuriesDuring practice, they wear shoes with padding to limit the wear and tear on their legs, i.e., the stress on their tendons and so on. Later in this chapter we will discuss how foot injuries during running might be controlled by how we land with our feet and the running shoes we use (Fig. 3.33) [85, 87].

RunningA good running track can return up $\sim$ 12 J per step (Table 3.11, later). Such a track has a stiffness (spring constant) about $3\times$ the lower leg stiffness (80,000 N/m), and can increase running speeds by $\sim$ 2%. Such an increase in speed decreases Sprinting Running100 m sprint times by $\sim$ 0.2 s and could decrease Marathon running Runningmarathon times by $\sim$ 80 s. (Of course, marathons are not run on such a track.) These tracks can be tuned to have resonant frequencies of $\sim$ 2 Hz, which matches the times needed for the recovery of the stored energy. This impact of the step is a type of collision that will be discussed later in this chapter. This is also analyzed in a problem at the end of Chap. 4 (Problem 4.39).

RunningStill, more of the energy lost per step is recoverable inWalking walking than during running, perhaps somewhat over 70% (Fig. 3.26). HoppingKangaroos recover about 50% during each hop.

Accelerating to Sprint Speed

TheSprinting Running average speed for world class runners at the 100 and 200 m distances is about the same, $\simeq$ 10.4 m/s (in 2014) (see Table 6.34). It is slower at longer distances, because the stores of energy that activate muscles at the fastest rate are depleted after $\sim$ 200 m (Chap. 6). It is also slower at shorter distances, because it takes several strides to achieve peak speeds, starting from a dead stop. We now model this acceleration process by examining the energetics after each stride [67, 68, 78, 90]. The stride is step (c) in Fig. 3.28, with one foot on the ground propelling the body forward. The body consists of the other leg, which has mass $m_{\mathrm {leg}}$ , and the body above the legs (upper body), which has a mass $m_{\mathrm {b}}-2m_{\mathrm {leg}}$ , where $m_{\mathrm {b}}$ is the total body mass. (In this model we are ignoring the other stages: free flight, the slowing down of the body during touchdown, etc., and we are effectively incorporating all into this step (c).)

Sprinting RunningDuring the first stride, the foot on the ground, say the right foot, propels the upper body and the left leg to a speed with a force F as they move a distance L, and they attain a speed of $v_{1}$ at the end of the stride. The work done FL equals the kinetic energy in the upper body, $(m_{\mathrm {b} }-2m_{\mathrm {leg}})v_{1}^{2}/2$ , and that in the left leg, $m_{\mathrm {leg} }v_{1}^{2}/2$ ; the right foot is still on the ground, so the kinetic energy of the right leg is small and can be neglected. Therefore, after the first stride

$\begin{aligned} FL=\frac{1}{2}(m_{\mathrm {b}}-2m_{\mathrm {leg}})v_{1}^{2}+\frac{1}{2}m_{\mathrm {leg }}v_{1}^{2}=\frac{1}{2}(m_{\mathrm {b}}-m_{\mathrm {leg}})v_{1}^{2} \end{aligned}$

(3.38)

and

$\begin{aligned} v_{1}^{2}=\frac{2FL}{m_{\mathrm {b}}-m_{\mathrm {leg}}}. \end{aligned}$

(3.39)

Sprinting RunningIn the second stride the left foot is on the ground, does work FL, and accelerates the upper body from $v_{1}$ to $v_{2}$ and the right leg from 0 to $v_{2}$ , so

$\begin{aligned} FL= & {} \frac{1}{2}(m_{\mathrm {b}}-2m_{\mathrm {leg}})(v_{2}^{2}-v_{1}^{2})+\frac{1}{2} m_{\mathrm {leg}}v_{2}^{2}\nonumber \\= & {} \frac{1}{2}(m_{\mathrm {b}}-m_{\mathrm {leg}})v_{2}^{2}- \frac{1}{2}(m_{\mathrm {b}}-2m_{\mathrm {leg}})v_{1}^{2} \end{aligned}$

(3.40)

and

$\begin{aligned} v_{2}^{2}=\frac{1}{m_{\mathrm {b}}-m_{\mathrm {leg}}}(2FL+(m_{\mathrm {b}}-2m_{\mathrm { leg}})v_{1}^{2}). \end{aligned}$

(3.41)

Using (3.39) for $v_{1}^{2}$

$\begin{aligned} v_{2}^{2}=\frac{2FL}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\left( 1+\frac{m_{\mathrm {b} }-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\right) . \end{aligned}$

(3.42)

Sprinting RunningSimilarly, after the third stride, with the right foot on the ground,

$\begin{aligned} v_{3}^{2}=\frac{1}{m_{\mathrm {b}}-m_{\mathrm {leg}}}(2FL+(m_{\mathrm {b}}-2m_{\mathrm { leg}})v_{2}^{2}) \end{aligned}$

(3.43)

and using (3.42)

$\begin{aligned} v_{3}^{2}=\frac{2FL}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\left( 1+\frac{m_{\mathrm {b} }-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}+\left( \frac{m_{\mathrm {b} }-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\right) ^{2}\right) . \end{aligned}$

(3.44)

Sprinting RunningAfter n strides, $v_{n}$ is obviously given by

$\begin{aligned} v_{n}^{2}&=\frac{2FL}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\nonumber \\&\quad \times \left( 1+\frac{m_{\mathrm {b} }-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}+\left( \frac{m_{\mathrm {b} }-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\right) ^{2}+\mathrm { }\cdots \mathrm { }+\left( \frac{m_{\mathrm {b}}-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}} \right) ^{n}\right) . \end{aligned}$

(3.45)

The terms in the brackets form a geometric series $1+x+x^{2}+\cdots + x^{n}=(1-x^{n})/(1-x)$ for $$0<x<1$$

, so

$\begin{aligned} v_{n}^{2}=\frac{2FL}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\frac{1-\left( \frac{m_{ \mathrm {b}}-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\right) ^{n}}{ 1-\left( \frac{m_{\mathrm {b}}-2m_{\mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}} \right) }=\frac{2FL}{m_{\mathrm {leg}}}\left( 1-\left( \frac{m_{\mathrm {b}}-2m_{ \mathrm {leg}}}{m_{\mathrm {b}}-m_{\mathrm {leg}}}\right) ^{n}\right) .\qquad \end{aligned}$

(3.46)

Sprinting RunningThe final speed is (with $n\rightarrow \infty$ )

$\begin{aligned} v_{\mathrm {final}}=\sqrt{\frac{2FL}{m_{\mathrm {leg}}}}. \end{aligned}$

(3.47)

Table 1.7 shows that $m_{\mathrm {leg}} = 0.161\,m_{\mathrm {b}}$ , so a 70 kg sprinter, with $$F= 560$$

N and

m stride, attains a final speed of 10.0 m/s.

3.4.4 Bouncing Ball/Pogo Stick ModelBouncing

Much as walking can be modeled as a pendulum, Runningrunning can be modeled using mechanical analogies. Figure 3.35 shows the bouncing ball model. This is fairly equivalent to the even simpler pogo stick model, which does not have the complication of rotational ball motion. Kinetic energy Potential energyWhen the ball or stick is at its highest point, the gravitational PE is maximized, the KE is zero, and there is no stored elastic energy EE in the ball due to compression of the ball or in the pogo stick due to compression of the spring (which is also a form of potential energy). When it is about to touch the ground, the PE is almost minimized, the KE is almost maximized and the EE is still zero. During the elastic collision with the ground, the EE increases while the KE goes to zero, and then all of the EE is recovered, initially mostly as KE. The total energy $\mathrm{TE}=\mathrm{KE}+\mathrm{PE}+\mathrm{EE}$ is always constant [99]. There is exchange between the sum of the kinetic and gravitational potential energies with the elastic energy in a bouncing ball.Bouncing

Fig. 3.35

BouncingBouncing Running ball model of running, showing that the sum of the ball kinetic energy (dotted line), gravitational potential energy (thick line), and elastic potential energy (thin line) is constant (very thick line) during bounces (from [99]. Used with permission)

Fig. 3.36

ThreeWalking Running Hopping Skipping Bouncing mechanical models of locomotion, the (inverted) pendulum (left, for walking), single pogo stick (middle, for hopping and running), and double pogo stick (right, for skipping), with tension in pogo stick spring and the modes of energy exchange (from [99]. Used with permission)

Bouncing RunningThere is an analogy between running and this model, but it is not very strong. The PE and KE are both maximized during stage (d) in Fig. 3.26, while EE is zero. During stage (b), PE and KE are both minimized, and presumably there is EE stored in the tendons, foot arch, etc. Consequently, the phase of KE is different from that in the model. However, the analogy is fine for the phases of $\mathrm{PE}+\mathrm{KE}=\mathrm{TE}-\mathrm{EE}$ and of EE. Unlike walking, there is relatively little stored energy returned in running (Fig. 3.26); it is not very elastic.

Bouncing RunningFigure 3.36 shows the pogo stick model for running and a double pogo stick model for Skippingskipping. Skipping is a third, and fairly uncommon gait for bipeds. It differsWalking from walking in that it has a significant flight phase, and from running in that both feet can be on the ground (Table 3.7). Skipping is energetically very efficient: about 50% of the energy is recoverable (Fig. 3.26). These models for biped locomotion can also be used to model quadruped locomotion (Fig. 3.36). Whereas there are two main types of locomotion in bipeds (walking, running), there are three for quadrupeds (walking, trotting, galloping). The types of quadruped motion are described in Table 3.7.

Table 3.7

ComparisonRunning Walking Skipping of motion in bipeds and quadrupeds

Biped type	Foot sequence $^\mathrm{a}$	Features
Walk	R, L, R, L, etc.	No flight phase
Run	R, flight, L, flight, etc.
Skip
Right unilateral skipping	L, R, flight; L, R, flight; etc.	Left unilateral is R, L, flight, etc.
Bilateral skipping	L, R, flight; R, L, flight; etc.

Quadruped type (biped analog)	Foot sequence $^\mathrm{b}$	Features
Walk (walk)	FL, HR, FR, HL	Stride: up to 3 hooves on ground
Trot (run)	Two running bipeds w/50% phase lag	Rack is trot with 0% phase lag
Gallop (none)
Slow gallop or canter	HL, HR, FL simul.; FR flight	Right cantor; left switches R and L
Transverse gallop (fast run)	HL, HR flight; FL, FR flight	Right transverse
Rotary gallop (fast run)	HL, HR flight; FR, FL flight	Clockwise or counterclockwise

$^\mathrm{a}$ R right foot, L left foot

$^\mathrm{b}$ FL fore left, HR hind right, FR fore right, HL hind left (from [99])

3.5 Jumping

WeJumping will first examine the vertical jump and pole vault using an energetics approach and then the high jump and long jump by analyzing their equations of motion. In the last three athletic endeavors, kinetic energy from runningRunning is converted into potential energy. TheVertical jump Jumping vertical jump is a half-collision with the floor. (Here the half-collision being the second half of a collision with the floor.) Other types of jumping, as on trampolines and in others common inGymnastics gymnastics, involve full collisions with surfaces; how much energy these surfaces return to the jumper becomes important. This will be addressed in the discussion of collisions.

3.5.1 Vertical Jump

How high can you jump? It depends on howVertical jump Jumping fast you can take off. The four stick diagrams in Fig. 3.37 show the four stages of the vertical jump.

Fig. 3.37

Stick diagramVertical jump Jumping of a vertical jump, including initial standing in (a), crouching in (b) and then extension to take-off in (c) and free flight in (d) (based on [113])

FirstVertical jump Jumping the person stands upright. TheCenter of mass center of mass is about 1.0 m from the ground, 5 cm above theHip hip joints. Figure 3.37b shows the body after the crouch, after which the center of mass is about 0.65 m above the ground. The extension phase is next, which ends in takeoff shown in the third diagram. At takeoff (Fig. 3.37c) the center of mass is about 1.05 m above the ground, a bit higher than in Fig. 3.37a since the person is on his or her toes. The center of mass rises by a distance s during extension, and attains a vertical speed of $v_{ \mathrm {TO}}$ at takeoff. After takeoff the person is in free flight for a distance H and then the vertical speed v is zero, as shown in Fig. 3.37d.

TheKinetic energy Vertical jump Jumping kinetic energy at takeoff is all converted intoPotential energy potential energy at the apex of free flight. Therefore

$\begin{aligned} \frac{1}{2}m_{\mathrm {b}}v_{\mathrm {TO}}^{2}+m_{\mathrm {b}}g(1.05\mathrm { \;m})= & {} 0+m_{ \mathrm {b}}g(1.05\mathrm { \;m}+H) \end{aligned}$

(3.48)

$\begin{aligned} \frac{1}{2}m_{\mathrm {b}}v_{\mathrm {TO}}^{2}= & {} m_{\mathrm {b}}gH, \end{aligned}$

(3.49)

so $H=v_{\mathrm {TO}}^{2}/2g$ and the faster the takeoff speed the higher the jump.

DuringVertical jump Jumping extension both legs generate a normal force N(t) that leads to a net vertical force $F_{\mathrm {V}}(t)=N(t)-m_{\mathrm {b}}g$ . Figure 3.38 shows this normal force in measurements made on several men in the early 1930s at Columbia University [113]. This vertical force increases and then decreases, and while it is applied the center of mass accelerates upward and travels a distance s. The work W done on the center of mass during this phase is

$\begin{aligned} W=\int _{0}^{s}F_{\mathrm {V}}(t)\mathrm{d}z. \end{aligned}$

(3.50)

(Note that to perform this integration $F_{\mathrm {V}}(t)$ needs to be converted to a function of z.) This gets converted into the vertical kinetic energy so

$\begin{aligned} \int _{0}^{s}F_{\mathrm {V}}(t)\mathrm{d}z=\frac{1}{2}m_{\mathrm {b}}v_{\mathrm {TO}}^{2}=m_{ \mathrm {b}}gH \end{aligned}$

(3.51)

and

$\begin{aligned} H=\frac{\int _{0}^{s}F_{\mathrm {V}}(t)\mathrm{d}z}{m_{\mathrm {b}}g}. \end{aligned}$

(3.52)

Fig. 3.38

Kinematic and dynamic data during a vertical jump, including a reaction force from the ground (left scale) and height of the center of mass (right scale), and b vertical speed of the center of mass (left scale) and applied power (right scale, this power is the product of the force in (a) and speedVertical jump Jumping in (b)) (based on [113])

ForVertical jump Jumping $W_{\mathrm {b}}=m_{\mathrm {b}}g = 140$ lb (64 kg, 620 N), an average normal force during extension $\langle N(t)\rangle$ of 300 lb (Fig. 3.38), and $$s =$$

1.4 ft (0.43 m), we find

$\begin{aligned} H= & {} \frac{\int _{0}^{s}F_{\mathrm {V}}(t)\mathrm{d}z}{m_{\mathrm {b}}g}=\frac{(N(t)-m_{ \mathrm {b}}g)s}{m_{\mathrm {b}}g} \end{aligned}$

(3.53)

$\begin{aligned}= & {} \frac{{(\mathrm {300\,lb }-\mathrm {140\,lb})} \mathrm {1.4\,ft}}{\mathrm {140\,lb}} =1.6 \mathrm {\;ft \; (0.49 \;m).} \end{aligned}$

(3.54)

How long is takeoff? LetVertical jump Jumping us assume that the vertical acceleration is constant and equal to a during extension. (This would be true only if the vertical normal force were constant—and it is not.) From $F_{ \mathrm {V}}=m_{\mathrm {b}}a$

$\begin{aligned} a=\frac{N(t)-m_{\mathrm {b}}g}{m_{\mathrm {b}}}. \end{aligned}$

(3.55)

IfVertical jump Jumping the duration of takeoff is $\tau$

$\begin{aligned} v_{\mathrm {TO}}=a\tau \end{aligned}$

(3.56)

and using

$\begin{aligned} s=\frac{1}{2}a\tau ^{2} \end{aligned}$

(3.57)

we see

$\begin{aligned} \tau ^{2}=2\left( \frac{m_{\mathrm {b}}g}{N(t)-m_{\mathrm {b}}g}\right) \left( \frac{s}{g}\right) =2\left( \frac{\mathrm {140 \;lb}}{\mathrm {160 \;lb}}\right) \left( \frac{\mathrm {1.4 \;ft}}{\mathrm {32.2 \;ft/s}^{2}}\right) \end{aligned}$

(3.58)

because $\langle N(t)\rangle - m_{\mathrm {b}}g= 300\,\mathrm{{lb}} - 140\,\mathrm{{lb}} = 160\,\mathrm{{lb}}\,(710\,\mathrm{{N}})$ . The acceleration occurs over a time $\tau = 0.28$ s.

How are the jump height and takeoff time related? DividingVertical jump Jumping (3.56) by (3.57) eliminates a to give

$\begin{aligned} v_{\mathrm {TO}}=\frac{2s}{\tau }, \end{aligned}$

(3.59)

$\begin{aligned} H=\frac{v_{\mathrm {TO}}^{2}}{2g}=\frac{(2s/\tau )^{2}}{2g}=\frac{2s^{2}}{g\tau ^{2}}. \end{aligned}$

(3.60)

IfVertical jump Jumping you can decrease the extension time, you can jump much higher! We will see that extension times are shorter in people with larger fractions of fast-twitch muscles in their legs. HoweverMuscles, there can be a trade-off with shorter extension times. When muscles achieve steady-state motion, they develop less tension the faster they contract, which is much less than the isometric (fixed length) force that they can generate. (In Chap. 5 we will consider these and other features of muscle activation.) Measurements of the force on the floor using force plates have shown that while the peak force on a representativeRunning runner’s foot might be as high as 1,710 N, the force on each foot during his high jump turns out to be much less, 715 N [5]. This surprising observation is explained by the faster muscle contraction during jumping. Average forces are also expected to be lower in the high jump because muscles develop less tension when they are much longer or shorter than their resting length. Also, the momentum arm about theKnees knee joint is smaller for the bent knee in the squatting position of a vertical jump.

Fig. 3.39

a A mechanicalVertical jump Jumping model of vertical jumping, with b a simulated jump and c a simulated countermovement jump, withMuscles muscle force, ground reaction force, and angle plotted versus time in (b) and (c) (from [7], as from [6]. Used with permission)

Figure 3.39 shows a model of twoVertical jump Jumping variations of the vertical jump, which includesMuscles muscle activation, tendon stretching, and body response [7]. The squat jump starts in a squatting position. Initially, as the muscles contract there is only slow movement of theKnees knees and consequently much tendon stretching. Due to fast contraction, muscle forces in the squat jump are $0.2\times$ the isometric level. Just before liftoff these muscle forces and tendon recoil produce forces $0.4\times$ the isometric level. The countermovement jump starts with an upright person first falling freely to the starting position of the squat and then jumping upward. The muscle forces initially generated are larger than the isometric values because the muscles are initially braking (eccentric contraction, Chap. 5), and this leads to even more tendon stretching and upward force generated, and a higher elevation jump than the squat jump. This countermovement jump is higher also because muscles are stretched immediately before they are contracted, leading to greater forces during this contraction [145] (as in the later phases of this jump). (This effect is not included in the Fig. 3.39 model.)

3.5.2 Pole Vault

The pole vaultPole vault is not really a jump, but a propulsion. Still, it is fairly similar to the long jump in that horizontalKinetic energy kinetic energy is converted to a propulsion. More specificallyPotential energy, the pole vaulter’s horizontal kinetic energy is mostly converted into elastic potential energy stored in the pole, which is then mostly converted to gravitational potential energy of the pole vaulter (Fig. 3.40). A fast runningRunning pole vaulter can propel above and beyond the bar if the two “mostly”s in the previous sentence are really “essentially all”s. This happens with good techniques and good poles. To propel over the bar there has to be some remnant horizontal kinetic energy, but this necessary amount is minimal.

Fig. 3.40

Schematic of the heights involved in the pole vaultPole vault (based on [66])

Figure 3.40 shows the relevant heightsPole vault: the height of theCenter of mass center of mass at takeoff (the takeoff height, $H_{1}$ ), the maximum increase in the height of the pole vaulter’s center of mass while on the pole (the swing height, $H_{2}$ ), the further increase in height due to vertical kinetic energy after release of the pole (the flight height, $H_{3}$ ), and maximum height of the center of mass above the cross bar (the clearance height, $H_{4}$ —which could be negative). The takeoff angle of world-class pole vaulters using fiberglass poles is $\simeq$ 13–15 $^{\circ }$ (relative to the vertical).

The ideal case of total conversion of one type of energy toPole vault another gives a surprisingly good prediction of actual pole vaulting performance. Initially the pole vaulter accelerates to a speed $v_{\mathrm {H,i}}$ and has a center of mass $h_{\mathrm {CM}}$ above the ground ( $H_{1}$ in Fig. 3.40). DuringRunning running $h_{\mathrm { CM}}$ is approximately 0.9 m, corresponding to the slightly crouched position during running. The goal is to convert the initial $E_{\mathrm {H,i}}=mv_{\mathrm {H,i} }^{2}/2$ into the potential energy. Over the bar, the pole vaulter has a center of mass $h_{\mathrm {bar}}+h_{\mathrm {min}}$ , where $h_{\mathrm {bar}}$ is the height of the bar and $h_{\mathrm {min}}$ ( $H_{4}$ in Fig. 3.40) is the distance the center of mass of the vaulter needs to be above the bar without touching it, which we will take as $\sim$ 0.1 m. This corresponds to an increase in the height of the center of mass by $h_{\mathrm { bar}}+h_{\mathrm {min}}-h_{\mathrm {CM}}$ . A good pole vaulter can achieve a speed $v_{\mathrm {H,i }}$ $\sim$ 9.5 m/s. (World-classSprinting Running sprinters in a 100 m race, who do not carry poles while running, can attain speeds of 10.4 m/s.)

If the initialPole vault kinetic energy is converted into potential energy over the bar:

$\begin{aligned} \frac{1}{2}mv_{\mathrm {H,i}}^{2}= & {} mg(h_{\mathrm {bar}}+h_{\mathrm {min}}-h_{\mathrm {CM}}) \end{aligned}$

(3.61)

$\begin{aligned} h_{\mathrm {bar}}= & {} h_{\mathrm {CM}}-h_{\mathrm {min}}+\frac{v_{\mathrm {H,i}}^{2}}{2g}. \end{aligned}$

(3.62)

With

m/s $^{2}$ , we seePole vault that $h_{\mathrm {bar}} = 5.4$ m. In a good pole vault $h_{ \mathrm {bar}} \sim 5.4$ m; the world record is 6.16 m (in 2014) (see Problem 3.50). Of course, the pole vaulter is still moving forward over the bar and so $v_{ \mathrm {H}}$ does not actually become zero. Also, the maximum height of a pole vault jump can be higher than predicted here because the vaulter can extend his or her arms toPulling, pushing push up from the pole when it is vertical, just before releasing it.

The polePole vault is very elastic. It stores and returns energy to the pole vaulter much more efficiently than do human body components during jumping and running. There is an optimal pole stiffness that maximizes performance [134]. If the pole is too stiff, it straightens before the athlete is at the maximum height and the polePulling, pushing pushes the athlete horizontally away from the bar. If the pole is not stiff enough, it straightens too slowly and the athlete passes the bar before attaining the highest position.

Fig. 3.41

Multisegment model of a high jump usingHigh jump Jumping the Fosbury flop method, showing that the center of mass (the closed circle) of the jumper is always below the bar (in this example). Different heights during a high jump are also shown, with $H_{1}$ being the initial height of the center of mass, $H_{2}$ the maximum elevation of the center of mass, and $H_{3}$ the distance that the final center of mass is above the bar. The arrows are shown assuming the highest center of mass is above the bar, so $H_{3}$ would really be negative for a successful Fosbury flop (based on [66])

3.5.3 High Jump and Long Jump

Multisegment modeling High jump Long jump JumpingWe analyzed the vertical jump and pole vault by considering the energetics at different stages of the motion. Analyzing jumping by following the equations of motion is more complex and requires the generalMultisegment modeling multisegment model presented later this chapter. We now examine a model of jumping from a running start that is simpler than this to learn about the differences in the techniques needed to be successful high and long jumpers [3]. The high jump using the now-standard Fosbury flop method is depicted in Fig. 3.41 and the long jump is illustrated in Fig. 3.42. This model [3] focuses on one joint (theKnees knee) and two segments (the upper and lower legs, which are assumed to be of equal length). Though this model is conceptually simple, it still needs to be solved numerically. The energetics of the high jump and the long jump are analyzed in the problems at the end of this chapter (Problems 3.51–3.54, and 3.55–3.57, respectively).

Fig. 3.42

Different lengths during a hang-style long jumpLong jump Jumping (based on [66])

Fig. 3.43

Schematic of a force model of the high jump and long jump, with torques applied about the knee joint byMuscles muscles (based on [3])

The jumperKnees runs and plants one foot, which is considered a point at the end of the lower leg and the origin of a two-dimensional system (Fig. 3.43); the center of mass of the body is at (x, y) and at the time of implant it is moving horizontally at the approach speed of v. The line from the center of mass to the origin makes an angle $\phi = \arctan (-y/x)$ with the horizontal. In this simple model the center of mass is assumed to be at the height of the hip joints, although it is really 5 cm above them as noted in Chap. 2. The angle of the knee joint is $\theta$ , with $(x^{2}+y^{2})^{1/2}=2L \sin (\theta /2)$ , where L is the length of the upper and lower legs. The center of mass is followed from the time the foot is planted until takeoff and then after takeoff. There is a normal or reaction force, N, on the foot until takeoff, which, along with gravity, are the only forces exerted on the center of mass.

Lift is provided by the torque $\tau$ that is developed by the knee extensorMuscles muscles, which power the upper and lower legs (see Fig. 3.2); it is assumed that there is no torque generated about the hip joints. This knee torque produces a force at the ground with magnitude N, where $\tau =NL \sin (\pi /2-\theta /2))= NL \cos (\theta /2))$ , so the reaction force from the ground has magnitude $N= \tau /(L \cos (\theta /2))$ ; it needs to exceed the body weight to enable liftoff. Because there are no torques about the hip joints in this model, which is also the center of mass here, this force is parallel to the vector from the planted foot to the hips and not actually normal to the surface, as is shown in Fig. 3.43.

In Chap. 5 we will seeHill force–velocity curve Muscles that the maximum forces developed by muscles depend both on their length and contraction speed (Fig. 5.29). If they contract slowly during the motion, we can use a model in which the muscle force depends only on its length (Fig. 5.19). We will soon use this to model throwing a ball (even though this assumption might not be perfectly valid), and will, in fact, pretty much ignore the effects of small changes in muscle length. If the muscle lengths vary little and remain near those needed for maximum force during the motion, we can use a model in which the muscle force depends only on the muscle contraction speed, which is given by the Hill force-velocity curve (5.21) (Fig. 5.26). WeHill force–velocity curve will use this in this model of jumping [3].

As is oftenHill force–velocity curve Muscles done, we will recast the Hill force-velocity curve by changing from muscle forces to joint torques and from muscle contraction speeds to the rates of change of joint angles ( $\omega = -d\theta /dt$ , so $\omega$ is positive when the knee angle is becoming smaller). (This is described in Problem 5.20.) For joint flexion ( $$$\omega > 0$$” src=”http://basicmedicalkey.com/wp-content/uploads/2017/06/A114622_2_En_3_Chapter_IEq313.gif”></SPAN>)<br /> <DIV id=Equ63 class=Equation><br /> <DIV class=EquationContent><br /> <DIV class=MediaObject><IMG alt=$

(3.63)

and in resisting joint extension (for $\omega < 0$ )

$\begin{aligned} \tau =\tau _{\mathrm {max}}, \end{aligned}$

(3.64)

where $\omega _{\mathrm {max}}$ and $G \sim 3$ are constants. (In the classic Hill force-velocity formulation, (3.64) would be $\tau =1.8 \tau _{\mathrm {max}}$ .) At the time of foot implant, $\omega$ is zero and the torque can be at its maximum value, and during knee flexion $\omega$ increases and the torque decreases until it becomes zero when $\omega =\omega _{\mathrm {max}}$ .

When the foot is still on the ground, the equations of motion of the center of mass of the body are

$\begin{aligned} \frac{d^{2}x}{dt^{2}}=\frac{N}{m}\cos \theta \end{aligned}$

(3.65)

and

$\begin{aligned} \frac{d^{2}y}{dt^{2}}=\frac{N}{m}\sin \theta -g. \end{aligned}$

(3.66)

The equations in this model need to be numerically integrated from the time of foot implant to takeoff for typical muscle properties, with takeoff beginning when the torque falls to zero. (Numerical analysis is beyond the scope of this book.)

After takeoff the center of mass moves in free flight, with these same equations of motion, but now with

; consequently, from takeoff to landing they are easily solved analytically. Free flight depends on the center of mass position ( $x_{\mathrm {TO}}$ , $y_{\mathrm {TO}}$ ) and velocity ( $v_{\mathrm {x,TO}}$ , $v_{\mathrm {y,TO}}$ ) at time of takeoff. By usual kinematics, at time t after takeoff $v_{\mathrm {x}}(t)= v_{\mathrm {x,TO}}$ and $v_{\mathrm {y}}(t)= v_{\mathrm {y,TO}}-gt$ . The maximum height of the center of mass is $y_{\mathrm {TO}} + v_{\mathrm {y,TO}}^{2}/2g$ (which is significant for the high jump) and the distance travelled by the center of mass during flight is $(v_{\mathrm {x,TO}}/g)(v_{\mathrm {y,TO}}+(v_{\mathrm {y,TO}}^{2}+2gY)^{1/2})$ (which is significant for the long jump), where Y is how much lower the center of mass is at landing than at takeoff (Fig. 3.42). See Problem 3.96 for a derivation of this result, which is also important in throwing and hitting a ball at one height and having it land or caught at a different height. $Y=y_{\mathrm {TO}}$ if the center of mass is at ground level at landing. This distance traveled is a bit different from the distance the implanted foot travels (from its takeoff to landing) (Fig. 3.42).

The most interesting variables in this jump model are the approach speed and the angle of the lower leg at foot implant. The most significant results are the maximum increase in the height $H_{\mathrm {2}}$ of the center of mass and the height itself $H_{\mathrm {1}}+H_{\mathrm {2}}$ (Fig. 3.41), and the distance it travels until landing $L_{\mathrm {2}}$ (Fig. 3.42). In the high jump the goal is to maximize $H_{\mathrm {2}}$ for a given $H_{\mathrm {1}}$ , and this occurs when the leg angle at implant is $\sim$ 45–50 $^{\circ }$ and the approach speed is $\sim$ 7 m/s. It is a bit surprising that smaller $H_{\mathrm {2}}$ is predicted (and also seen) for even faster approach speeds; such faster speeds are easily attainable by runners, but are not desirable in the high jump.

In the long jump the goal is to maximize landing $L_{\mathrm {2}}$ , and the model shows that this occurs with steeper leg angles at implant of $\sim$ 60–70 $^{\circ }$ and the fastest possible approach speeds. In contrast to the high jump, long jumps are always better with faster speed, to the $\sim$ 11 m/s speed of sprinters.

For both jumps, the model analysis of best strategies is relatively insensitive to the specific choice ofMuscles muscle parameters and (our) ignoring of drag, and agrees with the guidance jumpers are given: High jumpers should not try to attain speeds as fast as sprinters, and should have relatively shallow leg angles at foot implant (which occurs when they have low centers of mass and their implant feet are well in front their centers of mass). Long jumpers should run as fast as possible, with their implant lower legs at a steeper angle.

3.6 Throwing a Ball

SportsThrowing Pitching Baseball Golf Football involving balls and other projectiles involve several of the same basic components: (1) human motion to throw the ball or swing a bat or racket, (2) hitting the ball (collision of the bat with the ball), (3) the flight of the ball, and (4) theBouncing bouncing andSkidding ball skidding orRolling ball rolling of the ball after it contacts the ground. Throwing a baseball or football involves 1, 3, and sometimes 4, while hitting a baseball or golf ball involves all four. We will address the throwing part of component 1 now and then the swinging motion part, contact with the ball during swinging, and ball flight, bouncing and rolling later in this chapter. We will then tie up this discussion in Appendix G, which gives an overview of the physics of sports, where we will also provide general references on the physics of various sports.

The biomechanics ofThrowing Pitching Baseball throwing a ball and, of course, of pitching in baseball are quite complex. They involve the coordinated motion of manyMuscles muscles about several body joints (Table 3.8), as is also true forRunning running, jogging, etc. We will model throwing a ball by first greatly simplifying the problem and subsequently reassessing many of our initial assumptions. The first assumption is that only the arm is involved in throwing. This is a major assumption.

Table 3.8

Joints and muscles involved in the acceleration phase of the overarm throw

Joint	Joint action moved	Segment	Major muscles
Shoulder joint	Horizontal abduction	Upper arm	Posterior deltoid
			Infraspinatus
			Teres minor
	Medial rotation		Subscapularis
			Latissmus dorsi
			Pectoralis major
Shoulder girdle	Abduction	Scapula	Serratus anterior
Shoulder girdle	Abduction	Scapula	Pectoralis minor
Elbows	Extension	Forearm	Triceps brachii
Radioulnar	Pronation	Forearm	Pronator teres
Radioulnar	Pronation	Forearm	Pronator quadratus
Wrists	Flexion	Hand	Flexor carpi radialis
			Flexor carpi ulnaris
			Flexor carpi digitorum
			Superficialis

Each joint and its action are shown, along with the segment being moved and the muscles undergoing concentric actionExercise Throwing Pitching

Using data from [64]

Fig. 3.44

Models of throwing a ballBaseball Throwing Pitching with a motion of the shoulder and elbow (extension), b motion of the shoulder only with a fixed arm, c extension of the elbow and a stationary shoulder, and d flexion of the elbow and a stationary shoulder

The diagramsBaseball Throwing Pitching in Fig. 3.44 shows four models of throwing a ball overhand. One can consider throwing by:

(a)

TheShoulders Elbows entire motion with motion in the shoulder and elbow joints (Fig. 3.44a).

(b)

Motion of a stiff arm (elbow fixed)—motion only in the shoulder joint, using theMuscles Deltoids deltoid muscles (Fig. 3.44b).

(c)

Motion of only the forearm (lower arm) (shoulder fixed)—motion only in the elbow joint using theExtensor muscles extensor tricepsTriceps brachii brachii (Fig. 3.44c).

(d)

Motion of only the forearm (shoulder fixed)—motion only in the elbow joint using theFlexor muscles flexorBiceps brachii biceps brachii (Fig. 3.44d).

Similar diagramsThrowing Pitching Baseball can be drawn showing several analogous ways of throwing a ball underhand, which involve these same (a), (b), and (d) models.

Each of theseBaseball Throwing Pitching models represents a fairly unnatural motion for throwing (try them!), but they represent good first-order models. Model (a) involves two-body joints, and consequently is aTwo-segment model two-segment model (because two segments can move). Models (b)–(d) involve one-body joint and areOne-segment model one-segment models. More sophisticated models require aMultisegment modeling multisegment model.

We will analyze throwing with the single-segment model (d) where theBaseball Throwing Pitching forearm moves due to theFlexor muscles Muscles flexor biceps brachii contraction, which causes the forearm to rotate about the elbow joint. This can be a model of either overhand or underhand throwing. Although it is very simple, this model has several complexities and we will analyze each one. Our goal is to see how fast we can throw a ball, specifically a baseball. We will first assume that the biceps brachii are the only muscles used to flex the forearms; later we will examine other muscles that can contribute to form an “effective biceps” muscle.

Throwing Pitching Torques BaseballWe have analyzed the torque in this system already, in Chap. 2. As seen in Fig. 2.9d (Case 3)

$\begin{aligned} \sum \tau _{z}= & {} Md_{\mathrm {M}}\sin \theta -W_{\mathrm {F}}d_{\mathrm {F}}\sin \theta -W_{\mathrm {B}}d_{\mathrm {B}}\sin \theta \nonumber \\= & {} (Md_{\mathrm {M}}-W_{\mathrm {F}}d_{ \mathrm {F}}-W_{\mathrm {B}}d_{\mathrm {B}})\sin \theta , \end{aligned}$

(3.67)

with the weight in the hand now called a ball (B). For simplicity, we first assume that the humerus is vertical. In a static situation the torques add up to zero. Here they do not. For our desired arm motion, the sum must be positive.

Figure 3.45 explains a small subtletyThrowing Pitching Baseball concerning the defined angle $\theta$ , which is the angle the moving forearm makes with the fixed humerus (upper arm). The forearm makes an angle $\theta ^{\prime }$ with the fixed x-axis. In relating the net torques to the second derivative of the angle, as in (3.67), the dynamical variable is actually what we have called $\theta ^{\prime }$ here, so

$\begin{aligned} \sum \tau _{z}=I\frac{\mathrm{d}^{2}\theta ^{\prime }}{\mathrm{d}t^{2}}. \end{aligned}$

(3.68)

Fig. 3.45

Angles involved in throwingBaseball Throwing Pitching a ball, include the angles of the rotating forearm relative to the x-axis ( $\theta ^{\prime }$ ) and the fixed upper arm ( $\theta$ )

Baseball Throwing PitchingBecause $\theta +\theta ^{\prime }=\pi /2 \;({=}90^{\circ })$ , a fixed quantity, we see that $\mathrm{d}\theta /\mathrm{d}t+\mathrm{d}\theta ^{\prime }/\mathrm{d}t=0$ and $\mathrm{d}^{2}\theta /\mathrm{d}t^{2}+\mathrm{d}^{2}\theta ^{\prime }/\mathrm{d}t^{2}=0$ , so

$\begin{aligned} \frac{\mathrm{d}^{2}\theta ^{\prime }}{\mathrm{d}t^{2}}=-\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}} \end{aligned}$

(3.69)

and from (3.25)

$\begin{aligned} \sum \tau _{z}=-I\frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}} \end{aligned}$

(3.70)

$\begin{aligned} \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}=-\frac{\sum \tau _{z}}{I}. \end{aligned}$

(3.71)

Baseball Throwing PitchingNow we need to evaluate theMoment of inertia moment of inertia I and the torques. The moment of inertia about the elbow pivot equals the sum of the components due to the arm and the ball. Using (3.23) and (3.30) this is

$\begin{aligned} I=\frac{1}{3}m_{\mathrm {F}}L^{2}+m_{\mathrm {B}}L^{2}, \end{aligned}$

(3.72)

where L is the length of the forearm ( $0.146\,H$ ) plus half the hand (0.108 H / 2), which is $L \sim 0.2\,H \sim 36$ cm. The weight of the forearm (with the whole hand) $\sim$ 0.022 $\,W_{\mathrm {b}}$ , so $m_{\mathrm {F}} \sim$ 2 kg for a 90 kg player. (We are modeling a 6 ft tall (1.8 m), 200 lb (90 kg) major leaguer.) A regulationBaseball major league baseball must weigh between 5 and 5 $\frac{1}{4}$ oz (Official Baseball Rules (Major League Baseball), Rule 1.09), which corresponds to a mass of 0.146 kg. This gives $$I = 1{,}053$$

kg-cm $^{2}$ . (These are mixed MKS/CGS units. This is often not the best practice, but is fine if we are careful.)

Baseball Throwing PitchingEvaluating the components of torque due to the forearm and ball is straightforward. Using $$W=mg$$

$\begin{aligned} W_{\mathrm {F}}d_{\mathrm {F}}= & {} (2\mathrm { \;kg})(9.8\mathrm { \;m/s}^{2})(18\mathrm { \;cm})=353 \hbox { N-cm} \end{aligned}$

(3.73)

$\begin{aligned} W_{\mathrm {B}}d_{\mathrm {B}}= & {} (0.146\mathrm { \;kg})(9.8\mathrm { \;m/s}^{2})(36\mathrm { \;cm} )=51.5\hbox { N-cm}, \end{aligned}$

(3.74)

for a total of 404 N-cm.

In evaluating the torque due to theMuscles muscle, we will initially make some assumptions. (1) We assume that the distance from the point of bicepsBaseball Throwing Pitching Points of insertion brachii insertion to the pivot axis, $d_{\mathrm {M}}$ , is 4 cm, independent of the arm angleMoment arm $\theta$ . Figure 3.46 shows that $d_{\mathrm {M}}=d_{\mathrm {M} }(\theta )$ and so including this variation would change our final answer a bit, but it will not change our overall conclusions. (2) We assume a fixed value of the maximum muscle force M, to see how fast we can throw a ball. As we will see in Chap. 5, this is an assumption because (a) M depends on muscle length ( $l_{\mathrm {M}}$ ), which depends on $\theta$ , and (b) M depends on the muscle contraction speed $v_{\mathrm {M}}=\mathrm{d}l_{\mathrm {M}}/\mathrm{d}t$ , which depends on $\mathrm{d}\theta /\mathrm{d}t$ . We will revisit several of these assumptions in Problems 5.25 and 5.26. For nowPhysiological cross-sectional area of muscle (PCA) we use the fact that most muscles exert a maximum force per unit cross-sectional area, which we will initially take as $\sim$ 20 N/cm $^{2}$ , a relatively modest value.

Fig. 3.46

VariationBaseball Throwing Muscles Pitching Moment arm Elbows Biceps brachii of the moment arm of biceps brachii versus elbow angle (from [23]. Reprinted with permission of Wiley)

Baseball Throwing Pitching MusclesFor biceps brachii with a 2 in diameter, the cross-sectional area is $\pi (1\,\mathrm{in})^{2} = 20.3$ cm $^{2}$ , so $$M = 405$$

N and $Md_{\mathrm {M}}= 1{,}620$ N-cm. So (3.67) becomes

$\begin{aligned} \sum \tau _{z}=({1{,}620\,{\text {N-cm}}} -{{404\,{\text {N-cm}}}})\sin \theta ={(1{,}216\,{\text {N-cm}})}\sin \theta \end{aligned}$

(3.75)

and

$\begin{aligned} \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}=-\frac{\sum \tau _{z}}{I}=-\frac{{(1{,}216\,{\text {N-cm}})}\sin \theta }{{1{,}053\,{\text {kg-cm}}}^{2}}. \end{aligned}$

(3.76)

Because 1 N $$=$$

1 kg-m/s $^{2}$ $$=$$

100 kg-cm/s $^{2}$ ,

$\begin{aligned} \frac{\mathrm{d}^{2}\theta (t)}{\mathrm{d}t^{2}}=-\mathrm {116/s}^{2}\sin \theta (t), \end{aligned}$

Only gold members can continue reading. Log In or Register to continue

Stay updated, free articles. Join our Telegram channel

Tags: Physics of the Human Body

Jun 11, 2017 | Posted by admin in GENERAL & FAMILY MEDICINE | Comments Off

Basicmedical Key

Fastest Basicmedical Insight Engine

Motion

3.2 Standing

3.2.1 StabilityStability

3.2.2 Forces on the Feet

3.3 WalkingWalking Running

3.3.1 Kinematics

3.3.2 Muscular Action

3.3.3 FrictionFriction

3.3.4 Energetics

3.3.5 Review of Harmonic Motion, Pendulums, and Moments of Inertia

3.3.6 Ballistic (or Pendulum) Model of Walking

3.3.7 Inverted Pendulum Model

3.4 Running

3.4.1 Kinematics

3.4.2 Muscular Action

3.4.3 Energetics

3.4.4 Bouncing Ball/Pogo Stick ModelBouncing

3.5 Jumping

3.5.1 Vertical Jump

3.5.2 Pole Vault

3.5.3 High Jump and Long Jump

3.6 Throwing a Ball

Like this:

Related

Stay updated, free articles. Join our Telegram channel

Full access? Get Clinical Tree

Basicmedical Key

Fastest Basicmedical Insight Engine

Motion

3.2 Standing

3.2.1 StabilityStability

3.2.2 Forces on the Feet

3.3 WalkingWalking Running

3.3.1 Kinematics

3.3.2 Muscular Action

3.3.3 FrictionFriction

3.3.4 Energetics

3.3.5 Review of Harmonic Motion, Pendulums, and Moments of Inertia

3.3.6 Ballistic (or Pendulum) Model of Walking

3.3.7 Inverted Pendulum Model

3.4 Running

3.4.1 Kinematics

3.4.2 Muscular Action

3.4.3 Energetics

3.4.4 Bouncing Ball/Pogo Stick ModelBouncing

3.5 Jumping

3.5.1 Vertical Jump

3.5.2 Pole Vault

3.5.3 High Jump and Long Jump

3.6 Throwing a Ball

Share this:

Like this:

Related

Related posts:

Stay updated, free articles. Join our Telegram channel

Full access? Get Clinical Tree