SAQMG2

In inverse dynamics the motion is measured and the resultant force iscalculated.

The three types of forces that act on a body segment are external, internal and inertial.

It is justifiable to treat a link segment model as quasi-static only when theaccelerations are small.

The results of this SAQ are dependent on your own dimensions. As an exampleI will use the following dimensions, height = 193 cm, measured upper limblength = 75 cm, and measured lower limb length = 100 cm.Length of upper limb is calculated as follows using the ratios given in Figure 9:LUPPER = (0.186 + 0.146 + 0.108) H = 0.440 × 193 = 85 cmLength of lower limb is calculated as follows:LLOWER = (0.186 + 0.146 + 0.108) H = 0.530 × 193 = 102 cmIf the calculated limb lengths are compared to measured limb lengths it can be seen that the calculated lower limb lengths are reasonable estimates. However, the calculated upper limb length is 10 cm out. This illustrates the diversity of human dimensions and why standardised data sets should only be used when actual measurements are not available.

To calculate the mass of the thigh, mTHIGH, the ratio reported in Table 1 is used:mTHIGH = 0.100mBODY = 0.100 × 80 = 8.0 kgTo calculate the position of the centre of mass of the thigh relative to itsproximal end, IPROXIMAL, the ratio, 0.433, reported in Table 2 is used:IPROXIMAL = 0.433ITHIGH = 0.433 × 390 = 169 mmTo calculate the position of the centre of mass of the thigh relative to itsproximal end, IDISTAL, the ratio, 0.567, reported in Table 2 is used: IDISTAL = 0.567ITHIGH = 0.567 × 390 = 221 mmThe mass of the thigh is 8.0 kg, and its centre of mass is located 169 mm distal to the greater trochanter and 221 mm proximal to the femoral condyles.Note that the centre of mass is closer to the proximal end than to the distal end. This is mainly due to the shape of the thigh which narrows towards its distal end so that more mass is distributed towards the proximal end. The position of the centre of mass is also dependent on the densities of the various tissues, such as bone, muscle and fat, and how they are distributed.

Calculating the mass of the lower leg and foot complex, m, using the ratio,0.061, in Table 1:m = 0.061 × 75 = 4.575 kgCalculating the radius of gyration about the centre of mass, kCM, using the ratio 0.416, given in Table 2:kCM = 0.416L = 0.416 × 450 × 10-3 = 0.1872 = 0.19 mCalculating the moment of inertia about the centre of mass, ICM:ICM = mAkCM = 4.575 × 0.1872 × 0.1872 = 0.1603 = 0.16 kg m2Calculating the radius of gyration, kP, using the ratio 0.735, and moment ofinertia, IP, about the proximal end:kP = 0.735L = 0.735 × 450 × 10-3 = 0.33075 = 0.33 mIP = mkP = 4.575 × 0.33075 × 0.33075 = 0.50048 = 0.50 kg m2Calculating the radius of gyration, kD, using the ratio 0.572, and moment ofinertia, ID, about the distal end:kD = 0.572 × L = 0.572 × 450 × 10-3 = 0.2574 = 0.26 mID = mkD = 4.575 × 0.2574 × 0.2574 = 0.3031 = 0.30 kg m2The radius of gyration about the proximal end is larger than the radius ofgyration about the distal end because more mass is distributed further from the proximal end than the distal end.

The insertion point of the muscle which connects the two segments does not change. However, the line of action of the muscle line changes as the two segments move relative to one another.

If the sign of the inter-segment force is positive then it means that it is acting to push the segments together.

The joint force is generally much larger than the inter-segment force because it includes muscle forces.

The inter-segment force is the resultant of all the forces crossing the jointbetween two segments.The joint force is the force acting between two segments at the joint.The muscle force is the force produced by the muscles crossing the joint.The bone-on-bone force is the force acting between the bones that form a joint.The ligament force is the force carried by ligaments, joint capsules and muscles that are not contracting.

The net external joint moment is equal and opposite to the net internal jointmoment.

If the mass was doubled then the quadriceps moment would need to bedoubled to maintain static equilibrium.

The interacting filaments in the sarcomeres are called the actin and myosinfilaments or the thin and thick myofilaments.

The maximum active muscle tension is produced when the muscle lengthis approximately equal to the resting length of the muscle.

In a concentric contraction the muscle is shortening and in an eccentriccontraction the muscle is lengthening.

For a certain level of activation, a muscle will produce a greater forceduring an eccentric contraction than during a concentric contraction.

Thus, as the pennation angle increases the physiological cross-sectional area decreases according to the cosine function.

Electromyography is the study of the electrical signal associated with thecontractions of muscles.

There is a relationship between EMG and muscle force. However, it is onlypossible to characterise this relationship under controlled conditions.

The distribution problem is the problem of how the total muscle force isdistributed between all the muscles crossing a joint.

The three segments of the lower limb are the thigh, leg and foot.

The three major joints of the lower limb are the hip, knee and ankle.

The two main functions of the joints of the lower limb are movement andweight bearing.

The hip joint is a ball-and-socket type synovial joint.

Synovial fluid lubricates the joint and provides nutrients to the articularcartilage.

The hip joint is intrinsically stable due to its shape, its strong joint capsuleand the surrounding ligaments and muscles.

The greatest range of hip joint motion occurs in the sagittal plane.

Typically around 110° of flexion-extension, 20° of abduction-adduction and15° of rotation is required to stand up and sit down.

The abductor muscle force is 1880 N and the subject's total body weight is 800 N thus the magnitude of the abductor muscle force is equal to 2.4 times the subject's total body weight.

Muscle activity generally increases the magnitude of the joint forces bypulling the two sides of the joint together. During unilateral stance thecontraction of the hip joint abductors acts in this way, causing a largeincrease in the joint force as compared to bilateral stance when there isusually no muscle activity.

The magnitude of the hip joint force is equal to 0.4 total body weightduring bilateral stance, and 3.2 total body weight during unilateral stance.

The two articulations which comprise the knee joint are the tibiofemoraland the patellofemoral.

The patella is a sesamoid bone.

The stability of the knee joint is derived from its ligaments, in particular thecruciate and collateral.