1. Advanced Biomechanics of Physical Activity (KIN 831) Anthropometry Material included in this presentation is derived primarily from: Winter, D.A. (1990). Biomechanical and motor control of human movement. (2 nd ed.). New York: Wiley & Sons Hamill, J. & Knutzen, K. (2003). Biomechanical Basis of Human Movement (2 nd ed.). Philadelphia: Lippincott Williams & Wilkins Hamilton, N. & Luttgens, K. (2002). Kinesiology – Scientific Basis of Human Motion (10 th ed.). Boston: McGraw-Hill.

2. What is Anthropology? Anthropometry?

3. Definitions Anthropology – the scientific study of the origin and of the physical, social, and cultural development and behavior of man Anthropometry – the study and technique of human body measurement for use in anthropological classification and comparison

4. What functions does anthropometry serve?

5. Uses of Anthropometry Studies of physical measurements of the human body Determining differences in individuals and groups –Age –Sex –Race –Body type (somatotype)

6. Emphases of Anthropometric Studies Evolution Historical Performance parameters Man-machine interfaces Past studies Associated with recent technological developments

7. What are some examples of anthropometric measurements?

8. Types of Anthropometric Measurements Static measurements –Length –Area –Volume –Breadths –Circumference –Skin fold thickness

9. Types of Anthropometric Measurements Static measurements (continued) –Ratios and proportions Body mass index – height/stature 2 Sitting height/stature Bicristal/biacromial Ponderal index – height/weight 1/3 Physique (somatotype – endomorphy, mesomorphy, ectomorphy)

11. Types of Anthropometric Measurements (continued) Kinetic measurements –Linear – using mass Force = mass x acceleration F = MA –Angular – using moment of inertia Torque or moment of force = moment of inertia x angular acceleration T or M = I 

12. Body Segment Lengths Vary with body build, sex, racial origin Dempster’s data (1955, 1959) – segment lengths and joint center locations relative to anatomical landmarks (see figure) Drillis and Contini (1966) – segment lengths/body height (see figure)

15. Body Segment Lengths Vary with body build, sex, racial origin Dempster’s data (1955, 1959) – segment lengths and joint center locations relative to anatomical landmarks Drillis and Contini (1966) – segment lengths/body height (see figure)

17. What is Mass? Inertia? Density?

18. Density, Mass, and Inertial Properties Kinematic and kinetic analyses require data on mass distribution, mass centers, and moments of inertia –Measured directly Cadaver Segment volume Segment density –Measured indirectly Density via MRI

19. Density, Mass, and Inertial Properties (continued) Whole body density Tissue density – varies with type of tissue Specific gravity – weight of tissue/weight of water of same volume –Cortical bone = 1.8 –Muscle  1.0 –Fat  0.7 –Lungs < 1.0

21. Density, Mass, and Inertial Properties (continued) Drillis and Contini (1966) expression for body density d = 0.69 + 0.0297c E answer in kilograms/liter where c E = (height in inches)/(weight in pounds) 1/3 ;this is the inverse ponderal index d = 0.69 + 0.9c m answer in kilograms/liter where c m = (height in meters)/(mass in kilograms) 1/3 Note that it can be seen that a tall thin person has a higher inverse ponderal index than a short fat individual and therefore greater density.

22. Proof of conversion from English to metric units in equations: Since d = 0.69 + 0.0297c E and d = 0.69 + 0.9c m 0.0297c E = 0.9c m  c m /c E = 0.0297/0.9 = 0.033 meters/kilograms 1/3 = 0.033 inches/pound 1/3 1 meter/1 kilogram 1/3 = 0.033 (disregarding units)  39.37 inches/2.2046 pounds 1/3 1  = 0.033 proof !!! 39.37/1.30

23. Example calculation of density: Person 70 inches and 170 pounds d = 0.69 + 0.0297c E c E = h/w 1/3 = 70/170 1/3 = 12.64 d = 0.69 + 0.0297 (12.64) = 1.065 kg/l

24. Density, Mass, and Inertial Properties (continued) Segment densities –Unique densities for each body segment –Each segment has a different combination of tissues (e.g., head vs. shank) –Individual segments increase in density with increase in total body density (see figure) –Upper and lower extremities more dense than whole body –Proximal segments less dense than distal segments

26. Density, Mass, and Inertial Properties (continued) Segment mass and center of mass –Center of mass and center of gravity used interchangeably –Total body mass vs. segment mass Increase in total body mass  increase in segmental mass (proportional increases) Possible to express mass of each segment as a proportion of the total body Proportions vary by age, gender, and other factors Location of center of mass determined as a percent of segment length (from proximal or distal end) Balance technique used in cadavers to determine center of mass

27. Total Mass of a Segment Mass =  m i m i = d i V i density = mass/volume M =  d i V i Note that if d is uniform or assumed to be uniform, M = d  V i Note that the center of mass creates the same net moment about any point along the segment axis as does the original distributed mass Mx =  m i x i

28. Why the same definition ? Head And Trunk

29. Example: If the greater trochanter has coordinates (72.1,98.8) and the lateral femoral condyle has coordinates (86.4,54.9), calculate the center of mass of the thigh.

30. Y X (72.1, 98.9) (86.4, 54.9) 72.1 98.9 54.9 86.4 center of mass 0.433(length) 0.567(length) 0.433(  Y) 0.567(  Y) 0.433(  X)0.567(  X)

31. Example: If the greater trochanter has coordinates (72.1,98.8) and the lateral femoral condyle has coordinates (86.4,54.9), calculate the center of mass of the thigh. Using information from the proximal end: x = 72.1 + 0.433(86.4 –72.1) = 78.3 y = 92.8 – 0.433(92.8 – 54.9) = 76.4

32. - step 1: determine the proportion of mass that each segment is of the entire multisegment system - step 2: multiply each segmental proportion times the x coordinate of the center of mass of that segment - step 3: multiply each segmental proportion times the y coordinate of the center of mass of that segment - step 4: add each of the x products - step 5: add each of the y products - step 6: the sums from steps 4 and 5 are the x and y coordinates of the center of mass of the multisegment system x 0 = m 1 x 1 + m 2 x 2 + m 3 x 3 +  + m n x n M y 0 = m 1 y 1 + m 2 y 2 + m 3 y 3 +  + m n y n M Calculate the center of mass of a multisegment system (e.g., lower extremity, entire body)

33. sample table SegmentProportion of total mass x value of center of mass x producty value of center of mass y product Segment 1 11 x1x1  x 1 y1y1  y 1 Segment 2 22 x2x2  x 2 y2y2  y 2 Segment 3 33 x3x3  x 3 y3y3  y 3          Segment n nn xnxn  x n ynyn  y n  = 1.00  = x value of the center of mass *  = y value of the center of mass * * Note that the calculated center of mass will be relative to the Cartesian coordinate system that was used for the center of masses used for the individual segments.

34. Example (not in book): Calculate the center of mass of the right lower extremity (foot, shank, and thigh) for frame 33 of the subject in Appendix A (use Table A.3(a-c))

36. - step 1: determine the proportion of mass that each segment is of the entire multisegment system mass of subject = 56.7kg Lower Extremity Mass mass of foot = 0.0145(56.7kg) = 0.82215kg = mass of shank =0.0465(56.7kg) = 2.63655kg = mass of thigh = 0.100(56.7kg) = 5.67kg = total mass of lower extremity = 9.1287kg Segmental Proportions of Lower Extremity Mass foot proportion = 0.8221/9.1287 = 0.09006 shank proportion = 2.6355/9.1287 = 0.28882 thigh proportion = 5.67/9.1287 = 0.62112 total mass proportion = 9.1287/9.1287 = 1.00 - step 2: multiply each segmental proportion times the x coordinate of the center of mass of that segment - step 3: multiply each segmental proportion times the y coordinate of the center of mass of that segment - step 4: add each of the x products - step 5: add each of the y products - step 6: the sums from steps 4 and 5 are the x and y coordinates of the center of mass of the multisegment system

37. SegmentProportion of total mass x value of center of mass x producty value of center of mass y product Foot0.090061.3420.120860.0690.00621 Shank0.288821.2560.362760.3310.0956 Thigh0.621121.1750.729810.6780.42112  = 1.0  = 1.21343  = 0.5229

38. What is inertia?

39. According to Newton’s first law of motion, inertia is an object’s tendency to resist a change in velocity. The measure of an object’s inertia is its mass. The more mass an object has the more inertia it has.

40. What is moment of inertia?

41.  F = ma and  = I , where F = force, m = mass, a = acceleration,  = torque or moment of force causing angular acceleration, I = moment of inertia =  m i x i 2, and  = angular acceleration  I of an object depends upon the point about which it rotates  I is minimum for rotations about an object’s center of mass I =  m i x i 2 =  m 1 x 1 2 +  m 2 x 2 2 +  m 3 x 3 2 +     +  m n x n 2 Moment of Inertia

42. The angular counterpart to mass is moment of inertia? It is a quantity that indicates the resistance of an object to a change in angular motion. The magnitude of an object’s moment of inertia is determined by its mass and the distribution of its mass with respect to its axis of rotation.

44. Hypothetical object made up of 5 point masses m1m1 m2m2 m3m3 m4m4 m5m5 y xx y 0.1m horizontal axis through center of mass vertical axis through center of mass 0.1m m 1 = m 2 = m 3 = m 4 = m 5 = 0.5 kg

45. Calculate the moment of inertia about y-y m1m1 m2m2 m3m3 m4m4 m5m5 y xx y 0.1m horizontal axis through center of mass vertical axis through center of mass 0.1m I y-y = (0.5kg)(0.1m) 2 + (0.5kg)(0.2m) 2 + (0.5kg)(0.3m) 2 + (0.5kg)(0.4m) 2 + (0.5kg)(0.5m) 2 = 0.275kgm 2

46. Calculate the moment of inertia about x-x m1m1 m2m2 m3m3 m4m4 m5m5 y xx y 0.1m horizontal axis through center of mass vertical axis through center of mass 0.1m I x-x = (0.5kg)(0.1m) 2 + (0.5kg)(0.1m) 2 + (0.5kg)(0.1m) 2 + (0.5kg)(0.1m) 2 + (0.5kg)(0.1m) 2 = 0.025kgm 2

47. Calculate the moment of inertia about vertical axis through center of mass m1m1 m2m2 m3m3 m4m4 m5m5 y xx y 0.1m horizontal axis through center of mass vertical axis through center of mass 0.1m I cg = (0.5kg)(0.2m) 2 + (0.5kg)(0.1m) 2 + (0.5kg)(0.0m) 2 + (0.5kg)(0.1m) 2 + (0.5kg)(0.2m) 2 = 0.05kgm 2

48. Moment of Inertia of Segments of the Human Body Segments of body made up of different tissues that are not evenly distributed or of uniform shape Moment of inertia of body segments determined experimentally Moment of inertia of body segments unique to individual segments and axes of rotation Calculation of moment of inertial of a body segment is based on the segment’s radius of gyration

49. Radius of Gyration - most techniques that provide values for segment moment of inertia provide information on the radius of gyration - moment of inertia can be calculated from the radius of gyration - radius of gyration denotes the segment’s mass distribution about an axis of rotation and is the distance from the axis of rotation to a point at which the mass can be assumed to be concentrated without changing the inertial characteristics of the segment I 0 = m  0 2 where I 0 = the moment of inertia about the center of mass, m = mass of object and  0 = radius of gyration for rotation about the center of mass

50. Radius of Gyration Denotes the segment’s mass distribution about an axis of rotation and is the distance from the axis of rotation to a point at which the mass can be assumed to be concentrated without changing inertial characteristics of the segment Moment of inertia (I) = m(ρl) 2 where m = mass of segment, l = length of segment, and ρ = radius of gyration as a proportion of the segment length

51. Parallel Axis Theorem Moment of inertia can be calculated about any parallel axis, given the: 1.moment of inertia about one axis, 2.mass of the segment, and 3.perpendicular distance between the parallel axes

52. Parallel Axis Theorem I = I 0 + mx 2 where I = moment of inertia about a parallel axis, I 0 = moment of inertia about the center of mass, m = mass of object (or segment), and x = distance from the center of mass and the center of rotation

53. SegmentCenter of Gravity Proximal End Distal End Shank0.3020.5280.643 Radius of Gyration/Segment Length in meters* (about a transverse axis) *(see Table 3.1 in class text)

54. Moment of inertia varies on the basis of axis of rotation: Proximal end Center of mass Distal end

55. 42 cm 20 cm 62 cm Example: a) A prosthetic shank has a mass of 3kg and a center of mass at 20cm from the knee joint. The radius of gyration is 14.1cm Calculate the moment of inertia about the knee joint.

57. 42 cm 20 cm 62 cm Example: a) A prosthetic shank has a mass of 3kg and a center of mass at 20cm from the knee joint. The radius of gyration is 14.1cm Calculate the moment of inertia about the knee joint. I 0 about the center of mass of the shank = m  0 2 = 3kg(0.141meters) 2 = 0.06kg meters 2 Using the parallel axis theorem:I k = I 0 + mx 2 = 0.06kg meters 2 + 3kg (0.2meters) 2 = 0.18kg meters 2 b) Calculate the moment of inertia for the prosthetic shank about the hip.

58. 42 cm 20 cm 62 cm Example: a) A prosthetic shank has a mass of 3kg and a center of mass at 20cm from the knee joint. The radius of gyration is 14.1cm Calculate the moment of inertia about the knee joint. I 0 about the center of mass of the shank = m  0 2 = 3kg(0.141meters) 2 = 0.06kg meters 2 Using the parallel axis theorem :I k = I 0 + mx 2 = 0.06kg meters 2 + 3kg (0.2meters) 2 = 0.18kg meters 2 b) Calculate the moment of inertia for the prosthetic leg about the hip. Using the parallel axis theorem: I h = I 0 + mx 2 = 0.06kg meters 2 + 3kg (0.62meters) 2 = 1.21kg meters 2 Note that I h  20 I 0

59. Calculate the moment of inertial of shank about its center of gravity and proximal and distal ends Given: mass of shank = 3.6kg, length of shank = 0.4 meters

60. Calculate the moment of inertial of shank about its center of gravity and proximal and distal ends Given: mass of shank = 3.6kg, length of shank = 0.4 meters I cm = I o = m(ρ cm l) 2 = 3.6kg[(0.302)(0.4m)] 2 = 0.0525kgm 2 I prox = m(ρ prox l) 2 = 3.6kg[(0.528)(0.4m)] 2 = 0.161kgm 2 I dist = m(ρ dist ) 2 = 3.6kg[(0.643)(0.4m)] 2 = 0.238kgm 2

61. In which twisting movement is the moment of inertia greater? Why is a layout flip worth more points than a tuck flip?

62. Why do we bend our knee during the swing phase of running?

63. Lower Extremity

64. Example: Calculate the moment of inertia of the leg about its center of mass, its distal end, and its proximal end for an 80kg subject who has a leg (shank) segment = 0.435meters.

66. mass of the leg is 0.0465 x 80kg = 3.72kg I 0 = 3.72kg(0.435 meters x 0.302) 2 = 0.064kg meters 2 I p = 3.72kg(0.435 meters x 0.528) 2 = 0.196kg meters 2 I d = 3.72kg(0.435 meters x 0.643) 2 = 0.291kg meters 2

67. Example: Calculate the moment of inertia of the leg about its center of mass, its distal end, and its proximal end for an 80kg subject who has a leg (shank) segment = 0.435meters. Using the Parallel-Axis Theorem:

69. Example: Calculate the moment of inertia of the leg about its center of mass, its distal end, and its proximal end for an 80kg subject who has a leg (shank) segment = 0.435meters. Using the Parallel-Axis Theorem: I p = I 0 + mx 2 = 0.064kg meters 2 + 3.72kg(0.433 x 0.435meters) 2 = 0.196kg meters 2 I d = I 0 + mx 2 = 0.064kg meters 2 + 3.72kg(0.567 x 0.435meters) 2 = 0.2903kg meters 2

70. Muscle Anthropometry Physiologic Cross-sectional Area (PCA) PCA is an experimental measure of maximum strength of contraction based on the cross-sectional area of the muscle PCA is based on a measure of the number of sarcomeres that make up the cross- sectional area of a muscle.

71. Logic Behind PCA The sarcomere is the basic contractile unit of the muscle. Because they are in series to each other in making up the length of the myofibril, the myofibril is limited in strength to the maximum force that can be generated by the weakest sarcomere in the series. Myofibrils are parallel to each other in a muscle cell. Therefore, the maximum force of contraction of the muscle cell is directly related to the sum of the force of maximum contraction of the each myofibril (Remember that the maximum force of contraction of the myofibril is the maximum force of the weakest sarcomere in its length.).

72. Logic Behind PCA Muscle cells are in parallel with each other. The maximum force of contraction of a muscle is directly related to the sum of the maximum forces of contraction of all the individual muscle cells that compose the muscle. Therefore, the PCA is directly related to the number of sarcomeres in this cross-section and directly related to the maximum force of contraction of the muscle.

75. PCA = m/d(l) where m = mass of the muscle fibers in grams, d = density of the muscle in grams/cm 3 (note that muscle mass  1.056g/cm 3 ), and l = length of the muscle fibers in cm

76. Based on the equation for PCA, in a fusiform muscle, the PCA is independent of length. Note that changes in length of a muscle are accompanied by the same proportional changes in muscle mass. Therefore, PCA is not changed. l m l/2 m/2 PCA = m/d(l) PCA = (m/2)/d(l/2)  PCA = m/d(l) Based on the architecture of muscle, why is there no change in PCA?

77. Shading represents PCA Example of PCA of fusiform and pennate muscle fiber arrangement: FusiformPennate

78. Example of PCA of fusiform and pennate muscle fiber arrangement: FusiformPennate Note the difference in the number of muscle fibers that occupy the same volume and the difference in the cross-sectional area required to cut across all the fibers in the muscle. In other words the PCA of the pennate is greater than the PCA of the fusiform of equal volume. Therefore, muscles with pennate fiber arrangements have an advantage for force of contraction. However, muscles with fusiform fiber arrangement have an advantage for length of contraction.

79. Pennation - angle of pull of the muscle fibers in relationship to the long axis of the muscle - as the pennation angle increases, the PCA increases - as the pennation angle increases, the force of contraction of a single muscle cell, in the direction of the long axis of the muscle, decreases - the combination of pennation angle and PCA influence the maximum force of contraction of a muscle FmFm FyFy FyFy θ FxFx F y = F m = force of contraction of each muscle cell in the direction of the long axis of the F y = F m cos  = force of contract of each muscle cell in the direction of the long axis of the muscle

80. Based in PCA, what would you suggest to be the strongest flexors and extensors?

81. PCA Force per unit cross-sectional area: - reports range from 20 –100N/cm 2 - higher values reported for pennate muscles

82. Example (not in book): Ignoring any influence pennation angle might have on force of contraction, calculate the collective peak force of contraction of the quadriceps (see Table 3.3 for PCA). Assume that these muscles exert a force per unit cross- section = 75N/cm 2. Calculate the answer in N and in pounds. Note that 1N = 0.2247lbs.

83. Example (not in book): Ignoring any influence pennation angle might have on force of contraction, calculate the collective peak force of contraction of the quadriceps (see Table 3.3 for PCA). Assume that these muscles exert a force per unit cross- section = 75N/cm 2. Calculate the answer in N and in pounds. Note that 1N = 0.2247lbs. Collective PCA of quadriceps muscle = 12.5cm 2 + 30cm 2 +26cm 2 +25cm 2 = 93.5cm 2 93.5cm 2 x 75N/cm 2 = 7012.5N 7012.5N x 0.2247lbs/N = 1575.71 lbs **impressive!!!

84. Mechanical Advantage of Muscle -mechanical advantage of a muscle at a joint is determined by the moment arm length (length of line from the joint center intersecting the line of long axis of the muscle at a perpendicular -mechanical advantage (and moment arm) change with changes in the angle of the joint - force of contraction of the muscle also changes with changes in joint angle (changes in muscle length) - maximum torque or turning force (moment) at a joint is therefore the dynamic product of two variables – length of the muscle (as determined by joint angle) and moment arm (also determined by joint angle)

85. Multijoint Muscles -individual muscles that cross two or more joints -muscle fiber arrangements of many multijoint muscles may be insufficient to engage in maximum shortening (simultaneously affecting multiple joints) - fusiform muscles have an advantage in length of contraction in multijoint muscles (e.g., sartorius muscle – longest muscle in body and two joint muscle), but disadvantage in PCA (and maximum force of contraction)