Advanced Biomechanics of Physical Activity (KIN 831)

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. (2nd ed.). New York: Wiley & Sons Hamill, J. & Knutzen, K. (2003). Biomechanical Basis of Human Movement (2nd ed.). Philadelphia: Lippincott Williams & Wilkins Hamilton, N. & Luttgens, K. (2002). Kinesiology – Scientific Basis of Human Motion (10th 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/stature2 • Sitting height/stature • Bicristal/biacromial • Ponderal index – height/weight1/3 • 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 + 0.0297cEanswer in kilograms/liter where cE = (height in inches)/(weight in pounds)1/3;this is the inverse ponderal index d = 0.69 + 0.9cmanswer in kilograms/liter where cm = (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 + 0.0297cE and d = 0.69 + 0.9cm 0.0297cE = 0.9cm cm/cE = 0.0297/0.9 = 0.033 meters/kilograms1/3 = 0.033 inches/pound1/3 1 meter/1 kilogram1/3 = 0.033 (disregarding units)  39.37 inches/2.2046 pounds1/3 1  = 0.033 proof !!! 39.37/1.30

23. Example calculation of density: Person 70 inches and 170 pounds d = 0.69 + 0.0297cE cE = h/w1/3 = 70/1701/3 = 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 = mi mi = diVi density = mass/volume M = diVi Note that if d is uniform or assumed to be uniform, M = dVi 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 =  mixi

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 (72.1, 98.9) 98.9 0.433(length) 0.433(Y) center of mass 0.567(length) 0.567(Y) (86.4, 54.9) 54.9 86.4 72.1 X 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. Calculate the center of mass of a multisegment system (e.g., lower extremity, entire body) - 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 x0 = m1x1 + m2x2 + m3x3 + + mnxn M y0 = m1y1 + m2y2 + m3y3 + + mnyn M

33. sample table Segment Proportion of total mass x value of center of mass x product y value of center of mass y product Segment 1 1 x1  x1 y1  y1 Segment 2 2 x2  x2 y2  y2 Segment 3 3 x3  x3 y3  y3             Segment n n xn  xn yn  yn  = 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. 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)) - 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. Segment Proportion of total mass x value of center of mass x product y value of center of mass y product Foot 0.09006 1.342 0.12086 0.069 0.00621 Shank 0.28882 1.256 0.36276 0.331 0.0956 Thigh 0.62112 1.175 0.72981 0.678 0.42112  = 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. Moment of Inertia  F = ma and  = I, where F = force, m = mass, a = acceleration,  = torque or moment of force causing angular acceleration, I = moment of inertia = mixi2, 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 = mixi2 = m1x12 + m2x22 + m3x32 +  + mnxn2

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 vertical axis through center of mass y horizontal axis through center of mass m1 m2 m3 m4 m5 0.1m 0.1m 0.1m 0.1m 0.1m 0.1m x x y m1 = m2 = m3 = m4 = m5 = 0.5 kg

45. Calculate the moment of inertia about y-y vertical axis through center of mass y horizontal axis through center of mass m1 m2 m3 m4 m5 0.1m 0.1m 0.1m 0.1m 0.1m 0.1m x x y Iy-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 = 0.275kgm2

46. Calculate the moment of inertia about x-x vertical axis through center of mass y horizontal axis through center of mass m1 m2 m3 m4 m5 0.1m 0.1m 0.1m 0.1m 0.1m 0.1m x x y Ix-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 = 0.025kgm2

47. Calculate the moment of inertia about vertical axis through center of mass vertical axis through center of mass y horizontal axis through center of mass m1 m2 m3 m4 m5 0.1m 0.1m 0.1m 0.1m 0.1m 0.1m x x y Icg = (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 = 0.05kgm2

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 I0 = m02 where I0 = 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