ANTHROPOMETRY (İnsan Vücudunu Ölçme Bilimi)

Transcription

1 ANTHROPOMETRY (İnsan Vücudunu Ölçme Bilimi) Dr. Kurtuluş Erinç Akdoğan

2 INTRODUCTION Anthropometry is the major branch of anthropology (insan bilimi) that studies the physical measurements of the human body to determine differences in individuals and groups. A wide variety of physical measurements are required to describe and differentiate the characteristics of race, sex, age, and body type. In the past, the major emphasis of these studies has been evolutionary and historical. More recently Anthropometry is needed for technological developments, especially man-machine interfaces workspace design, cockpits, pressure suits, armor, and so on. Human movement analysis requires kinetic measures masses, moments of inertia, and their locations

3 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

4 Segment Dimensions The most basic body dimension is the length of the segments between each joint. Lengths vary with body build, sex, and racial origin. These segment proportions serve as a good approximation It is preferred to measure directly from the individual.

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6 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

7 Whole-Body Density The human body consists of many types of tissue, each with a different density. Specific gravity weight of tissue/weight of water of same volume Cortical bone: 1.8 muscle tissue: 1.0 fat :1.0 Average density is a function of body build, called somatotype. d: body density c: ponderal index h: height (m) w: weight (kg) Calculate the whole-body density of an adult whose height is 1.78m and who weighs 77.3kg

8 Segment Densities Each body segment has a unique combination of bone, muscle, fat, and other tissue The density within a given segment is not uniform. Because of the higher proportion of bone, the density of distal segments is greater than that of proximal segments Individual segments increase their densities as the average body density increases.

9 Segment Mass and Center of Mass

10 The total mass M of the segment is: If the density d is assumed to be uniform over the segment, then

11 Calculate the coordinates of the center of mass of the foot and the thigh given the following coordinates: ankle (84.9, 11.0), metatarsal (101.1, 1.3), greater trochanter (7.1, 9.8), lateral femoral condyle (86.4, 54.9) The foot center of mass is 0.5 of the distance from the lateral malleolus (ankle) to the metatarsal marker. Thus, the center of mass of the foot is: The thigh center of mass is from the proximal end of the segment. Thus, the center of mass of the thigh is:

12 Calculate the coordinates of the center of mass of the thigh given the following coordinates: greater trochanter (7.1, 9.8), lateral femoral condyle (86.4, 54.9)

13 Y (7.1, 9.8) (length) 0.433(Y) center of mass 0.567(Y) 0.567(length) 54.9 (86.4, 54.9) (X) (X) X

14 Calculate the coordinates of the center of mass of the foot given the following coordinates: ankle (84.9, 11.0), metatarsal (101.1, 1.3),

15 Y (7.1, 9.8) (length) 0.4 (Y) center of mass 0.5 (Y) 0.5 (length) 1.3 (86.4, 54.9) (X) (X) X

16 Center of Mass of a Multisegment System step 1: determine the proportion of mass that each segment is of the entire multisegment system step : 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

17 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 = (56.7kg) = 0.815kg = mass of shank =0.0465(56.7kg) =.63655kg = mass of thigh = 0.100(56.7kg) = 5.67kg = total mass of lower extremity = 9.187kg Segmental Proportions of Lower Extremity Mass foot proportion = 0.81/9.187 = shank proportion =.6355/9.187 = thigh proportion = 5.67/9.187 = total mass proportion = 9.187/9.187 = step : 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

18 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 x 1 x 1 y 1 y 1 Segment x x y y Segment 3 3 x 3 x 3 y 3 y 3 Segment n n x n x n y n 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.

19 Segment Proportion of total mass x value of center of mass x product y value of center of mass y product Foot Shank Thigh = 1.0 = = 0.59

20 Mass Moment of Inertia What is inertia? 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. F=ma, F = force, m = mass, a = acceleration What is moment of inertia? 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. T=Iα, T=torque, I=moment of inertia and α=angular acceleration

21 Moment of Inertia T=Iα, T=torque, I=moment of inertia and α=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 n i1 m r i i m r 1 1 m r m n-1 r n-1 m r n n

22 Hypothetical object made up of 5 point masses y vertical axis through center of mass horizontal axis through center of mass x 0.1m m 1 m m 3 m 4 m 5 0.1m 0.1m 0.1m 0.1m 0.1m x y m 1 = m = m 3 = m 4 = m 5 = 0.5 kg

23 Calculate the moment of inertia about y-y y vertical axis through center of mass horizontal axis through center of mass x 0.1m m 1 m m 3 m 4 m 5 0.1m 0.1m 0.1m 0.1m 0.1m x y I y-y 5 i1 mir i m1r 1 mr m3r 3 m4r 4 m5r 5 I y-y = (0.5kg)(0.1m) + (0.5kg)(0.m) + (0.5kg)(0.3m) + (0.5kg)(0.4m) + (0.5kg)(0.5m) = 0.75kgm

24 Calculate the moment of inertia about x-x y vertical axis through center of mass horizontal axis through center of mass x 0.1m m 1 m m 3 m 4 m 5 0.1m 0.1m 0.1m 0.1m 0.1m x y I x-x 5 i1 mir i m1r 1 mr m3r 3 m4r 4 m5r 5 I x-x = (0.5kg)(0.1m) + (0.5kg)(0.1m) + (0.5kg)(0.1m) + (0.5kg)(0.1m) + (0.5kg)(0.1m) = 0.05kgm

25 Calculate the moment of inertia about vertical axis through center of mass y vertical axis through center of mass horizontal axis through center of mass x 0.1m m 1 m m 3 m 4 m 5 0.1m 0.1m 0.1m 0.1m 0.1m x y I cg 5 i1 mir i m1r 1 mr m3r 3 m4r 4 m5r 5 I cg = (0.5kg)(0.m) + (0.5kg)(0.1m) + (0.5kg)(0.0m) + (0.5kg)(0.1m) + (0.5kg)(0.m) = 0.05kgm

26 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

27 Radius of Gyration Radius of gyration of a body segment is provided measured data Using radius of gyration, moment of inertia of an segment about a given axis rotation can be calculated. 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 = the moment of inertia about the center of mass m = mass of object and ρ 0 = radius of gyration for rotation about Body Segment

28 Parallel Axis Theorem Moment of inertia can be calculated about any parallel axis, given the: moment of inertia about one axis, mass of the segment, and perpendicular distance between the parallel axes In vivo measures of the moment of inertia can only be taken about a joint center. The moment of inertia of a body segment about the axis at COM is body segment joint center

29 Parallel Axis Theorem Most body segments do not rotate about their mass center but rather about the joint at either end. The relationship between this moment of inertia and that about the center of mass is given by the parallel-axis theorem. x can be any distance in either direction from the center of mass as long as it lies along the same axis as I 0 was calculated on.

30 Radius of Gyration/Segment Length in meters (about a transverse axis) Segment Center of Gravity Proximal End Distal End Shank

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

32 Example: a) A prosthetic shank has a mass of 3kg and a center of mass at 0cm from the knee joint. The radius of gyration is 14.1cm. Calculate the moment of inertia about the knee joint. 4 cm 0 cm 6 cm I 0 about the center of mass of the shank = m 0 = 3kg(0.141meters) = 0.06kg meters Using the parallel axis theorem :I k = I 0 + mx = 0.06kg meters + 3kg (0.meters) = 0.18kg meters b) Calculate the moment of inertia for the prosthetic leg about the hip. Using the parallel axis theorem:i h = I 0 + mx = 0.06kg meters + 3kg (0.6meters) = 1.1kg meters Note that I h 0 I 0

33 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 Center of Gravity I cm = I o = m(ρ cm l) = 3.6kg[(0.30)(0.4m)] = 0.055kgm Proximal I prox = m(ρ prox l) = 3.6kg[(0.58)(0.4m)] = 0.161kgm Distal

34 Use of Anthropometric Tables and Kinematic Data

35 Calculation of Segment Masses and Centers of Mass

36 Calculation of Segment Masses and Centers of Mass

37 Calculation of Total-Body Center of Mass For an n-segment body system, the center of mass in the X direction is

38 Calculation of Total-Body Center of Mass

39 Calculate the total-body center of mass at a given frame 15. The time for one stride was 68 frames. Thus, the data from frame 15 become the data for the right lower limb and the right half of HAT, and the data one-half stride (34 frames) later become those for the left side of the body. All coordinates from frame 49 must now be shifted back in the x direction by a step length. An examination of the x coordinates of the heel during two successive periods of stance showed the stride length to be = cm. Therefore, the step length is 70.7 cm = m. Table shows the coordinates of the body segments for both left and right halves of the body for frame 15. The mass fractions for each segment are as follows: foot = , leg = , thigh = 0.10, 1/ HAT = The mass of HAT dominates the body center of mass, but the energy changes in the lower limbs will be seen to be dominant as far as walking is concerned