Models and Anthropometry

Transcription

1 Learning Objectives Models and Anthropometry Readings: some of Chapter 8 [in text] some of Chapter 11 [in text] By the end of this lecture, you should be able to: Describe common anthropometric measurements Choose an appropriate frame of reference to analyze a movement Understand the basic properties of forces and torques Draw link segment models and kinetic diagrams Calculate forces, moment arms, and torques (moments) using anthropometric data Calculate net torque (net moment) List the simplifications you used when performing the calculations above 1 2 Link Segment Models These model the body as a series of rigid segments linked by hinge joints (in two dimensions) They can be used to predict: muscle force joint reaction force reach and space requirements Link Segment Models To develop a link segment model, we need to know: segment lengths segment masses centre of gravity locations (and possibly the segment moment of inertia) joint centres of rotation locations force magnitude and direction 3 4 1

2 Anthropometry This is the science that deals with the measure of size (length, breadth, volume, etc.), mass, shape, and inertial properties of the human body. Anthropometric measurements are statistical data describing human populations. 5 Measuring Segment Volume, Mass, Density and Centre of Mass Length and volume from living subjects or cadavers (volume can be measured using fluid displacement) Mass from cadavers Density from cadavers, or indirectly: density = mass volume Centre of mass and moment of inertia from cadavers, mathematical modeling, or radioisotope scanning 6 Centre of Mass Centre of mass Sum of moments (torques) on all sides of a segment s centre of mass equals zero i.e. the mass is equally distributed about the centre of mass This is not the same as having equal mass on each side! Example: The distance from the elbow to the forearm centre of mass is about 43% of the length of the forearm: 60 N 20 N 0.5 m 1.5 m 7 8 2

3 Anthropometric Data Use data from the population you are working with. Examples: year old Canadian females adult females and males year old children soldiers subject-specific We will use data from the 50% male. Example: According to Webb Associates, in 1978 the average height of female Swedish civilians was cm, and the average height of female Japanese citizens was cm. 9 Force A force is a push, pull, rub (friction), or blow (impact). A force causes or tends to cause motion or a change in the shape of an object. We draw forces as arrows: the size of the arrow indicates the magnitude of the force the direction of the arrow indicates the direction of the force 13 Properties of Forces 1. Magnitude Vector Components 2. Direction point of application sin θ = y component magnitude line of application angle of application θ magnitude θ x component y component cos θ = x component magnitude

4 Quadrants Direction Conventions y axis y axis Quadrant II sin is cos is - Quadrant I sin is cos is Quadrant III sin is - cos is - Quadrant IV sin is - cos is x axis x axis Frames of Reference Absolute the frame of reference is stationary relative to the environment the object (person) moves relative to the frame of reference Frames of Reference Relative the frame of reference is stationary relative to the object (person) the environment moves relative to the frame of reference y axis y axis used to describe the motion of an object (person) relative to the environment used to describe the movement of body segments relative to each other x axis x axis

5 Three-Dimensional Movement Sometimes we need to describe movement in three dimensions. Three-Dimensional Coordinate System x axis horizontal (sagittal) axis y axis vertical axis z axis transverse (medial-lateral) axis 2D sagittal plane analysis 3D analysis 23 (sometimes z is used for the vertical axis) 24 Torque (moment) Properties of Torques Muscles move body segments by generating torque, causing rotation of the segments about joints. 1. Axis of rotation 2. Force 3. Moment arm perpendicular distance between force line of action and axis of rotation T = F x d moment arm

6 Properties of Torques Like force, torque is a vector: 1. Magnitude = (force) x (moment arm) 2. Direction counterclockwise = positive clockwise = negative y axis x axis moment arm 27 Moment Arms Make sure you use the perpendicular distance between the force line of action and the axis of rotation. NOT the distance between the force point of application and the axis of rotation! F F 28 Mechanical Advantage All lever systems have: 1. Effort force 2. Resistive force 3. Axis of rotation (fulcrum) resistive force mechanical advantage (MA): MA = effort force moment arm resistive force moment arm effort force axis of rotation 30 Mechanical advantage Types of lever systems: 1 st class lever 2 nd class lever favours effort force (MA > 1) 3 rd class lever favours range and speed of movement (MA < 1) 31 6

7 Mechanical Advantage Most musculoskeletal systems have a mechanical advantage less than 1. i.e. the moment arm of the effort force (muscle) is less than the moment arm of the resistive force This makes them 3 rd class levers. 3 rd class levers favour range and speed of movement, but require a larger effort force to move the same load Static Equilibrium If there is no linear acceleration: Sum of forces = zero (ΣF = ma = 0) If there is no angular acceleration: Sum of torques = zero (Στ = Iα = 0) ΣF x = 0 ΣF y = 0 ΣM = Static Equilibrium Example: Calculate the: mass of the total arm (hand, lower and upper arm). horizontal distance from the shoulder to the centre of gravities of the three segments. moment at the shoulder in this position Example Continued What is the net shoulder muscle moment? Answer: 9.51 Nm positive because it is counter-clockwise 30 o 34 Direction Conventions: Clockwise is negative Counter-clockwise positive 36 7

8 Muscle Force and Torque What if we want to find the force generated by a muscle? In addition to muscle moment (torque), we need to know how the line of action of the force relates to the axis of rotation: We need to know the moment arm of the muscle Muscle Force and Torque Example: What is the force generated by the muscle? (assume static equilibrium) F J F m given: F gf = -15N F gw = -50N moment arms are 0.03m, 0.15m, 0.3m 37 F gf F gw 38 Example continued: Why can t we solve directly for F m using ΣFy = 0? because we have two unknown forces Instead, we have to use ΣM = 0, taking moments about the elbow. Because F J passes through the elbow joint, we can ignore it that means we only have one unknown force, which we can solve for. Kinetic Diagram It is OK to draw stick figures on a Kinetic diagram (this is not a FBD). τm -15N Moment arms = 0.03m, 0.15m, and 0.3m ΣF x = 0 ΣF y = 0 ΣM = 0-50N

9 Muscle Torque Calculation Muscle Force and Torque τ = Fd τ = ± τ = 2.25 ± 15 Which direction are the torques? Both are counter clockwise τ = =17.25Nm This method of just eyeballing torques removes the need for cross product of vectors. 41 Example continued: If there is minimal co-contraction, the calculated muscle torque will be very close to the forearm flexor torque. If we make some simplifying assumptions, we can move on to calculate muscle force. 42 Simplifications To calculate muscle force from muscle torque, we usually need to make some simplifying assumptions, such as: 1. Assume only one muscle (single equivalent muscle) 2. Estimate the moment arm 3. The segments are rigid structures Example continued: 575N -15N T = F x d ΣF x = 0 ΣF y = 0 ΣM = 0-50N 43 Moment arms = 0.03m, 0.15m, and 0.3m 44 9

10 Simplifications We have also assumed that all the muscles crossing the joint have the same insertion point and line of action. Many people don t use these simplifying assumptions, and create more complex models than we will look at in this course. 46 Sample midterm question The diagram on the next slide shows our 50 th percentile male in the cross position on the rings (he is stationary). Assume the long axes of the arms are horizontal and the force distribution is the same in both hands (i.e. symmetrical). The external forces at the hands (F R & F L ) act at the hands centre of mass. a) Draw a free body diagram of one total arm system (upper arm, forearm and hand). Try to proportion the force vectors somewhat realistically. Name the forces you put in your diagram. [4] 47 θ = 80 o Sample question continued F R θ = 80 o F L b) What is the value of F R in the diagram (remember force is a vector)? [4] c) What is the muscle moment across his right shoulder joint? [8] d) Name two muscles that would be large contributors to this muscle moment. [2] e) Briefly explain why you cannot accurately calculate the amount of force these muscles exert. [2]