KIN 3309 1nd Edition Lecture 9 Outline of Last Lecture I. Human Movement AnalysisII. KinematicsIII. Linear KinematicsIV. Collection of Kinematic DataV. Spatial Reference SystemsVI. Cartesian Coordinate SystemVII. 2-D Cartesian Coordinate SystemVIII. Two-Dimensional Reference SystemIX. Quadrants in a Two-Dimensional Reference SystemX. Three-Dimensional Reference SystemXI. Three-Dimensional Coordinate SystemXII. Vectors and ScalarsXIII. Vector MagnitudeXIV. Distance from OriginXV. Vector ComponentsXVI. Vector DirectionXVII. Vector OrientationXVIII. Vector Orientation StandardXIX. Vector Arithmetic XX. Example 1aXXI. Example 1bXXII. Example 1cXXIII. Example 1dXXIV. Example 2aXXV. Example 2bXXVI. Example 2cXXVII. Example 3XXVIII. Example 4XXIX. Position and DisplacementXXX. Example 5XXXI. QuizOutline of Current Lecture These notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a substitute.I. Human Movement AnalysisII. Movements Occur Over TimeIII. Important ParametersIV. Velocity (v)V. Velocity (slope)VI. Velocity EstimationVII. Finite DifferentiationVIII. Instantaneous VelocityIX. Average vs. Instantaneous VelocityX. Graphical ExampleXI. Constant VelocityXII. AccelerationXIII. Instantaneous AccelerationXIV. Constant AccelerationXV. IntegrationXVI. Finite IntegrationXVII. ExampleXVIII. Riemann SumXIX. Kinematic AnalysisXX. Gait KinematicsXXI. How to Solve Kinematics ProblemsXXII. QuizCurrent Lecture*Constant Velocity: the xf is very importantI. Human Movement Analysisa.II. Movements Occur Over Timea. Knowledge of the temporal patterns of a movement is critical in a kinematic analysis since changes in position occur over timeb.III. Important Parametersa. Position? WHERE?i. Location in space relative to some reference pointii. Linear and angular position (s, theta)b. Displacement & Distance. HOW FAR?i.c. Velocity and Accelerationi.IV. Velocity (v)a.V. Velocity (slope)a.VI. Velocity Estimationa. Velocity is the slope of a displacement-time graphb. Approximating velocity from displacement data employs finite differentiationi. Differentiation gives the average velocity between displacement pointsc.VII. Finite Differentiationa. Differentiation over one time interval calculates the velocity of a point midway between the framesb.c. First central differences method reduces sensitivity (accuracy), but aligns velocities with the same time frame as displacementsd.e.VIII. Instantaneous Velocitya. Instantaneous velocity is the apparent velocity at any moment (every point of time)b. The instantaneous velocity can be positive, negative, or zeroc. The instantaneous speed: the magnitude of the instantaneous velocityd. The instantaneous speed has no direction associated with ite.IX. Average vs. Instantaneous Velocitya.X. Graphical Examplea.XI. Constant Velocitya. Indicates the instantaneous velocity at any instant during a time intervali. Same as the average velocity during that time intervalXII. Accelerationa.b. Acceleration is used in everyday terms as a scalar c. It is, strictly speaking, a vectord.XIII. Instantaneous Accelerationa. The instantaneous acceleration is the limit of the average acceleration as change in t approaches 0b.XIV. Constant Accelerationa. Indicates the instantaneous acceleration at any instant during a time intervali. Same as the average acceleration during that time intervalXV. Integrationa. There are times when it is more convenient to collect velocity (or acceleration) data than position data for biomechanical analysisb. To compute displacement from velocity we use integrationXVI. Finite Integrationa. Finite differentiation calculates the slope of the curveb. Finite integration calculates the area under the curvec.XVII. Examplea.b. Area A equals 3 m/s x 6 s = 18 mc. This is change in position from 0 – 6 sd. Area B equals 7 m/s x 2 s = 14 m e. Total change in position from 0 – 8 s is 32 mXVIII. Riemann Suma. Finite integration approximates the area under curves as a series of rectanglesi. This is called the Riemann sumii. If change in t is small enough, this is an accurate approximationiii.XIX. Kinematic Analysisa. There are many uses of kinematic analysisb. Sports scientists and coaches often use kinematics to characterize elite performancei. E.g., analyzing a movement pattern, golf club head speedc. Ergonomists use kinematics to assess injury riski. E.g., assessing poor postures, high task repetitiond. Doctors and physiotherapists also use kinematicsi. E.g., assessing walking gait with prostheses, range of motionXX. Gait Kinematicsa. Gait Cyclei. Single sequence of functions by one limbii. Begins when reference foot contacts the groundiii. Ends with subsequent floor contact of the same footb. Stance Phasei. Reference limb in contact with the floorc. Swing Phasei. Reference limb not in contact with the floord. Single Supporti. Only one foot in contact with the floore. Double Supporti. Both feet in contact with floorf. Cadencei. Number of steps per unit time (e.g., 100-115 steps/min)g. Step Lengthi. Distance between corresponding successive points of heel contact of the opposite feeth. Stride Length i. Distance between successive points of heel contact of the same footi. Support and non-support (swing) phases are also of interestj. In walking, the time in the support phase is approximately 66-60% (swing phase 34-40%)k. The relative time (% of cycle) spent in the support phase decreases with increasing speed:i. Jogging – 59-30%ii. Full Sprint – 25-20% l. Changes associated with pathological function/surface conditionsi. Increased support phase  decreased swing phaseii. Shorter step lengthiii. Increased time in double supportXXI. How to Solve Kinematics Problemsa. Step 1i. Identify clearly what the problem is askingb. Step 2i. Identify the information you are givenc. Step 3i. Identify relationshipsd. Step 4i. Combine the given information and the relationshipsXXII. Quiza. A sprinter is observed to run 50 meters in 7.5 seconds. Calculate the speed of thesprinter in m/sec.i. 50 m / 7.5 s = 6.67 m/sb. An individual runs 20 km in 89 minutes. What was the average speed in meters per second?i. 0.27ii. 0.90iii. 3.75: 20/89 = 0.23 km / min then .23x (1000/60) = 3.75 m/siv. 225c. During running 100 m, a runner runs from the 20- to 30- meter mark in 60 frames of a video record. If the video camera recorded data at 50 fps, how fast was she running during this interval?i. 6.5 m/sii. 36