Kinematics Kinematics Role of Kinematic Analysis in Biomechanics ...

KinematicsKinematicsMechanicsPhysicsOther areasof physicsStaticsDynamicsKinematicsKineticsKinematics: branch of dynamics which describes themotion of objects without reference to the forceswhich caused, or were generated by, the motionRole of Kinematic Analysisin Biomechanics• Kinematic data by themselves may provideuseful information about a human movement– Initial description of a previously unstudiedmovement pattern– Assessment of the coordination pattern of aparticular movementFrames of ReferenceFixed inertial (global)reference frameY• Kinematic data may be required as part of amore complete analysis of a movement– Kinematic and EMG data– Inverse dynamics (kinematic & kinetic data)ZXAbsolute position ofthe pelvis in a fixedinertial reference frameFrames of ReferenceFrames of ReferenceLocal or segmentalreference framesJoint-basedreference framesX TAbd/AddY TFlex/ExtZ TInt/Ext Rotreference frame fixedto the thigh segmentreference frame fixedin the knee joint1

Kinematic variablesLinear kinematics of particles (points)– Position - location at a given time– Displacement - change in position over aperiod of time (compare with distance)– Velocity - rate of change in position withrespect to time (compare with speed)– Acceleration - rate of change in velocity withrespect to time– Jerk - rate of change in acceleration withrespect to time*All are vector quantities(except distance and speed)Position & DisplacementDegrees of freedom - the minimum number ofindependent parameters necessary to specifythe configuration of a systemA point has 3 degreesof freedom, so itsposition is defined by 3coordinates (x, y, and z)ZYXPosition & DisplacementIn two-dimensional analysis, only two quantitiesare needed to completely describe the positionof a point or particleY5r3PXThe position of pointP can be convenientlydenoted by a vector r,with r defined asr = 3 i + 5 ji and j are unit vectorsin the directions of Xand Y, respectivelyPosition & DisplacementYr 1P 1P 2r 2XSay that the point movesfrom position 1 (P 1 ) toposition 2 (P 2 ) such thatr 1 = 3 i + 5 j andr 2 = 8 i + 7 jHow do we represent thedisplacement as a vector?Position & DisplacementYr 1d (r 1 -r 2 )P 1P 2r 2∆x∆yXThe displacement vector (d)is defined asd = r 2 - r 1 , sod = (8-3) i + (7-5) j = 5 i + 2 jThat is, the position haschanged 5 units in the Xdirection (∆x) and 2 units inthe Y direction (∆y)VelocityYd (r 1 -r 2 )P 2P 1 ∆y∆xr 1r 2XVelocity is the rate at whichthe X and Y coordinates arechangingThe average velocity in theX direction is given by_v X = ∆x∆tand the instantaneousvelocity in the X direction isgiven by.v X = lim ∆x = dx = x∆t→0 ∆t dt2

VelocityVelocityWhereas the position vectorto point P was given byYr 1d (r 1 -r 2 )P 2P 1 ∆y∆xr 2XThe average velocity in theY direction is given by_v Y = ∆y∆tand the instantaneousvelocity in the Y direction isgiven by.v Y = lim ∆y = dy = y∆t→0 ∆t dtYr = x i + y jv 1the vector representation ofv 2the velocity of point P isP 2written as. . .P 1 v = r = x i + y jXThe velocity vector v willalways be tangent to thepath followed by point PVelocityVelocityXtangent linex ix i-1x i+1When data are expressedin digital form, a trueinstantaneous velocitycannot be calculatedIt can be approximated,however, using finitedifference techniques.v X(i) = x ≈ x i+1 − x i-12∆tThe equation v X(i) = (x i+1 − x i-1 ) / 2∆tis a first order central difference equation, and can beapplied to all data points except the first and lastTo estimate the velocity for the first point use a secondorder,forward difference equationv X(1) = -3x 1 + 4x 2 − x 32∆tand for the last point use a second-order, backwarddifference equation∆t∆ttAs ∆t gets smaller, theapproximation gets betterv X(n) = x n-2 − 4x n-1 + 3x n2∆tVelocityGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s)1 0.0000 1.73 ?2 0.0083 1.82 ?3 0.0167 1.88 ?4 0.0250 1.79 ?5 0.0333 1.83 ?First point, use forward differencev X(1) = (-3×1.73 + 4×1.82 − 1.88) / (2×0.0083) = 12.65VelocityGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s)1 0.0000 1.73 12.652 0.0083 1.82 ?3 0.0167 1.88 ?4 0.0250 1.84 ?5 0.0333 1.85 ?Points 2 - 4, use central differencev X(2) = (1.88 − 1.73) / (2×0.0083) = 9.04v X(3) = (1.84 − 1.82) / (2×0.0083) = 1.20v X(4) = (1.85 − 1.88) / (2×0.0083) = -1.813

VelocityGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s)1 0.0000 1.73 12.652 0.0083 1.82 9.043 0.0167 1.88 1.204 0.0250 1.84 -1.815 0.0333 1.85 ?YAccelerationP 1P 2∆vThe vector change invelocity (∆v) is given by∆v = v 2 − v 1v 2v 1It can be represented interms of the change ofvelocity in the X and YdirectionsLast point, use backward differencev 1 v 2∆v∆v Xv X(5) = (1.88 − 4×1.84 + 3×1.85) / (2×0.0083) = 4.22X∆v YAcceleration∆v XAcceleration∆v XYX∆v Yv 1 v 2the X direction is given by_a X = ∆v X∆vThe average acceleration inP 2∆tP 1and the instantaneousacceleration in the Xdirection is given by.a X = lim ∆v X = dv X = v X∆t→0∆t dtYX∆v Yv 1 v 2the Y direction is given by_a Y = ∆v Y∆vThe average acceleration inP 2∆tP 1and the instantaneousacceleration in the Ydirection is given by.a Y = lim ∆v Y = dv Y = v Y∆t→0∆t dtYAccelerationthe vector representation ofv 1 P 2the acceleration of point Pa T(2)is written asP 1 . .. .. ..v 2a = v = r = x i + y ja N(2)XWhereas the velocity vectorof point P was given by. . .v = r = x i + y jThe acceleration vector, a,will have two components,tangent to and normal tothe path followed by point PAccelerationA similar finite difference approach as for velocity can beused to estimate accelerationA first-order, central difference equation for acceleration,written in terms of position data, isa X(i) = x i+1 − 2x i + x i-1∆t 2The second-order, forward and backward differenceequations for the first and last points area X(1) = 2x 1 − 5x 2 + 4x 3 − x 4 a X(n) = -x n-3 + 4x n-2 − 5x n-1 + 2x n∆t 2 ∆t 24

AccelerationGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s) X accel (m/s 2 )1 0.0000 1.73 12.65 ?2 0.0083 1.82 9.04 ?3 0.0167 1.88 1.20 ?4 0.0250 1.79 -1.81 ?5 0.0333 1.83 4.22 ?AccelerationGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s) X accel (m/s 2 )1 0.0000 1.73 12.65 1306.42 0.0083 1.82 9.04 ?3 0.0167 1.88 1.20 ?4 0.0250 1.79 -1.81 ?5 0.0333 1.83 4.22 ?First point, use forward differencea X(1) = (2×1.73 − 5×1.82 + 4×1.88 − 1.79) / (0.0083) 2 = 1306.4Points 2 - 4, use central differencea X(2) = (1.88 − 2×1.82 + 1.73) / (0.0083) 2 = -435.5a X(3) = (1.84 − 2×1.88 + 1.82) / (0.0083) 2 = -1451.6a X(4) = (1.85 − 2×1.84 + 1.88) / (0.0083) 2 = 725.8AccelerationGiven the following position data for the X coordinate ofa marker on the hip joint, calculate the velocityFrame Time (s) X coord (m) X vel (m/s) X accel (m/s 2 )1 0.0000 1.73 12.65 1306.42 0.0083 1.82 9.04 -435.53 0.0167 1.88 1.20 -1451.64 0.0250 1.79 -1.81 725.85 0.0333 1.83 4.22 ?Last point, use backward differencea X(5) = (-1.82 + 4×1.88 − 5×1.84 + 2×1.85) / (0.0083) 2 = 2903.2Velocity & AccelerationNote that if global polynomials or splinefunction were used to fit and/or smooth thedata, then velocity and acceleration can bedetermined analyticallyExample:x(t) = 3 + 7t − 4t 2 + 8t 3 + 5t 4 − 2t 5.v(t) = dx = x = 7 − 8t + 24t 2 + 20t 3 − 10t 4dt.a(t) = dv = v = -8 + 48t +60t 2 − 40t 3dtBack to SmoothingWhy were we so careful to properly smooth outhigh frequency noise from our data?Smoothing & DifferentiationDifferentiation amplifies high-frequency noise5

Smoothing & Differentiation• For the 1 st derivative (velocity) signalamplitude increases proportional tofrequency• For the 2 nd derivative (acceleration) signalamplitude increases proportional tofrequency squared• This is why it is so important to eliminatesources of high frequency noise before datacollection, and suppress the remaining highfrequency noise through low-pass filteringAngular Kinematic VariablesThe following is applicable to rigid bodies inplanar motion (3-D is more complicated)– Angular Position - angle at a given time– Angular Displacement - change in angularposition over a period of time– Angular Velocity - rate of change in angularposition with respect to time– Angular Acceleration - rate of change inangular velocity with respect to timeAngular Position (Orientation)Degrees of freedom - a rigid body in 3-D spacerequires six quantities to completely describe itsposition and orientationAngular Position (Orientation)In two-dimensional analysis, only two linearcoordinates (x, y) and one angle (θ) are neededto completely describe the position andorientation of a rigid bodyCould use the x, y, andz coordinate of thecenter of mass, plusthe angular rotationsrelative to the globalreference frame(other coordinate setsare also possible)(x,y,z)Yθ 2θ 3Zθ 1XY(x,y)θXSo in planar analyses,you must know the xand y coordinates of atleast one point on eachbody, plus the anglerelative to some fixedreferenceAngular Position (Orientation)Segment angles, relative to the right horizontal,can be calculated using the coordinates ofmarkers at the ends of the segmentAngular Position (Orientation)Y(x 1 ,y 1 )The segment angle canbe calculated anywherein the first two quadrantsusing the law of cosines:Y(x 1 ,y 1 )First, create a third,imaginary point, to theright along the x-axisθ(x 2 ,y 2 )a 2 = b 2 + c 2 − 2bc × cos θ( )θ = cos -1 b 2 + c 2 − a 22bcθ(x 2 ,y 2 )(x 3 ,y 3 )The value of y 3 is thesame as y 2 , the value ofx 3 simply needs to belarger than x 2XX6

Angular Position (Orientation)Angular Position (Orientation)Y(x 1 ,y 1 )b(x 2 ,y 2 )aθc(x 3 ,y 3 )XLabel the sides a, b, andc as shown, and solvethe equation for θθ = cos -1 b 2 + c 2 − a 2where( )2bca = √ (x 3 − x 1 ) 2 + (y 3 − y 1 ) 2b = √ (x 2 − x 1 ) 2 + (y 2 − y 1 ) 2c = √ (x 3 − x 2 ) 2 + (y 3 − y 2 ) 2Y(3,6)b(5,2)aθc(8,2)XWith point 1 (3,6), point2 (5,2), and point 3 (8,2)a = √ (8−3) 2 + (2−6) 2 = 6.4b = √ (5−3) 2 + (2−6) 2 = 4.5c = √ (8−5) 2 + (2−2) 2 = 3.0θ = cos -1 4.5 2 + 3.0 2 − 6.4 2θ = 115.7°( )2(4.5)(3.0)Angular Position (Orientation)Once segment angles are known, relative jointangles can be easily calculatedYθ THIGHθ SHANKXJoint angles are typicallycalculated as the angle ofthe proximal segmentminus the angle of thedistal segmentθ KNEE = θ THIGH − θ SHANKThis would make kneeangle 0 at full extension,pos for flexion, and negfor hyperextensionAngular Kinematics• For planar (2-D) motion, the relationshipsbetween angular displacement, angularvelocity, and angular acceleration areperfectly analogous to linear displacement,linear velocity, and linear acceleration• Once segment or joint angles are known,angular velocities and angular accelerationscan be calculated using the same finitedifference approach as we used for linearvelocity and linear accelerationAcknowledgementI want to acknowledge Brian Umberger, PhDfor the organization of much of the informationpresented on these slides7