Alpine Skiing And Models In Biomechanics
Alpine Skiingand Models in BiomechanicsCenter for Sensory-Motor InteractionSports BiomechanicsUwe Kersting – MiniModule 08 - 20111 Uwe Kersting, 2011Objectives Review fundamental concepts on skimovement on a slope Provide technical solutions for generatingand controlling forces to ‘turn’ skis Review the concept of inverse dynamicsmodeling Apply inverse dynamics to skiing Determination of muscle forces usingmodels Uwe Kersting, 201121
Contents1.Fundamental biomechanics of parallelskiing2.Body actions to initiate ski movement3.Inverse dynamics priciples4.Muscle forces by optimization5.Ankle joint loading and shoes6.Knee joint loading in skiing Uwe Kersting, 20113I. Fundamental biomechanics of parallel skiing The logics of the procedurehttp://www.youtube.com/watch?v LtMHcLgUFo4&NR 1 The specific motion of the skis Which are the principles the skis have tofollow? The four principal activities A necessary requirement: Balance Provisional result: Few simple technicalelements Important consequences Uwe Kersting, 201142
The logics of the procedurehttp://www.youtube.com/watch?v 3sSwo1-SGpY&feature list related&playnext 1&list SP143CF92A7E71E11A Skiers perform the ski-motion by means of specificactions. This implies the question: How do the skis move onthe slope? Then one has to ask: By which principles is thismotion created? Then one can explore: Which activities produce therequired effects? http://www.youtube.com/watch?v aofTCdhlyyY5 Uwe Kersting, 2011The specific motion of the skis In any turn the skis changefrom the outside- to theinside-edges. At the onset of edgerelease the skis drift inwardover the outside edges(faster at the tips):Inward-Drifting. Following the edge changethe skis drift outward overthe inside edges (faster atthe tails): Outward-Drifting Uwe Kersting, 201163
Investigation 13 skiers Carrying outstandard skischooldemonstrationtechniques High/lowoffloading, etc.7 Uwe Kersting, 2011Results Ski movement ‘low’Arrows are velocitiesMain result:Any offloading happensbefore the turning of theskis set in Uwe Kersting, 201184
Which effect are needed for this tohappen? lateral motion inwardonly needs lateral forcesinward. lateral motion outwardonly needs lateral forcesoutward. The faster inwarddrifting of the tips andfaster outward-driftingof the tails needparticular lateral forces. Uwe Kersting, 20119The four principal activitiesThe first principal activity: Falling-inwards (I) By means of falling-inwards the skis become able toslip sideways.We need to distinguish two questions: What is the effect of the lateral support of the bodyon the skis? What is the effect of falling-inwards itself on theskis? Uwe Kersting, 2011105
The four principal activitiesThe first principal activity: Falling-inwards (II) Only the lateral support of the skier FSt produces alateral force inward FS for the lateral motion, whendrifting inward (a). Only the lateral support of the skier FSt produces alateral force outward FS for the lateral motion, whendrifting outward (b). Uwe Kersting, 201111The four principal activitiesThe first principal activity: Falling-inward (III) Falling inwards itself induces a reaction force FSbeyond/behind the centre/midpoint of the skis because the bindings are located behind the center. When drifting inward (a) FS brakes the slipping of thetails and therefore the tips are slipping faster. When drifting outward (b) FS enhances drifting outwardof the tails. Uwe Kersting, 2011126
The four principal activitiesThe first principal activity: Falling-inward (IV)Conclusion: Drifting inwards and drifting outward - and thusparallel-turns - can be effected in principle only bymeans of falling-inward. All essential conditions are fulfilled to perform theski motion:Edge-change, inclination, ability to slip laterally andlateral forces.13 Uwe Kersting, 2011The four principal activitiesThe second principal activity: Ski-change (I) Ski-change only meanslifting the inside ski –without shifting thebody weight to theoutside ski. This activity isdifferent from the socalled stepping. Effect is: immediateinward falling Uwe Kersting, 2011147
The four principal activitiesThe second principal activity: Ski-change (II)Conclusion: Because the skichange immediatelycauses falling inward,all essential conditionsare fulfilled toperform the skimotion. Uwe Kersting, 2011 Parallel turns can beeffected in principleonly by means of skichange.15The four principal activitiesThe third principal activity: Angulation-change (I) Angulation-changemeans the lateralmotion changing fromone angulation to theother. What is the effectof this activity? Uwe Kersting, 2011 The lateral turningof the upper part ofthe body occurs incoincidence with acounter movement ofthe legs.168
The four principal activitiesThe third principal activity: Angulation-change (II)Conclusion: Angulation changeobviously affectsedge-change, abilityto slip, inclinationand lateral forcesin a similar way asfalling-inward. Uwe Kersting, 2011 Thus parallel-turnscan be effected inprinciple only bymeans ofangulation-change. 17The four principal activitiesThe fourth principal activity: Leaning-forward /backward (I)Graphical Explanation: After initiation of the ski-turn the centre of gravity(KSP) has the tendency to maintain the direction ofthe motion (v). Because thecentre of gravityis behind thecentre of theskis (MS), thedrifting of thetails is enhanced(M). Uwe Kersting, 2011189
A necessary requirement: Balance (I) When the skis perform a turn the skier will adopt adynamic balance; i.e. despite inclination the skierdoes not fall. In order to deviatefrom a linear directionof motion we need alateral force FZ(centripetal force). This force simplyresults from lateralsupport –FSt and weightFG . The skier does notfall, because hecontinuously needs FZ .19 Uwe Kersting, 2011A necessary requirement: Balance (II) The centripetal forceFZ has to varycontinuously becauseFZ depends on thecurrent radius of turnand the currentvelocity. The only thing theskier has to do andcan do for balancecontrol is to vary thelateral support -FSt. Uwe Kersting, 20112010
4 simple key elements (actions) for parallel skiingBy means of falling-inward, ski-change and angulation-changewe produce the specific motion of the skis ( inward-driftingand outward-drifting ).By means of leaning-forward and leaning-backward we controlthe turn of the skis.By means of varying lateral support we maintain the dynamicbalance for the turn.21 Uwe Kersting, 2011Important consequences:No mechanisms of turningFlattening of skis is in contradiction to the position of theedges, when performing turn.Unweighting does not fit with the requirement of lateralsupport of the skier.Leg rotation is not practicable, because the edges are alwaysdirected against the slope.Stepping ( shifting the body weight to the outside ski ) is notconceivable, because falling-inward ( inclination ) is required. Uwe Kersting, 20112211
A simple solution The sum of the friction forces FR ( external force )eccentrically have an effect on the skis and causesexternal torques from the slope to the skis. It is unmistakenly reality: The slope turns the skis!.23 Uwe Kersting, 2011A new view – A new understanding The slope turns the skis, but we have to establish theappropriate contact between the skis and the slope.We do this by means of the technical elementsdiscussed : by falling-inward, skichange and angulationchange we vary the edgeangle and thereby thefriction force FR. Uwe Kersting, 2011 Through leaning forwardand backward we vary thedistance of the centre ofgravity (KSP) from thefriction force and therebythe torques from the slopeto the skis.2412
ConclusionIn short:When performing parallel-turns- we always do the same !(G. Kassat, 1985, 1997) Uwe Kersting, 201125Lever arms and musclesEffort force and load force are applied on the sameside of the axis of rotationEffort force applied closer to axis than the load(ie, d effort d load)Effort and load force act in opposite directionsGood for moving load quickly or through large range ofmotion; poor for strengthd effortFeffortFloadFeffortaxisd loadFload Uwe Kersting, 201113
Static equilibriumA system is at rest and will remain at restNo translation or rotation is occurring or will occurConditions for static equilibrium(from Newton’s 1st law):Net external force in x-directionequals zeroNet external force in y-directionequals zeroNet torque produced by allexternal forces and all externaltorques equals zeroΣFx 0ΣFy 0ΣΤ 0Can use any point as the axis of rotationCan solve for at most three unknown quantities Uwe Kersting, 2011Lecture problem 1During an isometric (static) knee extension, atherapist measures a force of 100 N using ahand dynamometer in the position shown below.Find the resultant knee joint force and torque.Does the dynamometer position affect themeasured force?60 24cm30cmm 4.5kgFdynamometer 100N Uwe Kersting, 201114
Dynamic equilibriumApplies to rigid bodies that are acceleratingConditions for dynamic equilibrium(from Newton’s 2nd law):Net external force in x-directionequals mass times x-accelerationNet external force in y-directionequals mass times y-accelerationNet torque produced by allexternal forces and all externaltorques equals moment of inertiatimes angular accelerationΣFx m axΣFy m ayΣΤ Ι αNet torque must be computed about COM orfixed axis of rotationCan solve for at most three unknownquantities Uwe Kersting, 2011Computing joint forces and torquesIt is possible to measure:joint position (using video/motion capture)ground reaction forces (using force platform)centre of pressure (using force platform - point of application ofground reaction forces)From joint position data, we can compute:absolute angle of each segmentlocation of centre of mass of each segmentCan use central differences method to compute:angular velocity of segmentangular acceleration of segmentx- and y-velocity of segment COMx- and y-acceleration of segment COMFinally, use general equations of motion to compute joint forces andtorques Uwe Kersting, 201115
General equations of motionproximaljointFpyFrom dynamic equilibrium: ΤpFpx yΣFsegment msegment asegment Τcm ax Fdx - Fpx x ay θm ay Fdy - Fpy - FWαΣΤjoint/COM Ιjoint/COM αjoint/COMaxΙCOM α - Τd ΤpFW- (L - c) sinθ FdxL segment length- c sinθ Fpxc proximal end to COM (L - c) cosθ FdyFdxdistal c cosθ Fpyjoint ΤΤdFdy Uwe Kersting, orsiflexionExtension Uwe Kersting, 201116
How to solve the dynamic case?Knee extensor: Kinematic data – no externallymeasured forces60 24cmm 4.5kg30cm InverseDyn 01.xls Uwe Kersting, 2011So far we have only talked about net joing torque and force:The ‘many muscles’ problem?Solve the muscle recruitment problem: Uwe Kersting, 20113417
The effect of ski boot modifications onjoint loading during mogul skiingUwe G. KerstingPaul McAlpine, Nico KurpiersCENTER FORSENSORY-MOTORINTERACTIONDEPARTMENT OFHEALTH SCIENCEAND TECHNOLOGY Uwe Kersting, 2011BackgroundFreestyle skiing growingremarkably over thepast decade(Babic 2006, Fry 2007)Injuries affect mainly the knee(Langran, 2008; Hunter 1999) especially in mogul skiing(landings following aerials) Uwe Kersting, 201118
Mechanisms (?) Uwe Kersting, 201137Intervention possibilitiesEffect of equipment on knee joint loading infree-style skiingSkier-Shoe-Binding-Ski System? Focus: boot shaft Uwe Kersting, 201119
ApproachDetailed mechanical assessment of skiingtechnique!segmental movementexternal forces and momentsbiomechanical models .http://www.youtube.com/watch?v scLlZ5E-zCQOutdoors! Uwe Kersting, 2011Just move the lab outdoors?Kinematics are manageable by camerasForce measurement systems have beendescribed – few intervention studiespublished(Niessen et al., 1999)6 DOF force sensor(Kiefmann et al., 2006) Uwe Kersting, 201120
The ModelDetailed model repository includedKinematic optimisation/scalingInverse dynamicsOptimisation for muscle activationVL: 93%AnyBodyModelingSystemVL: 100% Uwe Kersting, 2011Adaptation to skiingUpper trunk and arms fixed(in a skiing position)Subtalar joint axis fixed inneutral postiionForceplates ‘attached’ to thefeetBoot stiffness adSTIFF vs. FLEX SHAFTded as angledependentjoint torque Uwe Kersting, 201121
ApplicationStiffness function addedto ankle modelApplication to two bootinterventions1) Wave course:(Kurpiers et al. ISSS, 2009)2) Mogul skiing run:(Kurpiers et al., 2011) Uwe Kersting, 2011A hypothetical application3000R SOR GL2500R GM2000F [N]1 subject, 1 trialSame kinematicsSame reaction forcesBoot specific stiffness1500100050000.313000 0.3100.51t [s]t [s]0.710.71R SOR GL2500-10002000F [N] F [N]Stiffness removedJoint compressionforce0.51R GM-20001500-30001000-4000500Fy ank 0 ST-500000.31-6000Fy ank BOOT0.710.51t [s] Uwe Kersting, 201122
In real life9 subjects, 3 trials each, 2 boot conditionsflexible shaft (FL) – standard (stiff) shaft(ST)Average muscle activation forceAt GRF maximum69 – 81% muscleforce reduction40% reduction ofankle joint compression Uwe Kersting, 2011The knee joint!10-10RightLeftKneeanteriorposteriorforceF [N/kg]-30-50-70FLST-90-110 Uwe Kersting, 201123
Results - Muscles Uwe Kersting, 2011ResultsExampleRight GRF with ap-tibia force1400Joint force-4855 NJoint force-1793 N1200F [N]1000800600GRF FL400GRF ST20000.20.30.40.50.60.70.8time [s] Uwe Kersting, 201124
Summary of findings Change in the body positiono Greater range of movement, particularly at ankle jointo Forward shift of the CoM at force maximum Reduced GRF Change in net forces and moments in the ankle and kneejointo Reduced ap tibia force (from 11 BW to 8 BW) Reduced muscle activation in the lower extremities General acceptance of the modified ski boot by freestyleskiers Uwe Kersting, 2011SummaryCommonly accepted technique descriptions maybe misleading implications for teaching ofsports (skiing) techniqueInverse dynamics to ‘inversed’ inverse dynamicsOutdoor skiing test setupestablishedesentedAnkle flexion stiffness of bootsalters joint loading at ankle andkneeFuture: Inclusion of landing afteraerials most criticalPerspectives for the generalskiing community Uwe Kersting, 201125
Thank you!AcknowledgementsPeter SpitzenpfeilVeit SennerDAADObels FamiliefondetSpar Nord FondAnyBody TechAndi KiefmannEllen HildFrédérik MeyerJürg BinerAnyBody GroupBergbahnen:Zermatt, Garmisch-P.HeadSkiOaseNovel GmbH Uwe Kersting, 201126
(from Newton’s 2 nd law): Net external force in x-direction equals mass times x-acceleration Net external force in y-direction equals mass times y-acceleration Net torque produced by all external forces and all external torques equals moment of inertia times angular acceleration Net to
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