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1
© Uwe Kersting, 20111
The ‚use‘ of Energy
in Sport Biomechanics
Center for Sensory-Motor Interaction
Anvendt Biomekanik
Uwe Kersting – MiniModule 07 - 2011
© Uwe Kersting, 20112
Objectives• Quickly review some history and methods
• Review general energy equations
• Discuss their potential role for sports biomechanics
• Go through examples on equipment, sport techniques, muscle mechanics, etc.
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© Uwe Kersting, 20113
Contents
1. Introduction
2. Fundamentals of photogrammetry
3. Planar kinematics
4. Calibration
5. Extension to 3D video techniques
6. (Examples: energy calculations for jumping and running)
7. Summary
© Uwe Kersting, 2011
Where do ‘we’ come from?
Sports biomechanics used to be rather descriptive
- But that was not Unexpected...
4
3
© Uwe Kersting, 20115
Historical perspective
Technical realisation: camera obscura
Also described and used by Aristotle, Kepler
First technical drawings by Leonardo DaVinci (1490)
© Uwe Kersting, 20116
Multiple exposure
Eadweard Muybridge
(1830 – 1904)
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© Uwe Kersting, 20117
Conceptual Models in Biomechanics
Partial heights model
(transferable to many movements)
(Hay, 1978)
© Uwe Kersting, 20118
Partial heights model: factors
5
© Uwe Kersting, 2011
Resolve the contributors
Kinematic descriptors:- knee angle, hip angle etc. � ?- at key instances during execution ?- segmental velocities (angular andtranslatory)
- anatomical model � implications formuscle action
Rough but easy: energies – segmental or ‘just’ the whole body
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© Uwe Kersting, 201110
Energy in a purely ‘physics’ sense
Mechanical energy:
Ekin = 0.5 m v2
Epot = m g h
Erot = 0.5 I ω2
Eel = 0.5 k ∆x2
Total body energy:
E tot = Σ Ekin i + Epot i + Erot i : n = # of segmentsi = 1
n
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© Uwe Kersting, 201111
Example: Running with and without shoes
Oxygen consumption: + 4 – 5 % when running with shoes(6 – 7 min in a marathon!)
Total work 107 JShoe mass: 100 gLifting height 0.2 mMaximal speed of foot during swing: 10 m/sStep length: 2 m => 20 ksteps
Additional work due to gravityAdditional work to accelerate the shoe
© Uwe Kersting, 201112
Additional Energyneeded
5
10
5 10 15
100g
200g
300g
400g
Add. Work [%]
Max foot speed [m/s]
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© Uwe Kersting, 2011
Application to the whole body
What do we need to know or have?
___________ ?
___________ ?
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© Uwe Kersting, 201114
COM calculation from segment positions
x0 = (Σ mi * xi)
M
y0 = (Σ mi * yi)
M
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© Uwe Kersting, 201115
E kin tot = Σ mi vi2
+ potential Energy
Epot = Σ mi g hi
Question: Is the total Energy equal to CoM
based calculations?
CoM velocity?
h
© Uwe Kersting, 201116
Rotational Energy contribution?
E rot tot = Σ Ii ωi2
Example
Leg: mleg = 0.0465 BM
Vleg sprint = 20 m/s
Ileg = 0.064 kg m2
ωleg sprint = 500 °/s
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© Uwe Kersting, 201117
Energy in a purely ‘physics’ sense
Mechanical energy:
Total body energy:
E tot = Σ Ekin i + Epot i + (Erot i ): n = # of segmentsi = 1
n
© Uwe Kersting, 201118
Energy of a pole vaulter (after putting down the pole)
(Schade et al., 2000)
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© Uwe Kersting, 201119
Application to high jump
In high jump the initial energy has to be transformed into jump
energy.
© Uwe Kersting, 201120
There is no linear relationship between approach velocity and vertical take-off velocity. (Dapena et al., 1990)
Transformation is coupled with a decrease in total energy during the last ground contact. (Brüggemann & Arampatzis, 1991)
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© Uwe Kersting, 201121
Purpose
• To examine the approach and take-off strategies of high jumpers at the world class level.
• To determine how to estimate the optimal take-off behavior from given initial characteristics.
© Uwe Kersting, 201122
Methods
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© Uwe Kersting, 201123
Data Aquisition and Analysis
• four stationary PAL video cameras (50 Hz), two for each side, synchronised
• calibration with 2 x 2 x 3 m cube
• digitisation with Peak Motus (19 points)
• calculation of body angles, CM position, CM velocity and total energy by „fast information program“ using DLT
© Uwe Kersting, 201124
Digitised Frames
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© Uwe Kersting, 201125
Calculations
(1)H = H1 +2 g
V2 sin2 α
(2)H = +2 g
E k2 sin2 αE p2
g
© Uwe Kersting, 201126
E k 2 =m
E kin 2; E p 2 =
m
E pot 2where:
T in =
And: ET = total Energy
E dec
α(3)
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© Uwe Kersting, 201127
(4)
H = +g
(E T1 - Edec - E p2) sin2 (Τin Edec)E p2
g
The effective height (H) can be expressed as a function of:initial energy, energy loss and
transformation index.
© Uwe Kersting, 201128
RESULTS
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© Uwe Kersting, 201129
2 3 4 5 6 7 8 9 10 11 12 13 14
22
24
26
28
30
32
34
36
38
40
42
ET1
42 J/kg
40 J/kg
38 J/kg
36 J/kg34 J/kg32 J/kg
Edec [J/kg]
ET2 [J/kg]
Group1 (n=16)
Group2 (n=10)
Cluster Analysis
© Uwe Kersting, 201130
Parameters Group1 (n=16) Group2 (n=10)
Initial energy [J/kg] 35.15 (1.38) 39.04 (2.06) *
Energy decrease [J/kg] 4.26 (0.99) 9.23 (2.06) *
Transformation index [°/J/kg] 12.05 (3.34) 5.47 (1.01) *
Final energy [J/kg] 30.88 (1.14) 29.81 (0.74)
Take-off angle [°] 48.83 (1.99) 48.67 (2.94)
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© Uwe Kersting, 201131
2 3 4 5 6 7 8 9 10 11 12 13 14
4
6
8
10
12
14
16
18
46
50
α [°]
Edec [J/kg]
TIn [°/J/kg]
Group1 (n=16)
Group2 (n=10)
© Uwe Kersting, 201132
Two groups were identified that showed both varying initial
conditions and jumping strategies.
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© Uwe Kersting, 201133
Group1 showed a lower horizontal velocity and therefore, a lower initial energy prior to the last ground contact.
Goup2 had a markedly increased initial energy and demonstrated a lower transformation index.
© Uwe Kersting, 201134
2 4 6 8 10 12 142
4
6
8
10
12
14
16
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Edec [J/kg]
TIn [°/J/kg]
R2 = 0.98
Formula: Tin = exp(-c * Edec)
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© Uwe Kersting, 201135
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1,6
1,7
1,8
1,9
2,0
2,1
2,2
2,3
2,4
2,5
2,6
2,7
2,8
Edec [J/kg]
Height [m
] Group1 (n=16)
Group2 (n=10)
With that:
© Uwe Kersting, 201136
Parameter [m/s] Group1 (n=16) Group2 (n=10)
Horizontal TD velocity 7.11 (0.29) 7.83 (0.43) *
Decrease in h. velocity 3.22 (0.30) 4.05 (0.56) *
Horizontal TO velocity 3.89 (0.27) 3.77 (0.22)
Vertical TO velocity 4.44 (0.11) 4.29 (0.26)
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© Uwe Kersting, 201137
Discussion
© Uwe Kersting, 201138
Both of the identified strategies resulted in comparable effective heights.
The faster approaching athletes could not benefit from their higher energy produced during the run-up.
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© Uwe Kersting, 201139
• Since body orientation at touch-down shows no significant differences a possible explanation could be an different muscle stiffness.
• The influence of changes in muscle stiffness has to be further investigated.
© Uwe Kersting, 201140
• For both groups the optimum
energy loss that leads to the
maximum vertical velocity after
the take-off can be estimated.
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© Uwe Kersting, 201141
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
1,6
1,8
2,0
2,2
2,4
2,6
2,8
3,0
Edec [J/kg]
Jump height [m
] Sotomayor E
T1=41.07 J/kg
Partyka ET1=36.53 J/kg
Forsyth ET1=35.82 J/kg
Matusevitch ET1=36.65 J/kg
© Uwe Kersting, 201142
Training concepts may be developed to increase
performance at both the world class level and in sub-elite
jumpers.
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© Uwe Kersting, 201143
© Uwe Kersting, 201144
Summary - IRelationships between kinematic parameters and jump performance are usually not very strong.
Therefore, predictions are almost impossible.
Energy approach provides much simpler analysis and can be used on a daily basis (?).
Muscle stiffness/joint stiffness has to be included.
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© Uwe Kersting, 2011
Force Platforms UseTo identify the magnitude and direction of ground reaction force (i.e., force and moments applied to the platform.
4545
Calculation of the point of force
application
© Uwe Kersting, 201146
Vertical Jump
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© Uwe Kersting, 201147
Predictors of Jump HeightJump type – CMJ > SJ
Leg Strength – Moderate effect
• Explains less than 30% of variation (Dowling & Vamos, 1993)
Peak force late in movement sequence (Dowling, 1992)
Max power output in propulsion phase –strong predictor (Dowling & Vamos, 1993)
• Dependant on knee and ankle power (Arampatzis et al, 2001)
Overall goal – increased jump height (Ford el at, 2005)
© Uwe Kersting, 201148
Jumping
Primary muscles• Hip (40%), knee (24.2%) and ankle (35.8%)
musculature (Hedrick & Anderson, 1996)
• Activation increases stiffness – utilization of elastic energy (optimal level? Non linear relationship)
Arm Swing – increase GRF, increase muscle loading
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© Uwe Kersting, 201149
Countermovement Jump
Four Common Theories to explain increased jump height:
• Storage and re-utilization of elastic energy(Komi, 1992)
• Increased time for force generation(Chapman & Anderson, (1990)
• Potentiation of contractile elements (through stretch)
• Spinal reflexes
© Uwe Kersting, 201150
Jump PerformanceCommon tests:- Jump and reach
- Assumptions: …
- Jump belt test:
Assumptions: …
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© Uwe Kersting, 201151
Determining Jump Height
- Determine flight time:(contact mats or foot switches, video??)
Assumptions: …- Error of 0.5 – 2 cm
(Linthorne, 2001)
V
time
signalflight time
© Uwe Kersting, 201152
Countermovement jump force trace
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© Uwe Kersting, 201153
Impulse-Momentum Relationship
Momentum
(of a moving object):Mom = m * v(t)at any point intime t
Impulse: I = F * t;
actually: I = F * dt
mv
© Uwe Kersting, 201154
Impulse-Momentum Relationship
Resulting vertical impulse (GRI):M * V = v – u – w;
M = body massV = TO velocityv, u, w = areas under force curve
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© Uwe Kersting, 201155
In other terms
BWimpulse: BWI = tcontact * BW
Total Impulse: TI = F(t) dt = Σ Fi * ∆t
Resulting
momentum:
Mom = TI - BWI
© Uwe Kersting, 201156
Realisation in excel
To calculate from measured force signal:
Acceleration of CoM
Velocity of CoM
Displacement of CoM
(all in vertical (z) direction; however applies for horizontal forces as well)
Example
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© Uwe Kersting, 201157
Viscoelasticity of tendon material
Stiffness from cadaver experiments: εTendon = 0.08 – 0.2 GPa
© Uwe Kersting, 201158
Oscillating systems
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© Uwe Kersting, 201159
Achilles tendon anatomy
© Uwe Kersting, 201160
Stiffness testing
Experiment by Shorten & Kerwin (1986)Setup:
Damped oscillation at 3 - 6 Hz
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© Uwe Kersting, 201161
ResultsEMG is constantduring oscillations� activation is const.
Stiffness increases curvilinearly with increasing load
Exp. on isolated muscle show a linear increase with number of cross bridges (Λ = ks P)
� change of the model
© Uwe Kersting, 201162
Mechancial muscle modelkt (kp + ks P)
K = ; overall stiffnesskt + kp + ks P
ks P
kT
kp
kp
kT
ks P
kt = 1091 Nm rad-1
kp = 14.7 Nm rad-1
ks = 13.5 rad-1
Ankle extension moment [Nm]
Angular equivalent stiffness [Nm/rad]
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© Uwe Kersting, 2011
SummarySports biomechanics covers more than the mere description of movement
Energy approaches can be used to explain effects of equipment
Total body (i.e., center of mass) information is a useful performance indicator and relatively easy to access
Energy or basic mechanical principles allow to estimate internal characteristics from external measurements
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