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1© Uwe Kersting, 2007
Energy and Power in
(Sports) Biomechanics
Department of Department of Sport Sport and Exercise Scienceand Exercise Science
SPORTSCI 306 SPORTSCI 306 –– Technique AssessmentTechnique Assessment
Uwe Uwe Kersting Kersting –– LectureLecture 0404 -- 20072007
Center for SensoryCenter for Sensory--Motor InteractionMotor Interaction
Anvendt BiomekanikAnvendt Biomekanik
Uwe Uwe Kersting Kersting –– MiniModuleMiniModule 0606 -- 20082008
2© Uwe Kersting, 2007
Objectives
• Review and learn basic energy formulae
• Determine total body energy by centre of mass calculations
• The use of power equations in performance assessment
• Apply energy estimations to an example of changes to sport equipment
• Learn about basic physics principles of rotational movements
3© Uwe Kersting, 2007
Contents
1. Calculation of total body energy
2. Estimates of rotational energy
3. Estimates of energy consumption with varied running footwear
4. Energy calculations in high jump
5. Rotational movements
6. Power calculations in jump tests
2
4© Uwe Kersting, 2007
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
5© Uwe Kersting, 2007
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
6© Uwe Kersting, 2007
COM calculation fromsegment positions
x0 = (Σ mi * xi)
M
y0 = (Σ mi * yi)
M
3
7© Uwe Kersting, 2007
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
8© Uwe Kersting, 2007
Hook‘s law:Force is proportionate to deformation
deformation: s [m]force: F [N]stiffness: k [N/m]
Hook‘s: Fs = - k * s
Mechanics
Fext
s
k
9© Uwe Kersting, 2007
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
4
10© Uwe Kersting, 2007
Additional Energyneeded
5
10
5 10 15
100g
200g
300g
400g
Add. Work [%]
Max foot speed [m/s]
11© Uwe Kersting, 2007
Applicationto highjump
In high jump the initial energy has to be transformed into jump
energy.
12© Uwe Kersting, 2007
There is no linear relationship between approach velocity andvertical take-off velocity. (Dapena et al., 1990)
Transformation is coupled with a decrease in total energy during thelast ground contact. (Brüggemann & Arampatzis, 1991)
5
13© Uwe Kersting, 2007
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.
14© Uwe Kersting, 2007
Methods
15© Uwe Kersting, 2007
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
6
16© Uwe Kersting, 2007
Digitised Frames
17© Uwe Kersting, 2007
Calculations
(1)H = H1 +2 g
V2 sin2 α
(2)H = +2 g
E k2 sin2 αE p2
g
18© Uwe Kersting, 2007
E k 2 =m
E kin 2; E p 2 =
m
E pot 2where:
T in =E dec
α(3)
7
19© Uwe Kersting, 2007
(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 andtransformation index.
20© Uwe Kersting, 2007
RESULTS
21© Uwe Kersting, 2007
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
8
22© Uwe Kersting, 2007
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)
23© Uwe Kersting, 2007
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)
24© Uwe Kersting, 2007
Two groups were identified that showed both varying initial
conditions and jumping strategies.
9
25© Uwe Kersting, 2007
Group1 showed a lower horizontal velocity and therefore, a lower initial energy prior to the last groundcontact.
Goup2 had a markedly increased initialenergy and demonstrated a lower transformation index.
26© Uwe Kersting, 2007
2 4 6 8 10 12 142
4
6
8
10
12
14
16
18
Edec [J/kg]
TIn [°/J/kg]
R2 = 0.98
Formula: Tin = exp(-c * Edec)
27© Uwe Kersting, 2007
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:
10
28© Uwe Kersting, 2007
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)
29© Uwe Kersting, 2007
Discussion
30© Uwe Kersting, 2007
Both of the identified strategies resulted in comparable effective heights.
The faster approaching athletes could notbenefit from their higher energy produced during the run-up.
11
31© Uwe Kersting, 2007
• Since body orientation at touch-down shows no significant differences a possible explanation could be an altered muscle stiffness.
• The influence of changes in muscle stiffness has to be further investigated.
32© Uwe Kersting, 2007
• For both groups the optimum
energy loss that leads to the
maximum vertical velocity after
the take-off can be estimated.
33© Uwe Kersting, 2007
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 ET1=41.07 J/kg
Partyka ET1=36.53 J/kg
Forsyth ET1=35.82 J/kg
Matusevitch ET1=36.65 J/kg
12
34© Uwe Kersting, 2007
Training concepts may be developed to increase
performance at both the world class level and in sub-elite
jumpers.
35© Uwe Kersting, 2007
36© Uwe Kersting, 2007
SummaryRelationships 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.
13
37© Uwe Kersting, 2007
Initial knee angle, incoming speed and jump height
Results from simulation study (Alexander, 1980)
38© Uwe Kersting, 2007
Some basic rules on rotation
1. Gathering angular momentum
2. Regulating rotations
3. AerialsSummersaults with twist
4. Assessment of jump performances- simple tests- using a force platform
39© Uwe Kersting, 2007
Aerials
Off-centre force during take-off creates torque. The amount of angular momentum in the airborne phase cannot be changed.
FBoard Resulting in:
Momentum: M = m v
(Impulse)
Angular Mom.: A = I ωFRot
FCom
14
40© Uwe Kersting, 2007
How does a snowboarder/skier produce angular momentum?
Virtually no friction to generate Frot!
From TO angular
momentum is
preserved; CoM
follows
parabolic path
(i.e., projectile
motion)
41© Uwe Kersting, 2007
Regulation of rotationAthletes have to change body
configuration to alter I (Moment of inertia); Angular Momentum unchanged when airborne: A = I ω � reducing I increases ω.
Used for slowing down or accelerating rotations.
How is forward rotation generated in ski jumps?
42© Uwe Kersting, 2007
Aerials with twistIf body position changes body rotates about a different local axis. Angular velocity is a vector along the rotational axis (right hand rule!)
ωω
ω||
ω⊥
15
43© Uwe Kersting, 2007
Result (Yeadon 2000, etc.)a) in athlete’s coordinate systemb) in laboratory coordinate system
44© Uwe Kersting, 2007
Jump Performance
Common tests:- Jump and reach
- Assumptions: …
- Jump belt test:
Assumptions: …
45© Uwe Kersting, 2007
Determining Jump Height
Common tests
- Determine flight time:(contact mats or foot switches, video??)
Assumptions: …
V
time
signalflight time
16
46© Uwe Kersting, 2007
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
47© Uwe Kersting, 2007
In other terms
BWimpulse: BWI = tcontact * BW
Total Impulse: TI = F(t) dt = S Fi * Dt
Resulting
momentum:
Mom = TI - BWI
48© Uwe Kersting, 2007
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)
..\Labs\Jumps_template.xls
17
49© Uwe Kersting, 2007
What else to get from GRF
F = m * a
v(t) = v0 + a dt
s(t) = s0 + v dt
s(t) = s0 + v0 (t-t0)
+ a dt2
50© Uwe Kersting, 2007
Example: Result
Counter Movement Jump
�
Force [N]
0
500
1000
1500
2000
2500
3000
0 0.2 0.4 0.6 0.8
vCM [m/s]
-2
0
2
4
0 0.2 0.4 0.6 0.8
W [J/kg]
-5
0
5
10
0 0.2 0.4 0.6 0.8
P [W/kg]
-100
-50
0
50
100
0 0.2 0.4 0.6 0.8
Msubj [kg] 81.00 Caine
hBox [m] 0.00
tcont [s] 0.70
Fmax [N] 2474.45
vTD [m/s] 0.00
vTO [m/s] 3.37
maxP [W/kg] 74.00
hj [m] 0.57
neg Work[J/kg] -2.60
pos Work[J/kg] 7.11
gain [J/kg] 9.71