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Quiz
Kinematics
30 januari 2017
1
1. How do you calculate velocity?
A) Final position / initial position
B) Final position – initial position
C) Final position – initial position
Final time – initial time
D) Initial position – final position
Initial time – final time
30 januari 2017
2
2. How do you calculate acceleration?
A) Final velocity / initial velocity
B) Final velocity – initial velocity
C) Initial velocity – final velocity
Initial time – final time
D) Final velocity – initial velocity
Final time – initial time
30 januari 2017
3
3. Which movement can describe a person
that is moving in an AZ+ direction?
A) A person riding on a bike
B) A person doing a backward summersoult
(kullerbyte)
C) A person doing a pirrouette
D) A person doing a forward summersoult
30 januari 2017
4
4. Which movement can described a person
that is moving in an TX+ and TY+ direction?
A) A person riding on a bike
B) A person doing a summersoult (kullerbyta)
C) A person doing a pirrouette
D) A person walking up the stairs
30 januari 2017
5
5. Think about the experiment we did yesterday.
Which movement parameter is not a temporal
parameter?
A) Time to peak%
B) Total movement time
C) Peak (maximal) displacement
D) Shoulder velocity
30 januari 2017
6
6. Think about the experiment we did yesterday.
Which movement parameter can be used for
quantitatively describing motion smoothness?
A) Number of changes in direction of velocity
B) Number of peaks in displacement
C) Area under the acceleration curves
D) Shoulder velocity
30 januari 2017
7
7. Think about the experiment we did yesterday.
Is it possible to describe the shoulder flexion
angle with the two markers?
No
Yes, by making a vector between the hand
and shoulder marker against the
A) Y-axis in the sagittal plane
B) X-axis in the sagittal plane
C) Z-axis in the sagittal plane
30 januari 2017
8
8. What curve could this be?
1
0,8
A)
B)
C)
D)
0,6
0,4
0,2
1
16
31
46
61
76
91
106
121
136
151
166
181
196
211
226
241
256
271
286
301
316
331
346
361
376
391
406
421
436
451
466
481
496
511
0
Displacement X
Velocity X
Acceleration X
Mina aktier/
Wims stockings
-0,2
-0,4
-0,6
-0,8
Grooten
30 januari 2017
9
Answers
1.
2.
3.
4.
5.
6.
7.
8.
C
D
B
D
C (and D)
A
No – but perhaps also C
B (and D)
31 januari 2017
10
KINETICS
Wim Grooten
Disposal of the lecture
*
*
*
*
*
Newtonian laws
Force plates
Vectors
Center of Mass
Momentarm and Torque
(external and internal torque
* Strength measurements
* Work and power
Forces




Gravity
Inertia
Centripetal
Friction
 Internal forces
 Muscle force
 Joint force
 Shear
 Compression
 Ground reaction force
30 januari 2017
12
Newtonian law I
I.
If a body is at rest it remains at rest
or if it is in motion it moves with uniform velocity
until it is acted on by an other force.
Newtonian law II
II. The acceleration of an object as produced by a
net force is directly proportional to the
magnitude of the net force, in the same direction
as the net force, and inversely proportional to the
mass of the object. Force = mass x acceleration
50N
100N
Newtonian law III
III. For every action, there is an equal and
opposite reaction.
Force plates
Piezoelectric force plates
KISTLER
…cover a wide temperature range, are overload-protected
and offer long-term stability as well as freedom from
fatigue. Piezoelectric sensors are ideal for almost all areas
of application, particularly for the type of dynamic and
highly sensitive processes encountered in biomechanics.
The operating principle of quartz crystal sensors in
Kistler force plates means that compared with sensors with
strain gages they offer decisive advantages, most of which
are attributable to their comparatively high rigidity
Force plates
Force plates
AMTI
AMTI’s platforms use strain gages mounted on precision metal sensing elements located
within the platform to perform the force and moment measurement task. The strain
gages are electrically wired in full four arm bridge arrangements to provide thermal
stability and to isolate the strains caused by forces applied in the several directions.
In order to function a strain gage bridge requires a source of stable excitation voltage. When
this voltage is applied across two terminals of the bridge the alternate two terminals of
the bridge will be balanced and no signal will be present on those terminals. When a
load is applied to the sensing element small mechanical strains will subtly change the
resistance of the bridge arms and the bridge will become unbalanced. When this occurs
a very small electrical signal will be observed across the bridge.
The output signals from the strain gage bridges must be amplified in order to produce
signals of sufficient strength to be useful. Typically a well designed transducer (or
force platform) will need an amplifier gain of between 1000 and 4000 to produce a
usable output signal.
How does Kistler calculate CoP?
Like this!!!
Computation of the CP
Generally, the true origin of the strain gauge force-plate is not at the geometric center of the plate
surface. This is due to problems in the manufacturing process. The manufacturers usually go through a
series of calibrations and estimate the position of the true origin. Here, we assume that the true origin
(O' shown in Figure 2) is at (a, b, c). The Z component of the CP position is always 0. The moment
measured from the plate is equal to the moment caused by F about the true origin plus Tz:
or:
Therefore, the position of the CP
can be computed from the moment caused by the ground reaction force about the true origin, Mx, My &
Mz, the ground reaction force, Fx, Fy & Fz, and the location of the true origin, a, b & c. Mx, My, Mz, Fx, Fy
& Fz can be directly measured from the 6 channels of the AMTI plates while the position of the true
origin can be found in the calibration data sheet.
The Kistler plates provide a different channel configurations: F1z, F2z, F3z, F4z, F1x + F2x, F3x + F4x, F1y +
F4y, & F2y + F3y. In Figure 3a, the location of the sensors are described by three distance factors: a, b &
g. Among these g is the depth of the sensor center from the surface. The sum of the moments caused
by the four forces is equal to the moment caused by the ground reaction force (F) plus the free vertical
torque (Tz) as shown in Figure
3b.
Vectors
 Point of application
 Direction
 Magnitude

ADDITION

DIVISION
Free body diagram
Centre of Mass
The terms "center of mass" and "center of gravity" are used synonymously in a
uniform gravity field to represent the unique point in an object or system which
can be used to describe the system's response to external forces and torques.
The concept of the center of mass is that of an average of the masses factored by
their distances from a reference point. In one plane, that is like the balancing of a
seesaw about a pivot point with respect to the torques produced.
If you are making measurements from the center of mass point for a two-mass
system then the center of mass condition can be expressed as
where r1 and r2 locate the masses. The center of mass lies on the line connecting the two
masses.
Body segments
Momentarm and Torque
 Forces
(Newton)
 Momentarm
(Meter)
 Torque / Moment of force
(Newtonmeter)
Equilibrium
What happens if person 8 and 5 shift places?
Thus, not only the force is important, also the
distance to the axis of rotation!
Lever arm !!
Torque:
M=Fxd
Moment of Force (Nm) = Force (N) x distance (m)
Moment arm
the perpendicular distance from the axis of rotation
to the force(vector)
Force vector
Moment arm
Not only for external forces, also for internal forces
produced by muscles
WHY DO WE NEED THIS?
We can use [M = F x d] for the estimation of the external
torque and when we know this, we can estimate the
internal torque
External torque = Internal torque
External forces:
Bowling ball: 10 kg = 100N
Lower arm: 5% of Body Weight
0.05x800N = 40N
Moment arms:
Bowling ball: 33 cm = 0.33m
Lower arm: 15 cm = 0.15m
M=Fxd
External torque = Mball + Marm
(100N x 0.33m) + (40N x 0.15m) = 33Nm + 6Nm
External torque = 39Nm
External torque = Internal torque
39Nm = Muscle force x 0.04m
Muscle force = 39Nm/0,04m = 975N
Strength measurements
 ISOMETRIC
 DYNAMIC
 ISOKINETIC
 Kincom
 Biodex
 Cybex
Strength measurements
ISOKINETIC BENCHPRESS
Dysfunction – leg extension
ACL
Menisc
Shoulder
 Some different types of diseases
of the supraspinatus muscles
Work and Power
 WORK
 Force produced during the movement (Nm)
 area under the curve
 Power
 Work produced over time (W = J/s)
WORK
POWER
= Force x displacement Nm = Joule
= WORK / time
Joule/sec = Watt
Example
Angles
Single-leg hop
testing following
fatiguing exercise:
reliability and
biomechanical
analysis
J. Augustsson, R.
Thomee, C. Linde,
M. Folkesson, R.
Tranberg,
J. Karlsson
Scand J Med Sci
Sports 2006: 16:
111–120
Moment
Power
GRF
Free Body Diagrams
Effects of a Force at an Instant in time
 F  ma
 If acceleration is zero, biomechanical study of the human is referred
to as a static analysis
 If acceleration is not zero (and is significant), biomechanical study of
the human is referred to as a dynamic analysis
Static Analysis
 If the acceleration of a system is zero, then we can perform a static
analysis
 Fsystem  0
 Fx  0
 Fy  0
Static Analysis (cont.)
Dynamic Analysis
 Static analysis is often used when acceleration occurs but is not
significant (like slow lifts)
 If acceleration is significant, a dynamic analysis must be performed
 Fsystem  ma
 Fx  max
 Fy  ma y
Dynamic Analysis (cont.)
Vertical Ground Reaction Force
Vertical Jump
Effects of a Force Applied over a Distance
Work  F  cos  s





F = the applied force
s = the displacement
 = the angle between the force vector and line of displacement
Units of work are joules (J)
Work is a scalar quantity
Mechanical Work on a Block