Download 13 Inverse Dynamics.pptx

Learning Objectives:
By the end of this lecture, you should be able to:
Inverse Dynamics
•  Summarize the process of inverse dynamics
•  List the information needed
•  Describe how to obtain the needed information (e.g.
data collection, analysis of data)
•  Describe the general process
•  List and interpret the information that can be obtained
from an inverse dynamic analysis
Readings:
Chapter 10 p. 387 [course text]
Chapter 11 p. 442 [course text]
Whittlesey, Chapter 5 [on Canvas]
•  Use inverse dynamics to calculate net forces and
net moment at joints, when given the necessary
kinetic, kinematic, and anthropometric data
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Overview
Uses of Inverse Dynamics
Biomechanics
Inverse dynamics is the process of
calculating kinetic information (forces and
moments) from measured kinematic
information (positions, velocities, and
accelerations).
•  calculate net muscle moments
•  determine which muscle groups may be active
•  estimate the amount of force produced by muscles
•  calculate mechanical work and power
produced at body joints
Sometimes measured kinetic information is
also included.
•  determine why muscle groups may be active
(concentric or eccentric contractions)
This is different than forward dynamics,
which is when we calculate the acceleration
resulting from a given force.
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Robotics
•  calculate the amount of torque that
motors must generate to achieve a
given motion.
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Free-Body Diagrams Review
Free-body Diagrams Review
Example: Find the force and net moment at the ankle.
You already know how to use free-body
diagrams to find forces and moments.
Example: Midterm 1
Fabductors
FJ
∑F=ma
FJ
Fg,hand
Fmusc
we can’t usually
measure Fmusc or FJ
Fg
∑M=Iα
FF
Fg,forearm
Fg,upper arm
FGRF
FGRF
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FN
we can measure
FF and FN with a
force plate
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Replacing Forces with Moments
Replacing Forces with Moments
We can “move” a force somewhere else, as long
as we account for the moment that it causes.
Example continued:
Fmusc
FJ
Fmusc
FJ
Fnet,ankle,y
Fnet,ankle
Fnet,ankle,x
Fg
Fg
FF
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Fg
FF
FN
Sure it is cheating – but it works! 
Mmusc
Mmusc
FF
FN
FN
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Moving moments to different
centers of rotation
Moving moments to different
centers of rotation
Moments can be moved to different centers
of rotation.
Fnet,ankle,y
Fnet,ankle,y
Fnet,ankle,x
Mankle
Fg
M
M
Fnet,ankle,x
Mankle
Fg
FF
Taking moments about the center of mass is good,
because it makes dealing with accelerations easy.
FF
FN
FN
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Finding net force and moment
Finding net force and moment
Example continued:
Example continued:
We can measure ground
reaction force with a force plate.
We can measure segment
position with motion capture,
and use this to determine
angular and linear acceleration.
Fnet,ankle,y
Fnet,ankle,y
Fnet,ankle,x
Mankle
Fg
FF
moment
arms
FN
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x direction:
∑Fx = max
FF + Fnet,ankle,x = m ax
force plate
anthropometry
motion capture
y direction:
∑Fy = may
FN + Fnet,ankle,y + Fg = m ay
Mankle
Fg
Therefore, we can solve for the unknown net
ankle force, and the unknown NET ankle
moment using
F=ma M=Iα
Fnet,ankle,x
unknown
FF
FN
angular:
∑Mcom = Iα
MGRF + Mnet,ankle + Mankle = I α
Note that Fg does not create a moment about the center of mass.
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Finding net force and moment
16.6°
a = 7.28
Example question:
You measure a normal force of 735 N and a friction
force of -75 N when a subject is walking over a force
plate.
From motion capture data, you calculate a linear
acceleration of the foot segment of 7.28 m/s2 at
16.6°, and an angular acceleration of -20 rad/s2.
α = -20 rad/s2
Fnet,ankle,y
Fnet,ankle,x
16.6°
a = 7.28 m/s2
FF = -75 N
m = 1 kg
FF
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a = 7.28 m/s2
m = 1 kg
angular:
∑Mcom = Iα
Example continued:
ay = (7.28 m/s2)(sin16.6°)
= 2.08 m/s2
y direction:
∑Fy = m ay
FN + Fnet,ankle,y + Fg = m ay
(735) + Fnet,ankle,y + (1)(-9.81) = (1)(2.08)
Fnet,ankle,y = -735 – (-9.81) + 2.08
= -723.11 N
Fg
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FF
FN = 735 N
16.6°
Fnet,ankle,y
Fnet,ankle,x
x direction:
∑Fx = max
FN
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x direction:
∑Fx = m ax
FF + Fnet,ankle,x = m ax
(-75) + Fnet,ankle,x = (1)(6.98)
Fnet,ankle,x = 75 + 6.98
= 81.98 N
dx
Fg
Example continued:
y direction:
∑Fy = may
dy
dy
Example continued:
ax = (7.28 m/s2)(cos16.6°)
= 6.98 m/s2
Mankle
dx
What is the magnitude of the net force and moment
at the ankle?
Assume that the ground reaction force and net ankle
force act at the ends of the segment. Use the
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anthropometry provided on the next slide.
m/s2
FN = 735 N
FF = -75 N
m = 1 kg
I = 0.04 kg m2
dx = 0.06 m
dy = 0.0575 m
FN
Net force at ankle:
|Fnet,ankle| = √(Fnet,ankle,x2 + Fnet,ankle,y2)
= √(82.982 + (-723.11)2)
= √(529,774) = 727.9 N
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Example continued:
FN = 735 N
FF = -75 N
α = -20 rad/s2
I = 0.04 kg m2
dx = 0.06 m
dy = 0.0575 m
Mankle
dx
dy
dy
Note that Fg does
not create a moment
about the center of
FN
mass.
Proximal segments
Fnet,ankle,x = 81.98 N
Fnet,ankle,y = -723.11 N
angular:
∑Mcom = I α
FFdx + FNdy + Fnet,ankle,xdx + Fnet,ankle,ydy
+ Mankle = I α
Fnet,ankle,y
Fnet,ankle,x
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dx
Mankle
FF-4.5 + 42.26 – 4.92 + 41.58 + Mankle =
-0.8
Mankle = -75.22 Nm
Proximal segments
FGRF
Fnet,knee,y
Fnet,knee,x
Fg
Fnet,ankle,y
-Fnet,ankle,x
-Mankle
•  When adding moments, you must take into account
the net distal moment, as well as the other three
sources of moments that we used for the most
distal segment (net distal force, net proximal force,
and net proximal moment).
Fnet,ankle,x
Fg
FF
FN
Calculations for proximal segments are very
similar to those for distal segments, except that:
•  The net force on the distal end of the segment is
equal and opposite to the net force on the proximal
end of the adjacent distal segment (Newton’s 3rd
Law).
-Fnet,ankle,y
Mankle
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Proximal segments
Mknee
FGRF
Fnet,ankle
-(75)(0.06) + (735)(0.0575) - (81.98)
(0.06) + (723.11)(0.0575) + Mankle =
(0.04)(-20)
Once you have analyzed the
most distal segment (e.g.
foot), you can use the net
joint force and net moment to
calculate joint force and
moment at the adjoining
proximal segment.
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Proximal segments
Considerations
Fnet,knee,y
x direction:
∑Fx = max
-Fnet,ankle,x + Fnet,knee,x = m ax
Fnet,knee,x
Mknee
y direction:
-Fnet,ankle,x
∑Fy = may
-Fnet,ankle,y + Fnet,knee,y + Fg = m ay
-Mankle
The calculated moment at the joint is not
necessarily the moment of any given muscle.
-Fnet,ankle,y
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Co-contraction of agonists and antagonists, as
well as moments created by other joint
structures, are included in the calculated NET
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moment.
Summary (Known Variables)
Summary (Known Variables)
Motion capture gives us
information about segment end
points.
From our measured position data,
we can calculate the angular and
linear velocities of the segment
α
center of mass.
Force plates give us
information about ground
reaction force.
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We usually begin by analyzing the most distal
segments, because they have only one linear
and one angular unknown.
Fg
angular:
∑Mcom = Iα
M-net,ankle + Mnet,knee - Mankle + Mknee = I α
If the foot is in swing phase, the ground reaction
force is zero.
Anthropometry gives us
information about segment
mass, center of mass location,
and moment of inertia.
a
Then, we can calculate segment
linear and angular accelerations
using finite differentiation.
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Summary (Unknown Variables)
Summary
Each segment has four
unknowns – these are what we
want to determine:
The most distal segment does not have a
distal joint reaction force or distal joint moment,
but may have a ground reaction force.
i.e. distal segments have only one linear and
one angular unknown
•  net proximal joint force
•  net proximal joint moment
Therefore, we usually begin by analyzing distal
segments.
•  net distal joint force
•  net distal joint moment
These are net forces and moments due
to all of the connective tissue, muscle,
etc. at each end of the segment.
The forces and moments calculated allow us to
analyze more proximal segments.
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Summary
We can use acceleration information to
calculate the net force and moment that
must be acting on the segment:
F = ma
M = Iα
Then we combine this with information about
the force (ground reaction force or joint
reaction force) and moment acting at the
distal end of the segment, to calculate the
net force and moment acting at the proximal
end of the segment.
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