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 1
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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. 3
Robotics • calculate the amount of torque that motors must generate to achieve a given motion. 4
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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 6
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 8
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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 11
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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 15
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 13 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 26 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. 29
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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. 31
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