Transcript Inverse Dynamics - University of Ottawa

Inverse Dynamics
D. Gordon E. Robertson, PhD, FCSB
School of Human Kinetics
University of Ottawa
Inverse Dynamics
(definition)
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The process of deriving the kinetics (i.e.,
forces and moments of force) necessary to
produce the kinematics (observed motion) of
bodies with known inertial properties (i.e.,
mass and moment of inertia).
Typically the process is used to compute
internal forces and moments when external
forces are known and there are no closed
kinematic chains (e.g., batting, shoveling).
Two-dimensional Derivation
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The following slides outline the derivation of
the equations for determining net forces and
moments of force for the two-dimensional
case.
The three-dimensional case follows the same
procedure.
Inverse Dynamics
Kinematic Chains, Segment & Assumptions
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First, divide body into kinematic chains
Next, divide chains into segments
Assume that each segment is a “rigid body”
Assume that each joint is rotationally frictionless
Space Diagram
Segments
Ordering of Segments
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Start with the terminal segment of a kinematic
chain
The ground reaction forces of the terminal
segment must be known (i.e., measured) or
zero (i.e., free-ended)
If not, start at the other end of the chain (i.e.,
top-down versus bottom-up)
If external forces are unknown, measure them,
otherwise, you cannot proceed
Free-body Diagram
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Make a free-body diagram (FBD) of the
terminal segment
Rules:
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Add all known forces that directly influence the
free-body
Wherever free-body contacts the environment or
another body add unknown force and moment
Simplify unknown forces when possible (i.e., does
a force have a known direction, can force be
assumed to be zero, is surface frictionless?)
1.
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Draw free-body
diagram of terminal
segment
2.
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Add weight vector
to free-body
diagram at centre
of gravity
weight
centre of gravity
(xfoot, yfoot)
3.
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If ground contact
add ground
reaction force at
centre of pressure
F ground
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
4.
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Add all muscle
forces at their
points of
application
(insertions)
force from
triceps surae
force from
tibialis anterior
F ground
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5.
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force from
triceps surae bone-onforce
from
bone
forces
tibialis anterior
Add bone-on-bone
and ligament forces
and the frictional
joint moment of
ligament
force
F ground
force
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
Equations are Indeterminate
Too many Unknowns, Too few Equations
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In two dimensions there are three equations of
motion. In three dimensions there are six
equations.
But there are more unknown forces (two or
more muscles per joint, several ligaments,
skin, joint capsule, bone-on-bone or cartilage
forces, etc.) then there are equations.
Thus, equations of motion are indeterminate
and cannot be solved.
Solution:
Reduce number of unknowns to
three (2D) or six (3D)
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The solution is to reduce the number of
unknowns to three (or six for 3D)
These are called the net force (Fx, Fy) and the
net moment of force (Mz) for 2D
or (Fx, Fy, Fz) and (Mx, My, Mz) for 3D
5a.
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Consider a single
muscle force (F)
force from
triceps surae bone-onforce
from
bone
forces
tibialis anterior
F
ligament
force
F ground
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5b.
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Move muscle force
to joint centre
(F*)
force from
triceps surae bone-onfrom
bone
forces
F* force
tibialis anterior
F
ligament
force
F ground
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5c.
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Add balancing
force (–F*)
force from
triceps surae bone-onforce
from
bone
forces
F* tibialis anterior
ligament
force
F
F ground
weight
–F*
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5d.
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Force couple
(F, –F*) is equal to
free moment of
force (MFk)
force from
triceps surae bone-onforce
from
bone
forces
F* tibialis anterior
ligament
force
F
F ground
= MF k
weight
–F*
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5e.
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Replace couple
with free moment
(MF k)
force from
triceps surae bone-onforce
from
bone
forces
F* tibialis anterior
ligament
force
MFF k Fground
= MF k
weight
–F*
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
5.
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Show all forces
again
force from
triceps surae bone-onforce
from
bone
forces
tibialis anterior
F ground
ligament
force
weight
centrecentre
of gravity
of pressure
(xfoot, (yxfoot) , y
ground
ground )
6.
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Replace muscle
forces with
equivalent joint
forces and free
moments
force from
force and
moment from
triceps surae
bone-ontriceps
surae
force
from
bone
forces
tibialis anterior
F ground
ligament
force
weight
force and
moment
from
centre
of
gravity
centre
of pressure
tibialis
anterior
(xfoot, (yxfoot) , y
ground
ground )
7.
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Add all ankle
forces and
moments to obtain
net ankle force and
moment of force
Mankle k
Fankle
Fground
8.
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Show complete
free-body diagram
Mankle k
Fankle
(xankle, yankle)
Fground
mfoot g j
centre of pressure
(xground, yground)
9.
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Show position
vectors (rankle ,
rground)
Mankle k
Fankle
(xankle, yankle)
rground
rankle
Fground
mfoot g j
centre of pressure
(xground, yground)
Three Equations of Motion for the
Foot
Σ Fx = max:
Fx(ankle) + Fx(ground) = max (foot)
Σ Fy = may:
Fy (ankle) + Fy (ground) – mg = may (foot)
Σ Mz= I a:
Mz (ankle) + [rankle × Fankle] z
+ [rground × Fground] z = Ifoot a (foot)
Moment of Force as Cross Product
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the moment of a force (M) is defined as the
cross-product (x) of a position vector (r) and
its force (F). I.e., M = r x F
Mz = [ r × F ]z = rx Fy – ry Fx
rankle = (xankle – xfoot , yankle – yfoot)
[ ... ]z means take the scalar portion in the zdirection
Equations of Motion for Foot
Solve for the Unknowns
Σ Fx = max:
Fx(ankle) = max(foot) – Fx(ground)
Σ Fy = may:
Fy(ankle) = may(foot) – Fy(ground) + mg
Σ Mz = I a:
Mz(ankle) = Ifoot a(foot) – [rankle × Fankle] z
– [rground × Fground] z
Note, moment of inertia (Ifoot) is about the centre of the
gravity of the foot, not the proximal or distal end)
Apply Newton’s Third Law to Leg:
Reaction = – Action
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Net force and moment of force at proximal end
of ankle causes reaction force and moment of
force at distal end of the leg (shank)
Reactions are opposite in direction to actions
I.e., reaction force = – action force
reaction moment = – action moment
10.
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Mknee k
Draw free-body
diagram of leg
using net forces
and moments of
force
–Fankle
–Mankle k
Fknee
Equations of Motion for Leg
Σ Fx = max:
Fx(knee) – Fx(ankle) = max(leg)
Σ Fy = may:
Fy(knee) – Fy(ankle) – mg = may(leg)
S Mz = I a:
Mz(knee) + [rknee × Fknee]z – Mz(ankle)
+ [rankle × – Fankle]z = Ileg a(leg)
Equations of Motion for Thigh
Σ Fx = max:
Fx(hip) – Fx(knee) = max(thigh)
Σ Fy = may:
Fy(hip) – Fy(knee) – mg = may(thigh)
S Mz = I a:
Mz(hip) + [rhip × Fhip]z – Mz(knee)
+ [rknee × – Fknee]z = Ithigh a(thigh)
Interpretation
Mathematical concepts not anatomical kinetics
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These forces and moments are mathematical
constructs NOT actual forces and moments.
The actual forces inside joints and the
moments across joints are higher because of
the cocontractions of antagonists.
Furthermore, there is no certain method to
apportion the net forces and moments to the
individual anatomical structures.
Computerize the Process
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Examples:
2D: Biomech MAS, Ariel PAS, Hu-m-an
3D: Visual3D, Polygon, KinTools, KinTrak,
Kwon3D, Simi