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EOM with Newton-Euler Equations
Concurrent Dynamics International
December 2017
copyright © Concurrent Dynamics International 12/7/2017
Content
• Introduction
• Notations
•
• O(N) Recursive Algorithms
• Acceleration Equations
• Dynamics Equations
• Forward & Inverse Dynamics
• Constraint Forces
• Computing System Mass Matrix
• O(N) Solution of EOM
• Examples
• Summary
• References
12/7/2017
Influence Matrices and T
copyright © Concurrent Dynamics International
Introduction
•
•
12/7/2017 copyright © Concurrent Dynamics International
Equations motion of a rigid-body system using the Newton-Euler equations is
considered. The influence matrix and its companion operators are used to
derive them. The acceleration solution by both
A
N
3order( ) and order( ) methods
are presented. Notably, Lagrange multipliers are not needed for the derivations given
here.
N N
3Much has been published on the order( ) method in the 60's and 70's
6 to 10 . The dynamics equations derivation shown here is post 1990's
similar to the work by Radriguez-Jain-Kreuze 1 , except for a di
N
fference
in the definition of the influence matrix operator .
•
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1: 1:
Generalized coordinates and system rates for these are
,
where inboard joint angle of
i ii N i N
i i
MBS
q q
b
Ensuing discussions cover 1) dynamics equations of an , 2) solving the
dynamics equations for the system accelerations, .
MBS
q
copyright © Concurrent Dynamics International
Tree-structured rigid multibody systems that have only single axis rotational
motion between joint connected bodies are considered to expedite the discussion.
The root body is connected to the grou
MBS
nd in all cases considered. These systems
encompass robots, mechanical arms, multi-fingered hands, cranes, gimbaled
antennas, transmission and suspension systems.
Fig. 1 Notations
12/7/2017
Inertial frame
bj
bi
x
z
y
dj sj
si
joint1
jointj
ir
i
cmi
yi
external force
external torque
i
i
f
if
i body
inertial position of joint
inertial position of cm
impact position from joint
joint position from joint
cm position from joint
angular rate of
i
i i
i i
i i i
i i ip
i i i
i i
b i
r
y f
d
s
b
copyright © Concurrent Dynamics International
jointi
i
• The root body b1 is the reference body whose position and attitude serve as the starting value to compute the same for other bodies in the system in a hierarchical manner. In the following, b1 is connected to the ground
• Body indexing rule used here is the Parent-First order meaning that the index of a body is always a lower integer number than the indices of its children.
• The chain of bodies between b1 and bj shall be denoted as The less-than-or-equal relation over body indices is a topological order and not a
numerical order.
• The set of bodies branching from bj shall be denoted as The greater-than-or-equal relation over body indices is a topological order and not a
numerical order. • All vectors in the following discussion are given in the format xi
j . The subsript i denotes the body that x belongs to and the superscript j denotes the coordinate frame that the vector is in.
• Vectors with no superscript are given in inertial coordinates unless defined otherwise
12/7/2017
{ | } or just .i i j i j
{ | } or just .i i j i j
copyright © Concurrent Dynamics International
•
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1 2 3
A stacked vector y is a column vector whose elements are also vectors.
The latter can be of different sizes. We define this stacked vector as
[ , , , , ], is a vector
where size( )
N iy y y y y y
y
1
1 2 3
1
1 2 3 2
3
( ) 1, ( )= ( )
For example:
Let , ,
Then [ , , ]
N
j
j
x
x
x y
y
z
x
x
y
x
y
z
len y len y len y
ws
y v y y ws
w
v
sy
sy y y y y
wy
w
w
copyright © Concurrent Dynamics International
•
•
•
•
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0
Skew matrix notation: 0 , if , ,
0
z y
z x x y z
y x
a a
a a a a a a a
a a
The relations { , , , } shall mean topological order in the following
expressions when used to group bodies, unless stated otherwise.
{ | ( ) } , indices of children of jj parent j b
identity matrix, depends on context e k k k
copyright © Concurrent Dynamics International
( ) , index of parent of iip parent i b
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,
6 6
Influence matrices , and , are defined by the parent child relations of the
considered mechanism and some block diagonal matrix :
if parent of [ ]
0 otherwise
T
i j i
i j
b b
,
6 6
Def. 1
if parent of [ ] Def. 2
0 otherwise
where
T
T j i j
i j
b b
1: 1 diag{ } , a block diagonal matrix, with R , 0
integer > 1
k k
j j N i
k
Forward influence matrix, is strictly lower triangular.
Influence Matrices and T
copyright © Concurrent Dynamics International
Backward influence matrix, is strictly upper triangular.T
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1: 1Let { } , , =0, and [ ], [ ], with
, , 1.
is a column vector representation of the forward element-
by-element (parent-to-child) calculations
Prop1.
k k
i i N i i i
k
i i
diag R x col x z col z
x z R k
z x
for 1: (P1)
is a column vector representation of the backward element-
by-element (child-to-p
Prop2.
i i ip
T
z x i N
z x
arent) calculations
for 1: (P2)T
i j j
j i
z x i N
1Each body in a tree-configured mechanism has one parent body except for .
Thus, each row of has only one non-zero submatrix entry and the first
row is all zeros.
b
.
1
Given that is square and strictly triangular, it is nilpotent of degree , i.e.
0
where, length of the longest link in the system from , and .
m
m
m b m N
copyright © Concurrent Dynamics International
•
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1
2 1
If is square and strictly lower triangular, then ( ) exists
and can be expressed as
(P3)
where ni
m
e
e
m
lpotency of A
copyright © Concurrent Dynamics International
1
,2 , 1
It follows that is square and strictly upper triangular, and ( ) exists
and can be expressed as
T T T
T T T T m
e
e
(P4)
1 1
6
The power series expansion of and serve to show the existence of
( ) and ( ) . More efficient way of computing and
for any R is shown shortly.
T
T T
N
e e x x
x
Fig. 2 Example Body Sets
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2 1
4 3
6
5
8 7
10 9
branch1={j ≥ 1}={1:10} branch2={j ≥ 2}={2} branch3={j ≥ 3}={3:8} branch4={j ≥ 4}={4,7,8}
1 {2,3,9}
2 6 8 10 { }
3 {4,5}
4 {7}
empty
{j ≤ 2}={2,1} {j < 5}={3,1} {j ≤ 6}={6,5,3,1} {j ≤ 10}={10,9,1}
10 body example
copyright © Concurrent Dynamics International
For Fig. 2 Example
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A
2
3
4
5
6
7
8
9
10
0
0
0 0 0
0 0 0
0 0 0 0
0 0 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0
copyright © Concurrent Dynamics International
For Fig. 2 Example
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2
3
4 3 4
1 5 3 5
6 5 3 6 5 6
7 4 3 7 4 7
8 7 4 3 8 7 4 8 7 8
9
10 9 10
If is a square submatrix for this example, then by (P3)
0 0
0
0 0
0 0
0 0 0
0 0 0
0 0 0 0 0 0 0
0 0 0 0 0 0 0
i i
e
e
e
e
ee
e
e
e
e
e
nilpotency( )=5
copyright © Concurrent Dynamics International
Fig. 3 Nilpotency 2 System
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Inertial frame
x
z
y
hinge1 1
b1
b2
b3
b4
bN d2
d3
d4
dN
copyright © Concurrent Dynamics International
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2
3
By Defs. 2 and 3, if is a square matrix, then for nilpotency 2 systems per (P3) are
with
0 0 0 0
0 0 0
0 0 0
0 0 0N
e
copyright © Concurrent Dynamics International
2:
1
2
3
A nilpotency 2 tree configured system has all bodies as immediate
children of . See Fig. 3. For these systems the forward recursive operator per (P3) is
0 0 0
0 0
0 0
0 0
N
A
N
b
b
e
A e
A e
A e
O(N) Recursive Algorithms
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Examples presented earlier showed the lower-triangular nature of given the
parent-child relation of bodies in a tree-configured system.
Computing expressions involving or by direct matrix-vector
multiplication (i.e. using (P3) and (P4) for and ) is inefficient.
T
T
x x
copyright © Concurrent Dynamics International
The next two algorithms are order( ) computations of those expressions.N
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A stacked vector given by is the result of solving for z,
where is square and strictly lower triangular per Def 1. Thus, given ,
the elements of are obtained by the algorithm bel
z x z z x
x
z
1 1
ow :
1. set since the first row of is zero
2. for 2 :
;
end
i i ip i
z x
i N
z z x
The above computation of given is order( ). Thus is an order( ) operator.z x N N
copyright © Concurrent Dynamics International
Algorithm for z x
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A stacked vector given by is the result of solving for ,
where is square and strictly upper triangular. Thus, given , the elements
of are obtained by the algorithm below :
T T
T
z x z z x z
x
z
1. set since the first column of is zero
2. for : 2
: ;
end
T
T
ip ip i i
z x
i N
z z z
copyright © Concurrent Dynamics International
Algorithm for Tz x
The above computation of given is order( ). Thus is an order( ) operator.T
z x N N
Acceleration Equations
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Relation between the two accelerations is
(2)
0where
( )
i i i i
i
i i i
v S v A
As
Relation between and is (See Fig. 4 )
(1)
where [ , ], [ , ] and to
j j
i i i
i i i i i i i
v v
v S v
v r v
1 1
tal angular rate of
velocity of , velocity of joint
0 since joint is attached to ground in the considered system
0 , displacement f
i
i i i i
i i
i
b
r cm
eS s
s e
rom joint to i i
cm
copyright © Concurrent Dynamics International
•
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1:
Total accelerations of (2) in stacked column form is
(3)
where [ ] , [ ]i i N i
v Sv A
v col v S diag S
1: 1: 1:
, [ ] , [ ]i N i i N i i N
v col v A col A
copyright © Concurrent Dynamics International
•
•
12/7/2017
1 1
Relation between and the relative rate is
(4)
0where , 0, 0
i i
i i ip i i
i p
i
v q
v D v G q
eD D v
d e
3 1 [ , 0 ] for 1:
i iG g i N
copyright © Concurrent Dynamics International
Given (4), the relation between the total acceleration and the relative
acceleration is
i
i
i i ip i i i
v
q
v D v G q A
1
(5)
where , 0 (5a).( )
ip i i
i i ip i i
ip ip i
gA D v G q a
d
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12/7/2017 copyright © Concurrent Dynamics International
1:
Given Def. 1 and Prop. 1, Eq. (5) can be put in column vector form as
(6)
where, [ ]
D
i i
v v Gq A
v col v
1: 1: 1:
1:
, [ ] , [ ] , [ ]
[ ]
N i i N i i N i i N
i i N
D diag D G diag G q col q
A col A
1
Total acceleration in (6) can be solved to be
( ) (7)
where ( )
D
D D
v Gq A
e
Given (7), Equation (3) becomes
( ) (8) D
v S Gq A A
Fig. 4 Forces and Torque on bi
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1
Inertial frame
cmi
x
z
y
dα
si
hingeα
τα
fα
fi
τi
ωi
yi
ir
i
bi
hinge1
if
i
body
hinge position from hinge
y impact positionfrom hinge
f interbody force from to
external force on
interbody torque from to
external force on
i
i
i i i
i ip i
i i
i ip i
i i
i
b i
d
f
b b
f b
b b
b
angular rate of ib
copyright © Concurrent Dynamics International
hingei
Dynamics Equations
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Newton-Euler equations for each body of an per Fig. 4 are
(9)
( ) ( ( ) )
r
i
i i i i i i i i i i i i
j
i
i i i i
i
b MBS
s f I I y s f
d s f
f m f f
(10)
Eqs. 9 and 10 can be made into a 6 x 1 equation to become
(11)
0where , ,
00
T T T
i i i i i i i
i
i iT
i i
i i
F D F S M v B Y F
Ie dF D M
f me
, ,0
, (see Fig. 4 for def. of variables.)0
i i i i
i i
i
iiT
i i
i
Iv B
e r
e yY F
fe
copyright © Concurrent Dynamics International
12/7/2017
, joint torque-force pair
, parent-to-child influence matrix0
0, cm centered mass matrix
0
, acceleration pair
, nonlinear force 0
i
i
i
T
i
i
i
i
i
i
i i i
i
Ff
e dD
e
IM
m e
vr
IB
pair
, backward moment-force influence matrix0
, external torque-force pair
iT
j
i
j
i
e yY
e
Ff
copyright © Concurrent Dynamics International
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1: 1:
Given Def . 2 and Prop. 2, 11 can be put in column vector form as
(12)
where [ ] , [ ] ,
T T T
D
T
i i N i i N
F F S Mv B Y F
F col F D diag D S
1: 1:
1: 1: 1: 1:
[ ] , [ ] ,
[ ] , [ ] , [ ] , [ ]
T
i i N i i N
T T
i i N i i N i i N i i N
diag S M diag M
v col v B col B Y diag Y F col F
1
in (12) is solved to be
( ) (13)
where ( )
T T T
D
T T
D D
F
F S Mv B Y F
e
Use (8) in (13) to get
( ( ( ) ) ) (14)T T T
D DF S M S Gq A A B Y F
f is the actuation force/torque from to along the moving
joint axis. The complement of that force in are the constraint forces/torque
that prevent motion along the constrained axes of
T
i i i ip i
i
G F b b
F
joint( ).i
copyright © Concurrent Dynamics International
•
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1:
After some rearrangement, 14 becomes
( ) (15)
where
{ }
T T T
D D D D
T T
i i i i i N
i i i i i
i
F M Gq M A B Y F
M S MS diag S M S diag M
I m s s m sM
1:
(16) , joint centered mass matrix
( ), nonlinear force
( )
i
i i i
T
i i N
i i i i i i
i
i i i i
m s m e
B S MA B col B
I m s sB
m s
(17)
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•
12/7/2017
By projecting of 15 onto the free joint axes, the dynamics equations
becomes
f ( , ) (18)
where f , actuatioT
F
q C
G F
n force/torque along free axes (18a)
( , ) ( ) (18b)
, system mass matrix
T T T
D D
T T
D D
C G M A B Y F
G M G
(18c)
copyright © Concurrent Dynamics International
•
12/7/2017
1
1
Acceleration state vector in (18) can be solved to be
(f ( , ) ) (19a)
[ ] [ ](f ( , )T T T T
D D
q C
e K G R e G K C
1
1:
1:
1:
) (19b)
(See [1][2] on the factorization of . Note that the operator in [2] is the
operator here. )
where [ ]
[ ]
[ ] , w
i i N
T
i i i i N
i i N
q
diag e G K D
D diag D
E
1 6 6
1 1
1:
1
1:
1:
ith 0
[ ] ,
inertia tensor
[ ] ,
[ ]
T
i i N i i i i
i i
i i N i i i i
i i N
D
R diag R R G J G
J branch
K diag K K J G R
G diag G
copyright © Concurrent Dynamics International
•
12/7/2017
1:
1
Backward recursive computation of branch inertia tensors :
a) for 1: ; ; end
b) for : 2
ˆ
ˆ
end
(see Ref
i i N
i i
T
i i i i i i i
T
ip ip i i i
J
i N J M
i N
J J J G R G J
J J D J D
. [1] and [2].)
copyright © Concurrent Dynamics International
•
•
•
•
12/7/2017 copyright © Concurrent Dynamics International
Note that (18a) follows from the definition of f. Each f of f is the net
force/torque from the control and the spring/damper forces transmitted along
the free axis of joint .
i
i
Eqs. 19a and 19b [2] are two acceleration state solution options.
Eq. 19a requires the assembly of using 18c . While that matrix is
fractored is an advantage in the assembly process. However, acceleration
solution of 18 by any linear equation solvers involving is3
an order( )
process.
N
Eq. (19b) requires that (f ( , ) ) be operated from right to left using the
indicated operators. Every one of those serial operations is an O( ) process.
Therefore, the solution process by (19b) is O
C
N
( ). N
Forward & Inverse Dynamics
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Forward dynamics regarding (18) is the problem of solving for the system
accelerations given the interbody forces along the free joint axes and the
external forces, i.e. {f , }. This is effectivelyi i
q
F (19a, or 19b).
*
Inverse dynamics regarding (18) is the calculation of the required interbody
forces along the joint free axes {f } given the system accelerations and
the external forces, i.e. { , }.
i
i iq F
copyright © Concurrent Dynamics International
Constraint Forces
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12/7/2017
6 5
Proposition 5. Given that f is the actuation force/torque along the
ˆfree axis of ' inboard hinge, then f is the constraint forces/torque
at the same hinge, where
ˆ R , c
T
i i i
c c
i i i i
i
G F
b s G F
G
1 5 5 5
ˆ ˆ ˆonstraint matrix; 0 , 1, (20)
ˆMore specifically, if is a particular column of , then f is the
constraint torque (or force) along .
T T T
i i i i i i
c c cT
i i i i i
c
i
G G G G G G e
G G G F
G
Given that the system accelerations are solved by 19a, or 19b , then all the
interbody forces can be determined by 15 .i
F
copyright © Concurrent Dynamics International
•
12/7/2017
3 1
Example: Let - be the free axis of joint frame and the request is for the
constraint force along the - of that frame. Given that has been computed
by (15), then
, 0 ,
i
i
x
i i
x axis
z axis F
G e
3 1
3 1
0 ,
Ans: f 0 ,
where unit - of joint frame in inertial coordinates.
c z
i i
Tc z
i i i
z
i i
G e
e F
e z axis
copyright © Concurrent Dynamics International
Computing System Mass Matrix
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Procedure to compute per (18c) :
1. Compute 2. Compute
with [ (:,1) (:, 2) (:, )] with [ (:
DJ G Q MJ
J J J J N Q Q
,1) (:, 2) (:, )]
[ (:,1) (:, 2) (:, )] { ( ), 1: }
for 1: for 1:
(:, ) (:, ) ; per AlD
Q Q N
G G G G N M diag M i i N
i N i N
J i G i
g.1 ( ,:) ( ) ( ,:) ; row block matrix mult.
end end
3. , end
Q1: what is the order of the above pr
T
Q i M i J i
J Q
ocedure?
Q2 : can it be made order( )?N
copyright © Concurrent Dynamics International
• Ans1: Steps 1, 2 and 3 are each order(N2), therefore the process is order(N2)
• Ans2: No. Computation of is order(N2) per Ans1.
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International
O(N) Solution of EOM
•
12/7/2017
0 1 2
1 1:
1:
For { , , , }, do {
Step 1: Given [ , ] at compute :
a. compute { } given { , }
b. compute { , , , , , , , , , , , , , }
end
k
i i i N
i i i i i i i i i i i i i i i N
t t t t t
q q t q
C q q
I g d s r y D G M J K R
1:
1:
c. compute { , , } given #a, #b and
d. compute { , , f } , application specific input
e. compute { , } per (5a) and (17)
i i i i N
i i i i N
r q
f
A B
3
f. solve (18) for
O( ) opt1: compute (f ( , ) ) per (18a,b,c) and use (19a)
O( ) opt2: compute (f ( , )) per (18a,b) and use (19b)
Step 2: Given [ ,
q
N C
N C
q
1, ] at , integrate [ , ] to }
k kq q t q q t
copyright © Concurrent Dynamics International
Examples
• The following examples illustrate how to use the methodology presented earlier to derive the eom of specific cases. These derivations can then be compared with other hand derived eom for the same cases for verification.
• It is recommended that the O(N) solution procedure shown earlier be used to develop a generic program that solves all tree-configured N-body dynamics. That program can easily solve the examples presented and produce identical dynamic responses as those produced by solving the hand derived eom.
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International
Example 1 Simple Pendulum
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θ cm1: I1, m1
s1
Inertial frame
x
y
z(out)
hinge1:ez pendulum angle
gravity accelerationg
yge
1 0
copyright © Concurrent Dynamics International
Model Parameters
12/7/2017
1 1
1
1
1
1 1
1 1
0,
(sin( ) cos( ) ) ; cm position from hinge
0 ; external force
f ; hinge torque
, with 0
z
x y
y
T
z
e
s L e e
Fm g
G F
eG G G
6 6
1
6
(M1)
0 (M2)
( )
D
D De e
6 (M3)
copyright © Concurrent Dynamics International
•
•
12/7/2017
1
2
1 1
1 1 1
1 1
Let ( , ) ( ) per (18b)
with 0 and for this example.
Thus, given (17) and (M3), we have
0
[0, ]
[ , ]
( ) [
T T T
D D
D
T
T
y y
T T
D
C G M A B Y F
A e
e sY
e
B m s
Y F s m ge m ge
B Y F s m
2
1 1 1
1
, ]
f ( , ) sin( )
y yge m s m ge
C m gL
2
1 1 1
per 18 for this pendulum is
( ) g sin( ) (M4)zz
EOM
I m L m L
copyright © Concurrent Dynamics International
Example 2 Pendulum with a Moving Base
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Inertial frame x
θ cm2: I2, m2
b1: I1, m1
s2
x
y
z(out)
(0,0) hinge1
hinge2:ez
xfe
cart force
pendulum angle
gravity acceleration
f
g
yge
copyright © Concurrent Dynamics International
Model Parameters
12/7/2017
2 1 1 2
1 2
1 2
1 2
0 , , 0,
0, (sin( ) cos( ) ) ; cm position from hinge
0 0, ; external forces
Let f [0,0] ; joint transmitted
x z
x y
x y y
T
d xe e
s s L e e
F Ffe m ge m ge
G F
1 2 1 2
6 6
2
1
force/torque
0{ , }, with , (N1)
0
0 0 0 0, where 0 0 nilpotency( ) 2 (N2)
0 0
( )
z
x
D D
D D
eG diag G G G G
e
D e
e e
6 6
6 6 6 6
0 (N3)
D
e
e e
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•
12/7/2017
1
2
2 2
2
1 6 6 2
1 2 2
Let ( , ) per (18b), and we have 0
and 0 for this example. Given (17) and (N3), we have
[[0,0],[0, ]]
0 ,0
[[0, ],[
T T T
D D
T T
T
x y
C G M A B Y F A
B m s
e sY Y
e
Y F fe m ge s m g
2
2
2 2 1 2 2 2
2
2 2 2 2 2
2
2 2
, ] ]
[[ , ],
[ , ] ]
( , ) [ ( ) , g ( )]
Given (18a) and f [0,0] for this cas
y y
T T
D y x y y
y y
e m ge
B Y F s m ge fe m ge m s m ge
s m ge m ge m s
C f m Ls m Ls
2
2 2
e,
f ( , ) [ ( ) , g ( )]C f m Ls m Ls
copyright © Concurrent Dynamics International
•
12/7/2017
21 2 2 2
2
2 2 2 2
EOM for this pendulum on a moving base is
cos( ) sin( ) = (N4)
cos( ) g sin( )
The system mass matrix in (N4) was deriv
zz
m m m L x f m L
m L I m L m L
ed using Eq. (18c)
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Summary
12/7/2017
• EOM for a tree configured multibody system using generalized coordinates and Newton-Euler equations has been presented.
• EOM formulation here requires no joint constraints to enforce. One can extract the interbody constraint force or torque at any joint from Eq. (15), once the system accelerations have been determined.
• O(N3) and O(N) solution to the EOM, Eq. (18), have been presented. (See Eq. (19a) and Eq. (19b).)
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References
1. Rodriguez, G., Jain, A., Kreutz-Delgado, K., ‘Spatial Operator Algebra for Manipulator Modeling and Control’, The International Journal of Robotics Research, Vol. 10, No. 4, August, 1991.
2. Tong, M. ,’Inverse Mass Matrix Factorization Using Momentum Equations of a Rigid Multibody System’, AIAA Guidance, Navigation and Control Conference, 13-16 August, 2012, Minneapolis, Minnesota, Paper No. 1359569
3. Greenwood, D., ‘Classical Dynamics’, Prentice-Hall, Inc., Englewood Cliffs, New Jersey, 1977.
4. Goldstein, H., ‘Classical Mechanics’, Reading, Mass: Addison-welsley Press, Inc. 1950.
5. Pars, L.A., ‘A Treatise on Analytical Dynamics’, London: Heinemann, 1965.
6. Hooker, W. W., and Margulies, G. 1965. The Dynamical Attitude Equations for an n-Body Satellite. J. Astronautical Sciences, vol. 12, no. 4, pp. 123-128.
7. Likins, P.W., 1975. Point-Connected Rigid Bodies in a Topological Tree. Celestial Mechanics, Vol. 11, No.3, 301-317..
8. Wittenburg, J., 1977. Dynamics of Systems of Rigid Bodies. Stuttgart: B. G. Teubner.
9. Ho, J.Y.L, 1977. Direct Path Method for Flexible Multibody Spacecraft Dynamics. Journal of Spacecraft and Rockets, Vol. 14, pp. 102-110.
10. Orlandea, N., Chace, M. A., and Calahan, D. A. 1977. A Sparsity-Oriented Approach to the Dynamic Analysis and Design of Mechanical Systems|Part1. Trans. ASME J. Engineering for Industry, vol. 99, no. 3, pp. 773-779.
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