4.0 (STUDY 2) UNDERSTANDING THE ‘YIPS’ IN SPORT: A GROUNDED THEORY INTERVIEW STUDY
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8.1: Constraint Estimation and Residuals
151
101
102
103 0
2 4
Angular Alignment Error and 1 Sigma Bounds (1 Contact)
Sample Size
Angular Error (deg)
Figure 8.2: Angular alignment accuracy of the constraint estimation procedure with 1 contact as a function of the number of samples. Mea-surements where positions and forces compensated for friction. Note the signicant improvement in performance over the stiness approach.
The completely free and xed case are t without any error (indicated by NA), since there is nothing for the estimator to estimate. The partial constraint cases are also signicantly improved.
If accurate estimates are required for assembly or other mating tasks, it may be worth the additional computational cost to use the constraint based approach. The eigenvalue computations are easier in the constraint approach since two k k sys-tems have to be solved, versus one 2k 2k system. However there is additional computational cost in the numerical iterations. In the test cases, this computation generally converged in two to three steps from the initial estimate, but each step requires searching for a minimum in three dierent directions.
152
Chapter 8: Constraint Estimation will be represented by the vectorr
. With these conventions, the planar rotation is given byR
() =cos() ;sin() sin() cos()=c ;s s c
; (8:30)
where c and s are abbreviations for cos and sin. This representation for the congu-ration space will make some of the derivations easier.
For polygons any contact constraint consists of contacts between one or more points and lines. Let
p
be the homogeneous coordinates of a point in space. Letn
= [n
;;d]be the homogeneous coordinates of a hyper-plane (line in 2D) with
n
a unit vector, and d the distance to the plane measured along the normal. With this notation the basic constraint that a point lie on the plane is0
n
T 0p
= 0: (8:31)This general constraint becomes a function of the conguration of the polygon. There are only two basic types of contact that can occur with polygons. Type B contact is contact between a vertex on the moving polygon and an edge on the xed obstacles.
Type A contact is contact between an edge on the moving polygon and a vertex on the xed obstacles. Type B constraints are most easily expressed in the xed frame, and by symmetry type A constraints are most easily expressed in the moving frame.
However, for uniformity, all the constraints will be given in the xed frame.
For type B contacts the general constraint becomes
0
n
T 0T
m mp
= 0: (8:32)In this equation m
p
is the coordinates of the contact vertex expressed relative to the moving frame, 0T
m is the transform taking the local coordinates to the global coordinates, and 0n
is in the xed frame. This equation can be expanded to0
n
T (0R
mp
+r
);d = 0: (8:33) For type A contact the constraint ism
n
T 0T
;m1 0p
= 0: (8:34)In this equation 0
p
is the coordinates of the contact vertex expressed relative to the xed frame, mn
is in the moving frame, and 0T
;m1 is the transform taking global coordinates to local coordinates. It is the inverse of 0T
m. This equation expands tom
n
T 0R
Tm (p
;r
);d = 0: (8:35)8.1: Constraint Estimation and Residuals
153
The dierence between the two constraint equations is whether 0
R
m is applied or not applied tor
.Both constraint equations involve a normal. To simplify the following notation we will drop the frame superscript. The choice of frame should be clear from the contact type.
8.1.2.1 Constraint Tangents and Cotangents
This section derives the constraint tangents and cotangents for both type A and type B contacts. It is shown that all the tangent bundles can be written as a projection of the product of a 4x4 curvature matrix with a constant 4 vector. This unifying view of the problem allows us to derive a single general form for the constraint estimator in the next section. In all of these equations
R
=0R
m to simplify the notation.8.1.2.2 Type B
Taking the derivative of equation 8.33 with respect to
x
gives the cotangent vectorY
B(x
) =n n
TXRp
(8:36)
where
X
=0 ;1 1 0and is the cross product matrix.
X
has the properties
y
TXy
= 0 for any vectory
X
T =;X
and
XR
=RX
, i.e.X
commutes with rotations.Expanding out the term
n
TXRp
results in the expressionn
TXRp
=n
TXp
cos();n
Tp
sin() (8.37)= [1 0]
Ra
; (8.38)154
Chapter 8: Constraint Estimation where a1 =n
TXp
and a2 =n
Tp
. This means thatY
B(x
) can be rewritten asY
B(x
) =P
Id
00
R
n a
=
P C
YBY
B; (8:39) whereId
is the identity matrix andP
is a projection matrix which keeps the rst three components.Y
B will refer to the constant four vector andY
B(x
) will refer to the function. The matrixC
YB encodes the curvature of the cotangent vectors and plays a critical role in the estimation problem.The tangent vectors
J
B1 andJ
B2 are reciprocal toY
BJ
B1(x
) =Xn
0
J
B2(x
) =RX
Tp
1
; (8:40)
and these can also be written as products of the projection matrix and curvature matrices as
J
B1(x
) =P
Id
0 0Id
Xn
0=
P C
JB1J
B1 (8.41)J
B2(x
) =P
R
0 0Id
2
6
4
X
Tp
10
3
7
5=
P C
JB2J
B2: (8.42)8.1.2.3 Type A
Taking the derivative of equation 8.35 with respect to
x
gives the cotangent vectorY
A(x
) = ;Rn n
TX
TR
T(p
;r
)(8:43)
Expanding out the term
n
TX
TR
Tp
results in the expressionn
TX
TR
Tp
=n
TX
Tp
cos();n
Tp
sin(): (8:44) Using the previous denitions of a1 and a2 shows thatY
A(x
) can be rewritten asY
A(x
) =P
2
4
R
0r
TXR
0
R
3
5
;
n a
=
P C
YAY
A: (8:45) The tangent vectorsJ
A1 andJ
A2 are orthogonal toY
AJ
A1(x
) =RX
Tn
0
J
A2(x
) =;X
(p
;r
) 1; (8:46)
8.1: Constraint Estimation and Residuals
155
and these can also be written as products of the projection matrix and curvature matrices as
J
A1(x
) =P
R
0 0Id
X
Tn
0
=
P C
JA1J
A1 (8.47)J
A2(x
) =P
Id
[Xr
0]0
Id
2
6
4
;
Xp
10
3
7
5=