‚ Part 5. Hakodate 1998 
ree dimensions. 
rithms of stereo 
ensional frames 
ability to follow 
es information is 
h more difficult 
ie line segments 
je the state and 
f our coordinate 
M=M,V(M)= 
M (2) 
£2 are known 
n (2) appears as 
equation in M. 
>neral,exists for 
is not a function 
nstant angular 
solution is given 
By (s)ds (3) 
nmetric matrix 
t with Q(Qxz —9 
A 2. e*75? and e“ 22 are rotation matrices. 
We can obtain slightly more detailed 
results that are useful in practice by making 
assumptions about the functional form of 
V (s). For example ,we may assume that it is a 
polynomial in s : 
VG) m ys 
= 
Using Rodrigues’ equation we can write 
sin(@— ss 12D 
I| e l| 
1 — cos((t — s) || 9 || )—, 
& 
p | 2 | 2 
498 — I + 
  
From this it is clear that,in order to compute 
the integral [ e€*-9?y (s) ds , which appears in 
0 
equation (3) , we need to compute the integrals 
L, = [ ssinC(t — s) || & || dds 
0 
M, = | eR od 
0 
It is simple to show that can be done in closed 
form. For the special cases a —0,1 (constant 
velocity and constant acceleration) we have 
the following result: 
proposition 
when V (s) —V -4-sA ,the trajectory of the point 
M is given by 
M(t) = U,M(t,) — UV +U,A (D 
with the following values for the matrices U ;,? 
=0,1,2 
_, 1 sin(@ —#) 9/5 y L=cos(C—#) 91) 
Wines El em FEE 2 
— cos((t — t9) || 9 |. 
(16-6) ^ LOL PD 
( — to) || 9l, —sinCG — t9 [L9 la 
* Tals S 
(—uV?, t—t)5| 9l —sincct —t0] 914 
2 
ET EE 
( — t9) || 9 I? — 20 —cos(¢t —t) 21 57 
201% 
Up=1 
+ 
2. CHOOSING A REPRESENTATION OF 
  
THREE-DIMENSIONAL 
LINE SEGMENTS 
we know a representation of 3-D lines by four 
numbers (a ,b,pand q) such that the equations 
of the line are (z —az--p,y —bz —4) in the 
first map g» With the two other maps 
obtained by exchanging the roles played by z , 
y and z. In g lines perpendicular to the z-axis 
cannot be represented , while in 9» and @ it is 
the lines perpendicular to the x-and y-axis, 
respectively that cannot be represented. From 
this representation, it is not too difficult to 
compute the weight (covariance) matrix of the 
endpoints M; and M; of a line segment ,a 6X 6 
matrix. 
One possible representation of a line 
segment is therefore the six-dimensional 
vector [ M7, M2 I”. Just as in the two- 
dimensional case,this assumes that segments 
are oriented. If they are not, we can use the 
representation of direction, midpoint and 
length which is the vector [a 5, M^ ,L]" ,where 
M is the representation of the midpoint and / 
is the length of the segment. Of course , weight 
(covariance)matrices for these representations 
can be computed up to first order. In what 
follows ,we will denote by r the representation 
vector. and we will let C be the corresponding 
covariance matrix. 
3. THE PLANT AND MEASUREMENT 
EQUATIONS 
Contrary to the two-dimensional case, the 
state of the segment is directly related to the 
kinematic screw of the solid to which we 
assume it is attached. we define it to be the 
vector 
QQ 
V 
a= VA) (5) 
V (n)