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Likely one gets 
m32 * 7(94n254*395n55*03n53) (12-2) 
Mag = -(r4n54*nron55*n3n53) (12—3) 
where eq.(12-3) is identical to eq.(10). 
3.4 Evaluation of ky 
Writing the first six expressions of 
eqs. (4) in the form of 
ü 
m» 1^31 p4*m34n^241: 
m341^32. 7? PBotma4n55, 
m21"33 ?. P3*m3 423. 
m32n341 = d4*m355n» 1: 
Maas = d5*m32n22* 
m32^n33 * d3t*n3»5n»3. 
multiplying the first with the forth, the 
second with the fifth and the third with 
the sixth of each side of the above ex- 
pressions and summing up them, one can 
calculate the right side of it. And the 
left side becomes 
m534m55(n342*n352*n332) = Mo4Moo 
--$ink4cosk4- -1/2sin2k,. 
This procedure produces four candidates 
for k4- 
Then n34.n32 and n are evaluated for 
each candidate for Kq - They are evaluated 
from following different equations for 
better precision. 
a) for -3/4% <k4<-M/4 or T/4<k,4<3/4T 
VB 4) 
n32* (potmg4n22)/C-=in ky), 
n33 *(pgtma34no3)/cos k4 (13—1) 
b) fon -T/Ask4«T/4 or 3/AT«k4«5/4tr 
nee msanaq37coskq: 
n327(05*m35022) 76093. 
n337(83*ma45n53)/cosk4 (13-2) 
3.5 Evaluation of £5, wo 
From eqs.(8-2) 
sin W4COS d = Nog, 
COS w5,COS gy =nag. (14) 
Since n33 >0, which means cos @, # 0, 
cos Bo = ros? + nas * (15) 
There are four candidates for £5. And for 
each candidate for £5, angle wy, is evalu- 
ated by 
sin w4- no5g3/cos gs, 
203 
COS W5 * n33/COS PB». (16) 
3.8 Evaluation of ka 
From eqs.(8-2); 
(-cos w5)sin ks 
+ (sin wo sin £g5)cos ko * n1 
( cos W5)cos ko 
+ (sin Wo sin $5)sin Ko = noo 
( sin wa)sin ko 
+ (cos Wo sin $5)cos Ka, 5 Naz (17) 
(-sin W5)coS ko 
+ (cos wasin $5)sin Ko * ngo, 
one solves the first two equations to get 
gin ks and cos kp. They are always solva- 
ble, even df sin g is zero. And this Ko 
is tested by substituting it into the 
third and forth equations. Any sets of 
candidates for do and Wo that do" not 
satisfy both are abandoned. 
3.1 Strict relative orientation and deter- 
mination of the sign of a base length 
Since the precision of approximations 
evaluated above is usually not suffi- 
cient, one should execute relative orien- 
tation again using those approximations. 
An independent model is thus obtained, 
which ‘igs either Fig.1-1 or 1-2. 
Next the sign of a base length is deter- 
mined the way that if Zp coordinates of 
objects in the independent model coordi- 
nate system are lesser than O0, it is set 
plus, and if Zp coordinates are greater 
than 0, it is set minus. 
4. EVALUATION OF ORIENTATION PARAMETERS 
IN THE OBJECT SPACE COORDINATE SYSTEM 
4.1 Model connection in the global model 
coordinate system 
Independent models thus produced are 
linked to make a global model by usual 
successive orientation. Scales of succes- 
sive models are adjusted by scaling base 
lengths. As a result exposing positions 
and rotation matrices associated with the 
global coordinate system XMYmZM are deter- 
mined. 
e global model! 
object space 
4.2 Transformation from t 
coordinate system to th 
coordinate system 
When an object space coordinate system XYZ 
is given, global model coordinates XyYwZw 
are further transformed to the object 
coordinates. Here let us consider the case 
the object space coordinate system is 
implicitly given in the form of a few of 
3-D control points. In most industrial 
measurements this is common. And in this 
case one can calculates orientation param- 
eters automatically in the following way. 
Similar transformation