Full text: XVIIth ISPRS Congress (Part B5)

  
    
  
  
     
    
   
  
  
     
   
   
   
  
      
    
    
    
     
     
   
  
  
    
    
   
    
   
   
    
   
    
     
       
    
     
  
    
      
      
    
    
vector f become equivalent. This means that N and f are 
parallel but have different ES (the value of 
y 
> 
components of the vectors N and f are multiple). 
     
X 
Figure 2.3.1 Normal vectors in object and image space. 
The vector f can be defined by the vector product: 
0 n -m Xc - X1 
$-3xp6-4I-l1]-5 0 1 [| Ye —Mi 14232) 
m -l 0 Zc - Z1 
The relation between their components can be 
analytically described introducing a scale factor A, 
which leads to: 
N= aR? (2.3.3) 
Combining equations 2.2.8, 2.3.2 and 2.3.3 the 
model can be stated more explicitly: 
f.cose] ril ri2 ri3 0 n -m||Xc - Xi 
f.sinO| = A.|r21r22 r23].|-n 0 1{.]Yc - Y1| (2.3.4) 
-p r31 r32 r33 m -l 0| IZc - Z1 
In equation 2.3.4, the scale factor A can be stated 
=> 
as the ratio of f and N modulus. 
In order to eliminate A in equations 2.3.4 the 
first and the third equations are divided by the second 
one, resulting: 
rii.fx + r12.fy + rı3.fz 
  
  
  
cotgo = 
r21.fx + r22.fy + r23.fz (2.3.5) 
-p SU q31.fx t. r3s2.fy * r33.fz 
f.senO . r21.fx * r22.fy * r23.fz 
Equations (2.3.5) can be rewritten as: 
rii.fx + r12.fy + ri13.f: 
T21.1x + r22.1y + r23.fı (2.3.6) 
r31.fx + r32.fy + r33.fz 
‘r21.fx + r22.fy + r23.fz 
Equations (2.3.6) are indefinite when © = 0° (or 
b=-f 
  
x = 90°) and by this reason it can be used only for 
45°< 8 < 135° or 225°< 6e < 315°. 
Defining a new representation for the straight 
line 5 . 
x=a.,y+b (2.3.7) 
where: a = —tan® = cotana b = p/cos@ = -p/sina 
a complementary set of equations can be derived from 
2.3.4 dividing the second and the third equations by 
the first one, which is applied for 315°< 6 < 45° or 
135°< 8 « 225°. 
à f21.fx * ra2.fy * r23.fz 
ril.fx + rT12.Íy * r13.Íz 
  
(2.3.8) 
bzr rise r32.fy + r33.fz 
US Tli.fx t ri2.fTy * r13.1z 
  
3. SYSTEM MODEL 
3.1 Introduction 
The problem of locating an object or calibrating a 
camera in Robot Vision is depicted in Figure 3.1.1. In 
this specific example it is assumed that the object is 
moving over a conveyor belt (a linear movement) and 
that the camera is attached to the robot wrist. In the 
eye-in-hand configuration the camera state vector is 
given by the kinematics of the robot plus a known 
transformation (a priori calibration) between wrist and 
camera coordinate systems. Due to cumulative errors in 
the robot joints the camera state vector can present 
some uncertainty. 
3.2 Transform equations 
The problem to be solved using eye-in-hand Vision 
can be better stated by defining the following 
homogeneous transformations: 
T, describes the station frame with respect to the 
base of the manipulator; 
T. describes the object frame with respect to 
station frame. The model of the object is 
specified in the object frame; 
T. describes the goal frame with respect to the 
object frame. The goal frame defines the 
position and orientation which must be reached 
by the manipulator wrist or an end effector; 
T., describes the wrist frame with respect to the 
base frame. This transformation is also known as 
the kinematics of the manipulator and is 
obtained by successive transformations over the 
  
  
  
  
" links; 
T. describes the camera with respect to the wrist 
Ü frame; 
T. describes the goal frame in camera coordinates; 
Te 
Ye 
Zp Yw 
Tw 9 
YB 
y 
2 ?n 2 
AA 
CM 
$»| y 5 
Zs Ys ^" 
Xo 
(* — He 9 
: Xs € 
Figure 3.1.1 Frames and transforms in  eye-in-hand 
problem. 
The general problem of transformation between two 
reference frames can be stated by the transform graph 
(Paul, 1981) presented in Figure 3.1.2. 
  
  
v 
à d 
x 
A 
N 
Figure 3.1.2 Transform graph. 
In the eye-in-hand approach two cases can be 
stated: Ww 
first case: the object is static and transforms  T,, 
B, C 
To To and T are known. Transform TG is 
computed Jsing space resection techniques (in fact, 
transform Te ). Then, transform T w can be obtained 
using the following transform equation: 
  
  
  
  
  
  
  
  
  
   
e+ pe fra ea 
pm set AE pe 
P e* © TH 
-—— ff aa MD 
f^
	        
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