Master Thesis - Fachbereich Informatik

4.2. MODEL KNOWLEDGE AND ASSUMPTIONS 57
Figure 4.5: The image intensity of transparent tubes is not uniform as for black tubes.
Depending on how much light can pass through a tube, regions appear darker or brighter.
One characteristic of transparent tubes under back light are two dark horizontal stripes at
the top and the bottom of a tube indicated by the arrows. The printing also reduces the
translucency and thus appears darker in the image.
4.2.6. Tube Orientation
The tube orientation is highly constrained by the guide bars as introduced in Section 3.1.
Thus, an approximately horizontal orientation can be assumed throughout the design of
the inspection algorithms.
In practice, the distance between the guide bars is slightly larger than the outer diameter
of a tube to prevent a blockage, since tubes may not be ideally round. This means, the
cross-section of a tube can be elliptical instead of circular. Let dspace denote the vertical
distance between the guide bar distance dGB, and hmax the maximum expected tube
extension in vertical direction with respect to the image projection. The remaining spacing
distance can be expressed as dspace = dGB − hmax ascanbeseeninFigure4.6(a).
The maximum possible rotation is reached if the tube hits both guide bars at two points
(see Figure 4.6(b)). The maximum angle of rotation θmax can be defined as the angle
between the longitudinal axis of the tube and the x-axis. One can define an unrotated
version of the tube with the longitudinal axis parallel to the x-axis and shifted so that the
two axis intersect at the center of gravity of the rotated tube. In Figure 4.6(b) this virtual
tube is visualized as dashed rectangle. The distance between the measuring points of the
rotated and the ideal horizontal tube can be also seen in the Figure and are denoted as
dL and dR for the left and right tube side respectively. Both dL and dR are ≤ dspace/2. If
thetubeisnotbent,dL = dR. The maximum error between the ideal distance l and the
rotated distance l ′ can be estimated as follows:
errθ = l ′ − l
�
(4.1)
= l2 + d2 space − l
For example, in a typical setup for 50mm tubes of 8mm diameter one tube has a length of
approximately 415 pixels and dspace = 15. This leads to an error of errθ =0.27pixel. Thus,
with one pixel representing 0.12mm in the measuring plane, the acceptable maximum error
due to orientation would be about 0.03mm. On average this error will be even smaller.
Based on these estimation, the orientation error is neglected in the following, i.e. all tubes
are assumed to be oriented ideally horizontal.