2016 International Conference on Artificial Intelligence and Computer Science (AICS 2016)
ISBN: 978-1-60595-411-0
Recovering Projected Centers of Circle-pairs with Common Tangents
Qian CHEN1,*, Hai-Yuan WU1,*, Ding CHEN2 and Ryuuki SAKAMOTO2
1
Faculty of Systems Engineering, Wakayama University, Wakayama City,
Wakayama 640-8510, Japan
2
Yahoo Japan Corp, Tokyo, Japan
E-mail: *{chen,wuhy}@sys.wakayama-u.ac.jp
[email protected], [email protected]
*Corresponding author
Keywords: Circle; Center; Computer Vision.
Abstract. This paper describes a closed-form algorithm of recovering the projected centers of a
parallel circle-pair from one image. By using the common tangents of a circle-pair as the key helper
for solving the problem, we have developed an algorithm that only uses real, visible and concrete
geometrical features for describing itself and for representing the intermediate results. This makes
the jobs of checking the behavior or verifying the intermediate and the final results easy. This
characteristic greatly improves the reliability of our algorithm when applied to noisy images. We
have checked the behavior of our algorithm in various camera settings and the influence of image
noise using simulated images and real images, and compared ours with the algorithm that uses two
concentric circles and the algorithm that uses three circles.
Introduction
Estimating the accurate locations of reference points or markers in images is a fundamental
requirement for camera calibration, 3D shape modeling, motion recovery and many other computer
vision applications. A circle has many unique features: 1) It is big. Compared with a corner or a
common point of two lines, a circle has much more pixels on its locus. Thus its center can be
estimated accurately. 2) A circle has a full symmetrical shape. Because of this, it can be recovered
even when only part of it is visible.
JS.Kim et al.[1] used concentric circles for camera calibration. They used the rank properties of
degenerate conics to recover the affine and Euclidean structures, including the projected center of
circles. A.Datta et al.[2] used the circles for camera calibration. They used the centers of ellipses in
image as the projected centers of circles, but this only an acceptable approximation when the
ellipses are small enough. Z.Zhao et al.[3,4] estimated the projected centers of n  3 co-planar
circles. Q.Chen et al.[5,6] estimated the projected centers of two co-planar circles and recovered the
Euclidean structure of the plane. They assumed that all the intrinsic camera parameters except the
focal length are known.
In this work, we proposed a straight forward and non-trial method for recovering the projected
centers of parallel circle-pairs in a single image taken by an un-calibrated camera. As far as we
have investigated, we have not discovered any paper that estimates the projected centers of two
non-concentric parallel circles. In our method, we only use real and visible geometrical features,
such as points, lines since they have intuitive meaning and can be visualized easily.
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Recovering the Projected Centers of Parallel Circle-pair
In our algorithm,

the common tangents are used as the key to solve our problem.

the points that the common tangents touch the circle-pairs are used to estimate a vanishing
point, which is then used to determine the line passing through the centers of the circles.

another vanishing point is determined by using the line passing through the centers. The line at
infinity is then determined and used to calculate the centers of the circles.
Determining the Outer Coincident Tangents of Two Circles
Considering two co-planar or parallel circles that they do not enclose each other. there are at least
two common tangents of the two circles. Let Q1 and Q 2 be the quadric form of two circles, either in
the real world or on the image plane. The common tangents, which are the tangent lines of both
circles simultaneously, can be determined by solving the simultaneous equations (eq.1),
t T Q11t  0
 T 1
t Q 2 t  0
,
(1)
where t is the parameters of the common tangent line in homogeneous coordinates. For
simplicity, we use degenerate conics to solve eq.1. The degenerate conics are the members of the
pencil Q11  Q 21 where  is a scalar that satisfies | Q11  Q 21 | 0 . As shown in Fig.1, the
degenerate members are the pairs of the common points on the common tangent-pairs: (AB,CD),
(AC,BD) and (AD,BC), which can be obtained by decomposing degenerate conics[9] into pairs of
column vectors (l, m) :
Q11  Q 21  lmT  ml T .
(2)
Figure 1. Degenerate conics of conics loci and conic envelopes.
The common tangents t1 , t 2 , t 3 , t 4 can be calculated with any two degenerate members. For
example, if we use the first two, (AB,CD) and (AC,BD), then the common tangents can be
calculated by
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 t1  AB  AC
t  AB  BD
 2
.

t 3  CD  AC
t 4  CD  BD
(3)
Figure 2. Outer and inner tangents of two circles.
Figure 3. Recovering the centers of two circles.
As shown in Fig.2, if we use pij to denote the point that the tangent t i touches circle Q j , then
pij can be calculated with
p ij  Q j 1t i ,
(4)
If two circles do not enclose each other, there are always two common tangents that the both
circles are on the same side ( t1 , t 4 in Fig.2). We call the common tangents of this kind as outer
tangents. The outer tangents can be obtained by selecting the ones among (t i , i  1,2,3,4) that
1.
t i only contains real elements, and
2.
 k  i  2j 1 t Ti p ki  0 .


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Determining the Line Passing Through the Centers of Two Circles
In order to recover the centers of the two circles, we want to determine the line that passes through
the two centers of them. As shown in Fig.3, since the mid-point m1 of the line segment p11p 41 and
the mid-point m 2 of the line segment p12p 42 are on the line that passes through the two centers, we
use m1 and m 2 to determine that line. Since p11p 41 and p12p 42 are parallel in 3D space, the
vanishing point n1(  ) can be calculated with
n1( )  (p11  p 41 )  (p12  p 42 ) .
(5)
Here, n1(  ) is described and calculated in homogeneous coordinates, thus it can be computed
numerically even when it is really at infinity. The distance p 41m1 and p 42m 2 can be calculated
according to the invariance of the cross ratio of four co-linear points,
 p 4ip1i  p 4in1(  )

p 4im i   2p 4in1(  )  p 4ip1i

p 4ip1i / 2

; if | n1(  ) | 
, i  1,2 ,
(6)
; if | n1(  ) | 
here the distance between two point a and b is denoted by ab and the calculations in eq.6 is
carried out in Cartesian coordinates. The mid-points are then calculated with
m i  p 4i 
p 4i m i
(p1i  p 4i ), i  1,2 .
p 4i p1i
(7)
Then the line that passes the two centers is obtained with
l cc  m1  m 2 .
(8)
Determining the Centers of Two Circles
We first find out the common points between line that passes through the two centers and the two
circles ( c11 , c12 , c 21 , c 22 ) by solving the following equations.
cTij Q i c ij  0
; i  1,2; j  1,2 .
 T
 l ij c ij  0
(9)
Without losing generality, we number i, j to let (c11  c12 )T (c 21  c 22 )  0 . Since the lines c11p11
and c12 p 12 are parallel, we can calculate another vanishing point n 2() with them.
n 2()  (c11  p11 )  (c21  p12 )
(10)
Then the line at infinity can be determined:
l   n1()  n 2() .
(11)
Since the center and the line at infinity are in the pole-polar relationship, the two centers can be
calculated as follows.
ci  Qi1l  , i  1,2
(12)
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Experiments
We have tested our algorithms (TwoCircles) using both simulateFigmages and real images and have
compared the performance of ours with the algorithm that uses concentric circles[1]
(ConcentricCircles) and the one[3] that uses n  3 circles (Three Circles). We generated simulated
images while changing the tilt angle from 10 to 80 degree, the pan angle from 0 to 360 degree, the
size and the location of each circle. We also added Gaussian noise to the edge points of the ellipses.
Fig.4 shows some example images (a) and the average errors in pixel due to the change of the
radius of the smallest circle (indicated by the ration between it and the biggest one) (b) and the  of
Gaussian noise (c).
(a)
(b)
(c)
Figure 4. Some example images (a), the error due to the change of radius (b) and due to the Gaussian noise (c).
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We printed the test pattern on to papers and took images with a Nikon D90 camera while
changing the pan and tilt angle while taking the images. The errors were calculated by comparing
the estimated center with the ''cross'' marker at the center (See Fig.5). Fig 6 and 7 show two
example results of the comparative experiments. The ration after each name of the algorithm
indicates the radius ration between the smallest and the biggest circle in the test pattern.
Figure 5. The center marker for indicating the ground truth.
Figure 6. The experimental results radius ratio=0.8:1.
Figure 7. The experimental results radius ratio=0.5:1.
Conclusions
In this paper, we have described a straight forward, stable and easy to implement method for
recovering the projected centers of parallel circles. We have shown that by using common tangents
as the key, the line at infinity and the centers of circles could be recovered easily. The experiments
using both the simulated images and the real images have shown that our algorithm was stable and
could give competitive results compared with the other tested algorithms. Moreover, compared with
concentric circle based algorithm and the one that requires more than two circles, the input of our
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algorithm is less restrictive. This makes that our algorithm be applied to a wider range of
applications where strong constraints such as concentric circles or more than two circles are not
available.
Acknowldgements
This work was partially supported by JSPS KAKENHI Grant Number 26330195.
References
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