ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
ECCENTRICITY ON AN IMAGE CAUSED BY PROJECTION
OF A CIRCLE AND A SPHERE
R. Matsuoka a, *, S. Maruyama a
a
Geospatial Technology Department, Kokusai Kogyo Co., Ltd., 2-24-1 Harumi-cho, Fuchu, Tokyo 183-0057, Japan (ryuji_matsuoka, shintaro_maruyama)@kk-grp.jp
Commission V, WG V/1
KEY WORDS: Eccentricity, Projection, Circle, Sphere, Ellipse, Formula
ABSTRACT:
Circular targets on a plane are often utilized in photogrammetry, particularly in close range photogrammetry, while spherical targets
are sometimes utilized in industrial applications. Both a circle and a sphere are projected as an ellipse onto an image. There is an
eccentricity on an image between the centre of the projected ellipse and the projected location of the centre of a circle or a sphere.
Since only the centre of the projected ellipse is measured, the correction of the eccentricity is considered to be necessary for highly
accurate measurement. This paper shows a process to derive general formulae to calculate an eccentricity of a circle and a sphere
using the size and the location of a circle or a sphere, and the focal length, the position and the attitude of a camera. Furthermore the
paper shows methods to estimate the eccentricity of a circle and a sphere from the equation of the projected ellipse of a circle or a
sphere on an image.
1. INTRODUCTION
There are many factors affecting the accuracy of image
measurement. Some of them should be considered when we
execute image measurement. We have conducted a camera
calibration in order to correct radial and decentering lens
distortions since the era of analytical photogrammetry using
analogue films. In recent years the correction of colour
aberrations becomes indispensable for highly accurate
measurement (Kaufmann and Ladstädter, 2005, Luhmann et al.,
2006a, Mallon and Whelan, 2007, Hattori et al., 2011). In the
era of digital photogrammetry the fact that sampling and
quantization in creating a digital image affect the accuracy of
image measurement using digital images was reported
(Matsuoka et al., 2011).
Furthermore there is an eccentricity of projection unless an
object to be measured is a planar object lying on a plane
parallel to the image plane of a camera. The eccentricity is a
disparity on an image between the centre of the projected object
and the projected location of the centre of an object. The
correction of the eccentricity is considered to be necessary for
highly accurate measurement.
In the meanwhile, circular targets on a plane are often utilized
in photogrammetry, particularly in close range photogrammetry.
Since a circle is radially symmetrical, circular targets are well
suited for photogrammetric use such as camera calibration and
3D measurement. Although circular targets offer advantages of
high image contrast, minimal size and high precision in image
measurement, they have some drawbacks such as restricted
visibility (typical: ±45°) depending on light source and
reflective properties (Luhmann, 2014).
On the other hand, in industrial and other technical applications
spherical targets are used in order to maximize visibility and
enable tactile probing by other systems, for example coordinate
measurement machines (CMM) (Luhmann, 2014). Spherical
targets are used in the German guideline VDI/VDE 2634 Part 2
for acceptance testing and re-verification of optical 3D
measuring systems (Verein Deutscher Ingenieure, 2002).
Both a circle on a plane and a sphere in space are projected as
an ellipse onto an image. If a circle does not lie on a plane
parallel to the image plane of a camera, or if the center of a
sphere does not lie on the view vector of a camera, there is an
eccentricity of projection to some extent. Figure 1 shows an
eccentricity of projection of a circle.
oblique image
e
pE
pC
O
O: projection center
P C : centre of a circle
p C : projected centre of circle
PC
p E : centre of projected ellipse
e : eccentricity of projection
Figure 1. Eccentricity of projection of a circle
Regarding the eccentricity of a circle, Matsuoka et al. (2009)
showed a formula calculating an eccentricity of a circle using
the size and the location of a circle, and the focal length, the
position and the attitude of a camera only when the circle lies
on the horizontal plane. Moreover they presented merely the
finally derived equation, although it would be convenient to
show a process to derive the equation.
Luhmann (2014) reported the results of numerical simulations
conducted in order to evaluate the effect of eccentricities to
stereo and multi-image intersections. He quantified and
investigated the eccentricities of circular targets and spherical
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
19
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
targets. Unfortunately he did not show a general formula
calculating an eccentricity of a sphere.
This paper shows a process to derive general formulae to
calculate an eccentricity of a circle and a sphere using the size
and the location of a circle or a sphere, and the focal length, the
position and the attitude of a camera. Furthermore the paper
shows a method to estimate the eccentricity of a circle or a
sphere from the equation of the projected ellipse of a circle or a
sphere on an image.
 X S   a11
  
 YS    a12
 Z S   a13
 
a31   X S  X O 
a32   YS  YO 
a33   Z S  Z O 
(4)
The projected location pS  xS , yS  in the image coordinate
system xy of the centre PS  X S , YS , ZS  in the camera coordinate
system XY Z can be expressed by the following form:

XS
 xS   f Z

S

Y
y   f S
 S
ZS
2. PROJECTED LOCATION OF THE CENTRE OF A
CIRCLE AND A SPHERE
The image coordinate system is assumed a planar rectangular
coordinate system xy, while the object space coordinate system
is assumed a spatial rectangular coordinate system XYZ. The
camera coordinate system is a spatial rectangular coordinate
system XY Z as well.
a21
a22
a23
(5)
3. GENERAL FORMULA TO CALCULATE AN
ECCENTRICITY OF A CIRCLE
3.1 From the Size and the Location of a Circle
Consider that an image of a circle C or a sphere S is taken by a
camera with focal length f. Let O(XO, YO, ZO) and R(, , ) be
the position and the attitude of the camera in the object space
coordinate system XYZ respectively. The position and the
attitude of the camera in the camera coordinate system
XY Z are O(0, 0, 0) and R(0, 0, 0) respectively as well.
The rotation matrix R = [aij] in the object space coordinate
system XYZ can be represented by the following equation:
 a11
a
 21
 a31
a12
a22
a32
a13  1
0


a23   0 cos 
a33  0 sin 
 cos 
  sin 
 0
  cos 
 sin    0
cos     sin 
 sin 
cos 
0
0
0 sin  
1
0 
(1)
0 cos  
0
0 
1 
a31   X C  X O 
a32   YC  YO 
a33   Z C  Z O 
(2)
The projected location p C  xC , yC  in the image coordinate
system xy of the centre PC  X C , YC , Z C  in the camera coordinate
system XY Z can be expressed by the following form:

XC
 xC   f Z

C

Y
y   f C
 C
ZC
(6)
l 2  m2  n 2  1
(7)
where
The normal vector v of the plane H in the object space
coordinate system XYZ can be represented as follows:
v  l
m n
t
(8)
H : l  X  X C   m Y  YC   n  Z  Z C   0
in the camera coordinate system XY Z can be expressed by the
following equation.
a21
a22
a23
H : l  X  X C   m Y  YC   n  Z  Z C   0
Here let the plane H be expressed by the following equation in
the camera coordinate system XY Z .
Consider that the centre of a circle C is PC(XC, YC, ZC) in the
object space coordinate system XYZ. The centre PC  X C , YC , Z C 
 X C   a11
  
 YC    a12
 Z C   a13
 
Let a circle C with radius RC in the object space coordinate
system XYZ be on the following plane H:
(3)
Similarly consider that the centre of a sphere S is PS(XS, YS, ZS)
in the object space coordinate system XYZ. The centre
PS  X S , YS , ZS  in the camera coordinate system XY Z can be
(9)
where
l 2  m2  n 2  1
(10)
The normal vector v of the plane H in the camera coordinate
system XY Z can be expressed as follows:
v   l
m n 
t
(11)
The following equation can be obtained as well.
 l   a11
  
 m    a12
 n   a13
 
a21
a22
a23
a31   l 
a32   m 
a33   n 
(12)
Consider that the following transformation T from  X , Y , Z  to
 X , Y , Z  makes the normal vector v parallel to the Z
 X   1
0
  

Y
0
cos


  
 Z   0 sin 
 
0   cos 
 sin    0
cos     sin 
0 sin    X 
 
1
0   Y 
0 cos    Z 
axis:
(13)
expressed by the following equation.
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
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ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
Because the transformation T makes the normal vector v
parallel to the Z axis, the following equation is obtained:
m n     cos  sin 
t
 l
t
cos  cos   (14)
sin 
Finally  and  are obtained as follows:
  cos 
 sin    0
cos     sin 
0
b13   X 
 
b23   Y 
b33   Z 
0 sin  
(16)
1
0 
0 cos  
(17)
 X   b11 b21 b31   X 
  
 
 Y   b12 b22 b32   Y  
 Z  b13 b23 b33   Z  
 
 
(18)
the circle C in the camera coordinate system XY Z is projected
to q  xQ , yQ  in the image coordinate system xy. Then the
following equation is obtained:
yQ
t
 f   kQ  X Q YQ
Z Q 
t
(19)
where
kQ  
f
ZQ
(20)
Substituting Equation (18) for Equation (19) and rearranging
gives the following equation:
 X Q 
 b11 b12
  1 
Y

 Q  k b21 b22
Q
 Z Q 
b31 b32
 
b13   xQ 
 
b23   yQ 
b33    f 
(21)
where
Z Q
1

kQ b31 xQ  b32 yQ  b33 f
(22)
Substituting Equation (22) for Equation (21) gives the
following equation:
 X Q 
Z Q
 
Y
b
x
b

31 Q
32 yQ  b33 f
 Q
 X C   b11 b12
  
 YC   b21 b22
 Z C  b31 b32
 
b13   X C 
 
b23   YC 
b33   Z C 
(25)
Substituting Equation (23) for Equation (24) gives the
following equation:
 b11 xQ  b12 yQ  b13 f

 X C 
 Z C

 b31 xQ  b32 yQ  b33 f

2
(26)
2
 b x b y b f

  Z C 21 Q 22 Q 23  YC   RC2
 b x b y b f

33
31 Q
32 Q


By rearrange Equation (26) as to
x
Q
, yQ , f  the following
quadratic equation is obtained:
Consider that the point Q  X Q , YQ , Z Q  on the circumference of
 xQ
(24)
where
Then the transformation T and the inverse of the transformation
T are expressed as follows:
 X   b11 b12
  
 Y    b21 b22
 Z   b31 b32
 
 X   X  2  Y   Y  2  R 2
Q
C
Q
C
C



 Z Q  Z C
(15)
Let bij be as follows:
b13  1
0
b23   0 cos 
b33  0 sin 
system X Y Z  , and is perpendicular to the Z  axis. Because
the point Q  X Q , YQ , Z Q  in the coordinate system X Y Z  is on
the circumference of the circle C, the following equation is
obtained:


m
1 
  tan  2

2

l
n





1  l 
    tan  n 
 

 b11 b12
b b
 21 22
b31 b32
The circle C transformed by the transformation T is a circle
with radius RC and centre PC  X C , YC , Z C  in the coordinate
pxx xQ2  p yy yQ2  p ff f 2
(27)
2 pxy xQ yQ  2 p xf xQ f  2 p yf yQ f  0
where pxx , p yy , p ff , p xy , p xf and p yf are as follows:
 pxx



 p yy



 p ff



 pxy



 pxf



 p yf


 Z C2  2  b11b31 X C Z C  b21b31YCZ C 
  X C2  YC2  Z C2  RC2  b312
 Z C2  2  b12b32 X C Z C  b22b32YCZ C 
  X C2  YC2  Z C2  RC2  b322
 Z C2  2  b13b33 X C Z C  b23b33YCZ C 
  X C2  YC2  Z C2  RC2  b332
   b11b32  b31b12  X C Z C   b21b32  b31b22  YCZ C 
(28)
  X C2  YC2  Z C2  RC2  b31b32
   b11b33  b31b13  X C Z C   b21b33  b31b23  YCZ C 
  X C2  YC2  Z C2  RC2  b31b33
   b12b33  b32b13  X C Z C   b22b33  b32b23  YCZ C 
  X C2  YC2  Z C2  RC2  b32b33
Here substituting Equations (25), (16) and (14) for Equation
(28) and rearranging gives the following equation:
 b11 xQ  b12 yQ  b13 f  (23)
b x  b y  b f 
 21 Q 22 Q 23 
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
21
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
 pxx

 p yy

 p ff


 pxy

 pxf

 p yf
   l 2  m 2  YC2   n 2  l
 Z  2mnY Z  l R
   l  m  X   m  n  Z  2nlZ X  m R
   n  l  X   m  n  Y  2lmX Y  n R
  lmZ  nlY Z  mnZ X   l  m  X Y  lmR
 nlY  lmY Z   n  l  Z X  mnX Y  nlR
 mnX   m  n  Y Z  lmZ X  nlX Y  mnR
2
2
2
2
2
C
2
C
2
C
2
2
C
2
2
2
C
2
2
C
2
C
C
C
2
C
C
2
C C
2
C
C C
2
C
2
C
C
C C
2
C
C
2
C
C
2
C
2
C
2
2
C
2
C
C
C
C C
2
C
2
C
2
C
2
C
2
C
(29)
Equation (27) shows an ellipse as which the circle C with radius
RC and centre PC (XC, YC, ZC) is projected onto the image. The
centre p E  xE , yE  , the semi-major axis a, the semi-minor axis b
and the rotation angle  of the major axis of the projected
ellipse on the image take the following forms:
p yy p xf  p xy p yf

 xE   f
p xx p yy  pxy2


 y   f pxx p yf  pxy p xf
 E
p xx p yy  pxy2


2w
a  
2

 pxx  pyy    pxx  p yy   4 pxy2


2w
b  

2
 pxx  pyy    pxx  p yy   4 pxy2

2 p xy
1
  tan 1
2
p xx  p yy
(30)
If we can derive  X C , YC , Z C  from pxx , p yy , p ff , p xy , p xf and
p yf by using Equation (29), we can calculate an eccentricity (x,
y) of the circle C from the projected ellipse E by using
Equation (34). Since it would be unable to derive  X C , YC , Z C 
from pxx , p yy , p ff , p xy , p xf and p yf analytically, we will
obtain  X C , YC , Z C  numerically by Newton-Raphson method.
Now the following equations from f1 to f 7 as to  X C , YC , Z C  ,
RC and  l , m, n  are to be solved by Newton-Raphson method:
 f1 :  l 2  m 2  YC2   n 2  l 2  Z C2  2mnYC Z C  l 2 R 2  pxx  0

 f 2 :  m 2  n 2  Z C2   l 2  m 2  X C2  2nlZ C X C  m 2 R 2  p yy  0

 f 3 :  n 2  l 2  X C2   m 2  n 2  YC2  2lmX CYC  n 2 R 2  p ff  0

 f 4 :  lmZ C2  nlYC Z C  mnZ C X C   l 2  m 2  X CYC  lmR 2  pxy  0

 f5 : nlYC2  lmYC Z C   n 2  l 2  Z C X C  mnX CYC  nlR 2  pxf  0

 f 6 : mnX C2   m 2  n 2  YC Z C  lmZ C X C  nlX CYC  mnR 2  p yf  0

2
2
2
 f 7 : l  m  n  1  0
(36)
(31)
Here, since
(32)
, YC , Z C  and RC are linearly dependent,
To secure the convergence of Newton-Raphson method we
adopt the following initial guess of  X C0 , YC0 , Z C0  , RC0 and
l , m , n  :
0
w  p x  p yy y  p ff f
2
E
C
consider that the centre of the circle is on the plane Z   f ,
that is Z C   f .
where w is as follows:
2
xx E
X
0
0
2
2 pxy xE yE  2 pxf xE f  2 p yf yE f
(33)
Hence the eccentricity (x, y) of the circle C with radius RC
and centre PC(XC, YC, ZC) on the plane H can be expressed as
follows:

 p yy pxf  pxy p yf X C 

 x  xE  xC    f  

2
Z C 
 pxx p yy  p xy


 pxx p yf  p xy pxf YC 

 y  y E  y C    f   p p  p 2  Z 
 xx yy
C 
xy


(34)
p yy pxf  pxy p yf

 X C0  xE   f
pxx p yy  pxy2


pxx p yf  pxy pxf
 YC0  yE   f
pxx p yy  pxy2

Z f
 C0

RC0  ab
3.2 From an Ellipse Projected onto an Image
From now on we are going to derive an eccentricity of a circle
from an ellipse projected onto an image. First consider that a
projected image of a circle C on an image is an ellipse E
represented by the following equation:
E : p xx xQ2  p yy yQ2  p ff f 2  2 p xy xQ yQ  2 p xf xQ f  2 p yf yQ f  0
(38)

 pxx  X  R   p Y  p ff Z  2 p yf YC0 Z C0
 l0 2 
2
RC0

 X C02  YC02  Z C02  RC02 
(39)

2
2
2
2
 2  pxx X C0  p yy YC0  RC0   p ff Z C0  2 pxf Z C0 X C0
m0 
2
RC0
 X C02  YC02  ZC02  RC02 


2
2
2
2
 2  pxx X C0  p yyYC0  p ff  Z C0  RC0   2 pxy X C0YC0
 n0 
2
2
2
2
2
RC0  X C0  YC0  Z C0  RC0


2
C0
Finally Equations (34), (29), (12), (2), and (1) provide the
eccentricity (x, y) of the circle C with radius RC and centre
PC(XC, YC, ZC) on the plane H represented by Equation (6).
(37)
2
C0
2
yy C0
2
C0
The solution of Equation (36) by Newton-Raphson method
gives the normal vector v of the plane H on which the circle C
places in addition to the eccentricity (x, y) of the circle C. If
the radius RC of the circle C is known, we can know the centre
PC(XC, YC, ZC) of the circle C.
(35)
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
22
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
Because OPC Q and QPC PS are similar, the following
equation is obtained:
4. GENERAL FORMULA TO CALCULATE AN
ECCENTRICITY OF A SPHERE
r Dd

d
r
4.1 From the Size and the Location of a Sphere
O
O (X0, Y0, Z0)
PS (XS, YS, ZS)
PC (XC, YC, ZC)
Q (XQ, YQ, ZQ)
Rearranging Equation (45) gives the following equation:
OP S = D
OP C = d
QP S = R S
QP C = r
r2  d D  d 
RS
r
PS

R2 
r 2  RS2 1  S2 
 D 
P CC
S
 X , Y , Z  makes the axis of the cone V parallel to the Z
Consider that a right circular cone V with vertex O  X O , YO , Z O 
in the camera coordinate system XY Z is tangent to the sphere
S. The tangent line to the sphere S is a small circle C of the
sphere S. The circumference of the ellipse on the image is the
projected circumference of the small circle C. The small circle
C is perpendicular to the axis of the cone V. Let r and
PC  X C , YC , Z C  in the camera coordinate system XY Z be the
radius and the centre of the small circle C respectively.
Consider that a point Q  X Q , YQ , Z Q  in the camera coordinate
system XY Z is on the circumference of the small circle C.
The centre PC  X C , YC , Z C  of the small circle C is represented
t
d
 X S YS
D
ZS 
t
(43)
0 sin    X 
  (48)
1
0   Y 
0 cos    Z 
0   cos 
 sin    0
cos     sin 
0 sin  
(49)
1
0 
0 cos  
Then the transformation T and the inverse of the transformation
T are expressed as follows:
(41)
(50)
(51)
Because the transformation T makes the axis of the cone V
parallel to the Z axis, the following equations are obtained:
 X S   b31 D   D cos  sin  
  
 

 YS    b32 D     D sin  
 ZS   b33 D    D cos  cos  
(52)
 X C   b31d   d cos  sin  
  
 

 YC    b32 d    d sin  
 Z C   b33 d    d cos  cos  
(53)
Finally  and  are obtained as follows:
Substituting Equation (43) for Equation (40) gives the
following equation:
XC 
 XS 
2
   RS   


1
Y
 C   D 2   YS 
Z 
 ZC  
 
 S
b13  1
0
b23   0 cos 
b33  0 sin 
b31   X 
 
b32   Y  
b33   Z  
Rearranging Equation (42) gives the following equation:
 R2 
d  D 1  S2 
 D 
 b11 b12
b
 21 b22
b31 b32
 X   b11 b21
  
 Y   b12 b22
 Z  b13 b23
(42)
axis:
Let bij be as follows:
(40)
Because OPSQ and QPS PC are similar, the following
equation is obtained:
RS D  d

D
RS
  cos 
 sin    0
cos     sin 
0
b13   X 
 
b23   Y 
b33   Z 
where
 D  X 2  Y 2  Z 2
S
S
S

2
2
2
 d  X C  YC  Z C
 X  1
0
  
 Y   0 cos 
 Z   0 sin 
 
 X   b11 b12
  
 Y    b21 b22
 Z   b31 b32
 
as follows:
Z C  
(47)
Consider that the following transformation T from  X , Y , Z  to
view vector
Figure 2. Sphere S and its small circle C
 X C YC
(46)
Substituting Equation (42) for Equation (46) gives the
following equation:
V
Q
(45)
(44)




YS
YC
1 
   tan 1 
   tan 
2
2
2
2


X
Z
X
S
S 
C  ZC





1  X S 
1  X C 
    tan 
   tan 

 ZS 
 ZC 

This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016



(54)
23
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
Consider that the point Q  X Q , YQ , Z Q  on the circumference of
the small circle C is projected to q  xQ , yQ  on the image. Then
the following equation is obtained:
 xQ
yQ
t
 f   kQ  X Q
YQ
Z Q 
t
(55)
where

c11 xQ  c12 yQ  c13 f
 d 
c31 xQ  c32 yQ  c33 f





c x c y c f
  d  21 Q 22 Q 23
c31 xQ  c32 yQ  c33 f

2
(64)
2

2
 r

x
By rearrange Equation (64) as to
kQ  
f
ZQ
(56)
 xQ 
 a11
 

 yQ   kQ  a12
 f 
 a13
 
a21
a22
a23
a31   b11
a32  b12
a33  b13
b21
b22
b23
b31   X Q 
 
b32   YQ 
b33   Z Q 
(57)
Here let cij be as follows:
c13   b11 b12
c23   b21 b22
c33  b31 b32
b13   a11
b23   a21
b33   a31
a12
a22
a32
a13 
a23 
a33 
(58)
Substituting Equation (58) for Equation (57) gives the
following equation:
 X Q 
 c11
  1 
 YQ   k  c21
Q
 Z Q 
 c31
 
c12
c22
c32
c13   xQ 
 
c23   yQ 
c33    f 
(59)
where
kQ  
, yQ , f  the following
quadratic equation is obtained:
pxx xQ2  p yy yQ2  p ff f 2
Substituting Equation (50) and Equation (51) for Equation (55)
gives the following equation:
 c11 c12
c
 21 c22
 c31 c32
Q
c x c y c f
f
 31 Q 32 Q 33
c13 X Q  c23YQ  c33 Z Q
Z Q
(60)
Substituting Equation (60) for Equation (59) gives the
following equation:
 X Q 
Z Q
 c11 xQ  c12 yQ  c13 f  (61)
 
c x  c y  c f 

 YQ  c31 xQ  c32 yQ  c33 f  21 Q 22 Q 23 
The small circle C transformed by the transformation T is a
circle with radius r and centre PC (0, 0, −d) in the coordinate
system X Y Z  , and is perpendicular to the Z  axis.
Accordingly the small circle C in the coordinate system
X Y Z  is represented as follows:
 X 2  Y 2  r 2

 Z   d
(62)
Because the point Q  X Q , YQ , Z Q  is on the circumference of the
small circle C, the following equation is obtained:
2
2
2
 X Q  YQ  r

 Z Q  d
(63)
(65)
2 pxy xQ yQ  2 pxf xQ f  2 p yf yQ f  0
where pxx , p yy , p ff , p xy , p xf and p yf are as follows:
 pxx

 p yy

 p ff


 pxy

 pxf

 p yf
2
  d 2  c31
d 2  r2 
2
  d 2  c32
d 2  r2 
(66)
2
  d 2  c33
d 2  r2 
 c31c32  d 2  r 2 
  c31c33  d 2  r 2 
  c32 c33  d 2  r 2 
Here substituting Equations (58), (53) and (41) for Equation
(66) and rearranging gives the following equation:

 pxx



 p yy



 p ff



 pxy


p
 xf


p
 yf


Y
 Z C2  X C2  YC2  Z C2   r 2 X C2
2
C
 X Y  Z 
 Z  X  X  Y  Z   r Y

 X Y  Z 
 X  Y  X  Y  Z   r Z

 X Y  Z 
X Y  X Y  Z  r X Y

 X Y  Z 
Z X X Y  Z  r Z X

X Y  Z 
Y Z X Y  Z  r Y Z

X Y  Z 
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
2
C
C
C
2
C
2
C
2
C
2
C
C
C
2
C
2
C
2
C
2
2
C
2
2
C
2
C
C
2
C
2
C
2
C
(67)
C C
2
C
2
C
2
C
2
C
2
2
C
2
C
2
C
2
C
2
C
2
C
2
C
C C
2
C
2
C
C
2
C
Equation (65) shows an ellipse as which the sphere S with
radius RS and centre PS (XS, YS, ZS) is projected onto the image.
The centre p E  xE , yE  , the semi-major axis a, the semi-minor
axis b and the rotation angle  of the major axis of the projected
ellipse on the image take the following forms:
p yy p xf  pxy p yf

 xE   f
pxx p yy  p xy2


 y   f pxx p yf  p xy pxf
 E
pxx p yy  p xy2

(68)
Substituting Equation (61) for Equation (63) gives the
following equation:
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
24
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic

a  

 pxx  pyy  


b  

 pxx  pyy  


2w
p
 p yy   4 pxy2
2
xx
(69)
2w
Moreover substitution Equation (75) for Equation (46) and
rearranging gives the following equation:
 pxx  p yy   4 pxy2
2
2 p xy
1
tan 1
p xx  p yy
2
(70)
where w is as follows:
2
E
2
(71)
Hence the eccentricity (x, y) of the sphere S with radius RS
and centre PS (XS, YS, ZS) can be expressed as follows:

 p yy p xf  p xy p yf X S 

 x  xE  xS    f  

2
Z S 

 p xx p yy  p xy

 pxx p yf  pxy pxf YS 

 y  yE  yS    f   p p  p 2  Z 
 xx yy
xy
S


(72)
Finally Equations (72), (67), (47), (41), (44), (4), and (1)
provide the eccentricity (x, y) of the sphere S with radius RS
and centre PS (XS, YS, ZS).
4.2 From an Ellipse Projected onto an Image
E : p xx xQ2  p yy yQ2  p ff f 2  2 p xy xQ yQ  2 p xf xQ f  2 p yf yQ f  0
(73)
If we can derive  X C , YC , Z C  from pxx , p yy , p ff , pxy , pxf and
p yf by using Equation (67), we can calculate the eccentricity
(x, y) of the sphere S from the projected ellipse E by using
Equation (72). Fortunately  X C , YC , Z C  is able to be derived
from pxx , p yy , p ff , pxy , pxf and p yf analytically. Since
, YC , Z C  and r are linearly dependent,  X C , YC , Z C  and r
are represented using a scale factor s as follows:
2


pxf p yf 
 Z   s 1   pxx  p yy  p ff  pxf p yf 



C
2
2
2

pxy 
3  p p    p p    p p 
xy yf
xf
yf

 xy xf


pxy

ZC
 X C  s
p yf


p
 YC   s xy Z C
pxf



(74)
2
(75)
r  2d   p  p  p 
xx
yy
ff
where
d 2  X C2  YC2  Z C2
(76)
d
2d   p
xx
 p yy  p ff
3d   p
2
d
xx
 p yy  p ff

2
(77)
Then the centre PS  X S , YS , ZS  in the camera coordinate system
XY Z of the sphere S is obtained as follows:
 XS 
XC 
  D 
 YS   d  YC 
 ZS 
 Z C 
(78)
Equations (76), (77) and (78) contain the scale factor s. If the
radius RS of the sphere S is known, we can estimate the scale
factor s and finally we can know the centre PS(XS, YS, ZS) of the
sphere S.
5. TRIAL CALCULATIONS OF ECCENTRICITIES
From now on we are going to derive an eccentricity of a sphere
from an ellipse projected onto an image. First consider that a
projected image of a sphere S onto an image is an ellipse E
represented by the following equation:
C
3d 2   pxx  p yy  p ff 
2
RS 
2 pxy xE yE  2 pxf xE f  2 p yf yE f
X
D
The radius RS of the sphere S can be obtained as follows:
w  p x  p yy y  p ff f
2
xx E
Now we can calculate an eccentricity (x, y) of the sphere S
from the projected ellipse E by substituting Equation (74) for
Equation (72).
We conducted trial calculations of eccentricities of circular
targets and spherical targets mostly according to Luhmann’s
simulation (2014). Since some errata were found in his paper,
we modified some of imaging conditions. Therefore obtained
results in the study were not necessary the same as Luhmann’s
simulation results.
5.1 Imaging Scenario
The assumptions of an imaging scenario by a digital single-lens
reflex (DSLR) camera are as follows:
(1) Camera
a) Image size: 4288 x 2848 pixels
b) Unit pixel size: 5.5μm
c) f = 16mm
d) (XO, YO, ZO) = (0 mm, 0 mm, 0 mm)
e) (, , ) = (0, 0, 0) and  varies from -35 to +35
(2) Targets
a) RC and RS vary from 1 mm to 10 mm.
b) A circular target and a spherical target are located at P4
(0 mm, 0 mm, -500 mm).
c) A circular target and a spherical target are located at P6
(360 mm, 240 mm, -600 mm).
d) The circular targets lie on a plane tilted relative to the
image plane by angles = 20° and  =10°.
Figures 3 and 4 show the semi-major axes a and the semi-minor
axes b of ellipses on the acquired images. Figure 3 shows a and
b against view angles  of the camera where the circular target
with RC = 5mm and the spherical target with RS = 5mm are
located at P4. Figure 4 shows a and b against radii RC, RS of the
circular target and the spherical target located at P6 where  =
0°.
(41)
This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
25
ISPRS Annals of the Photogrammetry, Remote Sensing and Spatial Information Sciences, Volume III-5, 2016
XXIII ISPRS Congress, 12–19 July 2016, Prague, Czech Republic
60
40
30
20
10
0
-40
sphere a
circle a
sphere b
circle b
-20
0
20
40
 
a / b (pixel)
a / b (pixel)
50
Figure 3. Semi-major axis a
and semi-minor axis b
vs. view angle 
where RC = RS = 5mm
40
resolution of a digital camera becomes higher, the influence of
eccentricities becomes larger. Finally the correction of
eccentricities would become unavoidable.
circle a
circle b
sphere a
sphere b
6. CONCLUSIONS
20
0
0
2
6
8
4
R c / R s (mm)
10
Figure 4. Semi-major axis a
and semi-minor axis b
vs. radius RC, RS
where  = 0°
5.2 Calculated Eccentricities
eccentricity (pixel)
Figure 5 shows calculated eccentricities (x, y) against view
angles  of the camera where the circular target and the
spherical target are located at P4 and RC = RS = 5mm. Figure 6
shows calculated eccentricities (x, y) against radii RC, RS of
the circular target and the spherical target located at P6 where 
= 0°.
0.4
circle y
0.3
sphere y
0.2
0.1
0.0
-0.1
-0.2
circle x
-0.3
sphere x
-0.4
-40 -30 -20 -10 0 10 20 30 40
view angle (°)
Figure 5. Eccentricity vs. view angle at P4
where RC = RS = 5 mm
eccentricity (pixel)
0.5
circle x
circle y
sphere x
sphere y
0.4
0.3
0.1
0.0
0
2
6
8
4
radius (mm)
Moreover the general formulae calculating an eccentricity of a
sphere by using the focal length, the position, the attitude of the
camera, and the radius, the location of a sphere was derived. As
for a sphere the paper shows general formulae to calculate the
eccentricity of a sphere from the equation of the projected
ellipse of the sphere on an image.
REFERENCES
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Aberration Compensation for Industrial Vision Metrology,
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0.2
-0.1
The general formulae calculating an eccentricity of a circle by
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The paper shows a numerical method to estimate the
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of the circle on an image as well.
10
Figure 6. Eccentricity vs. radius at P6 where  = 0°
5.3 Discussion on Eccentricities on Digital Camera Images
Figures 5 and 6 indicate that an eccentricity on an image
acquired by the DSLR camera reaches almost a half pixel,
which is considerably larger than the measurement accuracy of
the centre of an ellipse on image (Luhmann et al., 2006b).
Although the obtained formulae demonstrate that a smaller
target produces a smaller eccentricity, a smaller target reduces
the measurement accuracy of the centre of an ellipse on image
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effect of eccentricities on an image acquired by a digital camera
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This contribution has been peer-reviewed. The double-blind peer-review was conducted on the basis of the full paper.
doi:10.5194/isprsannals-III-5-19-2016
26