1
S1 Text. Appendix.
2
A rotation by an angle  about an axis defined by the unit vector k  (k x , k y , k z ) is geometrically
3
described by the rotation matrix [1]
4
 cos   (1  cos 2  )k x2

Rk (  )   (1  cos  )k x k y  k z sin 
 (1  cos  )k x k z  k y sin 

(1  cos  ) k x k y  k z sin 
cos   (1  cos 2  ) k y2
(1  cos  )k y k z  k x sin 
(1  cos  ) k x k z  k y sin  

(1  cos  ) k y k z  k x sin  
cos   (1  cos 2  ) k z2 
5
Let the 0xyz reference frame rotate with the camera. The optical axis defines the y-axis, with the
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7
8
camera at the origin. The pan axis is vertical and fixed. The x-axis coincides with the tilt axis, which
is horizontal, pans with the camera and is always perpendicular to the camera's optical axis. The
camera is horizontal at the starting position, as depicted in Fig. 2 in the main text.
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10
Consider the case where the camera is vertically tilted by an angle   (the minus sign means a
downward tilt) and simultaneously rotated by a pan angle  . From the point of view of the camera its
11
orientation is described by a rotation Ru ( ) followed by a rotation RRu (  ) v ( ) . The unit vector
12
u  (1,0,0) defines the tilt axis, v  (0,0,1) is the z-axis unit vector and the unit vector Ru ( ) v
13
14
defines the vertical direction, i.e. the pan axis, after the tilt rotation. For example, point A, on the
seafloor, has initial coordinates (l1 , y A0 , h) ; after the camera is rotated its coordinates become
15
16
17
sin 
 xA 
 l1   cos 
 

 
 y A   RRu (  ) v ( ) Ru ( )  y A0     sin  cos  cos  cos 
z 
 h    sin  sin  cos  sin 
 A

 
0   l1 
 

 sin     y A0 
cos    h 
Proceeding analogously for points B, C and D, after the complete rotation the coordinates of the four
seafloor points become
 x A  l1 cos   y A0 sin 

 y A  l1 cos  sin   y A0 cos  cos   h sin 
 z  l sin  sin   y sin  cos   h cos 
1
A0
 A
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 xB  l1 cos   ( y A0  L1 )sin 

 yB  l1 cos  sin   ( y A0  L1 ) cos  cos   h sin 
 z  l sin  sin   ( y  L )sin  cos   h cos 
1
A0
1
 B
 xC  l2 cos   yC 0 sin 

 yC  l2 cos  sin   yC 0 cos  cos   h sin 
 z  l sin  sin   y sin  cos   h cos 
C0
 C 2
 xD  l2 cos   ( yC 0  L2 )sin 

 yD  l2 cos  sin   ( yC 0  L2 ) cos  cos   h sin 
 z  l sin  sin   ( y  L )sin  cos   h cos 
C0
2
 D 2
1
(A1)
19
Relative to equation (1) in the main text, equation (A1) accounts for the more general case where y A 0
20
may not be equal to yC 0 – both coordinates are equal to f  d in (1). The distances AB  L1 and
21
CD  L2 are also not necessarily the same. Since the image plane I remains at y  f , the coordinates
for the projected points on the picture are obtained in a similar fashion as for equation (2) in the main
text:
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23
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
 x A 


z 
 A

 xB 


z 
 B

 xC 


z 
 C

 xD 


z 
 D
f
l1 cos   y A0 sin 
l1 cos  sin   y A0 cos  cos   h sin 
f
l1 sin  sin   y A0 sin  cos   h cos 
l1 cos  sin   y A0 cos  cos   h sin 
f
l1 cos   ( y A0  L1 ) sin 
l1 cos  sin   ( y A0  L1 ) cos  cos   h sin 
f
l1 sin  sin   ( y A0  L1 ) sin  cos   h cos 
l1 cos  sin   ( y A0  L1 ) cos  cos   h sin 
f
l2 cos   yC 0 sin 
l2 cos  sin   yC 0 cos  cos   h sin 
f
l2 sin  sin   yC 0 sin  cos   h cos 
l2 cos  sin   yC 0 cos  cos   h sin 
f
l2 cos   ( yC 0  L2 ) sin 
l2 cos  sin   ( yC 0  L2 ) cos  cos   h sin 
f
l2 sin  sin   ( yC 0  L2 ) sin  cos   h cos 
l2 cos  sin   ( yC 0  L2 ) cos  cos   h sin 
(A2)
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The camera’s rotation moves the vanishing point of the projected laser lines relative to the image
center to a new position ( xV  , zV  ) , as depicted in Fig. 4a in the main text. The vanishing point V’ is by
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definition the intersection point of the straight lines [A’B’] and [C’D’], described respectively by
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z  ( x  xA ) / cot 1  z A
z  ( x  xC  ) / cot  2  zC 
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where the “perspective” angles 1 and  2 (see Fig. 4a in the main text) are obtained by generalizing
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equation (3) in the main text to the present case,
cot 1  
x A  xB l1 cos   h sin  sin 

z A  z B
h cos 
cot  2  
xD  xC  l2 cos   h sin  sin 

z D  zC 
h cos 
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(A3)
(A4)
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From (A3) it follows that ( xV   xA ) / cot 1  z A  ( xV   xC  ) / cot  2  zC  applies at V’. Solving for
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xV  and using (A2) together with (A4), it results that
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xV   f
2
tan 
cos 
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From equation (11) in the main text, inserting f  W / 2 tan  H / 2  into the last equation yields
36
equation (15) in the main text:
xV  
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zV  
W
 tan 
2 tan( H / 2)
(A6)
Next, by adding the two equations in (A4) and solving for h , the general expression for the camera's
height relative the ground, equation (22) in the main text, is obtained:
h
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(A5)
The z-coordinate of the vanishing point is obtained by inserting (A5) into either one of the equations
in (A3):
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W
tan 

2 tan( H / 2) cos 
cos 
1
 cot 1  cot  2  l
cos 
If z A  zC  applies (as in Fig. 1c in the main text), it follows from (A2) that
y A0  yC 0  l tan 
(A7)
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L1  L2
(A8)
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Thus, let L  L1  L2 . From equations (A7) and (A8) follows that the trapezoid [A’B’D’C’] in Fig. 1c
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corresponds to a parallelogram [ABDC] on the seabed with sides l / cos  and L (see Fig. 4b in the
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main text) and inner angles 90º  . Accordingly, the area of [ABDC] is S  L  l .
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As for the special case of equation (8) in the main text, by similarity of triangles [A0C] and [A’0’C’]
it follows that
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l / cos  y A

l
f
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Analogously, if z B   z D then
(A9)
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Note that f  d in equation (8) corresponds to y A in (A9) and l is exchanged for l / cos  .
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From (A1) yB  y A  L cos  cos  applies, thus, by similarity of triangles [B0D] and [B’0’D’], the
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general case of equation (9) is
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l / cos  y A  L cos  cos 

l 
f
Combining (A9) and (A10) with (11) yields equation (17) in the main text:
l
 l  W
L    1  
 l   l  2 tan( H / 2)cos  cos 
3
(A10)
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In many cases the vanishing point V’ lies outside the picture, as in Fig. 1c in the main text. In such a
situation, the coordinates xV  and zV  must be calculated, so that the cosines of the tilt and pan angles
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63
may be used in equation (19) in the main text. Therefore, consider the case depicted in Fig. 4a in the
main text. At the vanishing point V’ equation (A3) becomes
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xA  xV   ( zV   z A )cot 1
xV   xC   ( zV   zC  )cot  2
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Adding the two equalities in (A11) and solving for zV  yields
zV   z A 
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l
d
s1  s2
zV   z A 
(A14)
Let H be the picture's height. Analogously to equation (11) in the main text, H obeys
f 
H
2 tan(V / 2)
Thereby,
W
H

2 tan( H / 2) 2 tan(V / 2)
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(A13)
it follows that (A12) is equivalent to
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(A12)
cot 1  s1 / d
cot  2  s2 / d
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l
cot 1  cot 2
Since (see Fig. 4a in the main text)
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(A11)
Inserting this result into equation (A6) and solving for cos  yields
2
 
 
zV 
cos   1  
 
  H / 2 tan(V / 2)  
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1/2
(A15)
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Let M be the vertical size of the picture in pixels, so that H / M is the height per pixel. Hence,
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d  mH / M
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l  /  s1  s2   n /  n1  n2  , equation (A14) becomes
and
z A    M / 2  m0  H / M
(see
Fig.
4a
in
the
main
text).
Since
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zV  m0 1
n
m

 

H M 2 n1  n2 M
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Inserting this equality into (A15), the cosine of the tilt angle is expressed in terms of pixel quantities
directly measurable from the picture,
4
2
  m 1
 
n
m
0
cos   1  
 
   2 tan(V / 2)  
  M 2 n1  n2 M 
 
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2
 x
 
cos   1   V   2 tan( H / 2)cos   
 
  W
xV   x A 
cot 1
cot 1  cot 2
xV   xA 
s1
l
s1  s2
Using quantities in pixels (see Fig. 4a in the main text), this last equation becomes
xV  1 n0
n1
n
  

W 2 N n1  n2 N
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(A17)
which, by the use of (A13), is equivalent to
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90
1/ 2
Subtracting the two equalities in equation (A11) and solving for xV ' , it follows that
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(A16)
which is equation (20) in the main text. Similarly, cos  may be obtained from equation (A5):
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1/2
Substituting xV  / W into (A17), equation (21) in the main text is obtained:
2

 
n1
n 
  1 n0

cos   1    
   2 tan  H / 2  cos   
2 N n1  n2 N 
 

 

93
1/2
(A18)
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Note that the procedure for obtaining cos  and cos  (equations (A16) and (A18), respectively)
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remains valid when V’ is inside the image.
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References
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1. Elliott JP, Dawber PG. Symmetry in Physics. Vol.1: Principles and Simple Applications.
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MacMillan; 1990.
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5