ISPRS Commission III, Vol.34, Part 3A „Photogrammetric Computer Vision“, Graz, 2002
From the single view observation ofthe cylinder, (19, no, sin(% $ Whe depth of the pipes can also be recovered directly: 5; =
we can deduce:
IM x lo — Aonol| — ho sin(5) (20)
This results in a second degree equation in ho:
(1—sin* 2 3) hà—2 n5(Mxlo) ho4-|[M x1o| 2 0 (21)
Solving this equation results in two possible values for ho.
The larger one is the correct depth. This is because the
point is visible in the image. The smaller solution would
correspond to a point on the other side of a transparent
cylinder.
Iftwo 3D points are measured on the cylinder, we can com-
pute the depth of cylinder without using the viewing angle
a. This could result in a more accurate estimate since the
viewing angle is also sensitive to noise. If we have two
points M, and M», ho will be defined uniquely and ex-
plicitly as:
IM x Io]? — ||M; x lof?
Ds 22
o3 (n5 (M; x 1o) — n$(M3 x 19)) ad
4 CYLINDER BASED POSE ESTIMATION
In the case we have a single view of a set of known cylin-
ders, the canonical representation makes it easy to formu-
late the pose estimation problem. Let us suppose that we
observe a few cylinders in the scene. We represent them in
the world coordinate system by (1;, N;, rj). Their images
could be respectively represented by: (12, n;, sin( m). We
have:
R03 Ou 1; I
E Ri 01 Nia = hin;
0153 0153 1 Tj hj sin(5&)
(23)
The unknowns are the motion parameters (R, T), and the
depth of different cylinders ^;. We could solve this in two
consecutive steps. First we could recover the camera ori-
entation by solving:
RL =I (24)
The optimum estimation of orientation is the unit quater-
nion, which is the eigenvector of the following 4 x 4 matrix
Q associated to its largest eigenvalue (Horn, 1987, Besl
and McKay, 1992):
Ei
Q= “la ;x Ip
We need at least two independent, i.e, non parallel cylin-
ders, 3D/2D cylinder correspondences in order to recover
the orientation of the camera. The more correspondences
we have, the more accurate the result will be.
1; X If
LIE + 11 es)
A - 222
sin( AL &-. The translation can then be estimated by EE.
>
T x RI; an h;Rnj N; (26)
This provides two equations per cylinder correspondences.
We therefore need two independent pairs of cylinder cor-
respondences to recover all the pose parameters. Note that
we can also solve for h; and T; simultaneously by solving
one system of linear equations combining the last two con-
straints. This would provide more accurate pose estimates.
5 CYLINDER BASED STRUCTURE FROM MOTION
In this section we try to recover motion and structure of
cylinders from two or more views. In the case two views
are available, the equations are quite similar to the pose es-
timation case. The only difference is that in the first frame
the pipe is defined up to one parameter, i.e, depth or radius.
The main constraint is the following:
lo R 053 05, li
hono = E R 03.1 hi In
ho sin() 0153 0153 1 hi sin( 94
(27)
The relative orientation can be computed in the same way
as in the pose estimation case. Once again we need at least
two indepeaden cylinders viewed in two images. Replac-
ing hy by ho = eS , we have:
sin(%)
TxRlı+h =
gt ol n(%)
This equation is a homogeneous one. It is known that in
structure from motion shape or translational motion can
be only recovered up to scale. Once again we need two
independent cylinders in order to recover the translational
motion up to scale.
Rn; — no) = 0 (28)
In the same way if we have three views, we can write the
following constraints:
Ri a 0
BY
n(%) :
T x Rlı +4o( 2 Rn; — no) zu) (31)
sin( 5)
ao
U x SI + en — no) = 0)
sin(72)
The Ao can be eliminated from the last two constraints.
This leaves us with a set of equations defining the rela-
tionship between the three perspective views. One could
estimate the motion parameters followed by the depth of
cylinders using the correspondences between two indepen-
dent cylinders.
6 CYLINDERS AND TRILINEAR TENSORS
In this section we present a new formulation for the pro-
jective geometry of cylinders in three views. This also
