Multiple View Geometry: Solution Sheet 4
Prof. Dr. Florian Bernard, Florian Hofherr, Tarun Yenamandra
Computer Vision Group, TU Munich
Link Zoom Room , Password: 307238
Exercise: May 14th, 2020
Part I: Theory
1. Image Formation
(a) Compute λ and show that (2) is equivalent to
u =
fX
Z
+ ox , v =
fY
Z
+ oy .
Performing the matrix multiplication in (2), one obtains
λuλv
λ
=
fX + oxZfY + oyZ
Z
From the third row, it directly follows that λ = Z. Substituting Z for λ and dividing the
equation by Z, one immediately obtains the result.
(b) A classic ambiguity of the perspective projection is that one cannot tell an object from
another object that is exactly twice as big but twice as far. Explain why this is true.
Let X̃1 = (X1 Y1 Z1)> be a point on the smaller object and X̃2 = (X2 Y2 Z2)> a point
on the larger object. Since X̃2 is twice as far away, we have Z2 = 2Z1, and since it is
twice as big we have X2 = 2X1 and Y2 = 2Y1. Thus,
u2 =
fX2
Z2
+ ox =
2fX1
2Z1
+ ox =
fX1
Z1
+ ox = u1
and analogous for v2 = v1.
(c) For a camera with f = 540, ox = 320 and oy = 240, compute the pixel coordinates u and
v of a point X̃ = (60 100 180)>.
u =
fX
Z
+ ox =
540 · 60
180
+ 320 = 500
v =
fY
Z
+ oy =
540 · 100
180
+ 240 = 540
Explain with the help of (b) why the units of X̃ are not needed for this task.
Using different units (mm, cm, m, etc.) can be interpreted as scaling the point coordinates
by a constant factor (10, 100, …). The argument of (b) for a factor of 2 can easily be
generalized to any factor α.
Will the projected point be in the image if it has dimensions 640× 480?
No, the point (u, v) = (500, 540) is not in [−0.5, 639.5]× [−0.5, 479.5].
1
https://tum-conf.zoom.us/s/62772800235?pwd=SUpZN2QrV0JpeXJyR2R1TWx5cHEwdz09
(d) Using the generic projection π, show that (3) — and therefore also (1) and (2) — is equiv-
alent to
uv
1
= K (π(X̃)
1
)
.
Insert in the RHS of the equation:
uv
1
= K (π(X̃)
1
)
=
f 0 ox0 f oy
0 0 1
X/ZY/Z
1
=
fX/Z + oxfY /Z + oy
1
2. Radial Distortion
(a) Can this model be used for lenses with a field of view of more than 180◦?
No, it can only model points for which the viewing ray intersects the image plane. Since
points are first projected on the canonical iamge plane with π(X̃) there is a singularity at
Z = 0 (which is 180◦).
Note: It is possible to rewrite the FOV model to avoid the division by Z and apply it to
lenses with more than 180◦ FOV.
(b) Derive a closed form solution for f in the undistortion formula
π(X̃) = f
(
‖πd(X̃)‖
)
· πd(X̃)
using (6) and g(r) = gATAN(r).
Define r := ‖π(X̃)‖ and rd := ‖πd(X̃)‖. The norms of (9) and (6) are:
r = f(rd)rd and rd = g(r)r
Inserting g = gATAN yields
rd =
1
ωr
arctan
(
2r tan
(ω
2
))
r =
1
ω
arctan
(
2r tan
(ω
2
))
⇒ tan(rdω) = 2r tan
(ω
2
)
⇒ r =
tan(rdω)
2 tan
(
ω
2
) = f(rd)rd ⇒ f(rd) = tan(rdω)
2rd tan
(
ω
2
)
2