224
The second one is more complicate but numerically more
accurate. In such a method an object with a side originally
parallel to a principal axis is transformed in the required
position. A simple perspective transformation is then applied.
The final concatenate matrix is then used to transform the
points to the infinity on the principal axis. The co-ordinates
resulting are those of the vanishing points.
[PF]-[T]
10 0 0
0 10 0
0 0 10
0 0 0 0
[T]
(60)
where:
7”n
7* 12
7Ì3
7*14
T' 21
7*22
T' 23
7*24
T' 31
T' 32
r 33
7*34
7**41
T' 42
7*43
7*'44
The results can be projected on a plane, for example z = 0,
using the following matrix:
T"=
10 0 0
0 10 0
0 0 0 0
0 0 0 1
(65)
17. Photographs and perspective transformations
A photograph is a perspective projection (we neglect the
deformations due to the camera and to the film). The
projection centre is the focal point of the lenses of the
camera. It is convenient to consider the creation of the
original photographic negative and of the print of the
negative as two separated cases. For convenience to negative
z = 0 is assigned with the projection centre and the scene in
[T n ] =
supplies the transformation:
lOO 0 "
0 10 0
0 0 0 1//
0 0 0 1
where / is the focal lenght. Note that the image on the
negative is inverted. In phase of print, the negative is
projected agam, and if s is the distance between the focal
point of the lenses on the enlarger and the sheet of
photographic paper, the transformation is:
The concatenation of the two matrices produces:
T = [T"]-[T] =
The transformation can be written as:
Tu
712
0
7*14
7*21
7*22
0
7*24
7*31
7*32
0
7*34
7*41
7*42
0
7*44.
(66)
e z. The perspective projection
- ,
,
0
.
where the negative is placed)
7*11
T \2
7*14
[x;y;z; !]•
7*21
7*22
0
7*24
Tii
7*32
0
7*34
7*41
7*42
0
7*44
= [x;y ;0;/z] =
(67)
Note that the x
= h[x ;y ;0;1]
and y are the co-ordinates of
the perspective projection on the plane z = 0. In a completely
similar manner we can proceed for the planes x = 0 and y = 0.
By explicating the equations before written in a matricial
form, we get:
[T p \ =
10 0 0
0 10 0
0 0 0 -1/s
0 0 0 1
(62)
The projection operated by the enlarger rectifies again the
image.
T\\X + T 2 ^y + 731 z + 741 = hx
7j 2 x + T 22 y + T 22 z + T 42 = hy (68)
Tj 4 x + T 24 y + T M z + T 44 =h
By substituting the third one in the first two, we get two
equations:
(Til - T u x*)x + (T 2l -T 24 x*)y +
18. Reconstruction of three-dimensional images
While the techniques of reconstruction of a three-dimensional
object from orthogonal projections are well known,
reconstruct from perspective views, like the photographs, is a
less common problem. As we will see the suitable techniques
to reconstruction of an object from a perspective projection
are the general cases of those used in the simpler
orthographic projections. Before to face the general problem,
let's see the particular and very simple case of reconstruction
of a point from the orthogonal projections. Let’s consider the
views high, front, right as in the figures:
Any projection brings the following information:
high
front
right
*■ *
▼ ?
(a K')
X
The problem is determining the three unknowns of the pomt.
The problem is iperdetermined, having six equations. Let’s
come back now r to the problem of the perspective projection,
and remind that in general such a transformation and
presentable with a 4x4 matrix. So:
[x;y;z;l\[T\=[x’;y’;z ;h] (63)
+ (T31 ~ 7’ 34 x*)z + (T 41 - 7’ 44 x*) = 0
(69)
(7*12 ~T\ 4 y )x + (T 22 -T 24 y )y +
+ (732 ~ 7/4 y )z + (T 42 - T u y ) = 0
This couple of equations can be used in three different
modes. The first one assumes T, x andy known; we have two
equations in the unknowns X and y . The problem is
solvable and supplies the solution of the perspective
projection. The way is to use the two equations and to assume
*
v *
T and y to be known. We have two equations in the
three unknowns x, y, z. The problem is not solvable. At least
two projections are needed, then four equations, two for any
projection:
