Roto-translations
Plane roto-translation
y
e2
y’
e '2
P
e '1
x’
ω
O’
O
x
e1
Figure 1
Consider a plane body at two different epochs. The position of a point on the body is given with respect to a
cartesian system of coordinates oxy by its coordinates P=P(x,y). Assume that between the two epochs the
body has been rotated with respect to the point O and then translated.
We want to express the coordinates of a fixed point on the body after the roto-translation with respect to
same reference system as a function of its coordinates at the initial epoch (in other words we want to find
the displacement field).
To this aim (see Figure 1), we consider the two following reference systems: o’x’y’ attached to the body
while the body is moving, and the original fixed system oxy coincident with o’x’y’ before the movement.
Let us call e1 , e2 the unit vectors associated to the fixed reference system and e1′ , e2′ the vectors associated
to the moving system.
Obviously, the coordinates of the point P after applying a roto-translation do not change with respect to the
system o’x’y’ but change with respect to the fixed one.
The displacement field can be retrieved in the following way (see Figure 1):
OP = OO ' + O 'P = x O ' e1 + y O ' e2 + x P′ e1′ + y P′ e2′
x P = OP ⋅ e1 = x O ' e1 ⋅ e1 + y O ' e2 ⋅ e1 + x P′ e1′ ⋅ e1 + y P′ e2′ ⋅ e1 = x O ' + x P′ ( e1 ⋅ e1′ ) + y P′ ( ⋅e1 ⋅ e2′ )
y P = OP ⋅ e2 = x O ' e1 ⋅ e2 + y O ' e2 ⋅ e2 + x P′ e1′ ⋅ e2 + y P′ e2′ ⋅ e2 = y O ' + x P′ ( e2 ⋅ e1′ ) + y P′ ( e2 ⋅ e2′ )
⎛ x P ⎞ ⎛ x O ' ⎞ ⎛ ( e1 ⋅ e1′ )
⎜ ⎟=⎜
⎟ + ⎜⎜
⎝ y P ⎠ ⎝ y O ' ⎠ ⎝ ( e2 ⋅ e1′ )
(⋅e1 ⋅ e2′ ) ⎞ ⎛ xP′ ⎞
⎟
( e2 ⋅ e2′ ) ⎟⎠ ⎜⎝ yP′ ⎟⎠
1
x P = x O ' + R x P′ = t + R x ′
the translation vector t contains the coordinates of the origin of the moving system with respect to the fixed
one:
⎛x ⎞
t = xO' = ⎜ O' ⎟
⎝ yO' ⎠
R is the rotation matrix; it depends on the rotation angle ω.
⎛ ( e ⋅ e′ )
R = ⎜⎜ 1 1
⎝ ( e2 ⋅ e1′ )
⎛
⎛π
⎞⎞
cos ω
cos ⎜ + ω ⎟ ⎟
⎜
( e1 ⋅ e2′ ) ⎞ = ⎜
⎝2
⎠ ⎟ ⎛ cos ω − sin ω ⎞
=
⎟⎟ ⎜
⎟ ⎜⎝ sin ω cos ω ⎟⎠
( e2 ⋅ e2′ ) ⎠ cos ⎛ π − ω ⎞
cos ω ⎟⎟
⎜⎜
⎜2
⎟
⎝
⎠
⎝
⎠
Note that anticlockwise rotations are positive, according to mathematical standards.
Rotation matrix properties
1. R is orthogonal:
R +R = I = RR +
2. R is regular
R λ = 0 ⇒ λ = 0 ⇒ ∃ R −1
3. The inverse of R is equal to its transpose
R −1 = R +
1.
⎣⎡R ⎦⎤ij = ei ⋅ e ′j
⎡⎣R + ⎤⎦ = ⎣⎡R ⎦⎤ ji = e j ⋅ ei′ = ei′ ⋅ e j
ij
⎡⎣R +R ⎤⎦ = ∑ k ⎡⎣R + ⎤⎦ ⎡⎣R ⎤⎦kj = ∑ k ( ei′ ⋅ ek ) ( ek ⋅ e ′j ) = ei′ ⋅ ∑ k ek ( ek ⋅ e ′j ) = ei′ ⋅ e ′j
ij
ik
i= j
⎧1
ei′ ⋅ e ′j = δij = ⎨
⎩0
i≠ j
or,
⎡⎣RR + ⎤⎦ = ∑ k ⎡⎣R ⎤⎦ik ⎡⎣R + ⎤⎦ = ∑ k ( ei ⋅ ek′ ) ( ek′ ⋅ e j ) = ei ∑ k ek′ ( ek′ ⋅ e j ) = ei′ ⋅ e ′j
ij
kj
i= j
⎧1
ei ⋅ e j = δij = ⎨
⎩0
i≠ j
2.
Rλ = 0 ⇒ (Rλ ) (Rλ ) = 0
+
(Rλ ) (Rλ ) = λ +R + (Rλ ) = λ + (R +R ) λ = λ + λ = 0
+
λ+λ = 0 ⇒ λ = 0
3.
RR + = I ⇒ R −1 RR + = R −1I ⇒ R −1R R + = IR + ⇒ R −1 = R +
(
)
(
)
From the rotation matrix properties it follows that
2
⎛ x P′ ⎞ ⎛ ( e1′ ⋅ e1 ⋅)
⎜ ′ ⎟ = ⎜⎜ ′
⎝ y P ⎠ ⎝ ( e2 ⋅ e1 )
( e1′ ⋅ e2 ) ⎞ ⎛ xP − x O ' ⎞ = ⎛ cos ω
⎟
( e2′ ⋅ e2 ) ⎟⎠ ⎝⎜ yP − y O ' ⎠⎟ ⎜⎝ − sin ω
sin ω ⎞ ⎛ x P − x O ' ⎞
⎟
⎟⎜
cos ω ⎠ ⎝ y P − y O ' ⎠
R + ( x P − x O ' ) = x P′
R + x P − R + x O ' = x P′
x P′ = −R + x O ' + R + x P = t ′ + R + x P
which can be seen as the formula to get back to the original position from the displaced one.
Note that if the opposite rotation −ω is applied the corresponding rotation matrix R ( −ω) = R + ( ω) .
3d rotation
The 3d displacement field due to a 3d rotation can be obtained by applying in sequence three rotations
around the three axes of the cartesian reference system oxyz which is kept fixed in the rotation.
Conventionally, the rotations around the z and x axes are anti-clockwise, while the rotation around the y axis
is clockwise.
By applying the same reasoning shown for the 2d case we have
OP = x P′ e1′ + y P′ e2′ + zP′ e3′
x P = OP ⋅ e1 = x P′ e1′ ⋅ e1 + y P′ e2′ ⋅ e1 + zP′ e3′ ⋅ e1
y P = OP ⋅ e2 = OP ⋅ e2 = x P′ e1′ ⋅ e2 + y P′ e2′ ⋅ e2 + zP′ e3′ ⋅ e2
zP = OP ⋅ e3 = OP ⋅ e3 = x P′ e1′ ⋅ e3 + y P′ e2′ ⋅ e3 + zP′ e3′ ⋅ e3
⎛ x P ⎞ ⎛ ( e1 ⋅ e1′ )
⎜ ⎟ ⎜
⎜ y P ⎟ = ⎜ ( e2 ⋅ e1′ )
⎜ ⎟ ⎜
⎝ zP ⎠ ⎝ ( e3 ⋅ e1′ )
( e1 ⋅ e2′ ) ( e1 ⋅ e3′ ) ⎞ ⎛ x P′ ⎞
⎟
( e2 ⋅ e2′ ) ( e2 ⋅ e3′ ) ⎟ ⎜⎜ yP′ ⎟⎟
( e3 ⋅ e2′ ) ( e3 ⋅ e3′ ) ⎠⎟ ⎝⎜ zP′ ⎠⎟
x P = Rx P′
If we perform the rotation around one axis at a time, this axis does not change its position with respect to
the fixed system and we get respectively for the three rotations:
anti-clockwise rotation around the z axis, e3′
⎛ ( e1 ⋅ e1′ )
⎜
R ( ωZ ) = ⎜ ( e2 ⋅ e1′ )
⎜ 0
⎝
( e1 ⋅ e2′ )
( e2 ⋅ e2′ )
0
0 ⎞ ⎛ cos ωZ
⎟ ⎜
0 ⎟ = ⎜ sin ωZ
1 ⎟⎠ ⎜⎝ 0
clockwise rotation around the y axis; e2′
⎛ ( e1 ⋅ e1′ ) 0
⎜
R ( −ωy ) = ⎜ 0
1
⎜ ( e ⋅ e′ ) 0
⎝ 3 1
( e1 ⋅ e3′ ) ⎞
e3 , e3′ ⊥ e1 , e3′ ⊥ e2
− sin ωZ
cos ωZ
0
0⎞
⎟
0⎟ ,
1 ⎟⎠
e2 , e2′ ⊥ e1 , e2′ ⊥ e3
⎛ cos ωy
⎟ ⎜
0
⎟=⎜ 0
′
e
e
⋅
( 3 3 ) ⎟⎠ ⎜⎝ − sin ωy
sin ωy ⎞
⎟
1
0 ⎟,
0 cos ωy ⎟⎠
0
3
anti-clockwise rotation around the x axis, e1′ e1 , e1′ ⊥ e2 , e1′ ⊥ e3
⎛1
⎜
R ( ωx ) = ⎜ 0
⎜0
⎝
0
( e2 ⋅ e2′ )
( e3 ⋅ e2′ )
⎞ ⎛1
0
⎟ ⎜
( e2 ⋅ e3′ ) ⎟ = ⎜ 0 cos ωx
( e3 ⋅ e3′ ) ⎟⎠ ⎜⎝ 0 sin ωx
0
0
⎞
⎟
− sin ωx ⎟ .
cos ωx ⎟⎠
It can be shown that every 3d rotation can be expressed as follows:
x P = R ( ωx ) R ( −ωy ) R ( ωz ) x P′ .
It is easy to show that a 3d roto-translation is given by:
x P = t + R ( ωx ) R ( −ωy ) R ( ωz ) x P′ .
Infinitesimal 3d rotations
When dealing with deformation problems, we have to manage residuals rotations, usually due to changes in
the reference system between the two epochs under consideration.
The residual rotation is characterised by infinitesimal rotation angles. Under this assumption the following
approximations can be done:
sin ω ω
cos ω 1
This yields:
⎛1
⎜
R ( ωZ ) = ⎜ ωZ
⎜ 0
⎝
−ωZ
1
0
0⎞
⎟
0⎟
1 ⎟⎠
⎛ 1
0 ωy ⎞
⎜
⎟
R ( −ωy ) = ⎜ 0 1 0 ⎟
⎜ −ω 0 1 ⎟
⎝ y
⎠
0 ⎞
⎛1 0
⎜
⎟
R ( ωx ) = ⎜ 0 1 −ωx ⎟
⎜0 ω
1 ⎟⎠
x
⎝
⎛1 0
⎜
R ( ωx ) R ( −ωy ) R ( ωz ) = ⎜ 0 1
⎜0 ω
x
⎝
⎛1 0
⎜
= ⎜0 1
⎜0 ω
x
⎝
0 ⎞⎛ 1
⎟⎜
−ωx ⎟ ⎜ ωZ
1 ⎟⎠ ⎜⎝ −ωy
−ωZ
1
ωy ωz
⎛
−ωZ
1
⎜
= ⎜ ωZ + ωx ωy 1 − ωx ωy ωz
⎜ω ω − ω
ωx + ωy ωz
y
⎝ x Z
0 ⎞⎛ 1
⎟⎜
−ωx ⎟ ⎜ 0
1 ⎟⎠ ⎜⎝ −ωy
0 ωy ⎞ ⎛ 1
⎟⎜
1 0 ⎟ ⎜ ωZ
0 1 ⎟⎠ ⎜⎝ 0
−ωZ
1
0
0⎞
⎟
0⎟ =
1 ⎟⎠
ωy ⎞
⎟
0 ⎟=
1 ⎟⎠
ωy ⎞
⎟
−ωx ⎟
1 ⎟⎠
4
by disregarding the product of two or more infinitesimal rotations, we finally get
⎛ 1
⎜
R ( ωx ) R ( −ωy ) R ( ωz ) = ⎜ ωZ
⎜ −ω
⎝ y
−ωZ
1
ωx
ωy ⎞
⎟
−ωx ⎟ = I + ⎡⎣R ( ωx ) − I ⎤⎦ ⎣⎡R ( −ωy ) − I ⎦⎤ ⎡⎣R ( ωz ) − I ⎤⎦
1 ⎟⎠
Eventually, the displacement field due to a 3d roto-translation (with an infinitesimal rotation) is given by:
⎛ 1
⎜
x P = t + ⎜ ωZ
⎜ −ω
⎝ y
−ωZ
1
ωx
ωy ⎞
⎟
−ωx ⎟ x P′ .
1 ⎟⎠
Least squares roto-translation parameters estimation
We have seen that a roto-translation does not imply any deformation in the body under analysis. No strain
means no stress. This is why we are interested in the displacement field after a roto-translation removal.
The displacement field of a body can be estimated by interpolating the displacements of a certain number of
points of the body itself. The position of those points is evaluated at two different epochs with respect to the
same reference system. The displacement of each point is the difference between the final position and the
initial one.
The position can be expressed in terms of geographical or cartesian coordinates, depending on the
observing methods.
For instance, by using a GPS positioning, we have the three cartesian components (X,Y,Z) in the WGS84
system or the corresponding ( ϕ, λ ,h ) on the ellipsoid associated to this system.
In the SAR case, we deal with the deformation affecting e point along the so called SAR line of sight; the
point being defined by the same kind of coordinates given by the GPS system.
In this case the deformation of a point is detected straightforwardly by the SAR interferometric phase
analysis.
If we use a total station we will get the displacement with respect to a local cartesian coordinates system.
In the following we will describe the way of estimating the roto-translation parameters (with an infinitesimal
rotation) when dealing with a set of 3d cartesian coordinates at two epochs. We assume that a previous
transformation of the geodetic coordinates to the cartesian coordinates has been performed, this
transformation and its inverse being well-known.
For each point in which the displacement is observed, we can write 3 equations that are linear in the 6
parameters of the sought roto-translation when we assume that the rotation is infinitesimal, as we do in
deformation analysis.
We will therefore solve the linear system by using the least squares technique, by modelling the
displacement field as a deterministic roto-translation plus an unknown stochastic field, the residual
displacement field whose strain tensor is different from zero, that we model here as a white noise.
LS observation equations
For each point P whose position is determined at the two different epoch, we can write 3 equations linear
with respect to the 6 roto-traslation parameters:
5
⎛ 1
⎜
x P = t + ⎜ ωZ
⎜ −ω
⎝ y
ωy ⎞
⎟
−ωx ⎟ x P′ + ν
1 ⎟⎠
−ωZ
1
ωx
⎛ xP ⎞ ⎛ t x ⎞ ⎛ 1
⎜ ⎟ ⎜ ⎟ ⎜
⎜ y P ⎟ = ⎜ t y ⎟ + ⎜ ωZ
⎜ ⎟ ⎜ ⎟ ⎜
⎝ zP ⎠ ⎝ zP ⎠ ⎝ −ωy
−ωZ
1
ωx
ωy ⎞ ⎛ x P′ ⎞ ⎛ ν x ⎞
⎟⎜ ⎟ ⎜ ⎟
−ωx ⎟ ⎜ y P′ ⎟ + ⎜ ν y ⎟
1 ⎟⎠ ⎝⎜ zP′ ⎠⎟ ⎜⎝ ν z ⎟⎠
zP′
0
− x P′
⎛ xP ⎞ ⎛ 1 0 0 0
⎜ ⎟ ⎜
⎜ y P ⎟ = ⎜ 0 1 0 −zP′
⎜ z ⎟ ⎜ 0 0 1 y′
P
⎝ P⎠ ⎝
⎛ tx ⎞
⎜ ⎟
⎜ ty ⎟
− y P′ ⎞ ⎜ ⎟ ⎛ x P′ ⎞ ⎛ ν x ⎞
⎜ ⎟ ⎜ ⎟
⎟ z
x P′ ⎟ ⎜ P ⎟ + ⎜ y P′ ⎟ + ⎜ ν y ⎟ =
⎜ω ⎟
0 ⎟⎠ ⎜ x ⎟ ⎜⎝ zP′ ⎟⎠ ⎜⎝ ν z ⎟⎠
⎜ ωy ⎟
⎜ω ⎟
⎝ Z⎠
and , eventually we get
⎛ x P ⎞ ⎛ x P′ ⎞ ⎛ 1 0 0 0
⎜ ⎟ ⎜ ⎟ ⎜
⎜ y P ⎟ − ⎜ y P′ ⎟ = ⎜ 0 1 0 −zP′
⎜ z ⎟ ⎜ z′ ⎟ ⎜ 0 0 1 y ′
P
⎝ P⎠ ⎝ P⎠ ⎝
zP′
0
− x P′
⎛ tx ⎞
⎜ ⎟
⎜ ty ⎟
− y P′ ⎞ ⎜ ⎟ ⎛ ν x ⎞
⎜ ⎟
⎟ zP
x P′ ⎟ ⎜ ⎟ + ⎜ ν y ⎟ =
⎜ω ⎟
0 ⎠⎟ ⎜ x ⎟ ⎝⎜ ν z ⎠⎟
⎜ ωy ⎟
⎜ω ⎟
⎝ Z⎠
By considering M points we have:
⎛ x1 ⎞ ⎛ x1′ ⎞
⎜ ⎟ ⎜ ⎟ ⎛1
⎜ y1 ⎟ ⎜ y1′ ⎟ ⎜ 0
⎜ z1 ⎟ ⎜ z1′ ⎟ ⎜
⎜ ⎟ ⎜ ⎟ ⎜0
⎜ .. ⎟ ⎜ .. ⎟ ⎜
⎜ ⎟ ⎜ ⎟ ⎜ ..
⎜ ⎟−⎜ ⎟ =⎜
⎜ ⎟ ⎜ ⎟ ⎜
⎜ ⎟ ⎜ ⎟ ⎜
⎜ ⎟ ⎜ ⎟ ⎜1
⎜ x M ⎟ ⎜ x M′ ⎟ ⎜
⎜ ⎟ ⎜ ⎟ ⎜0
⎜ y M ⎟ ⎜ y M′ ⎟ ⎜⎜
⎜ z ⎟ ⎜ z′ ⎟ ⎝ 0
⎝ M⎠ ⎝ M⎠
0
1
0
..
z1′
0 0
0 −z1′
1 y1′
.. ..
0
− x1′
..
0 0 0
1 0 −zM′
0 1 y M′
zM′
0
− x M′
⎛ ν x1
⎜
′
− y1 ⎞
⎜ ν y1
⎟
x1′ ⎟
⎜
⎛ t x ⎞ ⎜ ν z1
0 ⎟⎜ ⎟ ⎜
⎟ ⎜ t y ⎟ ..
.. ⎟ ⎜ ⎟ ⎜
⎟ ⎜ zP ⎟ + ⎜
⎟ ⎜ ωx ⎟ ⎜
⎟⎜ ⎟ ⎜
⎜
ω
⎟
− y M′ ⎟ ⎜ y ⎟ ⎜
⎜ω ⎟
ν
x M′ ⎟ ⎝ Z ⎠ ⎜ xM
⎜
⎟
ν
0 ⎠⎟
⎜ yM
⎜ νz
⎝ M
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟.
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
By using the common vector LS notation:
6
y o = Ax + ν;
⎛ x1 ⎞ ⎛ x1′ ⎞
⎛1 0 0 0
⎜ ⎟ ⎜ ⎟
⎜
⎜ y1 ⎟ ⎜ y1′ ⎟
⎜ 0 1 0 −z1′
⎜ z1 ⎟ ⎜ z1′ ⎟
⎜ 0 0 1 y1′
⎜ ⎟ ⎜ ⎟
⎜
⎜ .. ⎟ ⎜ .. ⎟
⎜ .. .. .. ..
⎜ ⎟ ⎜ ⎟
o
⎜
⎟
⎜
⎟
y =
; A=⎜
−
⎜
⎜ ⎟ ⎜ ⎟
⎜
⎜ ⎟ ⎜ ⎟
⎜1 0 0 0
⎜ ⎟ ⎜ ⎟
⎜
⎜ x M ⎟ ⎜ x M′ ⎟
⎜ 0 1 0 −zM′
⎜ ⎟ ⎜ ⎟
⎜⎜
⎜ y M ⎟ ⎜ y M′ ⎟
⎝ 0 0 1 y M′
⎜ z ⎟ ⎜ z′ ⎟
⎝ M⎠ ⎝ M⎠
z1′
0
− x1′
..
zM′
0
− x M′
− y1′ ⎞
⎟
x1′ ⎟
0 ⎟
⎟
.. ⎟
⎟;
⎟
⎟
− y M′ ⎟⎟
x M′ ⎟
⎟
0 ⎟⎠
⎛ tx ⎞
⎜ ⎟
⎜ ty ⎟
⎜z ⎟
P
x = ⎜ ⎟;
⎜ ωx ⎟
⎜ ⎟
⎜ ωy ⎟
⎜ω ⎟
⎝ Z⎠
⎛ ν x1
⎜
⎜ ν y1
⎜
⎜ ν z1
⎜ ..
⎜
⎜
ν=⎜
⎜
⎜
⎜
⎜ ν xM
⎜ν
⎜ yM
⎜ νz
⎝ M
0
∑ z′
⎞
⎟
⎟
⎟
⎟
⎟
⎟
⎟.
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎟
⎠
Least squares solution
The LS solution is given by
(
x̂ = N−1 ⋅ tN = A + A
)
−1
A+ yo
ˆ
ˆ
y = Ax
y
ν = yo − ˆ
ˆ
.
ν+ˆ
ν
ˆ
σ
ˆ =
n−m
n = 3M
2
0
m=6
In this case the N matrix is given by:
⎛
⎜
⎜
⎜
⎜
⎜
⎜
⎜
N = A+ A = ⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎜⎜
⎝
M
0
0
M
i =1
M
M
−∑ zi′
0
i
0
i =1
M
M
∑ y′
M
i =1
M
∑ z′
i =1
i
2
−∑ x i′
i
+ y i′2
i =1
M
−∑ y i′x i′
i =1
M
∑ z′
i =1
i
2
+ x i′2
⎞
⎟
i =1
⎟
M
⎟
x i′ ⎟
∑
i =1
⎟
⎟
0
⎟
⎟;
M
⎟
−∑ zi′x i′ ⎟
i =1
⎟
M
⎟
−∑ zi′y i′ ⎟
⎟
i =1
⎟
M
y i′2 + x i′2 ⎟⎟
∑
i =1
⎠
M
−∑ y i′
The normal vector of the known term is given by:
7
M
⎛ M
⎞ ⎛ M
⎞
⎜ ∑ x i − x i′
⎟ ⎜ ∑ x i − ∑ x i′ ⎟
i =1
⎜ i=1
⎟ ⎜ i=1
⎟
M
⎜ M
⎟ ⎜ M
⎟
⎜ ∑ y i − y i′
⎟ ⎜ ∑ y i − ∑ y i′ ⎟
i =1
⎜ i=1
⎟ ⎜ i=1
⎟
M
⎜ M
⎟ ⎜ M
⎟
⎜ ∑ zi − zi′
⎟ ⎜ ∑ zi − ∑ zi′ ⎟
i =1
i =1
⎟ = ⎜ i=1
⎟
tN = ⎜ M
⎜
⎟ ⎜ M
⎟
⎜ −∑ zi′ ( y i − y i′ ) − y i′ ( zi − z′ ) ⎟ ⎜ −∑ zi′y i − y i′zi ⎟
⎟
⎜ i=1
⎟ ⎜ i=1
⎟
⎜ M
⎟ ⎜ M
⎜ ∑ zi′ ( x i − x i′ ) − x i′ ( zi − z′ ) ⎟ ⎜ ∑ zi′x i − x i′zi ⎟
⎟
⎜ i=1
⎟ ⎜ i=1
⎟
⎜ M
⎟ ⎜ M
⎜⎜ −∑ y i′ ( x i − x i′ ) − x i′ ( y i − y i′ ) ⎟⎟ ⎜⎜ −∑ y i′x i − x i′y i ⎟⎟
⎠
⎝ i=1
⎠ ⎝ i=1
By looking at the above normal terms, one realizes soon that it is convenient to make a translation in the
barycentre of the points at the initial epoch, before entering the LS procedure.
Such a translation produces a simplification of the normal matrix, which becomes block diagonal; moreover,
it produces a better conditioning of the N matrix producing a stable inversion.
The barycentre coordinates are:
1 M
X ' = ∑ x i′
M i=1
1 M
Y ' = ∑ y i′
M i=1
1 M
Z ' = ∑ zi′
M i=1
We translate the system of coordinates oxy in the barycentre evaluated at the initial epoch by applying the
following translation formulas:
1 M
∑ x i′
M i=1
1 M
ηi′ = y i′ − Y ' = y ′ − ∑ y i′
M i=1
1 M
ζ i′ = zi′ − Z ′ = zi′ − ∑ zi′
M i=1
1 M
ξi = x i − X ′ = x i − ∑ x i′
M i=1
1 M
ηi = y i − Y ' = y i − ∑ y i′
M i=1
1 M
ζ i = zi − Z ′ = zi − ∑ zi′
M i=1
ξi′ = x i′ − X ′ = x i′ −
This yields:
8
M
M
M
M
∑ ξ′ =∑ ( x ′ − X ′) =∑ x ′ − ∑ X ′ = MX ′ − MX ′ = 0
i
i
i
i =1
i =1
i =1
M
M
M
i =1
M
∑ η′ =∑ ( y ′ − Y ′) =∑ y ′ − ∑ Y ′ = MY ′ − MY ′ = 0
i
i
i
i =1
i =1
i =1
M
M
M
i =1
M
∑ ζ′ =∑ ( z′ − Z′) =∑ z′ − ∑ Z′ = MZ′ − MZ′ = 0
i =1
i
i =1
i
i =1
i
i =1
and
⎛
⎜
⎜
⎜
⎜
⎜
N = A+ A = ⎜
⎜
⎜
⎜
⎜
⎜
⎜
⎝
M
0
M
0
0
M
0
0
0
M
∑ ζ′
i =1
2
i
0
0
0
+ ηi′2
0
0
0
M
−∑ ηi′ξi′
i =1
M
∑ ζ′
i =1
2
i
+ ξi′2
⎞
⎟
⎟
⎟
⎟
M
−∑ ζ i′ξi′ ⎟
⎟
i =1
⎟
M
⎟
−∑ ζ i′ηi′ ⎟
i =1
⎟
M
2
2 ⎟
′
′
η
+
ξ
∑
i
i ⎟
i =1
⎠
M
⎛ M
⎞ ⎛ M
⎞ ⎛ M
⎞ ⎛ M
⎞
⎜ ∑ x i − x i′
⎟ ⎜ ∑ x i − ∑ x i′ ⎟ ⎜ ∑ x i − MX ′
⎟ ⎜ ∑ ξi
⎟
i =1
⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟
M
⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟
⎜ ∑ y i − y i′
⎟ ⎜ ∑ y i − ∑ y i′ ⎟ ⎜ ∑ y i − MY ′
⎟ ⎜ ∑ ηi
⎟
i =1
⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟
M
⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟
⎜ ∑ zi − zi′
⎟ ⎜ ∑ zi − ∑ zi′ ⎟ ⎜ ∑ zi − MZ ′
⎟ ⎜ ∑ ζi
⎟
i =1
i =1
i =1
i =1
i =1
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
=
=
=
tN =
⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟
⎜ −∑ zi′ ( y i − y i′ ) − y i′ ( zi − z′ ) ⎟ ⎜ −∑ zi′y i − y i′zi ⎟ ⎜ −∑ ( zi′ − Z ′ )( y i − Y ′ ) − ( y i′ − Y ′ )( zi − Z ′ ) ⎟ ⎜ −∑ ζ i′ηi − ηi′ζ i ⎟
⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟
⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟
⎜ ∑ zi′ ( x i − x i′ ) − x i′ ( zi − z′ ) ⎟ ⎜ ∑ zi′x i − x i′zi ⎟ ⎜ ∑ ( zi′ − Z ′ )( x i′ − X ′ ) − ( x i′ − X ′ )( zi′ − Z ′ ) ⎟ ⎜ ∑ ζ i′ξi − ξi′ζ i ⎟
⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟ ⎜ i=1
⎟
⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟ ⎜ M
⎟
⎜⎜ −∑ y i′ ( x i − x i′ ) − x i′ ( y i − y i′ ) ⎟⎟ ⎜⎜ −∑ y i′x i − x i′y i ⎟⎟ ⎜⎜ −∑ ( y i − Y ′ )( x i′ − X ′ ) − ( x i′ − X ′ )( y i − Y ′ ) ⎟⎟ ⎜⎜ −∑ ηi′ξi − ξi′ηi ⎟⎟
⎝ i=1
⎠ ⎝ i=1
⎠ ⎝ i=1
⎠ ⎝ i=1
⎠
Note that the parameters do not change with respect to the original systems.
Once verified that all the estimated parameters are significantly different from zero we directly compute the
deformation field as the residuals vector of the LS adjustment.
Test on the LS parameters
The estimated parameters significance can be evaluated by using the standard test on Least squares
parameters. The H0 hypothesis for each parameter is that the mean of the corresponding estimator is equal
to zero. By assuming a normal distribution of the observed data, a student test can be set up at a fixed
level α
The H0 hypotheses and the corresponding statistics to set up the test on the 6 estimated parameters are the
following:
9
H0 : E[TX ] = tX = 0
TX − tX
∑ (N )
−1
0
t oss =
TX
=
∼ t 3M− 6
∑ (N )
2
−1
0
11
11
t̂ X
σ
ˆ0
(N )
−1
11
Note that we denote here with capital letters the estimators, i.e. the random variables, and with the
corresponding small letters the estimates, which are numbers. t 3M− 6 is the Student distribution with the
corresponding degree of freedom.
if
t oss < t lim ( α, 3M − 6 ) then the hypothesis is verified and the parameter is not significantly different from
zero.
t lim is the value at which the Student distribution function is equal to α (degrees of freedom equal to 3M-6).
H0 : E[TY ] = tY = 0
TY − tY
∑ (N )
0
t oss =
TY
=
−1
∑ (N )
2
0
22
∼ t 3M− 6 ,
−1
22
t̂ Y
σ
ˆ0
(N )
−1
22
H0 : E[TZ ] = tZ = 0
TZ − tZ
∑ (N )
−1
0
t oss =
TZ
=
∑ (N )
2
0
33
∼ t 3M−6
−1
33
t̂ Z
σ
ˆ0
(N )
−1
33
H0 : E[Ω X ] = ωX = 0
Ω X − ωX
∑ (N )
0
t oss =
ΩX
=
−1
∑ (N )
2
44
∼ t 3M−6 ;
−1
0
44
ω
ˆX
σ
ˆ0
(N )
−1
44
H0 : E[Ω Y ] = ωY = 0
Ω Y − ωY
∑ (N )
−1
0
t oss =
ΩY
=
∑ (N )
2
0
55
−1
∼ t 3M−6
55
ω
ˆY
σ
ˆ0
(N )
−1
55
10
H0 : E[Ω Z ] = ωZ = 0
Ω Z − ωZ
∑ (N )
=
−1
0
t oss =
ΩZ
∑ (N )
2
0
66
∼ t 3M− 6
−1
66
ω
ˆZ
σ
ˆ0
(N )
−1
66
If only a subset of parameters results different from zero, then the corresponding residual displacements are
obtained by applying just the corresponding translation and or rotation components, e.g by applying the
same LS residuals evaluation where the significantly equal to zero parameters are set equal to zero.
11