Download SMART CALIBRATION OF EXCAVATORS AN ESGI54 PROJECT

SMART CALIBRATION OF EXCAVATORS
AN ESGI54 PROJECT FROM MIKROFYN
Abstract. Excavators dig holes. But where is the bucket? The
purpose of this report is to treat different aspects of excavator
calibration encountered on the construction site in some detail.
Four problems are considered - all related to the question of how
to determine the position of excavator and/or bucket.
1. Participants
secParticip
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secTODO
Lars Overgaard, Mikrofyn - [email protected]
Marie Bro, Dept. of Math. DTU - [email protected]
Mikael S. Hansen, Dept. of Math. DTU - [email protected]
Steen Markvorsen, Dept. of Math. DTU - [email protected]
David Spence Dept. of Math. DTU - [email protected]
Mathias Stolpe, Dept. of Math. DTU - [email protected]
Katja Skaanning, Inst. Math. Sci. KU - [email protected]
Kasper Døring, MIP, SDU - [email protected]
Lars-Peter Ellekilde, MIP, SDU - [email protected]
Dorthe Sølvason, MIP, SDU - [email protected]
2. Date/Who/Comments/Revisions/Todos
(4) 25.08.05 (Dorthe and Steen): Notation conc. matrices and parameters in problem A has been decided as below. New reference [1] added to reference list and homepage.
(5) 28.08.05 (Mikael): Added to problem A, section on Magnetic
compass calibration, magnetic fields model and appendices on
magnetic fields models. Reconstructed references as a BibTeX
file. This file is now in the REFERENCES file at the home
page site and called Calib.bib . Figure (softhardeffects.pstex t
, softhardeffects.pstex) included and files added into FIGURES.
Package colors and new commands for controlling appendices
added into heading.
(6) 30.08.05 (Steen): To determine the Earths magnetic field assuming no extra error terms: Keep h fixed at 0 and let r and
p vary as much as possible. Next, given the Earths magnetic
coordinates from this setting, then the value of h in any other
orientation is determined from 3 trigonometric equations, much
as in problem D. The general strategy (with soft and hard iron
1
2
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effects added) is similar. The attack continues.
(7) 31.08.05 (Steen): The famous (figure of speech) ellipsoid seems
to be only a shadow of the real general problem? See equation
(4.11) and subsection 4.3.3.
(8) next comment, input, ...
(9) next comment, input, ...
3. Introduction
secIntro
4. Problem A: Magnetic Compass Calibration
secA
In this section we consider the question of how to determine the
heading h of a rotating object E given pitch p, roll r and (distorted)
components of Earths magnetic field in a magnetic compass frame situated on E . This is a ”classical” problem in the context of navigation
systems [2, 1] encountered in many different fields e.g. aviation [3, 4, 5],
robotics [6], and space engineering [7].
4.1. Magnetic fields model. [Mikael]
To a good approximation the direction of Earths magnetic field BE
only depends on the latitude which can be used to determine the orientation of an object. Assuming that E is free to move in all three dimensions BE will induce a sphere of radius kBE k centered at origo in the
compass coordinate system. However, there are two local effects distorting the field components measured by a magnetic compass known
as hard and soft iron effects in the navigation systems literature. Both
are properties of ferromagnetic materials and are collectively known as
ferrous effects.
4.1.1. Hard iron effects. In ferromagnetic materials the constituent atoms
each have a magnetic moment and quantum mechanical interactions
between electrons give rise to an alignment of these moments. This
permanent magnetization is the hard iron effect resulting in a translation
(4.1)
Bef f = BE + δBhard ,
of the magnetic field measured by the compass.
4.1.2. Soft iron effects. However not all ferromagnetic materials exhibit
a permanent magnetization. The quantum mechanical interactions alluded to above are of a local nature and on a larger scale dipole-dipole
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3
Bef f
Bef f
Bef f
δBhard
δBhard
No perturbation
Hard iron effects
Hard and soft iron effects
Figure 1. Due to the presence of materials with various
magnetic properties a magnetic compass will on measure
a distorted version of Earth’s magnetic field. This should
be included in a calibration scheme to obtain a precise
determination of heading.
fig:softhard
interactions tend to anti-align macroscopic magnetic domains1. This
explains why the introduction of an external field may increase the
magnetization. Material characteristics such as anisotropies introduce
a directional dependency of the measured magnetic field.
Taking both hard and soft iron effects together we have a distorted
magnetic field
(4.2)
Bef f = Csof t (BE + δBhard ).
as shown in Fig. 1. We assume that the materials in question have no
memory i.e. we neglect hysteresis effects. In this case Csof t is simply
a 3 × 3 matrix describing the angular dependency of the measured
magnetic field.
For more information on the physics of ferromagnets see Appendix
A.
4.2. Coordinate transforms. [Dorthe]
We consider coordinate systems with a fixed common origin O:
(1) The World coordinate system:
C1 = {O, x1 , y1 , z1 , ex1 , ey1 , ez1 } ,
where the z1 -axis is considered vertical, i.e. directed opposite
to the force of gravity, so that the (x1 , y1 )-plane is horizontal.
The x1 -axis is assumed to be directed in along a fixed direction
which has been defined and set up by the surveyors on site
location. This direction is NOT necessarily true North or any
approximation thereof.
1Regions
where the magnetic moments are aligned.
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Every inclinometer which is mounted on a vertical plane and
which is pointing in a given direction d is able to measure and
display the angle between d and the horizontal plane, i.e. the inclinometer measures the vertical component (or rather its arcsin
of its ’director’ d.
(2) The magnetometer coordinate system:
C2 = {O, x2 , y2 , z2 , ex2 , ey2 , ez2 } ,
whose origin is coinciding with the origin of the world C1 .
The relation between the two systems are obtained as follows. We
specify the orientation of C2 with respect to C1 :
(1) Rotate by the angle r around the x1 -axis, then
(2) Rotate by the angle p around the x2 -axis, and finally
(3) Rotate by the angle h around the x3 -axis.
Remark 4.1. It is important to note, that the value of p is directly
read off from an inclinometer (first inclinometer), whereas the value of
r is NOT directly an inclinometer reading. The second inclinometer
reads the value:
eqInclin
(4.3)
r̃ = arcsin(sin(r) cos(p)) .
The transform between the coordinate systems C2 and C1 are thus
determined by the following rotation matrices, using the the parameters
r, p, and h:


1
0
0
M1 (r) =  0 cos(r) − sin(r) 
0 sin(r) cos(r)


cos(p) 0 − sin(p)

1
0
M2 (p) =  0
sin(p) 0 cos(p)


cos(h) − sin(h) 0
M3 (h) =  sin(h) cos(h) 0 
0
0
1
eqTransf
Then the following holds true concerning the matrices giving the
resulting orientation of the magnetometer in relation to world coordinates:




x2
x1
 y1  = M (r, p, h)  y2  ,
(4.4)
z2
z1
where
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M (r, p, h) = M3 (h) M2 (p) M1 (r) =

cos (h) cos (p) − sin (h) cos (r) − cos (h) sin (p) sin (r)

 sin (h) cos (p)
cos (h) cos (r) − sin (h) sin (p) sin (r)

sin (p)
cos (p) sin (r)
5
sin (h) sin (r) − cos (h) sin (p) cos (r)

− cos (h) sin (r) − sin (h) sin (p) cos (r) 

cos (p) cos (r)
M −1 (r, p, h) = M1−1 (r) M2−1 (p) M3−1 (h) =

cos (h) cos (p)
sin (h) cos (p)

 − sin (h) cos (r) − cos (h) sin (p) sin (r)
cos
(h)
cos
(r)
− sin (h) sin (p) sin (r)

sin (h) sin (r) − cos (h) sin (p) cos (r)
− cos (h) sin (r) − sin (h) sin (p) cos (r)
sin (p)
subsubsecNoPerturb
eqMpure


cos (p) sin (r) 

cos (p) cos (r)
where, of course, the latter matrix is just the transpose of the former
(since both of them are orthogonal matrices).
subsecCombEq

4.3. Equations from combined model and transform. [Steen] We
first consider the simplest possible model for determining the heading
from a given measurement. Then the hard and soft iron effects are
taken into account using a most general linear model for their influences.
4.3.1. The pure Sphere Case. First we consider the non-disturbed case:
no hard and no soft iron, only the ambient constant magnetic field B
from the Earth is present. The coordinates of this field in the coordinate
system C1 are
 
a
BC1 =  b 
c
We next assume, that the magnetometer is perfect, so that it measures the exact coordinates of this vector in the magnetometer coordinate system C2 :
 m 
a

bm 
BC2 =
cm
Then
 m 
 
a
a
 bm  = M −1 (r, p, h)  b 
(4.5)
cm
c
Note that for each measurement we know am , bm , cm from the magnetometer displays, and r and p from inclinometer displays. The value
of r follows from both of the two inclinometer displays via equation
(4.3).
The unknown values here are thus h and a, b, c. The latter 3 could
be measured by some other device for comparison, but we will develop
a procedure to actually find the B-field as well as the heading h.
For each measurement we thus have 3 equations (one for each coordinate, see below) and 4 unknowns. Among the 4 unknowns, however,
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3 of them are constants, i.e. they do not vary from one measurement
to the other.
For each measurement the 3 equations alluded to are the following,
which are directly equivalent to equation (4.5):
am =
cos (h) cos (p) a
+ sin (h) cos (p) b
+ sin (p) c ,
b
eqAmA
m
=
cos (h) (− sin (p) sin (r) a + cos (r) b)
+ sin (h) (− cos (r) a − sin (p) sin (r) b)
(4.6)
+ cos (p) sin (r) c ,
cm =
cos (h) (− sin (p) cos (r) a − sin (r) b)
+ sin (h) (sin (r) a − sin (p) cos (r) b)
+ cos (p) cos (r) c .
Note that if we knew the values of a, b, and c (the constant ambient magnetic field), then we would now have (generically speaking) 3
equations to determine the heading h. Each of the equations is solved
in h in a similar way as in problem D. The system is over-determined.
On the other hand, suppose we let h = 0 then for this particular
heading (given by some fixed reference direction on site) we get values
of a, b, and c by just one reading of the magnetometer:


 m 
a
a
 b  = M (r, p, 0)  bm 
c
cm


cos (p) am − sin (p) sin (r) bm − sin (p) cos (r) cm


cos (r) bm − sin (r) cm
= 

m
m
m
sin (p) a + cos (p) sin (r) b + cos (p) cos (r) c
With a constant heading 0 we may do several measurements, by
varying the values of p and r. Each measurement gives rise to stabilize
the determination of a, b, and c.
With these values of a, b, and c we then change the heading to some
unknown value h. The value of h is then determined by one or a series
of measurements: With the fixed heading change first r and then p and
always read off the display of corresponding values of r, p, am , bm , and
cm .
The value of h is then determined by any one of the 3 equations in
(4.6). For N measurements at the heading h we have generically 3N
equations to determine the value of h.
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7
Each one of these equations is of the following type:
eqCosSin
subsubsecGeneral
(4.7)
α cos(h) + β sin(h) = γ
√
Let x = sin(h) so that cos(h) = 1 − x2 , assuming first that the
heading is ’forward’, i.e. h ∈ [−π/2, π/2] with respect to the fixed
direction of the x1 -axis in the World system C1 . Then
p
βγ ± α α2 + β 2 − γ 2
x =
α2 + β 2
Again, as in problem D, generically there will be 4 solutions, 4 heading values satisfying equation (4.7). However, only one of them will
be in the intersection of all sets of solutions stemming from the 3N
equations.
This then finishes the strategy in case of no perturbation terms in
the magnetic field, i.e. the ’pure sphere case’.
4.3.2. The General Case. The strategy is much the same as above:
First we determine the constants in the problem, and then for each
single measurement we apply these constants to determine the heading.
The constants in the general setting are: The hard iron contribution
H (constant coordinates in C2 ), the constant ambient field (constant
coordinates in C1 ) and the distortion matrix C (constant coordinates
in C2 ). The matrix C is assumed to be highly regular, since it will
usually be a perturbation of the identity matrix. But it is important
to note, that unlike several of our (best) references we do NOT assume
any structural simplicity (like e.g. symmetry) of this matrix. Thus our
model will take care of any first order perturbation effects resulting
from soft iron and scaling mismatch etc.
The intermixed variables are still r, p and h. We have in total for
the model:
B m = C (B + H) ,
or equivalently the coordinate expression with respect to system C2 :
eqC1
i.e.
eqGeneral
BCm2 = CC2 (BC2 + HC2 )
(4.8)
,

   


a
κ
c11 c12 c13
am
−1
m 







c21 c22 c23
b
M (r, p, h) b + λ 
=
(4.9)
m
c31 c32 c33
c
c
µ

The problem is to find the constants cij , the constants a, b, c, κ, λ,
µ and eventually the variable h (for each measurement).
The idea is first to tell as much about the constants without knowing
h except for the zero directional case: For h = 0 we get (for each choice
of r and p) 3 equations for the 15 unknown constant values. Thus we
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need only (generically) 5 measurements (with h = 0) to determine the
15 constants. Once found, we insert these constants into equation (4.9)
and then solve for h (as in the pure case discussed above) for all other
orientations.
eqGeneralINV
This method does not use any ellipsoid! Nor does it use the information hidden in M −1 (r, p, h), that this coordinate shift matrix is
orthogonal, so that it preserves lengths. We apply this now. If we let
C −1 = D , then D is also a regular matrix being close to the identity,
and we have from (4.9):

 m   
 
d11 d12 d13
a
κ
a
(4.10)  d21 d22 d23   bm  −  λ  = M −1 (r, p, h)  b 
d31 d32 d33
cm
µ
c
Let us denote the left hand side as follows:


¤1 (dij , am , bm , cm , κ, λ, µ)
 ¤2 (dij , am , bm , cm , κ, λ, µ) 
¤3 (dij , am , bm , cm , κ, λ, µ)
,
where each box ¤k , k = 1, 2, 3, is a quadratic function of the 15 variables
dij , am , bm , cm , κ, λ, µ (where i = 1, 2, 3, j = 1, 2, 3). Note, however,
that when inserting known measured values of am , bm , and cm , then
¤k is a linear function of the remaining 12 constant coefficients.
Then independently of r, p and h we have:
eqSquare
subsubsecEllipsoid
(4.11)
¤21 + ¤22 + ¤23 = a2 + b2 + c2
.
For each measurement we only know the values of am , bm , and cm , so
each measurement gives 1 equation, equation (4.11), with 15 unknown
constants (note that a, b, and c are not known). In short, generically
we need 15 measurements to determine the 15 constants. However,
with a significantly larger number of measurements we need to fit the
15 constants so that (4.11) is optimally satisfied. This is similar to,
but NOT precisely the ellipsoidal fitting problem.
4.3.3. The Ellipsoidal Shadow. To (re-)approach the general case from
the point of view of ellipsoidal fitting we now express C as a product
of 3 matrices, two of which, Q1 and Q2 , are orthogonal and one, Λ,
which is diagonal - this is the so-called SVD decomposition of C:
C = Q1 ΛQ−1
2
,
so that we now have:
eqCSVD
(4.12)
where
B m = Q1 Λ Q−1
2 (B + H)


λ1 0 0
Λ =  0 λ2 0 
0 0 λ3
.
,
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eqGeneralSVD
9
Each λi (the singular values) is the square root of an eigenvalue for
the symmetric matrix CC T (where T means ’transpose’). The corresponding unit orthogonal eigenvectors of CC T are the column vectors
in Q1 .
The Ellipsoid now appears as follows:
(4.13)
 m 



   
a
λ1 0 0
a
κ
−1
 bm  = Q1  0 λ2 0  Q−1




M (r, p, h) b + λ 
2
m
c
0 0 λ3
c
µ
In short, the measurements are positioned on an ellipsoid with semi
axes determined by λi and with an orientation determined by Q1 . Thus
we may read off geometrically these 6 constants from the ellipsoid construction. The values of κ, λ, and µ cannot be directly read off from
the position of the ellipsoid in C2 because this positioning involves the
unknown orientation matrix Q2 .
Still in short, this seems to suggest, that the geometric ellipsoid is
only delivering 6 out of the 9 constants to be determined to fix C. And
since the ellipsoid is the result of a fitting algorithm anyways, it seems
most appropriate to attack the problem (4.11) and find the (best fit) 9
constants in C directly (together with the remaining 6 constants a, b,
c, κ, λ, and µ). We therefore return now to that equation and line it
up for application to the data sets:
subsubsecFit
4.3.4. The Fit.
4.4. Description and pictures from on site experiments. [Katja]
4.5. Ellipsoidal data fitting. [Mathias]
4.6. Implementation on data set. [All]
secB
5. Problem B: Inclinometer Calibration
5.1. Robot data and inclinometry. [Kasper and Lars-Peter]
5.2. Calibration from tip motion. [Kasper and Lars-Peter]
5.3. Sensitivity analysis. [Kasper and Lars-Peter]
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5.4. Implementation on data set. [Kasper and Lars-Peter]
secC
6. Problem C: Center finding from Circular GPS motion
6.1. GPS and (r,p,h) coordinates. [Marie and David]
6.2. Center finding techniques. [Marie and David]
6.3. Implementation on data set. [Marie and David]
secD
7. Problem D: Distant Pitch generation via Rotation
We consider a fixed (world) Cartesian coordinate system
C1 = {O, x1 , y1 , z1 , ex1 , ey1 , ez1 } ,
where the z1 -axis is considered vertical, i.e. directed opposite to the
force of gravity, so that the (x1 , y1 )-plane is horizontal.
Another Cartesian coordinate system C2 is then constructed by translating and orienting the system C1 :
C2 = {P, x2 , y2 , z2 , dx2 , dy2 , dz2 } .
The relation between the two systems is determined partly by the translation vector, which in (C1 )−coordinates are, say:
 
a
~

b  ,
OP (C1 ) =
c
and partly by the orientation given by values of roll r, pitch p, and
heading h as developed in section A. The coordinates of the following
(selected two) basis vectors of system C2 with respect to system C1 are
then:


∗
dx2 (C1 ) =  ∗  and
sin(p)


∗
dy2 (C1 ) =  ∗  ,
sin(r)
where the ∗ denote some not important (for the present task) functions of r, p, and h. (The complete transformation is given in section
4, by M (r, p, h) equation (4.4), modulo the use of parameter r. In the
present section we do use r to denote the actual inclinometer display.)
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11
The resulting coordinate transformation then reads in total:




  
x1
∗
∗
∗
x2
a
 y1  =  ∗
∗
∗   y2  +  b  .
z1
sin(p) sin(r) ∗
z2
c
y2
x2
z1
y1
x1
z2
Figure 2. An oriented link to a rotating device (bucket
with (x2 , y2 , z2 ) system). Now rotate the bucket around
the z2 -axis. What happens then to the angle between
the horizontal plane and the x2 -axis (the bucket pitch) ?
figRPH1
Suppose now that the coordinate system C2 is rotated the angle α
around the z2 -axis of the coordinate system C2 . Then the pitch value
(the angle to the horizontal (x1 , y1 )-plane) of the new resulting 1’st axis
(to be named the x3 -axis below) becomes some angle θ.
probPitch
Problem 7.1. The problem is the inverse: For any given θ find - if
possible - the values of rotation angles α which will give the pitch value
θ by the above mentioned rotation.
7.1. Conditions and number of solutions.
Statement 7.2. There are solutions (rotational α-values) to problem
7.1 if and only if the given value of θ satisfies:
eqCond
(7.1)
sin2 (θ) ≤ sin2 (r) + sin2 (p) .
Suppose that θ, r, and p are given values satisfying this condition.
Then:
For θ = 0, r = 0, and p = 0: Every value of α is a solution, α ∈ [−π, π].
For θ 6= 0 or r 6= 0 or p 6= 0: If equality occurs in the condition
12
eqSolut
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above, then there are 2 solutions; otherwise there are 4 solutions. These
solutions are given by those α ∈ [−π, π] which satisfy:
p
sin(r) sin(θ) ± sin(p) sin2 (r) + sin2 (p) − sin2 (θ)
(7.2) sin(α) =
.
sin2 (r) + sin2 (p)
Remark 7.3. The solutions do not depend on the parameters h, a, b, c,
which was (maybe?) to be expected.
How to see this: The rotation by α around the z2 -axis in system C2
results in a new system:
C3 = {P, x3 , y3 , z3 , wx3 , wy3 , wz3 } ,
which is related to system C2 via the coordinate transformation:





x2
cos(α) − sin(α) 0
x3
 y2  =  sin(α) cos(α) 0   y3  .
z2
0
0
1
z3
In total, the transformation



x1
∗
∗
 y1  =  ∗
∗
z1
sin(p) sin(r)
from system C1 to system C3 is therefore:


  
∗
cos(α) − sin(α) 0
x3
a
∗   sin(α) cos(α) 0   y3 + b  .
∗
0
0
1
z3
c
Hence the first basis vector wx3 of the system C3 has coordinates with
respect to the system C1 which are obtained as the first column in the
product matrix



∗
∗
∗
cos(α) − sin(α) 0
 ∗
∗
∗   sin(α) cos(α) 0 
sin(p) sin(r) ∗
0
0
1


∗
∗ ∗
∗
∗ ∗ 
= 
sin(p) cos(α) + sin(r) sin(α) ∗ ∗
Thus

wx3 (C1 )

∗

∗
= 
sin(p) cos(α) + sin(r) sin(α)
.
Since sin(θ) is, by definition of the pitch θ, the projection of wx3 on
the z1 -axis, we have:
eqSinCos
(7.3)
sin(θ) = sin(p) cos(α) + sin(r) sin(α) .
√
Now let x = sin(α) so that cos(α) = ± 1 − x2 and let S = sin(θ), A1 =
sin(p), and finally A2 = sin(r). Inserting these shorthand notations into
equation (7.3) then gives:
√
A1 (± 1 − x2 ) + A2 x = S ,
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13
or equivalently:
eqX
(7.4)
(S − A2 x)2 − A21 (1 − x2 ) = 0 ,
x2 (A21 + A22 ) + x (−2A2 S) + (S 2 − A21 ) = 0 ,
p
A2 S ± A1 A21 + A22 − S 2
x =
,
A21 + A22
which is in accordance with equation (7.2) in the statement above.
Clearly the condition in equation (7.1) is necessary for the existence of
solutions because otherwise the discriminant of the second order equation would be negative. It is also sufficient, because the values of the
x-solutions are numerically less than 1, so that x = sin(α) really does
give α-solutions. Indeed (for ease of mind) this follows from inserting
x = 1 and x = −1 into equation (7.4) and observe that the left hand
side is positive in both cases.
¤
secCoda
secAppendices
8. Coda and Conclusion
9. List of appendices
(1) Codes for procedures ?
(2) Raw data ?
(3) What else ?
app:magnetism
Appendix A. Magnetism
For a fuller treatment of this subject see any graduate textbook on
classical electrodynamics e.g. [8].
The magnetic properties of materials are primarily given by the orbital angular momentum of electrons in the constituent atoms, changes
in their orbital angular momentum, and their spin.
Absence of both permanent spin and orbital angular momentum contributions gives rise to diamagnetism. Here an external magnetic field
Bext perturbs the motions of electrons around the atomic nuclei leading
to an induced field Bind opposite of Bext (Lenz’ law).
When atoms have permanent magnetic moments these will tend to
align with Bext leading to an increase in the effective magnetic field.
This phenomenon is known as paramagnetism.
However, both dia- and paramagnetism are weak effects compared
to that of ferromagnetism. While the former can be understood from a
classical treatment of noninteracting atoms ferromagnetism is a quantum mechanical, collective effect.
14
ESGI54
A.1. Ferromagnetism. Spin is a quantum mechanical property of elementary particle that intuitively can be understood as an internal
orbital angular momentum.
Electrons are fermions and according to Pauli’s exclusion principle [9]
two electrons of identical spin cannot have the same position in space.
It is therefore energetically favorable for two electrons two have parallel
spins since this will lower the electrostatic repulsion experienced by
each particle. This is known as the exchange interaction.
This effect can give rise to a permanent magnetization even in the
absence of an external field. In the navigation systems literature this
is known as hard iron effects. To understand why not all ferromagnetic
materials have a permanent magnetization we need one further concept.
The exchange interaction is short-ranged and for sufficiently large regions of aligned spins the magnetic dipole-dipole interaction makes it
favorable to anti-align such regions known as magnetic domains.
Applying an external field will now tend to align domains. How and
to what extent such an alignment actually takes place depends strongly
on the characteristics - e.g. anisotropy - of the material in question.
This field dependent phenomenon is known as soft iron effects.
References
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gebre01
caruso1
caruso2
caruso3
miller04
merayo
elbek97
bransden00
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