Analytical solution of the spacecraft magnetometer calibration problem using the method of least squares

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In this paper, an analytical method is proposed for solving the problem of magnetometer calibration for the model considered in [1]. When solving the problem of determining the calibration parameters of the magnetometer unit, it is taken into account that for measurements with any spatial orientation of the magnetometer unit, the value of the measured magnetic induction vector is preserved and is a known model value. A penalty function of 12 variables equal to the sum of the squares of the residuals is introduced into consideration. The algorithm for solving the problem of calibrating the measuring axes of the magnetometer unit is reduced to searching, by the method of least squares, for such values of the variables of this function that, for a given set of magnetometer measurement vectors, provide it with a minimum. For this purpose, the specified function is examined for a conditional extremum in the presence of three equality constraints. The Lagrangian function is compiled and, based on the necessary condition for the extremum of this function, the system of 15 equations in the 15 variables is formed. It is proved that the system has four solutions. Formulas are derived that make it possible to obtain the components of each of these four solutions. As a solution to the magnetometer calibration problem, it is recommended to choose a solution to the specified system that provides a minimum of the Lagrangian function.

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Introduction

Magnetometers are part of the orientation and stabilization system of low-orbit small-sized spacecraft. They are widely used due to the fact that they are lightweight, inexpensive and reliable. However, due to the physical properties of the sensitive element, modern magnetometers require mathematical calibration. At present, various methods for calibrating magnetometers have been proposed, and a considerable number of scientific papers are devoted to these methods [1–12], in particular, article [9], which provides an overview of various methods for performing such operations.

In the above-mentioned works, the problem of calibrating the magnetometer of a spacecraft was solved using numerical methods. In this article, an analytical method for solving this problem for the model considered in [1] is proposed.

1. Model of errors in measurements of magnetic induction vector

We denote by h the value of the measured magnetic induction vector at a certain spatial position of the magnetometer unit (MU). We use the simplified measurement model considered in  [1]:

h=h1h2h3=QPB+b+n.                                                                                       (1)

The following notations [1] are used in(1):

B = (B1, B2, B3)T is a true magnetic induction vector;

b = (b1, b2, b3)T is a constant vector corresponding to zero offsets for each of the measuring axes of the MU;

n is a random vector corresponding to uncorrelated noise for each of the measurement axes;

Р is a matrix, the rows of which are the unit vectors of the measuring axes of the MU, written in the “base” MU system;

Q is a diagonal matrix containing on the main diagonal the scaling factors k1, k2, kfor measuring axes MU i.e

.Q=k1000k2000k3.

In other words, the matrix P describes the non-orthogonality of the MU measurement axes, and the matrix Q corresponds to scaling along these axes.

The task of calibrating the MU measuring axes comes down to finding the elements of the matrices P and Q, as well as the zero offset vector b.

2. Development of an algorithm for determining the calibration parameters of the MU

 When solving the problem of MU determining the calibration parameters, we will use the fact that for measurements with any  MU spatial orientation, the value of the measured magnetic induction vector B is preserved and is a known model value.

We will assume that in flight, as a result of magnetometer measurements at discrete moments in time, a set of vectors is obtained h(l) = (h1(l), h2(l), h3(l))T, l = 1, …, N. Provided there is no measurement noise, from formula (1) we obtain:

h(l)=QPB(l)+b,l=1,,N,                                                                          (2)

where B(l) = (B1(l), B2(l), B3(l))T is the true vector of magnetic induction at the same point in space as the measured vector h(l), l = 1, …, N. We express vectors   (l = 1, …, N) from equalities  (2):

B(l)=Sh(l)b,                                                                                               (3)

l = 1, …, N, где

S=QP1=P1Q1.                                                                                    (4)

As noted in [1], without loss of generality, the non-orthogonality matrix P can be represented with a minimum number of unknown elements as follows:

P=100sin ε1cos ε10sin ε2cos ε2sin ε3cos ε2cos ε3,

where ε1, ε2, ε3 are small angles. Then the inverse matrix P – 1 will take the form

P1=100α1β10α1α3α2β3β1α3β2β3,

where αi = tg εi, βi = sec εi, i = 1, 2, 3, and since

Q1=k11000k21000k31,

then in accordance with (4) we obtain:

S=γ100α1γ1β1γ20α1α3α2β3γ1β1α3γ2β2β3γ3,                                        (5)

where γi = ki – 1, i = 1, 2, 3.

We rewrite equality (3) taking into account (5):

B1(l)B2(l)B3(l)=γ100α1γ1β1γ20α1α3α2β3γ1β1α3γ2β2β3γ3h1(l)b1h2(l)b2h3(l)b3,               (6)

l = 1, …, N. We write each of the N vector equalities (6) as a system of three scalar equalities:

γ1h1(l)b1=B1(l),α1γ1h1(l)b1+β1γ2h2(l)b2=B2(l),α1α3α2β3γ1h1(l)b1β1α3γ2h2(l)b2+β2β3γ3h3(l)b3=B3(l),                 (7)

l = 1, …, N. Due to the first equality of system (7), we rewrite the second equality of this system in the form

α1B1(l)+β1γ2h2(l)b2=B2(l),

β1γ2h2(l)b2=α1B1(l)+B2(l),

as a result of which the third equality of system (7) is written as follows:

α2β3B1(l)α3B2(l)+β2β3γ3h3(l)b3=B3(l),

l = 1, …, N.

We introduce into consideration a function of twelve variables αi, βi, γi, bi (i = 1, 2, 3)

Φ=l=1NB1(l)γ1h1(l)b12+B2(l)+α1B1(l)β1γ2h2(l)b22++B3(l)+α2β3B1(l)+α3B2(l)β2β3γ3h3(l)b32,

where the variables  αi и βi are related by the trigonometric identity

tg2φsec2​φ1

by the equalities

αi2βi2+1=0,                                                                         (8)

i = 1, 2, 3.

The algorithm for solving the problem of calibrating the MU measuring axes is reduced to searching by the least squares method [13] taking into account (8) such values of variables αi, βi, γi, bi (i = 1, 2, 3), which, for a given set of measurement vectors,   {h(l)} (l = 1, …, N) provide a minimum of the function F. For this purpose, it is necessary to investigate the function F for a conditional extremum [14] in the presence of three constraint equations (8).

We compose the Lagrange function

F=Φ+i=13λiαi2βi2+1,                                                                                  (9)

fifteen variables  αi, βi, γi, bi, λi (i = 1, 2, 3) and we write down the necessary condition for the local extremum of this function:

    

It is required to find the stationary points of the function F, i.e. the solution of the system of equations (10). Let us introduce the following notations:

Ci=l=1NBi(l),  Hi=l=1Nhi(l),  Fi=l=1Nhi(l)2,  Dij=l=1NBi(l)hj(l),  Gij=Gj ​​​i=l=1NBi(l)Bj(l),  i,j=1,2,3.

Taking into account these notations, for convenience we will divide the system of equations (10) into three nonlinear systems – a system of equations for two unknowns b1, γ1:

D11C1b1+F1+2H1b1Nb12γ1=0,C1+Nb1H1γ1γ1=0,                                                       (11)

a system of equations for five unknowns  b2, α1, β1, γ2, λ1:

Ci=l=1NBi(l),  Hi=l=1Nhi(l),  Fi=l=1Nhi(l)2,  Dij=l=1NBi(l)hj(l),  Gij=Gj ​​​i=l=1NBi(l)Bj(l),  i,j=1,2,3.

C1b2D12β1γ2+G11+λ1α1+G12=0,D22C2b2+D12C1b2α1+F2+2H2b2Nb22β1γ2γ2+λ1β1=0,D22C2b2+D12C1b2α1+F2+2H2b2Nb22β1γ2β1=0,C2+C1α1+Nb2H2β1γ2β1γ2=0,β1=α12+1       (12)

and a system of equations for eight unknowns.  b3, α2, α3, β2, β3, γ3, λ2, λ3:

G13+G11α2β3+G12α3+C1b3D13β2β3γ3β3+λ2α2=0,D33C3b3+D13C1b3α2β3+D23C2b3α3+F3+2H3b3Nb32β2β3γ3β3γ3+λ2β2=0,D33C3b3+D13C1b3α2β3+D23C2b3α3+F3+2H3b3Nb32β2β3γ3β2β3=0,C3+C1α2β3+C2α3+Nb3H3β2β3γ3β2β3γ3=0,G23+G12α2β3+G22α3+C2b3D23β2β3γ3+λ3α3=0,G13+G11α2β3+G12α3+C1b3D13β2β3γ3α2λ3β3D33C3b3+D13C1b3α2β3+D23C2b3α3+F3+2H3b3Nb32β2β3γ3β2γ3=0,βi=αi2+1,   i=2,3.(13)

неравенства Коши

(13)

In writing the expressions βi through αi in the last equations of systems (12) and (13), we used the fact that βi = sec εi > 0 due to the obvious inequalities–π/2 < εi < π/2, i = 1, 2, 3. We solve each of the systems of equations (11), (12), (13).

First, we note that the inequality is valid

Fi+2HibiNbi20,                                                                                    (14)

i = 1, 2, 3. Indeed, the discriminant of the quadratic equation is – Fi + 2HibiNbi2 = 0 with respect to the unknown bi, being equal to

δi=4Hi24NFi=4l=1Nhi(l)2Nl=1Nhi(l)2,

is negative due to the obvious consequence

l=1Nhi(l)2<Nl=1Nhi(l)2

of the Cauchy–Bunyakovsky inequality [15]  (here the sign “<” is used instead of “≤”, since at least one of the terms hi(1), …, hi(N) is obviously not equal to zero), therefore the quadratic equation under consideration has no real roots, i = 1, 2, 3.

We solve the system of equations (11). Taking into account inequality (14), we express γ1 through b1 from the first equation of this system:

γ1=D11C1b1F12H1b1+Nb12.

Substituting the obtained expression for γ1 into the second equation of system (11) and taking into account the inequality γ1 = k1– 1 ≠ 0, after simple transformations we arrive at an equality,

b1=C1F1D11H1C1H1ND11,

taking into account which the equality written above for γ1 is reduced to the form

b1=C1F1D11H1C1H1ND11,

Thus, the solution to the system of equations (11) is found.

We consider the system of equations (12). Since  γ2 = k2 – 1 ≠ 0, the second equation of this system can be written as

D22C2b2+D12C1b2α1+F2+2H2b2Nb22β1γ2+λ1β1γ2=0,                 (15)

since β1 = sec ε1 ≠ 0, the third equation of system (12) is equivalent to the following equation:

D22C2b2+D12C1b2α1+F2+2H2b2Nb22β1γ2=0.                                (16)

From (15), taking into account (16), we obtain that λ1β1γ2– 1 = 0 and then, by virtue of the inequality β1 ≠ 0, we have

λ1 = 0                                                                         (17)

The inequality D12 – C1b2 ≠ 0 holds, since a direct check shows that the value b2 = D12C1– 1, which is the root of the equation D12C1b2 = 0, does not satisfy the equations of system (12)

First, we consider the case H2Nb2 ≠ 0, i.e. b2H2N– 1. We express β1γ2 from (16), as well as from the first (taking into account equality (17)) and fourth (taking into account the inequality β1γ2 ≠ 0) equations of system (12):

β1γ2=D12C1b2α1+D22C2b2F22H2b2+Nb22, β1γ2=G11α1+G12D12C1b2, β1γ2=C1α1+C2H2Nb2.        (18)

As proven, the denominator of the fraction in the second of the equalities (18) is not equal to zero, the denominator of the fraction in the first of the equalities (18) is different from zero by virtue of (14). From the first and third equalities (18) follows the equation

C1α1+C2H2Nb2=D12C1b2α1+D22C2b2F22H2b2+Nb22,

we express b2 through α1:

b2=D12H2C1F2α1+D22H2C2F2ND12C1H2α1+ND22C2H2.                                                                          (19)

Similarly, from the second and third equalities (18) follows the equation

C1α1+C2H2Nb2=G11α1+G12D12C1b2,

we express b2 through α1:

b2=G11H2C1D12α1+G12H2C2D12NG11C12α1+NG12C1C2.                                                                        (20)

Equating the right-hand sides of equalities (19) and (20), after simple transformations we obtain a quadratic equation with respect to α1:

C12C1F2D12H2+C1G11H22NF2+C1D12ND12C1H2α12++2C1C2C1F2D12H2+ND12C1H2C1D22+C2D12+H22NF2C1G12+C2G11α1++C22C1F2D12H2+C2G12H22NF2+C2D22ND12C1H2=0.  (21)

The discriminant of equation (21) is equal to

and the roots of this equation are calculated using the formulae

α1=C2C1,α1=C2C1F2D12H2+G12H22NF2+D22ND12C1H2C1C1F2D12H2+G11H22NF2+D12ND12C1H2.                              (22)

 We note that the values (22) of the unknown α1 are not the roots of the denominators of the fractions in equalities (19) and (20). Substituting the first of the values (22) of the unknown α1 into the fourth equation of system (12), taking into account the inequality β1γ2 ≠ 0, we arrive at the equality H2Nb2 = 0, which contradicts the case under consideration.

 Now H2Nb2 = 0, i.e. b2 = H2N– 1. Taking into account the inequality β1γ2 ≠ 0, from the fourth equation of system (12) we obtain that α1 = – C2C1– 1, i.e. the found value of α1 coincides with the first of the roots (22) of equation (21). It is easy to show that equalities (19) and (20), proven under the assumption b2H2N– 1, are also valid forb2 = H2N– 1. In the case b2 = H2N– 1, the first two equalities from (18) also hold, since no restrictions on the value of the unknown b2 were taken into account when deriving them.

From the above reasoning it follows that the system of equations (12) has two solutions. The components of both solutions of the specified system are obtained in the following order:

  • – the values of the unknown α1 are found by formulae (22);
  • – the values of β1 corresponding to the found values of α1 are calculated by the formula written as the last equation of the system (12);
  • – the values of b2 – by any of the formulae (19), (20);
  •  – the values of γ2 – by any of the two formulae

γ2=D12C1b2α1+D22C2b2β1F22H2b2+Nb22, γ2=G11α1+G12β1D12C1b2,   

which follow respectively from the first and second equalities (18);

– the value of the unknown λ1 in both solutions of the system of equations (12) due to (17) is taken to be equal to zero.

We move on to finding solutions to the system of equations (13). Due to the inequalities γ3 = k3– 1 ≠ 0, β2 = sec ε2 ≠ 0 and β3 = sec ε3 ≠ 0 the second and third equations of this system can be written as

D33C3b3+D13C1b3α2β3+D23C2b3α3+F3+2H3b3Nb32β2β3γ3+λ2β2β3γ3=0,    (23)

D33C3b3+D13C1b3α2β3+D23C2b3α3+F3+2H3b3Nb32β2β3γ3=0                  (24)

respectively. From (23), taking into account (24), we obtain: λ2β23γ3)–1 = 0. Consequently, the equality is valid

λ2=0.                                                                                                         (25)

 Then the first equation of system (13), taking into account the inequalityβ3 ≠ 0, can be written in the form

G13+G11α2β3+G12α3+C1b3D13β2β3γ3=0.                                      (26)

 From the sixth equation of system (13), taking into account (24) and (26), we find – λ3β3 = 0, from which the equality follows

λ3=0.                                                                                                         (27)

The inequalities H3Nb3 ≠ 0, Di3Cib3 ≠ 0 take place, since a direct check shows that neither the root b3 = H3N– 1 of the equation H3Nb3 = 0, nor the roots b3 = Di3Ci– 1 of the equations Di3Cib3 = 0 satisfy the equations of system (13), i = 1, 2.

Taking into account equalities (25), (27) and inequalities β2 ≠ 0, β3 ≠ 0, γ3 ≠ 0, we express β2β3γ3 from the first, third, fourth and fifth equations of system (13):

β2β3γ3=G13+G11α2β3+G12α3D13C1b3, β2β3γ3=D33C3b3+D13C1b3α2β3+D23C2b3α3F32H3b3+Nb32,     β2β3γ3=C3+C1α2β3+C2α3H3Nb3, β2β3γ3=G23+G12α2β3+G22α3D23C2b3.                         (28)

As proven, the denominators of the fractions in the first, third and fourth equalities (28) are not equal to zero, the denominator of the fraction in the second equalities (28) is different from zero by virtue of (14). From the first and third equalities (28) follows the equation

G13+G11α2β3+G12α3D13C1b3=C3+C1α2β3+C2α3H3Nb3,

from the second and third equalities (28) follows the equation

D33C3b3+D13C1b3α2β3+D23C2b3α3F32H3b3+Nb32=C3+C1α2β3+C2α3H3Nb3,

from the third and thorthequalities (28) follows the equation

C3+C1α2β3+C2α3H3Nb3=G23+G12α2β3+G22α3D23C2b3.

From the last three equations we express α2β3 through α3 and b3:

α2β3=G12H3Nb3C2D13C1b3α3+G13H3Nb3C3D13C1b3C1D13C1b3G11H3Nb3,           (29)

α2β3=H3Nb3D23C2b3C2Nb322H3b3+F3α3+H3Nb3D33C3b3C3Nb322H3b3+F3C1Nb322H3b3+F3H3Nb3D13C1b3, (30)

α2β3=C2D23C2b3G22H3Nb3α3+C3D23C2b3G23H3Nb3G12H3Nb3C1D23C2b3.           (31)

By equating the right-hand side of equality (29) to the right-hand sides of equalities (30) and (31), after simple transformations we obtain the equations

C2C1b3D132+G11Nb3H3C2b3D23G12Nb3H3C1b3D13C1C1b3D13C2b3D23++C1G12C2G11Nb322H3b3 +F3α3Nb322H3b3+F3C3G11C1G13C1C1b3D13C3b3D33   G13Nb3H3C1b3D13+G11Nb3H3C3b3D33+C3C1b3D132H3Nb3=0,

  C2G12C1G22C1b3D13+G11G22G122Nb3H3+C1G12C2G11C2b3D23α3++C3G12C1G23C1b3D13+G11G23G12G13Nb3H3+C1G13C3G11C2b3D23H3Nb3=0,

from which we express α3 through b3, having previously divided both parts of each of them by (H3Nb3):

α3=H3b3F3C1G13C3G11+Nb3H3D33G11D13G13+C1b3D13C3D13C1D33H3b3F3C2G11C1G12+Nb3H3D13G12D23G11+C1b3D13C1D23C2D13,  (32)

α3=D23C2b3C1G13C3G11+Nb3H3G12G13G11G23+C1b3D13C1G23C3G12D23C2b3C2G11C1G12+Nb3H3G11G22G122+C1b3D13C2G12C1G22. (33)

Having equated the right-hand sides of equalities (32) and (33), after the simplest transformations we arrive at a quadratic equation with respect to b3

Γb32+Λb3+Ω=0,                                                           (34)

the coefficients of which are determined as follows:

Γ=G13C1H3ND13+C3C1D13G11H3+D33NG11C12G22NG11C12+G12C1C2NG12++C2C1G12C2G11+G23NG11C12+C3C1G12C2G11+G13C1C2NG12D23C12NG11+            +C2G11H3C1D13+G12ND13C1H3,    

Λ=G13C1H3ND13+C3C1D13G11H3+D33NG11C12G22C1D13G11H3+G12G12H3C2D13++D23C2G11C1G12+F3C3G11C1G13+D33C1D13G11H3+D13G13H3C3D13G12C1C2NG12++G22NG11C12+C2C1G12C2G11+G23NG11C12+C3C1G12C2G11+G13C1C2NG12××F3C1G12C2G11+D13C2D13G12H3+D23G11H3C1D13+G12G13H3C3D13+G23C1D13G11H3+    +D23C3G11C1G13C2G11H3C1D13+G12ND13C1H3+D23C12NG11,

Ω=F3C3G11C1G13+D33​​C1D13G11H3+D13G13H3C3D13G22C1D13G11H3+G12G12H3C2D13++D23C2G11C1G12G12C3D13G13H3+G23G11H3C1D13+D23C1G13C3G11F3C1G12C2G11+           +D13C2D13G12H3+D23G11H3C1D13.

The discriminant of equation (34) is equal to

  Δ~=C22D13G13D33G11+G12H3C3G12C1G23+C2G11G23H3G11H3C1D13+F3G11××C2G23C3G22+F3G12C3G12C1G23+F3G13C1G22C2G12+C1D23D33G12D23G13+C2D23××D13G13D33G11+C3D23D23G11D13G12C12NG11+C2H3D23G11D13G12+G22H3C1D13G11H3C1C2D13D23C3G11C1G13+C2G11D13G23D33G12+C3G11D23G12D13G22+C1G13D13G22D23G12+   +C1G12D33G12D13G23ND13C1H3+C2G12C1D13C3D13C1D33+G11H3C1D33G13H32,

and the roots of this equation are calculated using the formulae

b3=C1D13G11H3C12NG11,b3=C3G11D232F3G22+D33G11G22H3C2D23+G11G23C2F3D23H3+D13G12G23H3C3D23++D13D23C1G23C3G12+G13H3D23G12D13G22+D13D33C2G12C1G22+F3G12C3G12C1G23++D33G12C1D23G12H3+D23G13C2D13C1D23+D132C3G22C2G23+F3G13C1G22C2G12××ND23G12G13G11G23+ND33G11G22G122+ND13G12G23G13G22+C3G11C2D23G22H3++C2G11G23H3C2D33+C3D13C1G22C2G12+C3G12G12H3C1D23+C2G13C2D13G12H3++C1G23C1D23G12H3+C1G13G22H3C2D23+C1D33C2G12C1G22+C1C2D33G12D13G231.(35)

Since equation (34) has two solutions, the system of equations (13) also has two solutions. The components of both solutions of this system are obtained in the following order:

  • – the values of the unknown b3 are found using formulae (35);
  • – the values of the unknown α3 corresponding to the found values of b3 are calculated using any of the formulae (33), (34);
  • – the values of β3 – using the formula written as the last equation of system (13);
  • – the values of α2 – using any of the three formulae

α2=G12H3Nb3C2D13C1b3α3+G13H3Nb3C3D13C1b3β3C1D13C1b3G11H3Nb3,

α2=H3Nb3D23C2b3C2Nb322H3b3+F3α3+H3Nb3D33C3b3C3Nb322H3b3+F3β3C1Nb322H3b3+F3H3Nb3D13C1b3,

α2=C2D23C2b3G22H3Nb3α3+C3D23C2b3G23H3Nb3β3G12H3Nb3C1D23C2b3,

which follow from equalities (29), (30) and (31),

respectively;

  • – the values of β2 – by the formula written as the last equation of system (13);
  •  – the values of γ3 – by any of the four formulae

γ3=G13+G11α2β3+G12α3β2β3D13C1b3, γ3=D33C3b3+D13C1b3α2β3+D23C2b3α3β2β3F32H3b3+Nb32,

γ3=C3+C1α2β3+C2α3β2β3H3Nb3, γ3=G23+G12α2β3+G22α3β2β3D23C2b3,

which follow from (28);

  • – the values of the unknowns λ2 and λ3 in both solutions of the system of equations (13) are taken to be equal to zero due to equalities (25) and (27)

Thus, the system of equations (11) has a single solution, and each of the systems (12), (13) has two solutions. Consequently, the number of solutions of the original system of equations (10), defined as the product of the number of solutions of systems (11), (12) and (13), is four. Of the four solutions obtained,

α1j, α2j, α3j, β1j, β2j, β3j,γ1j,γ2j,γ3j,b1j,b2j,b3j,λ1j,λ2j,λ3j                      (36)

(j = 1, 2, 3, 4) the systems (10), which are also stationary points of the Lagrange function F, defined by equality (9), we are interested in the solution

α~1, α~2, α~3, β~1,β~2,β~3,γ~1,γ~2,γ~3,b~1,b~2,b~3,λ~1,λ~2,λ~3,

satisfying the equality

   Fα~1, α~2, α~3, β~1,β~2,β~3,γ~1,γ~2,γ~3,b~1,b~2,b~3,λ~1,λ~2,λ~3,==min1j4Fα1j, α2j, α3j, β1j, β2j, β3j,γ1j,γ2j,γ3j,b1j,b2j,b3j,λ1j,λ2j,λ3j.

Since in each of the solutions (36) of the system of equations (10) the last three components are equal to zero, i.e.

λ1j=λ2j=λ3j=0,

j = 1, 2, 3, 4, by virtue of (9) we have:

Fα1j, α2j, α3j, β1j, β2j, β3j,γ1j,γ2j,γ3j,b1j,b2j,b3j,λ1j,λ2j,λ3j=  =Φα1j, α2j, α3j, β1j, β2j, β3j,γ1j,γ2j,γ3j,b1j,b2j,b3j,

j = 1, 2, 3, 4, therefore, we determine the desired solution  α~1, α~2, α~3, β~1,β~2,β~3,γ~1,γ~2,γ~3,b~1,b~2,b~3​​,

using the formula

Φα1,α2,α3, β1, β2, β3,γ~1,γ~2,γ~3,b~1,b~2,b~3​​==min1j4Φα1j, α2j, α3j, β1j, β2j, β3j,γ1j,γ2j,γ3j,b1j,b2j,b3j.

The required values of the scale factors k~1,k~2,k~3 for the measuring axes of the MU are calculated using the formulae k~i=γ~i1,

i = 1, 2, 3, required angle values ε~1,ε~2,ε~3 are calculated using the formulae ε~i=arctg α~i,

which follow from the equalities α~i=tg ε~i

and obvious inequalities π/2<ε~i<π/2, i = 1, 2, 3.

Conclusion

Thus, we have obtained an analytical solution to the problem of calibrating the magnetometer of a spacecraft for the model considered in [1]. The procedure for calculating the calibration parameters of the MU using the derived formulas has a number of obvious advantages over numerical methods for solving this problem:

  • – the number of arithmetic operations is significantly reduced;
  • – the problem of possible instability of the method disappears.
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作者简介

Kirill Kirillov

Reshetnev Siberian State University of Science and Technology

编辑信件的主要联系方式.
Email: kkirillow@yandex.ru
ORCID iD: 0000-0002-3763-1303

Dr. Sc. (Phys. and Math.), Associate Professor, Professor of the Department of Applied Mathematics

俄罗斯联邦, 31, Krasnoyarskii Rabochii prospekt, Krasnoyarsk, 660037

Svetlana Kirillova

Reshetnev Siberian State University of Science and Technology

Email: svkirillova2009@yandex.ru
ORCID iD: 0000-0003-3779-2825

Cand. Sc. (Technical Sciences), Associate Professor, Associate Professor of the Department of Applied Mathematics and Data Analysis

俄罗斯联邦, 31, Krasnoyarskii Rabochii prospekt, Krasnoyarsk, 660037

Alexander Kuznetsov

Reshetnev Siberian State University of Science and Technology

Email: alex_kuznetsov80@mail.ru
ORCID iD: 0000-0003-0944-1817

Dr. Sc., Professor, Head of the Institute of Space Research and High Technologies

俄罗斯联邦, 31, Krasnoyarskii Rabochii prospekt, Krasnoyarsk, 660037

Denis Melent'ev

Siberian Federal University; JSC “Information Satellite Systems” Academician M. F. Reshetnev Company”

Email: denes.2000@mail.ru
ORCID iD: 0009-0009-6187-4098

Graduate Student, Engineer

俄罗斯联邦, 79, Svobodny Av., Krasnoyarsk, 660041; 52, Lenin St., Zheleznogorsk, Krasnoyarsk region, 662972

Konstantin Safonov

Reshetnev Siberian State University of Science and Technology

Email: safonovkv@rambler.ru

Dr. Sc. (Phys. and Math.), Professor, Head of the Institute of Informatics and Telecommunications

俄罗斯联邦, 31, Krasnoyarskii Rabochii prospekt, Krasnoyarsk, 660037

参考

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