CN101923169A - Atmospheric model for high-precision orbit confirmation of phased array radar - Google Patents

Abstract

The invention relates to an atmospheric model for high-precision orbit confirmation of a phased array radar, which aims at providing an atmospheric model applicable to the high-precision orbit confirmation of a phased array radar. Mathematic singularities in DTM 94 atmospheric model density calculation are removed. The invention has the technical scheme that the expression form of the atmospheric density Rho in the atmospheric model is shown in the specifications, wherein subscript i equals to 1, 2......5 which respectively correspond to five gas atoms or molecules of H, He, O, N2 and O2.

Description

Atmospheric models in the phased-array radar high-precision orbit

Technical field

The present invention relates to the Atmospheric models field, be specifically related to the Atmospheric models in the phased-array radar high-precision orbit.

Background technology

In Satellite Orbit Determination research, near-earth satellite particularly, atmospherical drag are a kind of perturbation sources that must consider.We relate to atmospheric density ρ and the atmospheric density partial derivative with respect to satellite position to the consideration of atmospherical drag in the phased-array radar high-precision orbit

Calculating.Atmospheric density ρ is provided by Atmospheric models, and therefore the research of Atmospheric models receive much concern.The people such as C.Berger of France are in the article " Improvement of theempirical thermospheric model DTM:DTM94-a comparative review of various temporalvariations and prospects in space geodesy app1ications " of Journalof Geodesy (1998) 72:161-178, the Atmospheric models of the still comparatively general level that is in a leading position, the i.e. hot Atmospheric models of DTM94 have in the world been proposed now.Domestic Jiang Hu, yellow alkali also has article " DTM94 Atmospheric models and application thereof " that these Atmospheric models are introduced in Yunnan Observatory platform 1999 the 4th phases of periodical, and its effect is affirmed.The hot Atmospheric models of DTM94 have characteristics such as precision height, computing velocity be fast, easy to use.

Yet we find in the process of exploitation phased-array radar high-precision orbit software, and in the use, the hot Atmospheric models of DTM94 are owing to adopt spherical coordinates as basic variable, so its atmospheric density is calculated ball the two poles of the earth and had incalculable singular point.Simultaneously, if directly the partial derivative of DTM94 atmospheric density with respect to the solid square coordinate in ground calculated in differentiate, then can on denominator, occur

The factor, the molecule of overhead partial derivative and denominator can appear as zero situation simultaneously at the two poles of the earth, cause near the partial derivative in the sky, the two poles of the earth to calculate and may collapse.Because the existence of this singular point causes correlation computations to lose efficacy near the two poles of the earth, has had a strong impact on the versatility and the robustness of software.Therefore we need seek a kind of improving one's methods, and can make phased-array radar high-precision orbit software be unlikely to singular point problem partial failure because of Atmospheric models.

Consider that from physical angle it is continuous that atmospheric density distributes, the singular point that do not have exists.And the hot Atmospheric models of DTM94 are when calculating atmospheric density and atmospheric density with respect to the partial derivative of the solid square coordinate in ground and since adopt be spherical coordinates as basic variable, so, have incalculable singular point in the terrestrial pole sky.This is mainly reflected in following several:

Sunday item:

D＝{a ₂₁P ₁₁+a ₂₂P ₃₁+a ₂₃P ₅₁+(a ₂₄P ₁₁+a ₂₅P ₂₁)cos[Ω(d-a ₁₈)]}cosωt

+{a ₂₆P ₁₁+a ₂₇P ₃₁+a ₂₈P ₅₁+(a ₂₉P ₁₁+a ₃₀P ₂₁)cos[Ω(d-a ₁₈)]}sinωt

Half cycle day item:

SD＝{a ₃₁P ₂₂+a ₃₂P ₃₂cos[Ω(d-a ₁₈)]}cos2ωt

+{a ₃₃P ₂₂+a ₃₄P ₃₂cos[Ω(d-a ₁₈)]}sin2ωt

1/3 Sunday item:

TD＝a ₃₅P ₃₃cos3ωt+a ₃₆P ₃₃sin3ωt；

The geomagnetic activity item:

Wherein

λ in the top formula,

Be the sub-satellite point longitude and latitude, t is the local true solar time of sub-satellite point, a _NmBe normal value coefficient, can from the hot Atmospheric models coefficient table of DTM94, look into and get, ω=2 pi/2s, 4 (hour ^-1), Ω=2 π/365 (day ^-1) also be constant, P _NmBe about Association rein in allow polynomial expression.As seen, in the sky, the two poles of the earth, the local true solar time t of substar longitude λ and substar all loses definition, becomes above several mathematics singular point, causes the DTM94 mathematical model can't calculate at the two poles of the earth.But it is not an essential singularity, and the appropriate mathematical method just should be able to be overcome and solve by reference.

Summary of the invention

The object of the present invention is to provide a kind of Atmospheric models that are applicable in the phased-array radar high-precision orbit, eliminate the mathematics singular point in the density calculation of DTM94 Atmospheric models.The present invention is based on the hot Atmospheric models of DTM94,, the atmospheric density of any can be calculated, realize that the no singular point of atmospheric density calculates improving of atmospheric density calculating method in the DTM94 Atmospheric models.Further goal of the invention of the present invention is to provide the no singular point computing method of atmospheric density with respect to ground solid square coordinate partial derivative.

The present invention uses for reference (Pines, 1973) and eliminates the used method of the terrestrial gravitation humorous expression formula singular point of gesture ball, with satellite the earth's core square r and direction cosine θ ₁, θ ₂, θ ₃The spherical coordinates r that replaces it, λ, As the basic variable of the hot Atmospheric models of DTM94, make λ and t no longer to occur in the formula.Make square coordinate and direction cosine that definition all be arranged in the arbitrfary point, therefore calculate meaning point in office and can realize, do not have mathematics singular point problem through the atmospheric density of transforming.Make that simultaneously atmospheric density can not be zero the collapse because of the molecule denominator with respect to the calculating of ground solid square coordinate partial derivative simultaneously.

The technical scheme that realizes the object of the invention is:

Atmospheric models in the phased-array radar high-precision orbit, the expression-form of atmospheric density ρ is:

Subscript i=1,2 ... 5, the corresponding H of difference, H _e, O, N ₂, O ₂This five kinds of gas atoms or molecule.

In the following formula,

A _VR=1/ (6.022 * 10 ²³) be the inverse of avogadros constant;

M _iBe the molecular weight of neutral gas composition i, M ₁=1, M ₂=4, M ₃=16, M ₄=28, M ₅=32;

n _i(h)=A _iExp (G _i(L)) f _i(h), h is the floor level of satellite;

Following formula the right symbol description is as follows:

$f_i\left(h\right)={(T_{120}/T\left(h\right))}^{1+{\mathrm{\α}}_i+{\mathrm{\γ}}_i}\exp (-\mathrm{\σ}{\mathrm{\γ}}_i\mathrm{\ζ}),$

This wherein,

T ₁₂₀=380K represents 120 kilometers highly temperature at place,

The temperature at expression floor level h place,

T _∞=A _T(1+G _T(L)), the expression Outer Atmospheric Temperature,

α _iBe the thermal diffusion coefficient of neutral gas composition i, α ₁=-0.40, α ₂=-0.38, α ₃=α ₄=α ₅=0.0,

γ _i＝M _ig ₁₂₀/(σRT _∞)，

R=831.4 is a gas law constant,

g ₁₂₀=944.6626 (cm/s ²), the gravitational acceleration at 120 kilometers places of expression floor level,

σ=T ' ₁₂₀/ (T _∞-T ₁₂₀), represent relative vertical temperature gradient,

T ' ₁₂₀=14.348, the thermograde at 120 kilometers places of expression,

Expression terrestrial gravitation gesture height,

R _ρ=6356.77km, expression earth polar radius,

Expression satellite ground height,

e ²=2 ε-ε ², expression earth meridian circle excentricity,

ε=1/298.25, expression earth meridian circle ellipticity.

And other coefficients:

A ₅＝4.775×10 ¹⁰(cm ^-3)，

A ₁, A ₂, A ₃, A ₄And A _TValue is looked in can the coefficient table by DTM94 model (Berger et al, 1998) and is got, and promptly is respectively the α of each neutral gas composition and temperature T correspondence in the table ₁Value;

For G _i(L), be described as follows:

G ₅(L)＝0，

G ₁(L), G ₂(L), G ₃(L), G ₄(L) and G _T(L) unified below with G (L) expression, because their difference is with top A ₁, A ₂, A ₃, A ₄And A _TSimilar, only be the value of when looking into DTM94 model coefficient table, looking into each neutral gas composition and temperature T correspondence respectively.

G(L)＝LAT+F+M+β(AN1+SAN1+AN2+SAN2+D+SD+TD)；

Wherein,

The latitude item:

LAT＝a ₂H ₂₀+a ₃H ₄₀+a ₃₇H ₁₀；

The solar activity item:

$F=a_4(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})+a_5{(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})}^2+a_6({\stackrel{\‾}{F}}_{10.7}-150)+a_{38}{({\stackrel{\‾}{F}}_{10.7}-150)}^2;$

The geomagnetic activity item:

Symmetry anniversary item:

AN1＝(a ₉+a ₁₀H ₂₀)cos[Ω(d-a ₁₁)]；

Symmetry item half a year:

SAN1＝(a ₁₂+a ₁₃H ₂₀)cos[Ω(d-a ₁₄)]；

Asymmetric anniversary item:

AN2＝(a ₁₅H ₁₀+a ₁₆H ₃₀+a ₁₇H ₅₀)cos[Ω(d-a ₁₈)]；

Asymmetric half a year item:

SAN2＝a ₁₉H ₁₀cos[2Ω(d-a ₂₀)]；

Sunday item:

D＝(-B ₁cosλ _s+B ₂sinλ _s)θ ₁-(B ₁sinλ _s+B ₂cosλ _s)θ ₂；

Half cycle day item:

$\mathrm{SD}=(B_3\cos 2{\mathrm{\λ}}_S-B_4\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3\sin {2\mathrm{\λ}}_S+B_4\cos {2\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2;$

1/3 Sunday item:

$\mathrm{TD}=({-B}_5\cos 3{\mathrm{\λ}}_S+B_6\sin 3{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{3\mathrm{\θ}}_2^2){\mathrm{\θ}}_1+(B_5\sin {3\mathrm{\λ}}_S+B_6\cos {3\mathrm{\λ}}_S)({\mathrm{\θ}}_2^2-3{\mathrm{\θ}}_1^2){\mathrm{\θ}}_2;$

Coefficient is described as follows in the top formula:

B ₁＝a ₂₁H ₁₁+a ₂₂H ₃₁+a ₂₃H ₅₁+(a ₂₄H ₁₁+a ₂₅H ₂₁)cos[Ω(d-a ₁₈)]，

B ₂＝a ₂₆H ₁₁+a ₂₇H ₃₁+a ₂₈H ₅₁+(a ₂₉H ₁₁+a ₃₀H ₂₁)cos[Ω(d-a ₁₈)]，

B ₃＝a ₃₁H ₂₂+a ₃₂H ₃₂cos[Ω(d-a ₁₈)]，

B ₄＝a ₃₃H ₂₂+a ₃₄H ₃₂cos[Ω(d-a ₁₈)]，

B ₅＝a ₃₅H ₃₃，

B ₆＝a ₃₆H ₃₃；

a _i(i=2,3 ... 39) value can look into DTM94 model coefficient table (Berger et, al.1998);

The serve as reasons day of year number of year first calculation of d;

F _10.7Be sun 10.7cm radiant flux, with d-1 days values;

Be sun 10.7cm radiant flux mean value, average in three solar rotation cycles (81 days) before the d;

K _ρBe 3 hours geomagnetic indicess, get calculate constantly before

Hour value, wherein

Being the geocentric latitude of sub-satellite point, is unit with the degree;

Ω＝2π/365(day ^-1)；

H ₁₀＝θ ₃，

$H_{20}=\frac{3}{2}{\mathrm{\θ}}_3^2-\frac{1}{2},$

$H_{30}=\frac{1}{2}(5{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3,$

$H_{40}=\frac{1}{8}(35{\mathrm{\θ}}_3^4-30{\mathrm{\θ}}_3^2+3),$

$H_{50}=\frac{1}{8}({63\mathrm{\θ}}_3^4-70{\mathrm{\θ}}_3^2+15){\mathrm{\θ}}_3,$

H ₁₁＝1

H ₂₁＝3θ ₃，

$H_{31}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1),$

$H_{51}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15),$

H ₂₂＝3，

H ₃₂＝15θ ₃，

H ₃₃＝15

About being described as follows of coordinate amount:

Satellite the earth's core square r and direction cosine θ ₁, θ ₂, θ ₃With the pass of its spherical coordinates be:

X, y, z are the square coordinate of satellite in body-fixed coordinate system, and they can calculate in the sky, the two poles of the earth.

With The expression square coordinate of the sun in body-fixed coordinate system, the pass of it and spherical coordinates is:

As a further improvement on the present invention, in the described Atmospheric models, further comprise the computing method of atmospheric density with respect to ground solid square coordinate partial derivative, atmospheric density with respect to the computing method of ground solid square coordinate partial derivative is:

If X ₁=x, X ₂=y, X ₃=z represents the square coordinate of satellite in body-fixed coordinate system, then has:

$\frac{\∂\mathrm{\ρ}}{\∂X_j}=\frac{1}{r}\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}+(\frac{\∂\mathrm{\ρ}}{\∂r}-\frac{1}{r}{\mathrm{\Σ}}_{i=1}^3\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_i}{\mathrm{\θ}}_i){\mathrm{\θ}}_j,(j=\mathrm{1,2,3}),$

$\frac{\∂\mathrm{\ρ}}{\∂r}=-({\mathrm{\ρ}}_2\mathrm{\σ}+(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{T_{120}^{\′}}{T\left(h\right)}\exp (-\mathrm{\σ\ζ})){\left(\frac{R_{\mathrm{\ρ}}+120}{R_{\mathrm{\ρ}}+h}\right)}^2;$

For j=1,2:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_j}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3};$

For j=3:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_3}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$-R_{\mathrm{\ρ}}{e}^2{(1-e^2+e^2{\mathrm{\θ}}_3^2)}^{-\frac{3}{2}}{\mathrm{\θ}}_3\frac{\∂\mathrm{\ρ}}{\∂r};$

In the top formula,

${\mathrm{\ρ}}_1=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\α}}_i$

${\mathrm{\ρ}}_2=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\γ}}_i$

$\frac{\∂G\left(L\right)}{\∂{\mathrm{\θ}}_3}=\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}+\frac{\∂M}{\∂{\mathrm{\θ}}_3}+\mathrm{\β}(\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂D}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3})$

$\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}=a_2{H}_{20}^{\′}+a_3{H}_{40}^{\′}+a_{37}{H}_{10}^{\′}$

$\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}=a_{10}{H}_{20}^{\′}\cos \left[\mathrm{\Ω}\left({d-a}_{11}\right)\right]$

$\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}=a_{13}{H}_{20}^{\′}\cos \left[2\mathrm{\Ω}(d-a_{14})\right]$

$\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}=(a_{15}{H}_{10}^{\′}+a_{16}{H}_{30}^{\′}+a_{17}{H}_{50}^{\′})\cos \left[\mathrm{\Ω}\left({d-a}_{18}\right)\right]$

$\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}=a_{19}{H}_{10}^{\′}\cos \left[2\mathrm{\Ω}\left({d-a}_{20}\right)\right]$

$\frac{\∂D}{\∂{\mathrm{\θ}}_3}=(-B_1^{\′}\cos {\mathrm{\λ}}_S+B_2^{\′}\sin {\mathrm{\λ}}_S){\mathrm{\θ}}_1-(B_1^{\′}\sin {\mathrm{\λ}}_S+B_2^{\′}\cos {\mathrm{\λ}}_S){\mathrm{\θ}}_2$

$\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3}=(B_3^{\′}\cos 2{\mathrm{\λ}}_S-B_4^{\′}\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3^{\′}\sin 2{\mathrm{\λ}}_S+B_4^{\′}\cos 2{\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2$

B′ ₁＝a ₂₂H′ ₃₁+a ₂₃H′ ₅₁+a ₂₅H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₂＝a ₂₇H′ ₃₁+a ₂₈H′ ₅₁+a ₃₀H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₃＝a ₃₂H′ ₃₂cos[Ω(d-a ₁₈)]

B′ ₄＝a ₃₄H′ ₃₂cos[Ω(d-a ₁₈)]

H?′ ₁₀＝1

H′ ₂₀＝3θ ₃

H′ ₂₁＝3

$H_{30}^{\′}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1)$

H′ ₃₁＝15θ ₃

H′ ₃₂＝15

$H_{40}^{\′}=\frac{5}{2}(7{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3$

$H_{50}^{\′}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15)$

$H_{51}^{\′}=\frac{15}{2}(21{\mathrm{\θ}}_3^2-7){\mathrm{\θ}}_3.$

The coefficient that does not directly illustrate in the above formula all has explanation in atmospheric density is calculated.

Beneficial effect of the present invention is the hot Atmospheric models in conjunction with DTM94, provided its atmospheric density and atmospheric density no singular point computing formula with respect to ground solid square coordinate partial derivative, for versatility and the robustness that improves low orbit satellite orbit determination software provides a practical tool, computing method of the present invention are expressed the no singular point of other Atmospheric models has reference function too.

Embodiment

Be described further below in conjunction with embodiment.

The present invention is successful Application in phased-array radar high-precision orbit software, and the method for the no singular point transformation of process makes total system to move continuously incessantly.

Atmospheric models in the phased-array radar high-precision orbit, the expression-form of atmospheric density ρ is:

Subscript i=1,2 ... 5, the corresponding H of difference, H _e, O, N ₂, O ₂This five kinds of gas atoms or molecule.

In the following formula,

A _VR=1/ (6.022 * 10 ²³) be the inverse of avogadros constant;

M _iBe the molecular weight of neutral gas composition i, M ₁=1, M ₂=4, M ₃=16, M ₄=28, M ₅=32;

n _i(h)=A _iExp (G _i(L)) f _i(h), h is the floor level of satellite;

Following formula the right symbol description is as follows:

$f_i\left(h\right)={(T_{120}/T\left(h\right))}^{1+{\mathrm{\α}}_i+{\mathrm{\γ}}_i}\exp (-\mathrm{\σ}{\mathrm{\γ}}_i\mathrm{\ζ}),$

This wherein,

T ₁₂₀=380K represents 120 kilometers highly temperature at place,

The temperature at expression floor level h place,

T _∞=A _T(1+G _T(L)), the expression Outer Atmospheric Temperature,

α _iBe the thermal diffusion coefficient of neutral gas composition i, α ₁=-0.40, α ₂=-0.38, α ₃=α ₄=α ₅=0.0,

γ _i＝M _ig ₁₂₀/(σRT _∞)，

R=831.4 is a gas law constant,

g ₁₂₀=944.6626 (cm/s ²), the gravitational acceleration at 120 kilometers places of expression floor level,

σ=T ' ₁₂₀/ (T _∞-T ₁₂₀), represent relative vertical temperature gradient,

T ' ₁₂₀=14.348, the thermograde at 120 kilometers places of expression,

Expression terrestrial gravitation gesture height,

R _ρ=6356.77km, expression earth polar radius,

Expression satellite ground height,

e ²=2 ε-ε ², expression earth meridian circle excentricity,

ε=1/298.25, expression earth meridian circle ellipticity.

And other coefficients:

A ₅＝4.775×10 ¹⁰(cm ^-3)，

A ₁, A ₂, A ₃, A ₄And A _TValue is looked in can the coefficient table by DTM94 model (Berger et al, 1998) and is got, and promptly is respectively the α of each neutral gas composition and temperature T correspondence in the table ₁Value;

For G _i(L), be described as follows:

G ₅(L)＝0，

G ₁(L), G ₂(L), G ₃(L), G ₄(L) and G _T(L) unified below with G (T) expression, because their difference is with top A ₁, A ₂, A ₃, A ₄And A _TSimilar, only be the value of when looking into DTM94 model coefficient table, looking into each neutral gas composition and temperature T correspondence respectively.

G(L)＝LAT+F+M+β(AN1+SAN1+AN2+SAN2+D+SD+TD)；

Wherein,

The latitude item:

LAT＝a ₂H ₂₀+a ₃H ₄₀+a ₃₇H ₁₀；

The solar activity item:

$F=a_4(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})+a_5{(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})}^2+a_6({\stackrel{\‾}{F}}_{10.7}-150)+a_{38}{({\stackrel{\‾}{F}}_{10.7}-150)}^2;$

The geomagnetic activity item:

Symmetry anniversary item:

AN1＝(a ₉+a ₁₀H ₂₀)cos[Ω(d-a ₁₁)]；

Symmetry item half a year:

SAN1＝(a ₁₂+a ₁₃H ₂₀)cos[Ω(d-a ₁₄)]；

Asymmetric anniversary item:

AN2＝(a ₁₅H ₁₀+a ₁₆H ₃₀+a ₁₇H ₅₀)cos[Ω(d-a ₁₈)]；

Asymmetric half a year item:

SAN2＝a ₁₉H ₁₀cos[2Ω(d-a ₂₀)]；

Sunday item:

D＝(-B ₁cosλ _s+B ₂sinλ _s)θ ₁-(B ₁sinλ _s+B ₂cosλ _s)θ ₂；

Half cycle day item:

$\mathrm{SD}=(B_3\cos 2{\mathrm{\λ}}_S-B_4\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3\sin {2\mathrm{\λ}}_S+B_4\cos {2\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2;$

1/3 Sunday item:

$\mathrm{TD}=({-B}_5\cos 3{\mathrm{\λ}}_S+B_6\sin 3{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{3\mathrm{\θ}}_2^2){\mathrm{\θ}}_1+(B_5\sin {3\mathrm{\λ}}_S+B_6\cos {3\mathrm{\λ}}_S)({\mathrm{\θ}}_2^2-3{\mathrm{\θ}}_1^2){\mathrm{\θ}}_2;$

Coefficient is described as follows in the top formula:

B ₁＝a ₂₁H ₁₁+a ₂₂H ₃₁+a ₂₃H ₅₁+(a ₂₄H ₁₁+a ₂₅H ₂₁)cos[Ω(d-a ₁₈)]，

B ₂＝a ₂₆H ₁₁+a ₂₇H ₃₁+a ₂₈H ₅₁+(a ₂₉H ₁₁+a ₃₀H ₂₁)cos[Ω(d-a ₁₈)]，

B ₃＝a ₃₁H ₂₂+a ₃₂H ₃₂cos[Ω(d-a ₁₈)]，

B ₄＝a ₃₃H ₂₂+a ₃₄H ₃₂cos[Ω(d-a ₁₈)]，

B ₅＝a ₃₅H ₃₃，

B ₆＝a ₃₆H ₃₃；

a _i(i=2,3 ... 39) value can look into DTM94 model coefficient table (Berger et, al.1998);

The serve as reasons day of year number of year first calculation of d;

F _10.7Be sun 10.7cm radiant flux, with d-1 days values;

Be sun 10.7cm radiant flux mean value, average in three solar rotation cycles (81 days) before the d;

K _ρBe 3 hours geomagnetic indicess, get calculate constantly before

Hour value, wherein

Being the geocentric latitude of sub-satellite point, is unit with the degree;

Ω＝2π/365(day ^-1)；

H ₁₀＝θ ₃，

$H_{20}=\frac{3}{2}{\mathrm{\θ}}_3^2-\frac{1}{2},$

$H_{30}=\frac{1}{2}(5{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3,$

$H_{40}=\frac{1}{8}(35{\mathrm{\θ}}_3^4-30{\mathrm{\θ}}_3^2+3),$

$H_{50}=\frac{1}{8}({63\mathrm{\θ}}_3^4-70{\mathrm{\θ}}_3^2+15){\mathrm{\θ}}_3,$

H ₁₁＝1

H ₂₁＝3θ ₃，

$H_{31}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1),$

$H_{51}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15),$

H ₂₂＝3，

H ₃₂＝15θ ₃，

H ₃₃＝15

About being described as follows of coordinate amount:

Satellite the earth's core square r and direction cosine θ ₁, θ ₂, θ ₃With the pass of its spherical coordinates be:

X, y, z are the square coordinate of satellite in body-fixed coordinate system, and they can calculate in the sky, the two poles of the earth.

With

The expression square coordinate of the sun in body-fixed coordinate system, the pass of it and spherical coordinates is:

As a further improvement on the present invention, in the described Atmospheric models, further comprise the computing method of atmospheric density with respect to ground solid square coordinate partial derivative, atmospheric density with respect to the computing method of ground solid square coordinate partial derivative is:

If X ₁=x, X ₂=y, X ₃=z represents the square coordinate of satellite in body-fixed coordinate system, then has:

$\frac{\∂\mathrm{\ρ}}{\∂X_j}=\frac{1}{r}\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}+(\frac{\∂\mathrm{\ρ}}{\∂r}-\frac{1}{r}{\mathrm{\Σ}}_{i=1}^3\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_i}{\mathrm{\θ}}_i){\mathrm{\θ}}_j,(j=\mathrm{1,2,3}),$

$\frac{\∂\mathrm{\ρ}}{\∂r}=-({\mathrm{\ρ}}_2\mathrm{\σ}+(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{T_{120}^{\′}}{T\left(h\right)}\exp (-\mathrm{\σ\ζ})){\left(\frac{R_{\mathrm{\ρ}}+120}{R_{\mathrm{\ρ}}+h}\right)}^2;$

For j=1,2:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_j}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3};$

For j=3:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_3}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$-R_{\mathrm{\ρ}}{e}^2{(1-e^2+e^2{\mathrm{\θ}}_3^2)}^{-\frac{3}{2}}{\mathrm{\θ}}_3\frac{\∂\mathrm{\ρ}}{\∂r};$

In the top formula,

${\mathrm{\ρ}}_1=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\α}}_i$

${\mathrm{\ρ}}_2=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\γ}}_i$

$\frac{\∂G\left(L\right)}{\∂{\mathrm{\θ}}_3}=\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}+\frac{\∂M}{\∂{\mathrm{\θ}}_3}+\mathrm{\β}(\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂D}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3})$

$\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}=a_2{H}_{20}^{\′}+a_3{H}_{40}^{\′}+a_{37}{H}_{10}^{\′}$

$\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}=a_{10}{H}_{20}^{\′}\cos \left[\mathrm{\Ω}\left({d-a}_{11}\right)\right]$

$\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}=a_{13}{H}_{20}^{\′}\cos \left[2\mathrm{\Ω}(d-a_{14})\right]$

$\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}=(a_{15}{H}_{10}^{\′}+a_{16}{H}_{30}^{\′}+a_{17}{H}_{50}^{\′})\cos \left[\mathrm{\Ω}\left({d-a}_{18}\right)\right]$

$\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}=a_{19}{H}_{10}^{\′}\cos \left[2\mathrm{\Ω}\left({d-a}_{20}\right)\right]$

$\frac{\∂D}{\∂{\mathrm{\θ}}_3}=(-B_1^{\′}\cos {\mathrm{\λ}}_S+B_2^{\′}\sin {\mathrm{\λ}}_S){\mathrm{\θ}}_1-(B_1^{\′}\sin {\mathrm{\λ}}_S+B_2^{\′}\cos {\mathrm{\λ}}_S){\mathrm{\θ}}_2$

$\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3}=(B_3^{\′}\cos 2{\mathrm{\λ}}_S-B_4^{\′}\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3^{\′}\sin 2{\mathrm{\λ}}_S+B_4^{\′}\cos 2{\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2$

B′ ₁＝a ₂₂H′ ₃₁+a ₂₃H′ ₅₁+a ₂₅H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₂＝a ₂₇H′ ₃₁+a ₂₈H′ ₅₁+a ₃₀H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₃＝a ₃₂H′ ₃₂cos[Ω(d-a ₁₈)]

B′ ₄＝a ₃₄H′ ₃₂cos[Ω(d-a ₁₈)]

H′ ₁₀＝1

H′ ₂₀＝3θ ₃

H′ ₂₁＝3

$H_{30}^{\′}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1)$

H′ ₃₁＝15θ ₃

H′ ₃₂＝15

$H_{40}^{\′}=\frac{5}{2}(7{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3$

$H_{50}^{\′}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15)$

$H_{51}^{\′}=\frac{15}{2}(21{\mathrm{\θ}}_3^2-7){\mathrm{\θ}}_3.$

The coefficient that does not directly illustrate in the above formula all has explanation in atmospheric density is calculated.

Claims (2)

1. the Atmospheric models in the phased-array radar high-precision orbit is characterized in that, the expression-form of atmospheric density ρ is:

Subscript i=1,2 ... 5, the corresponding H of difference, H _e, O, N ₂, O ₂This five kinds of gas atoms or molecule;

In the following formula,

A _VR=1/ (6.022 * 10 ²³) be the inverse of avogadros constant;

M _iBe the molecular weight of neutral gas composition i, M ₁=1, M ₂=4, M ₃=16, M ₄=28, M ₅=32;

n _i(h)=A _iExp (G _i(L)) f _i(h), h is the floor level of satellite;

Following formula the right symbol description is as follows:

$f_i\left(h\right)={(T_{120}/T\left(h\right))}^{1+{\mathrm{\α}}_i+{\mathrm{\γ}}_i}\exp (-\mathrm{\σ}{\mathrm{\γ}}_i\mathrm{\ζ}),$

This wherein,

T ₁₂₀=380K represents 120 kilometers highly temperature at place,

The temperature at expression floor level h place,

T _∞=A _T(1+G _T(L)), the expression Outer Atmospheric Temperature,

α _iBe the thermal diffusion coefficient of neutral gas composition i, α ₁=-0.40, α ₂=-0.38, α ₃=α ₄=α ₅=0.0,

γ _i＝M _ig ₁₂₀/(σRT _∞)，

R=831.4 is a gas law constant,

g ₁₂₀=944.6626 (cm/s ²), the gravitational acceleration at 120 kilometers places of expression floor level,

σ=T ₁₂₀/ (T-T ₁₂₀), represent relative vertical temperature gradient,

T ' ₁₂₀=14.348, the thermograde at 120 kilometers places of expression,

Expression terrestrial gravitation gesture height,

R _ρ=6356.77km, expression earth polar radius,

Expression satellite ground height,

e ²=2 ε-ε ², expression earth meridian circle excentricity,

ε=1/298.25, expression earth meridian circle ellipticity;

And other coefficient:

A ₅＝4.775×10 ¹⁰(cm ^-3)，

A ₁, A ₂, A ₃, A ₄And A _TValue is looked in can the coefficient table by the DTM94 model and is got, and promptly is respectively the α of each neutral gas composition and temperature T correspondence in the table ₁Value;

For G _i(L), be described as follows:

G ₅(L)＝0，

G ₁(L), G ₂(L), G ₃(L), G ₄(L) and G _T(L) unified below with G (L) expression, because their difference is with top A ₁, A ₂, A ₃, A ₄And A _TSimilar, only be the value of when looking into DTM94 model coefficient table, looking into each neutral gas composition and temperature T correspondence respectively;

G(L)＝LAT+F+M+β(AN1+SAN1+AN2+SAN2+D+SD+TD)；

Wherein,

The latitude item:

LAT＝a ₂H ₂₀+a ₃H ₄₀+a ₃₇H ₁₀；

The solar activity item:

$F=a_4(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})+a_5{(F_{10.7}-{\stackrel{\‾}{F}}_{10.7})}^2+a_6({\stackrel{\‾}{F}}_{10.7}-150)+a_{38}{({\stackrel{\‾}{F}}_{10.7}-150)}^2;$

The geomagnetic activity item:

Symmetry anniversary item:

AN1＝(a ₉+a ₁₀H ₂₀)cos[Ω(d-a ₁₁)]；

Symmetry item half a year:

SAN1＝(a ₁₂+a ₁₃H ₂₀)cos[Ω(d-a ₁₄)]；

Asymmetric anniversary item:

AN2＝(a ₁₅H ₁₀+a ₁₆H ₃₀+a ₁₇H ₅₀)cos[Ω(d-a ₁₈)]；

Asymmetric half a year item:

SAN2＝a ₁₉H ₁₀cos[2Ω(d-a ₂₀)]；

Sunday item:

D＝(-B ₁cosλ _s+B ₂sinλ _s)θ ₁-(B ₁sinλ _s+B ₂cosλ _s)θ ₂；

Half cycle day item:

$\mathrm{SD}=(B_3\cos 2{\mathrm{\λ}}_S-B_4\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3\sin {2\mathrm{\λ}}_S+B_4\cos {2\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2;$

1/3 Sunday item:

$\mathrm{TD}=({-B}_5\cos 3{\mathrm{\λ}}_S+B_6\sin 3{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{3\mathrm{\θ}}_2^2){\mathrm{\θ}}_1+(B_5\sin {3\mathrm{\λ}}_S+B_6\cos {3\mathrm{\λ}}_S)({\mathrm{\θ}}_2^2-3{\mathrm{\θ}}_1^2){\mathrm{\θ}}_2;$

Coefficient is described as follows in the top formula:

B ₁＝a ₂₁H ₁₁+a ₂₂H ₃₁+a ₂₃H ₅₁+(a ₂₄H ₁₁+a ₂₅H ₂₁)cos[Ω(d-a ₁₈)]，

B ₂＝a ₂₆H ₁₁+a ₂₇H ₃₁+a ₂₈H ₅₁+(a ₂₉H ₁₁+a ₃₀H ₂₁)cos[Ω(d-a ₁₈)]，

B ₃＝a ₃₁H ₂₂+a ₃₂H ₃₂cos[Ω(d-a ₁₈)]，

B ₄＝a ₃₃H ₂₂+a ₃₄H ₃₂cos[Ω(d-a ₁₈)]，

B ₅＝a ₃₅H ₃₃，

B ₆＝a ₃₆H ₃₃；

a _i(i=2,3 ... 39) value can be looked into DTM94 model coefficient table;

The serve as reasons day of year number of year first calculation of d;

F _10.7Be sun 10.7cm radiant flux, with d-1 days values;

Be sun 10.7cm radiant flux mean value, average in three solar rotation cycles before the d;

K _ρBe 3 hours geomagnetic indicess, get calculate constantly before

Hour value, wherein Being the geocentric latitude of sub-satellite point, is unit with the degree;

Ω＝2π/365(day ^-1)；

H ₁₀＝θ ₃，

$H_{20}=\frac{3}{2}{\mathrm{\θ}}_3^2-\frac{1}{2},$

$H_{30}=\frac{1}{2}(5{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3,$

$H_{40}=\frac{1}{8}(35{\mathrm{\θ}}_3^4-30{\mathrm{\θ}}_3^2+3),$

$H_{50}=\frac{1}{8}({63\mathrm{\θ}}_3^4-70{\mathrm{\θ}}_3^2+15){\mathrm{\θ}}_3,$

H ₁₁＝1

H ₂₁＝3θ ₃，

$H_{31}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1),$

$H_{51}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15),$

H ₂₂＝3，

H ₃₂＝15θ ₃，

H ₃₃＝15

About being described as follows of coordinate amount:

Satellite the earth's core square r and direction cosine θ ₁, θ ₂, θ ₃With the pass of its spherical coordinates be:

X, y, z are the square coordinate of satellite in body-fixed coordinate system;

With

The expression square coordinate of the sun in body-fixed coordinate system, the pass of it and spherical coordinates is:

2. the Atmospheric models in the phased-array radar high-precision orbit according to claim 1, it is characterized in that, described Atmospheric models further comprise the expression-form of atmospheric density with respect to ground solid square coordinate partial derivative, and atmospheric density with respect to the expression-form of ground solid square coordinate partial derivative is:

If X ₁=x, X ₂=y, X ₃=z represents the square coordinate of satellite in body-fixed coordinate system, then has:

$\frac{\∂\mathrm{\ρ}}{\∂X_j}=\frac{1}{r}\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}+(\frac{\∂\mathrm{\ρ}}{\∂r}-\frac{1}{r}{\mathrm{\Σ}}_{i=1}^3\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_i}{\mathrm{\θ}}_i){\mathrm{\θ}}_j,(j=\mathrm{1,2,3}),$

$\frac{\∂\mathrm{\ρ}}{\∂r}=-({\mathrm{\ρ}}_2\mathrm{\σ}+(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{T_{120}^{\′}}{T\left(h\right)}\exp (-\mathrm{\σ\ζ})){\left(\frac{R_{\mathrm{\ρ}}+120}{R_{\mathrm{\ρ}}+h}\right)}^2;$

For j=1,2:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_j}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_j}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3};$

For j=3:

$\frac{\∂\mathrm{\ρ}}{\∂{\mathrm{\θ}}_3}=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^4{M}_i{n}_i\frac{\∂G_i\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$+({\mathrm{\ρ}}_2(\frac{A_T{T}_{120}}{T_{\∞}\left(T_{\∞}{-T}_{120}\right)}\ln \left(\frac{T_{120}}{T\left(h\right)}\right)+\frac{A_T\mathrm{\σ\ζ}}{T_{\∞}})-(\mathrm{\ρ}+{\mathrm{\ρ}}_1+{\mathrm{\ρ}}_2)\frac{A_T(1-\exp (-\mathrm{\σ\ζ}))}{T\left(h\right)})\frac{\∂G_T\left(L\right)}{\∂{\mathrm{\θ}}_3}$

$-R_{\mathrm{\ρ}}{e}^2{(1-e^2+e^2{\mathrm{\θ}}_3^2)}^{-\frac{3}{2}}{\mathrm{\θ}}_3\frac{\∂\mathrm{\ρ}}{\∂r};$

In the top formula,

${\mathrm{\ρ}}_1=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\α}}_i$

${\mathrm{\ρ}}_2=A_{\mathrm{VR}}{\mathrm{\Σ}}_{i=1}^5{M}_i{n}_i{\mathrm{\γ}}_i$

$\frac{\∂G\left(L\right)}{\∂{\mathrm{\θ}}_3}=\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}+\frac{\∂M}{\∂{\mathrm{\θ}}_3}+\mathrm{\β}(\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}+\frac{\∂D}{\∂{\mathrm{\θ}}_3}+\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3})$

$\frac{\∂\mathrm{LAT}}{\∂{\mathrm{\θ}}_3}=a_2{H}_{20}^{\′}+a_3{H}_{40}^{\′}+a_{37}{H}_{10}^{\′}$

$\frac{\∂\mathrm{AN}1}{\∂{\mathrm{\θ}}_3}=a_{10}{H}_{20}^{\′}\cos \left[\mathrm{\Ω}\left({d-a}_{11}\right)\right]$

$\frac{\∂\mathrm{SAN}1}{\∂{\mathrm{\θ}}_3}=a_{13}{H}_{20}^{\′}\cos \left[2\mathrm{\Ω}(d-a_{14})\right]$

$\frac{\∂\mathrm{AN}2}{\∂{\mathrm{\θ}}_3}=(a_{15}{H}_{10}^{\′}+a_{16}{H}_{30}^{\′}+a_{17}{H}_{50}^{\′})\cos \left[\mathrm{\Ω}\left({d-a}_{18}\right)\right]$

$\frac{\∂\mathrm{SAN}2}{\∂{\mathrm{\θ}}_3}=a_{19}{H}_{10}^{\′}\cos \left[2\mathrm{\Ω}\left({d-a}_{20}\right)\right]$

$\frac{\∂D}{\∂{\mathrm{\θ}}_3}=(-B_1^{\′}\cos {\mathrm{\λ}}_S+B_2^{\′}\sin {\mathrm{\λ}}_S){\mathrm{\θ}}_1-(B_1^{\′}\sin {\mathrm{\λ}}_S+B_2^{\′}\cos {\mathrm{\λ}}_S){\mathrm{\θ}}_2$

$\frac{\∂\mathrm{SD}}{\∂{\mathrm{\θ}}_3}=(B_3^{\′}\cos 2{\mathrm{\λ}}_S-B_4^{\′}\sin 2{\mathrm{\λ}}_S)({\mathrm{\θ}}_1^2-{\mathrm{\θ}}_2^2)+2(B_3^{\′}\sin 2{\mathrm{\λ}}_S+B_4^{\′}\cos 2{\mathrm{\λ}}_S){\mathrm{\θ}}_1{\mathrm{\θ}}_2$

B′ ₁＝a ₂₂H′ ₃₁+a ₂₃H′ ₅₁+a ₂₅H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₂＝a ₂₇H′ ₃₁+a ₂₈H′ ₅₁+a ₃₀H′ ₂₁cos[Ω(d-a ₁₈)]

B′ ₃＝a ₃₂H′ ₃₂cos[Ω(d-a ₁₈)]

B′ ₄＝a ₃₄H′ ₃₂cos[Ω(d-a ₁₈)]

H′ ₁₀＝1

H′ ₂₀＝3θ ₃

H′ ₂₁＝3

$H_{30}^{\′}=\frac{3}{2}({5\mathrm{\θ}}_3^2-1)$

H′ ₃₁＝15θ ₃

H′ ₃₂＝15

$H_{40}^{\′}=\frac{5}{2}(7{\mathrm{\θ}}_3^2-3){\mathrm{\θ}}_3$

$H_{50}^{\′}=\frac{1}{8}(315{\mathrm{\θ}}_3^4-210{\mathrm{\θ}}_3^2+15)$

$H_{51}^{\′}=\frac{15}{2}(21{\mathrm{\θ}}_3^2-7){\mathrm{\θ}}_3.$