CN103017766A - Rapid coarse alignment method for large course - Google Patents

Abstract

The invention relates to a coarse alignment method for serial inertial navigation, and particularly relates to a rapid coarse alignment method suitable for a large course for a serial inertial navigation system. The rapid coarse alignment method disclosed by the invention comprises the following steps of: measuring an east initial attitude angle at time zero; measuring a north initial attitude angle at time zero; constructing a transformation matrix from a carrier coordinate system to a platform intermediate coordinate system, and executing a serial inertial navigation update calculation under a condition of a static base; obtaining the accelerations fn(1),..., fn(m) at a series of time points in a navigation coordinate system; determining an intermediate speed parameter vn of coarse alignment under the condition of a static base at a time end of 60 s; calculating the trigonometric function value of a coarse course; and constructing a transformation matrix from the platform intermediate coordinate system n' to the navigation coordinate system n, and determining the initial attitude matrix of the large course. In the method disclosed by the invention, the step of course alignment for the large course can be finished only by performing a group of serial calculations, thus greatly shortening the time of course alignment in case of no increase of hardware cost, and providing a high accuracy for fine alignment in the next step.

Description

A kind of large course Fast Coarse alignment methods

Technical field

What the present invention relates to is a kind of Fast Coarse alignment methods that is applicable to large course of a kind of coarse alignment method, the especially strapdown inertial navitation system (SINS) of inertial navigation.

Background technology

The main task of initial alignment is exactly the initial attitude battle array of given inertial navigation.The inertial navigation initial alignment generally can be divided into coarse alignment and two stages of fine alignment.The main task of coarse alignment be fast with the coarse value of attitude to estimating, be about to attitude error and be controlled in the smaller scope, create conditions for next step adopts linear model to carry out fine alignment.Requirement according to the main task of coarse alignment then is that the requirement to precision is not very strict in this one-phase, but that the rapidity of speed is it is desired.Therefore provide a kind of and determine that fast the coarse alignment method of the rough initial attitude battle array of inertial navigation is necessary.

But in prior art, as in " Chinese inertial technology journal " the 3rd phase of the 11st volume, employed coarse alignment method is the leveling aligning at horizontal attitude angle and two steps of azimuthal error estimation in Cong Li, Li Hanjie, Li Xuyou " comparative studies of two kinds of strapdown inertial navitation system (SINS) Alignment Method ", and designed different systematic parameters according to the characteristics in each stage and aimed at, result from its emulation, all be at 200s～300s two kinds of used times of self aligned coarse alignment stage, this time is long to coarse alignment.Summer man and, Qin Yongyuan, dragon auspicious in " rapid alignment method that resolves based on many strapdowns under the Large azimuth angle " that " firepower with commander control " delivered many strapdowns that are based on of employing resolve multi-model parallel computation alignment methods with many Kalman filtering algorithms.The method is carried out interval division to possible azimuth coverage, and take the central value in interval as the position angle initial value, many strapdowns resolve simultaneously and carry out, and each strapdown resolves all to having one based on the Kalman filtering algorithm of linear small angle error model; Constantly choose optimum calculation result according to certain criterion to aiming at finally to finish.Result from emulation, it is effective based on many strapdowns calculation method to the time that shortens coarse alignment, reach about 100s, but above-mentioned alignment methods is carried out interval division for the requirement of satisfying low-angle linear error model with possible course angle scope, as take course angle as 60 ° as example, be divided into 6 intervals, and the strapdown that carries out simultaneously 6 groups take the central value in interval as the course angle initial value resolves with Kalman filtering and calculates, increase considerably calculated amount, thereby just required the calculated performance of computing machine to want high.

More than the coarse alignment of inertial navigation is narrated and is probed into, but do not reach simultaneously rapidity and the little requirement of calculated amount of coarse alignment, therefore study a kind of calculated amount little and fast coarse alignment method have novelty and Practical Project is worth.

Summary of the invention

The object of the present invention is to provide a kind of strapdown inertial navitation system (SINS) that is applicable to, can more accurate, the quicker coarse alignment method of finishing large course.

The object of the present invention is achieved like this:

Step of the present invention is as follows:

(1) measure the east orientation accelerometer at initial time, i.e. the east orientation initial attitude angle in 0 moment:

$\mathrm{\θ}\left(0\right)=-f_E^b\left(0\right)/9.78049,$

Wherein

The output of east orientation accelerometer initial time;

(2) initial time of measurement north orientation accelerometer, i.e. the north orientation initial attitude angle in 0 moment:

$\mathrm{\γ}\left(0\right)=f_N^b\left(0\right)/9.78049,$

Wherein

The output of north orientation accelerometer initial time;

(3) the carrier construction coordinate is tied to the transition matrix of platform middle coordinate system

$C_b^{n^{\′}}=\left[\begin{array}{ccc}\cos \mathrm{\γ}\left(0\right)& \sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ 0& \cos \mathrm{\θ}\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)& -\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right],$

Wherein b is carrier coordinate system, and n ' is the platform middle coordinate system;

(4) under quiet pedestal condition, carry out more new Algorithm of inertial navigation:

$T_b^n\left(k\right)=T_b^n(k-1)\left\{\right[{\mathrm{\ω}}_{\mathrm{ib}}^b\left(k\right)-\left[T_n^b(k-1)\right]{\mathrm{\ω}}_{\mathrm{ie}}^n\left(k\right)]\×\},$

Wherein the execution time is 60s, and the strapdown matrix is

K is the number of times of sampling, k=1, and 2 ..., m; M is the total degree of sampling in 60s, i.e. m=60/ Δ t, and Δ t is the sampling interval time, Mathematical platform initial value as this stage

Be the renewal of the k time strapdown matrix, [] * be the antisymmetric matrix of [],

Be the output of gyroscope the k time,

Be engraved in the projection that navigation coordinate is fastened during for rotational-angular velocity of the earth k, in the situation of quiet pedestal A normal value, namely

In the formula It is the local latitude value; What n represented is navigation coordinate system;

(5) obtain the acceleration f of a series of moment point in the navigation coordinate system ⁿ(1) ..., f ⁿ(m):

$f^n\left(k\right)=T_b^n(k-1)f^b\left(k\right)+g^n\left(k\right),$

F in the formula ^b(k) be the output of accelerometer the k time, f ⁿBe engraved in the acceleration that navigation coordinate is fastened when (k) being k, g ⁿBe engraved in the projection that navigation coordinate is fastened when (k) being earth surface acceleration of gravity k, g in the situation of quiet pedestal ⁿ(k) be normal value, i.e. a g ⁿ(k)=[0 0 9.78049];

(6) determine the 60s finish time, the midrange speed parameter v of coarse alignment under the quiet pedestal ⁿ:

$v^n=\underset{k=1}{\overset{m}{\mathrm{\Σ}}}\left[f^n\left(k\right)\mathrm{\Δt}\right],$

In the formula $v^n={\left[\begin{array}{ccc}{v}_E^n& v_N^n& v_U^n\end{array}\right]}^T,$

Be respectively under navigation coordinate system east orientation, north orientation, day to speed;

(7) trigonometric function value in the rough course of calculating is:

$\cos H\left(0\right)={2v}_E^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right]+1$

$\sin H\left(0\right)=-2v_N^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right],$

H in the formula (0) is unknown course initial estimate,

In the gyrostatic output valve 60s finish time;

(8) construction platform middle coordinate system n ' is the transition matrix of n to navigation coordinate

$C_{n^{\′}}^n=\left[\begin{array}{ccc}\cos H\left(0\right)& \sin H\left(0\right)& 0\\ -\sin H& \cos H\left(0\right)& 0\\ 0& 0& 1\end{array}\right],$

(9) determine the initial attitude matrix in large course

That is:

$C_b^n=C_{n^{\′}}^n{C}_b^{n^{\′}}=\left[\begin{array}{c}\cos \mathrm{\γ}\left(0\right)\cos H\left(0\right)-\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ -\cos \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos H\left(0\right)+\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\end{array}\right.$

$\left.\begin{array}{cc}\cos \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)+\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ \cos \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)-\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right].$

So far just obtained the initial attitude matrix in rough large course Finished the Fast Coarse in the large course of strapdown inertial navitation system (SINS) and aimed at, for next step fine alignment provides initial value.

Beneficial effect of the present invention is:

Problem for the strapdown inertial navitation system (SINS) coarse alignment, the invention provides a kind of again little coarse alignment method of the fast calculated amount of alignment speed that is applicable to large course, method of the present invention only need be carried out one group of strapdown and be resolved the coarse alignment step that can finish large course, calculated performance to computing machine does not require, in the situation that does not increase hardware cost, significantly shorten the time of coarse alignment, and provide higher precision for next step fine alignment.

Description of drawings

Fig. 1 is the large course of the strapdown inertial navigation system Fast Coarse alignment methods process flow diagram that is applicable to of the present invention.

Embodiment

For example the present invention is described in more detail below in conjunction with accompanying drawing 1: the large course Fast Coarse alignment methods of strapdown inertial navitation system (SINS) comprises the steps:

Step 1. is utilized the initial time of east orientation accelerometer, i.e. 0 constantly output obtains east orientation (pitching) initial attitude angle; That is:

$\mathrm{\θ}\left(0\right)=-f_E^b\left(0\right)/9.78049---\left(1\right)$

Wherein

The output of east orientation accelerometer initial time.

Step 2. is utilized the initial time of north orientation accelerometer, i.e. 0 constantly output obtains north orientation (rolling) initial attitude angle; That is:

$\mathrm{\γ}\left(0\right)=f_N^b\left(0\right)/9.78049---\left(2\right)$

Wherein

The output of north orientation accelerometer initial time.

The angle calculation that step 3. utilizes step 1, step 2 to obtain obtains carrier coordinate system to the transition matrix of platform middle coordinate system That is:

$C_b^{n^{\′}}=\left[\begin{array}{ccc}\cos \mathrm{\γ}\left(0\right)& \sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ 0& \cos \mathrm{\θ}\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)& -\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right]---\left(3\right)$

Wherein b representative is carrier coordinate system, n ' representative be the platform middle coordinate system.

Step 4. utilizes gyrostatic output to carry out more new Algorithm of inertial navigation, and execution time 60s so just can obtain strapdown matrix on a series of moment point

Under the condition of quiet pedestal, carry out more new Algorithm of inertial navigation, that is:

$T_b^n\left(k\right)=T_b^n(k-1)\left\{\right[{\mathrm{\ω}}_{\mathrm{ib}}^b\left(k\right)-\left[T_n^b(k-1)\right]{\mathrm{\ω}}_{\mathrm{ie}}^n\left(k\right)]\×\}---\left(4\right)$

Wherein k is the number of times of sampling, k=1, and 2 ..., m; M is the total degree of sampling in 60s, i.e. m=60/ Δ t, and Δ t is the sampling interval time; In the formula with in the step 3 Mathematical platform initial value as this stage

Represent the renewal of the k time strapdown matrix, [] * what represent is the antisymmetric matrix of [],

Representative be the output of gyroscope the k time,

Be engraved in the projection that navigation coordinate is fastened when being rotational-angular velocity of the earth k, in the situation of quiet pedestal

A normal value, namely

In the formula

It is the local latitude value; What n represented is navigation coordinate system.

Step 5. is utilized a series of strapdown matrixes of obtaining in the step 4 and the output of accelerometer, obtains the acceleration f of a series of moment point in navigation coordinate system ⁿ(1) ..., f ⁿ(m), that is:

$f^n\left(k\right)=T_b^n(k-1)f^b\left(k\right)+g^n\left(k\right)---\left(5\right)$

F in the formula ^b(k) be the output of accelerometer the k time, f ⁿBe engraved in the acceleration that navigation coordinate is fastened when (k) being k, g ⁿBe engraved in the projection that navigation coordinate is fastened when (k) being earth surface acceleration of gravity k, g in the situation of quiet pedestal ⁿ(k) be normal value, i.e. a g ⁿ(k)=[0 0 9.78049].

Step 6. utilizes the result who obtains in the step 5 can calculate the finish time at 60s, the midrange speed parameter v of coarse alignment under the quiet pedestal ⁿThat is:

$v^n=\underset{k=1}{\overset{m}{\mathrm{\Σ}}}\left[f^n\left(k\right)\mathrm{\Δt}\right]---\left(6\right)$

In the formula $v^n={\left[\begin{array}{ccc}{v}_E^n& v_N^n& v_U^n\end{array}\right]}^T,$

Be respectively under navigation coordinate system east orientation, north orientation, day to speed.

The midrange speed parameter of coarse alignment under the quiet pedestal that step 7. utilization obtains in step 6 With

And formula (7) and formula (8), the trigonometric function value that then can try to achieve rough course is:

$\cos H\left(0\right)={2v}_E^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right]+1---\left(7\right)$

$\sin H\left(0\right)=-2v_N^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right]---\left(8\right)$

H in the formula (0) is unknown course initial estimate,

In the gyrostatic output valve 60s finish time.

It is the transition matrix of n to navigation coordinate that step 8. calculates platform middle coordinate system n '

That is:

$C_{n^{\′}}^n=\left[\begin{array}{ccc}\cos H\left(0\right)& \sin H\left(0\right)& 0\\ -\sin H& \cos H\left(0\right)& 0\\ 0& 0& 1\end{array}\right]---\left(9\right)$

Step 9. is utilized the result of step 4 and step 8 to calculate and is tried to achieve rough attitude matrix

That is:

$C_b^n=C_{n^{\′}}^n{C}_b^{n^{\′}}=\left[\begin{array}{c}\cos \mathrm{\γ}\left(0\right)\cos H\left(0\right)-\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ -\cos \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos H\left(0\right)+\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\end{array}\right.$

$\left.\begin{array}{cc}\cos \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)+\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ \cos \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)-\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right]---\left(10\right)$

So far just obtained the initial attitude matrix in rough large course

Finished the Fast Coarse in the large course of strapdown inertial navitation system (SINS) and aimed at, for next step fine alignment provides initial value.

In order to verify the beneficial effect of described method, the present invention carries out emulation, and with the technical background technology in mention method and contrast, its result such as table one: A group is the simulation result of " comparative studies of two kinds of strapdown inertial navitation system (SINS) Alignment Method ", the B group is the simulation result of " rapid alignment method that resolves based on many strapdowns under the Large azimuth angle ", and the C group is simulation result of the present invention.

Table one

Illustrate: represent and can not aim at.

Be applicable to as can be known the large course of strapdown inertial navigation system Fast Coarse alignment methods by contrast and can finish rapidly coarse alignment, and can provide precision higher initial value for next step fine alignment.

Claims (1)

1. a large course Fast Coarse alignment methods is characterized in that, comprises the steps:

(1) measure the east orientation accelerometer at initial time, i.e. the east orientation initial attitude angle in 0 moment:

$\mathrm{\θ}\left(0\right)=-f_E^b\left(0\right)/9.78049,$

Wherein

The output of east orientation accelerometer initial time;

(2) initial time of measurement north orientation accelerometer, i.e. the north orientation initial attitude angle in 0 moment:

$\mathrm{\γ}\left(0\right)=f_N^b\left(0\right)/9.78049,$

Wherein

The output of north orientation accelerometer initial time;

(3) the carrier construction coordinate is tied to the transition matrix of platform middle coordinate system

$C_b^{n^{\′}}=\left[\begin{array}{ccc}\cos \mathrm{\γ}\left(0\right)& \sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ 0& \cos \mathrm{\θ}\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)& -\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right],$

Wherein b is carrier coordinate system, and n ' is the platform middle coordinate system;

(4) under quiet pedestal condition, carry out more new Algorithm of inertial navigation:

$T_b^n\left(k\right)=T_b^n(k-1)\left\{\right[{\mathrm{\ω}}_{\mathrm{ib}}^b\left(k\right)-\left[T_n^b(k-1)\right]{\mathrm{\ω}}_{\mathrm{ie}}^n\left(k\right)]\×\},$

Wherein the execution time is 60s, and the strapdown matrix is

K is the number of times of sampling, k=1, and 2 ..., m; M is the total degree of sampling in 60s, i.e. m=60/ Δ t, and Δ t is the sampling interval time,

Mathematical platform initial value as this stage

Be the renewal of the k time strapdown matrix, [] * be the antisymmetric matrix of [],

Be the output of gyroscope the k time,

Be engraved in the projection that navigation coordinate is fastened during for rotational-angular velocity of the earth k, in the situation of quiet pedestal

A normal value, namely In the formula

It is the local latitude value; What n represented is navigation coordinate system;

(5) obtain the acceleration f of a series of moment point in the navigation coordinate system ⁿ(1) ..., f ⁿ(m):

$f^n\left(k\right)=T_b^n(k-1)f^b\left(k\right)+g^n\left(k\right),$

F in the formula ^b(k) be the output of accelerometer the k time, f ⁿBe engraved in the acceleration that navigation coordinate is fastened when (k) being k, g ⁿBe engraved in the projection that navigation coordinate is fastened when (k) being earth surface acceleration of gravity k, g in the situation of quiet pedestal ⁿ(k) be normal value, i.e. a g ⁿ(k)=[0 0 9.78049];

(6) determine the 60s finish time, the midrange speed parameter v of coarse alignment under the quiet pedestal ⁿ:

$v^n=\underset{k=1}{\overset{m}{\mathrm{\Σ}}}\left[f^n\left(k\right)\mathrm{\Δt}\right],$

In the formula $v^n={\left[\begin{array}{ccc}{v}_E^n& v_N^n& v_U^n\end{array}\right]}^T,$

Be respectively under navigation coordinate system east orientation, north orientation, day to speed;

(7) trigonometric function value in the rough course of calculating is:

$\cos H\left(0\right)={2v}_E^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right]+1$

$\sin H\left(0\right)=-2v_N^n/\left[9.78049t^2\left({\mathrm{\ω}}_{\mathrm{ib}}^b\left(m\right)\right)\right],$

H in the formula (0) is unknown course initial estimate,

In the gyrostatic output valve 60s finish time;

(8) construction platform middle coordinate system n ' is the transition matrix of n to navigation coordinate

$C_{n^{\′}}^n=\left[\begin{array}{ccc}\cos H\left(0\right)& \sin H\left(0\right)& 0\\ -\sin H& \cos H\left(0\right)& 0\\ 0& 0& 1\end{array}\right],$

(9) determine the initial attitude matrix in large course

That is:

$C_b^n=C_{n^{\′}}^n{C}_b^{n^{\′}}=\left[\begin{array}{c}\cos \mathrm{\γ}\left(0\right)\cos H\left(0\right)-\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ -\cos \mathrm{\θ}\left(0\right)\sin H\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos H\left(0\right)+\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\sin H\left(0\right)\end{array}\right.$

$\left.\begin{array}{cc}\cos \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)+\sin \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& -\sin \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\\ \cos \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \sin \mathrm{\θ}\left(0\right)\\ \sin \mathrm{\γ}\left(0\right)\cos \mathrm{\γ}\left(0\right)-\cos \mathrm{\γ}\left(0\right)\sin \mathrm{\θ}\left(0\right)\cos H\left(0\right)& \cos \mathrm{\γ}\left(0\right)\cos \mathrm{\θ}\left(0\right)\end{array}\right].$

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