• Infrared and Laser Engineering
  • Vol. 51, Issue 7, 20210784 (2022)
Xuerui Zhai, Yuan Ren, Lifen Wang, Ting Zhu, and Chen Wang
Author Affiliations
  • Space Engineering University, Department of Aerospace Science and Technology, Beijing 101416, China
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    DOI: 10.3788/IRLA20210784 Cite this Article
    Xuerui Zhai, Yuan Ren, Lifen Wang, Ting Zhu, Chen Wang. Full parameter rapid field calibration method for regular tetrahedral redundant inertial navigation[J]. Infrared and Laser Engineering, 2022, 51(7): 20210784 Copy Citation Text show less

    Abstract

    Regular tetrahedral Redundant Inertial Navigation System (RINS) has the characteristics of high reliability and high precision, and error calibration is a necessary means to realize high-precision navigation solution. At present, the error calibration of regular tetrahedral RINS needs to be realized by using the high-precision turntable, which not only has high calibration cost and long calibration time, but also cannot calibrate full error parameters under the condition of insufficient hardware conditions such as external field. Address this issue, a rapid field calibration method for full error parameters of regular tetrahedral RINS without high-precision turntable was proposed. Firstly, the error model of regular tetrahedral RINS was established. Then, according to the relationship between the analytic coarse alignment attitude error matrix and the bias of regular tetrahedral RINS, a bias calibration method based on six positions was proposed. Then, a three-position rotation scheme was designed to calibrate the scale factor and installation error of the gyroscope. Finally, the six-position scheme of bias calibration was used to calibrate the scale factor and installation error of the accelerometer. Simulation and experiment show that this method can effectively calibrate full error parameters. In the 1h static base navigation experiment, the north positioning accuracy are improved from 61.065 5 km to 0.476 7 km, and the east positioning accuracy are improved from 161.202 7 km to 4.842 2 km.
    $ {{\boldsymbol{\omega}} _{m}}{ = }{\boldsymbol{H}}{{\boldsymbol{\omega}} _{b}} $(1)

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    $ {\boldsymbol{H}}{\text{ = }}\left[ {001sinα0cosαsinαcosβ3sinαsinβ3cosαsinαcosβ4sinαsinβ4cosα} \right] $(2)

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    $ {\boldsymbol{H}}{\text{ = }}\left[ {0010.942800.33330.47140.81650.33330.47140.81650.3333} \right] $(3)

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    $ {\hat{\boldsymbol{ \omega}} _b} = {\left( {{{\boldsymbol{H}}^{\rm{T}}}{\boldsymbol{H}}} \right)^{ - 1}}{{\boldsymbol{H}}^{\rm{T}}}{{\boldsymbol{\omega}} _m} $(4)

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    $ {{{\boldsymbol{B}}}_g} = {\left[ {bg1bg2bg3bg4} \right]^{\rm{T}}} $(5)

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    $ \Delta {{\boldsymbol{m}}_b} = {{\boldsymbol{B}}_g} $(6)

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    $ \delta {{\boldsymbol{K}}_g} = \left[ {δKg10000δKg20000δKg30000δKg4} \right] $(7)

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    $ \Delta {{\boldsymbol{m}}_k} = \delta {{\boldsymbol{K}}_g}{{\boldsymbol{\omega}} _m} $(8)

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    $ {{\boldsymbol{H}}_g} = \left[ {Hg1xHg1yHg1zHg2xHg2yHg2zHg3xHg3yHg3zHg4xHg4yHg4z} \right] $(9)

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    $ \left\{ {Hg1x=δug1sinβcosβ1+δvg1cosβsinβ1Hg1y=δug1sinβsinβ1δvg1cosβcosβ1Hg1z=1 + δug1cosβHg2x=sinαδug2sinαcosβ2+δvg2cosαsinβ2Hg2y=δug2sinαsinβ2δvg2cosαcosβ2Hg2z=cosα + δug2cosαHg3x=sinαcosβ3δug3sinαcosβ3+δvg3cosαsinβ3Hg3y=sinαsinβ3δug3sinαsinβ3δvg3cosαcosβ3Hg3z=cosα + δug3cosαHg4x=sinαcosβ4δug4sinαcosβ4+δvg4cosαsinβ4Hg4y=sinαsinβ4δug4sinαsinβ4δvg4cosαcosβ4Hg4z=cosα + δug4cosα} \right. $(10)

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    $ \Delta {{\boldsymbol{m}}_h} = \left( {{{{{\boldsymbol{H}}}}_g} - {\boldsymbol{H}}} \right){\left( {{{\boldsymbol{H}}^{\rm{T}}}{\boldsymbol{H}}} \right)^{ - 1}}{{\boldsymbol{H}}^{\rm{T}}}{{\boldsymbol{\omega}} _m} $(11)

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    $ {{\boldsymbol{\omega}} _m} = \left( {{\boldsymbol{I}} + \delta {{\boldsymbol{K}}_g}} \right)\left( {{{\boldsymbol{H}}_g}{{\boldsymbol{\omega}} _b} + {{\boldsymbol{B}}_g}} \right) + {{\boldsymbol{\eta}} _g} $(12)

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    $ {{{{\boldsymbol{N}}}}_{\omega m}} = {{{{\boldsymbol{K}}}}_g}\left( {{{{{\boldsymbol{H}}}}_g}{{\boldsymbol{\omega}} _b} + {{{{\boldsymbol{B}}}}_g}} \right) + {{\boldsymbol{\eta}} _g} $(13)

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    $ {{{\boldsymbol{f}}}_m} = \left( {{\boldsymbol{I}} + \delta {{\boldsymbol{K}}_f}} \right)\left( {{{\boldsymbol{H}}_f}{{{\boldsymbol{f}}}_b} + {{\boldsymbol{B}}_f}} \right) + {{\boldsymbol{\eta}} _f} $(14)

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    $ {{\boldsymbol{N}}_{fm}} = {{\boldsymbol{K}}_f}\left( {{{\boldsymbol{H}}_f}{{\boldsymbol{f}}_b} + {{\boldsymbol{B}}_f}} \right) + {{\boldsymbol{\eta}} _f} $(15)

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    $ {\boldsymbol{C}}_b^n = {\left[ {(\boldsymbolgn)T(\boldsymbolωien)T(\boldsymbolgn×\boldsymbolωien)T} \right]^{ - 1}}\left[ {(\boldsymbolgb)T(\boldsymbolωieb)T(\boldsymbolgb×\boldsymbolωieb)T} \right] $(16)

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    $ {{\boldsymbol{E}}} = {\delta {\boldsymbol{C}}}_b^n{\left( {{{\boldsymbol{C}}}_b^n} \right)^{\rm{T}}} $(17)

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    $ {{\boldsymbol{E}}} = {{{\boldsymbol{E}}}_{ss}} + {{{\boldsymbol{E}}}_s} $(18)

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    $ {{{\boldsymbol{E}}}_{ss}}{\text{ = }}\frac{{{{\boldsymbol{E}}} - {{{\boldsymbol{E}}}^{{\rm{T}}}}}}{2} = \left[ {0φUφEφU0φNφEφN0} \right] $(19)

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    $ {{{\boldsymbol{E}}}_{s}}{\text{ = }}\frac{{{{\boldsymbol{E}}} + {{{\boldsymbol{E}}}^{{\rm{T}}}}}}{2} = \left[ {ηEoUoNoUηNoEoNoEηU} \right] $(20)

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    $ \left\{ δfU=gηUδωN=Ω[ηEηU+(δfNtanL)/g]cosLδωU=Ω(2oEηUtanL+δfN/g)cosL \right. $(21)

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    $ \left\{ {δω^xb=(δω^x1b + δω^x2b)/2δω^yb=(δω^y1b + δω^y2b)/2δω^zb=(δω^z1b + δω^z2b)/2} \right. $(22)

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    $ \left\{ {δf^xb=(δf^x1b + δf^x2b)/2δf^yb=(δf^y1b + δf^y2b)/2δf^zb=(δf^z1b + δf^z2b)/2} \right. $(23)

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    $ {{{\hat {\boldsymbol{E}}}}_s} = \frac{1}{2}\left[ {{{\hat {\boldsymbol{C}}}}_b^n{{\left( {{{\hat {\boldsymbol{C}}}}_b^n} \right)}^{\rm{T}}} - {{{\boldsymbol{I}}}}} \right] $(24)

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    $ \left\{ {δω^xb=Ω[(o^E1o^E2)+(η^U2η^U1)tanL2]cosLδω^yb=Ω[(o^E3o^E4)+(η^U4η^U3)tanL2]cosLδω^zb=Ω[(o^E5o^E6)+(η^U6η^U5)tanL2]cosL} \right. $(25)

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    $ \left\{ {δf^xb = g(η^U1η^U2)/2δf^yb = g(η^U3η^U4)/2δf^zb = g(η^U5η^U6)/2} \right. $(26)

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    $ \left\{ {bg1=δω^zbbg2=δω^xbsinα+δω^zbcosαbg3=δω^xbsinαcosβ3+δω^ybsinαsinβ3+δω^zbcosαbg4=δω^xbsinαcosβ4+δω^ybsinαsinβ4+δω^zbcosαbf1=δf^zbbf2=δf^xbsinα+δf^zbcosαbf3=δf^xbsinαcosβ3+δf^ybsinαsinβ3+δf^zbcosαbf4=δf^xbsinαcosβ4+δf^ybsinαsinβ4+δf^zbcosα} \right. $(27)

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    $ [ωxωyωz]=[1000cosθ(t)sinθ(t)0sinθ(t)cosθ(t)]×[ΩsinLΩcosL0]+[±ωix(t)00] $(28)

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    $ {K_{g1}}{H_{g1x}} = \frac{{\displaystyle\int_0^T {N_{\omega 1}^{x + }{\rm{d}}t - \displaystyle\int_0^T {N_{\omega 1}^{x - }{\rm{d}}t} } }}{{2\displaystyle\int_0^T {{\omega _{ix}}\left( t \right){\rm{d}}t} }}{\text{ = }}\frac{{\displaystyle\int_0^T {N_{\omega 1}^{x + }{\rm{d}}t - \displaystyle\int_0^T {N_{\omega 1}^{x - }{\rm{d}}t} } }}{{4n\pi }} $(29)

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    $ {{K}_{gm}} = \sqrt {{K^2}_{gmx} + {K^2}_{gmy} + {K^2}_{gmz}} $(30)

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    $ {H_{gmn}}{\text{ = }}\frac{{{K_{gmn}}}}{{{K_{gm}}}} $(31)

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    $ Nfm1=Kfm(Hfmxfx+Hfmyfy+Hfmzfz)+KfmBfm = KfmHfmxg+KfmBfm $(32)

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    $ Nfm2=Kfm(Hfmxfx+Hfmyfy+Hfmzfz)+KfmBfm = KfmHfmx(g)+KfmBfm $(33)

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    $ {{K}_{fm}}{H_{fmx}}{\text{ = }}\frac{{N_{fm}^1 - N_{fm}^2}}{{2g}} $(34)

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    $ {K_{fm}} = \sqrt {{K^2}_{fmx} + {K^2}_{fmy} + {K^2}_{fmz}} $(35)

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    $ {H_{fmn}}{\text{ = }}\frac{{{K_{fmn}}}}{{{K_{fm}}}} $(36)

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    Xuerui Zhai, Yuan Ren, Lifen Wang, Ting Zhu, Chen Wang. Full parameter rapid field calibration method for regular tetrahedral redundant inertial navigation[J]. Infrared and Laser Engineering, 2022, 51(7): 20210784
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