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    Journals >Spacecraft Recovery & Remote Sensing >Volume 45 >Issue 2 >Page 1 > Article
    • Spacecraft Recovery & Remote Sensing
    • Vol. 45, Issue 2, 1 (2024)
    Design and Analysis of a New Reusable Lander
    Shan JIA1,3,4, Rujie HU1,2, Xianghua ZHOU1, Shaoyang LIU1..., Mingyang WU1 and Jinbao CHEN1,3,4|Show fewer author(s)
    Author Affiliations
  • 1Academy of Astronautics, Nanjing University of Aeronautics and Astronautics, Nanjing 211106, China
  • 2Shenyang Aircraft Design and Research Institute, Shenyang 110035, China
  • 3Laboratory of Aerospace Entry, Descent and Landing Technology, Nanjing 211106, China
  • 4Key Laboratory of Deep Space Star Catalog Detection Mechanism Technology, Ministry of Industry and Information Technology, Nanjing 211106, China
    • show less
    DOI: 10.3969/j.issn.1009-8518.2024.02.001 Cite this Article
    Shan JIA, Rujie HU, Xianghua ZHOU, Shaoyang LIU, Mingyang WU, Jinbao CHEN. Design and Analysis of a New Reusable Lander[J]. Spacecraft Recovery & Remote Sensing, 2024, 45(2): 1 Copy Citation Text show less

    Abstract

    To improve the shortcomings of the current landing cushion device that requires high landing terrain and is not reusable, a reusable four-legged lander based on a bionic structure is designed with reference to the composition of insect landing leg structures in nature. The support leg adopts the composition of “rigid body control thigh structure plus circular arc structure leg structure”, using a friction brake as a design method for a cushioning and energy absorbing device, which makes the landing impact load as thermal energy by friction work. The friction brake is applied to dissipate the landing impact load in the form of thermal energy to achieve cushion energy absorption. The D-H parametric method is applied to derive the chi-square matrix transformation of the support leg, and the forward and reverse kinematics analysis of the support leg is carried out for the bionic leg type cushioning structure. The theoretical model of the support leg is established and dynamics simulation is performed. The result shows that the reusable lander has good landing buffer performance and can provide a feasible solution for China’s subsequent deep space exploration missions.

    Keywords

    $_i^{i-1}\boldsymbol{T}=\left[cos⁡θi−sin⁡θi0ai−1sin⁡θicos⁡αi−1cos⁡θicos⁡αi−1−sin⁡αi−1−disin⁡αi−1sin⁡θisin⁡αi−1cos⁡θisin⁡αi−1cos⁡αi−1dicos⁡αi−10001\right]$(1)

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    $ {}_1^0{\boldsymbol{T}} = \left[ {cos⁡θAB−sin⁡θAB00sin⁡θABcos⁡θAB0000100001} \right] $(2)

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    $ {}_2^1{\boldsymbol{T}} = \left[ {cos⁡θBD−sin⁡θBD0L1sin⁡θBDcos⁡θBD0000100001} \right] $(3)

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    $ {}_3^2{\boldsymbol{T}} = \left[ {100L2010000100001} \right] $(4)

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    $ {x_{i - 1}} = {}_i^{i - 1}{\boldsymbol{T}}{x_i} $(5)

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    $ _i^0\boldsymbol{T}= _0^1 \boldsymbol{T}_2^1\boldsymbol{T}\cdots\,_i^{i-1} \boldsymbol{T} $(6)

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    $ 30\boldsymbolT=10\boldsymbolT21\boldsymbolT32\boldsymbolT=[cos⁡(θAB+θBD)−sin⁡(θAB+θBD)0L1cos⁡θABsin⁡(θAB+θBD)cos⁡(θAB+θBD)0L1sin⁡θAB00100001][100L2010000100001]=[cos⁡(θAB+θBD)−sin⁡(θAB+θBD)0cos⁡(θAB+θBD)L2+cos⁡θABL1sin⁡(θAB+θBD)cos⁡(θAB+θBD)0sin⁡(θAB+θBD)L2+sin⁡θABL100100001] $(7)

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    $ {[{}_1^0{\boldsymbol{T}}({\theta _{\rm{AB}}})]^{ - 1}}{}_3^0{\boldsymbol{T}} = {}_2^1{\boldsymbol{T}}({\theta _{\rm{BD}}})_3^2{\boldsymbol{T}}(0) $(8)

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    $ [cos⁡θABsin⁡θAB00−sin⁡θABcos⁡θAB0000100001]30\boldsymbolT=21\boldsymbolT(θBD)32\boldsymbolT(0)=[R1R20pxR3R40py001pz0001]=[cos⁡θBD−sin⁡θBD0L1sin⁡θBDcos⁡θBD0000100001][100L2010000100001]=[cos⁡θBD−sin⁡θBD0L2cos⁡θBD+L1sin⁡θBDcos⁡θBD0sin⁡θBDL200100001] $(9)

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    $\left.R1=cos⁡(θAB+θBD)cos⁡θAB+sin⁡(θAB+θBD)sin⁡θAB；R2=−sin⁡(θAB+θBD)cos⁡θAB+cos⁡(θAB+θBD)sin⁡θAB；R3=−cos⁡(θAB+θBD)sin⁡θAB+sin⁡(θAB+θBD)cos⁡θAB；R4=sin⁡(θAB+θBD)sin⁡θAB+cos⁡(θAB+θBD)cos⁡θAB；\right\} $()

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    $ px=cos⁡(θAB+θBD)cos⁡θABL2+sin⁡(θAB+θBD)sin⁡θABL2+L1py=−(cos⁡θAB+θBD)sin⁡θAB+sin⁡(θAB+θBD)cos⁡θABL2pz=0}。 $()

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    $ ^1\boldsymbol{P}_{\text{C}_1}=l_1\hat{\boldsymbol{X}}_1=\left[l100\right] $(10)

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    $ {}^2{{\boldsymbol{P}}_{{{\text{C}}_2}}} = {l_2}{\hat{\boldsymbol{ X}}_2} = \left[ {l200} \right] $(11)

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    $ {}^{{{\text{C}}_1}}{{\boldsymbol{I}}_1} = 0 $(12)

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    $ {}^{{{\text{C}}_2}}{{\boldsymbol{I}}_2} = 0 $(13)

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    $ {{\boldsymbol{f}}_3} = \left[ {−F00} \right] $(14)

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    $ n_3=0 $(15)

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    $ \boldsymbol{\mathit{\omega}}_0=0 $(16)

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    $ \dot{\boldsymbol{\mathit{\omega}}}_0=0 $(17)

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    $ ^0\boldsymbol{\dot v}_0=g_{\text{n}}\hat{\boldsymbol{Y}}_0=\left[0gn0\right] $(18)

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    $ _2^1\boldsymbol{R}=\left[cos⁡θBD−sin⁡θBD0sin⁡θBDcos⁡θBD0001\right] $(19)

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    $ _1^2\boldsymbol{R}=\left[cos⁡θBDsin⁡θBD0−sin⁡θBDcos⁡θBD0001\right] $(20)

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    $ {}^1{{\text{ω}} _1} = {}_0^1{{\boldsymbol{R}}}{{\;}^{0}{\omega}} _0 + {\theta _{\rm{AB}}}{}^1{\hat {\boldsymbol{Z}}_1} = \left[ {00θAB} \right] $(21)

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    $ { }^{1} {\text{ω}}_{1}={ }_{0}^{1} \boldsymbol{R}{\;}^{0} \omega_{0}+{ }_{0}^{1} \boldsymbol{R}{\;}^{0} \omega_{0} \times \theta_{\mathrm{AB}}{ }^{1} \hat{\boldsymbol{Z}}_{1}+\theta_{\mathrm{AB}}{ }^{1} \hat{\boldsymbol{Z}}_{1}=\left[00θAB\right]$(22)

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    $ 1\boldsymbolv˙1=01\boldsymbolR[0ω˙0×0\boldsymbolP1+0ω0×(0ω0×0\boldsymbolP1)+0\boldsymbolv˙0]=[cos⁡θABsin⁡θAB0−sin⁡θABcos⁡θAB0001][0gn0]=[gnsin⁡θABgncos⁡θAB0]$(23)

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    $ 1\boldsymbolv˙C1=1ω˙1×1\boldsymbolPC1+1ω1×(1ω1×1\boldsymbolPC1)+1\boldsymbolv˙1=[00θ¨AB]×[l100]+[00θ˙AB]×([00θ˙AB]×[l100])+[gnsin⁡θABgncos⁡θAB0]=[−l1θ˙AB2+gnsin⁡θABl1θ¨AB+gncos⁡θAB0] $(24)

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    $ ^{1}\boldsymbol{F}_{1}=m_{1}{}^{1}\boldsymbol{\dot v}_{{\mathrm{C}}_{1}}=[−m1l1θ˙AB2+m1gnsin⁡θABm1l1θ¨AB+m1gncos⁡θAB0]$(25)

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    $ ^1{\boldsymbol{N}}_1=^{\text{C}_1} {{\boldsymbol{I}}}_1{}^1\dot{\boldsymbol{\omega}}_{\text{1}}+^1\boldsymbol{\omega}_{\text{1}}\times^{\text{C}_1} {{\boldsymbol{I}}}_1{}^1\boldsymbol{\omega}_{\text{1}}=0 $(26)

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    $ ^2{\text{ω}}_2={}_1^2\boldsymbol{R}^1{\text{ω}}_1+\dot{\theta}_\mathrm{AB}{}^2\hat{\boldsymbol{Z}}_2=[cos⁡θBDsin⁡θBD0−sin⁡θBDcos⁡θBD0001][00θ˙AB]+[00θ˙BD]=[00θ˙AB+θ˙BD]$(27)

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    $ 2ω˙2=12\boldsymbolR1ω˙1+12\boldsymbolR1ω˙1×θ˙BD2\boldsymbolZ^2+θ¨BD2\boldsymbolZ^2=[cos⁡θBDsin⁡θBD0−sin⁡θBDcos⁡θBD0001][00θ¨AB]+[cos⁡θBDsin⁡θBD0−sin⁡θBDcos⁡θBD0001][00θ˙AB]×[00θ˙BD]+[00θ¨BD]=[00θ¨AB+θ¨BD]$(28)

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    $ 2\boldsymbolv˙2=12\boldsymbolR[1ω˙1×1\boldsymbolP2+1ω˙1×(1ω˙1×1\boldsymbolP2)+1\boldsymbolv˙1]=[cos⁡θBDsin⁡θBD0−sin⁡θBDcos⁡θBD0001][00θ¨AB]×[l100]+[00θ˙AB]×([00θ˙AB]×[l100])+[gnsin⁡θABgncos⁡θAB0]=[l1θ¨ABsin⁡θBD−l1θ˙AB2cos⁡θBD+gnsin⁡(θAB+θBD)l1θ¨ABcos⁡θBD+l1θ˙AB2sin⁡θBD+gncos⁡(θAB+θBD)0] $(29)

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    $ 2\boldsymbolv˙C2=2ω˙2×2\boldsymbolPC2+2ω˙2×(2ω˙2×2\boldsymbolPC2)+2\boldsymbolv˙2=[0l2(θ¨AB+θ¨BD)0]+[−l2(θ˙AB+θ˙BD)200]+[l1θ¨ABsin⁡θBD−l1θ˙AB2cos⁡θBD+gnsin⁡(θAB+θBD)l1θ¨ABcos⁡θBD+l1θ˙AB2sin⁡θBD+gncos⁡(θAB+θBD)0]$(30)

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    $ ^2\boldsymbol{F}_2=m_2^2\dot{\boldsymbol{v}}_{{\rm{C}}_2}=[m2l1θ¨ABsin⁡θBD−m2l1θ˙AB2cos⁡θBD+m2gnsin⁡(θAB+θBD)−m2l2(θ˙AB+θ˙BD)2m2l1θ¨ABcos⁡θBD+m2l1θ˙AB2sin⁡θBD+m2gncos⁡(θAB+θBD)+m2l2(θ¨AB+θ¨BD)0] $(31)

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    $ ^2\boldsymbol{N}_2={}^{{\rm{C}}_2}\boldsymbol{I}_2{}^2{\text{ω}}_2+{}^2{\text{ω}}_2\times{}^{{\rm{C}}_2}\boldsymbol{I}_2{}^2{\text{ω}}_2=0 $(32)

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    $ 2\boldsymbolf2=32\boldsymbolR3\boldsymbolf3+2\boldsymbolF2=[m2l1θ¨ABsin⁡θBD−m2l1θ˙AB2cos⁡θBD+m2gnsin⁡(θAB+θBD)−m2l2(θ˙AB+θ˙BD)2−Fcos⁡θBDm2l1θ¨ABcos⁡θBD+m2l1θ˙AB2sin⁡θBD+m2gncos⁡(θAB+θBD)+m2l2(θ¨AB+θ¨BD)−Fsin⁡θBD0]$(33)

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    $^2\boldsymbol{n}_2=[00m2l1l2θ¨ABcos⁡θBD+m2l1l2θ˙AB2sin⁡θBD+m2l2gncos⁡(θAB+θBD)+m2l22(θ¨AB+θ¨BD)−Fl2sin⁡θBD] $(34)

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    $ 1\boldsymbolf1=21\boldsymbolR2\boldsymbolf2+1\boldsymbolF1=[cos⁡θBD−sin⁡θBD0sin⁡θBDcos⁡θBD0001][m2[l1θ¨ABsin⁡θBD−l1θ˙AB2cos⁡θBD+gnsin⁡(θAB+θBD)−l2(θ˙AB+θ˙BD)2]m2[l1θ¨ABcos⁡θBD+l1θ˙AB2sin⁡θBD+gncos⁡(θAB+θBD)+l2(θ¨AB+θ¨BD)]0]+[−m1l1θ˙AB2+m1gnsin⁡θABm1l1θ¨AB+m1gncos⁡θAB0]$(35)

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    $ 1\boldsymboln1=[00m2l1l2θ¨ABcos⁡θBD+m2l1l2θ˙AB2sin⁡θBD+m2l2gncos⁡(θAB+θBD)+m2l22(θ¨AB+θ¨BD)]+[00m1l12θ¨AB+m1l1gncos⁡θAB]+[000m2l12θ¨AB−m2l1l2(θ˙AB+θ˙BD)2sin⁡θBD+m2l1gnsin⁡θBDsin⁡(θAB+θBD)+m2l1l2(θ¨AB+θ¨BD)cos⁡θBD+m2l1gncos⁡θBDcos⁡(θAB+θBD)] $(36)

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    $ τ1=m2l22(θ¨AB+θ¨BD)+m2l1l2(2θ¨AB+θ¨BD)+(m1+m2)l12θ¨AB−m2l1l2θ˙AB2sin⁡θBD−2m2l1l2θ˙ABθ˙2sin⁡θBD+m2l2gncos⁡(θAB+θBD)+(m1+m2)l1gncos⁡θBD$(37)

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    $ \tau_{2}=m_{2}l_{1}l_{2}\ddot{\theta}_{\mathrm{AB}}\cos\theta_{\mathrm{BD}}+m_{2}l_{1}l_{2}\dot{\theta}_{\mathrm{AB}}^{2}\sin\theta_{\mathrm{BD}}+m_{2}l_{2}g_{\rm{n}}\cos(\theta_{\mathrm{AB}}+\theta_{\mathrm{BD}})+m_{2}l_{2}^{2}(\ddot{\theta}_{\mathrm{AB}}+\ddot{\theta}_{\mathrm{BD}})$(38)

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    Shan JIA, Rujie HU, Xianghua ZHOU, Shaoyang LIU, Mingyang WU, Jinbao CHEN. Design and Analysis of a New Reusable Lander[J]. Spacecraft Recovery & Remote Sensing, 2024, 45(2): 1
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