NASA Technical Reports Server (NTRS) 20130003227: Supersonic Pitch Damping Predictions of Blunt Entry Vehicles from Static CFD Solutions PDF
Preview NASA Technical Reports Server (NTRS) 20130003227: Supersonic Pitch Damping Predictions of Blunt Entry Vehicles from Static CFD Solutions
∗ a S B t C t mα φ C C V mα(BS) mα C C X mα(FB) mα C ∂Cm mα˙ ∂qd 2V C C α mo m d α o E γ f α o I λ yy M ξ BS q ρ ∗ φ ∞ ω ω ˙ o ω ¨ 1 ω 2 ≈ CO 2 V ∞ +C m +α V ∞ Forebody Backshell 1 d 1 I α¨ ρV2SdC α˙ ρV2SdC α=0 yy − 2 ∞ mα˙ 2V − 2 ∞ mα ∞ α¨+2ξα˙ +ω2α=0 o C qSd2 ξ = mα˙ − 4I V yy ∞ C qSd ω2 = mα o − I yy α=eλt λ2+2ξλ+ω2 eλt =0 λ2+2ξλ+ω2 =0 o ⇒ o ξ ξ C =C +C mα mα(FB) mα(BS) α¨(t)+ω2α(t t )+ω2α(t)=0 2 − φ 1 C qSd ω2 = mα(BS) 2 − I yy C qSd ω2 = mα(FB) 1 − I yy α(t tφ)=eλ(t−tφ) =eλte−λtφ − λ2+ω22e−λtφ +ω12 eλt =0 ⇒ λ2+ω22e−λtφ +ω12 =0 X2 X3 eX =1+X+ + +... 2! 3! λt φ λ2+ω2(1 λt )+ω2 =0 2 − φ 1 ω ω 2 1 λ2 ω2t λ+ω2 =0 − 2 φ o 2ξ = Cmα˙qSd2 = ω2t = Cmα(BS)qSdt − 2I V − 2 φ I φ yy yy ∞ 2C V t C = mα(BS) ∞ φ mα˙ − d λ = ξ iω 1,2 − ± ω ω = ω2 ξ2 = CmαqSd 2Cmα(BS)qSdt 2 o − − Iyy − Iyy φ C t C mα φ mα(BS) ω C qSd/I ≈ − mα yy ω f = 2π φ 1 φ t = = φ 2π f ω C = 2Cmα(BS)V∞φ = 2Cmα(BS) 2Iyy φ mα˙ − dω − −Cmα ρ∞Sd3 · M qSd BS damping −Iyy =2ξα˙ =−Cmα(BS)φ−CmαIyyα˙ C m(BS) C C mα(BS) m α=α + o α˙ =˙ α¨ =¨ α o C C α+C =C +C α +C m(bs) ≈ mα(bs) mo mα(bs) mα(bs) o mo 1 d 1 I ¨ ρV2SdC ˙ ρV2Sd[(C +C )+C α +C ]=0 yy − 2 ∞ mα˙ 2V − 2 ∞ mα(FB) mα(BS) mα o mo ∞ C mα˙ ¨(t) Cmα(bs)qSd(t t ) Cmα(FB)qSd(t)= CmαqSdα + CmoqSd φ o − I − − I I I yy yy yy yy ¨(t)+ω2(t)+ω2(t t )+ω2α +γ =0 1 2 − φ o o qSd γ = C −I mo yy ω 1 =eλt+B+E (t tφ)=eλte−λtφ +B+E − λ2 ω2 t λ+ω2 +ω2 B+ω2 E =ω2 α +γ − 2 |αo φ o |αo o |αo o |αo o |αo o B E B = α o − γ C E = = mo ω2 C o mα λ2 ω (α )2t λ+ω (α )2 =0 2 o φ o o − Cmα˙(αo)=−2ω2(αoq)S2Vd∞Iyytφ =−2Cmα(bs)d|αo V∞tφ C (α )= 2Cmα(BS) |αo 2Iyy φ mα˙ o − −Cmα ρ∞Sd3 · ξ λ2+(2ξ ω2 t )λ+ω =0 − 2 |αo φ o |αo 2C V t C =C mα(BS) |αo ∞ φ mα˙(total) mα˙(Mod.Newtonian) − d 1.000d 0.732d 0.247d 0.217d 62.18° 43.4° 0.365d 40° 0.469d 9.3° 0.566d 0.333d 20° 20° R0.007d R0.025 R0.250d R0.250d 1.000d Viking MER d (m) 0.07 S (m2) 0.00385 I (kg m2) 1.55E 04 yy − − ρ (kg/m3 1.23 ∞ a(m/s) 348 C mα˙ CO 2 α α=0 ◦ α C m(bs) 2 4 6 8 11 12 14 16 18 21 26 ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ 15 ◦ φ = 15 ◦ ωd St= 2V ∞ St = 0.015 Teramoto St =0.016 St =0.011 MER,M=2 MER,M=3 St = 0.032 Viking,M=3 15 ◦
The list of books you might like
