Meccanica del volo - Esercizi / prove esame

1) 8 MAGGIO 2006

Eq L=W

M_G=0 C_MG=0

C_MG=0

C_Me+C_Mαα+C_Mδδ+C_Mθθ=0

L=W

L=1/2 ρ V_cas S C_L

C_L=C_Lαα+C_Lδδ+C_Lθθ=0,4506

α=−(C_Me+C_Mδδ+C_Mθθ)/C_Mαg=0=0,09895 rad

L = W = 35160,236 N

2) 13 SETTEMBRE 2006

STABILITÀ

HP M_AC=0

SM=3_G+3_N

ξ₀=SM−3_N

M_PL^wbM_ACL^wub(x_AC−x) − W(x_L−x₀) − L^t(x_t−x_AC)

M_P,10=0 L^wbM_ACL^wb(x_wubx_wbx_t)

L_E=a+b δ

α' =α−ξ α

[ξ=ξ₀+ξ_E α]

L^wb=L^wb

M_E,10=9SMC_ME=9SM ξξ)9α9( 0−9ξ

C_Mp,10=9

3_N= ... = −0.057

−ξ_G=SM−3_N=−0.187

3) 15 Settembre 2003 Equilibrio e Stabilità

SM = 3_CL - 3_N = -C_MGx/C_L = 0,4833

C_La = C_L0 + C_L (α - α₀) + C_L/S (β - β₀)

C_MGy = C_MG0.1 + C_mg (β - β₀) + C_m δ - δ₀

C_mα = C_m0.1 + C_x (1 - ε_α ε_β)

S_3C3 = 5.924 lg M^-1

C_l = C_A + C_B + C_C + C_D

C_l = C_MGx (3 - ε) C_L + C_mg δ = 0

(3_C3 - 3_{C6 Cx0) = 0.03 δ}

CMG = 0 = C_MGx (1 - ε_αε_β)

3_S3 - 3_N ≥ C_L/C_LS

E_q → L − W ≥ LD = 2/3 S ÷

= (L/W) = 2,888 = 0,905 lα

C_L = C_cm (n - α) + C_CL4 (β - β₀) = 0,2422

4) Equilibrio

• Q= GM (α + CL/S + CL/C)=0
•  
• S C_MGMa' = 6.8522 rad^-1
• C_L0 = - 0.01133 rad^-1 α
• C_L = S^1/2/2 V_el

∑ &langer; C_MGy &chai; = = ε> √ 3 V_el

A m[ b C 3₃ C 3_{6 alpha} = 0. 1 C_c. C_lad

G_cmy _3-s

c/d = 8.800 N / m

-_Cl/S C_MG = N_q &false; "Lor"

8 GM/2

30 GIUGNO 2010

1. PRESTAZIONI

W/S = 38740 N/m²   b₂ = 8,735

k₂ = 1   T_cL = 0,0251

V₁ = 2W/S

ξ_max

V_a emax 17.68

2W/S

C_L → V_max → _D

L = 2 T_e C_Lmax

L₃ = 2S_VC_Lmean

T₅₀₀₀ → D

L₂ →

2 W/S → C_Lmax

m = L/W = 2,41

T = 2 2 8,25

T_m →

L_2S

m = W/S

L → C_L^∞

• V → L
• V_max → 316
• T = D
• S²&SubL;D

Q =

Ω = V/Ω

Ω = 0,166 rad/s

R = 5000 m    T_UNN

V_max = 560 W/S

m = 25

T_TUNN =

T₅₀₀₀ V_max T_TUNN

2. Equilibrio e stabilità

1. SM = sG - sN → sG = SM + sN = 0,027

   CLα = CL0 + CLICG + CLOG

   CNα = CNY0 + CNH / 8 + CNBCG

   CMf = CNH + CLα (sG - ξCG)

   CMt / sG = CL • SM = 0,06584

   β = CMN / sB = 0,68°

   β → sG - sN = CMN / sB

   CMN / sB (ξC - ξN ) = - 2,506

   CLα = 2 WI / S

   Sθ VS B = 0,2056

2. CL1α = CL1α0 (1 - ^{sG - sN}) _{sG - sC} = 5,61 nod^-1

1. Det

   CLp = 5,82 nod^-1

   CLβ = 0,63 nod^-1

   CL0 = 0,032

   sN = -0,123

   ξC = -0,098

   CMN0 = 0,096

   SM = 12%

2. Vess = 142 mtrs

   WI / S = 9100 N / m²

3. β ?

   SX

ESE 6

1. 25 GENNAIO 2012

1. • EG → D = Lw = ¹/₂ρ V_{eq, st}² S C_L,man = 2 W/S²_V²ρVeq,st = 1,168
   • D = ¹/₂ρV_eqSC_D = ¹/₂ρV_eq,stS (C_D + k C_L²)
   • D = ¹/₂ ρV_eq,st²S(C_Do + kC²L_,maxcas)
   • D_st = C_Do → V_eq = ^2W/S/_δₒ C_L
   • C_L,min ^ρ/₂ = ^3C_Do/_V₁ = 1,4132 → V_eq,min = 139, 7 m/s
   • C_L,min = 0,6427 → V_eq,min = 184,1 m/s
2. Φ=45° → M =¹/_cosφ = 1,414
3. V_eq,st = 63,26 mm/s
4. V_eq,max = 218,3 mm/s
5. D_st = 23830 N
6. D_{max, st} = 53098 N
7. W = 37396 N
8. S = 92 m²

2) EQUILIBRIO E STABILITÀ

SM = S_c - S_N → S_n = S_c - SM = -0,3

S_n = S_ac + mΘ(1 - ξ₁) σ′/σ a′ S_ac

+ mΘ(1 - ξ₁) σ′/σ a′ S_ac

S_N + mΘ(1 - ξ₂) σ′/σ a′ S_N + mΘ(1 - ξ₂) σ′/σ S_ac =

mΘ(1 - ξ₁) a′/a S_ac = S_ac - S_N

S⁺ = σS = ^{S_ac - S_N}a′ η(1 - ξ₁) [S_N - S_ac]

¼ S = 13,945 m²

DATI TRADIZIONALI

• S = 9,7 m²
• ω = 6,85 rad^-1
• σ′ = 3,40 rad^-1
• S_AC = -0,035
• S_AC = 0,430
• ξ₁ = 0,278
• m = 0,928
• S_C = 0,190

η

1S⁺ per STABILITÀcon SM = 12%

10 Febbraio 2016

1) Equilibrio e Stabilità

Cl = Clα0wb + Clwb S + Clo

Clmα = Clmαwb S + Clwb b ^M_c√ + Clα S

Clα = 5.50749

Clwb = 0.655

Clδ = 0.044

Clt = 0

Clαwb ^M_c√ (C_mαwb) = 0.4066

Clo = 2.58

Cl _ew = (2 Ws) / (b V_BAS) = 0.306

Cl₀ = 0

• [ 1 S ] [ α δ ] [ Clwb - Clo ]
• [ Clwb -Clmα Clγ0 / Clwb ] [ Clo - Clwb ]
• [ 1 0.655 ] [ Clmα Clwb - Clmαwb Clγ0 / Clt ] [ Clwb - Clo ]

α = 0.05c1 = 3.21°

δ = -0.9646 = -6.84°

2)

Cd = Cdo ^+M_cCd_b = 0.0291

Cdwb = Cdo^wb k (Cdo^wb)² = 0.0243

Cdo^wb = α^wb (α + i^wb) = 0.369

Cd^T = Cdo + k (Cd^T)² = 0.025

Ct^T = α^t (at+i - ξα - ξαα) + b^t δ = -0.283

• α = β_t
• Cd - conditional minima
• Eo = 0.35 = 9.064 / m²
• CDmot = 0.205
• Clwb = 0.655
• k = 0.0575