helikopteri i: osnove aerodinamike i mehanike leta...
TRANSCRIPT
![Page 1: Helikopteri I: Osnove aerodinamike i mehanike leta ...titan.fsb.hr/~mvrdolja/heli/heli_dinamika_talk.pdf · Dinamika rotora i upravljanje dr. sc. Milan Vrdoljak milan.vrdoljak@fsb.hr](https://reader030.vdocuments.pub/reader030/viewer/2022021602/5c8c5c8509d3f2a66a8cde29/html5/thumbnails/1.jpg)
Helikopteri I:Osnove aerodinamike i mehanike leta
Dinamika rotora i upravljanje
dr. sc. Milan [email protected]
Fakultet strojarstva i brodogradnjeZavod za zrakoplovstvo
Studeni 2011.
Milan Vrdoljak (FSB) Helikopteri I 1 / 19
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Uvod
raspodjela brzina po azimutnom polozaju kraka
Ω
ψ = 0
ψ = 90
ψ = 180
ψ = 270
V
ΩR
+V
ΩR−V
ΩR
ΩR
ψ
Milan Vrdoljak (FSB) Helikopteri I 2 / 19
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Uvod
raspodjela brzina na elementarnom kraku
Uz
Ut
U
φ
α
θdFx
dFz
dL
dD
dFp
ravnina rotacije
osro
taci
je
Milan Vrdoljak (FSB) Helikopteri I 3 / 19
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Uvod
Milan Vrdoljak (FSB) Helikopteri I 4 / 19
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Uvod
zglobni rotor, engl. articulated
Milan Vrdoljak (FSB) Helikopteri I 5 / 19
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Uvod
bez-zglobni rotor, engl. hingeless
Milan Vrdoljak (FSB) Helikopteri I 6 / 19
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Uvod
potpuno bez-zglobni rotor, engl. bearingless
Milan Vrdoljak (FSB) Helikopteri I 7 / 19
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Mahanje kraka
Sile na elementarnu masu kraka koje cine moment u zglobu mahanja
os
rota
cije
y
dFC
(dLdy
)dy
dI
mdy
β
zglob mahanja
Milan Vrdoljak (FSB) Helikopteri I 8 / 19
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Mahanje kraka
centrifugalni moment
dMC = my2Ω2βdy ,
inercijski moment
dI = (mdy) y2β ,
i aerodinamicki moment
dMA = −(dL
dy
)ydy .
Milan Vrdoljak (FSB) Helikopteri I 9 / 19
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Mahanje kraka
Jednadzba gibanja mahanja — suma ukupnih momenatau zglobu mahanja:∫ R
0
dMC +
∫ R
0
dI +
∫ R
0
dMA = 0 ,
∫ R
0
mΩ2βy2dy +
∫ R
0
mβy2dy −∫ R
0
(dL
dy
)ydy = 0 .
Milan Vrdoljak (FSB) Helikopteri I 10 / 19
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Mahanje kraka
moment tromosti kraka oko zgloba mahanja
Ib =
∫ R
0
my2dy ,
jednadzba mahanja
Ibβ + IbΩ2β =
∫ R
0
(dL
dy
)ydy ,
Ibβ + IbΩ2β = −MA .
Milan Vrdoljak (FSB) Helikopteri I 11 / 19
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Mahanje kraka
kako je ψ = Ωt
β =∂β
∂t=∂ψ
∂t
∂β
∂ψ= Ω
∂β
∂ψ= Ω
∗β ,
β =∂2β
∂t2= Ω2∂
2β
∂ψ2= Ω2
∗∗β ,
jednadzba mahanja kraka
∗∗β + β =
−MA
IbΩ2.
Milan Vrdoljak (FSB) Helikopteri I 12 / 19
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Mahanje kraka
−MA
IbΩ2= γMβ , γ =
ρclαcR4
Ib
Mβ =θ
(1
8+µ
3sinψ +
µ2
4sin2 ψ
)+
+ θtw
(1
10+µ
4sinψ +
µ2
6sin2 ψ
)− λ
(1
6+µ
4sinψ
)−
−∗β
(1
8+µ
6sinψ
)− βµ cosψ
(1
6+µ
4sinψ
)
Milan Vrdoljak (FSB) Helikopteri I 13 / 19
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Mahanje kraka
komponente brzine na elementarnom kraku
Milan Vrdoljak (FSB) Helikopteri I 14 / 19
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Mahanje kraka
jednadzba mahanja kraka
∗∗β + β = γMβ .
rjesenje za kut mahanja raspisom u Fourierov red(1. harmonici)
β(ψ) = β0 + β1c cosψ + β1s sinψ .
Milan Vrdoljak (FSB) Helikopteri I 15 / 19
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Mahanje kraka
jednadzba mahanja kraka
∗∗β + β = γMβ .
rjesenje za kut mahanja raspisom u Fourierov red(1. harmonici)
β(ψ) = β0 + β1c cosψ + β1s sinψ .
Milan Vrdoljak (FSB) Helikopteri I 15 / 19
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Mahanje kraka
rjesenje jednadzbe mahanja
β0 = γ
[θ0
8
(1 + µ2
)+θtw10
(1 +
5
6µ2
)+µ
6θ1s −
λ
6
]β1s − θ1c =
−43µβ0
1 + 12µ
2
β1c + θ1s =−8
3µ[θ0 − 3
4λ+ 34µθ1s + 3
4θtw
]1− 1
2µ2
Milan Vrdoljak (FSB) Helikopteri I 16 / 19
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Mahanje kraka
Za lebdenje µ = 0:
β1s − θ1c = 0 → β1s = θ1c
β1c + θ1s = 0 → β1c = −θ1s
Postavni kut: raspis u Fourierov red (1. harmonici)
θ(ψ) = θ0 + θ1c cosψ + θ1s sinψ .
odziv kuta mahanja je
β = β0 + θ1c cos(ψ − π
2
)+ θ1s sin
(ψ − π
2
).
Milan Vrdoljak (FSB) Helikopteri I 17 / 19
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Mahanje kraka
Za lebdenje µ = 0:
β1s − θ1c = 0 → β1s = θ1c
β1c + θ1s = 0 → β1c = −θ1s
Postavni kut: raspis u Fourierov red (1. harmonici)
θ(ψ) = θ0 + θ1c cosψ + θ1s sinψ .
odziv kuta mahanja je
β = β0 + θ1c cos(ψ − π
2
)+ θ1s sin
(ψ − π
2
).
Milan Vrdoljak (FSB) Helikopteri I 17 / 19
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Mahanje kraka
Za lebdenje µ = 0:
β1s − θ1c = 0 → β1s = θ1c
β1c + θ1s = 0 → β1c = −θ1s
Postavni kut: raspis u Fourierov red (1. harmonici)
θ(ψ) = θ0 + θ1c cosψ + θ1s sinψ .
odziv kuta mahanja je
β = β0 + θ1c cos(ψ − π
2
)+ θ1s sin
(ψ − π
2
).
Milan Vrdoljak (FSB) Helikopteri I 17 / 19
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Mahanje kraka
0 90 180 270 3600
2
4
6
8
ψ, [deg]
β, [d
eg]
a0=4° , a
1=1.3° , b
1=0.8°
ψ=90°
ψ=0°
ψ=180°
ψ=270°
Milan Vrdoljak (FSB) Helikopteri I 18 / 19
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Upravljanje rotorom
Upravljacka velicina: postavni kutraspis u Fourierov red (1. harmonici)
θ(ψ) = θ0 + θ1c cosψ + θ1s sinψ .
Milan Vrdoljak (FSB) Helikopteri I 19 / 19