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Helicopter rotor dynamics
1. HELICOPTER PERFORMANCE STABILITY AND CONTROL
COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Question 1. Flapping blade angle variation for the advancing blade for one complete revolution
Answer : My chosen helicopter is MBB Bo105
Locks inertia number 𝛾 =
𝜌𝑎𝑐𝑅4
𝐼 𝛽
= 5.07
where ρ = 1.225
𝑘𝑔
𝑚3 [sea level condition] ; a = lift curve slope = 6.113 /rad ; c = chord = 0.27 m ;
R = radius = 4.91 m ; 𝐼𝛽= 231.7
𝑘𝑔
𝑚2 [ data source : Helicopter Flight Dynamics –Padfield]
Induced Velocity in hover 𝜈𝑖 = √
𝑊
2𝜌𝜋𝑅2 = 10.8 m/sec ; where W = 2200 x 9.81 = 21582 N
Induced inflow ratio 𝜆𝑖 =
𝜈 𝑖
Ω𝑅
=
10.8
218
= 0.05
Flapping solution in hover β (t) = 𝜷 𝒉𝒐𝒎(t) + 𝜷 𝒑𝒂𝒓𝒕(t)
Particular solution 𝛽 𝑝𝑎𝑟𝑡(t) =
𝛾
8
( θ -
4
3
𝜆𝑖 ) =
5.07
8
(8 -
4
3
0.05) = 5.03 degree; where θ = 8 degree(given)
[𝛽 𝑝𝑎𝑟𝑡(t) = 5.03 degree is constant coning angle ]
Homogenous solution 𝛽ℎ𝑜𝑚(t) = 𝛽0 𝑒
−𝛾Ω𝑡
16 [cos(Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾
16
√1−(
𝛾
16
)
2
sin(Ω√1 − (
𝛾
16
)
2
. 𝑡)]
MS Excel was used to calculate and plot β (t) = 𝛽ℎ𝑜𝑚(t) + 𝛽 𝑝𝑎𝑟𝑡(t) ,
were azimuth angle ψ =Ω t , 𝛽0 = 𝛽 𝑝𝑎𝑟𝑡 𝑤𝑎𝑠 𝑐ℎ𝑜𝑠𝑒𝑛
Fig. 01 Flapping blade angle β (ψ) = βhom(ψ) + βpart(ψ) variation for one revolution
0
2
4
6
8
10
12
0 5 10 15 20 25 30
beta(degree)
PSI ( x 15 degree)
Beta_hom +
Beta_parti
coning angle
(Beta_parti)
2. HELICOPTER PERFORMANCE STABILITY AND CONTROL
COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Question 2. Angle of attack variation for the advancing blade for one complete revolution at different radii
positions on the blade varying from 0 to R
Answer : Angle of attack of blade element located at a distance r from the rotation axis
𝜶 𝒓 = θ -
𝝂 𝒊+𝜷𝒓̇
Ω𝒓
Now to calculate 𝛽̇ from the relation 𝛽 (𝑡̇ ) = 𝛽ℎ𝑜𝑚
̇ (t) + 𝛽 𝑝𝑎𝑟𝑡
̇ (t) = 𝛽ℎ𝑜𝑚
̇ (t) [as 𝛽 𝑝𝑎𝑟𝑡(t) is a constant term]
Writing 𝛽 (𝑡̇ ) = 𝛽0[A+B]
Were A = (
−𝛾Ω
16
) 𝑒
−𝛾Ω𝑡
16 [cos (Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾
16
√1−(
𝛾
16
)
2
sin (Ω√1 − (
𝛾
16
)
2
. 𝑡)]
And B = 𝑒
−𝛾Ω𝑡
16 [– (Ω√1 − (
𝛾
16
)
2
)sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾Ω
16
cos (Ω√1 − (
𝛾
16
)
2
. 𝑡)]
Simplifying [A+B] , cosine terms from both A and B cancel each other
[A+B] = (−)𝑒
−𝛾Ω𝑡
16 sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) [
(
𝛾
16
)
2
Ω
√1−(
𝛾
16
)
2
+ Ω√1 − (
𝛾
16
)
2
]
Simplifying the [ ] term further
[A+B] = (−)𝑒
−𝛾Ω𝑡
16 sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) [
Ω
√1−(
𝛾
16
)
2
]
or 𝛃 (𝐭̇ ) =
(−)𝛃 𝟎Ω 𝐞
−𝛄Ω𝐭
𝟏𝟔
√ 𝟏−(
𝛄
𝟏𝟔
)
𝟐
𝐬𝐢𝐧 (Ω√ 𝟏 − (
𝛄
𝟏𝟔
)
𝟐
. 𝐭)
using the β (ṫ ) value , 𝛼 𝑟 at each radial location was calculated using MS EXCEL and are plotted below.
Fig. 0.2 Angle of attack variation with radial and azimuth
5
5.5
6
6.5
7
7.5
8
0 10 20 30 40 50
Angleofattack(degree)
PSI (x 15)
r 0.1 r 0.2
r 0.3 r 0.4
r 0.5 r 0.6
r 0.7 r 0.8
r 0.9 r 1
3. HELICOPTER PERFORMANCE STABILITY AND CONTROL
COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Question 3. Coning angle , longitudinal and lateral disc tilt angles.
Answer : Coning angle a0 = βpart(t) = 5.03 degree
Longitudinal disc tilt angle a1 = 0 degree
Lateral disc tilt angle b1 = 0 degree
Assume that the helicopter is fixed to the ground but has a body pitch rate q and roll rate p.
Question 4. Advancing blade angular velocities
Blade angular velocities: i Ω sinβ + q sinψ cosβ – p cosψ cosβ
j q cosψ + p sinψ - 𝛽̇
k Ω cosβ + p cosψ sinβ – q sinψ sinβ
Question 5. Flapping equation when the helicopter is having body rates p and q and is fixed to the ground
Kinetic Energy:
T =
1
2
𝐼 [(𝑞 𝑐𝑜𝑠𝜓 + 𝑝 𝑠𝑖𝑛𝜓 − 𝛽̇)
2
+ (Ω 𝑐𝑜𝑠𝛽 + 𝑝 𝑐𝑜𝑠𝜓 𝑠𝑖𝑛𝛽 − 𝑞 𝑠𝑖𝑛𝜓 𝑠𝑖𝑛𝛽)2
] eq. 01
[Note: neglecting the i axis contribution of blade angular velocities towards Kinetic Energy calculation]
Writing the above eq. 01 for simplicity T =
1
2
𝐼[𝐴 + 𝐵]
Lagrange equation of motion:
𝑑
𝑑𝑡
(
𝜕𝑇
𝜕𝛽̇ ) −
𝜕𝑇
𝜕𝛽
+
𝜕𝑉
𝜕𝛽
= 𝑄 𝛽 were 𝑄 𝛽 is the aerodynamic moment about the flapping hinge 𝑀 𝑎(𝑡)
Evaluating the first term Lagrange EoM
Now
𝜕𝑇
𝜕𝛽̇ =
1
2
𝐼 [2𝛽̇-2(𝑞 𝑐𝑜𝑠Ω𝑡 + 𝑝 𝑠𝑖𝑛Ω𝑡)] and
𝑑
𝑑𝑡
(
𝜕𝑇
𝜕𝛽̇ ) = 𝐼[𝛽̈ + 𝑞Ω 𝑠𝑖𝑛𝜓 − 𝑝Ω 𝑐𝑜𝑠𝜓] with ψ = Ωt
Now expand the B term of eq. 01
B = [(Ω 𝑐𝑜𝑠𝛽)2
+ (𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )2
𝑠𝑖𝑛2
𝛽 + 2Ω 𝑐𝑜𝑠𝛽𝑠𝑖𝑛𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )]
= [(Ω 𝑐𝑜𝑠𝛽)2
+ (𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )2
𝑠𝑖𝑛2
𝛽 + Ω 𝑠𝑖𝑛2𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )]
= [(Ω 𝑐𝑜𝑠𝛽)2
+ 0 + Ω 𝑠𝑖𝑛2𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )]
Middle term became 0 as 𝛽 is small and neglecting 𝑝2
, 𝑞2
, 𝑝. 𝑞 terms
Now evaluating the second term of Lagrange EoM
𝜕𝑇
𝜕𝛽
=
1
2
𝐼[−2Ω2
𝑐𝑜𝑠𝛽𝑠𝑖𝑛𝛽 + 2Ω𝑐𝑜𝑠2𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 ) ]
= 𝐼[−Ω2
𝛽 + Ω(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 ) ] assuming 𝛽 is small
Third term of the Lagrange EoM is zero
![HELICOPTER PERFORMANCE STABILITY AND CONTROL
COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Question 1. Flapping blade angle variation for the advancing blade for one complete revolution
Answer : My chosen helicopter is MBB Bo105
Locks inertia number 𝛾 =
𝜌𝑎𝑐𝑅4
𝐼 𝛽
= 5.07
where ρ = 1.225
𝑘𝑔
𝑚3 [sea level condition] ; a = lift curve slope = 6.113 /rad ; c = chord = 0.27 m ;
R = radius = 4.91 m ; 𝐼𝛽= 231.7
𝑘𝑔
𝑚2 [ data source : Helicopter Flight Dynamics –Padfield]
Induced Velocity in hover 𝜈𝑖 = √
𝑊
2𝜌𝜋𝑅2 = 10.8 m/sec ; where W = 2200 x 9.81 = 21582 N
Induced inflow ratio 𝜆𝑖 =
𝜈 𝑖
Ω𝑅
=
10.8
218
= 0.05
Flapping solution in hover β (t) = 𝜷 𝒉𝒐𝒎(t) + 𝜷 𝒑𝒂𝒓𝒕(t)
Particular solution 𝛽 𝑝𝑎𝑟𝑡(t) =
𝛾
8
( θ -
4
3
𝜆𝑖 ) =
5.07
8
(8 -
4
3
0.05) = 5.03 degree; where θ = 8 degree(given)
[𝛽 𝑝𝑎𝑟𝑡(t) = 5.03 degree is constant coning angle ]
Homogenous solution 𝛽ℎ𝑜𝑚(t) = 𝛽0 𝑒
−𝛾Ω𝑡
16 [cos(Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾
16
√1−(
𝛾
16
)
2
sin(Ω√1 − (
𝛾
16
)
2
. 𝑡)]
MS Excel was used to calculate and plot β (t) = 𝛽ℎ𝑜𝑚(t) + 𝛽 𝑝𝑎𝑟𝑡(t) ,
were azimuth angle ψ =Ω t , 𝛽0 = 𝛽 𝑝𝑎𝑟𝑡 𝑤𝑎𝑠 𝑐ℎ𝑜𝑠𝑒𝑛
Fig. 01 Flapping blade angle β (ψ) = βhom(ψ) + βpart(ψ) variation for one revolution
0
2
4
6
8
10
12
0 5 10 15 20 25 30
beta(degree)
PSI ( x 15 degree)
Beta_hom +
Beta_parti
coning angle
(Beta_parti)](https://image.slidesharecdn.com/assignment04deepakpaultirkey-160719123312/85/Helicopter-rotor-dynamics-1-320.jpg)
![HELICOPTER PERFORMANCE STABILITY AND CONTROL
COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Question 2. Angle of attack variation for the advancing blade for one complete revolution at different radii
positions on the blade varying from 0 to R
Answer : Angle of attack of blade element located at a distance r from the rotation axis
𝜶 𝒓 = θ -
𝝂 𝒊+𝜷𝒓̇
Ω𝒓
Now to calculate 𝛽̇ from the relation 𝛽 (𝑡̇ ) = 𝛽ℎ𝑜𝑚
̇ (t) + 𝛽 𝑝𝑎𝑟𝑡
̇ (t) = 𝛽ℎ𝑜𝑚
̇ (t) [as 𝛽 𝑝𝑎𝑟𝑡(t) is a constant term]
Writing 𝛽 (𝑡̇ ) = 𝛽0[A+B]
Were A = (
−𝛾Ω
16
) 𝑒
−𝛾Ω𝑡
16 [cos (Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾
16
√1−(
𝛾
16
)
2
sin (Ω√1 − (
𝛾
16
)
2
. 𝑡)]
And B = 𝑒
−𝛾Ω𝑡
16 [– (Ω√1 − (
𝛾
16
)
2
)sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) +
𝛾Ω
16
cos (Ω√1 − (
𝛾
16
)
2
. 𝑡)]
Simplifying [A+B] , cosine terms from both A and B cancel each other
[A+B] = (−)𝑒
−𝛾Ω𝑡
16 sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) [
(
𝛾
16
)
2
Ω
√1−(
𝛾
16
)
2
+ Ω√1 − (
𝛾
16
)
2
]
Simplifying the [ ] term further
[A+B] = (−)𝑒
−𝛾Ω𝑡
16 sin (Ω√1 − (
𝛾
16
)
2
. 𝑡) [
Ω
√1−(
𝛾
16
)
2
]
or 𝛃 (𝐭̇ ) =
(−)𝛃 𝟎Ω 𝐞
−𝛄Ω𝐭
𝟏𝟔
√ 𝟏−(
𝛄
𝟏𝟔
)
𝟐
𝐬𝐢𝐧 (Ω√ 𝟏 − (
𝛄
𝟏𝟔
)
𝟐
. 𝐭)
using the β (ṫ ) value , 𝛼 𝑟 at each radial location was calculated using MS EXCEL and are plotted below.
Fig. 0.2 Angle of attack variation with radial and azimuth
5
5.5
6
6.5
7
7.5
8
0 10 20 30 40 50
Angleofattack(degree)
PSI (x 15)
r 0.1 r 0.2
r 0.3 r 0.4
r 0.5 r 0.6
r 0.7 r 0.8
r 0.9 r 1](data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7)