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

𝐼𝛽 = 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 = 𝜈𝑖

Ω𝑅 =

10.8

218 = 0.05

Flapping solution in hover β (t) = 𝜷𝒉𝒐𝒎(t) + 𝜷𝒑𝒂𝒓𝒕(t)

Particular solution 𝛽𝑝𝑎𝑟𝑡(t) = 𝛾

8( θ -

4

3𝜆𝑖 ) =

5.07

8(8 -

4

30.05) = 5.03 degree; where θ = 8 degree(given)

[𝛽𝑝𝑎𝑟𝑡(t) = 5.03 degree is constant coning angle ]

Homogenous solution 𝛽ℎ𝑜𝑚(t) = 𝛽0𝑒−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) + 𝛽𝑝𝑎𝑟𝑡(t) ,

were azimuth angle ψ =Ω t , 𝛽0 = 𝛽𝑝𝑎𝑟𝑡 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

be

ta (

de

gre

e)

PSI ( x 15 degree)

Beta_hom +Beta_parti

coning angle(Beta_parti)

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

𝜶𝒓 = θ - 𝝂𝒊+𝜷

Ω𝒓

Now to calculate from the relation 𝛽 (𝑡 ) = 𝛽ℎ𝑜𝑚 (t) + 𝛽𝑝𝑎𝑟𝑡

(t) = 𝛽ℎ𝑜𝑚 (t) [as 𝛽𝑝𝑎𝑟𝑡(t) is a constant term]

Writing 𝛽 (𝑡 ) = 𝛽0[A+B]

Were A = (−𝛾Ω

16) 𝑒

−𝛾Ω𝑡

16 [cos (Ω√1 − (𝛾

16)

2

. 𝑡) +𝛾

16

√1−(𝛾

16)

2sin (Ω√1 − (

𝛾

16)

2

. 𝑡)]

And B = 𝑒−𝛾Ω𝑡

16 [– (Ω√1 − (𝛾

16)

2

)sin (Ω√1 − (𝛾

16)

2

. 𝑡) +𝛾Ω

16cos (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)

2Ω

√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 (𝐭 ) = (−)𝛃𝟎Ω 𝐞

−𝛄Ω𝐭𝟏𝟔

√𝟏−(𝛄

𝟏𝟔)

𝟐𝐬𝐢𝐧 (Ω√𝟏 − (

𝛄

𝟏𝟔)

𝟐. 𝐭)

using the β (t ) value , 𝛼𝑟 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

An

gle

of

atta

ck (

de

gre

e)

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

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 - 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

2𝐼 [(𝑞 𝑐𝑜𝑠𝜓 + 𝑝 𝑠𝑖𝑛𝜓 − )

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

2𝐼[𝐴 + 𝐵]

Lagrange equation of motion:

𝑑

𝑑𝑡(

𝜕𝑇

𝜕) −

𝜕𝑇

𝜕𝛽+

𝜕𝑉

𝜕𝛽= 𝑄𝛽 were 𝑄𝛽 is the aerodynamic moment about the flapping hinge 𝑀𝑎(𝑡)

Evaluating the first term Lagrange EoM

Now 𝜕𝑇

𝜕 =

1

2𝐼 [2-2(𝑞 𝑐𝑜𝑠Ω𝑡 + 𝑝 𝑠𝑖𝑛Ω𝑡)] and

𝑑

𝑑𝑡(

𝜕𝑇

𝜕) = 𝐼[ + 𝑞Ω 𝑠𝑖𝑛𝜓 − 𝑝Ω 𝑐𝑜𝑠𝜓] with ψ = Ωt

Now expand the B term of eq. 01 B = [(Ω 𝑐𝑜𝑠𝛽)2 + (𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )2𝑠𝑖𝑛2𝛽 + 2Ω 𝑐𝑜𝑠𝛽𝑠𝑖𝑛𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )] = [(Ω 𝑐𝑜𝑠𝛽)2 + (𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 )2𝑠𝑖𝑛2𝛽 + 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

2𝐼[−2Ω2 𝑐𝑜𝑠𝛽𝑠𝑖𝑛𝛽 + 2Ω𝑐𝑜𝑠2𝛽(𝑝 𝑐𝑜𝑠𝜓 − 𝑞 𝑠𝑖𝑛𝜓 ) ]

= 𝐼[−Ω2𝛽 + 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

Now evaluating the complete Lagrange EoM:

𝐼 + 𝐼𝑞Ω 𝑠𝑖𝑛𝜓 − 𝐼𝑝Ω 𝑐𝑜𝑠𝜓 + 𝐼Ω2𝛽 − 𝐼Ω𝑝 𝑐𝑜𝑠𝜓 + 𝐼Ω𝑞 𝑠𝑖𝑛𝜓 = 𝑀𝑎(𝑡) Retaining the 𝛽 terms on LHS and dividing throughout by 𝐼

+ Ω𝟐𝜷 = −𝟐Ω𝒒 𝒔𝒊𝒏𝝍 + 𝟐Ω𝒑 𝒄𝒐𝒔𝝍 + 𝑴𝒂(𝒕)

𝑰 eq. 02

Question 6. Disc tilt angles a0 , a1 and b1 Angle of attack of a blade element

𝛼 = 𝜃 −𝜈𝑖+𝛽 𝑟−𝑞 𝑟 𝑐𝑜𝑠𝜓−𝑝𝑟𝑠𝑖𝑛𝜓

Ω𝑟

Evaluate aerodynamic moment about flapping hinge:

𝑴𝒂

𝑰=

1

𝐼∫ 𝑎 (𝜃 −

𝜈𝑖+𝛽 𝑟−𝑞 𝑟 𝑐𝑜𝑠𝜓−𝑝𝑟𝑠𝑖𝑛𝜓

Ω𝑟)

1

2𝜌(Ω𝑟)2𝑐𝑟 𝑑𝑟

𝑅

0

=𝜌𝑎𝑐

𝐼.

Ω2

2∫ (𝜃0 −

𝜈𝑖

Ω𝑅𝑥−

𝛽

Ω+

𝑞 𝑐𝑜𝑠𝜓

Ω+

𝑝𝑠𝑖𝑛𝜓

Ω) 𝑥3𝑅4 𝑑𝑥

1

0 taking 𝑥 =

𝑟

𝑅 ; 𝑟 = 𝑥𝑅 ; 𝑑𝑟 = 𝑅 𝑑𝑥

=𝜌𝑎𝑐𝑅4

𝐼.

Ω2

2∫ (𝜃0𝑥3 − 𝜆𝑖𝑥

2 −𝛽

Ω𝑥3 +

𝑞 𝑐𝑜𝑠𝜓

Ω𝑥3 +

𝑝𝑠𝑖𝑛𝜓

Ω𝑥3) 𝑑𝑥

1

0

= 𝛾.Ω2

2[

𝜃0

4−

𝜆𝑖

3−

𝛽

4Ω+

𝑞 𝑐𝑜𝑠𝜓

4Ω+

𝑝𝑠𝑖𝑛𝜓

4Ω]

𝑴𝒂

𝑰=

𝛾Ω2

8(𝜃0 −

4𝜆𝑖

3) −

𝛾𝛽 Ω

8+

𝛾 𝑞Ω 𝑐𝑜𝑠𝜓

8+

𝛾 𝑝Ω 𝑠𝑖𝑛𝜓

8 eq. 03

Now rewriting eq. 02 using eq. 03

+ Ω2𝛽 = −2Ω𝑞 𝑠𝑖𝑛𝜓 + 2Ω𝑝 𝑐𝑜𝑠𝜓 +𝛾Ω2

8(𝜃0 −

4𝜆𝑖

3) −

𝛾𝛽 Ω

8+

𝛾 𝑞Ω 𝑐𝑜𝑠𝜓

8+

𝛾 𝑝Ω 𝑠𝑖𝑛𝜓

8

+𝛾𝛽 Ω

8+ Ω2𝛽 =

𝛾Ω2

8(𝜃0 −

4𝜆𝑖

3) +

𝛾 𝑞Ω 𝑐𝑜𝑠𝜓

8+

𝛾 𝑝Ω 𝑠𝑖𝑛𝜓

8 − 2Ω𝑞 𝑠𝑖𝑛𝜓 + 2Ω𝑝 𝑐𝑜𝑠𝜓

+𝜸Ω

𝟖 + Ω𝟐𝜷 =

𝜸Ω𝟐

𝟖(𝜽𝟎 −

𝟒𝝀𝒊

𝟑) + Ω (

𝜸 𝒒

𝟖+ 𝟐𝒑) 𝒄𝒐𝒔Ω𝒕 + Ω (

𝜸 𝒑

𝟖 − 𝟐𝒒 ) 𝒔𝒊𝒏Ω𝒕 eq. 04 (using 𝜓 = Ω𝑡)

Assuming solution for the above eq. 04 of the form 𝛽 = 𝑎0 − 𝑎1 𝑐𝑜𝑠Ω𝑡 − 𝑏1 𝑠𝑖𝑛Ω𝑡

= 𝑎1Ω 𝑠𝑖𝑛Ω𝑡 − 𝑏1Ω 𝑐𝑜𝑠Ω𝑡

= 𝑎1Ω2 𝑐𝑜𝑠Ω𝑡 + 𝑏1Ω2 𝑠𝑖𝑛Ω𝑡

Now substituting the values of 𝛽, , on the LHS of eq. 04 LHS of eq.04

≡ 𝑎1Ω2 𝑐𝑜𝑠Ω𝑡 + 𝑏1Ω2 𝑠𝑖𝑛Ω𝑡 +𝜸

𝟖𝑎1Ω2 𝑠𝑖𝑛Ω𝑡 −

𝜸

𝟖𝑏1Ω2 𝑐𝑜𝑠Ω𝑡 + Ω2𝑎0 − 𝑎1Ω2 𝑐𝑜𝑠Ω𝑡 − 𝑏1Ω2 𝑠𝑖𝑛Ω𝑡

≡ Ω2𝑎0 + (𝑎1Ω2 −𝜸

𝟖𝑏1Ω2 − 𝑎1Ω2) 𝑐𝑜𝑠Ω𝑡 + (𝑏1Ω2+

𝜸

𝟖𝑎1Ω2 − 𝑏1Ω2) 𝑠𝑖𝑛Ω𝑡

≡ Ω2𝑎0 −𝜸

𝟖𝑏1Ω2 𝑐𝑜𝑠Ω𝑡 +

𝜸

𝟖𝑎1Ω2 𝑠𝑖𝑛Ω𝑡

Equating the above expression with the RHS of eq. 04 and comparing constant, sine and cosine terms Constant term

Ω2𝑎0 =𝛾Ω2

8(𝜃0 −

4𝜆𝑖

3) or 𝒂𝟎 =

𝜸

𝟖(𝜽𝟎 −

𝟒𝝀𝒊

𝟑)

HELICOPTER PERFORMANCE STABILITY AND CONTROL COURSE CODE: AE4314

Name: DEEPAK PAUL TIRKEY

ASSIGNMENT NUMBER 04 Student number: 4590929

Sine term

𝛾

8𝑎1Ω2 = Ω (

𝛾 𝑝

8 − 2𝑞 ) or 𝒂𝟏 = (

𝒑

Ω −

𝟏𝟔𝒒

𝜸Ω )

Cosine term

−𝛾

8𝑏1Ω2 = Ω (

𝛾 𝑞

8+ 2𝑝) or 𝒃𝟏 = (

− 𝒒

Ω−

𝟏𝟔𝒑

𝜸Ω)

Assume that the helicopter is in forward flight and has body rate q Question 7. Perform analytically the integral of blade aerodynamic moment and compare with given expression Aerodynamic moment is given as:

𝑀𝑎 = ∫ 𝐶𝑙𝛼 𝛼1

2𝜌(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)2𝑅

0𝑐 𝑑𝑟. 𝑟 Where 𝛼 = 𝜃 −

𝑉 𝑠𝑖𝑛𝛼𝑐+𝜈𝑖+𝑟−𝑞𝑟 𝑐𝑜𝑠𝜓+𝛽𝑉 𝑐𝑜𝑠𝛼𝑐 𝑐𝑜𝑠𝜓

Ω𝑟+𝑉 𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓

Performing integration with the 𝜽 part

𝑀𝑎1 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ 𝜃(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)2𝑅

0 𝑑𝑟. 𝑟

= 𝜌𝐶𝑙𝛼 𝑐1

2 𝜃 ∫ (Ω2𝑟3 + 𝑉2𝑟 cos 𝛼𝑐

2 𝑠𝑖𝑛𝜓2 + 2Ω𝑉 𝑟2𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)𝑅

0 𝑑𝑟

= 𝜌𝐶𝑙𝛼 𝑐1

2 𝜃 [

Ω2𝑅4

4+

Ω2𝑅4

2 (

𝑉𝑐𝑜𝑠𝛼𝑐

Ω𝑅)

2

𝑠𝑖𝑛𝜓2 +2 Ω2𝑅4

3(

𝑉𝑐𝑜𝑠𝛼𝑐

Ω𝑅) 𝑠𝑖𝑛𝜓] Where

𝑉𝑐𝑜𝑠𝛼𝑐

Ω𝑅= 𝜇

= 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 𝜽 [

𝟏

𝟒+

𝟏

𝟐𝝁𝟐 𝒔𝒊𝒏𝝍𝟐 +

𝟐

𝟑𝝁 𝒔𝒊𝒏𝝍]

Performing integration with 𝑽 𝒔𝒊𝒏𝜶𝒄+𝝂𝒊+𝒓−𝒒𝒓 𝒄𝒐𝒔𝝍+𝜷𝑽 𝒄𝒐𝒔𝜶𝒄 𝒄𝒐𝒔𝝍

Ω𝒓+𝑽 𝒄𝒐𝒔𝜶𝒄 𝒔𝒊𝒏𝝍 part

𝑀𝑎2 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (𝑉 𝑠𝑖𝑛𝛼𝑐 + 𝜈𝑖 + 𝑟 − 𝑞𝑟 𝑐𝑜𝑠𝜓 + 𝛽𝑉 𝑐𝑜𝑠𝛼𝑐 𝑐𝑜𝑠𝜓)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟

Now integrating each term:

𝑀𝑎3 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (𝑉 𝑠𝑖𝑛𝛼𝑐)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟 = 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 [

𝝀𝒄

𝟑+

𝝀𝒄 𝝁

𝟐𝒔𝒊𝒏𝝍]

𝑀𝑎4 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (𝜈𝑖)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟 = 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 [

𝝀𝒊

𝟑+

𝝀𝒊 𝝁

𝟐𝒔𝒊𝒏𝝍]

𝑀𝑎5 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (𝛽)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟= 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 [

𝟒Ω+

𝝁

𝟑Ω𝒔𝒊𝒏𝝍]

𝑀𝑎6 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (−𝑞𝑟 𝑐𝑜𝑠𝜓)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟= 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 [

−𝒒

𝟒Ω𝑐𝑜𝑠𝜓 +

−𝒒 𝝁

𝟑Ω𝒔𝒊𝒏𝝍𝑐𝑜𝑠𝜓]

𝑀𝑎7 = 𝜌𝐶𝑙𝛼 𝑐1

2 ∫ (𝛽𝑉 𝑐𝑜𝑠𝛼𝑐 𝑐𝑜𝑠𝜓)(Ω𝑟 + 𝑉𝑐𝑜𝑠𝛼𝑐 𝑠𝑖𝑛𝜓)

𝑅

0𝑑𝑟. 𝑟= 𝝆𝑪𝒍𝜶 𝒄𝑹𝟒Ω𝟐 𝟏

𝟐 [

𝜷𝝁

𝟑𝑐𝑜𝑠𝜓 +

𝜷𝝁𝟐

𝟐𝒔𝒊𝒏𝝍𝑐𝑜𝑠𝜓]

Now 𝑀𝑎/𝐼𝑏𝑙=𝑀𝑎1- (𝑀𝑎3 + 𝑀𝑎4 + 𝑀𝑎5 + 𝑀𝑎6 + 𝑀𝑎7) /𝐼𝑏𝑙

= 𝛾Ω2 1

2 𝜃 [

1

4+

1

2𝜇2 𝑠𝑖𝑛𝜓2 +

2

3𝜇 𝑠𝑖𝑛𝜓] −𝛾Ω2 1

2 [

𝜆𝑐

3+

𝜆𝑐 𝜇

2𝑠𝑖𝑛𝜓] − 𝛾Ω2 1

2 [

𝜆𝑖

3+

𝜆𝑖 𝜇

2𝑠𝑖𝑛𝜓]

−𝛾Ω2 1

2 [

4Ω+

𝜇

3Ω𝑠𝑖𝑛𝜓] +𝛾Ω2 1

2 [

𝑞

4Ω𝑐𝑜𝑠𝜓 +

𝑞 𝝁

3Ω𝑠𝑖𝑛𝜓𝑐𝑜𝑠𝜓] −𝛾Ω2 1

2 [

𝛽𝜇

3𝑐𝑜𝑠𝜓 +

𝛽𝜇2

2𝑠𝑖𝑛𝜓𝑐𝑜𝑠𝜓]

Now flapping equation is given as: + Ω2𝛽 = −2𝑞Ω 𝑠𝑖𝑛𝜓 + 𝑀𝑎/𝐼𝑏𝑙

+ 𝜸

𝟖Ω [𝟏 +

𝟒𝝁

𝟑𝒔𝒊𝒏𝝍]+Ω𝟐𝜷 [𝟏 +

𝜸 𝝁

𝟔𝒄𝒐𝒔𝝍 +

𝜸 𝝁𝟐

𝟖𝒔𝒊𝒏𝟐𝝍] = 𝜸Ω𝟐 𝟏

𝟖 𝜽(𝟏 + 𝝁𝟐) − 𝜸Ω𝟐 𝟏

𝟔(𝝀𝒄 + 𝝀𝒊)

+𝜸Ω𝟐 𝟏

𝟖 𝒔𝒊𝒏𝝍 (

𝟖

𝟑𝜽𝝁 − 𝟐𝝁 (𝝀𝒄 + 𝝀𝒊))

+𝜸Ω𝟏

𝟖𝒒 𝒄𝒐𝒔𝝍+𝜸Ω

𝟏

𝟏𝟐𝝁 𝒒 𝒔𝒊𝒏𝟐𝝍

−𝜸Ω𝟐 𝟏

𝟖 𝜽𝝁𝟐𝒄𝒐𝒔𝟐𝝍 − 𝟐𝒒Ω 𝒔𝒊𝒏𝝍 eq. 05

Question 8. Deduce disc tilt angles Assuming solution for the above eq. 05 of the form 𝛽 = 𝑎0 − 𝑎1 𝑐𝑜𝑠𝜓 − 𝑏1 𝑠𝑖𝑛𝜓

= 𝑎1Ω 𝑠𝑖𝑛𝜓 − 𝑏1Ω 𝑐𝑜𝑠𝜓 by using 𝜓 = Ω𝑡

= 𝑎1Ω2 𝑐𝑜𝑠𝜓 + 𝑏1Ω2 𝑠𝑖𝑛𝜓

HELICOPTER PERFORMANCE STABILITY AND CONTROL COURSE CODE: AE4314

Name: DEEPAK PAUL TIRKEY

ASSIGNMENT NUMBER 04 Student number: 4590929

Now substituting the values of 𝛽, , on the LHS of eq. 05 LHS of eq.05

≡ 𝑎1Ω2 𝑐𝑜𝑠𝜓 + 𝑏1Ω2 𝑠𝑖𝑛𝜓 +𝛾

8Ω (𝑎1Ω 𝑠𝑖𝑛𝜓 − 𝑏1Ω 𝑐𝑜𝑠𝜓 +

4𝜇

3𝑎1Ω 𝑠𝑖𝑛2𝜓 −

4𝜇

3𝑏1Ω𝑠𝑖𝑛𝜓𝑐𝑜𝑠𝜓)

+Ω2 (𝑎0 − 𝑎1 𝑐𝑜𝑠𝜓 − 𝑏1 𝑠𝑖𝑛𝜓 +𝛾 𝜇

6𝑎0 𝑐𝑜𝑠𝜓 −

𝛾 𝜇

6𝑎1 𝑐𝑜𝑠2𝜓 −

𝛾 𝜇

6𝑏1𝑠𝑖𝑛𝜓𝑐𝑜𝑠𝜓 +

𝛾 𝜇2

8𝑎0 𝑠𝑖𝑛2𝜓 −

𝜸 𝝁𝟐

𝟖𝒂𝟏𝒄𝒐𝒔 𝝍𝒔𝒊𝒏𝟐𝝍 −

𝜸 𝝁𝟐

𝟖𝒃𝟏𝒔𝒊𝒏𝝍 𝒔𝒊𝒏𝟐𝝍) eq. 06

Simplifying the last two terms of eq. 06 :

− 𝜸 𝝁𝟐

𝟖𝒂𝟏𝒄𝒐𝒔 𝝍𝒔𝒊𝒏𝟐𝝍 ≡ −

𝛾 𝜇2

4𝑎1𝑐𝑜𝑠 𝜓 𝑠𝑖𝑛𝜓 𝑐𝑜𝑠𝜓 ≡ −

𝛾 𝜇2

4𝑎1 𝑠𝑖𝑛𝜓 𝑐𝑜𝑠2𝜓

≡ − 𝛾 𝜇2

4𝑎1 (

1

4𝑠𝑖𝑛𝜓 +

1

4𝑠𝑖𝑛3𝜓 )

≡ − 𝛾 𝜇2

16𝑎1𝑠𝑖𝑛𝜓 −

𝛾 𝜇2

16𝑎1𝑠𝑖𝑛3𝜓

−𝜸 𝝁𝟐

𝟖𝒃𝟏𝒔𝒊𝒏𝝍 𝒔𝒊𝒏𝟐𝝍 ≡ −

𝛾 𝜇2

4𝑏1𝑠𝑖𝑛𝜓 𝑠𝑖𝑛𝜓 𝑐𝑜𝑠𝜓 ≡ −

𝛾 𝜇2

4𝑏1𝑠𝑖𝑛2𝜓 𝑐𝑜𝑠𝜓

≡ − 𝛾 𝜇2

4𝑏1 (

1

4𝑐𝑜𝑠𝜓 −

1

4𝑐𝑜𝑠3𝜓 )

≡ − 𝛾 𝜇2

16𝑏1𝑐𝑜𝑠𝜓 +

𝛾 𝜇2

16𝑏1𝑐𝑜𝑠3𝜓

Rewriting eq. 06

≡ 𝑎1Ω2 𝑐𝑜𝑠𝜓 + 𝑏1Ω2 𝑠𝑖𝑛𝜓 +𝛾

8Ω2𝑎1 𝑠𝑖𝑛𝜓 −

𝛾

8Ω2𝑏1 𝑐𝑜𝑠𝜓 +

𝛾

8Ω2 2𝜇

3𝑎1 (1 − 𝑐𝑜𝑠2𝜓) −

𝛾

8Ω2 2𝜇

3𝑏1Ω𝑠𝑖𝑛2𝜓

+Ω2𝑎0 − Ω2𝑎1 𝑐𝑜𝑠𝜓 − Ω2𝑏1 𝑠𝑖𝑛𝜓 + Ω2 𝛾 𝜇

6𝑎0 𝑐𝑜𝑠𝜓 − Ω2 𝛾 𝜇

12𝑎1 (1 + 𝑐𝑜𝑠2𝜓) − Ω2 𝛾 𝜇

12𝑏1𝑠𝑖𝑛2𝜓 +

Ω2 𝛾 𝜇2

8𝑎0 𝑠𝑖𝑛2𝜓 − Ω2

𝛾 𝜇2

16𝑎1𝑠𝑖𝑛𝜓 − Ω2 𝛾 𝜇2

16𝑎1𝑠𝑖𝑛3𝜓 − Ω2

𝛾 𝜇2

16𝑏1𝑐𝑜𝑠𝜓 + Ω2 𝛾 𝜇2

16𝑏1𝑐𝑜𝑠3𝜓

Rearranging the above expression in terms of free, sine, cosine terms etc..

≡ (𝛾

8Ω2 2𝜇

3𝑎1 + Ω2𝑎0 − Ω2 𝛾 𝜇

12𝑎1) + (𝑏1Ω2 +

𝛾

8Ω2𝑎1 − Ω2𝑏1 − Ω2

𝛾 𝜇2

16𝑎1) 𝑠𝑖𝑛𝜓

+ (𝑎1Ω2 −𝛾

8Ω2𝑏1 − Ω2𝑎1 + Ω2 𝛾 𝜇

6𝑎0 − Ω2

𝛾 𝜇2

16𝑏1) 𝑐𝑜𝑠𝜓 + ⋯ 𝑡𝑒𝑟𝑚𝑠 𝑐𝑜𝑛𝑡𝑎𝑖𝑛𝑖𝑛𝑔 ℎ𝑖𝑔ℎ𝑒𝑟 ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐𝑠

Comparing free terms with the RHS free term of eq.05

(𝛾

8Ω2 2𝜇

3𝑎1 + Ω2𝑎0 − Ω2 𝛾 𝜇

12𝑎1) = 𝜸Ω𝟐 𝟏

𝟖 𝜽(𝟏 + 𝝁𝟐) − 𝜸Ω𝟐 𝟏

𝟔(𝝀𝒄 + 𝝀𝒊)

Or 𝒂𝟎 = 𝜸𝟏

𝟖 [𝜽(𝟏 + 𝝁𝟐) −

𝟒

𝟑(𝝀𝒄 + 𝝀𝒊)]

Comparing sine terms with the RHS sine term of eq.05

(𝑏1Ω2 +𝛾

8Ω2𝑎1 − Ω2𝑏1 − Ω2

𝛾 𝜇2

16𝑎1) = 𝛾Ω2 1

8 (

8

3𝜃𝜇 − 2𝜇 (𝜆𝑐 + 𝜆𝑖)) − 2𝑞Ω

𝛾

8Ω2𝑎1 (1 −

𝜇2

2) = 𝛾Ω2 1

8 (

8

3𝜃𝜇 − 2𝜇 (𝜆𝑐 + 𝜆𝑖)) − 2𝑞Ω

Or 𝒂𝟏 = (

𝟖

𝟑𝜽𝝁− 𝟐𝝁 (𝝀𝒄+𝝀𝒊))−

𝟏𝟔𝒒

𝜸Ω

(𝟏− 𝝁𝟐

𝟐)

Comparing cosine terms with the RHS cosine term of eq.05

(𝑎1Ω2 −𝛾

8Ω2𝑏1 − Ω2𝑎1 + Ω2 𝛾 𝜇

6𝑎0 − Ω2

𝛾 𝜇2

16𝑏1) = 𝛾Ω

1

8𝑞

(1

8𝑏1 +

𝜇2

16𝑏1) = −

𝑞

Ω

1

8+

𝜇

6𝑎0

Or 𝒃𝟏 =−

𝒒

Ω +

𝟒 𝝁

𝟑𝒂𝟎

(𝟏+ 𝝁𝟐

𝟐)