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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 πΎ =ππππ 4
πΌπ½ = 5.07
where Ο = 1.225 ππ
π3 [sea level condition] ; a = lift curve slope = 6.113 /rad ; c = chord = 0.27 m ;
R = radius = 4.91 m ; πΌπ½= 231.7 ππ
π2 [ data source : Helicopter Flight Dynamics βPadfield]
Induced Velocity in hover ππ = βπ
2πππ 2 = 10.8 m/sec ; where W = 2200 x 9.81 = 21582 N
Induced inflow ratio ππ = ππ
Ξ©π =
10.8
218 = 0.05
Flapping solution in hover Ξ² (t) = π·πππ(t) + π·ππππ(t)
Particular solution π½ππππ‘(t) = πΎ
8( ΞΈ -
4
3ππ ) =
5.07
8(8 -
4
30.05) = 5.03 degree; where ΞΈ = 8 degree(given)
[π½ππππ‘(t) = 5.03 degree is constant coning angle ]
Homogenous solution π½βππ(t) = π½0πβπΎΞ©π‘
16 [cos (Ξ©β1 β (πΎ
16)
2
. π‘) +πΎ
16
β1β(πΎ
16)
2 sin (Ξ©β1 β (
πΎ
16)
2
. π‘)]
MS Excel was used to calculate and plot Ξ² (t) = π½βππ(t) + π½ππππ‘(t) ,
were azimuth angle Ο =Ξ© t , π½0 = π½ππππ‘ π€ππ πβππ ππ
Fig. 01 Flapping blade angle Ξ² (Ο) = Ξ²hom(Ο) + Ξ²part(Ο) 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
πΆπ = ΞΈ - ππ+π·
Ξ©π
Now to calculate from the relation π½ (π‘ ) = π½βππ (t) + π½ππππ‘
(t) = π½βππ (t) [as π½ππππ‘(t) is a constant term]
Writing π½ (π‘ ) = π½0[A+B]
Were A = (βπΎΞ©
16) π
βπΎΞ©π‘
16 [cos (Ξ©β1 β (πΎ
16)
2
. π‘) +πΎ
16
β1β(πΎ
16)
2sin (Ξ©β1 β (
πΎ
16)
2
. π‘)]
And B = πβπΎΞ©π‘
16 [β (Ξ©β1 β (πΎ
16)
2
)sin (Ξ©β1 β (πΎ
16)
2
. π‘) +πΎΞ©
16cos (Ξ©β1 β (
πΎ
16)
2
. π‘)]
Simplifying [A+B] , cosine terms from both A and B cancel each other
[A+B] = (β)πβπΎΞ©π‘
16 sin (Ξ©β1 β (πΎ
16)
2
. π‘) [(
πΎ
16)
2Ξ©
β1β(πΎ
16)
2+ Ξ©β1 β (
πΎ
16)
2
]
Simplifying the [ ] term further
[A+B] = (β)πβπΎΞ©π‘
16 sin (Ξ©β1 β (πΎ
16)
2
. π‘) [Ξ©
β1β(πΎ
16)
2]
or π (π ) = (β)ππΞ© π
βπΞ©πππ
βπβ(π
ππ)
ππ¬π’π§ (Ξ©βπ β (
π
ππ)
π. π)
using the Ξ² (t ) value , πΌπ 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 = Ξ²part(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 Ξ© sinΞ² + q sinΟ cosΞ² β p cosΟ cosΞ²
j q cosΟ + p sinΟ - k Ξ© cosΞ² + p cosΟ sinΞ² β q sinΟ sinΞ²
Question 5. Flapping equation when the helicopter is having body rates p and q and is fixed to the ground Kinetic Energy:
T = 1
2πΌ [(π πππ π + π π πππ β )
2+ (Ξ© πππ π½ + π πππ π π πππ½ β π π πππ π πππ½)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
2πΌ[π΄ + π΅]
Lagrange equation of motion:
π
ππ‘(
ππ
π) β
ππ
ππ½+
ππ
ππ½= ππ½ were ππ½ is the aerodynamic moment about the flapping hinge ππ(π‘)
Evaluating the first term Lagrange EoM
Now ππ
π =
1
2πΌ [2-2(π πππ Ξ©π‘ + π π ππΞ©π‘)] and
π
ππ‘(
ππ
π) = πΌ[ + πΞ© π πππ β πΞ© πππ π] with Ο = Ξ©t
Now expand the B term of eq. 01 B = [(Ξ© πππ π½)2 + (π πππ π β π π πππ )2π ππ2π½ + 2Ξ© πππ π½π πππ½(π πππ π β π π πππ )] = [(Ξ© πππ π½)2 + (π πππ π β π π πππ )2π ππ2π½ + Ξ© π ππ2π½(π πππ π β π π πππ )] = [(Ξ© πππ π½)2 + 0 + Ξ© π ππ2π½(π πππ π β π π πππ )] Middle term became 0 as π½ is small and neglecting π2, π2, π. π terms Now evaluating the second term of Lagrange EoM
ππ
ππ½=
1
2πΌ[β2Ξ©2 πππ π½π πππ½ + 2Ξ©πππ 2π½(π πππ π β π π πππ ) ]
= πΌ[βΞ©2π½ + Ξ©(π πππ π β π π πππ ) ] assuming π½ 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:
πΌ + πΌπΞ© π πππ β πΌπΞ© πππ π + πΌΞ©2π½ β πΌΞ©π πππ π + πΌΞ©π π πππ = ππ(π‘) Retaining the π½ terms on LHS and dividing throughout by πΌ
+ Ξ©ππ· = βπΞ©π ππππ + πΞ©π ππππ + π΄π(π)
π° eq. 02
Question 6. Disc tilt angles a0 , a1 and b1 Angle of attack of a blade element
πΌ = π βππ+π½ πβπ π πππ πβπππ πππ
Ξ©π
Evaluate aerodynamic moment about flapping hinge:
π΄π
π°=
1
πΌβ« π (π β
ππ+π½ πβπ π πππ πβπππ πππ
Ξ©π)
1
2π(Ξ©π)2ππ ππ
π
0
=πππ
πΌ.
Ξ©2
2β« (π0 β
ππ
Ξ©π π₯β
π½
Ξ©+
π πππ π
Ξ©+
ππ πππ
Ξ©) π₯3π 4 ππ₯
1
0 taking π₯ =
π
π ; π = π₯π ; ππ = π ππ₯
=ππππ 4
πΌ.
Ξ©2
2β« (π0π₯3 β πππ₯
2 βπ½
Ξ©π₯3 +
π πππ π
Ξ©π₯3 +
ππ πππ
Ξ©π₯3) ππ₯
1
0
= πΎ.Ξ©2
2[
π0
4β
ππ
3β
π½
4Ξ©+
π πππ π
4Ξ©+
ππ πππ
4Ξ©]
π΄π
π°=
πΎΞ©2
8(π0 β
4ππ
3) β
πΎπ½ Ξ©
8+
πΎ πΞ© πππ π
8+
πΎ πΞ© π πππ
8 eq. 03
Now rewriting eq. 02 using eq. 03
+ Ξ©2π½ = β2Ξ©π π πππ + 2Ξ©π πππ π +πΎΞ©2
8(π0 β
4ππ
3) β
πΎπ½ Ξ©
8+
πΎ πΞ© πππ π
8+
πΎ πΞ© π πππ
8
+πΎπ½ Ξ©
8+ Ξ©2π½ =
πΎΞ©2
8(π0 β
4ππ
3) +
πΎ πΞ© πππ π
8+
πΎ πΞ© π πππ
8 β 2Ξ©π π πππ + 2Ξ©π πππ π
+πΈΞ©
π + Ξ©ππ· =
πΈΞ©π
π(π½π β
πππ
π) + Ξ© (
πΈ π
π+ ππ) πππΞ©π + Ξ© (
πΈ π
π β ππ ) πππΞ©π eq. 04 (using π = Ξ©π‘)
Assuming solution for the above eq. 04 of the form π½ = π0 β π1 πππ Ξ©π‘ β π1 π ππΞ©π‘
= π1Ξ© π ππΞ©π‘ β π1Ξ© πππ Ξ©π‘
= π1Ξ©2 πππ Ξ©π‘ + π1Ξ©2 π ππΞ©π‘
Now substituting the values of π½, , on the LHS of eq. 04 LHS of eq.04
β‘ π1Ξ©2 πππ Ξ©π‘ + π1Ξ©2 π ππΞ©π‘ +πΈ
ππ1Ξ©2 π ππΞ©π‘ β
πΈ
ππ1Ξ©2 πππ Ξ©π‘ + Ξ©2π0 β π1Ξ©2 πππ Ξ©π‘ β π1Ξ©2 π ππΞ©π‘
β‘ Ξ©2π0 + (π1Ξ©2 βπΈ
ππ1Ξ©2 β π1Ξ©2) πππ Ξ©π‘ + (π1Ξ©2+
πΈ
ππ1Ξ©2 β π1Ξ©2) π ππΞ©π‘
β‘ Ξ©2π0 βπΈ
ππ1Ξ©2 πππ Ξ©π‘ +
πΈ
ππ1Ξ©2 π ππΞ©π‘
Equating the above expression with the RHS of eq. 04 and comparing constant, sine and cosine terms Constant term
Ξ©2π0 =πΎΞ©2
8(π0 β
4ππ
3) or ππ =
πΈ
π(π½π β
πππ
π)
HELICOPTER PERFORMANCE STABILITY AND CONTROL COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Sine term
πΎ
8π1Ξ©2 = Ξ© (
πΎ π
8 β 2π ) or ππ = (
π
Ξ© β
πππ
πΈΞ© )
Cosine term
βπΎ
8π1Ξ©2 = Ξ© (
πΎ π
8+ 2π) or ππ = (
β π
Ξ©β
πππ
πΈΞ©)
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:
ππ = β« πΆππΌ πΌ1
2π(Ξ©π + ππππ πΌπ π πππ)2π
0π ππ. π Where πΌ = π β
π π πππΌπ+ππ+πβππ πππ π+π½π πππ πΌπ πππ π
Ξ©π+π πππ πΌπ π πππ
Performing integration with the π½ part
ππ1 = ππΆππΌ π1
2 β« π(Ξ©π + ππππ πΌπ π πππ)2π
0 ππ. π
= ππΆππΌ π1
2 π β« (Ξ©2π3 + π2π cos πΌπ
2 π πππ2 + 2Ξ©π π2πππ πΌπ π πππ)π
0 ππ
= ππΆππΌ π1
2 π [
Ξ©2π 4
4+
Ξ©2π 4
2 (
ππππ πΌπ
Ξ©π )
2
π πππ2 +2 Ξ©2π 4
3(
ππππ πΌπ
Ξ©π ) π πππ] Where
ππππ πΌπ
Ξ©π = π
= ππͺππΆ ππΉπΞ©π π
π π½ [
π
π+
π
πππ πππππ +
π
ππ ππππ]
Performing integration with π½ ππππΆπ+ππ+πβππ ππππ+π·π½ ππππΆπ ππππ
Ξ©π+π½ ππππΆπ ππππ part
ππ2 = ππΆππΌ π1
2 β« (π π πππΌπ + ππ + π β ππ πππ π + π½π πππ πΌπ πππ π)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π
Now integrating each term:
ππ3 = ππΆππΌ π1
2 β« (π π πππΌπ)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π = ππͺππΆ ππΉπΞ©π π
π [
ππ
π+
ππ π
πππππ]
ππ4 = ππΆππΌ π1
2 β« (ππ)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π = ππͺππΆ ππΉπΞ©π π
π [
ππ
π+
ππ π
πππππ]
ππ5 = ππΆππΌ π1
2 β« (π½)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π= ππͺππΆ ππΉπΞ©π π
π [
πΞ©+
π
πΞ©ππππ]
ππ6 = ππΆππΌ π1
2 β« (βππ πππ π)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π= ππͺππΆ ππΉπΞ©π π
π [
βπ
πΞ©πππ π +
βπ π
πΞ©πππππππ π]
ππ7 = ππΆππΌ π1
2 β« (π½π πππ πΌπ πππ π)(Ξ©π + ππππ πΌπ π πππ)
π
0ππ. π= ππͺππΆ ππΉπΞ©π π
π [
π·π
ππππ π +
π·ππ
ππππππππ π]
Now ππ/πΌππ=ππ1- (ππ3 + ππ4 + ππ5 + ππ6 + ππ7) /πΌππ
= πΎΞ©2 1
2 π [
1
4+
1
2π2 π πππ2 +
2
3π π πππ] βπΎΞ©2 1
2 [
ππ
3+
ππ π
2π πππ] β πΎΞ©2 1
2 [
ππ
3+
ππ π
2π πππ]
βπΎΞ©2 1
2 [
4Ξ©+
π
3Ξ©π πππ] +πΎΞ©2 1
2 [
π
4Ξ©πππ π +
π π
3Ξ©π ππππππ π] βπΎΞ©2 1
2 [
π½π
3πππ π +
π½π2
2π ππππππ π]
Now flapping equation is given as: + Ξ©2π½ = β2πΞ© π πππ + ππ/πΌππ
+ πΈ
πΞ© [π +
ππ
πππππ]+Ξ©ππ· [π +
πΈ π
πππππ +
πΈ ππ
ππππππ] = πΈΞ©π π
π π½(π + ππ) β πΈΞ©π π
π(ππ + ππ)
+πΈΞ©π π
π ππππ (
π
ππ½π β ππ (ππ + ππ))
+πΈΞ©π
ππ ππππ+πΈΞ©
π
πππ π πππππ
βπΈΞ©π π
π π½πππππππ β ππΞ© ππππ eq. 05
Question 8. Deduce disc tilt angles Assuming solution for the above eq. 05 of the form π½ = π0 β π1 πππ π β π1 π πππ
= π1Ξ© π πππ β π1Ξ© πππ π by using π = Ξ©π‘
= π1Ξ©2 πππ π + π1Ξ©2 π πππ
HELICOPTER PERFORMANCE STABILITY AND CONTROL COURSE CODE: AE4314
Name: DEEPAK PAUL TIRKEY
ASSIGNMENT NUMBER 04 Student number: 4590929
Now substituting the values of π½, , on the LHS of eq. 05 LHS of eq.05
β‘ π1Ξ©2 πππ π + π1Ξ©2 π πππ +πΎ
8Ξ© (π1Ξ© π πππ β π1Ξ© πππ π +
4π
3π1Ξ© π ππ2π β
4π
3π1Ξ©π ππππππ π)
+Ξ©2 (π0 β π1 πππ π β π1 π πππ +πΎ π
6π0 πππ π β
πΎ π
6π1 πππ 2π β
πΎ π
6π1π ππππππ π +
πΎ π2
8π0 π ππ2π β
πΈ ππ
ππππππ ππππππ β
πΈ ππ
πππππππ πππππ) eq. 06
Simplifying the last two terms of eq. 06 :
β πΈ ππ
ππππππ ππππππ β‘ β
πΎ π2
4π1πππ π π πππ πππ π β‘ β
πΎ π2
4π1 π πππ πππ 2π
β‘ β πΎ π2
4π1 (
1
4π πππ +
1
4π ππ3π )
β‘ β πΎ π2
16π1π πππ β
πΎ π2
16π1π ππ3π
βπΈ ππ
πππππππ πππππ β‘ β
πΎ π2
4π1π πππ π πππ πππ π β‘ β
πΎ π2
4π1π ππ2π πππ π
β‘ β πΎ π2
4π1 (
1
4πππ π β
1
4πππ 3π )
β‘ β πΎ π2
16π1πππ π +
πΎ π2
16π1πππ 3π
Rewriting eq. 06
β‘ π1Ξ©2 πππ π + π1Ξ©2 π πππ +πΎ
8Ξ©2π1 π πππ β
πΎ
8Ξ©2π1 πππ π +
πΎ
8Ξ©2 2π
3π1 (1 β πππ 2π) β
πΎ
8Ξ©2 2π
3π1Ξ©π ππ2π
+Ξ©2π0 β Ξ©2π1 πππ π β Ξ©2π1 π πππ + Ξ©2 πΎ π
6π0 πππ π β Ξ©2 πΎ π
12π1 (1 + πππ 2π) β Ξ©2 πΎ π
12π1π ππ2π +
Ξ©2 πΎ π2
8π0 π ππ2π β Ξ©2
πΎ π2
16π1π πππ β Ξ©2 πΎ π2
16π1π ππ3π β Ξ©2
πΎ π2
16π1πππ π + Ξ©2 πΎ π2
16π1πππ 3π
Rearranging the above expression in terms of free, sine, cosine terms etc..
β‘ (πΎ
8Ξ©2 2π
3π1 + Ξ©2π0 β Ξ©2 πΎ π
12π1) + (π1Ξ©2 +
πΎ
8Ξ©2π1 β Ξ©2π1 β Ξ©2
πΎ π2
16π1) π πππ
+ (π1Ξ©2 βπΎ
8Ξ©2π1 β Ξ©2π1 + Ξ©2 πΎ π
6π0 β Ξ©2
πΎ π2
16π1) πππ π + β― π‘ππππ ππππ‘ππππππ βππβππ βππππππππ
Comparing free terms with the RHS free term of eq.05
(πΎ
8Ξ©2 2π
3π1 + Ξ©2π0 β Ξ©2 πΎ π
12π1) = πΈΞ©π π
π π½(π + ππ) β πΈΞ©π π
π(ππ + ππ)
Or ππ = πΈπ
π [π½(π + ππ) β
π
π(ππ + ππ)]
Comparing sine terms with the RHS sine term of eq.05
(π1Ξ©2 +πΎ
8Ξ©2π1 β Ξ©2π1 β Ξ©2
πΎ π2
16π1) = πΎΞ©2 1
8 (
8
3ππ β 2π (ππ + ππ)) β 2πΞ©
πΎ
8Ξ©2π1 (1 β
π2
2) = πΎΞ©2 1
8 (
8
3ππ β 2π (ππ + ππ)) β 2πΞ©
Or ππ = (
π
ππ½πβ ππ (ππ+ππ))β
πππ
πΈΞ©
(πβ ππ
π)
Comparing cosine terms with the RHS cosine term of eq.05
(π1Ξ©2 βπΎ
8Ξ©2π1 β Ξ©2π1 + Ξ©2 πΎ π
6π0 β Ξ©2
πΎ π2
16π1) = πΎΞ©
1
8π
(1
8π1 +
π2
16π1) = β
π
Ξ©
1
8+
π
6π0
Or ππ =β
π
Ξ© +
π π
πππ
(π+ ππ
π)