Download - 03-2 - Directional Control
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FLIGHT DYNAMICS & STABILITY
Lecture 03-2: Directional Control
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Rudder
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Yawing Moment Coefficient
ππ+
π΅β
π = βππ£πΆπΏπ£ππ£ππ£
π = βππ£ππ£
ππ£ = πΆπΏπ£ππ£ππ£
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Rudder Control Effectiveness
πͺπ = πΆππΏππΏπ = βππ£ππ£ππΆπΏπ£ππΏππΏπ
πͺππΉπ = βππ£ππ£π πͺπ³ππ πΉπ
ππΆπΏπ£ππΏπ=ππΆπΏπ£ππΌπ£
ππΌπ£ππΏπ= πΆπΏπΌπ£π
πͺππΉπ = βππ£ππ£πΆπΏπΌπ£π π€ππππ ππ£ =
ππ£π
ππ£ππ€
πͺππΉπ = βπππ»π πͺπ³ππ πΉπ= βπππ»πΆπΏπΌπ‘
π Compare with:
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Flap Effectiveness Parameter
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Rudder Design Requirements
β’ Adverse Yaw
β’ Cross Wind Takeoff/Landing
β’ Asymmetric Power Condition
β’ Spin Recovery
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Adverse Yaw
The rudder must be able to overcome the adverse yaw so that a coordinated turn can be achieved. The critical condition for yaw occurs when the airplane is flying slow (high CL).
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Cross Wind Takeoff/Landing
The rudder must be powerful enough to permit the pilot to trim the airplane for the specified cross-winds. For transport airplanes, landing may be carried out for cross-winds up to 15.5 m/s or 51 ft/s.
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Asymmetric Power Condition
The rudder must be powerful enough to overcome the yawing moment produced when one engine fails at low flight speed
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Example Obtain the minimum control speed in the event of an engine failure for the following airplane: ππ€ = 65 π2, ππ£ = 6.5 π2
, ππ£ = 10.5 π, π΅π»π = 880 ππ (πππ ππππππ),
πππππππππ ππππππππππ¦ = ππ = 75%, π¦π = 4.2 π, ππΆπΏπ£ ππΏπ = 0.02 πππ πππ,
πΏπ πππ₯ = 250.
πππ€πππ ππππππ‘ ππ’π π‘π ππππππ = π β π¦π = ππ β π΅π»π β π¦π π
= 0.75 β 880 β 1000 β4.2
π=2.772 β 106
π
πππ€πππ ππππππ‘ ππ’π π‘π ππ’ππππ
= βππ£ππ£ = βππ£πΆππ£ππ£ππ£ = βππ£ππΆπΏπ£ππΏππΏπππ£ππ€ππ£
= β10.5 β 0.02 β 25 β 1.0 β1
2β 1.225 β π2 β 6.5
= β20.9 β π2
Solution:
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πππ =πππ€πππ ππππππ‘ ππ’π π‘π ππππππ + πππ€πππ ππππππ‘ ππ’π π‘π ππ’ππππ = 0
Under equilibrium condition:
2.772 β 106
πβ 20.9 β π2 = 0
π =2.772 β 106
20.9
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= 0.51 β 102 = 51π π = 183.6 ππ ππ