03-2 - directional control
DESCRIPTION
STABILITYTRANSCRIPT
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FLIGHT DYNAMICS & STABILITY
Lecture 03-2: Directional Control
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Rudder
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Yawing Moment Coefficient
𝒀𝒗+
𝑵−
𝑁 = −𝑙𝑣𝐶𝐿𝑣𝑄𝑣𝑆𝑣
𝑁 = −𝑙𝑣𝑌𝑣
𝑌𝑣 = 𝐶𝐿𝑣𝑄𝑣𝑆𝑣
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Rudder Control Effectiveness
𝑪𝒏 = 𝐶𝑛𝛿𝑟𝛿𝑟 = −𝜂𝑣𝑉𝑣𝑑𝐶𝐿𝑣𝑑𝛿𝑟𝛿𝑟
𝑪𝒏𝜹𝒓 = −𝜂𝑣𝑉𝑣𝒅𝑪𝑳𝒗𝒅𝜹𝒓
𝑑𝐶𝐿𝑣𝑑𝛿𝑟=𝑑𝐶𝐿𝑣𝑑𝛼𝑣
𝑑𝛼𝑣𝑑𝛿𝑟= 𝐶𝐿𝛼𝑣𝜏
𝑪𝒏𝜹𝒓 = −𝜂𝑣𝑉𝑣𝐶𝐿𝛼𝑣𝜏 𝑤𝑒𝑟𝑒 𝑉𝑣 =
𝑙𝑣𝑏
𝑆𝑣𝑆𝑤
𝑪𝒎𝜹𝒆 = −𝜂𝑉𝐻𝒅𝑪𝑳𝒕𝒅𝜹𝒆= −𝜂𝑉𝐻𝐶𝐿𝛼𝑡
𝜏 Compare with:
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Flap Effectiveness Parameter
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Rudder Design Requirements
• Adverse Yaw
• Cross Wind Takeoff/Landing
• Asymmetric Power Condition
• 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: 𝑆𝑤 = 65 𝑚2, 𝑆𝑣 = 6.5 𝑚2
, 𝑙𝑣 = 10.5 𝑚, 𝐵𝐻𝑃 = 880 𝑘𝑊 (𝑝𝑒𝑟 𝑒𝑛𝑔𝑖𝑛𝑒),
𝑝𝑟𝑜𝑝𝑒𝑙𝑙𝑒𝑟 𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 = 𝜂𝑝 = 75%, 𝑦𝑝 = 4.2 𝑚, 𝑑𝐶𝐿𝑣 𝑑𝛿𝑟 = 0.02 𝑝𝑒𝑟 𝑑𝑒𝑔,
𝛿𝑟 𝑚𝑎𝑥 = 250.
𝑌𝑎𝑤𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑒𝑛𝑔𝑖𝑛𝑒 = 𝑇 ∗ 𝑦𝑝 = 𝜂𝑝 ∗ 𝐵𝐻𝑃 ∗ 𝑦𝑝 𝑉
= 0.75 ∗ 880 ∗ 1000 ∗4.2
𝑉=2.772 ∗ 106
𝑉
𝑌𝑎𝑤𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑟𝑢𝑑𝑑𝑒𝑟
= −𝑙𝑣𝑌𝑣 = −𝑙𝑣𝐶𝑙𝑣𝑄𝑣𝑆𝑣 = −𝑙𝑣𝑑𝐶𝐿𝑣𝑑𝛿𝑟𝛿𝑟𝜂𝑣𝑄𝑤𝑆𝑣
= −10.5 ∗ 0.02 ∗ 25 ∗ 1.0 ∗1
2∗ 1.225 ∗ 𝑉2 ∗ 6.5
= −20.9 ∗ 𝑉2
Solution:
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𝑁𝑐𝑔 =𝑌𝑎𝑤𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑒𝑛𝑔𝑖𝑛𝑒 + 𝑌𝑎𝑤𝑖𝑛𝑔 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑟𝑢𝑑𝑑𝑒𝑟 = 0
Under equilibrium condition:
2.772 ∗ 106
𝑉− 20.9 ∗ 𝑉2 = 0
𝑉 =2.772 ∗ 106
20.9
3
= 0.51 ∗ 102 = 51𝑚 𝑠 = 183.6 𝑘𝑚 𝑟