02-7 - Longitudinal Control
32
Flight Dynamics AAE 6710 Lesson 02-7 Longitudinal Control
description
FLIGHT CONTROLS
Transcript of 02-7 - Longitudinal Control
Flight Dynamics
AAE 6710
Lesson 02-7
Longitudinal Control
Factors Affecting Design
• Control Effectiveness: Measure of how effective the control is in producing the desired control moment
• Hinge Moments: The aerodynamic moments that must be overcome to rotate the control surface
• Aerodynamic & Mass Balancing: Techniques to keep the control forces within an acceptable range
Effect of Elevator Deflection
𝑪𝑳𝒕 = 𝑪𝑳𝜶𝒕𝛼𝑡 = 𝑪𝑳𝜶𝒕 𝛼𝑤 − 𝑖𝑤 − 휀 + 𝑖𝑡
𝑪𝒎𝒄𝒈𝒕= 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 𝜖0 + 𝑖𝑤 − 𝑖𝑡 − 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 1 −
𝑑휀
𝑑𝛼𝜶𝒘
𝛿𝑒 = 0
For arbitrary 𝛿𝑒
𝑪𝑳𝒕 = 𝑪𝑳𝜶𝒕𝛼𝑡 = 𝑪𝑳𝜶𝒕 𝛼𝑤 − 𝑖𝑤 − 휀 + 𝑖𝑡 + 𝜏 𝛿𝑒
𝑪𝒎𝒄𝒈𝒕= 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 𝜖0 + 𝑖𝑤 − 𝑖𝑡 − 𝜏 𝛿𝑒 − 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 1 −
𝑑휀
𝑑𝛼𝜶𝒘
Increment in Lift When the elevator is deflected, it changes the lift of the airplane.
𝑳 = 𝐿𝑤 + 𝐿𝑡
𝑪𝑳1
2𝜌𝑉2𝑆 = 𝐶𝐿𝑤
1
2𝜌𝑉2𝑆 + 𝐶𝐿𝑡
1
2𝜌𝑉𝑡
2𝑆𝑡
𝑪𝑳 = 𝐶𝐿𝑤 + 𝜂𝑆𝑡𝑆𝐶𝐿𝑡
𝐶𝐿𝑡 = 𝐶𝐿𝛼𝑡𝛼𝑡 = 𝐶𝐿𝛼𝑡
𝛼𝑤 − 𝑖𝑤 − 휀 + 𝑖𝑡 + 𝜏 𝛿𝑒
= 𝐶𝐿𝛼𝑡𝛼𝑤 − 𝑖𝑤 − 휀 + 𝑖𝑡 + 𝐶𝐿𝛼𝑡
𝜏 𝛿𝑒
∆𝑪𝑳 = ∆𝑪𝑳𝒕 = 𝜂𝑆𝑡𝑆𝐶𝐿𝛼𝑡
𝜏 𝛿𝑒 𝒅𝑪𝑳𝒕𝒅𝜹𝒆
= 𝐶𝐿𝛼𝑡𝜏 𝝉 =
𝑑𝛼𝑡𝑑𝛿𝑒
Elevator Effectiveness
𝒅𝑪𝑳𝒕𝒅𝜹𝒆
= 𝑪𝑳𝜶𝒕𝝉
• 𝒅𝑪𝑳𝒕/𝒅𝜹𝒆 is the elevator effectiveness
• The elevator effectiveness is proportional to the size of the flap being used as an elevator and can be estimated from the equation: where 𝝉 is the change in tail angle of attack per unit deflection of the elevator and is called the elevator effectiveness parameter
𝝉 =𝑑𝛼𝑡𝑑𝛿𝑒
Elevator Effectiveness Parameter
Elevator Effectiveness is proportional to the size of flap being used as an elevator
Increment in Pitching Moment
If ∆𝐶𝑚 = 𝐶𝑚𝛿𝑒𝛿𝑒 𝑤ℎ𝑒𝑟𝑒 𝐶𝑚𝛿𝑒
=𝑑𝐶𝑚𝑑𝛿𝑒
∆𝑪𝒎 = ∆𝑪𝒎𝒕= −𝜂𝑉𝐻𝐶𝐿𝛼𝑡
𝜏 𝛿𝑒 = −𝑉𝐻𝜂𝑑𝐶𝐿𝑡𝑑𝛿𝑒
𝛿𝑒
𝑪𝒎𝒄𝒈𝒕= 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 𝜖0 + 𝑖𝑤 − 𝑖𝑡 − 𝜏 𝛿𝑒 − 𝜼𝑽𝑯𝑪𝑳𝜶𝒕 1 −
𝑑휀
𝑑𝛼𝜶𝒘
𝑪𝒎𝜹𝒆= −𝑉𝐻𝜂
𝒅𝑪𝑳𝒕𝒅𝜹𝒆
= −𝑉𝐻𝜂𝐶𝐿𝛼𝑡𝜏
The derivative 𝑪𝒎𝜹𝒆 is called the elevator control power
Influence of Elevator on 𝐶𝑚 vs 𝛼 curve
𝑪𝒎 = 𝐶𝑚0+ 𝐶𝑚𝛼
𝛼 + 𝐶𝑚𝛿𝑒𝛿𝑒
Elevator Angle to Trim
An airplane is said to be trimmed if the forces and moments acting on the plane are in equilibrium
𝑪𝒎𝒕𝒓𝒊𝒎= 𝟎 = 𝐶𝑚0
+ 𝐶𝑚𝛼𝛼𝑡𝑟𝑖𝑚 + 𝐶𝑚𝛿𝑒
𝛿𝑒𝑡𝑟𝑖𝑚
𝜹𝒆𝒕𝒓𝒊𝒎 = −𝐶𝑚0
+ 𝐶𝑚𝛼𝛼𝑡𝑟𝑖𝑚
𝐶𝑚𝛿𝑒
𝑪𝑳𝒕𝒓𝒊𝒎 = 𝐶𝐿𝛼𝛼𝑡𝑟𝑖𝑚 + 𝐶𝐿𝛿𝑒𝛿𝑒𝑡𝑟𝑖𝑚
𝜶𝒕𝒓𝒊𝒎 =𝐶𝐿𝑡𝑟𝑖𝑚 − 𝐶𝐿𝛿𝑒𝛿𝑒𝑡𝑟𝑖𝑚
𝐶𝐿𝛼
𝐶𝐿𝑡𝑟𝑖𝑚 =𝑊
1
2𝜌𝑉2𝑆
Elevator Angle to Trim
If we substitute the value of 𝜶𝒕𝒓𝒊𝒎 in the 𝜹𝒆𝒕𝒓𝒊𝒎 equation, we get:
𝜹𝒆𝒕𝒓𝒊𝒎 = −𝐶𝑚0
𝐶𝐿𝛼 + 𝐶𝑚𝛼𝐶𝐿𝑡𝑟𝑖𝑚
𝐶𝑚𝛿𝑒𝐶𝐿𝛼 − 𝐶𝑚𝛼
𝐶𝐿𝛿𝑒
𝜶𝒕𝒓𝒊𝒎 = 𝐶𝑚0
𝐶𝐿𝛿𝑒 + 𝐶𝑚𝛿𝑒𝐶𝐿𝑡𝑟𝑖𝑚
𝐶𝑚𝛿𝑒𝐶𝐿𝛼 − 𝐶𝑚𝛼
𝐶𝐿𝛿𝑒
Elevator Angle to Trim
𝑪𝒎𝒕𝒓𝒊𝒎= 𝟎 = 𝐶𝑚0
+ 𝐶𝑚𝛼𝛼𝑡𝑟𝑖𝑚 + 𝐶𝑚𝛿𝑒
𝛿𝑒𝑡𝑟𝑖𝑚
𝐶𝐿𝑡𝑟𝑖𝑚 =𝑊
1
2𝜌𝑉2𝑆
∴ 𝜹𝒆𝒕𝒓𝒊𝒎≈ 𝛿𝑒0 −𝐶𝐿𝑡𝑟𝑖𝑚𝐶𝑚𝛿𝑒
𝒅𝑪𝒎
𝒅𝑪𝑳 𝑓𝑖𝑥 where 𝛿𝑒0 = 𝛿𝑒 𝑎𝑡 𝑧𝑒𝑟𝑜 𝐶𝐿 = −
𝐶𝑚0
𝐶𝑚𝛿𝑒
= 𝐶𝑚0+
𝑑𝐶𝑚𝑑𝐶𝐿 𝑓𝑖𝑥
𝐶𝐿𝛼𝛼𝑡𝑟𝑖𝑚 + 𝐶𝑚𝛿𝑒𝛿𝑒𝑡𝑟𝑖𝑚
0 ≈ 𝐶𝑚0+
𝑑𝐶𝑚𝑑𝐶𝐿 𝑓𝑖𝑥
𝐶𝐿𝑡𝑟𝑖𝑚 + 𝐶𝑚𝛿𝑒𝛿𝑒𝑡𝑟𝑖𝑚
Elevator Angle for Trim
Remarks
• For a given c.g. location 𝛿𝑡𝑟𝑖𝑚 is a linear function of 𝐶𝐿𝑡𝑟𝑖𝑚
• When an airplane is trimmed at a chosen 𝐶𝐿 by setting the elevator at corresponding 𝛿𝑡𝑟𝑖𝑚, to fly at a lower speed which implies higher 𝐶𝐿 the pilot would need to apply more negative elevator deflection or the incremental lift on the tail would be negative or Δ𝐿𝑇 would be negative
Remarks
• In trim (balanced) condition, the setting of the elevator determines the airplane's trim speed - a given elevator position has only one lift coefficient (and one speed at a given altitude) at which the aircraft will maintain a constant (un-accelerated) condition
Remarks
• Military airplanes which are highly maneuverable, sometimes have the following features:
– An all movable tail in which the entire horizontal tail is rotated to achieve higher Δ𝑀𝑐𝑔
– Relaxed static stability wherein may have a small positive value 𝐶𝑚𝛼
Example 1
An airplane has elevator power 𝐶𝑚𝛿𝑒 of -0.010 per degree.
The c.g. is placed such that the static margin is 10% of
m.a.c. Further, the tail setting , 𝑖𝑡 ,is such that the airplane is
in trim, with zero elevator deflection, at 𝐶𝐿 = 0.5. Plot the
curves of 𝐶𝑚𝑐𝑔 vs 𝐶𝐿 for constant elevator angles of
𝛿𝑒 = −200, −100, 00 , 100, 200. Cross plot these curves to
obtain the curve corresponding to 𝛿𝑡𝑟𝑖𝑚 vs. 𝐶𝐿𝑡𝑟𝑖𝑚. Note
𝐶𝐿𝑚𝑎𝑥= 1.5.
𝐶𝑚𝑐𝑔= 𝐶𝑚0
+ 𝐶𝑚𝛼𝛼 + 𝐶𝑚𝛿𝑒
𝛿𝑒 = 𝐶𝑚0+ 𝑑𝐶𝑚𝑐𝑔
𝑑𝐶𝐿 𝐶𝐿 + 𝐶𝑚𝛿𝑒𝛿𝑒
Since the airplane is in equilibrium with zero elevator deflection at 𝐶𝐿 = 0.5
0 = 𝐶𝑚0− 0.1 ∗ 0.5 + 0.0 𝐶𝑚0
= 0.05
𝐶𝑚𝑐𝑔= 0.05 − 0.1𝐶𝐿 − 0.01𝛿𝑒
𝛿𝑒 = −200, 𝐶𝑚𝑐𝑔= 0.05 − 0.1𝐶𝐿 − 0.01 −200
= +0.25 − 0.1𝐶𝐿 𝛿𝑒 = −100, 𝐶𝑚𝑐𝑔
= +0.15 − 0.1𝐶𝐿
𝛿𝑒 = 00, 𝐶𝑚𝑐𝑔= +0.05 − 0.1𝐶𝐿
𝛿𝑒 = +100, 𝐶𝑚𝑐𝑔= −0.05 − 0.1𝐶𝐿
𝛿𝑒 = +100, 𝐶𝑚𝑐𝑔= −0.15 − 0.1𝐶𝐿
Static Margin= 0.1 𝑑𝐶𝑚𝑐𝑔𝑑𝐶𝐿 = −0.1
Solution
𝐶𝑚𝛿𝑒= −0.01
𝑪𝒎𝒄𝒈 𝒗𝒔 𝑪𝑳
𝛿𝑡𝑟𝑖𝑚 vs 𝐶𝐿
𝛿𝑡𝑟𝑖𝑚 = 𝛿𝑒0 − 𝑑𝐶𝑚𝑐𝑔𝑑𝐶𝐿 𝐶𝐿𝑡𝑟𝑖𝑚 𝐶𝑚𝛿𝑒
𝛿𝑡𝑟𝑖𝑚 = 50 − −0.1 𝐶𝐿𝑡𝑟𝑖𝑚 −0.01 = 50 − 10 ∗ 𝐶𝐿
Note that 𝛿𝑡𝑟𝑖𝑚 = 0 𝑎𝑡 𝐶𝐿𝑡𝑟𝑖𝑚 = 0.5
∴ 0 = 𝛿𝑒0 −−0.1
−0.01∗ 0.5 𝛿𝑒0 = 50
Flight Measurement of XNP
Procedure
𝑤ℎ𝑒𝑛 𝐶𝑚𝛼= 0,
𝒅𝜹𝒆𝒕𝒓𝒊𝒎𝒅𝑪𝑳𝒕𝒓𝒊𝒎
𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝑧𝑒𝑟𝑜
𝒅𝜹𝒆𝒕𝒓𝒊𝒎𝒅𝑪𝑳𝒕𝒓𝒊𝒎
= −𝐶𝑚𝛼
𝐶𝑚𝛿𝑒𝐶𝐿𝛼 − 𝐶𝑚𝛼
𝐶𝐿𝛿𝑒
𝜹𝒆𝒕𝒓𝒊𝒎 = −𝐶𝑚0
𝐶𝐿𝛼 + 𝐶𝑚𝛼𝐶𝐿𝑡𝑟𝑖𝑚
𝐶𝑚𝛿𝑒𝐶𝐿𝛼 + 𝐶𝑚𝛼
𝐶𝐿𝛿𝑒
Neutral Point
Ground Effect
– The slope of lift curve of the wing, i.e. 𝐶𝐿𝛼𝑤 increases slightly
– The downwash due to wing decreases considerably
– Result: The airplane becomes more stable and requires more negative elevator deflection
Limitation on Fwd Movement of CG
• In Free Flight: Forward c.g. location at which the maximum negative elevator deflection would be just able to permit trim at 𝐶𝐿𝑚𝑎𝑥
• Near Ground: Requirement of more negative elevator deflection due to increased stability in ground effect
Limitation on Fwd Movement of CG
• Flare at Landing: Additional elevator deflection to compensate for gusts and to flare the aircraft. This requirement dictates a 𝑑𝐶𝑚 𝑑𝐶𝐿𝑚𝑎𝑥
less than the value required for
𝐶𝐿𝑚𝑎𝑥only
Restrictions on C.G. Movement from Stick Fixed Stability Considerations
Example 2.3 (Nelson)
Solution
H.W. Assignment # 3
Solve problems 2.4, 2.6, 2.7, 2.9 & 2.10 from Nelson’s book Submission date: 14 Apr. 2015 Submit at the start of class on due date (even if you plan to be absent). No credit afterwards. Do not copy any assignment.
