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FLIGHT MECHANICS
SIDESLIP
TOPIC WISE SOLVED QUESTIONS
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
1
2
FLIGHT MECHANICS - BASICS
Answer:
(C) is always negative
Explanation:
• Yawing Moment (+N & -N).
o Right Wing going back/ clockwise turn: Positive Yaw
o Left Wing going back/ anti-clockwise turn: Negative Yaw
• Sideslip Angle (+Ve & -Ve)
o Relative wind from right side of aircraft’s nose: Positive Sideslip angle
o Relative wind from left side of aircraft’s nose: Negative Sideslip angle
GATE 2011 Sideslip 1 Mark
Q1. If an aircraft is performing a positive yawing manoeuvre, the side slip angle
(A) is always zero (B) is never zero
(C) is always negative (D) could be any value
- β
V∞
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
3
FLIGHT MECHANICS - BASICS
Answer: (B) 5.65
Solution:
• Ground speed = Aircraft's true airspeed + wind speed & direction;
Therefore
• Aircraft's true airspeed = Ground speed - wind speed and direction
• [u, v, w] = [100-0, 5+5, 5+10] = [100, 10, 15]
β = sin-1 𝑣
𝑉
; 𝑉 = 𝑢2 + 𝑣2 + 𝑤2 = 1002 + 102 + 152 = 10325 = 101.612 𝑚/𝑠
GATE 2015 Sideslip 2 Mark
Q2. An aircraft is flying with inertial ground and wind speeds of 𝑣𝑔
𝑏= (100, 5, 5) 𝑚/𝑠 and 𝑣 𝑤
𝑏
= (0, −5, −10)
𝑚/𝑠, respectively, as expressed in the body frame. The corresponding sideslip angle (in degrees) is
(A) 0 (B) 5.65
(C) 8.49 (D) 9.54
β = sin-1 10
101.612
= 5.647°
i.e. sinθ =
𝑂𝑝𝑝
𝐻𝑦𝑝
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
4
5
FLIGHT MECHANICS - BASICS
Answer:
(D) [100.995, 5.71, 5.68]
Solution:
• [𝑢, 𝑣, 𝑤] = [100, 10, 10]
𝑉 = 𝑢2 + 𝑣2 + 𝑤2 = 1002 + 102 + 102 = 10200 = 100. 995 𝑚/𝑠
GATE 2016 Sideslip 2 Mark
Q3. The aircraft velocity (m/s) components in body axes are given as [𝑢, 𝑣, 𝑤] = [100, 10, 10]. The air velocity
(m/s), angle of attack (deg) and sideslip angle (deg) in that order are
(A) [120, 0.1, 0.1] (B) [100, 0.1, 0.1]
(C) [100.995, 0.1, 5.73] (D) [100.995, 5.71, 5.68]
β = sin-1 10
101.995
= 5.68°
β = sin-1 𝑣
𝑉
i.e. sinθ =
𝑂𝑝𝑝
𝐻𝑦𝑝
;α = tan-1 𝑤
𝑢
i.e. tanθ =
𝑂𝑝𝑝
𝐴𝑑𝑗
α = tan-1 10
100
= 5.71°
Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
6

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Flight Mechanics Sideslip Questions

  • 1. FLIGHT MECHANICS SIDESLIP TOPIC WISE SOLVED QUESTIONS Question Analysis | Book Reference | Important Concepts | Formulae | Solutions 1
  • 2. 2 FLIGHT MECHANICS - BASICS Answer: (C) is always negative Explanation: • Yawing Moment (+N & -N). o Right Wing going back/ clockwise turn: Positive Yaw o Left Wing going back/ anti-clockwise turn: Negative Yaw • Sideslip Angle (+Ve & -Ve) o Relative wind from right side of aircraft’s nose: Positive Sideslip angle o Relative wind from left side of aircraft’s nose: Negative Sideslip angle GATE 2011 Sideslip 1 Mark Q1. If an aircraft is performing a positive yawing manoeuvre, the side slip angle (A) is always zero (B) is never zero (C) is always negative (D) could be any value - β V∞ Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
  • 3. 3 FLIGHT MECHANICS - BASICS Answer: (B) 5.65 Solution: • Ground speed = Aircraft's true airspeed + wind speed & direction; Therefore • Aircraft's true airspeed = Ground speed - wind speed and direction • [u, v, w] = [100-0, 5+5, 5+10] = [100, 10, 15] β = sin-1 𝑣 𝑉 ; 𝑉 = 𝑢2 + 𝑣2 + 𝑤2 = 1002 + 102 + 152 = 10325 = 101.612 𝑚/𝑠 GATE 2015 Sideslip 2 Mark Q2. An aircraft is flying with inertial ground and wind speeds of 𝑣𝑔 𝑏= (100, 5, 5) 𝑚/𝑠 and 𝑣 𝑤 𝑏 = (0, −5, −10) 𝑚/𝑠, respectively, as expressed in the body frame. The corresponding sideslip angle (in degrees) is (A) 0 (B) 5.65 (C) 8.49 (D) 9.54 β = sin-1 10 101.612 = 5.647° i.e. sinθ = 𝑂𝑝𝑝 𝐻𝑦𝑝 Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
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  • 5. 5 FLIGHT MECHANICS - BASICS Answer: (D) [100.995, 5.71, 5.68] Solution: • [𝑢, 𝑣, 𝑤] = [100, 10, 10] 𝑉 = 𝑢2 + 𝑣2 + 𝑤2 = 1002 + 102 + 102 = 10200 = 100. 995 𝑚/𝑠 GATE 2016 Sideslip 2 Mark Q3. The aircraft velocity (m/s) components in body axes are given as [𝑢, 𝑣, 𝑤] = [100, 10, 10]. The air velocity (m/s), angle of attack (deg) and sideslip angle (deg) in that order are (A) [120, 0.1, 0.1] (B) [100, 0.1, 0.1] (C) [100.995, 0.1, 5.73] (D) [100.995, 5.71, 5.68] β = sin-1 10 101.995 = 5.68° β = sin-1 𝑣 𝑉 i.e. sinθ = 𝑂𝑝𝑝 𝐻𝑦𝑝 ;α = tan-1 𝑤 𝑢 i.e. tanθ = 𝑂𝑝𝑝 𝐴𝑑𝑗 α = tan-1 10 100 = 5.71° Question Analysis | Book Reference | Important Concepts | Formulae | Solutions
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