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Exercises Lecture 6 Stability & Control Prof. dr. ir. Jacco Hoekstra Tim van Leeuwen Delft University of Technology A. Sifre - CC - BY - NC - ND
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  • Exercises Lecture 6 Stability & Control

    Prof. dr. ir. Jacco Hoekstra Tim van Leeuwen

    Delft

    Univ

    ers

    ity o

    f Tech

    nolo

    gy

    A. Sifre - CC - BY - NC - ND

  • AE1110x - Introduction to Aeronautical Engineering

    Exercise 1According to who was the issue of stability the final hurdle to overcome before flight could takeplace?

    A) Samuel LangleyB) Theodore von KarmanC) The Wright brothersD) The Montgolfier brothers

    Exercise 2Why do we see modern planes still having difficulty with stability and control?

    A) The pilots may do manoeuvres the designers did not think of.B) It is impossible to make a perfectly controllable and stable aircraft.C) Control surfaces are very hard to design properly, due to complex aerodynamic effects.D) The control system of the aircraft does not match the pilots sensory and processing capabilities.

    Exercise 3What was the main difference between the controls of the Wrights Flyer I and the controls ofmost European early planes?

    A) The Flyer had a horizontal tail in the front, rather than at the backB) The Flyer had larger aileronsC) The Flyer had a smaller vertical tailD) The Flyer was less stable than European planes

    Exercise 4Below you see a picture of a Condor Boeing 767. Three of its control surfaces have been highlightedin a yellow triangle, a green rectangle and a pink circle.

    Figure 1: A Condor aircraft, with control surfaces highlighted. Image courtesy of Aero Icarus, CC- BY - SA

    1) What is the name of the control surface in the yellow triangle?2) What is the name of the control surface in the green rectangle?3) What is the name of the control surface in the pink circle?

    Exercises Lecture 6 - Stability & Control 1

  • AE1110x - Introduction to Aeronautical Engineering

    Exercise 5Early aircraft used wing warping to bank, modern aircraft use ailerons. Why dont modern aircraftuse wing warping?

    A) Banking using ailerons is more aerodynamically efficientB) Banking using ailerons allows for faster bankingC) Modern aluminium wings are not capable of warpingD) Ailerons are easier to control than wing warping mechanisms

    Exercise 6A pilot wants to bank his aircraft to the left. What will he do?

    Figure 2: A Delta aircraft banking to the left. Image courtesy of Pieter van Marion, CC - BY - SA

    A) Push the control column awayB) Pull the control column towards himC) Turn the control column clockwiseD) Turn the control column counter-clockwiseE) Push his right pedalF) Push his left pedal

    Exercise 7Modern aircraft usually have two sets of ailerons: one inboard aileron (close to the fuselage) andone outboard aileron (further up the wing). Why is this?

    Figure 3: A set of double ailerons. Image courtesy of Nelson Cunnington, CC - BY - NC - ND

    2 Exercises Lecture 6 - Stability & Control

  • AE1110x - Introduction to Aeronautical Engineering

    A) The outboard ailerons serve as a backup, to still achieve banking during for example gusts.B) Two sets of ailerons allow for redundancy: if one set fails the other set is used.C) The two sets of ailerons are used for different weather conditions.D) The two sets of ailerons are used for different flight speeds.

    Exercise 8For most conventional aircraft the line of action of the engine thrust is situated below the aircraftcentre of gravity. This means that in case the engine thrust is altered, the pitching moment ofthe aircraft will be affected. To prevent the aircraft from pitching, the elevator can be deflected abit.

    Suppose the pilot of a conventional aircraft decides to increase the thrust of this aircraft, but wantsto keep the pitch angle (and angle of attack) the same. What kind of elevator deflection shouldhe give to prevent his aircraft from pitching: a positive or negative one?

    Exercise 9Suppose an aircraft flies under an angle of attack of 5.6 degrees, and at a pitch angle of 8 degrees.Determine the climb angle of this aircraft, in degrees.

    Figure 4: A British Airways Boeing 747. Image courtesy of Aero Icarus, CC - BY - SA

    Exercise 10Suppose a passenger aircraft flies with a course of 80 degrees and a heading of 84 degrees (bothaccording to the compass). Determine the angle of sideslip it then experiences.

    Exercise 11Below you see a picture of a deflected rudder. If the pilot would deflect his rudder in this wayduring flight, what yaw angle will the aircraft get: a positive or a negative one?

    Figure 5: A British Airways Boeing 747. Image courtesy of Mark Jones Jr, CC - BY

    Exercises Lecture 6 - Stability & Control 3

  • AE1110x - Introduction to Aeronautical Engineering

    Exercise 12As an aircraft flies across the globe, it must continuously pitch (with respect to an inertial frameof reference) to keep following the curvature of the Earth.

    From that point of view, what must the sign of this continuous (though extremely small) pitchrate be?

    Exercise 13Below you see a picture of a mass m attached to a spring. By what kind of stability is this systemcharacterised?

    Figure 6: A mass-spring system.

    A) Positive static stabilityB) Neutral static stabilityC) Negative static stability

    Exercise 14Consider a statically stable aircraft in the yaw direction. What direction will the aircraft straightaway yaw to after the pilot has pushed his left foot pedal?

    A) The positive yaw directionB) The negative yaw direction

    Exercise 15In a car, you basically have two controls in the forward direction: the throttle and the brake pedal.Consider this type of control system.

    How many integrators are there in this control system?

    Exercise 16Which of the following actions is not a way to make an aircraft (more) longitudinally staticallystable?

    A) Increasing the horizontal tail surface areaB) Increasing the wing lift coefficient gradientC) Placing the horizontal tail out of the downwash of the wingD) Decreasing the wing surface area

    4 Exercises Lecture 6 - Stability & Control

  • AE1110x - Introduction to Aeronautical Engineering

    Exercise 17Below you see a figure of a flying wing, so an aircraft without a tail. Its centre of gravity is situatedbehind the aerodynamic centre of the wing, with lcg = 0.05c .

    Figure 7: Image courtesy of indianadinos, CC - BY - NC - SA

    If you are given that the wing flies with CL = 0.8 and given that Cm = 0.02 , determine themoment coefficient of the moment around the aerodynamic centre.

    Exercise 18Again consider the flying wing of the problem above. Is this flying wing longitudinally staticallystable?

    Exercise 19Below you see a picture of an Airbus A340-600.

    Figure 8: Image courtesy of indianadinos, CC - BY - NC - SA

    Of this aircraft you are given the following parameters: lwing = 34.9mltai l = 67.8mS = 437m2

    SH = 93m2

    c = 8.35ma = 4.8/radat = 5.2/rad

    Furthermore we will assume that the downwash angle is equal to 10% of the angle of attack ofthe wing. Given that the centre of gravity of this A340-600 is situated 39.2 metres from the nose,find the static margin (in metres) of this A340.

    Exercises Lecture 6 - Stability & Control 5

  • AE1110x - Introduction to Aeronautical Engineering

    Exercise 20In the picture below you see a King Air 200, an aircraft with a so-called T-tail. One of the primaryreasons of placing the tail higher up is to keep the tail out of the wake and downwash of thewing.

    Figure 9: Image courtesy of Mark Jones Jr., CC - BY

    For this aircraft, we will in this exercise investigate the required position of the centre of gravityto guarantee static, longitudinal stability. The first step in this analysis is to set up the momentequation around the centre of gravity of this aircraft.

    1) Derive the moment equation around the centre of gravity of this King Air 200, as a functionof the total lift L, the tail lift LH, the centre of gravity distance lcg, the tail arm lH and theaerodynamic centre moment Mac .Tips: Use the sign convention "Pitch up = positive" and take a positive moment in the aerodynamiccentre.

    2) Non-dimensionalise the equation just derived and give the result below. You are allowed to usethe quantities c , lcg, lH, S, SH, Cmac , CL and CLH .Tip: Dont introduce the tail volume yet.

    3) Derive this equation with respect to the angle of attack and give your result for Cm . You areallowed to use the quantities c , lcg, VH (the tail volume), a (the wing lift coefficient gradient) andat (the tail lift coefficient gradient).

    4) What is the requirement on the position of the centre of gravity? You are allowed to use thequantities c , lcg, VH (the tail volume), a (the wing lift coefficient gradient) and at (the tail liftcoefficient gradient).

    6 Exercises Lecture 6 - Stability & Control