70 + Interview Questions in Stability in Aircraft Design Page 2 InterviewSolution

51.	Determine the net velocity if sideslip angle is 3.8 degree and [v] = 4.2 unit.(a) 63.23 unit(b) 100 unit(c) 100.23 unit(d) 102.623 unitI got this question during an internship interview.The query is from Lateral-Directional Static Stability and Control in portion Stability, Control, and Handling Qualities of Aircraft Design
Answer» The CORRECT choice is (a) 63.23 UNIT The explanation is: NET VELOCITY V = vSINE (sideslip) = 4.2sine (3.8°) = 63.23 unit.

Discussion

52.	Which of the following is correct?(a) Elevator can float in stick free condition(b) Drag is always same as weight(c) Power loading is same as weight(d) Stability is same for every aircraftI got this question in examination.This intriguing question comes from Stick-Free Stability in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» The CORRECT answer is (a) ELEVATOR can float in stick free condition Explanation: Stick free condition allows floating of elevator. POWER loading is defined as the ratio of weight of the AIRCRAFT to the power produced. Stability of aircraft DEPENDS on the design requirements. Military aircraft is less stable whereas commercial are more stable.

Discussion

53.	Find the value of CHδe if tail angle of attack is 3 deg and CHɑt is -0.004. Consider elevator floats up by 2 degree.(a) -0.006(b) -7(c) 1.223(d) 1.225I have been asked this question in an interview for job.This interesting question is from Stick-Free Stability in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct option is (a) -0.006 To elaborate: GIVEN, tail AOA ɑ=3°, ELEVATOR float e =-2° (UPWARD deflection is NEGATIVE), CHɑt = -0.004 Now, CHδe = – ɑ * CHɑt / e = -3*-0.004/-2 = -0.006.

Discussion

54.	Longitudinal stability means __________(a) stability about pitching axis(b) stability about yawing axis(c) stability about lateral axis(d) stability about negative yawing axisThe question was asked at a job interview.This key question is from Longitudinal Static Stability and Control-1 topic in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» The CORRECT option is (a) stability about PITCHING axis To explain I WOULD say: Longitudinal stability MEANS stability about pitching axis. Pitch up or pitch down moments will be included in this type of stability. Stability about yawing axis is CALLED directional stability and similarly stability about lateral axis is called Lateral stability.

Discussion

55.	Which of the following is not the type of aircraft dynamic mode?(a) Flap(b) Dutch roll(c) Phugoid(d) Lateral modeI have been asked this question in an internship interview.My query is from Dynamic Stability in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» The correct option is (a) Flap The EXPLANATION: Flap is a high lift DEVICE. Flap is USED to INCREASE lift produced by the wing. Flaps are used to increase camber and overall surface area of the wing. This results in increased lift. Dutch roll, phugoid, lateral mode etc. are typical aircraft dynamic MODES.

Discussion

56.	A wing alone arrangement has wing lift curve slope of 2.65 per rad. Find slope of pitching moment coefficient. Given Xcg = 0.3.(a) 0.025 per degree(b) 0.03 per rad(c) 45 per rad(d) 0.0035 per degreeI have been asked this question in an interview for job.My question comes from Longitudinal Static Stability and Control-1 topic in chapter Stability, Control, and Handling Qualities of Aircraft Design
Answer» Right option is (a) 0.025 per degree For EXPLANATION I would say: Considering ideal LOCATION of aerodynamic centre i.e. XAC = 0.25 Slope of pitching MOMENT curve = (wing LIFT curve slope*[Xcg-Xac]) = (2.56[0.3-0.25]) = 0.1325 per rad = 0.00253 per degree.

Discussion

57.	If CM0 is -0.052 and lift coefficient at zero angle CL0 is 0.92 then, find CMac. Consider rectangular wing.(a) -0.282(b) 0.98(c) 2.5(d) 7.89The question was posed to me during an online interview.My doubt stems from Longitudinal Static Stability and Control-2 topic in portion Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct CHOICE is (a) -0.282 The EXPLANATION: CMac = CM0 +CL0[Xcg-Xac] For a typical RECTANGLE wing, Xcg-Xac =0.25. Hence, CMac = CM0 + CL00.25 = -0.052+0.95*0.25 = -0.282.

Discussion

58.	An aircraft is consider to be over damped. What will be the appropriate value for damping ratio?(a) 1.45(b) 0.1(c) 0.001(d) 0.04This question was posed to me in final exam.I would like to ask this question from Dynamic Stability topic in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct option is (a) 1.45 Easiest explanation: For overdamped system, value of DAMPING ratio is greater than 1. Hence, AMONG the given options, 1.45 is correct ANSWER as it is > 1.

Discussion

59.	Which of the following is correct?(a) Aircraft static longitudinal stability will be dependent upon the arrangement of different components(b) Lift is always same as weight(c) Static stability is similar to dynamic stability(d) All the aircrafts are statistically stableThis question was addressed to me by my college professor while I was bunking the class.This question is from Stick-Free Stability topic in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct OPTION is (a) Aircraft STATIC longitudinal stability will be dependent upon the arrangement of different COMPONENTS Explanation: Static longitudinal stability of aircraft will be dependent upon the arrangement of different components such as wing placements, tail location etc. At cruise CONDITION or at trim position lift will be same as weight. Static stability and dynamic stability are different. System can be STATICALLY stable but that doesn’t mean that the system is dynamically stable as well.

Discussion

60.	How can we say that the aircraft has initial tendency to return to its original equilibrium position after being disturbed?(a) If aircraft generates some restoring force or/and moment without any external help(b) If restoring force is not generated to oppose the disturbance(c) If lift is same as weight always(d) If thrust loading is always unityI had been asked this question during an online interview.My question is from Longitudinal Static Stability and Control-1 topic in chapter Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct answer is (a) If aircraft generates some RESTORING FORCE or/and moment without any EXTERNAL help For explanation I would say: If some disturbance is given to the aircraft and aircraft has initial tendency to generate some restoring force or/and moments without any external means then, we say that the aircraft has initial tendency to return to its original equilibrium POSITION after being disturbed. Hence, it has static stability.

Discussion

61.	If CLɑ wing = 1.2 per rad then, determine CMɑ. Given Xcg=0.29.(a) 0.076 per rad(b) 45 per rad(c) 0.09 per rad(d) 1.45 per radI had been asked this question in an interview for job.I would like to ask this question from Longitudinal Static Stability and Control-1 in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» CORRECT ANSWER is (a) 0.076 PER RAD Easy EXPLANATION: CMɑ = CLɑ[Xcg-Xac] = 1.2[0.29-0.25] = 0.076 per rad.

Discussion

62.	0166 per rad(a) -12.33(b) 0.0015 per rad(c) 0.098 per °This question was addressed to me in a job interview.Enquiry is from Longitudinal Static Stability and Control-2 topic in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct option is (a) -12.33 To EXPLAIN I WOULD say: Elevator control POWER Cmδe = (Cm of 1 – Cm of 2) / (δe of 1 – δe of 2) = 0.02-0.08/(1.1-(-2.5)) = -0.0166 PER RAD.

Discussion

63.	Find sideslip angle if [u, v, w] = [100, 5, 2.5]. Consider steady level flight.(a) 2.86°(b) 4.5°(c) 4.7°(d) 2.1°I had been asked this question during an interview.The doubt is from Lateral-Directional Static Stability and Control topic in portion Stability, Control, and Handling Qualities of Aircraft Design
Answer» Right answer is (a) 2.86° The BEST EXPLANATION: GIVEN, v = 5, V = [u^2+v^2+w^2] ^0.5 = [100100+55+2.5*2.5] ^0.5 = 100.156. Sideslip ANGLE = ARCSINE (v/V) = arcsine (5/100.156) = 2.86°.

Discussion

64.	Negative stick force gradient is used to provide ___________(a) speed stability(b) thrust(c) propulsive force(d) gliding ratioI had been asked this question by my school teacher while I was bunking the class.Query is from Stick-Free Stability topic in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct answer is (a) SPEED stability The best explanation: Negative stick force gradient is used to provide speed stability. Thrust is a propulsive force. Thrust is GENERATED by AIRCRAFT engine. Propulsive force is a typical force which is responsible for driving the aircraft or a system in forward direction TYPICALLY. Gliding ratio is RELATED to glider.

Discussion

65.	Dynamic stability is more concerned about time.(a) True(b) FalseThe question was posed to me in my homework.My question is based upon Dynamic Stability topic in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» The correct choice is (a) TRUE Easy explanation: Yes, it is true. Dynamic stability is more concerned about time. If an aircraft has INITIAL tendency to RETURN to its equilibrium position after displaced then, it is called that the aircraft has STATIC stability. Dynamic stability accounts for the time required for returning to original equilibrium position. System can be dynamically STABLE but may or may not be statically stable.

Discussion

66.	Which of the following is incorrect?(a) Uncambered wing lift curve is not same as that of the Cambered(b) Stick free and stick fixed are same(c) Lift is not always same as weight(d) Thrust required is not always same powerThis question was posed to me during an interview.My doubt is from Stick-Free Stability topic in chapter Stability, Control, and Handling Qualities of Aircraft Design
Answer» Correct choice is (b) STICK free and stick fixed are same Easiest explanation: Wing lift curve is DIFFERENT for uncambered wing and Cambered wing. Stick free and stick fixed phenomena are not same. Lift is not always same as weight. During cruise lift and Weight will be EQUAL. Thrust REQUIRED is calculated based on DRAG of the system.

Discussion

67.	If aircraft continues to go farther away from equilibrium position after being disturbed then, the aircraft is called _______(a) unstable(b) stable(c) statically stable(d) neutrally stableThis question was posed to me during a job interview.This question is from Longitudinal Static Stability and Control-1 topic in chapter Stability, Control, and Handling Qualities of Aircraft Design
Answer» Right choice is (a) UNSTABLE Best explanation: If after being DISTURBED from the equilibrium position, aircraft tends to GO further away from the original equilibrium position then it is SAID that the aircraft is not stable or it is Unstable. It will be called neutrally stable if it attains NEW equilibrium position however it is not mentioned in the question so the correct answer will be unstable.

Discussion

68.	If damping ratio is unity then, system is _____________(a) critically damped(b) over damped(c) high lift device(d) thrust augmentedThe question was asked by my college director while I was bunking the class.The query is from Dynamic Stability in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» The correct answer is (a) critically damped The best explanation: Damping RATIO is defined as damping coefficient DIVIDED by critical damping coefficient. If damping ratio is unity then, the damping coefficient is EQUAL to the critical damping coefficient. Hence, for such SCENARIO SYSTEM is said to be critically damped.

Discussion

69.	Find critical damping coefficient if damping ratio is 0.5 and damping coefficient is 1.1.(a) 2.2(b) 1(c) 4(d) 1.1I got this question by my school teacher while I was bunking the class.My doubt stems from Dynamic Stability in division Stability, Control, and Handling Qualities of Aircraft Design
Answer» RIGHT OPTION is (a) 2.2 The explanation: Critical DAMPING coefficient damping coefficient/ damping RATIO 1.1/0.5 = 2.2.

Discussion

70.	Consider the 2d phugoid approximation with natural frequency of 0.4. Determine the approximate value of aircraft speed.(a) 34.68(b) 56(c) 12(d) 12.68This question was posed to me during a job interview.I would like to ask this question from Dynamic Stability in section Stability, Control, and Handling Qualities of Aircraft Design
Answer» RIGHT answer is (a) 34.68 To explain I would say: GIVEN, natural frequency w = 0.4 Approximate value of AIRCRAFT SPEED = 1.414g/w = 1.4149.81/0.4 = 34.68 m/s.

Discussion

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Which of the following is not the type of aircraft dynamic mode?(a) Flap(b) Dutch roll(c) Phugoid(d) Lateral modeI have been asked this question in an internship interview.My query is from Dynamic Stability in section Stability, Control, and Handling Qualities of Aircraft Design

0166 per rad(a) -12.33(b) 0.0015 per rad(c) 0.098 per °This question was addressed to me in a job interview.Enquiry is from Longitudinal Static Stability and Control-2 topic in section Stability, Control, and Handling Qualities of Aircraft Design

Dynamic stability is more concerned about time.(a) True(b) FalseThe question was posed to me in my homework.My question is based upon Dynamic Stability topic in section Stability, Control, and Handling Qualities of Aircraft Design

Find critical damping coefficient if damping ratio is 0.5 and damping coefficient is 1.1.(a) 2.2(b) 1(c) 4(d) 1.1I got this question by my school teacher while I was bunking the class.My doubt stems from Dynamic Stability in division Stability, Control, and Handling Qualities of Aircraft Design