Aircraft Design - Configuration Layout and Loft Quiz(MCQ)

1 .

A good sketch should include _______

A)

only volume

B)

only drag area

C)

only lifting area

D)

overall aerodynamic concepts, location of components etc

Correct Answer : overall aerodynamic concepts, location of components etc

Explanation : A good sketch is the one that highlights as much as detail as it can. It should include overall aerodynamic concepts, internal placements of components, landing gear location etc.

2 .

What do you mean by the inboard profile'?

A)

Inboard parts

B)

Outboard parts

C)

First design of sketch

D)

Most detailed drawing drawn to show greater details of internal components

Correct Answer : Most detailed drawing drawn to show greater details of internal components

Explanation : Before a design can be finalized it needs to go through number of iterations. After this iterative process we make a design which has most detailed information. This detailed layout is called inboard profile. Typically, it is far more detailed than the initial layout.

3 .

Process of defining external geometry of an A/C is _______

A)

lofting

B)

flutter

C)

drafting

D)

drawing

Correct Answer : lofting

Explanation : Lofting is defined as the process in which external geometry of an A/C is defined. It is nothing but the mathematical modelling the outer layer of the aircraft. Flutter is related to control surface. Drawing and drafting are related to each other.

4 .

Determine the corrections or otherwise of the following assertion [A] and reason [R]:
Assertion [A]: In modern A/C, we generally do not use spline lofting.
Reason[R]: Spline lofting is drawn through connecting different known points via spline line.

A)

[A] is false but [R] is true

B)

Both [A] and [R] are true but [R] is not the correct reason for [A]

C)

Both [A] and [R] are false

D)

Both [A] and [R] are true and [R] is the correct reason for [A]

Correct Answer : Both [A] and [R] are true but [R] is not the correct reason for [A]

Explaination : Now, in modern days we generally do not use spline lofting due to some disadvantages. Spline lofting requires more trial and error in order to achieve the smoothness as per requirement. Spline lofting is done by connecting different points via spline line.

5 .

Following diagram represents _____
Aircraft-Design

A)

lift curve slope

B)

development of conic lofting

C)

drag curve slope

D)

always circular shape

Correct Answer : development of conic lofting

Explaination : A typical intermediate step of conic lofting is shown in the figure. Here, ‘A’ and ‘B’ points are selected as per requirements and called end points. C is the tangent point and S is known as the shoulder point for given conic lofting development.

6 .

The first step of conic construction is ______

A)

selecting ellipse always

B)

selecting parabola always

C)

selecting desired end points at desired tangent angle

D)

selecting a circle to proceed

Correct Answer : selecting desired end points at desired tangent angle

Explanation : Conic lofting method is a step by step method to generate conic of desired cross-section. In general, first step of any conic development is to fixe two end points of the shape. And based on this endpoint we will fix the tangent point as per our requirement.

7 .

What will be the resulting curve if following cone is cut through plane as shown?

A)

Circle

B)

Ellipse

C)

Square

D)

Rectangle

Correct Answer : Circle

Explaination : Conic shape is dependent on the way we cut the cone. As shown in the figure, if conic is cut through a plane which is perpendicular to the axis of the cone then, the resultant shape will be a circle. Similarly, if plane is slanted then it will produce an ellipse.

8 .

What is conic lofting?

A)

Lofting done by using various circle only

B)

Lofting done by using various square curves only

C)

Lofting done by using various conic curves

D)

Lofting done by using various triangle curves only

Correct Answer : Lofting done by using various conic curves

Explanation : When lofting is done by using conic curves then, it is termed as conic lofting. Conic lofting may include only circles but it is not mandatory that if conic lofting is mentioned then, it will have circles only.

9 .

Which of the following is an advantage of conic lofting?

A)

Availability of simple lift drag

B)

Availability of diverse curves

C)

Availability of simple drag only

D)

Availability of lift and drag only

Correct Answer : Availability of diverse curves

Explanation : Conic lofting is done by using conic curves. A conic has number of different curves in family such as circle, ellipse etc. We can generate all such shapes via conic lofting. Hence, it is one of the advantages of it that it has availability of diverse curves.

10 .

The purpose of the longitudinal control lines is ________

A)

to change lifting line slope

B)

only to produce control of lifting lines of same curve

C)

to connect tangent points at half distance only of same conic curve

D)

to connect end points, shoulder points and tangent point of different conic curves

Correct Answer : to connect end points, shoulder points and tangent point of different conic curves

Explanation : Longitudinal control lines are used to develop smoothly – lofted fuselage. Longitudinal control lines will connect end points of different conic curves. It is also used to connect different shoulder points and tangent points to respective shoulder points and tangent points of different conic curves.

11 .

Define control station.

A)

Auxiliary lines

B)

Elliptic contour only

C)

Conic shape with circle only

D)

Original cross-section used for longitudinal control lines

Correct Answer : Original cross-section used for longitudinal control lines

Explanation : Control stations are original cross section which are used to develop longitudinal lines. Control stations are also known as control cross- section. Control stations can be any type of conic curve. It can be circular, elliptic or any c/s as per our requirement.

12 .

What do you mean by longitudinal control lines?

A)

Lines which go in horizontal direction only

B)

Lines which expands to vertical direction only

C)

Line which can be expanded only in lateral direction

D)

Smooth lines which controls overall shape of the conic cross-section

Correct Answer : Smooth lines which controls overall shape of the conic cross-section

Explanation : Longitudinal control lines are smooth lines through which conic curves are connected. Since, it connects all different conics, it will control the overall shape of the body itself during lofting.

13 .

As a designer we need to design a single engine fighter A/C having circular nozzle exhaust. Most appropriate control station for nozzle exhaust is ________

A)

triangular

B)

elliptic shape

C)

circle with nozzle exhaust diameter

D)

square of circle diameter

Correct Answer : circle with nozzle exhaust diameter

Explanation : Control station are the original cross-section which are required to complete the lofting process. Here, our main body is nozzle which has circular cross-section. Hence, original shape required to loft this nozzle will be circular also. Hence, control station should be circle with similar diameter as the nozzle.

14 .

Buttock-planes are used to _______

A)

evaluate the smoothness of A/C fuselage

B)

evaluate Lift

C)

find lift curve

D)

find lift curve slope

Correct Answer : evaluate the smoothness of A/C fuselage

Explanation : Buttock planes or butt planes are vertically oriented cuts which can be used to find smoothness of the A/C fuselage. Lift curve slope is found by using lift and AOA. Lift can be estimated by measuring pressure difference between upper and lower surface.

15 .

What is meant by 'butt-plane 40'?

A)

Buttock plane is at the 40� from fuselage centerline

B)

Buttock plane is at a distance of 40 unit from fuselage centerline

C)

Buttock plane is at a distance of 80 from fuselage centerline

D)

Buttock plane is at a distance of 20 unit from fuselage centerline

Correct Answer : Buttock plane is at a distance of 40 unit from fuselage centerline

Explanation : Butt-planes or buttock planes are defined as vertically oriented cuts which are used for fuselage lofting verification. It cuts from the horizontal intersection of the A/C with vertical planes. These planes are defined by their distance from centerline. Hence, butt-plane 40 can be interpreted as butt plane is at distance of 40 unit from aircraft centerline.

16 .

Following diagram shows ________

A)

Elliptic converter

B)

Circle to square adapter

C)

Triangle to ellipse

D)

Triangle to circle adapter

Correct Answer : Circle to square adapter

Explaination : Above diagram is showing a typical problem during lofting namely, ‘circle to square adapter’. In such cases, designer is required to loft body with different cross-section. For e.g. A nozzle of typical jet is square at inlet but at front face it is circular. Hence, such problems are required to be solved through various techniques of circle to square adapter.

17 .

Let's consider we want to loft fuselage by using flat wrap method then, _____

A)

longitudinal lines must be avoided

B)

always use cross shape parameter

C)

longitudinal control lines must be straight

D)

longitudinal control lines must be curved

Correct Answer : longitudinal control lines must be straight

Explanation : Typically, to use flat wrap method following 2 conditions are required. 1. Longitudinal control lines needs to be straight and 2. Tangent angles of conic should remain constant longitudinally.

18 .

Find the distance DS in the following diagram. Given contour is parabolic.

A)

0.5

B)

1.12

C)

2.47

D)

3.12

Correct Answer : 3.12

Explaination : Given, a parabolic contour.
A parabolic shape has conic shape parameter of 0.5.
From figure,
Conic shape parameter CSP = |DS|/|DC|
From Pythagorean Theorem, |DC| = 6.24.
Hence, |DS| = CSP*|DC| = 0.5*6.24 = 3.12.

19 .

If distance between shoulder point and tangent point is much higher then, it will produce _________

A)

flat C/S shape

B)

circle shape only

C)

circle with diameter as distance

D)

circle of radius as distance between points

Correct Answer : flat C/S shape

Explanation : Selection of the shoulder point plays vital role in conic curve. If shoulder point is too far from tangent point it will result in flat c/s shape. Similarly, if it is too close it will generate non-flat curve.

20 .

A wing is to be loft with span as 2m and reference area of wing is 4m². If we want to design the wing with taper ratio of 0.6 then what will be the value of root chord?

A)

2.5m

B)

2.60m

C)

2.65m

D)

25.6m

Correct Answer : 2.5m

Explaination : Given, span b=2m, area S = 4m², taper ratio t = 0.6
Now, root chord Cr is given by,
Cr = 2*s / b (1+t) = 2*4 / 2(0.6+1) = 8/2*1.6 = 2.5m.

21 .

Consider below wing. Calculate the value of span of wing marked by '?'.

A)

3.3m

B)

3.36m

C)

6.66m

D)

6.96m

Correct Answer : 3.3m

Explaination : Cr = 4m, Ct = 2m, S = 20m²
Taper ratio t = Ct/Cr = 2/4 = 0.5.
Hence, wing span of given wing b/2 = S/Cr*(1+t) = 20/4*(1+0.5) = 20/4*1.5 = 3.3m.

22 .

Following graph represents ____

A)

graphical method to find lift

B)

graphical method to produce drag

C)

graphical method to find MAC

D)

graphical method to find lift-curve

Correct Answer : graphical method to find MAC

Explaination : Above figure is showing the typical graphical method to find the MAC. Here, first we project the root chord from tip chord and vice-versa as shown by dashed lines from respective chords. After that we draw a line from which will connect midpoints of root and tip chord. Intersection of both lines will provide MAC which will be located at Y distance from root chord as shown.

23 .

In linear interpolation method, new airfoil is created as _________

A)

lower chord than tip always

B)

higher chord than root always

C)

heavier than the root airfoil always

D)

weighted average of root and tip airfoil

Correct Answer : weighted average of root and tip airfoil

Explanation : Linear interpolation method is used to create new airfoil station between root and tip airfoil. In general, root airfoil is based on performance and tip is selected for stall properties. Linear interpolation will give new airfoil as weighted average of the root and tip airfoil.

24 .

Which of the following is correct in terms of the wing layout with twist?

A)

Only untwist airfoils can be used for layout

B)

Only rotation of chord line is required at c/6 points

C)

Only chord line should be rotated by reducing length of chord

D)

Incidence at each span station must be considered and chord should be rotated accordingly

Correct Answer : Incidence at each span station must be considered and chord should be rotated accordingly

Explanation : In general, airfoils are drawn to find the complete wing layout. When wing has some twist then, we need to find incidence at each span station in order to include effects of twist. Also chord needs to be rotated accordingly before drawing an actual airfoil for lofting.

25 .

If wing has MAC of 8m then, what will be the location of aerodynamic centre?

A)

At l.e.

B)

At t.e.

C)

2m from l.e.

D)

2m from t.e.

Correct Answer : 2m from l.e.

Explanation : Given, MAC = 8M
Location of aerodynamic center = 25% of MAC = 25% of 8 = 0.25*8 = 2m from l.e.

26 .

For wing to be lofted as shown in below figure, what will be the approximated value of MAC?

A)

8m from t.e.

B)

8m from l.e.

C)

12m from l.e.

D)

12m from t.e.

Correct Answer : 8m from l.e.

Explaination : Given, Cr=12m, Ct=3m span = 2*20 = 40m
Taper ratio t = Ct/Cr = 3/12 = 0.25.
Location of MAC, γ = (b/6)*(1+2*t) / (1+t) = (40/6)*(1+2*0.25) / (1+0.25) = 40*(1+0.5)/6*1.25 = 8m from l.e.

27 .

As a designer, our task is to design wing layout such that the location of MAC from root chord or centre line is at 8ft. Find the appropriate value of the wingspan if wing is rectangular.

A)

32ft

B)

50ft

C)

28m

D)

32m

Correct Answer : 32ft

Explaination : Given, rectangular wing.
MAC location y = 8ft. As wing is rectangular taper ratio t =1.
Wingspan b = (6*y*(1+ t)) / (1+t+t²) = 6*8*(1+1) / (1+1+1²) = 6*8*2/3 = 32ft.

28 .

A fighter jet is flying with wing span of 90ft. If mean aerodynamic chord MAC is at 30ft from tip chord then, to design such wing which type of planform should I use?

A)

Square

B)

Triangle

C)

Delta

D)

Rectangular

Correct Answer : Delta

Explaination : Given, wing span b=90ft, location of MAC from tip = 30ft
Location of MAC from root y = b/2 – 30ft = 90/2 – 30 = 45-30 = 15ft.
To find planform let’s find taper ratio first from y.
Location of MAC is given as, y = (b/6)*(1+t+t2) / (1+t). …. (1)
From given options, let’s consider delta wing first.
Delta wing has taper ratio near to none. Let’s consider taper ratio to be zero for idle case.
Hence, by substituting value of taper ratio in eq (1),
Y = (90/6)*1+0+0/1+0 = 90/6 = 15ft.
This value is same as we found earlier from given data. Hence, wing planform is delta.

29 .

Following diagram represents?

A)

LEX

B)

Midline tip

C)

Blended tip

D)

Rounded wingtip

Correct Answer : Rounded wingtip

Explaination : Wings can have different wingtip shapes. Above figure is showing a typical rounded wingtip. A smoothly rounded tip allows air to flow easily around tip. LEX stand for leading edge extension.

30 .

Following diagram represents?

A)

T.e. kick

B)

Straight tip

C)

Rounded tip

D)

T.e. extension

Correct Answer : T.e. kick

Explaination : Above figure is often termed as trailing edge kick. A straightened out trailing edge will increase the flap chord. It also reduces induced drag. Rounded tip will permit air to flow around tip easily.

31 .

An unstable A/C configuration is a typical wing-aft tail body. IF MAC is 4m then, what would be the approximated location of the CG?

A)

2m

B)

5m

C)

6m

D)

7m

Correct Answer : 2m

Explanation : Typical location Cg for an unstable wing = 50% of MAC = 50% of 4 = 0.5*4 = 2m.

32 .

Wing planform is always elliptic.

A)

True

B)

False

C)

Can Not Say

D)

None of the above

Correct Answer : False

Explanation : Wing planform is shape generate when viewed from top. Elliptic wings are more efficient in terms of lifting properties. However to manufacture it is also difficult. Hence, wing planform is not always elliptic rather it will be selected according to mission requirements.

33 .

At maximum thickness point, fillet is in ____________

A)

purely in hp

B)

plane at 45� always

C)

purely vertical plane

D)

plane at an angle of less than 20� always

Correct Answer : purely vertical plane

Explanation : Fillet arcs will be normal to wing surface only at the maximum thickness point. At any other section it is not perpendicular to surface. Hence, at maximum thickness point, fillet is in pure vertical plane.

34 .

Following diagram represents?

A)

Wing drag

B)

Wing fillet

C)

Wing chamfer

D)

T-tail arrangement

Correct Answer : Wing filletWing fillets are arcs as shown in figure. Wing fillets are used to improve smoothness of the root of the airfoil as shown. This will result in improvement of aerodynamic efficiency. A typical fillet is a circular arc.

35 .

Total exposed area that would get wet if immersed in fluid is called ________

A)

length

B)

volume

C)

planform of wing

D)

wetted area

Correct Answer : wetted area

Explanation : Wetted area is the total exposed area which would get wet if we immersed body in fluid. Volume is product of area and length or depth. Length is measurement of how long an object is. Wing planform is shape of wing when viewed from top.

36 .

Find the wetted area of given body.

A)

65m²

B)

85m²

C)

75

D)

200

Correct Answer : 65m²

Explaination : For given diagram,
Wetted area Aw = area under the curve = area of triangle + area of rectangle
Aw = (0.5*6*5) + (10*5) = 15+50 = 65m².

37 .

Find the approximated volume for body as shown in below.

A)

77 cm

B)

7.7m

C)

77m

D)

77m³

Correct Answer : 77m³

Explaination : Volume = area under the curve
= area of triangle + rectangle area + 8 + 4 + 6 = 0.5*6*3 + 5*10 + 8 + 4 + 6 = 9+50+8+4+6 = 77m³.

38 .

Fuselage has wetted area of 20m². Given Aright = 5m². Find the value of area viewed from top

A)

6.8m

B)

6.8m²

C)

9.8m²

D)

68m

Correct Answer : 6.8m²

Explaination : Given wetted area Aw = 20m², Aright = 5m².
Now, area vied from top A is given by,
A = 0.588*Aw – Aright = 0.588*20 – 5 = 11.764-5 = 6.764 = 6.8m².

39 .

Wetted area of a rectangular body is 64.64m². Find the average projected area.

A)

15.85 m²

B)

16.16m²

C)

25.368 km

D)

45.45 cm

Correct Answer : 16.16m²

Explaination : Given, Aw = 64.64m².
Now, average projected area A = Aw/4 = 64.64/4 = 16.16m².

40 .

Wing has planform area of 2m². If wing is paper thin then, find wetted area of the wing.

A)

4m²

B)

5m²

C)

10m²

D)

20m²

Correct Answer : 4m²

Explaination : Given, planform area = 2m².
Since, wing is paper thin wetted are = 2*planform area = 2*2 = 4m².

41 .

If for a given fuselage S_top=10m², A_side=14m², L = 10m then, find the ratio of wetted area to the volume of fuselage.

A)

3.42 per m

B)

4.001

C)

6.68 per m³

D)

8.36 m

Correct Answer : 3.42 per m

Explaination : Given, S_top=10m², A_side=14m², L = 10m.
Now ratio of wetted area Aw and volume V is given by,
Aw/V = 2*L*(S_top + A_side) / (S_top*A_side)
Aw/V = 2*10*(10+14) / (10*14) = 20*24/140 = 3.42 per m.