HW3_Sol___AAE_421 (2)

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AAE421 HW3 Solution October 4, 2023 1 Problem 1 (20pt.) For the C L and C m relationship shown in the following plots: Figure 1: C m vs C L (a) Find the linear expressions of C m in term of C L for line A and B, re- spectively. (b) To obtain a trim condition, which line should be selected? (c) Assuming ¯ x cg = 0 . 7, how to relocate the a.c center (¯ x ac ) to obtain a new CL trim = 0 . 8? 1

1.1 Solution (a) Line A : C m = 0 . 5 + 0 . 33 C L Line B : C m = 0 . 3 − 0 . 5 C L (b) For trim condition, Line B should be selected because C mα < 0 implies a stable configuration in B. (c) At trim, C m,cg = 0 and CL trim = 0 . 8. Given that ¯ x cg = 0 . 7, from Line B, C m,ac = 0 . 3 C m,cg = C m,ac + C L,trim (¯ x cg − ¯ x ac ) C m,ac C L,trim = (¯ x cg − ¯ x ac ) ¯ x ac = 1 . 075 2 Problem 2 (20pt.) Consider a wing-body model with C m ac,wb = − 0 . 022 , ¯ x cg = 0 . 43, ¯ x ac,wb = 0 . 25, a wb = 0 . 08. The area and chord of the wing are 0 . 2 m 2 and 0 . 2 m , respectively. Now assume that a horizontal tail is added to this model. We have the flowing parameters: l t = 0 . 17 m , S t = 0 . 02 m 2 , i t = 2 . 5 ◦ , α t = 0 . 1 per degree, ϵ 0 = 0 and ∂ϵ ∂α = 0 . 4. (a) If the angle of attack α wb = 9 . 38 ◦ , calculate C M,cg for the airplane model. (b) Does this model has longitudinal static stability and balance? (c) Calculate the neutral point location. 2.1 Solution 2.1.1 Solution A If student interpreted C L αw as 0.08 per rad. (a) If the angle of attack α wb = 9 . 38 ◦ , calculate C M,cg for the airplane model. V H = S t l t SC = 0 . 02 ∗ 0 . 17 0 . 2 ∗ 0 . 2 = 0 . 085 C m 0 wt = C m acw − V H [ C L α t ( ϵ 0 − i t )] = − 0 . 00075 Convert C L αw from per radian to per degree by multiplying with 0.0175. C L α t is in per degree already. C α wt = C L αw (0 . 0175)( x cg ¯ c − x ac w ¯ c ) − V H C L α t (1 − dϵ dα ) = − 0 . 0048 perdeg C M,cg = C m wt = C m 0 wt + C α wt α = − 0 . 00075 − 0 . 0048(9 . 38) = − 0 . 0458 2

(b) Does this model has longitudinal static stability and balance? C α wt = − 0 . 0458 < 0 C m 0 = − 0 . 00075 < 0 C M,cg represents the change in moment (pitching moment) coefficient with changes in the aircraft’s center of gravity (CG) position. A negative C M,cg means that as the center of gravity moves forward, the aircraft becomes more stable in pitch. In other words, if the CG is shifted forward, the aircraft will have a natural tendency to return to its original pitch attitude, which is gen- erally desirable for stability. C m 0 is the pitching moment coefficient when the aircraft is at zero lift (i.e., when it is in a level, unaccelerated flight). A negative C m 0 means that the aircraft tends to pitch nose-down when it is flying at zero lift. This can be compensated for through the design of horizontal stabilizers and control surfaces. However, the configuration does not include stabilizers and control surfaces. Hence, overall we can say that that the configuration is longitudinally stable but the a/c is not balanced. (c) Calculate the neutral point location. x NP ¯ c = ( x ac wb ¯ c + a t a wb S t S (1 − ∂ϵ ∂α )) 1 + a t a wb S t S (1 − ∂ϵ ∂α ) = 0 . 8581 2.1.2 Solution B If student interpreted C L αw as 0.08 per deg. (a) If the angle of attack α wb = 9 . 38 ◦ , calculate C M,cg for the airplane model. V H = S t l t SC = 0 . 02 ∗ 0 . 17 0 . 2 ∗ 0 . 2 = 0 . 085 C m 0 wt = C m acw − V H [ C L α t ( ϵ 0 − i t )] = − 0 . 00075 C α wt = C L αw ( x cg ¯ c − x ac w ¯ c ) − V H C L α t (1 − dϵ dα ) = 0 . 0093 perdeg C M,cg = C m wt = C m 0 wt + C α wt α = − 0 . 00075 + 0 . 0093(9 . 38) = 0 . 0865 (b) Does this model has longitudinal static stability and balance? C α wt = 0 . 0865 > 0 C m 0 = − 0 . 00075 < 0 C M,cg represents the change in moment (pitching moment) coefficient with changes in the aircraft’s center of gravity (CG) position. A positive C M,cg means that as the center of gravity moves forward, the aircraft becomes less 3

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stable in pitch. In other words, if the CG is shifted forward, the aircraft will have a natural tendency to move away from its original pitch attitude, which is not desirable for a stable aircraft. C m 0 is the pitching moment coefficient when the aircraft is at zero lift (i.e., when it is in a level, unaccelerated flight). A negative C m 0 means that the aircraft tends to pitch nose down when it is fly- ing at zero lift. This can be compensated for through the design of horizontal stabilizers and control surfaces. However, the configuration does not include stabilizers and control surfaces, and the aircraft does not posses longitudinal stability with the addition of the tail. Hence, overall we can say that that the configuration is longitudinally un- stable and the a/c is not balanced. (c) Calculate the neutral point location. x NP ¯ c = ( x ac wb ¯ c + a t a wb S t S (1 − ∂ϵ ∂α )) 1 + a t a wb S t S (1 − ∂ϵ ∂α ) = 0 . 8581 3 Problem 3 (20pt.) Show that the angle of attack of the horizontal tail of the airplane model that we have adopted in this class can be expressed as: α t = α (1 − ∂ϵ ∂α ) − ( ϵ 0 + i t )[1 − a t a S t S (1 − ∂ϵ ∂α )] Start with α t = α wb − ( ϵ + i t ). 3.1 Solution α t = α wb − ( ϵ + i t ) = α wb − ( ∂ϵ ∂α α wb + i t ) = (1 − ∂ϵ ∂α ) α wb − ( ϵ 0 + i t ) Since, C L = aα = − ( a t ) S t S ( ϵ 0 + i t ) + aα wb α wb = α + a t a S t S ( ϵ 0 + i t ) Substituting α aw into the α t equation, α t = (1 − ∂ϵ ∂α ) α wb − ( ϵ 0 + i t ) = (1 − ∂ϵ ∂α )( α + a t a S t S ( ϵ 0 + i t )) − ( ϵ 0 + i t ) Algebra, α t = α (1 − ∂ϵ ∂α ) − ( ϵ 0 + i t )[1 − a t a S t S (1 − ∂ϵ ∂α )] 4

4 Problem 4 (20pt.) Wind tunnel test on a full-scale flying wing yielded the following database Figure 2: Database The configuration weight is 650 lb and has a rectangular wing with an area of 80 ft2. The aspect ratio is 12 and the chord length is 3 ft. The configuration c.g is located 0.58 ft from the leading edge of the chord. 4.1 Solution: (a). Because the plot of the data at the right is a straight line, any two points can be used to estimate the slope. 5

Figure 3: Lift Curve Slope C Lα = ∆ C L ∆ α = 0 . 08 per deg (b) Yes. A plot of moment coefficient w.r.t angle of attack is shown below. 6

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Figure 4: Moment Coefficient vs Angle of Attack The two conditions for pitch moment static stability are satisfied: 1. C m,ac = 0 . 05 > 0, which means that moment about aerodynamic center should be positive. 2. C m,α = ∆ C m ∆ α = − 0 . 008 per deg which means that with every increase of angle of attack, moment generated should be restoring moment. (c) We can use the data provided directly, along with simple algebra: C m = C m,ac + C mα α − 0 . 014 = C m,ac − 0 . 008 ∗ 8 C m,ac = 0 . 05 To find location of aerodynamic center: C mα C Lα = (¯ x cg − ¯ x ac ) x ac = 1 . 0599 ft. (d) Yes, this configuration, as tested, can be trimmed in a steady glide condi- tion because according to Fig. 4 , we can see the the aircraft trims at a positive angle of attack, and it is statically stable. C mα < 0 and C m,ac > 0 7