Aerodynamics help please!!! Given Conditions: W=15,000 lb S = 200 ft 2 CLMAX = 1

Question

Aerodynamics help please!!!

Given Conditions:

W=15,000 lb

S = 200 ft 2

CLMAX = 1.5 Aircraft Certified Aerobatic Category (-3, +6) G

1. Find Stall Speed (KEAS) at 1 G, 0 deg AOB.

2. Find Stall Speed (KEAS), level flight 70 deg Bank Angle (~2.9 G).

3. Find Positive Limit Load and Positive Ultimate Limit Load of Given Aerobatic Aircraft-Max Gross Weight = 15,000 lb.

4. Find Maneuvering speed (Va) of given aerobatic aircraft W=15,000 lb.

5. Find Turn Radius (r) and Rate of Turn (ROT) of aerobatic aircraft performing level sustained 4 G turn at 250 KTAS.

6. Using Figure 14.10 from Flight Theory and Aerodynamics, find Bank Angle to fly a standard rate turn (3 deg/sec) at 250 KTAS.

7. Name four different design features for supersonic aircraft.

8. What is rate of turn (ROT) and turn radius(r) of Pitts doing 3G turn at 110 imph (indicated statute miles per hour)? Assume Pressure Altitude of 6,000 ft Standard Temperature. Convert to imph to KTAS. Assume no compressibility issues and not knowing the position errors, assume KCAS = KIAS; also, not knowing the temperature, assume a standard day at 6000 ft; therefore, KEAS =KTAS Compare your mathematical answer of rate of turn (ROT) to that of the aircraft aviation video link (Time the 3G turn of 360 deg and then divide by 360 to get deg/sec)?

Explanation / Answer

A.

Vstall= sqrt(295*G*W/(CLmax*sigma*S) )
G = 1, sigma = 1, W =15000 lb, S = 200 ft^2, Clmax =1.5
Vstall = sqrt(295*1*15000/(1.5*200)) =121.44 KEAS

B.

G = 2.9
Vs =sqrt(295*2.9*15000/(1.5*200)) =206.82 KEAS
or simpler

Vs2 = Vs1*sqrt(G) =121.44*sqrt(2.9) =206.82 KEAS

C.

Positive limit load PL = 6GW = 6*15000 =90000 lb =90000*0.4536*9.81 N =400483.44 N
Ultimate load UL = 1.5*PL =1.5*90000 = 135000 lb = 600725.16 N

D.

Maneuvering speed Va is the stall speed at the positive limit load. It is specified that this load should not be less than 6G for aerobatic aircraft. In practice this load is taken as 7G (where the stuctural damage occurs).
Va =Vs1*sqrt(G) = 121.44*sqrt(7) =321.30 KIAS
if we take the positive limit load as 6G (the lowest possible value) the maneuvering speed is
Va= 121.44*sqrt(6) =297.46 KIAS

E.

Fi = arccos(1/G) = arccos(1/4) = 75.52 degree
turn radius is
r = V^2/[g*tan(fi)]
V = 250 KTAS = 250*0.5144 m/s =128.6 m/s
g = 9.81 m/s/s
r = (128.6)^2/(9.81*tan(75.52)) =435.35 m =435.35*3.28 ft = 1427.94 ft

total length of turn L = 2*pi*r =2*3.14*435.35 =2734 m (= 360 degree)
it means l = 7.594 meter/degree
V = 250 KTAS = 128.6 m/sec
ROT = l/V = 128.6/7.594 = 16.93 degree/sec

F.

1. The wing desing is different for supersonic flight. Airfoils generate lift in a diffenernt
manner than for the subsonic speeds.
2. The jet engine design is different for supersonic speeds. Since the drag is higher in
supersonic flights the engines need to provide incresed fuel efficiency at supersonic speed.
3. The design of the entire structure of the plane is different for supersonic speeds. This is
because in supersonic flight the the stresses on the wings and fuselage are greater and also the
cruise temperatre of the structure is increased due to air friction.

G.

Fi = arccos(1/G) = arccos(1/3) = 70.52 degree
speed V = 110 imph = 110 *1609.34 = 177027.4 meter/h = 49.17 m/s
turn radius
r = V^2/[g*tan(fi)] = (49.17)^2/(9.81*tan(70.52)) = 87.19 m = 286 ft

total length of turn is L =2*pi*r = 309 m (=360 degree)
l = 0.858 meter/degree
V = 49.17 m/s
ROT = V/l = 49.17/0.858 = 57.29 deg/sec

6th part

From the figure 14.10 for 3 deg/sec rate of turn and V= 250 KTAS one gets
AOB = 35 degree ans

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