# Slow speed flight…

Conditions:

W=10,000 lb    weight of aircraft
S = 100 ft 2      wings surface
CLMAX = 1.5   CL for stall speed
Aircraft Certified Aerobatic Category (-3, +6) G

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

$Vstall= \sqrt{295*G*W/(CLmax*\sigma*S}$

$G = 1$, $\sigma = 1$, $W =10000 lb$, $S = 100 ft^2$, $Clmax =1.5$

$Vstall = \sqrt(295*1*10000/(1.5*100)) =140.24 KEAS$

2. Find Stall Speed (KEAS), level flight 70 deg AOB (2.9 G).

$G = 2.9$

$Vs =\sqrt(295*2.9*10000/(1.5*100)) =238.82 KEAS$

or simpler $Vs2 = Vs1*\sqrt{G} =140.24*\sqrt{2.9} =238.82 KEAS$

3. Find Positive Limit Load and Positive Ultimate Limit Load of Given Aerobatic Aircraft-Max Gross Weight = 10,000 lb.

Positive limit load $PL = 6GW = 6*10000 =60000 lb =60000*0.4536*9.81 N =266989 N$

Ultimate load $UL = 1.5*PL =1.5*60000 = 90000 lb = 400483.5 N$

4. Find Maneuvering speed (Va) of given aerobatic aircraft W=10,000 lb.

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).

$Va =Vs1*\sqrt{G} = 140.24*\sqrt{7} =371 KIAS$

if we take the positive limit load as 6 G (the lowest possible value) the maneuvering speed is

$Va= 140.24*\sqrt{6} =343.51 KIAS$

(I hope I am not mistaken this answer since the pages 224,225 of the book Flight theory and aerodynamics are hidden for me.)

5. Find Turn Radius and Rate of Turn of aerobatic aircraft performing level sustained 4 G turn at 100 KTAS.

$\Phi = \arccos(1/G) = \arccos(1/4) = 75.52 degree$

$r = V^2/[g*\tan(\phi)]$

$V = 100 KTAS = 100*0.5144 m/s =51.44 m/s$

$g = 9.81 m/s^2$

$r = (51.44)^2/(9.81*\tan(75.52)) =69.66 m =69.66*3.28 ft = 228.47 ft$

rate of turn $ROT = 1.091*\tan(\phi)/Vk =1.091*\tan(75.52)/100 =42.2 deg/sec$

another way of computing this

total length of turn $L = 2*\pi*r =2*\pi*69.66 =437.7 m (= 360 degree)$

it means $l = 1.216 meter/degree$

$V = 100 KTAS = 51.44 m/sec$

$ROT = l/V = 51.44/1.216 = 42.3 degree/sec$

6. Using Figure 14.10 from Flight Theory and Aerodynamics, find AOB to fly a standard rate turn (3 deg/sec) at 200 KTAS.

From the figure 14.10 for 3 deg/sec rate of turn and V= 200 KTAS one gets

AOB = 30 degree

7. Name four different design features for supersonic aircraft.

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.

8. What is rate of turn and turn radius of Pitts doing 3g turn at 110 imph (indicated statute miles per hour)? Compare your mathematical answer of turn rate to that of the aircraft aviation video link (Time the 3 g turn of 360 deg and then divide by 360 to get deg/sec)?

$\Phi = \arccos(1/G) = \arccos(1/3) = 70.52 degree$

speed $V = 110 imph = 110 *1609.34 = 177027.4 meter/h = 49.17 m/s$

$r = V^2/[g*tan(\phi)] = (49.17)^2/(9.81*\tan(70.52)) = 87.19 m = 286 ft$
total length of turn is $L =2*\pi*r = 309 m (=360 degree)$
$l = 0.858 meter/degree$
$V = 49.17 m/s$
$ROT = V/l = 49.17/0.858 = 57.29 deg/sec$