Aircraft performance
1. Find Climb Angle at 250 KTAS at 100 % RPM
2. Find Rate of Climb at 300 KTAS at 100 % RPM.
3. Find Max Level Airspeed at 95% RPM.
4. Find Max Range Airspeed (VBR) and Max Endurance Airspeed (VBE).
Propeller Aircraft Performance (Questions 5-8)
Refer to Figure 8.4 in Flight Theory and Aerodynamics and assume:
• $W = 20,000 lb$
• $S = 500 ft^2$
• Density Sea Level Standard Day
5. Find Max Specific Range and Max Range Airspeed (KTAS).
6. Find Max Specific Endurance and Max Endurance Airspeed (KTAS).
7. Find Max Rate of Climb and Max Rate of Climb Airspeed.
8. Find new Max Range Airspeed after fuel has burned down and W = 15,000 lb.
Landing Performance (Questions 9-14)
W = 15,000 lb CLMax = 1.5
$S = 500 ft^2$ Thrust @ Idle = 500 lb
75% of Weight on Main Tires 25 % of Weight on Nose Tire
Average Drag = 500 lb Sea level standard day
9. Find Approach Speed if V approach = 1.2 * V stall (KTAS) (For landing and takeoff performance use KTAS. At sea level standard day KEAS=KTAS).
10. Find Average Rolling Friction (lb) on Nose Tire during Landing Rollout if Rolling Coefficient of Friction = 0.2.
11. Find Average Braking Friction (lb) on Main Tires during Landing Rollout on Dry Concrete. Use Figure 13.9.
12. Find Average Deceleration (ft/s2) during landing rollout. Assume Lift on wing is zero during landing rollout. Account for residual Thrust, Rolling Friction, Drag, and Braking Friction.
13. Find Landing Distance Ground Roll-out (ft)
14. Find Landing Distance (ft) if Runway is at 5,000 ft Density Altitude.
Answers
1. Climb angle occurs at (TA-Tr)max
TAmax = 4200 lb at 100% RPM from the figure
TRmin = 830 lb
angle of climb is
$sin(gamma) = (Ta-Tr)/W = (4200-830)/12000 =0.2808$
$gamma =16.31 degree$
2. At 300 KTAS TA =4200 lb and Tr =1000 lb
rate of climb is
$ROC(fpm) = 101.3*Vk*(TA-Tr)/W = 101.3*300*(4200-1000)/12000 =$
$=8104 ft/min$
3. At 95% RPM , max level airspeed Vmax =500 KTAS
4. Max endurance airspeed V(BE) is at Dmin = Trmin, it means at V(BE) = 240 KTAS
This is the intersection of the horizontal tangent with the graphic
Max range airspeed V(BR) is where the tangent line draw from the origin intersects the graph
V(BR) = 300 KTAS
5. Max range airspeed occurs V(BR) occurs where the tangent line from the origin intersects
the graph (figure 8.4)
V(BR) = 135 KTAS
Max specific range occurs where the ratio SR=(Airspeed/Fuel flow)= (V/FF) is maximum
in other words SRmax occurs at (FF/V) minimum = (Horsepower required/V) min =[HP(R)/V] minim
This occurs where the tangent to the HP(R) from the origin intersects the graph (figure 8.9)
V = 150 KTAS, FF = 300 lb/hr
The fuel flow at
SR = (V/FF)max =150/300 = 0.5 KTAS*hr/lb
6. Max endurance airspeed V(BE) occurs at the intersection of the horizontal tangent with the graph
(figure 8.4)
V(BE) = 100 KTAS
Maximum endurance = (Fuel Flow) minim = 240 lb/hr
this is where the horizontal tangent line intersects the graph of fig 8.9
7.
$TA = 325*Pa/Vk$, Pa is the full power at sea level, Pr is the Horsepower required, Vk is the speed
$Tr = 325*Pr/Vk$
Rate of climb
$ROC = v*sin(gamma) = Vk*(TA-Tr)/W (knots) = 101.3*Vk*(TA-Tr)/W (fpm) =$
$= 33000*(Pa-Pr)/W$
Maximum rate of climb occurs where there is maximum excess power. (figure 8.8)
Vk =150 KTAS, this is maximum rate of climb airspeed
Pa =3500 HP
Pr = 600 HP
ROC(max) = 33000*(3500-600)/20000 =4785 fpm
8.There is no graph in the book for the weight of the plane W =15000 lb
9.
$CL =W/(q*S)$
$q =sigma*Vk^2/295$
$sigma =1$
$w =15000 lb$
$S =500 ft^2$
it implies Vstall =76.81 KTAS
Vapproach = 1.2*VStall = 92.17 KTAS
10.
Average rolling friction on nose tire
$Ff = mu*0.25*W =0.2*0.25*15000 = 750 lb$
11.
Maximum friction coefficient is $mu=0.7$ at 15% percent slip
Minimum friction coefficient is $mu=0.5$ at 100% percent slip
average braking friction coefficient $mu =0.6$
Average braking friction on main tires (0.75 Weight on main tires)
$Ff = mu*0.75*W =0.6*0.75*15000 = 6750 lb$
12. Total Force = W*a
Total force = Residual thrust – Rolling Friction on nose tire- Drag – Braking Friction on main tires =
=500 – 750 – 500 – 6750 = -7500 lb =-7500*0.4536 = -3402 kg =-3402*9.81 = -33373.62 N
W =15000 lb =15000*04536 = 6804 kg
$a = F/W =-33373.62/6804 = -4.905 m/(s^2) = -4.905*3.28 = -16.09 ft/(s^2)$
13. $V^2 = -2*a*s$, where V is the initial speed, a is the acceleration, s is the distance
V=Vapproach=92.17 KTAS = 92.17*1.688 = 155.58 ft/s
$a = -16.09 ft/s^2$
$s = V^2/(2a) = 752.18 ft$
14. At 5000 ft density altitude the total drag is different.
At sea level total drag is
$D(0) = Cd(0)*q*S$ where $q =sigma*(V^2)/295 =92.17^2/295 =28.80$ and
$S = 500 ft^2$
$Cd(0) = 500/28.80/500 =0.0347$
at 500 ft altitude Cd =0.02 for angle of attack =0 degree
(see figure 1.13 from Aerodynamics for naval aviators (1965) )
$D(5000) = Cd(5000)*q*S$
$D(0)/D(5000) = Cd(0)/Cd(5000) = 0.0347/0.02 = 1.736$
$D(5000) =500/1.736 =287.97 lb = 288 lb$
Total force = residual thrust – drag – rolling friction on nose tire – braking friction on main tires=
=500-288-750-6750 = -7288 lb =-7288*0.4536 = -3305.84 kg = -3305.84*9.81 = -32430.26 N
W =15000 lb = 15000*0.4536= 6804 kg
acceleration $a = F/W =-32430.26/6804 kg =- 4.766 m/s^2 = -4.766*3.28 ft/s^2 =-15.634 ft/s^2$
initial speed $V = 92.17 KTAS =155.58 ft/s$
Total distance $s = V^2/(2a) = 155.58/(2*15.634) = 774.14 ft$
