Performance Analysis of the Diamond DA-40

Michael Ditzler

8 DEC 2014

AERO 2200: Introduction to Aerospace Engineering I

Dr. Cliff Whitfield – TA Jonathan Kratz

Executive Summary To reveal as many performance characteristics the Diamond DA-40 by expanding upon a

few known values of the aircraft, a performance evaluation of the four seated general aviation

aircraft was done using MATLAB. The analysis was based on only the geometry, aerodynamics,

and propulsion characteristics given in order to practice the process of determining important

parameters of aircraft design.

In Task I, the parasitic drag coefficient (CD0) and Oswald efficiency factor of the aircraft

(e) were estimated using the drag build-up method, which consisted of calculating and adding up

the parasitic drag component of each part of the aircraft. The drag polar for the DA-40 was

plotted, with CD0 marked, and a separate plot of L/D vs. CL was also created, which helped

determine (L/D)max.. In all later tasks, values of 0.0300 and 0.75 were used for CD0 and e,

respectively.

In Task II, power required as a function of velocity was plotted for sea level, 5000 feet,

and 10000 feet. The stall speeds, speeds for minimum power required, and speed for maximum

L/D were found on the plot and reported onto Table 1. In Task III, the maximum power available

was calculated for all three altitudes using the given engine data and propeller efficiency, and

was plotted on the same graph from Task II. With power available plotted, maximum airspeed

was acquired for each altitude by examining the intersection point between the power required

curve and power available curve for each altitude.

In Task IV, climb performance was analyzed. A graph of rate of climb vs free stream

velocity was created for sea level, 5000 feet, and 10000 feet. On the graph, the maximum

climbing rates were marked, and appeared to be very close in value, ranging around 80 knots.

Those maximum climbing rates were extrapolated to make an altitude vs R/C graph, with the

absolute and service ceilings marked. The time to climb from sea level to 10,000 feet was

calculated, and a hodograph for a sea level climb at maximum weight was made. A line from the

origin of the graph a point tangent on the curve was used to determine the angle and true airspeed

for the best rate of climb.

In Task V, Breguet Range and Endurance relations were used to calculated the maximum

range and endurance for the aircraft while cruising at 10,000 ft, using 90% of its fuel. In Task VI,

a glide hodograph with horizontal and vertical velocity for the DA-40 was developed at sea level

with maximum weight. The airspeed to maximize glide distance, and time aloft were calculated

along with the time it would take for the aircraft to descend from 5000 to 1000 feet, with the

conditions that optimum flight speed is maintained.

In Task VII, the turning flight characteristics for the DA-40 were analyzed by creating a

V-n diagram, displaying the maneuver point of the aircraft, and calculating the minimum turn

radius and maximum turn rate for the aircraft at sea level and maximum weight.

Lastly, in Task VIII, takeoff and landing performance was analyzed by calculating the

ground roll distance for landing at maximum takeoff weight at sea level, maximum takeoff

weight at 5000 feet, and a weight of 2400 pounds at sea level. The landing ground roll distance a

maximum weight at sea level was also calculated.

Task I: Drag Polar

The parasitic drag coefficient and Oswald efficiency factor were estimated to be 0.0381

and 0.7348, respectively.

Figure 1: Diamond DA-40 Drag Polar

-0.5 0 0.5 1 1.5 20.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0.18

0.2

CD vs C

L

CL

CD

Drag Polar

CD,0

Figure 2: L/D vs. CL for the Diamond DA-40

-0.5 0 0.5 1 1.5 2-15

-10

-5

0

5

10

15

L/D vs. CL

CL

L/D

L/D

L/Dmax

Task II: Power Required, and Task III: Power Available

Figure 3: Horse Power (HP) vs Velocity (knots) of the DA-40 for various altitudes.

Table 1: Airspeed Data for the Diamond DA-40 at various altitudes.

Minimum Speed Speed for (L/D)max Maximum Speed

Altitude V∞ PR PA V∞ PR PA V∞ PR PA

Sea Level 59.90 kts

39.04 HP

90.89 HP

78.84 kts

44.50 HP

112.8 HP

137.9 kts

131.4 HP

131.4 HP

5,000 ft 64.53 kts

42.05 HP

82.61 HP

84.93 kts

47.94 HP

97.13 HP

133.8 kts

109.0 HP

109.0 HP

10,000 ft 69.31 kts

45.17 HP

69.72 HP

91.22 kts

51.49 HP

79.81 HP

126.0 kts

86.38 HP

86.38 HP

40 60 80 100 120 140 16020

40

60

80

100

120

140

160

180HP at SL, 5k ft, and 10k ft between 40 and 150 kts

Velocity (kts)

Hors

e P

ow

er

(HP

)

Sea Level

5000 ft

10000 ft

Stall Speed, SL

Stall Speed, 5k

Stall Speed, 10k

Vel. min PR, SL

Vel. min PR, 5k

Vel. min PR, 10k

Vel. Max Efficiency, SL

Vel. Max Efficieny, 5k

Vel. Max Efficieny, 10k

PA

, Sea Level

PA

, 5k ft

PA

, 10k ft

Task IV: Climb Performance

Figure 4: Rate of Climb vs. Velocity for the Diamond DA-40

40 60 80 100 120 140 160-600

-400

-200

0

200

400

600

800

1000Rate of Climb vs. Velocity

Velocity (knots)

Rate

of

clim

b (

ft/m

in)

Sea Level

5000 ft

10000 ft

Max R/C at Sea Level

Max R/C at 5k ft

Max R/C at 10k ft

Figure 5: Rate of Climb vs Altitude for the Diamond DA-40

0 100 200 300 400 500 600 700 800 9000

2000

4000

6000

8000

10000

12000

14000

16000

18000Rate of Climb (ft/min) vs Altitude (ft)

Altitude (

ft)

Rate of Climb (ft/min)

Altitude vs. Rate of Climb

Rate of Climb Data Points

Absolute Ceiling

Service Ceiling

Figure 6: Hodograph of Climb Performance for the Diamond DA-40

Table 2: Best Rate of Climb and Climb Angle Data for the Diamond DA-40

Rate of Climb (ft/min) Climb Angle (degrees) Velocity (knots)

Best Rate of Climb 853.4 5.968 81.19

Best Climb Angle Condition

768.5 6.571 66.32

Task V: Range and Endurance

The maximum range of the aircraft was found to be 803.1451 nautical miles, and the

maximum endurance was found to be 10.3106 hours.

0 2000 4000 6000 8000 10000 120000

100

200

300

400

500

600

700

800

900

X: 6672

Y: 768.5

Vertical vs Horizontal Velocity

Horizonal Velocity (ft/min)

Vert

ical V

elo

city (

ft/m

in)

Task VI: Gliding Flight

Figure 7: Hodograph of Gliding Flight for the Diamond DA-40

To maximize glide distance, and time aloft, the aircraft should travel at 81.73 knots. It

would take the aircraft 0.1277 hours to descend from 4000 feet to 1000 feet at this airspeed.

50 100 150 200 250 300 350 400-140

-120

-100

-80

-60

-40

-20

0

X: 137.5

Y: -11.08

Vv vs. V

h in Gliding Flight

Horizontal Velocity Vh (ft/sec)

Vert

ical V

elo

city V

v (

ft/s

ec)

Task VII: Turning Flight

Figure 8: V-n Diagram for turning flight for the Diamond DA-40

At maximum weight and in sea level conditions, the minimum turn radius for the

Diamond DA-40 is 249.67 feet and the maximum turn rate is 40.11 degrees per second.

Task VIII: Takeoff and Landing Performance

At maximum weight and sea level conditions, the ground roll distance for takeoff was

calculated to be 1910 feet. At maximum weight and 5000 feet, the ground roll distance for

takeoff was calculated to be 2403 feet. At a weight of 2400 pounds and sea level conditions, the

ground roll distance for takeoff was calculated to be 1481 feet. At maximum weight and sea

level conditions, the ground roll distance for landing was calculated to be 1524 feet.

0 50 100 150 200-5

-4

-3

-2

-1

0

1

2

3

4

5V-n Diagram

V

(knots)

Load F

acto

r (n

)

n

Manuever Point

Maximum n limit

Minimum n limit

Maximum Airspeed

Appendix

MATLAB Code:

clc clear

%Aero 2200 Final %Michael Ditzler.3 %TA - Jonathan Kratz, 8:00 AM Monday %Constants: Wmax = 2645; %maximum weight, lbs S = 145.7; %ft2, wing area b = 39.17; %ft, wing span AR=(b^2)/S; CLmax = 1.90; %Maximum lift, clean

MaxHPshaft_SL = 180; %hp, maximum shaft hp for sea level MaxHPshaft_5k = 150; %hp, maximuim hp for 5000 ft MaxHPshaft_10k = 120; %hp, maximum hp for 10000 ft

rhoSL = .0023769; %slugs/ft3 rho5k = .0020482; rho10k = .0017756;

P_SL = 2116.2; %(lbs/ft2) P_5k = 1760.9 ; P_10k = 1455.6 ;

T_SL = 518.69; %degrees R T_5k = 500.86 ; T_10k = 483.04;

% Task I: Drag Polar % Estimate the parasitic drag coefficient (CD0) and Oswald efficiency factor

(e) for the aircraft, % using the drag build-up method. Tabulate the component drag coefficients.

Plot the drag polar % as CD vs. CL and clearly mark CD0 on the polar. Create a separate plot of

L/D vs. CL and % determine (L/D)max.

% Drag build-up method % Most areas and all C_Dpi values are estimated based on the class example % and the picture of the aircraft in the document. S_wing = S; %area of wing C_Dpi_wing = 0.005; %partial coefficient for the wing

S_horiz_tail = 22; %horizontal component of tail wing C_Dpi_horiz_tail = 0.008;

S_vert_tail = 0.05; %vertical component of tail wing C_Dpi_vert_tail = 0.008;

S_fuselage = 15; %fuselage C_Dpi_fuselage = 0.05;

S_landing_struts = 0.8; %landing struts, 3 of them C_Dpi_landing_struts = 1;

S_wheels = 1.5; %wheels S_nosewheel = 1; C_Dpi_wheels = 0.6;

Sum_Fpi =

(S_wing*C_Dpi_wing)+(S_horiz_tail*C_Dpi_horiz_tail)+(S_vert_tail*C_Dpi_vert_t

ail)+(S_fuselage*C_Dpi_fuselage)+(S_landing_struts*C_Dpi_landing_struts*3)+((

S_wheels+S_nosewheel)*C_Dpi_wheels);

CD0 = Sum_Fpi/S_wing

%Oswald efficiency factor (e)

% K and delta values estimated from Day 29 Drag Build-up and % Oswald Efficiency Factor Estimation slide show. K = 0.01; delta = .03;

e = 1/(K*pi()*AR+1+delta)

CLs = linspace(-0.5,1.90,25);

CD = CD0 + CLs.^2./(pi()*AR*e); %tabulated CD values

figure(1) plot(CLs,CD,'b',0,CD0,'k^') title('C_D vs C_L') xlabel('C_L') ylabel('C_D') legend('Drag Polar','C_{D,0}')

LD = CLs./CD; %Lift over Drag

[LDmax,LDmaxi]=max(LD) %finds maximum lift over drag

figure(2) plot(CLs,LD,'b',CLs(LDmaxi),LDmax,'k^'); title('L/D vs. C_L') xlabel('C_L') ylabel('L/D') legend('L/D','L/D_{max}','Location','SouthEast')

% For the following tasks, use CD0 = 0.0300 and e = 0.75. CD0 = 0.0300; %profile drag coefficient

e = 0.75; %oswald efficiency factor

% Task II: Power Required % Determine the Power required as a function of velocity for flight at three

altitudes: sea level, % 5000 ft, and 10000 ft. Present the data on a single graph.

Vinf_kts = linspace(40,150,1000); %Velocity in knots Vinf = Vinf_kts*((5280*1.15078)/(3600)); %converting knots to ft/sec

P_req_SL =

(CD0*0.5*rhoSL*(Vinf.^3)*S)+(Wmax^2)./(0.5*rhoSL*(Vinf)*S*pi()*AR*e); %power

required at sea level HP_req_SL = P_req_SL./550; %divided by 550 to convert to horse power

P_req_5k =

(CD0*0.5*rho5k*(Vinf.^3)*S)+(Wmax^2)./(0.5*rho5k*(Vinf)*S*pi()*AR*e) ;%power

required at 5k ft HP_req_5k = P_req_5k./550 ;%divided by 550 to convert to horse power

P_req_10k =

(CD0*0.5*rho10k*(Vinf.^3)*S)+(Wmax^2)./(0.5*rho10k*(Vinf)*S*pi()*AR*e);

%power required at 10k ft HP_req_10k = P_req_10k./550; %divided by 550 to convert to horse power

% Locate the stall speed, speed for minimum PR, and speed for maximum lift-

to-drag ratio on this graph for each altitude. The graph % should cover the speed range from 40 to 150 knots. Units of velocity must

be knots, and units % of power must be horsepower

%Stall Speed stallSL=sqrt((2*Wmax)/(rhoSL*S*CLmax)); stallSL_kts = stallSL*.5925; HP_stall_SL =

((CD0*0.5*rhoSL*(stallSL^3)*S)+(Wmax^2)/(0.5*rhoSL*(stallSL)*S*pi()*AR*e))/55

0

stall5k=sqrt((2*Wmax)/(rho5k*S*CLmax)); stall5k_kts = stall5k*.5925; HP_stall_5k =

((CD0*0.5*rho5k*(stall5k^3)*S)+(Wmax^2)/(0.5*rho5k*(stall5k)*S*pi()*AR*e))/55

0;

stall10k=sqrt((2*Wmax)/(rho10k*S*CLmax)); stall10k_kts = stall10k*.5925; HP_stall_10k =

((CD0*0.5*rho10k*(stall10k^3)*S)+(Wmax^2)/(0.5*rho10k*(stall10k)*S*pi()*AR*e)

)/550;

%Speed for minimum Power Required

V_minPR_SL = ((4*Wmax^2)/(3*rhoSL^2*S^2*pi()*AR*e*CD0))^0.25; V_minPR_SL_kts = V_minPR_SL*.5925; HP_minPR_SL =

((CD0*0.5*rhoSL*(V_minPR_SL^3)*S)+(Wmax^2)/(0.5*rhoSL*(V_minPR_SL)*S*pi()*AR*

e))/550;

V_minPR_5k = ((4*Wmax^2)/(3*rho5k^2*S^2*pi()*AR*e*CD0))^0.25; V_minPR_5k_kts = V_minPR_5k*.5925; HP_minPR_5k =

((CD0*0.5*rho5k*(V_minPR_5k^3)*S)+(Wmax^2)/(0.5*rho5k*(V_minPR_5k)*S*pi()*AR*

e))/550;

V_minPR_10k = ((4*Wmax^2)/(3*rho10k^2*S^2*pi()*AR*e*CD0))^0.25; V_minPR_10k_kts = V_minPR_10k*.5925; HP_minPR_10k =

((CD0*0.5*rho10k*(V_minPR_10k^3)*S)+(Wmax^2)/(0.5*rho10k*(V_minPR_10k)*S*pi()

*AR*e))/550;

%Speed for Max L/D %CL for max L/D is when CD0 equals CDi CL_maxeff = sqrt(CD0*pi()*e*AR);

V_maxeff_SL = sqrt((2*Wmax)/(rhoSL*S*CL_maxeff)); V_maxeff_SL_kts = sqrt((2*Wmax)/(rhoSL*S*CL_maxeff))*.5925; HP_maxeff_SL =

((CD0*0.5*rhoSL*(V_maxeff_SL^3)*S)+(Wmax^2)/(0.5*rhoSL*(V_maxeff_SL)*S*pi()*A

R*e))/550;

V_maxeff_5k = sqrt((2*Wmax)/(rho5k*S*CL_maxeff)); V_maxeff_5k_kts = sqrt((2*Wmax)/(rho5k*S*CL_maxeff))*.5925; HP_maxeff_5k =

((CD0*0.5*rho5k*(V_maxeff_5k^3)*S)+(Wmax^2)/(0.5*rho5k*(V_maxeff_5k)*S*pi()*A

R*e))/550;

V_maxeff_10k = sqrt((2*Wmax)/(rho10k*S*CL_maxeff)); V_maxeff_10k_kts = sqrt((2*Wmax)/(rho10k*S*CL_maxeff))*.5925; HP_maxeff_10k =

((CD0*0.5*rho10k*(V_maxeff_10k^3)*S)+(Wmax^2)/(0.5*rho10k*(V_maxeff_10k)*S*pi

()*AR*e))/550;

figure(3) plot(Vinf_kts,HP_req_SL,'b',Vinf_kts,HP_req_5k,'g',Vinf_kts,HP_req_10k,'r',st

allSL_kts,HP_stall_SL,'bo',stall5k_kts,HP_stall_5k,'go',stall10k_kts,HP_stall

_10k,'ro',V_minPR_SL_kts,HP_minPR_SL,'bx',V_minPR_5k_kts,HP_minPR_5k,'gx',V_m

inPR_10k_kts,HP_minPR_10k,'rx',V_maxeff_SL_kts,HP_maxeff_SL,'bd',V_maxeff_5k_

kts,HP_maxeff_5k,'gd',V_maxeff_10k_kts,HP_maxeff_10k,'rd') title('HPat SL, 5k ft, and 10k ft between 40 and 150 kts') xlabel('Velocity (kts)') ylabel('Horse Power (HP)')

hold on

% Task III: Power Available % Using the engine data provided, along with the propeller efficiency data,

determine the % maximum power available at the three altitudes. (This will be a “thrust

horsepower”, or the % power imparted to the air by the propeller). On the same graph as the one

developed in Task II, % plot PA as a function of velocity. In your data report include a table

replicating the table below, % with all values filled in. Units of velocity must be knots, and units of

power must be horsepower.

prop_eff = 0.78*(1-(35./Vinf_kts).^2);

HPa_SL = prop_eff.*MaxHPshaft_SL; HPa_5k = prop_eff.*MaxHPshaft_5k; HPa_10k = prop_eff.*MaxHPshaft_10k;

plot(Vinf_kts,HPa_SL,'b--',Vinf_kts,HPa_5k,'g--',Vinf_kts,HPa_10k,'r--') legend('Sea Level','5000 ft','10000 ft','Stall Speed, SL','Stall Speed,

5k','Stall Speed, 10k','Vel. min P_R, SL','Vel. min P_R, 5k','Vel. min P_R,

10k','Vel. Max Efficiency, SL','Vel. Max Efficieny, 5k','Vel. Max Efficieny,

10k','P_A, Sea Level','P_A, 5k ft','P_A, 10k ft','Location','BestOutside')

hold off

% Task IV: Climb Performance % a) Prepare a graph of rate of climb (ft/min) vs. V? (knots) for the three

altitudes. Note the % maximum rate of climb and the true airspeed (V?) that corresponds to the

maximum rate of % climb.

Pa_SL = HPa_SL*550; %converts HPa to Pa Pa_5k = HPa_5k*550; Pa_10k = HPa_10k*550;

excessP_SL = Pa_SL - P_req_SL; %Excess Power excessP_5k = Pa_5k - P_req_5k; excessP_10k = Pa_10k - P_req_10k;

%Rate of Climb RateClimb_SL = (excessP_SL/Wmax)*(60); %550/60 converts back to ft/min RateClimb_5k = excessP_5k/Wmax*(60); RateClimb_10k = excessP_10k/Wmax*(60);

[maxRC_SL,maxRC_SL_i] = max(RateClimb_SL); %Maximum Rate of Climb values maxRC_SL_Velocity = Vinf_kts(maxRC_SL_i);

[maxRC_5k,maxRC_5k_i] = max(RateClimb_5k);

maxRC_5k_Velocity = Vinf_kts(maxRC_5k_i);

[maxRC_10k,maxRC_10k_i] = max(RateClimb_10k); maxRC_10k_Velocity = Vinf_kts(maxRC_10k_i);

figure(4) plot(Vinf_kts,RateClimb_SL,'b',Vinf_kts,RateClimb_5k,'g',Vinf_kts,RateClimb_1

0k,'r') title('Rate of Climb vs. Velocity') xlabel('Velocity (knots)') ylabel('Rate of climb (ft/min)')

hold on

plot(maxRC_SL_Velocity,maxRC_SL,'bo',maxRC_5k_Velocity,maxRC_5k,'go',maxRC_10

k_Velocity,maxRC_10k,'ro') legend('Sea Level','5000 ft','10000 ft','Max R/C at Sea Level','Max R/C at 5k

ft','Max R/C at 10k ft','Location','BestOutside')

hold off

% b) Plot the results of (a) on an altitude vs. R/C graph and extrapolate the

results to estimate % the absolute and service ceilings.

RC_maxes = [maxRC_SL maxRC_5k maxRC_10k]; %vector of max rates of climb alts = [0 5000 10000]; %vector of the three altitudes

RC_coeff = polyfit(RC_maxes,alts,1); %making a theoretical curve for Altitude

vs Rate of Climb alt_curve = RC_coeff(1)*RC_maxes + RC_coeff(2); alt_curve(length(alt_curve)+1) = RC_coeff(2); RC_curve = RC_maxes; RC_curve(length(alt_curve)) = 0;

figure(5) plot(RC_curve,alt_curve,'k-'); title('Rate of Climb (ft/min) vs Altitude (ft)') ylabel('Altitude (ft)') xlabel('Rate of Climb (ft/min)')

hold on

plot(RC_maxes,alts,'bo'); axis([0 900 0 18000]);

%service ceiling is accepted to be 100ft/min, absolute ceiling is 0 ft/min abs_ceil = RC_coeff(2); serv_ceil = RC_coeff(1)*100+RC_coeff(2);

plot(0,abs_ceil,'r^',100,serv_ceil,'k^')

legend('Altitude vs. Rate of Climb','Rate of Climb Data Points','Absolute

Ceiling','Service Ceiling')

hold off

% c) Use the results from (b) to determine the time to climb from sea level

to an altitude of % 10,000 ft.

%we have to find the area under the curve from 0 to 10000 ft

alt_wrt_time = 0:1:18000; %altitude with respect to time RC_wrt_time = (alt_wrt_time-RC_coeff(2))/RC_coeff(1); %Rate of climb with

respect to time

alt_low = 1; alt_high = 10000;

climbtime_min = 0;

for n = alt_low:1:alt_high climbtime_min = (RC_wrt_time(n))^(-1) + climbtime_min; end climbtime = climbtime_min*60;

% d) Develop a climb hodograph for sea level climb at maximum takeoff weight

to determine the % true airspeeds (V?) for best angle of climb and best rate of climb.

Complete the table % below.

V_vert_ftmin = RateClimb_SL(23:888); V_vert_fts = V_vert_ftmin./60 V_horiz_fts = (Vinf(23:888).^2-V_vert_fts.^2).^0.5; V_horiz_ftmin = V_horiz_fts*60;

figure(6) plot(V_horiz_ftmin,V_vert_ftmin); axis([0 12000 0 900]); title('Vertical vs Horizontal Velocity') xlabel('Horizonal Velocity (ft/min)') ylabel('Vertical Velocity (ft/min)')

% Task V: Range and Endurance % Use the Breguet Range and Endurance relations to determine the maximum

range and % endurance for the DA-40 while cruising at 10,000 ft and using 90% of the

fuel on board. For this % cruise flight, consider the specific fuel consumption and propeller

efficiency to be constant: % SFC = 0.49 (lb/hr)/shp and ? = 0.78

% Note the airspeed(s) that the pilot must maintain during these flights in

order to maintain the % optimum conditions assumed in the Breguet equations. Range values should be

reported in % nautical miles and endurance values should be reported in decimal hours.

FuelVolume = 50; %gallons, usable fuel volume rho100LL = 6; %lbs/gal CruiseSFC = 0.49; %(lb/hr)/shp c = CruiseSFC/(550*3600) %converts SFC to consistent units n = 0.78;

%At cruising, CD0 = CDi, CL_maxeff will be used for Range

CD_maxeff = CD0*2 %

W0 = Wmax; W1 = Wmax - 0.9*FuelVolume*rho100LL;

R = (n/c)*(CL_maxeff/CD_maxeff)*log(W0/W1); %feet R_nm = R/6076 %converts to nautical miles

CL_PRmin = (3*CD0*pi()*e*AR)^0.5; CD_PRmin = 4*CD0;

E = (n/c)*(CL_PRmin^(3/2)/CD_PRmin)*((2*rho10k*S)^0.5)*((W1^-.5)-(W0^-.5))

%in minutes Ehr = E/3600

% Task VI: Gliding Flight % a) Develop a glide hodograph, with VH and VV in ft/s for the DA-40 at

maximum takeoff weight % at sea level

glide_angle = atand(1./(CLs(6:25)./CD(6:25))); V_des = (2*cosd(glide_angle).*Wmax./(S*rhoSL*CLs(6:25))).^.5

V_horiz_glide = V_des.*cosd(glide_angle);

V_vert_glide = -V_des.*sind(glide_angle); V_vert_min = max(V_vert_glide);

figure(7) plot(V_horiz_glide,V_vert_glide,'b') title('V_v vs. V_h in Gliding Flight') xlabel('Horizontal Velocity V_h (ft/sec)') ylabel('Vertical Velocity V_v (ft/sec)')

% b) Assuming engine failure at a pressure altitude of 5,000 ft at maximum

takeoff weight, % determine the maximum distance (in nautical mi) that the aircraft can glide

to reach an

% airfield situated at a pressure altitude of 1,000 ft. What indicated

airspeed (in knots) % should the pilot fly in order to maximize glide distance? If the pilot

instead wants to remain % over a certain location during the descent, what airspeed should be used % in order to maximize time aloft? How long would it take for the aircraft to

descend from 5,000 ft to % 1,000 ft if the flight speed is maintained at the optimum for maximum time

aloft?

%from the hodograph: V_vert_maxeff = -11.08; V_horiz_maxeff = 137.5; V_inf_maxeff = (V_vert_maxeff^2+V_horiz_maxeff^2)^.5; V_inf_maxeff_kts = V_inf_maxeff*3600/6076;

Rmax = 4000*(LDmax); %4000 ft differential times maximum L/D Rmax_nm = Rmax/6076

time_gliding = -4000/V_vert_min; time_gliding_hr = time_gliding/3600

% Task VII: Turning Flight % Create a V-n diagram for the DA-40 if the “never exceed” airspeed is 178

knots, the maximum % positive load factor is 3.8, and the maximum negative load factor is -1.52.

Clearly define the % maneuver point on the diagram. What is the minimum turn radius and the

maximum turn rate % for the DA-40 at maximum takeoff weight at standard sea level conditions?

max_pos_LF = 3.8; %maximum load factor max_neg_LF = -1.52; %maximum negative load factor max_speed = 178; %max air speed

pos_LF = .5*rhoSL*Vinf.^2*S*CLmax/Wmax; %positive load factor neg_LF = -pos_LF; %negative load factor

Vx = ((2*max_pos_LF*Wmax)/(rhoSL*CLmax*S))^.5; %corner velocity Vx_kts = Vx*(3600/6076);

figure(8) plot(Vinf_kts,pos_LF,'b',Vinf_kts,neg_LF,'b',Vx_kts,0,'k^') title('V-n Diagram') xlabel('V_\infty (knots)') ylabel('Load Factor (n)')

hold on

plot(([0 180]), ([max_pos_LF max_pos_LF]),'b--') %graphs positive limit plot(([0 180]), ([max_neg_LF max_neg_LF]),'b-.') %graphs negative limit plot(([max_speed max_speed]), ([-5 100]),'r--')

axis([0 200 -5 5])

legend('n','','Manuever Point','Maximum n limit','Minimum n limit','Maximum

Airspeed','location','BestOutside') hold off

g = 32.2; %gravity constant min_turn_r = (2*Wmax)/(rhoSL*CLmax*g*S) max_turn_rate = (g*180/pi())*((max_pos_LF*rhoSL*CLmax*S)/(2*Wmax))^.5

% Task VIII: Takeoff and Landing Performance % Estimate the following for the DA-40:

% Given values: CL_takeoff = 0.5; %Coefficient of lift for takeoff CL_landing = 1.0; %coefficient of lift for landing u_TO = 0.02; %coefficient of friction for dry concrete u_L = 0.25; %coefficient of friction for dry concrete with brakes on h = 3.5; %height of wings, measured in the picture phi = ((16*h/b)^2)/((1+(16*h/b)^2)); n=0.78;

% a) Takeoff ground roll distance at maximum takeoff weight at standard sea

level conditions. V_takeoff_SL_Wmax = 1.2*(2*Wmax/(S*rhoSL*CLmax))^.5; %1.2 times stall speed

L_takeoff_SL_Wmax = 0.5*rhoSL*(0.7*V_takeoff_SL_Wmax)^2*S*CL_takeoff; D_takeoff_SL_Wmax =

.5*rhoSL*(0.7*V_takeoff_SL_Wmax)^2*S*(CD0+(phi*CL_takeoff^2)/(pi()*e*AR));

T_takeoff_SL_Wmax = (n*MaxHPshaft_SL*550)/(0.7*V_takeoff_SL_Wmax); s_liftoff_SL_Wmax = (1.44*Wmax^2)/(g*rhoSL*S*CL_takeoff*(T_takeoff_SL_Wmax-

(D_takeoff_SL_Wmax+u_TO*(Wmax-L_takeoff_SL_Wmax))))

% b) Takeoff ground roll distance at maximum takeoff weight at a high

altitude airport (5000 ft).

V_takeoff_5k_Wmax = 1.2*(2*Wmax/(S*rho5k*CLmax))^.5; %1.2 times stall speed

L_takeoff_5k_Wmax = 0.5*rho5k*(0.7*V_takeoff_5k_Wmax)^2*S*CL_takeoff; D_takeoff_5k_Wmax =

.5*rho5k*(0.7*V_takeoff_5k_Wmax)^2*S*(CD0+(phi*CL_takeoff^2)/(pi()*e*AR));

T_takeoff_5k_Wmax = (n*MaxHPshaft_SL*550)/(0.7*V_takeoff_5k_Wmax); s_liftoff_5k_Wmax = (1.44*Wmax^2)/(g*rho5k*S*CL_takeoff*(T_takeoff_5k_Wmax-

(D_takeoff_5k_Wmax+u_TO*(Wmax-L_takeoff_5k_Wmax))))

% c) Takeoff ground roll distance at a takeoff weight of 2400 lbs at standard

sea level % conditions.

W = 2400; %new weight

V_takeoff_SL_W = 1.2*(2*W/(S*rhoSL*CLmax))^.5 %1.2 times stall speed

L_takeoff_SL_W = 0.5*rhoSL*(0.7*V_takeoff_SL_W)^2*S*CL_takeoff; D_takeoff_SL_W =

.5*rhoSL*(0.7*V_takeoff_SL_W)^2*S*(CD0+(phi*CL_takeoff^2)/(pi()*e*AR));

T_takeoff_SL_W = (n*MaxHPshaft_SL*550)/(0.7*V_takeoff_SL_W); s_takeoff_SL_W = (1.44*W^2)/(g*rhoSL*S*CL_takeoff*(T_takeoff_SL_W-

(D_takeoff_SL_W+u_TO*(W-L_takeoff_SL_W))))

% d) Landing ground roll distance at maximum weight and standard sea level

conditions.

V_landing_SL = 1.3*((2*Wmax)/(rhoSL*S*CLmax))^.5;

D_landing_SL = .5*rhoSL*(0.7*V_landing_SL)^2*S*CD0;

s_landing = (1.69*Wmax^2)/(g*rhoSL*S*CL_landing*(D_landing_SL+u_L*Wmax))