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PERFORMANCE ANALYSIS OF THE DIAMOND DA-40

Submitted To:

Dr.C A Whitfield

Created By:

Tejaswini Gautham

AERO 2200: Introduction to Aeronautical Engineering 1 The Ohio State University

Columbus, Ohio December 8th, 2014

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EXECUTIVE SUMMARY In the analysis of the diamond DA-40 aircraft, some important results were obtained are have been summarized here. The drag polar was estimated in Task 1, to be 0.03 with the help of the drag build up method. The Oswald efficiency factor e, was estimated in Task 1 to be 0.73 using the same drag build up method. The maximum lift to drag ratio was found to be 14.40. The power analysis and rate of climb analysis

results have been tabulated in the table provided under task 3 and 4 respectively. Some note-worthy

results are, the power available at maximum velocity at sea level, 5000 ft and 10,000 ft are 131hp,

109hp, and 86hp respectively. The absolute and service ceiling values can be verified by the graph and

the program. They are 17448.12 ft and 15403.52 ft respectively. The range and endurance of the aircraft

were found to be 803 nautical miles and 10.25 hours. The takeoff ground roll distance at maximum

weight at sea level was found to be 3272 ft and the landing ground distance at maximum weight at sea

level was 621.35 ft

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TABLE OF CONTENTS Introduction…………………………………………………………………………………………………………………………………………….4 Nomenclature…………………………………………………………………………………………………………………………………………5 Task 1: Drag Polar……………………………………………………………………………………………………………………………………6 Task 2: Power Required…………………………………………………………………………………………………………………………12 Task 3: Power Available…………………………………………………………………………………………………………………………15 Task 4: Climb Performance……………………………………………………………………………………………………………………16 Task 5: Range and Endurance……………………………………………………………………………………………………………….20 Task 6: Gliding Flight……………………………………………………………………………………………………………………………..21 Task 7: Turning Flight……………………………………………………………………………………………………………………………23 Task 8: Take-off and Landing Performance……………………………………………………………………………………………24 Conclusion……………………………………………………………………………………………………………………………………………27 References……………………………………………………………………………………………………………………………………………28 Appendix………………………………………………………………………………………………………………………………………………29

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INTRODUCTION

This paper is about the performance analysis of the DIAMOND DA40 aircraft. It is done on a geometric basis, with the help of the given information. It is divided into various sections, based on the handout given. In each section, there are equations, tables and graphs, to support the results that were found. In the appendix, the MATLAB code, used for obtaining the graphs and results is found.

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NOMENCLATURE S- Wing Area b- Wing span AR- Aspect Ratio Cd0- Coefficient of Drag Polar Cl- Coefficient of lift Cd- Coefficient of Drag L/D- Lift to Drag Ratio Cf- Coefficient of Friction S(wet)- Surface Area that can be wetted R- Radius A- Area H- Height e- Oswald Efficiency Factor pi= 22/7 W- Weight ρ(SL)- Density At sea level ρ(5000)- Density at 5000 ft ρ(10,000)- Density at 10,000 ft η- Efficiency V- Velocity V(stall)- Stall Velocity R/C- Rate of Climb r- turn radius ω- turn rate g- acceleration due to gravity V(TD)- Velocity at touch-down V(TO)- Velocity at take-off T-time E-Endurance R-Range S(G)- take-off ground distance roll S(L)- Landing ground distance roll F(avg)- Average Force

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TASK 1: DRAG POLAR

To estimate the drag polar by the drag build up method we need the following values: Cf (0.009), the wing span (5 and 1/16 in) and the wetted area. In this drag build up method, we multiply the skin friction by the entire wetted surface of the aircraft and then divide it by the surface of the wings, to normalize it. The coefficient of skin friction is estimated after comparing the values of skin friction of planes similar to DA 40 from the graph below.

The wetted surface, wing span and wing area are values derived from the given figure of the aircraft. The following scale was used: Scale: 39 and 1/6 ft (actual) = 5 and 1/16 in (observed from the diagram) For the body of the aircraft, the nose can be seen as a cylinder, the main body as a cylinder and the back as a cone For the nose: A= 2(pi)*h*r R=0.25 in

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H= 0.5 in Scaled-up values: R=1.93 ft H=3.87 ft A= 47 ft^2 For the main body: R= 0.25 in H= 1 in Scaled-up values: R= 1.93 ft H= 7.74 ft A=94 ft^2 The back of the fuselage: R= 0.1 in H= 1 in Scaled-up values: R=0.78 ft H=7.74 ft A= (pi)*R*H= 18 ft^2 The wing span was estimated by direct observation, the wing area was calculated by considering the two wings as trapezoids and using the area formula for the trapezoid. (0.5*(parallel side 1 + parallel side 2 )* distance between the two sides) A= (side1+side2)H/2 side 1= 0.4 in side 2= 0.7 in distance between sides : 2.25 in Scaled-up Values: Side 1= 3.09 ft Side 2= 5.42 ft Distance between sides: 17.41 ft Plugging into the formula, and multiplying it by 4, to include the upper and lower surface of both wings, we get

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A= 296 ft^2 Similar calculations are carried out for the rudder and horizontal stabilizer to calculate the total surface area The rudder and the horizontal stabilizer are also consider trapezoids. Rudder: Side 1: 0.5 in Side 2: 0.4 in Distance between the sides: 0.5 in Scaled-up values: Side 1: 3.87 ft Side 2: 3.09 ft Distance between the sides: 3.87 ft Plugging the values into the formula and multiplying by 2, to include all side, we get: Area: 21.54 ft^2 Horizontal Stabilizer: Side 1: 0.2 in Side 2: 0.3 in Distance between the two sides: 0.7 in Scaled-up values: Side 1: 1.55 ft Side 2: 2.32 ft Distance between the two sides: 5.42 ft Plugging into the formula and multiplying by 4, to include both surfaces, we get, A= 41.92 ft^2 S(wet) = 18+ 94+ 47+ 41.92+21.54+296 = 518.46 ft^2 S= area of the wing S= 145.7 ft^2 ( given ) Now plugging in all these values, we can estimate drag polar coefficient: Cd0=Cf*S(wet)/ S = 0.031 To estimate Oswald efficiency factor: e= 1/(k’’*AR*(pi)+1+δ) Taper Ratio= 0.62 AR= b^2/S b= 39 and 1/6 ft s= 145.7 ft^2

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AR= 10.53

Using the above graph, δ= 0.03 k’’= 0.01 Using e= 1/(k’’*AR*(pi)+1+δ) We get, e=0.73

To graph Cd vs Cl Cd= Cd0+ (Cl^2 / (pi)*AR*e) = 0.031+0.0403 Cl^2

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(L/D)max = (Cl/Cd)max Cl=(Cd0*pi*AR*e)^1/2 Cd=Cd0+Cl^2/(pi*AR*e) Plugging in the values for the above equations, we get, Cl=(0.03*pi*10.53*0.73)^1/2 =0.85 Cd=0.059 (L/D)max= 14.40

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TASK 2: POWER REQUIRED Given: Cd0= 0.0300 e=0.75 AR=10.53 S=145.7 ft^2 W=2645 lb At sea level, ρ=0.0023769 At 5000 ft, ρ=0.0020482 At 10,000 ft, ρ=0.0017556 To find the Power Required, we need an equation that gives power required only in terms of the density and the values we have.

PR=( 1

2ρSCd0𝑉3+

𝑊^21

2𝑝𝑖𝐴𝑅𝑒𝑉𝜌

)/ 550 <we divide by 550 to get the power required in horsepower>

To find the stall speed, Given: ClMax=1.9 V(stall)=(2W/SρClMax)^1/2 At sea level, V(stall)= 89.66 fps = 89.66 ft/s = 53.122 knots At 5000 ft, V(stall)= 96.59 ft/s = 57.228 knots At 10,000 ft, V(stall)= 104.33 ft/s = 61.81 knots Velocity at Minimum Power Required= ((4W^2)/(3*ρ^2*S^2*pi*AR*e*Cd0))^1/4 At sea level, 101.10 ft/s =59.90 knots At 5000 ft, 108.92 ft/s= 64.54 knots At 10,000 ft, 117.64 ft/s= 69.71 knots Power Required at stall at the 3 different heights can be obtained by plugging in the stall velocity into

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the Power Required equation. At sea level, 39.8 hp At 5000 ft, 43 hp At 10,000ft, 46.3 hp Also, if we plug in the velocity we get from the ‘velocity needed to find PR(min) equation’ into the power required equation, we will get the minimum power required value. At sea level, 39 hp At 5000 ft, 42 hp At 10,000 ft, 45.5 hp To find the power required for (L/D)max, we can use the equation that was derived in class: PR=((2*W^3*Cd^2)/(ρ*S*Cl^3))^0.5 Diving the above answer by 550, we get the power required in Horse Power And to find the velocity associated with that power required, V= ((PR*(Cl/Cd)max)/W) We know (Cl/Cd)max=14.14 from task 1. Therefore, At sea level, PR=45 hp and V= 133 ft/s = 78.8 knots At 5000 ft, PR=49 hp and V=143.34 ft/s= 85 knots At 10,000 ft, PR=51.8 hp and V=155 ft/s= 92 knots

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COMBINED GRAPH FOR TASK 2 AND 3

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TASK 3:POWER AVAILABLE PA=SHPxη Given η=0.78(1-(35/V)^2) SHP(SL)= 180HP SHP(5000)= 150HP SHP(10,000)=120HP

MINIMUM SPEED

ALTITUDE V(knots) PR(hp) PA(hp)

SEA-LEVEL 59.9 39 92

5000 FT 64.54 42 83 10,000 FT 69.71 45.5 70

SPEED FOR (L/D)Max

ALTITUDE V(knots) PR(hp) PA(hp)

SEA-LEVEL 78.8 45 109.5

5000 FT 85 49 97

10,000 FT 92 51.8 80

MAXIMUM SPEED

ALTITUDE V(knots) PR(hp) PA(hp)

SEA-LEVEL 137 131 131

5000 FT 132 109 109

10,000 FT 126 86 86

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TASK 4: CLIMB PERFORMANCE

R/C=(PA-PR)/W

We know :

PA= SHP(0.78(1-(35/V)^2)))X550 <In hp>

PR=1/2*ρ*S*Cd0*V^3+(W^2/0.5*V*ρ*pi*e*AR)

Specific Values to calculate R/C and (R/C)max for different altitudes and velocity are supplied from task

3 to the program and a graph is plotted using MATLAB

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To find the time of climb to 10,000 ft, we have to integrate the dH/(R/C) from 10,000 to 0. This is

basically the area under the line indicating R/C for 10,000 ft. Coding this into MATLAB, we get the

following output:

The time to climb sea level to 10,000 ft.is 17.3 minutes.

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MATLAB OUTPUT:

The maximum ascent rate at sea level is 853.374 ft/min

The maximum ascent rate at 5000 ft is 615.015 ft/min

The maximum ascent rate at 10000 ft is 364.282 ft/min

The airspeed at maximum ascent rate at sea level is 81.763 knots

The airspeed at maximum ascent rate at 5000 ft is 82.948 knots

The airspeed at maximum ascent rate at 10000 ft is 84.725 knots

The absolute ceiling of the aircraft is 17448.12 ft

The Service ceiling of the aircraft is 15403.52 ft

The Rate of climb for maximum angle of attack at sea level is 779.41 ft/min

The true airspeed for maximum angle of attack at sea level is 67.25 knots

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The horizontal airspeed for maximum angle of attack at sea level is 66.81 knots

thetamax = 6.5720

thetarcmax = 5.9373

Rate of climb (ft/min) Climb angle (degrees) Velocity (knots)

Best Rate of Climb Condition

853.3 3.53 81

Best Climb Angle Condition

779.41 6.6 66.81

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TASK 5: RANGE AND ENDURANCE

Fuel Available= 50 gallons

Fuel Used= 45 gallons = 270 lb

SFC= 0.49

η= 0.78

Take-off weight (w1)= 2645 lb

Weight after fuel is used (w2)= 2375 lb

Altitude= 10,000 ft

Breguet Equations

Range Equation:

R=(η*Cl*ln(w1/w2)/(c*Cd)

Endurance Equation:

E= (η*Cl^1.5*((2*ρ*S)^0.5)*((w2)^-0.5)-(w2)^-0.5)/(c*Cd)

Maximum range is obtained when Cl/Cd is maximum.

Cl=(2*W)/(ρ(10,000)*VTRMin^2*S)

Cd=Cd0+(Cl^2/pi*AR*e)

R(max)= 803 nautical miles

Velocity at get to R(max) was obtained in task 2, as 91.73 knots

Maximum endurance is obtained when Cl^1.5/Cd is maximum,

E=10.25 hours

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TASK 6: GLIDING FLIGHT

1) Glide angle = arctan(Cd/Cl)

Vh=V cos ϴ

Vv=V sin ϴ

2) R(max) = h*(L/D)max = (5000-1000)*(14.40) =57517 ft =9460 nautical miles

To Determine Optimum Airspeed for max glide distance

ϴ(min)=arctan(Cd/Cl) =arctan(1/14.40)=3.972 degrees

V=(2*cosϴ*W/ρ(SL)*S*Cl)^0.5

Cl=0.85

Cl is approximated from the plot in Task 1

V=78.8 knots

To Determine airspeed for maximum time aloft

Vv(min)=V*sin ϴ

Vv(min)=8.08 from the MATLAB program

V=8.08/sin (4.6) =59 knots

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To Determine the time required to descend from 5000 ft to 1000 ft at Vvmin

T=dh/Vvmin= (5000-4000)/8.08= 495 s = 8.25 min

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TASK 7: TURNING FLIGHT

We know:

N=(ρ*V^2*S*Clmax)/(2W)

In the graph, the blue and red lines indicate the maximum and minimum limits for loading factor

respectively. The region between the two lines is the safe area.

Velocity(corner point)= 118 knots

R(min)=2*W/g*ρ(SL)*Cl(max)*S

= 250 ft

ω(max)= g*(n(max)*Cl(max)*S*ρ(SL)/(2W))^0.5

n(max)=3.8

ω(max)=0.7 rad/s

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TASK 8: TAKE OFF AND LANDING PERFORMANCE

Given:

CL_TO = 0.5

CL_L = 1.0

Cf=0.02 for dry concrete runway on take-off

Cf=0.25 for braking on dry concrete runway on landing

a) V(TO)=1.2*(2*W/ρ(SL)*S*Cl_max)^0.5

=107.6 ft/s

=63.74 knots

T(0.707V(TO))=P(0.707V(TO))/V(TO)

P(0.707V(TO)) is estimated from the power graph in Task 2

P(0.707V(TO))=23,700 lb/s

T(0.707V(TO))=220.26 lb

F(avg)=T(0.707V(TO))-(12*ρ(SL)*(0.707*V(TO))^2*S*Cd)-Cf*(W-(0.5*ρ(SL)*(0.707*V(TO))^2*S*Cl_TO))

Assuming Cd= Cd0= 0.03

F(avg)= 145 lb

S(G)=(W*V(TO)^2)/(2*g*F(avg))

=3272 ft

b) V(TO)=1.2*(2*W/ρ(5000)*S*Cl_max)^0.5

=115.91 ft/s

=68.68 knots

T(0.707V(TO))=P(0.707V(TO))/V(TO)

P(0.707V(TO)) is estimated from the power graph in Task 2

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P(0.707V(TO))=23,278 lb/s

T(0.707V(TO))=200 lb

F(avg)=T(0.707V(TO))-(12*ρ(5000)*(0.707*V(TO))^2*S*Cd)-Cf*(W-(0.5*ρ(5000)*(0.707*V(TO))^2*S*Cl_TO))

Cd=Cd0=0.03

F(avg)= 125 lb

S(G) )=(W*V(TO)^2)/(2*g*F(avg))

= 4385 ft

c) Given:

W=2400 lb

Sea Level conditions

V(TO)=1.2*(2*W/ρ(5000)*S*Cl_max)^0.5

=102.5 ft/s

=60.72 knots

T(0.707V(TO))=P(0.707V(TO))/V(TO)

P(0.707V(TO)) is estimated from the power graph in Task 2

P(0.707V(TO))=21,481 lb/s

T(0.707V(TO))=209.5 lb

F(avg)=T(0.707V(TO))-(12*ρ(5000)*(0.707*V(TO))^2*S*Cd)-Cf*(W-(0.5*ρ(5000)*(0.707*V(TO))^2*S*Cl_TO))

Cd=Cd0=0.03

F(avg)= 141.5 lb

S(G) )=(W*V(TO)^2)/(2*g*F(avg))

= 2767 ft

d) Given:

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W=2645 lb

Sea Level conditions

V(TD)=1.3*(2*W/ρ*S*Cl(max))^0.5

V(TD)=116.57 ft/s

= 69 knots

F(avg)=-(0.5*ρ(SL)*(0.707*V(TD))^2*S*Cd)-Cf*(W-(0.5*ρ(SL)*(0.707*V(TD))^2*S*CL_L))

= -898.21 lb

S(L)=-W*V(TD)^2/2*g*F(avg)

=621.35 ft

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CONCLUSION

From the above analysis, it can be concluded that the diamond DA 40 is an aerodynamically efficient

aircraft. Also, the above detailed analysis shows that, all the aerodynamic characteristics of the aircraft

can found accurately with some basic information.

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REFERENCES

1) Dr.Clifford Whitfield , The Ohio State University

2) Anderson, John David Introduction to Flight.7th Edition New York: McGraw-Hill, 2012. Print

3) http://adl.stanford.edu/sandbox/groups/aa241x/wiki/e054d/attachments/0567c/performancea

nddrag.pdf?sessionID=1a24a21ebb24dd09e8c25e50de7db2aadeb05ac6 -DRAG BUILD UP

METHOD

4) http://www.adac.aero/linked/12.5_parasite__zero-lift__drag.pdf -DRAG BUILD UP METHOD

5) http://web.mit.edu/16.unified/www/FALL/Unified_Concepts/BreguetNoteseps.pdf

6) AOE 3104 website readings was used in this document for better understanding of the concepts

and to obtain the formulas.

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APPENDIX

TASK 1:

clc clear figure(1) Cl=[0:0.001:2]; % lift Coeff Cd0= 0.03; %drag polar estimate e=0.73; % Oswald efficiency estimate AR= 10.53; Cd=Cd0+ ((Cl.^2)./(pi*AR*e)); % Cd equation % Drag Polar Plot plot(Cl,Cd) hold on plot(0,Cd0,'x') hold off title('Drag Polar') xlabel('Lift Cofficient') ylabel('Drag Coefficient') legend('Drag Coefficient vs Lift Coefficient','Cd0','Location','NorthWest') % L/D figure(2) LD=(Cl./Cd); plot(Cl,LD) hold on xlabel('Lift Cofficient') ylabel('Lift to Drag Ratio') for i=1:length(LD) if LD(i)==max(LD) plot(Cl(i),LD(i),'x') end end legend('L/D vs Cl','(L/D)max') max LD; Task 2 and 3: % given values SHP_SL=180; SHP_5000=150; SHP_10000=120; D_SL=0.0023769;

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D_5000=0.0020482; D_10000=0.0017556; S=145.7; Cd0=0.03; AR=10.53; e=0.75; W=2645; ClMax=1.90; Vel=(88:0.1:350); % velocity range Vel_k=(Vel.*3600)./(5280*1.15078); E=0.78.*(1-(35./Vel_k).^2); %given % power available PA_SL=SHP_SL.*E; PA_5000=SHP_5000.*E; PA_10000=SHP_10000.*E; A=4*W^2; B1=3*D_SL^2*S^2*pi*AR*e*Cd0; B2=3*D_5000^2*S^2*pi*AR*e*Cd0; B3=3*D_10000^2*S^2*pi*AR*e*Cd0; %Velocity at power required minimum VPRMin_SL=(A/B1)^0.25; VPRMin_5000=(A/B2)^0.25; VPRMin_10000=(A/B3)^0.25; C1=0.5*D_SL*S*Cd0; C2=0.5*D_5000*S*Cd0; C3=0.5*D_10000*S*Cd0; D1=(W^2)/(0.5*D_SL*S*pi*AR*e); D2=(W^2)/(0.5*D_5000*S*pi*AR*e); D3=(W^2)/(0.5*D_10000*S*pi*AR*e); PR_SL=C1*Vel.^3+D1*Vel.^-1; PR_5000=C2*Vel.^3+D2*Vel.^-1; PR_10000=C3*Vel.^3+D3*Vel.^-1; %Power required in horse power PRHP_SL=PR_SL./550; PRHP_5000=PR_5000./550; PRHP_10000=PR_10000./550; %velocity at power required minimum in knots VPRMin_SL_k=(VPRMin_SL*3600)/(5280*1.15078);

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VPRMin_5000_k=(VPRMin_5000*3600)/(5280*1.15078); VPRMin_10000_k=(VPRMin_10000*3600)/(5280*1.15078); %power required minimum PRMin_SL=(C1*VPRMin_SL^3+D1*VPRMin_SL^-1)/550; PRMin_5000=(C2*VPRMin_5000^3+D2*VPRMin_5000^-1)/550; PRMin_10000=(C3*VPRMin_10000^3+D3*VPRMin_10000^-1)/550; %Maximum velocity at sea level N1=find((PRHP_SL-PA_SL)>0,1); p11=PRHP_SL(1,N1); p12=PRHP_SL(1,N1-1); v11=Vel_k(1,N1); v12=Vel_k(1,N1-1); VelMax_SL=((PA_SL(1,N1)-p11)/(p12-p11))*(v12-v11)+v11; %maximum velocity at 5000 ft N2=find((PRHP_5000-PA_5000)>0,1); p21=PRHP_5000(1,N2); p22=PRHP_5000(1,N2-1); v21=Vel_k(1,N2); v22=Vel_k(1,N2-1); VelMax_5000=((PA_5000(1,N2)-p21)/(p22-p21))*(v22-v21)+v21; % maximum velocity at 10,000 ft N3=find((PRHP_10000-PA_10000)>0,1); p31=PRHP_10000(1,N3); p32=PRHP_10000(1,N3-1); v31=Vel_k(1,N3); v32=Vel_k(1,N3-1); VelMax_10000=((PA_10000(1,N3)-p31)/(p32-p31))*(v32-v31)+v31; %conversion to knots VelMaxf1=(VelMax_SL*5280*1.15078)/(3600); VelMaxf2=(VelMax_5000*5280*1.15078)/(3600); VelMaxf3=(VelMax_10000*5280*1.15078)/(3600); %power required maximum PRMax_SL=(C1*VelMaxf1^3+D1*VelMaxf1^-1)/550; PRMax_5000=(C2*VelMaxf2^3+D2*VelMaxf2^-1)/550; PRMax_10000=(C3*VelMaxf3^3+D3*VelMaxf3^-1)/550; CLMax=sqrt(Cd0*pi*AR*e); CdMax=Cd0+((CLMax^2)/(pi*AR*e)); %power required PrLto_SL=sqrt((2*((W^3)/S))/(D1*((CLMax^3)/(CdMax^2)))); PrLto_5000=sqrt((2*((W^3)/S))/(D2*((CLMax^3)/(CdMax^2))));

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PrLto_10000=sqrt((2*((W^3)/S))/(D3*((CLMax^3)/(CdMax^2)))); L_D=(CLMax/CdMax); %L/D ratio %velocity VLto_SL=(PrLto_SL)/(W/(L_D)); VLto_5000=(PrLto_5000)/(W/(L_D)); VLto_10000=(PrLto_10000)/(W/(L_D)); VLto_SLK=(VLto_SL*3600)/(5280*1.15078); VLto_5000K=(VLto_5000*3600)/(5280*1.15078); VLto_10000K=(VLto_10000*3600)/(5280*1.15078); PrLtoDHP_SL=PrLto_SL/550; PrLtoDHP_5000=PrLto_5000/550; PrLtoDHP_10000=PrLto_10000/550; %stall velocity Vstall_SL=sqrt((2*(W/S))/(D_SL*ClMax)); Vstall_5000=sqrt((2*(W/S))/(D_5000*ClMax)); Vstall_10000=sqrt((2*(W/S))/(D_10000*ClMax)); %stall velocity in knots VstK_SL=(Vstall_SL*3600)/(5280*1.15078); VstK_5000=(Vstall_5000*3600)/(5280*1.15078); VstK_10000=(Vstall_10000*3600)/(5280*1.15078); PA=(0:200); %power available range PA_SL=(C1*Vstall_SL^3+D1*Vstall_SL^-1)/550; PA_5000=(C2*Vstall_5000^3+D2*Vstall_5000^-1)/550; PA_10000=(C3*Vstall_10000^3+D3*Vstall_10000^-1)/550; figure(1) plot(Vel_k,PRHP_SL,'k',Vel_k,PRHP_5000,'r',Vel_k,PRHP_10000,'b',Vel_k,PA_SL,'--k',Vel_k,PA_5000,'--r',Vel_k,PA_10000,'--b',VstK_SL,PA,':r',VstK_5000,PA,':b',VstK_10000,PA,':k',VPRMin_SL_k,PRMin_SL,'ob',VPRMin_5000_k,PRMin_5000,'or',VPRMin_10000_k,PRMin_10000,'og',VLto_SLK,PrLtoDHP_SL,'*b',VLto_5000K,PrLtoDHP_5000,'*r',VLto_10000K,PrLtoDHP_10000,'*k') xlim([40 150]) ylim([0 170]) xlabel('Velocity (Knots)'); ylabel('Power (Horsepower)'); title('Power v. Velocity'); legend('Power Required at Sea Level','Power Required at 5,000 ft.','Power Required at 10,000 ft.','Power Available at Sea Level','Power Available at 5,000 ft.','Power Available at 10,000 ft.','Stall Speed at Sea Level','Stall Speed at 5,000 ft.','Stall Speed at 10,000 ft.',...

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'Minimum Power Required at Sea Level','Minimum Power Required at 5,000 ft.','Minimum Power Required at 10,000 ft.','L/D(max) at Sea Level','L/D(max) at 5,000 ft.','L/D(max)at 10,000 ft.','Location','SouthOutside')

TASK 4:

% finding rate of climb

RC_SL=(((PA_SL*550)-PR_SL)/W)*60; % sea-level condition RC_5000=(((PA_5000*550)-PR_5000)/W)*60; % at 5000 ft RC_10000=(((PA_10000*550)-PR_10000)/W)*60; % at 10,000ft % sea level conditions RCMax_SL=max(RC_SL); % R/C maximum VRCMaxM_SL=find((RC_SL-RCMax_SL)==0,1); % velocity VRCMax_SL=Vel_k(1,VRCMaxM_SL); % 5000ft conditions RCMax_5000=max(RC_5000); VRCMaxM_5000=find((RC_5000-RCMax_5000)==0,1); VRCMax_5000=Vel_k(1,VRCMaxM_5000); %10,000ft conditions RCMax_10000=max(RC_10000); VRCMaxM_10000=find((RC_10000-RCMax_10000)==0,1); VRCMax_10000=Vel_k(1,VRCMaxM_10000); figure(2) plot(Vel_k,RC_SL,'k',Vel_k,RC_5000,'r',Vel_k,RC_10000,'b',VRCMax_SL,RCMax_SL,'ok',VRCMax_5000,RCMax_5000,'or',VRCMax_10000,RCMax_10000,'ob'); xlabel('Velocity (Knots)'); ylabel('Rate of Climb (ft. per. min.)'); title('Rate of Climb v. Velocity'); legend ('Rate of Climb sea level','Rate of Climb 5,000 ft.','Rate of Climb 10,000 ft.','Maximum Rate of Climb at Sea Level','Maximum Rate of Climb at 5,000 ft.','Maximum Rate of Climb at 10,000 ft.','Location','SouthOutside') xlim([40 150]) ylim([0 900]) range=(0:10000); rcb=[RCMax_SL RCMax_5000 RCMax_10000]; alt=[0 5000 10000]; p=polyfit(rcb,alt,1); f=polyval(p,range); abscei=f(1,1); fprintf('\n\n The absolute ceiling is %6.1f ft.\n',abscei); sercei=f(1,101);

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fprintf('\n The service ceiling is %6.1f ft.\n\n',sercei); figure(3) plot(RCMax_SL,0,'ok',RCMax_5000,5000,'or',RCMax_10000,10000,'ob',range,f,'-m') xlabel('Rate of Climb (ft/min)'); ylabel('Altitude (ft)'); title('Rate of Clime v. Altitude'); legend ('Max Rate of Clime sea level','Max Rate of Clime 5,000 ft.','Max Rate of Clime 10,000 ft.','Best Fit Line','Location','South') xlim([0 900]) ylim([0 18000]) p2=polyfit(alt,rcb,1); f2=polyval(p2,range); x=(0:10000); fun= 1./f2; inta=trapz(x,fun); fprintf('\n The time to climb sea level to 10,000 ft.is %1.1f minutes.\n',inta);

% Climb hodograph figure(3) Vh= sqrt(abs((v.^2)-(RC_SL.^2))); % Slope creation for theta max tangent= 0.1152*Vh; plot(Vh,RC_SL,Vh,tangent) xlabel('Horizontal Speed (ft/s)') ylabel('Vertical Speed (ft/s)') title(' Climb Hodograph') for j= 1:length(tangent) if (tangent(j)-RC_SL(j))<0.000000000000000000000000000000001 Vthetamaxinf= v(j)*0.592484; Vthetamax= Vh(j)*0.592484; rcthetamax= RC_SL(j)*60; end if RC_SL(j)==max(RC_SL) Vhrcmax= Vh(j)*0.592484; end end fprintf('\nThe Rate of climb for maximum angle of attack at sea level is %5.2f ft/min\n',rcthetamax) fprintf('The true airspeed for maximum angle of attack at sea level is %5.2f knots\n',Vthetamaxinf) fprintf('The horizontal airspeed for maximum angle of attack at sea level is %5.2f knots\n',Vthetamax) % Calculation of Angle of attack thetamax= atand((rcthetamax/60)/(Vthetamax/0.592484)) thetarcmax= atand(max(rc1)/(Vhrcmax/0.592484))

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TASK 5:

No Code

TASK 6:

clc clear Cl=linspace(0,1.90,300); % Coeff of lift Cd=0.03+((Cl.^2)./(pi*10.53*0.75)); %Coeff of Drag theta=atand(1./(Cl./Cd)); %glide angle theta1= atand(1/14.3794); slope=tand(theta1); v=sqrt((2.*2645.*cosd(theta))./(0.0023769.*Cl.*145.7)); vh= v.*cosd(theta); %horizontal vv= (-1).*v.*sind(theta); %vertical x= (-1)*slope*vh; plot(vh,vv,vh,x) ylabel('Descent Rate (ft/s)') xlabel('Lateral Speed (ft/s)') title('Glide Hodograph') legend('Vv vs Vh','Line for Optimum glide angle') for i=1:length(Cl) if abs(theta(i)-theta1)<0.001 Cl(i) Cd(i) end if vv(i)==max(vv) theta(i) end end max(vv)

TASK 7:

clc clear %given ClMax= 1.90; D_SL=0.0023769; S=145.7; W=2645; %velocity range Vel=[0:300]; Vel2= sqrt((2*3.8*W)/(D_SL*ClMax*S)); % corner velocity line=linspace(-1.52,3.8,300); for i=1:length(Vel) a(i)= D_SL*(Vel(i)^2)*S*ClMax./(2*W);

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if a(i)<3.8 B(i)=a(i); else B(i)= 3.8; end end plot(Vel*0.592484,B,'b',Vel2*0.592484,3.8,'x') hold on for j=1:length(Vel) n(j)= (-1)*D_SL*(Vel(j)^2)*S*ClMax./(2*W); if n(j)>(-1.52) M(j)=n(j); else M(j)= -1.52; end end plot(Vel*0.592484,M,'r') plot(178,line,'b-') hold off xlabel('Airspeed (knots)') ylabel('Load Factor (n)') title('V-n Diagram') legend('Positive n vs V','Corner point','Negative n vs V','Max speed limit','Location','East')

TASK 8:

No Code