PERFORMANCE ANALYSIS OF THE DIAMOND DA-40. Bryan Drescher. December 2, AERO 2200: Introduction to Aerospace Engineering 1. Autumn Semester 2013

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1 PERFORMANCE ANALYSIS OF THE DIAMOND DA-40 Bryan Drescher December 2, 2013 AERO 2200: Introduction to Aerospace Engineering 1 Autumn Semester 2013 Class Number 8267 For Dr. James W. Gregory 1

2 SUMMERY The Diamond DA-40 was examined and the following data points of its performance were found. The drag polar was found by solving for the parasitic drag which was and the span efficiency factor which was The required and available powers, in horsepower, were graphed as a function of velocity for the altitudes of sea level, 5,000 ft. and 10,000 ft.. The DA-40 was found to have max climb of feet per minute at a speed of knots. The absolute ceiling is ft. The service ceiling is a ft. The best angle of to climb at is 6.57 degrees this not the angle for the max rate of climb. The aircraft is capable of flying for 9.77 hours for a distance of nautical miles using 90% of its fuel at 10,000 ft. From it gliding characteristics the Diamond can glide for a max distance of 9.47 nautical miles or stay gliding for a max tile of 8.25 minutes. The DA-40 has a maneuver velocity of knots. The takeoff ground roll distance at maximum takeoff weight at standard sea level conditions is ft. 2

3 TABLE OF CONTENTS Summery 2 Nomenclature...4 List of Figures...8 List of Tables.9 Introduction..10 Task I: Drag Polar..11 Task II: Power Required Task III: Power Available...31 Task IV: Climb Performance. 34 Task V: Range and Endurance Task VI: Gliding Flight...43 Task VII: Turning Flight.46 Task VIII: Takeoff and Landing Performance...49 Conclusion..52 References..53 Appendix..54 3

4 NOMENCLATURE Parasitic drag coefficient Aerodynamic cleanliness factor Wetted area Wing area Area Base one Base two Height Pi s Slant height of cone Radius one Radius two Radius Oswald efficiency factor Aspect Ratio Wing span 4

5 λ Taper ratio Drag Coefficient Lift Coefficient Lift Drag Velocity ρ Air Density Power Available Weight Stall Velocity Trust Minimum Thrust Trust Required Max Lift to Drag Ratio Max Lift Coefficient to Drag Coefficient Ratio Power Available Maximum Shaft Horsepower 5

6 Ƞ Propeller Efficiency Rate of Climb Starting Height Ending Height True Speed Vertical Speed Horizontal Speed Climb Angle Endurance Range Initial Weight Final Weight Corner Velocity Load Factor Turn Radius Turn Rate Fore of Gravity 6

7 Takeoff ground roll distance Landing ground roll distance Take Off Velocity Average Force Force of Friction Ground Effect Height of Wings Above Ground 7

8 LIST OF FIGURES Figure 1: Drag Polar (C_D v. C_L).. 17 Figure 2: LD vs. CL...19 Figure 3: Power Required and Power Available...32 Figure 4: Rate of Climb.36 Figure 5: Rate of Climb v. Altitude.. 37 Figure 6: Clime Hodograph..39 Figure 7: Glide Hodograph...44 Figure 8: V-n Diagram. 47 8

9 LIST OF TABLES Table 1: Power Required and Power Available.33 Table 2: Best Climb Performance 40 9

10 INTRODUCTION From a picture and a few values that were given many different performance characteristics can be found. Some of them include gliding performance, climbing performance turning performance, the range and endurance. Others are also found using MATLab. The program helps to organize and arrange the data in an orderly fashion. Many other values are found later on for the DA

11 TASK I: DRAG POLAR Part A The equation to fine is : The for the DA-40 was choose to be because the DA-40 is matches the description of the PA-28 and the PA-28 has an aerodynamic cleanliness factor of The PA-28 is a single prop, low wing and fixed landing gear aircraft just like the DA-40. The wetted area was estimated to be This was found by taking the diagram of the DA-40 and measuring different parts of the diagram and finding the surface area. The DA-40 has an actual wing span of 39 ft. but form the diagram it measured 5 in. so have to scale up the numbers gathered to make them correct. The area of the wing was found to be This was found by: The wings are a trapezoid so use the equation for the area of a trapezoid: 11

12 There are four trapezoids because there are two winds and a top and bottom of them both.,, ( ) But have to proportion to the actual size so: ( ) ( ) ( ) The tail is also a trapezoid so the same steps were followed:,, ( ) ( ) ( ) The rudder is also a trapezoid but there is only one and it has two sides.,, ( ) ( ) ( ) 12

13 Next is the body or the back of the fuselage. It is a cone that is missing its tip. The surface area equation for it is:,, Need to scale it up so it becomes: ( ) (( ) ( )) The rear of the cockpit is also a cone missing its tip so the same steps are used.,, ( ) (( ) ( )) The front is also a cone but just a different dimension. Again find the surface area of a cone.,, ( ) (( ) ( )) 13

14 The nose of the aircraft is a shaped like a cylinder without the ends though. The surface area of this kind of cylinder is:, Again, need to scale it from the drawing scale to the real live measurements. ( ) ( ) All those different components make up the wetted area: The wing area is the area of the wing and part of the fuselage. This is a by splitting the area in half the area can be found by finding the area of two identical trapezoids. Doing the as when finding the rooter area:,, 14

15 ( ) ( ) ( ) Now that all the variables are found in the parasitic drag coefficient equation, can be calculated: Part B The equation for the Oswald efficiency factor is: K is picked from a range of numbers; the range of numbers is For the DA-40 the K was picked to be The equation to find the aspect ratio is:, 15

16 is found using the graph provided. The graph has the taper ratio plotted on the x axis and on the y axis. is a function of taper ration and the aspect ratio. By looking at what the taper ration is and this aspect ratio to use can be found by seeing that the value is for the corresponding taper ratio and aspect ratio. The taper ratio is found by dividing the length of the tip of the wing by the length of the wing s base. For the DA-40 the tip measures and the base measures from the diagram so: ( ) ( ) From the graph, = On the x axis find where then go up to the line that corresponds to the aspect ratio being There is no line for that but there is one that is equal to 8 so just go a little higher than that line. Now that all the variables are found for the Oswald efficiency factor, e can be calculated. Part C To graph v need to find. The equation is: 16

17 Drag Coefficient is a function of so it will vary depending on what is. From this when = Drag Coefficient v Lift Coefficient Drag Polar Parasitic Drag Lift Coefficient Figure 1: Drag Polar (C D v. C L ) Part D 17

18 The equation for the max lift to drag ratio is: is given to be

19 Lift to Drag Ratio 15 Lift to Drag Ratio v Lift Coefficient Lift to Drag Ratio as a fuction of Cl The maximum Lift to Drag Ratio Lift Coefficient Figure 2: L D vs. C L 19

20 TASK II: POWER REQUIRED Part A The equation for the power required as a function of velocity is: For each different altitude the density of the air will change. At sea level the density is , at 5,000 ft. the density is and at 10,000 ft. the density is All that is needed now is to find the aspect ratio because all the other variables are given. The equation to find the aspect ratio is:, From here all the values for the different variables are known so can be found. For sea level For 5,000 ft. 20

21 For 10,000 ft. Where can be any value. For this case the values of ranged from fps. However, the power is in foot pounds it needs to be in horsepower so divide the value of by 550. Part B The stall speed is where no matter how much power is given to the engine cannot fly any slower. The Equation for it is: All the values are known so the stall velocity for sea level is: Need to convert into knots. Fps to knots= 21

22 So the stall velocity at sea level is: The stall velocity for 5,000 ft. is The stall velocity for 10,000 ft. is To find the power required at that speed all that is needed is to do is plug in the stall speed in for the velocity value in the power available equation. The stall speed needs to be in fps for the units to work out right and then divide the power by 550 to get it in to horsepower. For sea level: 22

23 ( ) ( ) For 5,000 ft.: ( ) ( ) For 10,000 ft.: 23

24 ( ) ( ) Part C To find the speed at the minimum power required use the equation: Again it will be different for the different altitudes because of the change in density. At sea level: Need to convert to knots: At 5000 ft: 24

25 Need to convert to knots: At 10,000 ft: Need to convert to knots: Again to find the power required at these speeds plug them into the power required equation. For sea level: 25

26 ( ) ( ) For 5,000 ft.: ( ) ( ) For 10,000 ft.: 26

27 ( ) ( ) Part D In order to find the speed for the max lift to drag ratio, need to find the minimum trust required. From this equation the minimum required trust happens at the same power value as when the max lift coefficient to drag coefficient ratio. is given to be

28 The Power Required in terms of trust is: can be solved for form the lift equation as: Subsisting for and the power required equation becomes: differs with altitude because the density changes For Sea level: Divide by 550 to get power in horsepower For 5,000 ft.: 28

29 For 10,000 ft.: To find the velocity solve for it in the power required equation. Take the power found and divide it by the trust required. At Sea Level the velocity is: has to be in ft. lbs. Now covert from fps to knots. 29

30 At 5,000 ft. the velocity is At 10,000 ft. the velocity is 30

31 TASK III: POWER AVAILABLE Part A The equation for the power available is: ( ) ( ( ) ) For the graph the same range of number for V that was used for the Pr. To find a certain Pa find the velocity at that point then plug it into the equation to find the power. The shaft horsepower at sea level is 180 HP, at 5,000 ft. it is 150 HP and at 10,000 ft. it is 120 hp. For example to find the power available for the minimum speed: At sea level the velocity was (needs to be in knots). Then the power available in horsepower is: ( ( ) ) HP The same process is done at any different speed or altitude. 31

32 Power (Horsepower) Power v. Velocity Velocity (Knots) Figure 3: Power Required and Power Available 32

33 Minimum Speed Speed for (L/D)max Maximum Speed Altitude V (knots) PR (HP) PA (HP) V (knots) PR (HP) PA (HP) V (knots) PR (HP) PA (HP) Sea Level ,000 ft ,000 ft Table 1: Power Required and Power Available 33

34 TASK IV: CLIMB PERFORMANCE Part A The equation to find the rate of climb is the difference of power required from the power available divided by the weight: Subsisting in for the power available and power required the equation becomes: ( ( ( ( ) ))) ( ) At sea level it is the equation becomes: ( ( ( ( ) ))) ( ) At 5,000 ft.: 34

35 ( ( ( ( ) ))) ( ) At 10,000 ft. ( ( ( ( ) ))) ( ) The power available needs to be in ft. lbs. The rate of climb will change as a function of velocity. It will also be different because at different altitudes the density will be different. To find the maximum rate of climb use the built-in MATlab function to find the maximum. At sea level the max rate of clime is fpm and knots, at 5,000 ft. the max rate of clime is fpm and knots, and at 10,000 the max rate of clime is 36 35

36 Rate of Climb (ft. per. min.) Rate of Clime v. Velocity Rate of Clime sea level Rate of Clime 5,000 ft. Rate of Clime 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 Velocity (Knots) Figure 4: Rate of Climb 4.28 fpm and 84.37knots. Part B The absolute ceiling is at ft. The service ceiling is at ft. These are found by taking the maximum and plotting them on a graph that is altitude as a function of. Then find a best fit line for the points and extend it to the y 36

37 Altitude (ft.) axis. The best way to do this by hand would be to find the rise over run between the t o farthest points then solve for the y intercept for then the line is equal to 0 which happens at the max at 10,000 ft. the best way to do this is to use a computer program. It will give a more precise line. The equation is for a line so to find the absolute ceiling plug 0 fpm (the y intercept) in and for the service ceiling plug 100 fpm in Rate of Clime v. Altitude Max Rate of Clime sea level Max Rate of Clime 5,000 ft. Max Rate of Clime 10,000 ft. Best Fit Line Rate of Climb (ft. per. min.) Figure 5: Rate of Climb v. Altitude Part C The equation for the rate of time is: 37

38 Where is the best fit line which is To find the time it takes to climb to 10,000 ft. the equation becomes: This equals 17.3 minutes. Part D In order to form the hodograph the horizontal speed must be found. To do this solve for it using the Pythagorean Theorem. The horizontal speed is already known because it is simply the rate of climb and the true speed is the same range of speeds as used before. ( ( ( ( ) ))) ( ) This graph is for sea level so the density used is the standard density of sea level. The best rate of climb is already known because it is the max rate of climb as well as the true speed at that rate of climb. To find the angle at that rate of climb need to find 38

39 Vertical Velocity /Rate of Climb (knots) the horizontal speed and then take the inverse tangent of the vertical speed over the horizontal speed. ( ) ( ( ) ) To find the max climb angle use the max function of MATlab. The max angle is then use the find command to find the correlating values for rate of climb and the true speed. Figure 6: Clime Hodograph 10 Climb Hodograph Rate of Climb 1 Best Rate of Climb Best Climb Angle Horizontal Velocity (knots) 39

40 Best Rate of Climb Condition Best Climb Angle Condition Rate of Climb (ft/min) Climb Angle (degrees) Velocity (knots) Table 2: Best Climb Performance 40

41 TASK V: RANGE AND ENDERANCE The DA-40 can carry 50 gallons of fuel on board, if it consumes 90% on it while flying it will have used 45 gallons. Each gallon of fuel weighs 6 lbs. then that means that the aircraft is 270 pounds lighter than when it took off. The max weight for takeoff is 2,645 lbs. So taking away the weight of the fuel burned then the plan now weighs 2,375 lbs. The DA-40 at 10,000 ft. has a propeller efficiency of 0.78 and a Specific Fuel consumption of 0.49 (lb/hr)/. The Breguet endurance equation is: The optimum endurance happens when the term is at its max. After finding and plugging in for the rest of the variables and converting for like units, the endurance is found to be 9.77 hours. This speed in which it has to fly at to reach this optimum endurance is the same speed for the minimum power required which was found earlier to be knots. The Breguet range equation is: The optimum range happens when the term is at its max. After finding which was found earlier and plugging in for the rest of the variables and converting for like units, the range is found to be knots. This speed in which it has to fly at to 41

42 reach this optimum range is the same speed for the minimum thrust required which happens at the max lift to drag ratio this was found earlier to be knots. 42

43 TASK VI: GLIDING FLIGHT In order to construct the glide hodograph the first thing needed was a range of. Since the max was 1.9 a range of -1.9 to 19 was used. Using the the could be calculated: Using the and values can be found by: From here the true speed can be calculated with and values: With the true speed and the angle known the component speeds can be found (the sink rate and the lateral speed) 43

44 Sink Speed (fps) 0 Glide Hodograp Lateral Speed (fps) Figure 7: Glide Hodograph Part B If the engine were to fail at a pressure altitude of 5,000 ft. and at maximum takeoff weight, the distance the DA-40 can glide can be found by: Which can to rewritten as: 44

45 To get the farthest glide distance the angle that the plane falls at has to be the smallest possible. However the minimum angle is happens at ( ) ( ) The aircraft is falling 4,000 ft., at that height the DA-40 would cover 9.47 nautical miles. At this angle the plane would be traveling at a true speed of Found by finding the angle at ( ) 3.98, then plugging it and the ( ), found earlier, into the true speed equation. If the pilot was trying to stay up in the air the longest as possible then he or she would want to set the plain to the lowest sink speed as possible. The true speed at this this point is knots and it would take 8.25 minutes for the plane to descend the 4,000 ft. 45

46 TASK VII: TURING FLIGHT Part A V-d diagram is graph of what an aircraft is capable physically of going. During a turn a plane must increase its minimum speed because the stall speed increases. However, as the speed increases the force acting on the plane. If the plane were to exceed the max speed or max load factor then the plane would suffer structural damage. The never exceed airspeed or high speed limit is given to be 178 knots, the maximum positive load factor is given to be 3.8 and the maximum negative load factor is given to be The maneuver point which is also the corner velocity is found by: was calculated to be knots, after converting it to knots from fps. The corner velocity is the velocity where the stall limit is the same as the max load factor. The stall limits were found by using the same equation but having the load factor vary from 0 to the max and min. 46

47 Load Factor Stall Limit Negative Stall Limit Positive Load Limit Negative Load Limit High Speed Limit Corner Velocity V-n Diagram of DA Velocity (knots) Figure 8: V-n Diagram Part B The minimum turn radius is the smallest distance for the plane to make a 180 turn. The equation for it is: Solving, the minimum turn radius is ft. or converting it to nautical miles the minimum turn radius is 0.09 nautical miles. 47

48 The maximum turn rate is the fastest the plane is able to turn. The equation is: Solving for it gives a maximum turn rate value of 0.47 fps or 0.28 knots. This would take 6.66 seconds to do. 48

49 TASK VIII: TAKEOFF AND LANDING PERFORMANCE In order for a plane to get off the ground it first has to reach a speed that will produce enough lift to get it off the ground. To find the distance that the plane is on the ground trying to reach that speed the use: Where: And and are the trust and speed at, respectably. Also: 49

50 Trust is equal to power divided by velocity. The velocity is the power is the power available. It is power available because the pilot wants to get off the ground as soon as possible in order to not run out of runway. The fastest way is to give the engine as much power as possible which is the power available. The takeoff ground roll distance at maximum takeoff weight at standard sea level conditions is ft. The takeoff ground roll distance at maximum takeoff weight at a high altitude airport (5000 ft) is ft. Since the air is less dense the engine does not have as much power and therefor takes longer to liftoff. The takeoff ground roll distance at a takeoff weight of 2400 lbs at standard sea level conditions is ft. Because there is less weight, there is less force pushing down on the plane and therefore can get off the ground sooner. To find the Landing ground roll distance the equation is: 50

51 The landing ground roll distance at maximum weight and standard sea level conditions is ft. The friction between the wheels and the ground is much higher so it stops is a short amount of distance. 51

52 CONCLUSION This shows that with a little information about an aircraft a lot can be uncovered by it. From just a picture the geometry of the plane the and e can be found. Then from there the power needed and available can be determined ant different speeds and altitudes. Next, the climbing characteristics were found including the absolute and survive ceilings as well as the optimum angle to climb and the speed to climb. After that how far and how long at a given altitude can be found. Also, the max gliding time and max gliding distance were discovered. After that the tuning flight was found, this is how does a airplane have to fly in order not to rip its self apart. The last thing that was found was to see how far a plane rolls on taking off and landing. 52

53 REFERENCES Dr. James W. Gregory Matt McCrink Introduction to Flight by John D. Anderson, Jr., published by McGraw Hill 53

54 APPRNDIX MATLab Code Task 1 clc clear clear all Cdo=0.0300; Cl=(-2:0.01:2); AR=(39.17)^2/(145.7); e=0.75; Cp=((Cl).^2)./(pi*AR*e); Cd=Cdo+Cp; CdoM=(ones(1,length(Cl))*Cdo); figure(1) plot(cl,cd,cl,cdom,'k') title('drag Coefficient v Lift Coefficient') xlabel('lift Coefficient') ylabel('drag Coefficient') legend('drag Polar','Parasitic Drag','Location','North') LD=(Cl)./Cd; Clmax=sqrt(Cdo*pi*AR*e); Cpmax=((Clmax)^2)./(pi*AR*e); Cdmax=Cdo+Cpmax; LDmax=(Clmax/Cdmax); figure(2) plot(cl,ld,clmax,ldmax,'o') title('lift to Drag Ratio v Lift Coefficient') xlabel('lift Coefficient') ylabel('lift to Drag Ratio') legend('lift to Drag Ratio as a fuction of Cl','The maximum Lift to Drag Ratio','Location','NorthWest') clc clear clear all Task 2 %power from engine in hourse power SHPa1=180; %Sea level SHPa2=150; %5000 ft SHPa3=120; %10000 ft S=145.7; V=(88:0.1:350); Vk=(V.*3600)./(5280* ); 54

55 N=0.78.*(1-(35./Vk).^2); CDo=0.03; AR=10.53; e=0.75; W=2645; Clmax=1.90; D1= ; D2= ; D3= ; Pa1=SHPa1.*N; Pa2=SHPa2.*N; Pa3=SHPa3.*N; A=4*W^2; B1=3*D1^2*S^2*pi*AR*e*CDo; B2=3*D2^2*S^2*pi*AR*e*CDo; B3=3*D3^2*S^2*pi*AR*e*CDo; VPrmin1=(A/B1)^0.25; VPrmin2=(A/B2)^0.25; VPrmin3=(A/B3)^0.25; X1=0.5*D1*S*CDo; X2=0.5*D2*S*CDo; X3=0.5*D3*S*CDo; Y1=(W^2)/(0.5*D1*S*pi*AR*e); Y2=(W^2)/(0.5*D2*S*pi*AR*e); Y3=(W^2)/(0.5*D3*S*pi*AR*e); Pr1=X1*V.^3+Y1*V.^-1; Pr2=X2*V.^3+Y2*V.^-1; Pr3=X3*V.^3+Y3*V.^-1; Prh1=Pr1./550; Prh2=Pr2./550; Prh3=Pr3./550; VPrmink1=(VPrmin1*3600)/(5280* ); VPrmink2=(VPrmin2*3600)/(5280* ); VPrmink3=(VPrmin3*3600)/(5280* ); Prmin1=(X1*VPrmin1^3+Y1*VPrmin1^-1)/550; Prmin2=(X2*VPrmin2^3+Y2*VPrmin2^-1)/550; Prmin3=(X3*VPrmin3^3+Y3*VPrmin3^-1)/550; num1=find((prh1-pa1)>0,1); p11=prh1(1,num1); p12=prh1(1,num1-1); v11=vk(1,num1); v12=vk(1,num1-1); 55

56 Vmax1=((Pa1(1,num1)-p11)/(p12-p11))*(v12-v11)+v11; num2=find((prh2-pa2)>0,1); p21=prh2(1,num2); p22=prh2(1,num2-1); v21=vk(1,num2); v22=vk(1,num2-1); Vmax2=((Pa2(1,num2)-p21)/(p22-p21))*(v22-v21)+v21; num3=find((prh3-pa3)>0,1); p31=prh3(1,num3); p32=prh3(1,num3-1); v31=vk(1,num3); v32=vk(1,num3-1); Vmax3=((Pa3(1,num3)-p31)/(p32-p31))*(v32-v31)+v31; Vmaxf1=(Vmax1*5280* )/(3600); Vmaxf2=(Vmax2*5280* )/(3600); Vmaxf3=(Vmax3*5280* )/(3600); Prmax1=(X1*Vmaxf1^3+Y1*Vmaxf1^-1)/550; Prmax2=(X2*Vmaxf2^3+Y2*Vmaxf2^-1)/550; Prmax3=(X3*Vmaxf3^3+Y3*Vmaxf3^-1)/550; CLmax=sqrt(CDo*pi*AR*e); Cdmax=CDo+((CLmax^2)/(pi*AR*e)); PrLtoD1=sqrt((2*((W^3)/S))/(D1*((CLmax^3)/(Cdmax^2)))); PrLtoD2=sqrt((2*((W^3)/S))/(D2*((CLmax^3)/(Cdmax^2)))); PrLtoD3=sqrt((2*((W^3)/S))/(D3*((CLmax^3)/(Cdmax^2)))); LtoD=(CLmax/Cdmax); VLtoD1=(PrLtoD1)/(W/(LtoD)); VLtoD2=(PrLtoD2)/(W/(LtoD)); VLtoD3=(PrLtoD3)/(W/(LtoD)); VLtoDk1=(VLtoD1*3600)/(5280* ); VLtoDk2=(VLtoD2*3600)/(5280* ); VLtoDk3=(VLtoD3*3600)/(5280* ); PrLtoDHP1=PrLtoD1/550; PrLtoDHP2=PrLtoD2/550; PrLtoDHP3=PrLtoD3/550; Vstall1=sqrt((2*(W/S))/(D1*Clmax)); Vstall2=sqrt((2*(W/S))/(D2*Clmax)); Vstall3=sqrt((2*(W/S))/(D3*Clmax)); Vst1=(Vstall1*3600)/(5280* ); Vst2=(Vstall2*3600)/(5280* ); 56

57 Vst3=(Vstall3*3600)/(5280* ); PA=(0:200); PVS1=(X1*Vstall1^3+Y1*Vstall1^-1)/550; PVS2=(X2*Vstall2^3+Y2*Vstall2^-1)/550; PVS3=(X3*Vstall3^3+Y3*Vstall3^-1)/550; figure(1) plot(vk,prh1,'b',vk,prh2,'r',vk,prh3,'g',vk,pa1,'--b',vk,pa2,'--r',vk,pa3,'-- g',vst1,pa,':b',vst2,pa,':r',vst3,pa,':g',vprmink1,prmin1,'ob',vprmink2,prmin2,'or',vprmink3,prmin3,'og',vltodk1,prltodhp1,'*b',vltodk2,prltodhp2,'*r',vltod k3,prltodhp3,'*g') 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.',... 'Minimum Power Required at Sea Level','Minimum Power Required at 5,000 ft.','minimum Power Required at 10,000 ft.','maximum Lift to Drag Ratio at Sea Level','Maximum Lift to Drag Ratio at 5,000 ft.','maximum Lift to Drag Ratio at 10,000 ft.','location','suthoutside') Task 3 RC1=(((Pa1*550)-Pr1)/W)*60; RC2=(((Pa2*550)-Pr2)/W)*60; RC3=(((Pa3*550)-Pr3)/W)*60; RC1max=max(RC1); VRC1maxM=find((RC1-RC1max)==0,1); VRC1max=Vk(1,VRC1maxM); RC2max=max(RC2); VRC2maxM=find((RC2-RC2max)==0,1); VRC2max=Vk(1,VRC2maxM); RC3max=max(RC3); VRC3maxM=find((RC3-RC3max)==0,1); VRC3max=Vk(1,VRC3maxM); figure(2) plot(vk,rc1,'b',vk,rc2,'r',vk,rc3,'g',vrc1max,rc1max,'xb',vrc2max,rc2max,'xr', VRC3max,RC3max,'xg'); xlabel('velocity (Knots)'); ylabel('rate of Climb (ft. per. min.)'); title('rate of Clime v. Velocity'); legend ('Rate of Clime sea level','rate of Clime 5,000 ft.','rate of Clime 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','northeastoutside') xlim([40 150]) 57

58 ylim([0 900]) range=(0:10000); rcb=[rc1max RC2max RC3max]; alt=[ ]; 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); fprintf('\n The service ceiling is %6.1f ft.\n\n',sercei); figure(3) plot(rc1max,0,'*r',rc2max,5000,'*g',rc3max,10000,'*b',range,f,'-k') xlabel('rate of Climb (ft. per. 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','Northeast') xlim([0 900]) ylim([ ]) 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); Task 4 VSL=(50:0.1:350); VkSL=(VSL.*3600)./(5280* ); NSL=0.78.*(1-(35./VkSL).^2); PaSL=SHPa1.*NSL; PrSL=X1*VSL.^3+Y1*VSL.^-1; RCSL=(((PaSL.*550)-PrSL)./W).*60; Vv=(RCSL*60)/(5280* ); Vinf=VkSL; Vh=sqrt((Vinf.^2)-(Vv.^2)); Vvmax=max(Vv); VinfmaxM=find((Vv-Vvmax)==0,1); Vinfmax=VkSL(1,VinfmaxM); Vhmax=sqrt((Vinfmax^2)-(Vvmax^2)); theta=atand(vv./vh); thetamax=max(theta); thetamaxm=find((theta-thetamax)==0,1); 58

59 Vinftheta=VkSL(1,thetamaxM); Vvtheta=Vv(1,thetamaxM); Vhtheta=sqrt((Vinftheta^2)-(Vvtheta^2)); thetavvmax=theta(1,vinfmaxm); figure(4) plot(vh,vv,'b',vhmax,vvmax,'ok',vhtheta,vvtheta,'*k'); ylim([0 10]) xlabel('horizontal Velocity (knots)'); ylabel('vertical Velocity /Rate of Climb (knots)'); title('climb Hodograph'); legend ('Rate of Climb','Best Rate of Climb','Best Climb Angle','Location','South') Task 5 N10=0.78; c=0.49; ch=0.49/(550*3600); Wf=(W-(45*6)); R=((N10/ch)*(LtoD)*log(W/Wf))/(5280* ); CL=(0:0.01:2); Cd=CDo+((CL.^2)./(pi*AR*e)); CLCd32=(CL.^(3/2))/(Cd); CLCd32max=max(CLCd32); E=((N10/ch)*CLCd32max*sqrt(2*D3*S)*((Wf^(-0.5))-(W^(-0.5))))/(3600); Task 6 clc clear clear all W=2645; Cd0=.03; e=0.75; AR=10.53; S=145.7; D= ; D2= ; Cl=(-1.9:.001:1.9); Cd=Cd0+((Cl.^2)/(pi*AR*e)); The=atand((Cd./Cl)); Vinf=sqrt((2*W*cosd(The))./(D*S.*Cl)); Vh=Vinf.*cosd(The); Vv=-((Vinf).*sind(The)); figure(1) plot(vh,vv) title('glide Hodograp') xlabel('lateral Speed (fps)') ylabel('sink Speed (fps)') LDmax=14.379; h=4000; 59

60 Rmax=h*LDmax; Rmaxk=Rmax/(5280* ); CLmax=0.8627; themin=atand(1/(ldmax)); Vtmin=sqrt((2*W*cosd(themin))./(D2*S.*CLmax)); Vtmink=(Vtmin*3600)/(5280* ); Vvmin=min(Vv); VvminM1=find((Vv-Vvmin)==0,1); Vminv=Vinf(1,VvminM1); Vminvk=(Vminv*3600)/(5280* ); t=(h/-vvmin)/60; Task 7 Vmax=178; nmax=3.8; nmin=-1.52; Vcv=(sqrt((2*nmax*W)/(D*CLmax*S))*3600)/( *5280); Cel=(Vcv:0.0001:Vmax); nmaxm=(ones(1,length(cel))*nmax); wall=(nmin:0.0001:nmax); VmaxM=(ones(1,length(wall))*Vmax); n=(0:0.01:nmax); Vst=(sqrt((2*n.*W)/(D*CLmax*S))*3600)/( *5280); nneg=(nmin:0.01:0); Vstneg=(sqrt((abs(2*nneg.*W))/(D*CLmax*S))*3600)/( *5280); or=(0:200); orn=(ones(1,length(or))*0); Vcvl=(0:0.01:nmax); VcvM=(ones(1,length(Vcvl))*Vcv); Vcv=(sqrt((2*nmax*W)/(D*CLmax*S))*3600)/( *5280); Cel=(Vcv:0.0001:Vmax); Vcvmin=(sqrt((abs(2*nmin*W))/(D*CLmax*S))*3600)/( *5280); base=(vcvmin:0.0001:vmax); nminm=(ones(1,length(base))*nmin); figure(2) plot(vst,n,'b',vstneg,nneg,'--b',cel,nmaxm,'g',base,nminm,'-- g',vmaxm,wall,'r',vcv,nmax,'ok',vcvm,vcvl,'--k',or,orn,'k') title('v-n Diagram of DA-40') xlabel('velocity (knots)'); ylabel('load Factor') legend('stall Limit','Negative Stall Limit','Positive Load Limit','Negative Load Limit','High Speed Limit','Corner Velocity','Location','NorthWest') 60

61 Task 8 g=32.2; mu=0.02; b=39.17; wh=3.5; B=((16*(wh/b))^2)/(1+(16*(wh/b))^2); CL=(1/(2*B))*pi*AR*e*mu; CD=Cd0+((B*CL^2)/(pi*AR*e)); W2=2400; Vto=1.2*(sqrt((2*(W/S))/(D*CLmax))); V707Vto=0.707*Vto; Vtok=(Vto*3600)/(5280* ); P=180*550*(0.78*(1-(35/Vtok)^2)); T707Vto=(P/V707Vto); Favg=T707Vto-0.5*D*(V707Vto)^2*S*CD-mu*(W-(0.5*D*(V707Vto)^2*S*CL)); Roll=(W*Vto^2)/(2*g*Favg); Vto2=1.2*(sqrt((2*(W/S))/(D2*CLmax))); V707Vto2=0.707*Vto2; Vtok2=(Vto2*3600)/(5280* ); P2=150*550*(0.78*(1-(35/Vtok2)^2)); T707Vto2=(P2/V707Vto2); Favg2=T707Vto2-0.5*D2*(V707Vto2)^2*S*CD-mu*(W-(0.5*D2*(V707Vto2)^2*S*CL)); Roll2=(W*Vto2^2)/(2*g*Favg2); Vto3=1.2*(sqrt((2*(W2/S))/(D*CLmax))); V707Vto3=0.707*Vto3; Vtok3=(Vto3*3600)/(5280* ); P3=180*550*(0.78*(1-(35/Vtok3)^2)); T707Vto3=(P3/V707Vto3); Favg3=T707Vto3-0.5*D*(V707Vto3)^2*S*CD-mu*(W2-(0.5*D*(V707Vto3)^2*S*CL)); Roll3=(W2*Vto3^2)/(2*g*Favg3); mu2=0.5; Vtd=1.3*(sqrt((2*(W/S))/(D*CLmax))); V707td=(0.707*Vtd); Fl=-0.5*D*(V707td)^2*S*CD-mu2*(W-(0.5*D*(V707td)^2*S*CL)); Land=-((W*Vtd^2)/(2*g*Fl)); 61