george andrew analysis of da40 (3)

1

The Analysis of the DA-40

Andrew George

11/17/2013

Aerospace 2200

Dr.Gregory

2

Table of Contents

Summary……………………………………………………………………………………….3

Introduction……………………………………………………………………………………4

Nomenclature………………………………………………………………………………….5

Drag Polar……………………………………………………………………………………………7-8

Power Required………………………………………………………………………………………8-9

Power Available………………………………………………………………………………10

Climb Performance………………………………………………………………………11-13

Range and Endurance…………………………………………………………………………………13-15

Glide Performance……………………………………………………………………………….15-16

Turn Performance……………………………………………………………………………17

Take Off and Landing Performance……………………………………………………………………………….18-21

Conclusion…………………………………………………………………………………….22

List of Figures……………………………………………………………………………………..23-28

List of Tables…………………………………………………………………………………........28-29

Code………………………………………………………………………………………..30-35

References…………………………………………………………………………………….36

3

SUMMARY

The following report details and expounds on the DA-40 and how well it performs in certain aircraft conditions. The first part of the analysis involved estimating the drag polar, and building a graph based on that. Using values given, the lift to drag ratio was found as 14.36. The power required section then involved finding how the power varies with velocity and how efficiency is altered. It was found that the higher the altitude, the higher the altitude was needed to keep steady level flight. The power available section followed the same suit and the same conclusion was held that a higher altitude lead to a higher velocity. When the climb performance section was reached, the rate of climb was found, and the rate of climb vastly got lower as the altitude got higher. The rate of climb was extrapolated to find the absolute and service ceilings, and the integral was used to find the time to climb from 0 to 10000ft. Range and endurance had the Breguet equations involved, and using a variety of parameters, predicted how far the aircraft could go(range) and how long it can stay in the air(endurance). The following task analyzed glide performance and how an aircraft performs in a glide. A hodograph was made using a range of coefficient of lift and coefficient of drag values, and made a vertical and horizontal component of the glide. The best time aloft and the range was found through the minimum sink rate velocity found on the graph, and the lift to drag ratio. The next task analyzed turn performance and used a V-N diagram. The V-N diagram showed the relationship between velocity and the load factor, or how many “g”s an aircraft could sustain. Finally the last section analyzed takeoff and landing, and what parameters lead to it. The final results for takeoff and landing were that a higher altitude airport required more runway, and aircraft with less weight required less runway.

4

INTRODUCTION:

This report took the basics of an aircraft and implored on it. It took the values given and step by step created an overall performance characteristic of the DA-40. The estimations used in the following report were acquired with the highest accuracy, and were put in to allow the most accurate analysis of the DA-40

5

NOMENCLATURE

𝐶𝑑0: The coefficient of profile drag. It represents the drag due to the airplane’s geometry and includes skin friction and form drag.

𝑏: Wingspan. The wingspan of the aircraft, from one end to the other.

𝑆: Wing area. The planform wing area of the wing

𝑆𝑤𝑒𝑡: Wetted area. This area includes everything on the airplane that feels the airflow or if dropped into water, would experience contact with the water.

𝑓: Equivalent parasite area. The area that is considered “clean”

𝐶𝑓: Cleanliness factor. An estimate on how clean the aircraft is

𝑓𝜋: Equivalent parasite area for the specific aircraft part.

𝑒: Oswald efficiency factor. A correction factor due to the wing not being elliptical.

𝑘′′: The change in coefficient of profile drag in respect to the coefficient of lift. Is estimated based on historical data

𝐴𝑅: Aspect Ratio. The wing span squared divided by the wing area

𝜆: Taper ratio. The ratio of wing root to wing tip

𝛿: Factor derived from the taper ratio to estimate the Oswald Efficiency factor

𝐶𝐿: Coefficient of Lift. A numerical value that adjusts for the lift of an aircraft

𝐶𝐷: Coefficient of Drag. A numerical value that adjusts for the drag of an aircraft

𝑃𝑟: Power required. The power that is needed to fly at a given airspeed

𝑃𝐴: Power available. The amount of power available for the aircraft

𝐿𝐷� : Lift to drag ratio. The ratio of lift to drag

𝐿𝐷𝑚𝑎𝑥� : The maximum lift to drag ratio that corresponds to minimum thrust.

𝑉𝑠𝑡𝑎𝑙𝑙: The stall speed of the aircraft, the point where the airflow separates and lift can no longer be produced.

6

𝜌: Density, the ratio of mass per unit volume

𝐶𝐿𝑚𝑎𝑥: The maximum coefficient of lift an aircraft can experience.

𝑉∞: The freestream velocity, or the velocity that is far up the upstream with no deflections from the aircraft.

𝑊: The weight of the aircraft.

𝐸: Endurance. How long an aircraft can stay in flight

𝜂: Propellor efficiency. How efficient a propeller is

𝑐: Specific fuel consumption. The rate which fuel is burned

𝑅: Range. How far an aircraft can go

𝑉𝑡𝑜: Take off velocity, the velocity to take off

𝑆𝐺: Ground roll takeoff distance

𝐹𝑥𝑎𝑣𝑔: Average force of all acting forces

𝜙: Ground roll factor

𝜔: Turn rate, how much an aircraft turns

7

TASK 1: THE DRAG POLAR

The drag polar was found through two main ways of analysis. The first was the wetted area method, and the second was the component buildup method. To expound both ways found the coefficient of profile drag or 𝐶𝑑0. This is the drag from the geometry of the airplane and has no influences from lift. The first method uses an estimation of the wing span, 𝑏 wing area, 𝑆 and the wetted area, 𝑆𝑤𝑒𝑡. The wing area is the area of the

wing, while the wetted area is the area that experiences airflow. Using estimations, the wing span, wing area and wetted area were found to be 39.162𝑓𝑡, 136.11𝑓𝑡2and

630𝑓𝑡2 respectively. From there, the aircraft was then estimated a cleanliness factor. This factor is based on past historical data. The cleanliness factor that was picked was .0049, as the DA-40 had many similarities with the Beech V35. From there using the

formula: 𝑓 = 𝐶𝑓𝑆𝑤𝑒𝑡 = 𝐶𝑑0𝑆,

the coefficient of drag could be found. The 𝐶𝑑0 was found to be .022 from the equation 𝐶𝑓𝑆𝑤𝑒𝑡

𝑆= .0049∗630𝑓𝑡2

136.11𝑓𝑡2= .022.

The component built up method was a method based on historical data. A drag coefficient was associated with each part of the aircraft. The area for each part was also estimated, and then using both values, the equivalent parasite area was found for that part, or 𝑓𝜋. Summing all the equivalent parasite area values, and then adding a 25% for landing gear and 10% for interference drag 𝑓 was found to be 4.8365. Using the area and then using the equation

𝐶𝑑0 = 𝑓𝑆� = 4.8365

136.11𝑓𝑡2� = .035.

Adding the two values of these methods and then averaging them gave us our final coefficient of drag of .029 or .035+.021

2= .02 = 9. The Oswald efficiency factor then

needed to be calculated. The Oswald efficiency factor is noted by the symbol 𝑒, and the equation is 𝑒 = 1

𝑘′′𝜋𝐴𝑅+1+𝛿. 𝑘′′ was estimated from past historical data and was picked as

.01. The aspect ratio was defined as 𝑏2

𝑆. Putting the values into the equation it became

𝑏2

𝑆= 39.162𝑓𝑡2

136.11𝑓𝑡2.

To find delta the tip-chord ratio needed to be found, and by using the geometry of the aircraft, was estimated to be .7. Using the graph for tip chord ratio to delta, delta was

8

found to be .005. These values were put into the equation for 𝑒 and 𝑒 was found to be .71. By using the values given, and finding the drag polar, the maximum lift to drag ratio

was found. It was found to be 14.3365, at a coefficient lift of .81. The following can be seen in the upcoming graphs.

TASK 2: POWER REQUIRED

The velocity for the minimum power required for sea level was 59.111kts. For 5000ft it was 64.5365kts. For 10000ft it was 69.7108kts. To find 𝑉𝑠𝑡𝑎𝑙𝑙, the lift equation must be

-1 -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

→ CD0

Coefficient of lift vs Coefficient of drag

Coefficient of lift

Coe

ffici

ent o

f dra

g

-1 -0.5 0 0.5 1 1.5 2-15

-10

-5

0

5

10

15 → L/DmaxLift to drag ratio vs Coefficient of lift

Coefficient of lift

Lift

to D

rag

ratio

(L/D

)

9

used. The lift equation, will have 𝐶𝐿𝑚𝑎𝑥, which is given to be 1.9. Using each specific density and then using the weight of the aircraft as lift, due to steady level flight, the

velocity can be found. The 𝑉𝑠𝑡𝑎𝑙𝑙 for sea level is

𝑉𝑠𝑡𝑎𝑙𝑙 = �2𝑊

𝜌𝑆𝐶𝐿𝑚𝑎𝑥=�

2∗2645𝑙𝑏𝑠

.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗145.17𝑓𝑡2∗1.9=89.92𝑘𝑛𝑜𝑡𝑠. For 5000ft it’s 𝑉𝑠𝑡𝑎𝑙𝑙 =

�2𝑊


2∗2645𝑙𝑏𝑠

.0020482𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗145.17𝑓𝑡2∗1.9=96.76𝑘𝑛𝑜𝑡𝑠. For 10000ft it’s

𝑉𝑠𝑡𝑎𝑙𝑙 = �2𝑊


2∗2645𝑙𝑏𝑠

.0017556𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗145.17𝑓𝑡2∗1.9=104.52𝑘𝑛𝑜𝑡𝑠.

To find 𝐿 𝐷𝑚𝑎𝑥� , the point of minimum thrust had to be found. This point was found

through making a tangential line for each power required graph, and reading the results off graphically. The 𝐿 𝐷𝑚𝑎𝑥� speeds were found to be 87𝑘𝑛𝑜𝑡𝑠 for sea level, 91𝑘𝑛𝑜𝑡𝑠 for

5000ft, and 96 𝑘𝑛𝑜𝑡𝑠 for 10000ft.

40 60 80 100 120 140 16020

40

60

80

100

120

140

160

180

X: 150Y: 129

velocity in knots

pow

er in

hor

sepo

wer

Velocity vs Power graph

Power required at sea levelPower required at 5000ftPower required at 10000ft

10

TASK 3: POWER AVAILABLE

Minimum Speed Speed for 𝐿/𝐷𝑚𝑎𝑥 Maximum Speed

Altitud

e

𝑉∞ 𝑃𝑅 𝑃𝐴 𝑉∞ 𝑃𝑅 𝑃𝐴 𝑉∞ 𝑃𝑅 𝑃𝐴

Sea

Level

43𝑘𝑡𝑠 44.43ℎ𝑝 44.43ℎ𝑝 87𝑘𝑡𝑠 50.08ℎ𝑝 117.7ℎ𝑝 138𝑘𝑡𝑠 132.1ℎ𝑝 132.1ℎ𝑝

5,000

feet


10,000

feet

52𝑘𝑡𝑠 50.41ℎ𝑝 50.41ℎ𝑝 96

𝑘𝑡𝑠

54.43ℎ𝑝 81.16ℎ𝑝 127𝑘𝑡𝑠 87.41ℎ𝑝 87.41ℎ𝑝

40 60 80 100 120 140 16020

40

60

80

100

120

140

160

180

Velocity(kts)

THP

(hp)


Sea level5000ft10000ft

11

TASK 4: CLIMB PERFORMANCE

A)

B)

40 60 80 100 120 140 160-600

-400

-200

0

200

400

600

800

1000Rate of Climb VS Velocity

Velocity(knots)

Rat

e of

Clim

b(ft/

min

)

RC at sea levelRC at 5000ftRC at 10000ft

0 100 200 300 400 500 600 700 800 900 1000-5000

0

5000

10000

15000

20000

Rate of Climb(ft/min)

Alti

tude

(ft)

Extrapolated rate of climb graph vs altitude

12

Absolute Ceiling: 17400ft

Service Ceiling: 154200ft

C)

17.92 minutes from sea level to 10000ft

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100001

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3x 10

-3

Altitude(ft)

Rat

e of

clim

b- 1(ft

/min

)

Time to Climb graph

13

D)

Rate of climb(ft/min)

Climb angle(degrees)

Velocity(kts)

Best rate of climb angle condition

505.5 .003 47.85

Best climb angle condition

250.14 .0024 29.93

TASK 5: RANGE AND ENDURANCE

To analyze endurance and range the equations for them have to be first introduced. The equation for range is

𝐸 = 𝜂𝑐𝐶𝐿

32�

𝐶𝐷�2𝜌𝑆 (𝑊−1

2� 1 -𝑊−1

2� 0),

where the 1 in the W denotes final weight, and the 0 denotes starting weight. 𝜂 was given as .78, and 𝑐 had to be found by converting the SFC to standard units. SFC was

given as . 49𝑙𝑏

ℎ𝑟�𝑠ℎ𝑝

, and converting it becomes

𝑐 = .45 𝑙𝑏(ℎ𝑝)(ℎ)

1ℎ𝑝550𝑓𝑡𝑙𝑏 𝑠�

13600𝑠

=2.47 ∗ 10−7𝑓𝑡−1.

To find the coefficient of lift and the ocefficient of drag, the velocity for minimum power required needed to be known. The velocity for minimum power required allows for best

20 30 40 50 60 70 80 90-4

-2

0

2

4

6

8

10Climb Hodograph

Horizontal Velocity, VH (knots)

Ver

tical

Vel

ocity

, VV

(kno

ts)

14

endurance, and it was found previously from task 2 as 69.7108kts or 116.45𝑓𝑡 𝑠� . Using

this in the lift equation the coefficient of lift could be found. It became

𝐶𝐿 = 𝐿12𝜌𝑆𝑉

2.

Lift is equal to weight, and the weight is max take off weight-fuel burned or 2645𝑙𝑏𝑠-

270𝑙𝑏𝑠=2375𝑙𝑏𝑠. The density is . 0017556 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� at 10,000ft and the wing area is

given as 145.7𝑓𝑡2. The equation then became

𝐶𝐿 = 2375𝑙𝑏𝑠12 .0017556𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� 145.7𝑓𝑡2116.45𝑓𝑡 𝑠�

2.=1.37.

The coefficient of drag could be found through the drag polar equation or

𝐶𝐷 = 𝐶𝐷0 + 𝐶𝐷𝑖 = 𝐶𝐷0 + 𝐶𝐿2

𝜋𝐴𝑅𝑒. 𝐶𝐷0

is given as .0300, oswald efficiency factor as .75, aspect ratio as

𝑏2

𝑆= 39.172𝑓𝑡2

145.7𝑓𝑡2=10.53.

𝐶𝐷 = .0300 + 1.372

𝜋∗10.53∗.75=.1056.

The last things are the weight final and weight initial. The weight initial is max takeoff weight at 2645lbs. With 90% fuel used, 270lbs of fuel was burned, so the final weight

becomes 2375lbs. Putting all these values into the final equation, it becomes 𝐸 =.78

2.47∗10−7𝑓𝑡−11.37

32�

..1056 �2 ∗ .0017556 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗ 145.7𝑓𝑡2 (2645−1 2� 1 -

2375−1 2� 0)=10.27hours. The range equation follows the same procedure, and the range is found out by 𝑅 = 𝜂

𝐶𝐶𝐿𝐶𝐷𝑙𝑛 𝑊0

𝑊1. All the previous values are used except the coefficient lift

and coefficient of drag. For that the velocity for minimum thrust had to be used in the lift equation to find coefficient of lift and drag. The velocity for minimum thrust at 10,000ft is

96 knots or 162.02𝑓𝑡 𝑠� . 𝐶𝐿 = 2375𝑙𝑏𝑠12 .0017556𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� 145.7𝑓𝑡2162.02𝑓𝑡 𝑠�

2 = .707. Plugging in the

same drag polar equation: 𝐶𝐷 = .0300 + .7072

𝜋∗10.53∗.75= .0501. The final range equation

15

becomes 𝑅 = .782.47∗10−7𝑓𝑡−1

.77.0544

𝑙𝑛 26452375

=5225886ft or

491526𝑓𝑡 ∗ 1𝑚𝑖𝑙𝑒6076𝑓𝑡

= 860.08𝑠𝑡𝑎𝑡𝑢𝑡𝑒 𝑚𝑖𝑙𝑒𝑠

TASK 6: GLIDE PERFORMANCE

A)

B) The best range for a glide is defined as

𝑅𝑚𝑎𝑥=h(𝐿𝐷𝑚𝑎𝑥

).

𝐿𝐷𝑚𝑎𝑥

from task 1 is found out to be 14.36. The differential change in height is 4000ft.

Multiply the two values for it to be 𝑅𝑚𝑎𝑥=h(𝐿

𝐷𝑚𝑎𝑥)=4000𝑓𝑡 ∗ 14.36 = 57360𝑓𝑡 𝑜𝑟 9.44 𝑛𝑎𝑢𝑡𝑖𝑐𝑎𝑙 𝑚𝑖𝑙𝑒s

100 150 200 250 300 350 400

0

20

40

60

80

100

120

Glide Hodograph(sea level)

Vh(ft/s)

Vv(

ft/s)

16

To maximize glide distance, the velocity for 𝐿𝐷𝑚𝑎𝑥

needs to be found. This is found by

drawing a tangential line from the origin with 𝜃𝑚𝑖𝑛. Where the 𝜃𝑚𝑖𝑛 meets the graph is the velocity where 𝐿

𝐷𝑚𝑎𝑥 occurs and can be extrapolated from the graph. The velocity is

122.76 knots. For the maximum time aloft, the minimum 𝑉𝑉 needs to be found as that allows the lowest sink rate. Extrapolating from the graph again the minimum 𝑉𝑉 is 63.63knots. If the maximum time aloft velocity is used, and a distance of 4000ft is covered, it becomes 4000𝑓𝑡

107.4𝑓𝑡 𝑠�=37.24seconds

100 150 200 250 300 350 400

0

20

40

60

80

100

120

Glide Hodograph(5000ft)

Vh(ft/s)

Vv(

ft/s)

17

TASK 7: TURN PERFORMANCE

The minimum turn radius is defined by the equation

𝑅𝑚𝑖𝑛 =2(𝑊𝑆 )

𝑔𝜌𝐶𝐿𝑚𝑎𝑥=

2( 2645𝑙𝑏𝑠145.17𝑓𝑡2

)

32.2𝑓𝑡 𝑠2� ∗.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� 1.9=250.58𝑓𝑡

The maximum turn rate is defined by the equation:

𝜔𝑚𝑎𝑥 = 𝑔�𝑛𝑚𝑎𝑥𝜌𝐶𝐿𝑚𝑎𝑥

2𝑊 𝑆�= 32.2 𝑓𝑡 𝑠2� �

3.8∗.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗1.9

2∗2645𝑙𝑏𝑠 145.17𝑓𝑡2�= .69 𝑟𝑎𝑑𝑖𝑎𝑛𝑠/𝑠𝑒𝑐𝑜𝑛𝑑

0 50 100 150 200 250 300-2

-1

0

1

2

3

4

→ Corner velocity

Velocity(freestream)

load

fact

orV-n diagram for DA-40

18

TASK 8: TAKEOFF AND LANDING PERFORMANCE

Since takeoff and landing performance no longer assumes steady level flight, new 2parameters must be put in. The equation for ground take off distance is

𝑆𝐺 = 𝑊𝑉𝑡𝑜2

2𝑔𝐹𝑎𝑣𝑔.

The derivation for takeoff velocity is

𝑉𝑡𝑜 = 1.2𝑉𝑠𝑡𝑎𝑙𝑙 = 1.2�2𝑊/𝑆𝜌𝐶𝐿𝑚𝑎𝑥

.

To find the average force, all forces have to reach an equilibrium or

𝐹𝑥𝑎𝑣𝑔 = 𝑇𝑎𝑣𝑔 − 𝐷𝑎𝑣𝑔 − 𝑅𝑎𝑣𝑔, which is simplified to

𝐹𝑥𝑎𝑣𝑔 = 𝑇.707𝑉𝑡𝑜 −12𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐷 − 𝜇[𝑊 − 1

2𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐿].

The coefficient of lift and coefficient of drag have one more factor effecting them, and that’s the ground effect factor or 𝜙. This factor is due to wind tip vortices are interfered by the ground, and less drag occurs. The modified forms of coefficient of lift and coefficient of drag are:

𝐶𝐿 = 12𝜙𝜋𝐴𝑅𝑒𝜇 and 𝐶𝐷 = 𝐶𝐷0 + 𝜙𝐶𝐿2

𝜋𝐴𝑅𝑒.,

and these are considered the optimum lift and drag coefficients for minimizing ground roll The ground effect factor is defined as

𝜙 =(16ℎ 𝑏� )2

1+(16ℎ 𝑏� )2,

where h is the height of the wing off the ground and b is the wingspan.

A) The first part of the analysis is finding out the ground effect factor. The height the wing is above the ground can be estimated at 6ft, and the wing span is given as 39. 17𝑓𝑡2. The ground effect factor becomes

𝜙 =(16ℎ 𝑏� )2

1+(16ℎ 𝑏� )2=

(16∗6𝑓𝑡 39.17𝑓𝑡2� )2

1+(16∗6𝑓𝑡 39.17𝑓𝑡2� )2=.86.

B) The average velocity must be found. The take off velocity is shown as 𝑉𝑡𝑜 =

1.2𝑉𝑠𝑡𝑎𝑙𝑙 = 1.2�2𝑊/𝑆𝜌𝐶𝐿𝑚𝑎𝑥

=1.2�2∗( 2645𝑙𝑏𝑠

145.7𝑓𝑡2)

.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗1.9= 107.904ft/s.

19

Using the factor .707 to get the average velocity it becomes 76.288ft/s. To find the coefficient of lift, a coefficient of rolling friction of .02 can be used. The coefficient of lift becomes

𝐶𝐿 = 12𝜙𝜋𝐴𝑅𝑒𝜇 = 1

2∗.86𝜋 ∗ 10.53 ∗ .75 ∗ .02=.288.

The coefficient of drag is

𝐶𝐷 = 𝐶𝐷0 + 𝜙𝐶𝐿2

𝜋𝐴𝑅𝑒.,= .0300 + .86∗.2882

𝜋∗10.53∗.75=.03288.

The thrust can be found through the power curve with the given given velocity. Through the equation 𝑃

𝑉=T, and power being

𝑠ℎ𝑝𝑎 ∗ 𝜂 = 180 ∗ .78(1 − �35𝑉∞�2

= 56.18ℎ𝑝

thrust in standard units is 30898𝑓𝑡𝑙𝑏𝑠76.288𝑓𝑡 𝑠�

=T=405.02bs.

Now the average force can be calculated as 𝐹𝑥𝑎𝑣𝑔 = 𝑇.707𝑉𝑡𝑜 −

12𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐷 − 𝜇[𝑊 − 1

2𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐿]=405. 02𝑙𝑏𝑠 − 1

2∗

.0023769 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗ 76.288 𝑓𝑡 𝑠�2∗ 145.17𝑓𝑡2 ∗ .03288 − .02 �2645𝑙𝑏𝑠 − 1

2∗

.0023769 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗ 76.288 𝑓𝑡 𝑠�2∗ 145.17𝑓𝑡2 ∗ .288� = 345.05 .

Putting it into the final ground roll equation, it’s is

𝑆𝐺 = 2645𝑙𝑏𝑠∗107.904𝑓𝑡 𝑠�2

2∗32.2𝑓𝑡 𝑠2∗� 345.05𝑏𝑠=1385𝑓𝑡

C) The same analysis applies, except with the density changing at 5000ft, some values vary. The ground effect stays the same as

𝜙 =(16ℎ 𝑏� )2

1+(16ℎ 𝑏� )2=

(16∗6𝑓𝑡 39.17𝑓𝑡2� )2

1+(16∗6𝑓𝑡 39.17𝑓𝑡2� )2=.86.

The velocity take off becomes


=1.2�2∗( 2645𝑙𝑏𝑠

145.7𝑓𝑡2)

.0020482𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗1.9= 115.908ft/s,

and the average velocity is 115.908 ∗ .707 = 81.94𝑓𝑡𝑠

. The optimum coefficient of

lift and drag remain the same, while the average force varies with density. 𝑃𝑉=T,

and power being

𝑠ℎ𝑝𝑎 ∗ 𝜂 = 150 ∗ .78(1 − �35𝑉∞�2

= 56.16ℎ𝑝.

20

The thrust changes with this new velocity and becomes 30893𝑓𝑡𝑙𝑏𝑠

81,94𝑓𝑡 𝑠�=T=377.34lbs.

The average force becomes 𝐹𝑥𝑎𝑣𝑔 = 𝑇.707𝑉𝑡𝑜 −

12𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐷 − 𝜇[𝑊 − 1

2𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐿]= 642.04𝑙𝑏𝑠 − 1

2∗


2∗

.0020842 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗ 81.94 𝑓𝑡 𝑠�2∗ 145.17𝑓𝑡2 ∗ .288� =262.54lbs.

The final ground roll becomes

𝑆𝐺 = 2645𝑙𝑏𝑠∗115.908𝑓𝑡 𝑠�2

2∗32.2𝑓𝑡 𝑠2∗� 561.64𝑙𝑏𝑠=1726ft

D) Now with the weight changing the velocity and the average force changes. The ground effect stays the same, and the optimum coefficient of lift and drag stay the same. With the modified weight the new velocity is


=1.2�2∗( 2400𝑙𝑏𝑠

145.7𝑓𝑡2)

.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗1.9= 102.49𝑓𝑡/𝑠

and the average velocity is .707*102.49𝑓𝑡/𝑠=72.46ft/s.

Power is

𝑠ℎ𝑝𝑎 ∗ 𝜂 = 180 ∗ .78(1 − �35𝑉∞�2

= 47.07ℎ𝑝

Thrust becomes 25893𝑓𝑡𝑙𝑏𝑠

72.46𝑓𝑡 𝑠�=T=357.34lbs.

The average force is then

𝐹𝑥𝑎𝑣𝑔 =

𝑇.707𝑉𝑡𝑜 −12𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐷 − 𝜇[𝑊 − 1

2𝜌𝑉2.707𝑡𝑜𝑆𝐶𝐿]= 357.34 − 1

2∗


2∗

21

.0023769 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗ 72.46 𝑓𝑡 𝑠�2∗ 145.17𝑓𝑡2 ∗ .288�=280.511lbs. Putting in the

final ground take off equation𝑆𝐺 = 2400𝑙𝑏𝑠∗102.49𝑓𝑡 𝑠�2

2∗32.2𝑓𝑡 𝑠2∗� 280.511𝑙𝑏𝑠=1395.5ft

𝐃) Much of the analysis for landing is the same as takeoff, with a few key differences. First the equation for the ground roll is

𝑆𝐿 = − 𝑊2𝑔𝐹𝑎𝑣𝑔

𝑉𝑇𝐷2.

The average force equation is 𝐹𝑎𝑣𝑔 = −1

2𝜌𝑉.707𝑣𝑡𝑑

2𝑆𝐶𝐷 − 𝜇[𝑊 −−12𝜌𝑉.707𝑣𝑡𝑑

2𝑆𝐶𝐿]. The same optimum of coefficient of lift and drag from the takeoff performance can be used. The rolling friction while breaking can be estimated at .25. The

velocity while landing is

𝑉𝑇𝐷 = 1.3𝑉𝑠𝑡𝑎𝑙𝑙 = 1.3�2𝑊/𝑆𝜌𝐶𝐿𝑚𝑎𝑥

=1.3�2∗( 2645𝑙𝑏𝑠

145.7𝑓𝑡2)

.0023769𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗1.9=116.56ft/s,

and the average is .707*116.56kts=82.40ft/s. Putting the values into the equation, it’s

𝐹𝑎𝑣𝑔 = −12𝜌𝑉.707𝑣𝑡𝑑

2𝑆𝐶𝐷 − 𝜇[𝑊 −−12𝜌𝑉.707𝑣𝑡𝑑

2𝑆𝐶𝐿]= −12∗ 0023769 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗

139.07 𝑓𝑡 𝑠�2∗ 145.17𝑓𝑡2 ∗ .03288 − .25[2645𝑙𝑏𝑠 − − 1

2∗ 0023769 𝑠𝑙𝑢𝑔𝑠 𝑓𝑡3� ∗

82.402 ∗ 145.17𝑓𝑡2 ∗ .288]=-304.651lbs. Putting into the final equation

𝑆𝐿 = − 2645𝑙𝑏𝑠

2∗32.2𝑓𝑡 𝑠2� 304.651𝑙𝑏𝑠116.562=1831.62ft

22

Conclusion:

The DA-40 performed to expectations. Many of the DA-40 performance values matched how an aircraft of its configuration would perform in real life. The multi-faceted analysis allowed the student to take the DA-40 with very base information and expand on it and make a full detailed report on its performance. These aspects included the drag polar, power required, power available, climb performance, range and endurance, glide performance, turn performance and take off and landing performance. The DA-40 was tested on these very principles and showed and explained to the writer how an aircraft functions and the complexities behind it. The writer learned how an aircraft performs in real world conditions and how using analysis techniques can lead to results on aircraft performance.

23

LIST OF FIGURES:

Page 8:

Page 8:

-1 -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

→ CD0


Coefficient of lift

Coe

ffici

ent o

f dra

g

24

Page 9:

Page 10:

-1 -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

→ CD0


Coefficient of lift

Coe

ffici

ent o

f dra

g

40 60 80 100 120 140 16020

40

60

80

100

120

140

160

180

X: 150Y: 129

velocity in knots

pow

er in

hor

sepo

wer


Power required at sea levelPower required at 5000ftPower required at 10000ft

25

Page 11:

Page 11:

40 60 80 100 120 140 16020

40

60

80

100

120

140

160

180

Velocity(kts)

THP

(hp)


Sea level5000ft10000ft

40 60 80 100 120 140 160-600

-400

-200

0

200

400

600

800

1000Rate of Climb VS Velocity

Velocity(knots)

Rat

e of

Clim

b(ft/

min

)

RC at sea levelRC at 5000ftRC at 10000ft

26

Page 12:

Page 13:

0 100 200 300 400 500 600 700 800 900 1000-5000

0

5000

10000

15000

20000

Rate of Climb(ft/min)

Alti

tude

(ft)

Extrapolated rate of climb graph vs altitude

0 1000 2000 3000 4000 5000 6000 7000 8000 9000 100001

1.2

1.4

1.6

1.8

2

2.2

2.4

2.6

2.8

3x 10

-3

Altitude(ft)

Rat

e of

clim

b- 1(ft

/min

)

Time to Climb graph

27

Page 15:

Page 16:

20 30 40 50 60 70 80 90-4

-2

0

2

4

6

8

10Climb Hodograph

Horizontal Velocity, VH (knots)

Ver

tical

Vel

ocity

, VV

(kno

ts)

100 150 200 250 300 350 400

0

20

40

60

80

100

120

Glide Hodograph(sea level)

Vh(ft/s)

Vv(

ft/s)

28

Page 17:

LIST OF FIGURES:

100 150 200 250 300 350 400

0

20

40

60

80

100

120

Glide Hodograph(5000ft)

Vh(ft/s)

Vv(

ft/s)

0 50 100 150 200 250 300-2

-1

0

1

2

3

4

→ Corner velocity

Velocity(freestream)

load

fact

or

V-n diagram for DA-40

29

Page 10:

Minimum Speed Speed for 𝐿/𝐷𝑚𝑎𝑥 Maximum Speed

Altitud

e

𝑉∞ 𝑃𝑅 𝑃𝐴 𝑉∞ 𝑃𝑅 𝑃𝐴 𝑉∞ 𝑃𝑅 𝑃𝐴

Sea

Level


5,000

feet


10,000

feet

52𝑘𝑡𝑠 50.41ℎ𝑝 50.41ℎ𝑝 96

𝑘𝑡𝑠

54.43ℎ𝑝 81.16ℎ𝑝 127𝑘𝑡𝑠 87.41ℎ𝑝 87.41ℎ𝑝

Page 13:

Rate of climb(ft/min)

Climb angle(degrees)

Velocity(kts)

Best rate of climb angle condition

505.5 .003 47.85

Best climb angle condition

250.14 .0024 29.93

30

CODE:

Task 1:

% Take home project clc; clear; % Purpose: To find the relationship between coefficient of lift and drag % Author: Andrew George % Functions: No user defined functions CD0=.0300; e=.75; S=145.7; b=39.17; AR=(b^2)/S; CL=(-1:.1:1.9); CD=CD0+((CL.^2)/(pi*e*AR)); plot([CL],[CD]) text(0,.0300,' \rightarrow CD0','FontSize',10) title('Coefficient of lift vs Coefficient of drag') xlabel('Coefficient of lift') ylabel('Coefficient of drag')

31

LD=CL./CD; plot([CL],[LD]) text(.8,14.34,' \rightarrow L/Dmax','FontSize',14) maxLD=max(LD); title('Lift to drag ratio vs Coefficient of lift') xlabel('Coefficient of lift') ylabel('Lift to Drag ratio(L/D)') Task 2:

% Purpose: To find the relationship between power and speed % Author: Andrew George % Functions: No user defined functions shpa1=180; shpa2=150; shpa3=120; V=[40:1:150]; Vf=V.*(6076/3600); n=.78.*(1-((35./V).^2)); P1=shpa1.*n; P2=shpa2.*n; P3=shpa3.*n; S=145.7; b=39.17; AR=(b^2)/S; CD0=.0300; W=2645; e=.75; pi=3.14; rho1=.0023769; rho2=.0020482; rho3=.0017556; Pr1=((((.5*rho1*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho1*S*3.14*AR*e)).*Vf.^-1)./550); Pr2=((((.5*rho2*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho2*S*3.14*AR*e)).*Vf.^-1)./550); Pr3=((((.5*rho3*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho3*S*3.14*AR*e)).*Vf.^-1)./550); plot([V],[Pr1],[V],[Pr2],[V],[Pr3]) xlabel('velocity in kts') ylabel('power in horsepower') title('Velocity vs Power graph') legend('Power required at sea level','Power required at 5000ft','Power required at 10000ft',0) C1=(4*W^2)/(3*(rho1^2)*(S^2)*pi*AR*e*CD0); C2=(4*W^2)/(3*(rho2^2)*(S^2)*pi*AR*e*CD0); C3=(4*W^2)/(3*(rho3^2)*(S^2)*pi*AR*e*CD0); VPrmin1=C1^(1/4); VPrmin2=C2^(1/4); VPrmin3=C3^(1/4);

32

Vparmin1=VPrmin1*(3600/6076); Vparmin2=VPrmin2*(3600/6076); Vparmin3=VPrmin3*(3600/6076); display(Vparmin1) display(Vparmin2) display(Vparmin3) Task 3:

shpa1=180; shpa2=150; shpa3=120; V=[40:1:150]; Vf=V.*(6076/3600); n=.78.*(1-((35./V).^2)); P1=shpa1.*n; P2=shpa2.*n; P3=shpa3.*n; S=145.7; b=39.17; AR=(b^2)/S; CD0=.0300; W=2645; e=.75; pi=3.14; rho1=.0023769; rho2=.0020482; rho3=.0017556; Pr1=((((.5*rho1*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho1*S*3.14*AR*e)).*Vf.^-1)./550); Pr2=((((.5*rho2*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho2*S*3.14*AR*e)).*Vf.^-1)./550); Pr3=((((.5*rho3*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho3*S*3.14*AR*e)).*Vf.^-1)./550); plot([V],[P1],'r',[V],[P2],'b',[V],[P3],'c',[V],[Pr1],'r',[V],[Pr2],'b',[V],[Pr3],'c') xlabel('Velocity(kts)') ylabel('THP(hp)') title('Velocity vs Power graph') legend('Sea level','5000ft','10000ft',0) Task 4: shpa1=180; shpa2=150; shpa3=120; V=[40:1:150]; Vf=V.*(6076/3600); n=.78.*(1-((35./V).^2)); P1=shpa1.*n; P2=shpa2.*n; P3=shpa3.*n;

33

S=145.7; b=39.17; AR=(b^2)/S; CD0=.0300; W=2645; e=.75; pi=3.14; rho1=.0023769; rho2=.0020482; rho3=.0017556; Pr1=((((.5*rho1*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho1*S*3.14*AR*e)).*Vf.^-1)./550); Pr2=((((.5*rho2*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho2*S*3.14*AR*e)).*Vf.^-1)./550); Pr3=((((.5*rho3*S*CD0).*Vf.^3)/550)+(((W^2)/(.5*rho3*S*3.14*AR*e)).*Vf.^-1)./550); RC1=33000*(P1-Pr1)/W; RC2=33000*(P2-Pr2)/W; RC3=33000*(P3-Pr3)/W; %Velocity VS RC Graph plot([V],[RC1],[V],[RC2],[V],[RC3]) title('Rate of Climb VS Velocity') xlabel('Velocity(kts)') ylabel('Rate of Climb(ft/min)') legend('RC at sea level','RC at 5000ft','RC at 10000ft') maxRC1=max(RC1) maxRC2=max(RC2) maxRC3=max(RC3) display(maxRC1) display(maxRC2) display(maxRC3) %Rate of Climb extrapolated x=[853.2105,614.8650,364.0977]; y=[0,5000,10000]; new_x=linspace(0,1000); coeffs1=polyfit(x,y,1); new_y=polyval(coeffs1,new_x); plot(x, y, 'rx'); plot(new_x, new_y, 'k'); xlabel('Rate of Climb(ft/min)') ylabel('Altitude(ft)') title('Extrapolated rate of climb graph vs altitude') %Time to climb plot(y,x) Z=trapz(y,x); display(Z) xlabel('Altitude(ft)') ylabel('Rate of climb(ft/min') title('Time to Climb graph')

34

Vv = RC1*(60/6076.12); %knots Vh = (V.^2-(Vv*6076.12/3600).^2).^(1/2)*(3600/6076.12); %knots plot(Vh,Vv) title('Climb Hodograph') xlabel('Horizontal Velocity, Vh (knots)') ylabel('Vertical Velocity, Vv (knots)') grid on display(Vh) display(Vv) Task 6:

Cl=[.1:.1:1.9]; Cd0=.0300; pi=3.14; AR=10.56; e=.75; W=2645; S=145.17; WS=W/S; rho=.0023769; Cd=Cd0+((Cl.^2)/(pi*AR*e)); R=Cl./Cd; theta=atan(1./R); display(Cl) display(Cd) display(theta) c=cos(theta); display(c) Vstream=(((2.*c.*WS)./(rho.*Cl)).^.5); display(Vstream) Vh=Vstream.*cos(theta); Vv=Vstream.*sin(theta); plot(Vh,Vv) axis ij axis([80,400,0,125]) title('Glide Hodograph') xlabel('Vh(ft/s)') ylabel('Vv(ft/s)') %5000ft Cl=[.1:.1:1.9]; Cd0=.0300; pi=3.14; AR=10.56; e=.75; W=2645;

35

S=145.17; WS=W/S; rho=.0020842; Cd=Cd0+((Cl.^2)/(pi*AR*e)); R=Cl./Cd; theta=atan(1./R); display(Cl) display(Cd) display(theta) c=cos(theta); display(c) Vstream=(((2.*c.*WS)./(rho.*Cl)).^.5); display(Vstream) Vh=Vstream.*cos(theta); Vv=Vstream.*sin(theta); plot(Vh,Vv) axis ij axis([80,400,0,125]) title('Glide Hodograph(5000ft)') xlabel('Vh(ft/s)') ylabel('Vv(ft/s)')

Task 7:

Vex=178; Vex1=178*(6076/3600); V1=0:.1:Vex; nmax=3.8; nmin=-1.52; W=2645; S=145.17; WS=W/S; rho=.0023769; Cl=1.9; Cln=-1.9; CV=(((2*nmax*WS)/(rho*Cl))^.5); VC=(0:1:CV)*(6076/3600); npositive=(.5.*rho.*VC.^2.*S.*Cl)./W; nCV=(((2*-1*nmin*WS)/(rho*Cl))^.5); NVC=(0:1:nCV)*(6076/3600); nnegative=(.5.*rho.*NVC.^2.*S.*Cln)./W; plot(VC,npositive,'r',NVC,nnegative,'b',[VC(end),500],[10.81,10.81],'r',[NVC(end),500],[-4.272,-4.272],'b',[0,500],[0,0],'c',[295.4,295.4],[0,10.81]) axis([0 500 -6 12])

36

text(295.4,0,' \rightarrow Corner velocity','FontSize',12) xlabel('Velocity(freestream)') ylabel('load factor') title('V-n diagram for DA-40')

REFERENCES:

Anderson, John. Introduction to Flight. 7th ed. New York NY: McGraw-Hill Companies, 2007.

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george andrew analysis of da40 (3)

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