Design Lab- Aero

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Spring 2010

Cory Brabon | Nick Deweese | Dilruhan Weerasinha

UNIVERSITY

AT

BUFFALO

MAE 434: AIRCRAFT DESIGN| EXAM 3



MAE 434: Aircraft Design Exam #3

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Contents

Part 1: 3-View Drawing ................................................................................................................................. 2

Part 2: Class I Weight and Balance Method .................................................................................................. 4

Part 3: Stability ............................................................................................................................................ 10

Part 4: Drag Polars ...................................................................................................................................... 15

Part 5: V-n diagrams.................................................................................................................................... 20

Part 6: How Do Airplanes Fly? ..................................................................................................................... 24

References .................................................................................................................................................. 25




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Part 1: 3-View Drawing

X (F.S.)

Z (W.L.)

Y (B.L)

X (F.S)

Y (B.L)

Z (W.L)




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For the three view drawing, each reference axis was spaced 5 ft from the plane in order to leave space

for future changes on the plane without having to compute negative center of gravities.

Wing Horizontal Tail Vertical Tail

Area 353 71.70 66.37

Span 53.2 17.84 17.35

MGC 6 4.13 7.43

AR 8 4.2 1.2

Sweep Angle 0 20 2.5

Taper Ratio .5 .5 .6

Thickness Ratio Root .18 .12 .15

Tip .12 - -

Airfoil Root NACA 9278 NACA 0012 NACA 0015

Tip NACA9212 - -

Dihedral Angle 3 0 -

Incidence Angle 3.5 VARIABLE -

Aileron Chord Ratio .345 Elevator Chord Rudder Chord

Aileron Span Ratio .75-.98 Ratio .47 Ratio .39

Flap Chord Ratio .345 - -

Flap Span Ratio .2 - -

Fuselage Cabin Interior Overall

Length 78.6 40 78.6

Maximum Height 8 6 18.78

Maximum Width 8 7.2 53.2




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Part 2: Class I Weight and Balance Method

Step 1

To determine the component weight break down, first the preliminary values needed to be recorded.

WTO 31,775 lbs WF 8,252 lbs

WE 18,863 lbs WPL 4,300 lbs

The payload is a bit higher because it also includes the weight of the crew. Also the gross weight or GW

will be equal to the takeoff weight due to this being more common among regional turboprops. This is

supported by Table A6.1

Considering Tables A6.1a and A6.1b, our plane is most similar to the Grumman G-I with similar takeoff

weight and empty weight. Using the data to this aircraft, several weight fractions were determined for

major components.

Wight fractions:

Structure/GW .304

Power Plant/GW .129

Fixed Equipm’t/GW .097

Empty Weight/GW .624

Wing Group/GW .100

Empenn. Group/GW .025

Fuselage Group/GW .112

Nacelle Group/GW .032

Land. Gear Group/GW .034

Multiplying these fractions by our planes GW gave us the component weights for our plane. These

weights were characterized and then broken down into smaller components based on the three major

sections; structure, power plant and fixed equipment.




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Component Weights:

Number of Engines: 2

Weight Item, lbs Final Values

Wing Group 3,189 +515 3,704

Empennage Group 794 +80 874

Fuselage Group 3,579 +645 4,224

Narcelle Group 1,017 +120 1,137

Land Gear Group 1,080 +140 1,220

Nose Gear 198 +40 238

Main Gear 882 +100 982

Structure Total 9,659 +1500 11,159

Engines 2,318 +310 2,628

Air Induct. System -

Fuel System 123 +25 148

Propeller Install. 799 +84 883

Propulsion System 859 +104 963

Power Plant Total 4,099 +523 4,622

Avionics + Instruments 90 90

Surface Controls 431 431

Hydraulic System 335 335Pneumatic System

Electrical System 765 765

Electronics 95 95

APU 384 384

Air Cond. System 436 436

Anti-icing System 177 177

Furnishings 353 353

Auxiliary Gear 16 16

Fixed Equipm’t Total 3,082 3,082

Total 16,840Under by 2,023

Woil + Wtfo 300

Max. Fuel Capacity 8,252

Maximum Payload 3,775




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Due to the fractions not being perfectly similar to our plane along with round off errors the total empty

weight was less than our actual empty weight. To account for this, the difference for this weight was

divided up similarly to how it was done before and then distributed.

Based our mission specifications, our plane is expected to be used in potentially dangerous missions and

was expected to have some armor on the fuselage. This is why the fuselage weight was shifted to

account for more weight. This effected the weight of the wing, nacelle and empennage groups but not

as heavy of an effect on the landing gear. This was not changed since the aircraft would need, if anything

stronger landing gear to account for the heavy fuselage. Also our landing gear was designed to have

eight tires on the main gear which is why they have a much greater weight when compared to the front

landing gear.

Moving onto the power plant weights, these were determined using historical data of the Grumman G-I

similarly to how the main components were estimated. The weights were shifted based on engine

specifications from Test 2 as well which amped up the total weight of the power plant to 4,622 lbs. This

was the same for the fixed equipment, which followed very similar trends as the Grumman G-I. Weightwas not added onto the fixed equipment after redistributing accounted weight sense this component

was not expected to change much due to a small increase in total weight.

Step 2

Using the 3-view drawing prom part one, the centers of gravity for the major components are shown,

based on calculations in the next step.




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Wing Fuselage V.T. H.T. LG

Main

LG

Front

Engine Fixed

Equip

Fuel Payload

1 2 3 4 5 6 7 8 9 10

These numbers correspond with the diagram above.

Step 3-4

These steps encompass locating the center of gravity for each major component for each major axis.

This was done by first finding the centre of each part using data from part 1 and multiplying it by the

corresponding weight from step 1. These values have been tabulated below

Component Weight X Wx Y Wy Z Wz

Lbs ft ft-lbs ft ft-lbs ft ft-lbs

Wing 3,704 30.5 112,972 - - 15.9 58,708

Empennage H. T. 577 69.6 40,159 - - 12.4 7,172V.T 297 70.6 20,968 - - 18.2 5,405

Fuselage 4,224 35.4 149,529 - - 12.4 52,504

Landing Gear

Nose238 13 3,094 - - 8.4 1,999

Main 982 40.6 39,898 - - 8.4 8,248

Engines+nacelle 4,876 31 151,156 - - 14.4 69,970

Fixed EQ 3,082 31 95,542 - - 10.4 32,145

Propellers 883 27 23,841 - - 14.4 12,671

Empty Weight 18,863 349 637,161 - - 114.8 248,825

TFO 360 31 11,160 - - 14.4 5,166

Fuel 8,252 30.5 251,686 - - 15.9 130,794

Passengers 2x 350 27 9,450 - - 12.4 4,350

2x 350 29 10,150 - - 12.4 4,350

2x 350 31 10,850 - - 12.4 4,350

2x 350 33 11,550 - - 12.4 4,350

2x 350 35 12,250 - - 12.4 4,350

2x 350 37 12,950 - - 12.4 4,350

2x 350 39 13,650 - - 12.4 4,350

2x 350 41 14,350 - - 12.4 4,350

1x 175 43 7,525 - - 12.4 2,175Equipment 800 28 22,400 - - 12.4 9,944

Crew 525 12 6,300 - - 12.4 6,525

Take-off Weight 31,775 416.5 1,031,432 - - 167 438,234

Note: The reference point was determined in part one, each major axis is backed off the plane by 5ft.




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Step 5

Xcg WE 33.8 ft Xcg WE+crew 33.2 ft Zcg WE 13.2 ft

Xcg WTO 32.5 ft Xcg WE+crew+fuel 32.4 ft Zcg WTO 13.8 ft

The x, y and z center of gravity for our plane were calculated and tabulated in the table above using data

from step 4. The Xcg was determined for four situations based on the empty weight, takeoff weight,

empty weight + crew and empty weight + crew + fuel.

Step 6

From the cg diagram you can see that the main gear was a little overshot and didn’t need to be so far

back on the plane. Adjusting this forward would also improve the cg, the only issue is with matching

historical data in terms of distance between the main gear and front gear which would be off if the gear

was moved.




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Step 7

Most forward c.g. occurs at W = 28,000 lbs at 32.4 ft and .317cw

The most aft c.g. occurs at W = 18,863 lbs at 33.8 ft and .55cw

The c.g. range for our airplane is 1.4 ft or .233cw

Table 10.3 shows regional TBP’s with a range of 12-20in or 1-1.67ft. Our cg range falls right in-between.

Also the c.g. range in terms of chord length in table 10.3 is .14-.27cw which corresponds with our data.

Step 8

Overall the plane seems very feasible, the only thing that seems to be an issue is the distance from the

aft cg and the main landing gear. Also the wing is just forward of the cg range which could cause some

issues. It’s important to keep the landing gear wing and center of gravity close due to structural

integrity. The plane is the strongest at this joint and needs to balance all of its moment arms and weight

here. As far as room goes for the landing gear, there should be no issue with this since most of the carry

on equipment is stored with the passengers not under the plane. Also the fixed equipment is loaded

mostly towards the front of the plane leaving ample space for the landing gear.

As far as a tip over issue there is no issue due to the landing gear, in fact there is an abundant amount of

stability due to an aft main gear. This would cause an issue during landing as far as structural integrity,

putting too much force on the fuselage too far away from the center of gravity.

Therefore the new position of the main landing gear will be at 35 ft from the x reference.




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Part 3: Stability

The following data is given:

From the previous exam:

Sh = 71.70ft2

Sv = 66.37ft2

S = 353.06ft2

Sh/S = 0.203

ft

Vh = 1.10

Xh = 32.5

**Assume a.c. of tail is at the mean quarter chord of each surface and the wing-fuselage a.c. is at a point

75% of the distance from the nose to the intersection of the of the wing with the fuselage**




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Step 1:

Longitudinal X-plot

From Previous Part:

5%

0.317

-0.875

33.3




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*For tail-aft airplane, Sc = 0 and equation (11.1) reduces to:

[ ]

[ ]

Therefore:

{ [ ] }

Step 2

This aircraft is designed to be inherently stable because a feedback augmentation system would not be

cost effective for our airplane design.

Step 3

This aircraft is considered a regional turboprop aircraft and therefore is in category 6 listed in chapter 2.

Step 4 is not required




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Step 5

33.3ft2

Empennage area = 33.3ft2


Step 7

The calculated area for our empennage is not within the 10% range when compared to the V-method

value so a review of the airplane weight and balance must be made. A problem I had making the

longitudinal plot that may have attributed to this difference in area is that our cg calculated in theprevious section is much farther behind the aerodynamic center of the aircraft, which in turn makes the

aircraft unstable, this should also be reviewed.

Step 8:

Xv = 64.4ft-24.3ft = 40.1ft

b = 53.2ft

* taken from historical data = 3.0 per radian.

*Assume overall level of directional stability is:




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Directional X-plot

Step 9:

Again, inherent directional stability would be more cost effective for this aircraft design rather than an

augmentation system.

Step 10:

As shown in step 8, the previous calculated area of vertical tail is larger than what is needed, 54.3ft2

would be sufficient for the assumed directional stability.


Calculation of Yawing Moment due to sideslip coefficient:

-17.19

54.3




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Part 4: Drag Polars

Total Wetted Area

Wing

The wetted surface area of a planform can be found using the equation,

where

For the wing, the following dimensions have been calculated.

Exposed surface area = 353.06 ft2

Thickness ratio at root = 0.18

Thickness ratio at tip = 0.12

Taper ratio = 0.5

Using this data,

τ = 1.5

Therefore,

Wetted surface area of wing = 743.19 ft2

Vertical Tail

For the vertical tail,


Thickness ratio = 0.15 { Note: The thickness ratio is constant }

Taper ratio = 0.6

Using this data,

τ = 1




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Therefore, from the above equation,

Wetted surface area of vertical tail = 137.72 ft2

Horizontal Tail

For the horizontal tail,


Thickness ratio = 0.12 { Note: The thickness ratio is constant }

Taper ratio = 0.5

Using this data,

τ = 1

Therefore, from the above equation,

Wetted surface area of horizontal tail = 147.70 ft2

Fuselage

The wetted surface are of a fuselage can be calculated from the equation,

⁄ ( )

where

For the fuselage,

Max height = 10.5 ft

Max width = 8.0 ft

Length = 78.6 ft

Using this data,

= (10.5+8.0)/2

= 9.25 ft

= 8.50




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Therefore,

Wetted surface area of fuselage = 1936.37 ft2

Nacelles

Using the perimeter method, the following can be estimated,

Wetted surface area of nacelles = 425 ft2

Wing-Fuselage Intersection

Max width of fuselage = 8.0 ft

Length of wing root = 6.0 ft

Therefore

Wetted surface area of intersection = 8 * 6 ft2

= 48 ft2

Total Wetted Surface Area

Using the data above

Total wetted surface area = 743.19 + 137.72 + 147.70 + 1936.37 + 425 - 48 ft2

= 3341.98 ft2

Comparison with Historical Data

The maximum takeoff weight of the aircraft was found to be 31,775 lb. Accordingly, the expected

wetted surface area can be found using statistical correlations of historical data. Such data is presented

in Figure 3.22b, p.124 of Ref 1. From this chart,

Expected wetted surface area = 3600 ft2

This is a 7.2% deviation from the calculated data, and is therefore within the expected margin of

variation.




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Zero Lift Drag Coefficient

It was determined that a skin friction coefficient of 0.006 should be attainable for the aircraft. From

Figure 3.21b, p.119 of Ref 1, for the wetted surface area found,

Equivalent parasite area “f” = 20 ft2

Therefore,

Clean zero lift drag coefficient “CDo” = f/s

= 0.0567

From Figure 12.7, p.286 of Ref 2,

Compressibility drag increment = 0.0005

Comparison

It was found in Exam 1

Clean zero lift drag coefficient = 0.0594

Since there is only a slight change in cruise drag, it will have a negligible effect on take-off and landing

polars. Therefore, from Exam I,

Takeoff flap drag increment = 0.02 [ and A = 8 | e = 0.75 ]

Landing flap drag increment = 0.06 [ and A = 8 | e = 0.70 ]

Landing gear drag increment = 0.02

The drag coefficient is given by the equation

Using the above data, and from exam one at cruise A=8 and e=0.8, the following drag polars can be

found.

Low speed, clean = 0.0572 + 0.0497

Takeoff, gear up = 0.0772 + 0.0530

Takeoff, gear down = 0.0972 + 0.0530




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Landing, gear up = 0.1172 + 0.0568

Landing, gear down = 0.1372 + 0.0568

From this, it can be found,

( ⁄ ) = 9.42

According to the old data,

( ⁄ ) = 9.20

Therefore, L/D increases by 0.22. In Exam I, it was found

(

⁄ ) = - 2969.86 lb

Therefore,

Takeoff weight decrease = 2969.86 * 0.22 lb

= 653.37 lb




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Part 5: V-n diagrams

Determination of positive 1g stall speed (+Vs1)

The 1g stall speed can be determined using the equation,

⁄

where ( ) ⁄

For preliminary design, it can be assumed,

The following data is known for the aircraft.

= 1.7

(W/S) = 90psf

Therefore,

= 1.1 * 1.7

= 1.87

= 201.19 ft/s

= 119.20 kts

Determination of negative 1g stall speed (-Vs1)

Similarly,

= -1.20

Therefore,

= 1.1 * 1.2

= -1.32




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= 239.47 ft/s

= 141.88 kts

Determination of positive limit load factor (-Nlim,pos)

The positive limit load factor can be found from the equation,

[ ]

It was found that the takeoff weight is 31,775lb. Therefore,

= 2.68

This is between the accepted values.

Determination of design maneuver airspeed (VA)

The design maneuver airspeed can be found from the intersection of the 1g stall line and the 2.68

() load factor line. This can be determined after the V-n maneuver diagram is drawn.

From the V-n diagram shown below,

VA = 195 kts

Determination of design maximum gust intensity airspeed (VB)

The design maximum gust intensity airspeed can be found from the intersection of the 1g stall line and

VB gust line. This can be determined from the V-n gust diagram.

From the V-n diagram shown below,

VB = 170 kts

Determination of design cruise airspeed (VC)

The design cruise airspeed can be found from the equation

Vc = VB + 43 kts

= 238 kts




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Determination of design dive airspeed (VD)

The design dive airspeed can be found from the equation

VD = VC *1.25

= 298 kts

Determination of gust load factor lines

The gust load factor lines can be found using the equation,

where (gust alleviation factor)

(airplane mass ratio)

For the airplane, it is known, (W/S) = 90, = 6 and = 5. Therefore, the following values are found.

= 78.40

Kg = 0.82

At 10,000ft, the derived gust values are as follows,

Ude for VC gust line = 66 ft/s



Using the above data, the gust load factor lines are

nlim,gust = 1 + 0.006069V for VB gust line

nlim,gust = 1 + 0.005498V for VC gust line

nlim,gust = 1 + 0.002299V for VD gust line




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V-n diagrams

Using the above data, v-n diagrams can now be plotted for the airplane.

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

3

3.5

0 50 100 150 200 250 300 350

L o a d

F a c t o r " n "

Speed "V" (KEAS)

V-n Maneuver Diagram

Cn,max = 1.87

119

142

A D

298238195

Cn,max = -1.32

0

0.5

1

1.5

2

2.5

3

0 50 100 150 200 250 300 350

L o a d F

a c t o r " n "

Speed "V" (KEAS)

V-n Gust Diagram

B C

D

170 238 298




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Part 6: How Do Airplanes Fly?

How does an airplane fly? In the simplest sense, it is all just a balancing act of the four main forces on an

aircraft: the lift force, the gravitational force, the thrust force and the drag force. The main force that

engineers or designers must deal with is, obviously, the lift force because if an airplane cannot create

lift, it will fall out of the air, or not even get off the ground in the first place. An airplanes lift must

counteract the gravitational force of the Earth, or whatever gravity is creating a force on it, to lift off

from the ground, reach higher elevations, or when it is equal, to stay in a trim flight condition. The

gravitational force is managed through the weight of the aircraft; it is ideal to keep the weight as low as

possible so that not as much lift and thrust must be created. The lift force is created mainly from wing

designs and is a function of the square of relative wind velocity.

This relative wind velocity brings up the next force which moves the plane and creates velocities suitable

for lift: the thrust force. Thrust is created or maintained in correlation to the engine of a fixed-wing

aircraft. There are different types of engines ranging from turboprops and turbofans to various jet

engines. Each creates varying levels of thrust with their own ramifications like fuel usage, weight and

efficiency.

The counterforce of thrust is drag. Drag is the consequence of velocity and lift forces as well as viscous

effects of the atmosphere. The coefficient of drag is proportional to the square of the lift force. It is a

very important force to study when in the design stages of creating an aircraft because designers must

take into account the stall characteristics of the aircraft and also must have enough thrust or power

from the engines to counteract the drag at high lift coefficients. This force becomes even more critical at

higher velocities ranging into the supersonic Mach numbers because wave drag must be considered

more intently.

In all, for an aircraft to fly the four main forces of the aircraft must be manipulated from the design

stage so that the aircraft can produce enough thrust and lift to counteract the weight and drag forces

and remain in a stable flight. Fairy dust just does not cut it anymore.




References

Ref 1: Roskam, Jan. Airplane Design Part I. DAR Corporation, 2003.

Ref 2: Roskam, Jan. Airplane Design Part II. DAR Corporation, 2003. Ref 3: Roskam, Jan. Airplane Design Part V. DAR Corporation, 2003.

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