November 18, 2003 Aero PDR #2 1

Team 5 Aerodynamics PDR #2Team 5 Aerodynamics PDR #2

• Scott Bird• Mike Downes• Kelby Haase• Grant Hile• Cyrus Sigari• Sarah Umberger• Jen Watson

November 18, 2003 Aero PDR #2 2

PreviewPreview

• Airfoil Sections and Geometry

• Aerodynamic mathematical model

• Launch Conditions• Endurance

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Walk-AroundWalk-Around

Data Boom

Conformal Pod

Engine

Landing Gear

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Aircraft 3-viewAircraft 3-view

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Wing GeometryWing Geometry

• Wing– Airfoil NACA 2412– Aspect Ratio 7– Span 14 (ft)– Chord 2 (ft)– Planform Area 28 (ft^2)– Taper Ratio 0 (deg)– Dihedral 0 (deg)– Sweep Angle 0 (deg)

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Horizontal Tail GeometryHorizontal Tail Geometry

• Horizontal Tail– Airfoil NACA 0012– Aspect Ratio 5– Span 7 (ft)– Chord (average) 1.38 (ft)– Planform Area 9.64 (ft^2)– Taper Ratio 0.8– Sweep Angle 5 (deg)

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Vertical Tail GeometryVertical Tail Geometry

• Vertical Tail– Airfoil NACA 0012– Aspect Ratio 2– Span 2.8 (ft)– Chord (average) 1.38 (ft)– Planform Area 3.9 (ft^2)– Taper Ratio 0.73 – Sweep Angle 10 (deg)

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Lift MethodologyLift Methodology

• Used XFOIL to get NACA 2412 2-D data

– Converted 2-D data to 3-D data for wing CLo and CLα

• Used XFOIL to get NACA 0012 horizontal tail data

– Used method described in Roskam Part VI, CH8 for CLδe

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Aerodynamic Mathematical Model of LiftAerodynamic Mathematical Model of Lift

• CL= CLo + CLα * α + CLδe * δe

• CLα = Cl α / (1 + (Cl α / (π * e * A))) [rad-1]

• CL = Cl * (CLα/Cl α)

• CLδe = CLα_ht * ηht * Sht / Sref * τe [rad-1]

– ηht = f (Sht_slip, Sht, Pavailable, U, Dp, qbar)

– (Roskam Part VI, CH8)

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Aerodynamic Mathematical Model of DragAerodynamic Mathematical Model of Drag

• Summation of individual components for CDo

– Wing, fuselage, horizontal tail, vertical tail

• CD=CDo + k * CL * CL

• CDo = Σ CDo(i)

– CDo(i) = (Cf * FF * Q * Swet) / Sref

• k = 1 / (pi * AR * e)

CDo(i)

CDo_wing 0.0104

CDo_horizontal tail 0.0039

CDo_vertical tail 0.0013

CDo_fuselage 0.0204

CDo_total 0.036

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Moment MethodologyMoment Methodology

• Used method described in Roskam, Chapter 3 of AAE 421 Book to find individual components of moment equation

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Aerodynamic Mathematical Model of MomentsAerodynamic Mathematical Model of Moments

• CM = CMo + CMα * α + CMδe * δe– (α and δe in degrees)

• CM0 = Cmac_wf + CL0_wf*(xbar_cg - xbar_acwf) + CLalpha_h*eta_h*(S_h/S)*(xbar_ach - xbar_cg)*epsilon

• CMalpha = (x/cw)*CLalpha_wf*(xbar_cg - xbar_acwf) - CLalpha_h*eta_h*(S_h/S)*(xbar_ach - xbar_cg)*(1 – depsilon/dalpha)

• CMdele = -CLalpha_h*eta_h*Vbar_h*tau_e

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Aerodynamic Mathematical Model SummaryAerodynamic Mathematical Model Summary

• CL= 0.1764 + 7.941 (rad-1)* α + 1.925 (rad-1)* δe– α and δe are in radians

• CD = .036 + .0522 * (.1764 + 7.941 (rad-1) * α)2

• Cm= - 0.0280 - 0.1562* α - 0.0898* δe

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Launch ConditionsLaunch Conditions

Known Values• W = 49.6 lbs• S = 28 ft2

• ρ = 0.0023328 slug/ft3 (at 600 ft)

• CLmax:To = 0.8*CLmax = 1.44

•

VTO = 39 ft/s

sft

CLS

WV

TO

TOTO 39

max

122.1

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Launch Conditions from EOMLaunch Conditions from EOM

ThrustDrag

Friction

Normal

Weight

WSCxx

Bhpxm

FrictionDragThrustmaF

Dop

horizontal

2

21550

Our Airplane

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Launch Conditions from EOM integrationLaunch Conditions from EOM integration

• STO = 76 ft– Constriant

of 127 ft

• tTO = 3.7 s

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EnduranceEndurance

Known Values• Bhp = 3.7 hp

• ηp = 0.40

• Cbhp = 0.0022 lb/s-hp

• Wi = 49.6 lbf

• Wf = 43.6 lbf – Using 1 gallon @ 6 lbf

• L/D = 10.39 • 12*.866 (86.6% of L/Dmax when flying @ Vmin power)

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EnduranceEndurance

• V = [2*W/( ρ*S)*(K/(3*CDo))1/2] ½

• (Raymer, eq 17.33)

– K = 1 / ( pi * AR * e )– Vmin@min power = 32.5 ft/s

• E = (L/D)*[(550*ηp)/(Cbhp*V)]*ln(Wi/Wf) • (Raymer eq 17.31)

E = 75.5 minutes

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Future WorkFuture Work

• Resize flaps for landing

• Verify known values

– Cbhp

• Weight reduction from extra fuel (Endurance)

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Review of Today’s PresentationReview of Today’s Presentation

• Wing Geometry and Size

• 3-view of aircraft

• Aerodynamic mathematical model

• Launch Conditions

• Endurance

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QuestionsQuestions

Questions??

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ReferencesReferences

• Roskam– AAE 421 Book

• Raymer– Aircraft Design: A Conceptual Approach

• AAE 251 Book– Introduction to Aeronautics: A Design Perspective

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Putting EOMs into MATLABPutting EOMs into MATLAB

• Put state space EOMs into MATLAB• Used ode45 command to integrate EOMs• Inputs to the code are initial conditions for position

and velocity – Used x(0) = 0.1ft and v(0) = 1ft/s to eliminate singularities

• Code output position and velocity as a function of time

xy

xy

2

1

WSCyy

Bhp

mxy

yxy

Dop

2221

2

55012

21

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EOM file in MATLABEOM file in MATLAB

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