Team 5 Aerodynamics PDR #2 Scott Bird Mike Downes Kelby Haase Grant Hile Cyrus Sigari Sarah Umberger Jen Watson November 18, 2003 Aero PDR #2

Preview Airfoil Sections and Geometry Aerodynamic mathematical model Launch Conditions Endurance November 18, 2003 Aero PDR #2

Walk-Around Data Boom Conformal Pod Engine Landing Gear November 18, 2003 Aero PDR #2

Aircraft 3-view November 18, 2003 Aero PDR #2

Wing 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) November 18, 2003 Aero PDR #2

Horizontal Tail Geometry 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) November 18, 2003 Aero PDR #2

Vertical Tail Geometry 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) November 18, 2003 Aero PDR #2

Used XFOIL to get NACA 2412 2-D data Lift 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 November 18, 2003 Aero PDR #2

Aerodynamic 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) November 18, 2003 Aero PDR #2

Aerodynamic 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 November 18, 2003 Aero PDR #2

Moment Methodology Used method described in Roskam, Chapter 3 of AAE 421 Book to find individual components of moment equation November 18, 2003 Aero PDR #2

Aerodynamic 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 November 18, 2003 Aero PDR #2

Aerodynamic 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 November 18, 2003 Aero PDR #2

Launch 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 November 18, 2003 Aero PDR #2

Launch Conditions from EOM Normal Drag Thrust Our Airplane Friction Weight November 18, 2003 Aero PDR #2

Launch Conditions from EOM integration STO = 76 ft Constriant of 127 ft tTO = 3.7 s November 18, 2003 Aero PDR #2

Endurance 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) November 18, 2003 Aero PDR #2

E = 75.5 minutes Endurance 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 November 18, 2003 Aero PDR #2

Resize flaps for landing Verify known values Cbhp Future Work Resize flaps for landing Verify known values Cbhp Weight reduction from extra fuel (Endurance) November 18, 2003 Aero PDR #2

Review of Today’s Presentation Wing Geometry and Size 3-view of aircraft Aerodynamic mathematical model Launch Conditions Endurance November 18, 2003 Aero PDR #2

Questions Questions?? November 18, 2003 Aero PDR #2

References Roskam Raymer AAE 251 Book AAE 421 Book Aircraft Design: A Conceptual Approach AAE 251 Book Introduction to Aeronautics: A Design Perspective November 18, 2003 Aero PDR #2

Putting 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 November 18, 2003 Aero PDR #2

EOM file in MATLAB November 18, 2003 Aero PDR #2

November 18, 2003 Aero PDR #2