NASA Technical Memorandum 102610 Hypersonic Vehicle ... · P NASA Technical Memorandum 102610 Hypersonic Vehicle Simulation Model: Winged-Cone Configuration John D. Shaughnessy, S

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NASA Technical Memorandum 102610

Hypersonic Vehicle Simulation Model:Winged-Cone Configuration

John D. Shaughnessy, S. Zane Pinckney, John D. McMinn,Christopher I. Cruz, and Marie-Louise Kelley

NOVEMBER 1990

(NASA-T_-IO_6]O}

SIMULATION MOOEL:

(NASA) 142

H YP:_RSON[C

WINGED-CONE

VEHICLE

CONFIGURATION

CSCL 0IC

c3/o8

N91-12105

Unclas

0312550

National Aeronautics andSpace Administration

Langley Research CenterHampton, Virginia 23665-5225

https://ntrs.nasa.gov/search.jsp?R=19910003392 2019-02-08T19:55:50+00:00Z

Hypersonic Vehicle Simulation Model:

Winged-Cone Configuration

John D. Shaughnessy, S. Zane Pinckney, John D. McMinn,

Christopher I. Cruz, and Marie-Louise Kelley

NASA Langley Research Center

Hampton, Virginia

SUMMARY

Aerodynamic, propulsion, and mass models for a generic, horizontal-takeoff, single-stage-to-orbit (SSTO) configuration are presented which are suitable for use in point mass as well asbatch and real-time six degree-of-freedom simulations. The simulations can be used to investigateascent performance issues and to allow research, refinement, and evaluation of integrated guidance/flight/propulsion/thermal control systems, design concepts, and methodologies for SSTOmissions. Aerodynamic force and moment coefficients are given as functions of angle of attack,Mach number, and control surface deflections. The model data were estimated by using asubsonic/supersonic panel code and a hypersonic local surface inclination code. Thrust coefficientand engine specific impulse were estimated using a two-dimensional forebody, inlet, nozzle codeand a one-dimensional combustor code and are given as functions of Mach number, dynamic

pressure, and fuel equivalence ratio. Rigid-body mass moments of inertia and center of gravitylocation are functions of vehicle weight which is in turn a function of fuel flow.

INTRODUCTION

The basic objective of the National Aero-Space Plane (NASP) Program is the developmentof a manned, horizontal takeoff and landing, single-stage-to-orbit (SSTO), airbreathing launchvehicle. Realization of this concept is theoretically possible through advanced technology in theareas of aerodynamics, materials, structures, flight control, and propulsion (ref. 1). Critical to thedevelopment of the vehicle is the integration of these disciplines to arrive at a technology mix

which maximizes propulsive efficiency while minimizing aerodynamic heating and structural loads.

Thestudiesreportedin references1-3for aconicalacceleratorconfiguration,testedexperimentallyfrom low subsonicto highhypersonicMachnumbers(refs.4-6),showthatsystemintegrationinvolvingaerodynamics,propulsion,andguidanceandcontrolearlyin thedesigncansignificantlyreducethefuel fractionto orbit. Themathematicalmodelsdevelopedfor theseinvestigationscontainedsufficientdetailto allowstudyof trim dragreductionconcepts,guidanceandcontrolstrategies,andvehicleperformanceissues.In reference2themodelsweremodifiedslightly to allow thrustvectoringandactivecenter-of-gravity(c.g.)locationcontrolto beinvestigatedaswaysto reduceexcessivefuel consumptioncausedbydragduetoelevonsusedtotrim thevehicle.

Thepurposeof thisreportis to presenttheaerodynamic,propulsion,andmassmathematicalmodelsusedin theinvestigationsreportedin references1-3. Aerodynamiclongitudinalandlateral-directionalforceandmomentcoefficientsaregivenasfunctionsof angleofattack;Machnumber,andelevon,rudder,andcanarddeflections.Thesemodeldatawereestimatedusingasubsonic/supersonicpanelcodeandahypersoniclocalsurfaceinclinationcode.Thrustcoefficientsandspecificimpulsearegivenasfunctionsof Machnumber,dynamicpressure,andfuelequivalenceratio,andwereestimatedusingatwo-dimensionalforebody,inlet,andnozzlecodeandaone-dimensionalcombustorcode.Themomentsof inertiaandcenterof gravitypositionaregivenasfunctionsof vehicleweightwhich in turnvarieswith fuel flow.

SYMBOLS AND ABBREVIATIONS

b lateral-directional reference length, span, ft

longitudinal reference length, mean aerodynamic chord, ft

CD D , n.d.total drag coefficient, _ Srel

CDa drag increment coefficient for basic vehicle, drag increment , n.d.Sref

CD,da drag increment coefficient for right elevon, drag incrementSref , n.d.

drag incrementCD,de drag increment coefficient for left elevon, Srer

, n.d.

drag increment

CD,dr drag increment coefficient for rudder, _ Sref, n.d.

drag incrementCD,dc drag increment coefficient for canard,

Sref, n.d.

C L total lift coefficient, _..k_ n.d.Sr_l '

lift incrementcoefficientfor basicvehicle,lift increment , n.d.

CL,da lift increment coefficient for right elevon, lift increment , n.d.Srcf

CL,delift increment coefficient for left elevon, lift increment , n.d.

Sref

CL,dclift increment coefficient for canard, lift increment , n.d.

S_cf

Cy total side force coefficient, _ Y_el ' n.d.

side force with sideslip derivative for basic vehicle, -- , n.d,

Cy,da side force increment coefficient for right elevon, side force increment , n.d.S_f

Cy,deside force increment coefficient for left elevon, side force increment , n.d.

S_ef

Cy,dr side force increment coefficient for rudder, side force increment, n.d.Sref

C 1L n.d.

total rolling moment coefficient, _ Srefb '

OClrolling moment with sideslip derivative for basic vehicle, _, n.d.

ot_

Cl,da rolling moment increment coefficient for right elevon, moment increment n.d.Srefb

Cl,de rolling moment increment coefficient for left elevon moment increment n.d.' q Srefb '

Cl,dr rolling momentincrementcoefficientfor rudder,moment increment, n.d.Srefb

C 1P

rolling moment with roll rate dynamic derivative, --3pb ,n.d.

2V

C 1r

rolling moment with yaw rate dynamic derivative, -- C1 n.d.

orb'2V

Cm

Mtotal pitching moment coefficient, _ Sref c

_, n.d.

Cma

pitching moment increment coefficient for basic vehicle, moment increment , n.d.S,cfc

Cm,da pitching moment increment coefficient for right elevon, moment increment , n.d.Stere

Cm ,de pitching moment increment coefficient for left elevon, moment increment , n.d.

S,efc

Cre,dr pitching moment increment coefficient for rudder, _moment increment, n.d.

SrefC

Cm,dc

pitching moment increment coefficient for canard, moment increment, n.d.S,cfc

Cm

qpitching moment with pitch rate dynamic derivative, --

0 Cm

oq c2V

, n.d.

Cn

D

N , n.d.total yawing moment coefficient, _ Srcfb

Cnl3yawing moment with sideslip derivative for basic vehicle, -- , n.d.

4

Cn,da yawingmomentincrementcoefficientfor rightelevon,moment increment n.d.Srefb

Cn,de

yawing moment increment coefficient for left elevon, moment increment n.d.Srefb

Cn,dr yawing moment increment coefficient for rudder, moment increment n.d.Srefb

Cn

P

yawing moment with roll rate dynamic derivative, --3 Cn

3p b , n.d.2V

Cn

r

yawing moment with yaw rate dynamic derivative,3rb

2V

• n.d.

CT

Iso

L

M

N

Mmrc

L,D,Y

X,Y,Z

M

thrust coefficient, thrust, ft 2

engine specific impulse, sec

aerodynamic rolling moment at center of gravity, ft-lb

aerodynamic pitching moment at center of gravity, ft-lb

aerodynamic yawing moment at center of gravity, ft-lb

aerodynamic pitching moment at moment reference center, ft-lb

total aerodynamic lift, drag, and side forces respectively, lb

total aerodynamic x, y, and z body axes forces respectively, lb

angle of attack, deg

sideslip angle, rad

engine fuel equivalence ratio, n.d.

Mach number, n.d.

dynamic pressure, lb/ft 2

Sref referencearea,theoreticalwing area,ft2

T enginenetthrust,lb

V vehiclefreestreamvelocity,ft/sec

W fuel flow rate, lb/sec

W vehicle weight, lb

W0 initial value of vehicle weight, lb

WCON weight of fuel consumed, lb

Xcg

longitudinal distance from moment reference center to vehicle c.g., positive aft, ft

Ixx, Ivy, Izz roll, pitch, and yaw moments of inertia respectively, sig-ft 2

APAS Aerodynamic Preliminary Analysis System

e.g. Vehicle center of gravity

G&C Guidance and Control

HABP Hypersonic Arbitrary Body Program

MACH Mach number, n.d.

mrc wind tunnel moment reference center

n.d. nondimensional

POST Program to Optimize Simulated Trajectories

SSTO Single-Stage-To-Orbit

UDP Unified Distributed Panel program

VEHICLE DESCRIPTION

The planform and side view of the study vehicle are shown in figure 1, and the geometriccharacteristics are given in table I. This aircraft is a full scale version of the wind tunnel testconfiguration reported in reference 4. A sizing analysis discussed below yielded a full-scale gross

weight of 300,000 lbs and an overall fuselage length of 200 ft.

The fuselage-centerline-mounted wing has convention'd, independently controllable,

trailing edge elevons with their hinge line perpendicular to the fuselage centerline. Deflections ofthe elevons are measured with respect to the hinge line, and positive deflections are with the trailing

edge up. The fuselage is modelled as an axisymmetric conical forebody, a cylindrical enginenacelle section, and a cone frustrum engine nozzle section. A fuselage-centerline-mounted vertical

tail has a full span rudder with its hinge line at 25 percent chord from the trailing edge. Deflectionsof the rudder are measured with respect to its hinge line, and positive deflections are with the

6

trailingedgeleft. Thesmallcanardshavea6percentthick symmetrical65A seriesairfoil andaredeployedonly at subsonicspeeds.Deflectionsof thecanardaremeasuredrelativeto thefuselagecenterline,andpositivedeflectionsarewith thetrailingedgedown.

AERODYNAMICS MODEL

A subsonic/supersonic/hypersonic analysis code developed jointly by NASA Langley andRockwell International Inc., referred to as the Aerodynamic Preliminary Analysis System (APAS),

was used to predict the longitudinal-and lateral-directional force and moment coefficients of thestudy configuration. The APAS program is described conceptually in Appendix A.

APAS estimated the basic vehicle lift and drag increment coefficients and the side forcesideslip derivative as functions of angle of attack and Mach number. The program estimated thelift, drag, and side force increment coefficients for the right and left elevons, the rudder, and thecanard as functions of angle of attack, surface deflection, and Mach number. Changes in lift, drag,and side force due to body angular rates and aerodynamic coupling between control surfacedeflections and sideslip are assumed negligible.

APAS estimated the basic vehicle roll and yaw moment sideslip derivatives, the pitchmoment increment coefficient, and the roll, pitch and yaw dynamic derivatives (for roll, pitch, andyaw rates) as functions of angle of attack and Mach number. The roll and yaw moment incrementcoefficients for right and left elevon and rudder and the pitch increment coefficients for right andleft elevon, rudder, and canard were estimated as functions of angle of attack, surface deflections,and Mach number. These quantities are given relative to the moment reference center. The totalmoments relative to the c.g. are obtained by adding those caused by lift, drag, and side force to the

above quantities. Aerodynamic moment coupling between control surface deflections, sideslip,and body angular rates is assumed negligible.

Data were computed at Mach numbers of 0.3, 0.7, 0.9, 1.5, 2.5, 4.0, 6.0, 10.0, 15.0,

20.0, and 24.2 and angles of attack of -1.0 °, 0.0 °, 1.0 °, 2.0 °, 4.0 °, 6.0 °, 8.0 °, 10.0 °, and 12.0 °.At each Mach number and angle of attack combination, coefficients were generated for a range of

deflections of the fight elevon, the left elevon, and the rudder, each taken separately. Deflections

of -20.0 °, -10.0 °, 0.0 °, 10.0 ° and 20.0 ° were used for each surface. At the three subsonic Mach

numbers and nine angles of attack above, data were estimated for canard deflections of - 10.0 °,

-5.0 °, 0.0 °, 5.0 ° and 10.0% The APAS Unified Distributed Panel (UDP) code was used at Mach

numbers of 0.3, 0.7, 0.9, and 1.5, and the Hypersonic Arbitrary Body Program (HABP) codewas used at Mach 2.5 and above. The results of the computations are shown in figures 2-32.

Model Implementation

Implementation of the aerodynamic model in a computer simulation requires a numericalalgorithm to extract the increment coefficients and derivatives. The drag, lift, and side force, androlling, pitching, and yawing moment increment coefficients and derivatives for the basic vehicle

are determined as functions of angle of attack and Mach number. The increment coefficientscaused by control surface deflections are determined as functions of angle of attack, surfacedeflections, and Mach number, and are added to the basic vehicle increments to form the total

aerodynamic force and moment coefficient. These coefficients are used to compute the forces andmoments relative to the moment reference center. These moments are then corrected to the c.g.The following sections discuss the computation of these quantities.

Drag Force Computation

The basic vehicle drag increment coefficient, CDa, is shown in figure 2 as a function of

angle of attack and Mach number. The drag increment coefficients for the right elevon, CD,da, left

elevon, CD,de, rudder, CD,dr, and canard, CD,dc, are shown in figures 3-6, respectively, as

functions of angle of attack, deflection angle, and Mach number. These coefficients are summed to

obtain the total drag coefficient as

C D = CDa + CD,de + CD,da + CD,dr + CD,dc

and the drag is given by

D= srefco

Lift Force Computation

The basic vehicle lift increment coefficient, CLa, is shown in figure 7 as a function of

angle of attack and Mach number. The increment coefficients for the right elevon, CL,da, left

elevon, CL,de, and canard, CL,dc, are shown in figures 8-10, respectively, as functions of angle

of attack, deflection angle, and Mach number. Changes in lift due to rudder deflections areneglected. These coefficients are summed to obtain the total coefficient as

C L = CLa + CL,de + CL,da + CL,dc

The lift is then given by

L = _ Sre f C L

Side Force Computation

The basic vehicle side force due to sideslip derivative, Cyl _ , is shown in figure 11 as a

function of angle of attack and Mach number. The increment coefficients for the right elevon,

Cy,da, left elevon, Cy,de, and rudder, Cy,dr, are shown in figures 12-14, respectively, as

functions of angle of attack, deflection angle, and Mach number. The total side force coefficient is

given by

Cy [_ + + += Cy[_ Cy,da Cy,de Cy,dr

where the side slip angle [3 is in radians. The side force is then given by

Y = _ Sref Cy

8

Rolling Moment Computation

The basic vehicle rolling moment with sideslip derivative, CII3 , is shown in figure 15 as a

function of angle of attack and Mach number. The rolling moment increment coefficients for fight

elevon, Cl,da, left elevon, Cl,de, and rudder, Cl,dr, are shown in figures 16-18 respectively as

functions of angle of attack, deflection angle, and Math number. The rolling moment dynamic

derivatives for roll rate, C 1 , and yaw rate, C 1 , are shown in figures 19 and 20, respectively, asp r

functions of angle of attack and Mach number. The total rolling moment coefficient is given by

C 1 _ + Cl,da + Cl,de + Cl,dr + Clp= ell 3 (2_) + Clr {2r-r-_V)

where 13is in radians, and the terms and are the computed nondimensional roll and

yaw rates. The rolling moment is then given by

L = _ b Sre f C1

For this configuration there is no rolling moment caused by the z-axis force or the y-axis forcebecause the e.g. and the moment reference center lie on the x-axis.

Pitching Moment Computation

The pitching moment increment coefficient relative to the moment reference center for the

basic vehicle, Cma, is shown in figure 21 as a function of angle of attack and Mach number. The

pitching moment increment coefficients for the fight elevon, C_n,d a, left elevon, Cm,de, rudder,

Cm,dr, and canard, Cm,dc, are shown in figures 22-25, respectively, as functions of angle of

attack, deflection angle, and Mach number. The pitch rate dynamic derivative, C m , is shown inq

figure 26 as a function of angle of attack and Mach number. The pitching moment coefficientrelative to the moment reference center is given by

C m = Cma + Cm,da + Cm,de + Cm,dr + Cm,dc + C m (q_-)q

The term _s the computed nondimensional pitch rate. The pitching moment relative to the

moment reference center is given by

Mmr c = _ c Sref C m

For the study configuration the aerodynamic pitching moment about the c.g. is equal to the sum of

the pitching moment relative to the moment reference center plus the pitching moment caused bythe z-axis force acting through the x-distance from the c.g. to the moment reference center as

9

M = Mmrc - XcgZ

wherethez-axisforceis givenby

Z = - D sin 0t - L cos ¢x

For the study configuration there is no pitching moment caused by the x-axis force because the c.g.and the moment reference center lie on the x-axis.

Yawing Moment Computation

The yawing moment with sideslip derivative relative to the moment reference center for the

basic vehicle, Cnl 5, is shown in figure 27 as a function of angle of attack and Mach number. The

increment coefficients for the right elevon, Cn,da, left elevon, Cn,de, and rudder, Cn,dr, are

shown in figures 28-30, respectively, as functions of angle of attack, deflection angle, and Mach

number. The yawing moment dynamic derivatives for roll rate, C n , and yaw rate, C n , arer

shown in figures 31 and 32, respectively, as functions of angle of aPck and Mach number. The

yawing moment coefficient relative to the moment reference center is given by

C n = Cn[ _ 13+ Cn,da + Cn,de + Cn,dr + Cnp(2_) +Cn (rbl2v!

r

and the aerodynamic yawing moment relative to the moment reference center is given by

m

Nrnrc = _ b Sre f C n

For the study configuration the total aerodynamic yawing moment about the c.g: is equal to thesum of the yawing moment relative to the moment reference center plus the yawang moment causedby the side force acting through the x-distance from the c.g. to the moment reference center as

N = Nmr c + Xcg Y

where the y-axis aerodynamic force is given above. For this configuration there is no yawingmoment caused by the x-axis force because the c.g. and the moment reference center lie on thex-axis.

MASS MODEL

The vehicle mass model is based on the assumption of rigid structure and time varyingmass, center of gravity, and moments of inertia. The total mass of the vehicle, its c.g. location,and its moments of inertia vary as fuel is consumed. Fuel slosh is not considered, and the

products of inertia are assumed negligible.

10

Model Generation

The vehicle takeoff weight and size were determined based on an iterative method (refs.1,7). This method involves assuming vehicle takeoff mass, vehicle size, propulsion system, andaerodynamics and running the three degree-of-freedom Program to Optimize Simulated Trajectories(POST) simulation (ref. 8) to orbit to obtain a final mass fraction. The mass fraction is then usedas input to an empirically based vehicle sizing program which outputs new values for mass and

size. This process is repeated until no significant change is output from the sizing program.

The sizing analysis yielded a vehicle gross weight of 300,000 lbs and an overall fuselagelength of 200 feet. The c.g. location and the moments of inertia were estimated at ten values ofvehicle weight ranging from 295,000 lb to 140,000 lb which corresponded to takeoff and engineshut down conditions, respectively. The moments of inertia are consistent with values given inreference 9 for several low aspect ratio military and research aircraft and the Concorde supersonic

transport. The precomputed c.g. location, Xcg, measured positive aft from the wind tunnel

moment reference center, and the vehicle moments of inertia IXX, Iyy, and IZZ are shown in

figure 33 as functions of vehicle weight.

Model Implementation

As in the aerodynamic model, implementation of the mass model in a computer simulation

requires an algorithm to extract the mass data. The weight of fuel consumed, WCON, is obtainedby integrating the fuel flow rate over the time, tT, the engines are thrusting as

tT

WCON = W dt

The vehicle weight is then given by

W = W0 - WCON

where W0 is the initial weight of the vehicle.

The c.g. position, x , and the moments of inertia are then determined as a function of thecg

vehicle weight. It is assumed that the c.g. moves only along the body x-axis as fuel is consumed.

PROPULSION MODEL

The propulsion model for the study configuration was developed using a two-dimensionalforebody, inlet, and nozzle code with a one-dimensional combustor code. The code analyzes avehicle nose-to-tail with the forebody, inlet, and nozzle flows computed by a two-dimensionalperfect gas code. The combustor performance is computed assuming one-dimensional flow withhydrogen combustion, utilizing the equilibrium chemistry combustion model of reference 10. Thecombustor code is based on an update of the cycle analysis method described in reference 11. Theforebody, inlet, and nozzle forces are determined using a two-dimensional finite difference codebased on the shock fitting method of reference 12. The boundary layer effects are determined bythe methods described in references 13 and 14.

The thrust coefficient (thrust per unit dynamic pressure) and specific impulse were

estimated at specified values of Mach number, dynamic pressure, and equivalence ratio. The thrust11

asdeterminedfrom thismodelis netthrust,andit actsalongthebodyx-axis. Theeffectsof angleof attackandsideslip,bodyangularrates,andcontrolsurfacedeflectionson thrustandspecificimpulseareconsiderednegligiblefor thethepresentconfiguration.

Thrustcoefficientswerecomputedat Machnumbersof 0.0,0.3, 0.5,0.7, 0.9,0.95, 1.0,1.5,2.0, 3.0, 3.5, 4.0, 6.0, 8.0, 10.0, 15.0,20.0,and25.0; five dynamicpressuresrangingfrom 0 to 5000lb/ft2; andfive equivalenceratiosranging from 0.0 to 10.0. Similarly, specific

impulse values were estimated at Mach numbers of 0.0, 0.3, 0.5, 0.7, 0.9, 1.0, 1.5, 2.0, 3.0,3.5, 4.0, 6.0, 8.0, 10.0, 15.0, 20.0, and 25.0 ; five dynamic pressures ranging from 0 to

5000 lb/ft2; and six equivalence ratios ranging from 0.0 to 100.0. The results of the computations

are shown in figures 34 and 35 respectively.

Model Impl¢mentation

As above, implementation of the propulsion model in a computer simulation requires analgorithm to extract the precomputed data. Using a linear interpolation/extrapolation scheme, thethrust coefficient and specific impulse are determined as functions of fuel equivalence ratio,

dynamic pressure, and Mach number. The following sections discuss the computation of thethrust, specific impulse, and fuel flow rate.

Thrust Computation and Throttlin_

The thrust coefficient, CT, as shown in figure 34, is a function of equivalence ratio,

dynamic pressure, and Mach number. The thrust is calculated as

T=qC T

Engine throttling is simulated by varying the equivalence ratio. Equivalence ratio of unitycorresponds to maximum fuel efficiency, and values greater than unity give more thrust but usedisproportionately more fuel.

Specific Impulse and Fuel Flow Computation

The specific impulse, lsp, is shown in figure 35 as a function of Mach number, dynamicpressure, and equivalence ratio. Once the thrust is computed and the specific impulse isdetermined the fuel flow rate is computed as

and is integrated to determine the weight of fuel consumed, WCON.

MODEL VALIDATION AND VERIFICATION

The aerodynamic model presented is based on established conceptual design estimationmethods. Good comparisons with available experimental aerodynamic data have been obtained

(for example, in ref. 4, 5, 6, and 15.)

The APAS program predictions compare well with Space Shuttle orbiter data book valuesas shown by C. I. Cruz at the Langley Research Center. There do not exist experimental data oncontrol effectiveness for the study configuration; however, the APAS estimates of control

12

effectivenesshavebeenvalidatedin reference15for theSpaceShuttleorbiter, the X-15, and ahypersonic research airplane (ref. 16). These comparisons indicate that APAS estimates of controleffectiveness are generally satisfactory for conceptual design studies.

The codes used to generate the propulsion model have been validated independently bycomparisons with estimates from computational fluid dynamics codes and with limitedexperimental data. The sharp conical forebody and axisymmetric inlet, combustor, and nozzle ofthe study configuration improve the accuracy of the assumptions used in developing the model.

For example, the cross flow around the forebody at angles of attack and side slip tends to keep thedifference in thrust small between the windward and leeward sides of the vehicle.

The vehicle mass characteristics and size were determined based on an established iterative

method that involved the trajectory, the aerodynamics, and the propulsion system (refs. 1, 7). Thesizing program uses empirical data, and the vehicle moments of inertia were made a function of

vehicle weight based on the trends given in reference 9.

The models presented have been used extensively in single-stage-to-orbit, three degree-of-freedom and six degree-of-freedom simulations (refs. 2, 3). In these cases active guidance andcontrol systems were simulated and vehicle performance issues were studied. In reference 2 the

model was modified slightly to allow thrust vectoring and active c.g. location control to beinvestigated as ways to reduce excessive fuel consumption caused by drag due to elevons used totrim the vehicle. In all cases the models gave reasonable and consistent results.

Based on the above considerations, the models presented are suitable for use in point massas well as batch and pilot-in-the-loop, six degree-of-freedom, horizontal takeoff and landing,

ascent-to-orbit analyses and simulations. The models can be used to investigate trajectoryoptimization, guidance algorithms, stability augmentation system issues, drag reduction concepts,and to allow research, refinement, and evaluation of integrated guidance/flight/propulsion/thermal

control system concepts and design methodologies for horizontal-takeoff single-stage to orbitmissions.

CONCLUDING REMARKS

Aerodynamic, propulsion, and mass mathematical models for a generic, horizontal takeoffand landing, single-stage-to-orbit, conical accelerator configuration are presented. The models canbe used to represent the vehicle in point mass as well as batch and pilot-in-the-loop, sixdegree-of-freedom, horizontal takeoff and landing, ascent-to-orbit analyses and simulations.These models can be used to investigate trajectory optimization, guidance algorithms, stabilityaugmentation system issues, drag reduction concepts, handling qualities, and to allow research,

refinement, and evaluation of integrated guidance/flight/propulsion/thermal control system conceptsand design methodologies for horizontal-takeoff, single-stage to orbit missions. Aerodynamic and

propulsion forces and moments are given as functions of local flow conditions, aerodynamiccontrol surface deflections, body rates, and propulsion parameters. Mass properties are given interms of vehicle weight which in turn varies with fuel flow. The models are based on acceptedconceptual design methods and have been validated and verified separately and together insimulations.

t3

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16.Dillon, J.L.; andPittman,J.L.: AerodynamicCharacteristicsatMach6 of aWingBodyConceptfor aHypersonicResearchAirplane,NASA TP 1189,1978.

17.Sova,G.; andDivan, P.: AerodynamicPreliminaryAnalysisSystemII, PartII,User'sManual,NASA CR-182077,1990.

18.Bonner,E.; Clever,W.; andDunn,K.: AerodynamicPreliminaryAnalysisSystemII,PartI, Theory,NASA CR-182076,1990.

19.Cruz,C. I.; andWilhite, A. W.: Predictionof High-SpeedAerodynamicCharacteristicsUsingtheAerodynamicPreliminaryAnalysisSystem(APAS)AIAA-89-2173,1989.

APPENDIX A

Aerodynamic Preliminary Analysis System (APAS)

The FORTRAN-based Aerodynamic Preliminary Analysis System (APAS) (ref. 17)utilizesthe Unified Distributed Panel (UDP) program for analysis in the subsonic/supersonic speed rangeand an enhanced version of the Hypersonic Arbitrary Body Program (HABP) in the highsupersonic and hypersonic speed range. APAS is contained in an interactive shell which allowsthe user to quickly generate, modify, and analyze arbitrary vehicle geometries.

The UDP program (ref. 18) estimates fuselage contributions to vehicle aerodynamics usingslender body theory. The wing and tail contributions are evaluated using a system of vortex panels(non linear vortex effects are estimated using the Polhamus suction analogy.) The components ofviscous drag, wave drag, and base drag are predicted by a combination of theoretical and empiricalmethods in UDP. The implementation of the vortex panel theory in UDP requires streamwisepanel edges; thus, engineering approximations had to be made when defining the geometry at thewing/body and tail/body junctures, particularly in the aft region of the vehicle where the wing andtail join the truncated cone section of the fuselage.

The HABP module (ref. 19) of APAS computes inviscid and viscous solutions on each

finite element of the APAS model and integrates the pressure and shear forces to obtain total forceand moment coefficients for the configuration. Pressures on the impact surfaces (surfaces whosenormal vector has a component pointing into the flow) of the APAS model were approximatedusing the tangent-cone (fuselage component) and tangent-wedge (wing and tail components)methods. Prandtl-Myer expansion was assumed for all shadow surfaces. The viscous shearforces were estimated using the reference enthalpy method.

15

TABLE l.- GEOMETRICatARACTERISTICS OFCONFIGURATION

Wing:

Referencearea (includes area projected to filselage centerline), ft 2 ....................... 3(g)3

Aspect ratio ....................................................................................... 1.00

Span, ft ............................................................................................ 60.0

l.eading edge sweep angle, deg ............................................................... 75.97

Trailing edge sweep angle, deg .................................................................. 0.0

Mean aerodynamic chord, ft .................................................................... 8(I.0

Airfoil section ................................................................................. diamond

Airfoil thickness to chord ratio, percent ......................................................... 4.0

Incidence angle, deg ............................................................................... 0.0

l)ihedral angle, deg ................................................................................ 0.0

Wing flap (devon):

Area each, ft 2 .................................................................................... 92.3

Chord (constant), ft ............................................................................. 7.22

Inboard section span location, ft ............................................................... 15.(}

()utboard section span location, ft ........................................................... 27.78

Vertical tail, body centerline:

Exposed area, ft 2 ............................................................................... 645.7

Theoretical area, ft 2 ........................................................................... 1248.8

Span, ft .......................................................................................... 32.48

Leading edge sweep angle, deg ................................................................ 71}.0

Trailing edge sweep angle, deg ............................................................... 38.13

Airfoil section ................................................................................. diamond

Airfoil thickness to chord ratio, percent ......................................................... 4.1}

16

TABLE I.- CONCLUDED

Rudder:

Area, ft 2 ......................................................................................... 161.4

Span, ft ............................................................................................ 22.8

Chord to vertical tail chord ratio, percent ..................................................... 25.0

Canard:

Exposed area, ft 2 ............................................................................... 154.3

Theoretical aspect ratio .......................................................................... 5.48

Span, ft ............................................................................................ 33.6

Leading edge sweep angle, deg ................................................................ 16.0

Trailing edge sweep angle, deg ................................................................. (I.0

Airfoil section ....................................................................... NACA 65A006

Incidence angle, deg .............................................................................. (1.0

Dihedral angle, deg ................................................................................ 0.0

Axisytumetric filselage:

Theoretical length, ft ........................................................................... 200.0

Cone half angle, (leg .............................................................................. 5.0

Cylinder radius (maximum), ft ................................................................ 12.87

Cylinder length, ft .............................................................................. 12.88

Boattail half angle, (leg ........................................................................... 9.0

Boattail length, ft ................................................................................. 40.0

Moment reference cenler, ft .................................................................. 124.(11

17

b

70

C

0

EO

12DcO

°_

L,,

°_

cO

oI°

t.k

18

r..)

0.14

0.12

O. !

(LOg

0.06

0.04

0.02

0 -..! ._J__

-2 0 2 4 6 g I 0 12 14

Angle of attack, (t, degrees

Figure 2.- l)rag increment c(_fficienl for basic vehicle as a function of

angle of attack and Math numlmr.

M

-- (>--- 0.3

m- 0.7

o O.O

1.5

2.5

...... 4.O

6.0

10.0

15,0

20.O

24.2

__l.J

rj

O.05

(I.04

0.03

0.(12

0.01

0

-0.01

-(I.(12

-0.03 _LI ,,I , I * * !.1 t ,-! 1 J , , I , , I ,

-2 0 2 4 6 8 I0 12

Angle of altack, or, (leg

(a) Mach number = 0.3

..._ ...-M

.-q.T*J

_.__.___._O...._._.__..__ .............

9. o o o

l)efleclion

angle, ¢le_

I) - -20

- -o .... 10

.....0- 0

---_4---10

t 20

.i . t. ]

14

Figure 3.- Drag increment coefficient fi_r right elevon as a function of angle ofattack, deflection angle, and Mach number.

19

CO

0.05

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

Deflection

1._..--FI angle, deg1-t --{ q- - -20

.r'" ..... - -o -- -10

• __..r ---_ - 10

X

-2 0 2 4 6 8 10 12 14

Angle of attack, ct, deg

(b) Math nmnher ---:-0.7

drO

0.05

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

- -<..-.._ __-<>J_<>f / -_- o/--+-,o

_o-o---_-__o o (, o [ , 20

_L_, l .I._,__,...L.._I_..L_._I_I__I_,_I t__l .,__J___l__l_,_, ,._1 L_.., J_l..J__J J__/

-2 0 2 4 6 g 10 12 14

Angle of attack, or, (leg

(c) Mach number = 0.9

Figure 3.- Continued.

20

_5co

0.05

0.04

0.03

0.02

0.0 l

0

-0.0 l

-0.02

-0.03

._ .o--.fjO

_0._7. 0 --J

__i?- <>........o

---_,+

Deflection

angle, deg

-- u- -20

---o--- -10

--O- 0

----_-- 10

---)-- 20

-2 0 2 4 6 8 I0 12 14

Angle of attack, or, deg

(d) Mach number = 1.5

cicO

0.012

0.01

0.008

0.(X)6

0.004

0.002

0

-0.002

-2 0

...........................................................

_.......__ angle, deg

_.._.o--*, -j --o- 0

2 4 6 8 10 12 14

Angle of attack, et, (leg

(el Mach number = 2.5

Figure 3.- Co.tim)e(t.

21

-8¢:5

L)

0.012

0.01

().flOg

0.006

(1.004

0.002

0

-0.002

......................................................................__--7_-]........Deflection

angle, deg

---r_}-- -20

--0 -. -10

--o- 0

--_- I0

.... _- 20

-2 0 2 4 6 g I0 12 14

Angle of attack, _t,deg

(f) Mach nmnber = 4.0

-8_5

ro

0.(R)8

0.006

0.004

0.002

0

-O.002

:'/ Deflection

,rr// angle, deg

j --cJ----20

- --o ...... 10

/t -''_ r_ o-0

-_" t:'J' t'- o ....._-lO[.r"_'_v---_ o---J- , 20

_L_--a_ I_.a _t..l_ I_ .L L._t___l_t_..,_,__l__.L.L_L I_.L.L --,..J. _,_,..___J__._t__a__j....

-2 0 2 4 6 g 10 12 14

Angle of attack, {z, deg

(g) Mach number = 6.0

Figure 3: Continued.

22

mr,.)

0.008

0.006

0.004

0.002

0

-0.(_)2

S Deflection

a ngle, deg

---D- -20

--o-- -10

--O-- 0

+ 10

--4-- 20

-2 0 2 4 6 8 10 12 14

Angle of attack, o_, deg

(h) Mach number = 10.0

0,008

0.006

0.004

0.002

0

-0.002 .--L__L __I____X__LL_L__L__J__.L__L.__t__,_ J__L,

-2 0 2 4 6 8 10


(i) Mach number = 15.0

, I

12

l)eflection

angle, deg

---C]-- -20

--o-- -10

---o- 0

-_ 10

.... t-- 20

i1.........14


23

-2 0 2 4 6 8 10 12 14

Angle of attack, 0t, deg

(j) Mach number = 20.0

0.008

[0.006 ] angle, deg

I --o--20

0.004 [ --o-- -10

/ ----o-- 0

ro J --x-- 10

-0.002 ' ' ' I ' t , I , , , I , , , I t , , I , , i I , , , I ,

-2 0 2 4 6 8 10 12 14


(k) Mach number = 24.2

Figure 3.- Concluded.

24

0.05

0.04 -

0.030.02 1

c5 0.01

0

-0.01 ...... '.... _.

-0.02

-0.03 ,,, _,,, i,,, I,,, _,,, _ , ,, I,,, _ , , ,

....... ......

Deflection

angle, deg

--43--- -20

--o-- -I0

--O-- 0

---)6- I0

20

-2 0 2 4 6 8 10 12 14



0.05 ]

0.04

0.03

_1 0.02

cf 0.01 --O--

0 --+--

-0.01

-0.02

-0.03 ,,, I,,,, ......

Deflection

angle, deg

---G-- -20

--o-- -10

L)

-2 0 2 4 6 8 10 12 14

0

10

20

Angle ofattack, a, deg

(b) Mach number = 0.7

Figure 4.- Drag increment coefficient for left elevon as a function of angle ofattack, deflection angle, and Mach number.

25

O.O5 ]....

0.04 Deflection

angle, deg

0.03 _ -20

•8 0.02 ----o-- -10. --o-- 0

¢9 0.01 _ I0

0 ----4--- 2O

-0.01 ]-0.02

-0.03 ,,, i,,, I,,, I,,, t,, ,_ I,,, I,,, I , , ,

-2 0 2 4 6 8 10 12 14


(c) Mach number -- 0.9

0.05

0.04 _(_o!O.O3

.8 0.02

--0- 00.01

----g--- 10

O ---4-- 20

-0.01

-0.02

-0.03 ,,, I,,, I , , , , ,

-2 0 2 4 6 8 10 12 14

Angle of attack, a, deg



26

-8c;

O

0.012

0.01

0.008

0.006

0.004

0.002

0

-0.002

-2

, ,., , I , _ , I , , , I , , , I , , , I , _ t ,

0 2 4 6 8 i0 12


(e) Mach number =2.5

Deflection

angle, deg

r3--20

--<>-- -10

---O-- 0

+ 10

---4-- 20

a !

14

0.012 }

0.01 Deflection

angle, deg

0.008 _ .20

-8 0.006 ---o-- -I0

--o-- 0

0.004 ---->¢- 10---4-- 20

0.002

0

-0.002 ......

-2 0 2 4 6 8 10 12 14


(f) Mach number = 4.0


27

o.oos IDeflection

angle, deg0.006

---[3- .... 20

--o- -10

0.004 --o- 0

.... _-- !0u

0.002 ..... _-- 20

0

-0.002 , , , j , , , n , , , i , , , I , ,, _ , , ,_L_----, v , , ,

-2 0 2 4 6 8 10 ! 2 14

Angle of attack, cx, deg

(g) Mach number -- 6.0

0.008 [

0.006

-8 0.004c5

0.002

0

-0.002 , , , 1 , , , I , , , I , , , I , , , I , , , I , , , I , , ,

Deflection

angle, deg

--0--20

---o-- -10

--o- 0

-----_ - 10

--+-- 20

-2 0 2 4 6 8 !0 12 14




28

0.008 I

d

0.006

O.(X)4

0.002

0

-0.002

Deflection

angle, deg

+ -20

- ,o .... 10

--(3- 0

---_ .... 10

----¢--- 20

-2 0 2 4 6 g I0 12 14

Angle of attack, _, deg


0.008/

Deflection

0.006 angle, deg

+-20

0.004 --o-- -10

d -o--0

0.002 __q__ 20

o J-0.002 ..... ,, .L___r._,_,_LL_L

-2 0 2 4 6 R 10 12 14




29

Deflectionangle,deg--o- -20---_- -10

--O-- 0

0.008 I

0.006

_1 0.004d

_ 10

0.002 --4--- 20

0

-0.002 ,,, i,,, t,,, I,,, i,,, I,,, t,,, t , , ,

-2 0 2 4 6 8 10 12 14

Angle of attack, ix, deg



dL_

0.008

0.006

0.004

0.002

0

-0.002

Deflection

angle, deg

--o-- -20

---o-- -10

--O- 0

10

---4 .... 20

i,, I, ,,I, I,ll I , I , ,, I , , L_I ,, , I i I ,

-2 0 2 4 6 8 10 12 14Angle of attack, or, deg


Figure 5.- Drag increment coefficient for rudder as a function of angle ofattack, deflection angle, and Mach number.

30

co

0.008

0.006

0.004

0.002

-0.002

..... L ......

Deflection

angle, deg

--c3-- -20

--o- 0

20---I

i I , , , I,,, I,, , I x, , I , , , I , , ,_l_x__

0 2 4 6 8 10 12 14



c/L)

0.008

0.006

0.004

0.002

0

-0.002

-2

Deflection

angle, deg

--or-- -20

---o-- -10

--O- 0

+ 10

--+-- 20

J I I , t I , , , I , , , I , , j J | i i I I i t i i___

0 2 4 6 8 10 12 14




31

c/L)

0.008

0.006

0.004

0.002

0

-0.002

1Deflection

angle, deg

--<3- -20

---o-- -10

--o-- 0

--_-- 10

--4-- 20

J _ I _ J _ I , _ , I__a_._a_._t_

0 2 4 6 8 10 12 14


(d) Mach number = !.5

ro

0.0035

0.003

0.0025

0.002

0.0015

0.001

0.0005

0

-0.0005

-ffl-_._

Deflectionangle, deg

---0---20

---o-- -If}

--o-- 0

--_-- 10

--4-- 20

-2 0 2 4 6 8 10 12


14

(e) Mach number = 2.5


32

r..)

0.0035

0.003

0.0025

0.002

0.0015

0.001

0.0005

0

-0.0005

_---o-----------o

-2 0 2 4 6 8 10 12


Deflection

angle, deg

---Or-- -20

--o-- -10

--O- 0

+ 10

--4-- 20


14

0.0035

0.003

0.0025

0.002

0.0015

0.001

0.0005

0

-0.0005 , , , I , , , I I i , I , , , I , I i I j i J I , , , I , i f

1Deflection

angle, deg

--o-- -20

--o-- -10

---O-- 0

+ 10

--+-- 20

]-2 0 2 4 6 8 10 12


14

(g) Mach number =6.0


33

c5c.)

0.0009

0.0008

0.0007

0.0006

0.0005

0.0003

0.0002

0.0001

0

-0.0001

-2 0 2 4 6 8 10 12


(h) Mach number= 10.0

Deflection

angle, deg

-- -t3 20

---o--- -10

--O-- 0

---x-- I0

---4-- 20

14

ci

0.0009

0.0008

0.0007

0.0006

0.0005

0.0003

0.0002

0.0001

0

-0.0001

Deflection

angle, deg

--0--20

---o-- -10

--O-- 0

---_ -- I0

----4---- 20

0 2 4 6 8 10 12 14


(i) Mach number - 15.0


34

0.0009

0.0008

0.0007

0.0006

0.0005

0.0003

0.0002

0.0001

0

-0.0001

.......

Deflection

angle, deg

-20

--o-- -10

--O- 0

10

--4--20

-2 0 2 4 6 8 10 12



14

0.0009

0.0008

0.0007

0.OO06

0.0005

0.0003

0.0002

0.0001

0

-0.0001 ' ' ' I ' , , i , , , I , , , I , , L I , ,,i f s i i I ] i i

Deflection

angle, deg

-43-- -20

--_-- -10

--0-- 0

10

--4--- 20

I-2 0 2 4 6 8 10 12


14



35

-8c5

r.)

0.02

0.015

0.01

0.005

0

-0.005 ,,, I,,, I,,, I I t i I m i t I,, J I,, , I , , ,

-2 0 2 4 6 8 10 12 14



]Deflection

angle, deg

---t3-- -10

--o--- -5

---C_ 0

+ 5

--¢--- 10

0.02

De_ection

0.015 angle, deg

---o-- -10

-8 0.01 ---o-- -5

c5 ----<>- 0

0.005 10

o-0.005 ,,, I,,, I,,, u,,, n,,, t,,, i,,, _ , ,

-2 0 2 4 6 8 10 12 14



Figure 6.- Drag increment coefficient for canard as a function of angle ofattack, deflection angle, and Mach number.

36

0.02 l_Deflection

0.0.15 angle,deg-10

-- o .... 5-8 0.01 --c_-- 0

n" -)_-- 5

0.005 ---4--- l0

0

-0.005 ,,, _ , , , i,,, _ , , I , , i , , ,

-2 0 2 4 6 8 10 12 14




0.4 t

0.3 y -----4-3-- 0.7+ 0.9

_l 0.2 _ 1.5-- - 2.5

0.1 ..... 6.0

..... 10.0

0 -- - - 15.0..... 20.0

..... 24.2

-0.| l , , I , , , 1 , , , I , j , I , , , I .... I ,., 11 * , j I

-2 0 2 4 6 8 10 12 14

Angle of attack, or, degrees

Figure 7.- Lift increment coefficient for basic vehicle as a function of

angle of attack and Mach number.

37

0.15

0.1

0.05

0

-0.05

-0. I

-0.15

0----0---0----0_--0----0------0- ..... 0-- ....... 0

I


-43---20

--o-- -10

----<>- 0

---_--- 10

-2 0 2 4 6 8 10 12 14

Angle of attack, (x, deg


dU

0.15

0.I

0.05

0

-0.05

-0.1

-0.15

_J__Deflection

angle, deg

---Ek- -20

---o-- -!0

--0-- 0

10

---+--20

-2 0 2 4 6 8 10 12 14

Angle of attack, ¢t, deg


Figure 8.- Lift increment coefficient for right elevon as a function of angle ofattack, deflection angle, and Mach number.

38

d

0.15

0.I

0.05

0

-0.05

-0.1

-0.15

+---+-_ :

-2

'' _ I' I t J,,, I t t t I t t i I t _ t I _,, I , j ,

0 2 4 6 8 10 12

Angle of attack, cz, (leg


1

Deflection

angle, deg

---4:3-- -20

--o--- -10

--<>-- 0

10

-----+--20

I14

d

0.15

0.1

0.05

0

-0.05

-0.1

-0.15

-2

+_.-+ 0 0 0_

IDeflection

angle, deg

---q_-- -20

---o-- -10

---o-- 0

+ 10

--¢--- 20

o , l _ , _ I , , t I , _ , I , , , } , _ , I , w , I , i J

0 2 4 6 8 10 12



14


39

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-2

X X X X X X X X

I

Deflection

angle, deg

---ck- -20

--o-- -10

--c>- 0

10

---4-- 20

I ' I

0 2 4 6 8 10 12 14



d

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

X X X : ,-,v X X X X X

Deflection

angle, deg

---CJ---20

--o-- -10

--0- 0

---_- 10

---+--20

I I I II

-2 0 2 4 6 8 10 12



14


40

-0

0

0.015

0.01

0.OO5

0

-0.005

-0.01

......4}.........

I_-.4']- Ff--[}..............[.I-......

O- -0- -o---0 .........o .........0 .......0

i___.4.....4....I--...........-I...........I--........4 - -

-2 0 2 4 6 R

Angle of attack, ct, (leg

(g) Math numl_er = 6.0

..........r-3........471[l.

....... 0

!...... I

If) 12

Deflectkm

angle, deg

f3 -20

o -l(I

O- 0

-×- 10

.... _ -- 20

I .

14

,_./

0.015

0.01

0.005

0

-0.005

-0.01

....................................................................................i-I...........M [)eflecti(m

..........ct 1.......I-}........

__----L]-.........__--.F]......

[-}--_[]./-I_ -_-_/

__.O.___.O_.___O ...........-0............0"...........O ......0<>---"0

0---0 -0---0 ...........0 ..............0 ...........0 ...........0 ........0

....+ ...........-4........i...... !.........

+---_.+........4.....4=......

n,gle,deg

....t-}---20

- o .....I0

--o-(l

--_<----10

-4-- 20

k.,, I , ,., _l_, ,., I , , , .I , , , I , , , I _ , , I , , ,

-2 0 2 4 6 $I if) 12 14

Angle of attack, ¢x, deg

(h) Mach number = 10.O


41

0.015

O.0 !

0.005

0

-0.005

-0.01

......._ ...............CI

----[ }............. 4"_.....

0 ...........

o___o._.._o____o-_ -_o---

o.......o.... -o__---d ........ ! .......... t ........ F

Deflection

angle, deg

[3 .... 20

--o ..... 10

....o-- 0

----x--- I0

---_-- 20

-2 0 2 4 6 8 10



14

G)

0.015

0.01

0.005

0

-0.005

-0.01

......[-t ......

c__-__ ......... ..........._>__---.----0-----

_.__0--.--0 -----0---

o---o- o-o_-_.___--__ .... o-- o .... o

p..14.___..4___.__._ _ I--+-.........+ .......}.........:+:

.___L_LJ .t._L_l_.l._t l L..I J__,. I._l., _l ,.I J f I .I J , L I.

-2 0 2 4 6 g 10 12

Angle of allack, eL, (leg


Deflection

angle, (leg

--r3-- -20

-- -_ -.-

14

-I0

0

10

20

Figure _.- Continued.

42

0.015 1

U

0.01

O.005

0

-O.(X)5

-(}.01

...........4"1 Deflection..... t-_....

i[] ..... angle, deg

_j.4-t-" r_ -20

[2_.[ 3......Ct-f_3-"" ......@ ..........o --o ......1(1....... o ..... --O-- 0

o___.o---_ --------'-°--_- + 10

_.l_J__t_l.._l.J__.L_].l_-c-J I J_._l._t I .,. I.i I ,., , I_J , x. I

-2 0 2 4 6 8 !0 12 14


(k) Mach number -- 24.2


-8

0

0.15

0.I

0.05

0

-0.05

-0.I

-0.15

____D_____.Ct_ ....... t--3------------40

O___O_____O______O--- ..... O- .......... O-- ....... -O

0---0_--0-_-0------0

>_-- ->(---_-X-- ---X-_ -.--)4-----_- ......... N ........ _._ ........ X

-2 0 2 4 6 g 10 12 14


(a) Math number ---1}.3

Figure 9.- l,ift coefficient increment for left elevon as a flmction of angle ofattack, deflection angle, and Math number.

................ ]

!

Deflection

angle, deg

-- {-3..... 20

-- o- .... 10

--O-- 0

-_- 10

..... _ . 20

43

0.15

0.1

0.05

0

-0.05

-0.1

-0.15

................................... .........

Deflection

.____----L-J-.... 49"-----q] angle, degM _._--U1-

Er_t_4[__Y_ _ --43-- -20

_ . .. _,.__..______----o ....... o ........ o- ..... ---o --o-- -10

--o-0

0----(_ ...... -0------0-----0 + 10

/, I , , , I , , , I , , , I__L_a__t__L_L___ _t__L_L._L__.J___L_, , I

-2 0 2 4 6 8 10 12 14



.8d

0.15

0.1

0.05

0

-0.05

-0.1

-0.15

.__--.--43---..... t-3--------{3_._.---c]----

0_____0___.0__ ..... -0----_---0

o_---o------o------o

.... .........

Deflection

angle, deg

--[3-- -20

.... o .... 10

--O--- 0

---_ ..... I0

---+--20

.j_j.._a._ I , , , I ,

-2 0 2

,_a_l , , , l__L__t__L...l_.t_ t_J.A+__t__t__l_L_a.____k_

4 6 8 10 12 14


(el Mach number = 0.9


44

,.5r..)

0.15

0.1

0.05

0

-().()5

-0.1

-0.15

Et__tr_T____C]_4.j______.O.____..O___ ....... UI--.... _ ....... 42

O-- O----_---O- .... "-'O-_<>- ......... O .......... O ........ --O

O-O---O ......O ........ O .......... -(3 ........ O --- O .......... O

.... -X--- _ ....... )<--.... _ ...... -9_ ......... )¢,_....... X

-F.... -t...... t.... t .......... t............ f...... -4-.............. t............. F

IllllJlJll

-2 0 2

.___Lt_a_J___l___L_t_J.._ht_ L___l__J__a__t__J__J_ L.a__

4 6 8 10 12 14


ii i-_ill ........

Deflection

angle, deg

--{3-- -20

.....o - -10

----C_- 0

---_--l(I

---t .... 20

(d) Mach number = i.5

L.)

0.015

0.01

0.005

0

-0.005

-O.OI

-0.015

o---o---.o--,ff- .... -o-- ----o ..... _- ...... O ...... -_

O -O -O-O ..... O ....... O O O • ()

X _ ...... )(-....... )< ........ -)<X x x

Deflection

angle, deg

---cr-- -20

--_> .... 10

('_ 0

- _ 10

.... _.... 20

, ....... ,........ ........_ ..L I ,__L.a__l_J_._J___.I__L-L _LJ__ L-L-_L--L-L__t__a__

-2 0 2 4 6 8 10 12 14

Angle of allack, or, deg



45

().015

0.01

0.005

0

-0.005

-0.01

..................................... ._] 1 ........

U1---O---4]_ ..... -C3---------CX- ....... f3--

...... _ ........ --[]

O-- --O---O--O---- ....... O--------O- ....... _ ........ O ......... O

0 ...... 00---0--0- -0 ......... 0-----0 - -(3

X -'><--..... )_-_-XX X X X X

l)efleclion

angle, (leg

- -_ - -20

--o--- -10

.....O-- 0

---_ 10

2O

....... _t__-4L ......... t....... t ........ 4-_.,__t,,, F ..... ,__1,,, _, ._,.._LL_.,.__,__I___,.,J._L,_..,_

-2 0 2 4 6 8 I0 12 14


(O Mach number = 4.0

.8J

0.015

0.01

0.005

0

-0.005

-0.01

.... 4_]-------------E3_.------O---

..--.--.El"-'---t-j_._.4_--43_

0_0__________ 0 ......... -O

O -O--- O--O--- .... O-- ..... O ..... O .......... O .... O

__..] .........

Deflection

angle, deg

....E1 -20

---o -- -10

---0-- 0

.... _.- 10

--_ .... 20

X-----)_ X X X - ------X_ -- --)<- _- -)<-..... X ....... r ........./

. _.a__________4______l-- ------I ..... l ...... -i |

-2 0 2 4 6 8 10 12 14




46

0.015

0.01

O.O05

0

-0.005

-O.O1

....................................... I.t_LL...........Deflection

...... t ) ......... (1 angle, (leg_.t'} ....

._x:) ....... -_n.... 20......rT ......

it .... f) 'r_'xl ........ o -10

.... o ............ o .......... o ........... o .... O .... 0

o_..__o___o--_--o-_- _ 10

o--o---O--O ....... O-----o ........ O ......... o-----o --+-- 2(1

X----_-04----)4- .... )4 ...... _ ...... )_ .......... X ........... Y7 ................

i-+ ....... -4 ............ t ......... t ............ t-

i.._l J..i ,_ I. J ,., I I J , I , , , I ,

-2 0 2 4 6 g



I() 12 14

,_5

0.015

0.0 i

0.005

0

-0.005

-0.01

.......................................5 77.........

.__.._-X-tI

t_.} .... ..

...... .--I.q..... l }....

l)efleclion

angle, (leg

- -CJ.... 20

.... o - -I0

- 0-- 0

0 _-_ ..........

....... O" ............ 0 ....

o---o .... o--o---- -- _-- I0

o---(>--o--o .... o .... _x.os__ ..... o ....... o -- o -_.- 20>(____)_..... _)_ ...... _ ................. x ............ _ .......... _( ..........q"...........

4 ............ t .......... t ..... t- /...........-"4......... [

/

/_L_X__a_] i__L--L_L--L..i---t__JJ--J-__i._-L..t L I .l t_.i...l I I_l.__l.._[ .l_ i J

-2 (I 2 4 6 R I 0 12 14

Angle of attack, 0t, (leg

(i) Mach number = i 5.0

Figure 9.- C0nlimled.

47

(}.015 --

.8

U

O.O!

0.005

-0.005

-0.01 __t.__L--t__l__L_L--t__l__l__L.k I---a_..L__t__l__k_L+, .k.,_ .i _J I k_l _t--I-J-_,.--, ....

-2 0 2 4 6 8 I 0 12 14


(j) Mach number =- 20.0

.8,2

0.015

O.OI

0.005

0

-0.005

-0.01

...... .......

__- u3 Deflection

..._.--tr-----_ angle, deg

----c3- -2o,

D_,..._..-/3--J .... ---o ---o-- -i0

---_-- 1(I0---0- ------ 0 ........ 0

"" __+________+____--4...........,--:----.-------_--] ......

_x__x__r_ t_a__L_.t__L_..t_ L_-L--t__L_l___---__Ld_a___L __J

-2 0 2 4 6 8 10 12 14

Angle of attack, m deg

(k) Math number = 24.2


48

0.01

0.005

0

-0.005

-0.01 ..t .U.._+t_I_-J_L.-L__t._.+_ L. L-L +t.._ t . I ..U ,_ ,_ I

-2 0 2 4 6 8

i+_::1........ (_ - t J ::::.....+:r+-:+:....ii!i

-+ 0 ........... 0 ....... -0

I

.)I )efleclion

angle, (leg

[ ] -10

-+-o _ - -5

---O-- 0

+ 10

10 12 14



,-5(.)

0.01

0.005

0

-0.005

-0.01

-_ --->(

..... - ............

Deflection

angle, (leg

-r__t- -10

--O ..... 5

--o 0

× 5

- _ I0

.--L--L--J---LL--J-__L_I_.+I+.J_+.',_+I_L_LU.__.I_ t__i_+ J .t_+t. _ +l., _ , +] '_,..I._

-2 0 2 4 6 8 I 0 12 4

Angle of attack, 0t, (leg


Figure 10.- Lift increment coefficient for canard as a function of angle ofattack, deflection angle, and Math number.

49

0.01 ]

-8

U

0.005

0

-0.005

-0.01

-2

__ Deflectionangle, deg

"fl ---_ ...... 10

o.._,,;,._.-o--o--___ -o _ .___----_ -- o---0

---[ .... I0

0 2 4 6 g I0 12 14


(c) Mach number =0.9


-0.3

>.,{J

-0.4

-0.5

-0.6

-0.7

-0.8

................. _ _. : _.7 2._.__.. 7"-,-......

--.....,............... . . ° . .

O--O--- O-- O--- O----- O--- ..... O ........ C_ - -:-O

O-=--O-=-_---_ _ _ _ O-...... --O- ........ <> ...... ---O

M--o-- 0.3

--<v- 0.7

---o-- 0.9

--._- - 1.5

2.5

.... 4.0

..... 6.0

..... 10.0

.... 15.0

..... 20.0

...... 24.2

J__LJ__! , , , I , , , I t t I I I__.LLJ__L._LJ__LLLJLL.I__IJ_J

-2 0 2 4 6 8 I0 12 14

Angle of attack, ix, degrees

Figure 11.- Side force with sideslip derivative for basic vehicle as a function of

angle of attack and Mach number.

50

r..)

o.ool

0.0(}o8

0.(x)o6

0.0004

0.0002

0

-0.0(x)2

-O.O(X)4

-0.0006

...... I

H I)cllcclion

' _ angle, deg

/ / \ "_o--_ ^ -o-0

d--o--_-_>..........o.........o .....o o o 2o

-X-.........N .........><..... X ..........-)<

"'_-___ ._............._.......-' I

-2 0 2 4 6 8 I 0 12 14

Angle of altack, or, deg


O.0008

0.0006

O.O004

0.0002

0

-0.0002

-0.0004

................................................................./ -(1........../._ l)efleclion+ \ angle, deg

_. _t_r-______)......c_ --_- -I0

"/__ -........... _>....... _+......... rJ ---o-- 0

d/ \ \"- .... o------o- ....... _ ........... o ......... o --× ..... 10/ 'k

/ k, _ ).-2o....... ......o ,> o ,,\_\

_.__.__.__(........... -X......... X ....... X__q_............ -k

._J___L_J__.l__t__J__a___l._.t_J__..t_l __t_____l. I .t_ J _.l_1 l...., __ I.._t__J_L__

-2 0 2 4 6 8 I0 12 14

Angle of atlack, ct, deg

(b) Mach ,unlber = 0.7

Figure 12.- Side force increment coefficient for righl elevon as a funclion ofangle of attack, deflection angle, and Mach number.

51

rj

0.0(X)8

0.0006

0.0004

0.0002

0

-O.O(X)2

-0.0004 I

-2

......o........o o.......o"--.. 'K.

-.....-.._ ........ _............. !

"l- , , , I , J _ I

.... L_7.-]............

Deflectk)n

angle, deg

i-} -20

---o -10

--O-- 0

--_---10

..... t-- 20

I I 1

0 2 4 6 g 10 12 14



>:L)

O.(X)08

O.O(X)6

0.0004

0.00()2

0

-0.0002

-0.0004

1

Deflection

angle, deg

- (7_.... 20

-o-10

x----- o 0

" ---t3----- t.1- × 10

...... i

_l_ . _L_l .__L I __LJ_._L_±_ LJ_iL _ _L._lm_ ant __LI _l_ i _l .l__l_ i I _l_ ,., .

-2 0 2 4 6 8 10 12 14




52

I.(X) 10 .5

0.00 10°

- i.00 10 .5

-2.(X) 10-5

-5100 10.5

............................. -21zl................-_ .......... _I- . l)eflecti(m

O---O---O---(0 7_: .... O ---III-_ angle, deg

---_ ---_*----f --C__--- -20

--o- 0

---o-...... ---o -.-×- 10

-t"f_ -_°_k_ .... ,- 20

f _ ..... [

i- - .......'El .......... f3- ..... /__]

i]__t_._i._l__--t_a___t._l__c. I. l- .t _a_.t_a._. I_.t. i t . l i i .l 1 .,._ t .J ALL t _t_--I

-2 0 2 4 6 g 10 12 14



¢..)

1.(X) 10-5

0.00 10°

-I.(90 10-5

-2.00 10-5

-3.(X) 10 5

-4.00 i 0 -5

-5.00 10 -5

-2 0 2 4 6 8 10 12 14

Angle of attack, ft, cleg

(0 Mach number = 431

Figure 12.- Contimled.

53

_z

1.00 i 0 .5

o.oolo°

- 1.00 10 -5

-2.00 10 .5

-3.0O I0 5

-4.(X) 105

-5.00 10 -5

.............................................. j-___

Deflection

angle, deg

--43-- -21)

+ -10

----O-- 0

--_--- 10

2t)

I [ I__LLJ_ l I ! l L A__L_J__.._I __ J__l_ J.__ i __L.2 _J__I__L__I__

-2 0 2 4 6 g 10 12 14



>:r..)

1.00 10-5

0.00 10 0

-1.00 10 -5

-2.00 I0 .5

-3.(X) 10 .5

-4.00 10 .5

-5.00 10 -5

O----O-

, , , I , , , I , , , I , , , I , , , I , L_a__L..__t_a__L___t__L_

-2 0 2 4 6 Iq 10 12 14



Deflection

angle, deg

----_ - -20

----o-- -10

---o-- 0

--_ .... 10

---t- 20


54

U

I.(X) I0 5

0.(_) 10o

- I .(X) !0 5

-2.00 105

4.O0 10 -5

-4.(X) 10-5

-5.(}0 10 -5

:7a<--7-_Z773.......Deflection

angle, (leg

--M -20

-. o---10

-- 0-- 0

• - )_ 1(1

.... _.... 20

-2 0 2 4 6 tl 10 i 2 14

Angle of attack, et, (leg


>

1.00 10.5

O.(X) I0 °

- 1.00 10 .5

-2.(X) 105

-3.(X) 10.5

-4.(}0 10 -5

-5.00 10 -5

Deflection

angle, deg

....E) -- -20

- o---10

0 ()

•-x--- 10

t - 20

--2

1. I I 1 I

0

t._l. , t_,_ I , , _, ..I ,. _ I ,

2 4 6 g

Angle of allack, or, deg

(j) Mach ntmlbcr = 20.0

IIII.

12 14

Figure 12.- (?()rilhiued.

55

I .(X) 10 .5

().(X) 10II

- 1.00 I0 -5

-2.(X) 10-5

-3.00 I0-5

-4.(X) !0 -5

-5.(X) 10 5

o - o o--o ...... ._-------8 ...... c) .-,,_

o_.......,1_,..___ ..... .4-X ....... 0-.-._._, ....._t"_-_ '

_""_'0"-_

........o...........o

--11--._

_j/"

0

l )cflecl ion

angle, (leg

r} 20

O -10

---O-- 0

....._ .... 10

...._--- 20

I ....

II I I I

12 14

fl .....

, , ,. I t ,., I , _ , I , , _ I t _, I , , , I ,_,

-2 0 2 4 6 g 10

Angle of altack, o_, deg


Fig,re 12.- C(mchuled.

o

>:L)

O.O006

0.0004

O.0002

0

-0.O0O2

-0.0004

-0.0006

-0,0008

-0.001

............... 2 :7722-....i I

O .

1 471........... [t ........ { } ..... { I-.... _-l

Deflection

angle, deg

-.-iT- -20

--o .... I0

() 0

-_. 10

_ 20

-2 0 2 4 6 g I{}

Angle (if ;llla(.'k, Ix, (leg

(a) Mach nunlber : 0.3

Figure 13.- Side force incremerfl coefficienl fi}r lefl elevnn ;is a filnclirm ofangle of altack, defleclion angle, and Math numher.

56

(..)

0.00()4

0.0002

0

-0.0002

-0.0004

-0.0006

-0.0008

__/,,- .... _ ......... × . ×. . .×

-_--- ['--_/-- .... o-------o ....... o - -- o ........ o

E /._0-- ........... 0 ............... 0 0 0

.....I} ............. 41t ......... I-1 ........ F} ........ I 1...... I]

J__,_.J_ L____.__..l_.t_L___ I_ _ _L_ I ._._ _ I __._ .._ 1 _..............

-2 0 2 4 6 g 10 ! 2

Angle of attack, c_, (leg

(h) Math numlx'r _-0.7

l)eflection

angle, deg

- 4_1.--20

---_-. -10

--O--0

---_- 10

2(I

14

Q)

0.0004

0.0002

0

-0.0002

-0.(R)04

-0.0006

-0.0008

j.-4 .... --t ...../ ....... 4

o" x.........................---O--O () - • -O

._.___.<,--.... o .... <,[ ....

4_t- .... "'[} ......... [} ....... f 1" - (l-- I1

.I._I_L_}_L.I._|.I.I L I l .V t.-,_ I.*., , I , t , I , , , t ,

-2 0 2 4 6 g I() 12

A,gle of allack, (Z, (leg


Figt,re 13.- Ccmtimved.

k. 11

Deflection

angle, (leg

--t_ -20

•o -10

- o 0

34 10

20

I

14

57

>:rj

O.O(X)4

0.0002

0

-0.0002

-0.0004

-0.0006

-O.(X)08

...... vW-71.

---0 ............... 0 - - "

" " 4 ........ t....... X ....... X

-2

J. I. .1 ] I. I.. L

0

I L.._J _, I t t. , I J , , I

2 4 6 R


(d) Mach number--- 1.5

I()

, I

12

- ] _I)efleclion

angle, (leg

-.r] -20

-o -10

o 0

....._- 10

.... t- 20

1 I I

14

,g

5,0 10.5

4.0 10.5

3.0 10 .5

2.0 10 .5

1.0 10.5

0.0 10°

- I .() 10 .5

Angle of attack, (x, (leg

(e) Math number--- 2.5

Figure ! 3.- (_ontinued.

I)eflecli_m

,'ingle, deg

_t -20

-- <> lf)

---o 0

10

i 20

I I

14

514

r,.)

5.0 1()-5

4.O 10 -5

3.0 !0 .5

2.0 10 -5

1.0 I(1.5

0.0 I0 °

- 1.0 10 .5

-2 0 2 4 6 g I 0 12 ] 4



--B>:

L)

3.O 10 -5

2.0 10 -5

0.0 10°

- !.0 104

-2 0 2 4 6 g I0 12 14

Angle of altack, or, (leg

._

l)effect ion

angle, deg

--_- -20

- --o-- -I0

....o-- (}

--_ I0

_, 20

(g) Math number = 6.0


59

-8

u

5.0 1(1-5

4.0 10.5

3.0 10-5

2.0 10-5

1.0 10 -5

0.0 I0 °

-1.0 10 -5 i

-2

......................... 7¸::1Deflection

_.._...__3 angle, deg

,____..___-__ --c} .... 20

I , I i__J__J__L_.l_ LL _ L_J._J_._L__IJ__ I __[_£_ L _LI _L

0 2 4 6 g I0 12 14

Angle of a/lack, o_, deg


-8

r,.)

5.0 10 .5

4.0 10 -5

3.0 10 -5

2.0 10 -5

1.0 10 .5

0.0 10 °

-I.0 10 .5

-2

i i

0 2 4 6 8 10 12


(i) Mach number= 15.0


Deflection

angle, deg

---G---20

---o----10

--O- 0

10

20

i i

14

6O

5.0 I0 5

4.0 10-5

3.0 10-5

2.0 10-5

1.010-5

o.o lo°

-1.0 10 -5 i

-2 0 2 4 6 8 10 12



Deflection

angle, deg

----0---20

---o-- -10

---O-- 0

10

--_---- 20

5.0 10 5

4,0 10 -5

3.0 10 -5

2.0 10 -5

1.0 10 -5

0.0 10 °

-1.0 10 -5

-2

Deflection

angle, deg

--c3-- -20

--o-- -10

--O-0

--+-- 20

i k_._l___t__

0 2 4 6 8 10 12 14




61

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

I I I I I t I I I

vr, X X X X X X X X

Q--C>_OIQ-_-QIQI-O

-2 0 2 4 6 8 l0 12



Deflection

angle, deg

-20

+ -10

--O-0

10

---4---20

14

0.06

0.04

0.02

0

-0.02

-0.04

-0.06-2

I I I I I I I I

X X X X X X X X X

0 2 4 6 8 10 12



Figure 14.- Side force increment coefficient for rudder as a fimction ofangle of attack, deflection angle, and Mach number.

62

0.06 I

0.04

0.02

-0.02

-0.04

-0.06

-2

. t I t I t I : ,_

X X X X X X )( XV

o.---o-

0 2 4 6 8 10 12



Deflection

angle, deg

--43---20

----o-- -10

--O-- 0

+ 10

--+--20

0.06

_D

0.04

0.02

>: 0

-0.02

-0.04

I I I I ,_ I I I I

X X X-X X X X X X

_--------o

Deflection

angle, deg

--o-- -20

-I0

--O--0

+ 10

--4-- 20

, J _ I t I , I , , , I ,j__j I , , , I J , , I _ t I , , ,

0 2 4 6 8 l0 12 14




63

0.015 I

0.01

0.005

0

-0.005

-0.01

-0.015

-2 0 2 104 6 8

Angle of attack, t_, deg


I I I

12

Deflection

angle, deg

-------0-- -20

+ -10

--O-0

10

---+--20

14

(.9

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

X X X X X X _ X ,

0----0---0---0--------0_ ...... 0-'------ <)- ...... 0-- ...... 0

Deflection

angle, deg

-4:y- -20

---o-- -10

--O-0

10

---_-- _ 20

-2

,,, I, _, I , J, I,, , I t , , I, ,, I,, , I J , ,

0 2 4 6 8 lO 12

Angle of attack,c_,deg


14


64

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

X X ,-.v X × × ×

f>----O---<3----O__

. _ .O

-2

J , , f , , , J j j , I , , , I J , , I I i J I J J t I , , t

0 2 4 6 8 10 12



__J_

Deflection

angle, deg

---Or- -20

---o-- -10

---<>- 0

+ 10

--+--20

14

r..)

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

Deflection

angle, deg

--43-- -20

---o-- -I0

--O--0

+ 10

+ 20

l , J I _ _ i I i , _ I , , , I , , , I , J i I , , , l._____j_

0 2 4 6 8 10 12


14



65

c,.)

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

yr. X X ..v X X X X _K

O---O--i_ ,0 0----------0 , O

Deflection

angle, deg

--o-- -20

--o-- -I0

--O-0

10

--+--20

-2

I I I I I I I I , I I I I ] I I I I I I ' , , I i j _ I , , ,

0 2 4 6 8 I0 12


(i) Math number = 15.0

14

4_

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

X X X X X X X X _-X

OiO---Oi@ _ _ ,0

.____---_

Deflection

angle, deg

--c}-- -20

---o-- -10

--O-0

10

---_.-- 20

0

, , I , , , I i i i [ i i ] I i i i I , I i [ , i ,

2 4 6 8 10 12

Angle of attack, at, deg


14


66

0.006 I

>:L_

0.004

0.002

0

-0.002

-0.004

-0.006

X X X X X X X X------_

t _, I,, _ I t t t I, l, I,,, I t t __A__,,, I,, t

-2 0 2 4 6 8 10 12 14




Deflection

angle, deg

---D- -20

---o-- -10

+0

---x-- I0

---4---20

0

-0.02

-0.04

-" -0.06

-0.08

-0.1

-0.12

-0.14

-_-Z-2" ___.'----" ----------_:--: =--: .__.: _ :-- ...... _ __.<_-_: _--_-"---: =-:-.

M

----<>-- 0.3

--o-- 0.7

--o-- 0.9

1.5

2.5

- - - 4.0

..... 6.0

..... 10.0

-- - - 15.0

..... 20.0

..... 24.2,,, I,, _ I,,, I, t , I t, ,I j ,._ I _ , _ I ....

-2 0 2 4 6 8 10 12 14


Figure 15.- Rolling moment with sideslip derivative for basic vehicle as afunction of angle of attack and Mach number.

67

.g

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-0.04

-2

, , , I , , , t , , , I , , , I , , , I , , , I , , , I , , ,

0 2 4 6 8 10 12

Angle of attack, tz, deg


Deflection

angle, deg

---o- -20

---o-- -10

--<>- 0

+ 10

+20

14

ro

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-0.04

IDeflectionangle, deg

--O- -20

---o-- -10

--o-0

+ 10

--¢-- 20

, , , I i , , I , i i I J , , I , , , I , , , I , , j I , , ,

-2 0 2 4 6 8 10 12



Figure 16.- Rolling moment increment coefficient for right elevon as afunction of angle of attack, deflection angle, and Mach number.

14

68

,....t

r..)

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-0.04

_.._..---4-

-2 0 2 4 6 8 10 12 14



Deflection

angle, deg

--cr--- -20

---o-- -10

--<>- 0

+ 10

+20

d

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-0.04

.I---+

I ___..+__._-.----b--"-_

-2 0 2 4 6 8 10 12




Deflection

angle, deg

--o-- -20

+ -10

---O-- 0

+ 10

-----4---20

14

69

0.006 I

P

0.004

0.002

0

-0.002

-0.004

-0.006

•I I i I I I i I I

X X X X X X X X X

-2

, , , I , , , l , , , I , _ , t , _ , I , , , _ _

0 2 4 6 8 10 12



Deflection

angle, deg

---cy- -20

---o-- -10

--O-0

+ 10

+20

14

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

I I l l

X X X X X X X X X

Deflection

angle, deg

--cy-- -20

--o-- -10

--O-0

+ 10

---_ 20

, _ _ I _, , I , _ i I,,, I, _ J I,, , I, , , I , , ,

0 2 4 6 8 10 12


(f) Macb number = 4.0

14


7O

r..)

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

-0.004

-0.005

-2

x----x----x ×_ ×,. x __--------x__

, t , I,,, I,,, I, i , I i i, I i | i LJ._l_l_J I,I I

0 2 4 6 8 10 12



Deflection

angle, deg

--oF- -20

---o-- -10

--O--0

10

+20

14

0.003

@

r..)

0.002

0.001

0

-0.001

-0.002

-0.003

-0.004

-0.005

-2

x × -*

, , , I i i i I i i i I i i i I i i i I , , I I I I I I , , ,

0 2 4 6 8 10 12 14

Angle of attack, tt, deg

Deflection

angle, deg

--z3---20

_-I0

--<y- 0

+ 10

20



71

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

-0.004

-0.005 , , , I , , , I , , , I , , , I , , , I , , , J j , , I , , ,

-2 0 2 4 6 8 10 12



I

Deflection

angle, deg

+-20

--¢-- -10

---O- 0

---)6- 10

+20

14

d

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

-0.004

-0.005

-2

; I- 4

v

0 2 4 6 8 10 12


Deflection

angle, deg

--cI-- -20

-----<>-- -10

--O-- 0

---x-- I0

20

14



72

..-t(..)

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

-0.004

-0.005-2

__ I -4-

, , , I , , , I ! , i I t , i l i , , I , t , l , i i I i J a

0 2 4 6 8 l0 12




Deflection

angle, deg

----t3-- -20

---o-- -10

---o- 0

10

---¢-- 20

14

d

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-O.O3

---43

_¢_._.-.4_I a __

Deflection

angle, deg

---o- -20

---o-- -10

---<>- 0

+ 10

-----+--20

-2 0 2 4 6 8 10 12



14

Figure 17.- Rolling moment increment coefficient for left elevon as afunction of angle of attack, deflection angle, and Mach number.

73

O

ro

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-2

]

Deflection

angle, (leg

---cF- -20

--o-- -10

---o- 0

+ I0

+20

, , , I , , , I , , , I , , , I , , , I , , , I , , , I , , ,

0 2 4 6 8 I0 12



14

d

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-2

IDeflection

angle, deg

--0--20

--o-- -10

--O--0

+ 10

---+-20

i

_ _ _ I,, t I, _ ] I _, , I,, _ I,, , I,, J I , , ,

0 2 4 6 8 10 12

Angle of attack, 0_, deg

14



74

0.04

0.03

0.02

0.01

0

-0.01

-0.02

-0.03

-2

,, , I, ,, I,, , I,,, I, ,, I,,, I,,, I , , ,

0 2 4 6 8 10 12

Angle of attack, (z, deg


Deflection

angle, deg

---o- -20

--+- -10

---O- 0

---x-- 10

+20

14

d

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

o----o----.o----0 o -o-------o o------o

X X X ,%1 X X X X X

Deflection

angle, deg

---t3-- -20

+ -10

----O-- 0

+ 10

--_-- 20

I t t I t ..... I t - t I

0 2 4 6 8 10 12


(e) Macb number = 2.5


14

75

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

x X x X X X x x_

-2

, , , I , , , I , , , I , j , I , _ _ I J _ , I , , , I , ,

0 2 4 6 8 10 12



Deflection

angle, deg

+-20

+ -10

--O-- 0

+ 10

---+--20

14

0.005

0.O04

0.003

0.002

0.001

0

-0.001

-0.002

-0.003-2

Deflection

angle, deg

----cv- -20

---o-- -10

--<9-- 0

+ 10

----4-- 20

x X × x x x ×

0 2 4 6 8 10 12



14


76

¢.)

0.005

0.004

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

x

_-------O

.--.---O----

-2 0 2 4 6 8 10 12

Angle of attack, (_, deg


Deflection

angle, deg

--O- -20

+ -I0

--<>- 0

+ I0

--+--20

14

,,._-

u

0.005

0.004

0.003

0.002

0.001

(1

-0.001

-0.002

-0.003

-2

.______<>----

o--_----o---o

_i - i ------I-

,,, I,,, I,, , I,,, I t t, I, ,, I,, t I t I t

0 2 4 6 8 10 12


Deflection

angle, deg

---q3- -20

----o-- -10

---O- 0

+ 10

--+-20

14



77

u

0.005

0.004

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

-2

f..q21----Z

Zf

J

Deflection

angle, deg

-q3-- -20

--o-- -10

--O-0

--g-- 10

+20

Z

0 2 4 6 8 l0 12



14

,..¢

0.005

0.004

0.003

0.002

0.001

0

-0.001

-0.002

-0.003

J..-cl----

__---o------o---o_

___1__

Deflection

angle, deg

--z3-- -20

--o-- -10

--O-0

10

----4---20

-2 0 2 4 6 8 10 12


14



78

4_

0.02

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-0.02

-2

; a,, _ ' I I I I --H !

× × × X X X X _-------X

O_ 0-------<>--------0 ,0-------0

Deflection

angle, deg

--0--20

---0-- -10

---<y- 0

----x-- 10

+20

0 2 4 6 8 10 12



14

_.r¢..)

0.02

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-0.02

-2

I I I I t-- I } I I

X X ..._ X X .×. X X X

• - - 0---_<>_--------0

Deflection

angle, deg

---u-- -20

--o-- -10

--O--0

l0

---4--- 20

,,, I I I I I I,, I,,, I i i I I, , , I , .a • I I _ _

0 2 4 6 8 10 12



Figure 18.- Rolling moment increment coefficient for rudder as aflmction of angle of attack, deflection angle, and Mach number.

14

79

L)

0.02

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-0.02

I I i I I I I -I I

X X X X X X X , X X

-2 0 2 4 6 8 10 12



Deflection

angle, deg

--_ -20

---o-- -10

--o--0

10

--+--20

14

0.02

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-0.02

I I I I I " • I' I t I

X X _ X X X X X X

e_

Deflection

angle, deg

--0---20

--o-- -10

---O-0

10

---4--20

-2

, , , I , t , I , , , I , , , I , , , I , , , I i , , I, , ,

0 2 4 6 8 10 12




14

80

0.006 l

c)

0.004

0.002

0

-0.002

-0.004

-O.006

-2 0 2 4 6 8 10 12



Deflection

angle, deg

----cy- -20

--<>-- -10

---O- 0

---x--- 10

+20

14

d

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

Z X X X I PX I X X X 2 X

0----0------0----0_--<>__0 ....... -0 _---0

Deflection

angle, deg

---c3-- -20

--o-- -10

--O-0

---x--- 10

--+-- 20

, , , I , , , I , , , I , , ., I , , , I , , , I , , , I , , ,

0 2 4 6 8 l0 12


(0 Mach number = 4.0


14

81

0.002

0.0015

0.001

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

X--'---X----X ×

-2

.___..42p._-.------C1---

I I l I I I ' i I i i i I I i I i I t ! I I ! I I I i i i

0 2 4 6 8 10 12


(g) Mach number -- 6.0

Deflection

angle, deg

+-20

--o-- -10

--O-- 0

----4¢-- 10

+20

14

0.002

0.0015

0.001

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

-2

_, _. x x x

o-_ --_ _-_-------_

Deflection

angle, deg

---o-- -10

--O--0

+ 10

----+--20

, , , I , I I I I I ' I | I I I ' ' ' I ' ' ' I I I I I , I '

0 2 4 6 8 10 12




14

82

0.002

0.0015

0.001

0.0005

0

-O.OOO5

-0.001

-0.0015

-0.002

-. X X X .." X )K X

-o------o o _ o

-2 0 2 4 6 8 10 12



Deflection

angle, deg

---o- -20

---o-- -10

--(y- 0

+ 10

---4--20

14

d

0.002

0.0015

0.001

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

-2

X X X X X X X X X

Deflection

angle, deg

--q:F- -20

---o-- -10

--O--0

---x--- 10

----4---20

0 2 4 6 8 l0 12




83

14

rf

0.002

0.0015

0.0OI

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

-2 0 2 4 6 8 10 12




I

Deflection

angle, deg

----_- -20

-10

---<3--0

10

-----+--20

14

-0.02

-0.04

-0.06

-0.08r,.)

-0.1

-0.12

-0.14

-0.16

-2

- 2.-.

c+_ O

M

-----<3-- 0.3

----D- 0.7

---o-- 0.9

1.5

2.5

- - - 4.0

..... 6.0

..... 10.0

- - 15.0

..... 20.0

..... 24.2_L.a_J__l J , _ I J J , I , J _ I , _ _ I , , , I _ , , I _ , , ,

0 2 4 6 8 10 12 14

Angle of attack, _, degrees

Figure 19.- Rolling moment with roll rate dynamic derivative for basicvehicle as a function of angle of attack and Mach number.

84

r..)

0.35

0.3

0.25

0.2

0.15

0.I

0.05

0

-2

0----0----0---0 0 0--------0 0 0

M

--<:>- 0.3

--t:y- 0.7

--o-- 0.9

1.5

2.5

4.0

..... 6.0

..... I0.0

--- - 15.0

..... 20.0

24.2

0 2 4 6 8 lO 12 14

Angle of attack, cz, degrees

Figure 20.- Rolling moment with yaw rate dynamic derivative for basicvehicle as a function of angle of attack and Mach number.

0.02

0

-0.02

-0.04

-0.06

-0.08

M

---O--0.3

---zy-0.7

--0--0.9

1.5

__ _ _ _ 2.53,99

..... 6.0

..... 10.0

----15.0

..... _.0

, , , I _ , , I , , , I , J , I , I i I I j j Lx , ,N , ,__L_u

-2 0 2 4 6 8 10 12 14

Angle ofatmck, a, degrees

Figure 21.- Pitching moment increment coefficient for basic vehicle as afunction of angle of attack and Mach number.

85

gL_

0.06

0.04

0.02

0

-0.02

-0.04

-0.06-2

4----+ i 'I :_

XL X X X X X _____g(_ _ X

_,, ,

0 2 4 6 8 10 12

Angle of attack, ¢t, deg


,,,

Deflection

angle, deg

--Ck- -20

---o-- -10

--o-0

---x-- 10

---4--20

14

gL)

0.06

0.04

0.02

0

-0.02

-0.04

-0.06-2

-F----q-- t • + I '_

X---X X X X

O--q3--qD-'-K__qD-----_O

, , , I ,._ , I , J J I t , t I , , J I , J , I , , _ I

0 2 4 6 8 l0 12

Angle of attack, 04 deg


1Deflection

angle, deg

-20

---o-- -10

---o- 0

10

----4-- 20

14

Figure 22.- Pitching moment increment coefficient for right elevon as afunction of angle of attack, deflection angle, and Mach number.

86

ro

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

I

Deflection

angle, deg

--cy- -2O

+ -10

--<5-0

+ 10

---+--2O

, , , I , , , I , , , I---La---L-L--x---,---L--L2LL___j_LL_I__,_.,._.i....

-2 0 2 4 6 8 !0 12 14


(c) Mach number ---0.9

r,.)

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

X _P

, , , I , i , I , , , I , , , I , , , I , j I I , , , I , ,j

-2 0 2 4 6 8 10 12




__3__


+-20

----o-- -I0

---O-- 0

--x--- 10

----v-- 20

J14

87

I

0.015

d

0.01

0.005

0

-0.005

-0.01

-0.015

I I' I I' " I I I d " I"

X X X X v X X X _'

-2 0 2 4 6 8 l0 12



Deflection

angle, deg

--[3---20

---o-- -10

----O-- 0

--x-- 10

--+--- 20

14

d

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

-2

I I' ' I I I

X X X X X X X X X

_O_---O----O

Deflection

angle, deg

--43-- -20

-----o-- -10

--<>-0

---_---- 10

--+- 20

0 2 4 6 8 10 12 14




88

0.006

0.004

0.002

0

d -0.002¢3

-0.004

-0.006

-0.008

-0.01

-2

_ ,,xv

,, _ I , , , I , , , I I I |__L_L-___-L.-J,__J__a_._L [-_L_--L__L_J__J______

0 2 4 6 8 10 12 14


Deflection

angle, deg

---4z_ -20

--o--- -10

----O- 0

---_-- 10

+20


t,.)

0.006

0.004

0.002

0

-0,002

-0.004

-0.006

-0.008

-0.01

-2 0 2 4 6 8 I0 12 14

Angle of attack, et, deg

Deflection

angle, deg

--43-- -20

----o-- -10

--o-- 0

---x--- 10

---+-- 20



89

0.006

0.004

0.002

0

d -0.002

-0.004

-0.006

-0.008

-0.01

-2



0.006

dL)

0.004

0.002

0

-0.002

-0.004

-0.006

-0.008

-0.01

-2

, , , I , , , I , , , I , , , l. , I , I ) ) ) I i , J I J i L.

0 2 4 6 8 10 12 14


Deflection

angle, deg

--Z:F- -20

---O--- -10

--o-0

-_ 10

+20



90

0.006

0.004

0.002

0

j -0.002

-0.004

-0.006

-0.008

-0.01

_-¢3--...

_",,, I,,, I,, _ I, i, I j L_LI,,, I , t , I , , j

-2 0 2 4 6 8 l 0 12




1

Deflection

angle, deg

---o-- -20

--o-- -10

--O--0

--x-- 10

----+--20

14

U

0.06

0.04

0.02

0

-0.02

-0.04

-0.06-2

_-------o

'', I m J, I,,, I J i u i j, i ,__1, , , f , J , I , , ,

0 2 4 6 8 I0 12



Deflection

angle, deg

----o-- -20

---o-- -10

--Q- 0

--g-- 10

---+--- 20

l14

Figure 23.- Pitching moment increment coefficient for left elevon as a

function of angle of attack, deflection angle, and Mach number.

91

r

.8d

0.06

0.04

0.02

0

°0.02

-0.04

-0.06

-2

-I----q-- _ 4

X X X X .. X X ------X

IZ_O--------Q

Deflection

angle, deg

--{3-- -20

---o-- -10

---(3--O

I0

----+--20

, J , I _ _ , I t _ m I J _ J I , , J I l , i I _, , J I , ,._L_

0 2 4 6 8 10 12 14



dU

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

-2

X X X X X ,_----_--X----------X

o--o-

0 2 4 6 8 10 12



Deflection

angle, deg

--EF- -20

---O-- -10

_0

---_-- 10

---_-- 20

14


92

dc.)

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

X X X X X X :. _ X '--"_

,_>

-2 0 2 4 6 8 10 12



Deflection

angle, (leg

----Er-- -20

----o-- -10

---(>- 0

---x--- 10

--+--20

14

r.)

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

4- I I I l I I I t I

X _ _ _ Z X Z X I

o_0--o---o-----_Ol O1 .

[Deflection

angle, deg

-20

--o-- -10

--<>- 0

10

--+--20

, , , I , , , I , , I I J J , I I I J L.J__J_.._.l__l I I___l I I I

-2 0 2 4 6 8 lO 12

Angle of attack, ct,deg


14


93

.8d

0.015

0.01

0.005

0

-0.005

-0.01

-0.015

+-'_,_ ' I I I I' t I I

X X X X X X X X • ):.

o---o--o--_, o--_--o_-_____>.______


---O-- -20

--O-- -10

---O-- 0

+ 10

+20

-2 0 2 4 6 8 I0 12



14

0.006

0.004

0.002

0.8

j -0.002

-0.1104

-0.006

-0.008

-0.01-2

X .._'. X × >(

_O-------O_------(>-_-O

,} ,,

Deflection

angle, deg

---O-- -20

+ -10

--o-- 0

+ 10

---4---20

, _ J I, _ _ I,,, I t t t I t t t I, t t I t t t I t t t

0 2 4 6 8 10 12


14



94

0.006

0.004

0.002

0

g -0.002

-0.004

-0.006

-0.008

-0.01

X - X X - X _---x

2-x.a.--x-_, , , I , _ , l , , , I i i i I t _ j l__t._, , f

-2 0 2 4 6 8 10 12

Angle of attack, ¢x, deg


Deflection

angle, deg

--D- -20

---o-- -10

--O-0

---04-- 10

+20

I

I I I J

14

x:J

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-0.008

-0.01

-2

-t]-......

''' I, l, I,,, I I, I I f i t I i I L i,,, I t , i_

0 2 4 6 8 10 12


Deflection

angle, deg

---£_ -20

---o-- -10

--O--0

---g-- 10

.... ;--20

14



95

0.006

0.004

0.002

0

j -0.002

-0.004

-0.006

-0.008

-0.01

4______ 4-------+

, , , I _ , ,.I , , , I , , , I , , , I , , , I , , , I , , ,

-2 0 2 4 6 8 lO 12



Deflection

angle, deg

+-20

--o-- -10

+0

+ 10

+20

14

0.006

0.004

0.002

0

d -0.002

-0.004

-0.006

-0.008

-0.01

-2

" b-----o

t t _ I , t , I _ , _ I , t _ I t , , I _ , t I t t t I t ,_ t

0 2 4 6 8 10 12

Angle of attack, tx, deg

Deflection

angle, deg

+-20

---o---10

--O- 0

+ 10

--f.-- 20

14

(k) Mach number = 24.2.


96

0.003

0.0025

0.002

o.oo15J

0.001

0.0005

0

-0.0005

-2

Deflection

angle, deg

---t3-- -20

---o-- -10

---<>- 0

+ 10

+20

, , , I , , , I , , , I , , , I , , , I , , , .k., , , I , , ,

0 2 4 6 8 10 12



14

0.003

0.0025

0.002

0.0015

L)0.001

0.0005

0

-0.0005

Deflection

angle, deg

--C3-- -20

--o-- -10

---O-0

---x-- 10

-------+---20

-2

, , , I , , , I , , , I , , , I , , , I , , , I , , , I , , ,

0 2 4 6 8 l0 12



Figure 24.- Pitching moment increment coefficient for rudder as afunction of angle of attack, deflection angle, and Mach number.

14

97

L_.-;.

0.003

0.(1025

0.002

8

0.001

0.0005

0

-0.0005

nn rn n.s rvsw

--fl]---------Ir8..........{1:1.... ffl

m---_- _-_ --------_-- _I

Deflection

angle, deg

--o-- -20

--o-- -10

--O-0

10

--+--20

-2

O--O- _-O------q_------O

, , , I , , _ I , , , I , , , I , , _ I , , , I , , , l , , ,

O 2 4 6 g I0 12 14

Angle of attack,or,deg


0.003

0.OO25

0.002

'_ 0.0015

L)0.001

0.0005

0

-0.0005-2

._ _---_F-------- t_----------_

Deflection

angle, deg

---C_ -20

---o-- -10

--<y- 0

10

-_-- 20

0 2 4 6 8 I0 12 14


(d) Math number = 1.5


98

0.005

0.004

0.003

0.002

O.OO1

0

-0.001

t

, , , I , , , I , , , I , i, ,_J__a__,_L_L__t___L, , , I , , ,

-2 0 2 4 6 8 10 12



Deflection

angle, deg

--o-- -20

--o-- -10

----<3--0

---+--20

14

0.003

0.0025

0.002

0.0015

U0.001

0.0005

0

-0.0005

-2

Deflection

angle, deg

--C3---20

---¢-- -I0

--O--0

10

.... t--20

0 2 4 6 8 I0 12 14




99

0.003

0.0025

0.002

.8 0.0015g

0.001

0.0005

0

-0.0005

¢

Deflection

angle, deg

--{3- -20

--o-- -10

--O--0

--x--- 10

---+---20

-2

_-<>------<3

0 2 4 6 8 10 12

Angle of attack, tx, deg


14

0.0014

().0012

0.001

0.0008.8

d 0.OO06u

0.0004

0.0002

0

-O.OOO2

-2

...... A ...........

-m-.__..------m

l)efleclion

angle, deg

--o-- -20

---o-- -10

---O-- 0

10

---t--- 20

j_----------______ ...._______._

_0-------0 ...... 0--------0

, , , l , , , il , , , I , _ , l , , , i , , , I , , , I , , ,

0 2 4 6 8 10 12


(h) Mach number = lif0

14


100

0.0014

0.0012

0.001

0.0008

0.0006

0.0004

0.0002

0

-0.0002

m.._...

Deflection

angle, deg

-q3-- -20

---o-- -10

--<_0

10

_20

-2 0 2 4 6 8 10 12



14

0.0014

0.0012

0.001

0.0008

d 0.0006ro

0.0004

O.0002

0

-0.0002

Deflection

angle, deg

--D-- -20

-I0

--O-- 0

---x-- 10

20

-2

, , , I , , j I t , , I , , _ I J , _ } , , j I * J , I , , J

0 2 4 6 8 10 12



14


101

0.0014

0.0012

0.001

0.0008

g 0.0006

0.0004

0.0002

0

-0.0002

1

Deflection

angle, deg

---1:3--20

---o-- -10

---O- 0

+ 10

---+-20I

, , i , , , I , , , I , , , l , , , I , , , I , , , I , , ,

-2 0 2 4 6 8 10 12

Angle of attack, (t, deg


14


.8

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

Deflection

angle, deg

-10

---0---5

--O--0

--+-- l0

-2

, , , I , , , I , , , I , , , I , , , I , , , I , , , I , , ,

0 2 4 6 8 10 12



Figure 25.- Pitching moment increment coefficient for canard as afunction of angle of attack, deflection angle, and Mach number.

102

14

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

X "_, X X X X _ X

-2 0 2 4 6 8 10 12



Deflection

angle, deg

+ -10

+-5

--O-- 0

+5

----4-- 10

14

U

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-2

X X X X X X X X

Deflection

angle, deg

---z3-- -10

----o-- -5

---o- 0

+5

---+-- lO

, , , i , , , I , , , I _ _ _ I J t _ I , _ t , , ,

0 2 4 6 8 10 12



14


103

t

-0.5

-1

-1.5

-2

o-----o----o----o o--------o o o---------o

-2 0 2 4 6 8 10 12

M

--o-- O.3

0.7

---o-- 0.9

1.5

-- - 2.5

- - - 4.0

..... 6.0

..... 10.0

-- - - 15.0

..... 20.0

..... 24.2

14

Angle of attack, ct, degrees

Figure 26.- Pitching moment with pitch rate dynamic derivative as afunction of angle of attack and Mach number.

r,.)

0.2

0.15

0.1

0.05

0

-0.05

-0.1

o----o----o---o , o _ o

..................... ° . ° . . . . . .....

M

0.3

--0-- 0.7

---o--- 0.9

1.5

-- - 2.5

- - - 4.0

..... 6.0

..... 10.0

- - 15,0

..... 20.0

..... 24.2

-2 0 2 4 6 8 10 12 14

Angle of attack, a, degrees

Figure 27.- Yawing moment with sideslip derivative for basic vehicleas a function of angle of attack and Mach number.

104

0.0002

0.0001

o

-0.0001

-0.0002 I

-0.0003 ' ,, i,,, I,,, I,,, I,,, I,,,

-2 0 2 4 6 8 10 12 14

Angle of attack, t_, deg

(a) Mach number -- 0.3

Deflection

angle, (leg

--D-- -20

--o-- -10

--o--0

10

---+--20

L)

0.0002

0.0001

0

-0.0_1

-0.0_2

-0.0003

Deflection

angle, deg

----0--20

---o-- -10

---0-- 0

10

----_- 20

i * i I * , * I | I [ [ * * I I i i i I i * l I * * , I j t I

-2 0 2 4 6 8 I0 12


(b) Math number = 0.7

14

Figure 28.- Yawing moment increment coefficient for fight elevon as afunction of angle of attack, deflection angle, and Mach number.

105

O.OOO2 I

dL)

0.0001

0

-0.0001

-0.0002

-0.0003-2 0 2 4 6 8 10 12



Deflection

angle, deg

---D- -20

---o-- -10

--O--0

I0

+20

14

0.0002

0.0001

0

-0.0001

-0.0002

-0.0003-2 0 2 4 6 8 10 12

Angle of attack, it, deg

Deflection

angle, deg

--42k- -20

----o-- -10

--O-0

I0

--+--20

14



106

=-

0.003 ]

Deflection0.0025 angle, deg

0.002 --o-- -20--o-- -lO

0.0015 --o-- 0---x-- 10

0.001 _ 20

0.0005

0

-0.0005 ,,, i .... J,,, i,,, i,,, I,,, i , ,-2 0 2 4 6 8 10 12 14



0.003 1

Deflection0.0025 - angle, deg

0.002 ---t3-- -20---o-- -10

_ 0.0015 ---0-- 0

0.:100 ---x-- 10---+--20

0.0005

0

-0.0005 ,,, J,,, I,,, I,,, J,,, i,,, I,,, J , , ,-2 0 2 4 6 8 10 12 14




107

0.002 I

r..)

0.0015

0.001

0.0005

0

-0.0005 , , , I , , , I , , , I , , , I , J , I , , , I , , , I , , ,

-2 0 2 4 6 8 10 12 14



Deflection

angle, deg

---t3-- -20

---<>-- -10

--o-0

---x-- 10

----+--20

=-t,.)

0.002 l

Deflection

0.0015 angle, deg

-----0- -20

--o-- -10

o.ool --o- 0!0

0.0005 -- 4--- 20

0

, , , I , I Lu__, , j-0.0_5

-2 0 2 4 6 8 10 12 14




1(18

r..)

0.002

fl

0.0015 ;I

0.001 0

_5

0

,,, I,,, I,, i I i f t I t,, I,,, I , , , I , ,

Deflection

angle, deg

---t3-- -20

mob -10

0.0005

-0.0005

-2 0 2 4 6 8 10 12 14



0.002 1

Deflection

0.0015 angle, deg

--t3- -20

--o-- -10

._t. 0.001 --0- 0d

_ I0

0.0005" ---+-- 20

! "j ' 1

0

-0.0005 ' i J i

-2 0 2 4 6 8 10 12 14




109

0.002

Deflection

0.0015 angle, deg

--c3-- -20

--o-- -10

-B 0.001 --O-- 0=-

---x-- 10

0.0005 ----+--20

0

-0.0005 ....

-2 0 2 4 6 8 10 12 14




d0

0.0003

0.0002

.'0.0001

0

-0.0001

-0.0002-2

Deflection

angle, deg

---£v- -20

--o-- -10

--O-0

----x--- 10

--+-- 20

0 2 4 6 8 10 12 14



Figure 29.- Yawing moment increment coefficient for left elevon as afunction of angle of attack, deflection angle, and Mach number.

110

r.)

0.0003

0.0002

o.oool

o

-0.o001

-0.0002

Deflectionangle,deg

-20----o----10--o-0

10----+--20

-2 0 2 4 6 8 10 12 14



r..)

0.0003

0.0002

0.0001

0

-0.0001

-0.0_2

I

Deflection

angle, (:leg

--cv- -20

-10

---O-- 0

10

---4--20

-2 0 2 4 6 8 10 12 14




111

0.0003 I

d

0.0002

0.0001

0

-0.0001

-0.0002

-2

: ___.----+ I"

)(_ v v

0 2 4 6 8 10 12



Deflection

angle, deg

--qSr-- -20

----o-- -10

--O-- 0

+ 10

+20

14

0.0005

0

-0.0005

•_ _"-0.001d

-0.0015"

-0.002

-0.0025

1 ! !

L)

-0.003

-2 0 2 4 6 8 10 12


Deflection

angle, deg

--43-- -20

--o-- -10

--O-- 0

---g--- I0

.----4----20

14



112

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

-0.0025

-0.003

-2

, t , I., , , I , , , I , , ,.1 , , , I , , , I , , , 1 , , ,

0 2 4 6 8 l0 12



14

0.0005

0

-0.0005

-0.001

-0.0015

-0.002

!

-2 0 2 10 124 6 8




Deflection

angle, (leg

-----t3-- -20

---o-- -10

---O--0

+ 10

20

I I I

14

113

O.OOO5 l

0

-0.0005

-0.001

-0.0015

-0.002

-2 14

, , , I , , , I , , , f _ _ , I , _ , I , , , t , , , I , , ,

0 2 4 6 8 I0 12

Angle of attack, cz, deg


Deflection

angle, deg

+-20

---o-- -10

--O--0

+ 10

--+-20

0.0005 I

Deflection

0 angle, deg

--c}-- -20

---o-- -10

_. .:0.0005, ---x----°--I00

-0.001 ----4-- 20

-0.0015

-0.002 , , , t , , , t , , , _ , , , I ,_.J. , t , , ,

-2 0 2 4 6 8 10 12 14

Angle of attack, oc, deg



114

0.0005

0

.8=- -0.0005

rO

-0.001

-0.0015

-0.002

-2 140 2 4 6 8 10 12



Deflection

angle, deg

--0---20

---¢-- -10

----O-- 0

+ 10

+20

0.0005 _t_

Deflection

0 angle, deg

---43-- -20

--o-- -I0

.8 --0.0005- ---o- od

tO + I0

-0.00 1 _ 20

-0.0015

-0.002 ....

-2 0 2 4 6 8 10 12 14




115

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

-2

0----0---0----0

X X X ,"( X X X X X

I I I I I I I I I

, _ I _ _ _ I I i i I J , , I , , , I , , , l , , , I , , l

0 2 4 6 8 10 12

Angle of attack, oh deg


Deflection

angle, deg

--cF- -20

--o-- -10

---O--0

10

--+--20

14

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

o-----o-----o-----o O--------o o---------o-------O

X X v v v X X -----)_-------X

Deflection

angle, deg

-20

-10

--o-0

-----x-- 10

.... _-- 20

-2

I I I I I I I , I I

, , , I , , , I , , , I , , i I i i , I , , i I _ i i I i i i

0 2 4 6 8 10 12



Figure 30.- Yawing moment increment coefficient for rudder as afunction of angle of attack, deflection angle, and Mach number.

14

116

dr,)

0.08

0.06

0.04

0.02

0

-0.02

-0.04

-0.06

-0.08

_0------0

Deflection

angle, deg

--U-- -20

+ -10

--O--0

+ 10

--+--20

v,., X X ^_ X X X X ..v

PII1[111

-2 0

t I I I I I

I , , , I , J , I , , _ I , , , I j , , I , , ,

2 4 6 8 10 12



14

t..)

0.08

0.06

0.04

0,02

0

-0.02

-0.04

-0.06

-0.08-2

O---------O O O

X X X X -_v × X' X v..

Deflection

angle, deg

-20

---o-- -I0

---O-0

+ 10

---+-- 20

+----F- t t ..... t I I I

,,, I, ,, i,,, l, ,, I l t, I, , , I , i i l , , ,

0 2 4 6 8 10 12 14




ll7

0.02

0.015

0.01

•_ 0.005=-

0

-0.005

-0.01

-0.015

-0.02

-2 0 2 4 6 8 10 12



__1__Deflection

angle, deg

---c3-- -20

-----o-- -I0

--O-0

---x-- I0

------+---20

I I !

14

0.02

0.015

0.01

0.005

,0

-0.005

-0.01

-0.015

-0.02

X X X X X X X X

Deflection

angle, deg

--c_ -20

---o-- -10

--O-- 0

--x-- 10

---+--20

-2

, , , I , , , I , , _ I , , _ I , , , I , , , I , , , I , , ,

0 2 4 6 8 10 12




14

118

ro

0.02

0.015

0.01

0.O05

0

-0.005

-0.01

-0.015

-0.02

X X X _,v X X X X X

-2 0 2 4 6 8 10 12



__1_

Deflection

angle, deg

-----0---20

---o-- -10

--O-0

+ 10

+20

14

C

0.02

0.Ol

,0

-0.01

-0.02

O'---O---O---_. _ O

Deflection

angle, deg

--c3-- -20

---o-- -10

---O-- 0

---x--- I0

--+-20

, _ , I , i I I , t , I j j I I , _ , I , , , I I , , I i , t

-2 0 2 4 6 8 10 12


(h) Mach number = I0.0

14


119

0.O06 1

dL)

0.004

0.002

0

-0.002

-0.004

-0.006

-[3

i_O------ 0 ------xD

X----)_- X X X -----_( X X

-2

_ , , I k , , I , , i I , , , I , , , I , , , I , , , I _ n n

0 2 4 6 8 10 12



Deflection

angle, deg

---43- -20

---o-- -10

--O--0

10

--+--20

14

4

0.006

0.004

0.002

0

-0.002

-0.004

-0.006

-2

X X X X X X X X X

Deflection

angle, deg

--o--20

--o-- -10

--<y- 0

----_--I0

---+-- 20

.._.---4-

..------4-----.......4-------

, _ , I , , _ I , , J I , , , I , , , I , , _t__l____ J I J , j

0 2 4 6 8 10 12


(3) Mach number = 20.0

14


120

0.O06 l

U

0.004

0.002

0

-0.002

-0.004

-0.006

_'-"-£3

_o------O

X X X X X )( X X X

-2 0 2 4 6 8 10 12




Deflection

angle, deg

--o-- -20

---o-- -10

--O--0

+ 10

+20

14

0.25

t£

0.2

0.15

0.1

(}.05

0-2

, , , I , ,., I , , i I , , , I , , , l , , , I , , , I ,

0 2 4 6 8 10 12


Figure 31.- Yawing moment with roll rate dynamic derivative as afimction of angle of attack and Mach number.

M

---O-- 0.3

--{3-- 0.7

+ 0.9

1.5

2.5

4.0

..... 6.0

..... 10.0

-- - - 15.0

...... 20.0

..... 24.2i t I

14

12l

t_t_

u

-0.2

-0.4

-0.6

-0.8

-1

-1.2

-1.4

-1.6

..................... . ° . . . ........

, , _ i , , _ i , , , I J.i J f_ , _ , I , J _ I , _ , i i , J J

-2 0 2 4 6 8 l0 12 14

Angle of attack, ct, degrees

Figure 32.- Yawing moment with yaw rate dynamic derivative as afunction of angle of attack and Mach number.

M

---o-- 0.3

----or- 0.7

--o-- 0.9

1.5

2.5

4.0

..... 6.0

..... 10.0

-- - - 15.0

..... 20.0

..... 24.2

da

.9

O

_3

O

107

10 6

105 ' ' ' 9

1.0 105 3.0 105

g_-r_---o---_

---O- lxx I -

IyyIr= I _

.l l i i ., _ l i J i , I , , J ,

1.5 10 5 2.0 10 5 2.5 10 5

15

14

13

12

11

10

W, lb

Figure 33.- Moments of inertia and center of gravity (relative to aerodynamicmoment reference center) as a function of vehicle weight.

122

2500O

2OOOO

1500O

l(XXX)

5000

0

-5000

-2 0 2 4 6 8 10

Equivalence ratio, ¢


Dynamicpressure, psf

0

-----o-- 150

--O-- 1000

+ 2000

---+--5000

12

250O0

200OO

1500O

O 10000

5000

0

-50O0-2

1


--or- 0

---o-- 150

--O-- 1000

+2000

--+-- 5000

, , , I , , , I , , , I , , t I , , , I , , , I i , J

0 2 4 6 8 I0

Fuel equivalence ratio, ¢


Figure 34.- Thrust coefficient as a function of fuel equivalence ratio,dynamic pressure, and Mach number.

12

123

lOOOO i

8OO0

6OOO

4000

2OOO

0

-2OOO


--O-- 0

--o-- 150

--O- I000

--x-- 2000

----+--5000

-2

i , , I , _ , I , , , I , _ t I t , t I _ , , I t , _

0 2 4 6 8 I0

Fuel equivalence ratio, t_


12

10000

8OOO

60OO

0 4o0o

2OOO

0

-2OOO


0

---o-- 150

--O- 1000

_2000

---+--5000

-2

, , J I _ J t I , , , I t , _ I , , , 1 , J , I l I ,

0 2 4 6 8 I0

Fuel equivalence ratio, t_


12


124

50OO

400O

3O00

20oo

1000

0

-1000

__1


--Or- 0

--o-- 150

--o- 1000

+2000

----+--5000

-2 0 2 4 6 8 10



12

5OOO

4O00

• 3000

2o0o

1000

0

-1000

__1


--or-- 0

---o-- 150

----O-- 1000

+2000

---+-- 5000

-2

J , , I , , , f , , , I , , , I ,_. , , I _ t , I , , ,

0 2 4 6 8 10



12


125

3500

3000

2500

2OOO

_-. 1500

1000

500

0

-500

Dynamic

pressure, psf

0

---o-- 150

--O- 1000

+2000

---+--5000

t _ _ I J J , f _ _ , I , , , I , _ , i , , , I , , t

-2 0 2 4 6 8 10 12

Fuel equivalence ratio, qb


3500

3000

2500

2000

f-- 1500

1000

500

0

-500-2 0 2 4 6 8 10

Fuel equivalence ratio,



--cy- 0

----o-- 150

---O-- 1000

----x-- 2000

------4-- 500O

12


126

3500

3000

2500

2000

r.J" 1500

1000

500

0

-500-2

, , , I , , , I , , , I , , , I , , , I , , , I , , ,

0 2 4 6 8 10




-----or- 0

---o-- 150

---<>- 1000

---x--- 2000

-----+--5000

12

3500

3000

2500

2000

_ 1500

10013

500

0

-500

-2


---or-- 0

--¢-- 150

----<>- I000

2000

----4-- 5000

, , , I , , J I , , , I J , , I , , , I , , , [ I i i

0 2 4 6 8 10


12



_,127

3500

3000

2500

2OOO

_ 1500

1000

500

0

-500 , , , I ' , i I j i i I,, , , I , , , I , , , I , , ,

-2 0 2 4 6 8 lO




--o-- 0

---o-- 150

+996

----x-- 1968

---+-- 50OO

12

35OO

30O0

2500

_:: 2000

t._ 1500

1000

5OO

0

-500

-2

1


--o-- 0

---o-- 150

+996

---x-- 1968

+5000

0 2 4 6 8 lO


(1) Mach number = 4.0

12


128

35OO

3OOO

250O

2OOO

_-. 1500

1000

500

0

-500


---t:3-- 0

---o-- 150

---O-- 928

---x-- 1866

---¢-- 5000

-2 0 2 4 6 8 10 12


(m) Mach number = 6.0

350

300

250

._ 200

150

100

50

0

-50

-2 0 2 4 6 8 10


(n) Mach number = 8.0


---0-- 0

---<>-- 150

---O-- 1043

----x-- 2003

--+-- 50OO

12


129

350

300

250

200

150

100

50

0

-50

1

Dynamic

pressure, psf

--or- 0

---o-- 150

--<7- 1164

-_ 1992

+ 50OO

-2 0 2 4 6 8 10 12


(o) Mach number = 10.0

f-,

350

300

250

200

150

100

50

0

-50-2


0

---o-- 150

--O- 1101

+ 2049

---¢-- 5000

, , , I , , , 1 , , , I , , J I , , , I. , , , } , , ,

0 2 4 6 8 10


12

(p) Mach number= 15.0


130

350

300 Dynamic

250 I _ pressure,__o_oPSfe]_ 200 ---o-- 590

r_ 150 ---O-- 9672001

100 5000

50

0

-50

-2 0 2 4 6 8 I0 12

Fuel equivalence ratio, 0

(q) Mach number = 20.0

350

30O _ IDynamic

250 _ pressure, psf--43-- 0

--_ 200 -o- 150

615150

1008

100 5000

50

0

-50

-2 0 2 4 6 8 10 12


(r) Mach number = 25.0


131

4OOO

3500

3OOO

u 2500Oe,t}

2OOO

1500

I000

5OO

0

-20

I I I I

0


0

150

---0-- 1000

+2000

---+--5000

I , , , I , . . I , , , I , , , .z. , , ,

20 40 60 80 100

Fuel equivalenceratio,¢


120

4000

3500

3000

u 2500O

1500

1000

500

0-20

II

, , , I , "P' , I , , , I , , . I _ , i I i , . _" , , i

0 20 40 60 80 100



___1


+ 0

+ 150

---O-- 1000

+2000

----+-- 5000

120

Figure 35.- Engine specific impulse as a function of fuel equivalenceratio, dynamic pressure, and Mach number.

132

4OOO

3500

3OOO

u 2500

2OOO

1500

1000

5OO

0

-2O

i

0 20 40 60 80 100

Fuel equivalenceratio,



--t:3-- 0

--<>--- 150

--O-1000

---x--- 2000

----+--5000

120

4OOO

3500

3000

u 2500¢o

20o0

1500

1000

500

0

-2O

k

0 20 80 1O0


--or- 0

---o-- 150

--O-- 1000

+2000

--+--- 5_

40 60

Fuel equivalence ratio, 0


120


133

4000

3500

3000

2500

k_ 200O

1500

1000

500

0

k

-20 0 20 40 60 80 100

Fuel equivalence ratio, _b



---43- 0

----o-- 150

--o-- 1000

----x-- 2000

--+--5000

120

4000

3500

3000

2500

20o0

1500

I000

500

0

-20

, , J I , , , I _ _ , I _ , , i , J J I , J'7_'r-..sla , [ ,

0 20 40 60 80 100



---or-- 0

---<>-- 150

---O-- I000

--x-- 2000

---+-- 5000

120



134

40O0

35OO

3000

2500

_, 2000

1500

IO00

5OO

0


---o-- 0

---o-- 150

---O-- 1000

-----x-- 2000

---+-- 5000

-20 0 20 40 60 80 100



120

4000

3500

3000

"o 2500

_, 200O

1500

1000

500

0 I I I II

-20 0


----c3-- 0

---o-- 150

---0-- 1000

+2O0O

--¢--- 5000

: i : ! : ; : ! : ; .;.I , , , J .... _ I I I

20 40 60 80 I00



120


135

4000

3500

3000

2500

;_ 200o

1500

1000

500

0

-20


---CY- 0

---o-- 150

--O-- 1000

_- 2000

----+--5000

w

0 20 40 60 80 1O0 120



4000

3500

3000

"o 2500"O

Se 20oo1500

lO00

5OO

0-2 0 2 4 6 8 10 12




136

4ooo ", 73500 Dynamic

3000 pressure, psf---O-- 0

25oo _ 15o"_ 2000 996

1968

1500 5000

1000

500

0

-2 0 2 4 6 8 10 12



4000

3500 Dynamic

3000 pressure, psf

----cy- 0

2_. 2000 9281866

1500 5000

1000

500

0

-2 0 2 4 6 8 10 12


(!) Mach number--- 6.0


137

OO

4OOO

350O

3OOO

25OO

2OOO

1500

tOO0

5OO

0 I

-2 0 2 4 6 8 10


(m) Mach number = 8.0


--O-- 0

--o-- 150

--(>- 1043

+ 2003

+5000

I I I

12

0"0

4OOO

3500

3OOO

2500

2o0o

1500

I000

5OO

0

-2

I


---q3-- 0

---o-- 150

----<>- 1164

---x--- 1992

---+--5000

I I I I I I /

0 2 4 6 8 lO 12


(n) Mach number = 10.0


138

1400

1200

1000

8OO

6OO

4OO

2OO

0-2

_ , , I _ , , I , l _ 1 1 1 1 1 i i _ I , , l I l t J

0 2 4 6 8 10


---qzy-- 0

---o-- 150

----O-- 1101

2049

------+--5000

12

Fuel equivalence ratio, ¢p

(o) Mach number = 15.0

¢..)

c,o

1.--.4

1400

1200

1000

-800

600

400

200

0

-2

l i J I , J , I , , , I j I I I I I I I J ' I I I J J

0 2 4 6 8 10

Fuel equivalence ratio, _p

Dynamic

pressure, psf

---t3-- 0

---o-- 590

--O-967

2001

5000

12

(p) Mach number = 20.0


139

fd

1400

1200

I000

8OO

6OO

4OO

2O0

0

-2

, , , I , , , I , _ , I , t _ I _ J , I _ _ , I , , ,

0 2 4 6 8 lO


(q) Mach number = 25.0


---U-- 0

---o-- 150

--O-- 615

1008

5000

12


140

Report Documentation Page_,at_l t*_r0t_O)CS ant3

1. Report No.

NASA TM-t026]0

2. Government Accession No.

4. Title and Subtitle

Hypersonic Vehicle Simulation Model:Winged-Cone Configuration

7. Author(s)

John D. Shaughnessy, S. Zane Pinckney, John D. McMinn,Christopher I. Cruz, and Marie-Louise Kelley

9. Performing Organization Name and Address

NASA Langley Research CenterHampton, VA 23665-5225

12. Sponsoring Agency Name and Address

National Aeronautics and Space AdministrationWashington, DC 20546-0001

3. Recipient's Catalog No.

5. Report Date

November 1990

6. Performing Organization Code

8. Performing Organization Report No.

10. Work Unit No.

763-01-51-0511. Contract or Grant No.

13. Type of Report and Period Covered

Technical Memorandum14. Sponsoring/_gency Code

15. Supplementary Notes

16. Abstract

Aerodynamic, propulsion, and mass models for a generic, horizontal-takeoff, single-stage- to-orbit(SSTO) configuration are presented which are suitable for use in point mass as well as batch andreal-time six degree-of-freedom simulations. The sinmladons can be used to investigate ascentperformance issues and to allow research, refinement, and evaluation of integrated guidance/flight/propulsion/thermal control systems, design concepts, and methodologies for SSTOmissions. Aerodynamic force and moment coefficients are given as functions of angle of attack,Mach number, and control surface deflections. The model data were estimated by using asubsonic/supersonic panel code and a hypersonic local surface inclination code. Thrust coefficientand engine specific impulse were estimated using a two-dimensional forebody, inlet, nozzle codeand a one-dimensional combustor code and are given as functions of Mach number, dynamicpressure, and fuel equivalence ratio. Rigid-body mass moments of inertia and center of gravitylocation are functions of vehicle weight which is in turn a function of fuel flow.

17."Key Words (Suggested by Author(s))

Single Stage-To-OrbitNational Aero-Space PlaneHypersonicSimulationLaunch Vehiclesk/I._1_.1

18. Distribution StatementUnclassified - Unlimited

!Subject Category - 08

19"."et_"e'_u'rityClassif. (of this report)

Unclassified

20. Security Classff. (of this page]

Unclassified

21. No. of pages

141

22. Price

A07

NASA FORM 1626 OCT 86

Documents

NASA Technical Memorandum 102610 Hypersonic Vehicle ... · P NASA Technical Memorandum 102610 Hypersonic Vehicle Simulation Model: Winged-Cone Configuration John D. Shaughnessy, S