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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
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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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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
Angle of attack, or, deg
(i) Mach number = 15.0
, I
12
l)eflection
angle, deg
---C]-- -20
--o-- -10
---o- 0
-_ 10
.... t-- 20
i1.........14
Figure 3.- Continued.
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
Angle of attack, or, deg
(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
Angle of attack, or, deg
(a) Mach number = 0.3
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
Angle of attack, o_, deg
(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
(d) Mach number = 1.5
Figure 4.- Continued.
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
Angle of attack, a, deg
(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
Angle of attack, a, deg
(f) Mach number = 4.0
Figure 4.- Continued.
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
Angle of attack, ct, deg
(h) Mach number = 10.0
Figure 4.- Continued.
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
(i) Mach number = 15.0
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
Angle of attack, a, deg
(j) Mach number = 20.0
Figure 4.- Continued.
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
(k) Mach number = 24.2
Figure 4.- Concluded.
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
(a) Mach number = 0.3
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
Angle of attack, or, deg
(b) Mach number = 0.7
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
Angle of attack, or, deg
(c) Mach number = 0.9
Figure 5.- Continued.
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
Angle of attack, or, deg
(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
Angle of attack, ct, deg
14
(e) Mach number = 2.5
Figure 5.- Continued.
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
Angle of attack, ix, deg
Deflection
angle, deg
---Or-- -20
--o-- -10
--O- 0
+ 10
--4-- 20
(f) Mach number = 4.0
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
Angle of attack, or, deg
14
(g) Mach number =6.0
Figure 5.- Continued.
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
Angle of attack, or, deg
(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
Angle of attack, a, deg
(i) Mach number - 15.0
Figure 5.- Continued.
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
Angle of attack, or, deg
(j) Mach number = 20.0
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
Angle of attack, ct, deg
14
(k) Mach number = 24.2
Figure 5.- Concluded.
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
Angle of attack, or, deg
(a) Mach number = 0.3
]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
Angle of attack, or, deg
(b) Mach number = 0.7
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
Angle of attack, or, deg
(c) Mach number = 0.9
Figure 6.- Concluded.
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
Deflectionangle, deg
-43---20
--o-- -10
----<>- 0
---_--- 10
-2 0 2 4 6 8 10 12 14
Angle of attack, (x, deg
(a) Mach number = 0.3
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
(b) Mach number = 0.7
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
(c) Mach number = 0.9
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
Angle of attack, or, deg
(d) Mach number = 1.5
14
Figure 8.- Continued.
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
Angle of attack, or, deg
(e) Mach number = 2.5
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
Angle of attack, a, deg
(f) Mach number = 4.0
14
Figure 8.- Continued.
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
Figure 8.- Continued.
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
Angle of attack, or, (leg
(i) Mach number = 15.0
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
(j) Mach number = 20.0
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
Angle of attack, 0t, deg
(k) Mach number -- 24.2
Figure 8.- Concluded.
-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
Angle of attack, 0t, deg
(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
Angle of attack, 0t, deg
(b) Mach number = 0.7
.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
Angle of attack, 0t, deg
(el Mach number = 0.9
Figure 9.- Continued.
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
Angle of attack, or, deg
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
(e) Mach number = 2.5
Figure 9.- Continued.
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
Angle of attack, or, deg
(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
Angle of attack, or, (leg
(g) Mach number = 6.0
Figure 9.- Continued.
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
Angle of attack, 0t, deg
(h) Mach number = 10.0
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
Angle of attack, or, deg
(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
Figure 9.- Concluded.
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
Angle of attack, or, deg
(a) Mach number = 0.3
,-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
(b) Mach number = 0.7
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
Angle of attack, ix, deg
(c) Mach number =0.9
Figure 10.- Concluded.
-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
(a) Mach number = 0.3
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
Angle of attack, or, deg
(c) Mach number = 0.9
>: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
Angle of attack, ix, deg
(d) Mach number = 1.5
Figure 12.- Continued.
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
Angle of attack, ct, deg
(e) Mach number = 2.5
¢..)
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
Angle of attack, 0t, deg
(g) Mach number = 6.0
>: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
Angle of attack, or, deg
(h) Mach number = 10.0
Deflection
angle, deg
----_ - -20
----o-- -10
---o-- 0
--_ .... 10
---t- 20
Figure 12.- Continued.
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
(i) Mach number = 15.0
>
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
(k) Mach number = 24.2
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
(c) Mach number = 0.9
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
Angle of attack, or, deg
(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
Angle of attack, or, deg
(f) Mach number = 4.0
--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
Figure 13.- Continued.
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
(h) Mach number = 10.0
-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
Angle of attack, ct, deg
(i) Mach number= 15.0
Figure 13.- Continued.
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
Angle of attack, or, deg
(j) Mach number = 20.0
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 13.- Concluded.
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
Angle of attack, or, deg
(a) Mach number = 0.3
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
Angle of attack, _, deg
(b) Mach number = 0.7
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
Angle of attack, ct, deg
(c) Mach number = 0.9
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
Angle of attack, or, deg
(d) Mach number = 1.5
Figure 14.- Continued.
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
(e) Mach number = 2.5
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
(f) Mach number = 4.0
14
Figure 14.- Continued.
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
Angle of attack, a, deg
(g) Mach number = 6.0
__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
Angle of attack, o_, deg
14
(h) Mach number = 10.0
Figure 14.- Continued.
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
Angle of attack, or, deg
(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
(j) Mach number = 20.0
14
Figure 14.- Continued.
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
Angle of attack, o_, deg
(k) Mach number = 24.2
Figure 14.- Concluded.
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
Angle of attack, or, degrees
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
(a) Mach number = 0.3
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
Angle of attack, or, deg
(b) Mach number = 0.7
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
Angle of attack, o_, deg
(c) Mach number = 0.9
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
Angle of attack, o_, deg
(d) Mach number = 1.5
Figure 16: Continued.
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
Angle of attack, _, deg
(e) Mach number = 2.5
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
Angle of attack, or, deg
(f) Macb number = 4.0
14
Figure 16.- Continued.
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
Angle of attack, ct, deg
(g) Mach number = 6.0
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
(h) Mach number = 10.0
Figure 16.- Continued.
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
Angle of attack, ct, deg
(i) Mach number = 15.0
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
Angle of attack, ix, deg
Deflection
angle, deg
--cI-- -20
-----<>-- -10
--O-- 0
---x-- I0
20
14
(j) Mach number = 20.0
Figure 16.- Continued.
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 16.- Concluded.
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
Angle of attack, o_, deg
(a) Mach number = 0.3
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
Angle of attack, ct, deg
(b) Mach number = 0.7
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
(c) Mach number = 0.9
Figure 17.- Continued.
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
(d) Mach number = 1.5
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
Angle of attack, tz, deg
(e) Macb number = 2.5
Figure 17.- Continued.
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
Angle of attack, or, deg
(f) Mach number = 4.0
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
Angle of attack, _, deg
(g) Mach number = 6.0
14
Figure 17: Continued.
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
(h) Mach number = 10.0
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
Angle of attack, or, deg
Deflection
angle, deg
---q3- -20
----o-- -10
---O- 0
+ 10
--+-20
14
(i) Mach number = 15.0
Figure 17.- Continued.
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
Angle of attack, or, deg
(j) Mach number = 20.0
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
Angle of attack, _, deg
14
(k) Mach number = 24.2
Figure 17.- Concluded.
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
Angle of attack, or, deg
(a) Mach number = 0.3
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
Angle of attack, a, deg
(b) Mach number = 0.7
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
Angle of attack, ct, deg
(c) Mach number = 0.9
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
Angle of attack, o_, deg
(d) Mach number = 1.5
Figure 18.- Continued.
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
Angle of attack, o_, deg
(e) Mach number = 2.5
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
Angle of attack, tz, deg
(0 Mach number = 4.0
Figure 18: Continued.
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
Angle of attack, or, deg
(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
Angle of attack, or, deg
(h) Mach number = 10.0
Figure 18.- Continued.
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
Angle of attack, o_, deg
(i) Mach number= 15.0
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
Angle of attack, or, deg
(j) Mach number = 20.0
Figure 18: Continued.
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 18.- Concluded.
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
(a) Mach number = 0.3
,,,
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
(b) Mach number = 0.7
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
Angle of attack, or, deg
(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
Angle of attack, ct, deg
(d) Mach number = 1.5
Figure 22.- Continued.
__3__
Deflectionangle, deg
+-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
Angle of attack, _, deg
(e) Mach number = 2.5
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
Angle of attack, o_, deg
(f) Mach number = 4.0
Figure 22.- Continued.
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
Angle of attack, _, deg
Deflection
angle, deg
---4z_ -20
--o--- -10
----O- 0
---_-- 10
+20
(g) Mach number = 6.0
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
(h) Mach number = 10.0
Figure 22.- Continued.
89
0.006
0.004
0.002
0
d -0.002
-0.004
-0.006
-0.008
-0.01
-2
Angle of attack, or, deg
(i) Mach number = 15.0
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
Angle of attack, or, deg
Deflection
angle, deg
--Z:F- -20
---O--- -10
--o-0
-_ 10
+20
(j) Mach number = 20.0
Figure 22.- Continued.
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 22.- Concluded.
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
Angle of attack, tt, deg
(a) Mach number = 0.3
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
Angle of attack, _, deg
(b) Mach number = 0.7
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
Angle of attack, or, deg
(c) Mach number = 0.9
Deflection
angle, deg
--EF- -20
---O-- -10
_0
---_-- 10
---_-- 20
14
Figure 23.- Continued.
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
Angle of attack, _, deg
(d) Mach number = 1.5
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
(e) Mach number = 2.5
14
Figure 23.- Continued.
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_-_____>.______
Deflectionangle, deg
---O-- -20
--O-- -10
---O-- 0
+ 10
+20
-2 0 2 4 6 8 I0 12
Angle of attack, 0t, deg
(f) Mach number = 4.0
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
Angle of attack, 0t, deg
14
(g) Mach number = 6.0
Figure 23: Continued.
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
(h) Mach number = 10.0
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
Angle of attack, or, deg
Deflection
angle, deg
---£_ -20
---o-- -10
--O--0
---g-- 10
.... ;--20
14
(i) Mach number = 15.0
Figure 23.- Continued.
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
Angle of attack, or, deg
(j) Mach number = 20.0
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.
Figure 23.- Concluded.
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
Angle of attack, (x, deg
(a) Mach number = 0.3
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
Angle of attack, (x, deg
(b) Mach number = 0.7
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
(c) Mach number = 0.9
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
Angle of attack, _, deg
(d) Math number = 1.5
Figure 24.- Continued.
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
Angle of attack, or, deg
(e) Mach number = 2.5
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
Angle of attack, o_, deg
(f) Mach number = 4.0
Figure 24.- Continued.
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
(g) Mach number = 6.0
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
Angle of attack, a, deg
(h) Mach number = lif0
14
Figure 24.- Continued.
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
Angle of attack, _, deg
(i) Mach number = 15.0
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
Angle of attack, _, deg
(j) Mach number = 20.0
14
Figure 24.- Continued.
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
(k) Mach number = 24.2
14
Figure 24.- Concluded.
.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
Angle of attack, _, deg
(a) Mach number = 0.3
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
Angle of attack, or, deg
(b) Mach number = 0.7
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
Angle of attack, tz, deg
(c) Mach number = 0.9
14
Figure 25.- Concluded.
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
Angle of attack, _, deg
(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
Angle of attack, or, deg
(c) Mach number = 0.9
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
(d) Mach number = 1.5
Figure 28.- Continued.
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
Angle of attack, or, deg
(e) Mach number = 2.5
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
Angle of attack, or, deg
(f) Mach number = 4.0
Figure 28.- Continued.
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
Angle of attack, ct, deg
(g) Mach number = 6.0
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
Angle of attack, tt, deg
(h) Mach number = 10.0
Figure 28.- Continued.
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
Angle of attack, a, deg
(i) Mach number= 15.0
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
Angle of attack, a, deg
(j) Mach number = 20.0
Figure 28.- Continued.
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
Angle of attack, o_, deg
(k) Mach number = 24.2
Figure 28.- Concluded.
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
Angle of attack, or, deg
(a) Mach number = 0.3
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
Angle of attack, ct, deg
(b) Mach number = 0.7
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
Angle of attack, ct, deg
(c) Mach number = 0.9
Figure 29.- Continued.
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
Angle of attack, or, deg
(d) Mach number = 1.5
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
Angle of attack, o_, deg
Deflection
angle, deg
--43-- -20
--o-- -10
--O-- 0
---g--- I0
.----4----20
14
(e) Mach number = 2.5
Figure 29.- Continued.
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
Angle of attack, or, deg
(f) Mach number = 4.0
14
0.0005
0
-0.0005
-0.001
-0.0015
-0.002
!
-2 0 2 10 124 6 8
Angle of attack, or, deg
(g) Mach number = 6.0
Figure 29.- Continued.
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
(h) Mach number = 10.0
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
(i) Mach number = 15.0
Figure 29.- Continued.
114
0.0005
0
.8=- -0.0005
rO
-0.001
-0.0015
-0.002
-2 140 2 4 6 8 10 12
Angle of attack, o_, deg
(j) Mach number = 20.0
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 29.- Concluded.
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
(a) Mach number = 0.3
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
Angle of attack, o_, deg
(b) Mach number = 0.7
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
Angle of attack, or, deg
(c) Mach number = 0.9
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
Angle of attack, _, deg
(d) Mach number = 1.5
Figure 30.- Continued.
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
Angle of attack, _, deg
(e) Mach number = 2.5
__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
Angle of attack, or, deg
(f) Mach number = 4.0
Figure 30.- Continued.
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
Angle of attack, or, deg
(g) Mach number = 6.0
__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
Angle of attack, or, deg
(h) Mach number = I0.0
14
Figure 30.- Continued.
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
Angle of attack, or, deg
(i) Mach number= 15.0
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
Angle of attack, o_, deg
(3) Mach number = 20.0
14
Figure 30.- Continued.
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
Angle of attack, or, deg
(k) Mach number = 24.2
Figure 30.- Concluded.
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
Angle of attack, or, degrees
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, ¢
(a) Mach number = 0.0
Dynamicpressure, psf
0
-----o-- 150
--O-- 1000
+ 2000
---+--5000
12
250O0
200OO
1500O
O 10000
5000
0
-50O0-2
1
Dynamicpressure, psf
--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, ¢
(b) Mach number = 0.3
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
Dynamicpressure, psf
--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_
(c) Mach number = 0.5
12
10000
8OOO
60OO
0 4o0o
2OOO
0
-2OOO
Dynamicpressure, psf
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_
(d) Mach number = 0.7
12
Figure 34.- Continued.
124
50OO
400O
3O00
20oo
1000
0
-1000
__1
Dynamicpressure, psf
--Or- 0
--o-- 150
--o- 1000
+2000
----+--5000
-2 0 2 4 6 8 10
Fuel equivalence ratio, ¢
(e) Mach number = 0.9
12
5OOO
4O00
• 3000
2o0o
1000
0
-1000
__1
Dynamicpressure, psf
--or-- 0
---o-- 150
----O-- 1000
+2000
---+-- 5000
-2
J , , I , , , f , , , I , , , I ,_. , , I _ t , I , , ,
0 2 4 6 8 10
Fuel equivalence ratio, ¢
(f) Mach number = 0.95
12
Figure 34.- Continued.
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
(g) Mach number = 1.0
3500
3000
2500
2000
f-- 1500
1000
500
0
-500-2 0 2 4 6 8 10
Fuel equivalence ratio,
(h) Mach number = 1.5
Dynamicpressure, psf
--cy- 0
----o-- 150
---O-- 1000
----x-- 2000
------4-- 500O
12
Figure 34.- Continued.
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
Fuel equivalence ratio,
(i) Mach number = 2.0
Dynamicpressure, psf
-----or- 0
---o-- 150
---<>- 1000
---x--- 2000
-----+--5000
12
3500
3000
2500
2000
_ 1500
10013
500
0
-500
-2
Dynamicpressure, psf
---or-- 0
--¢-- 150
----<>- I000
2000
----4-- 5000
, , , I , , J I , , , I J , , I , , , I , , , [ I i i
0 2 4 6 8 10
Fuel equivalence ratio, ¢
12
(j) Mach number = 3.0
Figure 34.- Continued.
_,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
Fuel equivalence ratio, ¢
(k) Mach number = 3.5
Dynamicpressure, psf
--o-- 0
---o-- 150
+996
----x-- 1968
---+-- 50OO
12
35OO
30O0
2500
_:: 2000
t._ 1500
1000
5OO
0
-500
-2
1
Dynamicpressure, psf
--o-- 0
---o-- 150
+996
---x-- 1968
+5000
0 2 4 6 8 lO
Fuel equivalence ratio,
(1) Mach number = 4.0
12
Figure 34.- Continued.
128
35OO
3OOO
250O
2OOO
_-. 1500
1000
500
0
-500
Dynamicpressure, psf
---t:3-- 0
---o-- 150
---O-- 928
---x-- 1866
---¢-- 5000
-2 0 2 4 6 8 10 12
Fuel equivalence ratio, ¢
(m) Mach number = 6.0
350
300
250
._ 200
150
100
50
0
-50
-2 0 2 4 6 8 10
Fuel equivalence ratio, ¢
(n) Mach number = 8.0
Dynamicpressure, psf
---0-- 0
---<>-- 150
---O-- 1043
----x-- 2003
--+-- 50OO
12
Figure 34.- Continued.
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
Fuel equivalence ratio, ¢
(o) Mach number = 10.0
f-,
350
300
250
200
150
100
50
0
-50-2
Dynamicpressure, psf
0
---o-- 150
--O- 1101
+ 2049
---¢-- 5000
, , , I , , , 1 , , , I , , J I , , , I. , , , } , , ,
0 2 4 6 8 10
Fuel equivalence ratio,
12
(p) Mach number= 15.0
Figure 34.- Continued.
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
Fuel equivalence ratio,
(r) Mach number = 25.0
Figure 34.- Concluded.
131
4OOO
3500
3OOO
u 2500Oe,t}
2OOO
1500
I000
5OO
0
-20
I I I I
0
Dynamicpressure, psf
0
150
---0-- 1000
+2000
---+--5000
I , , , I , . . I , , , I , , , .z. , , ,
20 40 60 80 100
Fuel equivalenceratio,¢
(a) Mach number = 0.0
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
Fuel equivalence ratio,
(b) Mach number = 0.3
___1
Dynamicpressure, psf
+ 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,
(c) Mach number = 0.5
Dynamicpressure, psf
--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
Dynamicpressure, psf
--or- 0
---o-- 150
--O-- 1000
+2000
--+--- 5_
40 60
Fuel equivalence ratio, 0
(d) Mach number = 0.7
120
Figure 35.- Continued.
133
4000
3500
3000
2500
k_ 200O
1500
1000
500
0
k
-20 0 20 40 60 80 100
Fuel equivalence ratio, _b
(e) Mach number = 0.9
Dynamicpressure, psf
---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
Fuel equivalence ratio,
Dynamicpressure, psf
---or-- 0
---<>-- 150
---O-- I000
--x-- 2000
---+-- 5000
120
(f) Mach number = 1.0
Figure 35.- Continued.
134
40O0
35OO
3000
2500
_, 2000
1500
IO00
5OO
0
Dynamicpressure, psf
---o-- 0
---o-- 150
---O-- 1000
-----x-- 2000
---+-- 5000
-20 0 20 40 60 80 100
Fuel equivalence ratio,
(g) Mach number = 1.5
120
4000
3500
3000
"o 2500
_, 200O
1500
1000
500
0 I I I II
-20 0
Dynamicpressure, psf
----c3-- 0
---o-- 150
---0-- 1000
+2O0O
--¢--- 5000
: i : ! : ; : ! : ; .;.I , , , J .... _ I I I
20 40 60 80 I00
Fuel equivalence ratio,
(h) Mach number = 2.0
120
Figure 35.- Continued.
135
4000
3500
3000
2500
;_ 200o
1500
1000
500
0
-20
Dynamicpressure, psf
---CY- 0
---o-- 150
--O-- 1000
_- 2000
----+--5000
w
0 20 40 60 80 1O0 120
Fuel equivalence ratio,
(i) Mach number = 3.0
4000
3500
3000
"o 2500"O
Se 20oo1500
lO00
5OO
0-2 0 2 4 6 8 10 12
Fuel equivalence ratio,
(j) Mach number = 3.5
Figure 35.- Continued.
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
Fuel equivalence ratio,
(k) Mach number = 4.0
4000
3500 Dynamic
3000 pressure, psf
----cy- 0
2_. 2000 9281866
1500 5000
1000
500
0
-2 0 2 4 6 8 10 12
Fuel equivalence ratio,
(!) Mach number--- 6.0
Figure 35.- Continued.
137
OO
4OOO
350O
3OOO
25OO
2OOO
1500
tOO0
5OO
0 I
-2 0 2 4 6 8 10
Fuel equivalence ratio,
(m) Mach number = 8.0
Dynamicpressure, psf
--O-- 0
--o-- 150
--(>- 1043
+ 2003
+5000
I I I
12
0"0
4OOO
3500
3OOO
2500
2o0o
1500
I000
5OO
0
-2
I
Dynamicpressure, psf
---q3-- 0
---o-- 150
----<>- 1164
---x--- 1992
---+--5000
I I I I I I /
0 2 4 6 8 lO 12
Fuel equivalence ratio,
(n) Mach number = 10.0
Figure 35.- Continued.
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
Dynamicpressure, psf
---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
Figure 35.- Continued.
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
Fuel equivalence ratio,
(q) Mach number = 25.0
Dynamicpressure, psf
---U-- 0
---o-- 150
--O-- 615
1008
5000
12
Figure 35.- Concluded.
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