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    NASA NACA-RM-A53H17-1953 Comparison of theoretical and experimental zero-lift drag-rise characteristics of wing-body-tail combinations near the speed of sound《在声速附近 翼身尾翼组合理论和实验零升力阻.pdf

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    NASA NACA-RM-A53H17-1953 Comparison of theoretical and experimental zero-lift drag-rise characteristics of wing-body-tail combinations near the speed of sound《在声速附近 翼身尾翼组合理论和实验零升力阻.pdf

    1、SECURITY INFORMATION.;RESEARCH MEMORANDUM -COMPARISON OF THEORETICAL AND EXPERIMENTAL ZERO-LIFTDRAG-RISE CHARACTERISTICS OF WING- BODY-TAILCOMBINATIONS NEAR THE SPEED OF SOUNDBy George H. HoldawayAmes Aeronautical.LaboratoryMoffett Field, Ca.lif.i.e., psrsllel to the yz plane.)angle between the y ax

    2、is and the intersection of the cuttingplanes X with the xy plane, arc tan (ml Cos e)first derivative of the projected cross-sectional area, d2Ssecond derivative of the projected cross-sectional area, _any local shock or separation effects which might occur dueto shape modification were not evaluated

    3、 in the development of the theory.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-NACA RMA53H17Concepts Leading to the DTag Equation *.The derivation of the drag equation for a wing-body-tail combination ais based on the theory that the configuration

    4、may be represented by aseries of equivalent bodies of revolution. This theory is dependent ona siinplifiedrelationship between source strength and cross-sectionalarea. This relationship is used in planar wing and slender body of revo- lution problems. .Specifically,as was pointed out in reference 9,

    5、 the source strength is assumed to be proportional to the normal component ofthe stream velocity at the body surface. The theory also assuuws thatthe configuration is of a conventional type with thin symmetrical airfoilsurfaces and a high fineness ratio body. Further exceptions and limi- tations to

    6、the theory are given in reference 9.The development of equivalent bodies of revolution will be illus-trated by using the configuration shown in figure 1. The wing-body-tailcombination is cut by a series of planes which always intersect the longi-tudinal axis at the Mach angle p. In other words, thes

    7、e planes aretangent to Mach cones. The plane identified im figure 1 as Xl repre-sents one plane of a series of parallel planes which cut the configurationalong the entire longitudinal axis. Each plane of this series interceptsthe yz plane in a line which forms the angle e= = with the zaxis. Similarl

    8、y, planes Xa form the angles ez with the z axis andthe yz plane. For any one cutting plane of the series of planesX2 . f (e2, p) the oblique cross-sectionalarea is projected on a planeperpendicular to the x axis. This projected cross-sectionalarea isplotted as a function of x. The resulting plot may

    9、 be considered asrepresenting the longitudinal distribution of cross-sectionalarea S(x)of an equivalentbody of revolution for the series of planesX2 s f (e2, v). For any one value of K this process is repeated forother values of e ranging from e = O to e = 2n. However, if the con-figuration is symme

    10、trical with respect to the xy and xz planes, thenequivalent bodies for G from O to n ofiy need be obtained. Forbodies of revolution the area distribution is independent of e.-.b .With these concepts and with the use of the simplified relationshipbetween source strength and cross-sectionalarea, the e

    11、quation for thezero-lift drag rise as a function of the r-teof change of cross-sectionalarea can be derived from equation (h6) of reference 7 and written as:fiuo do dowhere x1 and x2 are two differents“(x) =locations along the x axis, Pd2S(x) (2)dx2 Provided by IHSNot for ResaleNo reproduction or ne

    12、tworking permitted without license from IHS-,-,-NACA RM A53H17 5This generalized equation can be simplifiedby solving the doubleintegral of the functions of x through a Fourier sine series in thesame nner as used in reference 5.-1-1 =NJv mhc= I An sti n9n=lP=arccoB2/2thenwhere the coefficients are a

    13、 function of e, sinceOf e. With this solution the simplified equation(4)(5)(6)S(x) is a functioncan mow be written as(7)The computing procedure followed in applying the foregoing equationsand theory to the determination of the zero-lift drag rise is presentedin the Appendix of thts report.CONFIGURAT

    14、IONSAND TESTSPlan-view sketches of the models tested, and also the axial distri-bution of cross-sectionalarea normal to the longitudinal axis, areshownin figure 2. The different configurationswill be referred to as modelsA, B, C, and D as follows:lbdel A: aspect ratio 4 triangular wing with fusekge

    15、and tailModel B: aspect ratio 3 straight wing with fuselage and tailProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-6Model C: aspect ratio 6, 450 sweptbacklbdel D: fuselage and tail (consistinghorizontal surfaces)NACA RMA53H17wing with fuselage and t

    16、ail .of two vertical and two .-bGeneral geometric data for all the models are presented in table 1, withgreater detail given for model D in figure 3. The fuselage and tail werethe same for all models. The fuselage ordinates from the 8-inch to the139. therefore, the data of figure weremultiplied by t

    17、he respective areas used in computing the drag coef.ficients and the results presented in figure 6. The apparent late bagrise for model A is attributed to possible error in fairlng the experi-ADmental data points which had scatter equal to = =hL d%g “radiano l.1 go o 1.5718.08 1.285 70 8.08 1.0928.0

    18、8 1.857a 70 8.08 2.05Q11.38 1.173 u. 38 .89011.38 1. g69a 60 n. 38 2.g5217.75 .946 y 17.75 017.75 2.196a p 17.75 3.14228.70 0 4528.70 3.142* 45I M = 1.02e,dg raio 1.5718.08 .7878.08 2.35511.38 011.38 3.142hA#9070702aAngles G are taken symmetrical about 19= m/2.Step 5.- Integrate the area under curve

    19、s similar to figure 9(c) andcompute the drag-rise coefficient from the following equation derivedfrom equation (7): Calculation for Mach Number Range, Configuration SymmetricalWith Respect to xy and xz Planes, Wingsand Tail Surfaces in Both PlanesThe essential differences from the prior method is th

    20、at the verti-cal surfaces are rotated 90 into the xy plane, and if the same valuesof $ used for the cutting of the horizontal surfaces are used to cutvertical surfaces then the cross sections for the vertical surfaces willcorrespond to different angles e than those used for the horizontalsurfaces. B

    21、ecause the areas must be couibinedto give an equivalent bodyof revolution for one value of G, cross plots of the areas should bemade or the vertical surfaces should be cut at different angles than thehorizontal surfaces. A satisfactory and simple procedure is to cut the”vertical surfaces for one or

    22、two angles more than the horizontal surfacesto ensure a uniform variation of angles from O to “, and then proceedas follows:Step 1 Multiply the cross-sectionalareas for the verticalsurfaces by cosy.1Step 2.- For each fuselage station plot the areas from step 1 as afWCt.iOn Of =.eH - “. Subscripts V

    23、and H refer to the verti-cal and horizontal surfaces, respectively.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-14of Step 3.- Read the areas.from step 2 thatNACA RMA53H17correspondto the valuesE!s:?Taper ratio . . . . . . . . . . . . a71 . . . a71

    24、 . c a71 * a71 a71 0.22Airfoil section . . . . . NACA 65009 perndiculsx to 0.25 chordSweepofO.25chord . . . . . . . . . . . . . . . . . . . 450See figure 2.See figure 3. =?5=.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-.NACA RM A53H17Equivalent b

    25、ody of revolutionformed by the planes X2/1718459Figure 1.- Illustration of the cutting ples X and the emglesQ and $, which are the intercepts of these planes with theyz and xy planes, respectively.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-Pco0

    26、20 40 60 80 100 120 140 160 18o 2CCI 220Fuselage stations, Inches TFigure 2.- Pleu views ofthe models tested shorn with thefi axial distributions of cross-sectionsl area normsl to the longitudinal. sxis.E!r , *Provided by IHSNot for ResaleNo reproduction or networking permitted without license from

    27、IHS-,-,- ,Plan view.Angle-of-attackvaneEquation for body ordinatesx=139.4 Is : r = 85 1StatIonoIStation-25.1,4.4?T62.5. . -between x= 8 and Station 1148.25#( )- +2_l 2 3/4StationStatIon 210.5150.5 IStation StationSide view -21.7 139.4450. St;ty.1 2*- . - -Angle-of -sidesllpjY 48.7vaneNote: All dimen

    28、sions are in Inches.Figure 3.- Geometryand MmenElons of model D.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-20 NACA RM A53H17a713-/.2 /,Mode 1 Area, sq ftA 30.10.- 1-B 21.68 SW. c 9.02.1 D 1.58(CrossSectional)o .06 -t.04 # -d“.02 / -=$s=0.8 a719

    29、1.0 1.1 1.2Mach number, MFigure 4.- Expertiental zero-lift drag coefficients,based upon winga?ea for models A, B, end C end upon maximum cross-sectionalareafor model D.Provided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-.NACA RM A53H17.04.020. Figure 5.-

    30、.020.04.020.2.10 Experiment-I I eory I,- .- -(b) Model B.- - ,- -(c) Model C.-, - -/ yq=21.8 .9 LO L1 1.2Mach number, M(d) Model D.Coqpsrison of eerimental drag rises with values calculatedutilizing linearized theory.Provided by IHSNot for ResaleNo reproduction or networking permitted without licens

    31、e from IHS-,-,-22.8.40Q)ii!tY4.m.8.40(a) Model A.(b) Model 13,(c) Model C.8 Expertimt-Theory.8 .9 1.0 1.1 1.2Mach number, M(d) Model D.Figure 6.- Comparison of theoretical and experimental increases in dragdivided by dynamic pressure.Provided by IHSNot for ResaleNo reproduction or networking permitt

    32、ed without license from IHS-,-,-b.04.oi a71 1Unmodified. . - - - Modlficatlon 1- Modification 2- Modiflcatlon 3- Modlflcatlon 4- Sears-Haack bodywith same Sam.i /.Model B.96 1.00 L04 1.08 1.12 1.16 1.20 1.24Mach number, M(a) ULlcul.ated drag rise, o.Figure 7.- Comparisonof the computeddrag rise of m

    33、cdel B with the computedvalues fs=imi!=FadimsFigure 9.- The three steps in the theoretical solution ofFor an unsymmetrical configuration, plot C shouldbee =OtOe=.NACA-ey-1O-16-M.325the drag rise.made fromProvided by IHSNot for ResaleNo reproduction or networking permitted without license from IHS-,-,-


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