On the wave drag of non anixymmetric bodies at supersonic speeds

ON THE WAVE DRAG OF NON ANIXYMMETRIC BODIES AT SUPERSONIC SPEEDS (0 VOLNOVOM

SPOROTIVLENII NEOSESIMMETRICHNYKH V SVERKHZVUKOVOM POTOKE)

PMM

Vol.23,

No.2,

1959,

TEL

376-378

pp.

MAIKAPAR

G. I.

(Moscow) (Received

The bodies polygonal of

such

the

under cross

the plane the

is

shock

edges

waves.

on the

December

belong

with

formed

outer

wave drag

design

consideration sections

a body

straight

24

the

For such

the

wave drag

therefore wave drag of

bodies,

shock

waves,

intersections

a simple is

having The surface and of

dependence

developed

for

of

given

conditions.

Heretofore

narrow

of

plane

as linear

characteristics

pyramids

corners.

behind

appear

a class

geometrical

of

or concave

surfaces

body

Fig.

exact

a family

reentrant

by stream of

to

1958)

of

interest

is

easily

and the

bodies

only

is

class

solutions

bodies

found here.

even for

for

which

been obtained

to consider

solutions

studied

of

have

1.

other

though

isolated

They are

families these “design”

pyramids,

flow,

relatively

were circular of

families

bodies

simple cones. for

which

is the

may be relatively

Mach numbers. with

and It

or without

Such a class internal

having cross-sections in the form of reentrant polygons (see Figs. of such a body is formed by the stream surface 1 and 4). The surface behind a configuration of plane shock waves which intersect along (with-

528

On the

out

wave

extending Let

drag

of

beyond)

us consider

reentrant

non

The velocity

the

a body

polygon

(Fig.

bodies

anixyrnetric

straight with

convex

at

edges

supersonic

of

a cross-section

the

in the

speeds

529

pyramids.* shape

of

a regular

1).

components

behind

27%= V [I -

the

(1 -E)

uy = V ctg y (I-

oblique

wave are:

shock

(sirPy-sin’o)]

E) (sin2 y -

(1) (2)

sinlo)

Here sino

1 =co = - = M 7’ cp-c

e=-.=

From (1) axis

and (2)

and the

we obtain

reentrant

The functional base

section

An “equivalent” 8,

such

at

circular

and the

of

the angle

between

the

pressure

pressure

J

r = T (y)

is

(3)

shown in Fig.

The area

2.

of

x = 1 is: Y tg

x

(I-

’ = lZ tg n

cone

with

1 _

E) (sin2 y - sin2 0) (I_

equal

2 tg Z

I-

E)

(sin2

base

y

_

area

sin2o)

has a half

nose

(l- E) (sin2 y - sin2 63) (I- E) (sin” y - sin2 0)

behind

the

p = (i-

C)P~ V2 ( sin2 y -

coefficient p=

shock

is

given -j-&

by sin2 w )

is

p---P,

= 2 (I-

c) (sin2 y -

sin” 0)

(6)

Qco From equations

(5)

and (6)

follows

of the equivalent cone and the pressure which does not depend on Mach number: l

body

that

tg e,= The air

tangent

1

8

ctg2 y (1-c) (sin2 y - sin%0) 1 - (1 -c) (sin2 y - sin2 CO) = ’

’ = ’ tg t tg

angle

the

dependence

cross

-

rib:

ta 6 Ig=

the

u

cp+ cu

a simple

relation

coefficient

between

(wave

drag

For constructing such bodies one may also use a group of for instance three shocks intersecting in a single point, suitable cylindrical shocks.

the

angle

coefficient)

plane shocks, and also

G.I.

Yaikapur

(7)

Equation (8) facilitates comparison between the wave drag of the present class of bodies and that of the circular cones. One must bear in mind, however, that for each Mach number the body shape is different.

Fig.

Fig.

2.

3.

The drag coefficients. referred to that of a circular cone at M = 00 as unity, are shown in Fig. 4. The upper set of curves labelled with the corresponding values of M refer to circular cones. The same figure displays a series of cross sections of the pyramidal bodies equivalent in area to the circular cone with 8, = 15’ at different Mach numbers. The case of an asymmetric pyramidal interest. velocity

body shown

in Fig.

It resembles a (delta) wing with empennage. components behind the shock wave are: vz = V [I-

(1 -

c) (sin2y - sin2o)]

vy = V etg a (Z- E) (sin2y - sin2w)

5.

In this

may be of case

the

(9) (10)

On

the

mave

drag

non

of

bodies

anixyrmctric

at supersonic

apccdr

531

1.5

I \I

M=lO

M=7

M=4

_

1.3

I.1

0.8

a7

I

I

I 16

I

Us =

I

I

I

Fig.

4.

Fig.

5.

V ctg @ (i-c)

I

I

32

I

I

I

OK

(sin*y -sinew)

WI

sin Y = (1 + ctg2 a f ctgs p)-‘ia

The area of one quarter Fig. 4) equals s=z

The equations (3) E + 0. the reentrant

i

of the cross

tga ( t8p+

and (6) corners

:B tg

section

at x = 1 (right

side

(I- c) (sine y - sifP 0) a > l-((l-~)(sin~y-sina~)

of

(12)

for tan 6 and iT remain valid. When M + m and of the polygon straighten out.

Translated

by

M.V.M.