Analysis for Calculating the Lift-to-Drag Ratio for Flying Plates Having Square or Circular Shapes

688

APPENDIX E ANALYSIS FOR CALCULATING THE LIFT-TO-DRAG RATIO FOR FLYING PLATES HAVING SQUARE OR CIRCULAR SHAPES

Most of the fragments from explosions are usually "chunky" in shape and have a lift coefficient (C^) of 0.0.

However, in some cases, where one

predicts a breakup pattern which involves a large number of plate-like fragments, lift on fragments can become an important consideration.

The

discussion which follows gives a technique for calculating the normal force coefficients and lift and drag force coefficients of plates having square or circular shapes. Consider Figure E-l which shows a square plate moving with velocity v from left to right at an angle of attack a^.

The lift area for the larg­

est surface is A^ and the normal force coefficient for this surface is C^ . This normal force coefficient is divided into a lift component CL in the T l U vertical direction and a drag component C nl in the horizontal direction. Likewise, examining the smallest surface, it has an angle of attack which is

Figure E-l.

+ 90° and an area A^.

Square Plate in Flight

The normal force coefficient is C

N

and

689

is divided into a lift component C

and a drag component CL . Note that 2 2 this surface has a negative lift component, but because A is much larger than A ^

T

the fragment will experience a net positive lift force.

drag and lift forces,

Thus, the

and F^ respectively, can be expressed by:

2 F

D

= (1/2) p V

L

- (1/2) P V

^C

2 F

Di A x

+

A

(c^ A x

+

A^

(E-l)

2

(E-2)

From Figure E-l, one can readily obtain the following relationships for the lift and drag coefficients:

CL = C , T x l l

cos a

D = C CL l l

sin a.

C

LT

2

D

When both equal 0. and C

N

=

CL

N

2

2

cos a

N

1

h

Also, when a

2 equals When

on and (from Table 3-2) CD l Likewise, when

and (from Table 3-2) C

(E-4)

(E-5)

20

sin a_ 2

(E-6)

0° and 180°, cos

and

must

equals 1 and

equals 90°, area A^ is traveling face-

equals 1.17 and

[from Equation (C-2)] C

equals 90°, area A

equals 2.05 and 2

(E-3)

1

equals 0° and 180°, cos ot^ equals 1 and

must both equal 0.

equals 1.17.

2

= C._ N

n

2

also

1 is traveling face-on

[from Equation (E-6)] C

also 2

690

equals 2.05.

Intermediate values for

Hoerner (1958).

and

can be derived from

The results are shown graphically in Figure E-2 and are

tabulated in Table E-l.

Table E-l also contains calculated values for

the drag and lift coefficients. In order to use Figure 6-12 for lifting fragments, it is necessary to determine the ratio C^A^/C^A^ or, more accurately,

A^ +

A^/

2.50

2.05

h

2.00

(
2

1.70 1.50

I-

1.00

U

1.17

0.50

0

10

L,™

180

170

180

170

| 0

Figure E-2.

1

20

i

30

i

10

i

160

150

140

160

150

140

i

. 1

40

1

20

I 30

i

i

40

50

i

130

i

130

i

50

60 120

i

120

i

60

70

i

80

i

90

I

110

100 1

\

110

100

90

1

1

70

80

90

1 90

Determination of the Normal Force Coefficient as a Function of Angle of Attack

(« )

2

691

Table E-l.

Tabulation of Normal Force, Lift and Drag as a Function of Angle of Attack

Attack

1

C

C„ [ h i 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 110° 120° 130° 140° 150° 160° 170°

C D l

(fi) -A 0. 00 0. 43 0. 85 1. 28 1. 70 1..17 1,,17 1. 17 1. 17 1. 17 1. 17 1. 17 1. 17 1. 17 1. 70 1. 28 0. 85 0. 43

90° 100° 110° 120° 130° 140° 150° 160° 170° 0° 10° 20° 30° 40° 50° 60° 70° 80°

A + ^

^2

MN cos

a, \

_2._.05 ?_

( i 0. 00 *)

2.,05 2..05 2..05 2,.05 1..70 1..28 0..85 0..43 0,.00 0..43 0..85 1..28 1.,70 2. ,05 2,.05 2..05 2, .05

0.,42 0.,80 1.,11 1..30 0..75 0..59 0..40 0..20 0.,00 -0..20 -0..40 -0,,59 -0..75 -1.,30 -1.,11 -0.,80 -0..42

< <

^D ^ 2 '

pression.

/ C

/C

Coefficients

1

XN T

2

sin a, \

( i 0.,00 )

/ C \ _ h

0. 075 0. 29 0.,64 1.,09 0..90 1.,01 1..10 1.,15 1.,17 1.,15 1.,10 1.,01 0.,90 1.,09 0..64 0.,29 0..075

M

cos a_ \

0. 00 - 0 . 36 - 0 .,70 -1. 03 -1.,32 -1..30 -1.,11 -0..80 -0,.42 0..00 0..42 0..80 1..11 1,.30 1..32 1..03 0..70 0,.36

/ C

1

ll (

t n h e oe wn o u

^1'

ld

normally use the simpler ex­

As an example calculation of the lift-to-drag ratio, consider

a square plate of dimensions 1.0 m x 1.0 m x 0.01 m (see Table E - 2 ) . is, A^ equals 1.0 m m) .

:

2

(l.Omxl.Om)

and A^ equals 0.01 m

2

That

(0.01 m x 1.0

Using the lift and drag coefficients in Table E-l and the appropriate

values for A^ and A^,

one can readily calculate the lift-to-drag ratio.

The complete and approximate forms of this expression are contained in Table E-2 for various angles of attack a^.

For this particular example,

one can readily see that the approximate values of the lift-to-drag ratio do not differ greatly from the complete values.

si

N

2 2.05 2.02 1.93 1.78 1.57 1.09 0.64 0.29 0.075 0.00 0.075 0.29 0.64 1.09 1.57 1.78 1.93 2.02

692 Table E-2.

Example Calculation for Determining the Lift-to-Drag Ratio

k [ \

l

CA

A C+ L

2

2)

+ ( V i

A %

2 )

(

+ A C2

l

L l +L 2

( V *

)

V > )

(degrees)

(square meters)

(square meters)

(• -)

(-)

0 10 20 30 40 50 60 70 80 90 100 110 120 130 140 150 160 170

0..00 0..42 0..79 1..10 1..29 0..74 0.,58 0..39 0..20 0.,00 -0..20 -0.,39 -0.,58 -0..74 -1,.29 -1.,10 -0.,79 -0.,42

0,.02 0..10 0..31 0..66 1..11 0..91 1..02 1..10 1..15 1..17 1..15 1..10 1.,02 0..91 1.,11 0..66 0. 31 0.,10

0,.00 4 .20 2 .55 1..67 1,.16 0,.81 0,.57 0..35 0..17 0..00 -0..17 -0..35 -0..57 -0..81 -1..16 -1..67 -2..55 -4..20

0.00 5.60 2.76 1.73 1.19 0.83 0.58 0.36 0.17 0.00 -0.17 -0.36 -0.58 -0.83 -1.19 -1.73 -2.76 -5.60