unclassified ad 40 2 019 - dticcda d -gv2 (1) v 2 /l, v 5cos 0 = -1 d 2w ap - 1f(v22) (2) d$ (l•...

61
UNCLASSIFIED AD 40 2 019 DEFENSE DOCUMENTATION CENTER FOR SCIENTIFIC AND TECHNICAL INFORMATION CAMERON STATION, ALEXANDRIA, VIRGINIA UJNCLASSIFIED

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Page 1: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

UNCLASSIFIED

AD 40 2 019

DEFENSE DOCUMENTATION CENTERFOR

SCIENTIFIC AND TECHNICAL INFORMATION

CAMERON STATION, ALEXANDRIA, VIRGINIA

UJNCLASSIFIED

Page 2: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

NOTICE: When government or other drawings, speci-fications or other data are used for any purposeother than in connection with a definitely relatedgovernment procurement operation, the U. S.Government thereby incurs no responsibility, nor anyobligation whatsoever; and the fact that the Govern-ment may have formilated, furnished, or in any waysupplied the said drawings, specifications, or otherdata is not to be regarded by implication or other-wise as in any manner licensing the holder or anyother person or corporation, or conveying any rightsor permission to manufacture, use or sell anypatented invention that may in any way be relatedthereto.

Page 3: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

SSD~TD.-~6Q eport No.SSD-~q86 -60TDR-169(8660-10)TN-1

7 '7

M PROXIMATE SOLUTIONS OF THE LATERAL MOTION OFC.) E-ENTRY VEHICLES, DURING CONSTANT ALTIT UDE GLIDE

H. E. Wang 'Prepared forCOMMANDER SPACE. SYSTEMS, DIVISION

25 February 1963 - UNITED STATES AIR FORCE

ContactNo. FO4695)169 Inglewood, California

J SYSTEMAS RESEARCH AND PLANNING DIVISION.

A S IfI A,

AEROSPACE CORPORATION

Page 4: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

SSD-TDR-63-60 Report No.TDR-169(3560-I0)TN- 1

APPROXIMATE SOLUTIONS OF THE LATERAL MOTION OF

RE-ENTRY VEHICLES DURING CONSTANT ALTITUDE GLIDE

H.E. Wang

25 February 1963

Contract No. AF04(695)-169

AEROSPACE CORPORATION2400 East ElSegundo Boulevard

ElSegundo, California

Prepared for

COMMANDER SPACE SYSTEMS DIVISIONUNITED STATES AIR FORCE

Inglewood, California

Page 5: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page iii

APPROXIMATE SOLUTIONS OF THE LATERAL MOTION OF

RE-ENTRY VEHICLES DURING CONSTANT ALTITUDE GLIDE

25 February 1963

IaL

Prepared by__________H. E. Wang

7Approved byf/.

S. T. Chu, Heac--Re-entry Studies Section

' / ,'/

A. Mager NActing Department Head,Gas DynamicsDirector, Spacecraft Sciences

AEROSPACE CORPORATION2400 East ElSegundo Boulevard

El Segundo, California

Page 6: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page v

ABSTRACT

Approximate analytic solutions have been obtained to the lateral

motion of re-entry vehicles during constant altitude glide. The con-

stant altitude glide is maintained by banking the vehicle about its

velocity vector at a fixed angle of attack. Such a glide mode is often

used in design studies to achieve lateral range but no simple analytic

solutions have been available up to now. The approach taken here is

to first treat the "flat earth" case and then include the spherical earth

correction. Simple expressions have been obtained for the turn angle,

lateral range angle and the down range angle. A comparison with exact

numerical solutions shows that the approximate solutions are sufficiently

accurate for preliminary design purposes.

The approximate solutions, although simple, contain all the charac-

teristics of the glide mode in question. The results show that the lateral

range, being a function of W/(CDA p), exhibits a maximum for a given

L/D and initial velocity for the spherical earth case. In the flat earth

case, the maximum exists for suborbital and orbital initial velocities

and ceases to exist above a certain initial velocity depending on L/D.

The down range also exhibits a maximum for a given L/D and initial

velocity provided that the initial velocity is suborbital. The spherical

earth correction on the lateral and down rangc has be en shown to be

small for low W/(CDA p), L/D and initial velocities and becomes signifi-

cant when these parameters are high. All these findings agree with

exact numerical solutions.

Page 7: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page vii

CONTENTS

Page

I. INTRODUCTION ........... .......................... . . ....

II. EQUATIONS OF MOTION ...... ....... ..................... 3

III. FLAT EARTH SOLUTION ......... ..... ..................... 7

IV. SPHERICAL EARTH SOLUTION - SPHERICAL EARTHCORRECTION ........... ............................ .... 28

V. CONCLUDING REMARKS ............................. .... 41

APPENDIX

A SUMMARY OF APPROXIMATE SOLUTIONS .......... .43

B DISCUSSION OF THE APPROXIMATE SOLUTIONSFOR VERY SMALL C .... .................... 45

NOMENCLATURE . ............. .................... 49

REFERENCES .................................. 50

Page 8: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page ix

ILLUSTRATIONS

Figure Page

la Definition of Bank Angle . ........... .................. 4

lb Definition of Turn Angle 1, Lateral Range Angle X,and Down Range Angle . ......... . .... .................. 4

2 Variation of sin 0 as a Function of V/V and C .... ...... 8S

3 Comparison of Final Turn Angle ..... .............. . i.11

4a Comparison of Final Turn Angle (V. = 20, 000 ft/sec). . . . 1z4b Comparison of Final Turn Angle (Vi =26,000 ft/sec).... 13

4c Comparison of Final Turn Angle (V.i 3 2,000 ft/sec) .... 14

5a Comparison of Lateral Range (Vi = 20, 000 ft/sec) ... 15

5b Comparison of Lateral Range (Vi = 26, 000 ft/sec) ........ 16

5c Comparison of Lateral Range (Vi = 32, 000 ft/sec) ........ 17

6 Schematic Diagram of the Exact and ApproximateFlight Path ............... .......................... .... 19

7 Comparison of C*, "Flat Earth" ....... ............... 22

8a Comparison of Down Range (Vi = 20, 000 ft/sec) .... ...... 24

8b Comparison of Down Range (Vi = 26, 000 ft/sec) ...... ... 25

8c Comparison of Down Range (V. = 32,000 ft/sec) ...... .. 26

9 Spherical Earth Correction on Turn Angle ............ ... 31

10 Spherical Earth Correction on Lateral Range Angle . . .. 34

11 Spherical Earth Correction on Down Range Angle ..... .. 39

B-I Vehicle Flight Path at Very Small C .... ............ ... 48

B-Z Schematics of the Variation of X and ý. withW/(CDA p) at Very Small C ..... ................ ... 48

Page 9: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 1

I. INTRODUCTION

In many cases, a space vehicle entering the earth's atmosphere requires

lateral range capability for landing at a specific site. To gain lateral range,

the vehicle must be steered out of the original orbit plane and as a result one

needs an understanding of the lateral motion of such a vehicle. Slye (Reference 1)

has presented an analytical method for studying the lateral motion of re-entry

vehicles flying equilibrium glide trajectories. While this class of trajectories

is of great interest and importance in designing a re-entry flight path, its

practical use is limited to re-entry from orbital speed. When the vehicle enters

the atmosphere at superorbital speeds, a transition phase is required to bridge

the gap between the superorbital initial phase and the suborbital equilibrium

glide phase. One type of trajectory capable of carrying out this transition is the

constant altitude glide. Constant altitude glide has drawn the attention of many

designers; for example, Smith and Menard (Reference 2) used a constant altitude

glide to maneuver for range near orbital speed. Ferri and Ting (Reference 3)

also studied constant altitude glide to reduce the second peak deceleration and to

control the vehicle from skipping. These authors, however, did not consider

the advantage of obtaining lateral range by banking the vehicle while maintaining

the constant altitude glide. Banking the vehicle not only produces lateral range

but also ease the problem of designing a thermal protection system due to the

fact that only one side of the vehicle is exposed to the free-stream. This sub-

ject has been discussed by Gervais, duPont and Lowe (Reference 3).

While the constant altitude glide is of great practical interest, it is not

without problems to get a vehicle onto such a glide from the initial phase of the

re-entry, especially when the speed is superorbital. It can be readily appre-

ciated that lift modulation in the initial phase is required. Wang and Chu

(Reference 4) have attempted to analytically solve the problem of re-entry

trajectory'with"vafiable lift of a simple type. They have demonstrated that by

varying the parameters in the lift program, one is indeed able to make a smooth

transition from the initial plunge to a constant altitude glide. This can be done

either by pitch control or by bank control. In the present paper, it will be

assumed that the vehicle is already at the beginning of a constant altitude glide

and the initial phase of the re-entry trajectory is beyond the scope of this

analysis.

Page 10: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 2

The particular maneuver to be dealt with here, namely, constant altitude

glide by banking at a fixed angle of attack, has been studied by Skulsky (Refer-

ence 5). He presented exact numerical solutions to the equations of motion.

Later, Wang and Skulsky (Reference 6) discussed the characteristics of the

lateral range obtained during such a glide. Their discussion was based on the

numerical solutions presented in Reference 5. The present paper presents,

however, analytic solutions to the problem using a basic approximation. The

approximation appears to be crude at first but the results turn out to be quite

satisfactory. The "flat earth" solution will be given in the first part of the

paper and in the second part the spherical earth case will be dealt with. A

summary of the approximate solutions is given in Appendix A.

Page 11: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 3

II. EQUATIONS OF MOTION

The three-dimensional equations of motion of a hypervelocity vehicle flying

over a spherical earth have been derived by London (Reference 7), by Love and

Neustadt (Reference 8), and by Wang (Reference 9). In Reference 9, the lateral

centrifugal force term has been discussed extensively with regard to its origin

and physical identity. It is not considered necessary to include the derivation

of these equations in this paper. For constant altitude glide these equations

reduce to

(2W__ dV2

CDA d -gV2 (1)

V 2 /L, 2W 1F(V22)v 5cos 0 = -1 -D Ap (2)

d$ (L• Vg sin 0 Vg cos p tan X (3)dt - =D- 1 2W V 2

(C*A p; Vs

Here 0 is the bank angle measured from the vertical, i is the turn angle

measured from the original heading and X is the lateral range angle defined

as the ratio of the lateral range to the radius of the earth (equal to 3440

nautical miles in the calculations shown later). These angles are shown in

Figure 1. The other symbols have their usual meaning. Together with these

three equations of motion, there are two kinematic equations:

dX _ Vg sin t (4)dt V2

d__ Vg cos • (5)dt Cos

s

Here .i is the downrange angle as shown in Figure 1. It is to be noted that

Equations (1), (2), and (3) are applicable for a point mass. A point mass

vehicle is assumed in this analysis. The orbital speed Vs is assumed

Page 12: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 4

LSINO

L COSO

mg

Figure La. Definition of Bank Angle 0.(Velocity vector pointing into the paper)

POINT

OF ENTRY

Figure lb. Definition of Turn Angle $, Lateral Range Angle X,and Down Range Angle ýL.

Page 13: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 5

constant and equal to 26, 000 ft/sec in the numerical calculations given later

in the paper.

If we write

B W (6)

and transform the independent variable from time t to velocity V through

Equation (1), we obtain

V = sin 0 - = cos tan X (7)V

d ZB (8cTV-= - sin (8)

V s

Sd•t ZB cos (9)-V : - cos X

Denoting In V = X, these equations become

d = L. ZB-d L sin 0 - B cos tan X (10)

s

dX 2BT- = -7- sin ii(1

V

d_ = ZB cos ýIJ (12)

dX R VZcos Xs

Equations (10), (11), and (1Z) determine the unknowns i$, X, and kt in terms

of V. The bank angle 0 can be readily written in terms of V by Equation (2)

and the velocity V can be found as a function of t by integrating Equation (1).

The main problem here is to solve the three nonlinear differential Equations

Page 14: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 6

(10), (11), and (12). One can readily appreciate the difficulties associated

with solving these equations analytically. The approach taken here is to treat

the "flat earth" case first and then take up the spherical earth case. Flat

earth means a condition when the second term on the right hand side of

Equation (10) can be neglected. This term is proportional to the lateral

centrifugal force. If the earth is truly flat, this term will of course drop out

but so will the (V/Vs)2 term in Equation (2). In the flat earth solution pre-

sented below, Equation (2) remains the way it is and only Equation (10) is

simplified by neglecting the second term on the right hand side. By so doing,

it does not result in a true flat earth as the name implies.

Page 15: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 7

"III. FLAT EARTH SOLUTION

For the flat earth case, although the equations to be solved are much

simplified, they are still difficult to handle analytically because of the trigono-

metric functions involved. One certainly can eliminate sin 0 by Equation (2)

and integrate Equation (10) and in principle it is integrable for the case of flat

earth. But the resultant expression for ý will be very complicated. In fact,

this has been done by Skulsky (Reference 10), and one still has to use high

speed computing machines to get a numerical value. As for X and 1±, numeri-

cal integration is unavoidable. What is done here is to further simplify Equation

(10) by making the following approximation:

sin 0 = 1 (13)

Let us investigate this approximation. First, rewrite Equation (2) as follows:

sine = ( I [ G ) 114)

where

C- 2B 2/CApV2 (15)C w( s = ZW/(GLA=BVz) 1(IL7 s

Equation (14) is plotted in Figure 2. It is seen in Figure 2 that for small values

of C, sin 0 is practically equal to unity for a wide range of V/Vs. For example,

when the initial velocity is Vi/Vs = 1.2 and C = 0. 1, sin 0 is practically unity

down to V/Vs = 0.5 and rapidly decreases to zero at Vi/Vs = 0. 3. Integrating

Equation (10) (when the second term on the right hand side is omitted for flat

earth) is effectively to find the area under each sin 0 curve for various C.

The approximation sin 0 = 1 will give the area of the rectangle bounded by

V/V and Vf/V where Vf is the final velocity that is at the end of the constant

altitude glide where 0 = 0. This area will not be much larger than that under

the actual sin 0 curve. For high C, the error should increase. For low initial

velocities, the error should also increase. For example, when Vi/Vs = 0. 77

and C = 0.8, there is no point where sin 0 is close to unity. Therefore,

Page 16: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 8

Z -

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0 P:

Io oZ 40

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CA NIS ~'NIS

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Page 17: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 9

the approximation is the worst at high C and low initial velocities. The

numerical examples given later in the paper bear this out but the inaccuracy

in X and ý is surprisingly low.

By this approximation sin 0 = 1, Equation (10) for the flat earth case

reduces to

di 0 L

-T (16)

Here the subscript 0 denotes flat earth solution. The initial condition is

*0 = 0 at Xi. Integrating Equation (15) gives the following simple result.

*0 = L (Xi - X) (17)

or

L (8= l• n (V./V) (18)

By Equation (17), Equation (11) can be integrated with the initial condition that

X0 = 0 at Xi and the result is

X0 = C (1 - cos ýf0) (19)

At the end of the constant altitude glide, the bank angle 0 vanishes; that is to

say, the vehicle has reached the saturation point and cannot maintain a constant

altitude unless by changing the angle of attack. Since the flight mode considered

here is constant altitude glide by banking at a fixed angle of attack, the final

velocity must be determined by the condition 0 = 0. Therefore, by Equation

(14), one has

VVf - (20)

l+-

Page 18: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 10

The final turn angle and the final lateral range angle are respectively

V.

L L 1 + 1 (21)OfD 5 -

S

k Cof C (1 - Cos W0 f) (22)

At this point, it is worthwhile to compare these simple solutions with

exact numerical solutions. The exact numerical solutions ar.e taken from

References 5 and 10. Reference 10 gives the flat earth solutions and should

serve as a check against the sin 0 = 1 approximation. Reference 5 gives the

spherical earth solution and should indicate the magnitude of the spherical

eartlh effect.

Figure 3 is a plot of CO0f/(L/D) as a function of W/(CLA p) for initial

velocities 32, 000, 26, 000, and 20, 000 ft/sec. Equation (21) predicts too high

a turn angle as expected. The percentage error increases with increasing

W/(CLA p) for a given initial velocity and decreases with increasing initial

velocity for a given WACLA p). These are the expected nature of the approxi-

mation. The absolute magnitude of the error in the quantity %Of/(L/D) is of

the order of 2 to about 6 degrees in the range compared. Generally speaking,

the accuracy of the simple approximate solution is satisfactory; it correctly

predicts the general trend and correctly establishes the functional, dependence

of w 0f upon B, L/D and V.. Figures 4a, 4b and 4c show the same comparison

plotted in a slightly different form. Included in these figures are the exact

numerical solutions for spherical earth, which will be discussed later in

Section IV.

A comparison of the lateral range is given in Figures 5a, 5b and 5c. For

all three initial velocities, the approximate solution, Equation (22), agrees

with the exact numerical solutions for flat earth well within 10 per cent. In

some cases when the initial velocity is high, the present approximate solution

is almost exact and so is it for low W/(CDA p). The general characteristics

of the lateral range as a function of B, L/D and Vi including the peaks are

Page 19: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 11

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Page 20: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 12

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Page 21: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page. 13

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Page 22: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 14

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Page 23: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 14

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Page 24: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

page 15

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Page 25: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 16

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0 OD w

M_ NI IINU IN3

Page 26: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 17

6

x

8

8U

CD

0

IlIID X

bi U) )

U:U

bi

OD . 4

Ito ~I 00

0

10

00u

Lf)

o 00 0 0

In 0 n 0I1N clN

IN N'!N k -IU.V

Page 27: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 18

well preserved in the approximate solution. The striking feature that one finds

in these comparisons is that the accuracies of predicting *Of and X0 f are not

at all the same. The approximate solution predicts the lateral range with a

much higher accuracy than it does the turn angle. This can probably be

explained in the following manner. For a given initial velocity and W/(GDA p),

the exact flight path might be represented by the solid curve of Figure 6 while

the approximate solution is indicated by the dashed line. Point A might repre-

sent the end of the turn. The lateral range in both computations might be very

close to each other while the approximate solution predicts a much higher turn+

angle than the exact numerical solution. This explanation is fully supported by

the downrange prediction. The approximate solution always predicts a shorter

downrange as will be seen later.

Let us now return to the development of the approximate solution. Equations

(21) and (22) can be used to determine the maximum lateral range and the con-

dition under which it occurs. Differentiating Equation (22) with respect to C

and setting it equal to zero, one gets

dX Of & Of= (1- cos I Of) + C sin 'Of-j- 0 (23)

where d¶of/dC can be found from Equation (21) as

d il1 Of I L 1

dC= - D C(l +C) (24)

Combining Equations (24) and (23), one gets

1*- 1 L sin l Ofcos 0 f = 7 D5 1+ (C *

Here the * quantities denote the condition where the maximum lateral range

occurs. Using the trigonometric relation of half angles and eliminating $0f

by Equation (21), Equation (25) finally reduces to

Dr. Frank Billet analytically investigated the errors and showed similarresults.

Page 28: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 19

REPRESENTING EXACT FLIGHT PATH

REPRESENTING APPROXIMATE FLIGHT PATH

wz V

4 '(TWO DIFFERENT TURNw ANGLES AT ABOUT

A SAME LATERAL RANGE)

POINT DOWN RANGEOF

ENTRY

Figure 6. Schematic Diagram of the Exact and ApproximateFlight Path.

Page 29: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 20

1~~ LLV. In , 1 - = arc tan (26)

The maximum lateral range can be readily found to be

L V. I - -•

sins L rIn, -1 TX 1 LS / . (27)XOf -2 1+-

These equations have been used to compute the maximum lateral range and

where it occurs. The results are listed in Table 1. Included in Table I are

also the exact numerical solutions. The comparison is extremely favorable.

It is found that for a given L/D there exists an initial velocity beyond which

there is no maximum for a flat earth. This is illustrated in Figure 7 in which

Equation (26) is plotted.

The downrange angle can now be integrated. Dividing Equation (12) by

Equation (11) one gets for flat earth,

d~o cot 0S = cos X 0 (28)

By Equation (19), one can write

C - X 0cot *0 = 0 (29)

-c C2- (C - X0)o

1 2.

The I/cos X term can be approximated by 1 + 1 X This approximation is

accurate within 5 per cent up to X0 = 40 degrees which means lateral range =

2400 nautical miles. In Figures 5a, 5b and 5c, the lateral range is smaller

than 2400 nautical miles for all the conditions presented. If the lateral range

Page 30: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 2 1

41

14r- 0 0 N Ln 10cn a, r- 1o Ln

0 w 4~ N a'

N;4 u 10 '.0 '.0 .0 '.0 '`00. C0 000 00CD0

N X X XCD o- 0 Lfn 0 00 Lfn

* N Ln~ Lfl a, r-

10 -4

0.

-4

~~4 S n N n N O'r

u '.0 '.00 '.0'.0'.0j 0 C C) 000 000D

u -

W 4.CD- 0 0 LLn 0 Lfn Lf

* N ifn -4 a, o'.- N

-4

o c, 00 o 00 Lfn 00

-4 0 4 N 0o N*

~ 00 0)00 CD

'4-4

Page 31: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 22

N

w 4-4

____ 0

0

00 0.p S

00

0~ ino0: X -

4 w 0:3 LL 0: >

Page 32: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 23

is of the order of 1000 nautical miles, the approximation has an inaccuracy of

about 1 per cent. Together with this approximation, Equations (28) and (29)

combine to yield

dý (C - X0) (I +½I X020 2 ( -(C-0) (30)

The initial condition is •0 = 0 at k0 = 0. One finally gets by integrating

Equation (30)

1+ -2-+ OG- G-3 + s i -si-n O (31)

where

G = _jC2 - (C - kXf)2 (32)

In terms of O0f, Equation (31) becomes

3 s 1 1 sin3 Of( )

I.Of = C sin *0f + C3 sin ,!0 f - 1 -12[ sin 2, - (33)

For practical purposes, Equation (33) can be approximated by

40f = C sin o0f (34)

Equation (34) agrees with Equation (33) within 1 per cent for lateral range up

to 1000 nautical miles and within. 3. 5 per cent up to 2000 nautical miles. Note

that Equation (34) would be the result of the integration if cos X0 = 1 were

assumed. The downrange is plotted in Figures 8a, 8b and 8c. Comparison

with the exact numerical solutions show that the approximate solution predicts

shorter downrange as depicted by the schematics of Figure 6, but the accuracy

is very high. In these plots, Equation (34) is used except for a few points where

Page 33: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 24

< mw ow zWu<~ w - 0

_ 04 j

o Z -

4- x -fi(nu

O0 ~ O ~ - - 1 0 Q u

W LI m-00zL0L

00

0 0

2w

00

0

00___ _ ___ _ ___ _ ___ __ 0 0

0 0

00

co

:J4

0 0 0 0 0 00 0 0 0 00 D40 vN

W 'N '39NVd NMoa

Page 34: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 25

0

xI0

w4

~ W

"r U

0 0Dz 2 _____ 0 0

2 CD

"" 00S- 0

O D

of Il -t J .0N

_ 1.0 00__0 0

~~ 00000S0 • 0

I IINN'9¥ M

000

N W N 03NkfN

Page 35: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 26

x

o • o

l~co,X I

PO -o00

0m

4"•,

0

W W

0d

o-02

a)

aN

0

0 ~~ 3t 000 IQ 0

00

0

o b

N

00 0 0 0 0 00 0 0 0 0 000 0 0 0 0 010 p*) -ot

INl N '3!0NV NMOO

Page 36: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 27

the lateral range is above 1500 nautical miles. It is observed that for a sub-

orbital initial velocity, there exists a maximum down range. No maximum

exists for orbital or superorbital initial velocities because the vehicle, at

these speeds, can always fly at a sufficiently high altitude and go around the

earth as many times as desired. This would produce no maximum on down-

range. It is also noted that the maximum down range and maximum lateral

range do not occur at the same altitude for a given vehicle. The maximum

down range occurs at a higher altitude. One can readily find the maximum

down range at suborbital initial velocities by differentiating Equation (34) and

following the usual procedures. The result is given in Appendix A.

For completeness, the total flight time is given below. This is obtained

by integrating Equation (1) with the initial condition t = 0 at Vi and the final

condition t = tf at Vf.

tf L C (35)

This essentially completes the development of the flat earth solution. It

must be pointed out, however, that the solutions presented above must be used

with caution when W/(CDA p) becomes very small. In the previous figures,

the lateral and down range have been extended to zero at W/(CDA p) = 0 by

smooth curves. This was done as a matter of convenience. Strictly speaking,

this is incorrect. As one can readily see that for a given L/D and initial

velocity, the down range and lateral range expressions Equations (22) and (34)

can be combined to yield an equation of a circle with its center at X = C and

radius equal to C . When C is extremely small, although this may never

become the case in practice, the vehicle can circle around the X = C location

without gaining much lateral range and down range. The lateral range and

down range curves must have an oscillatory nature at very small W/(CDA p)

A more detailed discussion is given in Appendix B.

Page 37: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 28

IV. SPHERICAL EARTH SOLUTION -

SPHERICAL EARTH CORRECTION

As can be seen from Figures 4, 5 and 8, the exact numerical solutions for

spherical earth deviate only slightly from those for flat earth when the initial

velocity is low. For example, when V. = 20, 000 ft/sec, one can hardly see any1

difference. This is true for the turn angle, the lateral range as well as thedown range. When V. = 26, 000 ft/sec, the spherical earth solution begins to

1

show some difference. The difference on lateral range is about 10 per cent at

WACDA p)= 1000 x 106 ft /sec and L/D = 2. The spherical earth solution

shows higher difference at higher initial velocities. For example, when V. =1

32, 000 ft/sec, the spherical earth solutions deviate from the flat earth solutionson lateral range by about 20 per cent at W pCDA p)= 1000 x 106 ft2/sec2 and

L/D = 2. This order of magnitude of the error produced by the flat earth solu-

tion is too high to be ignored and must be taken into consideration. On the other

hand, an error of 20 per cent or so can still be treated mathematically as a

second order. Thus, the second term on the right hand side of Equation (10)

can be treated as a small perturbation on the first term. Accepting this idea,

one is justified to speak of a spherical earth correction and treat it accordingly.

In the case of an equilibrium glide trajectory, Slye's solutions (Reference 1)

are for flat earth. The spherical earth correction in that case has been dis-

cussed by Jackson (Reference 11) and by London (Reference 12). London

advanced an argument that the proper criterion for determining the validity of

the flat earth solution is the relative magnitude of the integrated lateral

centrifugal force to the integrated lateral aerodynamic force, while the magni-

tude of the lateral range is not in itself a sufficient criterion. In the present

case, we also observe that it is not sufficient to say when X is small, the

lateral centrifugal force term can be dropped from Equation (10). If one tries

to do this, he will run into difficulties, because the first term (L/D) sin 0 tends

to zero toward the end of the glide while the second term (2B cos T tan X)/V25

can be very large. However, if one compares the integrated values of the term2(L/D) sin 0 and the term (2B cos * tan X)/V , he will find that for most cases

the latter will be smaller than the former. If the latter is so small that one can

neglect it, the flat earth solution becomes valid.

Page 38: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 29

The spherical earth correction for constant altitude glide has been dis-

cussed in Reference 6 where an empirical correlation was given for L/D = 2.

What will be done here is to find an analytic expression for this correction

within the present framework of approximation. Let us write

0 + 1 (36)

thus

ditdtlf - d~r0 (37)dUR _a + 7Y7

Combining Equations (10) and (37), one has

dt 0 dtl L ZBD =--" cos t tan X (38)

V s

By the sin 9 = I approximation, the first term on both sides of the equation

cancels each other and one gets

S= - cos tan X (39)Vs

With fairly good accuracy, one can replace cos * and tan k by cos *0 and

X0, respectively, and get

dIY1 0B X0S= V cos 0 (40)

or

d = - C (I - cos '0) cos *0 d* 0 (41)

Page 39: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 30

The initial condition is •I = 0 and 0= 0. Integrating Equation (41) yields

If = - C- (sin *of - 1 %0f - T sin 2 *of) (42)

Equation (42) relates the spherical earth correction and the flat earth solution.

It is noted that t1if is proportional to C and it is always negative. Equation

(42) is plotted in Figure 9 and also tabulated in Table 2. Applying this correc-

tion, the approximate spherical earth solution becomes

f= If C-2 (sin If 1 in2 ý of) (43)•f •Of -Of 40*f T1[ sn Of

where *of has been given by Equation (21). For comparison, Equation (43) is

plotted in Figures 4a, 4b and 4c as the dashed lines. The comparison is quite

favorable.

The spherical earth correction on lateral range is found as follows. With

Equation (36) one can write Equation (11) as

dX -2B

d -T sin (10 + *1 ) (44)V-X ZB v-

dk =7 (sin It cos t + cos It sin ( (44)S= v--z.(i o 'V1

0

Since ,I, is small, one is justified to assume

cos 1 and sin , 1 ( 'l (45)

Thus,

dX 2B Z B (46)-•- = -a si 40 + l4 cos (46

V V5 5

Page 40: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 31

0d

N

0z 0

U)U

0

2 -IN

w z

0

0 0 0

Pdd

a0 In

0r~

00

Page 41: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 32

Table 2. Spherical Earth Correction on Turn Angle.

Tof *of/C2

(deg) (deg)

0 0

10 0. 063

20 - 0.384

30 - 1.245

40 -2.728

50 4.710

60 - 7. 175

70 9.65

80 11.48

90 12. 33

Page 42: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 32

Table 2. Spherical Earth Correction on Turn Angle.

*Of *Of/Ic2

(deg) (deg)

0 0

10 - 0.063

20 - 0.384

30 - 1.245

40 - 2.728

50 - 4.710

60 - 7. 175

70 - 9.65

80 - 11.48

90 - 12.33

Page 43: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 33

Following Equation (36), one can write

X = X0 + (47)

and Equation (46) is reduced to

dX0 dX1 ZB 2B

- CX- - a v-z- sin ý0 + - * •l cos 0(48)Vs Vs

Cancelling the flat earth terms, i. e.,

dXo ZB

- aX %in *0 (49)V

s

one gets for the spherical earth correction

dk1 2B co

dk -B (50)a = *- 1 Cos *0

Ss

or

dX cos 0 sin C0 + -0 + 4 d*0 (51)

The initial condition is -I = 0 at *o = 0. One finally gets by integrating

Equation (51)

= 3 5 I 3 1 2 i 0( ZXf - c + Cos f- - cos - cos C OS* *f 0f s *0 (52)

Equation (52) is plotted in Figure 10 and also tabulated in Table 3. Note that3x1 f, being negative, is proportional to C . Applying this correction, the

approximate spherical earth solution becomes

Page 44: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 34

4)0d0<

0 4 i

Z

0 bC

j7- 0

0 0

w" • IN 0

I 1.-44

o0-00

I- ~ 0

00

_____o o___ o. _o .1

0 0

0 C-?

Page 45: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 35

Table 3. Spherical Earth Correction on Lateral Range Angleand Lateral Range.

Of x 1f/C3 Lateral Range/C3

(deg) (radians) (naut mi)

0 0 0

10 - 0.00075 - .2.58

20 - 0.00170 - 5.85

30 - 0.00325 - 11.18

40 - 0.00730 - 25.1

50 - 0.0157. - 54.0

60 - 0.0266 - 91.5

70 - 0.0365 - 125.5

80 - 0.0458 - 157.5

90 - 0.0483 - 166.0

Page 46: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 36

S= C 1( - COS *0f) - C 3 + cos - Cos - Cos - sin . 0f

(53)

Equation (53) is plotted in Figures 5a, 5b and 5c as the dashed lines and is

compared with the exact numerical solutions from Reference 5. The comparison

shows that the corrected X expression can predict the lateral range very closely

for most cases. One interesting thing is that whenever the spherical earth cor-

rection is negligible in the exact numerical solutions, it is also negligible in the

approximate solutions. When the spherical earth correction is large, the

approximate solution deviates from the exact numerical solution no more than

10 per cent in the range compared. As for the maximum lateral kange, it has

been noted in the flat earth solutions that there is no maximum beyond a certain

V.. With the spherical earth correction, the maximum lateral range at low V.1 1

is not affected but a maximum exists at high Vi due to the correction. This fact

has been observed in the exact numerical solutions. It can now be concluded that

the maxima at low initial velocities are caused by the integrated lateral aero-

dynamic force while those at high initial velocities are caused by the spherical

earth effect. This indicates the importance of the spherical earth correction at

high initial velocities.

The spherical earth correction on the down range angle 4 is found as

follows. Taking Equation (1Z) and substituting *r by 90 ÷+ and X by x 0 +l

one gets

- - ?B cos (*0 + $1)

- = V---Zcos (X 0 + %s

or

dý _ ZB Cos{0cos - tan*O sinv-z o•sK00 CosX - tan XO sin (55)

s

Since and X are small, one can write

cos *1 = cos X, = 1, sin 1 = I sin X = Xl (56)

Page 47: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 37

T and Equation (55) becomes

dii= ZB cO ios ( 01-i tan ,0" -d = 2 0 Co -1 tan X 0) (57)

Since X is always small and X0 very rarely reaches beyond 45 degrees, one

can assume

XI tan 10 << 1 (58)

thus

d4 _ 2B cos 0(- =cos-X 0 (1 -1 tan i 0 ) (59)

Writing ýi = F + •1 and substituting in the above, one gets

2B (*1sin0 (60-aX =v-"z\,Cos 0

and by Equation (16),

sin= - i cosk0 (61)

0

Using the approximation cos X = 1 which has been proven adequate in com-

puting the down range, one gets

dl= - C sin to d*o (62)

or

= C 3 (sin *- 0) sin *0 dp0 (63)

Page 48: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 38

The initial condition is 41 = 0 at *0 0. Integrating Equation (63), one gets

C sin ýOf) sin3

(4[= C3 (1 + cos J0f) (*0f sin 0f g sin (0'Sf Of Of

Equation (64) is plotted in Figure 11 and also tabulated in Table 4. It is noted

that 4 if is positive and proportional to C 3. Using this correction, the approxi-

mate spherical earth solution for down range is

= C sin tOf + C3 (l I + cos $of) (•Of - sin Odf) - 1 sin 3 (65)

The corrected down range is plotted in Figures 8a, 8b and 8c as the dashed

lines. The comparison with exact numerical solutions is quite satisfactory.

/

Page 49: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 39

00

0.1,

2-1. u

"-4

U)

Nb

0CO w 0

C! O&J C2...

Page 50: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 40

Table 4. Spherical Earth Correction on Down Range Angleand Down Range.

*Of ý'if/IC3 Down Range/C3

(deg) (radians) (naut mi)

0 0 0

10 0. 00001 0.0344

ZO 0.00022 0.7560

30 0.00119 4.10

40 0.0045 15.50

50 0.0126 43.40

60 0. 0276 95.00

70 0.0507 174.20

80 0. 0823 283.00

90 0.1187 407.50

Page 51: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 41

V. CONCLUDING REMARKS

An analytic study has been made of the lateral motion of a re-entry vehicle

gliding at a constant altitude. The constant altitude is maintained by banking the

vehicle about the velocity vector at a constiant angle of attack. The analytic

approximate solutions have been compared with numerical exact solutions. It

has been shown that the approximate solutions predict higher turn angle, higher

lateral range but lower down range. The accuracy of the prediction is generally

satisfactory. Among the three variables, the lateral and down range can be pre-

dicted well within 10 per cent while the turn angle, with lower accuracy.

There are two important features about these approximate solutions. First,

they are simple and can be readily applied to preliminary design work. Secondly,

they preserve the characteristics and provide a good understanding of a constant

altitude glide.

The results of the approximate solutions show (and the exact numerical

solutions agree) that the lateral range, being a function of W/(CDA p) at a

given L/D and Vi, exhibits a maximum. The maximum exists for all initial

velocities in the case of a spherical earth. For a flat earth it exists at low

initial velocities only. The spherical earth correction has been shown to be

small for low W/(CDA p), L/D and Vi and become significant only when these

parameters assume high values.

Page 52: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 43

APPENDIX A

SUMMARY OF APPROXIMATE SOLUTIONS

I. TOTAL FLIGHT TIME (EXACT)

Lc C • I v sýt, •Vs I + - I-7tf D g~ _

where

C 2C L Ap V s

II. FINAL VELOCITY (EXACT)

VVf - s

'I C

III. FLAT EARTH SOLUTIONS

Basic approximation:

sin9 = I

Final turn angle:

V1Of = 1 In

Final lateral range angle:

xOf = C (1 - cos *!Of)

Final down range angle:

ý'of = C sin $Of

Page 53: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 44

C* for maximum lateral range:

I LL: (n ' -,J +arc tan Z

Maximum lateral range

V. -sin LIn jl +

k~f 1 + --

C*: for maximum down range

(_lL

L( L+ -= arc tan5 VsC*/

Maximum down range

*L iL [D V

Cos in (@-Isf-*

of TL

IV. SPHERICAL EARTH CORRECTION

Correction on final turn angle

-Cz (sin sin 2 *of)if5 C2( Of - *Of -

Correction on final lateral range

3(+1 s3 *f_1 2 1lf + c **Oc - " cos T - T of sin *Of

Correction on final down range

= C3 (1 + cos *of) (*Of - sin *of) -sin f0"lf f

Page 54: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 45

APPENDIX B

DISCUSSION OF THE APPROXIMATE SOLUTIONS

FOR VERY SMALL C+

It has been mentioned in the text that the sin @ = I approximation becomes

more accurate when C gets smaller. The text also shows that the spherical

earth correction becomes negligible when C is very small. Furthermore, the

lateral range angle X must be small for small C (X is proportional to C) and

as a result the cos X = 1 approximation used in computing the down range

angle becomes more accurate as C gets smaller. Therefore, for very small

C , Equations (10), (11) and (1Z) can be written, with high degree of accuracy,

as

Ad JL =-n7 (B-l)

dX _ ZBdX 2B sin , (B-Z)Vs

-dp. _ ZB- 2 cos (B-3)

s

The solutions of these equations are:

I= . n (B-4)

X = C (1 - cos ýJ) (B-5)

S= C sin d (B-6)

The author is grateful to Dr. Frank Billet for his helpful suggestions in thisdiscussion.

Page 55: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 46

Combining Equations (B-5) and (B-6), one gets

(X -C) + 2 = C(B-7)

Equation (B-7) represents a circle with its center at X = C and its radius

equal to C . Such a circle is plotted in Figure B-i. The tangent at any point

on the circumference of the circle makes an angle jr with the X = 0 line. The

line of X = 0 is along the original heading of the vehicle and if; must be the

turn angle. Figure B-i clearly indicates that the vehicle, while making a

constant altitude turn, describes a circular path. Depending on the aero-

dynamics, the mass characteristics and the altitude, the vehicle may find

itself anywhere along the circumference of the circle at the end of the turn.

The final turn angle has been given by Equation (21) as

1ilf = + in (B-8)

Here we drop the subs~ript 0 because Equation (B-8) can now be applied to a

spherical earth also. In the main text, most computations were made foruf < 90 0 . When C is very small, however, the final turn angle can be greater

than 90 degrees. If *f = 1800 , the vehicle would have made a lateral displace-

ment of 2C with no forward displacement and would fly from that instant on in

the opposite direction to the original heading. If IIIf = 360 , the vehicle would

have made a full 360-degree turn and find itself back to the original location at

a much lower velocity. In such case, the vehicle would be dissipating its

kinetic energy through drag while turning and it would have no loss in altitude

and no gain in either the longitudinal or lateral displacement at the end of the

turn.

While maneuvers of 180-degree and 36 0-degree turn appear to be extremely

interesting, the obvious question that has to be answered is whether or not it is

practical. To answer this question, we examine a vehicle with L/D = 22 oW/CLA = 25 lb/ft , at V. = 32, 000 ft/sec . In order to have *if = 1801

Page 56: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 47

the turn must be performed at an altitude of 184, 000 feet. For a 360-degree

turn, the altitude must be 104, 000 feet. The corresponding value of C is

0. 0704 and 0. 00282, respectively. The radius of the turn becomes 242

nautical miles and 9. 7 nautical miles, respectively. The centrifugal force

created by these turns is in the tens of g's for = 1800 and in the hundreds

of g's for #f = 360 . For lower L/D , higher W/CL A and lower Vi , it

becomes more difficult to perform such turns in the sense that the altitude will

be lower and the centrifugal force will be higher. Thus, one is led to conclude

at this time that constant altitude turns at very small values of C are not likely

to be practical.

The above discussion does clarify the statement made in the text about

the incorrectness of the curves of lateral range and down range at very small

W/(CDA p) as shown in Figures 5 and 8. As one can see from Figure B-1,

the lateral range and down range must have the characteristics of a damped

oscillation as W/(CDA p) goes to zero. This is schematically shown in

Figure B-Z. Because of the smallness of the region in which the damped

oscillation character occurs, it was not shown in Figures 5 and 8 to avoid

confusion.

Page 57: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 48

q, ISO*

27d' ° e0

p.00, 3600

ORIGINAL HEADINGOF VEHICLE

Figure B-1. Vehicle Flight Path at Very Small C

/w

0 CoAp

CoAp

Figure B-Z. Schematics of the Variation of X and ýiwith W/(CDA p) at Very Small C

Page 58: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 49

"NOMENCLATURE

A = reference area

B = abbreviation for W/(CDA p)

C = abbreviation for 2 B/(L/D) V2

s

CD = drag coefficient

D = drag force

g = gravitational acceleration

L = aerodynamic force perpendicular to drag

t = time

V = velocity

Vs = orbital velocity (= 26,000 ft/sec for earth)

W = vehicle weight

X = abbreviation for ln V

X = lateral range angle

i = down range angle

p = atmospheric density

0 = bank angle

= turn angle

Subs cripts

0 = "flat earth'" solution

I = spherical earth correction

f = at the end of the constant altitude glide

i = initial condition

Superscripts

* = maximum

Page 59: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

Page 50

REFERENCES

1. Slye, R. E., "An Analytical Method for Studying the Lateral Motion ofAtmosphere Entry Vehicles", NASA TN D-325, September 1960.

2. Smith, R. H. and J. A. Menard, "Superorbital Entry and Recovery withManeuverable Manned Vehicles", Aerospace Engineering, p 12,October 1961.

.3. Ferri, A. and L; Ting, "Practical Aspects of Re-entry Problems",Polytechnic Institute of Brooklyn, PIBAL Report No. 705, July 1961.

4. Wang, H. E. and S. T. Chu, "Variable Lift Re-entry at Superorbital andOrbital Speeds", Presented at IAS Summer Meetings, Los Angeles,Paper No. 62-165, 19-22 June 1962.

5. Skulsky, R.S., "Study of a Particular Lateral Range Re-entry Maneuver",Martin Company, TM 1364-61-16, August 1961.

6. Wang, H.E. and R.S. Skulsky, "Characteristics of Lateral Range DuringConstant Altitude Glide", to be published in the Journal of AerospaceSciences.

7. London, H.S., "Change of Satellite Orbit Plane by Aerodynamic Maneuver-ing", Journal of Aerospace Sciences, Vol 29, No. 3, p 323, March 1962.

8. Love, J. A. and L. W. Neustadt, "Derivation of Three-Dimensional Trajec-tory Equations for a Lifting Re-entry Vehicle", Space TechnologyLaboratories, Interoffice Correspondence 7431.1-107, 9 November 1960.

9. Wang, H-.E.- --"-On th-e•-Three-Dimensional Equations of Motion of a Lifting.Re-entry Vehicle Over a Non-Rotating Spherical Earth", AerospaceCorporation Interoffice Correspondence A62-1761-225, 8 November1962.

10. Skulsky, R.S., Aerospace Corporation Report to be published.

11. Jackson, W.S., "An Improved Method for Determining the Lateral Rangeof a Gliding Entry Vehicle", Reader's Forum, Journal of AerospaceSciences, Vol 28, No. 11, p 910, November 1961.

12. London, H.S., "Comments on Lateral Range During Equilibrium Glide",Reader's Forum, Journal of Aerospace Sciences, Vol 29, No. 5,p 610, May 1962.

Page 60: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

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Page 61: UNCLASSIFIED AD 40 2 019 - DTICCDA d -gV2 (1) V 2 /L, v 5cos 0 = -1 D 2W Ap - 1F(V22) (2) d$ (L• Vg sin 0 Vg cos p tan X (3) dt - =D- 1 2W V 2 (C*A p; Vs Here 0 is the bank angle

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