ballistic missile trajectories

7/29/2019 Ballistic Missile Trajectories

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Ballistic Missile

Trajectories


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Ballistic Missiles

Ballistic missiles are used for transportation of a

payload from one point on the Earth (launch site) toanother point on the surface of the Earth (impact

point or target). They are accelerated to a high

velocity during a relatively short period.

Then a re-entry vehicle, containing the warhead,

is released and this vehicle then simply coasts in a

ballistic or free-fall trajectory to the final impact

point.


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Ballistic Missile Trajectory

The trajectory of a missile differs from a satellite

orbit in only one respect

it intersects the surfaceof the Earth. Otherwise, it follows a conic orbit

during the free-flight portion of its trajectory.


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General Ballistic Missile Problem

A ballistic missile trajectory is composed of three

parts: the powered flight portion which lasts from

launch to thrust cutoff or burnout,

the free-flight portion which constitutes mostof the trajectory, and

the re-entry portion which begins at some ill-

defined point where atmospheric drag

becomes a significant force in determining the

missile's path and lasts until impact.


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Since energy is continuously being added to the

missile during powered flight, we cannot use 2-body mechanics to determine its path from launch

to burnout. The path of the missile during this

critical part of the flight is determined by the

guidance and navigation system.

During free-flight, the trajectory is part of a conic

orbit

almost always an ellipse

which we alreadyanalyzed.

Re-entry involves the dissipation of energy by

friction with the atmosphere.


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Figure 1:Geometry of a Ballistic

Missile Trajectory


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Non-dimensional Parameter Q

Here we need to define a non-dimensional

parameter Q as

(1)

From the energy equation , we can

prove

(2)

22

cs

v v rQ

v

2

2 2

v

r a

22

r ra or Q

Q a


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Free-flight Range Equation

Since the free-flight trajectory of a missile is a

conic section, the general equation of a conic canbe applied to the burnout point.

(3)

Solving for ,we get

(4)

1 cosbo bo

p

r e

cosbo

cos bobo

bo

p r

r e


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Figure 2: Symmetrical Geometry


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Since the free-flight trajectory is assumed to be

symmetrical ( ), half the free-flight rangeangle, , lies on each side of the major axis, and

(5)

Therefore equation (4) can be written as

(6)

Equation (6) is an expression for the free-flight

range angle in terms ofp, e, and .

bo reh h

cos cos2 bo

cos2

bo

bo

r pr e

bor


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Since and , we can use the

definition of parameter Q to obtain

(7)

Now, since (8)

Substituting equations (2) and (7) in equation (8),

we get

(9)

2p h cosh rv

2 2 2

2coscos

r vp r Q

2 21 , 1 pp a e ea

2 21 cos 2e Q Q


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Now substituting equations (7) and (9) into

equation (6) we have one form of the free-flightrange equation:

(10)

Given a particular launch point and target, the

total range angle, , can be calculated (We will seelater). If we know how far the missile will travel

during powered flight and re-entry, the required

free-flight range angle,, also becomes known.

2

2

1 coscos

2 1 cos 2

bo bo

bo bo bo

Q

Q Q


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If we now specify and for the missile, what

should the flight-path angle, , be in order that themissile will hit the target?

In other words, it would be nice to have an

equation for in terms of , and .

So we need to consider a geometry shown in

Figure 3 to derive an expression for flight-pathangle equation.

bor

bov

bo

bo bor bov


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Flight path Angle Equation

Figure 3:

Ellipse Geometry


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From Figure 3, it can be proven that the angle

between and is bisected by the normal. Thissimply gives that the angle between and is

.

Let us concentrate on the triangle formed by

and the burnout point. Let us divide the

triangle into two right triangles by the dashed line,

d, as shown in Figure 4.

bor borbo

r bor

2 bo

,F F


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Figure 4:

Triangle formed

from Ellipse Geometry


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Now From Figure 4, we can express das

(11)

and also as

(12)

Combining the two equations (11) and (12), weget

(13)

sin2

bod r

sin 180 22

bo bod r

sin 2 sin

2 2

bo

bo

bo

r

r


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Since and from equation (2),

, we can write equation (13) as

(14)

Equation (14) is called the flight path angle

equation.

2bo bo

r r a

2bo bor a Q

2sin 2 sin

2 2

bo

bo

bo

Q

Q


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Maximum Range Trajectory

Figure 5: Range vs flight-path angle


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To derive expressions for the maximum range

condition, a simpler method is to see under what

conditions the flight-path angle equation yields a

single solution.

If the right side of equation (14) equals exactly 1,

we get only a single answer for . This must, then,be the maximum range condition.

(15)

(16)

for maximum range conditions only.

bo

2

sin 2 sin 12 2

12 90 180

2 4

bo

bobo

bo bo

Q

Q


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From equation (15) we can easily find the

maximum range angle attainable with a given .

(17)

for maximum range conditions.

Solving for , we get

(18)

for maximum range conditions.

boQ

sin2 2

bo

bo

Q

Q

boQ

2sin 2

1 sin 2bo

Q


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Time of Free-flight


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Time of Free-flight

The value of eccentric anomaly can be computed

by taking as

(19)

And the time of free-flight can be obtained from

(20)

1180 2

1

cos 2cos

1 cos 2

eE

e

3

1 12 sin

ff

at E e E


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Effect of launching errors

on Free-flight Range


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Effect of Launching errors on Range

Variations in the speed, position, and launch

direction of the missile at thrust cutoff will produceerrors at the impact point.

These errors are of two types errors in the

intended plane which cause either a long or a short

hit, and out-of-plane errors which cause the missile

to hit to the right or left of the target.

We will refer to errors in the intended plane as

"down-range" errors, and out-of-plane errors as

"cross-range" errors.


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Effect of Lateral Displacement of the Burnout

Point

If the thrust cutoff point is displaced by an

amount, DX, perpendicular to the intended plane of

the trajectory and all other conditions are nominal,

the cross-range error, DC, at impact can bedetermined from spherical trigonometry.


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Effect of Lateral Displacement of the Burnout

Point

arc length DC= cross-range error


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Effect of Lateral Displacement of the Burnout Point

Applying the law of cosines for spherical

trigonometry to triangle OAB in previous diagramand noting that the small angle at O is the same as

DX, we get

(21)

Since both DXand DCwill be very small angles,

we can use the small angle approximation,

to simplify equation (21) to

(22)

2 2cos sin cos cosC XD D

2cos 1 / 2x x

cosC XD D


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Cross-range Error due to Incorrect Launch Azimuth

If the actual launch azimuth differs from the

intended launch azimuth by an amount, Db, a cross-range error, DC, will result.


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Cross-range Error due to Incorrect Launch Azimuth

From the law of cosines for spherical triangles we

get(23)

If we assume that both

Dband

DC will be very

small angles, we can use the small angle

approximation, to simplify equation

(23) to

(24)

2 2cos cos sin cosC bD D

2cos 1 / 2x x

sinC bD D

Eff t f D R Di l t f th


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Effect of Down-Range Displacement of the

Burnout Point

An error in down-range position at thrust cutoffproduces an equal error at impact.

If the actual burnout point is 1 nm farther down-

range than was intended, the missile will overshoot

the target by exactly 1 nm.


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Effect of burnout flight-path angle errors on range

In the above graph D will represent a down-

range error causing the missile to undershoot or

overshoot the target.


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A good approximate value for D for very small

values of is given by

(25)

where is the slope of the curve at the point

corresponding to the intended trajectory.

boD

bo

bo

D D

bo


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The expression for may be obtained by

implicit partial differentiation of the free-flight

range equation.

The free-flight range equation can be converted

into an alternate form for the simple differentiation.

Recall the free-flight range equation

bo

2

2

1 coscos

2 1 cos 2

bo bo

bo bo bo

Q

Q Q


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Let us consider the numerator of equation (10) as

aand denominator asb.

Then .

Substituting for aandbwe get

But , therefore,

2 2cos cot

2 2and

a a

b b a

21 cos sin

bo bo

2

2

1 coscot

2 cos 1 cos

bo bo

bo bo bo

Q

Q

21 cos

cot

2 cos sin

bo bo

bo bo bo

Q

Q


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Since we can further

simplify to obtain

Now express the above equation in terms ofand ,

(26)

Now we can differentiate equation (26) implicitly

with respect to , considering as constants.

sin 2 2cos sin ,x x x

2cot csc 2 cot

2bo bo

boQ

,bo bor vbo

2

2cot csc 2 cot

2bo bo

bo bov r

bo ,bo bor v


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(27)

This partial derivative is called an influence

coefficient since it influences the size of the rangeerror resulting from a particular burnout error.

Therefore the free-flight range error due to

burnout flight-path angle error is given by

(28)

2sin 22sin 2

bo

bo bo

2sin 22

sin2

bo

bo

bo

D D

Down-Range Errors caused by Incorrect Burnout


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Down-Range Errors caused by Incorrect Burnout

Height

Again a good approximate value for D for very

small values of is given by

Again differentiating the equation (26) implicitly

with respect to , and solving for , we get

(29)

borD

bo

bo

r

r

D D

bor

bor

2

2 2

sin4 2

sin 2bo bo bo bor v r

Down-Range Errors caused by Incorrect Speed at


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Down-Range Errors caused by Incorrect Speed at

Burnout

A good approximate value for D for very small

values of is given by

Again differentiating the equation (26) implicitly

with respect to , and solving for , we get

(30)

bovD

bo

bo

vv

D D

bov

bov

2

3

sin8 2

sin 2bo bo bo bov v r

T l D R E


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Total Down-Range Error

Now the total down range-error is given by

TOTAL bo bo bo

bo bo bo

r vr v

D D D D

Eff t f E th R t ti


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Effect of Earth Rotation

The Earth rotates once on its axis in 23 hrs 56 min

producing a surface velocity at the equator ofapprox 465 m/sec (or 1524 ft/sec). The rotation is

from west to east.

The free-flight portion of a ballistic missile

trajectory is inertial in character. That is, it remains

fixed in the XYZ inertial frame while the Earth runs

under it. Relative to this inertial XYZ frame, both the

launch point and the target are in motion. Thus we need to compensate for motion of the

launch site and the motion of the target due to

earth rotation.

C ti f th I iti l V l it f th Mi il


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Compensating for the Initial Velocity of the Missile

We can express the speed of any launch point onthe surface of the earth as

Vo = 1524 cos L0 (ft/sec)

Compensating for the Initial Velocit of the Missile


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The north, east, and up components of the true

velocity vcan be obtained as

Now the true velocity, flight-path angle, and

azimuth can then be found from

0

cos cos

cos sin

sin

N e e e

E e e e

Z e e

v v

v v v

v v

b

b

2 2 2

sin ; tan

N E Z

Z E N

v v v v

v v v v b



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Figure 6:True velocity and Direction at burnout

Compensating for Movement of the Target


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Figure 7:Launch site and aiming point at the instant of launch



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The latitude and longitude coordinates of the

launch point are , respectively, so the arc

length OA in Figure 7 is just . If the coordinates

of the target are then the latitude and

longitude of the aiming point should be and

respectively. The term, , representsthe number of degrees the Earth turns during the

time . The angular rate, , at which the earth

turns is approximately 15 deg/hr. Arc length OB is simply , and the third side

of the spherical triangle is the ground trace of the

missile trajectory which subtends the angle

.

0 0L and N0

90 L

,t tL N

tL

tN t t

t

90t

L



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The angle formed at O is just the difference in

longitude between the launch point and the aiming

point, , where DN is the difference in

longitude between launch point and target.

By considering the launch azimuth, b in the

spherical triangle,

N t D

0 0cos sin sin cos cos cost tL L L L N t D

0 0

0

0

sin sin cos cos sin cos

sin sin coscos

cos sin

t

t

L L L

L L

L

b

b

T b k( )


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Textbook(s)

Roger R. Bate, Donald D. Mueller, Jerry E. White,

Fundamentals of Astrodynamics, Dover Publications, 1971.

ballistic missile trajectories

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