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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General Ballistic Missile Problem
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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General Ballistic Missile Problem
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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Free-flight Range Equation
Figure 2: Symmetrical Geometry
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Free-flight Range Equation
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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Free-flight Range Equation
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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Free-flight Range Equation
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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Free-flight Range Equation
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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Flight path Angle Equation
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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Flight path Angle Equation
Figure 4:
Triangle formed
from Ellipse Geometry
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Flight path Angle Equation
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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Flight path Angle Equation
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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Maximum Range Trajectory
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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Maximum Range Trajectory
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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Effect of burnout flight-path angle errors on range
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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Effect of burnout flight-path angle errors on range
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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Effect of burnout flight-path angle errors on range
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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Effect of burnout flight-path angle errors on range
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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Effect of burnout flight-path angle errors on range
(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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Compensating for the Initial Velocity of the Missile
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
Compensating for the Initial Velocity of the Missile
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Compensating for the Initial Velocity of the Missile
Figure 6:True velocity and Direction at burnout
Compensating for Movement of the Target
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Compensating for Movement of the Target
Figure 7:Launch site and aiming point at the instant of launch
Compensating for Movement of the Target
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Compensating for Movement of the Target
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
Compensating for Movement of the Target
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Compensating for Movement of the Target
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.