Strategic Missile Zarchan

7/27/2019 Strategic Missile Zarchan

1/75

AMITY UNIVERSITY, UTTAR PRADESH

AMITY INSTITUTE OF SPACE SCIENCE AND

TECHNOLOGY

SUMMER INTERNSHIP AT IIT BOMBAY

LAMBERT GUIDANCE FOR

MISSILES AND SPACECRAFTS

Submitted in partial fulfilment of Dual Degree

(B.TECH AEROSPACE+M.TECH AVIONICS)

NAVUDAY SHARMA, A4717210001

JAYRAJ INAMDAR, A4717210010

PRITAM KUMAR PRATIHARI, A4717210016

7/27/2019 Strategic Missile Zarchan

2/75

2 | P a g e

ACKNOWLEDGEMENT

Words would not be sufficient to express the gratitude that we feel towards Prof HARI B.

HABLANI, FACULTY, INDIAN INSTITUTE OF TECHNOLOGY BOMBAY, and project guide, for

guiding us, understanding our needs and being very considerate. We also thank him for his

constant intellectual guidance and support.

We also owe a debt of gratitude towards our faculty guide, Prof AK VERMA, FACULTY, AMITY

INSTITUTE OF SPACE SCIENCE AND TECHNOLOGY. Apart from his intellectual guidance and

expert opinion, he gave us freedom and encouragement to think creatively and perform

consistently.

We thank each other for staying together through rough times and for showing immense interest

in completion of the project.

We would also thank our individual families for the constant support and encouragement,

unconditionally at every phase.

Last, but not the least, we thank the lord almighty for his blessings in our life.

7/27/2019 Strategic Missile Zarchan

3/75

3 | P a g e

INDEX

7/27/2019 Strategic Missile Zarchan

4/75

4 | P a g e

LIST OF FIGURES

7/27/2019 Strategic Missile Zarchan

5/75

5 | P a g e

7/27/2019 Strategic Missile Zarchan

6/75

6 | P a g e

ABSTRACT

Chapter 1: Strategic Considerations

This chapter introduces strategic interceptor concepts from a tactical point of view. A

gravitational model, based on Newtons law of universal gravitation is developed for strategic

flight. Comparisons between flat earth gravitational model and earth centered coordinate

system are made. Although strategic engagement simulation models are presented in the

text in a Cartesian Earth-centered coordinate system, a polar coordinate system is also

introduced so that important closed-form solutions can be derived. Key formulas for velocity

and flight time for an impulsive ballistic missile to travel a fixed distance, given an initial flight

path angle are developed using the polar coordinate system. Cartesian Earth-centered

coordinate system is used to confirm the results through MATLAB simulations.

Chapter 2: Boosters

This chapter shows how preliminary strategic booster sizing can be done with rocket

equation. Rocket equation is also derived. Simplified booster sizing examples are presented

in order to clarify the concepts. The rocket equation is extended to the virtues of staging.

Finally gravity turn maneuver is introduced as the simplest possible steering method a

booster can employ in traveling from its launched point to the desired destination.

Chapter 3: Lambert Guidance

Lambert Guidance introduces different phases of flight and is devoted to two body problem.

Also lambert theorem is discussed in this chapter. Starting from the closed form solutions

derived in chapter 1, the concepts of lambert steering, are also developed in this chapter. Asimple to understand but numerically inefficient way of solving Lamberts problem is derived.

A numerical example is presented showing how the numerical solution to Lamberts problem

be implemented. A novel use of the secant method is demonstrated to speed up the solution

to Lamberts problem by more than two orders of magnitude. It is then shown how the

implemented solution can be modified with a simple feedback scheme to steer an

interceptor, during its boost phase, to its intended target. Another subset of Lambert steering

known as general engineering management (GEM) steering, is also derived and

demonstrated. A numerical example highlighting the similarities and differences between

Lambert and GEM is presented.

Chapter 4:Orbital Mechanics

This chapter includes brief details about space flight. It deals with orbital mechanics by

deriving the orbit equation. Then it discusses about the peri-focal and celestial frames. It also

gives some information about the time equation and orbital transfer. In the end, we discussed

about lamberts theorem and lamberts problem for space crafts.

7/27/2019 Strategic Missile Zarchan

7/75

7 | P a g e

PRE-REQUISITE

In a modern military, a missile is a self-propelled guided weapon system, as opposed to an

unguided self-propelled, referred to as just a rocket (weapon) . Missiles have four system

components: targeting and/or guidance, flight system, engine, and warhead. Missiles come in

types adapted for different purposes: surface-to-surface and air-to-surface

missiles (ballistic, cruise, anti-ship, anti-tank, etc.), surface-to-air missiles (anti-aircraft and anti-

ballistic), air-to-air missiles, and anti-satellite missiles. All known existing missiles are designed to

be propelled during powered flight by chemical reactions inside a rocket engine, jet engine, or

other type of engine. Non-self-propelled airborne explosive devices are generally referred to

as shells and usually have a shorter range than missiles.

Missiles are generally categorized by their launch platform and intended target. In broadest

terms, these will either be surface (ground or water) or air, and then sub-categorized by range

and the exact target type (such as anti-tank or anti-ship). Many weapons are designed to be

launched from both surface and the air, and a few are designed to attack either surface or air

targets. Most weapons require some modification in order to be launched from the air or surface,such as adding boosters to the surface-launched version.

Ballistic missile

A ballistic missile is a missile that follows a ballistic flight path with the objective of delivering one

or more warheads to a predetermined target. Shorter range ballistic missiles stay within

the Earth's atmosphere, while longer range ones are designed to spend some of their flight time

above the atmosphere and are thus considered sub-orbital.

Ballistic missiles can vary widely in range and use, and are often divided into categories based on

range. Various schemes are used by different countries to categorize the ranges of ballisticmissiles:

Tactical ballistic missile: Range between about 150 km and 300 km Battlefield range ballistic missile (BRBM): Range less than 100 km Theatre ballistic missile (TBM): Range between 300 km and 3,500 km Short-range ballistic missile (SRBM): Range 1,000 km or less Medium-range ballistic missile (MRBM): Range between 1,000 km and 3,500 km Intermediate-range ballistic missile (IRBM) or long-range ballistic missile (LRBM): Range between

3,500 km and 5,500 km

Intercontinental ballistic missile (ICBM): Range greater than 5500 km Submarine-launched ballistic missile (SLBM): Launched from ballistic missile submarines (SSBNs),

all current designs have intercontinental range.

7/27/2019 Strategic Missile Zarchan

8/75

8 | P a g e

Tactical Ballistic Missile:

A Tactical Ballistic Missile is a ballistic missile designed for short-range battlefield use. Typically,

range is less than 300 kilometers (190 mi). Tactical ballistic missiles are usually mobile to ensure

survivability and quick deployment, as well as carrying a variety of warheads to target enemy

facilities, assembly areas, artillery, and other targets behind the front lines. Warheads can include

conventional high explosive, chemical, biological, or nuclear warheads. Typically tactical nuclear

weapons are limited in their total yield compared to strategic rockets.

Short-range ballistic missile:

A Short-range Ballistic Missile (SRBM) is a ballistic missile with a range of about 1,000 km or less.

They are usually capable of carrying nuclear weapons. In potential regional conflicts, these

missiles would be used because of the short distances between some countries and their relative

low cost and ease of configuration. In modern terminology, SRBMs are part of the wider grouping

oftheatre ballistic missiles, which includes any ballistic missile with a range of less than 3,500 km.

Intercontinental ballistic missile

An Intercontinental Ballistic Missile (ICBM) is a ballistic missile with a range of more than 5,500

kilometers (3,400 mi) typically designed for nuclear weapons delivery (delivering one or

more nuclear warheads). Most modern designs support multiple independently targetable

reentry vehicles, allowing a single missile to carry several warheads, each of which can strike a

different target.

Similarly other types of missiles have other agendas.

In the simulations done in MATLAB in the report, ICBMs and SRBMs are considered. Also note

that the simulations are done in FPS UNITS.

INDIAN SRBM (PRITHVI) INDIAN ICBM (AGNI-V)

7/27/2019 Strategic Missile Zarchan

9/75

9 | P a g e

CHAPTER 1 Strategic Consideration

This section introduces strategic interceptor concepts from a tactical point of view. A

gravitational model, based on Newtons law of universal gravitation, is developed for strategic

flight. Comparisons between a flat-Earth gravitational model and a strategic gravitational model

are made. Although strategic engagement simulation models are presented in the text in a

Cartesian Earth-centered coordinate system, a polar coordinate system is also introduced so that

important closed-form solutions can be derived. Key formulas for velocity and flight time for an

impulsive ballistic missile to travel a fixed distance, given an initial flight-path angle, are

developed using the polar coordinate system. A Cartesian Earth-centered simulation is used to

confirm the analytic results.

For tactical interceptor missions, where speeds are less than 5000ft/s, altitudes under 100kft,and ranges covered under 100 nautical miles(n.mi.), missiles are designed based on the flat-

Earth constant-gravity model. However, in the strategic world where speeds are near orbital and

the distances covered are intercontinental (for ICBMs), the flat-Earth constant-gravity

assumption is inaccurate.

1.1) Flat-Earth Constant Gravity Model:

1.1) Flat-Earth Constant Gravity Model

In the tactical world, in the absence of thrust, drag and the lift, the flat-Earth constant-gravity

assumption is easy to understand. In this mathematical model the gravitational acceleration is

independent of altitude with value 32.2ft/s2, always in the downward direction. The tactical

missile inertial coordinate system is fixed to the surface of flat earth and is depicted in in FIG1.1.

7/27/2019 Strategic Missile Zarchan

10/75

10 | P a g e

Here the missile has velocityV and is at flight-path angle with respect to the surface of the

Earth. In addition the missile is in a constant-gravity field with acceleration level g. The missile

is at an initial location that is distance downrange from the origin of the coordinate systemand at an altitude from the surface of the Earth.The differential equations acting on the missile are:

= 0 = = =

Where V is velocity and R is range. The downrange component is denoted by 1, and the altitude

component is denoted by 2. The initial conditions for velocity and position are given by

0 = 0 =

0 =

0 =

Since the coordinate system is inertial, we can integrate directly in the downrange and altitude

directions to get the velocity from acceleration and position from velocity. In this model,

gravitational acceleration is always 32.2ft/s2 in the downward direction regardless of the altitude.

Therefore, this model is only valid for lower altitudes, since the actual gravitational acceleration

decrease with increase in altitude.

7/27/2019 Strategic Missile Zarchan

11/75

11 | P a g e

1.2) Earth-Centered Coordinate System:

1.2) Earth-Centered Coordinate System

In general, a body in a gravitational field can be depicted in an Earth centered coordinate systemshown in figure 1.2. In this system the Earth is non-rotating and the gravitational acceleration

acting on the missile is toward the center of the Earth. Also actual gravitational acceleration

decrease with increase in altitude.

The missile has velocity V with respect to a reference that is tangent to the Earth and

perpendicular to r(the line from center of Earth to missile). The radius of the Earth is denoted

by a in this figure.

According to Newtons law of universal gravitation, two bodies attract each other with a force

that acts along a line connecting the two bodies. The force is proportional to the product of the

masses of the two bodies and inversely proportional to the square of the distance between them.

If one of the bodies is the Earth and the mass of the second body is negligible compared to the

Earth, Newtons law of universal gravitation can be expressed as

=

Where r is a vector form the center of the Earth to the second body, and is known as thegravitational parameter with the value

= 1.4077 10 / For simulation purposes, Cartesian coordinates are required.

7/27/2019 Strategic Missile Zarchan

12/75

12 | P a g e

= Now Newtons law of gravitation in Cartesian coordinates are:

=

.

= . where and are component distances from the center of the Earth to the body or missile.From figure 1.2 the initial conditions for the preceding differential equation are:

0 = 0 = 0 = 2

0 = 2 where is the initial missile velocity, alt0 the initial missile altitude with respect to the surface ofthe Earth,

the angle the velocity vector makes with respect to the reference, and

the initial

angular location of the missile with respect to the x axis.

Velocity and position components, with respect to the center of the Earth, can be found from

repeated integration of the preceding differential equation. The instantaneous altitude of the

missile can be found by finding the distance from the center of the Earth to the missile and then

subtracting the Earths radius,

= .

We can find the distance traveled along the surface of Earth from FIG. 1.3. The initial location of

the missile can be expressed in vector notation as:

= And the future location of the missile at any arbitrary time can be expressed as:

7/27/2019 Strategic Missile Zarchan

13/75

13 | P a g e

= The angle between the two vectors and can be found from the defination of the vector dotproduct:

= c o s ||||

1.3) Total Distance Travelled

Therefore the distance traveled, which is projected on the surface of a circular Earth is given by

=

For comparitive purpose, a simulation is done between flat-Earth constant-gravity model and

Earth-centered coordinate system using Newtons law of universal gravitation. In the Earth-

centered coordinate system, the missile position is converted to a downrange and altitude so

that a trajectory comparison can be made with flat-Earth model.

7/27/2019 Strategic Missile Zarchan

14/75

14 | P a g e

1.4) Both the gravitational model yield the same result when the missile velocity is small

Above simulation was done for an impulsive missile at 45 degrees initial launch angle and 3000

ft/s initial velocity. The above figure shows that the Flat-Earth model (valid for tactical missile)

and the Earth-centered coordinate system model (valid for a strategic missile) yield the same

missile trajectories. The total range traveled in both cases is about 47 n.mi., and the maximum

altitude is about 12 n.mi.

0 5 10 15 20 25 30 35 40 450

2

4

6

8

10

12

Downrange (Nmi)

Altitude(Nmi)

3000 Ft/s INITIAL VELOCITY

45 DEG LAUNCH

FLAT -EART H CONST ANT -GRAVIT Y MODEL

EART H-CENTERED MODEL

7/27/2019 Strategic Missile Zarchan

15/75

15 | P a g e

1.5) Flat-Earth Model is still fairly accurate when missile speed is doubled

Again same simulation is carried out at same initial launch angle but initial speed of the impulsive

missile is doubled(6000 ft/s). As the speed is doubled, some differences in the resultant

trajectories occur. In this case the missile travels about 180 n.mi. and the maximum altitude

reached is about 50 n.mi.

The correct answers are the one given by Earth-centered coordinate system. However, in above

case, the flat-Earth approximation (i.e. constant gravity model) is fairly accurate.

0 20 40 60 80 100 120 140 160 180 2000

10

20

30

40

50

60

Downrange (Nmi)

Altitude(Nmi)

6000 Ft/s INITIAL VELOCITY

45 DEG LAUNCH

FLAT -EART H CONST ANT -GRAVIT Y MODEL EA RT H-CENT ERED MODEL

7/27/2019 Strategic Missile Zarchan

16/75

16 | P a g e

1.6) Flat-Earth Model is not accurate when the missile speed is doubled further

The above figure shows that when the impulsive missile speed is again doubled to 12000 ft/s, the

flat-Earth model yields large discrepancies in the resultant missile trajectory. The missile actually

travels much farther than the downrange indicated by flat-Earth model because the gravitational

acceleration is reduced at the higher altitudes according to Newtons law of universal gravitation.

In this case the distance traveled is more than 800 n.mi and the peak altitude is about 220 n.mi.

0 100 200 300 400 500 600 700 8000

50

100

150

200

250

300

350

400

Downrange (Nmi)

Altitude(Nmi)

t s

45 DEG LAUNCH

EA RT H-CENT ERED MODEL

FLAT -EART H CONST ANT -GRAVIT Y MODEL

7/27/2019 Strategic Missile Zarchan

17/75

17 | P a g e

1.3) Polar Coordinate Systems:In previous sections, we have simulated tactical and strategic missile trajectories in flat-Earth

constant gravity model and Earth-centered coordinate system. The differential equations

representing Newtons law of universal gravitation were first presented in vector form and then

converted for simulation purposes to an Earth-centered Cartesian coordinate system. Earth-

centered coordinate system is extremely useful for simulation work because all integration can

be done directly in the inertial frame. However to get closed form solutions, it is more convenient

to work analytically in a polar coordinate system. FIG 1.7 displays the polar coordinate system.

1.7) Polar Coordinate System with missile in a gravity field

In the above figure, we have defined a moving coordinate system that has the missile at the

origin. The new coordinate system has an i component along the distance vector and a

jcomponent perpendicular to r. The relationship between the inertial Earth-centered coordinate

system and the moving coordinate system is depicted in FIG 1.8.

The relationship between the fixed and moving coordinate frames can be expressed

mathematically as

=cos sin = sin cos Since the new coordinate system is moving, we can express its rate of change with respect to the

polar angle . Differentiating the preceding expressions with respect to the polar angle yields

7/27/2019 Strategic Missile Zarchan

18/75

18 | P a g e

=sincos =cossin

Now the rate of change of the new coordinate system as a function of time would be

= = = =

The distance vector rcan be expressed in the moving coordinate system as

=

1.8) Relationship between fixed and moving coordinate system

Taking the derivative of the previous equation yields

= Now substituting value of in above equation, we get

= Taking another derivative gives,

= 2

7/27/2019 Strategic Missile Zarchan

19/75

19 | P a g e

We know that gravitational acceleration is along i and there is no acceleration along j. Therefore,

the preceding vector differential equation can be expressed as the following two scalar

differential equations:

=

0 = 2

Since,

= 2 Therefore,

= 0Integrating yields a constant of integration that must be a moment arm times a tangential

velocity, or

= cosIn summary, the differential equations describing a missile in a gravity field can be expressed in

polar coordinates as

= 0And,

= cos Where the initial conditions are,

0 = 0 = 00 = sin

7/27/2019 Strategic Missile Zarchan

20/75

20 | P a g e

1.9) Polar and Earth-centered gravity field equations yield identical trajectories

A simulation is shown to demonstrate that both polar and cartesian Earth-centered differential

equations are equivalent. A impulsive missile with an initial velocity of 24,000 ft/s was launched

from the surface of the Earth at an angle of 45 degree with respect to the reference. Downrange

covered by the misslile is approximately 4500 n.mi. and reached at an altitude of 1700 n.mi.

1.4) Closed Form SolutionIn this section we will solve the previously derived differential equations of the Newtons law of

universal gravitation expressed in polar coordinates. In other words, we seek to find closed-form

solutions of the polar differential equations.

= 0 = cos For convenience, let us define constants r0 andp such that

0 =

0 500 1000 1500 2000 2500 3000 3500 40000

200

400

600

800

1000

1200

1400

1600

1800

Downrange (Nmi)

Altitude(Nmi)

24000 Ft/s INITIAL VELOCITY

45 DEG LAUNCH

POLAR

CART ESIAN

7/27/2019 Strategic Missile Zarchan

21/75

21 | P a g e

= cos = 0 cos

In addition, we will define an inverse range to be

= 1The goal is to convert both polar differential equations to one second-order differential equation

in terms ofu. First we know that from chain rule u varies with time according to

= = = An alternate way of seeing how u changes with respect to time is

= = 1 Equating both expressions yields,

= Next we define z to be

= Using the chain rule to see how z changes with respect to time yields

= = [] = [ ]Therefore, we can say that

= = Substitution allows us to rewrite the second-order differential equation in range as

= 0 = Simplification yields,

7/27/2019 Strategic Missile Zarchan

22/75

22 | P a g e

= Here a new constant is defined:

=

We can now summarize the transformed range polar differential equation to be

= 1 cos The original initial conditions on the polar differential equations were,

0 = 0

0 = sin Since we already know that

= 1 = 1cos We can say that,

0 = tan The other initial condition is simply,

0 = 10 = 1The solution to the preceding second-order differential equation is,

= sin cos 1 cos where A and B can be found from the initial conditions. After the complete algebra, we obtain

the complete solution in terms ofu as

7/27/2019 Strategic Missile Zarchan

23/75

23 | P a g e

= 1 c o s cos cos cos However since,

= 1

The solution in terms ofrbecomes

= 1 c o s cos cos cos Or more conveniently,

= cos

1 c o s c o s c o s = cos 1 s i n c o s s i n c o s 1 c o s Thus, given missile altitude(r0), velocity() and flight path angle , we find the missile location r

as a function of the central angle .The preceding closed form solution is also the equation of anellipse in polar coordinate system.

The equation of ellipse in polar form is given as,

= 1 1 c o s = 1 1 s i n s i n c o s c o s

Where is the semimajor axis, e is the eccentricity, and is the argument of apogee. Thetrajectory equation for an ellipse are equivalent if,

s i n = c o s s i n c o s = 1 c o s

7/27/2019 Strategic Missile Zarchan

24/75

24 | P a g e

On squaring and adding the above equations, we get an expression for eccentricity in terms of, = 1 2cos.

The trajectory equation yields a circle ife=0, an ellipse if 0

7/27/2019 Strategic Missile Zarchan

25/75

25 | P a g e

1.11) Setting =1.5 yields elliptical orbit

1.12) Setting =2 results in parabolic trajectory for missile

7/27/2019 Strategic Missile Zarchan

26/75

26 | P a g e

1.13) Setting to small results in orbit that intersects earth1.5) HIT EQUATION:

We can get closed-form solutions from the trajectory equation, which, given an initial missile flight pathangle, altitude, distance to be travelled (downrange on earth) will define the magnitude of missile

velocity required.

For the missile to travel a distance , total central angle is given by;=/

Where is radius of the Earth. The missile hits the earth when r= . Substituting r= and = intotrajectory equation solution yields;

=1 c o s

coscos

We know that; = Solving for velocity;

= 1cos cos cos cos This equation gives the velocity required to hit a target at a certain distance away from launch point,

given a flight-path angle (

). If initial

is too large, missile will never hit earth because resultant velocity

will exceed escape velocity ( = 2) and trajectory will not be elliptical. For longer distance we requirelarger missile velocities.

7/27/2019 Strategic Missile Zarchan

27/75

27 | P a g e

1.6) FLIGHT TIMETime to reach the target or time of flight . Based on the trajectory equation solution for, a close-form solution for can be derived. From original gravity field differential equation in polar coordinates,we know that;

= cos

Cross multiplying terms to set up the integrals;

= cos

Integration of the right-hand side of the equation and substitution of trajectory solution into left-hand

side yields the integral

=1

cos

1 cos c os

After integration and algebra; the closed-form solution assuming < 2 for the flight time simplifies to; = cos {

tan 1 c o s 1 sin 2 [1 c o s cos cos ]

2cos 2 1. tan

(

2 1coscot 2 sin)}

Flight time increases smoothly and monotonically with increasing values of flight-path angle.

7/27/2019 Strategic Missile Zarchan

28/75

28 | P a g e

CHAPTER 2 Boosters

2.1) Rocket Equation:

A tactical missile gets up to speed by burning propellant. If the missile is launched from the air, it

already has a large initial speed. However, if the missile is launched from the ground, it needs

more propellant to reach the same speed since it is starting from rest. In addition, a ground-

based missile needs additional propellant since it must travel through more of the denser

atmosphere. Figure 2.1 shows a typical weight and thrust profile for a boost-coast missile. The

initial or total weight of the missile is denoted by WT, and its final weight, after the propellant is

expended, is the glide weight WG. The glide weight consist of missile structure, electronics and

warhead. While propellant is being consumed, the thrust is assumed to be constant, with

magnitude T.

We can find the missile velocity after all the propellant is consumed from basic physics. Applying

Newtons second law yields

= = = The change in velocity with respect to time can be expressed in terms of thrust and weight as = =

2.1) Boost-Coast Thrust Weight Profile

7/27/2019 Strategic Missile Zarchan

29/75

29 | P a g e

As long as the missile is burning propellant (0

7/27/2019 Strategic Missile Zarchan

30/75

30 | P a g e

2.2) Increasing fuel mass fraction and specific impulse, increases missile speed

Using the preceding equation, Fig 2.2 shows how the change in missile velocity varies with fuel

mass fraction and specific impulse. We can see from the figure that, if the fuel mass fraction is

200s, the change in velocity is about 2300 ft./s. That means if the missile were launched from the

ground, its final speed, in the absence of drag and gravitational effects, would be 2300 ft./s. If

the missile with the same fuel mass fraction were launched from an aircraft traveling at 1000

ft./s, its final speed would be 3300 ft./s.

2.2) Tsiolkovsky Rocket Equation:

The Tsiolkovsky Rocket Equation, or Ideal Rocket Equation, describes the motion of vehicles that

follow the basic principle of a rocket: a device that can apply acceleration to itself (a thrust) by

expelling part of its mass with high speed and move due to the conservation ofmomentum. The

equation relates the delta-v (the maximum change of speed of the rocket if no other external

forces act) with the effective exhaust velocity and the initial and final mass of a rocket (or

other reaction engine).

For any such maneuver (or journey involving a number of such maneuvers):

= ln

Where:

is the initial total mass, including propellant,is the final total mass,

is the effective exhaust velocity,

7/27/2019 Strategic Missile Zarchan

31/75

31 | P a g e

is delta-v - the maximum change ofvelocity of the vehicle (with no external forces acting),In this section preliminary boosters designs are investigated so that speeds required for strategic

travel (20kft/s) can be achieved.

As per the rocket equation the change in velocity is related to specific impulse and fuel massfraction

.

Mathematically,

= ln 11 A strategic interceptor consist of two sections viz. booster and payload. A single-stage booster

consists of fuel and structure denoted by the weights and , respectively as shown in figurebelow:

2.3) Single Strategic interceptor model

Initially, it is assumed that the sole purpose of the single-stage booster is to get the payload up

to the speed. The payload denoted by , consists of structure, electronics, a divert engineand fuel. The purpose of the payload for strategic guided interceptors is to acquire the target and

maneuver, using divert fuel to hit the target.

If it is desired that the interceptor change its velocity by amount of , then the weight of thestructure fuel and payload must also follow the rocket equation which is given as:

=

exp

Where denotes the specific impulse of the booster fuel and is measured in seconds. The fuelmass fraction has been defined as the ratio of the propellant weight to the total weight.

To simplify mathematical computations, an approximate fuel mass fraction is considered and is

defined as the ratio of the propellant weight to the sum of the propellant weight and structure

or,

7/27/2019 Strategic Missile Zarchan

32/75

32 | P a g e

= For small payloads the approximate and actual fuel mass fraction are equivalent. On rearranging

the above equation we can relate the weight of booster structure to the propellant weight and

fuel mass approximate mass fraction as,

= 1 Putting this relationship in the rocket equation a formula for the propellant weight in terms of

the payload weight, velocity desired, approximate fuel mass fraction and specific impulse which

is given as,

= 1 1

The total interceptor weight

consists of the booster fuel and structure plus the payload,

or

= 2.3) Booster Staging:

We know that all the rockets use the thrust generated by propulsion system to overcome the

weight of the rocket. Moreover the weight of the payload is only a small portion of the lift-off

weight. Most of the weight of the rocket is the weight of the propellants. As the propellants are

burned off during powered ascent, a larger proportion of the weight of the vehicle becomes the

near-empty fuel tanks and the structure that was required when the vehicle was fully loaded. In

order to lighten the weight of the vehicle to achieve orbital velocity, most launchers discard a

portion of the vehicle in a process called staging.

2.4) Two-Stage booster

Figure above represents the schematic of two stage booster, in this figure, the propellant and

structural weights are indicated in each of the stages.

Therefore, the second stage propellant weight can be expressed as,

= 1 1

7/27/2019 Strategic Missile Zarchan

33/75

33 | P a g e

Where is the desired velocity change is attributed to the second stage, is the secondstage fuel mass fraction, and is the second stage specific impulse.First and second stage burn times are given by,

=

And

= 2.4) Gravity Turn:

If we attempt to align the thrust vector with the booster velocity vector, we will obtain a gravity

turn. The acceleration due to the booster thrusting is given by,

= Where T is the thrust magnitude in pounds, W the missile weight, and g is 32.2 ft/s2 .The booster

velocity V at any time could be found from the velocity components as,

= .Therefore, during a gravity turn at any time the components of acceleration acting on the booster

in our Earth-centered coordinate system are given by,

= . = .

Where the initial conditions on velocity are related to the initial flight path angle and location.

The initial launch characteristics could be either counterclockwise or clockwise depending upon

the location of the target. Figure below shows a typical case of counterclockwise travel,

7/27/2019 Strategic Missile Zarchan

34/75

34 | P a g e

2.5) Counterclockwise Travel

For counterclockwise travel, the velocity initial conditions are

0 =0cos2

0 =0sin2 A case for clockwise initial launch condition is given below

2.6) Clockwise Travel

The velocity initial conditions are,

0 =0cos 2 0 =0sin 2

7/27/2019 Strategic Missile Zarchan

35/75

35 | P a g e

The initial components of the booster location are given by,

0 = 0 = A simulation was performed for a given booster design parameters to find the appropriate thrust-weight profiles and in addition, the booster performs a gravity turn. Here the initial flight path

angle of the booster during gravity turn is 85 deg. During the trajectory the flight path angle will

start from 85 deg and gradually reduce to smaller values.

Cases were run with the nominal design, and the initial flight path angle was made a parameter.

2.7) Larger Flight-Path angles are required for initial booster design

The resultant shown in the above figure indicate that large flight path angles are required just to

get a trajectory for a gravity turn. If the flight path angle is less than 80 deg, the booster will

immediately crash into the earth. As the booster thrusts, the flight path angle rapidly decreasesdue to the small booster acceleration (about 4g at the beginning). Eventually the flight path angle

decreases to the point where the component of the booster acceleration which is perpendicular

to the surface of the earth is not sufficient to overcome gravity.

To remedy the situation so that we could get smaller flight path angles to yield long range

trajectories, the maximum axial booster acceleration during each stage was increased from 10g

to 20g. The resultant velocity and acceleration profiles due to this change was once again

simulated.

A comparison is done with the initial and the final graphs are shown in figure 2.8.

7/27/2019 Strategic Missile Zarchan

36/75

36 | P a g e

2.8) New booster design versus nominal configuration

Gravity turns were performed, via the simulation, for the new booster design, and results for

different flight path angles are shown in the figure below.

7/27/2019 Strategic Missile Zarchan

37/75

37 | P a g e

2.9) New booster design yields longer fly-out range

We can see that the larger axial booster acceleration allowed the booster to experience lower

flight path angles, which increased the booster range. With the nominal design, the maximum

range achieved with a flight path angle of 85 deg was about 2300 n.mi. The new design, which

permitted a lower flight path angle of 65 deg, increased the maximum range about 2600 n.mi

7/27/2019 Strategic Missile Zarchan

38/75

38 | P a g e

CHAPTER 3 Lambert Guidance

A ballistic missiles trajectory is composed of three segments.

1.Powered Flight: The portion, which lasts from the time of launch to missile motor thrust cutoff

or burnout and exit from the atmosphere (depending on cut off altitude). The terms cutoff and

burnout, define the conditions at the beginning of the free-fall, that is, the termination ofpowered flight. Therefore, they denote the initial conditions necessary to solve the differential

equations of motion. More specifically, this is the flight through the atmosphere and extending

into free space where the aerodynamic forces may be neglected. During this portion of the flight,

the greatest force acting on the vehicle is the thrust, which is derived from a rocket engine. The

acceleration of the missile from this thrust is usually about 1.1 to 1.5 gs at liftoff; it increases as

the mass of the vehicle decreases with fuel consumption and staging, until a final value in the

range of 5 to 10 gs will be reached. At the time of cut-off (or burnout), the vehicle will have

attained an altitude such that aerodynamic forces are no longer of major importance to the

trajectory.

However, the velocity and position of the vehicle must be controlled along the trajectory so as to

limit the aerodynamic loading of the structure and to place the vehicle on a free-fall trajectory,

which will carry it to its target. It should be pointed out that the guidance of a ballistic missile

occurs entirely during this powered portion (or phase) of the flight; consequently, its objective is

to place the missile on a trajectory with flight conditions that are appropriate for the desired

target. This is equivalent to steering the missile to a burnout point that is uniquely related, as

stated above, to the velocity and flight path angle for the specified target range. If there were no

restrictions on the maneuvers that the missile can make during the powered flight, the guidance

and control would be relatively simple, and the only major problem would be that of precision

guidance. Structural limitations and flight performance requirements will combine to restrict the

ascent trajectory such that only limited correction maneuvers may be employed. Typically, an

ICBM will burn out at about 264.4 nm (490 km) altitude and 420.9 nm (780 km) downrange from

its target.

2.Free-Flight (or Free-Fall): The portion that constitutes most of the trajectory. The free-flight

trajectory is a conic section (i.e., an ellipse). This is also called vacuum flight. For this phase of

the flight, the initial conditions determine the parameters of the orbit; in other words, these

parameters establish the trajectory to be followed.

After the termination of powered flight, the missile is in a free-fall condition under the influence

of gravitation alone. Above the thrust termination point (or cut-off point) the atmosphere is, in

general, almost nonexistent for missiles capable of attaining ranges on the order of 5,000 to 6,000

nautical miles (9,260 to 11,112 kilometers). As the missile converges on the target, it will reenter

the atmosphere.

7/27/2019 Strategic Missile Zarchan

39/75

39 | P a g e

The missile is then no longer in the free-fall condition; this, as we shall see below, is the reentry

phase of the missile trajectory. Many effects influence the free-flight trajectory. The main effects

are those arising from the assumption that the Earth is a homogeneous rotating sphere. This

gives rise to an elliptical trajectory passing through the cut-off point and target with one focus at

the center of the Earth. All the other factors that affect the free-flight trajectory can be

considered to cause only perturbations of the elliptic orbit. As in the powered flight trajectory,

there is also a broad selection of free-flight trajectories to choose from, for a given range.

The choice must be based on both technical and strategic factors.

It should be noted that the entry of a ballistic missile into its free-fall trajectory occurs abruptly

upon the event of thrust cut-off, but the termination of the free-fall trajectory is not similarly

well defined. For the definition of free-fall, it will be convenient to adopt the convention of a

reference sphere. The reference sphere is defined as the sphere with the center at the center

of the Earth having the thrust termination (i.e., burnout) point on its surface. The free flight will

be assumed to terminate when the missile returns to the reference sphere (see Figure 3.1). This

same convention may be employed to define the initial point for reentry. Consequently, the flight

conditions that obtain at the time of initiation of the free-fall phase of the flight have the greatest

influence on the impact point of the missile. The powered flight is designed to place the vehicle

in an appropriate trajectory so that upon thrust termination the missile will begin a free-fall orbit

to the target. As stated above, under powered flight, no guidance need be employed during this

free-fall, since the trajectory will be fully predictable.

3. Reentry: The portion that begins at some point where the atmospheric drag becomes a

significant force in determining the missiles path and lasts until impact (i.e., target on the surface

of the Earth). The reentry trajectory is determined to a great extent by the conditions of flight

that obtain as the missile approaches the effective atmosphere of the Earth. Frequently, it isconvenient to treat the reentry phase as terminal perturbation acting on the free-fall trajectory.

The reentry phase of the trajectory should begin at an altitude of about 100,000 ft (30,480 m),

where the dynamic pressure starts to significantly affect the motion of the missile. The

computation of this trajectory phase involves knowledge of aerodynamic stability derivatives of

the missile. It can be shown that the effects of reentry constitute only a perturbation to the free-

flight trajectory. The importance of this phase of flight to navigation and guidance arises from

the high accelerations that are experienced by the missile on reentry. In particular, the extremely

high heating rates that are obtained during this flight limit the reentry trajectories that are

permissible for any given missile configuration. While the transition from powered flight to free-fall is abrupt, the transition from free-fall to reentry flight is more gradual as a result of the built

up of air density as the missile penetrates the atmosphere. It should be noted here that the

reentry point is not defined precisely.

7/27/2019 Strategic Missile Zarchan

40/75

40 | P a g e

3.1) Geometry of a ballistic missiles trajectory

Definition:A conic section is the locus of points so situated that the ratio of the distance of each

point from a fixed point to its distance from a fixed line not through the fixed point is a constant.

The fixed point is called thefocus of the conic, the fixed line is called its directrix, and the constant

ratio, generally denoted bye, is called its eccentricity. Note that the directrix has no physical

significance as far as orbits are concerned. However, the focus and eccentricity are indispensable

concepts in the understanding of orbital motion.From Figure 3.2 we note that the family of curves called conic sections (i.e., circle, ellipse,

hyperbola, and parabola) represent the only possible paths for an orbiting object in the two-body

problem. The focus of the conic orbit must be located at the center of the central body.

The most important types of curves (i.e., conic sections) can be represented by the general

equation of the second degree in two variables as follows:

= 0 (3.1)An equation of this type may represent an ellipse (or circle), a parabola, or a hyperbola. The polarequation of a conic section is given by the equation:

= + (3.2)Where p is a geometric constant of the conic called the parameteror semi-latus rectum, e is

called the eccentricity, which determines the type of conic section represented by (3.2), and is

7/27/2019 Strategic Missile Zarchan

41/75

41 | P a g e

the polar angle known as the true anomaly, which is the angle between rand the point on the

conic nearest the focus.

Consequently, (3.2) is the expression for the polar conic sections (i.e., the equation of all curves

formed by the intersection of a complete conic surface and a plane, as shown in Figure 3.2).

3.2) The Conic Sections

It is the trajectory equation of a body expressed in polar coordinates. Equation (6.2) is also known

as a Keplerian ellipse.

The exact nature of the resulting curves depends only upon the absolute value of the constant e,

that is, the eccentricity. The origin of the r, coordinate system is located at the primary focus

of the conic sections. The following relations define the various paths:

|e| > 1 hyperbola,

|e| = 1 parabola,

|e|

7/27/2019 Strategic Missile Zarchan

42/75

42 | P a g e

mass, whose gravitational attraction is an inverse-square central force. For example, for an

artificial satellite moving around the Earth as its focal center, the gravitational attraction is:

= Where M and m are the masses of the Earth and satellite, G is the universal constant, and r is thedistance of m from the center of the Earth. Equation also applies to the EarthSun, the Moon

Earth, and the Earthmissile systems. From the above discussion, consider again the motion of a

particle of small mass (e.g., the missile) m that is attracted by a particle of large mass (the Earth)

M. The force of gravitational attraction between the masses is along the line joining them, so that

the resulting motion is called motion under a central force. The acceleration ofM is much smaller

than that ofm, so that without making too great an error we may consider M to be at rest, with

m moving about it. Any motion under a central force takes place in a plane. From the above

discussion, we can make the following assumptions:

(1) Assume an inverse square law of force between the missile and the Earth.(2) Assume that the gravitational acceleration is a constant.

(3) Assume that the missile follows a path described by a conic section. This implies:

(a) The dissipative forces of the system are negligible. This means that the system is conservative

and the sum of the kinetic and potential energies is constant.

(b) The only forces acting on the missile after engine cut-off is that of gravity (i.e., no guidance

forces).

(c) The path of the missile is in a single vertical plane.

In vector form, the equation of motion for the two-body problem is,

= (3.4)Where,=G(M +m)GM, G is the universal gravitational constant, M is the mass of the central

body, and m is the mass of the orbiting body. In order to derive the trajectory (or orbit) equation,

we will use scalar notation instead of vector notation given by (3.4).

7/27/2019 Strategic Missile Zarchan

43/75

43 | P a g e

3.3) Shows an ellipse where S (Sun) and F(Focus) are two foci, C Is the center, and AB is the major axis

Keplers first law states that the path, or orbit, of a planet around the Sun is an ellipse, the

position of the Sun being at one focus of the ellipse. Keplers first law is illustrated in Figure 3.3.

Furthermore, Keplers second law states that the radius vector SP sweeps out equal areas in equal

times. Now, from Figure 3.3 we let (x, y) be the coordinates of the planet referenced from these

axes. Therefore, using (3.4), the equations of motion in the orbital plane of the planet are:

= 0 3.5 = 0 3.5

where r =(x2 +y2)1/2. We wish now to transform these equations given in rectangular coordinates(x, y) into polar coordinates (r, ). Let,

x= = = 3.6

=

= cos 2 sin cos sin 3.6

7/27/2019 Strategic Missile Zarchan

44/75

44 | P a g e

Similarly,

= (3.7a)

= sin cos cos sin cos = sin 2 cos sin cos (3.7b)

Substituting (3.6) and (3.7) into (3.5) results in,

[ 2 = 0 (3.8a) 2 = 0 (3.8b)

Since (3.8) must hold for all values of , then a planets motion is governed by the following

equations of force:

=

2 = 0OrRadial force:

= (3.9a)Transverse force:

2 = (3.9b)The second equation, (3.9b), leads to the statement of conservation of moment of momentum

per unit mass r2(d/dt) =h. These are the polar equations of motion.

Since,

7/27/2019 Strategic Missile Zarchan

45/75

45 | P a g e

= 2 The function, = (3.10)Satisfies (3.9b) where h is the constant of integration (h is also called the angular momentum).

Equation (3.10) is simply the mathematical expression of Keplers second law. Now let us

introduce the variable

= (3.11)From (3.10) we have

= = (3.12)Taking the derivative of (3.11) yields

= = 1 and from (3.10) and (3.11),

= (3.14)Hence, (3.13) may be written as follows:

= = (3.15)Taking the second derivative of (3.15), we have

= (3.16)Substituting (3.10) into (3.9a), results in

1 1 =

7/27/2019 Strategic Missile Zarchan

46/75

46 | P a g e

Or = (3.18)

The differential equation represented by (3.18) is called the harmonic equation; its solution is

well known. The complementary solution of (3.18) is the general solution of = 0.

i.e. = ,or =1 2, (3.19)where C1 and C2 are constants of integration. The particular solution is readily found to be =/.Then the complete solution of (3.18) is

= =1cos 2 =11cos 2

or

= + . (3.20)This is the polar form of an ellipse with origin at one focus. In terms of Figure 3.3, the constantC1 is identified with the eccentricity e, and the constant C2 identified with . Therefore, we can

write (3.20) as

= 1 c o s

where e and are constants of integration. The initial conditions on the motion are the burnout

conditions of the ballistic missile or orbital vehicle (or the burnout conditions of the retrorocket

in the case of reentry). These conditions must, of course, exist at a point of zero aerodynamicforces. A statement of the initial conditions that appears natural from an engineering point of

view is

at = 0; = 0 = = , = ,Note that the polar angle has been set equal to zero at the initial conditions. This puts norestrictions on the solution, since the effect of having = rather than = 0 is simply torotate the reference for measurement of the polar angle.

From Figure 6.3 we note that the semilatus rectump is given by = = 1 , (3.22a)

7/27/2019 Strategic Missile Zarchan

47/75

47 | P a g e

or we can write = / (3.22b)so that = = 1 .The general equation of a conic section, general equation of a conic section, which may be

(i) an ellipse ife1.

Although case (i) is that with which we are closely concerned here, the extension of the

possibilities concerning the motion of a body under the gravitational attraction of the Sun should

be noted.

It is convenient to interpret the initial conditions in terms of, Vi, and i, rather than in termsof , / , and . Here i is the initial missile flight-path elevation angle, measured, ofcourse, in the plane of motion, and Vi is the magnitude of the initial velocity vector in inertial

space, or relative to the nonrotating Earth.

7/27/2019 Strategic Missile Zarchan

48/75

48 | P a g e

From Figure 3.4, we have = sin

, (3.23a)

= cos (3.23b)It is now convenient to introduce a parameter o defined by the relationship, / (3.24)

7/27/2019 Strategic Missile Zarchan

49/75

49 | P a g e

This parameter is termed the initial condition parameter, and it can be interpreted as the ratio of

twice the particles initial kinetic energy to its initial potential energy.

We can now obtain expressions for in terms of, , and i. From (3.10) and (3.23) we obtain, = / = (3.25)Squaring and introducing (3.24), we have

= (3.26)Figure 3.4 illustrates the geometry of the ellipse applicable to an elliptical orbit.

It can be verified that the equation of motion (3.2) reduces to the statement r=ri= [constant] by

using (3.25) and (3.22b) to write

=

1 c o s =

1 c o s = 1 c o s

or the circular orbit i =0, e=0, and 1, and hence r =ri . The discussion of the complete ellipticalorbit best proceeds by considering first the maximum and minimum values ofr, as given by (3.2),

taking the derivative ofrwith respect to and setting the result equal to zero. Thus,

= sin1 c o s = sin = 0sin =0, for =0, , 2, . . . , gives the extreme values for r. These values are:

Minimum (perigee) radius occurs for =0: = + (3.27a)Maximum (apogee) radius occurs for =: = (3.27b)

Dividing (3.27b) by (3.27a) gives the expression for e in terms ofraand rp: = + (3.28)Now introduce a, the semi-major axis of the ellipse defined by = + (3.29)

7/27/2019 Strategic Missile Zarchan

50/75

50 | P a g e

We can now add the two equations (3.27a) and (3.27b) and solve forp, which, as we saw earlier,

is often called theparameterof the motion:

= 2 = [ 11 11 ] = 1 1 1 or

= 1 (3.30)which allows the equation of motion to be written in the simple form = + , || < 1 (3.31)This is the equation of the elliptical orbit that is usually encountered in the astronomical

literature.

Lamberts Theorem

With the preliminaries complete, we will now discuss Lamberts theorem. The German

mathematician Johann Heinrich Lambert (17281777) showed in the eighteenth century (in

1761) that in elliptic motion under Newtonian law, the time required in describing any arc

depends only on the major axis, the sum of the distances from the center of force to the initial

andfinalpoints, and the length of the chord joining these points. Therefore, if these elements

are given, the time can be determined regardless of the form of the ellipse.

7/27/2019 Strategic Missile Zarchan

51/75

51 | P a g e

3.1) Statement of Lamberts Theorem:

In elliptic motion under Newtonian law, the time required in describing any arc depends only on

the major axis, the sum of distances from the center of force to the initial and final points, and

the length of the chord joining these points. Therefore, if these elements are given, the time can

be determined regardless of the form of ellipse. This was shown by Johann Heinrich Lambert.

Statement of Lamberts Problem:

A body in a gravity field satisfies newtons law of gravitation, or;

= . = .

Assume the initial location of body in the gravity field is given by;

0 =

0 = After tfseconds, location of body;

= = Lamberts problem is to find the initial velocity orientation of body in the gravity field so that the

preceding initial conditions and boundary values are satisfied, or;

0 =?0 =?Solution to Lamberts Problem:

Initial missile velocity;

= 1 cos cos cos Where

is the central angle separating the initial location of missile and it intended target,

is

initial flight path angle of missile, a is the radius of earth, and r 0 is the initial distance from the

center of earth to missile which can be expressed as;

= alt is the intial altitude of missile with respect to surface of earth.

7/27/2019 Strategic Missile Zarchan

52/75

52 | P a g e

If we desire to hit a target at any location r f, the preceeding velocity equation can be modified

to;

= 1 cos cos

cos

= Where is altitude of intended target.If velocity vector is oriented for counterclockwise travel; we can find the initial conditions on the

velocity components in earth-centered system by trigonometry as;

0 = 2

0=sin2

Where is the orientation of missile velocity with respect to a reference that is tangent to theearth and perpendicular to the vector from center of earth to the initial location of missile andis the initial angular location of missile with respect to x-axis of earth-centered Cartesiancoordinate system.

3.4) Counterclockwise travel

7/27/2019 Strategic Missile Zarchan

53/75

53 | P a g e

3.5) Clockwise travel

If the velocity vector is intended to travel clockwise as shown in figure; then the initial conditions

on the velocity components in the Earth-centered system can be shown as;

0 = 2 0=sin 2

The formula for the time required for the missile to reach its intended target (t f) , which is valid

for elliptical travel ( < 2), does not require the target to be on the surface of the earth and isgiven by;

= + + + . .Where is the required velocity to hit the object and is defined as;

= is the angular distance to be travelled. = .||.

7/27/2019 Strategic Missile Zarchan

54/75

54 | P a g e

3.6) Central angle between initial and final position

In this figure, denotes a vector from the center of the earth to the initial location of the object,

denotes a vector from the center of the earth to the final location of the object. The angle

between the vectors is the central angle .We now have sufficient information to numerically solve Lamberts problem.

With a central angle , , and a flight path angle , sufficient information is available to findthe required velocity from our closed-form solution. The resultant velocity can then be used to

solve for the flight time from our other closed-form solution. It is important to note that the flight

time and velocity obtained are exact solutions for the flight-path angle used. Statd

mathematically, we can say that given , and , we can use the following relationships whichare based on exact closed-from solutions:

= , = , ,, =,,In Lamberts problem we are given , and and we seek to find and . If we use thepreceding relationships, we do not know how to choose , nor are we guaranteed that aparticular value of will yield the desired flight time .We can solve the problem by the method of Brute Force. That is, we work out all solutions until

we find the one that satisfies the constraints of the problem. For example, we start with =90 , solve for the velocity, and then solve for the time of flight. If the flight time is less thanthe desired flight time, we repeat the procedure with a slightly larger value of . We stop theloop when the computed flight time is greater than the desired flight time. If the flight-path angle

that satisfies the preceding procedure is negative, we know that the solution must be rejected

since it requires the missile to travel through the Earth. This numerical method converges

7/27/2019 Strategic Missile Zarchan

55/75

55 | P a g e

because we know that the flight time is smooth and monotonically increasing with increasing

flight-path angle.

NUMERICAL EXAMPLE

This example represents sample code, using double-precision arithmetic, for finding the Lambert

solution, based on the procedure developed before. We can say that, given an initial angle and

altitude for the missile (XLONGMDEG, ALTNMM), an initial angle and altitude for the target

(XLONGTDEG, ALTNMT), and a desired flight time (TF), the program iterates on the flight-path

angle (GAMDEG) until a solution is found. From the listing we can see that the program consists

of two loops. The first loop iterates on the flight-path angle in units of 0.1 deg. When a flight time

is found that exceeds the desired flight time we exit the loop for another loop that increments

the flight-path angle (after decreasing the last flight path angle by o.15 deg) in very fine units of

0.0001 deg. This loop is required to get extremely precise answers. When the desired flight time

is achieved, we exit the loop and the routine. For most cases the program takes about 1.5 s to

execute on a 16 MHz, 32-bit microcomputer with a math coprocessor. The routine, as written, is100 times slower than more elegant lambert routines. By performing a more intelligent search it

is possible to find the correct solution to Lamberts problem in a few iterations, thus making this

approach very competitive with more elegant Lambert routines.

To demonstrate how the routine works, the nominal case, shown in the listing was run. In this

case the missile is on the surface of the Earth 45 deg away from the target. It is desired to find

the velocity orientation of the missile (VRX, VRY) so that the missile will hit the target in exactly

1000 s.

SPEED UP LAMBERT ROUTINE

We can speed up the Lambert routine by two orders of magnitude, by restricting the brute force

search and eliminating many iterations by recalling the previous velocity () formula. = 1 cos cos cos

Here we are only interested in trajectories of ballistic missiles, so we can rule out the case that

lead to escape velocity ( = 2) or; = 2 =

7/27/2019 Strategic Missile Zarchan

56/75

56 | P a g e

Substitution of the escape velocity condition into the velocity formula yields

2 = 1 cos cos cosWe can solve the preceding equation for the flight-path angle

. After much algebra we get two

solutions corresponding to the minimum and maximum flight-path angles as

=sin 2 1 c o s 1 c o s

= sin 2 1 c o s 1 c o s

Solution for the velocity and time of flight were smooth, well-behaved functions of the flight-

path angle. Based on the non-pathological nature of these solutions and the fact that the flight-

path angle is well-bounded, we do not have to evaluate each flight-path angle but can instead

perform a more efficient search in finding the flight-path angle that corresponds to the desired

flight time. For example, we can use algorithm known as secant method to perform the search

or

+ = We can see from the preceding equation that the new flight-path angle + is related toprevious values , . At each iteration the new computed value of flight-path angle is limitedto the minimum and maximum possible values of the flight-path angle derived from the escape

velocity condition. The search is terminated when the computed flight time is sufficientlyclose to the desired flight time

.

The new lambert routine is not only accurate than the previous one but it is also more than two

orders of magnitude faster.

7/27/2019 Strategic Missile Zarchan

57/75

57 | P a g e

Simulation Results:

3.7) Lambert Routine using brute force method to calculate flight time

3.8) Lambert routine using brute force method to determine velocity

7/27/2019 Strategic Missile Zarchan

58/75

58 | P a g e

3.2) Booster Steering:

Given that we have the initial position and final position (target destination) of the missile and

an arrival time, the Lambert subroutine will give the orientation of velocity vector for an

impulsive missile to satisfy the problem. Since we do not have impulsive missiles (missiles that

get up to speed immediately), it is desirable to find out if the Lambert subroutine could be of usein enabling a non-impulsive missile or booster to reach its target. If atmosphere is neglected then

the solution to this problem is quite simple and is known as Lambert Guidance.

Consider the vector Diagram shown below.

3.9) Basis of Lambert Guidance

At small time-increments; while the missile is boosting, we have to find desired velocity from

lambert subroutine VLAMBERT and subtract the current missile velocity VM. The difference in

velocities is known as velocity to be gained (V). If boosting missile thrust vector is aligned withvelocity to be gained vector, then desired velocity will be obtained in a feedback fashion. When

desired velocity is achieved, engine is cut off and missile flies ballistically to the intended target.

Mathematically, components of velocity to be gained are;

= = Total velocity to be gained;

= .If magnitude of current thrust acceleration is given by

, then the direction of thrust

acceleration at each instant of time should be aligned with the velocity to be gained vector, or; = / = /

7/27/2019 Strategic Missile Zarchan

59/75

59 | P a g e

NUMERICAL EXAMPLE:

Following numerical presents a simulation of a two-stage booster using lambert guidance during

the boost phase. Scenario is unrealistic since g loading and range safety considerations have been

ignored, but it is useful for demonstrating how Lambert guidance works. The booster considered

in this example has a capability of reaching a velocity of 20000 ft/s. The booster is assumed to

have two stages, each of which has a fuel mass fraction of 0.9 and specific impulse of 300 s. The

maximum acceleration in each stage is 20 g. One-third of the speed will be attained in the first

stage and rest of the speed will be attained in the second stage. Burnout of the second stage will

be completed at about 60 s. it is desired that the booster, which is initially at angular location = 30 (ANGDEG=30), reach a target at 45 deg (XLONGTDEG) in 500 s (TF=500).As the Lambert feedback loop runs, when the difference between the desired velocity and the

attained velocity is less than 500 ft/s, the simulation automatically sets the actual velocity to the

desired velocity to avoid making the integration interval very small in the simulation. At this time

the booster cuts off and coasts.

3.10) X component of achieved velocity reached Lambert Solution

7/27/2019 Strategic Missile Zarchan

60/75

60 | P a g e

3.11) Y component of achieved velocity reaches Lambert solution

Fig 3.6 displays the x-component of the achieved velocity along with the desired or Lambert

velocity. The two velocities converge at about 45 s.

Fig. 3.7 displays the y-component of the achieved and desired velocities. This component is much

larger than the x-component. The discontinuity in the y-component at about 15s is due to staging,

and the slight discontinuity near the end of the display is due to setting the achieved velocity to

the desired velocity when the velocity to be gained was less than 500 ft/s.

Lambert solution was reached in 45 s even though the missile was capable of burning fuel for

nearly 60 s. Thus, lambert guidance can be used to steer a strategic missile with a thrust

termination system in absence of atmospheric effects. The lambert guidance principal can be

used for interceptors that fly ballistically to hit the stationary targets. Lambert guidance can also

be used to for guided interceptors that must hit the moving and accelerating targets. In this case,

the purpose of the lambert guidance is to place the interceptor on a collision triangle at the end

of a boost phase.

7/27/2019 Strategic Missile Zarchan

61/75

61 | P a g e

SIMULATION:

3.12) New booster design velocity curve

3.13) Nominal booster design velocity curve

7/27/2019 Strategic Missile Zarchan

62/75

62 | P a g e

3.14) Booster simulation using Lambert Guidance

7/27/2019 Strategic Missile Zarchan

63/75

63 | P a g e

3.3) General Energy Management (GEM) Steering:

It is possible to steer a strategic interceptor to the intercept point using Lambert guidance and to

make it possible the thrust had to be terminated before the end of burn in order to achieve the

desired Lambert solution. Often there is a restriction, in the absence of a thrust termination

system, that all the booster fuel must be consumed. In this case a method other than Lambert

guidance must be employed to waste some of the boosters excess energy. A popular energy

wasting technique is known as General energy management (GEM) Steering.

Consider the figure given below to understand the energy wasting concept.

3.15) Basic angles in general energy management

VCAP denotes length of arc of circle which resembles velocity capability of booster, r denotes

radius of circle, 2 is the central angle and denotes the chord connecting two endpoints ofarc. It is the velocity to be gained. If the thrust vector is drawn tangent to the chord at the

beginning of the arc, it is easy to show from the geometry that the thrust vector is at an angle of

with respect to the chord. Finally, a perpendicular is dropped from the chord to the center of

the circle. It is also easy to show that the perpendicular bisects the chord and the central angle.

The arc length is related to the central angle as; =2Since the perpendicular bisects the central angle , so = 2 s i n Therefore, we can ratio the two velocity expressions, yielding

7/27/2019 Strategic Missile Zarchan

64/75

64 | P a g e

= 2sin2 = sin Expanding the sine term into a two-term Taylor series leads to

=

6

= 1

6

Solving for the angle yields;

= 61 Formula suggests that if at each instant of time, we ensure that the thrust vector is at an angle with respect to the velocity to be gained vector (), then we can still achieve Lambert solutionat the end of burn and hit the target.

3.16) Sign conventions

Above figure shows the proper relationship between the thrust and the velocity to be gained

vectors relative to the inertial Earth-centered coordinate system.

For counterclockwise travel, the components of thrust accelerations are given by;

7/27/2019 Strategic Missile Zarchan

65/75

65 | P a g e

= cos = sin represents angle between the thrust vector and velocity to be gained vector (). represents angle between and the x-axis.For clockwise travel, thrust acceleration components become; = cos = sinListing 14.4 presents a simulation of a booster intercepting a ground target using general energy

management guidance. The axial acceleration capability of the booster is continually being

computed according to;

= Where H is the integration step-size and is instantaneous axial acceleration of booster.NUMERICAL EXAMPLE:To avoid numerical problems, the GEM logic is terminated when the velocity to be gained drops

below 50 ft/s. It is still necessary to implement lambert subroutine in order to implement GEM

guidance technique.

3.17) GEM angle reaches steady state quickly

A nominal case was run to see how the GEM guidance logic performed. We can see from the fig

3.9 that although the booster burn lasts for nearly 60 s, the angle the thrust vector makes with

the velocity to be gained vector approaches steady state in slightly over 50 s. Plot shows how the

7/27/2019 Strategic Missile Zarchan

66/75

66 | P a g e

velocity capability of the booster diminishes during the burn. The discontinuity in the curve is due

to staging.

3.18) Lambert and GEM trajectories during boost phase are vastly different

Fig 3.14 displays the gem trajectory during the boost phase. It appears from the figure that the

booster will never hit the target because initially it appears to be heading in the wrong direction.

However, after wasting energy, the GEM-guided booster heads in the right direction.

Superimposed on the figure is the Lambert guidance trajectory during the boost phase for the

same case. Both trajectories are vastly different during the boost phase.

3.19) Both Lambert and GEM trajectories hit target at the same time

Fig 3.15 displays the GEM and Lambert trajectories for the entire flight (boost and coast phases).

Although both the trajectories are vastly different during the boost phase, they eventually

converge, and both hit the target at the same time.

7/27/2019 Strategic Missile Zarchan

67/75

67 | P a g e

SIMULATION:

3.20) Simulation to support the theory

7/27/2019 Strategic Missile Zarchan

68/75

68 | P a g e

CHAPTER 4 Orbital Mechanics

4.1 Introduction:

Space flight is much more predictable than either aircraft or rocket flight, since it is not subject

to substantially large disturbances at any given time. Being governed primarily by the

gravitational field of a spherical body (called orbital mechanics), the equations of motion are

amenable to simplified solution in several important cases. However, the small magnitude of

perturbations such as non-uniform gravity, atmospheric drag, solar radiation pressure can

appreciably modify the trajectory over a long duration. Since spacecraft flight times are orders of

magnitude larger than those of either aircraft or rockets, it is necessary to account for the

perturbations while guiding a vehicle in its space mission.

4.2 Orbital Mechanics:

The relative motion of two spherical bodies in mutual gravitational attraction is the fundamentalproblem of translational space dynamics, called orbital mechanics, and possesses an analytical

solution. The motion of a spacecraft under the influence of a celestial body is usually

approximated as a two-body problem by neglecting the small gravitational effects caused by the

other objects, as well as the actual, non-spherical shapes of the two bodies.

Consider spherical masses , min mutual attraction given by Newtons law of gravitation. Theequation ofmotion of each mass is given by Newtons second law of motion referred to an inertial

frame. Subtracting the two equations of motion from one another, we have

= 0

wherer(t) is the position of the center of mass relative to the center of mass and = G(+ ), with G = 6.6726 1011 m3/kg/s2 being the universal gravitational constant. Usually, thespacecrafts mass, , is negligible in comparison with that of the celestial body, , and we canapproximate G.Orbit Equation

Let us begin the solution of equation (6.1) by taking the vector product of both the sides with r:

= or = which implies that the specific angular momentum ofm2 relative to m1, defined by

7/27/2019 Strategic Missile Zarchan

69/75

69 | P a g e

= = ,is conserved. Here = is the velocity. Since h is a constant vector, we have the followingconsequences:

(a) The direction ofh is a constant. This implies that the vectors r and v are always in the same

plane called the orbital plane and h is normal to that plane.

(b) The magnitude ofh is constant. Writing h in polar coordinates, (r,)

= | | = | | = The trajectories in the orbital plane called orbits are classified according to the magnitude

and direction of a constant h. The case of h = 0represents rectilinear motion along the line

joining the two bodies, while h 0represents the more common orbits involving rotation of m2

about

.

We next take the vector product of both sides of equation (6.1) with h:

= 0Since h is constant, we have

= Furthermore, the following interesting identity can be derived by differentiating r2 with respect

to time: = Hence, the second term on the left-hand side of equation (6.6) becomes, =

=

=

= / =

7/27/2019 Strategic Missile Zarchan

70/75

70 | P a g e

On substituting, we have = Thus, a constant vectore, called the eccentricity vector, can be defined such that

= Since the eccentricity vector is normal to h, it lies in the orbital plane and is employed as a

reference to specify the direction of the relative position vector, r(t). The angle, (t), made byr(t)

with e (measured along the flight direction) is called the true anomaly, (t).

Taking the scalar product of e with r(t), we have

r + e r= (1/)r[ (v h)] = h2/

or

= [ ]1 c o s

The above equation called the orbit equation defines the shape of the orbit in polar

coordinates, (r, ), and indicates that the orbit is symmetrical aboute. Furthermore, the minimum

separation of the two bodies (called periapsis) occurs for = 0, implying thatepoints toward the

periapsis. From the orbit equation it is clear that the general orbit is a conic section the shape

obtained by cutting a right-circular cone in a particular way. For e < 1, the orbit is an ellipse. The

circle is a special ellipse with e = 0 and r = h2

/.

For e = 1, the orbit is a parabola. The rectilinear trajectory is a special parabola with h = 0. For e

> 1, the orbit is a hyperbola. In all cases, the focus of the orbit is at the center of the celestial body,

and the semi-major axis, a, is given by

= [ ]1

4.3) Perifocal and Celestial Frames:

A right-handed coordinate frame fixed to the orbital plane, with unit vectors ih= h/h, ie = e/e, andip = ih ie, is used to specify the relative position and velocity vectors of the spacecraft in the

orbitalplane. This frame, (ie, ip, ih), is called the perifocal frame. The position and velocity of the

spacecraftin the perifocal frame are given by

=

7/27/2019 Strategic Missile Zarchan

71/75

71 | P a g e

= sin cos sin cos where (t) is the flight-path angle (taken to be positive above the local horizon) shown in Figurebelow by

4.1) Orientation of the perifocal frame relative to the celestial plane given by Euler angles

t a n=sin

1 cos

On eliminating the flight-path angle from the velocity expression, we have

= / / .In most practical applications, the spacecrafts position and velocity are resolved in a stationary

celestial frame that is fixed with respect to distant stars and has its origin at the center of the

celestial body. Examples of Earth-based celestial frames are the geocentric frames oGxGyGzG,where the plane (xGyG) is either the equatorial plane, or the ecliptic plane, and the axisoGxGpoints toward the vernal equinox. Normally, the equatorial plane is employed as the reference

plane, (

xGyG), when an orbit close to a planet is of interest, while the ecliptic plane is used for

interplanetary trajectories. The intersection of the orbital plane with the reference plane,

((xGyG), yields the line of nodes, as shown in Figure. The ascending node is the name given to thepoint on the line of nodes where the orbit crosses the plane ((xGyG) from south to north. A unitvector, n, pointing toward the ascending node makes an angle with the axis oGxG. The angle is measured in the plane (xGyG) in an anti-clockwise direction, and is termed right ascension ofthe ascending node. The inclination, i, is the angle between the orbital plane and (xGyG), and isthe positive rotation about n required to produce ihfrom the axis oGzG. The angle represents

7/27/2019 Strategic Missile Zarchan

72/75

72 | P a g e

a positive rotation ofn about ihto produce iein the orbital plane, and is called the argument of

periapsis.

4.2) Orientation of celestial frame with vernal equinox

4.4) Time Equation:

The vectors h and e completely determine the shape and orientation of a two-body trajectory,

but do not provide any information about the location of the spacecraft at a given time. Thismissing data is usually expressed as an equation that relates the variation of true anomaly with

time, (t), beginning with the time of periapsis, , for which = 0. On substituting the orbit

equation (6.13) into the angular momentum magnitude, equation (6.5), we have

= 1 cos Integration of equation (6.20) provides , thereby determining the function (t), and completing

the solution to the two-body problem. However, such an integration is usually carried out by a

numerical procedure, depending upon whether we have an elliptical, parabolic, or hyperbolicorbit.

7/27/2019 Strategic Missile Zarchan

73/75

73 | P a g e

Elliptical Orbit (0