CONCEPTS OF ROCKETRY

The following articles on rocketry which are collected from

http://hometown.aol.co.uk/chrlswrphl/favorite.html
Are given in this document. They are arranged in the order shown below.

Physics of Rocketry An introduction to the physics which is used in rocketry.

Includes concepts of force, work and energy

Forces on a Rocket Analysis of the linear and rotational forces acting on a rocket.

Propulsion The principles of rocket motors and the ideas of enthalpy,

nozzle design, flow expansion, impulse and power

The Rocket Equation Derivation of the rocket equation, multi stage rockets.

Radio Links Basic link calculations for radio links from rockets

Wind and Rocket Dynamics A paper considering the effects of wind profile on flight
dynamics

Logarithms and Decibels A primer in logs and decibels to support the section on radio
links.

Keplers Laws An introduction to the principles of orbits, conic sections, and

Keplers laws

Orbital Manouevers The principle types of orbital manouevres for spacecraft

Interplanetary Trajectories Trajectories for interplanetary flight, patched conic

approximation

Relativity and Rocketry Einstein's theory of special relativity explained using rocketry
examples

Wind Drift Model A paper presented at BROHP 2006. This gives the detail of
the wind drift model for rocket dispersion including the basic
equations and data used in the software.
Wind Drift Software This is the Excel spreadsheet with VBasic macro that
implements the wind drift model. It is uncalibrated so results
will be unreliable. It is being distributed so that people can
have a play and provide feedback.
 Physics of Rocket Flight
In order to understand the behaviour of rockets it is necessary to have a basic grounding
in physics, in particular some of the principles of statics and dynamics. This section
looks at the relationships between distance, velocity, acceleration, force, work, impulse,
energy and power. These basic ideas are often encountered in rocket science, and it is
useful to have a knowledge of how they interrelate.

Distance, Velocity & Acceleration

We all understand the concept of distance. We can draw two points and measure the
distance between them. If we know the time it takes to travel from one point to the next,
and we know the distance, we can calculate our speed. Or can we?

What we haven’t taken into account is the path we follow. For example, if we travel
from London to Bath, about 100 miles, in 5 hours we would say that we travelled at an
average speed of 20 mph. This, however, is incomplete information. If the journey took
us via Birmingham it may be that we travelled 250 miles, so our average speed was 50
mph. Clearly we need to think a little bit more about what we mean by distance and
speed.

In physics it normal to cal the distance between two points in a straight line the
displacement. The term “displacement” not only considers the distance between the
points but also the direction, thus it is a vector quantity. Speed also has an associate
direction in physics, and is thus a vector quantity called velocity. A change in either the
magnitude or the direction of the velocity vector is called acceleration. Most people can
identify with a change in the magnitude of the velocity vector due to acceleration, but
why the direction?

Newton’s first law tells us that “a body will remain at rest or uniform motion in a straight
line unless acted on by a force”. If we have a force acting on the body it could change its
motion from “uniform motion in a straight line” by either changing the uniform motion
(magnitude), moving it off the straight line (direction) or both. At any instant of time a
body which is being accelerated has an instantaneous velocity.

Imagine a model rocket. In terms if Newton’s first law it is “at rest” immediately before
launch. On ignition of the motor it is acted on by a force (the thrust of the motor) and
accelerates upwards. During its flight it may be acted on by other forces as it pushes air
molecules aside (drag) or is pressed by crosswinds. These result in accelerations which
may deflect it from its path, or change the magnitude of its velocity.

Sometimes we use calculus to express displacement, velocity and acceleration. Calculus

was another of Newton’s contributions to science. If we consider travel over a distance r,
we can define velocity v as the rate of change of distance, and acceleration a as the rate of
change of velocity. In mathematical terms:

Distance: r

dr
Velocity: v=
dt

dv d 2 r
Acceleration: d= =
dt dt 2

Distance is always measured in meters (m), velocity in measured in meters/second (m/s)

and acceleration in meters/second/second (m/s2). Having grasped these basic concepts of
displacement, velocity, and acceleration, let’s go on and consider forces.

Equations of Motion

The equations of motion for a body under a constant acceleration are often encountered in
physics. If we consider a body travelling at a velocity v1 m/s for a time t sec, we know
that it will travel a distance of r meters. We calculate this using the equation:

r = v1t

If we apply a force to the body we will cause the body to accelerate. We can add this
factor into the equation:

1
r = v1t + at 2
2

If we differentiate this equation with respect to time we get:

dr d (v1t ) 1 d at 2
= +
( )
dt dt 2 dt

which simplifies to:

v 2 = v1 + at

So is we know the acceleration and initial velocity v1 we can find the velocity v2 at any
subsequent time. By rearranging the above equations we get the final equation:

2 2
v 2 = v1 + 2ar
These equations are very useful, but also very misleading. They only apply to motion
under a constant acceleration, whereas in rocketry the acceleration is seldom constant.
The equations become useful when using numerical methods to approximate the motion
of a body under varying accelerations. In these circumstances the acceleration is viewed
as “instantaneously constant” over a very short interval of time, typically thousandths of a
second.

They should NOT be used in general rocket analysis as the results can be highly
misleading.

Force

Newton’s second law defines force as:

Force = mass x acceleration

In its convenient mathematical form:

dv
F = ma = m .................... equation 1
dt

If we apply a force to a body with low mass (light), the one with high mass (heavy) we’ll
find that the light body will accelerate more. Another way of thinking about force is to
think about acceleration. If we want the two bodies to accelerate at the same rate we
must apply a larger force to the heavier body. Force has other more subtle definitions.
One used in rocketry is:

Force = rate of change of momentum

This is a variation of equation 1 in which neither mass nor acceleration are constant. This
is the case in rockets where the mass decreases as fuel is burned. Mathematically we can
express this as:

d (mv )
F=
dt

The correct unit for measuring force is the kilogram-meter/second/second. This is a bit
of a mouthful so, as is common in science the unit was named after a great scientist. The
unit of force is thus the “Newton” (N).

Weight is also a force, however there is a common misconception the unit of weight is
the kilogram. In fact this is the unit of mass, to convert this to weight (a force) the mass
must be subjected to some acceleration. On Earth, this acceleration is due to gravity and
has a value of 9.81 m/s2. A rocket with a mass of 2 kg thus has a weight of 19.62 N.
This is important when considering the rating of motors where the “weight” of the rocket
on the scales is in fact its mass. The weight is about 10 times more.

Example: to safely launch a rocket a thrust to weight ratio of at least 5:1 is required.
The rocket weighs 0.4 kg, and the motor has a thrust of 15N. Is this safe?

Answer: The weight of the rocket is 0.4 x 9.8 = 3.9 N. Thrust is 15 N. The thrust to
weight ratio is thus 15:3.9, or 3.8:1. This is less than 5:1 so the launch would be unsafe.

Momentum

From the previous section we can see that momentum can be defined as:

Momentum = mass x velocity

The momentum of an object is simply the product of its mass and velocity. A rocket with
a mass of 2 kg travelling at 100m/s has a momentum of 200 kgm/s. Similarly a car with a
mass of 1000kg travelling at 30 m/s has a momentum of 30,000 kgm/s. We often use P
to represent momentum, thus:

P = mv

From our definition of force above, we can see that:

d (mv) dP
F= =
dt dt

Momentum has an interesting property in that it is always conserved. It can thus be

transferred from one object to another. Imagine a red snooker ball of mass m travelling
along a table with a velocity v, where it hits a stationary blue ball of mass m. Before the
impact the total momentum is mv plus the momentum of the seconds ball (zero, since its
velocity is zero).

After the impact both balls are moving. There is an important principle of physics which
says that the total momentum in a system is constant, in other words it is conserved. If
we look at a diagram of the balls after the collision:
The red ball and the blue ball are travelling with velocities v1 and v2 respectively. The
momentum of the red ball is thus mv1 and that of the blue ball is mv2. As momentum is
conserved, the momentum in the “system” before the collision is equal to the momentum
in the system after the collision, thus:

mv = mv1 + mv 2

The derivation of the rocket equation makes use of the property of conservation of
momentum.

Energy

Energy, like momentum, is conserved. It cannot be created or destroyed (although

Einstein demonstrated that it can be converted into mass). If we look at a “system”
which has no energy entering or leaving it (called a “closed system” in physics), then the
total energy in the system will be constant. The energy within the system may change
form, but the total energy will be constant.

The unit of energy is the Joule.

In rocketry we are concerned with three main types of energy: kinetic energy (KE),
potential energy (PE) and chemical energy. Kinetic energy is the energy of a moving
object, and can be easily calculated if we know the mass m of the object and its velocity
v:

1 2
KE = mv
2

A rocket with a mass of 10kg travelling at a velocity of 100 m/s has a KE of ½ x 10 x

1002 Joules, or 500,000 Joules.

Potential energy describes the energy which a rocket has by virtue of its position. A
rocket 10 metres off the ground has energy which could be released if it were allowed to
fall. If the rocket was dropped, the PE would fall to zero when it landed. PE can be
calculated easily: If an rocket of mass m is held at a height h above the ground (we’ll
assume that PE is zero at ground level), then the PE can be calculated from:

PE = mgh

The value of g is the acceleration due to gravity (9.81 m/s2). We know that the mg term
is a force and is thus measured in Newtons, since it is mass multiplied by acceleration.
The height h is measured in meters. We can thus see that potential energy is measured in
Newton-meters, and that these are the same as Joules.

Another form of energy, chemical energy, is more complex. It is the energy “stored” in
the chemical bonds within the propellant, and released when the propellant is burned.
The chemical energy released for every molecule liberated by the chemical reactions in
the propellant can be approximately calculated as:

5
CE = kT
2

The term k is Boltzmann’s Constant (1.38 x 10-23 J/K) and T is the absolute temperature
of the chemical reaction. It can be seen that the hotter the reaction, the more chemical
energy is released.

Putting all this together, we can start to consider the energy in a rocket flight. When the
rocket is on the launch pad it is at height of 0 m, and is travelling at 0 m/s. It thus has no
PE or KE, but it does have a lot o CE stored in the propellant.

After ignition, and during the burn phase, the rocket loses some of this CE, but acquires
PE and KE as it gains height and speed. At burnout the CE goes to zero as there is no
propellant left to burn, and the KE is at a maximum as the rocket is travelling at its
maximum velocity. When the rocket reaches apogee it has no velocity for an instant, it
thus has no KE, no CE, but has the maximum amount of PE. In descent, at a constant
rate, the rocket has a constant KE and diminishing PE. Finally at landing the rocket has
no PE, no KE and no CE.
Hang on, didn’t we start by saying that energy was conserved? We’ve just seen a rocket
start with a load of CE and finish with nothing! So where did all the energy go?

The answer is that the rocket is not a “closed” system. Energy has left the rocket in many
forms:
• As heat, passed to the atmosphere by the hot exhaust gases
• As sound, the energy was passed to the atmosphere as pressure which we detected
as sound
• As turbulence, as the rocket and its parachute moved the air

If we flew the rocket in a large, closed, room and measured all the different energy
outputs we would find that the total energy in the room would remain constant. Energy
would be conserved.

Work

The concept of “work” gets relatively little attention in physics, and yet is provides an
essential link between force and energy. It also allows us to shortcut the analysis of
systems where the forces, and hence accelerations, are not constant. At its simplest level,
and assuming a constant force, we can define work by considering a force F acting on a
body and causing the body to be displaced by a distance r. The work done of the body is
the product of the displacement and the force in the direction of the displacement.

W = Fr

To understand the usefulness of work we need to consider more complex systems. The
forces acting on a rocket will vary with time. We can consider a force which varies with
position.

Force F(r)

r1 r2
Distance r

We denote the force at a position r as F(r), so the work done at position r is:

The work done by the force over the distance r1 to r2 can be approximated by adding up
all the work done over each element ∆r. As we make ∆r progressively smaller, this
addition becomes integration, thus we can calculate the work as:

r2

W = ∫ F (r )dr
r1

Work is measured in Newton-meters. We know from considering potential energy that it

is also measured in Newton-meters, and that this energy is measured in Joules. Are work
and energy similar? The answer is yes. We can readily show a relationship between
work and another form of energy, kinetic energy.

We know that:
 dv
F =m
dt

Thus we can say that:

dv dr
W = ∫m dr = ∫ m dv
dt dt

We saw earlier that velocity is the rate of change of distance, so that:

dr 1
W = ∫m dv = ∫ mv.dv = mv 2
dt 2

Work and energy are thus very closely related. When we “do work” due to a force in the
direction of travel we give the body energy. Conversely, forces opposing the direction of
travel rob the body of energy. Forces at right angles to the direction of travel do no work.

Impulse

Impulse is the product of Force and time. If we apply a force F to a body for a time t the
impulse is:

I = Ft

Of course this assumes a constant force. In rocketry the net force varies with time as the
mass and acceleration of the rocket are not constant. In these circumstances the total
impulse can be regarded as the area under the force/time curve:

The impulse is the shaded area. Mathematically,

I = ∫ Fdt

Sometimes in rocketry we use the term “specific impulse”. This is used to describe the
impulse provided by 1 kg of propellant. This term is used more in British text books than
American ones.

Power

Power is defined as:

Power = the rate of use of energy = the rate at which work is done

Power is measured in Joules/second, also known as Watts. Another unit named after a
great scientist. A common misconception is that a motor with “G” impulse is twice as
powerful as one with “F” impulse. In fact it has twice as much impulse, not power. If we
imagine a motor which gives a thrust of F Newtons for t seconds, it has an impulse of:

I = Ft Newton-seconds

If it moves the rocket r meters in that time, it does W Joules of work where:

W = Fr Joules

The power of the rocket P is the energy divided by the time, in other words the rate of
doing work, so that:

W Fr
P= = Watts
t t

From which we can derive a relationship between power and impulse:

r
P=I
t2

In other words, the power of a rocket depends on the impulse, the distance over which the
motor burns, and the square of the time of the burn. Impulse and power are clearly NOT
the same thing!
 Forces on a Rocket
In order to understand the behaviour of rockets it is necessary to have a basic grounding
in physics, in particular some of the principles of statics and dynamics. This section
considers the forces acting on a rocket, and how they affect its performance. In
particular, the section looks at the linear and rotational behaviour of a rocket.

Forces

Newton’s second law leads to the well known equation:

Force = mass x acceleration

Expressed mathematically it can take one of several equivalent forms:

dv d 2r
F = ma = m =m 2
dt dt

Most basic text books leave the concept of force at this point, possibly illustrating the
idea of acceleration as being a change of velocity over time. The conventional view of a
model rocket is to consider three forces acting on a rocket: thrust, drag and weight.

Forces on a rocket

Weight

Drag

Thrust

The forces on a rocket vary throughout the flight. At a basic level the weight reduces as
propellant is consumed, the thrust changes depending on the burn profile, and drag
increases with the square of the velocity. Most model rocket flights take place within a
few thousand meters of the Earth, however in higher altitude flights other “constants”
start to change. Air density, used to calculate drag, changes with temperature and hence
with altitude. Even gravity reduces slightly as a rocket moves away from the Earth. Our
Newton’s second law equation starts to look complex as it contains nothing which is
constant.

Motion in the direction of flight

The motion of a rocket can be predicted if we can understand how the acceleration varies
with time. Knowledge of the rocket’s acceleration allows us to calculate its velocity and
altitude through calculus. If we can arrive at a generalised equation for how acceleration
varies with all the other possible factors, we can apply this to any rocket, motor, flight
profile, atmosphere or even planet and predict how our rocket will behave.

Let’s consider a rocket in the atmosphere flying at some angle to the horizon and with an
angle of attack to the airflow. We will regard the rocket as having a fixed motor so that
thrust is always aligned with the axis of the rocket. The forces acting on the rocket, their
directions and apparent locations are shown on the diagram below.

Axis

Velocity v

Angle of Attack α

γ
Horizon

Drag D

Thrust F
Weight mg

Note that the thrust acts along the axis, aerodynamic forces act through the centre of
pressure, and weight acts through the centre of gravity. As acceleration acts through the
centre of gravity, the diagram shows velocity acting through that point. If we consider all
the forces, and components of forces, acting in the direction of the instantaneous velocity
vector (direction of flight):

dv
m = F cos(α ) + mg cos(90 + γ ) − D
dt

Rearranging this equation, and assuming that the angle of attack is small (less than
10degrees) we arrive at an equation for the acceleration in the direction of flight.

dv F D
= − g sin(γ ) − ........ equation 1
dt m m

Let’s consider this equation. The thrust force acts in a forward (positive) direction and
the gravity and drag terms act to oppose it (negative direction). The acceleration is
positive if the F/m component is greater than the other two, which happens when the
thrust is larger than the combination of weight & drag. If the thrust term is zero, then the
acceleration is negative, which makes sense as the rocket will be slowing down. In
horizontal flight the weight component is at right angles to the direction of travel, so it
introduces no acceleration in that direction.

This gives us an equation for the acceleration in general terms, but many of the terms are
also variable. Mass m decreases as fuel is burned. The thrust F changes throughout the
burn. Drag D changes with the square of the velocity v. To make use of this general
equation we need to know how each of the terms varies with time. In practice these
equations are extraordinarily complex to solve, and the general approach is to solve them
using numerical methods such as linear extrapolation and Runga-Kutta’s method.

Linear extrapolation treats the equations of motion as being linear over very short time
intervals. The values of distance, velocity, acceleration and all the forces are initially
zero. Over a very short time interval the acceleration, forces, air density and other factors
are assumed to be constant. This allows calculation of the values at the end of this time
interval, and establishes the values for the start of the next time interval. By repeating
this over many time intervals it is possible to approximate the values at any time in the
future. It will be obvious that this process results in accumulation of errors over a period
of time as small errors will be carried forward into all future calculations. In order to
increase the accuracy of results over long periods the time intervals must get smaller.
Eventually the method collapses under the weight of computational intensity. It is,
however, quite simple to understand and lends itself to spreadsheets.

Runga-Kutta’s method is a numerical technique for solving first and second order
differential equations. It is less computationally intensive that linear extrapolation, and is
generally more accurate, but requires a firmer grasp of mathematics. It is not normally
taught until second year on engineering degree courses.
Motion perpendicular to the direction of flight (Pitch)

Having considered the motion along the direction of flight, we’ll now take a look at the
motion at right angles to the direction of flight. The forces in this direction are in
equilibrium if the rocket is flying straight, thus accelerations are zero. The first gust of
crosswind soon knocks the rocket out of equilibrium and we then have to consider the
restorative forces from the fins, forces from the wind, and any components of thrust
which are no longer along the direction of flight.

The rocket will tent to rotate about its centre of gravity, with all the aerodynamic forces
acting through its centre of pressure. Forces which do not pass through the centre of
gravity, which is by definition the axis of rotation, are called torques. A torque is the
force multiplied by the distance at which it acts from the axis of rotation.

The forces are thus “torques” about the centre of gravity, and since they are trying to alter
the direction of flight (pitch) we call them moments of pitch. Physics purists will criticise
this simplistic explanation, however it will suffice for our purpose.

Newton’s second law allows us to analyse linear motion but needs to be adapted to
analyse rotational motion. As shown earlier, the basic equation that we use for linear
motion is:
d 2r
F =m 2
dt

Torques cause a rotational acceleration which is measure in degrees per second per
second or more usually in radians per second per second, where 360 degrees is 2π
radians. In linear motion we have mass, which has an inherent inertia which resists
acceleration. It requires more force to accelerate a heavy object than a light one. In
rotational motion the equivalent of mass is called the moment of inertia, denoted I.

There is an equivalent form of this equation when we consider rotational motion. We say
that:

Torque = moment of inertia x angular acceleration

Or mathematically:

d 2θ
τ =I
dt 2

The equivalence of the linear and rotational equations should be obvious. In place of
force F we have torque τ, in place of mass m we have moment of inertia I, and in place of
distance r we have angle θ. Moment of inertia is a relatively straightforward idea. We
define moment of inertia as the sum of all the masses which comprise a body multiplied
by the square of the distances:
 Axis

Velocity v

Angle of Attack α

γ
Horizon
Lift L
d1
Wind W

d2

Drag D

Thrust F
Weight mg
From the diagram it can be seen that the forces causing an angular acceleration about the
centre of gravity are aerodynamic: the lift due to the fins and any torque inducing forces
such as crosswinds. These act through the centre of pressure. These torques combine to
overcome the moment of inertia and cause an angular acceleration of the rocket.

It is useful to express this mathematically in a generalised form, that is an equation

involving all terms. This allows various forms of the equation to be derived, for example
versions with and without thrust or wind, positive or negative or zero angles of attack etc.
For convenience, we’ll define a clockwise rotation as being positive and the direction of
the velocity vector as the reference direction for all angles:

d 2α
I = Fd 2 sin(α ) − Ld 1 − Wd 2 cos(γ ) ........ equation 2
dt 2
This is an important equation, and we’ll have a look at each of the terms in this equation
before deriving a more informative version.

Moment of Inertia. The moment of inertia of a body is a function of its mass

distribution. In some texts it is referred to as the “rotational inertia” or as the “second
moment of mass distribution”. We define moment of inertia as:

I = ∑i =1 mi ri 2 ........ equation 3
k

If we have a body which comprises k pieces, and each of these has a mass of mi kg and is
located ri metres from the bodies centre of gravity, we can calculate I by adding up the
total of all the mi ri2 for the whole body. As an example, if a body comprises 3 blocks,
two of which have masses of 4 kg and one has a mass of 3 kg. They are linked by a
lightweight beam as shown below:

The moment of inertia is:

I = ∑i =1 mi ri 2 = m1 r12 + m2 r22 + m3 r32 = 4 × 5 2 + 4 × 3 2 + 1 × 8 2 =200 kg m2

3

It can be seen from the example what is meant by mass distribution. If we move the
masses around we don’t change the overall mass of the system, but we will change their
location and hence the moment of inertia. To calculate the moment of inertia of a rocket
we need to know the mass, centre of gravity location, and distance from the rocket’s
centre of gravity of all its constituent pieces. Armed with this knowledge we can easily
calculate I.

Lift. The section on aerodynamics shows that the lift of the fins can be calculated from:

1
L= ρAF C L v 2
2

For small angles of attack (less than 10 degrees) the coefficient of lift CL of a flat fin is
proportional to the angle of attack. We can thus define a constant kL for small angles of
attack such that:
C L = k Lα

Thus

1
L= ρAF k L v 2α ........ equation 4
2

Reconciliation

If we substitute equation 4 into equation 2 we get:

d 2α 1
I = ρAF k L v 2 d 1α − Wd 2 cos(90 − γ ) + Fd 2 sin(α )
dt 2 2

A useful device called the “small angles approximation” tells us that for small angles of
attack sin(α)~α , as long as the angle is measured in radians. We thus get:

d 2α 1
I 2
= ρAF k L v 2 d1α − Wd 2 cos(90 − γ ) + Fd 2α
dt 2

Or with minor rearrangement to collect terms containing α:

d 2α  1 
I 2 =  ρAF k L v 2 d1 + Fd 2 α − Wd 2 cos(90 − γ ) ........................ equation 5
dt 2 

Maths tyros will recognise this as a second order differential equation in terms of α. So
what, I hear you ask. Well the solution for a second order differential equation of this
form is a sinusoid, as long as the force is acting to restore the rocket to straight flight.
This means that if the rocket has an angle of attack the restoring force will cause it to
swing into positive and negative angles of attack. The frequency of this oscillation will
depend of the moment of inertia I of the rocket, and its amplitude will depend on the lift
generated by the fins, whether thrust F is applied and the distance d1 between the CP and
CG. This bit of maths has just explained why we sometimes see oscillating smoke trails
in marginally stable rockets and linked it to the factors which affect the frequency and
amplitude of the oscillation.

We’ve used the small angles approximation quite a bit here, but have never justified it. If
you’ve read any aerodynamics you’ll recall that the lift coefficient increases with
increasing angles of attack. When it increases beyond a critical angle, typically 12
degrees, the lift coefficient drops sharply as the airflow stalls. At this point the fins cease
to apply any aerodynamic force and the rocket becomes unstable. Not only is 10 degrees
a convenient mathematical upper limit beyond which sin(α)≠α, but it also conveniently
matches the aerodynamic limits of the fins.
 Rocket Propulsion
In the section about the rocket equation we explored some of the issues surrounding the
performance of a whole rocket. What we didn’t explore was the heart of the rocket, the
motor. In this section we’ll look at the design of motors, the factors which affect the
performance of motors, and some of the practical limitations of motor design. The first
part of this section is necessarily descriptive as the chemistry, thermodynamics and maths
associated with motor design are beyond the target audience of this website.

General Principles of a Rocket Motor

In a rocket motor a chemical reaction is used to generate hot gas in a confined space
called the combustion chamber. The chamber has a single exit through a constriction
called the throat. The pressure of the hot gas is higher than the surrounding atmosphere,
thus the gas flows out through the constriction and is accelerated.

Combustion chamber Throat Nozzle

It sounds simple, so why is rocket science so complex? Well, firstly there’s chemistry
and the selection of the right reagents from many thousands of possibilities. Then there’s
the design of the motor to make it capable of withstanding the temperatures and pressures
of the reaction while still being as light as possible. There’s also the design of the throat
and nozzle to ensure that the exhaust velocity is as fast as possible. Putting all these bits
together, the average rocket scientist needs (as a minimum) to understand chemistry,
mechanical engineering, thermodynamics, materials science and aerodynamics.

Propellants

The chemical reaction in model rocket motors is referred to as an “exothermal redox”

reaction. The term “exothermal” means that the reaction gives off heat, and in the case of
rocket motors this heat is mainly absorbed by the propellants raising their temperature.
The term “redox” means that it is an oxidation/reduction reaction, in other words one of
the chemicals transfers oxygen atoms to another during the reaction (OK chemists, I
know that this is not a comprehensive definition but it will suffice!). The two chemicals
are called the oxidising agent and the reducing agent.

The most popular rocket motors are black powder motors, where the oxidising agent is
saltpetre and the reducing agents are sulphur and carbon. Other motors include
Potassium or ammonium perchlorate as the oxidising agent and mixtures of hydrocarbons
and fine powdered metals as the reducing agents. Other chemicals are often added such
as retardants to slow down the rate of burn, binding agents to hold the fuel together (often
these are the hydrocarbons used in the reaction), or chemicals to colour the flame or
smoke for effects. In hybrid motors a gaseous oxidiser, nitrous oxide, reacts with a
hydrocarbon, such as a plastic, to produce the hot gas.

Energy Conversion

This reaction releases energy in the form of heat, and by confining the gas within the
combustion chamber we give it energy due to its pressure. We refer to the energy of this
hot pressurised gas as its “enthalpy”. By releasing the gas through the throat the rocket
motor turns the enthalpy of the gas into a flow of the gas with kinetic energy. It is this
release of energy which powers the rocket. So the energy undergoes two conversions:
• Chemical energy to enthalpy
• Enthalpy to kinetic energy

The conversion from chemical energy to enthalpy takes place in the combustion chamber.
To obtain the maximum enthalpy it is clearly important to have a reaction which releases
lots of heat and generates lots of high energy molecules of gas to maximise pressure.
There is clearly a limit to the temperature & pressure, as the combustion chamber may
melt or split if these are too high. The designer has a limitation placed on his choice of
reagents in that the reaction must not heat the combustion chamber to a point where it is
damaged, nor must the pressure exceed that which the chamber can survive.

Changing enthalpy to kinetic energy takes place in the throat and the nozzle. Our mass of
hot gas flows into the throat, accelerating as the throat converges. If we reduce the
diameter of the throat enough, the flow will accelerate to the speed of sound, at which
pint something unexpected occurs. As the flow diverges into the nozzle it continues to
accelerate beyond the speed of sound, the increase in velocity depending on the increase
in area. This type of nozzle is called a De Laval nozzle.

Nozzle

Throat

You will recall that the kinetic energy of a body can be calculated from:

1 2
KE = mv
2
If we consider a small volume of gas, it will have a very low mass. As we accelerate this
gas it gains kinetic energy proportional to the square of the velocity, so if we double the
velocity we get four times the kinetic energy. The velocity of the supersonic flow
increases proportional to the increase in area of the nozzle, thus the kinetic energy
increases by the fourth power of the increase in nozzle diameter. Thus doubling the
nozzle diameter increases the kinetic energy by 16 times! The De Laval nozzle make
rocket motors possible, as only such high velocity flows can generate the energy required
to accelerate a rocket.

In model rockets the reaction is chemical generally short lived, a few seconds at most, so
the amount of heat transferred to the structural parts of the motor is limited. Also, the
liner of the motor casing acts to insulate the casing from the rapid rise in temperature
which would result from a reaction in direct contact with the metal casing. Model rocket
motors also run at quite low pressure, well below the limits if the motor casing, further
protecting the casing. It can be seen that the enthalpy of a model rocket motor is thus
quite low. In large launch vehicles such as Ariane, the pressure and temperature are high,
the burn may last several minutes, and the mass budget for the designer is very tight.
Designing motors for these purposes is highly complex.

Thrust

The basic principles of a rocket motor are relatively straightforward to understand. In

rocketry the motor exists to accelerate the rocket, and thus it has to develop a force called
“thrust”. One of several definitions of force is that:

Force = rate of change of momentum

If we ignore (for a few paragraphs) any external effects we can say that the thrust is
entirely due to the momentum of the propellant, a force called the “momentum thrust”. If
we denote the thrust as F and the momentum as P, then mathematically:

dP
F=
dt

Sometimes for mathematical clarity we us the notation of P with a dot on top to denote
the first derivative of P, and with 2 dots for the second derivative. Thus, in this new
notation:

dP &
F= =P
dt

You may also recall from the section on the rocket equation that momentum is the
product of the mass and velocity. Thus we can say that the momentum of the flow from
the nozzle of the rocket has a momentum:
P = mve

Thus:

dP & d (mve )
F= =P=
dt dt

If the exhaust velocity remains constant, which is a reasonable assumption, we arrive at

the equation:

d ( m)
F = ve = m& ve ...... equation 1
dt

The term “m-dot” is known as the mass flow rate, in other words the rate at which mass
is ejected through the nozzle in kg/sec. In other words this is the rate at which the rocket
burns fuel. This is an interesting relationship, which can be expressed in words as:

Momentum Thrust = mass flow rate x exhaust velocity

Flow expansion

The propellant is accelerated into the atmosphere. As it leaves the nozzle the propellant
has an exit pressure Pexit and enters an atmosphere which has a pressure Patm. The
transition from one pressure to the other cannot happen instantaneously as any pressure
difference will cause a flow of high pressure fluid into the low pressure region. So what
does this do to the thrust?

Pexit Patm

We define pressure as:

force
pressure =
area

So the force (a component of thrust called “pressure thrust”) depends on the pressure
difference and the area of the nozzle. If the area of the nozzle is A, we can produce an
equation for the total thrust:

F = m& ve + A(Pexit − Patm )

This suggests that we should aim for a maximum pressure in the nozzle so that the
pressure thrust combines with the momentum thrust to produce a greater overall thrust.
In fact, this intuitively correct result is wrong! If Pexit > Patm the exhaust gases will
expand in all directions when they leave the nozzle, not only the direction of thrust. The
total thrust of such a motor is less than could be delivered by just momentum thrust. We
call this an “under expanded” flow as the propellant needs to expand more within the
nozzle.

So what if Pexit < Patm? In this circumstance the atmosphere will try to flow back into the
nozzle. This causes sudden transitions from supersonic to subsonic flow to occur in the
nozzle setting up shock waves. These shock weaves turn some of the kinetic energy of
the flow back into enthalpy, reducing the overall thrust. We call the flow “over
expanded” as the flow expands too much in the nozzle reducing the overall pressure.

The ideal situation is when Pexit = Patm which only occurs over a narrow range of
altitudes. This is not a major problem for modellers, as the burns tend to occur at low
altitudes and over a relatively narrow range of atmospheric pressures. It is easy to design
motors which are efficient over this range. It is a real problem for manufacturers of
launch vehicles as the motor may burn from sea level to several tens of miles above sea
level. It is normal practice on major launchers to tune the motor for an altitude around
the middle of the range of pressures and accept some loss of efficiency at the start and
end of the burn.

This effect is very pronounced on the Saturn V rocket. Next time you see any video of a
launch, watch the plume. At launch it is long and thin as the flow is over expanded. At
high altitudes the plume is very wide, exhibiting under expansion.
 At launch At altitude
over expanded under expanded

Propellant Grain

Solid propellants are the most common type used in model rocket motors. The propellant
is ignited at the end away from the nozzle. The only escape route for the hot gas is to
flow through the grain to the nozzle. As the gas flows through the grain it ignites all the
exposed surfaces of the grain. As the surface burns away it exposes more grain to burn
until it has all burned away.

The diagram shows a simple grain, a hollow cylinder. The area of the burning surface is
the sum of the area of the top disc and the cylinder through the grain. As the burn
progresses this surface area changes, thus the amount of hot gas changes. The amount of
hot gas produced is directly proportional to the surface area.
 1 2 3
The mass of hot gas produced per second is the mass flow rate, and thrust depends on
mass flow rate. We saw earlier in this section that thrust is directly proportional to mass
flow rate, so the thrust thus depends on the burning surface area. We can use this
property to change the thrust profile.

By arranging the grain so that the burning surface area increases with time we get a
profile where the thrust increases with time. This is called a progressive burn.
Conversely if the area decreases with time we get a reduction in thrust or regressive burn.
If the area stays constant we get constant thrust or a neutral burn.

Thrust

Regressive Progressive

Neutral

Time
In practice there is not a grain geometry which can give a truly neutral burn. Most
neutral grains will give a degree of regression or progression.
Some common propellant grains used on model rocketry are shown below. Most black
powder motors use an end burn. These are ignited from the bottom. Slotted tubes are
used in medium and high power rocketry, and these are ignited from the top end of the
motor.

End burn Slotted tubes Internal burning tube

(Neutral) (neutral) (Progresive)

Specific Impulse

How do we measure the effectiveness of a rocket motor? In cars we compare the

effectiveness of motors through measures like miles per gallon of fuel, and time for 0-60
mph. The equivalent in rocket motors is called the specific impulse (Isp). Isp is defined
as:
Ft
I sp = ........ equation 2
W

Where F is the thrust in Newtons, t is the duration of the burn in seconds, and W is the
weight of fuel in Newtons. Overall this gives a measure of the impulse Ft provided by a
weight of fuel W. If we think about this, both F and W are forces, thus SI has the units of
seconds. If we imagine rocket motor with an Isp of 300 seconds, then Newton of fuel (i.e.
1 kg under the acceleration due to earth’s gravity at sea level) will give 1 Newton of
thrust for 300 seconds. The same amount of fuel could also give 150 Newtons of thrust
for 2 seconds. It can be seen that the notion of Isp gives a measure of the effectiveness of
a motor and fuel combination which is independent of the rate at which the fuel burns.

Some typical values of Isp are:

Black powder: 20-40 seconds

Ammonium perchlorate and aluminium: 100-150 seconds
Liquid oxygen and liquid hydrogen: 400 seconds.

By considering the mass flow rate of the motor as instantaneously constant, we can
modify equation 1 to read:

m
F = ve
t

We also know that the weight of fuel W is the mass of fuel multiplied by the acceleration
due to gravity, so that
W = mg 0

Substituting for F and W in equation 2, and manipulating, we get:

ve
I sp = ........... equation 3
g0

Thus specific impulse is directly proportional to the exhaust velocity, ve. The constant of
proportionality is 1/g0, where g0 is the acceleration due to gravity at sea level.

Why is equation 3 significant? It shows that the higher the exhaust velocity the more
efficient the motor becomes. In theory, we can keep on increasing the exhaust velocity
and hence the efficiency of the motor. There are practical issues such as chamber
temperature, pressure and flow expansion which limit the efficiency of chemical motors.
Once outside the atmosphere we can accelerate ions to very high velocities in the vacuum
of space, and thus get ion propulsion motors with Isp of many thousands of seconds, but
that is another story.....
 Momentum and the Rocket Equation
The rocket equation gives an explanation of how the gas ejected from the nozzle is used
to propel the rocket forwards. In its classical form it has little practical use in model
rocketry, as it assumes no drag and no gravity. It does, however, provide a useful means
of understanding the relationships between the mass of the rocket, mass of fuel, exhaust
velocity and the “burn out” velocity of a rocket.

This short paper considers the idea of momentum, and examines how an understanding of
momentum can be used to derive the rocket equation. It then examines some of the
design “trade offs” in building a rocket.

Momentum

The behaviour of a rocket motor can best be explained by understanding the principle of
conservation of momentum.

We can consider a rocket at some point during its flight as having a mass of m with a
little bit of fuel of mass dm about to leave the rocket. At this point the rocket has a
velocity v. An instant later, the element of propellant has left the nozzle at the exhaust
velocity (relative to the rocket) of -ve. Why the minus sign? Well the bit of propellant is
travelling in the opposite direction to the rocket, and positive velocities are in the
direction of travel of the rocket so anything travelling in the opposite direction is
negative.

As a result of ejecting this small bit of propellant the rocket increases its velocity by a
small amount, +dv. This is positive because it’s in the direction of travel of the rocket.
Conservation of momentum tells us that the momentum before the propellant is ejected is
the same as the momentum after the propellant is ejected. Thus:

Initial momentum = (m + dm)v

Final momentum = m(v + dv) – dm ve

As conservation is conserved, the initial momentum and the final momentum are equal,
thus:

(m + dm)v = m(v + dv) – dm ve

If we rearrange this, ignore second order terms, and integrate over the duration of the
burn we find that:

m 
v final = ve ln initial 
m 
 final 

Where ln denotes the natural logarithm. This is the classic form of the rocket equation.
The initial mass of the rocket is the total structural, payload and fuel mass, whereas the
final mass assumes that all the fuel has been burned.

The Rocket Equation

How can we use this equation in practice? Imagine we have a rocket which has a “dry
mass” mr , a payload mass mp and starts with propellant of mass mf If the rocket burns
the propellant and ejects it from the nozzle at a rate ve, then the “burn out” velocity vb
obtained by burning all the propellant is:

 mr + m p + m f 
vb = ve ln  ……. equation 1
 m +m 
 r p 

So what does this tell us? Well for a start, if we have a way of calculating the final, or
“burn out” velocity of a rocket which depends on the exhaust velocity, the mass of fuel
and the mass of the rocket. The exhaust velocity depends on the design of the motor, and
the amount of fuel and mass of the rocket depend on the design of the rocket.

Lets explore this equation further. If the amount of fuel carried is very small compared to
the mass of the rocket, then the bit in brackets has a value very close to 1. Since the
natural logarithm of 1 is zero (try it on your calculator), we can see that the change in
velocity is zero. This makes sense, as a rocket with very little fuel is not going to go very
far. Let’s consider the other “limiting case”, a rocket which is almost entirely fuel. In
this case mf is much greater than mr so the bit within the brackets gets close to infinite.
As the natural log of infinite is a very large number, we end up with a velocity many
times the exhaust velocity. This also makes sense, as a long burn will continuously
accelerate the rocket to a very high velocity.

It can be seen that there is a relationship between the performance of a rocket and the
relative masses of the structure and fuel. To explore this further, we need to
We can define two useful terms which can be used to quantify this relationship: the
payload ratio and the structural ratio. The payload ratio, denoted π, is simply the mass
of any payload carried by the rocket to the total mass of the rocket:

mp
π=
mr + m p + m f

The structural ratio, denoted ε, is simply the ration of the structural mass of the rocket to
the total mass of the rocket:

mr
ε=
mr + m p + m f

Manipulating equation 1 we get:

vb = −ve ln(ε + (1 − ε )π )

There is thus a maximum velocity that can be obtained by a rocket, and that occurs when
there is no payload, in other words when π = 0. This result is intuitively correct. The
value of this maximum velocity increases as the structural ratio decreases. This, too, is
intuitively correct as the structural ratio with no payload present indicates how much of
the rocket is structural. As the remainder of the mass comprises propellant it is easy to
see why a low structural ratio results in a higher burnout velocity. A typical satellite
launcher comprises about 80% propellant at launch.

MULTISTAGE ROCKETS

So far we’ve considered single stage rockets, how can we apply this to multistage
rockets? The trick is to recognise that all the stages above the one which is burning are
its “payload”. Each stage thus has a separate payload ratio, and it can be shown that the
payload ratio for the whole rocket is simply the product of all the payload ratios. For
example, if we have a rocket with 3 stages the payload ratio for the whole rocket is:

π total = π 1 × π 2 × π 3

Similarly for a rocket with N stages, the payload ratio is:

π total = π 1 × π 2 × π 3 × K × π N

It is not intended to prove it in this essay, but the burn out velocity of a multistage rocket
can be maximised by making the payload ratio of all the stages identical. Practically,
such a rocket cannot be readily built as the designer will reach a state for low values of N
where the stages become so massive that the rocket ceases to be affordable. It can be
shown that, for any given exhaust velocity ϖε, and payload ratio π, the advantages of
more than 4 stages are negligible.
 A Guide to Radio
Introduction
Radio contains elements of both science and “black art”. The science allows you to determine
what should happen to a radio signal over a predictable radio path. The black art explains what
really happens because no radio path is wholly predictable. This section of the website explains
the science of predictable radio paths and introduces some of the factors which can be used to
take account of the lack of predictability.
This section does not cover the design or building equipment as most rocketry is based on
commercial equipment. I would refer anyone thinking of building their own transmitters,
receivers or antennae to the excellent range of publications by the Radio Society of Great Britain
(RSGB) or the American Radio Relay League (ARRL). The aim is to link the specifications of
equipment, such as transmitted power, receiver sensitivity, antenna gain, to be used to generate
the maximum range, antenna pointing accuracy, and other metrics which will allow the system
performance to be estimated.
Radio calculations require a working knowledge of decibels. This site contains a short primer on
logarithms and decibels for those who wish to re-acquaint themselves with decibels.
General Principles
Radio links from a rocket can be used for many purposes:
• Telemetry, used to carry information about the rocket or payload to the ground in real
time.
• Tracking, to locate the position of the rocket in the air
• Location of the rocket after it has landed
• Payload data, for example live video cameras
The majority of electronic devices in model rocketry are commercial devices which need to be
integrated and powered. The instructions for the use of these devices are normally quite adequate
to bolt together the bits and make them work in the workshop. It is more difficult to predict how
the radio system will perform between a rocket in flight and the ground, when the received signal
is much weaker. Why is it weaker? Here are a few considerations:
• The power of the signal decreases with the square of the distance: double the distance
you get a quarter of the signal, treble it and you get an eighth of the signal.
• The receiver antenna is often directional. If you’re not pointing directly at the rocket
you’ll lose some signal.
• Antennae on rockets are generally very inefficient and radiate power in all sorts of
unwanted directions.
• The rocket materials may absorb some of the transmitted signal, so that power will not
get radiated (this becomes more critical as you increase in frequency)
• The polarization of the signal will vary as the rocket’s orientation changes. This can
affect the signal strength very significantly (try receiving horizontally polarized TV
signals on a vertically polarized antenna).
• The receiver will not only pick up the signal, but also receives noise from the
environment. If there is too much noise the receiver won’t be able to segregate the signal
from the noise.
These, and other, effects can be calculated or estimated individually. The results of each
calculation can be used to estimate the overall system performance in a LINK BUDGET. This
section of the website concerns itself with the calculation of link budgets.
The components of a typical radio system are:
Electrical
Noise

Signal

Tx Rx

Feeder Transmit Receive Feeder

Transmitter Radio path Receiver
cable antenna antenna cable

Radio Path Components

Each of the components has properties which affect the performance of the whole system. These
properties can be summarised as follows:

Transmitter Puts power into the feeder. Normally this is measured in Watts,
although a more useful unit is the dB relative to one watt dBW
(0 dBW = 1 Watt) or one milliwatt dBm (0 dBm = 1 mW)

Transmit feeder Has a loss in dB per meter. This means that the signal power is
attenuated before reaching the aerial so less power is radiated.

Aerial Has gain in dBi, and beamwidth in degrees, which are related.
Efficiency of power transfer between feeder and radiated power
is typically 50%. The output signal from the aerial is the input
signal times the gain. The beamwidth dictates the accuracy
with which the aerial must be aligned, and also its tolerance of
movement (wind etc.).

Radio Path The signal losses in free space increase with distance and
frequency. There are many other factors, such as
meteorological effects and path geometry, which affect the
propagation of the radio signal through the atmosphere.

Receive Aerial Same properties of gain and efficiency as transmit aerial. Sky
noise is treated as being an equivalent temperature presented at
the aerial output.
 Receive feeder Receive feeder has loss in dB/m. The longer the feeder the
weaker the signal.

Receiver The receiver will have a stated sensitivity, usually a power level
in dBm below which the signal will be too weak to be received
reliably.

Transmitters
The purpose of the transmitter is to take the baseband signal, convert it efficiently to a radio
frequency, amplify it and present it to the aerial. We can view a transmitter as 4 blocks:
• An interface stage, which sorts out the voltage and timing requirements of the input
signal
• A modulation stage, which turns the input into a radio signal
• An oscillator/mixer stage which changes the frequency of the radio signal
• An amplifier stage, which boosts the power of the radio signal

Transmitter Block Diagram

Receivers
The function of the receiver is to take the weak received signal, amplify it and demodulate it. It
must do this reliably in the presence of noise, fading and other unwanted effects.
Receivers are, generally, more complex than transmitters as they have to perform more functions.
One of the more complex functions of a receiver is to cope with a range of signal levels without
degrading the signal quality. For example, the signal received from a tranmsmitter which is
100m from the receiver (a rocket on the pad) will be much stronger than a signal from an
identical transmitter at 2000m (rocket at 6000 ft). In this case the signal power will reduce to
about 1/3000th of the original signal.
The receiver copes with this by detecting the signal level and controlling how much the signal is
amplified. This technique, called automatic gain control (AGC), is best implemented at one
frequency so the receiver will change all radio signals to a common “intermediate frequency” (IF)
at which most of the amplification and signal processing takes place.
A typical receiver has several stages:
• The first stage amplifies the weak radio signal.
 • The second stage is an oscillator and mixer which convert the signal to the intermediate
frequency
• There are usually two or three stages of amplification at IF
• The demodulator stage recovers the signal from the noise (to the best of its ability) and
controls the AGC system; is the signal is too weak it increases amplification until an
adequate signal is obtained (sometimes the signal is too weak).
• Finally, the output stage presents the recovered signal in a form that the end system can
use, maybe audio, video or data for a PC.

First Second First Second

Aerial Stage - Stage - IF IF Output
Demodulator
Amplifier Mixer Amplifier Amplifier Stage

Local
AGC
oscillator

Radio Block diagram

Antennae
The purpose of a transmit antenna is to convert the currents in a cable to E-M waves radiated
through “free space”. A receive antenna reverses this process, converting E-M waves to
currents. Almost all antenna types behave identically whether being used for transmitting or
receiving E-M waves; this is known as reciprocity.
A perfect form of antenna, known as an isotropic radiator, will radiate its signals equally in all
directions. The isotropic radiator has a fundamental problem from an engineering perspective: it
is impossible to make on in practice. Nevertheless, the isotropic radiator has one use: it provides
a theoretical benchmark against which all other antenna can be compared.
One such comparison is gain, which is the apparent increase in signal power caused by the
focussing of the beam in one preferred direction. We generally compare the gain of an antenna to
an isotropic radiator and quote this value in dBi. Sometimes gain is quoted in dB with respect to
a dipole (dBd), however manufacturers generally don’t use this as a dipole has a gain of 2.1 dBi,
thus the value for gain in the sales literature is smaller and less impressive in dBd than dBi.
It can be seen that the gain of a directional antenna decreases as we move a small angle off the
boresight. The gain of the antenna is achieved at the cost of reducing the width of the main lobe.
We define the edge of the beam as being the direction in which the power has reduced to half the
value at the boresight. Halving the power is akin to a loss of -3dB in power, so this definition of
beamwidth is generally called the -3dB beamwidth.
An approximate equation for calculating the -3 dB beamwidth of an antenna was estimated by
Kraus:
228π
θh ≈ degrees …………………………..…………………….……….. equation 1
G
The value of G is the gain as a number, not in dBi. Beamwidth is important for pointing dishes.
The narrower the beamwidth the more accurately the dish needs to be pointed. This can give
problems when trying to point a high gain (narrow beamwidth) antenna at a distant rocket.

For small pointing errors, defined as angles which are less than half the -3dB beamwidth, the
pointing loss can be approximated to:

2
θ 
L p ≈ 12 e  dB ……………………………………………………………….Equation 2
θh 
Note that this calculation gives its answer in dB.

Radio Path
In a perfect environment, such as interplanetary space, the only losses are due to the spreading of
the signal. The received signal power on a perfect radio path can be calculated from the Friis
power equation:

Pt Gt Gr λ
2

Pr =
(4πr )
where Pr is the received power, Pt the transmitted power, Gt and Gr the gain of the transmit and
receive aerials, λ the wavelength and r the distance between aerials, all distances being in metres.
The received signal power clearly decreases with the square of the distance, and increase as the
gain of the aerials increases. It also increases as wavelength increases, (or conversely the
received signal strength decrease as frequency increases).
We can rearrange the Friis equation to show the how many Watts of received power we get for
every watt of transmitted power. We call this the “free space loss equation”, as it shows the loss
in power of the signal is “free space”, which is another term for a perfect environment.
 Pr Gt G r λ
2

L fs = =
Pt (4πr )
Where Lfs is the amount of loss in free space. A more convenient form of the free space loss
equation is the logarithmic form, where the loss is expressed in decibels (dB):

L fs = 32.4 + 20 log10 f + 20 log10 d dB …………………………………… equation 3

Note that the frequency is in MHz, and the distance in Km. The Friis equation holds true on all
radio paths. Additional effects due to imperfections in the path, for example when the path goes
through an atmosphere, can be factored in as additional losses. In the earth’s atmosphere it is
common to account for losses due to rain (Lrain) and the effects of Oxygen (LO2) and Water vapour
(LH2O). These generally don’t affect model rockets as the frequencies concerned are not available
to rocketeers, furthermore we tend not to fly in the rain!
The Friis and free space equations makes the important assumptions that the aerials are correctly
pointed. There are additional pointing losses due to the changing attitude of the rocket and the
pointing accuracy of the antenna on the ground. The rocket antenna is particularly difficult to
model; rather than try to work out the pointing loss I tend to assume it is omnidirectional with
very low gain (-3dBi). The pointing loss of the transmit antenna can be estimated using equation
1.
While the rocket is in flight the path is not obstructed by vegetation, but may be obstructed after
landing. Foliage losses can be a real pain to calculate. NASA have done extensive modelling of
foliage losses of signals from satellites, and concluded that the amount of loss depends on a load
of factors including:
• Frequency
• vegetation type (grass/crops, scrub, coniferous forest, broad leafed forest etc)
• season (amount of leaf cover, moisture content of plants)
• terrain profile (flat, rolling, irregular, hilly, mountainous)
For practical purposes it is assumed that most rocket flying takes place in flat terrain with
grass/crops and scrub. Moisture content is viewed as moderate. This gives the following
approximate losses in dB for each meter of foliage along the path:

Frequency Loss (dB/m)

(MHz)
200 0.05
500 0.1
1000 0.2
2000 0.3
3000 0.4
Cables
Feeder loss characteristics of a cable are usually quoted in terms of frequency and loss/100m in
dB. The data is available from the manufacturer or, in the case of common feeder types, it is
published in data books. If we have a cable of length l meters, and it has a loss of x dB/m, then
the total cable loss is simply:

……………………………………………………………………. equation 4

Link Budget
To calculate the performance of the link we build a link budget. Typically these are laid out in a
table, and look something like:

Parameter Symbol Value Notes

Transmit power (dBm) PT + 10.0 From Tx spec
Transmit cable loss (dB) LCTx - 0.5 equation 4
Transmit antenna gain (dBi) GTx - 3.0 Assume -3 dBi
Path Loss(dB) LFS - 85.2 equation 3
Other path losses: (dB) Rain, atmospheric gases,
foliage as required
Receive antenna pointing LP - 0.7 equation 2
loss (dB)
Receive cable loss (dB) LCRx - 0.6 equation 4
Signal presented at receiver PR - 80.0 Sum of all the above
input (dBm) numbers
Required signal at the PReq - 90.0 From the receiver spec.
receiver (dBm)
Margin + 10.0 the difference between
required and received
signals.

Worked Example
A rocket flies to 1300 m altitude from a short grass field. It’s transmitting a telemetry signal to
the ground, where it is received using a manually pointed antenna with 15 dBi gain; the tracker
reckons he can point to within 5 degrees. The rocket is transmitting 8 mW ( 9dBm) at 433 MHz
using a cheap wire antenna with – 5 dBi gain. The antenna is wired directly to the transmitter so
there is no cable loss.
The receiver has a stated sensitivity of -93 dBm, and is wired to the antenna using 2m of cable
with a loss of 0.1dB/m at 433 MHz. The rocket appears to land in a 300m diameter patch of
scrub.
Q1. can the signal be received all the way to apogee.
Q2. Will there be enough signal to locate the rocket from the edge of the scrub?

A1. First, draw the problem.

We know most of the data to assemble a link budget. We need to work out the path loss and
pointing loss.
Frequency is 433 MHz, distance = 1.3 km (1300m) if we put the receiver underneath the expected
apogee.
Path loss = 32.4 + 20 log (433) + 20 log (1.3) = 87.4 dB
Pointing loss requires that we know the -3 dB beamwidth θh to substitute this into equation 2.
We know the gain (in dBi) so we can convert this into a ratio and substitute it into equation 1 to
get the beamwidth.

Gain = 15 dBi, which is as a ratio = 31.6

228π 228π
θh ≈ =
G 31.6
θh = 127 degrees (quite a wide beam)
As the tracker can point within 5 degrees:
 2 2
θ   5 
L p ≈ 12 e  = 12 
θh   127 
LF = 0.01 dB. This is not a problem so it can be ignored.
Populating the link budget:

Parameter Symbol Value Notes

Transmit power (dBm) PT + 8.0 From Tx spec
Transmit cable loss (dB) LCTx - 0.0 equation 4
Transmit antenna gain (dBi) GTx - 5.0
Path Loss(dB) LFS - 87.4 equation 3
Other path losses: (dB) 0.0 Rain, atmospheric gases,
foliage as required
Receive antenna pointing LP - 0.0 equation 1 then 2
loss (dB)
Receive cable loss (dB) LCRx - 0.2 2m @ 0.1 dB/m
Signal presented at receiver PR - 84.6 Sum of all the above
input (dBm) numbers
Required signal at the PReq - 93.0 From the receiver spec.
receiver (dBm)
Margin + 8.4 dB the difference between
required and received
signals.
This is a comfortable margin, so the link should work OK.

A2. With the rocket on the ground, and from the edge of the scrub, there are 2 losses. The first is
the path loss, then we must add the loss from the foliage. From the previous calculations we can
assume that pointing loss is not a problem.
Frequency is 433 MHz, distance = 0.3 km (300m) from one side to the other. Assuming that the
searchers will stand at the front edge and look for a signal, the max path length is 0.3 km. (This
could be shortened by walking around the edge)
Path loss = 32.4 + 20 log (433) + 20 log (0.3) = 74.7 dB
From the table above the loss in the scrub at about 500MHz (close enough) will be 0.1 dB/m.
The foliage loss is thus 0.1 x 300 = 30 dB.
The total path loss is thus 20 + 71.2 = 91.2 dB
Populating a link budget:

Parameter Symbol Value Notes

Transmit power (dBm) PT + 8.0 From Tx spec
Transmit cable loss (dB) LCTx - 0.0 equation 4
Transmit antenna gain (dBi) GTx - 5.0
Path Loss(dB) LFS - 74.7 equation 3
Other path losses: (dB) - 30.0 Foliage loss
Receive antenna pointing LP - 0.0 equation 1 then 2
loss (dB)
Receive cable loss (dB) LCRx - 0.2 2m @ 0.1 dB/m
Signal presented at receiver PR - -101.9 Sum of all the above
input (dBm) numbers
Required signal at the PReq - 93.0 From the receiver spec.
receiver (dBm)
Margin - 8.9 dB the difference between
required and received
signals.
The margin is negative, so the path will not work. The searchers will need to get closer to the
rocket to pick up a signal. If they can get within 200m of the rocket they should get a good
enough signal.
 THE IMPACT OF WIND PROFILE ON HIGH ALTITUDE
FLIGHTS BY FIN STABLISED ROCKETS

Phil Charlesworth MSc CEng

UKRA 1232 L1 RSO

Summary

This paper reflects conclusions which the author has reached as a result of his
involvement in such an altitude attempt prior to leaving the team in early 2004.
It is based on approximately 18 months research and consultation with academic
and industrial colleagues.

The paper considers how the wind changes speed and direction with altitude, and
how these changes affect the various stages of flight of a fin stabilised rocket. It
concentrates on two main areas: Firstly it considers the initial few seconds of
flight, the causes of weathercocking and its impact on launch angle and ballistic
range. It then considers parachute drift through the high and low level winds and
a method of predicting likely touchdown areas.

The paper considers the relationship between these factors and the design of the
rocket, establishing that the process of rocket design and launch conditions are
closely interdependent. It links these ideas together by proposing a system for
establishing launch conditions and criteria for informed “launch/don’t launch”
decisions by the RSO.

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TABLE OF CONTENTS

Table of Contents ......................................................................................................... 2

Table of Figures............................................................................................................ 2
1. Introduction.................................................................................................... 3
2. The Wind ........................................................................................................ 5
3. Practical Considerations ............................................................................... 7
3.1 Issues of High Altitude Flight................................................................... 7
3.2 Ballistic Range.......................................................................................... 7
3.3 Drift......................................................................................................... 12
3.4 Sources of Weather Information............................................................. 16
4. A Zonal Approach ....................................................................................... 19
4.1 Habitation ............................................................................................... 20
4.2 Roads ...................................................................................................... 20
4.3 Landowners............................................................................................. 21
4.4 Inaccessible Areas................................................................................... 21
5. Responsibilities............................................................................................. 22

Table of Figures
Figure 1 - High and low level winds............................................................................ 5
Figure 2 - Ballistic Range Model................................................................................. 8
Figure 3 - Ballistic Range vs Launch Angle Example............................................... 10
Figure 4 - Angular Displacement by Wind................................................................ 11
Figure 5 - An Ideal Wind Layer................................................................................. 13
Figure 6 - A Practical Wind Layer............................................................................. 13
Figure 7 - Drift Through Multiple Wind Layers........................................................ 14
Figure 8 - Data for Example ...................................................................................... 15
Figure 9 - Wind Profile for Example ......................................................................... 15
Figure 10 - Drift Simulation ........................................................................................ 16
Figure 11 - Zones ......................................................................................................... 19

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1. Introduction
The majority of rocket flights within the UK are made by single stage rockets
with maximum attitudes of under 10,000 ft. Such flights can be expected, in
normal flying conditions, to land within a short walk of the launch site.
Experience indicates that a large farm, or contiguous set of fields, is adequate
for this type of recreational flying.
A minority of flights aspire to reach higher altitudes, whether for the personal
challenge or as part of campaigns to set new altitude records. Flights of this
type have the potential to travel greater distances, whether by parachute drift
from high altitudes or ballistically due to a failure of the deployment system.
There are many other means by which a rocket could land outside the range,
for example a motor malfunction which could cause it to ignite at some
random launch angle. It is probable that rockets designed for high altitude
would land outside of a normal club flying site.
The UKRA safety code considers these circumstances in section 3, when it
requires minimum site dimensions based on the motor impulse or the apogee
altitude, whichever gives the greatest result. The RSO for the flight can grant
concessions to the site dimensions. There is also a requirement that the worst
case scenarios of a stage misfire or failure of a recovery device should be
considered.
The safety code, as currently worded, leaves some ambiguity as to how
distances corresponding to worst case scenarios should be calculated and who
should calculate them. Section 4 places the RSO in the situation of making
the “launch/don’t launch” decision, possibly based on incomplete
information; it is unlikely that the RSO will have knowledge of the current
wind profile or how the rocket will respond to the changes in wind profile. In
such circumstances an RSO should refuse permission for the flight, not
because it is unsafe but because of lack of information on which to authorise
it.
The safety code also requires landowners permission to be obtained for the
recovery area. For normal rocket flights, whose recovery direction and
distance are dictated by low level winds, this is likely to be some distance
downwind. For high altitude flights the high level winds, which can be at
right angles to the low level wind, will influence the location of the recovery
area. It is likely that this will not be “downwind” from the launch site and
may be much further away due to the longer descent time.
Getting landowners permission over a large potential recovery area may well
be impractical. Furthermore the larger area may contain villages or farms,
increasing both the difficulty of getting multiple permissions and the
potential for damage to property.
Making the decision to permit or refuse a high level launch clearly requires a
detailed knowledge of prevailing conditions throughout the atmosphere, and
the ability to translate these conditions into potential landing sites associated
with the successful operation of the rocket, or failures of either the motor or
recovery systems. This paper examines some of the factors which contribute

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 to that decision and proposes a standard approach to determining flight
conditions. It expresses some of the self-evident principles of rocketry in
simple maths, then uses these mathematical models to link the design of the
rocket and launch system to a set of “safe to launch” criteria. It is intended to
supplement, rather than replace, the existing site dimension criteria.
This paper summarises the conclusions of the author’s research and
consultation into these issues over the period mid 2002 to early 2004. It is
based on analysis and simulation of the factors which affect flight dynamics,
resulting from consultation with academic and industrial colleagues. The
basic mathematical models can be found on the author’s website. These were
turned into simulations using Excel, Rocksim and MATLAB.

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2. The Wind
Wind systems which affect rocket flight can be considered in 3 layers:
• Surface wind
• Low level wind
• High level wind
This section briefly introduces the characteristics of each of these layers,
setting the scene for the subsequent discussions.
Wind is caused by differences in the local temperature of air. As air
increases in temperature its pressure increases, and as it cools its temperature
decreases. At a macro level, this causes high and low pressure regions in the
atmosphere. Air flows from the high pressure systems to low pressure
systems, however the flow is not constant in either direction or speed.
The Coriolis affect causes the air at lower levels to spiral into low pressure
systems, or out of high pressure systems. At high level the winds flow
directly into and out of these systems. Furthermore the speed of the wind
varies with the rate of change of pressure.

High A

High level wind

Low level wind

Figure 1 - High and low level winds

The effect of this is that low level winds will circulate around pressure
systems, and high level winds move directly into or out of such systems. the
wind speed and direction thus changes with altitude. The change from low
level to high level occurs in the region of 5,000 ft to 10,000 ft depending on
conditions. In this region both wind speed and wind direction will change,
and the region will contain some degree of turbulence.
A further factor is that hot air is less dense and will rise, whereas cool air is
more dense and will sink. Rockets will descend more slowly in rising air,
and faster in sinking air. Such effects can be very local.

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 As most rocket flight take place under the influence of low level winds, that
is less than 10,000 ft, simple 2-D models of the atmosphere are generally
sufficient. Once a rocket enters the high level winds the increased speed and
change of direction need to e taken into account.
The wind observed at, or near, ground level at the launch site is not a true
representation of the low level wind. This wind, sometimes referred to as the
“surface wind”, is not a wind per se but reflects the results of the interaction
between the low level wind and surface features. Its effects are generally not
felt more than a few hundred feet above the ground, but are very significant
at ground level.
Surface wind is affected by the shape of the terrain. When it encounters
rising ground the wind accelerates, an effect predicted by Bernoulli’s
equation. Close to the ground the wind forms eddies and turbulence due to
its interaction with obstructions such as trees, hedges and buildings. These
effects can cause local and often large variation in windspeed and direction
which strongly influence the first few seconds of flight. Weathercocking, and
the subsequent launch angle of the rocket, will be dictated by the surface
wind.

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3. Practical Considerations

3.1 Issues of High Altitude Flight

When a rocket flies to high altitude two main sets of criteria will dictate
where its lands. The first is its ballistic range, which is the distance it will
coast in the event of some failure. The second set of criteria is the distance
which the rocket will drift during parachute decent. Each of these distances
can be linked to a set of factors which include:
• High altitude windspeed and direction
• Low altitude windspeed and direction
• Rocket design (moment of inertia, susceptibility to wind torque)
• Launch tower design (length, stiffness)
• Failure modes of motors or deployment systems
• Rate of descent
• Apogee altitude
By analysing these factors it is possible to understand and “tune” the launch
criteria. It becomes possible to trade off aspects of the design, for example
minimum rail length against desired launch angle, surface wind speed against
fin area, location and moment of inertia. This reduces the dependence on
intuition, and allows informed decisions to be made during the design phase,
launch campaign and on the day.

3.2 Ballistic Range

If a rocket is launched at an angle other than vertical and the deployment
mechanisms fail, then it will follow a parabolic trajectory. It will land some
distance from the launch site, and this distance is called the ballistic range.
The classical approach to calculating ballistic range is to use the vertical
velocity component to calculate time of flight the use this flight time to
calculate the horizontal range. The assumptions in this method are that the
only retarding acceleration is due to gravity, so that the horizontal velocity
component is constant. The ballistic range thus depends on the angle at
which it is launched and its velocity.
Consider a simple textbook example of a particle launched at a velocity V
and a launch angle θ, as shown below.

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 Apogee

Altitude (m)

V
Vcos(θ)

Vsin(θ) Ballistic range (m)

Figure 2 - Ballistic Range Model

Resolving the initial velocity into vertical and horizontal components we get:
Vvertical = V cos(θ )

Vhorizontal = V sin(θ )
Solution of the equations of motion for the vertical component gives us the
time t taken to reach apogee. After apogee the projectile will fall for a further
time t until it lands. The total flight time is thus 2t.
The horizontal component of velocity is unimpeded, as the acceleration on
the rocket in that direction is negligible. The distance travelled in time 2t is
thus the ballistic range, which can be found from:
Ballistic range = rballistic = 2Vt sin(θ ) ………………………………………(1)
This simplistic analysis ignores the third force, drag, as it makes the
equations of motion practically insoluble by analytic means. We can,
however, take drag into consideration by using numeric solutions based on
Euler’s method or the Runge-Kutta method, both of which are used in
Rocksim for their computational simplicity. These can give us values of V
and t for any particular rocket flight. These values can be fed into equation 1
to produce an estimate of ballistic range.

3.2.1 Maximum Ballistic Range

It can be shown that the maximum ballistic range occurs when θ =45 degrees.
The maximum range can be shown to be twice that apogee from vertical

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 flight. This rule is often called “Tartaglia’s rule” after the mathematician who
discovered it while investigating the behaviour of artillery.
Such a situation would not occur in normal flight, but as a result of some type
of failure. It is not possible to consider all failure modes, but two
immediately spring to mind. The first is a structural failure of the launch
stand, which is uncommon and can be overcome by good design. The second
is a failure of the motor, for example it fails to fully ignite causing the rocket
to tumble as it leaves the launch rail. Ignoring the laxative effects of such an
event, a rocket expected to reach 15000 ft/5 km apogee has a maximum
ballistic range of 10km. As few launch sites are 20km across, such a flight
would undoubtedly land outside the range, and would have considerable
potential for adverse publicity. A rocket intended to reach 40000 ft/17 km
has a maximum ballistic range of 34 km, requiring a launch site 68km across!
While such a motor failure is uncommon it is not unprecedented. Most HPR
fliers will have seen launch failures of this type. In mitigation, the maximum
range would only be achieved by failure of both the motor and deployment
system on the same flight. It is more probable that a low angle flight would
be terminated by parachute deployment.
If CPR is used, this deployment would occur at about half the maximum
ballistic range and about half the apogee altitude. A rocket with a planned
apogee of 15,000 ft/5km would this deploy at a range of 5km would drift
from 7,500 ft. Rockets with timers or standard delay grains set for 15,000 ft
would deploy beyond 5km while travelling at high velocity. The site
dimensions table in the safety code assumes the greater of half this distance,
in this case 2.5 km radius, or half the site dimension corresponding to the
impulse of the rocket. This is reasonable for the current generation of high
altitude rockets, but may need to be reconsidered as designs improve and
records increase.

3.2.2 Expected Ballistic Range

The expected ballistic range for a rocket which launches successfully, but
suffers a failure of its deployment system, can be calculated from equation 1.
The graph below plots ballistic range against launch angle for a 10kg, 4 inch
diameter, rocket. The two altitudes, 15,000 ft and 40,000 ft, reflect the
current records for flights inside and outside the UK. Knowledge of a
particular rocket design would permit equivalent graphs to be produced quite
simply.

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 14000

12000

10000
Ballistic range (m)
8000

6000

4000
15000 ft
2000 40000 ft

0
0.0 5.0 10.0 15.0 20.0
Launch angle (degrees)

Figure 3 - Ballistic Range vs Launch Angle Example

As an example of how to use Figure 3; if the nearest inhabited area is 4km
from the launch site, then a launch to 15,000 ft would be acceptable if the
launch angle could be assured to be within 11 degrees, whereas a launch to
40,000 ft requires a launch angle of no greater than 3 degrees. Unless the
launch system and conditions could provide a high degree of confidence that
the launch angle would not exceed 3 degrees, a launch to 40,000 ft would not
be advisable from that site.
As seen earlier, the ballistic range is a function of initial velocity and launch
angle. If launch angle can be kept low, the ballistic range can be reduced.
This leads to the conclusion that any launch attempt needs to be supported by
analysis of the launch angle and the consequent ballistic range. As launch
angle is determined by weathercocking, and weathercocking is due primarily
to the torque applied from the wind pressure on the rocket and the moment of
inertia of the rocket about its CG, there is a clear relationship between launch
conditions and rocket design. This relationship allows anyone comfortable
with basic aerodynamics and differential equations to make a good estimate
of the maximum permissible low level windspeed for a launch.
The velocity at which the rocket leaves the launch rail is also significant. It is
self evident that a longer launch rail will support the rocket to a higher
forward velocity. The higher velocity airflow over the fins generates lift
which opposes any torque induced by the wind. The length of launch rail
required for any given rocket in any windspeed is thus estimable on a case-
by-case basis.

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 Wind Force

Time

Displacement
Angle

Peak

Average

Time

Figure 4 - Angular Displacement by Wind

If we consider a suddenly applied wind force, analogous to the force applied
when the rocket leaves the rail, the rocket will rotate about its CG. This
angular displacement is opposed by the lift from the fins, which increases as
the angle of attack increases. The opposition of the wind and lift forces
causes the rocket to oscillate about its CG, eventually converging on an
average displacement. Figure 4 shows the typical response to a sudden
change in crosswind.
The frequency of the oscillation about the average displacement angle, and
the time it takes to settle to the new angle, are strongly influenced by the
mass distribution of the rocket (moment of inertia) and the rocket’s
dimensions. In a poorly designed rocket this oscillation will never terminate,
or even increase. This corresponds to the definition of an unstable rocket.
Alternatively the rocket may settle very slowly, corresponding to an over
stable rocket. Ideally the rocket should be designed to settle to the average
angle as quickly as possible. Empirically, this leads to the rule of the CP
being 1 or 2 diameters behind the CG. This rule of thumb is not universally
applicable and starts to break down with tall slender rockets such as are used
to reach high altitudes. The behaviour of rockets with transitions, parallel
staging, and multiple fin sets can also diverge significantly from this rule of
thumb.
A well designed rocket will settle quickly at the average displacement angle,
and for practical purposes this angle can be considered as the launch angle θ .
The low level wind speeds, rail length, and the dynamics of a particular
rocket should be analysed by the rocket designer so as to ensure
understanding of launch conditions. It is strongly suggested that such

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 calculations are checked independently as a prerequisite to launch approval
by the RSO.

3.3 Drift
The troposphere is not a predictable environment, as shown earlier. The high
level winds flow from high pressure areas to low pressure areas, whereas the
Coriolis effect causes low level winds to circulate around these areas. The
lower and high level winds can be up to 90 degrees apart, so the region where
they interact is subject to some turbulence. Wind direction and speed in this
region can be very unpredictable, and its height and depth vary considerably.
Local wind effects also influence the air flow. These include rising and
falling currents of air due to ground heating, and air masses rising over hills.
The Met Office has difficulty predicting the behaviour of the wind and
airflows in the troposphere, so it is unreasonable to expect UKRA members
to do any better.
Drift of a relatively light object such as a spent rocket is thus a highly
complex subject, as the behaviour of the atmosphere is not regular and
predictable but statistical. Predicting exactly where a rocket will touch down
is impractical, however models may be derived which identify where a rocket
should touch down, and defining the area surrounding that point where you
have high confidence that it will touch down.
It is clear that establishing the touchdown point of a rocket requires
prediction of drift through the high and low level winds. Consequently a 3D
model of the atmosphere is required. Programmes such as Rocksim make the
simplifying assumption that wind changes in neither speed nor direction with
altitude, so drift can be plotted as a straight line from apogee. This
assumption is adequate for flights in the low level winds where direction is
usually within 10 degrees and the influence of changes in windspeed profile
is not significant for short descents.
Descent from high altitude requires that the direction and speed of both the
low and high level winds be considered. The longer descent time, higher
windspeeds at high level, and difference in direction between low and high
level winds result in a touchdown point which can be for from that predicted
by simple 2D models such as Rocksim.
A simple 3D model is described in the following paragraphs. The model is
based on that used for the recovery of stratospheric balloons and by freefall
parachutists.

3.3.1 Drift Model

If we cannot describe the path which a descending rocket will follow from
apogee, maybe we can consider its behaviour over short distances and apply
some statistics to its behaviour.
Imagine a rocket falling through a slice of the atmosphere which is not very
thick, maybe a few hundred feet. Over that distance it will be subjected to a
crosswind, so the true path will not be vertical but will follow the vector sum

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 of the descent velocity and wind velocity. If we consider the whole descent
from apogee as a sequence of n layers, where the descent velocity vector is
approximately constant (neglecting the atmosphere’s density profile) and the
wind velocity changes with each layer, we can establish the ideal touchdown
point.
Point of entry

Descent
Actual
velocity
velocity

Wind
Point of exit velocity

Figure 5 - An Ideal Wind Layer

In practice, the wind vector will vary in each layer. We can refine the model
by assuming that the magnitude of the vector has a mean value, obtained
from a reliable source and a standard deviation. The exit from the layer is
thus not a point but a probability distribution based around that point.

Point of entry

Descent
Actual
velocity
velocity

Region of Average wind

uncertainty of exit velocity

Figure 6 - A Practical Wind Layer

If we assume that the amplitude of the wind velocity vector is normally
distributed about the means speed, and assume a standard deviation, we can
link the radius of the circle to the probability of the exit point lying within
that circle. The linkage is:
Diameter Probability
1 SD 68.2%
2 SD 95.4%
3 SD 99.7%

Why select a normal distribution? Two reasons: firstly many natural

occurrences exhibit this distribution, and secondly because of its
mathematical convenience.

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 By considering the effect of descent through n layers we can establish an
ideal touchdown point, and the probability of the actual touchdown point
lying within a circle around this point.

Apogee

Layer n

Layer n-1

Layer n-2

Ideal
descent path

Layer 4

Layer 3

Layer 2

Ideal
touchdown
point

Layer 1
90% confidence
region

Figure 7 - Drift Through Multiple Wind Layers

3.3.2 Implementing the Drift Model

The model was written as an Excel workbook. The input data was
interpolated from the Met Office F214 for the Bristol area for 0600 to1200 on
a day which looked like a good flying day. The layers were each considered
to be 1000 ft thick. A drogue descent rate of 60 mph from an apogee of
40,000 ft was modelled. The F214 data input is shown below:

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 INPUTS
F214 Input
Alt (k ft) Heading Variable Speed
24 130 0 53
18 130 0 48
10 130 0 40
5 60 1 25
2 60 1 15
1 40 0 10
0 40 0 8 From local observation

Descent Input
Altitude 40 ,000 ft
Descent rate 60 mph

Figure 8 - Data for Example

The data was plotted onto heading and speed graphs. In order to extend the
F214 data to above 24,000 ft three assumptions were made:
1. The peak windspeed occurs at 40,000 ft, corresponding to the average
altitude of the peak over the UK.
2. The windspeed dies off linearly above this altitude until it reaches a
negligible speed in the tropopause, around 60,000 ft.
3. The high altitude wind has a constant heading above 24,000 ft
These three assumptions are meteorologically reasonable, and should not
introduce large errors into the model except in unusual met conditions. As
no sensible rocketeer would launch in those conditions, the assumptions are
viewed as practical. The wind profile on the day is plotted below.
Windspeed Profile Wind Direction Profile

7 7
Altitude (x1000 ft)

Altitude (x1000 ft)

6 6
5 5
4 4
3 3
2 2
1 1
0 0
0 1 2 3 4 5 6 7 0 10 20 30
Velocity (mph) Direction (degrees)

Figure 9 - Wind Profile for Example

It was assumed that the standard deviation of the windspeed was the mean
windspeed divided by 6. This allows the very unlikely probability of the
windspeed being 0 mph and the equally unlikely probability of gusts being
twice the mean speed. Confidence levels of 99.7%, or 3 SD, were expected.
The results of this simulation are shown below.

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 2

0
-1 0 1 2 3 4 5 6

-1

-2

-3

-4

-5

Center
-6 Scatter
All axes are miles
Point (0,0) corresponds to apogee Rocksim

Figure 10 - Drift Simulation

The plot on Figure 10 shows the scattering of potential touchdown points.
These are centred on a point 5.5 miles SE of the apogee point (NOT the
launch site) and scattered in a one mile long ellipse. The location of the
touchdown area, and orientation of the major axis of the ellipse, suggest that
the touchdown area location is strongly influenced by the direction and speed
of the high altitude wind. Only 25 points were plotted for clarity.
This scattering is intuitively correct, since 75% of the descent would be
through the high altitude wind which was, on the day of the example,
moderately strong and blew on a consistent heading of 130 degrees. This
wind would have influenced descent from 40,000 ft to 10,000 ft, over 75% of
the path.
Rocksim’s 2D model puts the touchdown point about 1.2 miles NE of the
apogee point, some 5 miles away from the predicted point and in the wrong
compass quadrant. It is possible that a recovery team would be incorrectly
deployed and watching the wrong part of the sky if a 2D model was used,
with consequent risk of loss of the rocket.

3.4 Sources of Weather Information

It is clear from the above that the low and high level wind profiles have a
very significant influence on the launch and recovery criteria for a high
altitude rocket. Practical sources of wind profile information tend to have
their limitations, the most significant of which are the timeliness and
granularity of the information. A few sources of data have been uncovered

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 while researching this subject. While not exhaustive, this list may help to
uncover the practical issues surrounding making a launch decision based on
weather conditions.

3.4.1 Broadcast Weather Forecasts (Low level wind)

Weather forecasts from the TV and radio only provide mean windspeed data
at 10m above ground level for a 12 hour period over a wide area. Such data
is of little value in planning a high altitude launch.

3.4.2 Local Weather Station (Surface and low level wind)

A weather station at the launch site would be an unreliable source of
information. The surface wind can vary a lot due to terrain profile and local
obstructions. A local weather station would need to measure wind at the
launch site at several altitudes. Typically this would be from ground level to
an altitude corresponding to that where any weathercocking tendency is
negligible. Building a weather station on a mast of this height may be
impractical.
There is some potential for near real time measurement using, perhaps,
lightweight model rockets and observers. By simultaneously plotting the
drift of a descending model rocket a 3-D picture of the wind speed and
direction could be established. By radioing the results of several observers
back to a suitable PC, the surface and low level profile could be quickly
computed. After several flights (to eliminate any rogue results) a good
picture could be established in near real time.

3.4.3 Met Office UK Low Level Spot Wind Chart (low and high level wind)
The Met Office provides a free lower wind service up to 24,000 ft at spot
heights of 1, 2, 5, 10, 18 and 24 kft. This service, known as the F214, is part
of its aviation services, and is intended for flight planning purposes.
Interpolation between the spot heights can give a good idea of the wind
profile below 24,000 ft, but extrapolation beyond these altitudes requires
caution. This service only covers a few points over the UK, and gives a 6
hourly mean value for horizontal wind components. It doesn’t cover
windspeeds and directions at ground level and through the critical first few
seconds of flight.
Typically the F214 can be used to give 6 to 12 hours notice of conditions,
which is adequate for its target audience of flight planners but may not be
adequate notice to make a decision as to whether to travel to a launch site and
prep a rocket.

3.4.4 Local Meteorology Sources (low and high level wind)

Local sources such as the met officer on civilian and military airfields may be
able to provide more comprehensive meteorological data. Like the F214,
they may not be able to offer much advance notice of particular wind
conditions, but their local experience may be able to assist with identifying

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 periods when conditions are likely to be come favourable. A launch team
could then be brought to a state of readiness for a possible launch window,
and preparations be made while watching the short term weather information.
Building a positive relationship with the met officer near the site is clearly
beneficial.

3.4.5 Atmospheric Research Centres (low and high level wind)

The Met Office, and a number of universities, conduct measurement of the
wind profile using radio techniques. Such information tends to be local to
the site of the instrument, but has the advantage that it is real time and
accurate. A launch site located near such an instrument could get high
quality and local wind information.
The Met Office sites are used as the source of F214 information so their
processed data is released every 6 hours. Also, these sites are constrained to
reporting at an altitude of 24,000 ft and below. Some of the university sites
look beyond the troposphere into the tropopause, and report on winds up to
100,000 ft. One university, Aberystwyth, has a mobile sounder but it only
looks at low level winds. It is worth monitoring this area of research as
universities may develop mobile sounders capable of looking at the high
level winds, and such a capability may be available for a “one off” altitude
attempt.

3.4.6 Radiosondes (low and high level wind)

There are a number of Met Office and university radiosonde launch sites
around the UK. These are used to generate a range of metrological data
including wind profile. Radiosonde flights are limited to a few per day, so
the timeliness of the data may be less than ideal.
Radio interference issues prevent the launching of radiosondes by private
individuals, otherwise this would be an ideal way of establishing local
conditions. The author has not investigated the idea of inviting an authorised
radiosonde launch to be made from the rocket launch site, but this may be
worth exploring.

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4. A Zonal Approach
The previous section established some criteria for establishing where a rocket
might touch down in the event of failure or due to drift. The next significant
issue is to assess the impact of either situation.
Conceptually, the models produce 4 touchdown zones:
1. A green zone, within which the rocket will touchdown if the flight
goes to plan, and the rocket launches within its intended launch
angle.
2. A blue zone, which is where the rocket will touchdown if the
deployment mechanisms fail and the rocket flies ballistically.
3. An amber zone, which is where the rocket will touchdown if the
motor misfires on launch but the deployment mechanisms work,
causing ballistic flight.
4. A red zone, which is where the rocket will touchdown if the motor
misfires on launch and the deployment mechanisms fail.
The zones are shown below.

Half max
ballistic range
Max ballistic
range

Expected
ballistic range

Launch site
RED ZONE
BLUE ZONE

Expected
AMBER ZONE Landing site

GREEN ZONE

Figure 11 - Zones
It is assumed that the blue zone lies entirely within the amber zone.

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 The green zone will move around with its radius and location being
dependent on the wind profile at the time of launch, apogee altitude and
descent rate. It is necessary to plot the location of the green zone as close to
the launch time as possible.
The remainder of this section considers launch constraints and how they
relate to the zones. In planning a high altitude flight, and subsequent
recovery, the impact of landing in each of the zones needs to be assessed.
Issues to be considered include:
• Are the zones inhabited?
• Do roads pass through the zone?
• Who are the landowners?
• Are there any inaccessible areas?

4.1 Habitation
Curiously, the safety code does not comment on the presence or absence of
houses in the areas in which a rocket could land. It is theoretically possible
to launch in the middle of a town, provided that all the landowners give
consent! In practice it would be unwise to launch if there is the possibility of
the rocket landing in a populated area, so some criteria need to be set. The
following criteria are proposed.
If there are any populated areas inside the amber zone then a flight should not
take place from that site. If there are a small number of houses on the edge of
the amber zone then the RSO may judge the risk to be acceptably small.
The blue zone should be entirely unpopulated. In writing this paper
consideration was given to a “mail shot” to anyone who lives within the blue
zone, however this may generate more issues than it solves.
A launch should only be permitted if simulation shows that the green zone
lies entirely inside a depopulated area. There may be a case for increasing the
size of the green zone to allow for random factors, such as a reduction of
descent rate due to thermals.

4.2 Roads
Consideration should be given to closing any roads within the amber, blue or
green zones for the duration of the flight, requiring Police involvement. A
site several miles from a motorway or major route may be OK for normal
flying, but could be unsuitable for high altitude flights due to the drift
distances. Practically, this requires that the zones may contain infrequently
used rural roads but no major routes.
The mobility of the green zone raises practical issues here: how likely is it
that the police will respond to close a road in response to the short timescales
associated with identifying good launch conditions and plotting the green
zone? In practice the green zone may have to contain no roads.

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4.3 Landowners
In normal club and event flying, all rockets will land in a contiguous set of
fields for which permission has been obtained. As the size of the zones
increases, the difficulty of obtaining all the landowners’ permissions
increases significantly. Some landowners may be relaxed about a rocket
landing on their plot, others may be hostile.
One approach may be to approach landowners in which the green zone may
fall, and use their responses to plot areas where the green zone may occur and
where it may not. If a landowner refuses permission then a launch which
shows their land in the green zone should not take place. The position
becomes ambiguous where no response is received as the landowner’s
intentions cannot be assumed. The situation becomes more complex for a 2-
stage rocket as there would be two green zones.
The ballooning community have had mixed experiences of landing without
permission; in some areas there is considerable hostility, though this may be
due to overuse. The British Balloon and Airship Club (BBAC) issue a
monthly update on sensitive areas, and have guidelines for landing. We
could learn from their experiences, and an approach to the BBAC may help
to establish guidelines for rocketry.

4.4 Inaccessible Areas

If the rocket comes down in any inaccessible areas such as dense forestry,
lakes or mud flats, then it’s tough luck! The rocket should be considered
lost. The landowner should be contacted to advise him of the event as the
rocket may still be armed.

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5. Responsibilities
Effectively, this proposal requires that the individual or group planning to
launch a high altitude rocket flight presents the RSO with a data pack. This
can be done in 2 stages.
Stage 1 considers the calculation of the launch conditions and plotting of the
red amber and blue zones based on calculation of the initial flight parameters
and failure modes. As this depends on the rocket and launch tower design it
cannot commence until that process is complete. It should include the
calculations used to determine the maximum surface and low level
windspeed for a launch within the launch angle imposed by the site. It should
also include a recce to establish occupied areas, and subsequent contacting
landowners. This information should be presented to the RSO some weeks
before the planned flight to allow it to be checked.
Stage 2 is considers the prevailing conditions on the day of an intended
launch. It is primarily concerned with measurement of the low level wind, to
establish safe launch conditions, and using data on high level winds to plot
the location of the green zone(s).
Once the RSO is satisfied that the launch and recovery conditions are within
the design limits, an informed launch/don’t launch decision can be made
relatively easily.

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6. Conclusions
The wind profile has a very significant effect on the flight of a fin stabilised
rocket to high altitude. An understanding of the interactions between wind
profile, rocket design, launch criteria and parachute drift can only be
achieved by analysis and modelling. Programmes such as Rocksim have a
role in this, but the inbuilt assumptions and simplifications of that package
will give unreliable results. Rocksim’s 2D model is unsuitable for estimating
parachute drift from high altitude.
Good analysis of the interactions of any high altitude rocket and the wind are
needed, particularly for the initial stages of flight and during recovery.
Suitable models will permit a trade off between rocket design factors, wind
conditions at all altitudes, and site limitations. It is proposed that
consideration of the outputs of such models should form an essential part of
the RSO’s decision to allow or refuse a launch.
A method of presenting the outputs of the models is proposed. This involves
plotting a number of zones on a map of the proposed launch and recovery
areas, and considering the land use in each of those zones. Such a method
would allow the RSO to make an informed decision as to the safety of the
launch and recovery.

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Note that all the above examples are to an arbitrary base a. These properties hold true for
any base, provided that we use the same base for all calculations. We can even do
complex examples:

 vw 
z= 
 x 

Then:

Loga z = yLoga v +y Loga w - yLoga x

Gain
Imagine a device, for example an amplifier, which has an input signal of power Pi and
amplifies this to an output power Po.

We can say that the device has a “gain” G such that:

P0 = GPi

Which can be rearranged to give a definition of gain as the ratio of output power to input
power:

Po
G=
Pi

We can express gain logarithmically as:

Po
G = Log 10 Bel
Pi

Historically the unit for power gain was the Bel, named after the founder of telephony
Alexander Graham Bell. Early telephone engineers found that, by multiplying the answer
by 10, they could calculate to fewer decimal places without loss of accuracy. This was
useful in the pre-calculator era, and gave rise to the unit ten-Bel or deci-Bel. By
convention we now use the following equation for gain:
 Po
G = 10 Log10 dB
Pi

Example: A radio amplifier has an input power of 1 watt and output power of 30 watts.
What is its gain in dB?

Po 30
G = 10 Log10 = 10 Log 10 = 14.8 dB
Pi 1

Imagine a chain of amplifiers each with different gains.

By considering the input to each amplifier as the output of the previous amplifier, it can
be shown that the total gain is:

Gtot = G1 × G2 × G3 × G4 L × Gn

In decibels, this equation becomes:

Gtot = G1 + G2 + G3 + G4 + LGn dB

We’ve changed multiplication into addition. If the gain of one of the amplifiers changes
we only need to add numbers rather than multiply them.

Loss
Not all devices have gain. Some devices lose power so that the output power is less than
the input power.

Example: A transmitter feeds 100W into a low quality radio cable but only 75W reaches
the antenna. What is the gain of the cable in dB?

Po 75
G = 10 Log10 = 10 Log10 = - 1.25 dB
Pi 100

Note that the gain in dB is negative. We refer to the device having “loss” rather than
“gain”.
If the cable had no loss, so that 100W of power was input to the cable and 100W was
delivered to the antenna, it would have a gain of 0dB. The following table contains
useful values of gains in dB and ratio form.

Gain (dB) Gain (ratio)

100 × 1010 (× 10,000,000,000)
50 × 105 (× 100,000)
4
40 × 10 (× 10,000)
30 × 103 (× 1000)
20 × 10 2
(× 100)
19 × 80
16 × 40
13 × 20
10 × 10
9 ×8
6 ×4
3 ×2
0 ×1
-3 × 1/2 (÷ 2)
-6 × 1/4 (÷ 4)
-9 × 1/8 (÷ 8)
-10 × 1/10 (÷ 10)
-13 × 1/20 (÷ 20)
-16 × 1/40 (÷ 40)
-19 × 1/80 (÷ 80)
-20 × 1/100 (÷ 100)
-30 × 1/1000 (÷ 200)

Mixing gains and losses

Imagine that we have a system which contains a mixture of devices, some with gain and
others with loss.

The total gain expressed as a ratio is:

G1 × G2 × G3
Gtot =
L1 × L2 × L3
Whereas I decibels it becomes:

Gtot = G1 + G2 + G3 − L1 − L2 − L3 dB

As systems become more complex the decibel option becomes simpler to use. In the
above example, if each of the gains and losses depended on a number of other factors, for
example the length of cables or the values of electronic components, calculation as a ratio
would become unwieldy. In these circumstances the advantage of decibels becomes
apparent: the contribution of each term can be calculated independently and the values of
each gain and loss added together.

Turning decibels to ratios

This is just like taking antilogs, with the exception that the ×10 factor needs to be
accounted for. If we say that X is the ratio value, and x is the value in dB:

ratio to decibel

x = 10 Log10 X

decibel to ratio

x
X = 10 10

When converting back to a ratio it is imperative to remember whether the dB value is a

gain or a loss, and include the sign. If x represents a device with loss, then it is really –x
and should be calculated as:

−x
X = 10 10

This point cannot be over emphasized, as it is the cause of many incorrect results.

Decibel Units
In all the above examples we’ve considered simple devices with power gain and power
loss. Decibels can be used in other contexts. We can define units of power and other
elements in decibels. For example we can take a reference power of 1 Watt and define a
unit of power in decibels called the “decibel watt” or dBW, such that 1 Watt is 0 dBW.

Example: A transmitter has an output power of 20W. What is its output power in dBW?

Po 20
Power = 10 Log10 = 10 Log10 = 13.0 dBW
Pref 1
Another useful decibel power unit is the dB referenced to 1 milliwatt (1/1000 watt),
denoted dBm. This is often used in low power radio transmitters or telephony. We can
relate dBW to dBm as long as we remember that 0 dBW = 30 dBm (or conversely 0 dBm
= -30 dBW).

In communications we sometimes use dB in relation to 1 Hz bandwidth (dBHz),

temperature in Kelvin above absolute zero (dBK), and the gain of an antenna referenced
to an isotropic (truly omnidirectional) antenna (dBi).
Orbits and Keppler’s Laws
This web page introduces some of the basic ideas of orbital dynamics. It starts by
describing the basic force due to gravity, then considers the nature and shape of orbits.
The next section considers how velocity change (∆V, pronounced “delta-vee”) is used to
initiate manoeuvres, and looks at sources of velocity change.
Having established these basic ideas it then looks at interplanetary manoeuvres, starting
with a simple Earth-Moon transfer and moving on to more complex interplanetary
manoeuvres. By the end of this section readers should be in awe of the guys at JPL who
do this for a living!

Gravitational Force
There is much discussion among physicists about the nature of gravity. For the purpose
of this website we’ll use Newtonian mechanics and ignore notions of gravity waves or
any other current theories.
In the Newtonian world any two lumps of matter will exert a gravitational force on each
other. Imagine we have two masses m1 and m2, and they are a distance r apart (masses
in kg, distance in metres).

F F
m1 m2

The attractive force between the two masses can be calculated from:
Gm1 m 2
F=
r2
where G is the universal gravitational constant, G = 6.672 × 10-11 m3 /kg sec2. If one of
the masses is fixed and very much greater than the other, for example a planet and
spacecraft, then it is sometimes written:
GMm
F=
r2
where M is the mass of the plane an m is the mass of the spacecraft. The product GM is
called the gravitational parameter, written as the Greek letter µ. The value of µ is
different for each planet, so remember to change value when working on interplanetary
manoeuvres. Some useful gravitational parameters are:
Body µ (m3/s2) µ (km3/s2)
Sun 1.327 × 1020 1.327 × 1011
Earth 3.986 × 1014 3.986 × 105
 Moon 4.902 × 1012 4.902 × 103
Mars 4.281 × 1013 4.281 × 104

If a spacecraft was stationary above the planet it would simply fall out the sky due to
gravity. In order to stay in orbit it needs to move in an orbit.

Basic orbits
Let’s debunk one myth straight away – a spacecraft is not kept in orbit by centrifugal
force. Let’s consider Newton’s first law:
“A body will remain at rest or uniform motion in a straight line unless acted on by a
force”
If spacecraft is travelling in empty space with its motors off. There are no forces on the
spacecraft so it will travel in a straight line, as predicted by Newton’s first law. Imagine
that a planet suddenly appears beneath the spacecraft. The only force introduced is
gravity, and the effect of this is to deflect the path of the spacecraft towards the planet. If
the spacecraft is travelling very fast its path will bend, but it will escape the gravitational
force of the planet. If it is travelling very slowly it will spiral into the planet and crash.
Over a narrow range of speeds it will fall towards the planet, but it will never land. It is
in orbit.
Spacecraft No planet

High velocity
r Low velocity (fly-by)
(spirals into planet)

Optimum velocity
(enters orbit)

Planet

One way of thinking about this is that the curvature of the planet is such that it falls away
at the same rate as gravity is pulling the spacecraft towards it. The spacecraft is in “free
fall” around the planet.
If the spacecraft is travelling with a velocity v m/s at a distance of r metres from the
centre of the planet, then a circular orbit will occur only when the following equation is
true:
µ
v=
r
If the spacecraft is travelling slightly faster than this it will just fail to escape the
gravitational pull and remain in an orbit which is not circular but elliptical. It can be
shown that, if the velocity is equal to:
2µ
v=
r
then it will just escape from the planet’s gravitational pull. The shape of the orbit will be
parabolic. If we exceed this velocity the shape of the orbit becomes hyperbolic and the
spacecraft escapes from the planet’s gravitational pull at a higher velocity.
We can summarise the shape of orbits in the following table:
Spacecraft Velocity Orbital shape Comments
µ Unsustainable Spacecraft spirals into planet
v<
r

µ Circular Closed orbit

v=
r

µ 2µ Elliptical Closed orbit

<v≤
r r

2µ Parabolic Open orbit, minimum escape velocity

v=
r

2µ Hyperbolic Open orbit, escape velocity

v >
r
The shapes of orbits are either circles, ellipses, parabolae or hyperbolae. These shapes
may appear to be very different but mathematically they all have something in common:
they are derived from the “cuts” of a cone. The orbital shapes are thus referred to as
“conic sections”.
 Hyperbola

Parabola

Circle

Circle Ellipse
Ellipse

Hyperbola Parabola

Later on we’ll see that these “conic section” orbit shapes have a mathematical
relationship to each other, but for now we’ll just accept this as fact.

Kepplers laws
Between 1609 and 1619 Johannes Keppler published his famous 3 laws of planetary
motion, based on observations made by the astronomer Tycho Brahe. Keppler’s laws
described the orbits of bodies to remarkable accuracy, and stated that all orbits would be
elliptical or circular. In 1687 Sir Isaac Newton supplied the theoretical explanation for
why the orbits were this shape and allowed calculation of the velocities which a satellite
would need to reach if it was to sustain an orbit.
Keppler’s laws state:
1. A body orbiting around a planet will describe an orbit that is an ellipse with the
planet at one of the foci.
2. If we draw a line from the planet to the body in orbit around it, the line will
sweeps out equal areas in equal intervals of time.
3. The square of the time taken for a body to complete one orbit is proportional to
the cube of the major axis of the orbit.
What do these mean in practice?

Keppler’s First Law

Keppler’s first law describes the shape of an orbit, based on observations of the motions
of planets.
 Satellite
(at some point on orbit)

r
b

θ
a

Planet
(at focus)
b

a(1+e) a(1-e)

a a

The geometry of an ellipse tells us that it is an “eccentric” circle, where the degree of
eccentricity is denoted by the letter e. The value of e is defined from the dimensions of
the ellipse a and b, by the equation:
b2 = a2 (1-e2)
From the geometry of an orbit it can be shown that, for a satellite at any point on the orbit
at a distance r from the planet, the following relationship is always true:
a(1 − e 2 )
r=
1 + e cos(θ )
Let’s think about these equations a bit. If the eccentricity is zero, then what do we get?
Substituting e=0 into the first equation we find that a=b. Form the second equation we
find that r=a. Both of these cases indicate that we have a circle when e=0, so a circular
orbit is just a special case of an elliptical orbit where the eccentricity is zero.

Keppler’s Second Law

Imagine our satellite is at point
in its orbit around a planet. A time t seconds later it
has moved on to point. The line between the planet and the satellite will sweep out an
area A in that time. Later on in its orbit it moves from point  to point  in a time t,
sweeping out an area B.
 Time = t seconds
4 3

2
B Time =t seconds
A
1
Planet

Keppler’s second law tells us that area A will always equal area B, regardless of where
we start to measure the time t.

Keppler’s Third Law

The diagram below shows an elliptical orbit with a major axis a. It takes a time, which
we’ll call T seconds, to complete one orbit of the planet. Instinct says that if we make
the orbit bigger, in other words increase a, the satellite will take longer to complete one
orbit.
Time = T seconds

Satellite
2a metres
1

Planet

Keppler’s third law tells us how T varies as we change a:

a3
T = 2π
µ
 New orbit

V2
Current orbit

V1

As the satellite is staying at the same altitude we know that the magnitude of the velocity
vectors v1 and v2 will be the same, but the direction will be different. To work out the
velocity vector ∆V to effect a change the direction through an angle θ we simply
construct a vector triangle.

∆V
V2 2
V2
∆V
∆V
2
2
2

V1
V1

Applying simple trigonometry to the velocity we can show that the magnitude of the
velocity vector ∆V is:
θ
∆v = 2V sin
2
The direction of the burn is in a direction of 90+θ/2 degrees from the direction of travel
of the satellite.
What does this meen in reality? Imagine our satellite has a mass of 1000kg, which is
quite small for a modern communications satellite, and is travelling at 8km/s. To change
its direction by 5 degrees we need to give it a ∆V of magnitude:
θ 5
∆v = 2V sin = 2 × 8000 × sin ≈ 700m / s
2 2
If we revisit the section on the rocket equation, and assume that our booster motor on the
satellite has a specific impulse of 250 seconds, we can calculate how much fuel we need
to burn to execute this orbit change:
 msat + m fuel 
∆v = ve ln 
 msat 
It can be shown that the rocket exhaust velocity, ve, can be found from the equation:
ve = g o I sp
Where go is the acceleration due to gravity at sea level. Thus:
 msat + m fuel   1000 + m fuel 
∆v = I sp g o ln  ⇒ 700 = 250 × 9.81 × ln 
 m sat   1000 
From which we can determine that mfuel = 330 kg. To put this into perspective, we have
to burn about a third of the mass of the satellite as fuel to execute a 5 degree change in
direction. It can be seen that orbit change manoeuvres are very fuel intensive, and are
thus carried out very infrequently.
Interplanetary Trajectories
Interplanetary trajectories build on the basic properties of orbits. In the article on
Kepler’s laws we saw how basic orbits take the shape of conic sections with the focus at
the centre of the planet. In the section on orbital manoeuvre we saw how it is possible to
move between orbits by simple velocity changes. Interplanetary manoeuvres exploit and
combine these basic ideas.

In this section we’ll start by considering a trivial example, and use this to identify some
of the difficulties of trajectory design. After this we’ll look at a more realistic design
method. Finally, we’ll look at the oft misunderstood slingshot manoeuvre.

A SIMPLE EXAMPLE

Let’s start with a simple trajectory, a transit from Earth to Mars. We’ll time our launch
so that, at the end of the transit, Mars is at a point in its orbit directly opposite the point at
which the Earth was when the probe was launched. The trajectory approximates to a
Hoffmann transfer with the Sun at one focus of the ellipse. There are two possible
directions for the launch: in the direction of travel of the earth (figure 1) and in the
opposite direction (figure 2). When the probe is launched in the direction of travel of the
Earth it gains an extra 29.7 km/s due to the motion of the earth, thus the total delta-v of
the launch is reduced by this amount. Less fuel is required at launch when the direction
of travel is the same as the Earth.

Earth’s orbit Earth’s orbit

Vprobe

Position of Mars
E when probe arrives E
Sun Sun
M M
Ve = 29.7 km/s
Vprobe Ve = 29.7 km/s

Mars’ orbit Mars’ orbit

Vm = 24.1 km/s
Vm = 24.1 km/s
M
M
Position of Mars when
Position of Mars when probe launched
probe launched

Mars is travelling around the Sun at approximately 24.1 km/s. In order to intercept Mars
at the correct point the probe has to be launched when Mars lags the intercept pint by a
distance corresponding to the flight time.

This trivial example treats the interplanetary trajectory as a simple Hoffman transfer. In
practice a trajectory based on such a simple model would miss Mars for several reasons:
 1. The trajectory would be inaccurate. The times and velocities need precise
calculation as as even a small error would accumulate into a very significant error
over the long flight times.
2. The gravity of other planets will affect the trajectory, notably the earth in the early
stages of the flight and mars in the latter stages. The gas giants, particularly
Jupiter, also exert a perturbing force on the trajectory.
3. The orbits of the planets in the solar system are not perfectly circular but are
elliptical, albeit with small eccentricity. This complicates the trajectory.

Point 1 is self explanatory, and dictates the need for high accuracy in the trajectory
design. As launches and burns are imperfect there is always some initial error in the
trajectory. One way of mitigating these errors is to allow a fuel margin onboard the
probe for a mid course correction. How much fuel to add is a matter of judgement: too
much fuel increases the probes weight and a compensating reduction in payload weight is
required – never popular with the scientists!

Point 2 can be managed by a good design method. A mathematical analysis of all the
individual gravitational forces on a probe is complex. It is known as the “n-body
problem” as there are “n” bodies (planets and the Sun), and each body exerts a force on
the probe. As all the bodies are in relative motion the resulting force vector is
continuously changing. An analytical solution to the n-body problem is mathematically
hideous, so numerical methods are usually employed.

Numerical methods set start point of the planets and the probe all the forces on the probe
are calculated as if the planets are stationary. A computer then calculates the resulting
motion of the probe due to all the gravitational forces for a small time interval, perhaps
the next few seconds of flight. The position of all the planets and the probe is then re-
plotted and the calculations repeated. By repeating these calculations many millions of
times the whole trajectory of the probe can be calculated and a “final” position of the
probe determined.

The resolution of Point 3 is fairly straightforward with numerical analysis as the

instantaneous position of the planets can be calculated.

A REALISTIC METHOD

The problem with numerical methods is that a long computer run will only tell you where
the probe ends up. It is very difficult to tune the variables to produce a realistic trajectory
that links a start time and point to an end time and point via a minimal set of manoeuvres.
The art of this method is to know which variables to correct (launch time, arrival time,
velocities, masses etc) to arrive at a satisfactory trajectory. Some supporting method is
clearly required that will allow the trajectory designer top get a simple approximation to
the variables, and then fine-tune this to arrive at a useable trajectory. Such a technique
exists, and it is called the “patched conic approximation” or PCA.
The PCA “patches” different segments of a trajectory into a sequence. Let’s revisit our
simple example of a space probe transiting from Earth orbit to Mars orbit, this time
giving the planets arbitrary start points. At the start of this sequence of manoeuvres the
space probe is in a circular Earth orbit. Earth and Mars are at different points in their
orbit around the Sun. The relative positions of the bodies are shown in Figure 3. The
probes orbit and subsequent trajectory are shown in red.

Mars

Mars’ orbit
Earth

Probe in circular
orbit around Earth

Earth’s orbit
Sun

Figure 3- Initial Positions

In the first manoeuvre we conduct a burn to inject the probe into an orbit that will allow it
to intersect the orbit of Mars. The manoeuvre is shown in Figure 4.

Mars

Mars’ orbit
Earth Injection burn

Probe in circular
orbit around Earth

Earth’s orbit
Sun

Figure 4 – Injection burn

Initially the probe moves under Earth’s gravity in an orbit that permits it to intercept the
orbit of Mars. At some point it will leave the gravitational pull of the Earth and its
trajectory will be dominated by the gravity of the Sun. For simplicity the PCA defines an
imaginary spherical surface around the Earth. Within this “sphere of influence” (SOI) we
only consider the effects of Earth’s gravity and outside it we only consider the Sun’s
gravity. This is slightly artificial as the transition from Earth’s to the Sun’s gravitational
fields is gradual; however it greatly simplifies the mathematics. We show this point at
which the trajectory leaves the Earth’s SOI as point 1 in Figure 5.

We can calculate the radius of a planet’s SOI very simply:

2
M  5
R SOI = RSP  P 
 MS 

Where:
RSOI = radius of the sphere of influence
RSP = radius of the planets orbit around the Sun
MP = mass of the planet
MS = mass of the Sun

Using this equation we can find the SOI for planets with an approximately circular orbit.
The RSOI for the Earth is approximately 927,000 km, or about slightly under 150 Earth
radii.

Mars
Earth SOI
1

Mars’ orbit
Earth

Probe in circular
orbit around Earth

Earth’s orbit
Sun

Figure 5 – Leaving Earth’s SOI

The probe is now subject only to the forces of gravity from the Sun and follows an
interplanetary trajectory. This will be an ellipse, parabola or hyperbola with the Sun at
one focus. The shape of the orbit will depend on the probes velocity, distance from the
Sun and µSun as described in the section on Keplers laws. Eventually the probe will enter
the SOI of Mars as shown in Figure 6
Mars SOI

2 Mars

Mars’ orbit

Earth

Earth’s orbit
Sun

Figure 6 – Interplanetary trajectory

Once inside the SOI of Mars the probe will be subject only to the gravity of Mars and
will execute an orbit with Mars at the Focus. Left unattended it will perform a fly-by,
leaving Mars SOI at the same velocity (relative to Mars) as it arrived. There are a lot of
misconceptions about this manoeuvre, which is called a “slingshot” in popular science,
and it will be analyzed in more depth later on.

In our trajectory we need to enter a circular orbit around Mars so we decelerate the probe.
This is normally done be rotating the probe by 180 degrees in either pitch or yaw, and
doing a burn to provide some negative delta-v. More recent Mars probes have conserved
fuel by using the Martian atmosphere to aerobrake. This takes longer than a burn, but
tends to produce a good circular orbit over a period of time. This last manoeuvre can be
seen in Figure 7.
 Injection burn
Mars SOI

2 Mars

Mars’ orbit

Earth’s orbit
Sun

Figure 7 – Mars orbit injection burn

In this analysis we’ve basically treated each leg of the trajectory as a restricted 2-body
problem. The restriction is that one of the bodies (the probe) is very much lighter than
the other bodies (the Earth, the Sun or Mars). As a result the gravitation attractions will
deflect the path of the lighter body but will not deflect the heavier body. Each leg has
thus been a simple conic section, and it has allowed us to calculate the velocity vector
(speed and direction) at which each leg ends and the next leg begins. You can see why
the method is called the patched conic approximation:
• Consecutive legs of the trajectory are patched together in sequence
• Each leg is a simple conic section as it is a solution to the restricted 2-body
problem
• The restriction of the problem to two bodies simplifies the maths and gives a fair
approximation to the actual trajectory.

This method has allowed us to produce an approximation to the trajectory, and gives
strong pointers to the values of the variables in a proper numerical analysis.

THE GRAVITATIONAL SLINGSHOT

There’s a lot of misunderstanding about the slingshot manoeuvre. We need to go back to

basics to understand the slingshot and where the additional velocity comes from.

Imagine a probe approaching Mars on a fly-by trajectory. As the probe enters the SOI of
Mars it has a velocity VP1 relative to Mars. By definition, the probe is approaching Mars
at a velocity greater than the escape velocity so it will follow a hyperbolic trajectory with
Mars at the focus. The probe will leave the SOI of Mars with a velocity VP2. The
distance from Mars at which the probe enters the SOI will be the same as the istance at
which it leaves the SOI. Basic orbital properties dictate that the magnitude of velocities
VP1 and VP2 will be the same, but the direction will be deflected. We can see the
geometry of the slingshot manoeuvre, referenced to Mars, in Figure 8.

VP2
Deflection
θ

Mars

SOI

VP1

Trajectory

Figure 8 – Slingshot referenced to the planet

The planet Mars is, of course, moving around the sun at a velocity VMars of around 24.1
km/s. If we consider the velocity of the probe relative to the Sun, we can add the velocity
vectors to see the initial and final velocities of the probe relative to the Sun. We’ll denote
the initial velocity Vin and the final velocity Vout.

VMars VP2
VP2
Vout

Mars
VMars

VP1

VMars
VP1
Vin

Figure 9 – Slingshot referenced to the Sun

The vector addition shows that magnitude of Vout is different to Vin. If the probe
approaches Mars from “behind” as shown in Figure 9 it will depart with a higher velocity
relative to the Sun but the same velocity relative to Mars.

Slingshots do not change the velocity relative to the planet, but only relative to a body
with some velocity relative to the planet.

The slingshot is thus very useful for changing the direction of a trajectory without
burning fuel, and is also useful for changing velocity relative to the Sun. This makes it
ideal for manoeuvres within the solar system, and it has featured in many missions
including the Voyager probes at Ulysses.

As the probe’s velocity has increased its kinetic energy must have changed relative to the
Sun. Energy must be conserved, so this energy must have come from somewhere. The
kinetic energy has come from the planet, deflecting its orbit by a tiny amount.
Relativity and Rocketry
By Phil Charlesworth
In 1905 a little known patent clerk and part-time scientist published three scientific
papers that laid the foundations of modern physics. The first paper explained Brownian
motion, and led to the acceptance of the existence of atoms. The second paper explained
the photoelectric effect and laid the foundations of quantum physics. The third paper was
the theory of special relativity, and the patent clerk who wrote it was Albert Einstein.
Any one of these papers would have assured Einstein’s fame. It was for his paper on
Brownian motion that he received the Nobel Prize in 1921, not his better-known work on
relativity. A century after it was published, relativity is widely regarded as either the
province of a few geeky scientists, or a convenient way of moving the plot along in
second-rate science fiction films. In fact relativity is a part of everyday life; it’s just that
the effects of relativity are so small we tend to ignore them.
This article introduces special relativity, explaining the theory by using examples based
on rocketry. It starts by introducing the basic ideas of relativity, leading to the slightly
disturbing conclusion that space and time are not quite as well behaved as we’d like them
to be. Later on it considers just how strange space and time really are. It finishes by
considering how Einstein arrived at his famous equation E=mc2 and what the equation
means.
Basic Relativity
We all understand the idea of velocity, or do we? When we launch a rocket we know that
it leave the launch rail at velocity, burns out at a higher velocity, and coasts until apogee
where it has no velocity. Einstein’s basic question, which led him to investigate
relativity, was to ask “velocity relative to what?”. To the observer on the ground the
rocket has one velocity, to an observer flying alongside the rocket, it appears to be
standing still.
A rocket has a velocity relative to an observer standing on the surface of the Earth, but
this is not the only possible reference point. The rocket has a different velocity relative to
a car driving past the launch site, another velocity relative to the aircraft infringing our
NOTAM, and many other possible velocities relative to other moving bodies such as the
planets and stars. Einstein questioned whether there was an absolute reference in a
universe where everything is in motion relative to everything else.
In the case of a rocket, it is convenient to measure the rocket’s velocity relative to a point
on the surface of the Earth. Consequently all people on the surface of the Earth who can
see the rocket will see it travelling at the same velocity v. If we call the Earth our “frame
of reference”, then the rocket’s velocity is measured relative to our frame of reference.
We can turn this concept around by imagining that the rocket is the frame of reference.
An observer on the rocket will see the rocket standing still while the Earth moves away at
the same velocity v. Both of these frames of reference are equally valid.
A two-stage rocket provides another consideration of relativity. It is reasonable to expect
that the final velocity of the sustainer stage (relative to the Earth) is the sum of the
velocities imparted by the booster and sustainer motors. In other words:
vfinal= vbooster + vsustainer.
If we consider the booster as our frame of reference, we find that vfinal= vsustainer. The final
velocity, relative to any frame of reference, can be found by simply adding or subtracting
velocities.
A “Thought Experiment”
Einstein liked to do “thought experiments”, which were experiments done entirely in the
mind and without any equipment. In our “thought experiment” we’ll take two torches,
one is held by a man on the ground and the other is attached to a rocket.
Light wave
travelling at a
Light wave velocity c
travelling at a
velocity c

Light wave
travelling at a
velocity c

Rocket
travelling at a
velocity v

If the man shines his torch straight upwards, then the torchlight will go straight up at a
velocity c, the velocity of light. This velocity can be measured relative to the Earth’s
frame of reference. If we turn on the torch attached to the rocket then the lightwaves
leaving that torch will also have a velocity c relative to our Earth frame of reference. We
now launch the rocket and it flies at a velocity v relative to our Earth frame of reference.
How fast is the light from the torch on the rocket travelling?
Relative to the rocket’s frame of reference the light is travelling at the velocity of light, c.
As the rocket is travelling at a velocity v relative to the earth, then our man on the ground
should “see” the lightwaves travelling at a velocity v+c relative to the Earth’s frame of
reference. Unfortunately the man sees all the lightwaves from both torches travelling at
the velocity of light, c, relative to the Earth’s frame of reference.
So how can the velocity of light be the same in both the Earth’s and rocket’s frames of
reference when one is moving relative to the other? According to the theory of special
relativity the velocity of light is constant when observed from any frame of reference, so
the man on the ground must see the light wave travelling at a velocity c. This is a rather
strange and unexpected result that contradicts our previous ideas about adding and
subtracting velocities.
Velocity, Distance and Time
Special relativity implies that all physical laws are the same in every frame of reference.
Consequently, when one frame of reference moves with respect to another the velocity of
light is identical in both frames of reference. We all learned at school that:
velocity = distance ÷ time
In Einstein’s universe, the velocity of light is constant, so time and distance must be
doing something strange. Special relativity shows that time and distance, which we rely
upon to be constant in our measurement of velocity, depend on the frame of reference
from which they are observed. Einstein suggested that the only reason that we don’t
notice this in everyday life is because we travel at such incredibly slow velocities,
relative to the velocity of light, that we fail to notice the effects of relativity.
Because of special relativity all measurements of distance and time depend on the frame
of reference in which they are measured. Experiments have shown this to be true, not
only in the measurement of one-dimensional distance, but also in three-dimensional
space.
Spacetime
It should be clear by now that our notions of space, which has 3 dimensions, and time, the
fourth dimension, depend on the frame of reference from which they are observed. The
universe in which we live is thus a strange place in which the four dimensions of space
and time, generally referred to as “spacetime”, don’t behave properly.
One of the consequences of the theory of special relativity is that the amount by which
spacetime “misbehaves” can be predicted. The amount of misbehaviour can be
calculated using Lorentz’s transformation, usually denoted by the Greek letter gamma γ.
Gamma is defined by the equation:
1
γ =
v2
1−
c2
We can use the value of gamma to calculate the values of length and time that we would
observe when looking at other frames of reference. In this equation v is the velocity of
the moving frame of reference, and c is the velocity of light (300,000,000 m/s).
We can do some more “thought experiments” in spacetime to illustrate the usefulness of
gamma. These experiments led Einstein to the conclusion that energy and matter were
interchangeable, and to some of the more interesting consequences of special relativity.
So What is Gamma?
Gamma allows us to take a measurement in the moving frame of reference and convert it
into the value that the observer in the other frame of reference would measure. Putting
this into a “word equation”:
Length as seen by observer on Earth = length in rockets frame of reference ÷ γ
L E = L R÷ γ
For example, if v is 90% of the speed of light (v/c=0.9) then we can calculate that γ=2.3.
If we want to find out how big a moving rocket would be to an observer on Earth, then
we simply divide the actual length of the rocket by 2.3. A pair of “thought experiments”
best explains this slightly bizarre and counterintuitive idea.
Spatial Dilation
In the first “thought experiment” we’ll consider a rocket being launched and observed by
someone standing on the Earth. If we measure the length of the rocket before launch, i.e.
while it is stationary in our frame of reference, and find it has a length L.

Rocket
travelling at a
Observer in Earth’s
velocity v
frame of reference

Length L’

Length L

After we launch the rocket it is travelling at a velocity v in the Earth’s frame of reference.
An observer in the rocket’s frame of reference would still measure the length of the
rocket as L, whereas an observer in the Earth’s frame of reference would measure the
length as LE. In the rocket’s frame of reference the length of the rocket LR is L.
Consequently in the Earth’s frame of reference we should see the rocket having a length
LE = LR ÷γ =L÷γ.
We call this change in length “spatial dilation” as the space occupied by the rocket seems
to reduce, or dilate. How big is spatial dilation? If our rocket was travelling at a velocity
v=0.9c, at which speed γ=2.3, then it would have a length LR =2 metres in the rocket’s
frame of reference. To the observer in the Earth’s frame of reference the rocket would
only appear to be 2÷2.3=0.87m long. It appears to have shrunk to less than half its
length.
If we consider velocities more likely to be encountered on the rocketry range, a 2 metre
long rocket travelling at v=300m/s would appear to shrink by 0.000000001 mm. This is a
very small amount, and we would not expect to notice it.
Time Dilation
The effects of relativity also impact on our measurement of time. In another of Einstein’s
“thought experiments”, called the “twins paradox”, one of a pair of twin brothers
journeyed to another star system on a spacecraft at a velocity close to c. When the twin
returned, he found that the brother who stayed behind had aged.
As with our previous thought experiment the two frames of reference are the Earth and
the spacecraft. Just like spatial dilation, the amount of ageing depends on the relative
velocities of the two frames of reference. If our planet was 10 light years away and the
spacecraft travelled at v= 0.8c, then the trip will take 25 years in the Earth’s frame of
reference, but only 15 years in the spacecraft’s. The twin who stayed behind will be 10
years older than his brother.
Why is this? The brother on Earth knows that the spaceship will travel at 0.8c relative to
him and cover a round trip of 20 light years. The time that should elapse is thus 20/0.8
years, or 25 years. The brother on the spacecraft experiences a time interval of 25/γ
years, which is only 15 years.
Mass Increase
Another unexpected effect of relativity is that the mass (inertia) of a rocket will increase
as it accelerates. All mentions of mass in this article, so far, have been the mass at zero
velocity, generally called the “rest mass”. When a rocket moves, it behaves as if it has a
mass M=γm. As the velocity increases, the value of gamma increases and the rocket
behaves as if it has more and more mass. As the rocket’s speed approaches the speed of
light its mass increases without limit until it has infinite mass.
I used this argument to try to convince one of the girls at work that she’d lose weight by
standing still, but that’s another story….
The Famous Equation
One other consequence of relativity is that our understanding of energy needs to be
refined. Our definition of kinetic energy, K=½mv2 only holds true at speeds which are
very much less than the speed of light. A more complete equation for kinetic energy
must take account of the frame of reference in which it has the velocity v:
K= γ mc2- mc2
A bit of mathematical manipulation (a series expansion of (1-x)-½ for those who enjoy
maths) allows us to establish that the kinetic energy possessed by a rocket approximates
to K=½mv2 at velocities very much lower than c. Further manipulation shows that the
total energy possessed by the rocket is:
mc 2
E = γmc 2 =
v2
1− 2
c
When the rocket is stationary in out frame of reference, in other words v=0, we get the
familiar equation:
E= mc2
Why is this equation so important? It implies that the energy and mass of the rocket are
interchangeable. They are different aspects of the same thing.
Consequences of Special Relativity
If mass and energy are interchangeable then accepted laws of physics such as the
conservation of mass and conservation of energy are no longer true. If a small amount of
mass ceases to exist, as happens in radioactive decay of an atom, then it must reappear in
the universe as energy. Mass and energy are not conserved, but “mass-energy” is
conserved.
Multiply mass by the square of the speed of light means that a little bit of mass is
equivalent to a lot of energy. Every Kg of mass is the equivalent of
100,000,000,000,000,000 Joules of energy. To put this number into perspective, if a
rocket weighing 1 kg sat on a table in the UKRA hut suddenly ceased to exist it would
release enough energy to power 25,000,000,000 electric fires for one hour; and HSE
worry about a few kg of AP….
Our fundamental ideas of time and distance, and everything we base on those ideas, need
to be revised. Newtons Laws are clearly incomplete descriptions of the universe, as both
space and time depend on the frame of reference from which they are observed. They
are, at best, approximations that are only true at low velocities.
So to summarise: When a rocket moves it will shrink in size, get younger than the person
launching it, and increase in mass. If it suddenly ceased to exist then the energy released
would make an H-bomb look like an Estes ejection charge.
If by now you think the world of special relativity is strange, you should try Einstein’s
theory of General Relativity. Perhaps I’ll wait until its centenary in 2016 before writing
that article. Meanwhile, if you want to read more about Special Relativity, try
“Relativity” by Albert Einstein, published by Routledge. Einstein wrote it in 1916 to
introduce the ideas of special and general relativity to non-scientific audiences, and
produced a book that can be read at many levels. At only £7.99 it’s a great read.
Predicting Wind Dispersal of Descending HPR Using a
Monte-Carlo Method
Phil Charlesworth MSc CEng
UKRA L2 RSO

Background
Last year saw a new altitude record in the UK with a flight exceeding 20,000 ft. As
altitudes increase the distance that rockets will drift during descent will also increase,
leading to the possibility of drift outside the range. A descending HPR, even under
parachute, has sufficient kinetic energy to cause damage to property or serious injury1.
It is thus highly desirable to keep rockets within the range in order to manage the risk
to people and property not associated with rocket flying.
A model for predicting the dispersal pattern of descending rockets would be a useful
too for managing this risk. It would allow both fliers and RSOs to make an informed
decision before a high altitude flight, and allow them to judge whether the flight could
be made safely. The only known model was “Splash” by Apogee, and this is only
available to US nationals. Splash is a full 6 degrees of freedom (6DOF) model of the
complete flight from launch to landing. It incorporates failure modes and their
probability in its estimate of likely landing sites, and is a comprehensive package for
predicting the probability of landing outside the range.
In 2003 the author decided to start researching how such a model could be produced
in the UK. A period of study and consultation on the technical issues of rocket drift
with colleagues in aerospace companies and academia helped the author to
discriminate between useful and impractical approaches. The current model, together
with an approach for matching it to launch criteria, was proposed on the author’s
website in 2004.
This paper reports on the first part of the author’s research. It proposes a model for
predicting the dispersal of HPR descending from high altitudes. The model is based
on readily available data and uses Monte-Carlo simulations of multiple descents to
establish the statistics of the dispersal of landing sites.
The model has been implemented as an Excel spreadsheet “front end” with the model
coded as a VBasic macro. This allows it to be used on any personal computer and, if
required, easily modified. It does not, however, lend itself to easy implementation of
more complex maths functions required to model the entire flight profile in 6DOF; the
author’s preferred language for calculations in 6DOF would be MATLAB. The paper
explains the principles of the model in sufficient detail to allow others to code it in the
language of their choice.
It must be emphasised that the proposed model is experimental and, as yet, untested.
The paper starts by describing the basic principles of the model before expressing the
equations of motion for the descending rocket. It continues by describing the

1
One source indicates that it takes about 20 Joules to fracture a skull; this corresponds to a 5kg rocket
descending at 3 m/s (about 6 mph)
software and how to use it, before concluding with a short section on the future
development of the model.

The Wind
Wind is caused by differences in the local temperature of air. At a macro level, this
causes high and low pressure regions in the atmosphere. Air flows from the high
pressure systems to low pressure systems; however the flow between pressure
systems is not constant in either direction or speed.2
The origins of winds can be illustrated by considering a low pressure system. In this
system the Sun heats the ground which, in turn, warms the air above it. The warm air
is less dense so it expands and rises. As the warm air rises it creates a pressure
gradient which causes low level air to be drawn in from the higher pressure
surrounding air.
The warm air continues to rise slowly, until it encounters air of the same density at an
altitude of between 15,000 and 20,000 ft. At this altitude, it starts to diverge until it
eventually sinks and completes the convection cycle. These effects can be seen in
Figure 1.

SUN

4. The warm air meets air

of the same density and
5. The airflow diverges stops rising 5. The airflow diverges

2. Warm air
expands and
3. Surrounding air is 3. Surrounding air is
rises
drawn in to replace drawn in to replace
the rising air the rising air

1. The ground heats up

Figure 1 - A Low Pressure System

The wind does not, of course, flow directly into or out of the low pressure system.
The Coriolis force causes the air to spiral into the system at low level and spiral out at
high level. The wind speed and direction thus change with altitude.
The wind observed at, or near, ground level at the launch site is not a true
representation of the low level wind. This wind, sometimes referred to as the “surface
wind”, is not a wind per se but reflects the results of the interaction between the low
level wind and surface features. Its effects are generally not felt more than a few
hundred feet above the ground, but are very significant at ground level.

2
Chapter 6 of “Atmosphere, Weather and Climate” by Barry and Chorley gives a good description of
winds and the forces that drive them.
Surface wind is affected by the shape of the terrain. When it encounters rising ground
the wind accelerates, an effect predicted by Bernoulli’s equation. Close to the ground
the wind forms eddies and turbulence due to its interaction with obstructions such as
trees, hedges and buildings. These friction effects can cause local and often large
variation in wind speed and direction which strongly influence the first few seconds
of flight. Weathercocking, and the subsequent launch angle of the rocket, will be
dictated by the surface wind.

The Wind Model

It is mathematically complex to describe the path that a descending rocket will follow
from apogee. A simpler approach is to consider its behaviour over short distances and
apply some statistics to its behaviour.3 By summing these effects over the whole
descent path we can obtains a reasonable model for the whole descent.
Imagine a rocket falling through a thin slice of the atmosphere. Over that short
distance the velocity of the wind and the rocket can be approximately represented as
constant values. The rocket will be subjected to a crosswind, so the true path will not
be vertical but will follow the vector sum of the descent velocity and wind velocity.
If we consider the whole descent from apogee as a sequence of n layers we can
establish the approximate touchdown point.
Point of entry

Descent
Actual
velocity
velocity

Wind
Point of exit velocity

Figure 2 - An Ideal Wind Layer

In practice, the wind vector will not be constant in each layer. We can refine the
model by assuming that the magnitude of the vector has a mean value, obtained from
a reliable source and a standard deviation. The exit from the layer is thus not a point
but a probability distribution based around that point.

Point of entry

Descent
Actual
velocity
velocity

Region of Average wind

uncertainty of exit velocity

Figure 3 - A Practical Wind Layer

3
SI Adelfang of Marshall Spaceflight Centre has published some interesting papers on the subject of
wind profiles and statistics. There are also some very interesting public domain papers on parachute
drift from the AIAA and US Dept of Defense.
If we assume that the amplitude of the wind velocity vector is normally distributed
about the means speed, and assume a standard deviation, we can link the size of the
ellipse to the probability of the exit point lying within that circle. The linkage is:
Diameter Probability
1 SD 68.2%
2 SD 95.4%
3 SD 99.7%
Figure 4 - Normal Distribution
Why select a normal distribution? Two reasons: firstly many natural occurrences
exhibit this distribution, and
Apogee
secondly because of its
mathematical convenience. Layer n

By considering the effect of

descent through n layers we can
L a y e r n -1
establish an ideal touchdown point,
and the probability of the actual
touchdown point lying within a
L a y e r n -2
ellipse around this point.
A decision was made to use 100ft
layers in the model. The
sensitivity of the model to different
layer thicknesses was tested for Id e a l
d e s c e n t p a th
descents from over 20,000ft. The
model was found to be relatively
insensitive to change once the Layer 4
thickness was below 500 ft. A
thickness of 100ft was chosen as
this corresponded to one flight Layer 3
level.

Layer 2

Id e a l
to u c h d o w n
p o in t

Layer 1
9 0 % c o n fid e n c e
r e g io n

Figure 5 - Drift Through Multiple Wind Layers

Mathematical Model
This section derives the equations of motion for a rocket descending through a layer.
It considers vertical and horizontal motion separately, and gives the general equations
that can be used in coding the model.
Forces
The drag force D on a body can be calculated from:
 1
D= ρAC D v 2
2
where ρ is the density of the air, A is the surface area presented to the airflow, CD is
the coefficient of drag and v is the velocity of the airflow. This equation can be used
to find the equilibrium descent velocity where the resistance due to the drag of the
parachute(s) and rocket body exactly balance out the weight of the rocket.
Vertical Motion
A rocket of mass m kg has a weight of mg Newtons. As the rocket descends under
parachute the weight and drag forces will reach equilibrium such that:
1
mg = ρAC D v 2
2
A rocket with CPR can be modelled as several different components, each with its
own surface area Ak and drag coefficient. CDk. As these components are linked they
will all descend at the same velocity v, and can be treated as a single mass m. The
equilibrium descent velocity of a rocket comprising n components can thus be
calculated from:
1 2 n
mg = ρv ∑ Ak C Dk
2 k =1

Thus the descent velocity v can be found from:

2mg
v= n
……………………………………………………………………(1)
ρ ∑ Ak C Dk
k =1

If we know, or can reasonably estimate, the mass of the rocket, the surface areas and
drag coefficients of the components, and the variation of air density with altitude, we
can calculate the equilibrium velocity at any altitude. Appendix 1 shows a
representative drag model of a rocket that was used for coding the model. It is
reasonably representative of a CPR rocket falling under a drogue parachute. Care
should be used when using this model for a rocket falling as two sections connected
by a shock cord; rockets in this mode tend to tumble rather than achieve a stable(ish)
attitude.
The model assumes that the atmosphere can be sliced into horizontal layers, and that
all the atmospheric parameters such as density, wind speed and wind direction are
constant within that layer. This allows us to calculate the descent velocity in that
layer and, since distance = rate x time, the time it takes to fall through that layer.
In the ith layer, where the air density is ρi and the layer is di meters thick, the descent
velocity vi can be calculated from equation 1:
2mg
vi = n
ρ i ∑ Ak C Dk
k =1

Hence the time ti it takes to descend through the ith layer is:
 n
ρ i ∑ Ak C Dk
di k =1
ti = = di
vi 2mg

Horizontal Motion
The horizontal velocity of the rocket is assumed to be that of the wind in that layer.
During time ti the rocket is carried by the wind in that layer and will move with that
layer. Analysis of the wind speed in each layer thus allows us to determine the
direction and distance that the rocket moves when transiting that layer.
The wind velocity vector is not assumed to be constant in each layer but is assumed to
vary in both amplitude and direction. He timescales for these variations are assumed
to be shorter than the total descent time for the rocket. Mean wind speed and direction
at altitudes of 1000, 2000, 5000, 10000, 18000 and 24000 ft can be obtained from the
Met Office F214. This data is linearly interpolated to produce mean wind speed and
direction for 100ft thick slices in the atmosphere. These mean values are used to
produce the normal distribution for speed and direction.
There are three main variables, each of which varies about a mean value with a
normal distribution:
• Mean wind speed
• Mean wind direction
• Variability
Mean wind speed and direction are subject to two effects: randomness and variability.
Randomness contributes small changes to both these variables and is associated with
the small-scale changes in wind profile, whereas variability contributes larger scale
changes and is associated with the turbulence in the boundary region between high
and low level winds.
Randomness for a mean wind speed v is implemented as a standard deviation of σ =
1.33m/s for up to FL180 and σ = 2m/s above FL180. These figures are Met office
data for 24 hour forecasts4. Similarly the standard deviation for a wind direction is σ
= ±15 degrees. Thus for normally distributed variations in wind we would expect that
99.7% of speeds would lie between 0 and 2v m/s and within ±45 degrees of the mean
direction. There would be a 0.15% probability of the randomness allowing the wind
speed to reverse.
Variability is a more complex phenomenon to model. In the software it is treated as a
logical variable inasmuch as it is either on or off. If the F214 states direction as
“VRB” then variability is “on” and the statistics of variability come into play.
Variability is treated in the model as randomness with a much greater standard
deviation. The variability for speed has σ = v, thus allowing wind speed to vary
between –2v and 4v. Similarly variability has a direction standard deviation of σ = ±
60 degrees, allowing a variation of σ = ± 180 degrees. Consideration was also given
to introducing a vertical component to variability by varying the descent rate, but this
was shown to have little overall impact on the scatter plot, possibly because the

4
Information on the accuracy of forecasts was obtained from The Met Office.
turbulent region is transited quite quickly as rockets tend to be descending under
drogue at that time.
The standard deviations for randomness and variability are two of the key parameters
that can be “tuned” in the model. As real data becomes available these parameter can
be adjusted to provide a closer match.
Air Density in the Troposphere5
Some models treat air density ρ as a constant. This is a reasonable assumption for
altitudes below 3000 ft as the decrease in density is only a few percent. For high
altitude flights the decrease in ρ can be significant, for example it decreases by 50%
by 24000 ft. The variation of the density of the atmosphere with altitude is one of the
key factors in calculating aerodynamic forces at different altitudes. It can be shown
that the air density at any altitude h, denoted ρh, varies according to the equation:
 g 
 −1 
 T   LR 
ρ h = ρ 0  h 
 T0 
The variables in this equation are:
ρ0 = Air density at mean sea level (kg/m3)
ρh = Air density at an altitude of h meters (kg/m3)
T0 = Air temperature at sea level (K)
Th = Air temperature at an altitude of h meters (K)
h = altitude AMSL (m)
g = acceleration due to gravity = 9.81 (m/s2)
L = lapse rate of temperature with altitude (K/m)
R = characteristic gas constant (J/kg-K)
Note that all altitudes and temperatures are above mean sea level (AMSL) not ground
level. This is significant as rocket altitudes are referenced to ground level. The
programme offsets the density calculations to launch site altitude. The value of Th can
be calculated from T0 and the lapse rate L:
Th = T0 − hL
Which leads to a usable equation linking air density to altitude:
 g 
 −1 
 T − hL   LR 
ρ h = ρ 0  0 
 T0 
Some of these constants may vary with location, so the air density spreadsheet allows
users to change the data. Typical values are:
ρ0 = 1.225 kg/m3 (may vary with location)
T0 = 288.15 K (may vary with location)
g = 9.81 m/s2
R = 287.05 J/kg-K
L = 0.008 K/m
5
It was a conscious decision to avoid the “standard atmosphere” model and allow this model to be
tuned.
Lapse rate, L, differs between dry air and moist air. It is higher in dry air and lower
in saturated, typical values are 0.01 K/m (dry) and 0.005 K/m (saturated). Moist air
generally produces cloudy weather, and rockets are not launched into cloud, so a
typical value of 0.008 K/m is suggested for the model.
The value of T0 can be measured at the launch site. If measured in centigrade (0C) the
value can be converted to Kelvin (K) by adding 273.15.

The Software
This section describes the software. It starts with a brief functional description of the
software, then describes the inputs and
outputs from the current stable version 1.5. Start

Flowchart
Read data and create
The basic functionality can be described wind profile

from a flowchart.
Y
The program commences by dimensioning Is this the first
run?
variables and arrays, reading in data from
Create mean path and
the spreadsheet using the “cells(x,y)” plot descent between
N

command, and initialising variables. It uses FL240-FL180

this information to calculate the mean wind

profile from mean sea level to FL240. This Create mean path and Create random
plot descent between number and plot
information is used to create the spreadsheet FL180-surface descent between
and graphs “plot 1 – wind profile”. FL240-FL180

The next step is to do a first pass through the Plot descent rate Create random
number and plot
program without adding any randomness or graphs
descent between
variability. This produces a mean descent FL180-surface

path which is used to create the spreadsheet

and graphs “plot 2 – descent”. Note that
apogee, main deployment and launch site
(landing) altitude are referenced to AGL but Is this the last N
run?
wind profile is referenced to mean sea level.
The program takes account of this Y
difference. Plot scatter graphs &
calculate scatter mean
The remaining passes add randomness and and SD

variability and are used to create the scatter

End
plot on the “input & output” spreadsheet.
Figure 6 - Flowchart
Inputs
The program gets its inputs by reading from the spreadsheet. The wind profile data is
typed into the “F214 Input” area of the spreadsheet. The data comes from the Met
Office aviation forecast, a free service which requires registration. From the met
office homepage www.met-office.gov.uk click on “Aviation”, log in, press “Launch”
and open the F214 for the period you require.
Note that the F214 covers a 6 hour period, and is normally available for 12-18 hours
ahead.
With the F214 open, locate the grid point nearest your launch site and you should see
a table similar to Figure 7. The first columnist ha altitude in thousands of feet, the
second gives the heading, the third gives the wind speed and the fourth give the air
temperature in degrees Centigrade. Copy the heading and speed columns into the
spreadsheet. If the wind heading is shown as “VRB”, which means variable, type in
the value of either adjacent layer (it’s not too critical which you use) and set the
corresponding variable column to “1”. This triggers variability within the program.

F214 Input
Alt (k ft) Heading Variable Speed
24 50 0 25
18 50 0 20
10 30 0 15
5 90 0 20
2 110 1 5
1 120 0 10
0 120 0 4

Figure 7 - F214 Transferred to Spreadsheet

In the “Descent Input” area of the spreadsheet you should put the apogee altitude.
Apogee is measured above ground level (AGL) so it can be referenced to the launch
site. For normal motor ejection set the main chute altitude to apogee, or whatever
altitude you expect the main chute to deploy. If the rocket uses CPR you should also
set the main chute altitude to match the altimeter settings. A CPR entry can be seen in
Figure 8.
Descent Input
Apogee altitude 24000 ft AGL
Main chute altitude 1000 ft AGL

Figure 8 - Rocket Descent Inputs

The “Rocket Data” area is used to calculate the weight and drag forces acting on the
rocket during descent. The rocket dry mass is used to calculate the weight of the
rocket. Rocket diameter, length and fin area are used to estimate the total drag from
the airframe after deployment, the method of estimation is described at Appendix 1.
The drogue and main chute diameters and drag coefficients are used to estimate the
drag added by each chute as it deploys. If the rocket uses motor ejection, or uses CPR
but without a drogue, you should set the drogue diameter to zero.
Rocket Data
Rocket dry mass 8 Kg
Rocket diameter 3 inches
Rocket length 70 inches
Area of one fin 18 sq inches
drogue chute diameter 0 inches
Drogue CD 1.5
Main chute diameter 60 inches
Main chute CD 2

Figure 9 - Rocket Data for Drag Model

The “Prog Data” area contains two variables. The first is the number of points to be
plotted which should be an integer between 2 and 200. Setting this to a larger
number requires some adjustment to both the program and the spreadsheet otherwise
misleading results and plots will be produced.
The second data input is the launch site altitude in feet AMSL. This is used to
calculate the apogee and landing points AMSL.
Prog Data
Number of points 200 (max 200)
Launch site altitude 750 ft ASL
Figure 10 - Other Program Data
Note that the wind model is only valid to 24,000 ft AMSL, so the total of the launch
site altitude and apogee altitude must not exceed 24,000 ft. A warning is displayed in
the spreadsheet if this condition occurs.
A final set of inputs can be found in the “Atmospheric Density Model”. These should
normally be left untouched, but may be adjusted if launch weather conditions are
abnormal or you plan to launch on another planet.
Sea level pressure 1013.25 mb NOTE
Sea level density 1.25 kg/m^3 The values of the atmospheric
Sea level temp 288.15 K parameters can be changed in this
Lapse rate 8 K/km spreadsheet, and they will
J kg-1 automatically be picked up and read
Gas constant (dry air) 287.04 K-1 by the model.
Acceleration due to
-2
gravity 9.80665 ms

Figure 11 - Atmospheric Density Model

Once you’re satisfied with the inputs, press the “Press to Run” button. The data in the
“Results” area should change and a new scatter graph produced in about 2-3 seconds.
Outputs
The outputs are written to the spreadsheets using the “cells” command in VBasic.
They are produced as graphs, although you can take the data from the cells for further
manipulation or analysis.
The main plot is the scatter plot. This shows the distribution of landing points around
the sub-apogee point NOT THE LAUNCH POINT. This data is analysed in the
“Results ±3σ” area to show the statistical mean, min and max values of range and
bearing.
A scatter plot can be seen in Figure 12.
The scatter plot has been set to a scale of ±5km N-S and ±5km E-W. The scale can be
adjusted by right clicking on the X and Y axes in turn, selecting “format axis” and
setting the max and min values under the “scale” tab.
 Scatter plot from apogee (distances in m)
5000

4000

3000

2000

1000

0
-5000 -4000 -3000 -2000 -1000 0 1000 2000 3000 4000 5000
-1000

-2000

-3000

-4000

-5000

Figure 12 - Scatter Plot

The plots of wind profile can be seen in the “Plot 1- Wind Profile” sheet. Typical
plots can be seen in Figure 13.
Wind Speed Wind Heading
30000
30000

25000 25000

20000 20000
Altitude (ft)

Altitude (ft)

15000 15000

10000 10000

5000 5000

0 0
0 5 10 15 20 25 30 0 90 180 270 360
Speed (MPH) Heading

Figure 13 - Wind Profile Plots

A further pair of graphs can be seen in the “Plot 2- Descent” sheet. These show the
descent time and descent velocity in seconds and m/s respectively. They are provided
mainly because the data was available, but may help to give an idea of how long the
rocket should be airborne after apogee. Typical graphs for a CPR flight can be seen at
Figure 14.
 Des cent Time Des cent V elocity
30000
30000

25000
25000

20000 20000
Altitude(ft)

ltitude(ft)
15000 15000

A
10000 10000

5000
5000

0
0 50 100 150 200 250 0
0 10 20 30 40 50 60
Time (sec, from apogee)
Des cent V elocity (m/s)

Figure 14 - Descent Plots

Next Developments
Having created the model the next step is to calibrate it. This comprises taking data
on real HPR flights to significant altitudes and using these to tune the variables. The
general approach will be to take F214 and launch data from real rocket launches and
adjust the variables to fit the data.
There are several variables that could be adjusted, most notably the standard
deviations for wind speed and direction above and below FL180 (4 variables) and the
scaling factors in the random number subroutines (2 variable). The standard
deviations were set based on information from the Met Office and should not require
much adjustment. The focus of the tuning the model should thus be on adjusting the
two scaling factors for the random number subroutines.
This parametric tuning may also account for some of the assumptions, in particular
the assumptions that a rocket moves with the air in every layer and descends at a
steady rate.
The model only considers descent from apogee, and does not consider any aspects of
the launch or ascent. Further developments may extend the scope of the model to
cover the whole flight. These extensions include:
• Integration of the scatter plot with map data to aid visualisation
• Flight analysis in 5 DOF (assuming the effect of rotation around the roll axis
is insignificant) based on a given weathercock angle.
• A detailed analysis of weathercocking in 6DOF, taking into account the
changing mass distribution of the rocket.
• Consideration of failure modes, their probabilities, and their impact on
landing site.
The end result would be a program that could predict the scatter plot for all
conditions, including rocket failures. This would be a powerful tool for assisting
RSOs to make go/no-go decisions on high attitude flights. It would also be useful for
clubs to determine the best launch point, and perhaps set altitude limits depending on
prevailing conditions. Ultimately, it will help to maintain the good safety record of
rocketry in the UK while we continue to push the limits.
APPENDIX 1 – A SIMPLE DRAG MODEL OF A DESCENDING ROCKET
If we ignore the drag from the shock cord, we can produce a simple model of a CPR
rocket. Such a rocket has the following components:
• Main chute
• Drogue chute
• Main body
• Fins
The layout of a typical rocket descending is shown in the diagram below. This
diagram serves to explain
d1 how the various components
of the rocket add to the total
drag.
The main and drogue chutes
have the area of circles of
diameters d1 and d2
respectively. This ignores
spill holes. Construction
techniques may result in
different drag coefficients so
Af these can be input via the
spreadsheet. It is possible to
“free fall” with no drogue by
making the diameter d2=0.
For simplicity, we assume
that the main body of the
d2 rocket approximates to a
cylinder of length l and
l diameter d3. The surface area
of this cylinder (ignoring
ends) is:
450
d 32
A3 = πl
4
d3 Of course, the whole surface
is not exposed to the wind.
Simple geometry shows that
only 1/π of the area contributes to drag. As the main chute provides more drag than
the drogue then the main body will be inclined to the horizontal. This inclination is
assumed to be 45 degrees, thus the area presented to the airflow is only 0.7071/π of
the total area.
The fin area represents the area and number of fins that are exposed to the airflow.
This is multiplied by 0.7071 as the fins are inclined at 45 degrees to the airflow.
A table summarizing the area contributions of each part of the rocket that are used for
calculating drag can be seen below.
k Component Assumed area Ak Assumed Remarks
CDk
1 Main chute
πd12 1 Diameter d1 m
A1 =
4 Note 1
2 Drogue chute
πd 22 1 Diameter d2 m
A2 =
4 Note 1
3 Main body 2 1 Treat as a cylinder
d
A3 = 0.7071 × l × 3 diameter d3 and length
4 l 3.

Note 2
4 Fins 1 There are 3 or 4 fins
A4 = 0.7071 × 1.66 A f and the area of each fin
is Af.

Notes 2,3

Note 1. The values of CD for the main and drogue parachutes can be varied as inputs
in the spreadsheet.
Note 2. Only 1/π of the surface is exposed to the airflow. It is also assumed that the
body does not hang vertically but at 45 degrees, thus only cos(45) = 0.7071 of the
exposed area contributes to drag.
Note 3. The coefficient 1.66 is derived from the aspect ratio of the fins. For 4 fins the
area presented lies between 2 Af and 2xcos45 =1.414 Af. For 3 fins this aspect lies
between 1+cos60 = 1.5 Af. and 2 cos(30) Af = 1.73 Af. The average of these is 1.66 Af.
APPENDIX 2 – GENERATING NORMALLY DISTRIBUTED RANDOM
NUMBERS
VBasic has the ability to generate random numbers between 0 and 1 with a flat
distribution. A technique was needed to generate normally distributed random
numbers with a mean of 0 and standard deviation of around 0.3.
Two techniques were considered:
• Generate the random numbers in Excel and read them into VBasic
• Write a routine in VBasic to generate the random numbers.
Normally distributed random numbers can be generated in any cell in an Excel
spreadsheet using the function NORMDIST. The form of the command is:
= Normdist (Rand(), mean, variance)
Since the variance is the square of the standard deviation it is easy to produce random
numbers on demand. The cell is refreshed whenever anything is written to the
spreadsheet, either from the keyboard or from VBasic. The VBasic function thus
needs two lines: the first to write to any cell to trigger a new random number and the
second to read the new random number into a VBasic variable.
The advantages of this approach are simplicity and the ability to easily tune the mean
and standard deviation of the random number. The big disadvantage is that it is very
slow.
An alternative approach was adopted. This exploits a useful property of statistics
called the central limit theorem. This theorem states that the distribution of an
average tends to be Normal, even when the distribution from which the average is
computed is non-Normal. In other words, if we take the mean of a number of random
numbers then the mean will lie on a Normal distribution.
A short subroutine was written to generate 30 flat-distributed random numbers in
VBasic and find their mean. Tests showed that the distribution of numbers generated
this way was normal, with a mean of 0.5 a variance of 0.083 and s standard deviation
of 0.288.
This distribution was centred on zero by subtracting 0.5 from all the generated
random numbers. This value was then returned as a normally distributed random
number. This subroutine reduced the run time of the programme by a factor of 12,
and was thus adopted.