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2
Contents
1. What is a Lunar Space Elevator………………………………3
2. Historical Overview………………………………………….4
3. Helium 3 – Rare Energy Source…………………………….5
4. Abstract……………………………………………………….6
5. Payload Capsule Trajectory…………………………………7
6. Tether Characteristics………………………………………..8
7. Elevator Design……………………………………………….13
8. Climber Movement…………………………………………..31
9. Tether Oscillations…………………………………………....34
10. Capsule Reentry to Earth……………………………………36
11. Bringing the Spaceship to L1 Point – Theory……………..40
12. Bringing the Spaceship to L1 Point………………………...43
13. Tether Deployment………………………………………….47
14. Navigation……………………………………………………53
15. Solar Panels…………………………………………………..55
16. Tether Mooring to the Moon……………………………….57
17. Anti-skid Climber’s Wheels System………………………..58
18. Conclusions………………………………………………….61
19. Appendix – Matlab Codes………………………………….62
20. Bibliography…………………………………………………83
3
?Lunar Space Elevatora What is
The Lunar Space Elevator is a cargo transportation system between the Moon and Earth
and vice versa. The system uses a very long tether, connected to the Moon, stretching
towards the Earth. An electrically powered climber is designed to move on the tether. The
electrical energy is supplied by solar panels, on both sides of the climber, always directed
to the sun. The climber is carrying a cargo capsule. When the climber reaches the
appropriate release point, between the Moon and Earth, the capsule detaches and enters
“free” orbit around Earth. The orbit brings the capsule into the atmosphere, due to the
strong gravitation field of the Earth. At entry into the atmosphere, the capsule decelerates
ballistically due to drag and at a set altitude, breaks up and releases its payload to land
without use of parachutes.
The space elevator is not a new idea nor is it the exclusive idea of one person. The idea of
space elevator was conducted and studied by several scientists and institutes. Today there
are even companies, dedicated to promotion and development of components for the
space elevator. We have used as baseline, the report by Jerome Pearson, containing a
detailed study of the lunar space elevator. Pearson’s idea was the use of the lunar space
elevator for transporting Helium 3 to the Earth, in order to use it in the process of Nuclear
Fusion process, capable of delivering huge amounts of energy.
4
Historical Review
In 1960, a Sunday edition of the Russian newspaper “Komsomolskaya Pravda”
published an article by Russian scientist, Yuri Artusanov, which proposed to build a
“Space Train”. The article contained a description of a long tether, stretched from a base
on Earth’s surface, to geostationary orbit, where a “satellite” was proposed to be located.
This orbit is unique, such that a satellite on this orbit rotate at such speed, that it appears
to be stationary compared to a point on Earth, and so the tether “stands still” and allows
transportation of cargo to geostationary orbit.
The tether’s tension is provided by the satellite mass by the free end, and mooring the
other end to the ground on Earth. This article was received as science fiction, since there
weren’t available technologies or materials to fulfill these requirements.
In 1979, Jerome Pearson and Yuri Artusanov (independently) suggested to build a lunar
space elevator, which will be based on tether from the lunar surface towards the Earth,
with moving climber on it, maintaining the following principles:
1. Same face of the Moon always faces the Earth
2. Lagrange (null gravity) point in Earth-Moon system, allows stretching the tether
3. No environmental effects (atmosphere, satellite/aircraft, etc.)
4. No political boundaries
5. Moon’s gravity is considerably smaller than Earth’s
In 2005, Jerome Pearson submitted a lunar space elevator proposal report to NASA with
several main parameters:
• 290,000km long tether, attached to the Moon’s ground, on the side facing
the Earth
• 100 robotic climbers, moving on the tether at 100kph velocity
• The tether would be made from M5 polymer
The goal for this lunar elevator, by Pearson’s vision, was to transport Helium 3 that will
be used for nuclear fusion energy process. Helium 3 is located in relatively large
quantities on the Moon’s surface, and thus there is justification for mining and refining
the Helium 3 on the Moon and therefore inexpensive, a reliable and effective
transportation system is needed.
5
Helium 3
Helium 3 is an isotope, consisting of two protons and a neutron. It can be used in the
process of nuclear fusion, and from all available substances for this process, it is the least
radioactive and the most energetic.
The source of Helium 3 on Earth is mainly from tritium decay, created from nuclear
warheads material. Its scarcity makes it very valuable – it is estimated that only about 500
kg of this substance exist on Earth, capable of providing enough energy for electricity for
one year. On the other hand, on the Moon, it is estimated to contain about a million tons,
enough to provide energy for 2 million years! The explanation for Helium 3 existence on
the Moon in relatively large quantity is due to the solar wind particles, created in nuclear
reactions in the Sun, which bombards the lunar surface. Solar wind particles don’t reach
Earth surface, since the charged particles, get caught by Earth magnetic field, but the
Moon isn’t protected such way, and so Helium 3 on the Moon is therefore a renewable
energy source.
Although the amount of lunar Helium 3 is relatively large, its concentration is only 0.01
ppm for each ton of regolith (lunar soil), meaning that for each ton we will get 0.01gr of
Helium 3. Thus there is need for a large investment for suitable automatic facilities and
machinery for extraction of significant amount of this isotope from the surface.
The process of extraction should contain an element for cargo transport to Earth, and so,
the lunar elevator could be an inexpensive and efficient alternative for cargo
transportation, compared to conventional rockets.
The mining machinery will load the Helium 3 in special containers, which will be
transported, by the solar powered climber, moving on the tether, to a release point, for
reentry free fall orbit back to Earth.
6
Abstract
In this project, we used as a baseline J. Pearson’s report to NASA, in order to design our
version of lunar space elevator, and address additional issues.
The system follows these several principles:
• Tether of 325,000km length, made of M5 polymer, moored to the Moon, on the
side facing the Earth, stretched through L1 point (Earth-Moon gravity system),
with free end near the Earth.
• Cargo capsule, weighing one ton, will contain the Helium 3 in pressurized
containers and will incorporate heat shield for Earth atmosphere reentry, electrical
and control systems.
• The climber, weighting about half a ton, will be powered by solar panels, have
electric powered motors for the wheels, control and stability systems and
communications equipment.
• Rocket like anchor, for mooring the tether to the Moon’s surface, by penetrating
the surface to the necessary depth.
For designing the system, we had to get through the tether design, materials and
vibrations, completed initial design of the cargo capsule and the climber and selecting the
right high pressure containers for the Helium 3 transport.
Movement of the climber of the tether and its effects was closely analyzed by numerical
simulations and final elements method. Earth and Moon gravities, reciprocal movements
of the system and Coriolis Effect on the tether, caused by the climber movement, were
taken into account.
Furthermore an analysis and design was performed of the climber’s wheels, anti-skid
system and general design of the climber structure and solar panels parameters. A
navigation system was proposed for desired operation of the climber.
Finally, we dealt with tether deployment, by bringing it to the Lagrange point and
anchoring it on the Moon’s surface. This was performed by using three-body approach
and linearization of the proper equations.
The project’s home website is: www.lunarjacobsladder.webs.com
7
Trajectory of the Cargo Capsule
Planning of the cargo capsule trajectory was according the following requirements:
• The capsule would be released on the tether, at the proper point to insert it to
Earth inbound trajectory (without using additional thrust).
• The perigee of this trajectory will be at the altitude of 80 km from Earth’s surface,
in order to enter the atmosphere
To find the parameters of the trajectory we use the fact that the velocity of the capsule at
the release point will be equal to the entry velocity for the desired trajectory:
It is equal for the velocity of the climber v R= Ω ⋅ , when Ω the angular velocity of the
Earth-Moon system and R is the radius from the release point to the center of the Earth-
Moon system.
RRdRdR p
Ω=++
µ−
+
µ EE 22
Rp is the radius of the Earth and d is the center of the Earth-Moon system,
By approximation (when Rp and d are very small compared to R), we will get:
4/1
2
E
2
E222~2
Ω
µ≈⇒
µ≈Ω
pp RR
R
RR
Giving us:
R=155,000 km
T= 66 hours
Ve= 10.9 km/sec
While T is the cycle time and Ve is the entry velocity to the Earth.
Additional information: Inertial velocity at the release point is 0.412 km/s and entry angle
to the Earth is 2.5 degrees under horizon.
1 1( ) 2
2e
v RR a
µ
= −
8
Tether Characteristics
In order to characterize the tether, we must first to specify the forces acting on it:
1. Earth’s gravity
2. Moon’s gravity
3. Earth-Moon system radial motion
It will give us the next equation for calculation of the acceleration acting at each moment
on the tether:
d
Gravity parameters of the Earth and the Moo
h
Lr
La
LE µµ ,
Ω
2
2 2( ) ( )
( ) ( )
E LL L
L L L
g h a r h da r h r h
µ µ= − − − ⋅Ω − +
− − +
9
This equation allows us to perform appropriate calculations in Matlab and receive the
next chart, which shows the acceleration on a body as a function of distance from lunar
surface:
As we expected, at Lagrange point (L1) we will get null forces and after that, the gravity
acceleration is negative and towards the Earth, meaning that we can “cruise” with the
climber, without using motors, after passing the Lagrange point.
For tether’s tension we’ll use the next equation:
Tργ σ=
2
2 2( ) - ( ) ( - - - ) - ( ) 0
( - - ) ( )
E LL L
L L L
T h dh T h a r h d h dha r h r h
µ µγ
+ + × Ω + × = +
10
Integrating the equation brings us to the next form:
T(0) is the required tension at the base of the tether on the Moon’s surface, and it is equal
to 3240[N], which is the product of multiplying of capsule and climber mass (1.5ton) by
Moon’s gravity acceleration and by safety factor of 1.33.
We will get the next chart, showing the tension as a function of distance from the Moon:
The maximum tension is at the Lagrange point, meaning that the tether should withstand
this maximum tension, of course considering its cross-sectional area.
If we take the maximum tension and divide it by maximum allowed stress, we will get
the minimal cross-sectional area. In order to keep the same ratio of tension to stress, we
decided to choose non-constant cross-sectional area of the tether:
( )
( )
22
22
exp2
( ) (0)
exp2
E LL L
L L L
E LL L
L L L
a r h da r h r h
T h T
a r da r r
ρ µ µ
σ
ρ µ µ
σ
Ω− − − − − − − − + = Ω
− − − − − −
11
Cross-section of the tether as a function of distance from Moon’s surface is shown in the
next chart:
Now, after we’ve found the desired cross-section for the tether, we must choose its
design. Pearson proposed in his article a round cross-sectional tether, but our choice is
rectangular cross-section, allowing more surface for contact with wheels and less chance
of rupture, due to micrometeorites impact.
To make the manufacturing easier, we’ve decided to leave the tethers thickness constant,
at 0.015mm, and changing width, from 12mm to 36.7mm.
maxσ=
TS
12
End mass:
An end mass is required to balance the tether, much like a pendulum. As we make the
tether longer, a smaller end mass is required, since the gravity influence of Earth rises
and so, the force acting on the end mass increases. In the next chart, we can show the
required end mass, as a function of tether’s length:
We chose the tether to be 325,000 km long, since there the difference in the total mass
(tether + end mass) start to become negligible.
In order to increase the system’s redundancy in case of rupture, we chose four parallel
tethers instead of one, using the recommendation from Pearson’s report.
13
Initial Goals–Capsule Design /Climber
During the project, we had a complicated task of designing the climber structure and
concept of operation. Our main missions were:
• Choosing the materials for:
o Climber
o Capsule (including heat shield)
o High pressure containers for the payload
• In the climber:
o High speed light wheels system
o Anti-skid design
o Stability and control
o Initial assessment of sub-systems weight
o Stress and strain calculations
• Design and analysis of the return capsule:
o Design of capsule’s structure and containers for the payload
o Calculations of reentry to atmosphere and related thermodynamics
Main characteristics of the climber
• Mass 1.5 ton (1 ton for the return vehicle and 0.5 ton for the climber itself)
• Main sub-systems – batteries, photoelectric power cells, wheels, 4 electric
motors, vibrations dumping system, anti-skid system, thermal control system,
computer, general structure.
• 4 wheels system (0.7m radius of each wheel).
• Maximal cruise speed of 0.7 km/sec.
• Required energy – 20,000W (first iteration 7,500W).
• Solar panels area – 100 square m. (first iteration 28.6sq.m.), with 20%
efficiency.
14
Main characteristics of the return capsule
• Mass 1 ton (230kg high pressure containers, 100kg He3, ~200kg heat shield, ~500kg
structure)
• Ten high pressure containers for payload
• Ballistic reentry to Earth (without parachutes)
Design of the Climber system – First Iteration
In this chapter, shown the first iteration, for the initial design of the climber (PDR phase).
The climber consists of several parts:
1. Two sets of four pairs of wheels (16 wheels total), at the front and at the back of
the climber – allowing the climber to move on the tether.
2. Four electric motors – each motor connected to four wheels.
3. Return capsule – details of the design in the relevant chapter.
4. Solar panel – with the required area to produce 7,500W.
5. The main structure of the climber – containing the passage for the tether and
various sub-systems for dynamically and thermally controlling climber.
6. General structure of the climber – built of carbon-epoxy for best strength to
weight ratio.
15
Iteration First–skid system -Anti
A control circuit is used in order to control the wheels, at its motion on the tether and
prevent skidding from the tether, by tilting relatively to each other. The idea can be
Return vehicle
Solar panel
Motor 1 Wheel system
Main structure
& Systems
Tether system
Motor 2
Motor 3
Motor 4
16
explained, by simulating two truncated cones, rolling each on other. When the skid from
the tether occurs, the wheels’ axis makes a turn relatively to each other, and by that,
allowing rolling on the conical cross-section of the wheels, instead on the central part,
causing a turn to the center of the roll (see picture):
Longitudinal roll control system
When the climber moves on the tether, a longitudinal roll can occur (around the
motion direction), around the tether. The proposed control approach to solve this
problem, is by a pair of momentum wheels, one for each pair of electric motors.
When the sensor notices longitudinal roll, the momentum wheels start spinning,
creating angular momentum at the opposite direction and returning the climber to the
initial angular position.
Momentum wheels use is widespread in the space industry (for example in the
Hubble space telescope), as they serve a stabilizing gyroscope controller. Momentum
wheels usually have a high RPM rate (around 5,000rpm) and relatively high weight.
Center of roll
17
Design of the return capsule – First Iteration
The return capsule was designed as such, that it could contain ten high pressure spherical
pressure containers, which would contain pressurized isotope He3. The basic design
included ten high pressure containers, upper access hatch to the containers, anchors for
connecting to the climber and space for additional systems for operating the capsule.
Additionally, the bottom of the capsule, including the heat shield, is intended to detach
itself pyrotechnically at the last stage of reentry to the Earth’s atmosphere, as described in
following chapters.
Motor + Momentum
wheel
Motor
Double wall pressure tanks
Hatch
Heat shield
Return vehicle structure
Connection to the elevator
Space for systems
R=0.371 m
Hatch
Heat shield
Pressure tank
Main structure
18
Second Iteration–Design of the return capsule
Second iteration of the preliminary design of the reentry capsule, presents a different
geometrical form of the capsule and different arrangement of the pressure containers
(tetrahedral form – for better usage of available space). More conical form of the return
capsule (thanks to the new rearrangement of the pressure containers), allowed us to
maximize the heat shield at the bottom and reduce the temperature effect on the
unprotected areas of the capsule. Other characteristics remained the same.
Calculations-High pressure containers
As mentioned earlier, the return capsule contains ten spherical high pressure containers,
for transporting the payload. In order to maximize the available payload which could be
taken and reduce structure weight, an analysis was performed to calculate the structure
and choose materials for the pressure containers. After several iterations, the chosen
configuration was a two-layered structure, in order to maximize strength, keeping the
weight as low as possible. Examples for such containers can be found on the Ariane
Double wall pressure tanks
Hatch
Return vehicle structure
Connection to the elevator
Capacity for systems
R=0.371 m
Heat shield
Hatch
Heat shield
Pressure tank
Main structure
19
vehicle of European Space Agency (ESA). The inner layer is made from titanium and the
outer layer made from carbon-fiber, with pre-tension between the two layers, in order to
prevent separation. The information about the materials was taken from web materials
library (www.matweb.com):
• Inner layer - Titanium Ti-6Al-4V (Grade 5), Annealed
Outer layer – Carbon Fiber T1000G
• Operating pressure of the high pressure container – 394 atmospheres
• Safety factor chosen (due to the nature of high pressure containers) is 2, so:
The initial pressure acting between the two layers in the empty container is between 6 to
33 MPa and can be neglected.
The mass calculation of the structure for one container:
( ) ( )( )( ) ( )( )33 33
_ 1000 1000 1000 1000
4 422.726
3 3press capsule T G Ti T G Ti T G T Gm R t R t t R R t kgπ ρ π ρ= ⋅ ⋅ − − − + ⋅ + ⋅ ⋅ − − ⋅ ≅
3
640 10 0.2037 39.8
8314.472 300
w
helium
P V Mm kg
R T
⋅ ⋅ ⋅ ⋅ ⋅= = ≅
⋅ ⋅
[ ] [ ] 3880 , 113.8 , 4430y
kgMPa E GPa
mσ ρ = = =
[ ] [ ] 33040 , 165 , 1792y
kgMPa E GPa
mσ ρ = = =
( )_ ._ tan
1000
0.371
0.003 30.2 0.8
2
out press k
Ti T G
R m
t m mmP R t
tσ σ
= ⇒ ≅ =⋅ −
⋅ + ⋅ = ⋅
10006 1 , 5tot Ti T G
t mm t mm t mm≅ → = =
20
And so, for ten containers (with He3 inside), we will get mass of 325kg total:
Normal force calculations and friction losses
The wheel system of the climber contains two sets of four pair of wheels, at the front and
at the back of the climber. Between each pair of wheels a normal force acts, allowing the
climber to stay on the tether without falling. This chapter is about the characteristics of
this force.
In the next chart we can see the deviation of weight of the climber, as a function of
distance from Moon’s surface. It is easy to see, that we’ll get the highest weight at the
Moon’s surface, logically where the gravity force is the highest along the tether. The
calculation is made by multiplying the full mass of the elevator (1.5 tons) by varying
gravity acceleration (Moon and Earth gravities changing along the tether). It can be seen
from the chart that the maximum weight of the elevator on the Moon’s surface is
2.435kN:
2.435kN max weight
_
_
2 2 7
9 8
s t r u c t u r e t o t
p a y l o a d t o t
m k g
m k g
=
=
21
Using this calculated weight, we can now easily calculate the normal force required to
hold the climber without slipping. We use the friction forces, between rubber wheels
cover to the polymer tether. The friction coefficient taken is 0.5, while this value is only
an approximation, due to lack of experimental data. Multiplying the changing weight
values by changing contact area of the tether (due to changing cross-section) and dividing
by friction coefficient, we’ll receive the normal force distribution, as a function of
distance from Moon’s surface. As before, the maximum normal force required, is on the
Moon’s surface, and its value is 23.56kN. The next chart presents the normal force
distribution:
When the climber is on the tether, we have inevitable wheels friction, using extra power
and requiring higher power input. In the next chart we can see the power waste, due to
wheels friction (with estimated wheels friction coefficient of 0.005). It can be easily seen,
that this lost power, is much smaller than the total power available (20kW) and can be
neglected:
23.56kN max normal force
22
Strength calculation of wheels system
The wheels system of the climber is complicated, and so, accurate calculations are
required, in order to determine the required wheels structure, without heavy penalty on its
mass. In the next chart the design choice is shown. At the upper part, the results by
various radii of the wheel can be seen, up-to the sizing of 2 meters. The maximum force
will act on the wheel’s rim, and so choosing the materials according their elasticity
modulus, would lead us to possible solution. The leading consideration is the lower
centripetal force at the larger radius wheel, but the mass will increase in this case, as well.
In our case the chosen radius size is 0.7m, with the required elasticity modulus that suits
carbon-epoxy structure. The mass of the wheel, with its unique structure, is about 2kg,
with slight variations due to materials choice.
23
Choosing the value of 0.7m for the preferred radius of the wheel was due to motor rpm
speed consideration. The climber moves at the speed of 0.7km/sec = the linear velocity of
the outer boundary of the wheel. Thus the calculated angular velocity required from the
electrical motor, spinning the wheel is about 9,550rpm. This is a high value, but
reasonable for a realistic electric motor, suitable for propulsion of the climber.
From considerations of sizing of the tether damping hole and safety factor (so that the
tether won’t be too close to the end of the wheel), the wheel’s width was chosen to be
110mm.
3: 70 , 1600
kgCarbon Epoxy E GPa
mρ− = =
Weight per wheel ~ 2kg
24
The chosen configuration for the climber wheel is a bicycle type wheel – first of all, due
to weight consideration, while keeping the required strength. The spokes used in this kind
of wheel, are a steel spoke, with pretension, so when stress is acting on the wheel, the
change is only at the tension of the spike and buckling wouldn’t occur.
In the next chart a change of stress on the wheel’s spokes can be seen, with acting of
required maximal normal force. This force acts on the wheel’s axis (as shown in the
drawing) and so, the tension of the lower spokes is getting lower, and the upper turning
higher (the contact area between the pair of wheels is at the bottom of the wheel in the
drawing). The change of the tension of the spokes is within stress limits for steel spokes
of 1mm radius, while this value can be lowered even more, by increased number of the
spokes or their cross-section.
In order to maximize the contact area between the pair of wheels (in compression), an
additional rubber coating is necessary on the wheels perimeter. Small contact area can
cause high stress concentration and local buckling and even can cause the tether to be cut.
The method of iterative calculation of the stress for given contact area is called Hertzian
change in stress in tension spokes
-800.00
-600.00
-400.00
-200.00
0.00
200.00
400.00
600.00
800.00
0 18 36 54 72 90 108
126
144
162
180
198
216
234
252
270
288
306
324
342
360
angle [deg]
delt
a s
tres
s [
MP
a]
25
Contact Stress method. The calculated results are for a pair of cylindrical wheels, with
parallel axes.
Length of the contact area:
( )( )
2 2(1 ) 422
4
RPB
l E R
ν
π
−=
Strain:
2 20.25R R B
d d
δε
− −= =
Maximum pressure applied: max
2Pp =
Blπ
B – length of the contact area
l – width of the wheel
R – radius of the wheel
d – thickness of rubber layer
P – applied pressure
ν - Poison coefficient of the material
E – Yang modulus of the material
δ - change in rubber thickness due to compression
If we choose maximum strain value of 30% (which is logical maximum
value for rubber), we will get the following results;
[ ] [ ]2
max0.0569 , A =0.0078 m , 1.917E GPa p MPa = = ,
Which are acceptable values for the material we have chosen (rubber with
such Yang modulus can be found and the applied pressure isn’t high for the
material).
R
B
l
26
Design of the climber system – Second Iteration
In this section, the updated preliminary design of the climber is presented. The major
changes since the first iteration:
1. Change of the location, size and form of the solar panels – the new form and
location, allows movement of the panels according the position relative to the sun,
the area is larger due to larger energy requirement.
2. Change of wheels configuration – at this stage (CDR) more detailed work has
been done for the climber’s wheels and their configuration changed compared to
the initial stages.
3. Addition of tether oscillation vibrations – at the front and back side of the climber
– this was developed at the CDR stage as well and did not appear in the first
iteration.
4. Addition of springs between the wheel’s axes – addition of wheels in each side of
wheel’s axes, for appliance of required normal force (between the wheels pairs) to
hold on the tether, with allowance overcoming small obstacles on the tether (like
fold or dirt).
5. Addition of communication and equipment pole – behind the main structure of the
climber – for clear view for the communication antenna and star-tracker cameras.
6. Geometrical change of the return capsule – for more efficient packing of the
pressure containers and maximizing the area of the heat shield (see the
appropriate chapter).
7. Addition of electrical servos for solar panels movement – at the joints of the
panels to the main structure.
27
28
29
Spring System
The spring system in the climber is intended to apply the required normal force between
the pair of wheels (23.5kN) and to overcome small obstacles, as folds or dirt on the
tether. The best method is to use a pair of springs, each at opposite end of the axis, in
order not to create a moment (such as in the case of using only one spring). We can use
springs as used in the automotive industry, because it is intended for a weight (on each
spring) of about 1,200kg, which is a typical weight of a common family car. Using this
off-the-shelf product, saves the need for special developing and manufacture of special
tailored solution.
30
systemdamping Tether guide and oscillation
The tether is threaded through a guide oscillation damping spring, leading its path
between the wheels pairs. In order to overcome the friction that might be created
touching walls of the hole we’ve designed a smart system intended to reduce the friction.
A set of rolling cylinders, are located at the perimeter of the hole – having conical ends,
they can roll freely around each axis. When the tether touches such a cylinder, the sliding
friction will be converted to rolling friction (because the cylinder will roll), which is
much smaller, thus reducing the overall friction of the tether. The conical ends between
the cylinders will be closely fit, leaving no gaps (so the tether won’t be caught between
them).
31
Movement of the climber
The velocity of the climber is a direct function of available power. This connection is
given by the following energy equation:
( )
( ) ( )( ) ( )
2
0
2
2 2
2
h
E LL L
L L L
d VM g h dh W
dt
M cable car mass
V cablecar velocity
g total forceonthecablecar
g h a r h da r h r h
µ µ
+ =
−
−
−
= − − − ⋅Ω − +− − +
∫
From the above equation we can deduce:
( )( )2 *
*
2,
0 ,
W MVg hdV V V
MVdh
V V
−<
= =
When the V* is the required cruise velocity.
Defining 3Vζ = we will get:
( )1
33 3d W
g hdh M
ζζ= −
Solving this deferential equation numerically, we will get the velocity of the climber as a
function of distance from Moon’s surface.
Apparently, if we have a lot of power, we can accelerate as fast as possible, but additional
restriction of velocity exists, the most important of them originating from oscillation
created on the tether due to climber movement, and Coriolis effect as a result (the
climber’s movement is linear, while the tether itself is in a rotating system with the
Moon, around the Earth. Coriolis force created in the rotation plane of the Moon around
the Earth and perpendicularly to the climber’s movement. This force will cause the tether
to oscillate).
32
So, the initial restriction on the velocity, is the wave propagation velocity, given in the
next equation:
2.36sec
kmc
σ
ρ
= =
Our desire is that the velocity of the climber will be lower than the above velocity, to
prevent a “shock wave” that might cause the tether to rupture.
We have examined the tether oscillation for several velocities, lower than the maximum
one and discovered that even at lower velocities; oscillations created, are large enough
and can possibly rupture the tether. After several iterations, the optimal velocity chosen
was * 0.7
sec
kmV
=
, which is about M0.3 (Mach speed is defined in our case, as a ratio
of this velocity to the maximum wave propagation velocity).
For this cruise velocity, the required power needed is about 7,500W (initial estimate,
apparently reasonable to achieve, though later in this project, we have came to a
conclusion that even larger power can be achieved, using reasonably sized solar panels).
The next charts present the velocity and the acceleration as a function of altitude:
33
The next chart shows how the required power changes as a function of altitude:
At the first stage, when the climber accelerates, the required power is constant at 7,500W.
Afterward, the cruise velocity stays constant, and the required power is diminishing. At
34
some point, this power value becomes negative, due to the more dominant gravity force
of the Earth (we gain power), this happens, of course, after we cross the LaGrange point
(L1).
An additional chart is obtained, showing the travel time as a function of altitude:
It can be seen here, that the total time needed to reach the release point on the tether is
about 200 hrs (marked with red dot on the chart). This travel time is calculated, from our
chosen required power (7,500W). In following chapters, we will show, how increasing
the power, reduces the travel time.
Tether oscillations
In the previous chapter we have mentioned, that the movement of the climber on the
tether creates Coriolis force, and this force makes tether oscillate.
These oscillations are given by the following equation:
35
( ) ( )
( )
2
2,
d u d duh T C t h
dt dh dh
u transversal displacement
h segment mass
T tension
C coriolis force
γ
γ
= +
−
−
−
−
Discretization of this equation gives us:
( ) ( )( ) ( )
( )( )
1 1 1
1 1
1
2,..., 1
:
2
i i i i i i
i i i
N N N
N N N
i ii
u u T u u Tm u C t i N
l
u u Tm u C t
l
where h il
T Tm hρ
σ
− − +
− −
−
− + −= + = −
−= +
=
+=
&&
&&
Additional development of the equations gives us the final equation of the oscillations:
( ) ( )( )
( )( )
( )
( ) ( )
( )( ) ( )
1 1 1
2
1 1
1
2
1
2 2,..., 1
2
1
0
i i i i i i i
i
i i i i
N N N
N
N
i
u u T u u T C tu i N
l T T l T T
u u C tu
l lT
m h when i l h t ilC t
otherwise
σ
ρ
σ
ρ
− − +
− −
−
−
− + −= + = −
+ +
−= +
Ω − < ≤=
&&
&&
&
Solving these equations, gives us the oscillations of the tether due to the climber’s
movement. As mentioned earlier, we have checked these oscillations for different
“Mach” numbers (1,0.6,0.3). We have presented the oscillations developing on the tether
in each of the cases in the PDR. From these simulations we have concluded that the upper
limit for our velocity is M0.3, equal to * 0.7
sec
kmV
=
.
36
The oscillations developing in such velocity, are reasonable and don’t present a serious
disturbance, but there is a problem of cumulative oscillations of the tether, due to several
journeys on it. The proposed solution is travel on the tether at the opposite direction, in
the same manner as before. This movement will create reversed oscillations, to these
created in the initial travel. This way, we will create each time reversed oscillations,
causing the tether to be relatively balanced. Simulation of this solution was presented in
the CDR presentation.
Return capsule reentry with its payload
After releasing from the tether, the return capsule will make its way back to Earth.
Every body entering the atmosphere has to be thermally protected. We have chosen to
protect the capsule, the same way as other space missions that include Earth reentry. The
chosen material for the heat shield is Phenolic Impregnated Carbon Ablator (PICA) (this
is the material used in NASA’s missions).
This is a relatively light material, with density of 31140[ / ]kg mρ = , ablative – meaning
that evaporating layers of the material, are cooling the next layers. We have chose the
heat shield’s thickness of 5cm, which is a typical for such heat shields for space reentry
missions.
Solving the reentry problem motion equations: 3
1
2DXm X SC X X
R
µρ= − −
r r r&& &&
Parameters of Earth atmosphere reentry:
Altitude – 80km from Earth’s surface
Reentry velocity – 10.46 km/s
Reentry angle – 2.5deg (we have checked a range of reentry degrees, in order to select the
respective reentry degree so we could get the minimal maximum load factor, so that the
capsule won’t break down during the reentry).
The maximum load factor received is 6.396.
At altitude of 20km (after most of the heat is dissipated by the heat shield), the return
capsule is designed to open, and the payload pressure spherical containers starting their
free fall, ending in splashing on water, at 20m/s velocity. The reentry period, from
entering the atmosphere to landing, is about 884 seconds.
37
In the next chart we can see the altitude above Earth’s surface, plotted against the reentry
time, where the marked points are the opening of the return capsule and landing of the
pressure containers respectively:
The next chart shows the velocity vs. time, showing the landing velocity:
38
The heat flux on the return capsule can be calculated by:
21
2DF V SC
W F V
W
S
ρ=
=
Φ =
The heat flux vs. time is shown in the following chart, when the marked point is the
break-up of the capsule and release of the pressure containers, at the 20km altitude. It can
be seen that in this point, the heat flux is lower by two orders of magnitude compared to
heat flux at atmosphere reentry.
drag force power heat flux
39
This chart is showing the load factor, as a function of the reentry angle. It can be seen that
for 2.5 degrees, we got the minimal load factor.
40
Bringing the spaceship to the L1 point – Theoretical
background – Three bodies problem
The three bodies’ problem is dealing with the motion of three bodies, each with a gravity
field. This problem doesn’t have a closed and explicit solution, but we are simplifying the
problem.
We are assuming, that the small mass (spaceship in this case), doesn’t affect the other
two big masses (Earth and Moon). For this kind of assumption, we can get a reduced
form of the three-body problem.
For these conditions we can define the kinetic and potential energy by the next equations:
( ) ( )2 2 2
3
1 23
1 2
1
2T m x y y x z
m mU fm
r r
ν ν = − + + +
= +
&& & &
1 2
'
, tan
int
trueanomaly of themoon sorbit
r r are thedis ces fromtheearth
and themoon to a certain po
ν −
Thus the Langragian will be defined by: L T U= +
The motion equation will be derived from Oiler-Lagrange equations:
0
, ,
i i
i
d L L
dt q q
q x y z
∂ ∂− =
∂ ∂
=
&
And the obtained equations are:
2
2
2
2
Wx y y x
x
Wy x x y
y
Wz
z
ν ν ν
ν ν ν
∂− − − =
∂
∂+ + − =
∂
∂=
∂
& && &&& &
& && &&& &
&&
41
After normalizing the equations by
, ,x r y r z rξ η ζ= = =
we will receive:
1 12
1 cos 1 cos
1 12
1 cos 1 cos
1 1
1 cos 1 cos
W
e e
W
e e
W
e e
ξ η ξν ν ξ
η ξ ην ν η
ζ ζν ζ
∂′′ ′− − =
+ + ∂
∂′′ ′+ − =
+ + ∂
∂′′ + =
+ + ∂
while:
( )
( )
2
1 2 1 2
2 2 2
1
2 2 2
2
1,
1
mW
r r m m
r
r
µ µµ
ξ µ η ζ
ξ µ η ζ
−= + =
+
= + + +
= + − + +
Defining the potential function for the problem:
( )2 2 21 1cos
2 2e Wξ η ν ζΩ = + − ⋅ +
The motion equations are reduced to the following:
12
1 cos
12
1 cos
1
1 cos
e
e
e
ξ ην ξ
η ξν η
ζν ζ
∂Ω′′ ′− =
+ ∂
∂Ω′′ ′+ =
+ ∂
∂Ω′′ =
+ ∂
Finding the equilibrium point:
If we take 0ζ η= = in order to find L1 point we will get from the motion equations the
next equation:
42
( )( ) ( )
1 1 2 2
3 3
1
11 0
1
W W
r r r r
ξ µ ξ µξ
ξ µ ξ µξ µ µ
ξ µ ξ µ
+ ∂ + − ∂+ = −
∂ ∂
+ + −− − − =
+ + −
This equation gives us the coordinateξ , which is the location of the L1 point. From
numerical analysis, the value is 0.849065ξ = in normalized units (the normalization is
by the half of the largest axis of Moon’s orbit around the Earth – meaning half of distance
between them).
Movement near the L1 point:
When the spaceship moves near the L1 point, the above motion equations become linear
(we will look at the problem additionally, as a radial movement, meaning we will
assume 0e = for orbit of the Moon around the Earth).
The equations will now be:
( )
( )
1 3 3
1 2
1 1 2 1
1
2 1 2 0
2 1 0
0
:
1
, 1
1
k
k
k
a
a
a
where
ar r
r r
locationof L
ξ η ξ
η ξ η
ζ ζ
µ µ
ξ µ ξ µ
ξ
′′ ′− − + =
′′ ′+ − − =
′′ + =
−= +
= + = + −
−
43
1Bringing the spaceship to L
As presented in the last chapter, the movement of the spaceship is dictated by the three-
bodies-problem. In order to bring the spaceship to the L1 point, in order to unroll the
tether, two methods were considered:
First method: Two momentary pulses
First pulse is to direct the spaceship to the Lagrange point and the second pulse is at the
proximity of the Lagrange point, in order to counteract its movement and to stay at this
point. This method isn’t good enough for several reasons, the main one is that the
Lagrange point is a non-stable equilibrium point, and as such, every deviation from
perfect conditions will make the spaceship drift away from the point.
Second method: Continuous control
It means reaching the Lagrange point by “asymptotic trajectory to L1”. This trajectory
brings us to the point with almost zero velocity, by using a continuous control law. This
method solves the instability problem of the equilibrium point, by continuous control and
diminishing velocity, up-to zero at the point. This method is also preferable from the
aspect of fuel consumption.
Eventually we chose the second method, the continuous control. In the next chart we will
present this asymptotic trajectory – a solution of non linear motion equations presented
earlier.
44
Developing the control law:
The control law is formulated from the last chapter’s linear motion equations (since it is
derived from linear equations it will be correct in the relative proximity of the Lagrange
point, a matter that will be explained later):
0 0 1 0 0 0
0 0 0 1 0 0
1 2 0 0 2 1 0
0 1 2 0 0 1
x
y
x Ax Bu
Fx A B u
Fa
a
a is the a fromthe linearized equations
ξ
η
ϕ
ψ
′ = +
= = = =
+
− −
r r r
r r
We are interested to bring the state vector to 0 (nullifying the location in order to get to
the Lagrange point and nullify the velocity in order to stay there).
It is a regulation problem, namely the LQR problem of the next form:
( )0
T TJ x Qx u Ru dν∞
= +∫r r r r
The maximum amplification for this problem is: 1 T
K R B P−=
r
When P is the solution of the algebraic Riccati equation:
1 0T TA P PA PBR B P Q
−+ − + =
For this amplification, the optimal controller is given by: u Kx= −r r
We have used this control law on our problem and showed that while we are close
enough to the Lagrange point (up-to 0.15 normalized units), this control law will work. It
can be seen in the next charts, in which are presented the trajectory of the spaceship for
different initial conditions for the location and velocity. It can be seen, that in each case,
the spaceship eventually reaches the Lagrange point.
45
46
Acceleration limitation
An important issue to be discussed is the limitation to the acceleration applied. The
spaceship has a high mass, so we couldn’t reach a high enough acceleration, using the
existing available engines.
Since we have solved the LQR problem and in such a problem, there is no limitation on
the controllers (meaning that from theoretical point of view, we can get unlimited or very
high controllers, which we won’t be able to provide). A realistic approximation for the
required acceleration for the control law, was conducted.
Our selected limitation (maximum) for the acceleration value is chosen to be 1[m/s2].
Now we will plot the contour lines of the natural acceleration and contour lines of the
control law acceleration – our desire is to be at the range in which the control law
acceleration is higher than the natural acceleration and lower than our maximum limit.
The control law acceleration is linear, since the control we have devised is linear.
And from a closer look:
47
The desired range for us, is between the white lines. The way to reach this range is by
regular ballistic trajectory (from Earth) and at the moment the spaceship enters this range,
the control law is activated and then we can converge the trajectory in to the Lagrange
point.
Tether deployment
After the spaceship successfully reaches the Lagrange point, as was shown in the last
chapter, it can now start to deploy the tether.
The tether deployment consists of two stages:
1. Free motion – the spaceship, the tether and the anchor all moving freely in
opposite directions
2. The tether is anchored to the lunar surface, the deployment continues in the other
direction
First stage – free motion
48
As previously mentioned, the tether is not moored and all motion is free in the space.
The goals at this stage are:
1. Moor at the lunar surface, at the closest location to the Earth.
2. Keep the tether tense
Our assumption is that the tether’s mass is much lower than the spaceship’s mass and
then we can get a physical model of two masses, connected to each other with a mass-less
tether. The motion equations for this model are:
)2,1(
)(2)1(
)(2)21(
3
3 =
−+−−=′
−+++=′
=′
=′
−
− i
yyDm
Tuyav
xxDm
Tvxau
vy
ux
ii
i
iii
ii
i
iii
ii
ii
D is the part of the tether already deployed.
1,2 are the indexes of the two end masses (the spaceship and the anchor).
If we’ll multiply the first equation by 1
1 2
m
m m+ and the second one by 2
1 2
m
m m+ , we’ll
receive the next set of equations:
(1 2 ) 2
(1 ) 2
x u
y v
u a x v
v a y u
′ = ′ =
′ = + + ′ = − −
x,y represent the coordinates of the mass center of the system.
Now we can mark 2 1x xξ = − , 2 1y yη = − in order to devise the next equations:
( )
( )
1 2
1 2
1 11 2 2
1 11 2
Ta
D m m
Ta
D m m
ξ ξ η ξ
η η ξ η
′′ ′= + + − +
′′ ′= − − − +
After multiplying the first equation byξ and the second one by η , we can now reach the
next form:
( )[ ] ϕ−=+ϕ′ 2sin2
31 22
aDDdt
d
49
The ϕ in the equation, is the angle between the tether and the x axis. We would prefer,
that his angle will be 0, when we will moor at the Moon, and that its derivative (the
angular rate) will be 0 as well.
Practically, this equation describes the motion around the center of mass of the system.
We can now derive the left sided part of the above equation and get the next equation:
( ) ϕ−=+ϕ′′+ϕ ′′ 2sin2
312 D
aDD
If we will assume that the tether release rate is constant, meaning
( ) ,D t ut u const= = and small angles present, then our equation becomes:
232 −=ϕ+ϕ′+ϕ′′ att
And its general solution will be:
2
cos sin 2( )
t tt
t t t
ρ ρϕ α β
ρ= + −
3aρ = Now we can develop the Taylor series for the sinus and the cosine:
2 3/21 2 2
2
2 3/22
2
2( ) ( )
2 6
2( ) ( )
2 3
t t t t o t
t t t o t
ρ ρ βϕ α ρβ α
ρ
ρ ρ βϕ α α
ρ
−
−
= − + − − +
′ = − − − +
From the next boundary conditions:
( ) ( )
( ) ( )
0 0
0L L
t t
ϕ ϕ
ϕ ϕ
′< ∞ < ∞
′= =
We can find the constants ,α β and the next equations are obtained for the angle and
angular rate:
[ ]
[ ]
2
2 2
2( ) cos 1
2 2( ) cos 1 sin
t tt
t t tt t
ϕ ρρ
ϕ ρ ρρ ρ
= −
′ = − − −
50
Because of our requirement for nullifying the angle and angular rate at the end of the
process, the deployment time will be at the next form:
2L
kt
π
ρ=
In this equation, k is an arbitrary constant that will determine the number of cycles till the
mooring (shown later in the charts).
Since the distance is known (our requirement that it will be the shortest distance, so
actually it is the distance from L1 point to the lunar surface - * 0.148D = in canonic
units.
The velocity of the tether deployment, maintaining all the requirements, is defined by the
next equation:
* *
2L
D DV
t k
ρ
π= =
In this chart, we can see the angle and angular rate as a function of time, while the tether
is deployed:
The required conditions are fulfilled, with nullifying of the angle and angular rate values
at the end of the process.
51
In the next chart, we can see, that the requirement for the tension of the tether during the
deployment, is fulfilled as well.
It can be seen, that the tension grow bigger almost linear (with small fluctuations), as
more of the tether is deployed.
Second stage – moored tether
At this stage, after the tether is anchored to the lunar surface, the spaceship continues to
deploy it farther from the L1 point, towards the Earth.
We assume that constξ = and then the dynamic equations become:
W
W
η ηη
ζζ
∂− =
∂
∂=
∂
&&
&&
While:
( ) ( ) 222222
1
1 ζ+η+ξ+µ
µ−+
ζ+η+ξ−µ−
µ=W
Assuming small angles, we can now write:
52
( ) ( )
( ) ( )ζξ−=ζ
ξ+µ
µ−+
ξ−µ−
µ−≈
ζ∂
∂
ηξ−=η
ξ+µ
µ−+
ξ−µ−
µ−≈
η∂
∂
)(1
1
;)(1
1
33
33
fW
fW
Placing it the above dynamic equations, we’ll get:
( );)(
;)(1
ζξ−=ζ
ηξ−=η
f
f
&&
&&
These are equations of two harmonic oscillators with different frequencies. For eachξ
value, we have oscillation at the plane ηζ . An example for such oscillations can be seen
in the next chart:
53
Navigation
The need for a navigation system for the climber is due to the fact that its location in
inertial frame is not known and this information is essential due to these two main
reasons:
1. We want to know the location of the climber, in order to release the return capsule
at the right timing, so it will enter the return trajectory to the Earth atmosphere.
2. We want to know the location of the climber all the times, in order to operate it
efficiently and safely and to know it is working properly.
The navigation system is based on two cameras located on the mast on the back of the
climber; one directed to the Moon and the other toward the Earth. By simultaneous
analysis of the pictures of the both celestial bodies, we can determine the location and the
angular position of the climber, when the goal is to conclude from the analysis the real
distances.
At the first stage we will determine the d distance on the x axis (between the centers of
the Earth and Moon). It can be from the Earth or from the Moon.
The angleα is derived from the following equation: κ
α apparantR= assuming, that the
climber deviates by small angles only, when apparantR is the radius shown in the picture
above and κ is the constant of the camera’s lenses. It is to be noted that if the climber
doesn’t deviate from x axis at all, we can find its location simply by counting its wheels’
revolutions on the tether.
In this situation, the combined picture from both cameras will look like this:
αd
R
d
αx
αsin
Rd =⇒
54
At the second stage, the location / angular position of the climber is derived, from both
photos and the information transformed into real values. Now we’ll demonstrate the
method, for the case that the climber deviates in the z axis (perpendicular to the orbit
plane of the Moon around the Earth) – in this case, the combined pictures will look like
this:
It is the distance seen on the focal plane of the camera lens and it has to be found in its
real apparentdz value. For this, we will use the distance Ed , that can be found from the
first stage (using the distance from the Moon, as described earlier).
f is the distance from the lens (camera) to the focal plane of the lens
dz is the real deviation distance of the climber
Gravity
apparentdz
focal plane
f
Ed
apparentdz
dz
Earth
Moon
55
Now we can find the real distance using simple geometric relation – triangles similarity:
Solar panels
One of the main features in the space elevator is the use of solar energy instead of fuel, in
order to transport cargo to and from the Moon. Solar energy is a freely available and
renewable source of power and as such, is preferred over chemical or organic fuel, which
requires large quantities, which is very expensive to manage at such distance from Earth.
Therefore, our goal is to use solar power, in order to operate the climber’s wheels, in
order to get the cargo capsule to the release point. The size of the solar panels will
determine the amount of available energy and by this, the achievable acceleration rate of
the climber. As the acceleration could be performed faster, the travel time will be shorter,
and more trips on the tether could be made annually – eventually allowing transporting
higher payload of He3 to Earth.
In the following chart, we can see travel time as a function of available power:
E
apparent
d
f
dz
dz= apparent
E dzf
ddz =⇒
56
Out initial power value was of 7,500W, which was increased iteratively to 20,000W.
In order to obtain the required energy, we’ve chosen the solar panels of type UTJ, which
can deliver up to 28.3% of the collected solar energy. These kind of panels, were already
used in space applications by several companies, so it can be used as a proven option.
From manufacturer datasheet, we estimated the mass of the solar panels, as a function of
the required energy:
For 20,000W power, the mass of the panels is about 118kg. Taking a conservative
approach, it can be estimated that the total mass of the solar panels system, will be larger,
around 150kg, allowing for power surplus, for climber systems and redundancy in case of
backup or malfunction of some part of the solar panels.
The solar panels therefore, constitute a reasonable 30% of the climber’s total 500kg mass.
Another aspect of using the solar energy is the period of blackout, during which the
climber is concealed from the solar rays, by Earth or Moon’s shadow. Since the Earth
eclipse is a rare event in the lunar cycle, we’ll analyze only to the lunar shadow.
57
Using the MATLAB simulation, we’ve estimated the darkness period, in which the
climber located in two cases:
1. The climber starts its motion, when the Earth facing side is unlit by the Sun.
2. The climber starts its motion, when the Earth facing side is lit by the Sun.
In both cases, during one month, we’ve receive only 3% of the time of darkness. If we
calculate for the main part of the travel, we’ll receive a higher value of 12% of the time.
It is important to remember, that the solar energy is required only during the acceleration
phase, so the system can be operated with minimal batteries required, especially if
specifically planned, that the climber won’t be in the darkness during its acceleration
phase.
Mooring the tether to the lunar surface
Returning to the subject of tether deployment, the subject of tether mooring to the lunar
surface needs to be explained.
We need a force of 03240[ ]F N= in order to hold the tether on the surface. For this
purpose, we are proposing to use an anchor, which would penetrate the lunar surface,
during the tether’s deployment.
Anchor initial drop altitude (over the Moon): 6km
Penetration velocity: 140m/s
In order to calculate the depth of the penetration, we’ll use a Yang equation:
0.7
6
018 10 ( 30.5)m
h S N VA
− = ⋅ ⋅ ⋅ ⋅ −
m
A
is the ballistic coefficient, taken by typical value, similar to bunker buster bombs
S is the soil penetration constant
N is the coefficient of the anchor’s tip
58
Proposed material for the anchor: HSS
Mass: 230kg
Penetration depth: 6m
In order for the anchor to penetrate the soil, it will be stabilized and achieve angular
momentum using two small rocket motors.
Non skid wheels system
In order to prevent skidding of the wheels off the tether, we will use a guide, through
which the tether will pass:
G
O
h
ez ΓΓΓΓ cable
G
O
h
G
O
h
ez ΓΓΓΓ cable
The moment the tether is sliding aside (skidding of the wheels off the tether) it hits the
walls of the guide and returns back to its initial position.
The motion equations in this case are:
1 zJ J H T h eω ω ω ω
ω
+ × + × = ⋅ ⋅ ×Γ
Γ = ×Γ
&
&
59
The system achieved is asymptotic and non-stable. In order to stabilize the system, we
can use cadmium-plated springs, which are eventually oscillations dampers of the tether,
inside the guide.
Now we can solve the new motion equations set:
1
1 2 0( ) ( )
J J H T h E
E c E c E E
ω ω ω ω
ω
+ × + × = ⋅ ⋅ × Γ
= Γ − + −
Γ = × Γ
&
&
&
When E it is the location on the end of the damper.
The result is that the system is stable asymptotically.
In the first chart, of vector motion E, it can be seen that the system is converging to 0
value (initial condition is 0):
guide
damper
60
Second char shows the vector motion of the velocity of the tether, when it is converging
to 0 as well:
61
Conclusions
• Payload transport from the Moon to Earth can be achieved in 6 days, using solar
energy only, without fuel.
• Transporting the payload can be done, using a climber, moving on long tether, by
friction force between the wheels and the tether.
• The tether will be anchored to Moon’s surface, using surface penetration anchor,
to depth of 6 meters, while the free end of the tether (towards the Earth) will have
an end mass of about 16 tons.
• Releasing the return capsule at the desired point, will insert it to a reentry orbit
towards Earth’s atmosphere.
• Landing the capsule will be done without parachute, using only aerodynamic drag
for slowing down, and splashing the helium filled pressure containers on the
water, floating eventually to be picked up.
• Adaptive control system will control the tether rewinding, bringing the tether to
the Lagrange point and the tether’s oscillations.
• There is no need for large batteries for storing the energy for the climber, since
the cruise phase doesn’t require much energy input.
• Climber wheels deviations from the tether can be controlled using damping
system.
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APPENDIX
MATLAB codes
Calculations and charts for velocity, tension, force, cross-section, tether mass,
acceleration, power, time of travel and tether oscillations:
clear all; close all; clc; defaults; %% % Data m=1500; % [kg] mass of tether-car+load W=7500; % [Watt] Power a_l=3.84*10^8; % [m] r_l=1738*10^3;% [m] d=4700*10^3;% [m] omega=2.6617*10^-6;% [rad/sec] miu_e=3.986*10^14; % [m^3/sec^2] miu_l=4.903*10^12; % [m^3/sec^2] T_0=3240; % [N] rho=1.7*10^3; % [kg/m^3] sigma=9.5*10^9; % [N/m^2] %% % Velocity Calculation v=importdata('v_vector.mat'); h=importdata('h_vector.mat'); figure(1) plot(h/1000,v/1000) Title('Velocity Vs. Height') Xlabel('Height [km]') Ylabel('Velocity [km/sec]') grid on axis([0 350000 0 0.8]) %% % Tension Calaulation T=T_0*exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d-h).^2-miu_e./(a_l-r_l-h)-
miu_l./(r_l+h)))/exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d).^2-miu_e./(a_l-r_l)-miu_l./r_l)); figure(2) plot(h/1000,T) hold on plot(h(1)/1000,T(1),'r*') hold on [y,j]=max(T); plot(h(j)/1000,T(j),'r*') Title('Tension Vs. Height') Xlabel('Height[km]') Ylabel('Tension [N]')
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grid on % just g h_lala=linspace(0,325000000,1000); g_lala=(a_l-r_l-h_lala-d)*omega^2-miu_e./(a_l-r_l-h_lala).^2+miu_l./(r_l+h_lala).^2; figure(3) plot(h_lala/1000,g_lala) title('g vs. Height(from the moon)') xlabel('Height [km]') ylabel('g [m/sec^2]') grid on %% % Cross Section Calculation S_F=1; % safety factor s=T*S_F/sigma; figure(10) plot(h/1000,s*10^6) % h in km, s in mm^2 Title('Cross Section Area Vs. Height') Xlabel('Height[km]') Ylabel('Cross Section Area [mm^2]') grid on %% % Tether Mass Calculation s_m=s(2:length(s)); m_tether_segment=rho*s_m.*(h(2:length(h))-h(1:length(h)-1)); m_tether=importdata('m_tether_vector_full.mat'); h_big=h; h_big(length(h_big)+1)=340000000; h_big(length(h_big)+1)=350000000; T_big=T_0*exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d-h_big).^2-miu_e./(a_l-r_l-h_big)-
miu_l./(r_l+h_big)))/exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d).^2-miu_e./(a_l-r_l)-miu_l./r_l)); s_big=T_big*S_F/sigma; s_m_big=s_big(2:length(s_big)); m_tether_segment_big=m_tether_segment; m_tether_segment_big(length(m_tether_segment_big)+1)=rho*s_m_big(length(s_m_big)-
1).*(h_big(length(h_big)-1)-h_big(length(h_big)-2)); m_tether_segment_big(length(m_tether_segment_big)+1)=rho*s_m_big(length(s_m_big)).*(h_big(length(
h_big))-h_big(length(h_big)-1)); m_tether_big=m_tether; m_tether_big(length(m_tether_big)+1)=sum(m_tether_segment_big(1:length(m_tether_segment_big)-1)); m_tether_big(length(m_tether_big)+1)=sum(m_tether_segment_big(1:length(m_tether_segment_big))); h_m=h_big(119664:length(h_big)); m_tether_tag=m_tether_big(119664:length(m_tether_big)); figure(4) plot(h_m/1000,m_tether_tag) title('Tether Mass Vs. Height') xlabel('Height [km]') ylabel('Tether Mass [kg]') grid on %% % End Mass+Total Mass
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% end mass if the length is 325000 km T_m=T_big(119664:length(h_big)); g_m=(a_l-r_l-h_m-d)*omega^2-miu_e./(a_l-r_l-h_m).^2+miu_l./(r_l+h_m).^2; end_mass=T_m./g_m; % h_final=350000[km] not 3250000[km] end_mass=abs(end_mass); total_mass=m_tether_tag+end_mass; figure(5) plot(h_m/100,total_mass) title('Total Mass Vs. Height') xlabel('Height [km]') ylabel('Total Mass [kg]') grid on figure(6) plot(h_m/1000,m_tether_tag,'b',h_m/1000,total_mass,'r--') title('Tether Mass and Total Mass Vs. Height') xlabel('Height [km]') ylabel('Mass [kg]') grid on legend('tether mass','total mass') %% % Acceleration Calculation a=importdata('a_vector.mat'); h_tag=h(1:length(h)-1); delta=a(2:length(a))-a(1:length(a)-1); delta=abs(delta); jumps=delta>0.009*10^(-3); ind=find(jumps); h_tagag=h_tag; h_tagag(ind)=[]; a_tag=a; a_tag(ind)=[]; figure(7) plot(h_tagag/1000,a_tag*1000) title('Acceleration Vs. Height') xlabel('Height [km]') ylabel('Acceleration [mm/sec^2]') grid on %% % Power Calculation v_const=700; h_p=(0:1000:40225006); W(1:1:length(h_p))=7500; % power at the acceleration phase h_p_all=(0:1000:325000000); h_v_const=h_p_all(length(h_p)+1:1:length(h_p_all)); g_p=(a_l-r_l-h_v_const-d)*omega^2-miu_e./(a_l-r_l-h_v_const).^2+miu_l./(r_l+h_v_const).^2; W(length(h_p)+1:1:length(h_p_all))=m*v_const*g_p; figure(8) plot(h_p_all/1000,W) title('Power Vs. Height') xlabel('Height [km]') ylabel('Power [Watt]') grid on axis auto
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%% % Time Calculation N=1000; l=325000000/N; v_ac=v(1:119600); h_ac=h(1:119600); p1=polyfit(h_ac,v_ac,5); h_pol=linspace(0,325000000,N); h_pol_a=h_pol(1:125); v_pol_a=p1(1)*h_pol_a.^5+p1(2)*h_pol_a.^4+p1(3)*h_pol_a.^3+p1(4)*h_pol_a.^2+p1(5)*h_pol_a+p1(6)
; v_pol=v_pol_a; v_pol(length(v_pol)+1:length(h_pol))=v_pol_a(length(v_pol_a)); T_pol=T_0*exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d-h_pol).^2-miu_e./(a_l-r_l-h_pol)-
miu_l./(r_l+h_pol)))/exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d).^2-miu_e./(a_l-r_l)-miu_l./r_l)); dh=h_pol(2:length(h_pol))-h_pol(1:length(h_pol)-1); dt=dh./v_pol(1:length(v_pol)-1); t_vec(1)=0; for i=2:1:length(h_pol) t_vec(i)=sum(dt(1:i-1)); end figure(9) plot(h_pol/1000,t_vec/3600) hold on plot(h_pol(708)/1000,t_vec(708)/3600,'r*') Title('Time Vs. Height') xlabel('Height [km]') ylabel('Time [hours]') grid on %% % Displacement Calculation N=100; l=325000000/N; p2=polyfit(t_vec,h_pol,7); h_pol_t=p2(1)*t_vec.^7+p2(2)*t_vec.^6+p2(3)*t_vec.^5+p2(4)*t_vec.^4+p2(5)*t_vec.^3+p2(6)*t_vec.^2
+p2(7)*t_vec+p2(8); slope=h_pol_t(15)/t_vec(15); rect = [100 50 800 500]; hf=figure('Position',rect); h_seg=linspace(0,325000000,N); T_seg=T_0*exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d-h_seg).^2-miu_e./(a_l-r_l-h_seg)-
miu_l./(r_l+h_seg)))/exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d).^2-miu_e./(a_l-r_l)-miu_l./r_l)); dt=10; k=1; x1(1:N)=0; % x1=u x2(1:N)=0; % x2=u_dot x1_mid(1:N)=0;
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x2_mid(1:N)=0; for t=0:dt:864000*2, if t<864000 cor=zeros(1,N); h=h_t(t,p2,slope,t_vec(15)); delta=abs(h-h_seg); [y,j]=min(delta); else j=1; end if t<864000, if j<=(40/325*(N-1)+1) % cor(j)=Cor(m,omega,700); cor(j)=Cor(m,omega,v_seg(h,p1)); else % cor(j)=Cor(m,omega,700); cor(j)=Cor(m,omega,v_seg(h_seg(13),p1)); end else cor=zeros(1,N); end c(k)=cor(j); k=k+1; % for i=2:1:N-1, % u(i)=u(i)+du(i)*dt; % du(i)=du(i)+ddu(i)*dt; % ddu(i)=2*sigma/rho*((u(i-1)-u(i))*T_seg(i-1)+(u(i+1)-u(i))*T_seg(i))/(i*l^2*(T_seg(i-
1)+T_seg(i)))+2*(sigma*cor(i))/(i*l*rho*(T_seg(i-1)+T_seg(i))); % end % ddu(N)=sigma/rho*(u(N-1)-u(N))/(N*l^2)+sigma/rho*cor(N)/(N*l*T_seg(N-1)); % du(N)=du(N)+ddu(N)*dt; % u(N)=u(N)+du(N)*dt; for i=2:1:N-1, x1_dot(i)=x2(i); x2_dot(i)=2*sigma/rho*((x1(i-1)-x1(i))*T_seg(i-1)+(x1(i+1)-x1(i))*T_seg(i))/(l^2*(T_seg(i-
1)+T_seg(i)))+2*(sigma*cor(i))/(l*rho*(T_seg(i-1)+T_seg(i))); % x2_dot(i)=2*((x1(i-1)-x1(i))*T_seg(i-1)+(x1(i+1)-x1(i))*T_seg(i))/(T_seg(i-
1)+T_seg(i))+3*cor(i)/(T_seg(i-1)+T_seg(i)); x1_mid(i)=x1(i)+x1_dot(i)*dt/2; x2_mid(i)=x2(i)+x2_dot(i)*dt/2; x1_dot_mid(i)=x2_mid(i); x2_dot_mid(i)=2*sigma/rho*((x1_mid(i-1)-x1_mid(i))*T_seg(i-1)+(x1_mid(i+1)-
x1_mid(i))*T_seg(i))/(l^2*(T_seg(i-1)+T_seg(i)))+2*(sigma*cor(i))/(l*rho*(T_seg(i-1)+T_seg(i))); % x2_dot_mid(i)=2*((x1_mid(i-1)-x1_mid(i))*T_seg(i-1)+(x1_mid(i+1)-
x1_mid(i))*T_seg(i))/(T_seg(i-1)+T_seg(i))+3*cor(i)/(T_seg(i-1)+T_seg(i)); x1(i)=x1(i)+x1_dot_mid(i)*dt; x2(i)=x2(i)+x2_dot_mid(i)*dt; end x1_dot(N)=x2(N); x2_dot(N)=2*sigma/rho*(x1(N-1)-x1(N))/(l^2)+2*(sigma*cor(N))/(l*rho*(T_seg(N-1))); % x2_dot(N)=2*(x1(N-1)-x1(N))/(N*l^2)+3*cor(N)/(T_seg(N-1)); x1_mid(N)=x1(N)+x1_dot(N)*dt/2; x2_mid(N)=x2(N)+x2_dot(N)*dt/2;
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x1_dot_mid(N)=x2_mid(N); x2_dot_mid(N)=2*sigma/rho*(x1_mid(N-1)-
x1_mid(N))/(l^2)+2*(sigma*cor(N))/(l*rho*(T_seg(N-1))); % x2_dot_mid(N)=2*(x1_mid(N-1)-x1_mid(N))+2*cor(N)/(T_seg(N-1)); x1(N)=x1(N)+x1_dot_mid(N)*dt; x2(N)=x2(N)+x2_dot_mid(N)*dt; if mod(k,700)==0, plot(h_seg,x1,'o') hold on; if t<864000, plot(h_seg(j),x1(j),'rs') end hold off; axis([0 350000000 -1000000 1000000]) % axis auto M(k/700)=getframe; % M(k)=getframe; end end -----------------------------------------------------------
Calculation of velocity as a function of altitude (from Moon):
% Velocity Calculation clear all; close all; clc; m=1500; % [kg] mass of tether-car+load W=7500; % [Watt] Power a_l=3.84*10^8; % [m] r_l=1738*10^3;% [m] d=4700*10^3;% [m] omega=2.6617*10^-6;% [rad/sec] miu_e=3.986*10^14; % [m^3/sec^2] miu_l=4.903*10^12; % [m^3/sec^2] i=1; h(i)=0; dh=1; zeta(i)=0; V(i)=0; while V(i)<700, g_vec(i)=g(h(i),a_l,r_l,d,omega,miu_e,miu_l); x(i)=f_v(W,m,zeta(i),g_vec(i)); zeta_mid(i)=zeta(i)+x(i)*dh/2; g_mid_vec(i)=g(h(i)+h(i)/2,a_l,r_l,d,omega,miu_e,miu_l); y(i)=f_v(W,m,zeta_mid(i),g_mid_vec(i)); zeta(i+1)=zeta(i)+y(i)*dh; V(i+1)=sign(zeta(i+1))*(abs(zeta(i+1)))^(1/3); if mod(i+1,1000)==0, V(i+1)=V(i+1); end if mod(i+1,10000)==0, dh=dh*2 h(i) end
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h(i+1)=h(i)+dh; i=i+1; end i=i-1; k=i; while h(i)<325000000, V(i+1)=V(i); h(i+1)=h(i)+1000000; i=i+1; end % V(i:length([0:(325000000-h(i))/1000:325000000]))=V(i); % h(i:length([0:(325000000-h(i))/1000:325000000]))=(h(i):(325000000-h(i))/1000:325000000); plot(h,V,'*')
-----------------------------------------------------------------------
:פונקציות עזר function f_v=f_v(h,zeta) global a_l r_l d omega miu_e miu_l W m g=g_fun(h,a_l,r_l,d,omega,miu_e,miu_l); f_v=3*W/m-3*g*sign(zeta)*(abs(zeta))^(1/3); function g=g(h,a_l,r_l,d,omega,miu_e,miu_l) g=(a_l-r_l-h-d)*omega^2-miu_e/(a_l-r_l-h)^2+miu_l/(r_l+h)^2; function Cor=Cor(m,omega,v) Cor=m*omega*v;
-----------------------------------------------------------------------
Oscillations simulation:
clear all; close all; clc; rect = [100 50 800 500]; hf=figure('Position',rect); % h_seg=linspace(0,300000000,100); x=importdata('NECKLACE_03.dat'); % i<=327 in row 18 in the code % x=importdata('NECKLACE_06.dat'); % i<=164 % x=importdata('NECKLACE_1.dat'); % i<=99 [m,n]=size(x); for i=1:1:m, % plot(h_seg,x(I,2:n),'o') if mod(I,3)==0, plot(x(I,2:n),'o') hold on; if i<=327, j=round(x(I,1)); if j==0, j=1; end plot(j,x(I,j),'rs') end axis([0 100 -100 100])
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axis off M(i/3)=getframe; end hold off; % plot(x(I,1)*10^6,x(I,j),'rs')
end
-----------------------------------------------------------
Velocity calculations for different power sets:
% Velocity Calculation clear all; close all; clc; defaults; global a_l r_l d omega miu_e miu_l W m m=1500; % [kg] mass of tether-car+load W=20000; % [Watt] Power a_l=3.84*10^8; % [m] r_l=1738*10^3;% [m] d=4700*10^3;% [m] omega=2.6617*10^-6;% [rad/sec] miu_e=3.986*10^14; % [m^3/sec^2] miu_l=4.903*10^12; % [m^3/sec^2] T_0=3240; % [N] rho=1.7*10^3; % [kg/m^3] sigma=9.5*10^9; % [N/m^2] %% Initial Conditions zeta0=0; %% Integration options=odeset('MaxStep',10^(-3)*a_l,'Events',@events); tic [h,zeta]=ode45(@f_v,[0,100000000],zeta0,options); toc %% Analysis v=zeta.^(1/3); i=length(h); k=i; while h(i)<325000000, v(i+1)=v(i); h(i+1)=h(i)+1000000; i=i+1; end figure(1) plot(h/1000,v/1000) grid on title('Velocity vs. Height') xlabel('Height [km]') ylabel('Velocity [km/sec]') %% Travel Time
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N=1000; l=325000000/N; v_ac=v(1:k); h_ac=h(1:k); p1=polyfit(h_ac,v_ac,5); h_pol=linspace(0,325000000,N); index=81; % 7500W - index=155 % 11000W - index=121 % 15000W - index=97 % 20000W -
index=81 h_pol_a=h_pol(1:index); v_pol_a=p1(1)*h_pol_a.^5+p1(2)*h_pol_a.^4+p1(3)*h_pol_a.^3+p1(4)*h_pol_a.^2+p1(5)*h_pol_a+p1(6)
; v_pol=v_pol_a; v_pol(length(v_pol)+1:length(h_pol))=v_pol_a(length(v_pol_a)); T_pol=T_0*exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d-h_pol).^2-miu_e./(a_l-r_l-h_pol)-
miu_l./(r_l+h_pol)))/exp((rho/sigma)*(-0.5*omega^2*(a_l-r_l-d).^2-miu_e./(a_l-r_l)-miu_l./r_l)); dh=h_pol(2:length(h_pol))-h_pol(1:length(h_pol)-1); dt=dh./v_pol(1:length(v_pol)-1); t_vec(1)=0; for i=2:1:length(h_pol) t_vec(i)=sum(dt(1:i-1)); end % -----------find 225,000 km----------------- distance=h_pol-225000000; distance(distance<0)=1000000000; [min,min_index]=min(distance); % ------------plotting------------------ % figure(2) % plot(h_pol/1000,t_vec/3600) % hold on % plot(h_pol(min_index)/1000,t_vec(min_index)/3600,'r*') % Title('Time Vs. Height') % xlabel('Height [km]') % ylabel('Time [hours]') % grid on % text(h_pol(min_index)/1000,t_vec(min_index)/3600,'\uparrow Release Point',... % 'VerticalAlignment','top') t_7500=importdata('t_7500.mat'); h_7500=importdata('h_7500.mat'); t_11000=importdata('t_11000.mat'); h_11000=importdata('h_11000.mat'); t_15000=importdata('t_15000.mat'); h_15000=importdata('h_15000.mat'); t_20000=importdata('t_20000.mat'); h_20000=importdata('h_20000.mat'); figure(2) plot(h_7500/1000,t_7500/3600,h_11000/1000,t_11000/3600,h_15000/1000,t_15000/3600,h_20000/1000,t_
20000/3600) hold on plot(h_7500(min_index)/1000,t_7500(min_index)/3600,'r*') text(h_7500(min_index)/1000,t_7500(min_index)/3600,'\uparrow Release Point',... 'VerticalAlignment','top') plot(h_11000(min_index)/1000,t_11000(min_index)/3600,'r*')
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% text(h_11000(min_index)/1000,t_11000(min_index)/3600,'\uparrow Release Point',... % 'VerticalAlignment','top') plot(h_15000(min_index)/1000,t_15000(min_index)/3600,'r*') % text(h_15000(min_index)/1000,t_15000(min_index)/3600,'\uparrow Release Point',... % 'VerticalAlignment','top') plot(h_20000(min_index)/1000,t_20000(min_index)/3600,'r*') % text(h_20000(min_index)/1000,t_20000(min_index)/3600,'\uparrow Release Point',... % 'VerticalAlignment','top') Title('Time Vs. Height') xlabel('Height [km]') ylabel('Time [hours]') grid on legend('W=7500 [Watt]','W=11000 [Watt]','W=15000 [Watt]','W=20000 [Watt]',2) -----------------------------------------------------------
% aid function: function f_v=f_v(h,zeta) global a_l r_l d omega miu_e miu_l W m g=g_fun(h,a_l,r_l,d,omega,miu_e,miu_l); f_v=3*W/m-3*g*sign(zeta)*(abs(zeta))^(1/3); function g=g(h,a_l,r_l,d,omega,miu_e,miu_l) g=(a_l-r_l-h-d)*omega^2-miu_e/(a_l-r_l-h)^2+miu_l/(r_l+h)^2; function [value,isterminal,direction] = events(h,zeta) v=(abs(zeta))^(1/3)*sign(zeta); value=v-700; isterminal = 1; direction=0; -----------------------------------------------------------
Bringing the spacecraft to L1 – controlled and uncontrolled orbit: clear all; close all; clc; defaults; %% Constants Definitions global a miu a_earth_moon r1_L1 K miu1=3.986e5; % earth gravitation constant miu2=4.903e3; % moon gravitation constant miu=miu2/(miu1+miu2); zeta=0; a_earth_moon=3.84*10^8; % [m] r1_L1=0.849065; % L1 distance from earth r2_L1=1-r1_L1; % L1 distance from moon a=(1-miu)/r1_L1^3+miu/r2_L1^3; [lamda1,lamda3]=roots(a); T_norm=406066; % 1month/(2*pi) %% Intial Conditions tet_0=[-0.005;0;0;0]; K=alt_control(a); %% Integration options=odeset('RelTol',10^(-6),'AbsTol',10^(-6),'MaxStep',10^(-3)); [ni,tet]=ode45(@dx_shifted,[0 10],tet_0,options);
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%% Plotting (one section at a time - other should be deactivated) plot(tet(:,1)+r1_L1,-tet(:,2)) title('Asymptotic Trajectory to L1') xlabel('\xi') ylabel('\eta') hold on plot(0,0,'x') text(r1_L1,0,'\leftarrow L1',... % to plot the asymptotic trajectory set F_x=F_y=0
in the dx_shifted function 'HorizontalAlignment','left') text(0,0,'\leftarrow Earth',... 'HorizontalAlignment','left') axis([-0.9 0.9 -0.9 0.9]) plot(tet(:,1),tet(:,2)) title('Controlled Trajectory to L1') xlabel('\xi') ylabel('\eta') text(0,0,'\leftarrow L1',... % to plot the controlled trajectory set F_x=F_y to
other than zero in the dx_shifted function 'HorizontalAlignment','left') text(tet(1,1),tet(1,2),'\leftarrow Start',... 'HorizontalAlignment','left')
-----------------------------------------------------------------------
% aid function function dx=dx(ni,tet) global miu r1_L1 K F=-K*tet; F_x=F(1); % F_x=0; F_y=F(2); % F_y=0; r1=((tet(1)+r1_L1-miu+miu)^2+tet(2)^2)^0.5; r2=((tet(1)+r1_L1-miu+miu-1)^2+tet(2)^2)^0.5; d_omega_ksi=(tet(1)+r1_L1-miu)-(1-miu)*(tet(1)+r1_L1-miu+miu)/r1^3-miu*(tet(1)+r1_L1-miu+miu-
1)/r2^3; d_omega_eta=tet(2)-(1-miu)*tet(2)/r1^3-miu*tet(2)/r2^3; dx(1)=tet(3); dx(2)=tet(4); dx(3)=2*tet(4)+d_omega_ksi+F_x; dx(4)=-2*tet(3)+d_omega_eta+F_y; dx=dx'; end function [value,isterminal,direction] = events(ni,tet) global a_earth_moon miu r1_L1 value=(tet(1)^2+tet(2)^2)^0.5-10^-3; isterminal = 1; direction=0;
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function [lamda1,lamda3]=roots(a) delta1=((2-a)+((2-a)^2-4*(1-a)*(1+2*a))^0.5)/2; % positive root delta2=((2-a)-((2-a)^2-4*(1-a)*(1+2*a))^0.5)/2; % negative root lamda1=i*sqrt(abs(delta2)); lamda3=abs(sqrt(delta1)); end -----------------------------------------------------------
% Control law:
function K=alt_control(a) A=[0,0,1,0; 0,0,0,1; 1+2*a,0,0,2; 0,1-a,-2,0]; B=[0,0; 0,0; 1,0; 0,1]; Q=eye(4); R=eye(2); [K,S,E] = LQR(A,B,Q,R) -----------------------------------------------------------
Acceleration calculations:
clear all; close all; clc; defaults; %% Constants miu1=3.986e5; % earth gravitation constant miu2=4.903e3; % moon gravitation constant miu=miu2/(miu1+miu2); zeta=0; a_earth_moon=3.84*10^8; % [m] r1_L1=0.849065; % L1 distance from earth r2_L1=1-r1_L1; % L1 distance from moon a=(1-miu)/r1_L1^3+miu/r2_L1^3; [lamda1,lamda3]=roots(a); T_norm=406066; % 1month/(2*pi) K=alt_control(a); %% Acceleration - Control (Only Position) and Natural x_vec=linspace(-0.3,0.1,500); y_vec=linspace(-0.3,0.3,500); for i=1:length(x_vec), u_mat_x(i,:)=-(K(1,1)*x_vec(i)+K(1,2)*y_vec); u_mat_y(i,:)=-(K(2,1)*x_vec(i)+K(2,2)*y_vec); r1(i,:)=((x_vec(i)+r1_L1-miu+miu)^2+y_vec.^2).^0.5; r2(i,:)=((x_vec(i)+r1_L1-miu+miu-1)^2+y_vec.^2).^0.5; d_omega_ksi(i,:)=(x_vec(i)+r1_L1-miu)-(1-miu)*(x_vec(i)+r1_L1-miu+miu)./(r1(i,:)).^3-
miu*(x_vec(i)+r1_L1-miu+miu-1)./(r2(i,:)).^3; d_omega_eta(i,:)=y_vec-(1-miu)*y_vec./(r1(i,:)).^3-miu*y_vec./(r2(i,:)).^3; end
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u_mag=(u_mat_x.^2+u_mat_y.^2).^0.5; a_nat=(d_omega_ksi.^2+d_omega_eta.^2).^0.5; a_nat_dim=a_nat*a_earth_moon/T_norm^2; a_cont_dim=u_mag*a_earth_moon/T_norm^2; a_cont_dim=a_cont_dim'; a_nat_dim=a_nat_dim'; figure(2) subplot(2,1,1) [c,h] = contourf(x_vec,y_vec,a_cont_dim*1000); clabel(c,h) colorbar title('Control Acceleration (in mm/sec^2)') xlabel('\xi') ylabel('\eta') text(0,0,'\coloryellow\leftarrow\coloryellow L1',... 'HorizontalAlignment','left') text(-r1_L1,0,'\leftarrow Earth',... 'HorizontalAlignment','left') % text(r2_L1,0,'\leftarrow Moon',... % 'HorizontalAlignment','left') subplot(2,1,2) [c,h] = contourf(x_vec,y_vec,a_nat_dim*1000); clabel(c,h) colorbar title('Natural Acceleration (in mm/sec^2)') xlabel('\xi') ylabel('\eta') text(0,0,'\coloryellow\leftarrow\coloryellow L1',... 'HorizontalAlignment','left') text(-r1_L1,0,'\leftarrow Earth',... 'HorizontalAlignment','left') -----------------------------------------------------------
Back and Forth simulation:
clear all; close all; clc; rect = [100 50 800 500]; hf=figure('Position',rect); x=importdata('NECKLACE_03.dat'); [m,n]=size(x); x(1:m,2:n)=2*x(1:m,2:n); % increasing amplitude by a factor of 3 (for graphic demonstration only) for i=1:1:327, % plot(h_seg,x(i,2:n),'o') if mod(i,3)==0, plot(x(i,2:n),'o') hold on; if i<=327, j=round(x(i,1)); if j==0, j=1; end
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plot(j,x(i,j),'rs') end axis([0 100 -100 100]) axis off M(i/3)=getframe; % M_forward(i/3)=getframe; end hold off; % plot(x(i,1)*10^6,x(i,j),'rs')
end for i=1:1:327, % plot(h_seg,x(i,2:n),'o') if mod(i,3)==0, plot(x(328-i,2:n),'o') hold on; if i<=327, j=round(x(328-i,1)); if j==0, j=1; end plot(j,x(328-i,j),'rs') end axis([0 100 -100 100]) axis off M((i+327)/3)=getframe; % M_backward(i/3)=getframe; end hold off; % plot(x(i,1)*10^6,x(i,j),'rs')
End -----------------------------------------------------------
Tether deployment – first and second stages:
clear all; close all; clc; defaults; %% Constants miu1=3.986e5; % earth gravitation constant miu2=4.903e3; % moon gravitation constant miu=miu2/(miu1+miu2); r1_L1=0.849065; % L1 distance from earth r2_L1=1-r1_L1; % L1 distance from moon a=(1-miu)/r1_L1^3+miu/r2_L1^3; rho=(3*a)^0.5; k=5;
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D_star=0.148; % --------------First Stage-------------------- %% Phi t_l=2*k*pi/rho; t=linspace(0,t_l,1000); phi=2./(rho^2*t).*(cos(rho*t)-1); omega=-2./(rho^2*t.^2).*(cos(rho*t)-1)-2./(rho*t).*sin(rho*t); figure(1) plot(t,phi,t,omega) title('\phi and \omega Vs. Time') xlabel('Time') legend('\phi','\omega') %% Tension u=D_star/t_l; D=u*t; T=D.*(omega.^2+1+a/2+3*a/2*cos(2*phi)+2*omega); figure(2) plot(t,T) title('Tension Vs. Time') xlabel('Time') ylabel('Tension') % --------------Second Stage--------------------- %% eta and ksi t_new=linspace(0,5*t_l,1000); ksi=0.7; f_ksi=miu/(1-miu-ksi)^3+(1-miu)/(miu+ksi)^3; r1=(f_ksi-1)^0.5; r2=(f_ksi)^0.5; eta=0.03*cos(r1*t_new)+0.03*sin(r1*t_new); zeta=0.03*cos(r2*t_new)+0.03*sin(r2*t_new); figure(3) plot(eta,zeta) title('Lissajous trajectory of the anchored spacecraft') xlabel('\eta') ylabel('\zeta')
------------------------------------------------------
Normal force calculations and friction losses:
clear all; close all; clc;
% Data m=1500; % [kg] mass of tether-car+load W=7500; % [Watt] Power a_l=3.84*10^8; % [m] r_l=1738*10^3;% [m] d=4700*10^3;% [m] omega=2.6617*10^-6;% [rad/sec] miu_e=3.986*10^14; % [m^3/sec^2]
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miu_l=4.903*10^12; % [m^3/sec^2] T_0=3240; % [N] rho=1.7*10^3; % [kg/m^3] sigma=9.5*10^9; % [N/m^2] miu=0.5; %static friction (Polystyrene on Polystyrene) Crr=0.002; %rolling friction (polymer on steel - softer surface=large coef.)
% Velocity Calculation v=importdata('v_vector.mat'); h=importdata('h_vector.mat'); % Weight Calculations h_lala=linspace(0,325000000,119675); g_lala=(a_l-r_l-h_lala-d)*omega^2-miu_e./(a_l-r_l-h_lala).^2+miu_l./(r_l+h_lala).^2; figure(3) weight=m.*g_lala; plot(h_lala/1000,weight) title('Weight vs. Height(from the moon)') xlabel('Height [km]') ylabel('Weight [N]') grid on
% Normal Force Calculation S_F=1; % safety factor s=10^6*T*S_F/sigma; %mm^2 tether_width=s/0.015; %mm ratio=110./tether_width; N=weight.*ratio./miu; %normal force figure(4) plot(h_lala(1:length(N)-20)/1000,abs(N(1:length(N)-20))); % h in km, normal force in N Title('Normal Force Vs. Height') Xlabel('Height[km]') Ylabel('Normal Force [N]') grid on P=N.*v.*Crr; figure(5) plot(h_lala(1:length(P)-20)/1000,abs(P(1:length(P)-20))) % h in km, P in watt Title('Power Needed to Overcome Rolling Friction Vs. Height') Xlabel('Height[km]') Ylabel('Power [Watt]') grid on;
-----------------------------------------------------------
Choosing the material and weight of the wheel:
close all; clear all; clc;
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ro0=7850; %steel alloy ro4=1600; %carbon/epoxy v=700; %m/sec omega=v/1; r_vec=linspace(0.01,2,1000); m_vec=(0.01*2*pi*0.110*0.001*ro4)+pi.*0.001^2.*ro0.*20.*r_vec... +(2.*pi.*r_vec.*0.110.*0.001.*ro4); E_vec4=(v^2)*ro4./r_vec/10e9; figure (1) subplot(2,1,1) plot(r_vec,E_vec4,'r'); title ('Needed Young Modulus (at wheel`s perimeter)'); xlabel ('radius [m]'); ylabel ('young modulus [GPa]'); grid on; subplot (2,1,2) plot (r_vec,m_vec,'b'); title ('Weight of the wheel vs. radius'); xlabel ('radius [m]'); ylabel ('weight [kg]'); grid on;
-----------------------------------------------------------
Reentry to Earth:
%Main
close all;
clear all;
clc;
R0=6378; %Earth radius
Mu=398600; %Earth gravity constant
V_e=0.400; %Earth velocity
V_a=10.86-V_e; %Relative velocity in the entering on the Earth
Cd=2; %Drag coefficient
S=4.12/10^6; %Area of cross section
S1=pi*(0.371/1000)^2; %Area of 1 tank
m=952.5;%Total mass
m1=32.5;%mass of 1 tank +helium
R_gas=287*10^-6; %Universal gas constant
gama=1.4;
X=[6458,0];
fi=2.5*pi/180; %Angle of entering
V=V_a*[-sin(fi),cos(fi)];
A0=[X,V];
options = odeset('Maxstep',1);
[t1,A1]=ode45(@sol,[0 270.4],A0,options);
[t2,A2]=ode45(@sol1,[270.401 900],A1(length(A1),:),options);
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t=[t1;t2];
A=[A1;A2];
H=sqrt(A(:,1).^2+A(:,2).^2)-R0;
V_fin=sqrt(A(:,3).^2+A(:,4).^2);
for i=1:1:length(H)
row(i)=row_k(H(i));
g_e(i)=9.81*(R0/(R0+H(i)))^2;
if i<=length(A1)
F(i)=row(i)*S*V_fin(i)^2*Cd/2;
n(i)=F(i)*10^3/(m*g_e(i));
W(i)=F(i)*V_fin(i);
H_F(i)=W(i)/S;
else
F(i)=row(i)*S1*V_fin(i)^2*Cd/2;
n(i)=F(i)*10^3/(m1*g_e(i));
W(i)=F(i)*V_fin(i);
H_F(i)=W(i)/S1;
end
end
figure
plot(t,n);
title('Overload Vs. Time in Atmospheric entry');
xlabel('Time [sec]');
ylabel('n');
grid on
figure
plot(t,H_F/10^6);
title('Heat flux Vs. Time in Atmospheric entry');
xlabel('Time [sec]');
ylabel('Heat flux [MJ/m^2*sec]');
grid on
figure
plot(t,H);
grid on
title('Height Vs. Time in Atmospheric entry');
xlabel('Time [sec]');
ylabel('Height [km]');
figure
plot(t,V_fin);
% plotyy(t,V_fin,t,Mach);
grid on
title('Velocity Vs. Time in Atmospheric entry');
xlabel('t[sec]');
ylabel('V[km/sec]');
function d = sol(t,A)
Mu=398600;
S=4.12/10^6;
m=952.5;
R0=6378;
Cd=2;
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X(1)=A(1,1);
X(2)=A(2,1);
V(1)=A(3,1);
V(2)=A(4,1);
R=sqrt(X(1)^2+X(2)^2);
v=sqrt(V(1)^2+V(2)^2);
h=R-R0;
c=[0.198341,-0.0871016,-0.00243284,-1.60352e-005,2.00856e-006,-2.96832e-008,1.31533e-010];
row=exp(c(1)+c(2)*h+c(3)*h^2+c(4)*h^3+c(5)*h^4+c(6)*h^5+c(7)*h^6)*10^9;
if h<0
row=1.22*10^9;
end
V_f=-(Mu/R^3)*X-(S/(2*m))*Cd*row*v*V;
d=[V,V_f];
d=d';
function row=row_k(h) %Density
c=[0.198341,-0.0871016,-0.00243284,-1.60352e-005,2.00856e-006,-2.96832e-008,1.31533e-010];
row=exp(c(1)+c(2)*h+c(3)*h^2+c(4)*h^3+c(5)*h^4+c(6)*h^5+c(7)*h^6)*10^9;
if h<0
row=1.22*10^9;
end
if h>85
row=0;
end
-----------------------------------------------------------
Solar energy – daylight and darkness on the elevator’s path:
clear all; close all; clc;
i=1; V0=384000; V01=0; a=1; S_counter=0; Sp_counter=0; for teta=0:0.0002:2*pi/13 i=i+1; a_m=384000; %% orbits % moon r=a_m*(1-0.05^2)/(1+0.05*cos(teta*13)); x=r*cos(teta*13); y=r*sin(teta*13); r1=1738; teta2=0:0.01:2*pi;
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x2=x+r1.*cos(teta2); y2=y+r1.*sin(teta2); % sun r_s=(1.5*10^8)*(1-0.01^2)/(1+0.01*cos(teta)); x_s=r_s*cos(teta); y_s=r_s*sin(teta); r1_s=696000; teta2_s=0:0.01:2*pi; x2_s=x_s+r1_s.*cos(teta2_s); y2_s=y_s+r1_s.*sin(teta2_s); % earth r_e=6378; x_e=r_e.*cos(teta2); y_e=r_e.*sin(teta2); % tether car w=365*24*3600/(13*2*pi); t=teta*13*w; V=V0-a*(t*0.7-V01); if V<50000 V01=V01+334000; a=-a; V0=50000; end; if V>384000 V01=V01+334000; a=-a; V0=384000; end; r_car=V*(1-0.05^2)/(1+0.05*cos(teta*13)); x_car=r_car*cos(teta*13); y_car=r_car*sin(teta*13); % tether r_tether=60000*(1-0.05^2)/(1+0.05*cos(teta*13)); x_tether=r_tether*cos(teta*13); y_tether=r_tether*sin(teta*13); %% sun light % shadow (umbra) alpha=atan((y_s-y)/(x_s-y)); x2_s1=x_s+r1_s.*cos(alpha+pi/2); y2_s1=y_s+r1_s.*sin(alpha+pi/2); x2_1=x+r1.*cos(alpha+pi/2); y2_1=y+r1.*sin(alpha+pi/2); x2_s2=x_s+r1_s.*cos(alpha-pi/2); y2_s2=y_s+r1_s.*sin(alpha-pi/2); x2_2=x+r1.*cos(alpha-pi/2); y2_2=y+r1.*sin(alpha-pi/2); m1=(y2_s1-y2_1)/(x2_s1-x2_1); m2=(y2_s2-y2_2)/(x2_s2-x2_2); b1=y2_1-x2_1*m1; b2=y2_2-x2_2*m2; x_shadow=(b1-b2)/(m2-m1); y_shadow=m1*x_shadow+b1; % part shadow (pan umbra) m3=(y2_s2-y2_1)/(x2_s2-x2_1); m4=(y2_s1-y2_2)/(x2_s1-x2_2); b3=y2_1-x2_1*m3;
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b4=y2_2-x2_2*m4; x3=-x2_s2; x4=-x2_s1; y3=m3*x3+b3; y4=m4*x4+b4; % Shadow counter; m_em=y/x; x_em1=b1/(m_em-m1); y_em1=m_em*x_em1; x_em2=b2/(m_em-m2); y_em2=m_em*x_em2; d1=((x_em1-x)^2+(y_em1-y)^2)^0.5; d2=((x_em2-x)^2+(y_em2-y)^2)^0.5; d_car=((x_car-x)^2+(y_car-y)^2)^0.5; if d1>d2 if d_car<d2 S_counter=S_counter+1; end else if d_car<d1 S_counter=S_counter+1; end end % part shadow counter; y_p1=m3*x_car+b3; y_p2=m4*x_car+b4; if y_p1>y_p2 if (y_p1>y_car) && (y_p2<y_car) Sp_counter=Sp_counter+1; end else if (y_p1<y_car) && (y_p2>y_car) Sp_counter=Sp_counter+1; end end %% ploting plot(x,y,x2,y2,'blue',[x,x_tether],[y,y_tether],x_car,y_car,'*',[x2_1,x_shadow],[y2_1,y_shadow],'black',[x2
_2,x_shadow],[y2_2,y_shadow],'black',x_e,y_e,[x2_1,x3],[y2_1,y3],'blue',[x2_2,x4],[y2_2,y4],'blue') axis([-420000 420000 -420000 420000]) if mod(i,20)==0 M(i/20)=getframe; end end Pre_s=S_counter/i*100; Pre_sp=Sp_counter/i*100;
83
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