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Question on the space elevator



 
 
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  #21  
Old April 7th 04, 01:34 PM
Bob Martin
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Default Question on the space elevator

So it would take about 16 kg of fuel+oxidizer, give or take, to send 1
kg up the elevator. But you also have to carry the fuel to lift the
fuel, and the fuel to lift the fuel to lift the fuel, and so on.
Mathematically, you would have to carry e^16= 9 million kg of fuel+ox.
(Factors of 2 or 1/2 at that point become very important.)

Anyway, beaming the power becomes economical at that point.



That number of 9 million kg seems a little large... we've sent much
larger things out of the gravity well on much less fuel, using less
efficient processes.
  #22  
Old April 7th 04, 03:47 PM
Dr John Stockton
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JRS: In article , seen in
news:sci.space.science, Gordon D. Pusch g_d_pusch_remove_underscores@xn
et.com posted at Tue, 6 Apr 2004 21:08:29 :


Your proposal is not _totally_ implausible. The energy required to climb
a beanstalk is only a small fraction of the energy required to accelerate
a payload into Low Earth Orbit; the fuel and oxygen tankage required
would be large, but not prohibitively so.




Earth radius is 7 Mm, GSO radius is 42 Mm, so in units of the Earth's
radius 1 and 6.

Potential energy in an inverse square field is inverse linear, so with
it being zero at infinity, it is 1/6 at GSO and 1 on the ground,
difference 5/6.

But the energy required for LEO is half that for escape, so climbing the
stalk to GSO requires 5/3 of the energy to LEO.

"Centrifugal force" will supply a significant part of that energy (by
lengthening the day), but not enough to leave only a small fraction.

Much of the fuel and oxygen will need to be lifted, though there is the
advantage that the exhaust need only carry a little energy other than
that of height.

--
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  #23  
Old April 7th 04, 07:21 PM
Poliisi
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Default Question on the space elevator

The current answer to the last question isn't magnets but .. lasers.
Free Electron lasers beam power to the climber, which converts the
energy into mechanical energy (wheels or treads). IIRC, a FEL has
been designed that can do the job.


Please give me a link to this design. If you are talking about simply a
beamer that could perhaps be used in an unknown design that incorporates
aiming (you know, to hit that elevator), suitable transfer medium, and
receiver system, which can operate on these lurdicous lenghts, then dont
bother.
  #24  
Old April 7th 04, 07:41 PM
Sander Vesik
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Default Question on the space elevator

Gordon D. Pusch wrote:

For any reasonable strength of material, the amount of additional energy
that can be stored by pressurizing the tanks is negligible compared to the
chemical energy stored in the propellants themselves. About all you will do


Consider tanks that contain solid hydrogen instead of liquid hydrogen -
or more simply, contain say supercoold hydrogen at twice the pressure while
weighting no more than present ones.

by pressurizing the tanks is to allow you to eliminate the mass of the
turbopumps and drive turbines, which is likely to be marginal compared to
the additional mass of high-pressure propellant tanks. Pressure-fed rockets


But as you used much better materials, the mass of the tanks went down or
at least didn't increase much.

_may_ be justifiable on the basis of lower cost or higher reliability,
but are =VERY= unlikely to provide significantly better performance than
pump-fed rockets.


But it is not really a question of replacing the pumps with a pressure-fed
system, its a question of using better materials whichmeans that either
you can leave the mass the same - while getting higher performance *or*
reducing the mass of all compoents while still getting same performance.


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'


--
Sander

+++ Out of cheese error +++
  #25  
Old April 7th 04, 11:39 PM
Makhno
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Default Question on the space elevator

There's a lot of traffic in this thread about powering the climber. Why
can't it simply have a diesel/gasoline engine with its own oxygen
supply?


just possibly something to this...

Or run electrical cables up the elevator to power an electric motor?


But this was a bad idea.

Why make things more complicated than they need to be?


Your proposal is not _totally_ implausible. The energy required to climb
a beanstalk is only a small fraction of the energy required to accelerate
a payload into Low Earth Orbit; the fuel and oxygen tankage required
would be large, but not prohibitively so.


I did a quick calculation assume modern diesels assuming a 10 tonne payload
starting with 10 tonnes of fuel. From my matlab script:

G=6.67300e-11; %gravitional constant
Mc=10e3; %Mass of cargo, 10tonnes
Me=5.95e24; %mass of earth
e=0.5; %efficiency of typical 4 stroke
Esp=40e6; %fuel energy per kg (assume air breathing)
r0=12760e3/2; %start point (radius of earth)
Mf=10e3; %mass of fuel

Equating the energy required to get out of the gravity well to the energy
available in the fuel tank:
GMe(Mf+Mc)(1/r0-1/r1) = e Esp Mf

This makes three assumptions, One is that we're air-breathing all the way,
which might not be too unreasonable because the hardest part (where g is
highest and the engine will be working its hardest) has plenty of air. Even
at high altititudes a super-charger or similar technology could be applied
to allow the engine to breath the air, only switching to a liquid oxygen
tank, or oxygen-less fuel when truely in space.
The second is that the mass of the fuel in the tank remains constant -
obviously it would be used up.
The third is that the 10tonnes of payload would of course include the heavy
engine.

I get from this about 1000km of altitude - only about 1/30th of the altitude
needed. So perhaps not entirely unreasonable, but difficult nonetheless.


Rest of matlab script:

r1=1/(1/r0 - (Esp*e*Mf)/(G*Me*(Mf+Mc))); %calculate height

altitude=(r1-r0)/1e6 %million meters (need about 30e6)



  #26  
Old April 9th 04, 02:48 AM
Rick Jones
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Default Question on the space elevator

Makhno wrote:
This makes three assumptions, One is that we're air-breathing all
the way, which might not be too unreasonable because the hardest
part (where g is highest and the engine will be working its hardest)
has plenty of air. Even at high altititudes a super-charger or
similar technology could be applied to allow the engine to breath
the air, only switching to a liquid oxygen tank, or oxygen-less fuel
when truely in space.


Could you be better off stopping part way up to fill the O2 tanks
using those superchargers/whatever instead of lugging them up filled
from the surface? After all, you could just hang there on the tether
right?

rick jones
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these opinions are mine, all mine; HP might not want them anyway...
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  #27  
Old April 9th 04, 03:19 AM
David M. Palmer
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Default Question on the space elevator

In article , Bob
Martin wrote:


That number of 9 million kg seems a little large... we've sent much
larger things out of the gravity well on much less fuel, using less
efficient processes.


The energetics are different.

If you are halfway up the beanstalk, then burning a kg of fuel will
give you the same amount of energy as it would at the base. think of
the energy wasted as you drop the ash from your fuel over the side and
let it just fall back to Earth. (If you could use it as a
counterweight to pull you up, then you'd be better off.)

In a rocket, the amount of kinetic energy you get from burning a
certain amount of fuel (thrusting along the velocity vector) is
proportional to your speed. Intuitively, this is because the fuel in
your tanks has kinetic energy that you partially recover--if your
exhaust velocity is equal to your speed (in some frame of reference)
then the burned fuel ends up with zero kinetic energy and the rocket
gets it all.

--
David M. Palmer (formerly @clark.net, @ematic.com)
  #28  
Old April 9th 04, 04:38 AM
Gordon D. Pusch
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Default Question on the space elevator

Dr John Stockton writes:

JRS: In article , seen in
news:sci.space.science, Gordon D. Pusch g_d_pusch_remove_underscores@xn
et.com posted at Tue, 6 Apr 2004 21:08:29 :

Your proposal is not _totally_ implausible. The energy required to climb
a beanstalk is only a small fraction of the energy required to accelerate
a payload into Low Earth Orbit; the fuel and oxygen tankage required
would be large, but not prohibitively so.



Earth radius is 7 Mm, GSO radius is 42 Mm, so in units of the Earth's
radius 1 and 6.

Potential energy in an inverse square field is inverse linear, so with
it being zero at infinity, it is 1/6 at GSO and 1 on the ground,
difference 5/6.

But the energy required for LEO is half that for escape, so climbing the
stalk to GSO requires 5/3 of the energy to LEO.

"Centrifugal force" will supply a significant part of that energy (by
lengthening the day), but not enough to leave only a small fraction.

Much of the fuel and oxygen will need to be lifted, though there is the
advantage that the exhaust need only carry a little energy other than
that of height.


The significant difference is that, when climbing a beanstalk, one only has
to supply the change in _energy_, not the change in _momentum_, which is
is instead provided by the lateral force exerted by the beanstalk, which
very efficiently extracts momentum from the angular momentum of the Earth.

By contrast, much of the energy stored in the propellant of a rocket is
uselessly wasted, because it goes into accelerating the _rocket exhaust_,
rather than the _payload_...


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

  #29  
Old April 9th 04, 12:39 PM
Gordon D. Pusch
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Default Question on the space elevator

Rick Jones writes:

Makhno wrote:
This makes three assumptions, One is that we're air-breathing all
the way, which might not be too unreasonable because the hardest
part (where g is highest and the engine will be working its hardest)
has plenty of air. Even at high altititudes a super-charger or
similar technology could be applied to allow the engine to breath
the air, only switching to a liquid oxygen tank, or oxygen-less fuel
when truely in space.


Could you be better off stopping part way up to fill the O2 tanks
using those superchargers/whatever instead of lugging them up filled
from the surface? After all, you could just hang there on the tether
right?


Please note that the density of the atmosphere is already negligible
at a mere 150 km of altitude, less than 0.4% of the way to GSO,
whereas the gravitational force at that altitude is still essentially
the same as that at "sea level." It will =NOT= be possible to use
"superchargers" to get oxygen from the air --- LOX tanks will be needed.


-- Gordon D. Pusch

perl -e '$_ = \n"; s/NO\.//; s/SPAM\.//; print;'

  #30  
Old April 9th 04, 02:48 PM
nafod40
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Default Question on the space elevator

Gordon D. Pusch wrote:
Dr John Stockton writes:
The significant difference is that, when climbing a beanstalk, one only has
to supply the change in _energy_, not the change in _momentum_, which is
is instead provided by the lateral force exerted by the beanstalk, which
very efficiently extracts momentum from the angular momentum of the Earth.


Does the beanstalk have to be straight up and down? If canted so as to
lag the Earth's rotation (at least during a climb maneuver)how would
that make a difference?

 




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