From: ederd@bcstec.ca.boeing.com (Dani Eder)
Subject: Re: Orbital Elevators
Date: May 16 1995
Newsgroups: rec.arts.sf.science

burton@het.brown.edu (Joshua W. Burton) writes:

>Dani Eder writes

>> The least mass structure combines a compressive tower up from
>> the ground and a cable in tension hanging from orbit, with the
>> two meeting tip to tip.

>And what keeps the compressive part of the tower from buckling?  Make
>it a tube, to increase its stability against Euler buckling, and you
>will eventually have a thin enough tubewall to have Brazier buckling,

You would build the compressive tower the same way that TV transmission
towers are built - as a truss of smaller elements.  A reasonable 
height-to-base ratio is 20:1.  So a 100 km tower would have 3 base
points 5 km apart, assuming you have a triangular cross section
for the tower as a whole.  Each principal column would in turn
be a truss with 3 sub-columns spaced 250 meters apart, which in
turn are made of tertiary columns 12 meters apart and 0.6 meters
in diameter each.  The tertiary columns have a wall thickness of
0.03 meters.

>And since the wind loading also goes as the square
>of the height (when is a column really a beam?), at least to the
>top of the troposphere, you are paying far more per vertical half-km

Wind loading is indeed the driving force on the tower in the first
20 km of height.  Up to 10 km, higher wind speeds dominate over
pressure drops.  Above that height, wind speeds no longer increase,
and reduced atmospheric pressure dominates.  When I did the studies
of this type of structure, I used pivoting airfoils around the
truss elements to minimize drag.  They can lower it quite a bit
below that of a circular tube.

>> Actually, a tensile strength for carbon fiber of 1.4 million psi
>> is sufficient, and 1.0 million psi is available.  Here's how:
>> 
>> Carbon fiber of that strength and a density of 0.68 lb/in^3 can
>> support it's own weight at 1 g over a length of 325 miles.  At
>> reduced load for some margin of strength, we use 200 miles.  
>> In compression, we get about 40% of the tensile fiber strength in
>> a composite strut.  We taper the structure up from the ground
>> and down from orbit by a factor of e (2.718...) for each 'scale
>> length', our 200 and 0.4x200=80 mile distances.
>> 
>> The Earth's gravity well is equal to 1 g times the radius of the
>> Earth = 4000 miles.  Using the tower and cable combined we have
>> 4000/280 = 14.3 scale lengths.  Thus the cross section changes
>> by a factor of e^14.3 = 1,600,000 from ground to tip of tower
>> and GEO to tip of cable.  If you have a 100 inch x100 inch structure
>> at the fat ends, you have a 1/160 square inch structure at the
>> narrow end capable of supporting 6,000 lb.  

>Okay, let's get this into Homo sapiens units so that we can check it
>against what I said; I left my knapped flint slide rule at home today.
>A million psi is about six billion Pa, and your carbon fiber density
>would come to 18,800 kg/m^3, which is ludicrous.  (The only things I

Sorry, I dropped a decimal point.  Carbon fiber is 0.068 lb/in^3.

>quintals and drams---and say 1880 kg/m^3.  This makes your fiber
>"3x piano wire" in my notation, which as I said lets you carry a
>thousand tons of tension with a half-teraton tower.  If you can get
>up to 5x, the mass of the tower goes down a thousandfold.

>The depth-of-gravity-well calculation you just did is actually
>slightly pessimistic, because you are not going to the top of the 
>earth's gravity well (escape speed), but only to GEO, which is
>8.8% closer.  Further, you are not starting from rest but from 
>equatorial ground speed, although this is only a 0.3% effect with 
>respect to kinetic energy.  However, the 325-mile scale length is 
>on the high side.  A 3x cable of a square meter cross-section can
>support h = 3000/g = 306 km of its own weight under one gee, at 

I was using 1.4 million psi carbon fiber in my assumptions.

>Carbon fiber comes under the heading of "exotic" engineering materials

I guess that means that we're building commercial aircraft out of
'exotic' materials.  The tail of the new 777 aircraft (and other
smaller parts) are made of composites.  Nowadays, graphite-epoxy
is competitive with aluminum for aerospace-grade construction.  And
I have assumed all along that a space elevator type structure is
an aerospace-grade project rather than a civil-engineering grade
project.

>in my view:  my bicycle's frame is made of it, but I haven't seen
>anyone rushing to use it in suspension bridges or radio-mast guylines

Actually, consideration is being seriously given to a graphite
epoxy suspension bridge.  Aside from being much lighter than a
steel bridge, you wouldn't have to do as much maintenance to keep
it from rusting.

>Isn't it fascinating that our gravity well is right at the point
>of exponentially diminishing returns for _both_ chemical rocketry
>and materials science?  If we lived on Mars, the orbital tower
>would be easy, but we wouldn't need it because so would SSTO.  And
>on a world with twice our gravity well, no one would even consider
>getting to orbit with chemical rockets or skyhooks.  Makes you
>sorta wonder if the quarantine is an accident....

No coincidence at all.   Both ultimate strength of materials and
chemical rockets are based on the strength of a chemical bond.

Dani Eder

From: ederd@bcstec.ca.boeing.com (Dani Eder)
Subject: Re: Orbital elevators
Date: May 19 1995
Newsgroups: rec.arts.sf.science

burton@het.brown.edu (Joshua W. Burton) writes:

>Dani Eder writes

>As I said, I have no problem with a few tens of kilometers.  But by
>this argument your halfway-to-orbit compression structure, even though
>it only goes up 6000 km or so (halfway in terms of scale height) would
>have a primary structure with a footprint covering most of a 
>banana republic.  And the inward-leaning elements of the truss have
>a larger side loading at each level of the hierarchy.  I'm not saying
>it can't be done...just that every extra kilometer of nice simple
>tension member you can get away with will save you billions of dollars
>worth of structural complication.  Further, in the levels-of-three
>structure you have specified, ONE failure of a quaternary member
>will bring the whole edifice down, by a `want of a nail' effect.

For minimum mass and the given assumptions, your tower would be
12 scale heights high, or 1800 km.  Therefore the base points would
be 90 km apart.  Note that the first cross beam is 90 km high, so
you have 3 columns 4.5 km wide each spaced 90 km apart at ground
level.  The areal fill of the tower is so low that agriculural
work would go on with no discernable impact.  The cross beams
resist the inward lean of the columns, and amount to a 20% overhead
(the inward lean only produces 5% of the force of gravity,
so the cross beams are sqrt(0.05) the mass).  The orbital tether
will not be a simple tension member.  If you have a single strand,
then a single piece of orbital debris can break the whole thing.

Given the orbital debris environment, you want a minimum of 
6 strands, any 4 of which can carry the design load, and cross
strapping every 10 km to re-distribute loads around a broken
strand.  Repair then means replacing a 10 km section of one strand.
The strands should be several times the largest piece of orbital
debris apart, so that a piece of orbital debris is likely to
only hit one strand at a time.  With empty upper stages and
such in the 10 meter range, I would place the strands 50 meters
appart in a hexagonal pattern.

As for the tower being liable to single point failure, No sane
civil engineer designs that way.  You build 'damage tolerant'
structure, in case an airplane or satellite (it's tall enough)
crashes into it, with adequate strength margin and redundant
load paths.  At a minimum, the tower should survive the removal
of any 4.5 km length of any of the 9 secondary columns.  This
requires the design load to be carried by the two remaining
secondary columns on that edge. 

The first scale height of tower reduces the tether from e^17 to
e^16 times the payload mass.  Unless the tower structure costs
1.7 x e^16 times as much as the tether, it makes sense to do this.
Using an equal number of scale heights gives you a minimum mass
structure, and assumes that the tower and tether are equal cost
per weight, which is reasonable since they both use the same
carbon fiber, and the epoxy is relatively cheap.
 
>> I was using 1.4 million psi carbon fiber in my assumptions.
>> 
>> >Carbon fiber comes under the heading of "exotic" engineering materials
>> 
>> I guess that means that we're building commercial aircraft out of
>> 'exotic' materials.  

>We're talking about billions of tons at least, so even high-tensile
>steel is a bit exotic.  What we need is a material that costs dollars
>per ton; pity that mild steel is about 0.05x in my notation.

For a 1 ton payload, you would need about 2 million tons of carbon
fiber for a full GEO tower.  At $40,000 per ton in bulk, we are
talking $80 billion.  This is one or two orders of magnitude
out of economic reach.  At 1 payload per hour, and a 12.5 year
payback time (8% rate of return), you can afford to deilver
100,000 tons of payload.  Since the GEO tower itself had to be
deilverd into space, you are too heavy to make sense.  Either
structural materials would need to get to the 2 million psi
range or you build a sub-GEO size system that weighs less.

Dani Eder

From: "Geoffrey A. Landis" <geoffrey.landis@lerc.nasa.gov>
Newsgroups: sci.space.tech
Subject: Re: Advantages of high-altitude launch (was: Re: Alternative means of 
	achieving orbit)
Date: 1 Apr 1996 15:03:38 GMT

>"Geoffrey A. Landis" <geoffrey.landis@lerc.nasa.gov> put forth:
>>I'm still greatly enamoured of the idea of reducing the delta-V to orbit
>>by launching from an altitude: a mountain-top or a tall artificial
>>structure, or preferentially a tall structure built *on* a mountain-top.

In article <4jb345$bi4@vixen.cso.uiuc.edu> Blair Patric Bromley,
bromley@students.uiuc.edu comments:
>   To be of any use, the structure would need to be at least a mile-high;
>no engineering firm that I'm aware of is capable of building such a structure.

I'd like several miles at least.  Ten would be nice.  The engineering
problem just isn't that hard; remember that Frank Lloyd Wright designed
mile-high skyscrapers back in the 30's.  Materials properties allow you
to make structures *hundreds* of kilometers tall, if you wanted.  I'd say
that comparing the engineering design of a single stage to orbit vehicle,
and the engineering design of a twenty mile high building, the building
is an easier problem.

>Remember:  after you build your rocket, you have to get it on top of this
>launch structure.  No easy task.

Well, yes, but easier than using a rocket to get the same altitude.

>>Some of the advantages of high-altitude launch:
>>(1) You start with a fraction of the potential energy needed to get to
>>orbit.
>
>    If you could launch at 10000 ft above sea level, you could reduce
>your velocity change to get into orbit by approx. 250 m/s.  However, you
>need about 8000 m/s to get into orbit.  A 3% improvement.

But three percent is a *tremendous* improvement.  A RL-10A has an Isp of
about 450 seconds; thus, exhaust velocity Ve is about 4400 km/sec. 
Structure & payload mass fraction is exp[deltaV/Ve]; a RL-10A powered
vehicle could achieve a maxium amount of structure plus payload to 8
km/sec of 16.3%.  Typically about 5% of this is actually payload.  A 3%
decrease in delta-V to orbit increases this to 17.3%.  This increases the
*payload* to 6% of the gross lift-off mass -- a 20% increase in payload.

>>(2) You start at a lower atmospheric pressure.
>>        a. reduced atmospheric drag loss

Which wasn't commented on by Bromley or by Pat, but is a significant
effect, at *minimum* equal to the potential energy gain.

>>        b. vehicle can be designed with less attention paid to
>>aerodynamics.  Lower aerodynamic design penalty means higher performance
>>designs (ie., smaller fineness ratio allows more efficient tanks)
>>        c. More optimum trajectory possible; you can curve toward
>>horizontal thrust much faster since you start out closer to out of the
>>atmosphere
>>        d. Max-Q occurs at a much lower pressure; lower aerodynamic
>>stress on the system means vehicle can be designed lighter.

Pat, prb@clark.net commented on this one:
>I think Max-Q is going to be at the same altitude or lower depending
>upon tank fineness.

To the contrary.  Max-Q is the product of air density, the square of
velocity, and a vehicle-dependent factor which depends on mach number. 
For a given acceleration profile, Max-Q occurs at the same altitude
*above the launch site* independent of how high the launch site is.  That
is, the actual value of dynamic pressure will decrease linearly with the
initial pressure.  

True, if you decrease the fineness ratio to gain tank fraction, you will
then increase the Max-Q again.  This makes the whole thing an engineering
trade-off-- how much do you gain in tank fraction and robustness, versus
how much of the decrease of Max-Q do you lose?  Doing such trade-offs is
why we pay engineers.

However, I don't have a good number which tells me how much of the
structural mass is due to making the vehicle robust to survive Max-Q. 
Anybody have a guess?

>>        e. Aerodynamic vibrations lower; allows less robust payload
>>(e.g., lighter)
>>        f. Wind loads on vehicle in flight much lower
>>        g. Acoustic loads much lower
>>        h. Cryogenic storage easier (lower conduction and convective
>>heating)

>   All true, but what is the impact of these changes?  Atm. pressure is
>about 10 psia at 10000 ft.
[Bromley discusses point (4)a here; for logical continuity I moved this
discussion down]

>>(3) Above the weather means no compromises needed for weather
>>        a. Fewer delays for weather
>>        b. Above lightning hazard
>>(4) Lower pressure means rocket nozzle operation is closer to vacuum
>>operation.
>>        a. Higher performance out of the rocket nozzle at launch

Blair Patric Bromley, bromley@students.uiuc.edu commented:
>Atm. pressure is about 10 psia at 10000 ft.
>This would result in approx. a 2% increase
>in exhaust velocity, which, under certain assumptions would result in 
>roughly a 3% increase in payload mass fraction.

How do you figure?  A 2% increase in exhaust velocity gives a 2% increase
in Isp, Structure & payload mass fraction is exp[deltaV/Ve], or roughly
exp[0.98 * 1.82].  I calculate this as a 3.6% decrease in *total*
mass-ratio.  Using the same  [structure+payload] mass, that comes to an
11.8% increase in *payload* mass.  

Adding this to the potential energy gain [factor of 20% payload increase]
and the drag gain [quick estimate says another factor of 20%] gives
(going back to the rocket equation, instead of adding the increases
separately) an increase in [structure plus payload] to 18.9% instead of
16.3%.  That increases the payload by 52%.

>>        b. No design compromise for both sea-level and vacuum performance
>
>    If you use aerospike engines, you avoid this issue altogether.

At a cost in complexity and design evolution.  Aerospike engines are not
a highly developed technology (yet).

>>(5) You start at a lower gravity
>>        a. Lower gravity losses (gravity loss = the fraction of rocket
>>thrust used in simply cancelling gravity)
>>        b. This means that your thrust/weight ratio is higher, meaning
>>you accelerate faster, again reducing gravity losses

>    True, but the acceleration due to gravity does not change much in
>at an altitude of 200 km.  

Yeah, about 0.09%.  I wanted to try to list everything.  This adds
roughly 1% to the payload, though; not worth carrying through the numbers.

Blair Patric Bromley, bromley@students.uiuc.edu concludes by saying:
>    I really should take back what I said about the significance of the 
>changes.  Launching near the equator is definitely the way to go, and
>launching in an area that doesn't have pesky weather problems is probably
>a big advantage as well; however, the best possible high-altitude, ground-
>based launch areas are most likely in Colorado or New Mexico (U.S.) or in
>Bolivia or Peru (South America).  Realistically, you could probably have
>a launch site at 10000ft, but not much more.

Which I mostly agree with.  I like Mt. Kenya myself, though-- tall
mountain on the equator, what more could you want?

>If you were to air-launch,
>as is done with Pegasus, then you have an advantageous system.  Why
>launch from one location at 10000 ft when you can launch at even a
>higher altitude, heading east, near the equator, and with an additional
>(albeit small), additional velocity increment from the carrier aircraft.

Because the design difficulty of making a carrier aircraft to launch an
orbital vehicle of any size is tremendous.

____________________________________________
Geoffrey A. Landis,
Ohio Aerospace Institute at NASA Lewis Research Center
physicist and part-time science fiction writer

From: ederd@bcstec.ca.boeing.com (Dani Eder)
Subject: Re: Tower Launch (final)
Date: Apr 22 1996
Newsgroups: sci.space.tech

Bruce Dunn <bruce_dunn@mindlink.bc.ca> writes:

[Discussion that points out that buckling is the primary failure mode for
tall slender structures under load]


Mr. Dunn is correct in pointing out this well known design constraint.
In the analysis I did on tall structures, I assumed that the tower
would be a tapering pylon shape with a 20:1 ratio of height to base
width.  Since most things you would put on top of such a tower,
even 2000 ton Space Shuttles, require nowhere near a solid structure,
you would build it from small truss elements that provide the needed
stiffness and load bearing capacity.  For an example, look at any
radio transmitting tower.  They are usually made of a triangular
truss of small pieces.

For a 15 km tower, you would build the main legs of the pylon 750
meters apart at the base, in a triangular pattern.  Each leg would
be horizontally braced to the other legs at 750 meter intervals,
and diagonally braced with tension wires.  Each leg would in turn be
made of 3 vertical struts spaced 40 meters apart, and each strut
would be made of 3 tubes spaced 2 meters apart, with a 10 cm diameter
for each tube, and a wall thickness of 0.5 cm in the tube.  These
numbers are driven by buckling conditions.  Now, high strength graphite
epoxy tubes have a compressive strength of 17.5 tons/cm^2.  Let us
work at no more than 10 tons/cm^2.  Then each tube can support
149 tons, each strut 447 tons, each leg 1343 tons, and the tower as
a whole 4,029 tons.

Graphite epoxy weighs 1.6 grams/cm^3, so the mass of the vertical
members is 966 tons.  The horizontal braces don't have to withstand
as much load as the vertical members, so they probably can be figured
at no more than equal to the vertical members in mass.  So roughly 
we would have 2000 tons of structure and thus 2000 tons of payload
for the tower.

Another consideration is wind loading for a structure this large.
Winds tend to get stronger up to 10 km altitude (the Jet Stream),
after which the lower air pressure overcomes higher wind
velocity.  In order to minimize the effects of wind on the tower,
you would want to surround the circular structural tubes with
airfoils pivoting on roller bearings so that they can pivot into
the wind.  This way the wind drag on the tower can be reduced by
a factor of 10-20 (i.e. the ratio of the drag coefficient of a
circular tube to that of a good airfoil).

The tower itself would not require dedicated land, since it occupies
a set of 3 points on the ground with a lot of room in between.
You could graze cattle or raise crops on the same land since the
tower would be a fairly open structure in total (perhaps 10% optical
density, i.e. even in the shadow of the tower a lot of light gets
to the ground.

Dani Eder

Date: 29 Apr 87 23:30:35 GMT
From: ssc-vax!eder@beaver.cs.washington.edu  (Dani Eder)
Subject: Re: The Bottom-Up Approach to Space Travel

In article <8704280047.AA22888@angband.s1.gov>, DIETZ@slb-test.CSNET ("Paul F. Dietz") writes:
> Hypothesis: building a tower from the ground up to about 100 km requires
> much weaker materials than dropping a line from GSO.
> 
> How strong do materials have to be? A tower 100 km tall made of material
> with a density of 2 gr/cc will exert a pressure of 20 kilobars at the base
> (actually, less, since the tower would be tapered). That's not impossible;
> carefully shaped diamonds can withstand pressures > 1 megabar. I wonder how
> tall a tower can be if it is made of, say, graphite composite trusses?
> 

The idea you are searching for is 'scale height'.  Imagine a constant
cross section column of a given material.  For some column height, the
weight of the column per unit area of its base will equal the compressive
strength of the material.  This height is called the scale height. Using
english units, for structural carbon steel, the strength is 36,000 pounds
per square inch, and the weight is 0.3 pounds per cubic inch.  Thus the
scale height is 36,000/0.3=120,000 inches, or 10,000 feet.  Thes best
graphite epoxies I know of (Amoco 'Thornel' type T40 carbon fiber + type
1962 epoxy) have a compressive strength of 250,000 psi and a density of
0.06 lb/cubic inch.  Thus their scale height is 4.16 million inches (106 km).

The minimum mass tower to support a 'payload' at a given height , in theory,
has an exponential taper in cross section by a factor of e per scale height.
Theory, however has little bearing on the design of a realistic tower. 
The two largest 'real world' considerations are (1) you must have a
design factor of safety >1.0, or in other words you must design for less
than ultimate strength, and (2) there are winds.

The factor of safety you use will depend on the use of the tower,
especially whether people will be on it.  Reasonable figures derived
from airplane design would be 2 for static loads and 4 for dynamic
loads.  This means you design as if your structure were 1/2 and 1/4
as strong as it really is, respectively.  An example of static loads
is the weight of the structure itself.  An example of dynamic loads
is winds.

Unfortunately for tower builders, winds generally get stronger with
altitude up to 10 km, where you encounter the 'jet stream'.  If you
want your tower to last in that environment, remember to account for
100 mph AVERAGE winds.

Very rough calculations of weight of guyed towers using graphite-epoxy
mast and fiberglass guy wires indicate the following ratios of tower weight
to 'payload at top' weight:
Height=2 km, 0.1lb/lb; 4 km, 0.43 lb/lb; 6 km 1.22 lb/lb; 8 km, 2.63
lb/lb; and 10 km; 5.13 lb/lb.  (Pardon my mixing of units, but the
source numbers are that way and I don't have time to convert everything to SI)

Above 10 km the winds become a smaller effect, since the atmosphere is
getting thinner. 

Dani Eder/Advanced Space Transportation/Boeing/ssc-vax!eder
(why, you ask, do I have information on wind loads for 10 km towers?
Its because I was looking into the idea of big towers in 1986, with most
of the same reasons Paul Dietz listed.  Maybe the time for big towers
has arrived, since multiple people are independantly thinking about
them)

Date: 5 Apr 93 18:58:33 GMT
From: Dani Eder <eder@hsvaic.boeing.com>
Subject: Elevator to the top floor
Newsgroups: sci.space

Reading from a Amoco Performance Products data sheet, their
ERL-1906 resin with T40 carbon fiber reinforcement has a compressive
strength of 280,000 psi.  It has a density of 0.058 lb/cu in,
therefore the theoretical height for a constant section column
that can just support itself is 4.8 million inches, or 400,000 ft,
or 75 Statute miles.

Now, a real structure will have horizontal bracing (either a truss
type, or guy wires, or both) and will be used below the crush strength.
Let us assume that we will operate at 40% of the theoretical 
strength.  This gives a working height of 30 miles for a constant
section column.  

A constant section column is not the limit on how high you can
build something if you allow a tapering of the cross section
as you go up.  For example, let us say you have a 280,000 pound
load to support at the top of the tower (for simplicity in
calculation).  This requires 2.5 square inches of column cross
sectional area to support the weight.  The mile of structure
below the payload will itself weigh 9,200 lb, so at 1 mile 
below the payload, the total load is now 289,200 lb, a 3.3% increase.

The next mile of structure must be 3.3% thicker in cross section
to support the top mile of tower plus the payload.  Each mile
of structure must increase in area by the same ratio all the way
to the bottom.  We can see from this that there is no theoretical
limit on area, although there will be practical limits based
on how much composites we can afford to by at $40/lb, and how
much load you need to support on the ground (for which you need
a foundation that the bedrock can support.

Let us arbitrarily choose $1 billion as the limit in costruction
cost.  With this we can afford perhaps 10,000,000 lb of composites,
assuming our finished structure costs $100/lb.  The $40/lb figure
is just for materials cost.  Then we have a tower/payload mass
ratio of 35.7:1.  At a 3.3% mass ratio per mile, the tower
height becomes 111 miles.  This is clearly above the significant
atmosphere.  A rocket launched from the top of the tower will still
have to provide orbital velocity, but atmospheric drag and g-losses
will be almost eliminated.  G-losses are the component of
rocket thrust in the vertical direction to counter gravity,
but which do not contribute to horizontal orbital velocity.  Thus
they represent wasted thrust.  Together with drag, rockets starting
from the ground have a 15% velocity penalty to contend with.

This analysis is simplified, in that it does not consider wind
loads.  These will require more structural support over the first
15 miles of height.  Above that, the air pressure drops to a low
enough value for it not to be a big factor.

Dani Eder

-- 
Dani Eder/Meridian Investment Company/(205)464-2697(w)/232-7467(h)/
Rt.1, Box 188-2, Athens AL 35611/Location: 34deg 37' N 86deg 43' W +100m alt.

Date: 3 Dec 1993 02:57:29 GMT
From: Jordin Kare <jtk@s1.gov>
Subject: Very tall launch towers
Newsgroups: sci.space

In article <schlegel.754286593@cwis> schlegel@cwis.unomaha.edu (Mark Schlegel) writes:
...
>If there was some way to launch your vehicle from say, a tower that was
>many kilometers high, much more of the vehicle fuel percentage could be
>allocated to building up the orbital speed of the vehicle.  This would
>allow a given launcher to put much more mass into LEO or for a given
>mass to be placed there with a smaller, cheaper booster.  Although
>the highest structures yet built are no where near the height needed
>to make this worth while (~1200 ft for current towers/skyscrapers),
>these structures have not pushed the limits of current exotic
>materials (graphite, kelvar, sapphire whiskers, etc) or even steel.

Actually, the tallest such structures currently are ~7 km (>20,000 ft) tall,
and are made of very modest-performance materials such as granite and 
basalt.  They are known as "mountains" :-)

Still higher-altitude launch platforms are constructed of aluminum and
polyethylene ("airplanes" and "balloons" respectively)

Seriously, the effort involved in providing a "high place" of any kind
to launch from, and getting a massive launch vehicle and its associated
infrastructure to said place, has generally been more trouble than the 
modest reduction in gravity and drag losses was worth.  It's been easier
just to provide a little more performance in the first stage.  

The only currently-flying exception to this is Pegasus, a very small
booster that's light enough to carry on an airplane.  Even that was
only feasible because an existing airplane (NASA B-52) was already 
configured to carry similar-sized rockets (X-15), thus saving 
development costs.  Pegasus' big brother, Taurus, will be ground-launched.

There is, to be sure, one extra problem with mountains -- they aren't
located on the East coast shoreline where (for safety reasons) most
US orbital launches must fly from.  On the other hand, would _you_ want
to build your tall tower right in Hurricane Alley?  

All that being said, you are correct that a tower tall enough to 
reduce/eliminate drag losses and reduce gravity losses is feasible.
There is an additional benefit, btw., which is that you get to design
your engines and vehicle aerodynamics to work only at high altitude; 
you don't need translating nozzle skirts, aerothermal design, etc. for
sea level operation.  

>A launch tower would be less loaded per unit length than a skyscraper
>of the same height because it would not need to support all the
>extraneous mass of furniture, floors, fixtures, etc 

This is a comparatively minor factor.  Mile-high skyscrapers have been
designed.  The problems, I gather, are less structural than some subtler
ones, for instance:
	Elevators (An adequate set takes the entire building cross section)
	Fire exits (How do you evacuate a 500-story building???)
	Water (You can't pump water a mile high; you have to have
		multiple stages of pumps with "reservoirs" every few
		dozen floors.  Bear that in mind when designing a 
		launch tower -- where do you get deluge/firefighting water??)

>and could also
>be built with a taper (it would get narrower with height, this
>would be advantageous in resisting wind too), this would allow towers
>of very large heights.
>
>My question, does anyone out there more versed in mechanical engineering
>and strength of materials know how deeply this has been looked into?
>
>Mark


	I know Dani Eder has looked into it in some detail; I have also
looked at tall towers as a way to put a launch laser above (most of) the
atmosphere and increase its useful range).  The biggest design issue
is probably wind loading on the first 5-10 km -- dynamic loads are much
harder to design for than static loads (look up Tacoma Narrows Bridge....)
The biggest practical issue is probably safety -- what if it falls down?
These days, even building it in the middle of the desert wouldn't
get you away from safety codes, insurance issues, environmental 
impact statements, etc., etc.  There are practical difficulties
in construction (How do you lift building materials to the top of
a half-completed 10 km tower?  They don't make 5 km crane cables....)
but they're probably not insurmountable.

	I almost got a 4 km-tall tower structure / laser launch
facility onto the cover of an SF book a couple of years back (The 
Quiet Pools, by Michael Kube McDowell -- had laser propulsion as
a key part of the plot, and Michael used me as his consultant) but
the artist and publisher chickened out and used an old-fashioned
rocket launch as the backdrop instead....

	BTW, the single easiest way to pay for construction of 
such a tall tower is probably to put a revolving restaurant on top :-)

		Jordin (acrophobia?  What's that?) Kare