Airship to orbit -- isn't the buoyancy enough?

I'm interested in revisiting something I read here, posted by John
Schilling:

http://groups.google.com/groups?q=g:... spock.usc.edu

The quote that interests me is this one:

But, by the time the vehicle reaches an airspeed of a few hundred
meters per second, it had better be operating in a purely aerodynamic
mode, using the envelope as an inflatable lifting body to hold the
vehicle at the highest possible altitude.

The article then goes on to discuss the impossibility of the scheme in
terms of lift-to-drag ratios, as lots of other articles do.

Can someone explain to me why the buoyancy can be ignored in this
fashion? I do understand that the buoyancy helps less and less as we
get closer to vacuum, but it's not at all obvious to me that the
buoyancy helps so much less that it ought to be ignored. Does
buoyancy drop faster that much faster than lift and drag increase?

--
Patrick Doyle

"Patrick Doyle" wrote in message
om...
I'm interested in revisiting something I read here, posted by John
Schilling:


http://groups.google.com/groups?q=g:...6ck1%24s03%241
%40spock.usc.edu

The quote that interests me is this one:

But, by the time the vehicle reaches an airspeed of a few hundred
meters per second, it had better be operating in a purely
aerodynamic
mode, using the envelope as an inflatable lifting body to hold the
vehicle at the highest possible altitude.

The article then goes on to discuss the impossibility of the scheme in
terms of lift-to-drag ratios, as lots of other articles do.

Can someone explain to me why the buoyancy can be ignored in this
fashion? I do understand that the buoyancy helps less and less as we
get closer to vacuum, but it's not at all obvious to me that the
buoyancy helps so much less that it ought to be ignored. Does
buoyancy drop faster that much faster than lift and drag increase?

--
Patrick Doyle

Consider this extreme and grossly simplified case. An inflated cylinder
ten kilometers long with an arbitrary frontal area, say 100m^2, with a
drag coefficient of say 0.2, (hypersonic, ignoring skin friction).
Further assume 4.5km/s, if memory serves this is near the point of
maximum required thrust.

Drag = 0.5*density*drag coefficient*Area*velocity^2

Lift = density*Area*length (Assuming hydrogen weight negligible)

Lift/Drag = length/(0.5*drag coefficient*velocity^2) = 0.0049

For a practical system to work, L/D has to be more like a thousand. Any
practical system would be much worse than this example, hence no one
thinks this is even close to possible.

Drag can be mitigated by going higher in to less dense air, but then
lift has to come from elsewhere. My best guess is that they are getting
very high L/D by inversely charging the entire top and bottom skins so
as to create a very large dipole. This dipole might then levitate on
the earth's magnetic pole and/or interact with the ionosphere so as to
achieve the very high effective L/D. This would seem coherent with
available information.

Perhaps this charging is done directly using solar energy via the
photoelectric effect over the entire skin, (in the hundred megawatt
range), overcoming charge dissipation. A very thin fabric coating would
sort of act as a thin film solar cell, but without the wires/weight,
though some of the skin charge might also be tapped for power/thrust.

Pete.

Can someone explain to me why the buoyancy can be ignored in this
fashion?
--
Patrick Doyle

For a practical system to work, L/D has to be more like a thousand.
Any
practical system would be much worse than this example, hence no one
thinks this is even close to possible.
[/quote:ae5e1b6d43]

Bouyancy can't be ignored, and the relevance of the L/D ratio is not
as significant as you might think.

For an aircraft to fly, it's thrust-to-weight ratio must exceed its
drag-to-lift ratio (inverse of L/D). If you look at a vector plot of
these forces, you'll see that this is just saying that the sum of the
forces pushing the craft forward and up must be greater than the sum
of the forces pulling it down and backward.

For the JP Aerospace ATO, these four vector forces are best thought of
as effective sums, because not all of the forces on the craft are
external. So, when figuring the weight, for example, you're not
really interested in the massive weight alone. Instead, you're
interested in the sum of the massive weight, bouyancy, centripetal
force, updraft/downdraft drag from global thermal waves, etc. When
you compute the effective weight instead of the massive weight,
you'll see that - unlike a heavier-than-air craft - JP Aerospace's
ATO has an equilibrium position at which its effective weight is zero
and the vehicle floats, even with the engines off. At that position,
the effective Thrust/Weight ratio is infinity.

It turns out that it's not hard to get less than that.

At it's equilibrium point, spitting off the tail at the right angle
will make the ATO "fly". A favorable ratio of forces is not
impossible. The only question is whether it can be maintained all
the way up.

Admittedly, it would be tricky. There are several obvious critical
points during the ascent - the transition to compressible flow and
supersonic velocities, crossing the terminator, the mesopause, etc.
However, it is still perfectly conceivable that this idea could work.


Posted Via Usenet.com Premium Usenet Newsgroup Services
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com

lid (C M Edwards)
wrote in :



Can someone explain to me why the buoyancy can be ignored in this
fashion?
--
Patrick Doyle

For a practical system to work, L/D has to be more like a thousand.
Any
practical system would be much worse than this example, hence no one
thinks this is even close to possible.
[/quote:ae5e1b6d43]

Bouyancy can't be ignored, and the relevance of the L/D ratio is not
as significant as you might think.

For an aircraft to fly, it's thrust-to-weight ratio must exceed its
drag-to-lift ratio (inverse of L/D). If you look at a vector plot of
these forces, you'll see that this is just saying that the sum of the
forces pushing the craft forward and up must be greater than the sum
of the forces pulling it down and backward.

For the JP Aerospace ATO, these four vector forces are best thought of
as effective sums, because not all of the forces on the craft are
external. So, when figuring the weight, for example, you're not
really interested in the massive weight alone. Instead, you're
interested in the sum of the massive weight, bouyancy, centripetal
force, updraft/downdraft drag from global thermal waves, etc. When
you compute the effective weight instead of the massive weight,
you'll see that - unlike a heavier-than-air craft - JP Aerospace's
ATO has an equilibrium position at which its effective weight is zero
and the vehicle floats, even with the engines off. At that position,
the effective Thrust/Weight ratio is infinity.

It turns out that it's not hard to get less than that.

At it's equilibrium point, spitting off the tail at the right angle
will make the ATO "fly". A favorable ratio of forces is not
impossible. The only question is whether it can be maintained all
the way up.

Admittedly, it would be tricky. There are several obvious critical
points during the ascent - the transition to compressible flow and
supersonic velocities, crossing the terminator, the mesopause, etc.
However, it is still perfectly conceivable that this idea could work.


Posted Via Usenet.com Premium Usenet Newsgroup Services
----------------------------------------------------------
** SPEED ** RETENTION ** COMPLETION ** ANONYMITY **
----------------------------------------------------------
http://www.usenet.com


There's an interesting discussion about this, especially the probelms
with going up to orbital speed in the atmosphere at:
http://www.universetoday.com/forum/i...showtopic=3315

Basically the thought was, if you're still using lighter than air boyuncy
or aerodynamic effects to keep your altitude, there's going to be WAY too
much drag on a large object...something you would need a massive amount
of thrust to overcome, not to mention a massive strain on your structure.

Tom