Burt Rutans plans for a manned mission to Mars

Fred J. McCall wrote:
"Alex Terrell" wrote:

:
:Fred J. McCall wrote:
: "Alex Terrell" wrote:
:
: :Who mentioned a space elevator?
: :
: :Steve was talking about a rotovator, which could mass as little as 200
: :tons, of which 3/4 is anchor mass.
:
: Show your work. How many ounces are you lifting per cycle?
:
:Not my work - the formulae are exceedingly complex, but check slide 6
f this
:
:http://www.niac.usra.edu/files/libra...9/355Bogar.pdf
:
r this for more detail:
:
:http://www.tethers.com/papers/HASTOLAIAAPaper.pdf
:
:Best to skip the bit about the plane.
:
:With a material twice as strong a Spectra, system mass is 131 times
ayload mass, of which 120 is anchor mass. The anchor mass would make a
:good use for old rocket stages.

So all we need is a hypersonic cargo plane that we can't build, a
material with a tensile strength twice what we have, and a willingness
to transfer 1 ton cargoes during Mach 12 rendezvous at 300,000 feet.

You said: "If we can lift enough mass to put up a space elevator we
don't need a
space elevator. "

I just pointed out that no one in this thread was propsing an elevator,
rather a rotovator which could mass just a few hundred tons.

Reaching Mach 12 is quite feasible, though I think a conventional
rocket is a nearer term solution.


Yeah, I see that happening in the next decade....

This is another one of those things. If we can build a hypersonic
transport with those capabilities, why not just push a little harder
and take it all the way to orbit? That has to be easier and cheaper
than all this other stuff (not to mention safer).

OK, now I see you don't know much about orbital mechanics, which
explains your earlier comments.

To reach orbit, you need to reach ~Mach 25. This requires more than 4
times the energy required to reach Mach 12.

Even more critically, plugging the numbers into the rocket equation,
assuming Isp = 350s, and V = 4,500m/s for rotovator rendez-vous, and
9,000m/s for orbit (because of gravity losses and friction losses),
gives a dry mass fraction of 26.9% and 7.3% respectively.

Assuming the rocket masses about 5%, then the cargo is about 22% and
2.3% respectively.

In other words, its about 10 times easier for a rocket to reach Mach 12
as it is to reach orbit.

So at some point, a rotovator will make a lot of sense.

Fred J. McCall wrote:

"Alex Terrell" wrote:

:
:Fred J. McCall wrote:
: "Alex Terrell" wrote:
:
: :Who mentioned a space elevator?
: :
: :Steve was talking about a rotovator, which could mass as little as 200
: :tons, of which 3/4 is anchor mass.
:
: Show your work. How many ounces are you lifting per cycle?
:
:Not my work - the formulae are exceedingly complex, but check slide 6
f this
:
:http://www.niac.usra.edu/files/libra...9/355Bogar.pdf
:
r this for more detail:
:
:http://www.tethers.com/papers/HASTOLAIAAPaper.pdf
:
:Best to skip the bit about the plane.
:
:With a material twice as strong a Spectra, system mass is 131 times
ayload mass, of which 120 is anchor mass. The anchor mass would make a
:good use for old rocket stages.

So all we need is a hypersonic cargo plane that we can't build, a
material with a tensile strength twice what we have, and a willingness
to transfer 1 ton cargoes during Mach 12 rendezvous at 300,000 feet.

What's insurmountable about a craft that can attain 100 km altitude and
mach 12?

During rendezvous the plane will be going nearly the same velocity as
the tether tip. Hell, my computer keyboard and I are both moving mach 88
around the sun. Doesn't make it harder to type.

(Although I admit the the formula for airplane velocity perplexes me:
Vo-Vt-470 m/s. Why the 470 meters/sec? Also I can't see how Vt is
calculated)


Yeah, I see that happening in the next decade....

This is another one of those things. If we can build a hypersonic
transport with those capabilities, why not just push a little harder
and take it all the way to orbit?

Let me get this straight. You're going mach 12 and you ask "why not push
a little harder and take it all the way to orbit?"

Hop

Hop David wrote:


(Although I admit the the formula for airplane velocity perplexes me:
Vo-Vt-470 m/s. Why the 470 meters/sec? Also I can't see how Vt is
calculated)

I assume 470m/s is the speed of the Earth at the equator.

Vt is simply the tether tip speed, relative to the centre of mass. The
peak launch velcoity of the cargo is therefore Vo+Vt, which tends to
put it into a very highly elliptical orbit.

wrote:
On 12 May 2006 12:28:53 -0700, "Alex Terrell"
wrote:

An interesting point about rotovators is that they're not good for
reaching their orbit. They're very good at reaching higher orbits.

Reaching their own orbit takes a little doing, but is possible.

Can you please describe how, or provide a reference?

My own view is that with a rotovator, LEO would be pretty much
abandoned.

Alex Terrell wrote:

Hop David wrote:


(Although I admit the the formula for airplane velocity perplexes me:
Vo-Vt-470 m/s. Why the 470 meters/sec? Also I can't see how Vt is
calculated)


I assume 470m/s is the speed of the Earth at the equator.

Makes sense. Should've thought of that.


Vt is simply the tether tip speed, relative to the centre of mass. The
peak launch velcoity of the cargo is therefore Vo+Vt, which tends to
put it into a very highly elliptical orbit.


I get an orbital period of about 6000 seconds. Looking at the
illustration, seems like the tether turns two full revolutions each
orbit with it's cg in the middle of the tether.

A tether length of 600 km gives a radius of 300 km

4*pi*300 km / 6000 secs is about .6 km/sec or 600 meters/sec.

The chart gives 3006 meters/sec.

Hop

Hop David wrote:
Alex Terrell wrote:

Hop David wrote:


(Although I admit the the formula for airplane velocity perplexes me:
Vo-Vt-470 m/s. Why the 470 meters/sec? Also I can't see how Vt is
calculated)


I assume 470m/s is the speed of the Earth at the equator.

Makes sense. Should've thought of that.


Vt is simply the tether tip speed, relative to the centre of mass. The
peak launch velcoity of the cargo is therefore Vo+Vt, which tends to
put it into a very highly elliptical orbit.


I get an orbital period of about 6000 seconds. Looking at the
illustration, seems like the tether turns two full revolutions each
orbit with it's cg in the middle of the tether.

A tether length of 600 km gives a radius of 300 km

4*pi*300 km / 6000 secs is about .6 km/sec or 600 meters/sec.

The chart gives 3006 meters/sec.

There are a lot of different parameters (constraints) to play with,
including the release velocity. Bear in mind the tether has a large
anchor mass, so it rotates about a C of G which is near the anchor
mass.

I'm not sure how many rotations it does per orbit. Other constraints
for reaching lunar orbit (or L4 / L5 / L1 include:
- The device should orbit in the plane of the moons orbit, and not
around the equator.
- The release point, and hence capture point, are determined by the
position of the moon.

Hence I prefer the rototvator concept to work with an air launched
rocket - say dropped from a 747, or "Custom Very Large Aircraft"

The configuration I had in mind was:

"The Rotovator orbits 7,160 km from the Earth's centre, or at an
altitude of 783 km. The tether length is 663 km, and the tip velocity
is 2,549m/s.

A rocket travelling at an altitude of 120 km, and at 4,452 m/s relative
to the Earth's surface is able to rendezvous with the tip and
transfer its cargo to the end of the tether. The cargo is then
accelerated (at about 1g) by the tether system as it is rotated about
the centre of mass. Some 14 minutes later, the cargo is released at an
altitude 1,413 km, with a velocity of 10 km/s. Following release, the
cargo travels to an apogee of some 360,000 km. At that point, an
additional boost of 900m/s will enable the cargo to arrive at the Earth
- Moon L1 point. The cargo could then be used at the L1 point, or
transported to the lunar surface on a lunar tug."

On 13 May 2006 13:34:46 -0700, "Alex Terrell"
wrote:

wrote:
On 12 May 2006 12:28:53 -0700, "Alex Terrell"
wrote:

An interesting point about rotovators is that they're not good for
reaching their orbit. They're very good at reaching higher orbits.

Reaching their own orbit takes a little doing, but is possible.

Can you please describe how, or provide a reference?

Sorry, I don't have a reference, but ISTR it involved "rolling up" the
tether over a certain interval prior to release of the payload.

My own view is that with a rotovator, LEO would be pretty much
abandoned.

It's possible. LEO is not that great a place anyway. Too much junk
and air drag. There's a lot of useful space below the Van Allen
belts.

-- Roy L

wrote:

:On Fri, 12 May 2006 14:00:36 GMT, Fred J. McCall
wrote:
:
wrote:
:
::On Fri, 12 May 2006 03:34:02 GMT, Fred J. McCall
wrote:
::
::If we can lift enough mass to put up a space elevator we don't need a
::space elevator.
::
::?? ROTFL!! "If we can send enough men and supplies and equipment and
::rails and crossties all the way across the continent to build a
::railroad, we don't need a railroad."
:
:Cute, but hardly the same thing.
:
:Calculate the mass of your space elevator. Now figure out what it
:costs to put up at current launch costs.
:
:Why? I have already described the bootstrap process that refutes that
:arugment.

Unfortunately for you, I don't buy into arguments based on handwaving,
pixie dust, and wishing real hard.

:If launch costs come down enough for a space elevator to be
:economically practical, launch costs are so low that the elevator is
:redundant.
:
:It's the elevator itself that brings them down, by building itself. I
:have already informed you of that. Can't you read?

And so you are postulating a large, sophisticated manufacturing plant
up there churning out stuff that's stronger than anything we can
currently make out materials that somehow magically get there (or you
are postulating a hell of a space population already existing, in
which case the elevator is probably unnecessary.

After all, if you can manufacture such a thing starting in space, what
the hell do you need to bring up from Earth?

--
"Some people get lost in thought because it's such unfamiliar
territory."
--G. Behn

"Alex Terrell" wrote:

:Fred J. McCall wrote:
: "Alex Terrell" wrote:
:
: :
: :Fred J. McCall wrote:
: : "Alex Terrell" wrote:
: :
: : :Who mentioned a space elevator?
: : :
: : :Steve was talking about a rotovator, which could mass as little as 200
: : :tons, of which 3/4 is anchor mass.
: :
: : Show your work. How many ounces are you lifting per cycle?
: :
: :Not my work - the formulae are exceedingly complex, but check slide 6
: f this
: :
: :http://www.niac.usra.edu/files/libra...9/355Bogar.pdf
: :
: r this for more detail:
: :
: :http://www.tethers.com/papers/HASTOLAIAAPaper.pdf
: :
: :Best to skip the bit about the plane.
: :
: :With a material twice as strong a Spectra, system mass is 131 times
: ayload mass, of which 120 is anchor mass. The anchor mass would make a
: :good use for old rocket stages.
:
: So all we need is a hypersonic cargo plane that we can't build, a
: material with a tensile strength twice what we have, and a willingness
: to transfer 1 ton cargoes during Mach 12 rendezvous at 300,000 feet.
:
:You said: "If we can lift enough mass to put up a space elevator we
:don't need a
:space elevator. "
:
:I just pointed out that no one in this thread was propsing an elevator,
:rather a rotovator which could mass just a few hundred tons.
:
:Reaching Mach 12 is quite feasible, though I think a conventional
:rocket is a nearer term solution.

All sorts of things are "quite feasible". We just can't DO them right
now.

: Yeah, I see that happening in the next decade....
:
: This is another one of those things. If we can build a hypersonic
: transport with those capabilities, why not just push a little harder
: and take it all the way to orbit? That has to be easier and cheaper
: than all this other stuff (not to mention safer).
:
:OK, now I see you don't know much about orbital mechanics, which
:explains your earlier comments.

No doubt you see all sorts of silly **** that isn't true.

:To reach orbit, you need to reach ~Mach 25. This requires more than 4
:times the energy required to reach Mach 12.

Quite right, but it's still probably easier than a Mach 12 rendezvous
in atmosphere and a cargo transfer. This isn't like being in orbit
where you can creep up on the other object. The tip of your rotovator
isn't holding a nice steady speed and course.

:Even more critically, plugging the numbers into the rocket equation,
:assuming Isp = 350s, and V = 4,500m/s for rotovator rendez-vous, and
:9,000m/s for orbit (because of gravity losses and friction losses),
:gives a dry mass fraction of 26.9% and 7.3% respectively.
:
:Assuming the rocket masses about 5%, then the cargo is about 22% and
:2.3% respectively.
:
:In other words, its about 10 times easier for a rocket to reach Mach 12
:as it is to reach orbit.

And about 1,000,000 times harder to do a Mach 12 rendezvous in
atmosphere with the tip of your rotovator and actually transfer cargo
than it is to get the rest of the way to orbit. That's disregarding
the requirement for unobtainium to build your rotovator out of.

There's more to the problem than mere energy required. The energy is
the CHEAP part. That's fortunate for your scheme, since you're going
to have to keep reboosting your rotovator, as well.

:So at some point, a rotovator will make a lot of sense.

I'm not holding my breath.

--
"The reasonable man adapts himself to the world; the unreasonable
man persists in trying to adapt the world to himself. Therefore,
all progress depends on the unreasonable man."
--George Bernard Shaw

wrote:
On 13 May 2006 13:34:46 -0700, "Alex Terrell"
wrote:

wrote:
On 12 May 2006 12:28:53 -0700, "Alex Terrell"
wrote:

An interesting point about rotovators is that they're not good for
reaching their orbit. They're very good at reaching higher orbits.

Reaching their own orbit takes a little doing, but is possible.

Can you please describe how, or provide a reference?

Sorry, I don't have a reference, but ISTR it involved "rolling up" the
tether over a certain interval prior to release of the payload.

This takes time, and gives problems with angular momentum, and requires
a reeling mechanism.

Once a tether is up, it makes sense use it as much as possible, and
rolling up makes that difficult.

If the tether use electric rather than electrodynamic thrust, then it
would probably be easier to bring the supplies in with an electric
propulsion device from a highly elliptical orbit.

My own view is that with a rotovator, LEO would be pretty much
abandoned.

It's possible. LEO is not that great a place anyway. Too much junk
and air drag. There's a lot of useful space below the Van Allen
belts.

There's even more above the Van Allen belts.