Sub-Orbital Earth Transits

On Dec 10, 10:51 pm, David Spain wrote:

When does the cabin start to experience zero 0? Is it as soon as the asce
nt
rocket cuts out or at apogee and the return leg until you hit the atmosph
ere?

Engine cutoff

On Dec 10, 10:51 pm, David Spain wrote:
"Jonathan Thornburg [remove -animal to reply]" -zebr
a.edu writes:

How much rocket fuel does it take to maintain 0.5g acceleration for
3 hours?

When does the cabin start to experience zero 0? Is it as soon as the asce
nt
rocket cuts out or at apogee and the return leg until you hit the atmosph
ere?

Ignoring cabin zero 0 for the moment, and say you're only trying to reduc
e
descent velocity (again relative to your airpseed), wouldn't you only nee
d to
maintain .5g during the descent leg, starting at apogee, 'airspeed' = 0
, which
does not exceed 1.5 hours?

By your prior estimate it would seem more like:

exp(1.5 hours * 3600 seconds/hour / 800 seconds) = exp(6.75) = 854.

That would yield, by your own estimate, 854 kilograms of fuel per kilogra
m of
structure + payload, correct?


Which still weighs more than TPS. The whole basis premise of your
idea is not viable.
The reason that the shuttle, Soyuz and all entry vehicles use the
atmosphere to reduce speed is because it is more efficient than using
propellant.

This is the line in your proposal where if goes wrong:
"glider to come
in far too fast for a safe rentry without a significant TPS that
would add so much to the weight and cost as to make it largely
uneconomical."

Using propellant to reduce speed would be worse

I asked
| How much rocket fuel does it take to maintain 0.5g acceleration for
| 3 hours?
and then calculated it to be a (very) impractically large amount.

David Spain asked
When does the cabin start to experience zero 0? Is it as soon as the ascent
rocket cuts out or at apogee and the return leg until you hit the atmosphere?

Free-fall (a.k.a. "zero g" starts as soon as the ascent rocket cuts out
(assuming this is above the atmosphere, which is the case for any practical
rocket & trajectory), and it lasts until you hit the atmosphere again on
the way back down.


Ignoring cabin zero 0 for the moment, and say you're only trying to reduce
descent velocity (again relative to your airpseed), wouldn't you only need to
maintain .5g during the descent leg, starting at apogee, 'airspeed' = 0, which
does not exceed 1.5 hours?

By your prior estimate it would seem more like:

exp(1.5 hours * 3600 seconds/hour / 800 seconds) = exp(6.75) = 854.

That would yield, by your own estimate, 854 kilograms of fuel per kilogram of
structure + payload, correct?

Yes, a 100%-fuel-fraction rocket with a specific impulse of 400 seconds
which maintains 0.5g acceleration for 1.5 hours has a mass ratio of
exp(6.75) = 850, i.e., the rocket needs 850 kg fuel per kilogram of
structure + payload.

[But that wouldn't solve the original poster's problem
of making it "more comfortable for the passengers of such
a 'business-class' flight to avoid zero-g altogether".]

However, due to the differing accelerations-with-respect-to-the-Earth,
a trajectory which is in free-fall going up and accelerating-with-respect-
-to-free-fall going down, isn't going to spend equal times in the up and
down parts. Actually it's going to spend more time in the down segment.

Let's see....
height = 1/2 * acceleration * time^2
and this has to match for the up and down parts, so a couple of lines
of algebra gives
a_up/a_down = (t_down/t_up)^2
and hence (with a_up/a_down = 2) with a couple more lines of algebra
we get t_up = 75 minutes and t_down = 105 minutes.

Still assuming a flat Earth and no altitude variation to the Earth's
gravitational field, the peak altitude is now a mere 98,000 km (!) and
the re-entry velocity a mere 31 km/second (!). [Again, these numbers
are bogus, because our assumption of no altitude variation to the Earth's
gravitational field is very wrong.]

Since t_down = 105 minutes, the mass ratio to maintain 0.5g for that
time (still for a 100%-fuel-fraction rocket with a specific impulse of
400 seconds) comes out as exp(7.9) = around 2700:1 for the "down" rocket
burn.


Because we're dealing with an exponential It could be even less assuming the
..5g burn is not required for the full 1.5 hours on the return leg, esp. after
you again reach the atmosphere and can return to being a glider. I'm not
saying its practical, but it seems we could do better than your original 700
metric tonnes estimate.

Yes, that would help things a bit. But it would probably still be
seriously impractical.

--
-- "Jonathan Thornburg [remove -animal to reply]"
Dept of Astronomy, Indiana University, Bloomington, Indiana, USA
"The first strike in the American Colonies was in 1776 in Philadelphia,
when [...] carpenters demanded a 72-hour week." -- Anatole Beck

David Spain wrote:

Not hover, migtigate. .5g will not keep you from re-entry, only
slow your retry airspeed. But the fuel fraction to weight ratio
might be way too high, I'm not an expert here, that's why I am
asking the question.

David,

1. .5g is still much too high a thrust level to be maintained for
hours waiting on the earth to rotate beneath you. Propellant
consumption is still prohibitive.

2. The mitigation is grossly insufficient. Thrust for .5g still means
you're accelerating downward a 4.9 m/s2. At that acceleration you'll
cover the first 100 km in 3.36 minutes. The next 100 km will be
covered in 1.4 minutes. Your flight time is measured in minutes, not
hours, unless you're very nearly hovering.

Sorry,

Jim Davis

I (Alain Fournier) wrote:

So at about a height of 100 km you will have a
deceleration of 1 g and a speed of 5.6 km/s. One second later, you
went down 5.6 km, the atmosphere is twice as dense and you now
have a 2 g deceleration. Another second, the atmosphere is again
twice as dense and you have a 4 g deceleration. A third second
and you have an 8 g deceleration. You are now going at about
5.5 km/s, so it now takes about 1.02 seconds to double the pressure
and the deceleration. So after 4.02 seconds of re-entry you now
have a deceleration of 16 g

I made a mistake when I wrote that. Because of the slower speed
you need to more than double the atmospheric pressure to double
the deceleration. The atmospheric drag is roughly proportional to
the cube of the speed. So to compensate the speed decrease from
5.6 km/s to 5.5 km/s, you need to increase the air density by
(5.6/5.5)^3 = 1,056. This will take you an extra 0.08 seconds.
So it would be only after 4.1 seconds that you have a deceleration
of 16 g. The difference isn't important in this case, but if
you try to do a similar calculation with a slower entry speed,
using the erroneous calculations I made above can lead to thinking
that the atmospheric entry is not survivable when it is. But
in any case for a vertical atmospheric entry to be practical
you need much slower entry speeds.


Alain Fournier

I want to thank Alain Fournier, Charlie Murphy, Jonathon Thornburg,
and Jim Davis for their helpful insights.

I would agree that it seems mitigation via ullage rocketry is impractical
given current technology.

As was also pointed out, you have to overcome the horizontal velocity
component as well, which means expending fuel to reduce it. It is
more efficient, if using a more traditional trajectory, to fly
eastward and use your fuel to add to the horizontal component, esp.
since you can 'fall forward' to dramatically increase your ground
speed as the space shuttle does today. However, that also adds so
dramatically to your airspeed as to require a TPS, something I was
trying to avoid by zeroing the horizontal velocity component by
flying anti-spin-ward (westward) and only having to deal in the
vertical, using Earth's rotation to handle the ground speed component.

To alter the discussion somewhat (hope that's ok mods), let's shift
the focus to recovery using a TPS. Once you've achieved a measure
of speed reduction in shuttlecock mode, perhaps you could deploy
either a parachute or perhaps a tethered balloon with a TP skin to
slow further to allow a return to glider configuration.

I think recovery options are somewhat orthogonal, if you assume a TPS,
to the discussion of whether or not it is practical to fly the anti-
spin-ward trajectory but I'm up for further discussion of that.

Dave

David Spain wrote:
I want to thank Alain Fournier, Charlie Murphy, Jonathon Thornburg,
and Jim Davis for their helpful insights.

I would agree that it seems mitigation via ullage rocketry is impractical
given current technology.

As was also pointed out, you have to overcome the horizontal velocity
component as well, which means expending fuel to reduce it. It is
more efficient, if using a more traditional trajectory, to fly
eastward and use your fuel to add to the horizontal component, esp.
since you can 'fall forward' to dramatically increase your ground
speed as the space shuttle does today. However, that also adds so
dramatically to your airspeed as to require a TPS, something I was
trying to avoid by zeroing the horizontal velocity component by
flying anti-spin-ward (westward) and only having to deal in the
vertical, using Earth's rotation to handle the ground speed component.

David, I think you're still not quite getting it.

The Earth's atmosphere, to first approximation, rotates with the Earth.
Therefore thrusting eastward to add to your inertial velocity does *not*
increase your velocity *relative to the atmosphere* any more than
thrusting westward to cancel your inertial velocity; therefore, the TPS
requirements are no different.

David Spain wrote:

To alter the discussion somewhat (hope that's ok mods), let's shift
the focus to recovery using a TPS. Once you've achieved a measure
of speed reduction in shuttlecock mode, perhaps you could deploy
either a parachute or perhaps a tethered balloon with a TP skin to
slow further to allow a return to glider configuration.

If I understand well what you are saying here, you want to deploy
a parachute at hypersonic speeds as a way to lower the the heat
load during atmospheric entry. I think it might be workable, but
it isn't necessarily easy and straightforward. It can't be a nylon
parachute, that would burn in an instant. But you could make it out
of some kind of asbestos fabric or some other heat resistant
material. This parachute wouldn't be very different from a heat
shield, except that you have the additional constraint that it
must be foldable. On the other hand, you do have the advantage
of increasing the surface area without increasing mass all that
much. So the idea might work, but the devil might be in the details.

In my opinion, using a tethered balloon instead of a parachute
just makes the thing more complicated, less tolerant to punctures
and rips without much advantage. The lift provided by the
balloon would be negligible compared to the drag it or the
parachute provides.


Alain Fournier

"Jorge R. Frank" writes:


David, I think you're still not quite getting it.

The Earth's atmosphere, to first approximation, rotates with the
Earth.

Yeah, I agree to think otherwise seems unreasonable,
read stupid if you prefer :-)...

I think what happened here, is that my friend and I got off
on a tangent, worrying about the difference between airspeed and
groundspeed. Aerodynamic heating is caused by airspeed, but as you
say, to first approximations, it doesn't really matter, they can
be considered nearly the same. If you cancel the horizontal component
the air will essentinally be speeding past you at nearly the same
speed (ignoring wind currents, etc) as the ground.

There really doesn't seem to be a good way to game the Scaled
Composites solution to enable hypersonic transport w/o going the
TPS route, unless you posit a rocket system powered by unobtainum,
or maybe a gravitic drive. At that point you might as well punt
and consider teleportation as well. :-)

Dave

Alain Fournier writes:

David Spain wrote:

To alter the discussion somewhat (hope that's ok mods), let's shift
the focus to recovery using a TPS. Once you've achieved a measure
of speed reduction in shuttlecock mode, perhaps you could deploy
either a parachute or perhaps a tethered balloon with a TP skin to
slow further to allow a return to glider configuration.

If I understand well what you are saying here, you want to deploy
a parachute at hypersonic speeds as a way to lower the the heat
load during atmospheric entry.

Not exactly. I was thinking more in terms of a speed brake in order
slow the vehicle enough to reduce aerodynamic forces enough to enable
re-deployment of the booms back to a glider configuration.

So the general idea would be, coat the high T surfaces with TP, to allow
the 'shuttlecock' configuration to be able to withstand the heating of
re-entry, maybe even deploying mechanical speed brake vanes during
descent. The idea being you're exchanging V for heat. Eventually when
your V is low enough (as is your T) you deploy the parachute (or perhaps
a para-sail) to slow further so that the booms can be re-deployed back
to the glider configuration, at which time you jettison the sail and
come down on your wings for a landing.

Dave