spaceship one

On 24 Jun 2004 01:35:58 GMT, (Regnirps) wrote:

(Steve VanSickle)

This particular *design* won't work, yes, but why not the "method" (i.e.
moving surfaces to make for a "hands off" reentry)?

I suppose it can be dome somehow, but you are talking 18,000 mph instead of
3,000. If there is a way to skim along and slowly loose energy I'd love to find
it. But as it now stands, as you lose energy you start to drop into more
atmosphere and more drag and loose it faster and drop faster and more heat
and....

Anyway, an ninformed quick calculation. Kinetic energy is proportional to the
square of the velocity. So, 18,000 mph is six times faster than 3,000 mph but
you will have 36 times as much kinetic energy, which will become heat (mostly I
think).

-- Charlie Springer

You've captured a large portion of the problem.

First let me say that Hypersonics and Reentry are not my field, so
what I say is necessarily of a general nature and may miss the
details.

The higher you go the more potential energy you've got to get rid of
in re-entry. As you move from sub-orbital through low earth orbit
(Shuttle) to high earth orbit to lunar return and beyond you end up
with more and more energy to get rid of. I believe it's significantly
worse than just the additional potential energy because of increasing
orbital velocity. I defer to our resident orbit wonk on the point,
but I seem to remember that the higher your circular orbit the higher
your orbital velocity. I don't even want to think about elliptic
orbits, they make my head hurt. But the higher you go you gain
kinetic energy as well as additional potential energy. All of this
has to go to heat if you want to end up at zero velocity by the time
you reach the dirt.

Reentry trajectories are often depicted on velocity-altitude plots.
These have velocity across the x-axis and altitude going up. A
vehicle arrives at the upper edge of the atmosphere with more velocity
the higher it comes from. so it comes in the top of the plot at a
greater entry speed. If you plot the trajectories we have data for
you see that Mercury and Gemini and Apollo move progressively to the
right on the plots. They begin with more and more energy. Apollo is
the case we have with the highest entry speed and the worst
atmospheric heating problem.

As you move to the right (move from sub orbital, to higher orbits,
etc) you move into regions where the thermochemistry get more and more
difficult. As you move to the right you enter regions where Oxygen
disassociates, then Nitrogen disassociates, then ionization occurs.
SS1's Mach 3 peak velocity avoids nearly all of that. At Mach 3 the
regime is arguably not Hypersonic at all. (The boundary between
supersonic and hypersonic is not a well defined line.)

As you move faster than SS1 and into the Hypersonic regime you end up
being far more concerned with heating problems than aerodynamic ones.
As a case in point, when the shuttle burned up over Texas the peak
dynamic pressure was 75 lbs per square foot. Not much. All the
trouble came from the viscous heating at those Mach numbers.

One additional factor which bears on the shuttlecock concept is that
in the hypersonic regime the heating problem gets worse as the leading
radius of curvature gets smaller. This is why vehicles come back
blunt side forward. Never say never, but I suspect that unless they
are made of pure unobtaneum the leading edges of the shuttlecock's
"fletching" would burn right off once you got up past Mach 5 or so.

With enough ablative material you might be able to counteract this,
but as you move up in speed other techniques of attitude control will
begin to look more attractive. The ESA is planning a vehicle which
will test aerodynamic control of attitude. It's called Cheops or
something like that, and they were testing models of it in the Mach 6
tunnel while I was at von Karman last year.

--
David Munday -
Webpage: http://www.ase.uc.edu/~munday
"Adopt, Adapt, and Improve" -- Motto of the Round Table

On Thu, 24 Jun 2004 04:50:24 GMT, (David
Munday) wrote:

The higher you go the more potential energy you've got to get rid of
in re-entry. As you move from sub-orbital through low earth orbit
(Shuttle) to high earth orbit to lunar return and beyond you end up
with more and more energy to get rid of. I believe it's significantly
worse than just the additional potential energy because of increasing
orbital velocity. I defer to our resident orbit wonk on the point...

Hmmfff. Weren't referring to me, I hope. :-)

I ain't really an orbit type, though I've picked up a bit over the years.

Anyway, you do have it backwards...orbital velocity decreases with circular
orbit altitude. ~25,200 FPS at 200 nm, ~10,100 FPS at geosynchronous
altitude (~19320 NM).

You're right about the potential energy, though. Dropping from
geosynchronous altitude to ground level, you'll hit the atmosphere at over
23,000 miles per hour. And if you're an old-timer like BOb, you'll have
the turn-signal flashing the entire way....

Ron Wanttaja

Ron Wanttaja wrote:

Anyway, you do have it backwards...orbital velocity decreases with circular
orbit altitude. ~25,200 FPS at 200 nm, ~10,100 FPS at geosynchronous
altitude (~19320 NM).

You're right about the potential energy, though. Dropping from
geosynchronous altitude to ground level, you'll hit the atmosphere at over
23,000 miles per hour. And if you're an old-timer like BOb, you'll have
the turn-signal flashing the entire way....

Ron Wanttaja

So?
To catch up with the guy in front of you, you first slow down?

On Thu, 24 Jun 2004 08:33:29 GMT, Richard Lamb
wrote:

Ron Wanttaja wrote:

Anyway, you do have it backwards...orbital velocity decreases with circular
orbit altitude. ~25,200 FPS at 200 nm, ~10,100 FPS at geosynchronous
altitude (~19320 NM).

You're right about the potential energy, though. Dropping from
geosynchronous altitude to ground level, you'll hit the atmosphere at over
23,000 miles per hour. And if you're an old-timer like BOb, you'll have
the turn-signal flashing the entire way....

So?
To catch up with the guy in front of you, you first slow down?

Precisely. Think of it in terms of angular rate. If you're both in the
same 90-minute equatorial orbit, you're traveling around the Earth at an
angular rate of four degrees of longitude per minute. To catch up to the
guy in front, you need to increase your angular rate, e.g., an orbit with a
shorter period.

The angular rate is inversely proportional to orbit altitude...LEO (Low
Earth Orbit) satellites are at a few hundred miles and go around the Earth
in 90 minutes or so, but Geosynchronous satellites are at ~19300 NM and
take a full 24 hours to orbit the Earth. The velocities needed to maintain
the orbit are lower, and the circumference of the orbits is longer...you're
not only flying slower to start with, but the distance you have to travel
for one orbit is longer.

So... if your buddy is a few hundred miles ahead of you in LEO, you slow
up. This lowers your average orbit altitude and decreases your orbit
period. You've now got a period, say, of 88 minutes and an angular rate of
4.1 degrees a minute. Every minute, you're 0.1 degrees (roughly 6 nautical
miles, in LEO) closer. You're also slightly below your buddy, but when you
catch up, you increase your speed, which raises your orbit.

Mind you, if your friend had been only a hundred feet in front of you, you
just would have popped your little thrusters and flown directly to him. It
will perturb your orbit, but would be minor compared to normal orbit
maintenance maneuvers. Compare it to having to climb 10 feet in an
airplane vs. 10,000 feet. You'll just tug the stick back slightly for the
first case, but add power for the second.

Otherwise, though, orbit mechanics is *definitely* non-intuitive to an
aircraft pilot. In an airplane, we're always concerned about how far we can
fly, and can easily change directions if we desire. The situation is
exactly opposite in orbit... the vehicle has nearly unlimited range, but a
change in compass course is prohibitively expensive.

Ron Wanttaja

On Thu, 24 Jun 2004 05:28:59 GMT, Ron Wanttaja
wrote:

Anyway, you do have it backwards...orbital velocity decreases with circular
orbit altitude. ~25,200 FPS at 200 nm, ~10,100 FPS at geosynchronous
altitude (~19320 NM).

You're right about the potential energy, though. Dropping from
geosynchronous altitude to ground level, you'll hit the atmosphere at over
23,000 miles per hour. And if you're an old-timer like BOb, you'll have
the turn-signal flashing the entire way....

Thanks. I've had an aversion to orbits ever since an undergraduate
dynamics final exam question: "There is an object over the pole. It's
polar coordinates and velocity are _______. Should we launch a
counterstrike?" A question from a world which is now mostly gone.
RWR, RIP.

An example of the velocity-altitude plot I mentioned is at:
http://www.ase.uc.edu/~munday/pics/trajectories.ppt

You can see Mach 3 is a long way from the STS (Shuttle) LEO return.

I assume AOTV is that winged orbit transfer trick you refered to, Ron.

--
David Munday -
Webpage: http://www.ase.uc.edu/~munday
"Adopt, Adapt, and Improve" -- Motto of the Round Table

Ron Wanttaja wrote:

You're right about the potential energy, though. Dropping from
geosynchronous altitude to ground level, you'll hit the atmosphere at over
23,000 miles per hour. And if you're an old-timer like BOb, you'll have
the turn-signal flashing the entire way....

Ron Wanttaja

Hey Ron, help me out some more here on
rec.aviation.homebuilt.spacecraft.

For the reentry phase from orbit...
For the sake of argument (and ignoring the increased fuel required)
wouldn't slowing down too much before reentry be a problem?

Steeper path, higher G load, and even more reentry heat?

Richard (air breathing, gravity bound) Lamb

On Fri, 25 Jun 2004 01:03:41 GMT, Richard Lamb
wrote:

Hey Ron, help me out some more here on
rec.aviation.homebuilt.spacecraft.

For the reentry phase from orbit...
For the sake of argument (and ignoring the increased fuel required)
wouldn't slowing down too much before reentry be a problem?

Steeper path, higher G load, and even more reentry heat?

Like I said on an earlier post, I don't have much background on re-entry
physics. But I think it's possible to deorbit going slowly at a fairly
shallow angle...you just have to time the deorbit burn properly.

But one thing you can't do is "slow fly" a satellite. For any given speed,
for any given velocity vector, there is only one possible orbit. Sure, you
can probably increase your angle of attack and do a "skip", but that just
means that on the other side of the world, you're going to come down at a
much steeper angle. Kinda like bouncing a landing without the ability to
add a burst of power to catch the bounce.

Ron Wanttaja

On Fri, 25 Jun 2004 03:41:16 +0000, Ron Wanttaja wrote:

On Fri, 25 Jun 2004 01:03:41 GMT, Richard Lamb
wrote:

Hey Ron, help me out some more here on
rec.aviation.homebuilt.spacecraft.

For the reentry phase from orbit...
For the sake of argument (and ignoring the increased fuel required)
wouldn't slowing down too much before reentry be a problem?

Steeper path, higher G load, and even more reentry heat?

Like I said on an earlier post, I don't have much background on re-entry
physics. But I think it's possible to deorbit going slowly at a fairly
shallow angle...you just have to time the deorbit burn properly.

But one thing you can't do is "slow fly" a satellite. For any given speed,
for any given velocity vector, there is only one possible orbit. Sure, you
can probably increase your angle of attack and do a "skip", but that just
means that on the other side of the world, you're going to come down at a
much steeper angle. Kinda like bouncing a landing without the ability to
add a burst of power to catch the bounce.

I'm having one of those moments...

I had always wondered why you couldn't dolphin in and out of the earth's
atmosphere, cooling down in between hops.

AC

On Fri, 25 Jun 2004 05:41:36 +0100, anonymous coward
wrote:

But one thing you can't do is "slow fly" a satellite. For any given speed,
for any given velocity vector, there is only one possible orbit. Sure, you
can probably increase your angle of attack and do a "skip", but that just
means that on the other side of the world, you're going to come down at a
much steeper angle. Kinda like bouncing a landing without the ability to
add a burst of power to catch the bounce.

I'm having one of those moments...

I had always wondered why you couldn't dolphin in and out of the earth's
atmosphere, cooling down in between hops.

I think you'd re-enter at steeper and steeper angles each time, since you
lose velocity at each encounter with the atmosphere. I suspect, at some
point, you can't "pull out" and may break up due to the overly steep
re-entry.

Just a guess, mind you.

Ron Wanttaja

Ron Wanttaja wrote:

Like I said on an earlier post, I don't have much background on re-entry
physics. But I think it's possible to deorbit going slowly at a fairly
shallow angle...you just have to time the deorbit burn properly.


Me either, Ron. And I don't think my Holiday Inn Express line is gonna
work on this one, either.

I thought I'd try to see if I could at least set the problem up.
Hey, it's only calculus, right? But I found nothing there I could get
hold of except the (obvious) arrogance of ignorance. Humbling...


For a body in motion, the first derivative gives the rate of change
in position per unit time (i.e.: speed = rate of change of position
per second)

If the object is accelerating, the second derivative gives the rate
at which the first derivative is changing, or the rate of change in
speed (acceleration = rate of change in position per second per second).

Third derivative gives the rate of change in acceleration (what the
physics guys call 'jerk' = rate of change of position per second per
second per second). Like the way an old car jerks if there is too
sudden a change is how it is accelerating.

Quoting Martin Gardner, "Beyond the third, higher order derivatives are
seldom needed. This testifies to the fortunate fact that the universe
seems to favor simplicity in it's fundamental laws".

BUT
Simplicity is relative.

On orbit, our ship is in steady state unaccellerated motion, right?
Well, not exactly.

Due to the curved path of the orbit there is an 'outward' centrifugal
force that is exactly opposed by the opposite 'inward' centripetal force
(of gravity).

So our steady state 'unaccelerated' motion is actually a _second_
derivative from the straight line path (ASSUMING the orbit path is
perfectly circular?).

therefore

Adding an acceleration to our _forward_ motion (second and third
derivatives) causes an immediate third derivative reaction of the
orbital
altitude, i.e.: motion inward (if slowing) or outward (if speeding up).

If I'm not too badly mistaken, we are up to the SIXTH derivative,
and still haven't accounted for any deviation that would result if the
acceleration vector is not EXACTLY aligned with the true orbital path
in both pitch and yaw.

Taking those into account, we are looking at the TWELFTH derivative
just to predict what's going to happen when we try to change speed.

If we are off in pitch, I think the end result would be an oscillation
in the the orbital path. Think about an AC electrical signal imposed on
a DC carrier.

If we were thrusting straight 'outward', the thrust pushes us to a highe
altitude that our orbital velocity will not be able to maintain.

As soon as the thrust is removed, and momentum decays, we will drop back
down, gaining inward momentum on the way, which will cause an
'undershoot'
of our previous altitude, which will again bleed off momentum until we
go back 'up', and over shoot again.

There is probably going to be a fairly strong damping effect that will
eventually (sota) stabilize at the original altitude, but I haven't a
clue
how to set THAT one up...



Sheesh! Rocket Science....