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Horizontal take off/landing



 
 
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  #1  
Old October 23rd 15, 12:14 PM posted to sci.space.policy
Jeff Findley[_6_]
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Posts: 2,307
Default Horizontal take off/landing

In article om,
says...

On 2015-10-22 05:54, Jeff Findley wrote:

Which would not be at all practical for a reentry vehicle, which is one
of the scenarios being proposed.


With very this atmosphere, high up, couldn't a winged vehicle be able to
"fly" at very high speed without breaking up, and bleed speed slowly,
enabling it to gradually lower altitude ? (and with very thin
atmosphere, wouldn't heating be far less of a problem ?)


Nope. The dynamic pressure will be far too high due to the very high
reentry velocities.

Think of the difference between the SR-71 and U-2. They both fly high,
but only one could fly fast while the other could stay up for a very
long time (lower speeds, lower drag forces, so lower fuel consumption).

It would look like a U-2 and be extremely difficult to handle during
takeoff and landing.


I was thinking more of a tiny fuselage with 80m wide wings, a bit like
what pedal powered bicycles are built.


Still in the same class as a U-2 in terms of overall design. But those
human powered planes are much more fragile than a U-2, so they're not
going to handle weather at all. Record setting planes like those pedal
powered planes only fly when there is essentially zero wind.

Would such wings really come right off at supersonic speeds when you
consider how high the craft would still be (and stay until it has bled
off enough speed) ?


Yep.

extreme comparison: the space station's solar panels don't get ripped
out despite the station travelling at 25,000km/h in "thin" atmosphere.
And without reboost, the station loses altitude due to drag.


Yes, extreme example, but ISS will never be able to reenter and "land on
a runway".

The other aspect to consider is that the wings might be adjustable,
changing angle from having wings alongside fuselage to having wings
spread way out during low speed period where more lift is required.


Now you're adding mass and complexity to an aircraft design that needs
to be as light as possible to land and take off from a runway. Add a
thermal protection system and enough structural strength to survive
reentry, and there is no way you'll be able to land the thing on a
runway on Mars. You'll never meet the mass fractions required.

Jeff
--
"the perennial claim that hypersonic airbreathing propulsion would
magically make space launch cheaper is nonsense -- LOX is much cheaper
than advanced airbreathing engines, and so are the tanks to put it in
and the extra thrust to carry it." - Henry Spencer
  #2  
Old October 24th 15, 11:58 AM posted to sci.space.policy
Alain Fournier[_3_]
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Posts: 548
Default Horizontal take off/landing

On 10/24/15 6:37 AM, Fred J. McCall wrote :
JF Mezei wrote:


But on mars, what if you did re-entry over say 5 days with a vehicle
that loses orbital speed but maintains altitude due to wings generating
lift ? Couldn't you arrange it so that before each drop in altitude, you
wait to lose enough speed/energy so plane can glide safely at the lower
altitude/air density ? (aka, not burn up, and not break up)


No, you can't. Physics - get some.


Nothing in physics says it can't be done. Getting the vehicle to be
sufficiently aerodynamic for that would be hum!, let's say a challenge.
But that is an engineering challenge, not physics. So basically, you
are right it can't be done.


Alain Fournier

  #3  
Old October 25th 15, 12:26 AM posted to sci.space.policy
Alain Fournier[_3_]
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Posts: 548
Default Horizontal take off/landing

On 10/24/15 8:47 AM, Fred J. McCall wrote :
Alain Fournier wrote:

On 10/24/15 6:37 AM, Fred J. McCall wrote :
JF Mezei wrote:


But on mars, what if you did re-entry over say 5 days with a vehicle
that loses orbital speed but maintains altitude due to wings generating
lift ? Couldn't you arrange it so that before each drop in altitude, you
wait to lose enough speed/energy so plane can glide safely at the lower
altitude/air density ? (aka, not burn up, and not break up)


No, you can't. Physics - get some.


Nothing in physics says it can't be done.


Wrong.

If you disagree, run the numbers and show your work.


Your the one making a claim, your the one who should run the numbers
if you wish so. If you don't want to do it but still want to convince
that physics says it can't be done, you should at least say which
law of physics you think would be violated.

As I already said. I don't think it can be done.


Alain Fournier

  #4  
Old October 25th 15, 02:31 AM posted to sci.space.policy
Alain Fournier[_3_]
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Posts: 548
Default Horizontal take off/landing

On 10/24/15 9:03 PM, Fred J. McCall wrote :
Alain Fournier wrote:

On 10/24/15 8:47 AM, Fred J. McCall wrote :
Alain Fournier wrote:

On 10/24/15 6:37 AM, Fred J. McCall wrote :
JF Mezei wrote:


But on mars, what if you did re-entry over say 5 days with a vehicle
that loses orbital speed but maintains altitude due to wings generating
lift ? Couldn't you arrange it so that before each drop in altitude, you
wait to lose enough speed/energy so plane can glide safely at the lower
altitude/air density ? (aka, not burn up, and not break up)


No, you can't. Physics - get some.

Nothing in physics says it can't be done.


Wrong.

If you disagree, run the numbers and show your work.


Your the one making a claim, your the one who should run the numbers
if you wish so.


"You're".


Thank you. I appreciate when someone helps me improve my English.

Prove God doesn't exist. That's what you're asking me to
do.


No. You seem to say that doing so violates some law of physics.
If so, you could say which one. It is possible to prove that something
is impossible to do because of laws of physics. I don't think
this is the case here.

I can come up with any number of configurations that DON'T work.
The trick is to come up with one that does. No one has been able to
do that. So if you disagree that any configuration that purports to
work must violate one or more laws of physics (built of unobtainium,
powered by handwavium), you need to show a configuration that works
that doesn't violate any of those laws.


If you don't want to do it but still want to convince
that physics says it can't be done, you should at least say which
law of physics you think would be violated.

As I already said. I don't think it can be done.


Why don't you think it can be done?


Too complicated. But it doesn't violate any laws of physics.


Alain Fournier

  #5  
Old October 25th 15, 01:46 PM posted to sci.space.policy
Alain Fournier[_3_]
external usenet poster
 
Posts: 548
Default Horizontal take off/landing

On 10/25/15 4:12 AM, JF Mezei wrote :
On 2015-10-25 02:58, Fred J. McCall wrote:

Handwavium is not a convincing argument. Run the numbers...



That is something NASA or JPL have expertise to do. Not me. however,
what i am saying is that the "no, can't be done" comes from current
experience with re-entry which has always been ballistic.

One big de-orbit burn causes vehicle to drop into elliptical orbit
with perigee low enough for vehicle to hit dense atmosphere at still
orbital speed. Entry into atmosphere is as a ballistic object with
aerodynamics playing a small part to slow descent rate.

And in the case of Apollo, just come in extra fast and aim for the right
atmosphere density to capture you and slow you down, but not destroy
you/burn you.

Obviously, in those scenarios, a "flying" vehicle cannot survive this
quick descent at orbital speed into dense atmosphere. It would break up
before it has a chance to burn up.

But what if you had no significant de-orbit burn, remain circular orbit,
and simply let drag in thin atmosphere slow you down, and use the wings
to maintain altitude until you are close to stalling at that altitude
and then drop at lower altitude where wings once again maintain altitude
until you bleed enough speed to drop again ?

The goal here is to never drop to an altitude at a speed that you can't
survive. And you can use the wings to stay at survivable altitude until
ready to drop.


Do you agree that a B787 can fly at mach 0.84 at 30k feet ?


Yes.

Do you agree that a B787 would survive at 25,000km/h at 400km altitude
(ISS orbit) ? There would be insufficient drag to cause the plane to
break up.


Yes.

As soon as the 787 would drop low enough to start to feel a bit of drag,
do you agree that such drag at very thin atmosphere but high speed could
exert forces well within the plane's structural capabilities ? Do you
agree that the wings would generate some lift at that point ? (not much,
but just enough to keep altitude which would otherwise decay slowly due
to orbital speed going down bit by bit).

Would you not agree that for such a plane, one could devise a table of
maximum speed for each altitude where the plane would neither break up
nor burn up ?


Certainly not for a 787. I don't think you understand the difficulty of
generating lift at very high speeds and in very thin atmosphere. You
will generate more heat than lift, so you will burn through your wings.
You will also likely generate enough drag to bring you down into denser
parts of the atmosphere too quickly.

When you are going 90% of speed needed to maintain orbit at altitude x,
the wings only need to generate a small amount of lift to keep plane at
that altitude since the orbital speed does most of that work.


For normal scenarios, that small amount of lift needs to be generated in
an ionised gas because the very high speeds generate so much heat. Not
generating too much drag when you are ionizing the gas around you is, to
put it mildly, challenging.


Alain Fournier

  #6  
Old October 25th 15, 03:31 PM posted to sci.space.policy
Jeff Findley[_6_]
external usenet poster
 
Posts: 2,307
Default Horizontal take off/landing

In article om,
says...

On 2015-10-24 06:37, Fred J. McCall wrote:

No, you can't. Physics - get some.



This "no can do" attitude is what closes many doors.

You're in orbit high and with lots of energy.

You should be able to lower to just the right altitude where the wings
generate just a bit drag, and as your orbital speed slowly drops, you
change angle of attack so wings generate enough lift to maintain
altitude. At that altitude air is extremely thin so orbital speeds are
not a structural nor heat problem, just as it is not for ISS.


Back to basics. So, you have Martian gravity acting on the craft,
atmospheric drag, and lift from your (currently swept back) wings. The
problem here is that L/D won't be high enough to let your altitude drop
at a small enough rate to let you bleed all of off the speed necessary
to deploy the wings before your altitude drops below zero (with respect
to the surface). At those extremely high velocities, you'll get
precious little lift to counter Martian drag. So, your craft will drop
"like a rock", despite the swept back wings.

That is, unless you add propulsion (remember that thrust is the fourth
force acting on an aircraft). But now you're generating thrust, and
that violates the assumption that you can do this without thrust. Even
if you do add thrust, where is it going to come from? The whole point
of this exercise was to try to land *without* engines.

You only descend to lower altitude once the speed has been reduced
sufficiently for your wings to survive at lower altitude (denser
atmosphere) (structurally and heat).

This is a very slow and calculated transformation from orbit to
aerodynamic flight. Not a 30 minute re-entry as we are used to.

is the Martian atmosphere density gradually increases without sudden
jumps, this should be easier to do.


Wishing you could do this won't make it so. Only running the numbers
will prove if it is possible. It is the responsibility of anyone
claiming this is possible to do this. Guess what, it's not possible.
If it were, NASA would not be investigating blunt body inflatables for
Mars atmospheric entry.

Jeff
--
"the perennial claim that hypersonic airbreathing propulsion would
magically make space launch cheaper is nonsense -- LOX is much cheaper
than advanced airbreathing engines, and so are the tanks to put it in
and the extra thrust to carry it." - Henry Spencer
 




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