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Jens
2008-Jan-17, 10:27 AM
Any ideas of what would happen if you tossed a paper airplane out of a space shuttle as it began to reenter the atmosphere? My assumption is that it would burn up, but I was just watching the news and apparently a research group at some university (maybe Tokyo University) was actually testing it in a wind tunnel. So it must be an interesting question.

astromark
2008-Jan-17, 10:49 AM
1/ Would its lack of mass and weight mean that it would loose velocity very quickly on contacting the Earths atmosphere and just slow down to a gentle descent.
2/ Or would it just incinerate shortly after entering any significant density of atmosphear....
I am suggesting option 2. The shuttle de-orbits from 39000km/hr. Puts new meaning to the 'Running Hot' idea a?

hhEb09'1
2008-Jan-17, 11:09 AM
That'd be a wonderfully cheap experiment, but a PR nightmare: spacewalking astronaut throws out a thousand handbills, all printed with "If you find this, contact NASA 1-800-548-8377"

G O R T
2008-Jan-17, 12:11 PM
I was just watching the news and apparently a research group at some university (maybe Tokyo University) was actually testing it in a wind tunnel.

Did they say how this wind tunnel would get air at ~30 Pa moving at Mach 29 ???

NEOWatcher
2008-Jan-17, 12:54 PM
Here's an old (short) discussion (http://www.bautforum.com/astronomy/10554-would-feather.html) about a feather.

a surface area of about 3 sq. in. (1/2" x 6") would "fight the wind". At 1 psi resistance (9 Newtons total) and a weight of 0.1 gram, I get a deceleration rate of about 10,000g's (.1 sec to nearly stop).

joema
2008-Jan-17, 03:18 PM
Extensive discussion about whether a party balloon would survive reentry:

http://www.bautforum.com/questions-answers/63473-balloon-orbit.html

A balloon has a greater surface area to mass ratio than a paper airplane, which makes me think it has a greater chance.

However it probably depends on the size and design of paper airplane, type of paper, etc.

If constructed of very thin paper for the lightest possible weight and greatest ratio of surface area to mass, it's conceivable it might survive.

E.g, the shuttle orbiter has a wing loading of about 100 lb per square foot. This gives an idea of how much mass (hence at orbital velocity, kinetic energy) is supported per unit area.

By contrast a paper airplane constructed from an 8.5 x 11" of thin paper weighs about 3 grams, and (depending on the design) might have wing loading of 0.01 to 0.027 pounds per square foot.

So at the low end, the wing loading is 1/10,000th the space shuttle.

Like a styrofoam cooler blown from a pickup truck, it would decelerate very rapidly at the slightest hint of atmospheric drag. It might never develop a hypersonic shock wave, hence might survive to reach earth.

hhEb09'1
2008-Jan-17, 03:39 PM
It might never develop a hypersonic shock wave, hence might survive to reach earth.might never? this is a piece of paper right? hard to imagine

Warren Platts
2008-Jan-18, 02:39 PM
Supposedly they found a couple of embroidered shoulder patches from the space shuttle Challenger astronaut uniforms that were in near perfect condition. So if a shoulder patch can survive reentry, a paper airplane might be able to as well.

Saluki
2008-Jan-18, 02:56 PM
Supposedly they found a couple of embroidered shoulder patches from the space shuttle Challenger astronaut uniforms that were in near perfect condition. So if a shoulder patch can survive reentry, a paper airplane might be able to as well.

Challenger did not reach orbit. It blew up at about 48,000 feet, well shy of orbital velocity.

Warren Platts
2008-Jan-18, 03:49 PM
Challenger did not reach orbit. It blew up at about 48,000 feet, well shy of orbital velocity.
You are correct. I should have written "Columbia" instead of "Challenger".

BISMARCK
2008-Jan-18, 03:59 PM
Even the videotape from one of the Columbia astronaut's camcorder survived and was playable, sort of. It had to have some repair work done to it.

hhEb09'1
2008-Jan-18, 04:02 PM
I was under the impression that the Columbia shuttle had already taken the brunt of the re-entry at the time of the accident. No? Maybe I should look it up.

PS: This wiki article (http://en.wikipedia.org/wiki/Space_Shuttle_Columbia_disaster) has a timeline, and it says the shuttle "entered a 10-minute period of peak heating" at around 8:51, and a transmission from the mission commander was recorded almost 9 minutes later.

joema
2008-Jan-19, 03:40 PM
I was under the impression that the Columbia shuttle had already taken the brunt of the re-entry at the time of the accident. No? Maybe I should look it up...
Columbia broke up at about 203,000 ft (61 km) altitude at a speed of Mach 18 (13,700 mph, 6.1 km/sec).

Small objects like shoulder patches, papers, etc were immediately disgorged into an intense thermal environment, however some survived to reach the ground.

This was a much MORE intense situation than similar lightweight objects reentering from orbit. In that case they'd more slowly decelerate as atmospheric friction gradually increased.

Cougar
2008-Jan-19, 09:36 PM
In that case they'd more slowly decelerate as atmospheric friction gradually increased.
As mentioned early on by astromark, and further detailed in your previous link (http://www.bautforum.com/questions-answers/63473-balloon-orbit.html), I would think that it largely depends on the initial relative velocity of the {paper airplane} and the molecules in the atmosphere. At high enough relative velocities, any interaction with even the "uppermost" atmosphere would turn your paper airplane to toast.

Grains of sand have different aerodynamics than paper airplanes, but they're also pretty light, and it may be instructive to recall that grains coming in at very high speeds spectacularly combust into fireballs visible from the ground.

A more "fun" question might be,

What initial conditions (location and x-y-z velocities relative to the earth) would allow a paper airplane to enter the atmosphere and float, one way or another*, down to the surface "unburned"?

* By the way, the design of paper airplanes (http://www.ncgraphicarts.com/ryan/other/planes.htm
) can vary greatly, with varying capabilities. Perhaps a less-complicated "probe" would focus the question to the orbital and atmospheric dynamics.

Cougar
2008-Jan-19, 09:47 PM
...x-y-z velocities...
...not to mention orientation...

joema
2008-Jan-19, 11:59 PM
...I would think that it largely depends on the initial relative velocity of the {paper airplane} and the molecules in the atmosphere. At high enough relative velocities, any interaction with even the "uppermost" atmosphere would turn your paper airplane to toast.

Grains of sand have different aerodynamics than paper airplanes, but they're also pretty light, and it may be instructive to recall that grains coming in at very high speeds spectacularly combust into fireballs visible from the ground...
The small "grain of sand sand" meteoroids illustrate an object:

(1) With a low ratio of surface area to mass
(2) On a steep reentry angle (30 degrees typical)
(3) At very high velocity (11 to 71 km/sec)

...will burn up

None of these apply to a paper airplane or a balloon reentering from earth orbit.

As opposed to a tiny meteoroid, a paper airplane:

(a) Has a high ratio of surface area to mass
(b) Is on a shallow reentry angle (1 degree)
(c) Is going much slower on average (7.8 km/sec)

For the airplane, the faint air drag at entry interface would cause aerobraking and rapid deceleration due to the large surface area, low mass and kinetic energy.

Compare a tiny "grain of sand" meteroid to a paper airplane:

Consider a reentering spherical meteoroid that is 1 cubic millimeter, specific gravity of 3 g/cm^3 (average), moving at 15 km/sec on a 30 degree entry angle (also average).

The cross-sectional area is 1.2 mm^2, and the mass is .003 grams. The ratio of cross-sectional area to mass is 4 square cm per gram, or 0.5 lb per square foot.

Kinetic energy is given by KE = 1/2 * m * v^2, so the sand grain's kinetic energy is 33,750 Joules.

Compare this to a paper airplane, mass 3 grams, wing area 500 square cm, reentering from orbital velocity of 7.8 km/sec.

The paper airplane's ratio of wing area to mass (wing loading) is 0.012 pounds per square foot, or 40 times lower than the "grain of sand" meteoroid.

The paper airplane's kinetic energy is 91,260 Joules, 3x the tiny meteor, but dispersed over 41,000 times the surface area.

The paper airplane must only dissipate 182 Joules per square cm, vs the meteroid's 2.75 MILLION Joules per square cm. That's why the tiny meteroid burns up.

I don't know for sure a paper airplane would survive, but it seems conceivable once you think about the physics involved.

KaiYeves
2008-Jan-20, 01:18 AM
Even the videotape from one of the Columbia astronaut's camcorder survived and was playable, sort of. It had to have some repair work done to it.
My gosh, you mean that cliche has a basis in reality?
(The cliche of the hero finding a camera in the ruins of a disaster, and the hero plays it and finds out...)

Kaptain K
2008-Jan-21, 01:36 AM
Make the "paper" airplane out of aluminum (aluminium if you're English) foil and track it with radar!

Torsten
2008-Jan-21, 02:35 AM
Make the "paper" airplane out of aluminum (aluminium if you're English) foil and track it with radar!

Wouldn't that just be space junk?

KaiYeves
2008-Jan-21, 08:07 PM
Wouldn't that just be space junk?
No, space junk is RCH's books.

Jens
2008-Jan-22, 01:28 AM
By the way, New Scientist has a story (http://space.newscientist.com/article/dn13208-origami-spaceplane-aims-for-space-station-descent.html)on the piece of news I mentioned.

John Mendenhall
2008-Jan-22, 06:03 PM
No, space junk is RCH's books.

Rocket wash, Space Cadet!

Yes, he wrote those too.

Ara Pacis
2008-Feb-02, 09:57 PM
Didn't the Mir solar panels come fluttering down after the rest of it burned up? I remember them saying it was expected to happen that way.

mugaliens
2008-Feb-03, 06:20 PM
Like a styrofoam cooler blown from a pickup truck, it would decelerate very rapidly at the slightest hint of atmospheric drag. It might never develop a hypersonic shock wave, hence might survive to reach earth.

At the first molecular collision, it's developing a supersonic shock wave (hypersonic merely refers to very fast supersonic velocities), since the speed of sound decreases with decreasing atmospheric density.

Whether hypersonic or not isn't the issue. The issue is whether the drag/weight ratio is great enough to slow it down before heating burns it up.

The critical figure involves the mass of O2 required to combust the paper airplane. The exosphere consists of gaseous particles which, above the exobase, are considered to be on pure trajectories (collisions between them are scarce (beyond the region where the Knudsen number is approximately equal to 1).

Thus, the paper airplane would not encounter sufficient atmosphere to slow it down as it descended to the exobase, the bottom of the exosphere.

Since the proportion of O2 in the thermosphere (lies below the exosphere) is significantly less than at sea level, there's not much O2 available to combust paper. However, the temperature of the thermosphere reaches 2,500 deg C during the day, which, if there are enough of them present around the descending paper airplane, would be enough to evaporate nearly all elements except carbon.

The question is whether there's enough mass of O2 in the thermosphere to actually unite with the carbon in the paper as it descends through the thermosphere.

As it turns out, there are no where near enough molecules in the thermosphere to heat the paper airplane, either directly, or by friction.

In fact, the ISS has a stable orbit in the upper regions of the thermosphere.

This leads us to the mesosphere. The temps are very cold (-100 deg C in the upper mesosphere), but this is also the region where most meteors burn up due to atmospheric drag. Most of these meteors range in size from dust to grains of sand to small pebbles. Literally millions burn up every day in the Mesosphere.

It's here where our paper airplane will either disintegrate in a flame of glory or slow to a reasonably gentle fall.

Consider the following: The atmospheric pressure at sea level is 14.7 lbs/square inch. The atmospheric pressure at the border between the stratosphere and the mesosphere (approximately 130,000 ft, aka 50-60 km (180,000 ft) according to some), is just 1 mb, about 1/1000th the pressure at sea level.

Thus, a 1 square foot sheet of paper falling through the mesosphere would encounter .0147 lbs of atmosphere, of which approximately 10% would be either O2 or O (ozone). The question now is, what does that sheet of paper weigh? What's it's carbon weight? How does that compare to the oxygen weight (mass, actually) it's likely to encounter?

In reality, the sheet will likely descend through the mesosphere with enough velocity to cause immediate oxidation upon contact with any molecular oxygen or ozone. And the mesosphere provides just enough oxygen to oxidize the sheet of paper without slowing it down significantly.

Thus, it's unlikely that the sheet of paper will ever reach the stratosphere intact.

joema
2008-Feb-03, 08:52 PM
...Thus, it's unlikely that the sheet of paper will ever reach the stratosphere intact.
As cited above, professional aerodynamicists have already examined this and it appears plausible a paper airplane of proper construction might survive reentry. In fact they've been successfully tested in wind tunnels to Mach 7:
http://space.newscientist.com/article/dn13208-origami-spaceplane-aims-for-space-station-descent.html

Similar effects and properties could apply to a sheet of paper.

For an object reentering from orbit, the first aerodynamic effects appear at entry interface, about 400,000ft (122 km).

The Columbia Accident Investigation Report included various data about this, based on STS-107 investigation.

Dynamic pressure at entry interface is almost zero, and ramps up to 0.5 psf (0.0024 Newtons/cm^2) over about 160 seconds.

Over this period, the space shuttle actually accelerates from Mach 24.57 to mach 24.66, because atmospheric drag is tiny relative to the vehicle mass and area.

However a sheet of paper (or paper airplane) would behave very differently over this period.

We can crudely calculate the initial deceleration of a 3 gram, 8.5" x 11" (603 cm^2) sheet of paper over this period, assuming its oriented perpendicular to trajectory. To simplify things, we'll assume a constant dynamic pressure of 1/2 the midpoint, or 0.25 psf (0.00112 N/cm^2)

603 cm^2 * 0.00112 N/cm^2 = 0.68 Newtons

F=m*a
a = f/m
a = 0.68 N / 0.003 kg
a = 223 m/s^2 (23 g)

That deceleration would totally halt the paper within 34 seconds, while it's still around 400,000 ft!!

In actuality, the initial deceleration would be much more gradual and ramp up slowly (over 10s of seconds).

There are obviously unresolved issues about orientation of the airplane or sheet of paper, structural strength, etc.

However because the mass and wing loading are very low, it appears the object would quickly decelerate in the extreme upper atmosphere -- just after entry interface.

This is probably one reason the above-mentioned aerodynamicists think a paper airplane might survive.

JohnD
2008-Feb-03, 11:26 PM
Didn't NASA do some school science experiments at one time?

1000 sheets of A4, folded by school kids into paper planes that must fold flat. Copier paper weighs about 100gms/m^2, and that's 16 A4 sheets. SO 1000 A4s weigh 6.25kgs
NO! 10,000 sheets! 60Kgs! Think of it, 10,000 school kids from all over the world, their planes go into space, and maybe, back again!
Opened in space and jettisoned as the Shuttle descends (that would be the difficult part)

How cheap could an experiment be?
How do we petition NASA to do this one?
John

Jens
2008-Feb-04, 08:41 AM
Opened in space and jettisoned as the Shuttle descends (that would be the difficult part)

Just roll down the window, and toss it out. :)

dgavin
2008-Feb-04, 03:54 PM
OK this discussion just gave me a vision of a person on the ISS shooting spit wads at the earth to see if they would still 'stick' to the ceiling of his old school:P

KaiYeves
2008-Feb-04, 09:05 PM
OK this discussion just gave me a vision of a person on the ISS shooting spit wads at the earth to see if they would still 'stick' to the ceiling of his old school
Can't... stop... laughing...

dgavin
2008-Feb-05, 04:30 PM
Glad you liked it:lol:

The really disturbing thing is it did just pop in there after a few seconds reading of the OP. Sometimes a wandering mind is a wonderful thing! Just wish mine would return some day. . .:doh:

samkent
2008-Feb-05, 04:47 PM
Why don't we settle this whole thing.

Someone call Adam and Jamie!

pzkpfw
2008-Feb-07, 12:12 AM
Today in the news here:

http://www.stuff.co.nz/4391467a4560.html

(Japanese paper planes tested at high heat.)

ravens_cry
2008-Feb-07, 07:43 AM
I am reminded of Jim Lowell's (or maybe it was just Tom Hanks's) explanation on going to the bathroom in space.
"Well, mam, I tell you it's a very complicated procedure that involves cranking down the window and looking for a gas station"

astromark
2008-Feb-07, 08:12 AM
1/ Would its lack of mass and weight mean that it would loose velocity very quickly on contacting the Earths atmosphere and just slow down to a gentle descent.
2/ Or would it just incinerate shortly after entering any significant density of atmosphear....
I am suggesting option 2. The shuttle de-orbits from 39000km/hr. Puts new meaning to the 'Running Hot' idea a?

And its worth the repetition ( that was post two.) If like the parachute jumper from the high altitude balloon you have zero orbital velocity then you have just gravity to exhilarate you down through the atmosphere. Unfortunately if you are too high your re-entry velocity will incinerate you. The paper dart has the same problem. If it were stationary at the commencement of its descent then yes I believe it would survive the fall to Earth. If its dropped at a point near to the top of Earths atmosphere. Drop it out of a orbiting spacecraft then its charcoal and ash... To think that money might be wasted actually attempting this is surly nonsence...

Ara Pacis
2008-Feb-07, 08:27 AM
I thought experimentation was part of the Scientific Method.

astromark
2008-Feb-07, 08:41 AM
Yes of coarse it is, but. I do not need to jump of a building to know it will hurt.
We know the answer to this. Its done. Adam and Jamie can rest.

JohnD
2008-Feb-07, 10:17 AM
astromark,
joema's calculation (post 25) contradicts you - something as light as a paper dart would be stopped by the atmosphere almost immediately, and would not be burnt up.
John

astromark
2008-Feb-07, 10:41 AM
John : re post 25. Yes I read it and promptly dismissed it. It completely ignores the fact of orbital velocity of 39500 km/hr to near to stationary requires a loss of energy. That would be turned into heat POOF ! end of theorie...and dart. Now if you coat that paper with a heat dispersant layer of thermal officiant heatable tephlon....:) Naa, its still going to burn., or maybe BUT! Whats the point of this. Are we planning a very grand ticker tape parade ? If its got to have so little mass what use is this to us?

Neverfly
2008-Feb-07, 10:59 AM
I have to agree here.
Those speculating about the paper release don't even seem to be taking it seriously. 20 pieces with no idea where they will land (Ocean most likely) and oh.. let's throw a calling card on there in a couple of languages just in case. Hmmm....

joema
2008-Feb-07, 01:15 PM
...Yes I read it and promptly dismissed it. It completely ignores the fact of orbital velocity of 39500 km/hr to near to stationary requires a loss of energy. That would be turned into heat POOF !..
The deceleration occurs in the tenuous upper atmosphere BEFORE reentry heating happens.

This is different from all other vehicles that have ever reentered, so we can't be guided by those. Rather we must examine the facts and what calculations show, not what our intuition says.

A paper airplane has extremely low wing loading which allows rapid deceleration in the extreme upper atmosphere. This happens at very low dynamic pressures, long before frictional heating is a factor.

This has been examined by professional aerodynamicists, who think a properly constructed paper airplane might survive:

http://space.newscientist.com/article/dn13208-origami-spaceplane-aims-for-space-station-descent.html

HenrikOlsen
2008-Feb-07, 03:47 PM
The deceleration occurs in the tenuous upper atmosphere BEFORE reentry heating happens.

This is different from all other vehicles that have ever reentered, so we can't be guided by those. Rather we must examine the facts and what calculations show, not what our intuition says.
Actually, it's similar to one vehicle that has reentered.
It's exactly the effect that made SpaceShipOne possible without advanced ceramics or ablative coatings.

To understand what's going on, you have to remember that it isn't friction that heats meteors, it's the compression of the air in front of them that heats the air, the hot air then heats the meteor, which burns up.

With SpaceShipOne, (and the paper airplanes), the drag is so high that the speed is slowed while the airpressure is still very low.
This means the air in front of the plane isn't compressed nearly as much, and subsequently doesn't heat it as much.

A paper airplane would have an even higher drag ratio that SpaceShipOne (depending on orientation) and would thus slow down with even less heat transferred to it.

joema
2008-Feb-07, 06:54 PM
SpaceShipOne wing loading is 49 lbs/ft^2, the space shuttle orbiter is about 70 lbs/ft^2 (CAIB modeling).

By contrast the wing loading of a paper airplane is about 0.012 lbs/ft^2. SpaceShipOne is much closer to the shuttle orbiter's wing loading than to a paper airplane.

Another way to examine this is the ratio of kinetic energy at reentry vs wing area.

KE=1/2 * m * v^2, where
KE=kinetic energy in joules
m=mass in kg
v=velocity in m/s

Paper airplane reentry mass, speed, wing area: 3 grams, Mach 25, 0.06 m^2
SpaceShipOne reentry mass, speed, wing area: 1200 kg, Mach 3, 15 m^2
shuttle orbiter reentry mass, speed, wing area: 94,800 kg, Mach 25, 260 m^2

Paper airplane ratio of reentry kinetic energy to wing area:

KE = 0.5 * .003 kg * 7500^2 m/s
KE = 84,375 joules
ratio = 84,375 joules / 0.06 m^2 = 1.4 mega joules / m^2

SpaceShipOne ratio of reentry kinetic energy to wing area:

KE = 0.5 * 1200 kg * 1020^2 m/s
KE = 624 mega joules
ratio = 624E6 J / 14.9 m^2 = 42 mega joules / m^2

Shuttle orbiter ratio of reentry kinetic energy to wing area:

KE = 0.5 * 94800 kg * 7500^2 m/s
KE = 2.6 tera joules
ratio = 2.6E12 J / 260 m^2 = 10 giga joules / m^2

So expressed as either wing loading or kinetic energy loading per wing area, the paper airplane is vastly different than any other vehicle that has reentered.

The reason SpaceShipOne survived reentry is largely due to its much lower performance and kinetic energy. It reenters at Mach 3, not Mach 25. Kinetic energy scales as the square of velocity, so SpaceShipOne has little energy to dissipate relative to the shuttle. However it has far more energy per unit area to dissipate than a paper airplane.

A paper airplane might survive reentry from orbital velocity, whereas SpaceShipOne would be incinerated.

a1call
2008-Feb-07, 08:21 PM
To understand what's going on, you have to remember that it isn't friction that heats meteors, it's the compression of the air in front of them that heats the air, the hot air then heats the meteor, which burns up.

Thank you for that HenrikOlsen.

In order to remember something you need to have known it at some point in the past. :)
I finally understand the source of the heat. So, am I correct to assume that the upper side of a vessel will experience a cooling effect due to the relative low pressure?

Ara Pacis
2008-Feb-07, 10:16 PM
Yes of coarse it is, but. I do not need to jump of a building to know it will hurt.
We know the answer to this. Its done. Adam and Jamie can rest.

Ah, but when good men disagree, sometimes it's worthwhile to just go out and find out what will happen. As for jumping off of buildings, did you mean with or without a parachute? :-)

astromark
2008-Feb-08, 06:23 AM
This thread has become closely related to the next one re; 're-entering the atmosphere.' and no I am not interested in base jumping...
So now you understand the heat from re-entry does come from that compression in front of the object.. look up any clear night and see those smaller than a gram objects being destroyed as they re-enter... and you think some sort of paper dart could fool this..? Not from an orbital velocity of 39500km/hr...It can not. Which is what was asked... OP. Loose that velocity and the chances improve... but that is not this question:);)

joema
2008-Feb-08, 01:19 PM
...you think some sort of paper dart could fool this..? Not from an orbital velocity of 39500km/hr...It can not. Which is what was asked... OP...
The OP asked what would happen as a paper plane "began to reenter the atmosphere".

That point is entry interface, at about 400,000 ft. The above calculations show deceleration would happen long before heat buildup, due to the plane's high ratio of wing area to mass and kinetic energy.

By comparison the space shuttle has 5,800 times greater wing loading, and 7,100 times greater kinetic energy per unit area.

This probably explains why professional aerodynamicists have also concluded the paper plane might survive.

astromark
2008-Feb-09, 10:52 AM
Loose that velocity and the chances improve... but that is not this question

You seem to have missed the simple " If it was dropped from the shuttle or ISS...

You also seem to have missed the point that was made in posts 38 & 39.:)

Dropping small pieces of paper into Earths upper atmosphere is a tad pointless... and a wast of paper. Use money:) yours, not mine.

joema
2008-Feb-09, 02:25 PM
Loose that velocity and the chances improve... but that is not this question..
The question is whether a paper airplane could survive reentry from orbit.

As you can see in the below graph, the heat buildup is very low for the first few minutes. During that interval the dynamic pressure is also low. This doesn't slow the space shuttle, because of its great mass vs wing area. Thus it's still moving at high velocity when deeper in the atmosphere and heat builds up.

By contrast a paper airplane would be quickly decelerated during this low heat period. By the time it's deeper in the atmosphere it would be moving very slowly, and flutter down as if dropped from a high-altitude ballon.

This was confirmed by the data analyzed from the STS-107 Columbia disaster, where the left wing was open and unprotected from a thermal protection breach. Despite this, the exposed wing spar inside showed no sign of reentry heating until about entry interface+270 sec. That's because heating is extremely low during the initial period after entry interface.

During that period the shuttle (or all other conventional vehicles) slow down very little, thus encounter great heat when still moving a high velocity in the denser atmosphere. However a paper airplane would rapidly decelerate during that period due to the low mass relative to large wing area.

Kaptain K
2008-Feb-09, 02:27 PM
Give up guys. No calculation is ever going to match the power of astromark's superior intuition! :wall:

Ara Pacis
2008-Feb-10, 09:57 AM
I've heard that NASA has a saying: "One test is worth a thousand expert opinions."

astromark
2008-Feb-10, 06:32 PM
"The question is whether a paper airplane could survive reentry from orbit."

"One test is worth a thousand expert opinions."

Tell me what can be achieved by this fruitless want to experoment?... We could not hope to build a bigger winged area to gain the advantage you amagine...
As you know, I see ash and carbon as the outcome of all this... mark.

Kaptain K
2008-Feb-10, 07:02 PM
As you know, I see ash and carbon as the outcome of all this...
In spite of all the calculations that say you're wrong!

mugaliens
2008-Feb-10, 10:01 PM
Astromark, Joema, I think you both have some serious things to consider.

Having read both your remarks, and diving into the engineering details, I do believe that a paper airplane might survive atmospheric reentry.

However, the proof lies in the pudding, and I'm eagerly awaiting the results of the Japanese test.

Ara Pacis
2008-Feb-11, 12:00 AM
Tell me what can be achieved by this fruitless want to experoment?

Data.

KaiYeves
2008-Feb-11, 12:28 AM
Tell me what can be achieved by this fruitless want to experoment?
Don't you adults ever just do anything for fun?

mugaliens
2008-Feb-11, 06:57 PM
As cited above, professional aerodynamicists have already examined this and it appears plausible a paper airplane of proper construction might survive reentry. In fact they've been successfully tested in wind tunnels to Mach 7:
http://space.newscientist.com/article/dn13208-origami-spaceplane-aims-for-space-station-descent.html

Similar effects and properties could apply to a sheet of paper.

For an object reentering from orbit, the first aerodynamic effects appear at entry interface, about 400,000ft (122 km).

The Columbia Accident Investigation Report included various data about this, based on STS-107 investigation.

Dynamic pressure at entry interface is almost zero, and ramps up to 0.5 psf (0.0024 Newtons/cm^2) over about 160 seconds.

Over this period, the space shuttle actually accelerates from Mach 24.57 to mach 24.66, because atmospheric drag is tiny relative to the vehicle mass and area.

However a sheet of paper (or paper airplane) would behave very differently over this period.

We can crudely calculate the initial deceleration of a 3 gram, 8.5" x 11" (603 cm^2) sheet of paper over this period, assuming its oriented perpendicular to trajectory. To simplify things, we'll assume a constant dynamic pressure of 1/2 the midpoint, or 0.25 psf (0.00112 N/cm^2)

603 cm^2 * 0.00112 N/cm^2 = 0.68 Newtons

F=m*a
a = f/m
a = 0.68 N / 0.003 kg
a = 223 m/s^2 (23 g)

That deceleration would totally halt the paper within 34 seconds, while it's still around 400,000 ft!!

In actuality, the initial deceleration would be much more gradual and ramp up slowly (over 10s of seconds).

There are obviously unresolved issues about orientation of the airplane or sheet of paper, structural strength, etc.

However because the mass and wing loading are very low, it appears the object would quickly decelerate in the extreme upper atmosphere -- just after entry interface.

This is probably one reason the above-mentioned aerodynamicists think a paper airplane might survive.

Joema - nicely done!

I've changed my position. Properly configured, I now believe that a paper airplane could survive reentry.

I guess this brings up another question - if a paper airplane could survive reentry, what about a bacteria, or similarly-sized form of life?

Noclevername
2008-Feb-11, 07:01 PM
Data.

Without any means to track it (and they have no plans for any), the only data they get will be whether an astronaut can let go of something.

Larry Jacks
2008-Feb-11, 07:57 PM
I believe someone here has proposed using some sort of radar reflective material (perhaps a type of mylar) that would increase the RCS of the airplane enough for it to be tracked by the space surveillance network. It isn't the physical size of the object that really matters, just the RCS. Even then, unless you have some idea of where it came down or unless you released a lot of them to increase the chances that one of them might be found, you won't know if the test was successful.

KaiYeves
2008-Feb-11, 09:18 PM
Without any means to track it (and they have no plans for any), the only data they get will be whether an astronaut can let go of something.
As I said before-
Don't you adults ever do anything just for fun? Fun is important!

Noclevername
2008-Feb-11, 10:24 PM
As I said before-
Don't you adults ever do anything just for fun? Fun is important!

Sure. But that's not why they (claim they're) doing this. If you're going to spend a billion and a half and risk six lives to do something, it better be a lot more fun than throwing a paper airplane.

joema
2008-Feb-12, 01:22 PM
...If you're going to spend a billion and a half and risk six lives to do something, it better be a lot more fun than throwing a paper airplane.
They're not spending \$1.5 billion nor risking six lives to throw a paper airplane.

It's similar to when Apollo 15 astronauts dropped a feather and hammer on the lunar surface to demonstrate Galileo's experiment: http://video.google.com/videoplay?docid=6926891572259784994

The astronauts are already there on their main mission, it's just a momentary diversion.

Re whether it can be tracked, they haven't revealed specifics on this. However a small metallic strip on the plane (think aluminium foil and adhesive tape) could be tracked from ground radar, especially if known beforehand.

Similar size items were tracked by Space Survellance Radar during the investigation for Space Shuttle Columbia (STS-107): http://en.wikipedia.org/wiki/Air_Force_Space_Surveillance_System

Ara Pacis
2008-Feb-12, 09:54 PM
Without any means to track it (and they have no plans for any), the only data they get will be whether an astronaut can let go of something.

If this particular experiment has flaws then so be it. But conducting an experiment for this particular theory can be valuable if properly conducted. Criticizing a proposed experiment because you don't think it will result in an answer is one thing, criticizing a proposed experiment because you claim to already know the answer is another.

Neverfly
2008-Feb-12, 09:58 PM
If this particular experiment has flaws then so be it. But conducting an experiment for this particular theory can be valuable if properly conducted. Criticizing a proposed experiment because you don't think it will result in an answer is one thing, criticizing a proposed experiment because you claim to already know the answer is another.

I'm criticizing it because of the lack of ability to track it.

Larry Jacks
2008-Feb-12, 11:07 PM
About the only way I can think of to do this as a meaningful experiment would be to put a bunch of paper airplanes into a minimal microsat so you can do a controlled deorbit. The airplanes would need RF reflectors on board so they could be tracked with radar through the deorbit and hopefully be located if they survive. Otherwise, you're basically just adding to space junk for whatever time it takes them to decay somewhere between N50 and S50 degrees latitude (ISS inclination). Under those conditions, the chances of finding them even if they survived reentry is so small as to be absurd.

NEOWatcher
2008-Mar-27, 04:08 PM
Update:
This project now has official JAXA 3 year funding of \$300k/yr to study its feasibility.
Scientists say a successful flight could advance spacecraft design (http://www.msnbc.msn.com/id/23827045/)

Toda and Suzuki first met about 10 years ago, when Suzuki and other scientists attended Toda's launching of a 6.6-foot-long giant paper craft from the top of a mountain. The successful flight impressed Suzuki, and Toda revealed his long-cherished dream.

How does this relate to reentry? Ok, the guy can make a good paper airplane... there's got to be more to this impression. :think:

The effort has been a labor of love. It's had no outside funding so far, relying on paper donated by the origami association and Suzuki's access to Tokyo University equipment.

That makes it sound like the paper cost was a big factor. Just how exotic was this paper?

JohnD
2008-Mar-28, 01:12 PM
It's on that linked webpage, "Suzuki and Toda use origami paper made of sugar cane fibers that are resistant to heat, wind and water. They spray a special coating onto the paper."
And the paper shuttles are 8x4inches, not the 6.6 foot version - which did not surprise me. As anyone knows who has folded paper planes, size does not scale. A large plane of the same construction will be much weaker thna a small one.

Jhn

EvilEye
2008-Mar-28, 07:23 PM
Wouldn't the paper airplane just collapse as soon as it entered the atmosphere...like hitting a brick wall, and fall as crumpled piece of paper?

NEOWatcher
2008-Mar-28, 07:32 PM
Wouldn't the paper airplane just collapse as soon as it entered the atmosphere...like hitting a brick wall, and fall as crumpled piece of paper?
No; there is no actual boundary. It is a gradual thing.

Kind of like the beginning of a rainstorm. A little fog at first, getting thicker until it becomes a mist which keeps getting thicker until a slight drizzle, to a light rain, slowly building up to somebody dumping a barrel of gatorade on your head.

All my life, I've heard the phrase "skipping off the atmosphere". That is very misleading.

mugaliens
2008-Mar-29, 03:07 PM
Did they say how this wind tunnel would get air at ~30 Pa moving at Mach 29 ???

Easy. When I was in school, we could do M 50. The push came from high-pressure tanks of air. The rest involved a very delicately designed set of valves and attenuators. If someone goofed (a PhD always reviewed our efforts), the reward was usually an M2 or less result.

It got noticed!

publiusr
2008-Jul-21, 09:08 PM
Interesting

mugaliens
2008-Jul-22, 07:01 PM
John : re post 25. Yes I read it and promptly dismissed it. It completely ignores the fact of orbital velocity of 39500 km/hr to near to stationary requires a loss of energy. That would be turned into heat POOF ! end of theorie...and dart.

KE = 1/2 * m * V2

Let's see... 500 sheets of the 24 lb paper I use in my printer weighs 6 lbs. Most people use 20 lb paper, so let's go with 5 lbs. Divided by 500 sheets, that comes to 0.01 lb, or 0.004536 kilograms. Velocity is 39,500 km/hr, or 10,972 m/s.

So, the KE = 273,043 joules, which must be shed over approximately 30 seconds, so that's approximately 9,101 Watts experienced throughout that 30 seconds. To put this into perspective, that's about 8.6 BTU/s, 2,173 cal/s (2.17 kcal/s - aka "food calorie"), 2.6 ton of refridgeration (most heat pumps are 3 ton units), and 12.2 HP.

12 horsepower, eh? If you could shunt 12 horsepower into a sheet of paper, it would probably reach 451 deg F in less than a second.

Sorry, folks, but that paper airplane is TOAST!

antoniseb
2008-Jul-22, 07:19 PM
I've closed this resurrected thread. Send me a link if you want it reopened.

cjl
2008-Jul-22, 07:50 PM
Did you remember to actually close it, or did someone want it reopened already?

Oh, and mugaliens, remember that all of the heat doesn't actually need to go into the paper. Some of the heat (or most of it) can go into the air around the paper instead (the energy would go to compressing the air).

joema
2008-Jul-22, 09:37 PM
KE = 1/2 * m * V2

Let's see... 500 sheets of the 24 lb paper I use in my printer weighs 6 lbs. Most people use 20 lb paper, so let's go with 5 lbs. Divided by 500 sheets, that comes to 0.01 lb, or 0.004536 kilograms. Velocity is 39,500 km/hr, or 10,972 m/s.

So, the KE = 273,043 joules, which must be shed over approximately 30 seconds, so that's approximately 9,101 Watts experienced throughout that 30 seconds....it would probably reach 451 deg F in less than a second...Sorry, folks, but that paper airplane is TOAST!
Your calculations are incorrect, but you nonetheless have a good point. Let's use a 4 gram sheet of 8.5"x11" paper, and orbital velocity is 17,318 mph (7,742 m/s).

KE = 1/2 * m * v^2
KE = 1/2 * .004 kg * 7823^2 m/s
KE = 119,877 Joules

119,877 J = 33.3 watt hours, or 3,960 watts over 30 sec.

Now 3,960 watts for 30 sec seems like a lot, but as already mentioned only a small fraction is delivered into the paper.

Also we're dealing with a paper airplane (not a sheet), which would likely decelerate slower, thus spreading the energy release over a longer period.

From the STS-107 data, we know reentry heating isn't even measureable until about 420 sec after entry interface.

If the paper airplane decelerated over 420 sec, that would equate to 286 watts for that period.

The paper plane need not absorb all of the 286 watts, no more than the shuttle or any other reentering body absorbs all their own kinetic energy as heat.

alainprice
2008-Jul-23, 03:48 AM
There's also the fact that this problem is not easily solved with algebra, you need calculus(integration) to account for the changing atmosphere.

The re-entry will be much longer than 30 seconds. Even 420 seconds seems short. After all, it have very low momentum compared to surface area. Consider it a glider more than a ballistic entry. I'd expect a paper airplane to last longer than that from 35,000 feet.

cjl
2008-Jul-23, 05:36 AM
It wouldn't behave anything like a glider, due to the high supersonic speeds and extremely low atmospheric density. I'd say a ballistic entry probably isn't far off.

joema
2008-Jul-23, 12:01 PM
...The re-entry will be much longer than 30 seconds...
The reentry will be longer, but the deceleration phase of a paper plane (or sheet) would be very short. See the calculations in the above post: http://www.bautforum.com/questions-answers/69247-paper-airplane-reentry.html#post1166097

This is sometimes called the "shuttlecock effect". It's used by Burt Rutan's SpaceShipOne and Two. The vehicle enters a very high drag configuration where surface area is large relative to mass (similar to a paper plane). It thus decelerates in the extreme upper atmosphere, avoiding extensive heat shielding.

If it works for SpaceShipOne, it could possibly work for a paper plane. See the subheading "Feathered Reentry" in this article: http://en.wikipedia.org/wiki/Atmospheric_reentry

mugaliens
2008-Jul-23, 06:53 PM
Did you remember to actually close it, or did someone want it reopened already?

Oh, and mugaliens, remember that all of the heat doesn't actually need to go into the paper. Some of the heat (or most of it) can go into the air around the paper instead (the energy would go to compressing the air).

Ok. 1HP for a duration of 30 seconds and an 8-1/2x11 sheet of 20# paper is still toast.

mugaliens
2008-Jul-23, 07:01 PM
Your calculations are incorrect...

Where? If you don't mind my asking.

...but you nonetheless have a good point. Let's use a 4 gram sheet of 8.5"x11" paper, and orbital velocity is 17,318 mph (7,742 m/s).

KE = 1/2 * m * v^2
KE = 1/2 * .004 kg * 7823^2 m/s
KE = 119,877 Joules

119,877 J = 33.3 watt hours, or 3,960 watts over 30 sec.

Now 3,960 watts for 30 sec seems like a lot, but as already mentioned only a small fraction is delivered into the paper.

By your calculations, that's the rough equivalent of holding the paper up to a hair dryer.

Also we're dealing with a paper airplane (not a sheet), which would likely decelerate slower, thus spreading the energy release over a longer period.

Nope. Not buying it. The Shuttle takes several minutes. I can only imagine it, but my rough wag says that 30 seconds is about right to terminal velocity, which, at 398,000 ft is probably still well into MACH velocities, but in such a thin atmosphere, terminal velocity is nevertheless "cool."

But what do you do when the paper passes through the thermosphere, where "Temperatures are highly dependent on solar activity, and can rise to 15,000°C." - Wikipedia (http://en.wikipedia.org/wiki/Thermosphere)

Oooops...

Toast again.

From the STS-107 data, we know reentry heating isn't even measureable until about 420 sec after entry interface.

So that firery red stuff seen in the shuttle's windows is meaningless...

As far as the "286 W" claim I think you need to re-work your math.

Van Rijn
2008-Jul-23, 08:00 PM
But what do you do when the paper passes through the thermosphere, where "Temperatures are highly dependent on solar activity, and can rise to 15,000°C." - Wikipedia (http://en.wikipedia.org/wiki/Thermosphere)

Oooops...

Toast again.

Not at all. From that same page:

Even though the temperature is so high, one would not feel warm in the thermosphere, because it is so near vacuum that there is not enough contact with the few atoms of gas to transfer much heat.

cjl
2008-Jul-23, 08:51 PM
That's the thing about seemingly high temperatures in gas that thin - it doesn't behave at all like you might expect.

mugaliens
2008-Jul-25, 04:04 PM
Perhaps not, but it's not cooling it off, any...

However, the paper would probably be able to radiate heat away far faster than the hot gas could impart it via conduction.

cjl
2008-Jul-25, 09:58 PM
That would be what I would think.

mugaliens
2008-Jul-26, 09:54 AM
Still, given the fact that even dust-sized particles glow brightly into ultraviolet and oxidize completely into micrometeorite ash, I still don't think a paper airplane has a snowball's chance of surviving reentry.

cjl
2008-Jul-26, 03:14 PM
That happens at a much lower altitude though, and dust particles have a much larger mass per cross section (just because their cross sectional area is so small).

mugaliens
2008-Jul-27, 02:25 PM
That happens at a much lower altitude though, and dust particles have a much larger mass per cross section (just because their cross sectional area is so small).

The smaller one gets, the less effect density has on atmospheric penatration. This is due to Reynolds numbers (http://en.wikipedia.org/wiki/Reynolds_number), which is the ratio of inertial forces to viscous forces (specifically, the dynamic pressure divided by the shearing stress).

It's why aircraft with a high Renolds number, such as a 777, will always outperform smaller jets, such as a Cessna Citation III, in terms of pounds of fuel burned per pound of cargo per nautical mile travelled.

It's also why mini-UAVs such as the BATMAV have a different form factor than larger UAVs like the Predator, and why micro-UAVs (a few inches across) swing fat props instead of the more slender ones found on larger engines.

To larger aircraft, Earth's atmosphere "feels" like a highly fluid medium, with very little viscosity. To micro-UAVs, however, it feels more like creamy soup. Just as normal props are slender, ships props are fat, as water is far more viscous than air. Similarly, when you cut the size way down, air starts behaving on that small scale more like the much more viscous water.

It's also why bees really can fly.

That claim arose because the "engineers" applied common lift/drag equations applicable to normal aircraft without scaling them appropriately using the correct Reynolds numbers for bees.

Back to the concept, at the micrometeorite scale, mass/frontal area plays far less of a role due to it's minute Reynolds number.

mugaliens
2008-Jul-27, 06:35 PM
Joema, I'm going to redo the math on the reentry calcs... In detail, this time, showing my work as I go.

PLEASE read all the way through to the end, as I do a lot of rabbit-trailing and discover a few things along the way. It all comes together towards the end, though, so mission accomplished.

Units: SI base (http://en.wikipedia.org/wiki/SI_base_unit)and derived (http://en.wikipedia.org/wiki/SI_derived_unit), thoughout.

Primer:

Force is measured in Newtons (N). It's m*kg/s.

Pressure is measured in Pascals (Ps). It's force distributed over an area, N/m[sup]2.

Energy/work/heat is measured in Joules (J). It's force done over a distance, power generated over time, and charge moved by a voltage. Thus, N∙m = C·V = W·s. You buy power from the electric company in kW-hrs (kilowatt hours). 1 kW-hr equals 3,412 BTU.

Power is given in Watts (W). It's the rate of energy/work/heat over time, or the current moved under a voltage. Thus, J/s = V·A. One horsepower is about 745 watts, and about 2,544 BTU/hr.

Temperature is given in Kelvins (K).

Kinetic energy is given in joules, and it's equal to 1/2 the mass times the square of the velocity, or:

KE=1/2 * m * v2[/suo]

The mass is in kg and the velocity is in m/s.

Paper thickness is measured in pounds, the weight of a ream (500 sheets). Standard paper is a ream that has been cut into 4 equal pieces. Thus, a 500-sheet stack of 8-1/2" x 11", 20 lb paper weighs 4 pounds. Thus, one sheet weighs 4/500 lbs.

That's 0.008 lb, or 0.003 628 738 96 kilograms. Let's round this to 3.63 x 10[sup]-3 for the sake of expedience (also can be written as 3.63E-03 for simplicity).

That's out mass.

As Astromark said, "the shuttle de-orbits at 39000km/hr." That's 10,833 m/s.

Thus, KE = 1/2 * 3.63E-03 * 10,8332 = 212,936 Joules.

This equates to the following:

201.8 BTU

50,871 calories (a calorie is the amount of energy required to raise the temperature of one gram of water by 1 deg C)

This is important, as we'll get to this later...

157,054 ft-lbs

1.6E-03 gallons of gasoline. That's .157 oz of gasoline

0.059 kilowat hours (59.15 watt hours)

Back to the calories.

It's 50,871 calories, which means that it's enough energy to raise the temperature of one gram of water by 50,871 degrees, or 50871 grams of water by 1 degree.

Our paper airplane weighs 3.628739 grams. Call it 3.62 grams. If instead of our paper airplane, if it were 3.62 grams of water, the KE has enough energy to raise it's temperature by 14,019 degrees (ignoring the latent heat of evaporation, for a moment).

This is why micrometeorites are white-hot incandescent streaks across the sky.

Now...

Different materials have different properties when it comes to heating them. Heat capacity (how much energy it takes to raise the material's temperature by a certain amount) is given by:

Cp = m * cs

Let's look at this further... Aluminum has a cs of 0.897 J/g*C, which is 0.214 cal/(g*deg C). Thus:

Cp = 3.62 grams * 0.214 cal/(g*degC) = 0.77468 cal/degC.

So, if instead of a paper airplane we have 3.62 grams of aluminum, we have:

Delta degC = 50,871 calories / 0.77468 cal/degC = 65,667 degC.

Now...

The specific heat capacity of paper is roughly equivalent to that of wood, or about 0.42 J/g*C, which comes to 0.1 cal/g*C.

Thus:

Cp = 3.62 grams * 0.1 cal/(g*degC) = 0.362 cal/degC.

And:

Delta degC = 50,871 calories / 0.1 cal/degC = 508,710 degC.

This is how much hoter that piece of paper would become if 100% of it's kinetic energy were absorbed into the paper itself.

Now, a bit about reentry physics:

During reentry, the energy is dissipated in one of five ways: Shock waves (via sound as well as compression heating), heating of the air (friction), heating of the paper (friction), and radiation (radiating heat away after it has been absorbed due to friction).

Furthermore, and counterintuitively, Allen and Eggers (NACA) discovered in 1951 that blunt objects make the best heat shields, as the air mass immediately in front of the object (bow wave) acts as a cushion to push the shock wave and heated air mass out of the way.

So, let's use this technology to our advantage and wad our paper airplane up into a ball in order to give it the greatest chance of survival. An added benefit is that the ball will tend to turn, which exposes more sides to the worst of it, spreading out the heating. Besides, most of the reentry data is for spheres and spherical sections.

From Wiki: "An approximate rule-of-thumb used by heat shield designers for estimating peak shock layer temperature is to assume the air temperature in kelvins to be equal to the entry speed in meters per second - a mathematical coincidence." This works for objects both large and small, high density and low density, because the peak shock layer temperature is dependant on velocity in the earliest stages of reentry, before the objects have slowed significantly.

Thus, the peak shock layer temperature for our little ball of paper, in kelvins, will be approximately 10,833 deg K. This compares with the Apollo 4's command module peak shock temp of 10,700 deg K.

Back to our paper "airplane" (aka, wadded up in "reentry mode").

What are the proportions of each of the energy dissipation modes? (sonic and shock heating, heating of paper due to friction, heating of the air due to friction, and radiation?)

Let's look at each:

Sonic - Any who's heard a baseball whishing into the catcher's mitt knows that it takes energy to produce sound. Thus, some of the KE is radiated away as sound. However, sonic radiation is most pronounced at sea level, and progressively less prominant as altitude increases. Throughout most of its deceleration, our paper ball only rids itself of a tiny fraction of it's energy via sonic radiation.

Shock - This is identical to sound, but I've broken the two apart for clarification. Sonics are the subsonic "shhhh" sounds produced by the airflow over the surface of the paper ball. Shock is the shock wave produced. Interestingly enough, as you can see from these photographs (http://en.wikipedia.org/wiki/Image:Blunt_body_reentry_shapes.png), no part of a blunt reentry vehicle pierces the shock wave. In other words, all airflows over the ball are subsonic. While creating shock waves does take energy, the amount of energy it takes at any given velocity is proportional to the density of the ambient air. At 400,000 feet, not a lot of energy is being absorbed into the production of the shock wave.

Friction - There are really two parts to friction. The first part is the friction due to air moving over the surface of the ball. This isn't very much, because, as noted before, the airflow over the surface is subsonic. The second is the friction imparted by the surface to the air. Amazingly enough, these are roughly equal. The bad thing, however, is that while the heated paper is radiating it's heat outward, the heated air is radiating some of it's heat back to the paper. However, only about 1/20th is radiated to the ball, as the rest is radiated where the ball isn't. Nevertheless, most of the friction imparted to the air remains with the warmer air streaming behind the ball, so not really a factor.

Compression - This is the bad boy. What happens when you pressurize air? It gets warm. In the case of reentry, it gets hot.

Let's go back to the total Kinetic Energy for a moment. One unit of measurement was the BTU, specifically, around 200 BTU. Since a BTU is the amount of heat required to raise the temperature of 1 lb of water by 1 deg F, 200 BTU can heat 200 lb of water by a degree, 1 lb of water by 200 deg, or 3.62 grams of water by 25,000 degrees.

Wow. That's a lot of heat.

And most of that heat is produced at the leading edge of the shock wave, where it's closest to the paper ball. But it's also present in the shock wave which wraps itself around about 3/8ths of the ball, as well.

So, to sum:

Sonics: Negligible.

Shock: Minimal.

Friction: So-so.

Compression: (occurs within the shock wave) Accounts for roughly 80% of the energy dissipation. Half of the heat of compression is radiated outward, the other half inward. Since about 3/8th's is exposed, that's about 3/16ths of the heat of compression, or roughly 3/16 * 80% * 200 BTU, which comes to 30 BTU.

That's a lot better than 200 BTU! However, that's still able to raise the temperature of 3.62 grams of water by 3,750 degrees.

And that's the energy that's imparted to our paper ball. If it occurred all at once, the paper would turn to ash in an instant.

As I said before: Toast.

To make matters worse, the entire paper isn't heated. Rather, just the outer layer of the ball is heated. Since that's perhaps 1/3 of the mass of the entire ball, we're back to temps of around 10,000 degrees.

However, it doesn't occur in an instant. Rather, it occurs over a period of time. How long?

Well, I'd have to break out some texts I haven't used in a long, long time.

Still, we can get to an approximately solution by looking at some of the other times. Micrometeorites burn up in a second or two, whereas the shuttle goes through several minutes. Let's say it's 3.5, and then err on the side that gives our paper ball the greatest chance for success by saying it'll take 5 whole minutes to slow down to terminal velocity.

Five minutes.

Now - how fast does paper shed (radiate) heat? To answer that we must delve into Kirchoff's law of thermal radiation, as well as the laws of Stefan-Boltzmann and Wien.

To keep it simple, let's assume the worst, which is actually the best, namely, that our paper ball gets to 450 deg F and no hotter, and that it's a sphere with a radius of .0381 m (it's a three-inch diamater ball of paper), thus having a surface area of 4*pi*r2, or 0.01824 m2. 450 deg F is 505 deg K.

The appropriate formulas are given here (http://en.wikipedia.org/wiki/Thermal_radiation#Formula).

Thus, at 505 deg K, the max wavelength is given by b/T, where b is Wien's displacement constant. The result is 5.7E-6, which is very low indeed.

Instead of crunching the numbers manually, however, I found an online calculator (http://infrared.als.lbl.gov/content/black-body-emission-calculator)... Although it's a black-body emission calculator, if you simply enter the temp over 4*pi steradians (the solid angle of a sphere) it very handily gives you the rate at which that sphere at that temp is radiating

Result: 269 Watts.

Recalling the beginning of this post, that's the rate at which our minimally pre-Toast ball of paper is dissipating energy. Since this is also 269 Joules per second, over 5 minutes (300 seconds) it can, at 505 deg K, shed a total of 80,700 Joules of energy.

And, since the total kinetic energy is just 212,936 Joules, of which just 31,652 joules actually makes it to the paper, under ideal conditions and with some key assumptions, the ball of paper looks like it might survive.

That's a factor of safety of 2.5.

Key assumptions:

1. That deceleration takes at least 5 minutes.

2. That deceleration is fairly constant over those five minutes.

3. That the ball tumbles randomly, resulting in it's entire surface being heated evenly.

4. That no "corners" from the wadding-up process are sticking out into the slip-stream.

However, as design criteria go, particular with respect to safety, a 2.5 factor of safety is pretty slim.

Guesses:

1. The deceleration is not constant, and that there will be a period where the heating is much faster than the paper's ability to shed heat.

I'd guess this cuts my factor of safety in half (50%). Let's be optimistic, though, and choose to believe it's only cutting it by 30%. we're still down to an SF of 1.75.

2. The surface isn't uniformly smooth, thus, there will be several "hot spots" where the temperature will be significantly higher than average.

I'd guess this cuts it by another 30%. It's getting slim, at just 1.225.

3. That some of the radiation (perhaps 30%?) is radiated inward, heating the entire mass of the ball, which then re-radiates that back through the surface.

Looks like another 30%. Uh-oh, that's 0.8575, which is less than 1, which usually indicates failure.

You can use these percentages on the 31,652 Joules, which raise that to 92,279 Joules (actually, not really, but localized and generalized heating produce the same effect). When compared to it's 80,700 Joule-shedding ability, the result is the same: Failure.

In other words, folks...

TOAST.

:whistle:

mugaliens
2008-Jul-27, 06:40 PM
That happens at a much lower altitude though, and dust particles have a much larger mass per cross section (just because their cross sectional area is so small).

Actually, most space dust/micrometeors are quite porous.

joema
2008-Jul-27, 09:14 PM
...
As Astromark said, "the shuttle de-orbits at 39000km/hr." That's 10,833 m/s...
Your initial number is incorrect, and since KE is a squared term, the error is magnified.

10,833 m/s is 24,232 mph, or roughly earth escape velocity. That's the velocity an Apollo capsule is traveling at entry interface when returning from the moon.

However we aren't talking about a paper airplane returning from the moon, but deorbiting from a low earth orbit.

It would be moving considerably slower, so KE (being a squared term) would be much less.

All calculations should be re-done based on a velocity of about 7,700 meters per second, which is velocity at entry interface returning from a typical shuttle orbit.

mugaliens
2008-Jul-29, 04:23 PM
Your initial number is incorrect, and since KE is a squared term, the error is magnified.

10,833 m/s is 24,232 mph, or roughly earth escape velocity. That's the velocity an Apollo capsule is traveling at entry interface when returning from the moon.

er...

As Astromark said, "the shuttle de-orbits at 39000km/hr." That's 10,833 m/s...

My initial numbers... vel=39,000 km/hr * 1,000 m/km / 3,600 s/hr = 10,833 m/s...

They look correct to me, unless Astromark's original estimate/figure was off.

However we aren't talking about a paper airplane returning from the moon, but deorbiting from a low earth orbit.

Last time I checked, the Shuttle doesn't return from the Moon...

...and checking Wikipedia...

"The five species model is only usable for entry from low Earth orbit where entry velocity is approximately 7.8 km/s. For lunar return entry of 11 km/s, the shock layer contains a significant amount of ionized nitrogen and oxygen."

Hmmm... You're right, but only to a point.

It would be moving considerably slower, so KE (being a squared term) would be much less.

That's the point, as 7.8 km/s isn't "considerably" slower than 10.8 km/s, and even the square of the difference still comes to a ratio of 52.2% (thus, 52% of the originally calculated KE).

All calculations should be re-done based on a velocity of about 7,700 meters per second, which is velocity at entry interface returning from a typical shuttle orbit.

So now it's 7.7 km/s, not 7.8? I'll assume some slop, then, as it's an unguided paper ball, after all, not a high-tech space shuttle controlled by 5 onboard computers and several supercomputers on the ground.

While I'm at it, I'm changing the 5 min timeline. Here's why: 7.8 km/s equates to 25,590 ft/s. As most of us know, the appreciable atmosphere of the Earth extends to between 300,000 ft and 400,000 ft, depending on who you talk to. Thus, straight down, that 25.6 kft/s disappears in about 15.6 seconds.

On a more appropriate 30 deg approach it goes by in about 30 seconds. 5 deg gets you 179 seconds, or three minutes.

However, that's to the surface of the Earth.

As the ball is falling, it's slowing rapidly, such that a lightweight paper ball will reach terminal velocity in about one minute, not five.

That, although we have 1/2 the KE of the previous calculations, we also have about 1/5th the time, and that's still assuming a steady deceleration, which it most certainly is not.

Like I said, folks - toast, particularly once we remove the very optimistic assumptions I included in the original assesment.

NOW:

Before you pick apart anything else, have the decency to throw out your own calculations.

galacsi
2008-Jul-29, 05:35 PM
I pick apart an other thing : This time for going down is ridiculously too short. IMO your paper plane is a glider , loosing may be 1 m/s by s . So if successful it will glide for two hours before reaching the surface.

samkent
2008-Jul-29, 05:47 PM
I don’t have the math to back up my assumptions but…..

You assumptions that the speed of the airplane will remain close to 7.8 km/s may not be correct, here why.

If the airplane contacts one molecule of atmosphere at 7.8 km/s it would not burn up since one molecule cannot consume (in flames) the entire paper. But what effect will it have on the speed of the plane while it pokes a molecule sized hole in the paper? I (and others) propose that the forward speed will diminish very quickly without burning.

Thought experiment:
Drop a baseball and a wadded up sheet of paper out the window of your car at 60 miles/hr. The distance traveled when each contacts the pavement will be vastly different.

It’s this difference that will allow the paper airplane to survive, albeit with several thousand microscopic holes in it. At some point these molecules will stop making holes and start creating lift. As we all know paper airplanes normally travel a few tens of feet before they loop, depending on speed. In my opinion a paper airplane released from the shuttle just after the re entry burn will still be looping it’s way to the ground long after the shuttle has landed.

korjik
2008-Jul-29, 06:07 PM
I don’t have the math to back up my assumptions but…..

You assumptions that the speed of the airplane will remain close to 7.8 km/s may not be correct, here why.

If the airplane contacts one molecule of atmosphere at 7.8 km/s it would not burn up since one molecule cannot consume (in flames) the entire paper. But what effect will it have on the speed of the plane while it pokes a molecule sized hole in the paper? I (and others) propose that the forward speed will diminish very quickly without burning.

Thought experiment:
Drop a baseball and a wadded up sheet of paper out the window of your car at 60 miles/hr. The distance traveled when each contacts the pavement will be vastly different.

It’s this difference that will allow the paper airplane to survive, albeit with several thousand microscopic holes in it. At some point these molecules will stop making holes and start creating lift. As we all know paper airplanes normally travel a few tens of feet before they loop, depending on speed. In my opinion a paper airplane released from the shuttle just after the re entry burn will still be looping it’s way to the ground long after the shuttle has landed.

A paper airplane will burn up just like anything else.

First, a molecule striking the plane wont just punch a hole in the plane. It will knock a few molecules around and deposit some heat. The heat is from the kinetic energy of the molecule's impact being dissipated.

Second, you wont hit thousands of molecules, you will hit trillions. Per meter travelled. That many hits will deposit more than enough energy to vaporize the plane.

Third, at some point your plane will have to transition from supersonic speeds to subsonic. I doubt a paper airplane can survive that.

samkent
2008-Jul-29, 06:41 PM
Second, you wont hit thousands of molecules, you will hit trillions. Per meter travelled. That many hits will deposit more than enough energy to vaporize the plane.
It will start with one per minute then one/sec and gradually build up. But at the same time the speed will be bleeding off quickly. Also not all of the energy of that one molecule will be transferred to the paper. Only a small fraction will transfer before it pops through the other side. Plus cannot vaporize paper without air. A few molecules of O2 per minute just won’t do it.

Third, at some point your plane will have to transition from supersonic speeds to subsonic.

I expect it will be subsonic long before the atmosphere gets thick enough to create a shock wave. Remember paper doesn’t have enough mass to sustain high rates of speed as compared to say a baseball. My feelings are that it would reach mach 1 while still being over 90 miles high. It wouldn’t take long before it would be equivalent to dropping it from a ladder with orbital reach.

joema
2008-Jul-29, 07:14 PM
...I (and others) propose that the forward speed will diminish very quickly without burning...
Yes, that is called the "shuttlecock effect", and it has already been tested (albeit at a lower velocity) by Burt Rutan's SpaceShipOne. That vehicle relies on a high-drag configuration to decelerate in the extreme upper atmosphere before reentry heating is a major factor. It manages to reenter without a heat shield, unlike the X-15.

A prototype paper airplane has survived Mach 7 in a wind tunnel. Pictures here: http://www.discoverychannel.ca/reports/article.aspx?aid=6514

Researchers plan on treating the paper airplane with silicon to improve its heat resistance. The prototype paper plane has already achieved nearly 300C (572F) without burning up.

I'm sure the final configuration will have drag characteristics calculated to best ensure its survival.

They'll calculate the sustainable heat pulse the special paper can tolerate, then fold the plane to provide a drag and deceleration profile that best leverages this. If they think a 30 sec or 5 min. deceleration period is more favorable, they'll fold the plane to achieve one or the other.

Regardless of what simplistic calculations indicate, we have actual reentry temperature data from STS-107. It indicates there's very little heating -- even for totally exposed, unprotected items -- for about 200 sec after entry interface. Multiple graphs and tables are available in the CAIB Working Scenario, available here: http://history.nasa.gov/columbia/CAIB.html

Columbia's left wing had a hole in it. Behind that (exposed unprotected to reentry heating) were wires, etc. They showed virtually no temperature rise for about 200 sec after entry interface. During that same low-heating period a paper plane with low mass and high surface area would decelerate.

NEOWatcher
2008-Jul-29, 07:15 PM
Plus cannot vaporize paper without air.
Why not?
I can understand that paper won't burn without air, but vaporization is just a state change. Certainly, the heat can excite the paper enough to start knocking molecules off, no?

mugaliens
2008-Jul-29, 07:57 PM
I pick apart an other thing : This time for going down is ridiculously too short. IMO your paper plane is a glider...

Not before it slows to terminal velocity, it's not. And since pointy gliders quickly become toast (which is why all NASA's designs are blunt body, to keep the red-hot shock wave well out in front), we're talking about a wadded-up paper ball, not a winged glider.

Winged glider === definate toast. Toast man. Toastmania. Post-Toasties. Toasted oats, rolled and toasted! Especially if it's got a pointy nose on it. But any non-spherical shape --> toast.

Ball of paper = possible/probable toast, but best chance for success.

ggchuck
2008-Jul-29, 07:58 PM
A couple of things bother me about this discussion. I don't have the math ability to fall back on, so I have to go by intuition.

The first is the statement that low mass meteors vaporize. Comparing them to a wad of paper released at orbital speed seems irrelevant as: 1. They tend to enter the atmosphere at a steep angle, 2. They are probably entering the atmosphere at the orbital speed of the earth around the sun, not the speed of orbiting the earth, 3. They are more dense than a wad of paper.

My major objection to the logic presented here is that the energy dissipation seems to be accounted for by heat. It seems to me that most of the energy loss of the paper ball (or airplane) would be accounted for by a direct transfer of KE to the air molecules that are encountered, not the absorption of heat.

i.e. shooting ping pong balls at bowling ball (the shuttle) vs. a styrofoam ball (the paper wad), the loss of energy would go changing the KE of the ping pong balls, not to heat. The bowling ball would not slow significantly until the density of them would cause them to crush (and release heat). The styrofoam ball would be slowed significantly and may not have the energy to crush a ping pong ball even as they became more dense.

mugaliens
2008-Jul-29, 08:00 PM
In my opinion a paper airplane released from the shuttle just after the re entry burn will still be looping it’s way to the ground long after the shuttle has landed.

I agree whole-heartedly!

But we were talking about a paper airplane surviving reentry. Without the Shuttle around it to take it through the tough parts.

mugaliens
2008-Jul-29, 08:05 PM
I expect it will be subsonic long before the atmosphere gets thick enough to create a shock wave.

MACH (beyond which we have a shock wave) decreases as altitude increases. It's highest at sea level.

The moment it encounters it's first molecule, it's way, way beyond MACH 1. By the times it begins to slow appreciably (say, 90% of its original entry velocity), it's still well beyond MACH 20.

Thus, once it begins interacting with the atmosphere, the lower it goes, the slower it goes, falling through MACH 19, then 18, then... ...then 7, 6,... ...2, 1, and finally subsonic.

Subsonic ash, that is...

mugaliens
2008-Jul-29, 08:14 PM
A couple of things bother me about this discussion. I don't have the math ability to fall back on, so I have to go by intuition.

Ok. I have some math, along with some tuition...

The first is the statement that low mass meteors vaporize. Comparing them to a wad of paper released at orbital speed seems irrelevant as: 1. They tend to enter the atmosphere at a steep angle, 2. They are probably entering the atmosphere at the orbital speed of the earth around the sun, not the speed of orbiting the earth, 3. They are more dense than a wad of paper.

1. Who says? Actually, they enter the Earth's atmosphere at all angles. Some, even the smaller ones, just skim the nether reaches before continuing on - sort of a glancing blow.

2. They enter the Earth's atmosphere at all numbers of speeds, from suborbital speeds to many times faster than does the shuttle.

3. I would say most are more dense. Some, however, are as nebulous as the dust bunnies under your bed.

My major objection to the logic presented here is that the energy dissipation seems to be accounted for by heat. It seems to me that most of the energy loss of the paper ball (or airplane) would be accounted for by a direct transfer of KE to the air molecules that are encountered, not the absorption of heat.

I covered this extensively, in excrutiating detail, in an earlier post (http://www.bautforum.com/questions-answers/69247-paper-airplane-reentry-3.html#post1290828). But if you want to pit your intuition against the math of hundreds of others and the results of 57 years of testing, be my guest.

mugaliens
2008-Jul-29, 08:26 PM
Yes, that is called the "shuttlecock effect", and it has already been tested (albeit at a lower velocity) by Burt Rutan's SpaceShipOne. That vehicle relies on a high-drag configuration to decelerate in the extreme upper atmosphere before reentry heating is a major factor. It manages to reenter without a heat shield, unlike the X-15.

Rutan himself acknowledges that the SS1 wouldn't survive the velocities of an orbital reentry. SS1 essentially goes straight up, with both kinetic and potential energies well below anything required to achieve LOE. Thus, it has a lot less energy to lose on the way back. It's shuttlecock design works well, as it's essentially travelling at zero velocity at it's apex, barely into space. Thus, it really doesn't get going very fast before the density builds up to begin checking it's velocity.

A prototype paper airplane has survived Mach 7 in a wind tunnel.

Cool! Although Mach 7 at 400,000 ft isn't terribly impressive. I guarantee you that if he tried that at sea level he'd be seeing temps on the order of 8,500 deg K.

Also, I'm very surprised he went with the pointy-nose configuration. All such tests resulted in meltdowns at higher MACHs because it concentrates the peak shock temp in a much smaller space, and that space is a lot closer to the object. Think of holding a white-hot pencil torch to it's pointy nose.

Researchers plan on treating the paper airplane with silicon to improve its heat resistance. The prototype paper plane has already achieved nearly 300C (572F) without burning up.

Hey! That's cheating... It says that it stiffens the paper quite a bit, too.

I'm sure the final configuration will have drag characteristics calculated to best ensure its survival.

Such as the spherical or semispherical shapes NASA figured out years ago?

Hey, I don't invent this stuff, nor do I pull it out of a hat. I look it up! Why should I repeat the mistakes of others when I don't have to?

mugaliens
2008-Jul-29, 08:30 PM
Columbia's left wing had a hole in it. Behind that (exposed unprotected to reentry heating) were wires, etc. They showed virtually no temperature rise for about 200 sec after entry interface. During that same low-heating period a paper plane with low mass and high surface area would decelerate.

To some extent, yes, just as the more dense Columbia continued to accelerate despite the resistance. There is a point during any reentry where acceleration changes from positive to negative. That point would occur sooner with a paper ball than with the Shuttle, but the general profile is the same.

I strongly disagree that the paper ball would decelerate to a terminal velocity (where acceleration again reaches it's second equilibrium (well, nearly so)) within that first 200 seconds. "Virtually no temperature rise" is the result of "virtually no friction/shock wave heating."

ggchuck
2008-Jul-29, 09:22 PM
1. Who says? Actually, they enter the Earth's atmosphere at all angles. Some, even the smaller ones, just skim the nether reaches before continuing on - sort of a glancing blow.

2. They enter the Earth's atmosphere at all numbers of speeds, from suborbital speeds to many times faster than does the shuttle.

3. I would say most are more dense. Some, however, are as nebulous as the dust bunnies under your bed.I say, by intuition. I visualize the earth like a flyswatter through a swarm of bees. Granted some bees may overtake it and hit it from the side or behind, but most will make strong contact on the face of the flyswater... and those are the ones that burn brightly. Do you contend that all meteoroids do flash as meteors when they encounter the earth's atmosphere?

I covered this extensively, in excrutiating detail, in an earlier post (http://www.bautforum.com/questions-answers/69247-paper-airplane-reentry-3.html#post1290828). But if you want to pit your intuition against the math of hundreds of others and the results of 57 years of testing, be my guest.I did read this post. The quote from it that I took interest in was "During reentry, the energy is dissipated in one of five ways: Shock waves (via sound as well as compression heating), heating of the air (friction), heating of the paper (friction), and radiation (radiating heat away after it has been absorbed due to friction)."

I assume 'shock waves' accounts for 2 of the five ways or I have to fall back on the lack of my math background to explain why I can't count more than four. As far as I can tell, none of these relate to a transfer of KE from the reentry vehicle to the air molecule. Is it your contention that this transfer has no bearing on the energy that must be absorbed by the paper? I did not see any reference to it in your list of energy dissipation.

I'm not disputing your math, I'm perplexed because I don't see a reference that addresses or refutes the idea that much of the KE energy can be dissipated without heat.

mugaliens
2008-Jul-29, 09:51 PM
I say, by intuition. I visualize the earth like a flyswatter through a swarm of bees. Granted some bees may overtake it and hit it from the side or behind, but most will make strong contact on the face of the flyswater... and those are the ones that burn brightly. Do you contend that all meteoroids do flash as meteors when they encounter the earth's atmosphere?

What does your intuition say about how the Sun rises in the East, sets in the West, and how you view most of your meteorites in the evening, in the area of the sky behind where the Earth just was, it's backside, so to speak?

Meteorites travel at all speeds, but the fastest travel around 42 km/s. By comparison, the Earth orbits the Sun at 29.7 km/s. So there is indeed some "sweeping" going on. But if you've ever seen them after sundown, you're watching the faster ones that are coming up behind us.

I assume 'shock waves' accounts for 2 of the five ways or I have to fall back on the lack of my math background to count more than four.

Yes, as mentioned carefully in the post, two: sonic and shock compression. No, these aren't the same, as sound travels, whereas shock compression becomes hot, luminescent gas, the heat of which is both radiated away, as well as carried around behind the reentry object.

As far as I can tell, none of these relate to a transfer of KE from the reentry vehicle to the air molecule. Is it your contention that this transfer has no bearing on the energy that must be absorbed by the paper? I did not see any reference to it in your list of energy dissipation.

I detailed all this in the post. You're just going to have to go back and read it more thoroughly (and don't get ticked - you missed it, and I'm not going to hog white space typing it again).

I'm not disputing your math, I'm perplexed because I don't see a reference that addresses or refutes the idea that much of the KE energy can be dissipated without heat.

Trebuchet. Converts potential energy (raised counterweight) into kinetic energy (flung boulder). It works in reverse, too, but your aim better be pretty good or you'll wind up with a pile of splinters.

And look, Ma - no heat!

Consider reading through some examples (http://en.wikipedia.org/wiki/Kinetic_energy#Some_examples), particularly where it says, "However it becomes apparent at re-entry when the kinetic energy is converted to heat."

And then consider that heat due to friction is only one way kinetic energy is dissipated when an object hurtles through the atmosphere. By the generation of sound waves (requires energy, otherwise your stereo system wouldn't need a plug), shock compression (that big bright glow at the leading edge, which wraps it's way around towards well behind the reentry object), and last but not least, direct contact frictional heating.

The SR-71 get's a lot of that, due to it's sharp leading edge and the fact that the root of the shock was is actually aft of the various leading edges. Thus, you have MACH+ airflow along the first several inches of the leading edges. That direct contact makes for a lot of friction.

By contrast, during reentry, spherical and semispherical bodies have a blunt shock wave that's well ahead of the object. At no point does the air travelling over a blunt body actually break MACH at the surface of the body. That only happens when the angle of a leading edge is sharper than the shock wave.

ggchuck
2008-Jul-29, 10:54 PM
What does your intuition say about how the Sun rises in the East, sets in the West, and how you view most of your meteorites in the evening, in the area of the sky behind where the Earth just was, it's backside, so to speak?I'm not sure what your question is. I think we both agree that meteors come from all angles, my point was that most will average in the speed range of earth's 29.7 km/s. I noticed you avoided my question however.

I detailed all this [heatless KE transfer] in the post. You're just going to have to go back and read it more thoroughly (and don't get ticked - you missed it, and I'm not going to hog white space typing it again).Either we are talking about different things or I missed it again. I'm not being deliberately obtuse, my confusion remains.

Consider reading through some examples (http://en.wikipedia.org/wiki/Kinetic_energy#Some_examples), particularly where it says, "However it becomes apparent at re-entry when the kinetic energy is converted to heat."I understand your example. What I'm trying to address is the very next example in the source you referenced: "Kinetic energy can be passed from one object to another. In the game of billiards, the player gives kinetic energy to the cue ball by striking it with the cue stick. If the cue ball collides with another ball, it will slow down dramatically and the ball it collided with will accelerate to a speed as the kinetic energy is passed on to it."

I visualize his KE transfer would take place before the paper encounters dense enough air to heat it and before shock waves become a dominant factor. As stated before, I don't have the math to know if it is significant, but I haven't seen it addressed either and I noticed you bypassed my second question which addressed it.

mugaliens
2008-Jul-30, 07:09 PM
I'm not sure what your question is. I think we both agree that meteors come from all angles, my point was that most will average in the speed range of earth's 29.7 km/s. I noticed you avoided my question however.

No.

Either we are talking about different things or I missed it again. I'm not being deliberately obtuse, my confusion remains.

No, yes. Yes, and yes.

I understand your example. What I'm trying to address is the very next example in the source you referenced: "Kinetic energy can be passed from one object to another. In the game of billiards, the player gives kinetic energy to the cue ball by striking it with the cue stick. If the cue ball collides with another ball, it will slow down dramatically and the ball it collided with will accelerate to a speed as the kinetic energy is passed on to it."

The reason I didn't reference the next example is that it's not germain to the discussion. The reason I addressed the sentance immediately prior is that it is germain to the discussion.

I visualize his KE transfer would take place before the paper encounters dense enough air to heat it and before shock waves become a dominant factor.

I visualize a petite brunette, but that's not going to happen, either. The energy (KE) has to go somewhere, and I've already exhaustively detailed the modes by which it's disippated and their relative contributions to that dissipation.

As stated before, I don't have the math to know if it is significant...

I do.

...but I haven't seen it addressed either and I noticed you bypassed my second question which addressed it.

Keep looking, and no. Keep re-reading, too.

Aside: This board is about discovery, discussion, theorizing, and learning. I'm one of the few who will argue the points of an argument to death, dig up the body, and argue some more.

But I do my best to stay on topic. I do not perceive either arguing that someone else didn't address something to my liking, or ignoring their posts or their responses to yours as staying "on topic." Nor do I view needlessly defending what I've previously stated, exhaustively, as stying on topic.

So - have a nice day.

ggchuck
2008-Jul-30, 08:25 PM
No.

No, yes. Yes, and yes.Cute.

The reason I didn't reference the next example is that it's not germain to the discussion.It related directly to what I was discussing. If you don't want to respond to my query or if you don't have the background to address it, then why bother responding to my original post?

I've already exhaustively detailed the modes by which it's disippated...And you deemed my question about that list not worthy of a serious response.

I visualize a petite brunette, but that's not going to happen, either.Perhaps if you improved your people skills...

Aside: This board is about discovery, discussion, theorizing, and learning. Which is why I'm puzzled about your response. I feel like someone who comes in on a discussion where it is shown in mathematical terms that a big object and a small object fall at the same speed. I wonder if air resistance would have an effect and I'm told that it's off topic and re-read the equations even though they don't address air resistance.

So - have a nice day.You too.

korjik
2008-Jul-30, 09:00 PM
It will start with one per minute then one/sec and gradually build up. But at the same time the speed will be bleeding off quickly. Also not all of the energy of that one molecule will be transferred to the paper. Only a small fraction will transfer before it pops through the other side. Plus cannot vaporize paper without air. A few molecules of O2 per minute just won’t do it.

Where are you getting your numbers? At 150km density is about 10^-12 g/cm3, giving about 3.5 million oxygen atoms (used for convienence) per cubic centimeter. Assuming a cross section of 1cm3, and a velocity of 8500 m/s, you should get about 32 billion collisions per second.

Also, assuming that each impact is such that it imparts an impulse of mv, where v is the velocity of the plane, and m is the air molecule mass, you get an accelleration of 7^10-5 m/s^2.

Assuming a forward velocity of 8500 m/s and a decent velocity of 1 m/s, by the time you reach 100km, your forward velocity will be reduced by 3.6 m/s. The number of collisions is up to 32 trillion per second. Acceleration has increased to .07 m/s^2. (acc numbers are low, did not integrate the increase in density).

I expect it will be subsonic long before the atmosphere gets thick enough to create a shock wave. Remember paper doesn’t have enough mass to sustain high rates of speed as compared to say a baseball. My feelings are that it would reach mach 1 while still being over 90 miles high. It wouldn’t take long before it would be equivalent to dropping it from a ladder with orbital reach.

Your paper is starting at orbital velocity. That is a bit higher than a baseball. I also dont think you can slow to lower than mach 1 without a shock forming. The shock is what does the decelleration. I would still like to see your figures.

Why cant you vaporize paper without air? Notice I did not say burn. I said vaporize, as in vaporizing water to make steam.

mugaliens
2008-Jul-31, 12:11 AM
Cute.

And accurate. I'm two for two, today. :lol:

It related directly to what I was discussing. If you don't want to respond to my query...

Remainder of post will remain unread as you're obviously not getting the rather large clue-bird which landed on this thread recently which says that I do not appreciate you or others claiming behavior on my part that's not true.

If you can't find it, have the grace to simply ask.

Perhaps if you improved your people skills...

Good-bye, ggchuck.

Oh, and congratulations. You're the first person I've ever placed on my ignore list.

YamatoTwinkie
2008-Jul-31, 01:19 AM
Compression: (occurs within the shock wave) Accounts for roughly 80% of the energy dissipation. Half of the heat of compression is radiated outward, the other half inward. Since about 3/8th's is exposed, that's about 3/16ths of the heat of compression, or roughly 3/16 * 80% * 200 BTU, which comes to 30 BTU.

That's where you lost me. What is your reasoning for assuming that fully 3/8 of the atmospheric particles that the plane encounters essentially "stick" to the compression area for the entire ride down, merrily re-radiating all of their energy as they go? Why haven't they escaped(relatively quickly, I might add), taking their heat load and kinetic energy with them?

ggchuck
2008-Jul-31, 02:25 PM
At 150km density is about 10^-12 g/cm3, giving about 3.5 million oxygen atoms (used for convienence) per cubic centimeter. Assuming a cross section of 1cm3, and a velocity of 8500 m/s, you should get about 32 billion collisions per second.From your numbers, I get (3,500,000 * 8500 * 100 = 3.2 trillion collisions per second.)

...by the time you reach 100km, your forward velocity will be reduced by 3.6 m/s.I don't know how to compute KE impulse deceleration, but if my number of 3.2 trillion collisions per second is correct, then your number of 3.6 m/s should be increased to 360 m/s.

The number of collisions is up to 32 trillion per second. Acceleration has increased to .07 m/s^2. (acc numbers are low, did not integrate the increase in density).If the drag increases by a thousandfold, then the deceleration should also increase significantly... obviously not (360 * 1000 = 360,000 m/s) as the speed that can bleed off is only 8500 m/s.

Please let me know if my logic is in error or if I misinterpreted your figures.

ggchuck
2008-Jul-31, 02:33 PM
I do not appreciate you or others claiming behavior on my part that's not true.I'm not sure what you think I'm claiming. I can clarify what I do feel about your behavior; you don't answer direct questions, you mock my lack of math background, you show arrogance rather than guidance when clarification is requested.

If you can't find it, have the grace to simply ask.I did and you responded with the ambiguous (not 'accurate') and snide: "No, yes. Yes, and yes."

Oh, and congratulations. You're the first person I've ever placed on my ignore list.Thanks for the honor.

joema
2008-Jul-31, 03:41 PM
We have actual flight data from STS-107 that shows heating is fairly low during the first 100 sec or so. Even measured on the shuttle wing leading edge just behind the RCC panels, it stays below 500F for roughly 120 sec (see below attachment).

As already stated, prototype paper planes have already survived higher temps than this in a hypersonic wind tunnel.

During the first 160 after entry interface, dynamic pressure (qbar) increases from 0 to about 0.5 psf (see below attachment).

That's easily enough to decelerate a paper plane during that interval. Let's assume 46.75 in^2 (301 cm^2), about 1/2 of a 8.5"x11" sheet, and mass of 5 grams.

Assume a constant dynamic pressure of 1/2 the midpoint, or 0.25 psf (0.00112 N/cm^2)

301 cm^2 * 0.00112 N/cm^2 = 0.337 Newtons

F=m*a
a = f/m
a = 0.337 N / 0.005 kg
a = 67.4 m/s^2 (6.9 g)

That's easily enough to decelerate from orbital velocity within 120 sec, before aerodynamic heating becomes severe.

While it's not certain, it appears from actual flight data a specially constructed paper plane might survive reentry. The "shuttlecock effect" would cause rapid deceleration before it was destroyed by aerodynamic heating.

This is a unique, non-intuitive effect that's only possible because of the extremely high ratio of surface area to mass in a paper plane.

As already stated, Rutan's SpaceShipOne uses this from lower velocities, but in real man-carrying vehicles it can't be scaled up to work from orbital velocities. However in a small paper plane it appears it might be possible.

korjik
2008-Jul-31, 04:02 PM
From your numbers, I get (3,500,000 * 8500 * 100 = 3.2 trillion collisions per second.)

I don't know how to compute KE impulse deceleration, but if my number of 3.2 trillion collisions per second is correct, then your number of 3.6 m/s should be increased to 360 m/s.

If the drag increases by a thousandfold, then the deceleration should also increase significantly... obviously not (360 * 1000 = 360,000 m/s) as the speed that can bleed off is only 8500 m/s.

Please let me know if my logic is in error or if I misinterpreted your figures.

I use momentum for impulse, not kinetic energy.

korjik
2008-Jul-31, 04:08 PM
From your numbers, I get (3,500,000 * 8500 * 100 = 3.2 trillion collisions per second.)

I don't know how to compute KE impulse deceleration, but if my number of 3.2 trillion collisions per second is correct, then your number of 3.6 m/s should be increased to 360 m/s.

If the drag increases by a thousandfold, then the deceleration should also increase significantly... obviously not (360 * 1000 = 360,000 m/s) as the speed that can bleed off is only 8500 m/s.

Please let me know if my logic is in error or if I misinterpreted your figures.

Oops, I didnt convert the cm to m. It should be 3.2 trillion.

However, 360 is stll small compared to 8500.

the thing is, properly done, this should show a very rapid increase in acceleration fairly low in the atmosphere. The airplane is not going to slowly drift down from orbit.

korjik
2008-Jul-31, 04:11 PM
We have actual flight data from STS-107 that shows heating is fairly low during the first 100 sec or so. Even measured on the shuttle wing leading edge just behind the RCC panels, it stays below 500F for roughly 120 sec (see below attachment).

As already stated, prototype paper planes have already survived higher temps than this in a hypersonic wind tunnel.

During the first 160 after entry interface, dynamic pressure (qbar) increases from 0 to about 0.5 psf (see below attachment).

That's easily enough to decelerate a paper plane during that interval. Let's assume 46.75 in^2 (301 cm^2), about 1/2 of a 8.5"x11" sheet, and mass of 5 grams.

Assume a constant dynamic pressure of 1/2 the midpoint, or 0.25 psf (0.00112 N/cm^2)

301 cm^2 * 0.00112 N/cm^2 = 0.337 Newtons

F=m*a
a = f/m
a = 0.337 N / 0.005 kg
a = 67.4 m/s^2 (6.9 g)

That's easily enough to decelerate from orbital velocity within 120 sec, before aerodynamic heating becomes severe.

While it's not certain, it appears from actual flight data a specially constructed paper plane might survive reentry. The "shuttlecock effect" would cause rapid deceleration before it was destroyed by aerodynamic heating.

This is a unique, non-intuitive effect that's only possible because of the extremely high ratio of surface area to mass in a paper plane.

As already stated, Rutan's SpaceShipOne uses this from lower velocities, but in real man-carrying vehicles it can't be scaled up to work from orbital velocities. However in a small paper plane it appears it might be possible.

quick question: is dynamic pressure the pressure against an object moving through the air?

second quick question: shouldnt we assume a nose first configuration?

quick comment: pressure should rise exponentally, you cant really use 1/2 max as the average (that requires linear change)

ggchuck
2008-Jul-31, 04:29 PM
I use momentum for impulse, not kinetic energy.Regardless of my confusion between momentum and kinetic energy, are my thoughts about your computations valid? Would not the paper decelerate drastically before encountering atmosphere dense enough to vaporize it?

joema
2008-Jul-31, 04:40 PM
quick question: is dynamic pressure the pressure against an object moving through the air?
http://en.wikipedia.org/wiki/Dynamic_pressure

second quick question: shouldnt we assume a nose first configuration?
We don't know the configuration or orientation. The prototypes tested look like conventional paper airplanes, but further testing might indicate a totally different design. E.g, a ball, cube, blunt cone, etc. Issues of stability are unknown. I'm sure based on further calculations and testing the researchers will use whatever configuration will best achieve survivability.

quick comment: pressure should rise exponentally, you cant really use 1/2 max as the average (that requires linear change)
We have limited data during that period, simply that qbar rises from 0 at entry interface to 0.5 psf at EI+160 sec. It appears from other reentry qbar graphs the early increase is mostly linear before encountering a knee in the curve. These calculations are simply rough approximations based on limited data.

ggchuck
2008-Jul-31, 05:04 PM
However, 360 is stll small compared to 8500.

the thing is, properly done, this should show a very rapid increase in acceleration fairly low in the atmosphere. The airplane is not going to slowly drift down from orbit.Granted, 360 is small compared to 8500. However, I assume from your comments that by the time the aircraft drops to 100 km [(150km-100km)*1000/(1m/s) = 50,000 seconds], it will be encountering 1000 times more collisions than at the initial state. This seems like more than enough time and distance for impulse deceleration to slow the paper (whether an aircraft or wad) to flying speed or terminal velocity.

YamatoTwinkie
2008-Jul-31, 09:22 PM
Not sure what happened to my post, so I'll try this again:

Pretty key to this discussion is The 1958 NACA Allen&Eggers Study, which can be found at http://en.wikipedia.org/wiki/Atmospheric_reentry#cite_note-2

Of note is Equation (34):

Q=m/4[Cf*S/(Cd*A)]Ve^2

where
Q is the total heat load of the reentry body
m is mass
Cf is the coefficient of skin friction
S is Surface area
Cd is the pressure drag coefficient
A is wetted area
Ve is entrance velocity

To reduce heat load, you've got to minimize skin friction, maximize pressure drag, and maximize wetted area. This is the reason most re-entry bodies are blunt-nosed, as the Cd can be a couple orders of magnitude higer than Cf.

How much can you reduce the heat load by? Well, using Figure (6) as an example, a blunt-nosed, 1000lb 30,000ft/s missile has its heat reduced about 140 times vs. its initial kinetic kinetic energy.

Using that (conservative) estimate for our paper airplane(in reality, the paper airplane is going to have a much better S/A ratio than a missile, and a higher Cd as well), the airplane will only receive about 0.75 Btusduring re-entry.

That's a *whole* lot more survivable than mugalien's 30 Btus. No way the paper is going to hit 451F...

publiusr
2008-Aug-01, 07:17 PM
Only .75?

That makes me wonder if there may be a way to soft land large articles from orbit using tethers.

Imagine a bola, Which allows a weight to be deposited on the surface. Another idea---if one art of the bola could be violently caught in the atmosphere--sacrificed, in other words, could a shuttlecock then descend with less heating?

I'm thinking that the tether would break prematurely. But a tether itself is thin and light--as a paper airplane. So the weight--the bob--would have to be shielded and multiple tethers used....

Any thoughts?

mugaliens
2008-Aug-01, 08:30 PM
This is a unique, non-intuitive effect that's only possible because of the extremely high ratio of surface area to mass in a paper plane.

True, if the density remains constant, as the radius of a ball is halved, it's surface area is reduced by a factor of two, while it's mass, which is proportional to it's volume, is reduced by a factor of four.

Thus, conceivably, when you get to particles the size of sand grains, you should have something which slows down almost immediately. Yet Wikipedia still mentions, "A meteor is the visible event that occurs when a meteoroid or asteroid enters Earth's atmosphere and becomes brightly visible. For bodies with a size scale larger than the atmospheric mean free path (10 cm to several metres) the visibility is due to the heat produced by the ram pressure (not friction, as is commonly assumed) of atmospheric entry. Since the majority of meteors are from small sand-grain size meteoroid bodies, most visible signatures are caused by electron relaxation following the individual collisions between vaporized meteor atoms and atmospheric constituents. The meteor is simply the visible event rather than an object itself."

Our three-inch diameter wadded up paper ball is a little shy of 10 cm. Thus, more of the heat is produced by friction, which is imparted to the ball, not by heat produced by ram pressure.

Why the difference based on size? Reynolds numbers. Just as they effect both subsonic and supersonic flows around wings, they affect subsonice and supersonic flows around balls (or anything, for that matter): "Reynolds number is important in the calculation of a body's drag characteristics. A notable example is that of the flow around a cylinder. Above roughly 3×106 Re the drag coefficient drops considerably. This is important when calculating the optimal cruise speeds for low drag (and therefore long range) profiles for airplanes." - Wikipedia

Also, please recall that due to the geometry of a shock wave around blunt bodies, the peak shock wave temp is dependant only on velocity, not on size.

However, the distance from the object to the peak shock wave temp layer is proportional to the object's linear size. This, half the size gives you a shock wave temp layer that's half the distance, or twice as close, and to the same temperature, I might add.

Back to Reynolds... Because of it's smaller size, that peak shock wave temp layer also appears much larger, thereby occupying a much greater field of view in front of the object.

Same temp, but closer, and with more of a thermal-blanket wrap-around?

Toast.

joema
2008-Aug-02, 03:13 PM
...conceivably, when you get to particles the size of sand grains, you should have something which slows down almost immediately...
This was answered in this post: http://www.bautforum.com/questions-answers/69247-paper-airplane-reentry.html#post1155545

Summary: a typical reentering "grain of sand" meteor has about 2.75 million Joules per square cm, whereas a paper plane only about 200 Joules per square cm.

The tiny meteor has 13,000 times the kinetic energy per unit surface area, which is why it burns up.

mugaliens
2008-Aug-02, 04:57 PM
This was answered in this post: http://www.bautforum.com/questions-answers/69247-paper-airplane-reentry.html#post1155545

Summary: a typical reentering "grain of sand" meteor has about 2.75 million Joules per square cm, whereas a paper plane only about 200 Joules per square cm.

The tiny meteor has 13,000 times the kinetic energy per unit surface area, which is why it burns up.

If that were the whole picture, I should be able reenter in a PEX ball, and survive. After all, the melting point of of PEX is just over 200 deg F.

If your calculations show it's a PEX ball with a diameter of 3 m is too small (too much mass/cross-sectional area), we'll simply increase the diameter to, say, 9 m.

joema
2008-Aug-03, 01:23 PM
If that were the whole picture, I should be able reenter in a PEX ball, and survive. After all, the melting point of of PEX is just over 200 deg F.

If your calculations show it's a PEX ball with a diameter of 3 m is too small (too much mass/cross-sectional area), we'll simply increase the diameter to, say, 9 m.
To match the kinetic energy per unit surface area of a paper plane, a PEX ball carrying a 180-lb human would have to be 1,250 square meters in cross-sectional area, or 40 meters in diameter.

The ball itself would be heavy, thus requiring even more surface area to disperse the energy, driving up the size requirement further. Structural strength would also be a problem.

Because of the cube-square law, small objects will often survive events that would destroy a large object: http://en.wikipedia.org/wiki/Square-cube_law

E.g, you can drop a tiny "matchbox" model car from a scaled height of hundreds of feet, yet it will be hardly dented. A real car dropped from hundreds of feet would be destroyed.

The cube-square law favors small objects, structurally. This in turn allows sufficient structural strength to mechanically support the surface area required for aerodynamic braking of the small mass. However it cannot be scaled upward for large objects.

If you had infinitely strong, zero-mass PEX, you could probably make a gigantic sphere that would allow a human inside to survive reentry.

mugaliens
2008-Aug-03, 08:58 PM
To match the kinetic energy per unit surface area of a paper plane, a PEX ball carrying a 180-lb human would have to be 1,250 square meters in cross-sectional area, or 40 meters in diameter.

The ball itself would be heavy, thus requiring even more surface area to disperse the energy, driving up the size requirement further. Structural strength would also be a problem.

We're using ThermaPEX with a wall thickness of 0.070 inches. In 1/2" pipe, it'll withstand 80 psi at 200 deg F. For 1" pipe, the wall thickness is 0.097 inches, and it'll withstand the same 80 psi at 200 deg F.

Naturally, we're using a large sphere of ThermaPEX, but we're not pressurizing it to 80 psi, either. Let's say 20 psi, with 15% O2, and the rest Nitrogen. Or put the astronaut in a spacesuit and use 100% helium.

As for the structure, let's build it like a giant soccer ball, and put the astronaut at the center, connected to the many anchoring lines each of which goes to a different intersection in the soccer ball panels. Let's assume there's enough structural support inside the PEX ball to evenly distribute the load.

Let's say the astronaut weighs 180 lbs, his suit weighs the same at 180 lbs (what does a self-contained EVA suit weigh, sans propulsion?), and that all the lines and supports to the interior of our giant ThermoPEX soccer ball weighs 100 lbs (very reasonable, as a typical parachute weighs only 40 lbs.

Let's also state (according to my quick curve plot and the manufacturer's info) that the PEX at 20 psi has a max temp of 230 deg F.

For safety's sake, let's rake 10% off that and say that 200 deg F is max temp at 20 psi.

Questions:

1. Given the above information, by your calculations, what size ThermaPEX sphere would be required for a successful reentry?

2. What would be the weight of the sphere itself? With the astronaut, suit, and internal support structure as given above?

3. What would be the terminal velocity of such a sphere at 100,000 ft MSL? 10,000 ft MSL? Sea level?

4. How much larger would the ThermaPEX need to be to support 7 astronauts, without EVA suits, but in a 7-person escape module that weighs 1,000 lb?

5. What would the terminal velocities be at the altitudes given in query 3?

Because of the cube-square law, small objects will often survive events that would destroy a large object: http://en.wikipedia.org/wiki/Square-cube_law

E.g, you can drop a tiny "matchbox" model car from a scaled height of hundreds of feet, yet it will be hardly dented. A real car dropped from hundreds of feet would be destroyed.

We're talking reentry, not impact after being dropped from the Eiffel Tower.

Different matter entirely, which is why micrometeorites burn up all the time.

The cube-square law favors small objects, structurally. This in turn allows sufficient structural strength to mechanically support the surface area required for aerodynamic braking of the small mass. However it cannot be scaled upward for large objects.

Agreed.

If you had infinitely strong, zero-mass PEX, you could probably make a gigantic sphere that would allow a human inside to survive reentry.

What about given the parameters I specified above?

joema
2008-Aug-04, 03:03 PM
We're using ThermaPEX with a wall thickness of 0.070 inches...let's build it like a giant soccer ball, and put the astronaut at the center, connected to the many anchoring lines each of which goes to a different intersection in the soccer ball panels...Given the above information, by your calculations, what size ThermaPEX sphere would be required for a successful reentry?...What would be the weight of the sphere itself? With the astronaut, suit, and internal support structure as given above?..
A 180-lb astronaut in a 225-lb EMU suit would mass 405 lb (184 kg). Excluding the sphere mass, to achieve the same ratio of kinetic energy to cross-sectional area as the above paper plane (182 Joules/cm^2), the PEX sphere would have to be about 68 meters in diameter. That's much bigger than the gigantic Echo II inflatable satellite: http://en.wikipedia.org/wiki/Image:Echo_II.jpg

However that's just considering the mass of the astronaut and suit. A 68-meter dia. PEX sphere with 0.070 wall thickness (1.78 mm) would comprise about 25.8 cubic meters of PEX material:

A = 4 * pi * r^2
A = 4 * 3.14 * (34m)^2
A = 14,519 square meters

Volume of materal = 14,519 m^2 * .00178 mm = 25.8 m^3

At a typical PEX specific gravity of 0.94, the sphere itself would mass about 25.3 metric tons. That in turn would drive the size further upward. IOW it's impossible for the reasons already stated -- runaway cube-square ratio.

YamatoTwinkie
2008-Aug-04, 04:34 PM
Our three-inch diameter wadded up paper ball is a little shy of 10 cm. Thus, more of the heat is produced by friction, which is imparted to the ball, not by heat produced by ram pressure.

Why the difference based on size? Reynolds numbers. Just as they effect both subsonic and supersonic flows around wings, they affect subsonice and supersonic flows around balls (or anything, for that matter): "Reynolds number is important in the calculation of a body's drag characteristics. A notable example is that of the flow around a cylinder. Above roughly 3×106 Re the drag coefficient drops considerably. This is important when calculating the optimal cruise speeds for low drag (and therefore long range) profiles for airplanes." - Wikipedia

For a sphere, friction drag is only going to be greater than pressure drag when Re<1. Obviously, our paper airplane never hits this value (at least, not when it matters). After Re>1000, pressure drage is 95% of the total drag, and it keeps increasing.

During transition (around Re=3x10^6 for a smooth sphere), pressure drag does dip around 5-fold, but it's still dominated by pressure drag.

Besides, you're still under the assumption that a "crumpled ball" is the best design. A flat plate(normal to the flow) has even higher pressure drag than a sphere, and does not experience transition.

Also, please recall that due to the geometry of a shock wave around blunt bodies, the peak shock wave temp is dependant only on velocity, not on size.

However, the distance from the object to the peak shock wave temp layer is proportional to the object's linear size. This, half the size gives you a shock wave temp layer that's half the distance, or twice as close, and to the same temperature, I might add.

Radiation is negligible(air does not radiate very well). Heat transfer is dominated by convection for this application.

Peak temperature of the compression area should located at the stagnation point(where the air velocity is zero with respect to the moving object), which is going to be pretty close to the surface of the reentry vehicle, regardless of it's size.

Anyway, absolute temperature of the gas shouldn't matter much anyway. What matters is localized heat transfer.

mugaliens
2008-Aug-04, 05:49 PM
Besides, you're still under the assumption that a "crumpled ball" is the best design.

For providing both a spherical, sem-spherical surface while rotating the maximum surface area in order to spread the heat out over the greatest surface area possible.

Show me a better design.

A flat plate(normal to the flow) has even higher pressure drag than a sphere, and does not experience transition.

And gets very hot at the edges of that plate. No thanks.

Radiation is negligible(air does not radiate very well). Heat transfer is dominated by convection for this application.

Shuttle astronauts who feel the heat from the incandescent air streaming past the windows, long before the Shuttle's tiles or windows ever begin to heat up, would disagree.

Peak temperature of the compression area should located at the stagnation point(where the air velocity is zero with respect to the moving object)...

Correct.

...which is going to be pretty close to the surface of the reentry vehicle, regardless of it's size.

Incorrect. For any given geometry, the distance between the peak compression temp and the leading edge is proportional to the object's size. Thus, if it's an inch for an object five inches across, it's a foot for an object five feet across.

YamatoTwinkie
2008-Aug-04, 07:36 PM
And gets very hot at the edges of that plate. No thanks.

round the edges then, if you insist. A flying dinner-plate.

Shuttle astronauts who feel the heat from the incandescent air streaming past the windows, long before the Shuttle's tiles or windows ever begin to heat up, would disagree.
.

Didn't say that radiation was non-existant, just that it was negligible compared to the convection transfer from a superheated, compressed gas blowing across the wetted area at sonic velocity.

For any given geometry, the distance between the peak compression temp and the leading edge is proportional to the object's size. Thus, if it's an inch for an object five inches across, it's a foot for an object five feet across

For your object five feet across, what is the velocity vector for the molecule of air located at a distance of 1 inch from the leading edge, if you believe V=0at D=1 foot?

mugaliens
2008-Aug-05, 09:00 PM
round the edges then, if you insist. A flying dinner-plate.

You mean a semi-spherical object? Sort of like these pictures (http://en.wikipedia.org/wiki/Image:Blunt_body_reentry_shapes.png)to which I linked eons ago? In which picture is the maximum shock heating the furthest from the body? Blunt body concept 1953, isn't it?

Naturally, there's no information given on their radii, the atmospheric pressure, or it's MACH.

Didn't say that radiation was non-existant, just that it was negligible compared to the convection transfer from a superheated, compressed gas blowing across the wetted area at sonic velocity.

Behind the shock wave, the velocity flow is subsonic. Yes, convection plays a large factor. And (you know this - just putting it out there for the others) unlike what people imagine from the pics of the shock waves, the air doesn't hit the shock wave then flow along the shock wave. It passes right through the shock wave then flows along the skin of the reentry vehicle/shape/body.

For your object five feet across, what is the velocity vector for the molecule of air located at a distance of 1 inch from the leading edge, if you believe V=0at D=1 foot?

I assume you're referring to "on-axis" measurements, that is, a blunt body, symmetrically aligned, center of area, in the direction of travel.

By "V=0at D=1 foot" are you saying that a given of this on-axis, blunt body measurement is at 1 foot immediately in front of the leading edge, the velocity is 0?

If so, then at 1 inch and all points between D0'-1' V=0.

In reality, if that's 1 inch forward, and inside the shock wave, it's close to 0. If it's ahead of the shock wave, it's equal to the velocity of the object, opposite direction. The shock wave is the transition zone, where the air is compressed, slowed to zero.

At subsonic speeds, the bow wave of a leading edge theoretically extends out to infinity. In practice, that's not quite the case, but dolphins still have a great time playing in them! In practice, for blunt bodies, such as the nose of an aircraft or the bulbuous subsurface "nose" of modern cargo carrying sealift, it's generally assumed to be zero at a forward distance equal to the mean radius of the "nose." The shape determines whether there's any practical justification for extending the practical boundary further (the more blunt the shape, the further the boundary.

YamatoTwinkie
2008-Aug-06, 01:57 AM
Behind the shock wave, the velocity flow is subsonic.

Just to pick nits, behind the bow/oblique shockwave, the flow can be either subsonic(stagnation region) or supersonic as you go farther out from the centerline. A google books search for "Space Vehicle Design" -Griffin, page 311 has a pretty good figure of this, if you're interested.

Yes, convection plays a large factor.

It's most likely the dominant factor from everything I've read (Granted, much of this stuff is new to me). Fay-Riddell doesn't even address it. True, at higher re-entry velocities radiation can be substantial, as it scales to the fourth power of temperature.

By "V=0at D=1 foot" are you saying that a given of this on-axis, blunt body measurement is at 1 foot immediately in front of the leading edge, the velocity is 0?

If so, then at 1 inch and all points between D0'-1' V=0.

Maybe I'm not being clear. I'm used to seeing stagnation points at the surface that the streamline is hitting. What you seem to be describing is multiple points up and down the streamline where the velocity is zero. This makes no sense to me.

matthewota
2008-Aug-08, 02:14 AM
Seems to me that you just cannot throw a paper airplane from a shuttle or the ISS with enough velocity to negate the 17,500 mph orbital speed to allow it to fall into the atmosphere.

cjl
2008-Aug-08, 04:50 AM
You don't need to negate all of it. Just enough so that the perigee dips down into the atmosphere. In fact, the ISS is in the upper part of the atmosphere, so if you weren't picky about when and where it would come down, you could just release it at the ISS, and its orbit would decay fairly quickly from the drag.

mugaliens
2008-Aug-08, 09:56 PM
Just to pick nits, behind the bow/oblique shockwave, the flow can be either subsonic(stagnation region) or supersonic as you go farther out from the centerline. A google books search for "Space Vehicle Design" -Griffin, page 311 has a pretty good figure of this, if you're interested.

WHAT is the problem with BAUT tonight? This is the third time I've responded to this (once yesterday) and my response keeps disappearing.

Arg!

For hemispherical objects of larger Reynolds numbers (capsule-sized and above) the surface flow is subsonic.

Maybe I'm not being clear. I'm used to seeing stagnation points at the surface that the streamline is hitting. What you seem to be describing is multiple points up and down the streamline where the velocity is zero. This makes no sense to me.

On-axis velocities within the bow wave of spherical objects is effectively zero out to approximately the radius of the object. With hemispherical objects it's less than the radius. However, the off-axis radius of this effect narrows rapidly the further in front of the object one goes.

Technically, the distant free-stream on-axis velocity is zero, and increases to the pseudo-asymptotic (tangent) velocity of the object by the time it reaches the object.

YamatoTwinkie
2008-Aug-11, 04:41 PM
For hemispherical objects of larger Reynolds numbers (capsule-sized and above) the surface flow is subsonic.

Well sure. I was just pointing out that flow can be supersonic behind oblique/bow shockwaves, once you get far enough away from the object, as only incoming flow normal to the angle of the oblique shockwave is affected.

On-axis velocities within the bow wave of spherical objects is effectively zero out to approximately the radius of the object. With hemispherical objects it's less than the radius. However, the off-axis radius of this effect narrows rapidly the further in front of the object one goes.

Technically, the distant free-stream on-axis velocity is zero, and increases to the pseudo-asymptotic (tangent) velocity of the object by the time it reaches the object.

And please tell me, what is the tangent velocity at the surface of a sphere/hemisphere at the middle (i.e, along the center axis)? Remember, velocity has both a magnitude and a *direction*. And no, you can't have multiple directions at the same point.

Please provide citations to back up your assumptions. What you seem to be saying (distant stagnation points, increasing velocity as you approach the surface), is contrary to every fluid mechanics textbook I've ever seen. See Figure 16 at: http://www.princeton.edu/~asmits/Bicycle_web/Bernoulli.html

How about some experimental data:

http://naca.central.cranfield.ac.uk/reports/1958/naca-tn-4354.pdf

Look at Table I. Notice how the velocity is still 0.44M-0.62M behind the bow shockwave(region 3)? Notice how both pressure and temperature increase as you go from region 3 to region 0, at the surface of the sphere(coincidentally labeled "stagnation point")?

mugaliens
2008-Aug-11, 10:17 PM
And please tell me, what is the tangent velocity at the surface of a sphere/hemisphere at the middle (i.e, along the center axis)?

Zero.

Please provide citations to back up your assumptions.

Go fish!

I'm merely recounting my education and experience, not writing a dissertation.

What you seem to be saying (distant stagnation points, increasing velocity as you approach the surface), is contrary to every fluid mechanics textbook I've ever seen.

Yeah, it's contrary to every textbook I've seen, too. How in the world did you conclude that? I was very specific in that the relative velocity (relative to the object - stationary compared to the rest of the medium) of the medium along the centerline is zero at the surface, and increase the further away you get from the moving object.

How about some experimental data...

...

Notice how both pressure and temperature increase as you go from region 3 to region 0, at the surface of the sphere(coincidentally labeled "stagnation point")?

Yes. Behind the shock wave the temperture and pressure both increase along the centerline and reach a peak at the stagnation point. That's standard subsonic flow stuff.

Quiz time: At what point is the temperature and pressure greater? The peak of the shockwave, on centerline, or at the stagnation point (the air, not the heat shield)?

YamatoTwinkie
2008-Aug-12, 02:53 AM
I was very specific in that the relative velocity (relative to the object - stationary compared to the rest of the medium) of the medium along the centerline is zero at the surface, and increase the further away you get from the moving object.

Sorry, that's my failure to parse your phrasing correctly. I apologize.

Quiz time: At what point is the temperature and pressure greater? The peak of the shockwave, on centerline, or at the stagnation point (the air, not the heat shield)?

To be clear, you're talking about within the shockwave itself? To be honest, I have no clue. They've always been treated as discontinuities, as they're so small in thickness (~.0001mm). Usually you just know the immediate upstream/downstream conditions. If you have a better understanding of this area, I'd be welcome to reading it.

mugaliens
2008-Aug-12, 10:26 PM
Sorry, that's my failure to parse your phrasing correctly. I apologize.

No worries. :)

To be clear, you're talking about within the shockwave itself?

Yep.

To be honest, I have no clue. They've always been treated as discontinuities, as they're so small in thickness (~.0001mm). Usually you just know the immediate upstream/downstream conditions. If you have a better understanding of this area, I'd be welcome to reading it.

Here's a copy of my understanding (http://en.wikipedia.org/wiki/Reentry#Real_.28non-equilibrium.29_gas_model).

The key here is in the following few sentences (bold, mine):

"An important aspect of modeling non-equilibrium real gas effects is radiative heat flux. If a vehicle is entering an atmosphere at very high speed (hyperbolic trajectory, lunar return) and has a large nose radius then radiative heat flux can dominate TPS heating. Radiative heat flux during entry into an air or carbon dioxide atmosphere typically comes from unsymmetric diatomic molecules, e.g. cyanogen (CN), carbon monoxide, nitric oxide (NO), single ionized molecular nitrogen, et cetera. These molecules are formed by the shock wave dissociating ambient atmospheric gas followed by recombination within the shock layer into new molecular species. The newly formed diatomic molecules initially have a very high vibrational temperature that efficiently transforms the vibrational energy into radiant energy, i.e. radiative heat flux."

When they start incorporating a Schrödinger equation into aerospace engineering, however, it's time for me to drink a warm glass of milk and go to bed.

YamatoTwinkie
2008-Aug-13, 12:05 AM
These molecules are formed by the shock wave dissociating ambient atmospheric gas followed by recombination within the shock layer into new molecular species. The newly formed diatomic molecules initially have a very high vibrational temperature.

Interesting. But it still kinda leaves your original question unanswered. Is the initial shock going to actually be higher in temperature than the rest of the shock layer in the stagnation region? For a lunar return vehicle(shock layer temp ~10,000K), how much higher? Where's the peak located, if it takes ~1ms to occur?

While I gather that radiative heat transfer is more than I initially considered for high entry velocities, it still appears to be more or less meaningless for return velocities of 30,000ft/s or less, if Figure 18.10 of "Hypersonic and High Temperature Gas Dynamics" is any indication.

(Edit: broken link. Try a Googlebook search for the above title) It'd be nice if I could figure out what Ref 213 is.

So our LEO paper airplane isn't going to have to worry much about it anyway.

mugaliens
2008-Aug-13, 06:24 PM
Interesting. But it still kinda leaves your original question unanswered. Is the initial shock going to actually be higher in temperature than the rest of the shock layer in the stagnation region? For a lunar return vehicle(shock layer temp ~10,000K), how much higher? Where's the peak located, if it takes ~1ms to occur?

The peak is located along the symmetrical axis, where the initial shock wave compression is the greatest. Essentially, it's in the dark area in the B&W photos I linked to several times.

The larger the body, or the greater the velocity, the greater the radiative heating due to this non-equilibrium effect.

I'm not sure why you're referencing the short time. While air flows through the peak shock region, the region itself, where the radiation originates, remains in the leading shock wave ahead of the object.

While I gather that radiative heat transfer is more than I initially considered for high entry velocities, it still appears to be more or less meaningless for return velocities of 30,000ft/s or less, if Figure 18.10 of "Hypersonic and High Temperature Gas Dynamics" is any indication.

Might you have an online link? I'm having some difficulty downloading Figure 18.10 from here... Could be the electric bill you used as a placemark is acting as a firewall...

I'm certainly not going to pay \$90.95 for it at Barnes and Noble. My days of buying megabuck college texts ended when I graduated.

YamatoTwinkie
2008-Aug-14, 03:34 AM
I'm not sure why you're referencing the short time.

My only reasoning was that, as an individual air molecule passes through the initial shock, 1ms is still quite a bit of time compared to the speeds involved. While it's true that the wave itself stays in the same position relative to the object, the air molecules themselves certainly don't.

Again, I'd be interested in knowing the actual peak temperatures vs. the rest of the shock layer along the central axis, if you can manage to find it. I'm curious as to why, if there is in fact a large discrepancy between the two, it's not covered more often.

Might you have an online link? I'm having some difficulty downloading Figure 18.10 from here...

Hmmm... I'm still somewhat unfamiliar with how GoogleBooks does things, as even I can't access that particular page (688) anymore. You might want to try searching for the book yourself in GoogleBooks and see how much text it brings up for you. Yeah, it'll take awhile to load :sad:

Essentially, it was a plot that broke down the convective vs. radiative loads on a large Earth reentry vehicle according to the entry speed. The radiative portion was pretty much nothing until ~25,000 ft/s, but it exponentially rose after that to overtake convection shortly after ~30,000 ft/s, IIRC.

mugaliens
2008-Aug-15, 06:09 PM
Again, I'd be interested in knowing the actual peak temperatures vs. the rest of the shock layer along the central axis, if you can manage to find it. I'm curious as to why, if there is in fact a large discrepancy between the two, it's not covered more often.

I'm not sure what you mean by a difference between the two, as the peak shock temperature occurs in the shock wave where the pressure is at a maximum. For all objects hemispherical to pointed, that occurs at the central axis.

I've mentioned this (http://en.wikipedia.org/wiki/Reentry#Shock_layer_gas_physics)a few times before:

It's a coincidence, but a handy one, in that the peak shock wave temperature in Kelvins is roughly equal to the reentry velocity in meters per second. Thus, at 6,000 m/s, the peak shock wave temperature in Kelvins is roughly 6,000 deg K.

Naturally, the temperature experienced by the reentering object is less than that. How much less depends on several factors, including geometry, Reynolds number, and the ablative properties, thermal mass, and thermal conductivity of it's heat shield.

Essentially, it was a plot that broke down the convective vs. radiative loads on a large Earth reentry vehicle according to the entry speed. The radiative portion was pretty much nothing until ~25,000 ft/s, but it exponentially rose after that to overtake convection shortly after ~30,000 ft/s, IIRC.

It's interesting that the reentry velocity from LOE is roughly 7.8 km/s, which equates to 25,590 ft/s. From the moon (11 km/s), it's 36,089 ft/s. Thus, the Apollo gents had to contend with more radiative heating than convective heating.

I'm confused by your use of the word "convection."

Radiative heating occurs when the gases in the peak shock area are compressed. The air is heated two ways. The first is via compression, according to the ideal gas model. The second is chemical, as under the heat and pressure (heat, mostly), the five main gases, N2, O2, NO, N and O, recombine into diatomic molecules which have a very high vibrational temperature. ie, they radiate.

Convection is simply the moving of air, usually referred to in ovens. Thus, air warmed by the shock wave chemistry flows over the object, and warms it.

Conduction is simply warmer air flowing over a cooler surface tends to warm that surface.

There's also friction heating, but at subsonic velocities, it's minimal. Interestingly, there is a correction factor that needs to be used when computing TAS from EAS/CAS. ICE-T (Indicated --> Calibrated --> Equivalent --> True). One correction is for altitude, another for temperature (thus the combined correction is for density altitude), and a third for the increase in temperature, which becomes noticeable at velocities above about .5 M. Still, the correction factor is only a couple of degrees. One doesn't start seeing significant temperature increases along leading edges until you're into the transonic and beyond regions, and then the effect is commensurate with density.

Of another note:

"The five species model is based upon 5 ordinary differential equations and 17 algebraic equations. Because the 5 ordinary differential equations are loosely coupled, the system is numerically "stiff" and difficult to solve. The five species model is only usable for entry from low Earth orbit where entry velocity is approximately 7.8 km/s. For lunar return entry of 11 km/s, the shock layer contains a significant amount of ionized nitrogen and oxygen. The five species model is no longer accurate and a twelve species model must be used instead. High speed Mars entry which involves a carbon dioxide, nitrogen and argon atmosphere is even more complex requiring a 19 species model." - Wikipedia

Thus, in the days of Apollo it probably took months of calculations by calculators (humans busy with slide rules) to map out these effects. Today, I could crunch the numbers to a very high degree of accuracy overnight on my home computer.

YamatoTwinkie
2008-Aug-16, 12:32 AM
I'm not sure what you mean by a difference between the two, as the peak shock temperature occurs in the shock wave where the pressure is at a maximum. For all objects hemispherical to pointed, that occurs at the central axis.

I've mentioned this (http://en.wikipedia.org/wiki/Reentry#Shock_layer_gas_physics)a few times before:

It's a coincidence, but a handy one, in that the peak shock wave temperature in Kelvins is roughly equal to the reentry velocity in meters per second. Thus, at 6,000 m/s, the peak shock wave temperature in Kelvins is roughly 6,000 deg K.

What your wiki reference is referring to is the shock layer, i.e, the subsonic area of compressed, heated gas between the bow shock wave and the object's surface. See Figure 9.1 (pg 365)in "Hypersonic and High Temperature Gas Dynamics", if you can manage to get Googlebooks to load it for you.

Have you ever seen the standard Fluid Mechanics equations for normal shock waves before? You can find them at http://www.grc.nasa.gov/WWW/K-12/airplane/normal.html There's even a calculator. Note that for high entry velocities (Mach 32, for example), you can't use a specific heat of k=1.4 anymore because of the chemical reactions/ionization of the air, making k effectively much different. If you use k=1.4, you'll get shock layer temperatures on the order of 58,000K!.

You can also use the formula Ts=To(1+(k-1)/2*M^2)

Where Ts = stagnation temperature of the air in the shock layer (you're only going to see this at the stagnation point near the surface of the object)
To= incoming air temperature ahead of the shock
M=incoming air Mach Number (relative to the object)
k=specific heat of the air (again, *not* going to be =1.4 at higher Mach numbers)

Which will give slightly *higher* tempeatures than T1 (the shock layer temperature immediately following the shockwave).

It's interesting that the reentry velocity from LOE is roughly 7.8 km/s, which equates to 25,590 ft/s. From the moon (11 km/s), it's 36,089 ft/s. Thus, the Apollo gents had to contend with more radiative heating than convective heating.

Sorry, i was going off of memory. I've since been able to look at that page again (using a different computer, I don't know why Googlebooks is so finicky)
, and the convection/radiation crossover for a 15ft nose radius appears to occur at ~37,000 ft/s. Extrapolating from the plot, radiation looks like it would cross the x-axis at about 32,000 ft/s. That gives you a good idea of how quickly it rises from nowhere.

Additionally, In another part of "Hypersonic and High Temperature Gas Dynamics", it states that radiative heating was approximately 30% of the total load for the lunar re-entry capsule.

Also, "Hypersonic Aerothermodynamics" (also available to preview at GoogleBooks) has an example problem (7.1) where they calculate heat loads for the Shuttle Orbiter during re-entry, and find the radiative component to be 0.1% of the total.

I'm confused by your use of the word "convection."

I mean convection in the standard Heat Transfer sense of the word. Hot, compressed gas, formed as air molecules transition past the shockwave, flows over the surface area of the re-entry object at high velocity (subsonic in the stagnation zone, supersonic as you get away from the centerline), heating the object. Convection. Or, if you want to be pedantic, "forced convection."

mugaliens
2008-Nov-16, 07:01 PM
Have you ever seen the standard Fluid Mechanics equations for normal shock waves before?

Uh, no, they completely skipped that when I got my aero eng degree...

:rolleyes: