# Challenging the Kármán line from above

The initial conditions of the thought experiment is (very) LEO / reentry. capsules, space shuttle and other spacecraft can generate lift in upper atmosphere during reentry, in order to reduce deceleration maximum G force loading and maximum heating.

Since capsules and space shuttle are not the best lift producing devices ever made, what would happen if we try to deorbit a hypothetical heat resistant/unbreakable high lift-to-weight ratio (let's say) Nimbus4 glider?

Subsidiaries: At which (maximum) altitude & speed would the variometer tell 0 m/s vertical speed? What would be max temperature reached? Have there ever been real tests of deorbiting high finesse / low wing loaded gliding devices? What would angle of attack be during the whole descent, until subsonic speed? How does Coanda effect work at hypersonic speeds?

Illustrations found here: https://www.quora.com/In-regards-to-atmospheric-reentry-what-exactly-is-a-ballistic-reentry

• This falls apart at "hypothetical heat resistant". The hypersonic heating is a huge thing and you need to lose velocity fast for it not to become an insurmountable issue. See related questions: 1 2
– SF.
Commented Apr 26, 2017 at 10:18
• @SF. I think OP mean that the glider is made of unoptainium; so heat is not an issue. Commented Apr 26, 2017 at 10:20
• It's an interesting question as a thought experiment; "Ignoring heating for a moment, what might a gradual re-entry trajectory of a high lift-to-weight ratio craft look like?" I'd say Gedankenexperimente are on-topic. It worked for Einstein - nobody poked fun at his sub-light train. Worked well for Schrödinger too! But less well for his cat, though the jury is still out on that.
– uhoh
Commented Apr 26, 2017 at 10:55
• Both lift and drag are proportional to the same v^2, so as long as you have good speed, you have good lift, even in very thin air, and you can stay high enough that drag won't be slowing you down excessively. But heating is also proportional to v^2 and with a very nasty multiplier factor. So, yes you can glide in very, very thin atmosphere; your lift-to-drag coefficient changes significantly around 1 mach but then remains quite stable at higher speeds; aerodynamics of gliding works about the same at 2 mach and at 20, at 0.1 bar and 0.001 bar, but heating becomes prohibitive.
– SF.
Commented Apr 26, 2017 at 11:23
• @qqjkztd: Yes, especially that with being cooked for a short time, you cook a lot of air and a little of ablator, which you then immediately leave far behind - heat doesn't get to penetrate deep. Cooking slowly, you cook yourself - heat penetrates into the craft.
– SF.
Commented Apr 26, 2017 at 11:54

For a starting point Falcon 9's Fairing is going 8700 kph (Mach 7.7 sea level) and it has no ablative shell. The returned fairings come back water logged but un-scorched. The interesting point about this is that the booster was going 8650 kph, but suffers scorching inspite of the re-entry burn since it does not slow down in the wispy upper atmosphere and Block 5 has protective covers. SpaceX plans to recover the upper stage with a ballute. Presumably without a re-entry burn.

If this happens then you will have a real answer to your question. I'm pretty sure the answer is yes if the craft is light enough and has enough drag/lift then yes it will survive re-entry, like a paper plane.

• This does not address the high l/d premise of the question at all. Commented Dec 29, 2018 at 22:14

Here is the reentry plan:

Control your altitude during re-entry such that

1. There is enough air density to generate enough lift to equal 1g at your current speed. At an angle of attack that isn't too close to stall. Earlier in the flight you will not need anywhere near 1g of lift since you are somewhat in orbit, but it will not hurt to have it: just lower your angle of attack to lower your lift.

2. There isn't enough air to get 2g of lift. This ensures aerodynamic forces don't grow too high.

Here are the issues:

1. Supersonic handling. The glider as-is wouldn't be controllable at supersonic speeds without delta-wings or swing-wings, etc.

2. Hypersonic lift:drag ratio is around 4 unlike a sailplane which can reach 50. At optimum L/D ratio the lift coefficient is very low but at lower L/D ratio (of around 2) we get a lift of around 0.3. Subsonic airfoils can have a lift coefficient of around 1.0 at a safe margin from stall. This would mean 3x the force of a slow-gliding sailplane (not too extreme), unless your delta-wing is 3x fatter. In which case the wind-force would be similar to the subsonic case. You have 0.5g or so of drag-deceleration thus less sight-seeing time but that means a peak g-force of about 1.1g. No big deal.

Even with these two issues you can could stick your (pressure-gloved) hand into the airflow all the way through reentry and the force is no worse than a car on a freeway. Although it would feel different since it is a hypersonic flow at very low air density instead of subsonic at high density. You could see the shock-wave glowing and enveloping your hand. It would look pretty.

1. Heat! Heat! Heat! The force of the wind is ~ρv^2 and is kept just below hurricane force. But heat is ~ρv^3. When v is mach 20 you can't have enough wind to generate the needed lift without getting incinerated. Your hull is made out of muileh which is a material that cannot melt no matter how hot it gets. But the payload and people aren't, so you have the issue of thermal soak.

Ablative heat-shields saturate in terms of heat flux: Above a critical temperature the heat-shield starts to sublimate and the gas created prevents further heat flux to the shield. This means time is more important than temperature and you want to go in quick.

The shuttle has a non-ablative system and indeed reenters at a lower force. But there is still some advantage in not drawing reentry out too much (time still "wins" against temperature) even if there isn't as much of a saturation effect.

With this plan in mind:

At which (maximum) altitude & speed would the variometer tell 0 m/s vertical speed?

No limit here. You could (for a short time) maintain 0 m/s speed at any altitude/speed on this reentry.

What would be max temperature reached?

Probably around 1500C as that is the what the space-shuttle's relatively gentle reentry is. But for an even longer time than the space-shuttle.

Have there ever been real tests of deorbiting high finesse / low wing loaded gliding devices?

Not that I am aware of. Supersonic wings aren't "high finesse" (low chord). Keeping tensile strength high anywhere near 1500C is a real challenge and a long wing exposed to the airflow would be hard to heat-shield. As is protecting the avionics.

What would angle of attack be during the whole descent, until subsonic speed?

Near the zero-lift point in orbit (for my reentry plan), to highest in hypersonic but much below orbital speed and then lower again for the subsonic portion.

How does Coanda effect work at hypersonic speeds?

I am guessing it would be reversed. At low speed airflow tries to follow the curve of the ball. Air flowing over the top will end up getting deflected downward (lifting the ball upward) when it passes the upper-backside of the ball. Visa-versa for air passing the bottom of the ball. The Coandă effect uses skin-friction: a pitched baseball with backspin will slow down air on the lower side and speed up air on the upper side. This asymmetry generates a net upward lift. At supersonic speeds the air doesn't have time to follow the curve of the ball. It slams into the front of the ball as a shock-wave and is flung outward at great speed. It will eventually be pushed back inward and fill the void left by the ball but (enough above mach 1) this is too far downstream to affect the ball. A back-spinning ball will tend to push the air ramming into the front of it up due to skin friction and thus push itself down which is opposite to the subsonic case.

Can't be done. The Karman line is defined as the point where you can't maintain enough lift without having orbital speed and thus aerodynamic flight is impossible.

• "and thus aerodynamic flight is impossible" do you have a source for this?
– user19132
Commented Nov 25, 2022 at 21:50
• @qqjkztd That's the definition of the Karman line--the point where you can't keep yourself up with wings. Commented Nov 25, 2022 at 21:54

If you could modify the glider's wings to fold to enter the atmosphere like an arrow then expand the wings a bit to gradually shed speed while creating lift then yes you could use a glider to enter from orbit to meet the Kármán line, glide and then land safely. The glider would still need a heat shield, modification and weigh more. As it is the wings would rip off to just to start.

This booster rocket kind of does this:

Baikal flyback booster with second stage
The flyback wing is stowed above and parallel to the fuselage

Source: Russian Foundation for Advanced Studies (FPI) via russianspaceweb

• It is more than obvious that gliders can return from space; the shuttle did it over 100 times. This doesn't answer the question. Commented Dec 29, 2018 at 22:09
• @OrganicMarble the booster is a glider on the return trip empty of fuel.
– Muze
Commented Dec 31, 2018 at 19:39