# What would a “Kármán plane” look like, a bird, or a plane?

If I understand correctly (which I might not), the Kármán line is roughly the altitude where a "Kármán plane's" upward lift force at the orbital velocity for that altitude would be equal in magnitude to the gravitational downward force.

A simple expression for lift force would be:

$$F_L = \frac{1}{2} \rho v^2 S C_L$$

where $$\rho$$ is the density at that altitude, S is the aircraft's wing area, and $$C_L$$ is the aircraft's coefficient of lift.

The gravitational force downward at an altitude $$h$$ above a given Earth radius $$R_E$$ would be

$$F_G = \frac{GM_Em}{(R_E+h)^2}$$

where $$GM_E$$ is Earth's standard gravitational parameter and numerically is about 3.986E+14 m^3/s^2.

Setting those equal gives:

$$v^2 = \frac{2 GM_E m}{\rho S C_L (R_E+h)^2}$$

Orbital velocity can be gotten from the vis-viva equation:

$$v^2 = \frac{GM_E}{(R_E+h)}$$

and setting those two expressions equal yields

$$\frac{m}{S} = \frac{1}{2} \rho C_L (R_E+h)$$

Plugging in nominal values for lift coefficient (unity), $$R_E+h$$ (6378 + 100 km), and an estimated density of 4.575E-07 * 1.225 kg/m^3 from an old NASA standard atmosphere (see the (currently unanswered) question Why does Earth's atmospheric density have a big “knee” around 100 km? Is there a good analytical approximation?), I get a mass to wing surface area of this "Karman plane" of about 1.8 kg/m^2.

This ratio is also called wing loading and a value this low is literally "for the birds" and for paragliders. Values in that article for commercial aircraft are in the low to mid hundreds.

Question: What would a Kármán plane look like, a bird, or a plane? In other words, have I done my maths right, and understood the concepts and definitions correctly, and if so, why would the object used to conceptually define the approximate altitude of the Kármán line have a wing loading of about 2 kg/m^2 rather than a realistic airplane?

So far, the only thing I've found within this site about the topic is in one of @MarkAddler's answers (always a good place to start), which says (in part):

von Kármán picked some representative values for $$m\over A$$ and $$C_L$$, which I don't know. But I don't need to know.

This may be discoverable in Theodore von Kármán's original calculation, which is likely in German. While that didn't lead to exactly 100 km originally, an analysis of that result may lead to an answer.

• Don't you need the "centrifugal force" in there, if the craft is traveling at near orbital speed? – Organic Marble Jul 27 '18 at 4:18
• Orbit is sort of a balance between that and gravity. Don't see how you can leave it out. It's good to start with a force diagram. – Organic Marble Jul 27 '18 at 4:24
• Well, I learned a lot about the Karman line from reading this and researching it a little. It's a lot more abstract than I thought, since as you say, it ignores the centrifugal force. Any real world vehicle flying at the Karman line would have to take that into account. So, +1 for making me learn something. This is probably why we never talked about the Karman line at all in shuttle. – Organic Marble Jul 27 '18 at 17:04
• The NASA explanation of the definition makes more sense to me than the Wiki one: "Somewhat later, aeronautical scientist Theodore von Kármán calculated that above an altitude of approximately 100 kilometers (62 miles, or 328,084 feet), a vehicle would have to fly faster than orbital velocity in order to derive sufficient aerodynamic lift from the atmosphere to stay aloft. (see e.g. nasa.gov/centers/dryden/news/X-Press/stories/2005/…) – Bob Jacobsen Aug 3 '18 at 3:15
• Note that small flyers like birds have a lower mass per wing surface because of the square-cube law: wing surface increase with the square of size, while mass increase with the cube of size. So for the same shape, a bigger plane has a higher mass per wing surface. As such, a plane with the same ratio than a bird needs to have much wider wings, or be made of lighter materials like aerogel. – Eth Oct 16 '18 at 11:07

This atmospheric model from NASA, states that above 25km altitude:

$$Temp = -131.21 + 0.00299*h$$,

$$pressure = 2.488*((Temp+273.1)/216.6)^{(-11.388)}$$ and

$$\rho = pressure / (0.2869*(Temp+273.1))$$.

So for 100km we have $$\rho$$ = 6.006E-06, which is one order of magnitude higher than the one used in the question.

By plugging this number into the final equation I get a wing loading of about 19.45 $$kg/m^2$$, which is still low compared to today's airplanes but still more reasonable than 1.8 $$kg/m^2$$. and close to the limits of birds according to Wiki.

Also consider that Kármán calculations did not yield 100km (see this), but a lower value, which was then rounded up to 100km because it was easier to remember. If we consider the Kármán line to be as low as 83.6km, as suggested in one of the comments to the question, we would get $$\rho$$ = 2.589E-05 and a wing loading of 83.62 $$kg/m^2$$. This is higher than a Piper Warrior light aircraft, so the Kármán plane is definitely starting to look like a plane, and not like a bird.

On the other hand, $$C_L = 1$$ might be high (e.g. a 747-200 is stated as having a $$C_L = 0.52$$ ) and reducing it would reduce the wing loading again.

However, using 83.6km and $$C_L = 0.52$$ , the result would still be a 40+ wing loading, well beyond a bird max wing loading of 20.

If Kármán used an atmospheric model which yielded similar density values to this and imagined a plane with $$C_L = 0.5$$ and a wing loading of 40, then indeed he could have drawn the line around 80km.

My answer: yes, the Kármán plane looks like a plane, although probably not like a fighter jet or a big airliner but more like a small light aircraft.

• Well, it is not in the answer, it is only in the comment. – BlueCoder Oct 17 '18 at 9:55
• I have changed the last sentence so that it doesn't convey the possibility that one or the other model is "wrong". We just need to know what values Karman used :) – BlueCoder Oct 17 '18 at 11:02
• I've been looking for original Karman calculations as well because my interim conclusion is that I'm completely confused by all this. No luck so far. In fact, I think that would be a good question to post! – Organic Marble Oct 17 '18 at 13:09
• @uhoh I have updated the answer by considering 83.6km as the original altitude.. the result is now safely in the plane realm, even with a more realistic C = 0.5 :) – BlueCoder Jan 7 '19 at 10:55
• Thanks for taking a walk on the Karman side i.stack.imgur.com/YApMx.png – uhoh Jan 7 '19 at 11:03

Based on your initial stipulations, and the wording provided by Wikipedia, the altitude Karman was calculating was the altitude where, at orbital velocity, the lifting effect of aerodynamic forces on an aerospace frame is just sufficient to hold it aloft against gravity. Ergo, a body with sufficient lift to stay aloft at any velocity below orbital speed could, in theory, maintain an orbit at less than the speed dictated by Newtonian physics.

However, where aerodynamics provide lift, they also provide drag. Hence a craft operating in this way would need to provide periodic or continuous thrust, vs. the occasional boost burns needed for some LEO satellites.

It's really not an easy question to answer, nor is why one would want to make such a craft. My estimation is that it would look something related to both the U-2 and the B-2 airframes, but probably lighter and larger than either.

From Wikipedia Kármán line: Kármán's comments:

In the final chapter of his autobiography Kármán addresses the issue of the edge of outer space:

Where space begins… can actually be determined by the speed of the space vehicle and its altitude above the earth. Consider, for instance, the record flight of Captain Iven Carl Kincheloe Jr. in an X-2 rocket plane. Kincheloe flew 2000 miles per hour (3,200 km/h) at 126,000 feet (38,500 m), or 24 miles up. At this altitude and speed, aerodynamic lift still carries 98 per cent of the weight of the plane, and only two per cent is carried by centrifugal force, or Kepler Force, as space scientists call it. But at 300,000 feet (91,440 m) or 57 miles up, this relationship is reversed because there is no longer any air to contribute lift: only centrifugal force prevails. This is certainly a physical boundary, where aerodynamics stops and astronautics begins, and so I thought why should it not also be a jurisdictional boundary? Haley has kindly called it the Kármán Jurisdictional Line. Below this line space belongs to each country. Above this level there would be free space

(Theodore von Kármán with Lee Edson (1967) The Wind and Beyond, page 343)

• Again, thank you for the invocation. I may come back to this in a handful of days after investigating Karman's equations. – Hunting.Targ Dec 3 '18 at 5:04
• I hope that is a more informed, if not more enlightened, response. – Hunting.Targ Dec 15 '18 at 4:35
• I can't up vote twice, looks good though. – uhoh Dec 15 '18 at 6:11
• A good answer provides proof, this answer does:space.stackexchange.com/questions/31738/… – Cornelisinspace Dec 15 '18 at 11:03
• as a small aside, the block quote starting "...final chapter of his autobiography Kármán addresses the issue..." was likely written by the other author; 1) "autobiography" was published four years after Kármán's death, 2) contains scientifically incorrectly worded statements like " there is no longer any air to contribute lift". – uhoh Dec 18 '18 at 1:30

This answer will show that the starting point of the question, namely that the lift force $$F_L$$ would equal the gravitational force $$F_G$$, is wrong !

The FAI defines the Kármán line as the altitude of 100 km, so a Kármán plane would fly at that altitude.
The lift force for that plane is:
$$F_L = \ 1/2. \rho v^2 S C_L$$ According to this talk page the lift coëfficiënt for a supersonic airplane is:

$$C_L = \frac{4\alpha}{\sqrt{M^2 - 1}}$$ where $$\alpha$$ is the angle of attack in radians and $$M$$ is the Mach number.
(According to one of the editors instead of $$4\alpha$$ the numerator could be 4sine($$\alpha$$), with $$\alpha$$ in degrees)

To look for the different forces acting on a supersonic Kármán plane we can take the North American X-15 as an example.
With 4$$\alpha$$ = 2 and $$M$$ = 25 (first line) the lift coëfficiënt becomes: $$C_L$$ = 0.08 .

With $$\rho$$ = 5.6 x 10$$^-^7$$, $$v$$ = 7.5 km/sec and $$S$$ = 18.6 the lift force(X-15) = 23.4

$$F_G(gravitational force) = \frac{G M_Em}{(R+h)^2}$$

With $$h$$ = 100 and $$m$$ = 7000 the gravitational force(X-15) = 66,667 so $$F_L$$ < 0.04 % of $$F_G$$.

So this example shows that the lifting force at that height is only a fraction of the gravitational force !

Because the Kármán plane is supposed to maintain the 100 km altitude with a speed near the orbital velocity, the acceleration downwards to the centre of the Earth has to be taken into account.

Whatever the Kármán plane looks like, there is always a speed near the orbital velocity that is sufficient to keep that plane in orbit.

• That's not an article, it's a talk page. the angle of attack for a Karman plane would not be moderate - closer to 1 radian. Orbital velocity is high hypersonic, not supersonic, so the formula wouldn't necessarily extend to that regime. – JCRM Jan 11 '19 at 22:20
• the editor who included that equation didn't know if $\alpha$ was radians or degrees – JCRM Jan 11 '19 at 22:34
• A Karman plane (my word) is an instantaneous, hypothetical construct used as a linguistic tool to address forces at one instant in time. Karman planes do not fly! They exist for a mere instant, then disappear into the same virtual world from which they were originally imagined. – uhoh Jan 12 '19 at 0:28
• @uhoh That's simply nonsense. Kármán planes use the lifting force to keep them aloft so they have to fly whether or not the Kármán plane is your word. – Cornelisinspace Jan 12 '19 at 10:42
• @JCRM You're right, it's a talk page so i've changed that. If the AoA of the plane would be close to $57^0$ it would get pretty "stalled" ! – Cornelisinspace Jan 12 '19 at 10:46