Don't you need the "centrifugal force" in there, if the craft is traveling at near orbital speed?

@OrganicMarble the definition is just when these two forces are equal. It's addressed in the Wikipedia article that when these two stated forces are equal, the actual trajectory would be a straight line, rather than a circle around the Earth. * The Karman line is therefore the highest altitude at which orbital speed provides sufficient aerodynamic lift to fly in a straight line that doesn't follow the curvature of the Earth's surface.*

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.

@OrganicMarble you will have to argue directly with von Karman then. This is the definition, it is not a calculation of actual forces.

@OrganicMarble the goal here is to deduce the nature of the representative lifting body that von Karman envisioned for his definition where he arrived at an altitude of approximately 100 km.

I'm not arguing with anyone. Just describing how I would approach it.

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.

@OrganicMarble If the data on the Wikipedia page is right the Shuttle would have a wing loading of 274 kg/m^2 (empty). 1% of what he's suggesting for a Karman plane, of course it was ignored. (The page lists wing area, I don't know if that includes the area underneath the orbiter..)

@LorenPechtel ignoring my deleted comment with a massive units problem, that wing loading seems to be in the ballpark.

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/…)

@OrganicMarble maybe it was never talked about re Shuttle can be understood per BobJacobsen's link: It is interesting to note that the U.S. government has never officially adopted either of these standards, because doing so would complicate the issue of overflight rights for surveillance aircraft and reconnaissance satellites. (The Department of Defense is the exception, which, for purposes of aeronautical ratings, does subscribe to the FAI definition.)

The only "line" we cared about was Entry Interface, defined as 400K feet.

Oddly, Karman himself mentions the centrifugal force when writing about this: "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."

@OrganicMarble "...no longer any air..." is surprising. If you're reading an English translation, I wonder if that was meant to be slightly different. x doesn't need to go to zero for y to prevail.

It appears to have been written in English, but published posthumously and with a collaborator, so perhaps not from the man himself.

@Conelisinspace I'll believe it when I see it; cite your source indicating that my starting point is wrong, because I've simply summarized what I believe that my cited source carefully explains. Have you read the question carefully and thoroughly, or are you just "comment-hopping"?

Where do you get 4.575E-07 * 1.225 kg/m^3 in the source exactly? By using grc.nasa.gov/www/k-12/airplane/atmosmet.html I get for 100km altitude a density of 6E-6 which is one order of magnitude higher (so it would 18 wing loading instead of 1.8...better but still a bit low I guess).

@BlueCoder page 68 i.stack.imgur.com/C2dlC.jpg

@uhoh Thanks. Shouldn't be the escape velocity 2*GM/(R+h)? en.wikipedia.org/wiki/Vis-viva_equation#Practical_applicatio‌​ns

@uhoh This paper suggests that a better lift coefficient might be no higher than 0.4, but that just makes it worse by cutting the wing loading even more. ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19710010231.pd‌​f

@BlueCoder orbital velocity for any time in an ellipse: $v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$, substitute $r=a$ (circular orbit) gives $v^2 = GM\left(\frac{1}{a}\right)$, substitute $a = R_E+h$ then gives the expression $ \left(\frac{GM}{R_E+h}\right)$. I don't see how escape velocity has anything to do with the question.

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.

The vis-viva equation states: >It is the direct result of the principle of [conversation of mechanical energy][2] which applies when the only force acting on an object is its own weight. Because you're also using the lifting force as an acting force you cannot use the vis-viva equation, and so you cannot set the velocity you derived from equalling lifting force with gravitational force, equal to the orbital velocity. [2]: en.wikipedia.org/wiki/Mechanical_energy#Cons

@OrganicMarble deleted in both places. I'll revise my comment there.

@uhoh About your bounty: can you explain what you exactly mean with "capture the summary" ? And don't you mean the 80 km number instead of the 100 km number ?

@Conelisinspace No I mean 100. The bounty is aligned with the topic of the question. It's a dense paper and before he starts in on 80 he walks us carefully through how we initially got to 100. That's what this question and this bounty are about.

@uhoh: I still wonder how Karman's original calculation of 83.6 km would influence the wing load.

@SF. you may have found the answer to my question, go for it !!!

@uhoh: No, I didn't, but I found something else. Talking about "Karman calculated", "Karman's equation" etc may be an overstatement. It's a mess. Karman sketched a rough graph. Haley added a couple extra lines to that graph, including one at a certain 'knee', marking it 'Karman's Primary Jurisdiction Line'. And it intersected the Y scale. There.

@SF. oh that is a really interesting read, I'm going to have to take some more time later to read through thoroughly. Still though, you did find at least a factor of 10, a sizable chunk of the bird/plane disparity.