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The "lift coefficient" $C_L$ can be very different for one specific aircraft at different speeds.

According to this article from NASA about the lift coefficient:

So it is completely incorrect to measure a lift coefficient at some low speed (say 200 mph) and apply that lift coefficient at twice the speed of sound (approximately 1,400 mph, Mach = 2.0). The compressibility of the air will alter the important physics between these two cases.

In 1963 the North American X-15 flew twice above the Kármán line altitude of 100 km with speeds around 6,000 km/h.

Joseph A. Walker flew those two spaceplane flights that qualified him as an astronaut under the rules of the U.S. Air Force and the FAI.

Could have been or has been the lift coefficient calculated from the collected flight data from those two events?

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  • $\begingroup$ they would be able to measure the $C_L$ for the specific altitude, Mach, and AoA, but that wouldn't be much use for extrapolating to the extreme conditions of a Karman plane experiment. $\endgroup$ – JCRM Jan 8 at 19:46
  • $\begingroup$ @JCRM If that $C_L$ could be compared with the $C_L$ at,,for instance, 10 km altitude and Mac 2, that could give an indication for the $C_L$ of the X-15 at the Kármán altitude $\endgroup$ – Conelisinspace Jan 8 at 20:05
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    $\begingroup$ @Conelisinspace again with the useless Karmin line +1 $\endgroup$ – Muze Jan 8 at 21:02
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The X-15 never flew at any reasonable speed close to the Karman line. The maximum speed was reached at a much lower height (around 50 km or lower) and then followed a purely ballistic trajectory for several minutes. The speed at 100 km height was just a few hundred meter per second and close to zero at its top height.

First, this is way too low to give any reasonable numbers (as stated in your quote). Second, it's pretty clear that the tiny wings of the X-15 are several orders of magnitude too small to produce any meaningful amount of lift at this height.

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  • $\begingroup$ Indeed not much speed near the Kármán line, but maybe at 50 km we could get an idea if 6,000 km/h would alter the lift coëfficiënt because there the ballistic trajectory would be influenced by the lift force. From that force the coëfficiënt could be calculated. $\endgroup$ – Conelisinspace Jan 8 at 18:45
  • $\begingroup$ No way. Air density is 500 times higher at 50km and speed is 4 times too low. As lift goes with $\rho v^2$, there is a factor 8000 in between. $\endgroup$ – asdfex Jan 8 at 19:06
  • $\begingroup$ Yes, that's the lift force, When from that force there the coëfficiënt could be calculated that would be a lot better than calculating the coëfficiënt from data collected at 10 km with Mac 2, for instance. $\endgroup$ – Conelisinspace Jan 8 at 19:19
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    $\begingroup$ @Conelisinspace when the magnitude force is so small compared to other forces (e.g. gravity) it becomes impossible to extract unambiguously from flight data. $\endgroup$ – uhoh Jan 9 at 0:50
  • $\begingroup$ @uhoh That would confirm that the lift force is small compared to the g-force. $\endgroup$ – Conelisinspace Jan 9 at 15:08
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Not sure what you're looking for — above the atmosphere the lift coefficient is going to be zero in practical terms.

NASA has a free ebook ("X-15 Research Results") which might get you pointed in the right direction. Interesting quote:

The data also confirmed another peculiar trend of hypersonic flight: the reduced importance of the wings for lift. At Mach 6 and 25° angle of attack, the large fuselage and side fairings on the X–15 contribute 70 percent of the total lift, enough to permit reentry from an altitude of 250 000 feet with fuselage lift alone.

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  • $\begingroup$ Interesting link ! Aren't you confusing the lift coëfficiënt with the lift force ? $\endgroup$ – Conelisinspace Jan 8 at 18:31
  • $\begingroup$ Is the lift coëfficiënt mentioned somewhere in that e-book ? $\endgroup$ – Conelisinspace Jan 8 at 19:20
  • $\begingroup$ I don't see how the two are being confused. Lift coefficient depends on dynamic pressure (q). Up in the mesosphere (X-15 territory), ambient pressure is something like 1/100,000 of sea level so q would be so small as to be meaningless. Same reason they had to use gas reaction jets instead of aerodynamic control surfaces. I'd have to wonder if the equipment at the time would've even been able to measure it. $\endgroup$ – PHChilly Jan 8 at 21:15
  • $\begingroup$ Because $L = C_L q S$, $C_L$ is the proportionality constant. So i think it does not depend on the dynamic pressure $q$. $\endgroup$ – Conelisinspace Jan 9 at 9:12

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