The veteran Space Shuttle commander Charlie Precourt writes, in the July 2022 issue of EAA's Sport Aviation, p. 38:

Another interesting reality about our flight controls was their working essentially in reverse early in entry. To effect a right turn, we actually deflected the right elevon down which created more drag on the right, yawing the vehicle right, increasing lift on the left wing and causing roll to the right. The downward deflected right elevon was not causing roll to the left as it would in an airplane.

In this situation, why wasn't the right elevon deflected up instead? Wouldn't that also add drag, just like deflecting down? Wouldn't that also somewhat reduce the right wing's lift, thereby encouraging the desired right roll? Even in this regime (hypersonic, low air density, high angle of attack) there's some lift going on: he explicitly mentions the left wing's lift. So some of the conventional behavior of wings and ailerons ought to still apply.


4 Answers 4


For an explanation let me first explain hypersonic flow a bit. In super- and hypersonic flow, when a body moves through a fluid what changes foremost is not speed but density. At the high angle of attack during re-entry, the flow over the upper side of the body is almost a vacuum and on the lower side the local density is much higher than the (very low) ambient density.

Therefore, any upward control surface deflection will make no difference as the control surface is moving essentially into a vacuum. For the downward moving control surface the picture is very different: the already denser flow gets compressed even more, producing a bit of lift and a lot of drag.

Early in re-entry, most of the lift is still provided by the centrifugal force of the orbital motion. The high angle of attack is only for braking and stability, and in the rarefied atmosphere little lift is produced by the wings. The high-g phase of descent is still to come.

Shuttle hypersonic reentry test of 1975

1975 test of a Shuttle model in a helium wind tunnel (the glow from the shock structures is a result of electron-beam fluorescence | picture source)

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    $\begingroup$ Helium wind tunnel? Oooh la la $\endgroup$
    – DKNguyen
    Jan 1, 2023 at 6:37
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    $\begingroup$ Is there a similar picture that plots the density field, instead of the pretty shock waves? $\endgroup$ Jan 1, 2023 at 18:30
  • $\begingroup$ @CamilleGoudeseune: I wish there were - I found this pic by googling, but was unable to find anything about density. $\endgroup$ Jan 1, 2023 at 20:21
  • $\begingroup$ For a cone at Mach 2, arxiv.org/pdf/2208.04280.pdf, Fig. 9, left column, shows the density field dropping to ambient, but hardly lower. Mach 25 must be a different beast entirely. $\endgroup$ Jan 1, 2023 at 20:35
  • $\begingroup$ @CamilleGoudeseune perhaps this computer generated image is helpful. At re-entry altitude of around 100 km "ambient" density will not be much. $\endgroup$ Jan 3, 2023 at 13:42

This observation is not unlike what F4 Phantom pilots experience using ailerons at higher AoA slow flight.

The downward elevon undoubtedly produces more (adverse yaw) drag. The Shuttle is very low aspect ratio, meaning the elevons are much closer to the centerline. In order to have the same roll torque as a higher aspect wing (ailerons further out), they must be larger.

As mentioned, extremely low air density$^2$ at re-entry altitude of around 100 km renders the upward elevon useless.

Primarily, at this point in flight, any resultant rolling into the downward elevon may not caused by increased lift on the opposite side as much as "dihedral effect" from its large vertical fin (and boxy fuselage).

So, the aerodynamic behavior of the Shuttle would be, overall, more drag$^1$ based until it reached denser atmosphere. Until then, right elevon down was almost as good as right rudder. There just isn't enough air available to form a higher pressure lifting "pocket" underneath the right wing.

$^1$ from wikimedia CFD computer generated image
$^2$ from source chart of atmospheric density vs altitude

  • $\begingroup$ Indicated airspeed at 300000 feet is almost nil. But the Shuttle using wing warping to turn? I never would have believed it. $\endgroup$ Jan 1, 2023 at 5:17
  • $\begingroup$ Don't forget the dihedral effect from the swept wing (quarter-chord line of the delta wing is highly swept.) $\endgroup$ Jan 2, 2023 at 15:05
  • $\begingroup$ @quietflyer the roll torques aerodynamicly are indeed the sum of all forces, including the wing. What is interesting here, even at re- entry speeds, is that air density is close to 0 except in shockwave areas. I think a modern Shuttle would feature twin rudders, more like a F-15. In those high AoA situations, rudder is your friend. $\endgroup$ Jan 2, 2023 at 16:08
  • $\begingroup$ A lengthy conversation has unfolded in chat, focussing mainly on how the side area of the fuselage may be affecting the roll torque developed during sideslip, and why-- read or join in at chat.stackexchange.com/rooms/141704/… -- $\endgroup$ Jan 2, 2023 at 16:47

To effect a right turn, we actually deflected the right elevon down which created more drag on the right ... The downward deflected right elevon was not causing roll to the left as it would in an airplane.

The reason why deflecting the elevon down created a lot of drag but almost no additional lift as in a conventional subsonic airplane, lies in the particular behaviour of hypersonic aerodynamics.

At the hypersonic speeds typical of space vehicles reentry (Mach number much bigger than 5) air becomes an incandescent plasma mixture of ions O$^+$ and N$^+$ and electrons and radios do not work anymore. It is a complicated chemical matter but, on the other hand, the aerodynamic behaviour becomes extremely simple. To explain subsonic and supersonic aerodynamics, at least a couple of books of hundreds of pages are needed. But to explain hypersonic aerodynamics a short chapter might be enough.

Indeed, at these extreme Mach numbers, the shock wave becomes so bent toward the body that it practically coincides with the body surface itself¹:

hypersonic shock wave

This is a huge simplification: if we follow one particle of air, it simply flies on a straight line until it bumps into the body; at that point it just bends its trajectory to follow the body surface:

hypersonic streamlines

The particles of air flowing beyond the surface of the body do not impact with it and they just keep on flying straight on: the "shadowed" surface do not experience any aerodynamic force and the air there just remains undisturbed: only the surface of the body being directly impacted by the air experiences an aerodynamic force.

That's the all story about hypersonic aerodynamics! Very simple indeed.

Since upon impact the particle has changed its speed (its speed perpendicular to the surface has become zero), then by Newton's second law ($F=ma$) all the particles of air bumping into the body generate a force ($N$ in the following picture) which is:

  1. perpendicular to the body surface and;
  2. with a value proportional to the mass flow times its perpendicular deceleration to zero:

hypersonic force

For a flat plate the maths becomes really easy and simply returns:

  • $C_l= 2sin² α \times cosα$
  • $C_d= 2sin³α$

which look something like that ($C_l$ in red, $C_d$ in blue, my own work):

hypersonic aerodynamic characteristics

A couple of comments are definitely worth noting:

  • $C_l$ is not linear as it is instead at subsonic speeds.
  • $C_l$ starts at 0 at 0°, reaches a maximum and then decreases going again to 0 at 90°; the fact that it posseses a maximum has nothing to do with viscosity or stall or delta wing but only with the fact that at hypersonic speed the aerodynamic force is always perpendicular to the surface; there's no stall at hypersonic speeds.
  • For the same reasons, $C_d$ increases from 0 at 0° to 2 at 90° with an increase that, at least for small AoA, is cubic with AoA i.e. proportional to "³"; at subsonic speeds $C_d$ is simply constant at small AoA.
  • $C_{l_{max}}=0.77$ @ 54.7°; for a standard subsonic NACA 0012, $C_{l_{max}}$ is some 1.6 @ 22°, i.e. twice as much @ half AoA.
  • Obviously in reality at zero AoA $C_d$ is not zero but equals the drag due to viscosity $C_{d_0}$ and the whole plot can be simply shifted upward of $C_{d_0}$.
  • Maximum efficiency $L/D$ is between 6 and 7 while for the same NACA 0012 of before it is more than 20 i.e. three time as much.
  • Albeit being deduced for a flat plate, the two equations for $C_l$ and $C_d$ show a very good agreement with the aerodynamic characteristics of 3D slender bodies.

Now that we have learned the whole hypersonic aerodynamics (as promised in just one short chapter), how elevons worked on the space shuttle is easy to be understood:

  • first of all just forget about subsonic aerodynamics and any subsonic phenomenon related to moving surfaces; as seen, aerodynamics does not work in that way at hypersonic speed! There's no stall, lift is half at best and not linear with AoA, drag is bigger and so on.
  • Second of all, in order to present its belly to the atmosphere at reentry, the space shuttle maintained an AoA of 36 to 40°.

This is important: from the previous plot it can be seen that, at those AoA, lift and drag at hypersonic speeds have basically the same magnitude; the subsonic NACA 0012 of before at the same $C_l$ of 0.65 has a $C_d$ of 0.03 i.e. 20 times smaller! Furthermore, $C_l$ is already close to its maximum value: deflecting the moving surface downward does not have a big impact on the total $C_l$ (it might actually even decrease it) while it does modify $C_d$ a lot.

So, due to the peculiar aerodynamic characteristics at hypersonic speeds and due to the high AoA of the reentry phase, manoeuvring the space shuttle was achieved acting more upon its drag than its lift. And this finally explains why:

To effect a right turn, we actually deflected the right elevon down which created more drag on the right, yawing the vehicle right.

¹ all pictures from here

  • $\begingroup$ Now that's a really good answer. In my opinion, this is the best answer here +1. $\endgroup$ Jan 6, 2023 at 11:57
  • $\begingroup$ @AdityaSharma: I also think that this is the best answer 😂 But you know, sometimes is enough to be the first one and not the best one to win the competition here 😉 $\endgroup$
    – sophit
    Jan 6, 2023 at 19:19

Whenever you're generating lift with an airfoil, you're also generating drag. Deflecting an aileron up reduces the lift generated by the wing, and therefore, reduces induced drag. This causes the nose of an airplane to yaw in the opposite direction, a phenomenon called adverse yaw.

It's also true that the outside wing of an airplane in a turn generates slightly more lift than the inside one, because it's going faster through the air. This effect is normally so small it's not noticeable, but in certain situations (e.g. an incipient spin), this asymmetrical lift becomes a huge factor in the behavior of the airplane. In point of fact, the behavior of ailerons does reverse for a normal airplane in a spin, for much the same reason that it reverses for the Shuttle.

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    $\begingroup$ I agree with what you say about adverse yaw, but about your second reason: how much faster the outer wing is in comparison to the inner wing is directly proportional to the turn radius and wingspan. Span is constant, but the turn radius depends on TAS, and the space shuttle in this situation is operating at quite high TAS, probably so high that this effect becomes negligible. $\endgroup$ Jan 1, 2023 at 6:41
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    $\begingroup$ Correction: Directly proportional to the wingspan, inversely proportional to the turn radius. $\endgroup$ Jan 2, 2023 at 2:21

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