Here is what I read about the X-15 spaceplane:

The X-15 had a thick wedge tail to enable it to fly in a steady manner at hypersonic speeds.[14] This produced a significant amount of drag at lower speeds;

A wedge shape was used because it is more effective than the conventional tail as a stabilizing surface at hypersonic speeds. A vertical-tail area equal to 60 percent of the wing area was required to give the X-15 adequate directional stability.

The X-15 does indeed have flat trailing edges on its stabilizers. Some photos show this.

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So why does this produce better stability at hypersonic speeds? I thought stabilizers were more about the total surface area anyway. It just seems so counter-productive to make a flat end like that giving huge drag, especially for hypersonic stuff.

  • $\begingroup$ You may wish to rationize it by looking at a "boat tailled" bullet. At hypersonic speeds and altitudes, the density difference between the compression wave and ambient air may be well, like air and water. So rather than a higher viscous drag longer tail, it can be truncated, using the relative near vacuum behind the surface as a stabilizer. The compressed airwave pattern behind it would resemble the wake of a high speed boat, crossing well behind the aircraft. $\endgroup$ Nov 3, 2019 at 4:14
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    $\begingroup$ @RobertDiGiovanni What does "boat tailed" bullet mean? If your talking about normal bullets, the muzzle velocity is usually between Mach 1 and 2.5. That is not hypersonic at all. $\endgroup$
    – DrZ214
    Nov 3, 2019 at 9:58
  • $\begingroup$ @RobertDiGiovanni Compare the range between regular and spitzer bullets. A blunt end does indeed cause a lot of drag. Having been developed some 120-150 years ago, bullet shapes are not the ideal example of a low drag body. $\endgroup$ Nov 3, 2019 at 14:54
  • $\begingroup$ @Peter Kampf But drag was needed to help stabilize the aircraft, notice this novel approach allows for greater stability, at glide, WITHOUT adding MASS to the tail of the aircraft (definite improvement over the 307, no?), as a larger tail would have affected the center of gravity. This may have been "stability insurance", but at those speeds, who could blame them, $\endgroup$ Nov 3, 2019 at 16:22

2 Answers 2


A converging shape at hypersonic speed in a low-pressure medium will produce close to vacuum pressure on its surfaces (hypersonic shielding). A small sideslip angle will only result in a very small pressure difference between both sides.

Contrast this with a diverging shape which produces higher than ambient pressure on both sides. In hypersonic flow this pressure grows approximately with the square of the inclination angle. The same sideslip angle, therefore, will result in a much higher pressure difference, rendering the diverging shape more effective in sideslip.

A more detailed discussion can be found here, page 11-6 to 11-8.

A wedge shape was chosen since it allowed to predict forces and temperatures with greater confidence than would had been possible with a more complex shape.

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    $\begingroup$ Any drag (including a streamer or parachute) adds to stability. The X-15 flew with what amounted to a speed brake on the tail to account for potentially disasterous destabilizing forces on the nose at those speeds, as one Neil Armstrong famously experienced barely making it back to the Mojave after overshooting the base and flying out over Los Angeles. The return glide was close. But yes, the wedge forces would be more predictable. $\endgroup$ Nov 3, 2019 at 16:12
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    $\begingroup$ @RobertDiGiovanni: Please read the linked paper. It explains the wedge effect very nicely and drag is not part of it. $\endgroup$ Nov 3, 2019 at 17:03
  • $\begingroup$ Page 11-8 bottom paragraph certainly is interesting. $\endgroup$ Nov 3, 2019 at 20:13

Thanks for @PeterKampf's answer and reference for the illumination. This is to complement the results drawn within its reference, which I don't think are completely accurate.

As shown in Anderson, Fundamentals of Aerodynamics, the Newtonian theory is an ok first cut at hypersonic aerodynamics: due to the vast difference between the speed of sound and the airspeed, air flow is basically deflected by whatever shape it encounters. For a flat plate at an angle of attack ($\alpha$) with the hypersonic airflow, the bottom surface (incident with the airflow) will have a coefficient of pressure:


while the upper surface will have no effect:


Combined together, this makes a coefficient of lift:

$$C_l=sgn(\alpha)*2\sin^2\alpha\cos\alpha\approx sgn(\alpha)*2\alpha^2$$

Therefore, for a flat plate, the lift is nonlinear with respect to the flow incidence. For a vertical stabilizer, the directional stability would vanish at small perturbation and would also be nonlinear.

With a wedge shape making a half angle $\theta$, and for flow incidence smaller than $\theta$, the lift is:


Therefore, for flow incidence smaller than $\theta$, a wedge shape has a linear lift with respect to flow incidence and a fairly constant stability derivative even at small perturbations.

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    $\begingroup$ I will add constructively that these answers do seem to validate the need for all moving control surfaces, as the trailing edge is effectively "blanked out" (to the point of vacuum) at hypersonic speeds. $\endgroup$ Nov 4, 2019 at 4:44

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