As an aircraft accelerates through the transonic region from full subsonic to to full supersonic flow its directional or yaw stability decreases. What causes this reduction?


It is a combination of several effects:

  1. Aeroelasticity: With the higher forces at high speed, the structure deforms such that the effective flow angle at the tail surface is reduced.
  2. The supersonic lift curve slope of the tail surface decreases with Mach while that of the fuselage stays roughly constant, so with higher Mach the tail contribution to stability shrinks while the destabilizing contribution of the fuselage stays the same.
  3. Tail location: At supersonic speed the density of air on the upper fuselage is reduced while that on the bottom is increased. Since the vertical sits on top of the fuselage, it flies in rarefied air. Ventral fins, on the other hand, become more effective.

The supersonic lift curve slope is approximately $$c_{L\alpha}=\frac{4}{\sqrt{Ma^2-1}}\cdot\left(1-\frac{\lambda}{2\cdot AR\cdot\sqrt{Ma^2-1}}\right)$$ Note that the second term is for the influence of the triangular surface on the tip where the pressure difference is reduced. Nevertheless, it shows that lift curve slope of the vertical decreases with increasing Mach number.

For the fuselage, the lift curve slope is that of a slender body:$$c_{L\alpha}=\pi\cdot\frac{AR}{2}$$ Note that this equation is independent of Mach, so the destabilizing fuselage contribution stays constant over Mach.

For the beneficial influence of ventral fins see the diagram below which has been published by Lockheed in The F-104 Strafighter Test Pilot's Notebook by Glenn L. Reaves (source):

F-104 ventral fin contribution to directional stability

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  • $\begingroup$ Thank you. That seems to clear it up nicely. $\endgroup$ – Guy Inchbald Jun 6 at 13:09
  • $\begingroup$ while the destabilizing contribution of the fuselage stays the same why stays the same? Doesn't the center of pressure move forward? $\endgroup$ – Abdullah Jun 6 at 14:26
  • $\begingroup$ @Abdullah: The fuselage behaves like a slender body, its center of pressure is already near the tip and the forces on it don't change that much with Mach number. $\endgroup$ – Peter Kämpf Jun 6 at 20:51
  • $\begingroup$ Are you saying sub vs supersonic or supersonic vs more supersonic? I mean, at subsonic speeds, we have viscosity effects that just don't work that much at Mach>1 $\endgroup$ – Abdullah Jun 7 at 6:35
  • $\begingroup$ I guess the sensitivity to aeroelasticity must vary greatly depending on the relative locations of engine/s and tail fin/s. For example if the tail were rigidly mounted above a single engine, the thrust line would tend to move with it. $\endgroup$ – Guy Inchbald Jun 7 at 15:35


enter image description here

Here you see a fuselage viewed from above. The horizontal line at the right end of the ellipse is the tailfin. On top is a subsonic plane, and at the bottom is a supersonic one. The thin lines are pressure, the thick ones are forces, and the arrows show flow direction and magnitude.

As you can see, at subsonic speeds suction is more dominant, therefore, the tail, where more suction occurs, produces more sideforce, stabilising the plane.

EDIT: However, this effect is not that significant.

The below diagram shows the tailcone of the same plane.

enter image description here

As you can see, the same thing happens to the tailfin. But the effect of this center-of-pressure movement can't be that significant. However, the back half of the stabilizer becomes less effective at supersonic speed due to being in the low pressure "wake" of whatever is ahead, (the fuselage, as Peter said, or even the front section of the fin itself) as wakes are more defined at these high speeds.

The X-15 had a wedge-shaped tail fin because of this

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  • $\begingroup$ So if side pressure increases so much at the nose, why does it not also increase in similar proportion at the tail fin? $\endgroup$ – Guy Inchbald Jun 6 at 12:21
  • $\begingroup$ Let me add another diagram. $\endgroup$ – Abdullah Jun 6 at 12:45
  • $\begingroup$ @GuyInchbald why does it not also increase in similar proportion at the tail fin? You're missing one point. The destabilizing effect of pressure on the nosecone increases while the stabilizing effect of pressure on the tailcone decreases. $\endgroup$ – Abdullah Jun 6 at 14:30
  • $\begingroup$ Of course I am missing something, that is why I am asking! Thank you anyway, but I find Peter's analysis to be the more convincing. $\endgroup$ – Guy Inchbald Jun 6 at 14:49
  • $\begingroup$ @GuyInchbald That's OK, even I upvoted him. I was just clarifying my answer because you misinterpreted it. $\endgroup$ – Abdullah Jun 6 at 15:17

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