From the NTSB's preliminary report, we read in the "Findings" section:

  1. Although the copilot made the required 0.8 Mach callout at the correct point in the flight, he incorrectly unlocked the feather immediately afterward instead of waiting until SpaceShipTwo reached the required speed of 1.4 Mach.

  2. The unlocking of the feather during the transonic region resulted in uncommanded feather operation because the external aerodynamic loads on the feather flap assembly were greater than the capability of the feather actuators to hold the assembly in the unfeathered position with the locks disengaged.

Taking into consideration that, as reported elsewhere:

The vehicle's two tail wings are supposed to move when the spacecraft hits a certain speed, a maneuver known as "feathering." This repositioning helps to increase drag in order to slow the vehicle down for reentry.

From the article above, let's also take note of the visual depiction of such configuration change:

SpaceShip2 Feathering

  1. How is possible that this configuration change, meant to increase drag, needs a "preventing" (for lack of a better word) mechanism?
  2. Since the repositioning helps to increase drag, how do the aerodynamic loads make the repositioning occour?
  3. Wouldn't the (additional) drag on the tail section make it go back down again?
  • $\begingroup$ Not a full answer as I don't have any technical info: my assumption is that the tail was either generating lift or vibrating in part of the transonic region, or that normal use of the control surfaces in pitch would have this effect. (I'm assuming the elevator surfaces are on the boom section though.) Also the report mentions inertial loads, which might imply vibration? $\endgroup$
    – Andy
    Jul 29 '15 at 10:18
  • 3
    $\begingroup$ Watching the video, it seems to me like it's more that the fuselage pitched up, the tail wings stay mostly level. $\endgroup$ Jul 30 '15 at 1:28
  • $\begingroup$ Good find by @2012rcampion - the boom section remains approximately level and the body starts to tilt up, quite slowly (at first). So lift from the boom's surfaces seems unlikely. I think they would be starting to pitch up at approximately this time in the flight profile, by the way. I wonder if they would expect to be established in the climb, and at lower air density, at the preferred unlocking point. $\endgroup$
    – Andy
    Jul 30 '15 at 16:54

How do aerodynamic loads explain the Spaceship2 accident?

Aerodynamic load overpowered the actuators holding it in place, forcing it to open and causing SpaceShip2 to break up in flight. Although the rudders were unlocked early when SpaceShip2 was going at Mach 1.0, this action alone should not have been enough to pivot the tails upright, because neither pilot took the further step of turning the feather handle to actually move them. However, the tails rotated upward anyway, and the increase in drag at this point in the flight caused it to crash.

How is possible that this configuration change, meant to increase drag, needs a "preventing" (for lack of a better word) mechanism?

The feather is ultimately there for re-entry. It acts as a hinge and can go up and down. This system increases atmospheric drag on the spaceship so that it naturally positions itself during re entry. To make sure the tail doesn’t move forward early, SpaceShip2 has a structural lock that holds feather in place. Around Mach 1.4, the lock is moved so that the feather is now mechanically free. It won’t move though because at Mach 1.4, aerodynamic forces keep it nailed back. In addition to the lock, the feather itself has a big actuator that drives it up and down. So just because it’s unlocked doesn’t mean it’s just flopping in the wind and it can go where it wants. In the (initial) boost phase, the aerodynamic forces can overcome that, which is why it is locked .

How do the aerodynamic loads make the drag naturally increase?

Aerodynamic forces are generated by the difference in velocity between the spacecraft and the air. There must be motion between the spacecraft and the air. If there is no relative motion, there is no lift and no drag. In order to maintain a constant airspeed, thrust and drag must remain equal, just as lift and weight must be equal to maintain a constant vertical speed. If in level flight, the engine power is reduced, the thrust is lessened, and the aircraft slows down. As long as the thrust is less than the drag, the aircraft continues to decelerate until its airspeed is insufficient to support it in the air.

Calculations can be found here

Aerodynamic forces are used differently on a rocket than on an airplane. On an airplane, lift is used to overcome the weight of the aircraft, but on a rocket, thrust is used in opposition to weight. Because the center of pressure is not normally located at the center of gravity of the rocket, aerodynamic forces can cause the rocket to rotate in flight. The lift of a rocket is a side force used to stabilize and control the direction of flight. While most aircraft have a high lift to drag ratio, the drag of a rocket is usually much greater than the lift.

Wouldn't the drag on the tail section make it go down again?

There are two conventional ways to transition from space back to Earth: One is to come in as a winged airplane, like NASA’s space shuttle. In this way the spacecraft can land where it wants, but hitting the perfect angle of attack is tricky. The second method is as a capsule, which is essentially dummy-proof, but has no control over where it lands and subjects its occupants to high G-forces. SpaceShip2 is a bit of both. Its tail folds up and down, changing the vehicle’s shape from something that falls like a capsule to one that flies like an airplane once it’s back in the atmosphere.

Each of the two tails acts as a rudder, and has a small, horizontal flap at the back, extending to the outside. When the plane is descending, both tails can pivot upright, together, from 0 to 90 degrees, so that they stand “vertically” behind the plane. In this configuration, the tails and flaps create drag. When pointed upward, the tails are at right angles to the direction of airflow, creating a huge amount of drag on the vehicle, which slows it down without overheating the spacecraft. This method works because SpaceShip2 is not coming back from orbit.If put in feathered configuration during powered flight, the drag vector and thrust vectors are horribly misaligned, resulting in extreme pitch rates and, since the aircraft wasn't built to withstand those stresses, it would break apart

  • $\begingroup$ 1. However, the tails rotated upward anyway, that's my question: how was that possible? 2. It won’t move though because at Mach 1.4, aerodynamic forces keep it nailed back. which ones? the same that made it move up at 0.8? 3. I think I have to rephrase the third question. 4. I do not feel you have answered the fourth. $\endgroup$
    – Federico
    Jul 29 '15 at 19:04
  • $\begingroup$ @ Federico: After the spaceship reaches apogee the vehicle's twin tails rotate upward, stabilizing and slowing SpaceShip2 as it reenters the atmosphere. It's not feather deployment at high speed that is the problem, it's deployment during powered flight. In order to fly during powered flight, the thrust vector must be aligned with the center of mass and center of pressure (drag vector). Otherwise, the vehicle will tumble due to offset thrust. $\endgroup$
    – DSarkar
    Jul 29 '15 at 20:21
  • $\begingroup$ I'm sorry, but I do not see how your comment addresses any of the points I raised. Adds an interesting factoid, but does not answer my doubts. $\endgroup$
    – Federico
    Jul 29 '15 at 20:38
  • $\begingroup$ Your original question was "How do aerodynamic loads explain the SpaceShip2 accident?" I answered it in the opening para itself. In your question you need to segregate the cause of the crash from the mechanism involved in operating the spacecraft. Based on the NTSB report, so far, there is no specific clue that a malfunction occurred. The crash has been attributed to pilot error. $\endgroup$
    – DSarkar
    Jul 30 '15 at 8:38
  • $\begingroup$ that's the title, the question is in the body of the post. $\endgroup$
    – Federico
    Jul 30 '15 at 13:44

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