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Edit for justification:

This question is different from If the EPU (Emergency Power Unit) on a modern fighter jet fails, will the aircraft drop out of the sky? for the following reasons:

The question in If the EPU (Emergency Power Unit) on a modern fighter jet fails, will the aircraft drop out of the sky? asks if an aircraft drops out of the sky if the EPU (Emergency Power Unit) fails, for a generic aircraft. The accepted answer only briefly touches on the F-16, and certainly not in a way to answer this question. This question is specifically about an aircraft with RSS (Relaxed Static Stability), and how this relates to loss of hydraulics and power.

Question:

The F-16 has Relaxed Static Stability (RSS) to enable a high degree of maneuverability.

This causes the aircraft to be inherently unstable and RSS is corrected for by the on-board computer in real time, as part of the fly-by-wire system.

In the event of loss of hydraulic power, the ailerons cannot be operated, the on-board computer/avionics will also fail, if there is no power. The EPU (Emergency Power Unit) will re-establish hydraulic and electrical power, but this doesn't happen instantaneously. It takes some seconds.

During this critical window, what allows the F-16 to remain in stable flight in the context of RSS?

Reading this PDF it seems that the F-16 can be unresponsive to pilot input, yet remain in stable flight.

My question is, how is this achieved? How is it possible?

My own reasoning:

Either the F-16 is not as inherently unstable as I'm given the impression it is, or, there is enough hydraulic power and electrical power left in the interim between power from main hydraulics to switching to the EPU to allow the fly-by-wire system to keep the aircraft stable.

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    $\begingroup$ Make a scale model of an F-16 and throw it. It is designed for supersonic flight, and may only need computer help under certain conditions. A few seconds in straight and level subsonic flight (most of the time) should be enough time. If your F16 is too unstable, see what can be done to make it more stable if you are thinking of using it for passenger service (without computers). $\endgroup$
    – Robert DiGiovanni
    Commented May 24, 2019 at 14:12
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    $\begingroup$ Possible duplicate of If the EPU (Emergency Power Unit) on a modern fighter jet fails, will the aircraft drop out of the sky? $\endgroup$
    – fooot
    Commented May 24, 2019 at 15:18
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    $\begingroup$ The battery would provide power to critical things like the FBW computers, and a pressurized hydraulic system bleeds down, rather than going to zero instantly. Perhaps this combination is enough to keep the aircraft from departing controlled flight while the EPU starts. $\endgroup$
    – Ralph J ♦
    Commented May 24, 2019 at 15:46
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    $\begingroup$ @Robert DiGiovanni I've heard this before, and I don't understand its relevance. The center of gravity highly influences stability on any aircraft. What exactly is the center of gravity on the F-16? Also, I'm not convinced that a tiny F-16 model exhibits the same aerodynamic properties as a full scale model. Aerodynamic forces change as a function of speed and air density, the model might exhibit stable flight, while the full scale model may not. I would need to see a reference for this -- that a scale model F-16 can be used to exactly simulate a real F-16. $\endgroup$
    – AlphaCentauri
    Commented May 24, 2019 at 15:48
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    $\begingroup$ @fooot I've edited my question to explain how this question is different. $\endgroup$
    – AlphaCentauri
    Commented May 24, 2019 at 15:50

1 Answer 1

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As I understand it (I'm neither an F-16 pilot nor an F-16 mechanic), Relaxed Static Stability doesn't mean the aircraft is unstable, just that it's less stable than the long-established standard.

To put it another way, if you build a free flight model airplane scaled from an F-16, you'll have to put the center of mass significantly further forward than its location in the full size, piloted F-16, in order for the model to have static stability. The same is true of many radio control models intended to be flown by an experience model pilot -- they're what's call "neutrally stable". The primary characteristic here is that these models will "go where you point them" and require that you "fly them all the time" -- unlike a trainer, or many/most scale models of piloted aircraft, you can't just set the throttle to cruise and take your eyes off the airplane, and expect to find it in a predictable location a minute later.

The same is (mostly) true of the F-16, in that, in most flight regimes, it's stable enough that as long as a violent maneuver wasn't in process when the flight computer restarts, the aircraft won't be badly out of shape by the time the computer returns to full function. Back in the 1980s, however, it wasn't unheard of for a pilot to have to eject due to a fly-by-wire system failure; my parents lived near Hill AFB and would hear by local rumor mill and newspaper of "another F-16 crash" every so often.

Bottom line (again, based on my understanding as an "interested bystander") is that the F-16 is just about stable enough that it can remain recoverable through a control system restart, if it was in a steady state flight condition beforehand.

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  • $\begingroup$ +1 for your answer. You can set up a model to be "staticly unstable (in pitch)" as long as it is still DIRECTIONALLY stable (aerodynamic nuetral point is usually well behind the CG). The elevator is trimmed down. These models have an interesting property of "altitude or speed stability" because the increasing aerodynamic pitch down is cancelled by increasing lift trying to make the plane climb with increasing speed. But if you go too slow, the cat will be waiting. $\endgroup$
    – Robert DiGiovanni
    Commented May 24, 2019 at 17:55
  • $\begingroup$ @RobertDiGiovanni But if you go too slow, the cat will be waiting Which is probably why most of the Hill AFB crashes I recall (from the period my parents lived in that area) were on landing or takeoff. Lose fly-by-wire at 50 feet at 20 kt above stall, and you've got a problem... $\endgroup$
    – Zeiss Ikon
    Commented May 24, 2019 at 18:05

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