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The F-16 has a FLCS that constantly holds 1 g. What happens on high pitch angles?

What is defined as 1 g? let us say that you climb with 60 degrees relative to earth, does that mean that the FLCS just registers 0.5g and trims the plane up?

What happens at 90 degree pitch? Does it trim a whole G?

This is what DCS and BMS currently do and it doesn't make sense to me.

(This is all with AOA of less than 10 degrees)

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  • $\begingroup$ Ask why an aircraft needs an angle of attack: to make lift against gravity. The four forces must be in balance to fly a straight line. In a steep climb AoA is reduced because gravity vector (WRT the aircraft) is now more aligned with drag, requiring the engine to do more work. In a 90 degree climb "lift" will only curve the flight path. In order to fly straight, the computer sets AoA to zero lift and the engine does all the work, essentially as an airbreathing rocket. $\endgroup$ Feb 25 at 11:48

2 Answers 2

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The FLCS holds 1 g in absence of stick input.

This 1 g is maintained in relation to Earth. The result is that, given enough thrust and lift, the airplane's nose will keep pointing where the pilot has last pointed it. This is maintained whether pitch is up or down, or when the plane is in a coordinated turn.

When climbing at a 60 degrees up pitch, the airplane is still at ~1 g, which simply means it's not rapidly accelerating in any direction. Climb isn't acceleration per se. Whether at 15 degrees or 90 degrees, it just requires the forces to balance out while maintaining some upward velocity component.

Here's a detailed explanation of climb physics: Does lift equal weight in a climb?

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    $\begingroup$ "When climbing at a 60 degrees up pitch, the airplane is still at ~1 g," It's not, or at least it depends on your definition of G. At 60 deg pitch up, in steady state, the IRUs would measure 0.5 vertical G, and sqrt(2)/2 longitudinal G. If measured in earth axis, the up/down component would indeed be 1. OP is asking if the FLCS feedback is based upon body-axis G or earth axis G. $\endgroup$
    – sbabbi
    Feb 25 at 16:34
  • $\begingroup$ @sbabbi It's earth axis G. $\endgroup$
    – Therac
    Feb 26 at 3:46
  • $\begingroup$ Thanks for the answer, but this was a question specific for the F-16 FLCS. Does the F16 also take this into account because it would have to subtract acceleration and braking forces to the accelerometers reading, making it slightly more complicated. $\endgroup$
    – Sandpatch
    Feb 26 at 8:41
  • $\begingroup$ @Therac From what I can gather from online sources, the commanded g-rate is independent from the pitch attitude, and only (in newer blocks) is dependent on the roll attitude. Consider this video from DCS or this source. This means that for example during a climb, the pilot would have to trim the aircraft to command less then 1 g. If I am not mistaken, your answer implies that this is done automatically, however this is not the case. For example in inverted attitude the plane pulls down with now 2 g! $\endgroup$
    – U_flow
    Feb 26 at 8:41
  • $\begingroup$ @U_flow You command any g you need with the stick, and while you're doing it, it's anything within the envelope. Once released, the plane stabilizes at 1 g in relation to Earth. At least that's the idea, implementations aren't all perfect. But DCS seems to do just that. $\endgroup$
    – Therac
    Feb 26 at 11:42
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The gravity by definition always exerts 1 g of accerleration on any given object. If this acceleration is compensated by a counterforce, the object is weightless. Therefore if no other forces are acting on the object, the object is unaccelerated and continues on its current trajectory in a straight line.

Consider the example of the F-16 in straight an level flight and plot the different forces you arrive at the following well-known picture: enter image description here The complete gravity of 1 g is compensated by the lift of the aircraft and drag is compensated by thrust.

During a climb the situation is a bit different: enter image description here As you can see, the gravity still pulls the plane down towards earth, but now the lift and thrust combined cancel out the gravity combined with the drag of the aircraft. However the magnitude of the gravity is still 1 g.

But the FLCS of the F-16 does not do this for you as it always commands 1 g of acceleration regardless of pitch attitude (according to this source). Therefore during a climb you would have to trim (reduce) the acceleration commanded by the FLCS such that a stable climb is achieved. For example in a $60^\circ$ climb, you would have to reduce the acceleration command to only $\cos(60^\circ)\cdot 1 g = 0.5g$ and adjust the throttle such that more thrust is generated. This is not automatically done by FLCS!

However in newer blocks, the FLCS automatically compensated for roll attitude. Therefore if you fly a curve with a bank angle of say $60^\circ$, the acceleration command is automatically increased to 2 gs to compensate for the centrifugal acceleration of said curve. However apparently this is only implemented in newer F-16 models.

With this in mind, the target acceleration formula for hands-off stick flight, looks something like this:

$n_{cmd} = 1g\cdot sin(\varphi)\cdot \delta_{trim}$

with $n_{cmd}$ being the target acceleration, $\varphi$ the roll angle, and $\delta_{trim}$ the trim input, normally being 1.

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  • $\begingroup$ Thanks for the answer. So yeah, it makes sense to me too. But your sources isn't really helping, it seems to be 2 comments from a forum where both are unsure. At high pitch angles, the accelerometers would need to compensate for thrust and drag, making the calculation slightly more complicated. Is that something the 50 year old F-16 could do? $\endgroup$
    – Sandpatch
    Feb 26 at 8:49
  • $\begingroup$ @Sandpatch I included the link to the forum because both of them are sure that the pitch attitude is not compensated. Therefore the FLCS will command 1 g regardless of pitch attitude. Again, I cannot give you a flight handbook as a source, but consider this video. Regarding the computational power: The flight law they are proposing is a simple cosine calculation. The FLCS has to perform a lot more calculation than this to achieve stable flight. Therefore, no this would not be a problem. $\endgroup$
    – U_flow
    Feb 26 at 9:28
  • $\begingroup$ @Sandpatch but to reitterate: The FLCS always commands 1 g of upwards acceleration (therefore upwards in regards to the pilot, regardless of orientation of the aircraft) unless the pilot trims otherwise! If you want the aircraft to climb say at 60°, the pilot has to manually trim the aircraft such that the target acceleration is only $cos(60^\circ)\cdot 1 g=0.5 g$ and adjust the throttle accordingly. Perhaps I should include this in my answer to make this perfectly clear... $\endgroup$
    – U_flow
    Feb 26 at 9:32
  • $\begingroup$ @Sandpatch updated my answer accordingly $\endgroup$
    – U_flow
    Feb 26 at 9:44
  • $\begingroup$ @U-flow Thanks for the answer. Seems like it is like that. $\endgroup$
    – Sandpatch
    Feb 26 at 20:47

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