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

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)

• 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. Commented Feb 25 at 11:48

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?

• "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. Commented Feb 25 at 16:34
• @sbabbi It's earth axis G. Commented Feb 26 at 3:46
• 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. Commented Feb 26 at 8:41
• @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! Commented Feb 26 at 8:41
• @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. Commented Feb 26 at 11:42

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: 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: 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 normal acceleration (acceleration vertical to the body of the aircraft, $$n_z$$) regardless of pitch attitude (according to this source). Therefore during a climb you would have to trim (reduce) the normal acceleration $$n_z$$ commanded by the FLCS such that a stable climb is achieved. For example in a $$60^\circ$$ climb, you would have to reduce the normal 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:

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

with $$n_{z,cmd}$$ being the target normal acceleration (acceleration vertical to the body of the aircraft), $$\varphi$$ the roll angle, and $$\delta_{trim}$$ the trim input, normally being 1.

• 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? Commented Feb 26 at 8:49
• @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. Commented Feb 26 at 9:28
• @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... Commented Feb 26 at 9:32
• @Sandpatch updated my answer accordingly Commented Feb 26 at 9:44
• @U-flow Thanks for the answer. Seems like it is like that. Commented Feb 26 at 20:47

It seems that I'm late to the party, but the key reason is that the F-16, F-18, F-35, etc. uses Normal Acceleration, or Normal Load Factor feedback in the cruise or Up & Away (UA) flight mode. It does not compensate for pitch or roll. (The F-18 and F-35 however, will shift from Nz command to pitch-rate command at lower speeds.)

Definition: The Normal Acceleration (an) or Normal Load Factor (Nz) is the acceleration in “g” along the aircraft body Z axis, not including the components of gravity.

If you're climbing vertically with a 90 deg pitch, your Nz will be 0g. So the Flight Control System will constantly pull the aircraft to maintain Nz=1g.

Just be aware of in which axis system the acceleration you're talking about.

The ultimate source of F-16 FLCS using Nz feedback: (F-16 Digital Flight Control System Functional Block Diagrams. Data obtained from Lt. Bruce Peet, F-16 System Program Office, Wright-Patterson OH.) Quoted in the paper DTIC ADA 189675.

• Re "including the components of gravity" -- did you maybe mean not including the component of gravity? If gravity is included, then you have to say the situation in straight and level flight for the Nz component is 0G not 1G. Commented Apr 16 at 13:25
• @quiet flyer Yep I missed that. Thanks. Commented Apr 17 at 6:06
• +1 for the control scheme. Could you perhaps also add the link to the source? Commented Apr 18 at 15:03
• @U_flow Sure, i've included it in the answer. Commented Apr 18 at 15:15