Ok, your answers seem to be valid, and especially useful for Real Airbus Pilots. I stay away from this discussion, I have no interest to interfere with advice for active pilots.

My personal interest is to try to understand the system behaviour. The lateral aspect of roll control got confirmed by you by and large. Thank you, guys.

Now the more difficult part: Vertical Control.

Airbus FBW seems to incorporate a mix of Pitch Rate and g loads control. Pitch is predominant at low speeds, g load control at high speed. The A320 Veq "equal Speed" or changeover speed is not a hard changeover, but the speed where the influence of both factors are mixed 50/50 (at least that's what I picked up)

I ask myself it this is an intended mix or just the result of physical basics.

How I understand it:

Pitch Control means sidestick controls the pitch rate for changing the pitch.
Sidestick at zero most probably means the pitch angle is constant. I'm not sure if all influences are countered. Certainly Speed change, Power Change. Turbulence? Shouldn't have a great influence anyway.
Push/Pull of the side stick will command the speed of pitch change, so the pitch rate.
This is predominant at low speeds.

g load or load factor control means sidestick controls the vertical acceleration of the plane and will adjust the pitch to get the result.
Sidestick at zero most probably means the current vertical speed is constant. If plane is in level flight, it remains at current altitude; V/S=0. In Climb or Descent the V/S remains unchanged = unaccelerated. So the reference of g load as a measure for acceleration. sidestick neutral means 1g load, or 0g deviation, no vertical acceleration.
Push/Pull of the side stick will command a deviation from 1g. Like 1.2g, that will exert an upward vertical acceleration on the plane, V/S builds up until sidestick is neutralized again.
If this will also resist to uncommanded external influences (symmetric vertical turbulence, vertical airstream), then it sounds as a dream. Again it would prevent a large deviation from assigned altitude, and reduce "bumby rides" and sudden drops. with some significant delay I assume, cause the pitch will take time to adjust lift
G Load is predominant at high speeds, I learned. Where it really could help in turbulent air (within limits)

But why not at low speeds? On Final approach is is also the main interest to remain on the flight path (GS). Pitch Rate control doesn't do that directly.
What is the intention? Or, as I asked, is it simply physical (aerodynamic) laws and no intention?

The basis of my question is this formula from user CVividasku :

Me trying to use known words, pilot commands

C* = Load factor + Veq*Pitch Rate/g

I don't really understand it. Is it a zero sum on sidestick neutral? What changes when the sidestick is deflected?

The predominance of one or the other, he expains (modified to my terms):

So it's not really trajectory stable.

I'm confused and hope for somebody smart to enlighten me (or some of us)

Again, it's academic and I accept that Real Pilots don't need to know,

“Conventional aircraft”?
 

if by that you mean Boeing, many of those pilots need to learn how their over controlling is a wasted effort.

There are many videos of constant extreme control input which are not necessary, yet they broadcast their workouts as if they saved lives! The aircraft’s inertia and built in stability can do most of the work of correction.

and I speak from lessons and experience that started with the C150 and continued to to B737-800 and all makes and models in between.

but without a demonstration, most aggressive over-controlling pilots will not believe and remain convinced that a rapid extreme frenzy of aileron and elevator is the only thing that prevents crashing.

This. A million times this. I often wonder whether the complete lack of retrospection that allows someone to post a video of such awful flying is linked to the need to take and publish the video in the first place.

It's not extremely clear, but I think it works as follows.
C* = Load factor + Veq*Pitch Rate/g
The basics seem that you're asking for C* with your stick.
So to start with, if you're stick neutral, you're asking for 0 = load factor + veq*pitchrate/g
If you have a gust that increases the pitch, you have positive pitch rate, so load factor becomes negative, it leads to stabilizing the trajectory.

Now let's assume you're moving the stick. You ask for C* = 2.5g. You're gonna ask for 2.5g load factor so the elevators will deflect. This will give a pitch rate. You will obtain a load factor less than C*.

I'm not keen on trying to criticise or understand further this law. For example you could say that the stick directly asks for load factor.
Very clever people did it, and to get oneself to the same level of understanding would require a test airplane, or a test pilot training.
This costs one million. If you gave me one million I will gladly (very gladly) follow the test pilot course, but that's not near from happening.

Also, Boeing is using the same law (only with speed added), which shows that the veq*pitchrate term is there for a good reason.

You can still continue the analysis for a bit if you want.
Cz is a linear function of AOA. The lift equation is 1/2 rho s v² Cz
At higher speeds, a degree of AoA will give a higher load factor change. Let's say you ask C* = 2g (total, so 1g in reality)
You can derive something useful that is pitch rate 1° per second is equivalent to 0.05g at 60kt (and twice at 120kt, etc, linearly)
You are solving for C* = load factor + veq * pitchrate/ g
So you're going to solve 1 = (speed * 0.05 / 60 + veq/g)*pitch rate
That gives
Pitch rate = (C*-1)/(speed*0.05/60+veq/g/57.3) (in degrees per second

This gives a very "balanced" result. If you pull C* = 1, at low speeds you're going to get a lower load factor (that the airplane can give you), at high speeds you're going to get a higher load factor. But you're going to get a decreasing pitch rate.
That means the airplane is not going to feel agressive at high speeds, but still reactive at low speeds.
https://i.gyazo.com/bbfe34eaff064aa0...1bb58db63c.png

Also, this equation surely works in a non steady state.
At first, you don't have the equilibrium load factor. So the pitch rate target is higher. That probably makes the airplane more reactive in the very short term (first second or so)
Then, on the medium term (pitch rate established), you get this "equilibrium" in the graph above.
On the long term, the airplane if you pulled on it is decelerating, and the equilibrium load factor reduces. It's also "stabilizing" (in the right direction).

The system is much more complex than this simple analysis.
Hence, I don't know for example where the design load factor of 2.5g comes into play. Is it a cap on the C* law ? Is it included in the design of the C* law ?
At 1.6 VS you can pull 2.5g, theoretically. Let's say you pull C* = 2.5g at 160kt. This gives a load factor term of 0.6. If you want a load factor term of 1.5 at 160kt, you need to pull a C* of 3.7.
Then of course the protection also come into play.
Conversely if you want -1g, that is -2g variation compared to level flight, you need C* = -5.
Does it mean that the full course of the stick is from C* from -5 to 3.7 ? It's possible, but it's not even sure that the stick is proportional to C* (it should be according to all this theory but we can't know for sure)

Let's explore other possibilities :
If the stick is proportional to C*, then you will have an increasing load factor but decreasing pitch rate when increasing speed.
If the stick is proportional to load factor, you will have constant load factor and decreasing pitch rate (but decreasing very quickly) when increasing speed
If the stick is proportional to pitch rate, you will have a constant pitch rate and an increasing load factor (but very quickly) when increasing speed.
If the stick is proportional to elevator deflection.. You will have a constant AoA, and you will have an increasing pitch rate when increasing speed, so a load factor that increases even quicker.

Qualitatively, it seems like C* is the most balanced solution. Constant or increasing pitch rate with speed gives an airplane behavior that changes too much with speed. Constant load factor gives an airplane that would feel heavy or slow to react at higher speeds.

I stand to be corrected on this, if anyone has the training and experience related to aircraft design.

​​​​CVividasku , THANK YOU for that extreme effort to dig deeper. How amazing is that!

I need some time to understand what you think and suggest. But I already got your hint that harmonic behaviour is the reason for this mix of parameters.

It's absolutely interesting. Have nice weekend.