TsAGI paper suggests that the success of the Cobra is the rapid attainment of the Cobra maneuver. Staying in a position of 30-40 degrees AoA creates large lateral instability and oscillations and causes departure. To Cobra you must surpass this and go to the point of flow separation over the wing with little lateral movements for the maneuver to be a success.
Now, to further your point… the Su-27 has been seen doing great lateral motions in its’ maneuvers and recovering with ease and full control. This would be a better example for comparison;
It is not my intention to argue, but to find the truth.
Sorry but the video is not of a Su 27, but a Su 35 with vectored nozzles, completely different digital FbW and a significantly different airframe. I am attaching the video and a picture of the aircraft for comparison.
The first Su 27, the one in WT and DCS, does not do this "Su 35 Super version of Cobra. “Classic Su 27” Cobra is a post stall maneuver. The point is that longitudinally the aircraft can get dynamically above 100 AoA (if the pilot has the balls to do it, that is with a very good pilot, because the safety constraints of AoA and G are turned off, the Cobra maneuver can kill weaker pilots).
But laterally, above 28AoA it is uncontrollable ( according to the manual for the SK version.
so…did the you give up on fixing the flanker stupid speed bleed? or the r73 going crazy at low speeds and not tracking a target even with afterburner on(gripen) just cause he is flaring like a maniac
It’s back to position because of inercia and because of coefficient drag to lift go sky hi above ~18 AoA and it’s not existent around ~30 AoA (basically 0 lift> movement Vector stay the same as inercia Vector) . It’s have nothing comon with control flying. Pilot is passenger above point where plane going into deep stall (according definition what some one post for you)
In the SK manual there’re rolling and yawing moment coefficient derivatives with regard to sideslip angle (beta). Also known as lateral and directional static stability derivatives, denoted as Cl-beta (mx-beta) and Cn-beta (my-beta) respectively.
At M=0.2, Cn-beta is dropped to zero at 29° AOA, meaning there’s no yaw restoring moment available to reduce the sideslip. And Cl-beta is dropped to zero at 34° AOA, so no rolling moment can be generated with sideslip (by rudder).
These are two graphs of lateral and track stability, and we are talking about controllability. So mx- by delta flaperons at 28 degrees is equal to 0, and mx by delta rudder is not equal to 0 and the control moment is preserved
Yes, I’ve seen the graph of mx by delta flaperons and my by rudder some times back but couldn’t find it right now. It shows the mx-delta flaperons is zero at 28° AOA, and my-delta rudder at 30° AOA is about half of that at 10° AOA, but not at zero yet.
However I’d like to point out that, mx-beta (Cl-beta with opposite sign) is also reduced to zero at 34 deg AOA at M=0.2, which means roll controllability by inducing sideslip with rudder is also not possible at this point.
For actual roll response taking into consideration the lateral-directional stability of the aircraft and the roll-raw crossfeed control, one can calculate the LCDP parameter, as explained by NASA:
In the case of Su-27 at AOA=34° and M=0.2, Cl-beta is zero, so LCDP = Cn-beta, which is already negative at 34° AOA. So the roll response will be reversed and nose slice is possible even with roll-yaw crossfeed enabled.
Likewisely, the LCDP of F-16 is also in its lowest at 35° AOA in the AOA > 25 region, but still above zero, augmented by Aileron-Rudder Interconnect. (a.k.a roll-yaw crossfeed)
F-16 is equipped with Aileron-Rudder Interconnect (ARI) so you only need to look at the upper curve. It has a local minima at 35 deg AoA when the aircraft go beyond the 25 deg AOA limit.
Without an ARI, the F-16 would have a reversed roll response at 25 deg AOA.
