Install Steam
login
|
language
简体中文 (Simplified Chinese)
繁體中文 (Traditional Chinese)
日本語 (Japanese)
한국어 (Korean)
ไทย (Thai)
Български (Bulgarian)
Čeština (Czech)
Dansk (Danish)
Deutsch (German)
Español - España (Spanish - Spain)
Español - Latinoamérica (Spanish - Latin America)
Ελληνικά (Greek)
Français (French)
Italiano (Italian)
Bahasa Indonesia (Indonesian)
Magyar (Hungarian)
Nederlands (Dutch)
Norsk (Norwegian)
Polski (Polish)
Português (Portuguese - Portugal)
Português - Brasil (Portuguese - Brazil)
Română (Romanian)
Русский (Russian)
Suomi (Finnish)
Svenska (Swedish)
Türkçe (Turkish)
Tiếng Việt (Vietnamese)
Українська (Ukrainian)
Report a translation problem
No no, I'm not asking for tips on how to get out of a stall. I held the stick back all the time just to demonstrate what I had previously observed.
What I'm trying to say is that the aircraft is capable of pulling itself out from the stall during that brief level flight simply because it is light and has a comparatively powerful engine for its weight, but this has not happened.
The mirage does it beautifully at much lower speeds, granted it has a slightly higer power to weight ratio.
What happened is not a unreasonable result. In cases like that, there is no scenario where using the afterburner will suddenly take a MiG-21 out of a stall and back into normal forward flight. The ONLY solution is to release the aerodynamic load on the plane by releasing the stick, let the nose drop, gain airspeed, and then pull out of the dive.
the Mig-21 isn't meant to fly at less than 320km/h, landing at less than that is already a death trap, the engine can not overcome the drag due to the high AoA, pulling the nose will just increase your AoA while it will kill any velocity, and the plane will just fall to the ground.
Mirage-2000 is a much more modern plane, you may compare with the Mirage-3 which suffered from the exact same problem.
Mirage-3 needed military power to not lose altitude at 370km/h, only the use of afterburner or a decrease of the AoA could prevent the plane from falling like a rock at 350km/h.
in your video, you are not trying to land, you don t apply small corections with your control surface while you keep high AoA, you are completely pulling the stick back with extremely high AoA, it is expected that your your plane fall from the sky.
And it seems plenty of people have already given you the pointers on how to fix that issue.
Yes, since the power to weight ratio is ~ 1, it is capable to get out of the stall using engine power alone. Thats what I'm trying to get across.
I did not say that I was expecting it to "suddenly" take it out of the stall, what I was saying is there was no increase in the forward momentum, there should be nothing to prevent its forward velocity vector to increase and still the aircraft did not do that.
https://imgur.com/a/lOIMf
This is what I'm talking about.
Ok,
1. I'm not trying to maintain a glide slope
2. Please look at the grapgh I've posted in the imgur link that I've included 2 comments above.
3.Modern or not, it shares a lot of features with the mig such as the delta shaped wings, the only difference would be the slats at the leading edges of the mirage and its comparatively more powerful engine.
4. I am not asking people to give me tips on how to fly nor am I asking people to tell me how to get out of a stall, I wouldnt still be playing this if I didnt know these basic things
Hhhh....it was an intentional stall, the video exists to demonstrate the limitations of the FM of the mig 21, I am not asking people to tell me how to get out of the stall. Kindly look at the imgur link I've included 3 comments above. It summarizes what I want to say.
Basically when you draw the force diagram on the aircraft and split it into it's x and y components with respect to the ground. (postive x lets say forward direction level flight and negative y points torwards the ground.
If we assume the aircraft is not accelerating i.e. is failing at constant velocity, then we know that the magnitude of the thrust has to equal the magnitude of the drag force in the x direction, and the magnitude of the weight has to equal the magnitude of the drag in the y direction. (since this is a stall we ignore lift.)
The problem seems to come in is that the drag in the x direction is so great that even full afterburner provides no acceleration at relatively low speeds. (even if stall the continues you would still expect acceleration in the forward direction unless the plane is at max speed.)
And we also know the plane is failing and drag forces act antiparrallel to the velocity vector, so most the drag forces have to be in the vertical direction in this case. (which is probably realistic)
I would like to see you reproduce this scenario however, and post of video of you letting go of the controls and see how it behaves. That way we can see the effect of the control surfaces on the areodynamic forces of the aircraft.
On a second note I am also surprised that you can hold that position in a stall like that. I would expect there to be a torque on the aircraft, that either results in a spin or the nose to fall forward. The tail section would induce a lot more drag falling vertically like that than the nose, therefore pushing the nose down, regardless of the elevator/stabliator's position.
https://forums.eagle.ru/showthread.php?t=179681&page=8
https://forums.eagle.ru/showthread.php?t=194892
https://forums.eagle.ru/showthread.php?t=146550
"Another and much greater issue is that when reaching 90 deg. of AoA, which can only be reached through stall spins (which is absolutely normal for such a highly stable aircraft), after reducing the yaw rate to 0 the aircraft trims itself to 90 deg. of AoA like if it's statically relaxed there and won't budge in pitch whatever you'd try. This is also abnormal..., normally the higher the AoA (after the stall occurs) the higher the stability margin and therefore the higher the stabilizing moment should be towards reducing the angle of attack, but it seems that at 90 deg. there is no more stability margin left at all."
In this statement, you mention that drag is being exerted in the x axis, what is causing this drag? This is the part that I dont get. Since the aircraft is falling down (along -ve y axis) the drag exerted on the aircraft is in the direction of +ve y axis which in LEVEL FLIGHT (wrt to horizon) is LIFT.
Where does the drag along x axis come from when the resultant vector of the aircraft is pointing ~80° towards the ground?
I'd be happy to do the video for you, but if you want to, I can describe it as I've already experimented the same thing.
What happens when you let go at that speed is as you have described it in the next paragraph, a torque acts on the aircraft resulting it to spin (almost whip lash) around the axis of its center of gravity and into the direction of the resultant vector (this applies to any point after the moment of exit from the loop as long as the aircraft is in stall speeds)
The elevators are pointing along (almost) the resultant vector during that brief level flight. It should result in producing a small bit of lift along the x axis of the aircraft IMO
I dont understand this, I'm pretty new to this aircraft so I'm having a hard time visualising this.
But thank you for showing it to me, i'll go check it out!
Drag and lift are not the same thing. Lift is caused by an exchange of momentum when air passes over a wing. In a stall there is no airflow over the wing and therefore there is no lift. What you experience in the video is falling at terminal velocity where the vertical componment of the drag force is exactly equal to the weight of the aircraft.
I would really like to see a video so we have info of air speed, AoA, and so forth.