5
$\begingroup$

I just got a question after reading my old Principles of Flight book. The equation for the rate of turn is $\textrm{ROT} = 1,091 \cdot \frac{\tan(\phi)}{V}$, with $\phi$ denoting bank angle and $V$ denoting airspeed

How would that change when the airplane is equipped with thrust vectoring like the F22 or SU35? Is there an equation for that?

$\endgroup$
4
  • 2
    $\begingroup$ If you want to get down to actual equations, you are going to specify HOW the thrust vectoring is being used. Say we are in a left turn. Is the nozzle pointed to the left, to drive the nose to the left and expose the right side of the fuselage to the airflow? Or is the nozzle pointed to the right, to provide centripetal force to increase the turn rate? If the latter, what (if anything) is providing the yaw torque needed to keep the nose from yawing to the right? The question as you are asking seems to be insufficiently constrained. $\endgroup$ – quiet flyer Feb 15 at 12:39
  • $\begingroup$ Above was for the Su-35 , w/ 3-D thrust vectoring. F-22 only has 2-D. $\endgroup$ – quiet flyer Feb 15 at 12:42
  • $\begingroup$ An interesting case that may provide some food for thought in improving the question is Northrop YA-9. en.wikipedia.org/wiki/Northrop_YA-9 . Airbrakes on wings could be asymmetrically deployed to facilitate using the rudder to generate aerodynamic sideforce. As an example, to the boost the turn rate in left turn, without increasing bank angle, the plane might deflect rudder to right to generate an aerodynamic sideforce to the left. Meanwhile the nose would be prevented from yawing to right, and kept aligned with the instantaneous flight path, by opening the airbrake on left wingtip. $\endgroup$ – quiet flyer Feb 15 at 12:47
  • $\begingroup$ So are you envisioning something like this that combines some sort of differential wingtip drag device, with the thrust vectoring nozzle? (Does the Su-35 actually do this? Not as far as I'm aware, but maybe-- perhaps there's a good basis for another ASE question here--) As an aside, my understanding is that the basic purpose of the system on the YA-9 was to allow gun-aiming and flight path control to be de-coupled. $\endgroup$ – quiet flyer Feb 15 at 12:57
10
$\begingroup$

The standard equation applies to any coordinated level turn (so that the G felt is "straight down" to an occupant of the aircraft, in an aircraft frame of reference). Matters not how it's all achieved - flaps, high-lift devices, helo rotor, thrust vectoring, whatever... a coordinated level turn at "this" bank will take "this" much radius. And it will have a calculable G load.

Actually achieving (and sustaining) that G load may take all sorts of interesting effects, such as thrust vectoring. But once you're there, the standard equations apply.

And, if you choose to fly an un-coordinated turn, with some amount of side-loaded G force (example, a car turning at high speed... no appreciable "bank angle" so the G force pushes you toward the outside of the turn), then all the standard assumptions are out the window. At that point, your radius would be a function of G & not bank... and it'd get uncomfortable really fast!

$\endgroup$
3
  • 1
    $\begingroup$ I know we aren't supposed to use comments to say "nice answer"-- but this is a nice answer. Nicely sidesteps (yet also illuminates) any issues due to question being insufficiently constrained. $\endgroup$ – quiet flyer Feb 15 at 19:30
  • $\begingroup$ Not to be pedantic but in circular motion there is no "force pushing you towards the outside of the turn" you are only accelerating inwards. $\endgroup$ – bruh_weed Feb 15 at 20:50
  • $\begingroup$ @bruh_weed: True, the actual force is pushing the plane inwards, not pushing you outwards. But it feels the same either way — you and your seat are still being pushed together. $\endgroup$ – Ilmari Karonen Feb 16 at 15:03
8
$\begingroup$

With thrust vectoring you no longer turn (as in: the wing creates the force that accelerates you in the desired direction) but you do post-stall maneuvering. Next, you need to distinguish between highest instantaneous turn rate (trading altitude for higher rate) and continuous turn rate (which is limited by the available thrust in most cases).

enter image description here

Turn rate diagram (picture source). It plots flight Mach number on the X axis over turn rate on the Y axis. The bold colored lines show the sustained turn performance of several airplanes. At low speed the turn rate grows in proportion to the maximum possible load factor that the maximum lift of the airplane is capable of. The kink in the lines at around 10 - 12 degree per second shows the thrust limit - in order to fly tighter turns at even higher load factors, more than the installed thrust is needed. Now the curves run almost horizontal along the thrust limit and decline again at high Mach numbers, first at the speed of sound and then when supersonic drag cuts down the possible load factor.

The thin colored lines show the instantaneous turn rate, when altitude loss is permitted. The sharp peak at the maximum load factor (for example 8g for the Su-27 and 9g for the F-15) marks the maximum turn rate when thrust is unvectored and wing lift is used to force the change of direction.

With thrust vectoring the airplane can fly a direction change in a totally different way. It will pull up to reduce speed, then rotate using vectored thrust when flying at low speed in a zero-g parabola. When the fuselage points at the desired direction, it will use the altitude gained in the pull-up to accelerate again, now in the new direction. Now the turn rate depends on how quickly the speed can be reduced and how long it takes to build up new speed. The rotation itself only needs a second or two.

Since a conventional turn also requires to slow down and missile engagement can start as soon as the fuselage points towards the adversary, post-stall turns using thrust vectoring give a decided advantage in a dogfight with missiles.

$\endgroup$
1
  • 3
    $\begingroup$ This paper suggests a 3–5% higher turn rate (sustained), but I don't presume to be able to put it into words. The author derives the turn rate equation based on the thrust vectoring angle and its effect on the load-factor. Perhaps it can be of use. $\endgroup$ – ymb1 Feb 15 at 16:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.