V in the equation is the corner velocity is affected by TWR and the STR is further affected by max G-load which pilot can sustain. The very high TWR of F-22 can give it much higher sustained G load but it gets restricted to just 6G due to pilot limitation. In the lift equation L=CL x 1/2 x density x wing area x V^2, the only factors which a pilot can vary is V & CL by way of AOA or high lift devices.
weight affects the stall speed but not the CL as CL is dependent on AOA only.
So if you want a high turn rate, you need to have a low speed, high Lift and ability to sustain that speed. as the AOA increases, the drag also increases. if you study the energy diagram/graph of any fighter you will find that their ITR is calculated for the lowest speed which allows structural G load to be reached and the STR is calculated for speeds where the thrust and Drag are equal.
The rate of onset of G and roll rate is a factor in dogfight but they too have limitations imposed by the human body. Very high rate of onset of G can result in G-LOC to the pilot. It is for this reason that most RSS and unstable platforms have a controlled rate of G onset.
As already explained, theoretically wing area is a factor but for ITR,STR & min radius turn you need low speeds and in that case its the CL max which is the determining factor.
It is for this reason that Mig-29, Su-27, SU-30MKI, SU-35 have much higher ITR and STR compared to anyother fighter in the world except F-22 and they all have sweepback of 42* and an effectively cranked arrow wing form.
LCA has tried to overcome its low CLmax by increasing the wing area which created more problems for it.
ITR is dependent more on ability to pull high G load for short duration than anything else.
M2K cud pull nearly same ITR as Mig-29 because the total lift of an aircraft Ltotal= Lwing + Lthrust. As the AOA increases, a part of thrust gets added to the total lift. In M2K, due to 58* sweep, it has very high stall AOA and when it pulls to its stall AOA, a large amount of thrust is added as lift. For a 30* AOA, up to 50% thrust forms the lift component.
So M2K is able to to turn at lower speeds for 9G turns. This is the reason for it to be able to match Mig-29. The downside is that at Higher AOA the thrust component available to over come drag also reduces which results in very poor STR.
Banked turn - Wikipedia, the free encyclopedia
When a fixed-wing aircraft is making a turn (changing its direction) the aircraft must roll to a banked position so that its wings are angled towards the desired direction of the turn. When the turn has been completed the aircraft must roll back to the wings-level position in order to resume straight flight.[4]
When any moving vehicle is making a turn, it is necessary for the forces acting on the vehicle to add up to a net inward force, to cause centripetal acceleration. In the case of an aircraft making a turn, the force causing centripetal acceleration is the horizontal component of the lift acting on the aircraft.
What this effectively means is the acceleration of any fighter in a sustained turn depends upon the "horizontal component of lift acting on the fighter".
In straight, level flight, the lift acting on the aircraft acts vertically upwards to counteract the weight of the aircraft which acts downwards. During a balanced turn where the angle of bank is θ the lift acts at an angle θ away from the vertical.
It is useful to resolve the lift into a vertical component and a horizontal component.
If the aircraft is to continue in level flight (i.e. at constant altitude), the vertical component must continue to equal the weight of the aircraft and so the pilot must pull back on the stick a little more. The total (now angled) lift is greater than the weight of the aircraft so the vertical component can equal the weight.
The horizontal component is unbalanced, and is thus the net force causing the aircraft to accelerate inward and execute the turn.
Because centripetal acceleration is:
A=v^2/R
Newton's second law in the horizontal direction can be expressed mathematically as:
L Sin θ=M*V^2/R
This formula clearly shows that fighters with higher lift(lower wing loading ) can achieve a quicker sustained turn with lesser bank angles.
Also it clearly shows that for the same speed fighters with higher lift (low wing loading) can have a smaller radius of turn.
What is the significace of the above two point in a turning fight?
where:
L is the lift acting on the aircraft
θ is the angle of bank of the aircraft
m is the mass of the aircraft
v is the true airspeed of the aircraft
r is the radius of the turn
In straight level flight, lift is equal to the aircraft weight. In turning flight the lift exceeds the aircraft weight, and is equal to the weight of the aircraft (mg) divided by the cosine of the angle of bank:
L= Mg/cosθ
where g is the gravitational field strength.
The radius of the turn can now be calculated
R=v^2/g*Tanθ
This formula shows that the radius of turn is proportional to the square of the aircraft's true airspeed. With a higher airspeed the radius of turn is larger, and with a lower airspeed the radius is smaller.
This formula also shows that the radius of turn decreases with the angle of bank. With a higher angle of bank the radius of turn is smaller, and with a lower angle of bank the radius is greater.
SO now it is clear while lift has nothing to do with radius of turn, it is the sole thing that determines the all important " net force causing the aircraft to accelerate inward and execute the turn".
SO if you have lesser lift your horizontal component of the lift force which determines your acceleration will be lower. It means the pursuing fighter with higher wing lift(low wing loading fighter) will shoot you down with his guns , because his acceleration is faster.
By giving the equation for radius of turn alone and ignoring the which gives the all crucial centrepetal acceleration in a sustained turn, you are missing the whole point of the advantage of having a lower loaded wing.
