Combat Aircraft technology and Evolution

[ersakthivel]

The factors on which aerodynamic lift is dependent are,

1. Wing reference (or planform) area
2. Number of the wings
3. Vertical position relative to the fuselage (high, mid, or low wing)
4. Horizontal position relative to the fuselage
5. Cross section (or airfoil)
6. Aspect ratio (AR)
7. Taper ratio
8. Tip chord (Ct)
9. Root chord (Cr)
10. Mean Aerodynamic Chord (MAC or C)
11. Span (b)
12. Twist angle (or washout)
13. Sweep angle
14. Dihedral angle
15. Incidence
16. High lifting devices such as flap
17. Aileron
18. Other wing accessories

The total Lift of an aircraft is further dependent on a combination of aerodynamic thrust and engine thrust.

Myth Bustng - The advantages of low Wing Loading | Forums - Page 4

The lower the wingloading, the lower the angle of attack you have to pull to carry out the same manoeuvre (other things being equal)

If you look at the lift and drag equations, you can see why that's so important.

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

Coefficient induced drag (CDi) = (CL^2) / (pi * aspect ratio * Oswald efficiency)

Drag = coefficient * area * density *.5 * velocity squared

If you look at the equations, induced drag increases with the square of CL, and proportionately with the increase in wing area. So double wing area = a quarter the CL = half the induced drag.

To plug some figures in, an example aircraft with weight 3000 kg and wing area of 10 m^2, then the same aircraft with 20 m^2 wings. (this assumes weight doesn't increase with the larger wings, of course)

Assuming lift = 4 times weight, sea level density, speed = 400 km/h

117,600 = CL * 10 * .5 * 1.225 * 111^2
CL = 1.56

CDi = (1.56^)/(pi*6*.8) = 0.16

Induced drag = 0.16 * 10 * 1.225 * .5 * 111^2

Induced drag = 12,074 N

Now the same thing but with double the wing area

117,600 = CL * 20 * .5 * 1.225 * 111^2
CL = 0.78

CDi = (0.78^)/(pi*6*.8) = 0.04 (note how doubling the wing area results in a quarter of the CDi, because CL is squared)

Induced drag = 0.04 * 20 * 1.225 * .5 * 111^2

Induced drag = 6,037 N

Of course, parasitic drag increases with a larger wing area, but basically lower wingloading = an increasing advantage the tighter the turn, and the lower the IAS you fly (and IAS of course is lower at high altitudes)

these equations need no explanation as you your self has said that you are very talented in maths and physics.

Even though the weight increase due to larger wing is not factored in this formulae, the general directions of calculations are valid forever.

For your kind info tejas too has three sections slats, flaps, dihedral, twist angle all employed in the air frame.

So, lower wing loading gives,

Lower stall speed, tighter turns (in general).

Usually wings will stall at the same angle of attack, however planes with low wingloading will reach this angle of attack at a lower speed.

huge wing and a low weight aircraft you will get a very very low stall speed,

for example and probably a pretty good turner.

Another quote below,

In general, aircraft with higher wing loadings tend to be faster but less manoeuvrable. Since speed is life, this has historically been a good betting proposition. However, once top speeds become supersonic the trend can break down due to the operational, aerodynamic and thermodynamic issues associated with supersonic flight.

As a result of this, current fighters are no faster than 1960s fighters, but generally have lower wing loadings, since if you can't go faster the next best thing is to turn harder (apart from which, low wing loadings are necessary for high supersonic L/D and therefore help towards supercruise). As such, wingloadings are likely to stay roughly constant or possibly even decrease in future designs until such time as the upward trend in speed reasserts itself.

An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v2 s CL

* L = Lift, which must equal the airplane's weight in pounds
* d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
* v = velocity of an aircraft expressed in feet per second
* s = the wing area of an aircraft in square feet
* CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

The lower your wingloading, the less AOA you have to pull to get the same lift vector.
It is you lift vector that turns the plane as well as allows it to climb and fly.

For turning I would rather have a wing that needs less AOA than one that allows me to use more
but really needs it. The latter will have worse drag.

Of course, parasitic drag increases with a larger wing area, but basically lower wingloading = an increasing advantage the tighter the turn, and the lower the IAS you fly (and IAS of course is lower at high altitudes)

these equations need no explanation as you your self has said that you are very talented in maths and physics.

Even though the weight increase due to larger wing is not factored in this formulae, the general directions of calculations are valid forever.


So, lower wing loading gives,

Lower stall speed, tighter turns (in general).

Usually wings will stall at the same angle of attack, however planes with low wingloading will reach this angle of attack at a lower speed.

huge wing and a low weight aircraft you will get a very very low stall speed,

for example and probably a pretty good turner.


An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v2 s CL

* L = Lift, which must equal the airplane's weight in pounds
* d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
* v = velocity of an aircraft expressed in feet per second
* s = the wing area of an aircraft in square feet
* CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

The lower your wingloading, the less AOA you have to pull to get the same lift vector.
It is you lift vector that turns the plane as well as allows it to climb and fly.

For turning I would rather have a wing that needs less AOA than one that allows me to use more
but really needs it. The latter will have worse drag.

There are some counter arguments too,

Basically wingloading usually defines the aircraft's stall speed (a least in combat, without lift assisting devices such as slats and flaps), and through it, how slow the aircraft can get in turn, and thus the turning circle.

Simply to put, planes with low wingloading are usually having smaller turning circles.

Low wingloading on the other hand is not so pronounced as far as turn rate, or in other words, turn times go. Turn times are basically defined by how high G the aircraft can hold up in a sustained manner. That is basically speaking a race between the thrust of the propeller vs. the drag of the aircraft in turn. Here low wingloading doesn't give much of an advantage, as - all things equal - you will have a larger area wing, and thus higher drag to achieve lower wingloading. You will need more thrust to overcome the greater drag, if you're pulling the same Gs, or rate of turn.

The effect of increasing thrust (ie. engine output, 'boosting') on turn is interesting. Even huge increases in excess thrust (either by increasing engine power or doing a descending turn - tactics, tactics!) have very small effect on turn radii, since the aircraft just can't go below it's given stall speed at a given G-loading. However, increasing excess thrust can have a VERY pronounced effect on turn times.

Simply to put, low wingloading planes don't neccesarily beat high wingloading planes in turn times.

Best examples are the Yakovlevs vs. Spitfires. Spitfires have very low wingloading, and high thrust, high drag; the Yakovlevs have a fairly high wingloading, low thrust (poor engines), and very low drag. Comparing their real life turn radii and time values reveals the Yakovlevs have larger turn radius (by about 50m - about as much as the 109G, differences in turn radii were not as pronounced as some may think), but at the same time they beat the Spits in turn time by about 2 secs for a 360 degree turn.

It's classic example of how drag, thrust and wingloading effects turn time and radii. Turn time is IMHO more important, not to mention the other benefits of high wingloading being still there - lower drag, effecting level speed, zoom climb, dive performance etc.

but these arguments dont stand the test of questions,

The lower the wingloading, the lower the angle of attack you have to pull to carry out the same manoeuvre (other things being equal)

 

[ersakthivel]

can you explain what are those other parameters which need to be same?

I once again ask you to please list the factors which effect LIFT and where does the wing loading fall into it as the sole criteria.

You also need to know that wing loading is used as an explanation for laymen and not for hard core aviation buffs.

the entire structure of an aircraft generates Lift but for easy understanding, wing area or wing loading is used as a bench mark. I explained to you that total lift is a combination of wing lift + fuselage lift+ Lift component of thrust.

nothing wrong with it, I also know that ,"total lift is a combination of wing lift + fuselage lift+ Lift component of thrust".

but wing lift makes the substantial portion of total lift.

I told you that "the other parameters being the same " is a lay quote by a poster on the forum, which I added to give a balance and fairness to the discussion.

In the formula there are -NO OTHER THINGS.

So if your formula is there wiki it is correct and credible.

but if the set of formula posted by me is also there in Wiki , it is dubious perhaps!!!!

Well for hardcore aviation buffs , the wing loading gets into those critical formulas in the form of wing area in sq meters.

If you are a harcore buff you can easily see by now where wing area is there in those formulae.

just look below,

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

Coefficient induced drag (CDi) = (CL^2) / (pi * aspect ratio * Oswald efficiency)

Drag = coefficient * area * density *.5 * velocity squared

If you look at the equations, induced drag increases with the square of CL, and proportionately with the increase in wing area. So double wing area = a quarter the CL = half the induced drag.

To plug some figures in, an example aircraft with weight 3000 kg and wing area of 10 m^2, then the same aircraft with 20 m^2 wings. (this assumes weight doesn't increase with the larger wings, of course)

Assuming lift = 4 times weight, sea level density, speed = 400 km/h

117,600 = CL * 10 * .5 * 1.225 * 111^2
CL = 1.56

CDi = (1.56^)/(pi*6*.8) = 0.16

Induced drag = 0.16 * 10 * 1.225 * .5 * 111^2

Induced drag = 12,074 N

Now the same thing but with double the wing area

117,600 = CL * 20 * .5 * 1.225 * 111^2
CL = 0.78

CDi = (0.78^)/(pi*6*.8) = 0.04 (note how doubling the wing area results in a quarter of the CDi, because CL is squared)

Induced drag = 0.04 * 20 * 1.225 * .5 * 111^2

Induced drag = 6,037 N

Of course, parasitic drag increases with a larger wing area, but basically lower wingloading = an increasing advantage the tighter the turn, and the lower the IAS you fly (and IAS of course is lower at high altitudes)

these equations need no explanation as you your self has said that you are very talented in maths and physics.



And take a good look at the formula again,

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

Aircraft with low wing loadings tend to have superior sustained turn performance because they can generate more lift for a given quantity of engine thrust. The immediate bank angle an aircraft can achieve before drag seriously bleeds off airspeed is known as itsinstantaneous turn performance. An aircraft with a small, highly loaded wing may have superior instantaneous turn performance, but poor sustained turn performance: it reacts quickly to control input, but its ability to sustain a tight turn is limited. A classic example is the F-104 Starfighter, which has a very small wing and high wing loading. At the opposite end of the spectrum was the gigantic Convair B-36. Its large wings resulted in a low wing loading, and there are disputed claims[who?] that this made the bomber more agile than contemporary jet fighters (the slightly later Hawker Hunter had a similar wing loading of 250 kg/m2) at high altitude. Whatever the truth in that, the delta winged Avro Vulcan bomber, with a wing loading of 260 kg/m2 could certainly be rolled at low altitudes.[10]

Like any body in circular motion, an aircraft that is fast and strong enough to maintain level flight at speed v in a circle of radius Raccelerates towards the centre at ["‹IMG]. That acceleration is caused by the inward horizontal component of the lift, ["‹IMG], where ["‹IMG] is the banking angle. Then from Newton's second law,
Mv^2/R= L *sin θ=1/2 * v^2 *P * A* Sin θ (P- atmospheric pressure)

SOlving for this R is

R= 2* Ws/ p* Cl * Sin θ

SO higher the wing lift(possible only with low wing loading fighter) lower the turning radius for the same speed V.

that means at any corner velocity speed lower wing loading fighter will turn into a higher wing loading fighter as the radius of turn for lower wing loading fighter will be lower. What is the significance of this?

 

[ersakthivel]

If only wing area matters which the S in the lift equation, than what is the role played by coefficient of Lift, Density and speed in generating lift? Why are they in the lift equation and how can they be same for every aircraft?

the book mechnics of flight by Warren,

These are those other parameters which need to be same for wing loading to become the decisive factor.

https://books.google.co.in/books?id...ding affects turn radius of fighters?&f=false

Max turn rate that can be achieved without stalling the wing has CL max in it.

Is Cl max dependent on wing loading or not?

The only design parameter that can be used to significantly change the turning radius is wing loading-

the statement is from Mechanics of flight-Warren .F. Philips.

he again categorically says that both minimum turning radius and maximum turning rate are functions of max lift co efficient!!!!

And the prime factor of lift co efficient for any fighter is wing lift . of course wing lift, fuselage lift and thrust that functions as lift are all there.

but the prime factor is wing area. And low wing loading is the solution to get more lift out of wings.

he again says that designer has no control over max lift co efficient (which is between 1.2 and 1.6)and air density.

So the primary feature the designer must use to control the lift performanace is "wing loading".

Three times with in a span of three pages.

 

[ersakthivel]

You tell me how is the CL varied. I know it and I can say that wing area has nothing to do with CL.

Warren .F. Philip again says that designer has no control over max lift co efficient (which is between 1.2 and 1.6)and air density.

So the primary feature the designer must use to control the lift performanace is "wing loading".

Three times with in a span of three pages.


Cl=2L/Psv^2.

In the above equation

L is lift force vector.
v is velocity,
s is the wing surface area in sq meters,
P is atmospheric pressure.

SO Cl has everything

 

[ersakthivel]

Can you pls quote his exact words. bcoz if this is what he has written than he seems to have pedestrian knowledge of aerodynamics. It appears that you are again misquoting him or hiding the crucial part.
The design of a wing starts by selecting a Reynolds number(Re) for it. And Re depends on chord length (MAC), velocity and kinematic viscosity.

The CL is dependent on Re, angle of body w.r.t airflow (AOA) and wing shape. In an aircraft you control the CL by changing your AOA as the MAC is fixed and the shape is fixed. Some aircraft designs which utilized swing wings, used to change the shape and vary the CL by changing the sweep in addition to AOA.

I had asked you a question yesterday - how do you change wing area in flight.

Modern aircraft use flaps called Fowler flaps- double slotted or triple slotted. As these flap move outward, they increase the wing area and also change the chord length resulting in much higher increase in lift as they are able to vary two important factors in lift equation-CL and wing area.

I have read it now in detail and once again you have decided to misquote him without understanding the subject and the pretext. what you failed to understand was that he is referring to GA aircraft where the aircraft is turning at max CL and where density is fixed.

I also quoted to you that if there are two aircraft which are operating in same density altitude, with same speed and same CL, than the aircraft with lower wing loading will have better turn performance. that is possible if you have two different LCA which are exactly same but one of them has higher wing area. as will be the case between LCA MK1 & LCA MK2. These two aircraft operating at same altitude, same speed and same AOA will see that MK2 which is supposed to have slightly bigger span, out performing MK1 for the same reason which you and I have quoted.

https://books.google.co.in/books?id...ding affects turn radius of fighters?&f=false

"the low wing loaded fighter's greatest performance advantages are assumed to be,

1. good instantaneous turn performance,
2.tight sustained turning radius,
3.slow minimum speed"

"In some cases the aircraft will also have significant sustained turn rate advantage"

You have the habit of quoting what suits your purpose. No one has denied what you are quoting provided other parameters are equal. But I gave you the example of many other fighters and you quickly blamed that on TWR while that is not the sole criteria. Just FYI, Jaguar can hold out even against a M2K in combat in spite of much higher wing loading. Do you know the reason for it?

Tell me,

Jaguar was designed for operation from grass/roads and had STOL capability. This was possible due to its full span double slotted fowler flaps and 3/4 span LE slats combined with 40* sweep wing. (Now it has poor perf due to increase in weight and poor engines+ Indian hot conditions)

The aircraft is equipped with AMF which have a speed limitation of 580 kts. Once Jaguar is down to about 60% fuel load, it has better turning performance than when it is fully loaded compared to many fighters of IAF. M2K has poor vertical performance and once its speed drops, it also has poor turn ability due to its Delta Planform.

Jaguars can't fight in vertical but their wing gets completely transformed once the AMFs get deployed in response to AOA and speed variations.

This causes a very high increase in lift due to high increase in CL max value. So Jaguar is able to turn in much shorter radius than M2K. You have habit of dropping names of various test pilots. Nearly all of them have flown Jaguars also. I have done DACTS with Jags, M2K and Mig-21s in Sea Harrier.

You can ask those test pilots about what I have posted here. So you must know that wing area alone is not the sole criteria and many aircraft can have a very different wing for cruise and a completely changed wing for combat. I hope you have heard about MAW-Mission Adoptive Wing.

Pls read about them and you will understand the whole gamut of fighter flying and air combat much better and how the problem of having low wing loading is resolved. as I told you, if wing loading was the sole criteria, all fighters in the world wud have been built like gliders.

key word --"once mirage-2000 drops its speed".

What will happen when there is no drop in speed.

thats what Greek airforce chief mentioned about Mirage-2000 vs F-16.

An experienced Mirage-2000 pilot will finish the enemy off in the pass itself with high ITR .

tejas has far higher TWR and far lower wing loading than Mirage-2000 as well.

 

[ersakthivel]

Every simulated combat has rules for announcing a kill. we do all aspect missile combat simulations and we also do Guns only combat simulations. If we are in Mig-29/Su-30MKI using missile like R-73, we don't need to turn the aircraft. we just need to turn our head and you get a missile lock.

But aircraft like M2K, Jaguar and others will need some amount of turn as they do not have HMDS. But in guns only fight, The ITR/STR and especially STR is of high value. M2K has to make a kill in first pass and every F-16 pilot knows that and they also have standard counters to it. when we acquired M2K in response to PAF getting F-16 was to use BVR potential of M2k more than its agility in combat.

PAF F-16 in those days did not have BVR capability.
Any pilot who is up against an adversary with better ITR but lower STR will initially play in vertical to over come the ITR advantage of enemy and once that guy is down to low speeds, force him into a combination of vertical and turning fight. Poor STR also means poor TWR so low speed for such a fighter is death knell.

This reminds me that even IAF Mig-29 have higher wing loading compared to M2K and yet have better ITR and STR.

As per your equations,XXX degree bank at ZZ speed would produce the same turn rate regardless of the aircraft weight or wing loading ..

Rate of turn and Radius of Turn which I reproduce here also.
Rate of turn T (*/sec)= 1091*tanbank/v
Radius of turn R (ft)=v^2/ 11.26 tanbank

wing loading / g force / turn rate - PPRuNe Forums

"By the way, it really doesn't matter if you are in an Eagle or Falcon, a 9G turn is the same radius of action in either airplane, this means there is no trade-off between aircraft weight and wing surface area, the basic lift equation resultant will be equal for both jets."

"Not quite. A 9G turn is a 9G turn, yes. However, the wing loading makes a huge difference in how a plane is able to maintain these conditions.

For example, we have two 40,000 pound planes. One has a wing loading of 80 lbs./sq. ft., the other has a wing loading of 160 lbs./sq. ft. In a 9-G turn, both planes' centrifugal force will increase ninefold. In order to maintain a 9-G turn, the lift (centripetal force) must also increase ninefold. To do this, the 160 lbs./sq. ft. wing-loaded plane must increase its angle of attack a certain amount, but the 80 lbs./sq. ft. wing-loaded plane doesn't have to increase its angle of attack as much as the first plane has to.

Because the second plane can make a 9-G turn at a lower angle of attack, it can increase it's G-forces (rate of turn), to turn inside the first plane. In other words, at the same angle of attack, the second plane is producing more lift proportional to G-forces than the first plane is.

Wing loading is one indicator of turning ability, but others (such as wing geometry, camber, aspect ratio, etc.) all have influence as well.

Also keep in mind that if lift acts perpendicular to the chord line then a higher angle of attack will creater a higher backwards vector and create a resultant which works against forward motion. "

An aircraft travelling at the SAME SPEED and pulling the SAME G and therefore at the SAME BANK ANGLE, will have the SAME RATE OF TURN and SAME RADIUS OF TURN . Doesn't matter if it is a Tiger Moth or a Foxbat.

this link explains it all in clear terms with no sophisticated equations!!!!


But in real world,


My tiger moth, in a 60 degree banked turn will pull 2 gs and whip around through 360 degrees in under a minute. Your Foxbat, travelling at Mach 3 will take a long long time to do 360 degrees in a 2 g turn.

Why?

Lets look at extremes again to see why an aircraft with a low wing loading can out turn an aircraft with a high wing loading. Lets say the low wing loading aircraft is the Tiger Moth and the high wing loading aircraft is an experimental aircraft very similar to the Tiger Moth except it has such stubby wings that at 85 mph it requires an angle of attack close to the stalling angle to generate enough lift to maintain height.

Both aircraft cruising along at 85 mph together and they both enter a gentle 15 degree bank to the right. Both aircraft increase angle of attack slightly to maintain altitude. This brings the experimental aircraft even closer to the stalling AoA but at the moment it's not quite stalling.

Both aircraft are now at the same speed, pulling the same G and getting the same rate and radius of turn.

But what happens when the Tiger Moth increase bank angle even further? The experimental with very high wing loading tries to match it but doesn't have any excess angle of attack to use and stalls. Meanwhile the Tiger happily putts around in a nice tight turn.

Wing loading and G loading are not the same thing. If both planes fly at the same speed and bank angle (and experience the same G loading as a result of this bank angle), they will have exactly the same radius of turn.
.
I'd say that when comparing two airframes of similar weight but one with twice the wing loading (ie half the wing area), the aircraft with the higher wing loading is most probably designed to fly significantly faster. Therefore, with both at their design speed at any given bank angle and hence G loading, the faster aircraft will have a greater radius of turn. If both do fly at their design speeds, they could well fly at a similar angle of attack and there is no reason why the faster aircraft would 'run out of spare angel of attack' sooner than the other.
.
Now, if both of them do fly at the same speed (the design speed for the slower aircraft), the aircraft with the higher wing loading will be flying closer to its stalling angle of attack, to compensate for the loss of speed.
.
If in this situation they both fly a turn at a given bank angle the radius is still the same. However, the faster aircraft will be the first to reach a bank angle (and associated G loading and angle of attack) that will make the wing stall.
.
So, at slow speeds the aircraft with the lower wing loading could out-turn the one with the higher wing loading. But only because the aircraft with the higher wing loading is flying at a speed it is not designed for (ie too slow) and would be the first to stall.

So while the Gs and speed were the same, the aircraft had the same turning performance. But the low wing loading on the Tiger Moth meant it had plenty of excess angle of attack to use to further increase bank and therefore rate, radius and G.

A rate one turn which is defined as being 360 degrees in two minutes, requires a higher angle of bank, and therefore higher g, the faster you are going. In a light twin, a rate one turn is generally less than 25 degrees bank, in a faster passenger jet, the rate one turn requires more angle of bank, to the point where passenger jets are limited by 25 degrees bank angle during instrument procedures and light twins are limited by rate one. The only reason for this is that the passenger jet is going faster, if it slowed down to the speed of the light twin, the rate would be the same.

SO higher wing loading plane has to fly closer to its peak AOA and closer to its stall speed (with higher drag due to higher AOA),to do the 360 deg turn in 120 seconds. Because of its lower wing area it has to fly faster to get a higher horizontal component of lift that will let it remain at the same altitude.

 

[ersakthivel]

I have read it now in detail and once again you have decided to misquote him without understanding the subject and the pretext. what you failed to understand was that he is referring to GA aircraft where the aircraft is turning at max CL and where density is fixed.

I also quoted to you that if there are two aircraft which are operating in same density altitude, with same speed and same CL, than the aircraft with lower wing loading will have better turn performance.

that is possible if you have two different LCA which are exactly same but one of them has higher wing area. as will be the case between LCA MK1 & LCA MK2. These two aircraft operating at same altitude, same speed and same AOA will see that MK2 which is supposed to have slightly bigger span, out performing MK1 for the same reason which you and I have quoted.

As per Rate of turn and Radius of Turn equations,XXX degree bank at ZZ speed would produce the same turn rate regardless of the aircraft weight or wing loading ..

Rate of turn and Radius of Turn which I reproduce here also.
Rate of turn T (*/sec)= 1091*tan(bank angle)/v
Radius of turn R (ft)=v^2/ 11.26 tan(bank angle).

So it may seem larger wing area(lower wing loading airfame) has no advantage over low wing area ( higher wing loading) plane in any sustained turn or Instantaneous turn because Any aircraft travelling at the SAME SPEED and pulling the SAME G and therefore at the SAME BANK ANGLE, will have the SAME RATE and SAME RADIUS. Doesn't matter if it is a low wing loading or high wing loading fighter!!!!

because the above equation use nothing but bank angle and , v(air speed ) to calculate rate of turn and radius of turn.

not as you explained,

What Warren .F. Philips meant was ,"if you want to design a fighter with better turn radius and turn rates , you should lower the wing loading , which is the only thing the designer can control".And during the design of tejas the same low wing loading principle was applied also resulting in larger wings.

Tejas has split-flaperon and elevons. For high lifts, Slats are made use of. Of course you see a flaperon in the flaps position, but those things won't get actuated during take-offs and landings. As I mentioned, the leading edge flaps or simply the "slats" will automatically get actuated during take-offs, landings,at high AOAs and during low speed flights (depending on the speed). Carefully observe some LCA airshow videos available here and some close up shots to know how they work.

The LCA Wing Planform: So, they decided to have a wing which will have both non-slender and slender delta planform to get vortex generation in all AOAs.

That is the reason for the compound delta LCA wing design, with it's both a low-wing-sweep (50deg, so non-slender delta) and high-wing-sweep (~63deg aka in the "slender delta" territory) as you move from inboard (wing root) to outboard of a wing.

Thus for the relatively lower part of the high-AoA flight regime (say from around 18deg to 22deg etc), the outboard slender delta part of the wing would dutifully contribute to the vortex lift while keeping the drag as low as possible.

And with further increase of AoA, as that part of the wing starts to stall due to vortex bursting etc, the inboard non-slender-delta part of the wing will come into play with it's flow-reattachment aspects and keep on further enhancing the lift co-efficient (while still keeping the drag down as low as possible).

So where is the need of any additional control surface like a canard (and thus without the weight and complexity penalty of an additional control surface etc),

Did you know that for the F-16 the horizontal stabilizers come in the wake of the wing at greater than 25 degrees AoA, i.e become ineffective . This is called a deep-stall (or super-stall) and is mostly unrecoverable. Therefore the F-16 is limited by its FBW to 25 degrees AoA. SO not having a tail in tejas is the right thing to do, because with shorter fuselage length than F-16 , it would have created more problems in tejas than the longer f-16.

As usual you are drawing your own special inference on whatever is written anywhere about the impact of low wing loading on turn rates,

The same points you raised were also raised by another person in another forum and he was conclusively rebutted, in the link below.

wing loading / g force / turn rate - PPRuNe Forums


this link explains it all in clear terms with no sophisticated equations!!!!

Another illustration. A rate one turn which is defined as being 360 degrees in two minutes, requires a higher angle of bank, and therefore higher g, the faster you are going. In a light twin, a rate one turn is generally less than 25 degrees bank, in a faster passenger jet, the rate one turn requires more angle of bank, to the point where passenger jets are limited by 25 degrees bank angle during instrument procedures and light twins are limited by rate one. The only reason for this is that the passenger jet is going faster, if it slowed down to the speed of the light twin, the rate would be the same.

SO higher wing loading plane has to fly closer to its peak AOA and closer to its stall speed (with higher drag due to higher AOA),to do the 360 deg turn in 120 seconds. Because of its lower wing area it has to fly faster to get a higher horizontal component of lift that will let it remain at the same altitude.


An aircraft travelling at the SAME SPEED and pulling the SAME G and therefore at the SAME BANK ANGLE, will have the SAME RATE and SAME RADIUS. Doesn't matter if it is a Tiger Moth or a Foxbat.

But in real world,


My tiger moth, in a 60 degree banked turn will pull 2 gs and whip around through 360 degrees in under a minute. Your Foxbat, travelling at Mach 3 will take a long long time to do 360 degrees in a 2 g turn.

Why?

Lets look at extremes again to see why an aircraft with a low wing loading can out turn an aircraft with a high wing loading. Lets say the low wing loading aircraft is the Tiger Moth and the high wing loading aircraft is an experimental aircraft very similar to the Tiger Moth except it has such stubby wings that at 85 mph it requires an angle of attack close to the stalling angle to generate enough lift to maintain height.

Both aircraft cruising along at 85 mph together and they both enter a gentle 15 degree bank to the right. Both aircraft increase angle of attack slightly to maintain altitude. This brings the experimental aircraft even closer to the stalling AoA but at the moment it's not quite stalling.

Both aircraft are now at the same speed, pulling the same G and getting the same rate and radius of turn.

But what happens when the Tiger Moth increase bank angle even further? The experimental with very high wing loading tries to match it but doesn't have any excess angle of attack to use and stalls. Meanwhile the Tiger happily putts around in a nice tight turn.

So while the Gs and speed were the same, the aircraft had the same turning performance. But the low wing loading on the Tiger Moth meant it had plenty of excess angle of attack to use to further increase bank and therefore rate, radius and G.

Lets look at extremes again to see why an aircraft with a low wing loading can out turn an aircraft with a high wing loading. Lets say the low wing loading aircraft is the Tiger Moth and the high wing loading aircraft is an experimental aircraft very similar to the Tiger Moth except it has such stubby wings that at 85 mph it requires an angle of attack close to the stalling angle to generate enough lift to maintain height.

Both aircraft cruising along at 85 mph together and they both enter a gentle 15 degree bank to the right. Both aircraft increase angle of attack slightly to maintain altitude. This brings the experimental aircraft even closer to the stalling AoA but at the moment it's not quite stalling.

Both aircraft are now at the same speed, pulling the same G and getting the same rate and radius of turn.

But what happens when the Tiger Moth increase bank angle even further? The experimental with very high wing loading tries to match it but doesn't have any excess angle of attack to use and stalls. Meanwhile the Tiger happily putts around in a nice tight turn.

Wing loading and G loading are not the same thing. If both planes fly at the same speed and bank angle (and experience the same G loading as a result of this bank angle), they will have exactly the same radius of turn.
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I'd say that when comparing two airframes of similar weight but one with twice the wing loading (ie half the wing area), the aircraft with the higher wing loading is most probably designed to fly significantly faster. Therefore, with both at their design speed at any given bank angle and hence G loading, the faster aircraft will have a greater radius of turn. If both do fly at their design speeds, they could well fly at a similar angle of attack and there is no reason why the faster aircraft would 'run out of spare angel of attack' sooner than the other.

Now, if both of them do fly at the same speed (the design speed for the slower aircraft), the aircraft with the higher wing loading will be flying closer to its stalling angle of attack, to compensate for the loss of speed.

If in this situation they both fly a turn at a given bank angle the radius is still the same. However, the faster aircraft will be the first to reach a bank angle (and associated G loading and angle of attack) that will make the wing stall.

So, at slow speeds the aircraft with the lower wing loading could out-turn the one with the higher wing loading. But only because the aircraft with the higher wing loading is flying at a speed it is not designed for (ie too slow) and would be the first to stall.

 

[ersakthivel]

Tigermoth and b747 will have same radius of turn if their speeds, bank angle and g loading are same-This is a correct statement as we can see that radius of turn is dependent on bank angle and speed only.
BUT
The maximum bank angle that can be pulled is dependent on the load factor which an aircraft can withstand.
AND
The lift gets divided into two parts in a turn, one part is horizontal which creates centripetal force, the other part is the vertical part which balances the weight.
SO
The maximum bank angle which can be sustained by an aircraft is directly dependent on its ability to generate lift and the structural strength to withstand the required G load factor.
THEREFORE
An aircraft which can maintain lowest possible speed during a turn, has highest bank angle and g load compared to the other aircraft will have higher turn rate and smaller radius of turn.
G load is dependent on structural strength, Speed is limited to stall speed, bank angle will get limited by Lift force needed to overcome weight and provide centripetal force.
THE Lift is dependent on CL+Density+Speed+wing area. Any aircraft can play around with these components of the lift to maximize the lift to get the required bank angle to have high turn rate and low radius. Wing area is just one of them and NOT THE ONLY FACTOR.

You first show equations and source to substantiate your point --"THE Lift is dependent on CL+Density+Speed+wing area. Any aircraft can play around with these components of the lift to maximize the lift to get the required bank angle to have high turn rate and low radius. ", of course with high loading wing design.


First of all stop getting any fancy ideas like I am trying to learn anything from you.

I know where to look, when I need to learn, you are the last one I am ever going to turn.

I know that lift gets divided into two parts and low wing loading fighters have the max horizontal lift component at any given bank angle compared to high wing loading fighter,

And the low wing loading fighter need not get near stall AOA with huge drag which reduces acceleration in turn approaching stall speeds that makes them loose altitude.

but the high wing loading fighter has to do endure all the above to go neck and neck with low wing loading fighter.


Get things first into your wooden head before abusing me again and again , right.

"An aircraft which can maintain lowest possible speed during a turn, has highest bank angle and g load compared to the other aircraft will have higher turn rate and smaller radius of turn.". the primary means of achieving this is by low wing loading. Please don't repeat the same set of lies that it can be substantially achieved by means other than low wing loading again and again. You can get marginal improvement here and there.

They dont count for anything unless the area of the wing is big enough to sustain high lift at lower AOA (to give higher component of lift force) with low drag and low enough bank angle , well above stall speeds.

Flaps , etc, etc all help. Know one denies it. They too have been used in tejas.

"THE Lift is dependent on CL+Density+Speed+wing area. Any aircraft can play around with these components of the lift to maximize the lift to get the required bank angle to have high turn rate and low radius. Wing area is just one of them and NOT THE ONLY FACTOR.".

Of course I have posted all the equations which say the same thing you said above with all the explanations.

 

[ersakthivel]

All your theory about low wing loading falls flat when we compare Mig-29 with LCA and M2K which have lower wing loading than Mig-29. And let us not even compare F-22, Typhoon, Rafale, Su-27/30/35,
Regarding your shit about higher TWR, let me tell you that Rafale has a wing loading of 310, empty weight of 9.5 tons, internal fuel of 4.8 tons and thrust of 15300kgs which gives it a loaded weight TWR of 1.07. LCA MK1 has wing loading of 242, empty weight of 6.3 tons, internal fuel of 2.4 tons, thrust of 9100kgs and has a loaded TWR of 1.05. yet Rafale beats LCA by miles in ITR and STR. The LSP series of LCA are lighter by 200 kgs. I hope you know that.
Also you must know how to calculate TWR. In every forum you have been wrongly calculating it The correct formula is given below
TWR= Engine thrust/Loaded weight and not reverse of it which you do.
TWR stands for Trust to weight ratio and not Weight to thrust ratio. That's why I stated that you even have problems with English language.
LCA is not under powered by any standards, it is over winged.

I have always maintained that LCA is very good aircraft considering the tech that it has and every penny spent on this project is worth it. this ac is neither underpowered nor unfit for induction into IAF. This ac is over winged and every member may recall my posts regarding the wing of the LCA. reducing the wing area to 30sqm will lower its weight by another 200-300 kgs, increase its aspect ratio to 2.24 and wing loading will still be lowest among all its contemporary fighters but smaller wing will drastically reduce the drag which will boost its over all perf.

Mig-29 has double the engine thrust of tejas and has a very low fuel fraction , which was built out an out as an interceptor, while fighters like mirage-2000 and tejas are built RSS , low wing loading multi role fighters that can dominate the himalayan heights.

Go ask your pilot friends about why Mirage-2000 was used in kargil for precision bombing and not Mig-29s. you will get the answer.

if the high wing loading non-RSS, Mig-29 was the be all and end all of fighter design why is IAF so impressed with Mirage-2000 and insisted on embedding its flight qualities in tejas as retd Air chief marsahl Krishnasamy has spoken?

Then Why is the Mig design team in Intensive care unit?

Why is Sukhoi designing PAKFA as the same low wing loading RSS fighter with almost the same compound delta wing structure of tejas(even the LEVCONS of PAKFA was emphasised on naval tejas design as long back as 2001 itself as per Air chief marshal MSD Woolen's words!!!)

Rafale has a far higher TWR than tejas mk 1 , in half fuel , with couple of WVR missiles only close combat conditions.

Give me an exact break up and you will realize your mistake.

tejas mk2 will close the gap.

And no one knows the "fancy STRs and ITRs " of rafales in hot indian atmospheric conditions.


LCA is not under powered by any standards, it is over winged?????

Tejas mk2 is designed with HIGHER THRUST engine, while keeping the same wing loading .

So the aim of tejas mk2 program as far as IAF was concerned is to keep the same Wing loading and get a higher engine thrust, exactly the reverse of your diagnosis!!!!

 

[ersakthivel]

TWR can be calculated for any weight which cud be loaded wt or combat wt.
Mig-29 is an ASF and not a multirole fighter. M2K was used for bombing for the reason that only it had atlis pod and for no other reason. read my post in that blog of the Prof, I have blasted him as he is wrong. But you take the cake.
I can't fool you at all as only intelligent persons can be fooled, you don't fool a fool.
IAF knows all the values of ITR & STR of each aircraft in every possible Indian condition and configuration and that includes LCA and they have stated that LCA does not match upto Rafale or Mig-29s. Go and ask those test pilots whose names you keep dropping in every forum and every post.

Smaller wings doesn't necessarily reduce drag in proportion. The calculations I posted from Myth busting forum, conclusively prove that. Meanwhile a smaller wing would have reduced the win lift and reduced the STR and ITR drastically,

since no point in ," fancy high wing loading non -RSS fighter which will " play around with these components of the lift (CL+Density+Speed+wing area)to maximize the lift to get the required bank angle to have high turn rate and low radius".

The higher wing loading will lead to higher stall speeds and AOAs in turns getting close to stall limit and would have led to massive drags at those higher AOAs.

The design goal given to tejas was not to match Mig-29 STR !!!! but to exceed Mirage-2000 and to have a good performance in high himallayan theatre of war.

As late as 2001 Air marshal MSD woollen mentions that Design STR given to tejas was just 17 deg. So what is the point of saying that tejas doesnot have rafale and Mig-29 specs?

The design aim of tejas was to fit into mig-21 dimensions with single engines and have a low operating and procuring cost. What is the point of comparing it with mig-29s and Rafales which are much larger twin engined , highly costly fighters?

So please don't try to fool the fool!!!!

Which fool will expect tejas to have a STR-ITR combo twenty percent higher than its ASR specs?

I never stated that we replace LCA wing with a very small wing. I stated a reversed wing of 30sqm which will still give it a wingloading at loaded weight of 290 which is still lower than most other fighters. But its impact on drag will be drastic and it will not effect lift as the lower sweep at the outer section will give higher CL compared to present wing.

The reason for the double delta with lower sweep in the wing root and three sections flaps ,is to "CREATE FLOW ENERGIZING VORTICES".,Look at the picture.

With this design tejas has achieved a 26 deg AOA within fly by wire limit already(and will cross the 28 deg as per award winning test pilot Suneeth krishna), while even f-16 due to its deep stall issues had its fly by wire AOA limited to 25 degs. That wing design was tested extensively in tunnels and proven over 2800 flights and cleared all weather trials, wake penetration trials, and winter trials at LEH, where 4 of the multi billion MMRCA birde failed.

Tell me with how many tons will mig-29 lif off from leh?

Now you and the dumb prof just took a pencil and rubber and found out two diametrically opposite solutions to the "non existing drag problems" in tejas.

how many wing tunnel tests you guys did with those two diametrically opposite wing designs to cure the "non existing drag problems" in tejas?

 

[ersakthivel]

What is the value of V in the equation and how is it calculated.

Refer to the Newtons second law part in my past answer and you will find it.
Clarify your position on the following passage first,

Rate of turn and Radius of Turn.
Rate of turn T (*/sec)= 1091*tanbank/v
Radius of turn R (ft)=v^2/ 11.26 tanbank


And for claims like ,that rate of turn and radius of turn has nothing to with wing loading , we Eng grads start laughing our ass at it!!!!

because what is pathetic is the guy does not have any understanding of how the speed V in the equation is achieved by the fighter in sustained turn

and what is relationship of V ,with horizontal component of lift vector , which in turn relies on wing area to generate that,

Also he does not know what is the impact of bank angle on rate of turn and turn radius,

and how the AOA affects the drag in any given bank angle,

and how fighters with different wing loading pull off different AOAs required to achieve the same lift in varying bank angles,

And how higher wing loading fighters have to totter at the edge of their peak AOAs and stall speeds to produce the same horizontal component of lift facing very high induced drag

in order to keep up with

much lower wing loading fighters which pull off the same horizontal component of lift at much lesser AOA and well above stall speeds with much lesser induced drag,

An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v2 s CL

* L = Lift, which must equal the airplane's weight in pounds
* d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
* v = velocity of an aircraft expressed in feet per second
* s = the wing area of an aircraft in square feet
* CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

So v^2=2L/d*s*CL


If you look at the equations, induced drag increases with the square of CL, and proportionately with the increase in wing area. So double wing area = a quarter the CL = half the induced drag.

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

Coefficient induced drag (CDi) = (CL^2) / (pi * aspect ratio * Oswald efficiency)

Drag = coefficient * area * density *.5 * velocity squared

If you look at the equations, induced drag increases with the square of CL, and proportionately with the increase in wing area. So double wing area = a quarter the CL = half the induced drag.

As per the above set of formulas Cl reduces when surface area increases,

And the co efficient of induced drag reduces by square of CL ,

SO even though it appears that plain drag increases with directly with surface area , since the co efficient of drag reduce by the square of CL(which reduces further directly with wing surface area )

So v^2=2L/d*s*CL

SO even though v decreases with increase in surface area this is even out by decrease in Cl with surface area which leads to higher v.

And most important of all the L (the all important lift thing which determines the higher v) is always higher for low wing loading fighter at any given bank angle.

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

To plug some figures in, an example aircraft with weight 3000 kg and wing area of 10 m^2, then the same aircraft with 20 m^2 wings. (this assumes weight doesn't increase with the larger wings, of course)

Assuming lift = 4 times weight, sea level density, speed = 400 km/h

117,600 = CL * 10 * .5 * 1.225 * 111^2
CL = 1.56

CDi = (1.56^)/(pi*6*.8) = 0.16

Induced drag = 0.16 * 10 * 1.225 * .5 * 111^2

Induced drag = 12,074 N

Now the same thing but with double the wing area

117,600 = CL * 20 * .5 * 1.225 * 111^2
CL = 0.78

CDi = (0.78^)/(pi*6*.8) = 0.04 (note how doubling the wing area results in a quarter of the CDi, because CL is squared)

Induced drag = 0.04 * 20 * 1.225 * .5 * 111^2

Induced drag = 6,037 N

Of course, parasitic drag increases with a larger wing area, but basically lower wingloading = an increasing advantage the tighter the turn, and the lower the IAS you fly (and IAS of course is lower at high altitudes)

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

Coefficient induced drag (CDi) = (CL^2) / (pi * aspect ratio * Oswald efficiency)

Drag = coefficient * area * density *.5 * velocity squared

If you look at the equations, induced drag increases with the square of CL, and proportionately with the increase in wing area. So double wing area = a quarter the CL = half the induced drag.

To plug some figures in, an example aircraft with weight 3000 kg and wing area of 10 m^2, then the same aircraft with 20 m^2 wings. (this assumes weight doesn't increase with the larger wings, of course)

Assuming lift = 4 times weight, sea level density, speed = 400 km/h

117,600 = CL * 10 * .5 * 1.225 * 111^2
CL = 1.56

CDi = (1.56^)/(pi*6*.8) = 0.16

Induced drag = 0.16 * 10 * 1.225 * .5 * 111^2

Induced drag = 12,074 N

Now the same thing but with double the wing area

117,600 = CL * 20 * .5 * 1.225 * 111^2
CL = 0.78

CDi = (0.78^)/(pi*6*.8) = 0.04 (note how doubling the wing area results in a quarter of the CDi, because CL is squared)

Induced drag = 0.04 * 20 * 1.225 * .5 * 111^2

Induced drag = 6,037 N

Of course, parasitic drag increases with a larger wing area, but basically lower wingloading = an increasing advantage the tighter the turn, and the lower the IAS you fly (and IAS of course is lower at high altitudes)

 

[ersakthivel]

WOW, That is why I say that you have pedestrian knowledge of aerodynamics and just eating my brains here. I will teach you how to calculate the turn rate and radius step by step here. learn it Kid.
step-1
Calculate the stall speed of the aircraft in the configuration in which you will be during any maneuver. if you are going to use slats and partial flaps than calculate the stall speed for that configuration.
step-2
based on the structural load limit of your aircraft calculate the maximum bank angle which you can have. If an aircraft has 9G limit, use 9G and Load factor needed for any bank angle is calculated by formula n=1/cosbank. so CosBank=1/Load limit. Inthis case let us take it as 9G so the maximum bank that ac can maintain without exceeding design load limit is 83.62*
Step-3
Now calculate the revised stall speed for the load limit which in this case is 9 so the stall speed under 9G load will be 3 times the stall speed under 1G load.
Step-4
Now calculate the Turn rate and turn radius by substituting this speed in place of V in the equations which I had posted and repeated by you.

As I had told you that for ITR we take the structural G limit and for STR we take 6G being the maximum which human can sustain using G-suits.

SO

Any aircraft which has low stall speed will have lower stall speed in any maneuver and lower the speed, higher the bank angle, higher the load limit, higher the rate of turn and lower the radius of turn.

Why do we need max bank angle. the key as per Newtons second law is for any given bank angle,
First you understand Newton's second law which gives the following equation

L sin(θ)=mv*2/R

A low wing loading air craft will always bring higher horizontal lift component into sustained turn turn for any given bank angle θ at a lower AOA (which means with lesser drag)and speeds much hihger than stall speeds.

So it naturally transpires the lower wing loading fighter entering sustained turn at any given bank angle will have much higher L sin(θ). So it basically means that either it will have higher speeds or lower turn radius which directly translates into higher sustained turn rate .

On the contrary a higher wing loading fighter entering the same sustained turn at the same bank angle will have to pull a higher AOA , which means it will have higher drag and and lesser horizontal lift component , so it naturally means that it should either have lower speeds or higher turning radius which means inferior turn rate.

An aircraft's lift capabilities can be measured from the following formula:
L = (1/2) d v2 s CL

* L = Lift, which must equal the airplane's weight in pounds
* d = density of the air. This will change due to altitude. These values can be found in a I.C.A.O. Standard Atmosphere Table.
* v = velocity of an aircraft expressed in feet per second
* s = the wing area of an aircraft in square feet
* CL = Coefficient of lift , which is determined by the type of airfoil and angle of attack.

So v^2=2L/d*s*CL

Where is the stall speed, max load limit and max bank angle in this formula,

if I go on doing all the calculations I will spend countless hours doing repeated calculation circles forever.

First you understand Newton's second law which gives the following equation

L sin(θ)=mv*2/R

A low wing loading air craft will always bring higher horizontal lift component into sustained turn turn for any given bank angle θ at a lower AOA (which means with lesser drag)and speeds much hihger than stall speeds.

So it naturally transpires the lower wing loading fighter entering sustained turn at any given bank angle will have much higher L sin(θ). So it basically means that either it will have higher speeds or lower turn radius which directly translates into higher sustained turn rate .

On the contrary a higher wing loading fighter entering the same sustained turn at the same bank angle will have to pull a higher AOA , which means it will have higher drag and and lesser horizontal lift component , so it naturally means that it should either have lower speeds or higher turning radius which means inferior turn rate.

 

[ersakthivel]

The steps that I have written for calculating the Turn rate & radius is the ONLY WAY it is done. There is no other way in the world.
You did not even know what the V in equation means and how it is calculated as in your opinion only wing loading matters.
can you explain me why LCA has such high landing speed compared to other fighters which have higher wing loading?
When we do a 60* bank turn we will need- 1/cos60=1/0.5=2G to maintain level turn but the stall speed will be underroot of 2=1.41 times the stall speed in wings level flight.

Newton's second law which gives the following equation

L sin(θ)=mv*2/R

A low wing loading air craft will always bring higher horizontal lift component into sustained turn turn for any given bank angle θ at a lower AOA (which means with lesser drag)and speeds much hihger than stall speeds.

So it naturally transpires the lower wing loading fighter entering sustained turn at any given bank angle will have much higher L sin(θ). So it basically means that either it will have higher speeds or lower turn radius which directly translates into higher sustained turn rate .

On the contrary a higher wing loading fighter entering the same sustained turn at the same bank angle will have to pull a higher AOA , which means it will have higher drag and and lesser horizontal lift component , so it naturally means that it should either have lower speeds or higher turning radius which means inferior turn rate.

well it seems even Newton's second law of motion has become a garbage for you,

if you have the guts rebut this with formula and arguments with credible source,

I will be waiting here , not running away

In fact you are the only one who is running away at top speeds from something as basic as Newton's second law!!!!

Even if I post it hundred times, you won't understand something as basic as newton's second law and ask me to calculate all the values in the world,

First you understand Newton's second law which gives the following equation

Lift (N) = CL * area (sq m) * .5 * pressure (kg/cubic m) * velocity (m/s) squared

As per the above equation lift increases with wing surface area .

This means a higher wing loading plane will have lesser lift at speed v.

the lower wing loading plane will have more lift at the same speed v.

use the same v in the following equation

L sin(θ)=mv*2/R

A low wing loading air craft will always bring higher horizontal lift component into sustained turn turn for any given bank angle θ at a lower AOA (which means with lesser drag)and speeds much hihger than stall speeds.

So it naturally transpires the lower wing loading fighter entering sustained turn at any given bank angle will have much higher L sin(θ). So it basically means that either it will have higher speeds or lower turn radius which directly translates into higher sustained turn rate .

On the contrary a higher wing loading fighter entering the same sustained turn at the same bank angle will have to pull a higher AOA , which means it will have higher drag and and lesser horizontal lift component , so it naturally means that it should either have lower speeds or higher turning radius which means inferior turn rate.

 

[ersakthivel]

Regarding higher or lower lift component in horizontal during a turn, you must know that weight of the ac is balanced by vertical component and centripetal force is provided by horizontal component. So an ac which generates larger amount of lift at slowest possible speed and has higher bank angle and load limit will have higher turn rate and lower radius.

v is speed and that v is very much dependent on wing loading which determines the horizontal component of lift vector. if the horizontal component of lift vector is higher automatically v in increases in sustained turns because Newton's second law can not go wrong in speeds below the speed of light.

Also low wing loading fighters need lesser AOA to get an equivalent horizontal component of lift vector to get the same v. And higher wing loading fighter has to pull a much higher AOA with much higher drag to achieve the same speed. SO as turns get tighter with increased AOA the high wing loading fighter starts losing altitude and loose the dog fight in horizontal plane.


i haven't even raised the even worse vulnerability of high wing loading fighter in high altitudes , as with low wing area they struggle to perform close combat turns.

I haven't even raised the failure of all high wing loading fighters in LEH high altitude himalyan airfield where all of them abjectly failed to even take off with any meaningful load.

What you call elementary physics seems all Greek and Latin to you. In my response to Professor's article I had written the formula for calculating the Rate of turn and Radius of Turn which I reproduce here also.
Rate of turn T (*/sec)= 1091*tanbank/v
Radius of turn R (ft)=v^2/ 11.26 tanbank
Can you please show me where in these equations you find TWR or wing loading as a component.
You need to study some good books on aerodynamics to educate yourself. Low wing loading does not result in good ITR/STR. had that been the case, gliders wud have been the best fighters in the world.

from this position now you are saying that

Regarding higher or lower lift component in horizontal during a turn, you must know that weight of the ac is balanced by vertical component and centripetal force is provided by horizontal component. So an ac which generates larger amount of lift at slowest possible speed and has higher bank angle and load limit will have higher turn rate and lower radius.

ofcourse you will never accept that the primary requisite for any fighter to have a higher horizontal lift component in a sustained turn is low wing loading.

three section Slats ,twist in wing root and many other techniques are laso employed in tejas mk1 besides the low wing loading.


bangalore is located 100 plus meters above sea level and temp is in high 30s.

it is nice of you to accept this after a long discussion.

"The program announced an intention to change performance specifications for the F-35A, reducing turn performance from 5.3 to 4.6 sustained g's and extending the time for acceleration from 0.8 Mach to 1.2 Mach by eight seconds," Gilmore's report stated. The F-35B and F-35C also had their turn rates and acceleration time eased. The B-model jet's max turn went from 5.0 to 4.5 g's and its acceleration time to Mach 1.2 was extended by 16 seconds. The F-35C lost 0.1 g off its turn spec and added a whopping 43 seconds to its acceleration.

The changes likely reflect higher-than-expected drag on the JSF's single-engine airframe, according to Bill Sweetman of Aviation Week. The implications for frontline pilots are pretty serious. Less maneuverability makes the F-35 more vulnerable in a dogfight. And the slower acceleration means the plane can spend less time at top speed. "A long, full-power transonic acceleration burns a lot of fuel," Sweetman explained.

This is not the first time the Pentagon has altered its standards to give the JSF a pass. In early 2012, the military granted the F-35 a longer takeoff run than originally required and tweaked the plane's standard flight profile in order to claw back some of the flying range lost to increasing weight and drag.

Why this high TWR,lower swept wing leading edge+ tail , high wing loading fighter F-35 is down rated for gs and trans sonic acceleration?

The velocity in this case is the stall speed. F-22, Rafale and Typhoon have wings which have much lesser sweep compared to LCA. F-22 has just 42* and that gives it a very high CLmax plus it uses AMF and has negative stability unlike the RSS of LCA. So its tail plane also adds to wing lift.

why the high CL max from lower swept wing angle hasn't helped f-35 in pulling gs?

Why the tail plane lift is are not helping in f-35?.

Why the tail planes are coming in the way of wing wake in F-16 , so that its fly by wire AOA was reduced to 25 deg to avoid deep stall?

The answer is high wing loading.

tejas with no tail planes , no lower swept wing leading edge, no high wing loading can pull 26 deg AOA now and as per deiing tests in wind tunnel, it can cross 30 deg as per the award winning test pilot Suneeth krishna.

Now tell me F-35 which has a lower sweep than tejas mk1 gets no help from ,"the high CL max from low wing swept angle " you said will come with lower swept high wing loading wing?

now I am convinced that you are a hopeless nutcase!!!

Both tejas and F-22 are unstable design . tail has nothing to do with unstable or negative stability or RSS design.

Whether there is an RSS or not all tail planes will give some lift , what is got to do with RSS?

All tail surfaces produce lift and RSS has nothing to with it.

Rss is not about adding tail lift to wing lift. it will automatically be added to the total lift , whether there is any RSS or not!!!

Just try to read up what is RSS!!!

Even without tail lift tejas has enough lift from its wing. It has one of the lowest wing loading design of all times!!!

So why does it need a tail lift???

 

[ersakthivel]

One above is F-16 Xl, the one below is tejas

tejas wing has much lower sweep back than F-16 Xl, which outscored the much higher wing loading normal f-16 on all counts.And had no canards and tail planes as well , and had very low wing loading as well.

now "Wing sweep Angle " experts who defy newtons second law of motion ,

can raise their hand and propound their theory that why f-16 Xl high wing sweep was wrong to the legendary designer of F-16 Xl, Harry Hillaker.

Do they know more about wing sweep angle than him?

F-16 Designer Harry Hillaker | Code One Magazine

s this difference represented by the F-16XL?
Yes. The F-16XL had a better balance of air-to-air and air-to-ground capability. In fact, when I first started going to the Air Force with plans for the F-16XL, some of the Air Force people were so enthusiastic about it that they accused me of holding the design back so that we could sell the airplane twice. If you know anything about the history of the lightweight fighter, you know that this was not the case.

With the F-16XL, we reduced the drag of the weapon carriage by sixty-three percent. The drag of the XL with the same fuel and twice as many bombs is a little over thirty percent less than today's F-16 when you load it up. This points up a fallacy that has existed for thirty years, and I'm concerned that it may still exist. Our designs assume clean airplanes. Bombs and all the other crap are added on as an afterthought. These add-ons not only increase drag but they also ruin the handling qualities. They should be considered from the beginning.

We ought to start with the weapon. That's really the final product. We ought to determine what the weapon is and what it will take to deliver it and then do the airplane. Now, we design the airplane and smash the weapon on it.

How F-16 Xl which had almost twice the wing area could manage a a drag of the XL with the same fuel and twice as many bombs is a little over thirty percent less than today's F-16 when you load it up?

The wing and rear horizontal control surfaces were replaced with a cranked-arrow delta wing 120% larger than the original wing. Extensive use of carbon fiber composites allowed the savings of 600 lb (270 kg) of weight but the F-16XL was still 2,800 lb (1,300 kg) heavier than the original F-16A.

Less noticeable is that the fuselage was lengthened by 56 in (1.4 m) by the addition of two sections at the joints of the main fuselage sub-assemblies. With the new wing design, the tail section had to be canted up 3°, and the ventral fins removed, to prevent them from striking the pavement during takeoff and landing. However, as the F-16XL exhibits greater stability than the native F-16, these changes were not detrimental to the handling of the aircraft.

These changes resulted in a 25% improvement in maximum lift-to-drag ratio in supersonic flight and 11% in subsonic flight, and a plane that reportedly handled much more smoothly at high speeds and low altitudes.

The enlargements increased fuel capacity by 82%. The F-16XL could carry twice the ordnance of the F-16 and deliver it 40% farther. The enlarged wing allowed a total of 27 hardpoints:

How can F-16 Xl manage this-"These changes resulted in a 25% improvement in maximum lift-to-drag ratio in supersonic flight and 11% in subsonic flight"?

In February of 1984, the Air Force announced that it had selected the McDonnell Douglas design in preference to the proposed production versions of F-16XL. The McDonnell Douglas proposal was later to enter production as the F-15E Strike Eagle. Had the F-16XL won the competition, production aircraft would have been designated F-16E (single-seat) and F-16F (two-seat). John G. Williams, lead engineer on the XL: "The XL is a marvelous airplane, but was a victim of the USAF wanting to continue to produce the F-15, which is understandable. Sometimes you win these political games, sometimes not. In most ways, the XL was superior to the F-15 as a ground attack airplane, but the F-15 was good enough."

Any explanations?

The wing planform was altered in a cranked-arrow delta wing (120% larger than the original F-16 wing), with extensive use of carbon composite materials (in the upper and lower layers of the skin) to save weight. Weight savings in the wings alone amounted to 600lbs. or 272kg.

The wing is of multi-spar design with the leading edge sweep angle ranging from 50º to 70º, and is 2,800lbs (1,179 kg.) heavier than the original.
The increase in internal volume (both by lengthening the fuselage and expanding the wing) allowed for a 82% increase in internal fuel capacity, while the increased wing area allowed the incorporation of up to 27 stores stations. Despite the apparent lengthening of the fuselage involved with the program, the new XL designation does NOT stand for "extra large".

Through wing planform improvements and camber optimizations, the final configuration offered a 25% improvement in maximum lift-to-drag ratio over the F-16 supersonically, and 11% improvement subsonic.

The handling of the F-16XL was reportedly quite different from that of the standard F-16, offering a much smoother ride at high speeds and low altitudes. The configuration had matured into a very competent fighter with a large wing that allowed low-drag integration of large numbers of external weapons.

All these flies in the face of guys arguing that large wing area , high wing sweep angle tejas suffers from excess drag.

it is all shitloads of lies.

f-16 Xl's lower swept wing leading edge is in the back portion of the wing.

For tejas it was in the front portion of the wing to take care of vortex generation.

But what is most important to note is that F-16 Xl too had a 40 deg plus lower sweep like tejas near wing root for wing leading edge!!!

So both designs are exactly similar for more than 80 percent of wing area.

Still people claim that f16 Xl and tejas had different wing deisgn!!!

Do they expect all of f-16 Xl's performance to come from the last portion of the cranked arrow wing?

F-16 Xl had wing sweeps from 50 deg to 70 deg!!!

tejas too has a wing sweep from 50 deg to 62 deg.

So whats wrong?

 

[ersakthivel]

The lesser swept wing leading edge angle is 50 deg for tejas,

It consists of a pure double delta configuration with leading edge angles of 50 deg and 62.5 deg and a trailing edge forward sweep angle of 4 deg . The CG lies about 33.5% of MACand the wing area is 38.5 sq meter.

In F-16 Xl also the leading edge wing sweep angle was 50 deg as per the following link.
http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.51.2771&rep=rep1&type=pdf

The F-16 Xl is called derivative or baseline configuration .

The derivative configuration showed the potential for increased sustained-g maneuvering capability in relation to the F-16C model because of the
improved lift/drag; the derivative configuration also maintained comparable lateral-directional stability.

The F-16C configuration has a higher lift-curve slope and a lower lift-dependent drag coefficient (i.e., CD CDmin) mainly because of its lower wing sweep
and higher aspect ratio. Minimum drag coefficient for the baseline configuration is lower than for the F-16C model, osetting the lower lift and higher liftdependent
drag coeffecients to give the baseline configuration a substantially higher lift-drag ratio than the F-16C model for CL < 0:4. As mentioned previously,each set of data was referenced on its own wing reference area. The F-16C model has other lifting surfaces (i.e., horizontal tails, fore body, and shelf areas) that are not included in the reference area. If the data were reduced about a weighted planform area reference for each conguration, the magnitudes of lift-curve slope and drag dierences discussed above would likely be smaller characteristics.

what is being referred here as baseline configuration is F-16 XL

The data reduced with the common reference geometry show that the F-16C configuration now has a lower lift curve slope and higher lift-dependent drag coecient
(i.e., CD CDmin) relative to the baseline configuration.Although the baseline conguration has a higher minimum drag coeffecient than that of the F-16C because of skin friction associated with the larger wing and shelf area, it generates greater lift and L=D than the F-16C for CL > 0:15 for M = 1:60


The baseline conguration exhibits slightly higher lift-curve slopes and slightly more stable pitching-moment characteristics than does the generic wing model. The lift-dependent drag curves show little difference between the congurations at M = 1:60; however, at higher Mach numbers the baseline conguration shows lower untrimmed lift-dependent drag than the generic wing model.


Trimmed comparisons. The trimmed aerodynamic characteristics for the baseline conguration and the generic wing model are shown in gure 36.
The baseline conguration exhibited higher trimmed lift-dependent drag because of the larger pitching moment increment required to trim at a given lift
condition. A larger trailing-edge ap de ection was required for the baseline conguration not only to counter the larger pitching moment, but also to compensate
for the baseline model having a 10-percent smaller ratio of trailing-edge ap area to wing area than the other model

Approach to High Angle of Attack Testing of Light Combat Aircraft [LCA] Tejas

The Tejas LCA has been designed to be aerodynamically unstable in the longitudinal axis to obtain improved maneuverability and agility over the entire flight envelope and hence, has to be stabilized artificially by the use of active control technology.

The region from 0.5M to 0.7M and from 3Km to 8 Km is the zone of the highest instability with time to double amplitude dropping to 200 milli secs. This implies that an ydisturbance in pitch would cause an increase in amplitude by 32 times in a sec.

The lift, pitching moment, and lift-dependent drag characteristics of the two congurations at the tested Mach numbers are compared in gure 35. A translation of
lift curves is evident there, as well as a zero-lift pitching-moment shift between the two models that is likely attributable to the wing twist and body
camber differences.

Cl max continues to improve till 35 deg AOA.

directional characteristics indicated the proverbial 'cliff' with a sudden drop in Cnp ,, CRM (Coefficient of Rolling Moment) and CYM (Coefficient of Yawing Moment) at approx 25 deg AoA as shown at fig-4 and 5. These phenomena require the High AoA trials to be limited to 24 deg (as shown in dotted line) until directional stability is bolstered and augmented by rudder control up to an expected 26 deg . Currently the Tejas is flying to AoA limits of 20 deg and 22 deg never exceed. Fortunately as shown in fig-6, the LCA hassignificant rudder authority (CYM-Del R) even up to 30
exceeded(now it has gone higher than 26 deg.)

Fortunately as shown in fig-6, the LCA has significant rudder authority (CYM-Del R) even up to 30 deg AoA that will allow artificial stabilization in yaw at high AoA
AoA

 

[gadeshi]

Zvezda reports on MAKS aerobatics (Su-30SM and T-50)
http://tvzvezda.ru/news/vstrane_i_mire/content/201509261822-dnfa.htm#

 

[ersakthivel]

https://books.google.co.in/books?id...0EyAJf#v=onepage&q=marut underpowered&f=false

Marut's wing was highly swept and thin and large - all three characteristics for an interceptor. The sweep and thickness together determine the planes ability to fly across the speed of sound - greater the sweep and thinner the wing the lower is the thrust to weight ratio needed to get the aircraft supersonic. However, on the flip side, the greater the sweep and thinner the wing the higher becomes the landing speed and the less stable and maneuverable is the aircraft at low speeds below 250 knots (450 kmph). The Marut wing is a well balanced compromise of adequate sweep to get supersonic (provided the engines develop the thrust) and the thickness was enough to maintain moderate landing speeds and low speed stability. The wing bestowed on the HF-24 an acceleration and low altitude speed that the Pakistani Sabres and Indian Hawker Hunters could not match. In fact the Marut was one of the few, if not the only, frontline aircraft that could cross Mach 1.0 without afterburners - albeit just about at high altitudes.

There is wide consensus about excellent handling characteristics of the aircraft. Most pilots who have flown the aircraft describe it as pleasant to fly and excellent for aerobatics with fine control responses. And its ability to out-accelerate the Sabre jet, especially at low levels, was a useful asset in 1971. The Marut offered a stable weapon delivery platform and packed a formidable punch. While the Marut's pilots expressed an understandable desire for more thrust than the Orpheus 703 offered, they were unanimous in their view that the aircraft proved itself a thoroughly competent vehicle for the low-level ground attack profile. One defect which, I believe, remained was malfunction of roll control aerodynamic surfaces and the canopy flying off when all four 30mm cannons were fired simultaneously and the impact the recoil had on the electrics of the aircraft. HAL, I believe, claimed to have cured the problem but the IAF decided to be safe and blanked off the two upper cannons and operating only with the lower two in squadron service. The Marut was a robust aircraft with extremely good visibility for the pilot, and was aerodynamically one of the cleanest fighters of its time.

Today, 50 years later, the IAF has no indigenously built aircraft of any worth. The enthusiam that was associated particularly with Marut died a natural death because of a combination of two factors: import pressures in general and under-powered engines for the aircraft.

Retired IAF officers told Deccan Herald that neither Air Headquarters nor the Ministry of Defence pursued the indegenisation programme beginning with Marut manufactured by the then Hindustan Aircraft Ltd, later christened as Hindustan Aeronatics Ltd (HAL), with gusto. According to Wg Cdr (retd) Praful Bakshi, Maruts Achilles heel was its engine.

This did not help because the frame was designed for Mach 2-3 speed and the engines were grossly under-powered, another retired IAF officer said, adding that with no significant help from western countries in developing the Marut's engine, the plan to manufacture more of the HF-24 was dropped.

According to Wg Cdr Bakshi, the Marut was the only aircraft which flew supersonic without an afterburner, an aspect which our planners never gave importance to.

After the GNAT started flying, Kurt Tank (a German who had earlier designed the Focke-Wolf) designed the HF-24 which was a remarkable aircraft but fell short because of the lack of a proper engine.

This did not help because the frame was designed for Mach 2-3 speed and the engines were grossly under-powered, another retired IAF officer said, adding that with no significant help from western countries in developing the Maruts engine, the plan to manufacture more of the HF-24 was dropped.

According to Wg Cdr Bakshi, the Marut was the only aircraft which flew supersonic without an afterburner, an aspect which our planners never gave importance to.

Besides, the defence establishment never thought that this was a great tactical advantage. Senior personnel did not want to fly this aircraft because the worksmanship of HAL was not up to the mark, he notes.

The IAF was happy because nobody wanted an indigenous programme even though the Marut could do 640 knots, fly low level with four tanks (comparable to the American F-22).

Most retired IAF officers Deccan Herald spoke to faulted the Maruts engine whose under-performance was the main reason why production of the aircraft was grounded.[/b]

Link From karan's post Bharathrakshak.

 

[gadeshi]

Gas detonation practical research details (for detonative engines):

 

[gadeshi]

A great KRET concern CEO interview with a bunch of chocking news:
http://ria.ru/interview/20151112/1319135062.html#ixzz3rGulVsAT
One of them is completely new, revolutionary technology - Radio Photonics and new type of radar - Radio Optical AESA (ROESA) which allows to make a radar a tool for technical vision.