Since this post is more relevant and deals with tejas, i am posting it here,
As I said, this very discussion is moot. You are not going to get it. But to put it concisely, you are going to invite
@p2prada back, all the while keeping the stuff which forced him out in the first place and expect him to not leave again?
. Yeah, good luck.
Tirupati temple disagrees
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His question
can you explain what are those other parameters which need to be same? You keep posting same stuff over and over again without even reading what other members have written. I had also posted the same formula which you copied from google-wiki. You need wiki to know about aerodynamics but I have studied it as a subject for over three decades.
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.
my answer,
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,
["‹IMG]
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:
["‹IMG]
where g is the gravitational field strength.
The radius of the turn can now be calculated
["‹IMG]
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,
["‹IMG]
Solving for R gives
["‹IMG]
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?
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His question,
ersakthivel, It is for this reason that I had advised you to please read some good books on aerodynamics. If we have two aircraft operating at same density altitude at same speed with same CL value and same weight, the ac with more wing area will produce more lift as it will have lower wing loading. The factors listed here as SAME are the other things being equal.
what is so great about this. I have also written the same formula which you wrote and what has been stated by you but you have grossly misunderstood the meaning of it. Now I had given you the example of F-22, Rafale, Typhoon and others which have higher wing loading than LCA but much higher ITR & STR.
You keep trolling and posting useless stuff just to somehow fool the members. it is not going to succeed here. You can take your theory of low wing loading else where.
My answer,
Lift from the wings is the oxygen of a fighter, not the sweep angle of presence of all those "lift enhancing stuffs" you listed. they help , no doubt , but low wing loading is the primary way to get lift.
A low wing loading fighter like tejas will have much higher horizontal component of thrust in any sustained turn and can pull a smaller turning radius(it can always turn inside the turning circle of the high wing loading fighter it purses in close combat).
Also it can get that higher thrust at lower AOA , meaning lesser induced drag (skin friction drag from the larger wing area is significantly smaller , than the induced drag ) for the same horizontal component of lift achieved in turns . What this means is lower wing loading fighter like tejas can have higher speeds in turns than the high wing loading fighter.
Also the lower stall speeds of low wing loading fighters like tejas makes them excel at slow speed manoeuvres.
And in high altitude were the skin friction drag is even lesser , it helps them to perform more.
The so called "other things being equal" was an ordinary user quote with no formulas to base it , from that forum I posted to give a sense of balance, which was immediately highlighted by some dude to give a completely different meaning!!!! which primarily meant that larger wing area will lead to more fighter weight.
the excess weight added by the larger wing area is doesn't substantially adds to the fighter weight. just look at the empty weight of gripen c and tejas. tejas despite having lower wing loading has lesser empty weight than gripen.
Also tejas which has a substantially lower wing loading than JF-17 has more or less the same empty weight.
but even the conditions "all other things being equal" has been met by designers of tejas.
Why?
because the contribution to fighter weight from wing area portion is significantly smaller compared to other things like engine, fuselage, weapons, fuel , landing gear etc, etc,
So in a turning fight tejas has nothing to fear from any fighter in subcontinent , and once TWR goes up in tejas mk2, it can compare with any fighter in 4.5th gen class.
I am posting stuff , that neither you nor any one else can refute with any authentic stuff. Got it?
this one paragraph you have written doesn't nullify all those equations. Got it?
You just posted two equations and claimed that turn rate has nothing to do with wing loading or wing lift and entirely dependent on bank angle and velocity.
IMHO I haven't misunderstood anything. You can't ignore Newtons second law of motion in real world.
I proved you wrong categorically.
For example , if Tejas gets the same TWR as typhoon and rafale and the same thrust vectoring as F-22 we can talk.
Sure we will talk about it in future because they will be implemented in future tejas versions.
-----------------------------------------------------------------------------------
My answer,
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"
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?
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.
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@ersakthivel, iT is no point trying to explain anything to you. Long back my grand father had taught me that it is a sin not to teach a person who wants to learn and it is a bigger sin if you try and teach a person who does not want to learn. You have your mind closed and in each repeating post you contradict yourself to somehow prove your shit theory about low wing loading. You have filled tons of pages in the other forum writing the same shit.
Now it is my last post to make things clear to you.
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.
NOW STOP EATING MY BRAIN AND TAKE YOUR SHITTY THEORY TO A FORUM WHERE IT SELLS. DONT TEACH YOUR POP HOW TO ****.
My answer,
As per your equations,XXX degree bank at ZZ speed would produce the same turn rate regardless of the aircraft weight or wing loading ..
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.
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.".
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this was his question
@ersakthivel, All your knowledge can be gauged from just your reply to this question,
What is the value of V in the equation and how is it calculated. Now don't run to wiki for it?
This was my answer
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
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This what decklander asked me
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.
If you read these equations properly, you will find the mention of velocity and bank angle. 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.
LCA has a large wing, tailless design with very high sweep outwards of 62.5* which gives it extremely high stall AOA & poor CLmax and a very high power off stall speed. This is the main reason for it not having as good ITR/STR as the ac I mentioned.
Now another question for you.
How do you calculate ITR & STR and what all factors play a part in calculating them?
I will teach you very good things here and you will be able to stop making a fool of yourself over and over again in future.
The following is my answer,
and what is relationship of V ,with horizontal component of lift vector , which in turn relies on wing area to generate that,
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,
and what is the effect of lesser and higher AOA on drag and stall speeds,
You will teach me how to run in endless circles!!!!!
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.
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.
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.
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)
My further posts,
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.
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,
["‹IMG]
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:
["‹IMG]
where g is the gravitational field strength.
The radius of the turn can now be calculated
["‹IMG]
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,
["‹IMG]
Solving for R gives
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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?
finally he admitted
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.
If you have any problem with the above stuff, you contact any mech eng student and get clarification all by yourself and tell me where I was wrong and abusive.
I am all for inviting p2p, Deck lander And in future I will never reply to any of their posts,
I too will be glad to know where I have gone wrong.
Because I have replied more than thousand times and wont do it even once if they post here.
Okay, that is what you want from Arjun brigade(?!?!?),
because I will be reducing my activity here, getting engaged in a new venture.
I will not engage in arguments with any member here ,
and will not quote both of them and start an argument.
It's a pledge.
So p2p2 and Decklander can be free to post what ever they want.
Lets see how it contributes to the authenticity and credibility of the forum.