F-16.net

Elite 1K Joined: Aug 19, 2004 Posts: 1092

I have to make a correction to the equations I wrote earlier. I corrected all my equations in my earlier posts, but just to be perfectly clear we have:

Rmin = Vsturn^2 / {g * (n^2 - 1)^0.5}

where

Vsturn = { (2*n*W) / (rho * S * CLmax) }^0.5.

This is a good place to address another one of bf-fly's assertions:

bf-fly wrote:

As the load factor increases, so does the stalling speed because now in essense the plane weights twice as much. In a 60 degree bank constant altitude the stall speed increases about 1.5 times (actually about 1.41 if I recall correctly). This is an unquestioned fact. Google a load factor chart, use any source you like. An aircraft that stalls at 100 knots wings level will now stall at about 150 in a 60 degree bank. This is called an Accelerated Stall.

Stall speed, Vs, in level flight (where n = 1) is given by:

Vs = { (2*W) / (rho * S * CLmax) }^0.5

In a level turn you have lift, L=nW and so your stall speed in a level turn, Vsturn, becomes

Vs = { (2*n*W) / (rho * S * CLmax) }^0.5

Thus, your stall speed in a level turn compared to steady, level flight is given by:

Vsturn = Vs * (n)^0.5

Stall speed in a level turn is proportional to stall speed in level flight by the square root of the load factor. Where do you see bank angle up in there? Sure, you could substitute an equation for bank angle in place of n, but you could also substitute L/W for n too. So what? Bank angle doesn't generate load factor. An aircraft's shape generates lift which, when combined with enough speed can overcome the aircraft's weight. n=L/W. The trigonometrical relationship relating bank angle to load factor in a level turn is an approximate one. And it's just that... a trigonometrical relationship! You can't start drawing direct relationships between bank angle and stall speed simply because there's an approximate trigonometrical relationship relating load factor and bank angle in a level turn. It just doesn't work that way. You're mixing up cause and effect, to put things very simply.

F-16.net Sponsor

Elite 1K Joined: Aug 19, 2004 Posts: 1092

bf-fly wrote:

How do you get the smallest overall radius? The slowest speed at the greatest bank angle. I have provided this link on several occiasions. You can cross reference this charts provided with any source you like to verify them. I am providing this because it has all 3 charts, load factor, stall speed/bank and minimum radius turns per speed an bank angle. Use any source you like to verify these charts

http://www.aerospaceweb.org/question/pe ... 0146.shtml

First off, it's not necessarily good practice to gain a clear understanding of flight mechanics from websites like these. I'm not saying they're wrong or that the person responsible for writing what's on the website is incompetent. I'm just saying that it's easy to misinterpret the implications of what you find on a single web page like the one you linked to. And that is, in fact, what you have done. Let me be specific. I'll use a few equations and graphs from the website you linked to above.

1)

2)

3)

Okay, Vs in the first equation is the stall speed for level flight. It's not the stall speed for turning flight which I wrote out already and is:

Vsturn = { (2*n*W) / (rho * S * CLmax) }^0.5

Notice the numerator now contains n. This is the same equation that you use to find Vs for level flight where n = 1. The second equation from the website gives you lift, L as a function of aircraft weight and bank angle for level turning flight. Great! The third equation, however, is what's confusing you. It's not actually telling you that an aircraft's minimum turn radius is obtained at, as you say, "The slowest speed at the greatest bank angle." First off, I don't know what the point of that last equation and the above graph are beyond showing you the mathematical implications of the equations themselves. The "slowest speed" (as you call it) in turning flight is the stall speed given by:

Vsturn = { (2*n*W) / (rho * S * CLmax) }^0.5

The minimum turn radius is:

Rmin = Vsturn^2 / {g * (n^2 - 1)^0.5}

or equivalently,

Rmin = (Vs^2 * n) / {g * (n^2 - 1)^0.5}

where Vs is the stall speed for level flight given by equation 1. Equation 3 doesn't tell you what Rmin is. It doesn't tell you what the bank angle will be when flying at Vsturn either. If you're wondering how you'd actually figure out Rmin based on the aircraft's aerodynamics for a given altitude (or density if you will), it's not actually as simple as the equations themselves appear.

I wrote an energy maneuverability diagram generator in MATLAB to analyze the performance of F-16 flight models I helped develop for the Falcon 4.0 combat flight simulator, I actually had to write code that calculated things like minimum turning radius and maximum turn rate (just for starters). Those are just two points on a much more involved graphical representation of an aircraft's level turning capabilities, however. But needless to say, I know what I'm talking about here. If you want, I will provide the link to the flight model manual I helped write with examples of the EM diagrams that my program outputs.

To determine the absolute minimum turn radius based on aerodynamics alone (i.e. you might not be able to sustain this minimum turn radius in level flight due to negative excess power), you first have to calculate your aircraft's maximum load factor, nmax, as a function of airspeed. This is given by maximizing the following equation which I wrote out back on page 3:

n = ( T * sin(α) + q*S*CL ) / W

T = thrust
α = aircraft angle of attack
q = 0.5 * density * V^2 (the dynamic pressure)
S = reference area used to calculate the aircraft's lift coefficient
CL= lift coefficient of the aircraft
W = aircraft weight

Ignoring thrust effects, you'd simply have n = L / W or...

n = (0.5 * density * V^2 * S * CL) / W

Maximizing this equation for a given velocity gives you:

nmax = (0.5 * density * V^2 * S * CLmax) / W

You graph the above equation with nmax on the y-axis (dependent variable) and V on the x-axis (independent variable). Density is given by the standard atmospheric model for a given altitude, and S, CLmax, and W are the given physical and aerodynamic properties of the aircraft. So you've graphed nmax as a function of V. Whether you realized it or not, the velocity associated with this graph of nmax is none other than

Vsturn = { (2*n*W) / (rho * S * CLmax) }^0.5.

You didn't actually have to calculate Vsturn to graph nmax as a function of velocity in general though. I'm just pointing out that the velocity associated with the nmax curve you've plotted on your n vs. V diagram is indeed Vsturn. With this in mind, you have your plotting program calculate the aircraft's turn radius at all the discrete points for which you plotted your nmax curve. In other words, you'd calculate the R for each discrete [nmax, V] coordinate using the following equation:

R = V^2 / {g * (nmax^2 - 1)^0.5}

You then have your plotting program find the minimum value of R associated with the nmax vs. V curve. The more points you use to plot your nmax vs. V graph, the closer you'll get to predicting the exact value of Rmin via graphical/numerical analysis. Anyway, you have your plotting program save the [nmax, V] pair that gave the smallest turning radius, Rmin. And from there you can determine what your bank angle, phi, would be using the relation

phi = arccos(1/n)

where n = nmax from the [nmax, V] pair you determined gave you the smallest possible turn radius, Rmin. Notice that the bank angle, phi, does not depend the velocity. It only depends on the load factor, n. But Rmin clearly depends on both the load factor AND the velocity, V. There is some combination of maximum load factor for a given velocity that results in Rmin. It's not a matter of some maximum bank angle for a given velocity. What is maximum bank angle for a minimum radius turn? How do you figure that out? You figure that out by going through the exact steps I just explained. However, I hope you see that bank angle is an afterthought in this calculation process.

To further hammer home this point, tell me how you would calculate the maximum load factor an aircraft could achieve in level turning flight at a given airspeed based on the aircraft's "greatest bank angle" as you put it. It shouldn't take you too long to realize that you don't know what the "greatest bank angle" for the aircraft will be because you don't know what the maximum possible load factor, nmax will be at the specified airspeed (for a given altitude, aircraft weight, etc.) ahead of time. The idea that Rmin is governed by how far an aircraft can bank is 100% backwards. The load factor at which Rmin happens to occur at DETERMINES your bank angle when performing a minimum radius turn.

If, after all this trouble I've gone through, you still insist that bank angle is an independent variable in determining an aircraft's ABSOLUTE minimum turn radius (i.e. a variable you put on the left hand side of an equation), then I don't know what else to tell you. Just try hard to think about the fact that the equation n = 1/cos(phi) cannot possibly tell you what nmax is at any velocity or any bank angle because there's no such thing as a maximum bank angle, even for level flight. As someone said earlier, it's even possible for an aircraft to perform a level turn at 90 degrees of bank because the whole bank angle formula is a simplified approximation. Regardless, n=1/cos(phi) just tells you, approximately, what load factor is required to prevent the aircraft from losing (or gaining) altitude while banked. It tells you NOTHING about whether you'd be able to achieve this load factor. I can't believe I've gone into such detailed simplicity. Please... just get it.

Active Member Joined: Aug 28, 2006 Posts: 191

Quote:

The idea that Rmin is governed by how far an aircraft can bank is 100% backwards.

I have said no such thing. Only speed and bank angle control min radius of turn. You cannot have one without the other in level flight. Neither is controlling. Only when each intercect does min radius occur.

From page 3 or so: R1-

Quote:

You may have misunderstood what I meant by a level turn. A level turn does not imply 90 degrees of bank. A level turn implies no loss of altitude throughout the turn. This means that for a constant speed level turn, the aircraft must maintain a bank angle of less than 90 degrees. Also, for an aircraft to turn at its minimum "sustained" turn radius, it will be flying at its stall speed

No it does not. Only when each is optmised.

http://selair.selkirk.bc.ca/aerodynamic ... adius.html

http://www.csgnetwork.com/aircraftturninfocalc.html

Elite 1K Joined: Aug 19, 2004 Posts: 1092

Only speed and bank angle "control min radius of turn"? Whatever that means...

Please tell me EXACTLY how you would find an aircraft's ABSOLUTE minimum turning radius by "optimizing" your bank angle and airspeed. Take us through the steps. It's easy to talk to the talk, but now you have to walk the walk. I'm calling you out.

Active Member Joined: Aug 28, 2006 Posts: 191

Quote:

You can't start drawing direct relationships between bank angle and stall speed simply because there's an approximate trigonometrical relationship relating load factor and bank angle in a level turn.

Yes you can.

From NASA

"In steady, turning flight the lift developed by the wing must balance not only the weight of the aircraft but the centrifugal force generated by the turn. (The term "balance" is used here in a vector sense; that is, the lift vector must equal the sum of the weight and centrifugal force vectors.) The load factor is defined as the ratio of the lift in the turn to the weight of the aircraft and is usually expressed in g units, where g is the acceleration due to gravity. Thus, a 2-g turn is one in which the wing must develop a lift force twice the weight of the aircraft. The value of the load factor is uniquely defined by the aircraft angle of bank. For example, 2-g and 5-g turns require bank angles of 60 and 78.5 respectively. Finally, for a given bank angle and thus load factor, the turning radius varies as the square of the speed; for example, doubling the speed of the aircraft increases the turning radius by a factor of 4. It would then appear that two different aircraft flying at the same speed would have the same turning radius; however, this conclusion is not necessarily correct. The maximum load factor and associated turning radius may be limited by wing stalling."

It does not matter if we are talking about a Piper cub, a 747, or a F-22. A 60 degree bank is a 2 g turn and it increases the stall speed 1.41. It is a direct relationship. You seem to think that these charts I've provided are from some crackpot web sites. No, they appear exactly the same in every aviation textbook in the world.

When you say load factor, you are saying a bank angle.

"The value of the load factor is uniquely defined by the aircraft angle of bank"

Not "an approximate trigonometrical relationship"

Deleted

Elite 1K Joined: Aug 19, 2004 Posts: 1092

bf-fly wrote:

http://selair.selkirk.bc.ca/aerodynamic ... adius.html

http://www.csgnetwork.com/aircraftturninfocalc.html

bf-fly,

The first website is totally convoluted. I'm not sure what the author had in mind, but that is NOT how you calculate minimum turning radius. The main problem with the table and associated graph is that it assumes the aircraft can attain the load factors associated with the various bank angles curves. In other words, nowhere does the author take into account the aircraft's maximum lift coefficient, CLmax. He has the variable CLmax in some of the equations on his site, but never gives it an explicit value. Look at this page, for example:

http://selair.selkirk.bc.ca/aerodynamic ... age11.html

Without giving CLmax a value, you can't calculate a stall speed let alone an aircraft's maximum available load factor, nmax. But wait... what's the blue line in his graph, reposted below for convenience?

It's nonsensical is what it is. But here is the guy's explanation:

Quote:

The blue line in the graph is for an airplane with a stall speed of 60 knots in straight and level flight. This line will be different for every airplane of course.

WHAT????!!!! The blue line in the graph is for an airplane with a stall speed of 60 knots in straight and level flight?????? This just makes no sense. My first questions would be, "Stall speed in level flight for what weight and altitude (which determines density)." I wouldn't stop there, but why go any further? But tell me, how did he come up with that line by simply assuming that the aircraft has a stall speed of 60 knots? And what does the line represent besides "the airplane"?

Back to my original point though, however. Nowhere does this guy indicate what the aircraft's maximum available load factor is as a function of airspeed, so he can't possibly know which portions of the various bank angle lines are even valid. I laid out for you exactly how ABSOLUTE minimum turn radius would be calculated, but you obviously didn't pay my detailed explanation any mind. So tell me, where on that graph of his does the aircraft achieve its minimum turn radius? The point he marks as Va is the "maneuver speed" (or so he claims) which is not the same thing as minimum turn radius. He doesn't seem to make any distinction between the two, however. He also says this:

Quote:

As mentioned in the previous section the radius of turn will become smaller as the bank is increased, but the stress on the aircraft will become infinite at 90 degrees of bank. Thus, we are limited to approximately 75 degrees of bank, at which angle the g force will be 3.8gs.

Again... WHAT? Where does he pull these numbers from? Out of a magic hat? How does he know that the aircraft in question can attain a load factor of 3.8g at 75 degrees of bank? The answer is, he doesn't!!! He's just calculated n = 1/cos(75). He doesn't even round off to the nearest 10th of a g! He should have put either 3.86g or 3.9g. Not 3.8g!!! This guy doesn't even know how to round up/down. This guy is obviously just as confused as you. Just because someone's got a website doesn't mean everything they put up on it is correct, complete, or stated in the proper context.

On to the second site you posted. Nowhere does that site's calculator actually calculate the minimum turn radius. How could it? Look at the damned equation for Rmin and tell me how this calculator determines whether the load factor implied by the bank angle is attainable. Once again, you need to read my directions for calculating an aircraft's Rmin. You keep spewing the same BS and finding bunk websites to back up your claims, but you still have no clue how to actually calculate a REAL aircraft's minimum turning radius. You just keep spouting stuff about bank angle. You're obviously confused by the actual mathematics, so you keep going back to your (false) qualitative reasoning and ridiculous talking points. If you can't grasp the implications of multiple equations that must be considered simultaneously, you won't ever understand the nature of things like minimum turn radius, corner velocity, etc.

Lastly, the load factor is well approximated by the bank angle in level turning flight if the aircraft can actually generate said load factor at a given speed and altitude. If it can't, then the relationship between bank angle and load factor is meaningless and purely hypothetical. I guess your problem is separating the hypothetical from the physically possible. To calculate minimum turn radius, you have to determine the most favorable physically possible solutions to purely hypothetical equations. If you lack the necessary mathematical skills (and you, bf-fly, most certainly do), you will inevitably fixate on the implications of a single equation instead of using a complete set of equations to numerically optimize a certain variable like turn rate or turn radius.

End of lesson. Peace.

ok, check this out... I bought a book on aerodynamics, and if what I read was what I read, bf-fly is right - in a way. There is this "general example" on load factors and how bank angle affects it. At 60 degrees bank, the load factor is 2g, BUT THIS IS FOR A CONSTANT ALTITUD T-U-R-N. So it isn't just the bank angle - it's also the centrifugal force. To make a turn with 60 degrees of bank AND NOT LOSE ALTITUDE, the load factor (basically the centrifugal force, or g that the pilot would feel through his seat) would be 2g. That's what I didn't understand, bf-fly was saying (from what I read) that at 60 degrees bank, the load factor is 2g, but this is not entirely true - the load factor IS (generally speaking) 2g at 60 degrees bank, as long as the plane is TURNING with enough force that it is not losing altitude. However, the turn and the g force go hand in hand- if the plane isnt turning, no matter the bank angle, the pilot will always feel one g in the direction of the ground. This also does not seem to take into account speed (though it might be that I'm not seeing it). Be aware I have not read the entire book yet, so there might be something else here, but from what I read, I think this is pretty much it. Remember, too, that these are numbers for a single engine, small propeller plane, so it might not be truly representative of a loaded F-22 and in fact might be way off.

How this plays into the maneuverability of a fighter, I do not know completely, but in another post I'll try to explain whatever I find that might be of interest.

Active Member Joined: Aug 28, 2006 Posts: 191

Tinto, we were talking about constant altitude turns

I know, but it's the centrifugal force that makes the pilot feel 2g, not the bank angle per se. At least, this is where I personally got all mixed up. Now I seem to understand it, and in that context, now I understand some of what you (bf-fly) were saying. Remember not all ppl know about aeronautics and aerodynamics (I'm only 19 Smile

).

Active Member Joined: Aug 28, 2006 Posts: 191

Quote:

it's the centrifugal force that makes the pilot feel 2g, not the bank angle per se

Yes, but you can't seperate the two. It's a what came first the chicken or the egg scenario.

But, You you can do a 60 degree bank and have more than 2G's or less than 2G's if you don't maintain your altitude. (You climb and it's more, decend and it's less) And certainly of course you can pull 2 G's (or any) without a bank.

By the way, the book you bought on aerodynamics has the same charts that I've shown, correct? Bank angle/G's and hopefully Bank angle/stall speed.

Elite 1K Joined: Aug 19, 2004 Posts: 1092

Tinito_16 wrote:

ok, check this out... I bought a book on aerodynamics, and if what I read was what I read, bf-fly is right - in a way. There is this "general example" on load factors and how bank angle affects it. At 60 degrees bank, the load factor is 2g, BUT THIS IS FOR A CONSTANT ALTITUD T-U-R-N. So it isn't just the bank angle - it's also the centrifugal force. To make a turn with 60 degrees of bank AND NOT LOSE ALTITUDE, the load factor (basically the centrifugal force, or g that the pilot would feel through his seat) would be 2g. That's what I didn't understand, bf-fly was saying (from what I read) that at 60 degrees bank, the load factor is 2g, but this is not entirely true - the load factor IS (generally speaking) 2g at 60 degrees bank, as long as the plane is TURNING with enough force that it is not losing altitude. However, the turn and the g force go hand in hand- if the plane isnt turning, no matter the bank angle, the pilot will always feel one g in the direction of the ground. This also does not seem to take into account speed (though it might be that I'm not seeing it). Be aware I have not read the entire book yet, so there might be something else here, but from what I read, I think this is pretty much it. Remember, too, that these are numbers for a single engine, small propeller plane, so it might not be truly representative of a loaded F-22 and in fact might be way off.

How this plays into the maneuverability of a fighter, I do not know completely, but in another post I'll try to explain whatever I find that might be of interest.

Tinito,

No offense here, but I've read numerous books on aerodynamics, aircraft performance, and so on. I also have a B.S. degree in aerospace engineering. I've written computer programs to calculate and visualize aircraft energy-maneuverability performance which includes minimum turn radius and maximum turn rate. See here for yourself:

http://www.checksix-fr.com/downloads/fa ... manual.pdf

There's a difference between determining what load factor is required to maintain a level turn at some selected bank angle and what a REAL aircraft's minimum turn radius is at a given weight and altitude. Take your aerodynamics text book and figure out what the minimum turn radius, Rmin is for an aircraft with the following data:

CLmax = 1.5
Reference area, S = 500 ft^2
Aircraft Weight, W = 10,000 lbs

Assume the following:

Density, ρ = 0.0023769 slug/ft^2 (this is the air density at sea level)
Acceleration due to gravity, g = 32.2 ft/s^2

With only this data, you can determine the aircraft's minimum turn radius, Rmin, based solely on aerodynamics. We'll assume the aircraft produces enough thrust to maintain stall speed in a level turn. OK? Get cracking.

A few words of advice though. You are dealing two unknowns at the same time. They are as follows:

1) Maximum load factor, nmax.
2) Velocity, V.

The minimum turn radius, Rmin depends on V and nmax. But nmax also depends on V. Thus, you have to calculate a bunch of values for nmax corresponding to a bunch of values of V. You then have to calculate the turn radius for all these [nmax,V] data points to determine which one actually gives you the lowest value for the turn radius R, which will be the aircraft's Rmin. I hope you know how to use Excel or, better yet, MATLAB. This is not the kind of problem you can solve for exactly, but the more data points you use, the closer your answer will be to reality. Make sure you don't include any velocity points that give you a maximum load factor, nmax <= 1.0. Oh... and to make your life easier, assume that the aircraft's structural limit is 7g (i.e. don't bother using any velocities which give you an nmax > 7.0).

If you want to attempt to solve this problem using bank angle as a variable, ask yourself whether you can use it to determine an aircraft's maximum available load factor at a given airspeed. I've written out all the necessary equations in previous posts, but here's the one for maximum load factor again:

nmax = (0.5 * ρ * V^2 * S * CLmax) / W

Think long and hard about that one above. Can you use the following bank angle-load factor relationship to determine nmax at a given airspeed?

n = 1/cos(Φ)

Φ = bank angle (in radians, not degrees)

It should be clear to you that the simple equation for load factor as a function of bank angle (in a level turn) can't tell you anything about an aircraft's maximum available load factor at a given airspeed V, density ρ, reference area S, maximum lift coefficient CLmax, and weight W. You can certainly approximate Φ for the various values of nmax you calculate and for the specific [nmax,V] pair that gives you the absolute minimum turn radius, Rmin. You CANNOT determine Rmin from the n = 1/cos(Φ) equation though. It tells you nothing about nmax as a function of V, and as such, tells you nothing about turn radius R as function of both nmax and V.

End of lesson for today.

Elite 1K Joined: Aug 19, 2004 Posts: 1092

bf-fly wrote:

Quote:

it's the centrifugal force that makes the pilot feel 2g, not the bank angle per se

Yes, but you can't seperate the two. It's a what came first the chicken or the egg scenario.

But, You you can do a 60 degree bank and have more than 2G's or less than 2G's if you don't maintain your altitude. (You climb and it's more, decend and it's less) And certainly of course you can pull 2 G's (or any) without a bank.

By the way, the book you bought on aerodynamics has the same charts that I've shown, correct? Bank angle/G's and hopefully Bank angle/stall speed.

Dude... I'm assigning you the same homework problem I assigned Tinito. Get cracking! And don't forget to show your work. You will want to use a spreadsheet for this. I'll accept answers for minimum turn radius based on velocity intervals of no more than 10 ft/s. Extra credit if you narrow your step size to 1 ft/s around point that gives you an Rmin based on the 10 ft/s step size. If you refuse to do this homework assignment, you automatically lose the debate. And if you think I can't answer my own problem in astounding visual and numerical clarity, just try me. I'll use MATLAB to solve this problem with a velocity step size of 0.1 ft/sec and graph both nmax vs. V and Rmin vs. V. I'll even throw in a graph of bank angle at nmax vs. V since you're so fascinated by it. No problem. Just say the word. Of course you still have to complete the problem yourself. I won't present the right answer until all homeworks have been turned in.

Active Member Joined: Aug 28, 2006 Posts: 191

R1, here is your problem

Quote:

Without giving CLmax a value, you can't calculate a stall speed let alone an aircraft's maximum available load factor, nmax. But wait... what's the blue line in his graph, reposted below for convenience

You are so completely enamored with your formulas that you ignore (or are unaware of) the basics. It's called a manual which tells you the speeds.

Quote:

Again... WHAT? Where does he pull these numbers from? Out of a magic hat? How does he know that the aircraft in question can attain a load factor of 3.8g at 75 degrees of bank? The answer is, he doesn't!!! He's just calculated n = 1/cos(75). He doesn't even round off to the nearest 10th of a g! He should have put either 3.86g or 3.9g. Not 3.8g!!! This guy doesn't even know how to round up/down. This guy is obviously just as confused as you.

First off, I'm not confused in the slightest. Secondly, if you had read what he wrote he used an aircraft for his example that was FAA certificated as a "normal" catagory aircraft. Which as you know (actually it seems not) is limited to 3.8 G's. (plus of course a 150% tolerance)

I provide charts which are available in every textbook in the world that show a direct relationship between these three values, yet you say

Quote:

You can't start drawing direct relationships between bank angle and stall speed simply because there's an approximate trigonometrical relationship relating load factor and bank angle in a level turn.

You are very good with calculations, much better than me. I give you that. But you seem to be completely unaware of the basic foundations of the principles of flight, SO MUCH SO, you say this, once again:

Quote:

You can't start drawing direct relationships between bank angle and stall speed simply because there's an approximate trigonometrical relationship relating load factor and bank angle in a level turn

There is the albatrosse around your neck

(section 333) http://www.hq.nasa.gov/pao/History/SP-468/ch11-6.htm

It's not a question of me "drawing direct relationships", the whole aviation community does, with the apparent exception of you.

Ask Tinto 16 what his new book says.

ok, it's like 3:30 am where I'm at and I gotta get some shuteye, but I'll close with this: basically, bf-fly seems to be right on one thing, and Raptor_One seems to be right in another. Right now (I havent finished the book it's quite long) I think fly's right (mostly) about the g's/bank angle thing, and Raptor's right (mostly) when he says bank angle does not necessarily give you the Rmin. Given the F-22's good wing loading and excellent thrust, I don't think the pilot's gonna think twice about lighting the burners and banking about 90 degrees and just turning the nose as fast as possible to get whatever threatens him or her on the Raptor's sights. Unless the guy is 100 feet off the ground.

F-16.net

Forum: F-22A Raptor

Wing Loading on the Raptor