Picard, why are the Russians and Chinese pursuing internal weapon bays then with the T-50 and J-20? Why are they heavily investing in X-band AESAs for those aircraft?

Everyone has got it wrong. They are all idiots because Maginot line.

Turn rate is hellishly complicated. Climb rate is far more straightforward by comparison. For turn rate you have to consider different aerofoils, different planforms, instability margin, AoA differences, canard effects, LERX effects….. Not touching that with a barge pole. All you can really say is that a lower wing loading and higher TWR won’t hurt any if everything else was the same, which it likely isn’t.

I don’t see flight mechanics there anywhere. Which is undoubtedly the root source of this problem (and the Breguet range equation issue in the Eurofighter thread).

Covered in both aerodynamics and propulsion. Learning is always applied to real world scenarios in questions.

A mechanical engineering degree with a few aero modules tacked on is good – but don’t assume it makes you infallible on all matters aeronautical. Assuming you did graduate 14 years ago, then you really should be wise enough to know that by now.

It’s not a Mechanical Engineering degree (M stands for Masters). The degree was purpose designed by a leading aerospace defence contractor in conjunction with the University specifically for the industry, although some of it is obviously useful outside the industry.

Unfortunately, I’m too busy at work to settle this once and for all (by building a workbook to demonstrate it all) – if I get time over the next few days, I’ll see if I can throw something together.

My calculation:

http://forum.keypublishing.com/showthread.php?130883-fighter-agility&p=2150724#post2150724

is by no means flawless but the fundamentals are there and the assumptions are reasonable. In the post above I’ve demonstrated that Andraxxus’s method is flawed and even the implementation of maths within that flawed method is wrong at a fundamental level.

100 years ago, the world’s most professional general staff, directing the world’s best trained and equipped army, had a plan to win a European war before Christmas.

So anyone who says “this is so, because the military experts say so” is failing to use the brains he was given, if he was given any.

Actually it was because they failed to invest sufficiently in new technologies pre-war. We sold the Germans more Maxims than we bought ourselves. Tanks in large number in 1914 and lots of Maxims would have ended it a lot sooner.

Some idiot thought that cavalry on horseback with swords was going to win it. Kind of like the people who think gun dogfights and the occasional WVR missile is going to be a telling feature in future air wars.

Indeed you are. and problem is, even in zoom climb, your energy gain will wont change. In my calculation you will gain X amount of PE and 0 amount of KE. If you go through long version, You will gain X+Y amount of PE and lose Y amount of KE; net result will still be X.

You need to show your working from first concepts because at the moment it just look plain wrong.

Yes! Finally we are talking in the same language. Its called instantenious because the instant aircraft gains 1 meter altitude, there will be change in air density, so all thrust drag and weight will be affected; as aircraft gains altitude, it will lose climb rate even though airspeed stays the same.

That much I get.

I NEVER said that. V is the velocity; Assuming same V (which pretty valid for max climb rate, always around M0,85 and S/L); if you ratio New vs Old at same speed;

NewClimb/OldClimb = New(F-D)/Old(F-D)*(OldW/NewW) and “V” will cancel out.

What I assumed is Cd will be same between New and old, so D relation will be dependent on wing area alone.

As I have no initial data; I used “(NewF/OldF)-(NewD/OldD)” instead of “New(F-D)/Old(F-D)” Inaccurate I am fully aware of it, (hence why i called it so inaccurate; it would have much greater accuracy othervise).

This is where I disagree, V will not be the same and cancel and must be recalculated based on some kind of force balancing or account of a maximum on a SEP graph. Assuming the same Cd, or same climb angle (AoC) is fine but V will change and will likely be greater for aircraft with a larger T-D value. There are also problems with the mathematics of how you’re doing the ratio besides that and because of it. Because the different optimal V will also lead to a different optimal D, so it won’t move linearly with A. As

The bit in bold. Assuming V is the axial velocity, you obviously can’t get climb rates >300m/s if V is limited to M 0.85 @SL and you aren’t even going directly up. Assuming you’re just using that as a starting point for the climb, calculating the increase in SEP at this speed from ratios is incorrect because if I take a simple situation where:

T = 10
D = 5
W = 5

(T-D)/W = 1

If I increase T by 50% and D and W by 10%

(T-D)/W = 1.72 = [(15-5.5)/5.5]

1.72 is nowhere close to (1.5-0.1)/1.1 = 1.27 out by >26%.

which is what you did here:

http://forum.keypublishing.com/showthread.php?130883-fighter-agility&p=2150284#post2150284

Rafale instantenious turn rate 30,26 deg/s (On a big assumption that Rafale uses same airfoil as Mirage 2000, which Soviet information booklet gives Clmax of 1.05; Assuming close coupled canards improve this by ~15%, and assuming Rafale would need ~800kg of fuel for 500 km range, 10300kg + 1,2Cl +45,7m2 wing area gives such result) (Accuracy 3/10)
Rafale sustained turn rate 24,2 deg/s (On the same assumption Rafale behaves similar to Mirage 2000, which has 19,26deg/s STR, increase in 11% wing area will increase the drag by same amount and 59% increase in thrust will improve STR, , So 19,26*1,59/1,11/1.14 is the overall ratio) (Accuracy 2/10)
Rafale climb rate 370 m/s (Same assumption, same ratios above, 285m/s is the climb rate for M2k taken from wiki; 285*(59%-11%)/14%) (Accuracy 1/10)

285 x [(1.59-1.11)/1.14] = 370

(NewF/OldF)-(NewD/OldD)????

You actually did (NewF/OldF)-(NewD-OldD) or (NewF/OldF)-([NewD/OldD]-1) neither is right.

same as this actually; but as dV/dt = 0 you have to ignore KE change. If you dont; aircraft will trade PE for KE, and you will go through long and useless calculations to find same dE/dt or dH/dt, more likely you will make a mistake in your assumptions or calculations.

No, that’s what I did here:

http://forum.keypublishing.com/showthread.php?130883-fighter-agility&p=2150724#post2150724

I assumed 60 degrees as best angle of climb (BAOC) because that seems to be roughly what fighters seem to use. You’d do well to go through that calculation.

[QUOTE=Andraxxus;2150905]WRONG WRONG WRONG. Instantenious climb rate is where of KE change is 0 and the ability of aircraft to gain altitude AT GIVEN SPEED.

Horsesh*t

WRONG; If you were corret, all aircraft would have better climb rate at M2.0 than it has at M0,8. And ALL aircraft would have same SEP. Neither is the case.

I’m referring to a zoom climb not a sustainable climb. The problem is that your calculations have been such bollox that I’m still trying to figure out what the hell you’re talking about.

WRONG. Instantenious climb rate is given FOR a specific speed. Just ignore “instantenious climb rate” and focus on “specific excess power” if the word “climb” troubles your mind.

Well the word is confusing because they’re actually talking about something that’s sustainable at that point in time.

The problem is that whilst max climb rate is given by max excess power, given by:

[(F-D) x V]/W

The V in the equation is not the climb rate. Even in the equation pertaining to a Mirage 2000, the V is not the climb rate, the result of the whole equation is. That V which delivers maximum climb rate will change from aircraft to aircraft in a none linear manner. SO plugging in 285m/s V is complete crap and deserves, Accuracy 0/10.

Rest of your post is nonsense, because what you post is correct, but you simply don’t know what you are calculating.

Oh trust me I do.

From MiG-29 Aerodynamics booklet;
Vy* = (P-X)/G * V = (P/G-(Y/KG)*V = nxV

[ATTACH=CONFIG]229907[/ATTACH]

Now explain how that applies. It’s just stating this with different symbology:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node100.html

I’ve also used this same equation to prove you wrong. You seem to think that arbitrarily taking the climb rate of a Mirage 2000 and substituting it for V and applying a ratio to it based on T, D and W will give the new climb rate. You don’t understand a damn thing.

I don’t know how one can be proven WRONG so many times and still try to be smart-a** without a face. But of course; you know more than the MiG engineers too right? And since you can never be wrong, its the MiG engineers must be wrong. Also, typhoon is the best !!!11 . Have a nice day…

I’ll have a great day thanks. I know what I’m talking about unlike some people.

@lukos Again you are spreading BS without slighterst idea of what you are talking about.

and this is the apex of it. Firstly you cannot add forces in different planes like that; Secondly Drag = Thrust-Weight ROFL; WTF?????

Thirdly; an aircraft at an equlibrium state has L=W and T=D. This is also equal energy state because as Fnet is 0, dE/dt = 0. This state does not change if aircraft climbs or dives; an aircraft gained 1000 meter in altitude will lose 140m/s airspeed, but total energy of aircraft will newer change. Only way to change this energy state is increaing the T.

A 10 ton aircraft with 300m/s climb rate has 10000kg*9,81*300m/s = 2,94 kW excess power. It can use it for climb, accelerate, or for ANY combination of two. It can even dive, and accelerate quickly, but ability of this gain wont change, period.

This excess power is provided ****ONLY**** by “Thrust-Drag” equlibrium.
You can calculate any sh*t you want, it doesnt change SEP; you waste your time calculating how in change in KE affects PE or vice verse. Energy is a SCALAR quantity that doesn’t give **** about your calculations involving climb angle or anything.

In order to calculate a energy state; you assume level flight where L=W, but (T-D)*V will give energy gain rate on level flight. Then you convert this to dH/dt or dV/dt to find acceleration or climb rate at that instant. THIS IS THE METHOD TO DO IT, as its shown in fluid dynamics II, turbomachinary, advanced aerodynamics, and thermodynamic I and II courses, PERIOD. Go educate yourself in a university or a library or just STFU; your ignorance combined with your tone is really disgusting me.

You can’t do that either for sustained climb or instantaneous climb.

Maximum instantaneous climb involves starting from maximum speed in level flight KE and going into a vertical climb. At this point T = D, so you need to work out level flight V based on T, A, Cd and air density.

[T/(0.5 x Cd x A x Air Density)]^(0.5) = V

You start with KE and Rate of change in KE = Rate of change in PE. So:

0.5 x m x d(V^2)/dt = m x g x dH/dt

Essentially this but T-D = 0 because you’re starting from maximum level flight speed and we’re assuming instantaneous transition to vertical climb.

If you want a sustained climb then you need to force balance, as per the first 2 equations on here but dV/dt and d(AoC)/dt = 0:

http://web.mit.edu/16.unified/www/FALL/thermodynamics/notes/node100.html

Perpendicular to travel
L = W cos(AoC)

Parallel to travel
T-D-Wsin(AoC) = 0

You use the second equation to solve for V, and then either use.

dH/dt = Vsin(AoC)

OR

[(T-D)/W] x V = dH/dt

as per here:

http://forum.keypublishing.com/showthread.php?130883-fighter-agility&p=2150724#post2150724

If you’d like to question my education I can quite happily scan it for you. I received 80-100% in all mechanics, maths, aerodynamics, propulsion and control modules and achieved a first class honours MEng and a PhD in Physics. I have the results slips for each year.

Okay so if I write that I think:

2 + 2

roughly equates to:

(Number of legs on a centipede)^(0.25) + [(Number of craps a honey badger takes per day)/2]

and rate it with accuracy (1/10), that would be okay with you?

Yip.

Till someone comes along and produces better data that indicate otherwise.

:confused: