There are other reasons why the plane might only carry Brimstones on 4 pylons besides CFTs, although it’s far from definite that that’s the case.

For instance, a Tornado only has 4 pylons capable of using triple ejectors:

http://www.thinkdefence.co.uk/2012/03/the-brimstone-missile/

Equally an A-10 only has two pylons capable of carrying Mavericks.

Lets assume continuous burn: how long before it runs out of fuel ?
We had this discussion before,
i got the impression from a video that it can burn 25 sec, while Mercurius claim around 15 sec IIRC.
Coasting and throttle back works on high alt, but at lower altitudes it will need constant burn to overcome pressure

It’s still variable even if burn is continuous. That’s the huge difference between a rocket and a ramjet besides effficiency. A ramjet can slow down or speed up. The Meteor decides how fast to fly and what flight profile to use based on optimising the intercept. So if it was up against a MiG-31 doing Mach 2.5+ 50km away from a rear aspect, it might do Mach 5. If it was fired against an orbiting AEW aircraft from 250-300km away, it might do just over Mach 2.

As regards an NEZ, it’s rated at 3 times the that of the AMRAAM it replaces for the RAF. I believe the RAF’s current AMRAAM version is the AIM-120C-5. So I guess if we approximated NEZ as 35-40km less than stated maximum range based on the longest AIM-120B combat intercept at 21.6nmi, that would work out to about 65km for the C5 and therefore ~200km for the Meteor.

The other problem is communicating through the plasma stealth and using your own radar through it and how do you do SIGINT or RWR functions? We’ll have to wait to see if any of the new Euro drones have it. I guess we’ll know by 2030.

I wonder if it would be possible to have a thin layer of gas between an outer radar transparent skin and the aircraft main body and then ionise that, or does it need to be a certain thickness? Could you employ radioisotopes to expedite the process?

and who would want to buy the Tranche 1 Typhoons when the user nations themselves are eager to get rid of them so soon due to obsolescence creeping in?

Some air forces are still flying MiG-21s.

Been reading picards blog huh? Every number you just posted is fabricated. There is no evidence the f-35 tops out at .9 mach on dry thrust. The typhoon number is pure fantasy from a poster on a forum. Take a look at the KPP’s of the f-22 then think real hard.

Let’s be careful what we post as “fact”. That’s what starts those unpleasant flame wars.

It’s funny you should say that because the “half” was the problem in what he originally wrote.

What I’m trying to convey is simple. SEP tells you what your best rate of climb is and the velocity you should start at, but it doesn’t tell you the axial velocity in the actual climb.

When you initially go into a climb from that speed you will accelerate both axially and vertically.

T – D – Wsin(AoC) = ma [1]

but

a = dV/dt and m = W/g where V is the axial velocity in the direction of the climb path.

and

sin(AoC) = (dH/dt) / V where dH/dt is climb rate of velocity in the y-axis.

Hence substituting into [1] gives:

T – D – W(dH/dt)/V = (W/g).(dV/dt)

OR

T – D = W(dH/dt)/V + (W/g).(dV/dt)

Multiply each side by V/W

–>(T-D).V/W = dH/dt + (V/g).(dV/dt) = dH/dt + (1/2g).[d(V^2)/dt] [2]

Which is the same as what it says here:

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

He’s trying to make out that the force balancing is different to the energy balance but they’re one and the same.

When you start off from the V that gives maximum SEP (see LHS in [2]) and start to climb, both terms on the RHS will start to increase. That is, initially dH/dt will increase and dV/dt or d(V^2)/dt will increase.

The reason is obvious, in the case of the MiG-29 starting off at 309m/s, which is the V that gives maximum SEP at sea level, you cannot instantly achieve a dH/dt of 345m/s straight away just by pulling back on the stick and changing direction. V will need increase in the climb and it will continue to increase until:

T – D = Wsin(AoC) where D = 0.5 x Cd x A x Density x V^2 [3]

The value of V at which that happens is given by:

{ [ T – Wsin(AoC) ] / [0.5 x Cd x A x Density] }^(0.5)

When V stops rising equation [2] simplifies to:

(T-D).V/W = dH/dt

But from [3] we know that when V stabilises T – D = Wsin(AoC) therefore:

[Wsin(AoC)].V/W = dH/dt = Vsin(AoC)

Which is intuitively obvious from the diagram at top of page.

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

In [2], if dH/dt = 0 in level flight:

(T-D).V/W = (V/g).(dV/dt) = SEP (NOTE: Not V/2g.dV/dt as Andraxxus said)

–> SEP x (g/V) = dV/dt

The LHS is known as Specific Thrust.

I can’t? Math 101;

That isn’t what you did.

http://forum.keypublishing.com/showthread.php?130883-fighter-agility&p=2151125#post2151125

To make it more scientificc; typically this is where this formula has its use;

Divide both sides by W; assume W doesn’t change with respect to time so move it out of differantial part; eqn becomes
(T-D)*V/W = dH/dt +d/dt(V^2/(2g)

You know MiG-29’s SEP (T-D)/W*V = 345 m/s and max current rate of climb is 309m/s as limited by Vsin90;

345m/s – 309m/s + (1/2G) dV/dt*V where dV/dt is acceleration, positive indicating MiG-29 is accelerating in a 90 degree climb; math will follow to find the exact value.

Look carefully, ignoring G, you have said:

(1/2)d(V^2)/dt = (V/2).dV/dt NOT (1/2)d(V^2)/dt = V.dV/dt

which is exactly what I said. If you do if properly you get the equation for specific thrust. I.e.

[(T-D)V/W] = (1/2g).d(V^2)/dt = (2V/2G).dV/dt = (V/g).dV/dt

Hence:

dV/dt = [(T-D)V/W].(g/V) = Specific thrust = acceleration in level flight

Its a matter of simplification has nothing to do physics behind it. v*dv/dt = 0 if a = 0

That much I agree on.

MiG; engineers say othervise. MiG-29;
[ATTACH=CONFIG]229939[/ATTACH]

Also General dynamics engineers;
F-16@22000 lbs has 1000 feet per second = 304,8m/s @ M0,72 = 247m/s. Note that I circled STD DAY conditions.
[ATTACH=CONFIG]229940[/ATTACH]

Or McDonnell Douglas engineers of F-18;
[ATTACH=CONFIG]229941[/ATTACH]

Not to mention Boeing engineers of F-15, Sukhoi engineers of Su-27? But they are ALL wrong and you the lukos the infallible is right, right?

They are indifferent from energy gain perspecive, I don’t know how one can be so thickheaded to comply with this.

They’re not wrong, you just don’t understand what you’re reading. Maximum SEP is calculated at a V where SEP is maximum, obviously. That gives the maximum climb rate. However, it does not mean that the V during that climb is the same V that gives the maximum SEP. How can is be? At maximum SEP you can accelerate and/or increase climb. If you put all that excess power into a rate of change of PE, then you are using it. After doing this you will have increased axial and vertical velocity. You don’t understand it properly. You can’t attain a climb rate of 345m/s at an axial velocity of 309m/s, that isn’t what maximum SEP is saying. I’ve already proved this if you could actually understand the fundamentals.

Look:

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

If dV/dt = 0

T-D = Wsin(AoC)

Again if V isn’t changing:

dH/dt = [(T-D)V/W] = Wsin(AoC) x (V/W) = Vsin(AoC) i.e. the vertical component of the axial velocity. The axial velocity in the vertical climb is therefore greater than the climb rate.

In maximum climb: V > dH/dt

V is static and the excess power is put into rate of change of PE.

The maximum SEP is an entirely different condition, where V is selected to maximise SEP, which tells you what the maximum climb rate is but it is a different flight condition. Yes, it’s confusing, it’s telling you what your maximum climb rate is but not actually giving you the axial velocity for it. It tells you the velocity you should start at but not the velocity you will finish at when max climb rate is achieved.

Still you are not getting the point AT INSTANT V=309m/s, MiG-29 achieves 345m/s climb rate. There is no “no longer” thingie about it. Its irrelevant if this is physically possible or not; this is not “MiG-29 can climb 345 m/s @V=309m/s” statement. It is “MiG-29 has excess power equivalent to potential energy gain of 345m/s climb rate”.

That is why you are completely wrong in your following calculations:

Even going completely vertical at 90 degrees; an object with airpseed of 309m/s cannot have greater than 309m/s climb rate; but MiG-29 is known to have 345. So Your climb angle calculations are invalid due to (T-D)sin(AoC) = W cannot be valid;

Since you like accurate numbers, let me explain this way;

(T-D)/W*V = 345 when V = 309m/s;
(T-D) = W*345/309;

(T-D) = 1.116*W

You have to take arcsin of 1.116 which you cant, and due to you ignore the validity for input/output range, you take the the arcsin of 1.116^-1 (0,896) which gives out your 63,64 degrees, which is WRONG.

SO YOU CLAIM TWO EQUATIONS PLUS THREE CLIMB RATE GRAPHS ON MIG-29 FLIGHT MANUAL AERODYNAMICS BOOKLET IS WRONG????

Andraxxus is wrong, Amiga500 is wrong, 800+ engineers in MiG design bureu are also wrong, they designed MiG just by pure luck, only lukos is right. Why? because he is right and he also has PhD.

Perhaps its you who is using wrong ratios and getting wrong answers?

You have to take arcsin of 1.116 which you cant, and due to you ignore the validity for input/output range, you take the the arcsin of 1.116^-1 (0,896) which gives out your 63,64 degrees, which is WRONG.

Dude, the angle was actually wrong, and I’ve now removed it, but not for the reasons you think. It’s wrong because I wrote:

T – D = ma

and then said:

(T – D)sin(AoC) = ma x sin(AoC) = mg THIS WAS WRONG!

It should have been:

(T – D)sin(AoC) + Lcos(AoC)= ma x sin(AoC) + Lcos(AoC) = mg

You don’t asume if you are in the game of producing hard figures.
In a typical flight manual, its called a “climb profile”. And for the same jet, it could be different angle, as how(much) the jet is loaded. If you don’t know the exact climb angle used for the whole duration of the climb, your figures will be inaccurat.

Seems like a pretty good assumption to me, see above.