No.
Try such a trick with almost any recent AAM or SAM and you are going to receive a phisical direct hit.

I would bet quite a sum against a BVR AAM skin-skin contact in an accurate evaluation. 🙂

Arrghh. Alright folks,I guess I do need to explain this a bit more.

As everyone knows, an AAM does not aim ‘for’ the aircraft. It leads the target, by sometimes quite a significant margin (depending on target speed and relative flight path).

Then, at a predetermined proximity to the aircraft (which will be a function of the 2 relative vectors), it will explode. The (now) fragments of the missile will then expand to increase Pk. The fragmentation cone will propagate along the same flight trajectory as the missile – but with a greatly increased forward speed*.

Therefore, without detonation, the missile will naturally pass behind the aircraft.

Thus, it is relatively safe to propose firing of rounds that have the warhead removed. Telemetry can then be used to determine whether they would actually have made the intercept or not.

*It should be noted that some weapons, such as the AIM-7, do not use blast fragmentation warheads, but continuous rods – with quite different blast characteristics. These are unsuited to exercise firings as the blast fragments retain approximately the same forward velocity as the missile did prior to detonation. AMRAAM, with its blast fragmentation warhead, does accelerate the fragmentation cone forward, so is suited to exercise firings.

Or maybe, just maybe, you haven’t got the slightest clue about what you’re talking about where as those professionals in the many airforces of the world do have a clue.

The statistics of BVR combat would strongly indicate otherwise.

But you are one of the last people on here I would expect to be able to wrap their head around it.

Lets just go back to discussing F-22’s and Typhoons and you’ll save yourself any further embarrassment.

Have you not got any crayons to go play with?

The final three Meteor test shots were conducted against drones protected by jamming and the release of flares. All three shots resulted in direct body-to-body impacts.

Now that’s what I call “thoroughly validated”. In your world, these three Meteors would have shot down three aircraft and almost certainly killed three pilots.

I call that verified [using the distinction from a numerical modelling background]

Verifying that the tracking system works and it can evade ECM/flares does not prove the doctrine.

Were launch positions ideal?
Was the target engaged in aggressive non-cooperative manoeuvring?
Was the launch aircraft having to manoeuvre to avoid counter-shots?

Oh, and you know fine well that body-to-body =/= dead pilot.
Missile to cockpit might, but hitting a wing, tailplane or fuselage does not.

“The article talks about accelerations that are similar (with a slight advantage for typhoon) hence if thrust/weight ratios are similar then drag coefficient are similar too.”

No – for acceleration to be equal, then the following would need to hold:

drag1/drag2 = weight2/weight1

where 1 and 2 are the aircraft.

Drag =/= drag coefficient.

No, Cd is a vague term that can mean different drag factors and especially on modern plane the Cd’s are largely influenced by upstream conditioning of flows.

Drag coefficient is not a vague term.

Cd = drag/(0.5*density*wingarea*velocity^2)

[if your talking about a ground vehicle, change wingarea for frontal cross section area]

You *might* have a case for argument on turbulence intensities – but in flight, for the vast majority of the time, turbulence is extremely low. It is also extremely inaccurate to suggest the main source of Cd differences on modern aircraft is upstream conditions.

Acceleration is thrust divided by mass that why you have to take into account the T/W ratios, and those T/W are pretty similar on both planes meaning the drag coefficient are equal.

Correction:

In the horizontal plane:
Acceleration = (thrust – drag)/mass

In the vertical plane:
Acceleration = (thrust – drag – weight)/mass

At higher speeds, drag is extremely relevant to comparing acceleration figures.

Drag coefficient is also, nominally, a function of wing area – so you’d have to compare the wing areas of the aircraft to see if Cd is equal.

LOL, good joke.

Do you really think that intelligence services use the same information that is in the public domain? Seriously?

Do you not remember the hack on BAe?

Aside from that, with the right software, it is indeed possible to derive quite detailed geometric details from photographs. Yes, it will not be perfect, but it will give reasonably indicative figures for shape performance, which is what LO is driving at.

Are we still playing this interceptor game? Wing area is approximately 68,69 m^2 (length 19.36m, satellite pictures and J-10 length without pivot tube). The aircraft, if scaled from the F-35C, is about 30,400kg loaded.

This comes out to 440kg/m^2 wing loading.

How are you getting dimensions from the satellite?

I assume you are scaling off a known nearby object?