The only fact is the one made by MBDA:
MICA: BVR
ASRAAM: WVR
I guess that alone should be sufficient for Lukos.
Note AIM-120C5 range in tail chase at low level.
http://defense-update.com/newscast/0309/asraam_loal_test_130309.html
This engagement simulated a “chase down” situation by an enemy fighter and successfully demonstrated the potential for an all-round self protection capability with the ASRAAM. The missile was fired at a target located behind the wing-line of the ‘shooter’ aircraft, flying at low level and typical fighter speed, at a target located behind the fighter at a range in excess of 5km. The result was a direct hit on the target.
Let’s see MICA IR make that shot at low altitude!
Now, I intentionally asked you what climb rate would plane X have at 187 m/s on the deck for 60 and 90° climb angle and you answered (as I thought you would) 162 and 187 m/s.
First of all, answering that question without knowing a single thing about planes technical characteristics (since we are talking about plane X) is retarded by itself.
Second, the real world example is undoubtedly showing that numbers you provided have no base in reality.The video shows that F-15 at 187 m/s on the deck has potential (climb potential or specific excess power – see NASA paper) to climb at much higher rate than 187 m/s at 90° climb angle. The plane has enough specific excess power to accelerate through the speed of sound in the vertical climb.
Now, if we had Boing 747 performing the same thing at 187 m/s, situation would be much different 😉First of all, he is not accelerating through the speed of sound 25 seconds after pulling into the climb, he is reaching that speed 25 seconds after the brake release.
Really, let me think…the plane didn’t reach its max climb rate while flying horizontally at 450 mph…it had to actually go in to the climb, accelerate in the climb and only then it reached its max climb rate!
Wow, I think you are genius :rolleyes:
On the serious note, this only shows that the plane has potential (climb potential) to accelerate in the climb (55 and 90° climb angle) at that particular speed.
And what is the climb potential?
“The rate of change of the total energy per unit time for the airplane is a measure of the climb potential (specific excess power) of the vehicle“.
It is the same thing as with F-15 in the video I posted. At that speed (306m/s), Mig-29 has energy potential (climb potential) of additional 39m/s (positve SEP), which can be used for acceleration on a vertical climb. Record breaking F-15 demonstrates that perfectly (only the numbers are different).
Now, if we assume that EF2000 has its best climb rate of 315 m/s at 0,9M, (same speed as Mig-29), that meens that it has energy potential (climb potential) of additional 9 m/s (positve SEP), which can be used for acceleration on a vertical climb.But somehow, you with your perfect understanding of the subject concluded that EF2000 has almost TWO TIMES higher climb rate than Mig-29.
I stated that the angle of climb varies but is usually ~60deg for a fighter class aircraft. I have several times posted a link detailing how V optimum and alpha optimum are calculated, so please don’t try pretend you’re clever.
http://www.dept.aoe.vt.edu/~lutze/AOE3104/climb.pdf
You asked for climb rate, not SEP, you are again conflating the two. SEP != climb rate. The answer I provided was for the speed mentioned and that speed only. SEP is NOT Climb Rate!
It says, “after release”, that could mean brakes off or take off. Either way, it’s well known that the Streak Eagle was a non-standard aircraft. Stripped of nearly all avionics, even paint and had uprated thrust. TWR was over 2:1 on empty weight.
Yes, but sadly you are not a genius, which is why I’ve been trying to explain this very point to both you and Andraxxus for several pages. SEP at a given speed does not equal climb rate at that speed or maximum climb rate. It is a measure of the ability to gain speed and altitude. Unless Vsin(alpha optimum) equals climb rate then the SEP at that V does not equal climb rate.
Nope, you’re still not getting it. The SEP at 450mph was not all converted to climb rate, some was converted to velocity (KE). As per the following:
SEP = [dPE/dt + dKE/dt]/mg = dh/dt + (V/g)*(dV/dt)
As speed increased, SEP decreased, but the V was then sufficient to achieve the optimum climb rate and the remaining SEP could be converted to climb rate.
No because as the speed of the MiG increases, drag increases and the SEP, given by:
V*([T-D]/W)
Therefore reduces.
Seriously how can V*([T-D]/W) be the same at M0.9 as say M1.2?
The graph clearly shows the SEP at SL is zero at M1.2, so how can it climb at say a 55deg angle at the latter speed? And even if it did, it would only achieve 334m/s climb rate by Vsin(alpha)*.
(*I apologise for using alpha, it’s confusing, as it’s usually used to refer to AoA but here I use it as the angle of ascent.)
I never said it had twice the climb rate, I said it had nearly twice the SEP at a given speed. There is a difference, as I’ve been trying to point out.
Come back when you really understand this post and the subject matter, because it’s clear to all that you don’t.
Do you even understand this diagram?
First, MICA VL is not MICA and is vertically lauched. Dunno if CAMM is. Second, CAMM is not ASRAAM. Finally, you will not use (at all) the same flight profile in a ground to air interception an in a long range air to air intrception. I doubt trajectory of VL MICA and CAMM to be parabolic. But i may be wrong. For Justin Bronk, just read that : https://hushkit.net/2015/12/18/typhoon-versus-rafale-the-final-word/
PS you were right on one point, ranges stated by MBDA
It’s a MICA launched from the ground. Same weight, same size.
http://www.mbda-systems.com/ground-based-air-defence/vl-mica/
CAMM is also vertically launched.
He’s likely right, under demanding conditions against a non-submissive, alert target at 20,000ft, it would struggle beyond that range. MICA is a short range AAM by modern standards. AIM-120C has 105km range at 50,000ft but 40km range at 20,000ft. MICA is about 60% of that.
Sign In
