posted on Feb, 7 2015 @ 02:43 PM
"Is it possible flying supersonic with this kind of triangle shape ?"
Yes, I think so, based on the wing sweep angle. I have no inside information on this particular program, but from standard aeronautical engineering
principles, there is a relationship between the wing sweep angle and the top speed the aircraft is capable of. I won't go into all the mathematical
derivations, but the basic principle comes from the idea that when an aircraft is flying along at some Mach number, the nose tip of the aircraft
represents a pressure disturbance source that is sending out pressure waves in all directions. The pressure waves propagate outward at the speed of
sound (Mach 1) but they are swept backward at whatever the flight Mach number is. Collectively, the pressure waves form a conical surface attached to
the aircraft nose, where the cone angle gets shallower and shallower as the flight Mach number increases. (For a mathematical discussion, just
Google on "Mach angle").
Standard design practice is to keep the wings of the aircraft entirely within the Mach cone, so you don't have shock waves cutting across the lifting
surfaces. At a flight speed of Mach 1, the Mach angle is exactly 45 degrees. Look at the wing sweep of a B-2, and you will see that the wing sweep
is 45 degrees. (The included angle between both wing leading edges is 90 degrees.) This allows the B-2 to fly very close to Mach 1, while keeping
all parts of the aircraft within the Mach cone.
A flight speed of Mach 2 would create a Mach angle of 30 degrees, so the wing sweep would have to be at least 60 degrees. That's one reason the old
F-102s and F-106s used classic 60 degree delta wings; it was aerodynamically the simplest way to get Mach 2 aerodynamics. To my eye, the wing sweep
of the mystery Texas aircraft is definitely more than 45 degrees and could be as high as 60 degrees. My guess it is a super-cruiser, intended to
fly somewhere between Mach 1 and Mach 2. Probably intended to operate in concert with the F-22.