If a jet is in equilibium while making a turn/loop – ie the drag + gravity is fully counterbalanced by thrust and lift and the jet is maintaining a constant turn profile. The turn will be circular in shape. For this to be the case that means the jets alpha is irrelavent – the wing will always point to the centre of the turn as it produces the force that turns the jet. Therefore the jet engines which are 90 degrees to the wing will always instantaneously be parallel to the velocity vector – which is the tangent of the circular path taken by the jet.
I was actually referring to the angle (a’) between thrust vector (t) and its velocity component (v).
According to Greek defence media websites, the first two EMB-145H Erieye have been officially accepted by HAF. Perhaps we will see them for the first time with the Greek AF roundels in a few days at Tanagra AB during HAF’s Annual Celebration (12-14 Sep. 2008)
today’s photos
What I’m getting at here about the 4.5g turn is purely a theoretical concept. You know those, uhm… “internal dampeners” from Star Trek…?
Lets say the plane is equipped with a device that halves the g-load an aircraft experiences by half. I know it’s completely impossible, since inertia is a core element of basic existence as we know it, but if it WERE possible (A wizard did it; there’s your explanation.) How would that work? A 9g equivalent turn, but the aircraft itself only experiences 4.5g due to that impossible technology. In other words, the planes doesn’t experience as much as a load factor as it normally should. Do you understand what I’m getting at here?
Let me try to answer…:confused::confused::confused::confused:
An impossible technology question, will get you an impossible answer.
There is absolutely no logical challenge in answering this type of questions, so there is absolutely no point in making them:diablo:
If you could turn without experiencing g-forces, why move at all.
😮 Beam yourself 😮
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