Aircrafts Staying Level in Upside Down Flight: How Is It Possible?

  • Thread starter Thread starter gunblaze
  • Start date Start date
AI Thread Summary
Some airplanes can maintain a level flight even when inverted by angling the plane to ensure the wings generate lift through a high angle of attack. The engines also contribute by pulling upward during this maneuver. Although the fuselage may appear horizontal, the wings still operate at an angle of attack greater than zero. Stunt planes typically feature symmetrical wing sections, allowing them to perform effectively in inverted flight. This combination of techniques enables sustained flight in an upside-down position.
gunblaze
Messages
187
Reaction score
0
How can some airplanes sustain their flight level even when they are upside down?
 
Physics news on Phys.org
Poorly. :)

But it can be done. They are not really in level flight, they angle the plane high (or low?) so that the wing - even inverted - provides lift due to a high angle of attack. The engine is also pulling upward.
 
Along the same lines, the fuselage may appear to be horizontal, but the wing still has an angle of attack greater than 0.
 
Stunt planes often have symmetrical wing sections. In this case, the wing itself will perform just as well upside down.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top