Cool777 From , joined Dec 1969, posts, RR: Posted (10 years 7 months 2 weeks 5 days 16 hours ago) and read 1061 times:
I wondered about the forces that affect the airplane when the pilot performs the "rotation" and causes to the airplane to leave the ground. It's quite clear that when the nose of the airplane is raised, the angle of attack is increased as well, what causes to the lift to increase. But, another thing that happens is that the horizontal thrust of the engines is titled somewhat downward, what creates a component of vertical thrust. My question is, which one of the described forces is the main reason that causes to the airplane to take-off and start ascending ? Is it only the increase of the lift, or the angle of engine thrust change has some effect as well ? If it does, which percent of the total "take-off force" it generates ? Please refer to jet airplanes as well as light propeller driven airplanes.
Ikarus From United Kingdom, joined Jan 2001, 3524 posts, RR: 3 Reply 1, posted (10 years 7 months 2 weeks 5 days 15 hours ago) and read 1036 times:
On a large airliner, it is mostly the effect of the wing that lifts up the plane. The engine thrust is employed to speed up the plane until it reaches 1.1 or 1.2 times the stall speed, then it rotates and climbs away. The vertical component of the thrust is small compared to the lift of the wings.
For VTOLs, VSTOls and STOLs and military fighters things might be more dependent on the design, however.
Ralgha From United States of America, joined Nov 1999, 1614 posts, RR: 6 Reply 2, posted (10 years 7 months 2 weeks 5 days 9 hours ago) and read 1008 times:
During rotation, it's the increase in AoA that brings the plane off the ground. After it is stabalized in climb, it is the engine thrust that is causing the climb. Contrary to popular belief, a climb is not caused by excess lift, but rather excess thrust, defined as thurst beyond that which is required to maintain level flight in a given configuration.
Delta-flyer From United States of America, joined Jul 2001, 2676 posts, RR: 9 Reply 3, posted (10 years 7 months 2 weeks 5 days 6 hours ago) and read 1000 times:
The above are correct, but I just want to add the following.
It helps to visualize things by crunching some numbers. Let's use a B777-200 with maximum thrust of 2 x 77,000 = 154,000 lb. and MTOW of 545,000 lb.
At zero AoA the lift is zero, while at take-off AoA (say 5 degrees - I'm just guessing here) the lift is 545,000 lb. We know this because the plane takes off. So changing the AoA from zero to 5 deg. changes lift from zero to 545,000 lb.
The vertical component of the 154,000 lb. thrust at 5 degrees is only 154,000*sin(5 deg) = 13,450 lb. This is small potatoes compared to the 545,000 lb. needed to lift the plane off the ground.
When analyzing such problems, it is best to construct a free body diagram of the plane and draw the vectors of all forces acting on it. The weight is always vertically down, while lift is up at right angles to the wing chord, drag is in line with the wing chord and thrust is in line with the engine centerline. The horizontal and vertical components of all these forces will equal zero whenever there is no acceleration, whether the plane is climbing, descending or flying straight and level.
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