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The question itself is open ended but, in the simplest terms, it depends on how you execute the turn - are you flying a constant airspeed or a constant angle of attack? In either instance, in a level turn, the load factor will increase with the inverse cosine of the bank angle. If flying a constant airspeed turn, then an increase in lift coefficient will ...


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After some thought, yes, this is good question! The answer, from the pilots perspective, is to use all tools at their disposal to land safely, first and foremost. First, know how much altitude your plane loses in a 360 degree turn. Use this to determine a plan ahead of time for emergency response, for example, in a 172, that loses 750 feet in a 360 degree ...


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You would want to turn into wind for two reasons: The wind will push you away from the runway centerline, if you turn away from the wind then you will get pushed farther off the centerline then if you turned towards the wind. Turning to the wind means you will be in a better position to make the runway once the turn is completed, then you can let the wind ...


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If your objective is to make a 180 and land on the runway you took off from, always turn into wind. Less of a reverse turn to line up will be required the stronger the crosswind is, and I can't see any case where a downwind turn would make that any better since you will always end up farther off the center line than if you turn upwind, unless the crosswind ...


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Your questions are fair and there is a unique optimal response (or family of responses) to each question. It will be quite lengthy to detail all the variations, but I can at least answer some of them. Goal state First, we have to define what we wish to accomplish. Is the goal to be over the approach end of the runway at a reasonable height or to be down and ...


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Instead of confusing yourself with crosswind and relative velocity, imagine a parallel runway you want to land on. The answer is then obvious. You always turn in the direction the runway is located In the case it is close by, it just makes the required teardrop slightly smaller, say from turning 210° to only 190° before turning back to line up. In case it is ...


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No. Total vertical force implies the sum of all vertical forces. In a steady climb, the sum of all vertical forces must be zero, or it wouldn't be a steady climb. I think what you really want to ask is whether all upwards-pointing forces are higher than weight, and then the answer is yes, because drag adds a downwards pointing component which also needs to ...


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Equilibrium means all forces sum to zero. Per Newton's law $\sum F=ma$ this means that the total acceleration is also zero. In a steady turn, the flightpath is constantly changing from a straight line. There is a centripetal acceleration that is nonzero: a continuous acceleration perpendicular to the direction of movement. Working through Newton's law the ...


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It appears that you are assuming that there is a human pilot (or autopilot) in the loop, making pitch inputs as needed to prevent the aircraft from accelerating up or down as the bank angle is changed. (You could also model what the aircraft tended to do on its own if the aircraft entered a bank but the elevator position remained constant. This would be a ...


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Yes. Vertical forces are forces in the earth axes reference frame. So remaining in this reference frame: the aircraft is climbing steadily; it has a vertical speed component, which causes a vertical component of aerodynamic drag $D*sin \gamma$; and since F=m*a, at constant velocity there must be an equal opposite force pointing upwards. Notes: Upwards ...


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It's an axes definition issue, it is important to realise that altitude and weight are earth axis variables. Centripetal force is provided by lift, not by gravity. Downward gravity force does not reduce when the aeroplane takes on a different attitude, it always pulls downward to earth with magnitude $m*g$. in the statement "..angle of attack should be ...


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It seems you forgot to add centrifugal acceleration. Lift must increase to produce the centripetal force that is needed for turning. I suggest you start with the desired turn rate and determine the amount of sideways lift from there. This in relation to the amount of lift for compensating gravitational weight will give you the bank angle. Now it is helpful ...


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In a tricycle-gear plane that sits in a level attitude until the aircraft is rotated for liftoff, the following is true during the portion of the takeoff run that precedes rotation-- Engine torque does not create a turning tendency. In general, engine torque creates a rolling (banking) tendency, but the aircraft can't roll (bank) when the wheels are on the ...


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Something about the Tibbets statement does not ring true. I have talked to dozens of B-29 pilots and not one of them mentioned having to do either of those things during takeoff: stepping on the brakes or advancing the left engines first. Stepping on the brakes seems like an especially horrible idea. The crews were flying overloaded planes that needed every ...


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Yes, thrust force vector must have a force equal to (but opposite) the combined gravitational force and aerodynamic drag force (from velocity) to climb at constant speed. In horizontal flight, thrust equals drag to fly at constant speed. But some of the drag comes from the wing creating lift. Since the larger wing moves "a lot of air a little", ...


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Imagine a line extending up from the center of the wing perpendicular to the chord line of the wing. This is the lift vector. It points more-or-less straight up in level flight. In this scenario, 100% of the aircraft's lift is pointing up. If we roll into a turn, now some of that lift is pointing sideways- that pulls the aircraft sideways, which pulls it ...


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Interesting example. MY INTERPRETATION OF THE QUESTION [NEW] Here is an image from GlobalSecurity.org which I have modified to illustrate my interpretation of the question. The angle of attack of the wing is exaggerated in the original image. My interpretation is that the question involves only vertical movement of the wing - e.g. along the vertical red ...


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