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This might be a dumb question. And I could probably work it out myself. But here goes! Assume a plane is flying along at some velocity $v_0$ at angle-of-attack $a_0$, using throttle setting $tr_0$. Forces are balanced. At time $t=0$ the pilot gives input to the elevators to pitch the aircraft up, changing nothing else. What happens to the aircraft as a result (in terms of its position and velocity)?

Note that this differs from the traditional "climb" free-body diagram because we are not necessarily in equilibrium.

In terms of just instantaneous forces the second after the pitch changes:

  1. The angle of attack is now increased. Therefore more lift is generated.
  2. However, now some weight is added to drag component, and since the pilot has not increased the thrust from the engine, the plane experiences a net force downwards and backwards.

But not sure how to add all these up to get a time course of position and velocities.

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    $\begingroup$ Does the elevator remain fixed or go back to an equilibrated state? Nothing else is touched? $\endgroup$
    – sophit
    Commented Apr 10, 2023 at 9:32
  • $\begingroup$ It's helpful to keep in mind-- any net force acting perpendicular to instantaneous direction of flight path will curve ("bend") the flight path up or down. And any net force acting parallel to instantaneous direction of flight path will cause airspeed to increase or decrease. With this in mind, it should not be hard to go through each step in the process. Obviously, any upward or downward bend in the flight path changes the magnitude of the tangential and perpendicular components of the weight vector... $\endgroup$ Commented Apr 10, 2023 at 12:58
  • $\begingroup$ What was your own initial Answer, and why did you not trust that? $\endgroup$ Commented Apr 11, 2023 at 21:12

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A step increase in elevator (which is then held) will first upset the pitching moment equilibrium of the aircraft -- the pitching moment will be non-zero. It will pitch nose up, increasing angle of attack. However, this increase in alpha will over-shoot the new equilibrium trim point and the short-period oscillation will occur. This will damp out quickly (on the order of a few seconds).

At that point, the aircraft will be flying at a new alpha and CL, but will not yet be at the equilibrium velocity for that CL. The aircraft will be in a slight climb, speed will quickly bleed off while it gains potential energy. This will over-shoot the new equilibrium velocity and the phugoid oscillation will occur. The aircraft will trade kinetic for potential energy until drag damps out the excess energy and a new equilibrium velocity corresponding to the lift coefficient is achieved. The phugoid will damp out over the next 30sec to several minutes. Aircraft with excellent L/D will take longer for the phugoid to damp.

Since nose-up elevator was input, this new CL will be higher than CL0, and the equilibrium airspeed will be lower. The drag coefficient will be lower, but whether the drag (or power) is less depends on whether the initial equilibrium point was above or below the point of minimum drag/power i.e. were you on the front or back side of the power curve?

Here, we will approximate propeller powered aircraft as having approximately constant propulsive power with velocity -- and we will work in terms of the power curve. We would approximate jet powered aircraft as having approximately constant thrust with velocity -- and we would work in terms of the drag curve.

If you are on the front side of the curve, your increase in CL will drop Velocity and will increase CD but will lead to a reduction in drag/power required. At constant throttle, the aircraft will start to climb (as thrust is now greater than drag).

However, if you are on the back side of the curve, your increase in CL will drop velocity and will increase CD -- but drag/power required will increase. At constant throttle, the aircraft will start to descend (as thrust is now less than drag).

If, instead of holding the new elevator input constant, you were to simply pulse the elevator and let it return to its neutral position (assuming you were flying stick-free trimmed at the start).

The initial reaction would be similar -- the short-period would be excited, but would quickly damp out -- this would excite the phugoid, which would take longer to damp. At the end of the phugoid, you would return to essentially your initial velocity in steady level flight. Your altitude might be slightly lower or higher than your initial altitude. You would not be in a significant climb or descent -- you should return to your initial equilibrium condition.

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The increase in angle of attack increases the lift and drag. The airplane will begin to climb. If the elevator control is small enough, and the pilot continues to hold it in without adjusting, the airplane will begin to oscillate around a new equilibrium in airspeed and climb rate. That is to say, the climb rate will increase, which slows the airspeed. The slower airspeed reduces the climb rate, which allows the airspeed to increase again, and the cycle starts again.

This repetitive cycle of cyclically varying pitch and airspeed is called "phugoid oscillations." Airplanes are intentionally designed to dampen phugoid oscillations, so the airplane will eventually reach a new equilibrium, corresponding to the traditional "climb" free-body diagram you are familiar with (or descent, depending on airspeed). How quickly it reaches the new equilibrium will depend on the plane.

Of course, if the elevator control is too large, the airspeed will continue to decay and the airplane will eventually stall.

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    $\begingroup$ This assumes the pilot holds the elevator input, right? If he makes the input & then releases it, everything should damp out at about the same equilibrium conditions (maybe slightly higher altitude) as before, yes? $\endgroup$
    – Ralph J
    Commented Apr 10, 2023 at 1:44
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    $\begingroup$ @RalphJ Yes, my assumption here is that the pilot holds the input. $\endgroup$
    – Chris
    Commented Apr 10, 2023 at 3:52
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No weight (mass) is added. Assuming the elevator deflection is constant, this is what will happen.

  1. AOA will increase, causing an increase in Lift and Drag.
  2. The increase in Lift, (greater than the weight), will cause the aircraft pitch flight path to increase, (it will begin to climb).
  3. As pitch attitude continues to increase, (Climbing at increasing climb angle), the airspeed will decrease.
  4. As airspeed decreases, the Lift will decrease.
  5. When the aircraft, (climbing), has slowed sufficiently to the point where the Lift at its reduced airspeed is again equivilent to the weight, the aircraft climb angle will stop increasing.
  6. The aircraft, still climbing, will slow even more, and lift will decrease even more, below the weight.
  7. Lift now less than weight the climb angle will begin to decrease.
  8. The airspeed will continue to decrease as the nose, (still climbing), pitches down.
  9. When the descending nose reaches that pitch attitude necessary to maintain level flight, the airspeed will be below that necessary to generate 1 G of Lift, and the nose will continue to fall below level flight.
  10. Nose below the horizon, the aircraft will descend, and airspeed will begin to increase.

Cycle will repeat, with smaller and smaller oscillations, until it stabilizes at a slower than original airspeed, (since you did not increase power, and the higher AOA will have increased induced drag). If you started this at an airspeed above L/D max, you will probably be in a slight climb. If you were below L/D max, you will probably be in a slight descent.

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