3
$\begingroup$

An EASA ATPL question/answer insists, counterintuitively, that a twin which loses an engine and must drift down will find its least rate of sink at Vxse. I've been trying to understand this, but haven't come up with an answer yet. Do Vy and Vx even exist as concepts when above the ceiling? Does "best rate of climb" become instead "least rate of descent"? Does "greatest distance between the power-required and power-available curves" become "least distance" between same? Would the speeds of Vx and Vy (with or without SE) change when going from below ceiling to above while at the same altitude and mass?

One way to determine Vx and Vy is to look at a ROC vs Speed chart, where ROC will be an inverted parabola. The peak of the parabola is Vy, the tangent of it to the origin is Vx. If the whole curve is shifted below the x-axis, however, the tangent will fall on the far side of the Vy peak. So I would expect Vx > Vy when above the ceiling. But that still doesn't help understand why EASA would say least sink is at Vxse.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

Vx increases with altitude while Vy decreases with altitude. When they coincide, you have reach the airplane’s absolute ceiling, which it can no longer climb and can only maintain altitude at that specific calibrated airspeed. Vx and Vy do exist above you maximum single engine ceiling, however, should an engine fail above the maximum single engine ceiling the airplane will begin to descend in order to maintain a given airspeed. Pitching the airplane for blue line, Vyse, and maintaining the correct bank angle in the direction of the good engine (3°-5° typ.) will result in the minimum descent rate during drift down.

I have no idea why EASA would state that the airplane should be flown at Vxse above the maximum single engine altitude to minimize drift down; every source of information I’ve ever encountered indicates your airspeed should be maintained at Vyse, in addition to maintaining proper coordinated flight with OEI, in order to minimize your drift down rate. This makes sense, since the airplane will have the most power available to it for climbing and will result in the least loss of altitude over time in this configuration vis a vis conservation of energy.

Improve this answer
$\endgroup$
5
  • $\begingroup$ this may be a play on minimum power for level flight vs Vy, but I definitely agree with you for 2 reasons: 1. Asymmetric flight is easier controlled at Vy. 2. Vy will get one to their safe landing field faster and more efficiently. The question may be a bit myopic. $\endgroup$
    – Robert DiGiovanni
    Apr 11, 2021 at 11:14
  • $\begingroup$ But is minimizing the drift-down rate the goal? Shouldn't it be minimizing the drift-down angle? $\endgroup$
    – Jan Hudec
    Apr 14, 2021 at 4:19
  • $\begingroup$ Why would you want to minimize the drift down angle? $\endgroup$
    – Carlo Felicione
    Apr 14, 2021 at 4:27
  • $\begingroup$ @CarloFelicione, because I want to land as soon as practical, and want to stay as high as possible on the profile in case there will be some additional problems. And shallower angle means being higher on the profile, while the flight is faster, so the landing is sooner. $\endgroup$
    – Jan Hudec
    Apr 14, 2021 at 5:02
  • $\begingroup$ If you want to remain as high as possible to conserve altitude, you’re going to want to descent at Vyse. That will result of the lowest possible loss of altitude per unit time, and keep you in the air as long as possible. $\endgroup$
    – Carlo Felicione
    Apr 14, 2021 at 17:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .