2
$\begingroup$

Most people know the standard, "At a 60 degree bank, an aircraft experiences 2G's" checkpoint for load-factor on an airplane.

I assume that this statement has an underlying, unspoken assumption that the aircraft is maintaining a constant altitude in the bank.

This leads on to matters of increased stall speed during turns.

My question is, if no pitch adjustment is made while in a coordinated turn, which I think means that no additional load-factor is acting on the aircraft, is the stall speed still affected?

$\endgroup$
2
  • 2
    $\begingroup$ Note that to maintain 1G in a turn you must be not only descending but accelerating downward. $\endgroup$ Commented Apr 5, 2019 at 5:59
  • 1
    $\begingroup$ @pericynthion good point. $\endgroup$ Commented Apr 7, 2019 at 16:17

1 Answer 1

6
$\begingroup$

To fly a 60-degree bank coordinated turn with 1 G on the aircraft, you'll be dropping like a rock.

But your stall speed with 1 G on the aircraft is what it is, regardless of how you obtain that 1 G -- straight & level, or coordinated turn with deceleration down to reduce the G loading, or inverted & pulling 1 G... the angle of attack that stalls your wing will be reached at the same speed in all of those cases.

$\endgroup$
1
  • 2
    $\begingroup$ That aircraft is accelerating at 50% rate compared to the rock, because the fraction of the lift that remains to counter gravity is cos 60° = 1/2. Otherwise fully agree! $\endgroup$
    – bogl
    Commented Apr 5, 2019 at 9:22

You must log in to answer this question.

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