PPL-A (Canada) wrote:
Mxsmanic wrote:
d&tm writes:

I should add that this calulation assumes all the lift is coming from the
wing , but that theory would imply that an aircraft cant hold altitude in a
90 degree bank, and of course we have all seen aerobatic aircraft do this.

No, you haven't. It's impossible to hold altitude in a 90° bank. In
fact, it's impossible to execute a coordinated turn with a 90° bank.
A 90° bank requires infinite speed, because the acceleration vector
would have to be perpendicular to gravity, which is never possible as
long as gravity is non-zero. With both vertical and horizontal
non-zero components, the net acceleration vector can never be
completely horizontal or vertical. You can eliminate the non-zero
horizontal component in level flight, but you cannot eliminate the
force of gravity, so a 0° "bank" (i.e., level flight) is perfectly
possible, but a 90° bank is not.

You can come infinitely close to 90°, but you can never reach it, in
any type of aircraft. In an aircraft that can withstand 9 Gs, you can
reach slightly less than an 84° bank, but no more.

--
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I believe you have neglected to take into account that many aircraft
wings incorporate positive dihedral (as well as wash-out {although a
very few have wash-in}), which would have the effect, even while the
aircraft is in a 90º banked condition, of producing non-90º lift
vectors from the airfoils. The upward wing will still have a lift
vector that is not perpendicular to the weight vector, and depending on
the thrust available (or airspeed upon establishing the 90º banked
attitude), will allow a 90º banked turn for a period of time without
loss of altitude.


The upward wing will still have a lift
vector that is not perpendicular to the weight vector ...

Forgive me ... I meant to type the downward wing ... this wing will
have a lift vector that is not perpendicular to the weight vector ...

PPL-A (Canada)