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 belive 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.


If the fuselage is at 90 degrees, the dihedral makes no difference as
the slight upward vector from the bottom wing is offset by the negative
vector from the upper wing and thus the wing lift is still zero for all
practical purposes. However, you get lift from the fuselage and also
from the thrust vector being inclined upwards as well if top rudder is
applied.

Matt