Airspeed and G at the Stall

---------------------------

The faster we go, the more G we can pull (or push) before the wing

reaches the stalling angle of attack (AOA). From the lift equation:

Lift = K x Coefficient of lift x Velocity Squared (1)

we can derive:

Vs(G) = Vs(1 G) x square root(G) (2)

Example: with a 1 G Vs of 60 mph, we can pull 4 G above 60 x sqrt(4)

or 120 mph. Slower than 120 mph, we will stall before we can develop

4 G's of lift. Above 120 mph, we can pull more than 4G.

An interesting application of the above is during a turn, or pullout

from a vertical downline. Intuitively we might want to develop maximum

lift, and this is obtained at the stalling angle where Clmax is obtained.

From first year university calculus we can derive:

Acceleration = Velocity Squared divided by Radius, or

G = V^^2 / R rearrange:

R = V^^2 / G (3)

This makes sense. With a constant velocity, as we increase G, radius

decreases. As you decrease G, radius increases.

Example: with a 1 G Vs of 60 mph, in a 120 mph vertical downline, we

can pull a maximum of 4 G at the stall AOA, as per (2) above.

Substituting into (3):

R = (120 x 5280 / 3600)^^2 / 4 x 32 feet per second per second

= 242 feet

Example: with a 1 G Vs of 60 mph, in a 180 mph downline, we can pull

a maximum of 9 G at the stall AOA, as per (2) above.

Substituting into (3):

R = (180 x 5280 / 3600)^^2 / 9 x 32 feet per second per second

= 242 feet

Isn't that interesting - and quite counter-intuitive! The radius of

a 120 mph maximum-performance turn is the same as the radius of a

180 mph maximum-performance turn.

Why? Because we have velocity squared on both the top and bottom

of (3), so they cancel each other out, giving us a constant radius,

as long as we are willing to operate at the stalling angle of attack,

which at high airspeeds can develop enormous G loads - perhaps more

than the pilot or aircraft can withstand.

There is one important detail we have omitted above, for simplicity -

the effect of earth's gravity. During the pullout, we are being

constantly accelerated downward at 32 foot per second squared. So,

the less time we take to fly the maneuver, the higher we will be

when it has completed.

Therefore the maximum allowable structural G at Clmax will result

in minimum loss of altitude.

--

aboyd 2005