h = horizontal component of acceleration = g tan b

[L = lift vector]

The position vector for a circle is **r** = (R
cos q)**i **+ (R sin q)**j
**where
R is the radius of the circle and q is
the angle with the positive x-axis. If v_{a} is the angular velocity
in radians per second, then q = v_{a}t
where *t* is time, so

**r** = (R cos v_{a}t)**i** + (R sin
v_{a}t)**j **Differentiating, we get:

**v** = d**r**/dt = (-Rv_{a }sin v_{a}t)**i**
+ (Rv_{a }cos v_{a}t)**j **and

**a** = d**v**/dt = (-Rv_{a}^{2}cos
v_{a}t)**i** + (-Rv_{a}^{2}_{ }sin v_{a}t)**j
= **-Rv_{a}^{2}[(cos v_{a}t)**i** + (sin v_{a}t)**j**]

Now the length of this vector is Rv_{a}^{2}
= R·v_{a}· v_{a}

If v_{a}is in radians per second, the aircraft
velocity v is v_{a}·R, so v_{a}^{ }= v/R

Substituting for one of the v_{a}^{ }factors,
we have the length of the acceleration vector is

|**a**| = v_{a}·v

This must be equal to g tan b from the figure above. Thus we have

g tan b = v_{a}·v
so tan
b =

Now v_{a} is in radians per second, g and
v are in feet per second. We must apply some conversion factors here. Let
D be the turn rate in degrees per second and let s be the aircraft’s speed
in knots. Then
and (1
NM = 6076 ft) giving

It must be remembered when using a computer that the
arctan function gives results in radians, so the result must be multiplied
by after
the arctan is computed to get the bank angle in degrees. A 3^{o}
per second turn requires a bank angle of 15.4^{o} at 100 knots.