BANK ANGLE FOR AIRCRAFT TURN

  h = horizontal component of acceleration = g tan b
  [L = lift vector, b is bank angle, the angle at the top of the triangle]

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_at where t is time, so

r = (R cos v_at)i + (R sin v_at)j     Differentiating, we get:

v = dr/dt = (-Rv_a sin v_at)i + (Rv_a cos v_at)j and

a = dv/dt = (-Rv_a²cos v_at)i + (-Rv_a² sin v_at)j = -Rv_a²[(cos v_at)i + (sin v_at)j]

Now the length of this vector is Rv_a² = R·v_a· v_a

If v_a is in radians per second, the aircraft's linear 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, v is in feet per second, and g is 32 feet per second ². 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.
 
 
 
 

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Date created: 3/10/01