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  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 va is the angular velocity in radians per second, then q = vat where t is time, so

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

v = dr/dt = (-Rva sin vat)i + (Rva cos vat)j and

a = dv/dt = (-Rva2cos vat)i + (-Rva2 sin vat)j = -Rva2[(cos vat)i + (sin vat)j]

Now the length of this vector is Rva2 = R·va· va

If va is in radians per second, the aircraft's linear velocity v is va·R, so va = v/R

Substituting for one of the va factors, we have the length of the acceleration vector is

|a| = va·v

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

g tan b = va·v so tan b

Now va 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 3o per second turn requires a bank angle of 15.4o at 100 knots.

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Date created: 3/10/01

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