Below are formulas for calculating airspeed, radius, rate, and bank angle with respect to turns. Simple formulas are presented which relate the items to each other, followed by examples. Each example demonstrates solving for a different unknown. The formulas have a conversion constant, k, which can be adjusted for the units which the reader desires by using the unit analysis following each formula.
Airspeed = Radius of Turn * Rate of Turn * k
k = .01034 or (1/96.7)
At 100 knots, a standard rate turn has a radius of
= 3223 feet, or ½ NM |
At 220 knots, an airplane turns about a point exactly 1 NM away. The rate of turn is
= 3.5º/second |
An airplane turns about a point 2 NM away at 3º/second. Its velocity is
(6076)(2)(3)(.01034) = 376.9 knots
, where g = gravitation acceleration due to gravity (this is included in k below). |
k = .088537 or (1/11.29)
At 180 knots, an airplane turns about a point exactly 1 NM away. The bank angle is
An airplane turns about a point 1 NM away with a constant 15º bank. Its airspeed is
= 135.6 knots |
At 120 knots, with a 30º bank, the airplane turns at a radius of
= 2208 feet |
By substituting the first formula into the second, we can relate turn rate, radius, and bank angle. However, keep in mind that a particular rate and radius of turn only occur at a particular airspeed. Any problem solved with this formula can be solved with both of the preceding ones.
tan(Angle of Bank) = Radius of Turn * Rate of Turn^2 * k / g(where g = gravitation acceleration due to gravity (this is included in k below))
k = .000009467 or (1/105,621.01)
In a standard rate turn with a constant 20º bank, any airplane turns with a radius of
= 4271.4 feet, or almost ¾ NM; |
however, a standard rate turn with 20º bank can only happen at an airspeed (using formula #1) of
(4271.4)(3)(.01034) = 132 knots
Turning about a point 1/3 NM from the airplane with a 10º bank will result in a turn rate of
= 3.03º/sec, a standard rate that is close enough for government work |
and, this will only happen at an airspeed (using formula #1) of
(6076/3)(3.03)(.01034) = 63.5 knots! Any more and either the radius or rate of turn (or both) would increase.
To maintain a standard rate turn around a point ½ NM away requires a bank angle of
= 14.5º |
and, we can find that this combination of radius and rate only happen at an airspeed (using formula #1) of
(6076/2)(3)(.01034) = 94 knots
Almost every commercial pilot has been introduced to this formula for determining pivotal altitude in eights-on-pylons:
Altitude
= Groundspeed^2 / 11.3
This is almost like our formula #2, except what happened to the bank angle? Consider the following:
An airplane, P, is turning about a point on the ground, G, at bank angle t.
Any physics handbook will contain the formula for the force of centripetal acceleration (F_{a}):
, or solving for R: | where W = weight; V = velocity, g = acceleration due to gravity, R = turn radius. |
(The text will have the formula in terms of mass, just remember m = W/g)
The plane is flying level, so the vertical component of lift equals the weight of the plane. Brushing off trigonometry, we know that F_{a}, the horizontal component of lift, will equal tan(t) * w. We also know that the the radius R of the circle flown equals the height / tan(t), by the same principle.
If we set these two equations for R equal, we get:
By canceling tan(t) and W, we’re left with the following basic formula, that works, regardless of the bank used:
To use the traditional units of feet and knots, we need to add a conversion constant:
k = .088537 or (1/11.29)
That's where the mysterious 11.3 originates.