Turn radius for a 25° bank angle = (nm/min)^{2} / 9

Computing the turn radius of your airplane under various conditions is a fundamental building block for much of instrument flight. You will use it for many of the concepts that follow. The formula is complicated; the rule of thumb is easy.

I first learned about this rule of thumb at the Air Force Instrument Instructor's Course (AFIIC) where we were told it is derived from the 60-to-1 concept. The rule taught at AFIIC was slightly different: your turn radius is equal to the (nm/min)^{2} / 10. I thought that was inaccurate and used (nm/min) / 3. But then I realized much of this was based on a standard rate turn but we can't do that at higher speeds because we want to keep our bank angle under 25° as instrument pilots. So it was back to the drawing board and I came up with the rule of thumb presented here. But first, we'll define the rule, provide an example of its application, and then we'll provide a mathematical proof of the rule.

Turn radius for a 25° bank angle = (nm/min)^{2} / 9

Note: dividing by 9 can be a pain, in most cases dividing by 10 is close enough.

Figure: Circle 90° to runway, from Eddie's notes.

Let's say you want to determine the minimum runway displacement for a downwind when circling to an unfamiliar runway. Knowing your turn radius will make this easy.

If you were to shoot this approach at 150 knots the math isn't so easy. So let's try 120 knots first. Then the math comes to:

$turn\; radius=\frac{{2}^{2}}{9}=\frac{4}{9}$or about a half mile, so your turn diameter is about a mile.

Now let's try 180 knots:

$turn\; radius=\frac{{3}^{2}}{9}=\frac{9}{9}$That's a mile, so your turn diameter is about two miles.

So if you are circling at 150 knots, split the difference and you come up with a diameter of 1-1/2 miles. (If you did the math at 150 knots it would have come to 1.39 miles, so we are pretty close.

Figure: Turning flight, from Eddie's notes.

- In coordinated turning flight, we know the centrifugal force pulling the airplane away from the turn is equal to the mass of the airplane times the acceleration force.
- We also know that the aerodynamic force, lift, resolves to two components that oppose the weight vertically and the centrifugal force horizontally.
- The three forces, lift, weight, and centrifugal force can be drawn as a triangle with the acute angle equal to the bank angle θ of the airplane.
- The tangent of the bank angle θ is equal to the centrifugal force CF divided by the weight W.
- The weight of the airplane W is equal to its mass m times the gravitational constant g
- Substituting:
- By definition acceleration a is equal to the velocity divided by time:
- Substituting:
- Multiplying both sides of the equation by gt makes things a bit neater:
- By definition velocity V is equal to the distance divided by time, in this case the distance is equal to the turn radius, r:
- Solving for t:
- Substituting t in #11 into the equation in #9 give us:
- Multiply both sides of the equation by V:
- Solving for turn radius, r:
- The gravitational constant, g, varies from the equator to the poles and whom you ask. For our purposes, we'll say it is equal to 32 ft/sec
^{2}. We would like to have that in nautical miles per minute^{2}, which comes to (32 ft/sec^{2}) (1 nm / 6076 ft) (3600 sec^{2}/ 1 min^{2}) = 18.96 nm/min^{2}: - If we assume a 25° bank turn, θ = 25, we get:

$\mathrm{CF}=ma$

This comes from Newton's Second Law of Motion, F=ma. More about this: Basic Aerodynamics / Mechanics / Newton's Second Law.

$tan\left(\theta \right)=\frac{\mathrm{CF}}{W}$

$W=mg$

This from Galileo's work with dropping various objects and hypothesizing about gravity. More about this: Basic Aerodynamics / Mechanics / Falling.

$tan\left(\theta \right)=\frac{ma}{mg}=\frac{a}{g}$

$a=\frac{V}{t}$

$tan\left(\theta \right)=\frac{V}{gt}$

$gttan\left(\theta \right)=V$

$V=\frac{r}{t}$

$t=\frac{r}{V}$

$g\left(\frac{r}{V}\right)tan\left(\theta \right)=V$

${V}^{2}=grtan\left(\theta \right)$

$r=\frac{{V}^{2}}{gtan\left(\theta \right)}$

$r=\frac{{V}^{2}}{18.96tan\left(\theta \right)}$

$r=\frac{{V}^{2}}{8.84}$

An engineer would call that bottom term 8.84. A pilot says 9 because the result doesn't hinge on the 0.16 difference. (Just 2 percent.)

Turn radius for a 25° bank angle = (nm/min)^{2} / 9

The Instrument school taught (nm/min)^{2}/10 which is close, and easy to do while flying an airplane. But the formula we've derived here is closer and if you precompute a few radii for various speeds, just as easy. For example:

- At 120 knots and 25° of bank: turn radius = (2)
^{2}/9 = 0.44 nm. - At 150 knots and 25° of bank: turn radius = (2.5)
^{2}/9 = 0.69 nm. - At 180 knots and 25° of bank: turn radius = (3)
^{2}/9 = 1.00 nm

For more about this rule of thumb, see: Basic Aerodynamics / Turn Performance.

Portions of this page can be found in the book Flight Lessons 1: Basic Flight, Chapter 24.

So what about the claims this rule of thumb is based on 60-to-1? My conclusion: No, it has more to do with trigonometry.

Rule of Thumb | 60-to-1? | Trigonometry | π |

Turn Radius | ✓ |

60-to-1 — Rule of thumb is based on the mathematical relationship of a 360° circle and/or 6076' to 1 nm.

Trigonometry — Rule of thumb is based on the relationship to a right angle and the derived trigonometric functions.

π — Rule of thumb is based on the relationship of a 360° circle, the number π, and/or 6076' to 1 nm.