# How to calculate angular velocity and radius of a turn?

user1876
• How to calculate angular velocity and radius of a turn? user1876

How can I calculate with known tangents of different angles and by rules of thumb?

Which formulae should I use for fps and NM?

• Your terminology is a little confusing, but I'm going to assume you're asking how to calculate turn radius and rate of turn based on airspeed and bank angle. These formulas can all be found in the FAA's Pilot's Handbook of Aeronautical Knowledge which is available for free online.

The Handbook gives the formulas for rate of turn and turning radius on page 4-34:

$$R=\frac{V_2}{11.26\tan\theta}$$

$$\omega=\frac{1,091\tan\theta}{V}$$

The variables used are:

• $V$ = true airspeed in knots
• $R$ = turning radius in feet
• $\theta$ = bank angle in degrees
• $\omega$ = rate of turn in degrees per second

For example, at 120 knots and a 30° bank angle, the turn radius and rate of turn are:

$$R=\frac{120^2}{11.26\tan30}=\frac{14,400}{11.26\times0.5773}=2,215 feet\approx\frac13nautical~mile$$

$$\omega=\frac{1,091\tan30}{120}=\frac{1,091\times0.5773\tan30}{120}=5.25°/sec$$

The "magic constants" in these formulas ($11.26$ and $1,091$) are conversion factors for the units involved (knots, feet, and degrees). Physicists would use unitless formulas involving $g$, acceleration due to gravity (roughly $9.8m/sec^2$).

You can also rearrange the formulas above using simple algebra to figure out the required bank angle given a desired rate of turn or turn radius.

Finally, note that things get a lot more complicated if you factor in winds aloft. Rate of turn will always be the same regardless of wind, but turning radius no longer applies because the aircraft will trace a spiral path along the ground, not a circle. The turn will be "sharper" on the upwind portion of the turn and "wider" on the downwind portion. This is why turns around a point is a complex maneuver taught in basic flight training: in order to fly a circular ground track, the pilot must constantly vary the aircraft's bank angle according to wind: lower bank angle upwind, higher bank angle downwind. The pilot must also use the rudder correctly to keep the turn coordinated at all times.

There are also some "Related Questions" over on the right-hand side of this page that might be useful.

• If you're going to do this in a cockpit, a good rule of thumb will help more than an exact formula:

Bank angle for rate 1 turn is $\frac{speed}{10}+7$.

and

Turn diameter is 1% of the speed.

eg. for a 120kts turn you need $\frac{120}{10}+7=19°$ of bank and have a $\frac{120}{100}=1.2$nm turn diameter

• After all these answers with Imperial Units, let me explain it with SI units, starting from first principles. R is the radius, v the flight speed, m the mass, g the gravitational constant, Φ is the bank angle and L the lift.

The lift needs to be equal to the weight (m·g) and the centrifugal force (m·ω²·R = m·$\frac{v^2}{R}$), so

$$L = \sqrt{(m{\cdot}g)^2 + (m{\cdot}{\omega}^2{\cdot}R)} = \frac{\rho}2{\cdot}v^2{\cdot}c_L{\cdot}S$$

with ρ the air density, $c_L$ the lift coefficient and S the surface area of the wing. Now convert so you get v:

$$v = \sqrt[4]{\frac{(m{\cdot}g)^2}{{(\frac{\rho}2{\cdot}c_L{\cdot}S)^2} - (\frac{m}{R})^2}}$$

Now you can see that the nominator cannot become zero or less, which gives you the minimal radius for a given speed and maximum lift coefficient $c_{L max}$:

$$R ≥ \frac{2{\cdot}m}{\frac{\rho}2{\cdot}c_{L max}{\cdot}S},$$ and generally: $$R = \frac{2{\cdot}m}{\frac{\rho}2{\cdot}c_{L}{\cdot}S}$$

This is like a "radius barrier": Turns cannot be flown tighter than that. This is due to the increase in centrifugal force which comes from flying steeper turns. The steeper the turn, the faster you must fly to create enough lift to compensate both weight and centrifugal force.

What does still increase is your angular velocity ω:

$${\omega} = \frac{v}{R} = \frac{g{\cdot}tan{\Phi}}{v}$$

Below I have plotted them for a glider. You can clearly see the radius barrier at 40 m. Trust me, it looks just the same for an airliner, only the numbers are bigger.

Now for the other extreme: Hypersonic aircraft need lots of space for maneuvering. I have here some values, just for fun:

The high speed makes this almost tolerable, after all, a half turn at Mach 6 and 2 g takes only 336 seconds, that's under 6 minutes. Airliners bank to only 30° or less, so the first column is valid if you fly your hypersonic vehicle like an airliner.

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• How can I calculate with known tangents of different angles and by rules of thumb? Which formulae should I use for fps and NM?

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