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

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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What do you mean by "tangents of angles"? – dvnrrs Mar 29 '14 at 14:32
math constant values – user1876 Apr 7 '14 at 15:31

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.

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Physicist would never ever use a unitless formula. They are very careful with their units. They would, however, use a consistent unit system (of which SI is the most widely used, but consistent unit systems based on US customary units exist too). – Jan Hudec Mar 31 '14 at 5:39
@JanHudec I think you misunderstood what I meant by "unitless" -- I meant that the formula would contain no unit conversion terms and would be unit-agnostic (you can mix and match units all you want as long as you do the proper dimensional analysis to figure out what units are coming out as a result of the inputs). – dvnrrs Mar 31 '14 at 14:15
dvnrrs, thanks for your answer. You don`t use those 2 rules of thumb, do you? – user1876 Apr 7 '14 at 15:32
@user1876 I don't do any of this in the cockpit. I use the turn coordinator, and I've also flown enough hours that I just know roughly what the bank angle should be at the airspeeds I fly. If you don't have an easy way to see standard-rate in the cockpit, in my opinion this calculation should be something you do on the ground, and just memorize the values. Save your brainpower in the cockpit for something else, like keeping up a good instrument scan. – dvnrrs Apr 7 '14 at 19:51
I can't directly edit the answer, but in the first formula it should be V^2 rather than V_2. – DaG Feb 8 at 11:51

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

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Out of curiosity have you ever had occasion to need to do this in the cockpit? I've always wondered, and never have myself. – dvnrrs Mar 29 '14 at 18:41
Bank angle maybe a couple of times in a G1000 where a rate 1 is kinda' hard to see as there is no proper turn coordinator so I'd rather aim for a specific bank and hold it. Turn diameter not so much... – Radu094 Mar 29 '14 at 19:26
Thanks Radu094 .. – user1876 Apr 7 '14 at 15:33

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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Finally a proper SI answer with a bit of background to it! There is no need to get the lift coefficient involved though, don't make it too complex. But if you want to do complex then consider that at high speeds, your apparent weight is less. The turn radii you have calculated are too low for Mach 6 (effectively the weight is about 10% less, so the radius is about 10% bigger if travelling eastwards at Mach 6 above the equator) – DeltaLima Apr 18 '14 at 14:53
@DeltaLima: You are right, but I did not want to make it too complex ;-) – Peter Kämpf Apr 18 '14 at 18:26