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

Which formulae should I use for fps and NM?

  • 1
    $\begingroup$ What do you mean by "tangents of angles"? $\endgroup$
    – TypeIA
    Mar 29, 2014 at 14:32
  • $\begingroup$ math constant values $\endgroup$
    – user1876
    Apr 7, 2014 at 15:31

5 Answers 5


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:



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:

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


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.

Other useful links:

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

  • 6
    $\begingroup$ 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). $\endgroup$
    – Jan Hudec
    Mar 31, 2014 at 5:39
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    $\begingroup$ @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). $\endgroup$
    – TypeIA
    Mar 31, 2014 at 14:15
  • $\begingroup$ dvnrrs, thanks for your answer. You don`t use those 2 rules of thumb, do you? $\endgroup$
    – user1876
    Apr 7, 2014 at 15:32
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    $\begingroup$ @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. $\endgroup$
    – TypeIA
    Apr 7, 2014 at 19:51
  • 1
    $\begingroup$ DaG, I changed it. You are right. CN somebody explain to me where the0.5773 comes from in the 'turn rate' example? It appears out of nowhere. Or should the 'tan30' be removed as it is the answer of that part? $\endgroup$
    – Jelmer
    Mar 24, 2016 at 6:41

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. Airplane in banking flight

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:

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

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 $c_{L max}$:

$$R ≥ \frac{2{\cdot}m}{\rho\cdot c_{L max}{\cdot}S}$$ when neglecting weight and more precisely $$R ≥ \sqrt{\frac{m^2}{(\frac{\rho}2{\cdot}c_{L max}{\cdot}S)^2 - (\frac{m\cdot g}{v^2})^2}}$$ when we account for the fact that lift not only needs to counteract the centrifugal forces but also needs to keep the airplane up in the air.

This was for the minimum radius. For an arbitrary circle we only need to switch the maximum lift coefficient for the actual one, like this: $$R = \frac{m}{\sqrt{(\frac{\rho}2{\cdot}c_{L}{\cdot}S)^2 - (\frac{m\cdot g}{v^2})^2}} = \frac{v}{\omega} = \frac{v^2}{g\cdot\sqrt{n_z^2-1}}$$

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:

$${\omega} = \frac{v}{R} = \frac{g{\cdot}tan{\Phi}}{v} = \frac{g\cdot\sqrt{n_z^2-1}}{v}$$

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. plot for speed, nz and omega over radius

If you want a quick formula for estimating the radius, you need to use the square of airspeed, so this is not a simple linear relationship. For a turn with 30° bank ($n_z$ = 1.15), the denominator of the radius equation is about 4, so to calculate the turn radius in meters, divide the square of airspeed by 4 or take the square of half your airspeed in meters per second.

For the turn rate in degree per second, divide 220 by your airspeed in meters per second. Flying slower allows for a higher turn rate.

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

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.

  • 4
    $\begingroup$ 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) $\endgroup$
    – DeltaLima
    Apr 18, 2014 at 14:53
  • 1
    $\begingroup$ @DeltaLima: You are right, but I did not want to make it too complex ;-) $\endgroup$ Apr 18, 2014 at 18:26
  • $\begingroup$ @Peter, Wouldn't it be simpler (or at least equivalent) to just use Pythagoras, and note that in any turn, the aircraft total G (T) is the hypotenuse, the radial G (R) is the horizontal component, and One (1) G is the vertical component, so you can use the formula $R^2$ + 1 = $T^2$ and Turn Radius = $V^2$ / Radial G = $V^2$ / ($T^2$ - 1)? $\endgroup$ Apr 30, 2018 at 9:17
  • $\begingroup$ @CharlesBretana: The very first formula in the answer applies Pythagoras already. I expressed R in terms of $c_L$, but of course I can add more ways to calculate it. $\endgroup$ Apr 30, 2018 at 15:35
  • $\begingroup$ @Peter, Yes I saw that. Didn't mean to imply that your answer omitted or ignored that fact, just that it might be a simpler derivation to start with the Pythagorean formula expressed in G- units rather than lift forces. I was guessing that by doing that you might get to the result with fewer steps. No biggy though. $\endgroup$ Apr 30, 2018 at 23:57

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 $\frac{speed}{10}+7$.


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

  • $\begingroup$ Out of curiosity have you ever had occasion to need to do this in the cockpit? I've always wondered, and never have myself. $\endgroup$
    – TypeIA
    Mar 29, 2014 at 18:41
  • 2
    $\begingroup$ 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... $\endgroup$
    – Radu094
    Mar 29, 2014 at 19:26

Another approach is to just note that in any level turn, the relationship between total aircraft G ($G_T$), radial G ($G_R$), and God's G must conform to Pythagoras.

$G_T^2 = G_R^2 + 1$
$G_R = \sqrt{G_T^2 - 1}$

and since turn radius is Velocity squared over radial G, $$R = \frac{V^2}{\sqrt{G_T^2 - 1}}$$

total aircraft G, of course is just the Lift divided by the aircraft weight. and if we are below maneuvering speed, (lowest airspeed at which we can generate placard G-Limit load factor) and turning at the maximum Angle of Attack (AOA), then aircraft Lift is $C_{Lmax} p V^2 S$ and the weight, of course, can be represented by the Lift generated at at $C_Lmax$ when at Stall speed or $C_{Lmax} p V_s^2 S$.

So total aircraft G, ($G_T$), which is simply Lift divided by Weight, can be represented as $\frac{C_{Lmax} p V^2 S}{C_{Lmax} p V_s^2 S}$ or $\frac{V^2}{V_S^2}$

Substituting into the turn radius formula, we get the turn radius formula for a level max performance turn, (BELOW MANEUVERING SPEED), expressed as a function of aircraft True airspeed and Aircraft stall speed (in True):

$$R = \frac{V^2 V_s^2}{g (\sqrt{V^4-V_s^4})}$$


  1. $R$ .... Turn radius
  2. $V$.... Aircraft true air speed
  3. $V_S$... Stall speed (TAS)
  4. $g$ .... 32.2 $ft/sec^2$ (necessary to convert from Earth - G units to $ft/sec^2$

Graphing, it looks like the below: This is for an aircraft with a stall speed of 58 kt (true) and a placard G-limit of 3.8 Gs. (The kink at 122 Kt is due to the fact that once we are faster than maneuvering speed, we are limited by placard G and can no longer reach $C_{Lmax}$ without breaking or over stressing the airframe.)

enter image description here

  • 1
    $\begingroup$ "God's g?"-- this question could be improved by explaining or deleting this $\endgroup$ Mar 26, 2020 at 13:53

A simple rule of thumb for turn radius for a standard-rate turn is to divide airspeed by 180. For example at 90 kts it is .5 nm, and at 120 kts it is .66 nm. The turn radius is useful to decide when to lead a turn when approaching a fly-by fix. So for example at 90 kts, when .5 nm from the fix, begin a 90 degree turn. This is calculated by considering the speed of 90 nm per hour = 90/60 nm per minute, for 2 minutes for a standard rate turn, gives a 3nm circumference circle, which is approximately pi, so the diameter of that circle is approximately 1, and the radius is 0.5 nm. Substituting x for 90 in the original formula gives pi*2x/60 approximately equals x/180. Note that this does not account for wind or true airspeed.


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