Is the turn radius at fixed bank angle really independent of aircraft type?

I'm curious about the relationship between an aircraft's turn radius and the bank angle. This formula describes the relationship between the two:

$$R = \frac{v^2}{g \tan(b)}$$

Where $$R$$ is the turn radius, $$v$$ is the speed, $$g$$ is acceleration due to gravity and $$b$$ is the bank angle.

This formula is found dotted around the internet (with variations depending on units used) examples include here and here.

My question is, why is there no reference in the formula to the physical characteristics of the aircraft? Does the relationship between turn radius and angle-of-bank really have nothing to do with aircraft weight, wingspan or aerofoil design? Does the same formula really hold true for both glider (sailplane) and large passenger jet?

Hoping for an explanation with no maths please!

• Oct 12 '20 at 15:34

If the turn is fully coordinated, turn radius at a given bank angle will depend on the speed of the aircraft and the local value of gravity.

This is because of simple vector and geometry: the vertical component of lift must still support the weight of the aircraft (whether that's a hang glider or an AN-225, a Helio Courier on the edge of stall or an SR-71), but the horizontal component, generated by the bank, produces a horizontal (=> centripetal) acceleration that is fixed in relation to gravity for a given bank angle.

The problem is that this centripetal acceleration is determined by both turn radius and speed. So, if your hang glider and your SR-71 want to make the same radius turn (within their usual flight envelope), they'll do it at wildly different bank angles.

• Wrong. It has nothing to do with the "local value of gravity" the g in the equation is not gravity, it is the radial acceleration - G-radial, or Gr. The actual formula is V2/Gr. The formula in the question uses bank angle because g tan(b), where b is the bank angle is a formula for G-Radial (the acceleration of the aircraft in the direction of the turn). And your hang glider and your SR-71 if they were flying in formation (If they could both fly at the same speed) would need to be at exactly the same bank angle to remain in formation with one another. Oct 14 '20 at 2:08
• @CharlesBretana Better draw your force diagram again. The vertical component of lift will always be the local gravity * mass of the aircraft (taken as fixed; we discount fuel or stores expenditure burn for this). Therefore centripetal acceleration at a given bank angle depends on local gravity -- if your local gravity is lower (as for instance flying on Mars), you'll have less horizontal acceleration for a given bank angle than you'd have on Earth. Oct 14 '20 at 11:06
• Yes, my comment is perhaps slightly misstated. What I was trying to say is that implying that the local force of gravity affects turn radius in a horizontal turn is incorrect. The formula, using radial G, (Gr), instead of g Tan(b), always applies, for anything on a curved path. The g in the g Tan(b) expression is there because Gr = g Tan(b), (a shortcut to calculate radial G from bank angle), not because the local force of gravity is relevant. Indeed, in uncoordinated flight, or if some other force is augmenting the turn, (or on another planet), it would be wrong. Using V2/Gr will always work. Oct 14 '20 at 13:59
• The force diagram is also irrelevant, as it only serves to justify the use of the shortcut to calculate radial G from bank angle. It has no relevance to the calculation of the curvature (radius), of the path traversed by a moving body. That is determined by velocity (squared) divided by radial G, - i.e., the normal acceleration (component of acceleration in the radial direction). Oct 14 '20 at 14:05
• Assuming your turn doesn't involve vertical acceleration (it shouldn't, if correctly coordinated), local gravity is still in the mix. You vertical component must still be the same as local gravity, which constrains the horizontal to be dependent on bank angle. Draw your force diagram until you understand what you're arguing. Oct 14 '20 at 14:24

Why is there no reference in the formula to the physical characteristics of the aircraft?

In level steady state flight Weight = Total Lift = Vertical Lift. The ratio of Total Lift to Vertical lift = 1. In a bank more lift must be generated to maintain sufficient vertical lift. The ratio of Vertical Lift to Horizontal Lift is Tangent (bank angle). The horizontal lift component is what changes the direction of the flight path. The Force moving it in a new direction is mass x acceleration. This is always proportional the lifting force regardless of weight.

As lateral motion accelerates from zero and direction is always changing, the side drag of the aircraft is not nearly as significant as the sideways acceleration compared to the energy of forward motion.

Here the "physical characteristics of the aircraft" are accounted for in Velocity: the sailplane has a much tighter turning radius than the jet airliner because of a much lower stall stall speed. This is where airfoil type and wing loading come into play.

The key to understanding this is to put mass "m" back into the equation:

R = mv$$^2$$/mg (tan bank angle)

We can now clearly see the energy of the object is proportional to the square of the velocity. Weight (therefor lift) is mg. One simply gets a slower rate of turn with a faster moving object for a given amount of sideways acceleration.

Now, if one cuts power and wants to turn, physical characteristics of the aircraft will matter. Broad, low aspect, draggy wings make biplanes more maneuverable in turns than their faster mono plane counterparts. This is one reason triplanes were all the rage before new tactics were developed.