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!
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 (=&gt; 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.