component that turns the plane.
If a plane is flying level at 960 km/h and the banking angle is not to exceed 35 degrees, what is the minimum curvature radius for the turn?
4 Answers

Fw is the aerodynamic lift force acting on the wings, perpendicular to their surface
Fg is the gravitational force from the Earth.
Newton’s 2nd law in the vertical direction:
Fw*cos(theta) = Fg
Newton’s 2nd law in the horizontal direction:
Fw*sin(theta) = m*a
Recall the origin of Fg, Fg = m*g:
Fw*cos(theta) = m*g
solve for Fw:
Fw = m*g/cos(theta)
Substitute:
m*g*sin(theta)/cos(theta) = m*a
Cancel m’s and combine trigonometry:
g*tan(theta) = a
acceleration is due to centripetal acceleration:
a = v^2/r
Substitute:
g*tan(theta) = v^2/r
Solve for r:
r = v^2/(g*tan(theta))
Data:
v:=266.67 m/s; g:=9.8 N/kg; theta:=35 deg;
Result:
r = 10363 meters
r = 10.363 km

Plane Banking

At 45 deg bank, the lateral forces on the airplane are Fg = mgsin(45 d) and Fcent = mv^2/r*cos(45 d). To avoid sideslip, these forces must be equal to keep the net force on the plane perpendicular to the wing. Note sin(45 d) = cos(45 d). Thus Fg = Fcent ==> r = v^2/g = (940/3.6)^2/9.8 = 6957 m.

Acceleration v^2/r for the circular movement must be provided by the lifting force horizontal component. Hence v^2/r = F/m* sin(theta).
Now balance the vertical components : F*cos(theta) = m*g
So v^2/r = g*tan(theta) If you want theta < 35 deg then r > v^2/(g*tan(35)) = 10352 m