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?
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)
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
g*tan(theta) = v^2/r
Solve for r:
r = v^2/(g*tan(theta))
v:=266.67 m/s; g:=9.8 N/kg; theta:=35 deg;
r = 10363 meters
r = 10.363 km
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