When a plane turns, it banks as shown in the figure to give the lifting force of the wings a horizontal?

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)


    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

  • 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

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