E. Linacre and B.Geerts |
1/'99 |
The wind speed (S, in m/s) needed to lift an object off the ground depends on its weight (W, in kg) and the area of the side facing the wind (A, in m^{2}). For many objects, S is given by this rule-of-thumb formula -
S = (7.7 x W / A)^{ 0.5}
This gives a rough idea of the wind speed required to lift a car into the air during the passage of a tornado. If the car weighs a tonne and has a side area of 5 m^{2}, for instance, the wind needs to be at least 39 m/s, or 141 km/h. This corresponds with the maximum wind in a F1 tornado. Conversely, a 1000 kg car, with a frontal area of 3 m^{2}, driving over 182 km/h (under calm conditions), could become airborne unless it is designed suitably.
The above equation assumes that the object has normal angles (i.e., it is square). The lifting force also depends on the actual shape of the object. For instance, a house with a low triangular roof experiences more lift than a flat-roof house (Note 14.B). The cross section of a wing (or airfoil) is designed to maximize the lift. This lift is the result of different wind speeds adjacent to the opposite sides of a wing. According to Bernouilli's principle, the lift is proportional to the difference of the square of the velocities on the wing's top and bottom. The reason why gulls and other gliding birds fly so close to the surface of a lake or sea is because the wind speed increases most rapidly with height very close to the surface (Fig 14.2 in the book). Therefore the wind differential (between top and bottom of the wings) and the lift is largest there.