The nature of wind storms and windinduced damage - Wind Loading of Structures - page 21

For sheet objects, Equation (1.14) shows that the wind speed for flight depends on the
thickness of the sheet, but not on the length and width. Wills et al. expressed Equation
(1.14) in a slightly different form:
(1.16)
The left-hand side of Equation (1.16) is the mass per unit area of the sheet. This indicates
the wind speed for flight for a loose object depends essentially on its mass per unit area.
Thus, a galvanized iron sheet of 1 mm thickness with mass per unit area of 7.5 kg/m 2 will
fly at about 20 m/s ( C F =0.3).
2 For 'rod'-like objects, which include timber members of rectangular cross-section, a
similar formula to Equation (1.16) can be derived from Equation (1.15), with the 't'
replaced by 'd', the equivalent rod diameter. Using this Wills et al. calculated that a
timber rod of 10 mm diameter will fly at about 11 m/s, and a 100 mm by 50 mm timber
member, with an equivalent diameter of 80 mm, will fly at about 32 m/s, assuming C F is
equal to 1.0.
1.5.2 Trajectories of compact objects
A missile, once airborne, will continue to accelerate until its flight speed approaches the
wind speed or until its flight is terminated by impact with the ground or with an object
such as a building. The trajectories of compact objects are produced by drag forces
(Section 4.2.2), acting in the direction of the relative wind with respect to the body.
Consider first the aerodynamic force on a compact object (such as a sphere) in a
horizontal wind of speed, U . Neglecting the vertical air resistance initially, the
aerodynamic force can be expressed as:
where υ m is the horizontal velocity of the missile with respect to the ground and A the
reference area for the drag coefficient, C D (Section 4.2.2).
Applying Newton's law, the instantaneous acceleration of the object (characteristic
dimension, ℓ ) is given by:
(1.17)
taking A equal to ℓ 2 .