# Wind Pressure reduced by cos²

In standards you will often see that wind pressure on a pole, conductor etc is reduced by cos² (or sin² depending on the orientation of angles) of the angle between wind direction and object direction but there is no explanation of why cos is squared. Here is the explanation.

When moving air (wind) is stopped by a surface the dynamic energy in the wind is transformed to pressure (equation 1).1 The pressure on the surface transforms to a force (equation 2).2 Bernoulli’s equation for fluid flow is

$$\large q=0.5 \rho \times v^{2} \tag{1}$$

where
$$q$$ = dynamic pressure
$$\rho$$ = mass density of the fluid (air in this case)
$$v$$ = wind velocity

$$\large F=q A \tag{2}$$

where
$$F$$ = force on the surface
$$A$$ = surface area

Air density $$\rho$$ may be taken as 1.225 kg/m³ 3 which leads to equation B4 in AS7000:2016

$$\large q_{z}=0.6 V^{2} \tag{3}$$

with $$V$$ in equation 3 being the same as $$v$$ in equation 1. The 0.6 is rounded from 1.225 ÷ 2.

When the wind is not perpendicular to the surface its velocity is reduced by cos of the angle between the wind and the perpendicular4, ie in equation 1 $$v$$ becomes $$v \cdot cos \varphi$$.

Substituting eqn 1 into eqn 2 means eqn 2 becomes

$$\large F=q \times cos^{2} \varphi \times A \tag{4}$$

showing that wind pressure $$q$$ (Pa in metric units) is reduced by the square of cos of the angle between wind and perpendicular to the surface.