This shows a parabolic dish of radius $R$ and depth $D$ pointed at normals to a wind with velocity $v$.

This shows a parabolic dish of radius $R$ and depth $D$ pointed at normals to a wind with velocity $v$.

The dish has spent its first winter down at the site and, despite the fact that its not attached to anything, the issue of wind loading has recently come to the forefront of our team meetings. For those not in the know, wind loading is the force imposed on the dish by wind that may be blowing into it. In this post, we’ll be going over how to do the wind loading calculations for yourself and what we did with ours. First off, we’ll visit an important equation:

\begin{equation}
F = A \times P \times C_d
\label{eq:wind-load}
\end{equation}

where $F$ is the force that the wind exerts on an object, $A$ is the area of the object on which in the wind is incident, $P$ is the pressure of the wind and $C_d$ is the coefficient of drag of the object. This equation is the central equation which we can use to calculate the wind loading of a structure. In order to calculate this quantity we first need to calculate some other important quantities. Let’s go through these now. First, we need to calculate the surface area of the dish that’s facing into the wind. For this we can be safe and assume the worst case, and that is the velocity of the wind is normal to the dish and that the maximum surface area is subject to the wind flow. For a 2D parabolic dish the surface area is given by:

$$
A = \frac{\pi R}{6 D^2} \left( \left[ R^2 + 4D^2 \right]^{\frac{3}{2}} – R^3 \right)
$$

where $R$ is the radius of the aperture of the dish and $D$ is the depth from the aperture to the very rear of the dish. In our case the dish has $R = 4.5$m and $D = 0.59$m. This gives a total surface area of $A = 64.7\text{m}^2$. The next thing to consider in \eqref{eq:wind-load} is the dynamical wind pressure $P$ which is given by the equation:

\begin{equation}
P = \frac{1}{2} = \rho v^2
\label{eq:dynamical-wind-pressure}
\end{equation}

where $\rho$ is the density of the fluid that’s passing the object and $v$ is the velocity of the fluid. At this point we could simply take the density of air, but I went a step further with this one by customising it to the location of the observatory. We can attain the density of air from meteorological archive data using:

\begin{equation}
\rho = \frac{p}{R_\text{specific} T}
\label{eq:density-of-air}
\end{equation}

where $p$ is the absolute pressure measured in Pascals, $T$ is the absolute temperature measured in Kelvin and specific gas constant for dry air $R_\text{specific} = 287.058 J kg^{-1} K^{-1}$. We can now use data from the Met Office’s UK Climate Data Archive to take long term maxima for the average pressure and wind speed and put them into this equation. We must be aware of the units of this data. The pressure is expressed in hPa, or hectopascals, and the conversion between this and Pa is simply $1hPa = 100Pa$. We also remember that the conversion between degrees celsius and Kelvin is simply $0^{\circ}\text{C} = 273.15K$, so we have to add 273.15 onto any temperature measurement to get the absolute temperature. In our case we took data from Pensilva, a location very close to the observatory, which gives us $p = 1016$ hPa $ = 1.016 \times 10^5$Pa and $T= 13.2^{\circ}$C $ = 286.35$K. We can now go back to \eqref{eq:dynamical-wind-pressure} and substitute in \eqref{eq:density-of-air}. There is one more measure we need for this equation and that is the velocity of the wind. We can once again visit the Met Office climate archive to get the mean maximum velocity. This quantity is measured in knots, which can be converted to SI using  that fact that 1kts = 0.514ms$^{-1}$. We took the maximum to be $v = 11.9$kts $= 6.12$mS$^{-1}$. Putting all of these quantities into \eqref{eq:dynamical-wind-pressure} we get $P = 23.14$ kg m$^{-2}$ s$^{-1}$. Now we’ve got the dynamical pressure and the surface area of which it’s exerted. The only thing left now is to work out the coefficient of drag $C_d$ and we’re done. The issue is that calculating this analytically is problematic. The coefficient of drag is defined as:

\begin{equation}
C_d = \frac{2 F}{\rho v^2 A}
\label{eq:coefficient-of-drag}
\end{equation}

Herein lies the problem. The analytical solution to this parameter relies on the unknown quantity that we’re trying to calculate. Not only that, but the quantity is a function of many other parameters. These include flow speed of the liquid, flow direction,  size and even viscosity. Some of these can be parametrised and neglected, however it’s still a very complex problem that is only really solvable with either numerical computer simulations or experimentation. Fortunately, the supplier of the dish, RF Hamdesign in the Netherlands, has already performed these experiments for us under precisely controlled conditions, giving a proper wind loading calculation as a function of wind speed.

Wind loading on the dish as a function of wind speed. © 2012 RF Hamdesign.

Wind loading on the dish as a function of wind speed. © 2012 RF Hamdesign.

If you could ascertain the drag coefficient $C_d$, then you would be able to substitute in and calculate this value for yourself, however we chose to simply side with the data from the professionals. In this post, I’ve given you a bit of a look at one of the things it’s very important to consider when planning and constructing your own radio telescope, or for that many nearly any long-term engineering project. I hope I’ve been able to give you a taste of the calculations involved and how they can be solved, for the most part, using readily available data. In the following posts we shall be continuing our coverage of construction of the motor electronics.

Written by Sam Morrell – Science, Outreach and Press & Publicity Teams