simulations are performed without gravity and represent water droplets in air at room temperature.

Static
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![The interface of various droplets at time $0.1 \si{\s}$ \[figure:contourEquil\]](fig1){width="60.00000%"}

Figure \[figure:contourEquil\] shows the interface ($\alpha = 0.5$) of various droplets with different Young’s angles at a resolution of $512 \times 256$ in a box of $1.5 \si{\milli\m} \times 0.75 \si{\milli\m}$. The half droplets in the figure have a surface area of about $0.2 \si{\milli\m^{2}}$ (i.e. a radius of $0.5 \si{\milli\m}$ when the Young’s angle is $\pi/2$). As initial condition these droplet where given their equilibrium shape, and it can be seen that after a simulation time of $0.1 \si{\s}$ the droplets have maintained their shape.

![Radius of the base of the droplet and droplet height as a function of different contact angles.\[figure:angleRadiusEquil\]](fig2){width="60.00000%"}

To further quantify these droplets, figure \[figure:angleRadiusEquil\] shows the height and radius of the base of the droplets as a function of the Young’s angle. The radius of the base of the droplet $r_{B}$ is calculated as: $$r_{b} =  R \sin{\theta_{0}},$$ and the height of the droplet is equal to: $$h_{d} = R {\left( 1 - \cos{\theta_{0}} \right)}.$$ In the above equations: $$R = \sqrt{\frac{2 A}{\theta_{0} - \cos{\theta_{0}} \sin{\theta_{0}}}}$$ is the 