along the N-S direction and that the position angle (measured East of North) of the precession axis is given by $$\theta \sim\, \theta_m \,\sin(\omega t) - \theta _0~,~~~~~~~~     (1)$$

where $\theta_m$ is the amplitude of precession, $\omega$ is the precession rate and $\theta_0$ is the position angle at $t=0$. We assume that the time scale of the observable outflow phenomenon is much shorter than the precession period. Hence, over a short period the precession position angle changes linearly, $$\theta \simeq \theta_m \omega t - \theta_0 = \beta t - \theta_0~.~~~~~~~~~~~~~~(2)$$

The coordinates in the plane of the sky at the time of our observations, $t$, for the location of gas ejected from the star at time $t_e$ will be $$x(t) = r(t) \, \sin\,\theta_e~~~~~~~~~~~~~~~~~~(3)$$ $$y(t) = r(t)\cos \theta_e~~~~~~~~~~~~~~~~~~~(4)$$

where $\theta_e$ is the precession angle at the time of emission, and $r$ is the radial position of the ejecta, $r = v~(t-t_e)$. We assume that the ejection velocity, $v$, is constant. Since $t-t_e = r/v$ we obtain $$x(t) = r\,\sin\,(\frac{\beta r}{v} - \theta_0)~~~~~~~~~~~~~~~~~~~(5)$$ $$y(t) = r\,\cos\,(\frac{\beta r}{v} - \theta_0)~.~~~~~~~~~~~~~~~(6)$$

The prediction of this simple model is that $\theta$ is linearly related to $r$, that is

$$\theta(r) = \beta~r/v - \theta_0. ~~~~~~~~~~~~~~~~~~~~(7)$$

