I have been looking through my old dynamics books and derived the acceleration for a particle on a circular orbit. To begin with, I assume the particle is on a circular orbit around some center $x_0$. With a vector to the moving particle $\overline{e}_r$ and a tangential vector $\overline{e}_{\phi}$. The simple location of the particle can be written as:
$$\overline{x}(t)=r(t)\cdot e_r$$
Differentiating this gives:
$$\dfrac{d\overline{x}(t)}{dt}=\dfrac{dr(t)}{dt}\cdot \overline{e}_r + r(t) \cdot \dfrac{d\overline{e}_r}{dr}$$
Using $\frac{d\overline{e}_r}{dt}=\dot{\phi}\cdot \overline{e}_{\phi}$ and $\frac{d\overline{e}_{\phi}}{dt}=-\dot{\phi}\cdot \overline{e}_{r}$ which comes from a simple geometrical idea, I can simplify to (ignoring all the $(t)$ which should be obvious for all the variables):
$$\dot{\overline{x}}=\dot{r}\cdot \overline{e}_{r} + r\cdot \dot{\phi} \cdot \overline{e}_{\phi}$$
Differentiating this once more yields the acceleration: $$ \ddot{\overline{x}} = \left(\ddot{r}-r\dot{\phi}^2 \right)\overline{e}_{r} + \left(\ddot{\phi} r+ 2\dot{\phi}\dot{r} \right) \overline{e}_{\phi} $$
The part I was wondering about is the Coriolis-acceleration:
$$ a_c =2\dot{\phi}\dot{r} \overline{e}_{\phi} $$
I was trying to apply this to the passat winds on the earth.
So regarding the "Nordost Passat", we have wind particles flowing from North to South towards the equator. Since the formula above is only for a disc, yet the acceleration into the 3rd dimension (in this case the rotational axis of the earth) can be disregarded, I came up with this set of assumptions:
Since we are moving away from the rotational axis (towards the equator, I assume that $\dot{r} > 0$
Since the derivative of the radius with respect to the latitude at the equator is 0, the second derivative of the radius has to be smaller than $0$: $\ddot{r} < 0$
The wind is moving together with the earth's surface (roughly) which means a positive rotation around the rotational axis if the rotation axis is supposed to be "upwards" which corresponds to: $\dot{\phi} > 0$
Since the earth does not accelerate: $\ddot{\phi} = 0$
Also it is important to note that $\overline{e}_{\phi}$ shows to the "right" in the image which equals the direction the earth is rotating. Furthermore $\overline{e}_{r}$ shows outwards from the rotational axis.
Applying all of this to the Coriolis-acceleration, I end up with $a_c=(\text{something positive}) \overline{e}_{\phi}$. Since, according to the image, the wind is accelerated towards the negative $\overline{e}_{\phi}$, I am curious where I have messed up.
I am happy for any hints or advice.