<s>[INST] The two causes for the factor 2 in Coriolis effect While reading this document on Coriolis effect http://empslocal.ex.ac.uk/people/staff/gv219/classics.d/Persson98.pdf, I saw the followig sentence

Two kinematic effects each contribute half of the Coriolis acceleration: relative velocity and the turning of the frame of reference. 

And this is the reason why Coriolis term has that factor $2$. Unfortunately it does not specify anything about this two causes. Does anyone have some further explanation for how "relative velocity" and "turning of the frame" actually give rise to the Coriolis term?
 [/INST] Take a free particle moving on a plane in polar coordinates
$$ \begin{pmatrix} x \\ y \end{pmatrix}  = \begin{pmatrix} r \cos \theta \\ r \sin \theta \end{pmatrix}$$
The velocity is found from the chain rule, with clear separation for radial and tangential components:
$$\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \begin{pmatrix} \dot{r} \\ r \dot{\theta}\end{pmatrix} $$
The acceleration is found again by differentiation
$$ \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = \frac{{\rm d}\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}}{{\rm d}t} \begin{pmatrix} \dot{r} \\ r \dot{\theta}\end{pmatrix} + \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \frac{{\rm d}\begin{pmatrix} \dot{r} \\ r \dot{\theta}\end{pmatrix}}{{\rm d}t}$$
$$ =\begin{vmatrix} 0 & -\dot{\theta} \\ \dot{\theta} & 0 \end{vmatrix}\begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}\begin{pmatrix} \dot{r} \\ r \dot{\theta}\end{pmatrix} + \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix}\begin{pmatrix} \ddot{r} \\ r \ddot{\theta}+\dot{r} \dot{\theta}\end{pmatrix} $$
$$ = \begin{vmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{vmatrix} \left[\begin{vmatrix} 0 & -\dot{\theta} \\ \dot{\theta} & 0 \end{vmatrix}\begin{pmatrix} \dot{r} \\ r \dot{\theta}\end{pmatrix}+ \begin{pmatrix} \ddot{r} \\ r \ddot{\theta}+\dot{r} \dot{\theta}\end{pmatrix}\right] $$
The above is first a rotation matrix by $\theta$, then the effect of the rotation on the (local) velocity and finally the (local) acceleration. Notice in the radial direction the local acceleration is just $\ddot{r}$, and in the tangential direction it has two terms. One is Euler's acceleration $r \ddot{\theta}$ and the other 1/2 the coriolis term. This part is due to the change in direction of the radial velocity.
$$ \begin{pmatrix} \ddot{x} \\ \ddot{y} \end{pmatrix} = ({\rm Rotation}) \left[ \begin{pmatrix} -r \dot{\theta}^2 \\ \dot{r} \dot{\theta} \end{pmatrix} + \begin{pmatrix} \ddot{r} \\ r \ddot{\theta} + \dot{r} \dot{\theta} \end{pmatrix} \right] $$
Now the first part $(-r \dot{\theta}^2, \dot{r} \dot{\theta} )$ contains the centrifugal acceleration in the radial direction and the change in orientation of the tangential acceleration which is the other half of the coriolis effect.
But I find all this confusing. I'd rather look at a picture:

The changes in the velocity vector in radial coordinates (where the center of rotation is in the -x direction. The two $\dot{r} \dot{\theta}$ terms in the coriolis term are from a) turning of $\dot{r}$ and b) extension of $r \dot{\theta}$. 
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