Coriolis Acceleration in Local Cartesian Coordinates

Question

Generally, the coriolis acceleration is given as

$-2\vec{\Omega}\times\vec{v}$

Just as $\vec{\Omega}$ or $\vec{v}$, the coriolis acceleration can be rewritten in local cartesian coordinates (edited):

$-(2\omega\Omega\cos{\Theta}-2v\Omega\sin{\Theta}, 2u\Omega\sin{\Theta}, -2u\Omega\cos{\Theta})$

with $\vec{\Omega}=(0,\Omega\cos{\Theta},\Omega\sin{\Theta})$

For the coordinate system see Wikipedia image:

It's not easy to understand ... Why does $u$, the x-component of the velocity, appear in the second and third component of the coriolis acceleration? Could you eventually help me?

Answer 1

I think your question needs further clarification, if you are interested in the mathematical description, then the reason for $u$ to appear in the second and third component of the resulting vector is a matter of the cross product's definition only.

In your case it is defined as:

$$ \vec{a} \times \vec{b} = \left( \begin{array}{c} a_2b_3 - a_3b_2\\ a_3b_1 - a_1 b_3\\ a_1b_2 - a_2 b_1\\ \end{array} \right)$$

If you now replace $\vec{b}$ with $\vec{v}$, you will have $b_1 = u$, $b_2=v$ and $b_3=\omega$. From here it is easy to see why $u$ appears in the second and third component of the resulting vector.