# Winds around the eye of a storm

Continuing “Saturdays are made for mathematics” and bearing in mind that the other time I talk to you about how false are the centrifugal forces and centripeta being a differential coordinate change, today I want to talk about the winds around the center of a storm as a practical case.

First a bit of basic atmospheric physics to put you in position and is just to understand that it is an anticyclone or a storm. Both elements refer to variations in atmospheric pressure at specific points. Thus an anticyclone is an area with high pressures with a point where the pressure is maximum and, a storm, is just the opposite, an area with low pressures.

Remember that “normal pressure” is considered at 1 atmosphere or at 1000 bars.

I imagine that, if you see the time in the news, you will have your bare ass to see how these elements move and how their arms rotate around the central point that I indicate. Now let’s see why.

First of all, I imagine that you will know that when there is a differential of something (a point where there is much and another where there is little close), nature tends to homogenize the matter going from where there is more to where there is less so that they balance (The opposite happens in this society with money thanks to capitalism).

If we consider the eye of a storm, which is simply a particular area of the storm where there are very low pressures, such as low pressures, it absorbs the surrounding air in order to increase the pressure inside.

Obviously, the direction of rotation of the Earth causes the angular velocity of this point to the north. Suppose that this angular velocity is $\Omega$ and, obviously, what angular velocity means is parallel to the axis of rotation.

At any point P, $\Omega$ we can express it with its tangential components to the surface of the Earth. Remember that the Earth is spherical and the best used coordinates are spherical. This is important.

So we have to:

$\Omega (P) = \Omega_{T} (P) + \Omega_{N} (P)$

Where $\Omega_ {T}$ is the tangential part and $\Omega_ {N}$ is its normal part. Simple and but, look at the drawing (the bad drawing).

As we have seen the other day, in P (which for convenience we assume in the northern hemisphere, nothing more) we have that Coriolis force is:

$C=-2\Omega(P) \times v$

And now, we will use the magic of physics which, as I have said many times, is that a physicist interprets the equations to know what is going on here.

The fact is that the Coriolis force has broken it down into two, the normal and the tangential. By simple observation, the normal one can be despised because the air does not move much of horizontally for the point (remember, that we are on the terrestrial surface and horizontal is “up”).

The tangential is another song. First we know, because it is a vector product that is oriented 90 degrees clockwise with respect to av -vector-), so the particles, what pille is being absorbed towards the center (same for a storm) but with A deviation to the right by Coriolis of form, that because of Coriolis, to reach the center they do it by a route in form of spiral. And even this is where a physicist interprets the equations.

And now a little silly definition. If we have the anticyclones that are the ones that expel the air by excess pressure and Coriolis (in each hemisphere for its place) otherwise, the storms, also they are called cyclones.

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