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I was trying to do an alternative general expression for Coriolis acceleration to the one explained on Wikipedia which I find not too intuitive and too simple calculation wise.

Supposing the $z$ axis is the instantaneous axis of rotation of the Earth, then $\mathbf\Omega=\Omega\mathbf{\hat k}$. To model the velocity of the wind at some latitude $\lambda$ and longitude $\phi$ we'll use the spherical unit vectors: $$\begin{aligned} \mathbf V=u(-\mathbf e_\theta)+v\mathbf e_\phi&=-u(\sin\lambda\cos\phi\mathbf{\hat x+\sin\lambda\sin\phi\mathbf{\hat y}-\cos\lambda\mathbf{\hat z}})+v(-\sin\phi\mathbf{\hat x+\cos\phi\mathbf{\hat y}})\\ &=-(u\sin\lambda\cos\phi+v\sin\phi)\mathbf{\hat x}+(v\cos\phi-u\sin\lambda\sin\phi)\mathbf{\hat y}+u\cos\lambda\mathbf{\hat z}\\ &=V_x\mathbf{\hat x}+V_y\mathbf{\hat y}+V_z\mathbf{\hat z}, \end{aligned}$$ where there's no $\mathbf e_r$ term because we neglected the vertical velocity of the wind since it is usually low with respect to the others and where the minus sign of $\mathbf e_\theta$ is chosen in order for the north to be the positive wind direction.

At last, now I compute the cross product and get that the acceleration should be $$\begin{aligned} \mathbf{a_C}=-2\mathbf\Omega\times\mathbf V&=2\Omega(V_y\mathbf{\hat x}-V_x\mathbf{\hat y})\\ &=2\Omega(v\cos\phi-u\sin\lambda\sin\phi)\mathbf{\hat x}+2\Omega(u\sin\lambda\cos\phi+v\sin\phi)\mathbf{\hat y} \end{aligned} $$ The thing is when I evaluate the acceleration at the equator ($\lambda=0$), the acceleration is not zero like it's supposed to according to theory. What did I do wrong? Is Coriolis supposed to have a dependence on longitude anyways?

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  • $\begingroup$ This might help with the intuition. Coriolis Force: Direction Perpendicular to Rotation Axis Visualization $\endgroup$
    – mmesser314
    Mar 13 at 13:14
  • $\begingroup$ the Coriolis acceleration is $~\vec a_C=-2\vec\omega\times\vec v~$ the coordinate of $~\vec\omega~,\vec v~$ must be given in the rotating system !. you can't obtain the coordinates in inertial system, this is wrong $\endgroup$
    – Eli
    Mar 13 at 14:44
  • $\begingroup$ Right, when I change again to spherical basis I get something which makes more sense (I got rid of the vertical component): $$\mathbf{a_C}=2\Omega\sin\lambda(v\mathbf{e_\theta}+u\mathbf{e_\phi})$$ $\endgroup$
    – Conreu
    Mar 13 at 18:11

1 Answer 1

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You ask specifically about the rotation-of-Earth-effect on wind.

To discuss the physics of air mass in motion:
For this answer I will leave out pressure gradient, because that is a transient factor. I begin with considering the forces that are always there.

Forces on an oblate spheroid

Diagram 1.
Forces on a buoyant object (animated GIF)

We have that due to its rotation the Earth is an oblate spheroid. (The Earth started out as a protoplanetary disk, and over time contracted to its current oblate shape.)

In the diagram the oblateness is exaggerated, of course, the actual oblateness is is about 1/300 (The equatorial radius is about 20 kilometer larger than the polar radius.)

The blue arrow represents newtonian gravity, the red arrow represents buoyancy force.

On an oblate spheroid the direction of newtonian gravity is not perpendicular to the local level surface. For example, at 45 degrees latitude the deviation from perpendicular is about 0.1 degree.

Since newtonian gravity and the buoyancy force are not exactly opposite in direction there is a resultant force, in the diagram indicated in green.

(Note that in the case of ballistic motion Diagram 1 is not applicable. In the case of ballistic motion there is no buoyancy force.)


In the interpretation of measurement results we must allow for the equivalence of inertial and gravitational mass. Let me expand on what it means to take into account that inertial mass and gravitational mass are equivalent.

When you are located at the Equator you are circumnavigating the Earth's axis; a centripetal acceleration is required to sustain that centripetal acceleration. That centripetal acceleration goes at the expense of the amount of gravity that you are subject to.

At the Equator: to remain co-rotating with the Earth requires a centripetal acceleration of about 0.0339 $m/s^2$. That goes at the expense of gravitational acceleration.

Because of the equivalence of inertial and gravitational mass: a gravimetric instrument cannot measure that effect directly. A gravimetric instrument measures a single gravitational acceleration.

The rotation of Earth does become apparent when a measuring instrument has a velocity relative to the Earth. This effect is called Eötvös effect, after the hungarian scientist Eötvös, who designed and operated gravimetric instruments of the highest sensitivity.

To illustrate the Eötvös effect I wil use the example of an airship. Let an airship be flying parallel to the Equator, in west-to-east direction. The airship is trimmed to neutral buoyancy. Next the airship makes a U-turn. After the U-turn the buoyancy needs to be re-trimmed.

Before the U-turn the airship was circumnavigating the Earth's axis a bit faster than the Earth itself is rotating. So the airship was experiencing a bit less gravitational acceleration than when stationary wrt the Earth. After the U-turn the airship is circumnavigating the Earth's axis a bit slower than the Earth, so then it is experiencing a bit more gravitational acceleration.


Air mass and the rotation of the Earth

If you scroll back to Diagram 1:
The green arrow represent a resultant force that is acting in centripetal direction. From here on I will refer to that as 'the poleward force'.

The poleward force provides the amount of force that is required for buoyant mass to be co-rotating with the Earth. At any latitude: if air mass is flowing west-to-east the air mass in a sense "speeding", the provided centripetal force is then not enough, and the air mass will swing wide (deviate towards the Equator). If air mass is flowing east-to-west the air mass is experiencing a surplus of centripetal force, and subsequently the air mass will move to the inside of the latitude line that is is moving along.

Next:
Quantitative description of the rotation-of-Earth-effect.

First a simpler situation: motion over a flat surface, subject to a centripetal force such that at any distance to the axis of rotation a buoyant object remains co-rotating with the system.

Motion subject to a centripetal force

Diagram 2.
Motion subject to a centripetal force (animated GIF)

Diagram 2 represents side by side a stationary point of view, and a co-rotating point of view. The circle represents a rotating disk. Along the rim quadrants are added to give the viewer a reference of orientation.

The arrow in the animation represents the centripetal force. From center to rim there is a linear increase of the centripetal force.

The motion of the black dot is according to the following parametric expression:

$x = a \cos(\Omega t)$
$y = b \sin(\Omega t)$

$a$ half the length of the major axis
$b$ half the length of the minor axis
$\Omega$ 360 degrees divided by the duration of one revolution

For emphasis: the above is valid only in the case that a force is present force such that at any distance to the axis of rotation a buoyant object remains co-rotating with the system. Note that rotating systems tend to settle down to such a state spontaneously. Example: set up a dish with fluid in it to rotate at a uniform angular velocity. After sloshing has subsided the cross section of the surface will be in the shape of a parabola. That is: the slope of the fluid provides the required centripetal force to remain co-rotating.

To set up for transformation to a rotating coordinate system: rearrange as follows:

$$ x = \frac{a+b}{2} \cos(\Omega t) + \frac{a-b}{2} \cos(\Omega t) $$ $$ y = \frac{a+b}{2} \sin(\Omega t) - \frac{a-b}{2} \sin(\Omega t) $$

After transformation of the motion to the rotating coordinate system the motion relative to the co-rotating coordinate system is as follows:

$$ x = \frac{a-b}{2} \cos(2 \Omega t) $$ $$ y = - \frac{a-b}{2} \sin(2 \Omega t) $$

That is, the motion relative to the rotating coordinate system is perfectly circular, and proceeds at a frequency of $2\Omega$, twice the frequency of the rotating system.

I will refer to the circle of the motion relative to the rotating coordinate system as the epi-circle.

I will use the lowercase $\omega$ for the angular velocity relative to the rotating coordinate system.

with:
$a_c$ centripetal acceleration wrt to center of epi-circle
$r$ radial distance to center of epi-circle

$$ a_c = \omega^2r \tag{1} $$

Let $v_r$ be the velocity vector relative to the rotating coordinate system. the following substitution results in (2): $\omega r=v_r$

$$ a_c = \omega v_r \tag{2} $$

We make the expression independent of the center of the epi-circle with the following substitution: $\omega = 2\Omega$:

$$ a_c = 2 \Omega v_r \tag{3} $$

The form of (3) is independent of the direction of the velocity vector relative to the rotating system.

General remarks

The motion pattern of Diagram 2 consists of two coupled oscillations. Oscillation of distance to the axis of rotation, and oscillation of the angular velocity. Note especially that both of those oscillations are independent of the choice of coordinate system.
-In both coordinate systems, inertial and rotating, the distance to the center of rotation is the same.
-In both coordinate systems, inertial and rotating, the rate of change in angular velocity is the same.

The relevant aspects here are those aspects of the motion that are in both coordinate systems the same.

The interplay of those two oscillations (radial and angulular velocity) is described by the relation $a_c = 2 \Omega v_r$


Of course, we have that in the equation of motion for motion with respect to a rotating coordinate system there is a term of the form $2\Omega v_r$, which is the same form as in equation (3).

There is of course a reason that the same form makes another appearance - discussion of that is interesting, but outside of the scope of this answer.


Expanding to motion relative to the Earth's surface

In the case of rotation-of-Earth-effect on wind:
The magnitude of the rotation-of-Earth-effect is proportional to the projection onto the latitudinal plane.

Hence close to the Equator the rotation-of-Earth-effect on wind is very small, whereas close to the poles the motion of the air mass (which is parallel to the local surface) is close to parallel to the latitudinal plane. The closer to the poles, the closer the ratio approaches 1:1.

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