# A guided tour from the motion on carousels to cyclones and the Gulf Stream, full of surprises

## An interactive material for secondary school students for assisted or individual learning

Questions, projects are set in italics. Those marked by an asterisk (*) are recommended for particularly motivated readers.

### I Introduction

What do you think of an effect like this, is it realistic?

To have a better insight, let us start with a more straightforward video

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You might want to carry out analogous table-top experiments with balls on rotating platforms (e.g. LP players or potter’s wheel). Using a web camera rotating with the platform, you can observe motions like this.

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Can you decide by comparing the last two videos, whether the platform was turned in a clockwise or counter-clockwise direction in the second case?

Why do you think the balls follow such a curved path and not a straight course after tossing? You might have had similar experiences e.g., in science centers.

### The basic effect: deflection of moving bodies

If interested, carry out a simple home-experiment with a pencil, a sheet of paper and a ruler. Fix the mid-point of the paper with a pin to the table. First draw a straight line with constant speed on the non-moving sheet across this center point. Repeat the same motion with someone rotating the sheet clockwise with an approximately constant angular velocity while you are drawing the line. You might obtain a curved path, deflecting from the straight line, similar to this.

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X It is clear that the motion in the carousel (the path traced out on the rotated paper sheet) is deflected from its initial direction. Note that the deflection is to the left if the rotation is clockwise. This same deflection is reflected in the ball’s motion in the second and third video.

Note: the straight line on the non-rotated paper sheet corresponds to a uniform motion visible from stationary reference frame. The deflection is thus a consequence of the fact that an otherwise simple motion is observed from a rotating frame.

The curved path traced out on the rotated paper sheet is similar to a parabola, at least for short distances. The motion on the carousel is thus similar to one with a constant acceleration.

Based on the parabola plotted in your experiment, measure the amount of deflection at a few time distances. Show that the dependence is parabolic for short times at least, and determine the acceleration. Result

Use the fact that a point of distance $r$ from the center is rotated along an arc of length $r \Omega t$, where $\Omega$ is the angular velocity of the sheet. Express the acceleration by means of $\Omega$ and the pencil’s velocity $v$. What force can cause this acceleration? Answer

The fictitious force used to describe the deviation is the Coriolis force. Its magnitude is $$\boldsymbol{F_C=m 2 v \Omega},$$ and points perpendicular to the direction of velocity $v$, to the left and right in the case of clockwise and counter-clockwise rotation, respectively.

*For a deeper insight, carry out a slightly more advanced experiment. Repeat the first experiment so that you start drawing at a distance of $r_0$ away from the center of rotation and carry it out radially outwards. Do you obtain a different expression for the acceleration or the force acting perpendicular to velocity? Result

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The force remains $F_C=m 2 v \Omega$ but $v$ turns out to be the velocity of the body relative to the rotating system. Another force also appears acting radially outwards, this is the centrifugal force $F_{cf}=m r_0 \Omega^2$. Both forces are important in the motion of a ball on the carousel, but only the Coriolis force is responsible for the deflection from the original velocity.

### A basic qualitative measure: the strength of deflection

Consider a body which would move in a non-rotating system with a velocity of $U$ and estimate the deflection $D$ from the direction of this initial velocity after traveling a distance $L$, provided the deflection is weak. The acceleration $2 v \Omega$ is then constant and can be approximated by $2 U \Omega$. The distance $D$ traveled with this approximtion perpendicular to the initial velocity is, after time $t$, $(1/2) 2 U \Omega t^2$. The time needed to pass distance $L$ is approximately $L/U$, and we obtain $$D=U \Omega L^2 /U^2 = \Omega L^2 /U.$$ The relative deflection, the distance $D$ compared to the total distance $L$, i.e. $D/L$ is also worth considering: $$D/L = \Omega L/U.$$ This is a useful simple expression which enables us to estimate the magnitude of deflection. Note that this is inversely proportional to $U$, since the time during which the Coriolis force acts is also important for the deflection.

Check your knowledge acquired: Test I

### II The rotating Earth as a carousel

Our paper sheet experiment clearly illustrates that deflection is due to a rotation the axis of which is perpendicular to the plane in which the deflection takes place. If the plane of deflection is not perpendicular to the rotational axis, the angular velocity component pointing perpendicular to this plane will determines the rate of deflection.

It must be surprising to learn that angular velocity may have components. In fact, it is worth considering angular velocity as a vector. Its direction is that of the axis of rotation (and the arrow of the vector corresponds to our right thumb for counterclockwise rotations). Besides directionality, angular velocity is a vector also in the sense that it can be decomposed into perpendicular components. This implies that a rotation according to an angular velocity vector ${\bf \Omega}$ is equivalent to carrying out rotation around another axis with angular velocity ${\bf \Omega_1}$ and, afterwards, another rotation around a perpendicular axis with angular velocity ${\bf \Omega_2}$, where $\Omega^2=\Omega_1^2+\Omega_2^2$. This feature is illustrated by means of a simple animation. (Note that the animation is not to scale since the feature holds true for low angles only which would not be visible otherwise.)

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When applied to the rotating Earth, the angular velocity ${\bf \Omega_E}$ of the planet pointing to the north can be decomposed into arbitrary perpendicular components ${\bf \Omega_1}$ and ${\bf \Omega_2}$, as shown here

When dealing with the deflection effect of the Coriolis force on Earth - which is always observed in the local horizontal plane - it is most natural to decompose ${\bf \Omega_E}$ into the horizontal and vertical component with magnitudes denoted as $\Omega_H$ and $\Omega_V$, respectively.

If you are interested to see that the resultant angular velocity is indeed the vectorial sum of the horizontal and vertical components in this particular decomposition, too, consult this page

At a location on latitude $\phi$, the angle between the horizontal plane and the rotational axis is also $\phi$, therefore the angular velocity of Earth, $\Omega_E$, projected onto the local vertical component is $$\boldsymbol{\Omega_V = \Omega_E \sin \phi} .$$

The deflection expression can safely be applied to any motion on Earth, you just have to consider $\Omega$ to be $\Omega_V$. Note that the Earth's angular velocity is 1 rotation / 1 day = $2 \pi/ 86400 \text{ s}$, i.e., $$\Omega_E=7.3\ 10^{-5} \text{ 1/s}.$$ At mid latitudes, at $\phi \sim 45 \text{ degree} (=\pi/4)$, $$\Omega_V=5\ 10^{-5} \text{ 1/s},$$ a value easy to remember. Note that in the Southern Hemisphere $\phi$ is negative, and thus $\Omega_V$ is negative, too.

Remark: A simplifying feature in motions on Earth compared to those on a carousel is that the centrifugal force need not be taken into account explicitly since it is included in the gravitational acceleration ${\bf g}$ which is the vectorial sum resulting from the gravitational and the (small) centrifugal force.

Estimate the deflection of a cannonball over a distance of an L=100 km shot

During World War I, the average horizontal velocity of the canonball from this gun was about 500 m/s. Estimate.

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$$\Omega_V = 5\ 10^{-5} \text{ 1/s},\ L=10^5 \text{ m},\ U=500 \text{ m/s},\quad \text{thus}\quad D=\Omega_V L^2/U = 1 \text{ km}.$$ Coriolis compensators letting the canon shoot a little left to the target in the Northern Hemisphere (equalling approximately 1/100 radian, about 0.6 degree in this case) were used during World War I.

### Does water drain in different directions in the Northern vs. Southern Hemisphere?

If the Coriolis force was acting alone, it would be reasonable to think that this is the case, since the sign of $\Omega_E$ is different on the two Hemispheres as seen above. When draining starts, water parcels start to move towards the sink, but become deflected due to the Coriolis force. The figure corresponds to the Northern Hemisphere where a counter-clockwise spin is expected while water is going down in a plughole.

Before becoming fully convinced, estimate the deflection at mid latitudes with realistic values for a plughole: $L=10 \text{ cm},\ U=10 \text{ cm/s}$. Result.

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$$\Omega_V = 5\ 10^{-5} \text{ 1/s},\ L=0.1 \text{ m},\ U=0.1 \text{ m/s},\quad \text{thus}\quad D=\Omega_V L^2 /U = 5\ 10^{-6} \text{ m}\ (D/L= 5\ 10^{-5} ).$$ This value equals 5 micrometers, which can not be perceived by the naked eye! The fact that this number is so small is satisfactory from the point of view that our laboratories can be considered to be inertial reference frames in which Newton’s law hold, since the Coriolis force is practically negligble.

*To see how strong other effects can be in the draining process, estimate the acceleration of a small cube of the size of $\Delta r=1 \text{ cm}$ of water caused by the difference in the hydrostatic pressure generated by a tiny height difference: $\Delta h=1\text{ mm}$ on the surfaces of the opposite vertical planes of the cube. Result.

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The pressure difference on the vertical planes of area $A=\Delta r^2$ is $$F = \rho g \Delta h A,$$ where $\rho$ denotes the density of water. The mass of the water cube is $\rho \Delta r A$, its horizontal acceleration is thus $$a=g \Delta h/\Delta r = 1 \text{ m/s}^2$$ This is to be compared with the Coriolis acceleration which is, when $U=0.1 \text{ m/s}$, $$a_C=2\ 5\ 10^{-5}\ 10^{-1} \text{ m/s}^2 = 10^{-5} \text{ m/s}^2$$ A difference of 5 orders of magnitude!

The strength of the Coriolis deflection is so small, due to the small magnitude of Earth’s angular velocity, that this effect is typically suppressed in our bath tubes by other effects (residual fluid motion, waves on the surface, asymmetry of the container, etc.) and any direction of the plughole vortex can be seen in practice on both Hemispheres. In carefully planned experiments, however, where all the disturbances are ruled out, the expected direction of spin can indeed be observed. The first such experiment was carried out by A. Perrot in Paris in 1859. Later experiments with sinks.

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In such experiments the water should be held at rest for days in exactly circular containers, and the plugholes should be precisely in the center.

1911: Vienna, O. Turmlitz

1961: Boston, A. Shapiro, Nature 196, 1080

1965: Sidney, Trefethen, Nature 207, 1084

Considering these, how can you explain the events seen in the very first video? Hint: What is the magnitude of $\Omega_V$ on the Equator (where the local horizontal plane is parallel to the rotational axis), which is responsible for the deflection. Result.

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Since the total angular velocity vector points to the North, it has no vertical component here: $\Omega_V=0$. Therefore, no deflection occurs at the Equator! The effect seen in the video must be due to a clever way of pouring the water into the vessel.

*Estimate the tilt angle of an aircraft moving at a speed of 1080 km/h by which the Coriolis deflection can be compensated. Result.

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On the Northern Hemisphere the aircraft should tilt to the left under some angle $\alpha$ (as if it intended to turn left) to compensate the Coriolis force pointing to the right of its direction of flight.
The modulus of the lift acting perpendiculat to the aircraft is practically the weight $mg$ of the craft under this weak tilt, too. Since the aircraft is tilted, the horizontal component of this force is $mg \sin\alpha$. This should compensate the deflecting force to the right: $$mg \sin\alpha = 2 m \Omega_V U.$$ From here $\sin\alpha \approx \alpha = 2 \Omega_V U/g$. When $U=300 \text{ m/s}$, $\alpha=3\ 10^{-3} \text{ rad} = 0.17 \text{ degree}$, hardly discernible.

Historical experiments: Foucault and Eötvös

Other weak deflection effects with occasionally long-lasting consequences.

On the Northern Hemisphere, the right bank of rivers is eroding faster than the left one. This rule was formulated (in a somewhat restricting form) by the Estonian scientist K. E. Baer in 1860.

On geological time scales, the Coriolis froce might cause a drift of hundreds of kilometers of river beds to the right of the overall direction of the river.

On the Northern Hemispheres, the wear on the right rail of railway tracks is stronger than that of the left rail.

In sports, like hammer or javelin throw deflection due to the Coriolis force can be on the order of a few cm-s.

In summary, on the human scale, from a few meters up to hundreds of meters, the Coriolis effect is weak: in certain situations it cannot even be perceived, while in others one has to use special equipments to prove it in experiments.

Check your knowledge acquired: Test II

### III Strong Coriolis deflections

A remarkable feature of the deflection effect is, however, that on geographical scales it can become strong. When the deflection $D$ is not much smaller than the total spatial extension $L$, the estimated $D$ cannot be considered to be a faithful value of deflection, but the ratio $$\boldsymbol{\Omega_V L /U}$$ (with $\Omega_V$ replaced by $\Omega$ when dealing with motions on carousels) can be considered as a number measuring the strength of the Coriolis deflection. Indeed, at a ratio of order unity, we can be sure that the deflection is comparable to the total displacement. The deflection effect proves to be strong if this ratio is larger than 1.

Estimate the strength of deflection a) for our paper sheet experiment and b) in the carousel video. Result.

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The ratio $\Omega L /U$ is $\pi/5=0.6$ in a), and $2\pi=6.3$ in b). The fact that we consider the deflection strong if the ratio is larger than unity is in harmony with the carousel video since deflection there might lead to a complete return of the ball. The effect is strong even in our paper sheet experiment (otherwise it would have been very difficult to see the difference between the two lines drawn).

*Consider the case when the ball returns to one of the kids in the carousel video, and assume that the path of the ball (friction is negligible) goes through the center of rotation when observed from outside the carousel. Assuming the distance of the kid is $r$ from the center and the angular velocity is $\Omega$, how large should be the tangential initial velocity on the carousel (in which direction?) in order to ensure such a motion? How should the radial velocity be chosen in order to ensure a full return of the ball, after half a revolution? Result.

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The tangential velocity of the ball should compensate the analogous velocity of the carousel, it should be $r \Omega$ to the right. A stationary observer can see a uniform motion of velocity $v$, which passes the diameter $2r$ in time $2r/v$. For full return after half a revolution, $t=\pi/\Omega$ and this yields $v=2/\pi r \Omega$.

A rather remarkable property of the expression for the strength is its increase with the length. With a given velocity, the strength is the stronger the larger region the motions extends to. This shows that in spite of the slow rotation of Earth, strong deflections are present on large, geographical scales. In the motion of the atmosphere and of the ocean Coriolis deflections are never negligible. In fact, the Coriolis force overcomes most of the other forces.

Estimate the strength of deflection on Earth at mid latitudes a) for a wind blowing with $U=10 \text{ m/s}$ in a cyclone of the size $L=1000 \text{ km}$ and b) for an oceanic eddy with a radius of $L=100 \text{ km}$ in which water moves with velocity $U=1 \text{ m/s}$. Result.

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The ratio $\Omega_V L /U$ is $5\ 10^{-5}\ 10^6/10=5$ in a cyclon (a), and $5\ 10^{-5}\ 10^5/1=5$ in an oceanic eddy (b). The fact that the numbers are identical, indicates that the effect of the Coriolis force is of equal importance in the atmosphere and in the oceans.

### Cyclones as atmospheric sinks

Mid-latitude cyclones are known to bring rain. This is due to a low pressure region in their center, causing the upwelling of wet air from the surface which then leads to precipitation.

Upwelling can only be present if there is a flow toward the center in the lower layers. Thus the figures

used when discussing water draining from a bathroom, is perfectly valid for the air flow towards the center of cyclones. Air parcels are exposed to the Coriolis effect and turn, on the Northern Hemisphere, to the right (on the Southern to the left). The consequence is a huge air vortex spinning in counter-clockwise (clockwise) direction. It can be considered as a gigantic plughole vortex draining upward.

All cyclones follow this rule without any exception! This is so since, as we saw, hardly any other effects can compete with Coriolis on large scales. What we have not found to hold in our plugholes, is robustly present in our every-day weather forecasts dominated by cyclones. The latter always rotate as the Coriolis force dictates. (Anticyclones rotate, of course, in opposite directions but - lacking the clouds to trace out the spiraling motion - they are not visible in weather maps.) At the Equator, without an $\Omega_V$ and, therefore, a deflection effect, cyclones do not exist! (Note that tropical cyclones, or hurricanes, are not equatorial cyclones: they occur above 15 degrees to the North or the the South only.) How could then the plughole vortices seen in the very first video be anything else but faked?

### The air does not blow in the direction it is pushed but at a right angle to it

There is a single force which can be compared to the horizontal component of the Coriolis force on large scales. This is the pressure force arising from horizontal pressure differences. The direction of this force (from higher to lower level) is the direction in which the material is pushed. The resultant force accelerates air (or fluid) masses. Since velocity changes are rather slow in the atmosphere and in the ocean, the acceleration can practically be neglected. Accordingly, the Coriolis force $\boldsymbol{F}_C$ and the pressure force $\boldsymbol{F}$ originating from horizontal pressure differences compensate each other. The vectors $\boldsymbol{F}_C$ and $\boldsymbol{F}$ are opposite to each other, as the figure shows for the Northern Hemisphere (the straight lines represent curves of constant pressure, isobars, and force $\boldsymbol{F}$ points from larger to lower pressure).

The velocity vector $\boldsymbol{v}$ is, however, perpendicular to the Coriolis force, and you can see that velocity is perpendicular to the pressure force, too. As a result, the air or the water masses do not blow and flow, respectively, in the direction they are pushed, move instead square to the direction of the pressue difference. This is in conflict with your every-day experience, e.g. flow in a pipe, where the fluid moves in the direction it is pushed. No wonder, the Coriolis force is very weak on small scales.

The tendency of moving perpendicular to the driving force can be seen in a simple ball experiment on a carousel. Take a slope with a low tilt angle and let a ball moving on it from a state of rest. Initially, only gravity acts (the analogue of the pressure force) and the ball starts moving downward the slope. With increasing velocity the Coriolis force increases and a motion evolves along a spiral-like path in which the resultant displacement is practically horizontal, i.e., perpendicular to the pressure force. In this video the slope is tilted towards the bottom of the picture.

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### The weather map

As a surprising consequence of the orthogonality of the velocity and the pressure force, the isobars of a weather map, lines of constant pressure, run parallel to the wind arrows so that higher pressure values are on the right of the arrows on the Northern Hemisphere. A historical predecessor: Buys Ballot rule.

For winds in the Northern Hemisphere, the Dutch meteorologist formulated the following rule in 1857: if you turn your back to the wind, the low pressure will be to your left.

On the left and along the upper edge of the figure, two cyclones can be seen: closed isobars with the lowest values in the middle, and wind blowing in counter-clockwise direction.

### Water surface is tilted above oceanic currents

For oceanic currents a surprising consequence of the balance between the Coriolis and the pressure forces is that the water surface is not horizontal above currents, it is slightly tilted. The reason for this is that in slow flowing motions, pressure at some depth basically equals hydrostatic pressure which, in turn, is proportional to the height of the water column. The horizontal pressure difference is thus accompanied by a height difference. If an oceanic current is present with a certain velocity, the height of water should be larger on the right side in the Northern Hemisphere.

The figure shows the average elevation of the sea level (over a month's period, during which waves are, of course, averaged out) above the Gulf Stream. Arrows indicate flow velocities and colors the elevation above sea level relative to a reference value (given in m above mean sea level or mamsl). On the right of the Gulf Stream (longest arrows) elevation is about 60 cm on the right, and -40 cm on the left. The total level difference is thus 1 m across an approximately 100 km wide current. Eddies turning counter-clockwise are clearly discernible as orange patches within the red region since they have a pressure minimum in their centers.

### Laboratory experiments with fluids to demonstrate strong Coriolis deflection in nature

We have seen that the strength of deflection in the plugholes of our homes is on the order of $10^{-4}$. The low value is derived from the slowness of the Earth's rotation which is just 1 turn over a full day. The motion on an equipment rotating about 10 000 times faster could thus be exposed to a strong deflection. Since one day is 1440 minutes, such a rotational speed is not irrealistic, you just have to produce a few revolutions per minute (a lot slower than a CD player). In fact, experiments carried out with rotating tanks filled with fluids not only show strong deflections but, with properly chosen parameters, can faithfully imitate the essence of large scale atmospheric or oceanic flows.

Show that if you see a fluid pattern of size $L=10$ cm due to a fluid velocity of $U=1 \text{ cm/s}$ in a tank rotating with 5 revolutions per minute (rpm), the strength of deflection is exactly the same value as in a cylone or in an oceanic eddy. Result.

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In such a case the rotation period is $T=12$ s. The angular velocity $\Omega = 2 \pi/T \approx 0.5$ 1/s. The ratio $\Omega L /U$ is thus $0.5 \ 0.1/0.01=5$.

To illustrate the strong analogy with nature, let us take a cylinder with a radius of 25 cm, fill it with water up to a height of 10 cm, and rotate it with a speed of a few rpm. When injecting blue dye abruptly from above, penetrating in a few cm depth, an interesting pattern evolves as this video shows.

X One sees a dye curtain evolving (non-existent in stationary fluids!). The identical motion of the dye at any height is a consequence of the Coriolis deflection which only depends on the horizontal dye velocity (the same for any dye particles right after injection, and at any time later), and is thus independent of the height. The dye curtain persists and becomes wavy (when observed from above) due to emerging vortices next to each other, rotating in counter-clockwise or clockwise direction, the analog of cyclones and anticyclones. The optical impression is rather similar to that of Polar Light (Aurora Borealis).

X This is no accident, the motion of ionized molecules is also governed by strong horizontal Coriolis deflection in the stratosphere. For other similar experimental observations see

Other experiments in rotating tanks can be designed to model the essense of the general circulation of the atmosphere, including its change due to a decrease of the temperature contrast between the Equator and the Polar Region (i.e. in the presence of a climate change).

### Summary

This short introduction to the Coriolis force demonstrated that on the scale of our buildings (meters) the rotation of Earth is hardly detectable without carefully designed experiments. This force has the peculiar feature of increasing the strength of the deflection it causes with the extension. On the planetary scales (100 to 1000 kilometers) the horizontal component of the Coriolis force is dominant and only the pressure force can compete with it. This leads to surprising effects, since nothing like it is experienced in every-day life (except carousels and rotating tank experiments). You can't comprehend the elements of the Earth System without knowing about the Coriolis force (let alone the great challenge of our age, climate change). You can get an impression of the global dynamics of the atmosphere and of the oceans by looking at these NASA videos:

note that no spiraling motion, no cyclones and no deflection effect are present at the Equator!

note that no eddies are present at the Equator!

Check your knowledge acquired: Test III

### Recommended further materials:

http://marshallplumb.mit.edu/experiments

http://www.karman.elte.hu/index_eng.php

Copyright: A. Gróf, Á. Szeidemann, T. Tél
This study was funded by the Content Pedagogy Research Program of the Hungarian Academy of Sciences.