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The Coriolis Force, Part 1

Part 1 of 2: A Qualitative Explanation

The Coriolis force is a fictitious force that deflects large-scale motion on any rotating planet (including Earth).

To put it in simpler terms: if something is moving on a very large scale (e. g. an air mass moving from Mexico to Canada), then it will not move in the direction that you think it will move in. Rather, it will be deflected as it moves, due to the rotation of the earth. If it’s in the northern hemisphere, it will be deflected to the right; if it’s in the southern hemisphere, it will be deflected to the left.

The effect of the Coriolis force on a moving body is called the Coriolis effect. But “Coriolis force” and “Coriolis effect” are basically interchangeable terms.

A few examples would help to illustrate the concept.

Let’s say that you live in San Antonio, Texas, and you want to fire a ballistic missile such that it travels some 2500 km directly north and ultimately lands in Devil’s Lake, North Dakota. (Physics problems often seem to involve gratuitous violence and bizarre actions for which there is no apparent motive.)

However, if you fire your missile directly to the north, it will not land in Devil’s Lake, ND. It will probably land somewhere in Minnesota or Wisconsin. That’s because of the Coriolis force. We are in the northern hemisphere, so motion is deflected to the right. So instead of travelling due north, the missile will be deflected to the east and will land somewhere in Minnesota or Wisconsin.

If the intended flight path goes due north, the actual flight path will curve towards the northeast, due to the Coriolis effect. The Coriolis force deflects things to the right in the northern hemisphere and to the left in the southern hemisphere.

Conversely, let’s say you live in Devil’s Lake, ND, and you want to fire a missile such that it travels directly south and ultimately lands in San Antonio.

Again, if you fire your missile due south, it will not land in San Antonio, as planned. Rather, it will be deflected to the west and will probably land somewhere in New Mexico or Arizona. Since this is still the northern hemisphere, motion is still deflected to the right, so instead of travelling due south, the missile will be deflected to the west.

In the southern hemisphere, it is the opposite: all motion is deflected to the left.

The Coriolis force is named after the French scientist Gaspard-Gustave de Coriolis, who derived the mathematical expression for it (see Part 2) in the 19th Century.

The Coriolis force matters only for large-scale motion. For small-scale motion, the Coriolis force is negligibly small. If you’re throwing a tennis ball across your back yard, you don’t have to worry about the rotation of the earth. It makes no significant difference. On the other hand, if you’re firing a ballistic missile over thousands of kilometers, then you have to take the Coriolis force into account.

There are two properties that affect the strength of the Coriolis force.

First, the Coriolis force is stronger at higher latitudes. The farther you are from the equator, the greater the deflection. So if you fire a missile from the equator to 10°N, it will only be slightly deflected, while if you fire a missile from 70°N to 80°N, it will be deflected to a greater extent.

Second, the Coriolis force is stronger when the motion is faster. That seems logical enough: if an object is not moving, then there’s no Coriolis force acting on it. But when it starts to move, the Coriolis force begins to act on it, and the faster it moves, the stronger the pull of the force.

The Coriolis force is an example of what’s known as a fictitious force. That means that it isn’t really a force; it just appears to be a force because you’re viewing the motion from a moving reference frame.

A common example of a fictitious force is when you’re riding in a car, and the car turns right, and you are pulled towards the left side of the car. There is not actually any force pulling you to the left. It’s just that the car is turning right while you (initially) were going straight, so it seems like there must be a force pulling you to the left …… even though in reality, there’s not. So you could say that from your perspective, there is a fictitious force pulling you to the left.

In the same way, if you fire a missile from Texas to North Dakota, and it gets deflected by the Coriolis force and lands in Minnesota, there was not actually any force pulling it to the east. It’s just that the earth was constantly rotating, and we live on the rotating earth, so from our perspective, it looks like there was a force pulling the missile to the east, even though in reality, there was not. That’s why the Coriolis force is a fictitious force.

In other words, the Coriolis force is an illusion. From our perspective, it looks like there is this force — the Coriolis force — pulling on the missile, but in reality, that’s just because we are living on a rotating planet. In reality, there is no such force.

But even if it’s an illusion, we will always have to deal with it, because we will always be seeing our environment from the perspective of our rotating earth. So even if it’s only a fictitious force, we still have to define it, understand it, and take it into account whenever we analyze large-scale motion.

The main reason why we care about the Coriolis force has nothing to do with people firing ballistic missiles for no apparent reason. Rather, the reason why we care about it is that it is extremely important for meteorology. If the earth were not rotating, the general circulation of the atmosphere would be very different than it is, and the general patterns of weather that we all know would also be very different. The effects of the Coriolis force include:

· The Coriolis force is the reason why there are three cells of atmospheric circulation in each hemisphere, rather than just one

· The Coriolis force is the reason why the trade wins move east-to-west

· The Coriolis force combines with the pressure gradient force to establish geostrophic balance

· The Coriolis force (through geostrophic balance) is the reason why the winds in a hurricane rotate counterclockwise around the eye (in the northern hemisphere)

· The Coriolis force (through geostrophic balance) is the reason why the winds in the upper troposphere tend to flow parallel to the lines of constant pressure.

The Coriolis force (through geostrophic balance) is the reason why the winds in a hurricane rotate counterclockwise around the eye (in the northern hemisphere)

Clearly, the Coriolis force is quite an important phenomenon in meteorology. In fact, it is one of the three major forces that affects the direction of the wind, along with the pressure gradient force and friction.

But my discussion of the Coriolis force has still left three questions unanswered:

1. Why does the Coriolis force deflect things to the right in the northern hemisphere and to the left in the southern hemisphere?

2. Why is it stronger when you’re farther from the equator?

3. Why is it stronger when the motion is faster?

In my next article, I will use math and physics to answer these three questions.


The Coriolis Force, Part 1 was originally published in ILLUMINATION on Medium, where people are continuing the conversation by highlighting and responding to this story.

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The Coriolis Force, Part 2

Part 2 of 2: A Quantitative Explanation

In my last article, I provided a qualitative explanation of the Coriolis force, which is a fictitious force that deflects large-scale motion on any rotating planet (including earth). But at the end of the article, I posed three questions about the Coriolis force which I had not yet answered:

1. Why does the Coriolis force deflect things to the right in the northern hemisphere and to the left in the southern hemisphere?

2. Why is it stronger when you’re farther from the equator?

3. Why is it stronger when the motion is faster?

In this article, I will use math and physics to answer these questions. In order to understand these answers, you need to have a familiarity with Newton’s Laws of Motion and with the concept of a cross-product.

Newton’s Second Law, one of the most famous and important equations ever discovered, holds that force equals mass times acceleration. The net force on a particle is equal to the product of its mass and its acceleration. It is actually a vector equation: F = ma, where a bold letter represents a vector.

In fluid dynamics, it makes more sense to divide both sides by m: a = F/m. The acceleration of some little parcel of air in the atmosphere is equal to the net force per mass that acts on it. Thus, if we want to figure out the path that the parcel will travel, we need to identify all the forces (expressed as force per mass) that act on it and add them up.

There are three main forces that act on a parcel in the atmosphere: the pressure gradient force, the Coriolis force, and friction. The pressure gradient force turns out to be (-1/ρ)grad(p) and friction is usually just written as F. But an expression for the Coriolis force does not come so easily.

(“grad” is supposed to be the upside-down triangle symbol, which means “gradient”. I am not tech-savvy enough to know how to write this symbol in a Medium article. Also, p is pressure and ρ is density.)

Recall that the Coriolis force is a fictitious force. It doesn’t actually exist; it just appears to exist because we are living in a rotating reference frame (the earth is rotating). Thus, if we want to derive an expression for the Coriolis force, we would start by defining a, the acceleration, to be the relative acceleration (also called the apparent acceleration, because it is the acceleration that we would observe from our rotating perspective), and we would then have to derive an expression for a fictitious force (the Coriolis force) that accounts for the fact that our reference frame is actually rotating.

The derivation of this expression is long and complicated. I went through the whole derivation once, and it took me several hours. It also involves several approximations. It is quite a difficult derivation. I will not waste your time by going through the entire derivation, but I will tell you the final result: — f x v. The Coriolis force is approximated as — f x v (notice the negative sign). The vector f is equal to (0, 0, 2Ωsinθ), where Ω is the rate of rotation of the earth (7.292 x 10^(-5) radians per second) and θ is latitude. Meanwhile, v is the parcel’s velocity (its relative velocity, to be exact), and the x refers to a cross product. (Again, in order to understand this article, you need to know what cross products are and how they work.)

Thus, we have our version of Newton’s Second Law for the earth’s atmosphere:

a = (-1/ρ)grad(p) — f x vF

Notice that the vector f will always point either straight up or straight down. In fact, it points straight up in the northern hemisphere and straight down in the southern hemisphere. The reason for this is that the sine of a number between 0 and 90 is positive, while the sine of a number between -90 and 0 is negative.

And that is the key to answering the first of the three questions. Recall the right-hand rule for cross products and visualize — f x v in your head. You can see that in the northern hemisphere, where f points upwards, the Coriolis force points 90° to the right of v, while in the southern hemisphere, where f points down, it points 90° to the left of v. Since v is the (apparent) velocity, this tells you how motion will be deflected by the Coriolis force in each hemisphere.

And that’s the answer to the first question: due to the right-hand rule, the Coriolis force will deflect a moving air parcel to the right in the northern hemisphere and to the left in the southern hemisphere.

The right hand rule explains why the Coriolis force deflects things to the right in the northern hemisphere and to the left in the southern hemisphere.

The answers to the second and third questions both lie in the magnitude of — f x v. Recall that the magnitude of a cross product a x b is given by absinφ, where a = |a|, b = |b|, and φ is the angle between a and b. Thus, if we increase either the magnitude of f or the magnitude of v, we will increase the magnitude of the Coriolis force.

I’ll start with answering the third question, because that’s easier. If you increase the magnitude of v, you will therefore increase the magnitude of the Coriolis force. So when an air parcel is moving faster, the Coriolis force that acts on it has a greater magnitude than when the parcel is moving more slowly. And if an air parcel is stationary (v = 0), the Coriolis force has no effect on it at all.

The answer to the second question lies in the formula for f: f = (0, 0, 2Ωsinθ), where Ω = 7.292 x 10^(-5) and θ is latitude, which ranges from -90° (south pole) to 0° (equator) to 90° (north pole). Recall that sinθ increases continuously as θ ranges from -90° to 90°. sin(-90) = -1, sin(0) = 0, and sin(90) = 1. Thus, when θ is farther away from 0, in either direction, the magnitude of sinθ increases. That means when an air parcel is farther away from the equator, the magnitude of f will be larger.

And that’s the answer to the second question. When an air parcel is farther away from the equator, in either hemisphere, the Coriolis force will have a greater magnitude. That’s why the deflection due to the Coriolis force is greater for an air mass near one of the poles than for an air mass near the equator. In fact, if an air mass is exactly at the equator, there is no Coriolis force acting on it at all, because sin(0) = 0.

The fact that the Coriolis force disappears at the equator presents a challenge for some approximations about the atmosphere. It would be easier for us if the Coriolis force were never 0. But unfortunately for us, the Coriolis force has a magnitude of 0 at the equator.

I have now answered all three questions, by using math and physics. That shows you the explanatory power of mathematics. Mathematics can explain things that a qualitative explanation just can’t.


The Coriolis Force, Part 2 was originally published in ILLUMINATION on Medium, where people are continuing the conversation by highlighting and responding to this story.

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