Today is Pi Day, so named because the first three digits of pi are 3.14 and the date is March 14—or 3/14 in the format used in the United States. Yes, on most other parts of Earth today is also March 14, but they write that as 14/3—for them, the best Pi Day is July 22, (or 22/7) which is a fairly nice fractional representation of pi.

You can't actually write down all of pi since it's an irrational number and it has digits that go on *forever*. You can either use a fraction or write it as a decimal—like 3.14. But that's only three digits. How about 3.14159 or 3.14159265359 or even a trillion digits—wouldn't that be better? How many do you really need?

What Is Pi?

Let’s start with defining pi, also written as π. The most basic definition is that it’s the ratio of the circumference and the diameter of a circle. That means that if you take a circle and measure the distance *across* it (the diameter, d) and the distance *around* it (the circumference, C), then C/d = π. It doesn't matter what circle you use—this ratio is the same for *all* circles. A period at the end of a sentence has the same C/d ratio as the Earth's equator. (You can verify this yourself.)

But it's not just for circles. Pi turns up in lots of other places. It's in a random walk, and it's in the time it takes for an oscillating spring to go up and down. You can find pi with a swinging pendulum or with just a bunch of random numbers. Finally, pi is in the Euler identity—which is just a simple (but almost magical) equation.

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Parts of the Euler identity pop up in the solutions to differential equations, like in oscillating circuits, and the solutions to Schrödinger’s equation in quantum mechanics.

Could We Just Use Part of Pi?

We already do. No one ever writes down *all* the digits of pi, because you can't. The question is how much of pi is good enough.

In just about every physics class, we use 3.14—two digits—to represent pi. But could we try shortening it to just the number 3? That sure would make calculations easier. Let’s see what happens if we pretend that pi = 3.

Pi and Your Speedometer

Let's start with the speedometer in your car—no, not the speed reading from your smartphone's map. You know, the actual one on the dashboard, the one that goes from zero to 120 miles per hour. This determines your speed using the rotation of the wheels. Similarly, your odometer measures the distance your car travels based on the rotation of the wheels.

Since one full rotation of the wheels would make the car move the circumference of a tire, we can get the following relationship for the odometer:

Here I am using *s* as the distance a wheel travels and *f* as the number of rotations. If a wheel goes through one complete rotation (*f* = 1), then the distance traveled would be 2πR (the circumference of the wheel). In this expression, *f* can represent partial rotations or multiple rotations. (It is possible to use an angle measured in degrees or radians, but let's stick with the simple count for now.)

Now, what about the speedometer? Now that we have distance traveled, the velocity is just the rate of change of distance. That gives us the following relationship:

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So, what we have is a way to get the linear velocity (v) by looking at how fast a wheel spins (Δf/Δt). All you need is the radius of a wheel (*R*) and the value of π.

OK, now for some fun. Suppose I have a car with a wheel radius of 25 centimeters that travels at a speed of 50 mph (22.352 meters per second). This would have a wheel rotation rate of 14.2297 rotations per second.

But suppose that we went the other way. Let's say the vehicle measured the same rotation rate but used a value of π = 3 to calculate the speed. This would give a speedometer reading of 47.7466 mph (21.3446 m/s). That's a speed error of 4.5 percent.

Pi isn't the only issue here, because speedometers aren't perfect anyway. There's another thing that you have to worry about—the size of your tires. If you use smaller-diameter wheels, then for each rotation of the tires, the car would travel a shorter distance. This would make your speedometer reading too high. If you use larger tires, your speed reading will be too low. Tires can also effectively change size when they wear down or are not inflated properly.

In fact, according to the US Department of Transportation, a speedometer doesn't have to be perfectly accurate. They only have “reasonable accuracy”—which apparently means a margin of error of plus or minus 5 mph. (In other words, an actual speed of 50 mph could read between 45 and 55 mph.) So, in this case, we are good with a π value of 3. That's nice.

Finding the Density of the Earth

Now let’s try using pi with the value of 3 for another calculation: finding the density of the Earth, which is a sphere.

Density is defined as the ratio of total mass to total volume (m/V). We can determine the mass of the Earth by looking at the gravitational force. (Here are all the details.) There are several methods to determine the Earth's diameter—I even did it with a lake. With that, the density just depends on the volume of a sphere.

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Of course, this just gives the average density for Earth. Parts of it, like the surface, have lower density than the core. But still, there it is: The Earth has a mass of 5.972 x 1024 kilograms and a radius of 6.3781 x 106 meters, which gives an actual density of 5,494.87 kilograms per cubic meter.

If you use a value of 3, then the density would be 5,754.21 kg/m3.

That might seem like a huge difference, but actually neither one of those answers is exact. That’s because the Earth isn't a perfect sphere—it's an oblate spheroid. Because of the Earth's rotation, it's a little bit wider across the equator than from the North to South pole. So really, in this case, a π value of 3 wouldn't be so terrible.

What About Trig Functions?

Tons of classic math problems use trigonometry, or the study of the lengths and angles of triangles, but I'm going to work with this classic shadow problem. It goes like this: A tall tree casts a shadow on the ground. The length of the shadow is 14.5 meters, and the sun is at an angle of 34 degrees above the horizontal. How tall is the tree?

Here is a picture:

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Since the ground is perpendicular to the tree, its shadow forms one side of a right triangle. Boom, there's your trig problem. We know an angle and the adjacent side of the triangle (the length of the shadow). Since we want the height of the tree, we need the length of the opposite side of this triangle. That leaves us with the tangent function. (Tangent = opposite/adjacent.)

If we use the single digit version in which π = 3, what would happen to our height calculation? The answer: nothing.

Remember that the basic trig functions (sine, cosine, tangent) are just ratios of sides of right triangles. If you have a triangle with an angle of 34 degrees, then the ratio of the opposite side to the adjacent side is *always* 0.6745. So if you change the value of π nothing happens. It's still a right triangle and still has the same ratio of sides.

But how do we find these values of sine, cosine, and tangent for different angles? The oldest way is to just look them up in a trig table. These are just printed lists with angles and their corresponding sine, cosine, and tangent values. Your pocket calculator does something similar—usually a combination of a look-up table and an approximation of a type to get you that value of tangent (34 degrees). But it doesn't depend on the value of π.

How Many Digits of Pi Does NASA Use?

Let’s see if the number of digits matters when you’re calculating something vast, like a distance in space. For most calculations, NASA uses 15 digits: 3.141592653589793. Is that enough? Well, here is the full answer from NASA’s Jet Propulsion Laboratory, but I will give you the short answer.

In the NASA answer, they describe the digits of pi with an example using the Voyager 1 spacecraft at a distance of 12.5 billion miles away from the Earth. (Actually, that answer was created in 2015, and Voyager is now more like 14.5 billion miles away.) But let's think of that as Voyager's distance from the sun—it's pretty close to the same thing.

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So we can imagine this enormous distance as the radius of a huge circle centered on the sun, as if Voyager was in circular orbit around the sun. We can calculate the circumference of this circle by using 2πR. (I’ll use R = 14.5 billion miles.) Using 15 digits of pi gives a circumference of something like 91 billion miles, which is very long. If you use *more* digits of pi—like, say, 21 digits—the circumference would actually be longer.

But here’s the important part: Even with 6 more digits, you only get a circumference that’s 5.95 inches longer. Could you imagine measuring 91 billion miles and only being off by less than half a foot? That’s super accurate. So there’s not much point in calculating beyond the fifteenth digit. The returns really diminish beyond that point.

But what about just using 1 digit? If you used a value of 3 for π, that would make a circumference that is 9.1 billion miles shorter. Yes, I think that makes a difference.

So, just to be clear—in this case, 1 digit is not enough, and 15 digits is good enough for everything you can imagine. It's even good enough for NASA.

*Update 3-15-2022 4:37 PM: This story was updated to correct the relationship between tire size and speedometer reading.*

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