Ever wonder how fast you're spinning around the imaginary rod connecting the North and South Poles? Wonder no more. By finding your location on this map and searching for where your line of latitude intersects with the bold, black curve, you can determine the speed at which you're currently zipping around Earth's axis.

"I was initially going to make the graphic for Mars," planetary geologist Seth Kadish tells io9, " but people always enjoy a data visualization more when they can relate it to themselves. As no one lives on Mars (yet), I opted to use the Earth."

The result was the visualization you see above â€“ a graph that depicts the speed of a point on Earth's surface for a given latitude due to the rotation of the Earth about its axis. Note that this visualization does *not* take our rotation around the Sun into account (those interested can factor in our roughly 67,000-mph orbital speed, if they're so inclined).

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The speed at which you circle the Earth's axis, of course, depends upon your location on the planet's surface. An object on the Earth's equator travels once around the Earth's circumference (about 40,075 kilometers, or 24,901 miles) each day. Divide that distance by 24 hours (or 23 hours 56 minutes, if you're going by sidereal days â€“ which we'll get to shortly) and you get a speed of about 1,670 km/h, or roughly 1,040 mph. After that, all you need to do to calculated speed due to rotation at any other point on the Earth is multiply the speed at the equator by the cosine (remember trigonometry?) of the latitude at the point. [Trigonometric diagram via]

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But hang on a second. If we're going to bring simple trig into this calculation, it helps to assume that the Earth is a perfect sphere â€“ but it's not. You'd never know it from images like this one, but our planet is actually an oblate spheroid (a sphere that's been smooshed a little bit). It's also not perfectly smooth; the radius of the Earth measured to the base of Mount Everest is smaller than it is measured to the mountain's peak. And what about the fact that a full rotation of the Earth doesn't actually take 24 hours, but roughly 23 hours and 56 minutes â€“ wouldn't that affect our calculations, as well? Well... yes and no. As Kadish explains:

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Though this is an approximation, in an effort to be as accurate as possible, I used the length of a sidereal day (23 hrs, 56 min, 4 sec), which is a full 360Â° rotation of Earth. Because Earth is an oblate spheroid rather than a sphere, I varied the radius as a function of latitude when calculating the tangential speed. The polar radius is 3950 miles and the equatorial radius is 3963 miles; I approximated the radius at other latitudes via a linear interpolation. This has no visible effect on the curve, though. Using the average radius of the earth (3959 miles) as a constant changes the global tangential speeds by <1 mph. Topography of the Earth is equally unimportant for this level of accuracy because the difference between a mountain peak and the bottom of the ocean is trivial compared to the radius of the Earth. If, hypothetically, Mt. Everest's peak (5.5 miles above datum) and the deepest part of the Mariana Trench (6.8 miles below datum) were both located along the equator, the difference in tangential speed caused by the 12.3 mile elevation difference would only be about 3 mph, or less than a third of a percent of the equator's 1040 mph tangential speed.

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As I told Kadish in an e-mail, I've seen this visualization done a couple of times in the past, but I've never seen anyone take the time to explain the considerations that one might take into account when making such a chart (i.e. elevation differences, the fact that Earth is an oblate spheroid, solar vs. sidereal time etc.), let alone explain why those considerations give rise to such negligible differences in one's calculations. Really great work all around.

Check out more of Kadish's visualizations on his utter time-suck of a tumblr, Vizual Statistix.

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