The Earth’s Shape and Size? You Can Measure it Yourself — Part 4

In the last three posts (1,2,3) I showed how to establish the spherical nature of the Earth without the use of geography, geometry or trigonometry. All I used was was the timing of pressure spikes seen in barometers around the world as a result of two volcanic explosions — the one earlier this month from the Kingdom of Tonga, and the Krakatoa eruption of 1883 — along with addition and subtraction. This method, unlike any other I’m aware of, is suitable for especially young students; its only difficulties are conceptual, and even these only involve simple demonstrations, such as can be accomplished with a ball and a rubber band.

The timing data showed that it takes 35-36 hours for a pressure wave to circle the globe. (I showed this for this month’s eruption in the first post, pointed out a logical loophole in the second post, and closed the loophole by showing the same was true for the Krakatoa eruption in the third post.) Next, to determine the size of the globe, all we need is to estimate the pressure wave’s velocity. This requires a bit more information; we need some limited amount of local geography, and timing for one pressure spike as it moves across a small region of the Earth. In brief, all we need is to learn how much time X it took the pressure wave to cross a region of known width W; then the speed of the wave is simply v = W/X.

Measuring the Speed of the Wave

Fortunately a number of people made this easy for us, creating animations in which pressure measurements are shown over a brief period while the pressure wave was crossing their home countries. The only hard part is to make sure that we not only measure timing (X) correctly but that we define the width ( W ) correctly. The width has to be measured perpendicular to the direction of the wave (or equivalently, it has to be the shortest distance between the wave as measured at some initial time t and the wave as measured at a later time t+ X). Otherwise, as you can see in the figure, we’ll overestimate W and thus overestimate v. The difficulty of getting this right, along with the intrinsic thickness of the pressure wave, will be our biggest sources of uncertainty in estimating v.

As the wave moves from lower left to upper right, its speed can be estimated by measuring the distance W traveled during a certain time period. But the line drawn for this measurement must be perpendicular to the wave (black arrow); if we let geometry or geography fool us into measuring in any other direction (as in the red arrows), we will overestimate W and thus the speed.

We already have some circumstantial evidence that v varied by less than 5% or so, based on the success of the method I used to check the Earth’s a sphere. (At the end of this post is some satellite evidence that the Tonga volcano’s pressure wave had a nearly constant v; the evidence seems otherwise for Krakatoa, based on the observed timing of pressure spikes.) But still, in order to be certain that v didn’t vary much, and to reduce uncertainties on our measurements, it would be best to estimate v in a few places. I found useful animations of the pressure wave from Germany, China, New Zealand, and the United States. These represent the wave’s motion in four very different directions: north (and over the pole), northwest, southwest and northeast. Here’s the example from New Zealand, which we’ll go through in detail.

The pressure wave from the Tonga volcano crossing New Zealand.

Below are two stills from the above animation, which allows us to see the wave as it first enters New Zealand’s north island and as it exits. The time between the two stills is 1 hour and 2 minutes. How far has the wave traveled in that time? The wave is less obvious in the final still, so while the distance across New Zealand from northeast to southwest is about 720 miles, give or take 10 miles (1140-1175 km), the distance the wave has actually traveled is a bit less certain, perhaps as little as 700 miles or as much as 740 (1125-1190 km). So our measurement of the speed across New Zealand is about 700-740 miles per hour (1125-1190 km per hour.) It would be hard to get a more precise measurement.

Two stills from the above animation, 1:02 apart, showing the pressure wave as it enters New Zealand from the northeast and as it exits to the southwest just a few hours after the explosion.

When I tried to make similar estimates using the other animations from Germany, China and the United States, I found it was challenging if I tried to determine travel distances over times much less than an hour; the uncertainties were too great. But if the time was much longer than that, it became more difficult to determine the wave’s trajectory– remember it’s important to measure the distance in a direction perpendicular to the wave, so as not to overestimate the distance. In the end, using multiple measurements in both China and the United States and one measurement in Germany, I found the following:

LocationSpeed Estimate (mph)Speed Estimate (kph)
New Zealand700 – 7401125 – 1190
China620 – 7001000 – 1125
United States720 – 7601160 – 1225
Germany720 – 8001160 – 1285

The significant spread seen here probably reflects the challenges of an imprecise measurement, rather than actual variation in the wave speed; the round trip times found in an earlier post suggested variation in the speed of no more than 5%. It’s not obvious how to combine these statistically if you really wanted to do this with sophistication, but the whole point of this exercise is to see how far you can get without being sophisticated. So let’s eyeball it: you can see there is a preference for the 680-750 mph range (1095-1205 kph), so let’s take that as our most likely range. Of course you are free to draw a different conclusion from these numbers if you prefer, and to repeat the exercise I’m about to do.

Now that we have an estimate of v, we can determine the Earth’s circumference C. If the pressure wave traveled at constant speed v in the range just suggested, the distance C that it covered in a round trip, which required time T = 35–36 hours, is

  • C = v T = 23800 – 27000 miles = 38300 – 43500 km

The uncertainty of order 15% is not surprising given the difficulty of determining v, and perhaps its small variation from one place to another, combined with the imperfect measurements of the round-trip time.

The true answer for the Earth’s circumference varies slightly; it is 24,901 miles (40,075 kilometers) around the equator and 24,859 miles (40,008 km) around a circle that passes through both the north pole and the south pole. Of course these precise numbers are measured with sophisticated equipment. They lie well within my estimate (and quite close to its central value of 25400 miles, 40880 km). It shows that with this method, someone with no expertise in atmospheric science or surveying techniques, sitting in a chair in his living room, can characterize the planet. The same is true of kids in a science classroom, given a little time and a lot of guidance.

Some Last Thoughts

Admittedly I have used sophisticated equipment too — the computers, servers and communication lines of the internet, barometers with electronic output, software for putting that output into various useful forms, and social media for its distribution. But what I haven’t needed is illumination, travel, or knowledge of anything other than local geography. This method would work even if the Earth were forever in darkness, if international travel was impossible, and if a large fraction of the Earth had never been mapped.

That’s interesting, because all of the other methods I know for showing the Earth’s a sphere and measuring its size rely on light and/or on travel. Aristotle’s method for inferring Earth’s shape, and Eratosthenes’ method for measuring its size, rely on shadows; Eratosthenes needed geometry, too. If you travel off the Earth you can see the Earth from outside, either in visible light or in other invisible forms of light, such as infrared light — but you need the light. Of course you can remain on the Earth and travel around it, and if you’re really very careful you can learn about the planet’s shape and size without doing a complete circuit of it. That, however, requires some sophistication, and in particular trigonometry.

Here we’ve let a pressure wave do all the travel, and whether in sunlight or in darkness it has left its trace in local atmospheric pressure. We just need the data on that pressure in a few places, mostly without even knowing where those places are. All we need, after that, is addition and subtraction (to find T), followed by a brief application of division (to find v) and multiplication (to find C). I don’t know of a simpler method.

We’re done; now what exactly was the point of all this? I’m sure that there are plenty of people wondering why someone with a Ph.D. in theoretical physics and dozens of papers on particle physics and string theory would spend time showing how to measure something that’s been well-understood for thousands of years. My reasons range from an general interest in history, epistemology and volcanology to a vague concern about how science is taught and understood in the modern world. But that’s a subject for a future post.

Postscript on the Wave Speed

By the way, there’s satellite evidence that the wave speed v was very close to a constant, at least on first half-trip around the Earth. Here’s an animation of the pressure wave on its way out from Tonga (I have not been able to find the original clip), and below is an animation of the pressure wave as it converges on the point exactly opposite Tonga, in southern Algeria. If the wave speed were constant, the converging wave would form a shrinking circle. It’s not quite that, but pretty close! The approximation of a constant speed, while not perfect, is really quite good. And that’s why the methods I used worked so well.

6 responses to “The Earth’s Shape and Size? You Can Measure it Yourself — Part 4

  1. blaisepascal2014

    I think that, given three or more recording sites, you could determine the direction and velocity of the wave. Given times for the wave passing over, say, Seattle, New York, and Miami, there’s only one direction and velocity that would match those timings. Having more stations (LA, London, Rio de Janeiro, etc) would allow you to quantify and minimize the uncertainty, but you wouldn’t have to assume a direction of travel.

    • To use a small number of sites assumes you know more geography and geometry, and have more precision available, than I was assuming; My goal was to use only arithmetic. But it’s a good question: what’s the minimum info that you need if the speed is constant?

      I think you need 4. Let’s start simple and assume the wave passes two sites at the same time t. Then the wave formed a circle passing through those two sites at time t, but you don’t know its radius; for all you know, it could be a great circle, or the two points could be on opposite sides of the circle. If you now add a third site where the wave passes at t, that gives you the radius and center of the circle. But you still don’t know its speed. A fourth site at a different time t’ then fixes the circle’s location at time t’, and thus the speed.

      Now I believe that if you relax the requirement that the first three sites be simultaneous, you can still write down equations that will allow you to determine everything. (But you won’t be able to explain it to a fifth grade!)

      Still, you will want to confirm that the speed is constant both in time and in angle. So you need a handful of additional points to gain confidence in that, and/or observation of the wave coming the other way (or going round-trip) at your original sites. I think there’s no real minimum number for that purpose, nor an optimum unless you specify your goals.

      Do you think I got that right?

      • I think you only need three recording sites. Lay down a coordinate system so that the explosion occurs at the origin. The three recording sites are at coordinates (r1, a1), (r2, a2), and (r3, a3), where the r-coordinate is the radius from the explosion and the a-coordinate is the angle from an arbitrary axis through the explosion. These r- and a-coordinates are unknown because the source of the explosion is unknown and the velocity of the wave is unknown. But, for each pair of recording sites, we get two equation. If T21 is the time difference between detection at sites 2 and 1, then T21 = (r2 – r1)/v. If D21 is the distance between sites 2 and 1, then D21 = sqrt((r2cos(a2) – r1cos(a1))^2 + (r2sin(a2) – r1sin(a1))^2). This gives us six equations total (two for each pair of sites) for seven unknowns: r1, a1, r2, a2, r3, a3, and v. But, our axis for measuring angles was arbitrary, so rotate it to pass through site 1 by setting a1=0. Now we only have six unknowns.

        Actually, I just realized that the formula for the distances assumes a flat Earth since it’s just the Pythagorean theorem. If you’re allowed to know the radius of the Earth, then use the formula for the great-circle distance between the recording sites for the three D equations. If the Earth’s radius is unknown, then you need a fourth site since there are now seven unknowns.

        But, yes, this is not fifth-grade math.

  2. Isn’t this just a bang going at the speed of sound (at whatever height the sensors were, apparently a bit higher in China)? Would we expect the wave to go at any other speed?

    Great series, thanks (but looking forward to more “fundamental” physics when you can).

  3. These decades old measurements allowing calculation of the Earth’s size and shape lead us by ironclad logic to only one astounding conclusion.

    The globe-Earth cabal were already worldwide in the 19th century. Does their power know no bounds?