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22 August 2003

On a Method to Measure the Size of the Earth

by Doug Strain

Executive Summary

A method to measure the size of the Earth by timing successive views of sunset at an ocean cliff or pier is evaluated. Experiment design and some sources of error are investigated. Data collection and reduction procedures are detailed and the results of an experiment are reported. The radius of the Earth is found to be 3722.2 miles, a —6.1 per cent error.

 

Background

In a private conversation (I was kind of eavesdropping) Shawn Carlson described a method (of unknown but probably ancient origin) to measure the radius of the Earth with a tape measure, two stopwatches and a cliff. This intrigued me and caused me to ponder the dynamics of the experiment and potential error sources and the sensitivities thereto.

 

The Method

Two observers, one at the base of a cliff with a view of the western horizon and the other on the cliff top watch the Sun set. As the lower observer sees the last ray disappear, he (she) gives a holler and both observers start their watches. When the higher observer sees the same thing, she (he, whatever) gives an answering shout and both observers stop their watches. A little math later and an estimate of the Earth’s radius happens!

A whole bunch of derivation and math and stuff comes next. If it’s getting really close to sunset, skip directly to "Short Version" at the end.

One (cheap, dirty, non-rigorous) way to evaluate an experimental method is to simulate it. So let’s set up a simulation of the Earth-Sun system and play around with it.

So, How Fast Does the Sun Set?

Figure 1. Click image to enlarge

Figure 1 shows a proposed experimental apparatus; namely a West-facing ocean-side cliff at sunset. The view is sighting down the Earth’s axis from the North Pole (P). The curved feature is the surface of the Earth along a parallel of latitude, where RL is the distance from the axis (not the center) of the Earth to the surface at that latitude; RL = R*cos(latitude), where R is the radius of the Earth. The tangent line is the last ray of the setting Sun, above which is glorious sunshine, while below is fast disappearing into even’s gloom. As the Earth rotates counter-clockwise, the cliff sets behind the limb of the Earth as seen from the Sun. The shadow on the cliff face (the radial lines) climbs higher and higher at successive times ti.

Although I speak of a shadow on the cliff face, one would be very hard pressed to see it. What with refractive and scattering effects of the atmosphere, diffraction at the horizon, the finite diameter of the Sun (about 0.5 deg), etc., all one sees is a gradual darkening ("penumbra" — "almost shadow"). But observers at different points up the cliff face would see the Sun disappear below the horizon at distinctly different times. It is this phenomenon that I’m describing. And "shadow" is much more compact than "point where an observer would see the extinction of the last ray of the setting Sun".

The height of the shadow at any time t can be seen to be

h(t) = RL / cos(a) — RL, (1)

where a is the angle the Earth rotated in the t seconds after T0, the moment the shadow hit the water line.

Figure 2. Click image to enlarge.

In the following discussion, the quoted results are based on an experiment being conducted at 34 degrees latitude with the observers located 5 feet and 30 feet above the water line. (I’ve also assumed it’s the equinox ("equal night"), when the Sun sets exactly in the West, but I don’t know if that’s important yet.) The method and equations, though are good almost anywhere (not the poles) on almost any planet (Jupiter would be a little dicey).

Figure 2 is a plot of the height of the shadow above sea-level during the 60 seconds after T0. For the values of H1 = 5 feet and H2 = 30 feet, the resulting values of T1 and T2 are 10.3920 and 25.4551 seconds. These points are shown in Figure 2 as the little circles.

As an aside, the Earth is pretty darned flat! Although the cliff is receding from the horizon at close to 15 miles per minute, it takes a while before it starts dropping into shadow at an appreciable rate. Even after the first five seconds (with your feet at the waters edge) your knees are still sunlit; it takes about ten seconds for the shadow to reach your eyes, but after a minute (and 15 miles) it’s 166 feet up the cliff.

Another aside: the atmosphere refracts light a lot!1 Especially near the horizon2; exactly where we are looking. This is a problem known to astronomers for centuries. We know that, on average, one Earth (solar) day is 24 hours ("star days" are slightly shorter), so to an observer on the equator, on the day of the equinox, sunset should happen exactly 6 hours after local noon. In fact it’s 6 hours plus several minutes due to the bending of the suns rays by the atmosphere. And, likewise, sunrise happens several minutes more than 6 hours before local noon. How many is "several minutes" depends strongly on local air temperature and pressure, so as long as our observations are within a few seconds of each other, conditions won’t have changed enough to affect our measurements.

 

Analysis of the Method

To find the radius of the Earth, it is necessary to know its spin rate, the heights of the observers and the times of the observations after the Sun first disappeared at the base of the cliff, (no fair estimating T0 from Figure 2, R is already in there).

The rotation rate of the Earth (with respect to the Sun), adot, is 2*pi / (24*60*60) = 7.272e-5 radians per second (all angles in radians unless specified). a(t), the angle through which the Earth moves during the interval T0 to time t, is simply t*adot.

(The rotation period of the Earth is actually 23 hours, 56 minutes and 4.099 seconds3, that is, the time between successive appearances of a distant star (not the Sun) overhead. During one full rotation of the Earth, it has also moved along in its orbit around the Sun, so from noon on one day to noon on the next, the Earth has rotated just a little bit more. The extra rotation is just under a degree, 360 degrees around the Sun in about 365 days.)

If we know the values of H1 and T1 for the height and time of the lower observer and of H2 and T2 for those of the upper, and plugging these into Eqn. 1, we get

H1 = RL / cos( T1*adot ) — RL, and (2)

H2 = RL / cos( T2*adot ) — RL. (3)

Using the notation

c1 = cos(T1*adot) and

c2 = cos(T2*adot),

and subtracting Eqn. 2 from Eqn. 3, we can solve for RL

RL = c1 * c2 * (H2 — H1) / (c1 — c2 ). (4)

We can then find the radius of the Earth

R = RL / cos( latitude ). (5)

It will be seen that T1 and T2 cannot be measured directly because of very imperfect knowledge of T0. We still need the watches, though.

 

Figure 3. Click image to enlarge

Error Sources and Experiment Design

We run into our first problem in Eqn. 4 . Both cosine terms are, for practical purposes, equal to 1.00000, so the difference between them is pretty much 0.00000 AND in the denominator. Small errors in t are going to make large errors in RL.

What happens if the cliff face is not vertical?

As was mentioned earlier, the Earth is pretty flat.4 Even a quarter mile East-West difference results in only a 1 second time error. Qualitatively, if you’re close enough to make accurate h measurements, East-West differences between the observers will not be significant.

What happens due to errors measuring h?

Figure 3 is a plot of the sensitivity in the estimate of RL due to errors in the measurements of h. Eqn. 4 was repeatedly evaluated with intentional errors in H1 (red curve with ‘X’s) or H2 (blue with ‘O’s). A one foot error in either h causes about 150 mile error in R. A one foot error in both will have a negligible effect.

Figure 4. Click image to enlarge.

What happens due to errors measuring t?

Figure 4 is a plot of the sensitivity in the estimate of RL due to errors in the measurements of T1 (red ‘X’s) or T2 (blue ‘O’s). A one second error in T1 causes a mis-estimation of R of around 150 miles and 350-400 miles for T2. The green curve is the error caused by not knowing T0.

Another source of error is that when the Sun sets is subjective. It would seem to be good practice for the observers, anticipating this, to go to the test site for several evenings prior to the experiment and sit on the beach and watch the sunset, then sit on the cliff and watch the sunset and then arrive at a mutually agreeable definition of when the moment is right. O! The sacrifices demanded of scientists!

For this ocean scenario, it’s not practical for the lower observer’s eyes to be half-submerged (the ideal case): the surf; the effects of salt water on the stopwatch; the breathing thing; stuff like that. It will be necessary for the lower observer to be up on the beach. In which case we don’t really have values for T1 and T2, only dW, the length of time between watch clicks.

Figure 5. Click image to enlarge.

What happens if we just assume that T0 is the time of observer one’s sunset? Figure 5 is a plot of errors in R based on that assumption, for various values of H1. OK, that’s not going to work.

So, we’re not quite done yet. It’s necessary to determine T1 and T2 absolutely and we have enough data to do that. It can be shown5 that for small values of a, the curve in Figure 2 (resulting from Eqn. 4) can be very closely approximated by the equation

h = A*t2 (6)

where the coefficient A is some unknown constant.

Substituting H1, T1, H2 and T2 into Eqn. 6 we get

H1 = A*T12, and (7)

H2 = A*T22. (8)

Solving Eqn. 7 for A,

A=H1 / T12,

and realizing that, while we don’t know T1 or T2, we do know that dW = T2 — T1, so that

T2 = T1 + dW, (9)

and substituting for A and T2 in Eqn. 8, we get

H2 = (H1 / T12) * (T1 + dW)2

Which expands to

(H1-H2)*T12 + 2*H1*dW *T1 + H1*dW2 = 0. (10)

Eqn. 10 is a second order ("quadratic") polynomial in T1, in the form, A*x2 + B*x + C = y, where

x represents T1,

y = 0,

A = (H1-H2), (11a)

B = 2*H1*dW and (11b)

C = H1*dW2. (11c)

Solving for x (that is, T1) leads to the so-called "quadratic equations"

T1 = (-B + sqrt(B2 — 4AC)) / 2A and (12)

T1 = (-B - sqrt(B2 — 4AC)) / 2A. (13)

(A quadratic equation can have zero, one or two "roots", that is y can be 0 at zero, one or two values of x, hence the two equations. It’s also possible, for some values of A, B and C, that (B2 — 4AC) is negative, so no square-root, so zero "roots" in that case. But not in this case. The formulation of Eqn. 6, (and 7 and 8) guarantees a single (feasible) solution for T1. Only one of these equations will give the right answer; we only care about that one.)

Remembering that H1 = 5 feet, H2 = 30 feet and dW = T2 - T1 = 15.0631 seconds, and evaluating Eqns. 10 and 12 with

A = (H1 — H2) = 5 — 30 = -25

B = 2*H1*dW = 2*5*15.0631 = 103.92

C = H1*dW2 = 5*15.0631 = 1134.5

results in an infeasible solution, T1 comes out negative. So Eqn. 12 is the "other" equation, let’s abandon it and stick with the right one, Eqn. 13. Evaluating Eqn. 13 with the same values of dW, A, B and C returns T1 = 10.3920, the correct value. Notice that T1 and T2 do not appear in Eqn. 13, only dW. Cool.

Observer reaction times and observer-to-observer communications-latency times are sources of error in the measurement of t. This problem arises when observer A shouts out "NOW", or similar, and there is a slight, but finite, pause before observer B starts (or stops) his (her, you know) stopwatch. A simple solution is to start both stopwatches simultaneously well before sunset then take each to its respective observation point. When the time arrives, observer one stops her watch and starts climbing up the cliff since that’s where the bikes are. Observer two stops his watch at the appropriate time and waits while observer one struggles up the stupid cliff. Subtracting the times on the watches gives dW, which is what we want. This eliminates any missed signals.

 

Experimental Procedure (Short Version)

Before sunset, find a West-facing cliff with an unobstructed view of the horizon, where two observers can be located at different elevations. (Good sites are the Eastern shores of oceans and very large lakes. Recent studies6 have shown that Wichita may be suitable! On (and under) a pier is an excellent site since it would be very easy to measure the heights of the observers above the water line.)

Measure the heights of the observation points above sea level and note the lower one as H1 and the other H2

Start two stopwatches simultaneously and take one to each of the observation points.

At the moment each observer sees the Sun disappear over the horizon, s/he stops the watch.

Calculate dW as the difference in the readings of the watches.

Plug the values of H1, H2 and dW into Eqns. 11 and 13 to get T1,

calculate T2 as T1 + dW (Eqn. 9),

plug H1, T1, H2 and T2 into Eqns.2, 3 and 4 to get RL,

RL and latitude go into Eqn. 5, and we’ve just measured the radius of the Earth!

 

Does It Work?

Observers one and two proceeded on 27 July 2003 to Redondo Beach, CA where there is 1) an ocean pier (33.84 N, 118.39 W) and 2) a nice place to pound crabs with mallets (Captain Kidds). Sunset was scheduled for 8:00 pm PDT. By dangling the weighted (1-1/2"x6" galvanized pipe) end of a tape measure, it was determined that observer two’s eyes were 24 feet above the water. (When another couple asked "Are you seeing how deep it is?", and I told them what we were doing, they were so impressed they were speechless!)

The observers synchronized their watches and observer one made his way back to the beach. (We could see, but not hear each other; the suggested timing method can be used with NO communication.) A rough estimate of where the "water line" was (there was about a 2 foot break) was made by looking up the beach and guessing.

Among the sources of error: 1) where the water line was, 2) although the horizon appeared clear and distinct, the Sun began disappearing significantly above it (~ 1/5 of a solar diameter); a distant fog bank? and 3) observer one was a little "trigger happy" and anticipated rather than clocked "last light". Observer two is of a more sober disposition and waited.

Using the values: H1 = 5.67 feet, H2 = 24 feet and dW = 12.12 seconds and evaluating Eqns. 11, 13 and 9, yielded T1 = 11.46 seconds and T2 = 23.58 seconds.

Evaluating Eqns. 2, 3 and 4 gave RL = 3091.6 miles. Plugging RL and latitude = 33.84 degrees into Eqn. 5 yielded R = 3722.2 miles.

A more generally accepted3 value for the equatorial radius of the Earth is 3964.0 miles, 3950.7 miles at the poles. Using the larger value, the measurement error is -241.8 miles or -6.1 per cent.

 

Recommendations

Be sure to weight the end of the tape-measure.

Prior sunset-viewing practice cannot be over emphasized.

Think of a really good cover story.

 

Acknowledgments

I would like to thank Martha Strain for the ingeniously simple simultaneous stopwatch solution and for being observer two.

 

References

[1] http://mintaka.sdsu.edu/GF/explain/atmos_refr/astr_refr.html

[2] http://www.atm.damtp.cam.ac.uk/people/mgb/refraction.html

[3] http://ssd.jpl.nasa.gov/phys_props_earth.html

[4] http://www.flat-earth.org/

[5] http://www.sosmath.com/calculus/tayser/tayser01/tayser01.html, and/or

http://www.efunda.com/math/taylor_series/trig.cfm

[6] http://www.improbable.com/airchives/paperair/volume9/v9i3/kansas.html