Part B: What happens if there IS a shadow at both sites?

Now, we know how Eratosthenes determined the circumference of earth by measuring the shadow cast at one city, Alexandria, and determining the central angle when compared with Syene, where there was no shadow. However, how would he have calculated the sun angle if both cities cast a shadow? Actually, the answer is easier than it sounds...

To the right is a model of a cross-section of the earth with a shadow at two cities, City #1 and City #2. The parallel yellow lines are the sun's rays. Notice that neither Sun Angle A or Sun Angle B equals the Central Angle. In this case when both cities cast a shadow, the central angle between the two cities equals the difference of the two sun angles as in Graphic A. However, when the two cities are on opposite sides of the sun's zenith, as in the Graphic B, the central angle equals the sum of the two sun angles. So, what's the significance of this discovery? This means that you can figure out the central angle if you can measure the sun's angles at two different positions on the earth at the same time.
(EXTRA: Open the Java sketchpad sketch and drag the sun rays so that the zenith of the sun is both directly overhead the towns and above and below the towns. While dragging the sun's rays, pay particular attention to the relationship between the Central Angle and the Sun Angles).

As it turns out, schools from all over the world have been using the Internet to repeat Eratosthenes' amazing method since 1997 by participating in the Noon Day project, an online telecollaborative project conducted by CIESE. Participants measure the shadow of the sun's rays cast from a meter-long stick at local noon on the same day in each of the locations and then compare the sun angle and their physical location with other schools. Using this information and the method reviewed above, they are able to estimate the circumference of the Earth using Eratosthenes' method!

For example, students from Manasquan, NJ measured the shadow length of a meter stick at local noon to be 80.5 cm. Using a protractor, they measured a sun angle of 38.9º as in Graphic C and sent this information along with their geographic location (40º N, 74º W) to an online database. From this database, they saw that students from San Juan, Puerto Rico (18º N, 66º W) measured a sun angle of 18.6º on the same day at local noon. From this information, they were able to estimate the circumference of the earth to be 43,170.3 km, only a 7.3% error using 40,000 km as a benchmark! (see below for calculations)

1. Central Angle: 20.3º
   38.9º (Manasquan sun angle) - 18.6º (San Juan sun angle) = 20.3º
2. Number of 'earth slices': 17.7 'earth slices'
   360º (circle)  / 20.3º (central angle btwn. Manasquan & San Juan)  = 17.7 'earth slices'
3. North/south distance (arc length): 2,439 km
   4,430 km (distance from Manasquan to Equator) - 1,991 km (distance from San Juan to Equator) = 2,439 km
   (NOTE: Click why north/south and not just the direct distance? for an explanation)
4. Circumference (estimate): 43,170.3 km
   2,439 km * 17.7 'earth slices' = 43,170.3 km - (see Graphic D for illustration)
5. Percentage Error: 7.3% (using 40,000 km as a benchmark)
   40,000 km / 43,170.3 km = 92.7%
   100% - 92.7% = 7.3%

Now you're ready to estimate the earth's circumference for other locations... Using Manasquan, NJ (location: 40º N, 74º W; sun angle: 38.9º) as your comparison location, estimate the circumference of the earth and calculate the percentage error using 40,000 kilometers as your benchmark for the following locations. Remember, you will need to use the information collected from the schools below to first measure the sun angle using a protractor, then determine how many "slices" fit into this circle, calculate the north/south distance (arc length). Repeat the same process for a each school.

HINT: Use the distance Calculator (back-up) to calculate the distance from each location to the equator. Enter the geographic location of the city (for example 40º N, 74º W for Manasquan, NJ) and the corresponding longitudinal location on the equator (0º N, 74º W) to determine the distance to the equator.

5. Bonus: Why do you think some of the cities had a greater percentage error as compared to others?