One of the questions that came out of trying to duplicate Eratosthenes' method for estimating the circumference of the Earth pertained to the data we collected on the Sun's angle of elevation.
Looking at the data we obtained through the US Naval Observatory (Data Services) web site, we tried to determine if we could find a prediction equation that would describe the Sun's angle of elevation as a function of time.
Below you will find students' descriptions of their exploration into this topic:
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Unfortunately, due to directives from the school system we are unable to give credit on the Web to the students who submitted this work. Congratulations on a job well done, class! Your efforts are appreciated! We recently gained access to a scanner, and hope to add some of the images that students obtained on their calculators. Please stay tuned to this page for those results.
Submission #1:
At first, I graphed the scatter plot of my data on the TI-82 calculator. Then I estimated the two x-intercepts. Next I found the axis of symmetry and the maximum point. From these guides I guessed around and finally arrived at a function that closely matched my data points. That function is f(x) = -1.43(x - 6.5)(x - 17.5).
Submission #2:
How was the approximated function found?
The scatter plot looks similar to a parabola. Assuming that there is a quadratic equation to approximate this graph at least on the given interval, we can find such an approximation. The best form of quadratic equations to approximate graphs is the vertex form:
f(x)=a(x-b)² +c (the vertex is the point (b,c))
The vertex of the scatter plot can be easily read as (12, 43), thus we get the function:
f(x)=a(x-12)² + 43
a has to be a negative number, as the parabola opens downward. A sufficiently close number can be found by trying some values. a = -1 is a good first guess, because the scatter plot looks very similar in shape to a normal parabola. Displaying the graph of f(x)= -(x-12)² + 43 in one window with the scatter plot shows that it is too shallow. So the next guess has to be a number with a higher absolute value. I chose a = -2. This graph turns out to be too thin. The next guess a = -1.5 is shown in the second graph above (please see our poster board display for the graphs as we have not yet been able to scan the images in yet) and is a very close approximation of the scatter plot in the given interval.
f(x)= -1.5(x-12)² + 43.
Submission #3:
I used information from the USNO web site and plotted points corresponding with the angle of elevation and the time of day. With my previous experiences with parabolas I began estimating values using the vertex form of a parabola to find an equation.
Results: The final result for the sun's angle of elevation was within ½% error as calculated by my classmates. My final answer for an equation resulted in
y= -1.39(x-12)^2+43.The final value we determined for the circumference of the Earth was not as accurate as our value for the sun's angle of elevation. Our results were off by 1354 miles or 5.4% error.
Conclusion: I found this activity made me appreciate the intelligence of the ancient scholars, who without any of the modern technology we have were able to attain such accurate results.
Works Cited - Web sites:
http://www.tripquest.com
http://www-groups.dcs.st-and.ac.uk
http://astrosun.tn.cornell.edu
http://www.eranet.gr/eratosthenes/[Return to Top]
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