Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-l] method for measuring the size of the earth



David's idea is a nice thought experiment--if it could work. Unfortunately,
the shadow is not sharp enough for the experiment to work over a short
height difference. From where I live, in northern Utah, the mountains
reach to 10,000 feet or so from the valley floor of 4500 feet. You can watch
the shadow climb up the mountain quite nicely in the evenings. The shadow
climbing up the side of a building doesn't work, even if you have a very tall
building. I have thought of David's method, and came to realize that you
need a much greater height than most buildings offer.

There are two other methods that I have used to determine the size of the earth.
The first was published in TPT (Nov. 1987, pp. 500-501) along with Richard Hills.
That was a double sunset observation from an airplane. The time between
the two sunsets was about 17-18 minutes, and allowed a calculation of the radius
of the earth. You have to take off near sunset for this to work. We were flying from
Chicago to Atlanta and took off exactly at sunset, so it worked nicely.

The second method requires two measurements separated by a north-south
distance of a few hundred miles. At first it sounded like David wanted a method
that he could do at a single location. Then he suggested someone going to
Miami. The second method I am suggesting would work
very nicely between Miami and any northern city. I did it between Tucson, Arizona
and Ogden, Utah. On successive nights, or the same night if you have two people,
determine the elevation angle of Polaris above the horizon. The difference in
angle is the difference in latitude between the two observing spots. If the two cities
are along nearly the same longitude, all you need is the distance between the cities.
If they are along different longitude lines you need to adjust the distance so only
the component north-south is used. This is really a variation of Eratosthenes'
method. When I did it I got the circumference within an acceptable 10%. Your hand
outstretched at arm's length makes a fair protractor for this determination. The angle
between the tips of the thumb and little finger is about 20 degrees at arm's length.

Rondo Jeffery
Weber State University
Ogden, Utah 84408

"David Bowman" <David_Bowman@georgetowncollege.edu> 02/26/09 7:17 AM >>>
I've come up with a fairly simple means of measuring/calculating
the size of the earth using only local measurements (not requiring
multiple sightings at far away locations like Eratothenes' method
needs).

The idea is to observe and time the motion of the terminator at
sunset/sunrise ascending or descending the face of a building,
pole, or other tall structure with an exposed vertical face. It
is a fairly simple exercize in trigonometry to realize that if one
is situated on the equator during an equinox that the terminator
ascends vertically at sunset and descends vertically at sunrise
with a constant acceleration that is related to the rotation rate
of the earth and the radius of the earth. In this situation it is
a high school-level calculation to see that if a is the vertical
acceleration of the terminator then the radius R of the earth is

R = a*(T/(2*[pi]))^2

where T is the duration of one solar day.

If the observing location is not on the equator then there is a
correction for the local latitude of the observer. And if the
time of observation is not during an equinox then there is another
correction involving the declination of the sun due to the
tilted path of the sunrise/sunset at the horizon. Figuring out
these corrections is *not* at the high school level, but they
don't involve anything more than a lot of complicated
trigonometry. The date/place corrected formula involving these
corrections is

R*F = a*(T/(2*[pi]))^2

where F is a fudge factor that depends on the complications.
According to my calculations, the explicit value of F is

F = cos^2(D) + sin^2(L)*(cos^2(D) - tan^2(D))

where D is the declination of the sun on the date of observation
and L is the local latitude of the observation.

The actual value of a, being only a few cm/s^2, is plenty slow
enough for easy observation of motion of the terminator up or down
the observing wall using a wristwatch. To get the acceleration only
a few timings are needed between some fiducial marks on the wall
whose separation distance is measured.

BTW, in order for the method to work properly the horizon needs
to be an unobstructed 'true' horizon that accurately represents the
effectively smoothed surface of the earth. This means that the
observation ought to be done along a seacoast or the coast of a
great lake that can't be seen across from the top of the observing
wall. For those whose budgets are unlimited this means that a field
trip to Miami Beach for the purpose of observing the sunrise on the
beachfront hotels would make a very nice educational experience.

David Bowman
_______________________________________________
Forum for Physics Educators
Phys-l@carnot.physics.buffalo.edu
https://carnot.physics.buffalo.edu/mailman/listinfo/phys-l