Re: [MATHEDCC] circumference earth problem

Vern Kays (vkays@RICHLAND.CC.IL.US)
Thu, 10 Apr 1997 10:28:33 -0500

Approximately 3.18 feet.

Circumference1 = Pi * D1
25000/Pi = D1
D1 = 55234723/6941
Add ten feet
C2 = 25000.001893939.... or 25000 1/528 miles
divide by Pi = D2
D2=22178243/2787
D2-D1 @= 6.028566 E -4 miles in calculator ease
or about 3.1831 feet

Phil if you include your email address I can send the answer directly to you.

Vern Kays

At 12:22 AM 4/10/97 -0500, r philip grizzard wrote:
>Hi, my name is Phil Grizzard and I am a Junior in Math and Math Ed at the
>University of Illinois at Urbana-Champaign. I hope I'm not the only one
>out there who REALLY wants to hear the explanation for this problem.
>>From other responses, I gather that the answer is not microscopically
>close like I envision the solution to be. So please enlighten me.
>
>Thanks,
>Phil
>
>
>On 9 Apr 1997, Jerry Thornhill wrote (among other things):
>
>> Another good one involves assuming the earth is a sphere with a circumference
>> of 25,000 miles. (I know this is not realistic, but it gets worse.) Stretch
>> a string around the equator so that it exactly touches the surface of the
>> earth at all points. In other words, take a 25,000 mile long string and put
>> it around the earth at the equator. Then add 10 FEET (emphasis intended) to
>> the string and form a new larger circle of string that is equidistant
from the
>> earth at every point. How far above the surface of the earth is the new
>> string?
>>
>> Jerry Thornhill
>> jerry_thornhill@sw.cc.va.us
>> Southwest Virginia Community College
>> Richlands, VA
>>
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