Rule of Three
Date: 01/23/2002 at 10:08:38 From: Roger Staiger Subject: Geometry My grandfather's grandfather's father (born 1811) was a land surveyor in Southeast Pennsylvania. I am fortunate to have the notebooks he kept when he was studying to be a surveyor. He created these handwritten note books in 1828-29 when he was 17 or 18. The books are interesting in that there were no calculators or slide rules back then and everything is done by hand - even roots. Most of the problems are geometry. He makes liberal use of formulas and what was called back then "The rule of three," but today might be called simple proportions (A:B::C:?). There are several problems for which I am unable to replicate the answer, so I thought I would ask you to assist on one: How high above the surface of the earth must a person be raised to see 1/3 (one third) of its surface? Answer: To the height of the earth's diameter. He does the problem using "The rule of Three." I can not understand his logic, replicate his solution, or obtain my own. Can you assist? Thanks. Roger Staiger, Scotland, MD
Date: 01/23/2002 at 12:32:12 From: Doctor Peterson Subject: Re: Geometry Hi, Roger. Fascinating story! You are right that the rule of three refers to simple proportion; it's incredible to me how recently such a simple process was considered special enough to have a name, but it goes back to medieval times, I believe. I would first ask how to find the area of the portion you can see, which would be a spherical cap: http://mathforum.org/dr.math/faq/formulas/faq.sphere.html#spherecap S = 2 Pi r h where r is the radius of the earth and h is the height of the cap. Thus the area is proportional to h. How is that related to your height d? P+ |\ | \\ | \ | \ d| \ | \\ | \ | \ | \\ *********** \ ***** | ***** \ **** | ****\ ** h| **\ * |Q *\ R **-----------------+-----------------** * | ---- * * r-h| ---- * * | ---- r * * | ---- * *---------------------+---------------------* * O * * * * * * * ** ** * * ** ** **** **** ***** ***** *********** Comparing similar triangles OPR and ORQ, we find that r+d r --- = --- r r-h Cross-multiplying, (r+d)(r-h) = r^2 r^2 + dr - hr - dh = r^2 d(r-h) = hr d = hr/(r-h) Now, to make the area 1/3 of 4 pi r^2, the area of the earth, h must be 1/3 of 2r. Plugging this in, d = (2r/3)r / (r - 2r/3) = (2r/3)r / (r/3) = 2r How does that compare to his solution? I have to assume that the formula for the spherical cap, or at least the knowledge that the area is proportional to the height, must have been assumed. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
Date: 01/24/2002 at 09:16:13 From: Roger Staiger Subject: Geometry/Thank you Thank you for the solution to the problem. I need to study it, but it seems to be similar to the 1828 solution. It would appear that the early solution assumed that the area is proportional to height. This is not an assumption that was obvious to me until your answer. I do not have the original textbook, only the handwritten notebook. There is a reference in the notebook to "The Remainder of Promiscuous Questions, in Bonacastles Mensuration (sic, I think)." This may have been his textbook. I believe his notebook answers were numbered consistent with the text. Someday when I have extra time, I intend to stop into the Library of Congress and see if such a book exists. Thanks again. Roger Staiger
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