"Squaring the circle
" may as yet be
an unattainable goal for even the best mathematicians, but the November 2012 edition of
The Family Handyman
magazine had a tip for how to use
a square (of the framing type) and two nails to draw a circle. This is what it said (emphasis added by me):
Make a circle with a square
"Here's a tip for laying out small circles or parts of
circles. Tack two nails to set the diameter you want, then rotate a framing square against the nails while you hold
a pencil in the corner of the square. You might need to rub a little wax or some other lubricant on the bottom of
the square so it slides easily. Don't ask us why this process works; all we
know is that it does.
They're either very honest or they don't think the average reader would understand the explanation.
The Pythagorean theorem is the key, of course, for explaining the reason. For any right triangle:
where 'a' and 'b' are the lengths of the two perpendicular sides,
and 'c' is the length of the hypotenuse. The same equation also happens to be (not by coincidence) the equation for
a circle of radius 'c,' with the center at point (0,0). So, it stands to reason that if all of the parameters are
met (three intersecting straight sides with a right angle between two of them), then the locus of points of all permissible
value pairs (a,b) will result in a circle. It does not matter whether your value of 'c' represents a radius or a diameter.
The hypotenuse will always be the length between the two nails and sides 'a' and 'b' will always be the distance between
each nail and the 90° vertex. QED
Out of curiosity, I dug out my father's old Audels Carpenters and Builders
Guide (printed in 1945) to see if it described the method and if it did, was there an explanation offered. The author did
show how to draw a circle with a framing square, and even described how to find the diameter of a circle whose
area is equal to the sum of the areas of two given circles (not sure why that would be need by a carpenter).
However, an explicit reason for why it all works out is never given.
Here is what is included in the manual:
FIG. 939.-Problem 1: To describe a semi-circle with given diameter. Outer heel method:
Drive brads at points L, F, extremities of the given diameter. With pencil held at the outer heel M, slide
square around with its sides in contact with L, and F, then with the pencil held at M, describe a semi-circle.
Inner heel method:
Obviously if the pencil be held at S, it will be better guided, than
at M. In this method, the distance L'F' should be taken to equal diameter, the inner edges of the square sliding on
the tacks-the same edges (in either case) that guide the pencil.
At the ends of the diameter LF (fig.
939) drive brads. Place the outer edges of the square against the nails and hold a lead pencil at the outer heel M,
any semi-circle can be described as indicated.
This is the outer heel method. but a better guide for
the pencil is obtained by the inner heel method also shown in the figure.
FIG. 940 - Problem 2: To find the
center of a circle. At the points MS, and LF, where the sides of the square cut the circle when placed in any position
with heel in circumference, draw diameter and then intersection will be the center of the circle. Why?
941 - Problem 3: To describe a circle through three points not in a straight line. Let L, M, and F, be the given points.
Join these points with lines LM, and MF, bisecting them at 1 and 2. Apply square with heel at 1 and 2 as shown and
the intersection of perpendiculars thus obtained at S, will be the center of circle which, with radius LS, may be
described through LM and F. To find the center of a circle.
Lay the square on
the circle so that its outer heel lies in the circumference. Mark the intersections of the body and tongue with the
circumference. A line connecting these two points is a diameter and by drawing another diameter (obtained in the same
way) the intersection of the two diameters is the center of the circle as shown in fig. 940.
To describe a circle through three points not in a straight line.
Joint points with straight
lines; bisect these lines and at the points of bisection erect perpendiculars with the square. The intersection of
these perpendiculars is the center from which a circle may be described through the three points as in fig. 941.
To find the diameter of a circle whose area is equal to the sum of the areas of two given circles.
Let O, and H, be the given circles (drawn with diameters LR, and RF at right angles). Suppose diameter
of O, be 3 inches, and diameter of H, 4 inches. Then points L, F, at these distances from the heel of the square will
be 5 inches apart as conveniently measured with a two-foot rule as shown. This distance LF, or 5 inches, is diameter
of the required circle. Proof: LF2
, that is 52
or 25 = 9+16. (this is as close as they come to explaining the phenomenon, but not really)
Posted on 12/25/2012
These items are an archive of past Topical Smorgasbord items that have appeared on the RF Cafe
homepage. In keeping with the "cafe" genre, these tidbits of information are truly a smorgasbord of topics. They
all pertain to topics that are related to the general engineering and science theme of RF Cafe.
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