CHAPTER 2
REVIEW OF RELATED LITERATURE
The Reuleaux triangle
Two of the round windows (IV and XII) contain the shape known as the Reuleaux triangle. Starting with an equilateral triangle, we draw three arcs each with centre at a vertex and radius the side-length of the triangle.
If the triangle has side-length 1, the area of the Reuleaux triangle is ( – 3)/2 and the perimeter is .
The Reuleaux triangle is a curve which has constant width: it can be rotated between two fixed parallel tangent lines. There are many such curves; the circle is the simplest and best-known.
Since a Reuleaux triangle has constant width, it can be rotated inside a square so that it always touches the four edges. This idea has been used to construct a drill which will drill square holes!
Two arcs of the Reuleaux triangle give the characteristic Gothic arch; surprisingly, the triangle itself seldom features in Gothic architecture.
Working Model of Rotary Combustion Some pages for Geometers
Eric W. Weisstein has a comprehensive set of math pages including the Reuleaux Triangle and Constant Width Curves. Alexander Bogomolny has a page on Constant Width Shapes. David Epstein’s Geometry Junkyard has a discussion of the Reuleaux Triangles. Ivars Peterson expounds on Reuleaux Polygons. Mathsoft tells us that the Reuleaux Triangle has the largest Magic Geometric Constant.
William Chui has written a few pages about Franz Reuleaux(f) which are mildly interesting to those with a geometrical bent, though there is a slight confusion about what constitutes a Wankel rotary engine (RVICE creeps in).
A Curve of Constant Width constructed by drawing arcs from each Vertex of an Equilateral Triangle between the other two Vertices. It is the basis for the Harry Watt square drill bit. It has the smallest Area for a given width of any Curve of Constant Width.
The Area of each meniscus-shaped portion is
(1)
where we have subtracted the Area of the wedge from that of the Equilateral Triangle. The total Area is then
(2)
When rotated in a square, the fractional Area covered is
(3)
The center does not stay fixed as the Triangle is rotated, but moves along a curve composed of four arcs of an Ellipse (Wagon 1991).
Kakeya Needle Problem
What is the plane figure of least area in which a line segment of width 1 can be freely rotated (where translation of the segment is also allowed)? When the figure is restricted to be convex, Cunningham and Schoenberg (1965) found there is still no minimum area, although Wells (1991) states that Kakeya discovered that the smallest convex region is an equilateral triangle of unit height. The smallest simple convex domain in which one can put a segment of length 1 which will coincide with itself when rotated by 180° is
(Le Lionnais 1983).
For a general convex shape, Besicovitch (1928) proved that there is no minimum area. This can be seen by rotating a line segment inside a deltoid, star-shaped 5-oid, star-shaped 7-oid, etc. Another iterative construction which tends to as small an area as desired is called a Perron tree (Falconer 1990, Wells 1991).
Shapes of Constant Width
Yes – there are shapes of constant width other than the circle. No – you can’t drill square holes. But saying this was not just an attention catcher. Now then let us define the subject of our discussion. First we need a notion of width. Let there be a bounded shape. Pick two parallel lines so that the shape lies between the two. Move each line towards the shape all the while keeping it parallel to its original direction. After both lines touched our figure, measure the distance between the two. This will be called the width of the shape in the direction of the two lines. A shape is of constant width if its (directional) width does not depend on the direction. This unique number is called the width of the figure. For the circle, the width and the diameter coincide.
The curvilinear triangle above is built the following way. Start with an equilateral triangle. Draw three arcs with radius equal to the side of the triangle and each centered at one of the vertices. The figure is known as the Reuleaux triangle. Convince yourself that the construction indeed results in a figure of constant width. Starting with this we can create more. Rotating Reuleaux triangle covers most of the area of the enclosing square. For the width=1 the following formula is cited in Eric’s Treasure Trove of Mathematics (Oleg Cherevko from Kiev, Ukraine kindly pointed out a misprint in the original quote)
which looks pretty close to 1, the area of the square.
Extend sides of the triangle the same distance beyond its vertices. This will create three 60o angles external to the triangle. In each of these angles draw an arc with the center at the nearest vertex. All three arcs should be drawn with the same radius. Connect these arcs with each other with circular arcs centered again at the vertices (but now the distant ones) of the triangle.
There are many other shapes of constant width. There are in fact curves of constant width that include no circular arcs however small.
Drilling Square Holes
Smith S.(1993). A bit that drills square holes … it defies common sense. How can a revolving edge cut anything but a circular hole? Not only do such bits exist (as well as bits for pentagonal, hexagonal and octagonal holes), but they derive their shape from a simple geometric construction known as a Reuleaux triangle (after Franz Reuleaux, 1829-1905).
To construct a Reuleaux triangle, start with an equilateral triangle of side s (Figure 1). With a radius equal to s and the center at one of the vertices, draw an arc connecting the other two vertices. Similarly, draw arcs connecting the endpoints of the other two sides. The three arcs form the Reuleaux triangle. One of its properties is that of constant width, meaning the figure could be rotated completely around between two parallel lines separated by distance s.
It was with this property of constant width that the Reuleaux triangle was introduced in a sidebar of our geometry text (Moise and Downs, Teachers’ Edition, p. 555). “This figure has constant width,” Smith S. (1993) lectured, “just like a circle.” Without thinking, He volunteered, “Imagine it as wheels on a cart.” “What sort of cart?” “Why, a math cart, to carry his board compass and protractor,” He replied, digging myself in deeper. This was the first of several impulsive misstatements he made about the Reuleaux triangle, only to admit after a little reflection that it wasn’t so. Not in twenty years of teaching had his intuition failed him so completely.
The constant width property can be used to transport loads, but not by using Reuleaux triangles as wheels. If several poles had congruent Reuleaux triangles as cross sections, bulky items could ride atop them (Figure 2). Movement would occur as poles were transferred from back to front, providing a moveable base of constant height.
But the Reuleaux triangle cannot be a wheel. The only conceivable point for the axle, at the triangle’s centroid, is not the same distance from the Reuleaux triangle’s “sides” (Figure 3). If the sides of the equilateral triangle are s, then
2 s sqrt(3) (1) AP = - – sqrt(3) = ——- s »0.577s, 3 2 3
while
sqrt(3) sqrt(3) PB = s – ——- s = s(1 – ——-) »0.423s. 3 3
Even if four Reuleaux triangle wheels were synchronized, the load would rise and fall continuously — you’d need Dramamine to ride this cart.
“And since it has constant width, it would just fit inside a square whose sides are that width,” He continued, trying to regain their attention. He carefully drew a square circumscribing the Reuleaux triangle (Figure 4). The triangle is normally tangent to two sides of the square with two vertices touching the square directly opposite the points of tangency (why?), as in Figure 4a. The exception is Figure 4b, where the Reuleaux triangle has one point of tangency and all three vertices on the square (one directly opposite that point of tangency).
“If the Reuleaux triangle just fits inside the square, no matter what position it’s in, couldn’t it rotate around the inside of the square?” They needed convincing
– a model would have to be built. “But if it did rotate around the inside, doesn’t that mean that a sharp Reuleaux triangle could carve out a square as it rotated?” He had them. “Drill a square hole?”, one countered. “No way!”
That night he cut a four inch Reuleaux triangle from a manila folder to take to class the next day. With a lot of effort, he was able to show the triangle rotate around the inside of a four inch square. “And if this was metal at the end of a rotating shaft, it would cut out a square”,he continued, racking up two more falsehoods. Firstly, it was implied that the center of the Reuleaux triangle would coincide with the center of a drill’s shaft; it cannot. And secondly, the corners of the holes are not right angles, but slightly rounded.
Trying to show the triangle should be centered at the end of a rotating shaft, I stuck a pen through the triangle’s center which, while a student manually rotated the triangle within the square, traced the center’s path on paper beneath (Figure 5). “It’s definitely not a single point,” I had to admit, holding up the traced curve, “but it sure looks like a circle!” Falsehood #4.
Just what is the path of the centroid of a Reuleaux triangle boring a square hole? Assume the square and the equilateral triangle have sides of length 1. Center the square about the origin and position the Reuleaux triangle so vertex A is at (-1/2,0), as in Figure 6a. Using (1), the triangle’s centroid will be P (-1/2+sqrt(3)/3,0). Now imagine rotating the triangle clockwise through the position in Figure 6b, and ending up in Figure 6c, where the centroid is P”(0,-1/2+sqrt(3)/3). The path from P to P” lies in quadrant I. Let a be angle MA’B', ß the angle formed by A’P’ and a horizontal line through A’, and c the y-coordinate of point A’. We are interested in the coordinates of P’. Note that cosa=1/2+c and that ß=270°+a+30°=300°+a. Also note that during this rotation, a goes from 60°to 30° Because A’P'=sqrt(3)/3, if we measure from the coordinates of A’(-1/2,c), the x and y coordinates of P’ can be found.
-1 sqrt(3) -3 + sqrt(3) cos(a) + 3 sin(a) (2) x = – + ——- cos(300?a) = ——————————, 2 3 6
and
sqrt(3) (3) y = c + ——- sin(300?a) 3 sqrt(3) = (cosa – 1/2) + ——- sin(300?a) 3 -3 + 3 cos(a) + sqrt(3) sin(a) = —————————— 6
as a goes from 60?to 30°. Finding the path of the triangle’s center in the other three quadrants is similar in procedure and produces equations symmetric to the origin and both axes.
3 – sqrt(3) cos(a) – 3 sin(a) Quadrant II: x = —————————– 6 -3 + 3 cos(a) + sqrt(3) sin(a) y = —————————— 6 3 – sqrt(3) cos(a) – 3 sin(a) Quadrant III: x = —————————– 6 3 – 3 cos(a) – sqrt(3) sin(a) y = —————————– 6 -3 + sqrt(3) cos(a) + 3 sin(a) Quadrant IV: x = —————————— 6 3 – 3 cos(a) – sqrt(3) sin(a) y = —————————– 6
But these equations do not describe a circle. In equations (2) and (3), when a=30? P is on the x-axis at approximately (0.07735,0). But when a=45?
-6 + sqrt(6) + 3sqrt(2) x = y = ————————, 6
which makes the distance from P’ to the origin about 0.08168. This non-circularity is also shown by graphing the four parametric equations above with a circle whose radius is slightly smaller or larger. In Figure 7, the circle is the outer curve. Notice that the centroid’s path is farther from the circle at the axes than mid-quadrant.
So the Reuleaux triangle’s centroid does not follow a circular path. How then is the Reuleaux bit contained within the square outline it’s to cut? Harry Watts designed a drill in 1914 with a patented “full floating chuck” to accommodate his irregular bits. Bits for square, pentagonal, hexagonal and octagonal holes are still sold by the Watts Brothers Tool Works in Wilmerding PA. The actual drill bit for the square is a Reuleaux triangle made concave in three spots to allow for unobstructed corner-cutting and the discharge of shavings (Figure 8).
Even the modified bit leaves slightly rounded corners. How rounded? Assume the starting position in Figure 9a, in which the Reuleaux triangle is just tangent at point C. As the triangle rotates counterclockwise, C leaves that edge of the square temporarily (labeled C’ in Figure 9b) only to rejoin another at position C” in Figure 9c. In Figure 9b, let a be angle MA’B', ?be the angle formed by A’C’ and the horizontal line through A’, and c the y-coordinate of A’. Then ß= a+60°-90°= a-30°and cosa = 1/2+c. To generate the corner by C, a starts at 30°in Figure 9a and ends up at 60°in Figure 9c. Using A’C’ = 1 and measuring from the coordinates of A’, the coordinates of C’ are described by
-1 -1 + sqrt(3) cos(a) + sin(a) x = – + 1 cos(a-30°) = ———————————–, 2 2
and
1 y = c + 1 sin(a-30°) = (cosa – -) + sin(a-30°) 2 -1 + cos(a) + sqrt(3) sin(a) = ————————————. 2
The equations for the other three corners are similar, and when graphed with the rest of the square yield Figure 10.
Not only does the Reuleaux triangle has practical and interesting applications, and is easy to describe geometrically, but it generates a lot of discussion due to its nonintuitive properties. With this background, you can avoid the blunders I made. Further explorations into the topic might include other figures of constant width (see Gardner and Rademacher/Toeplitz); further identifying the curve of the Reuleaux triangle’s center as it cuts a square; and the shapes of bits for pentagonal, hexagonal and octagonal holes.
Credit:ivythesis.typepad.com
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