CALIFORNIA STATE UNIVERSITY, NORTHRIDGE
CURVES OF CONSTANT HIDTH AND A-CURVES ti
.A thesis submitted in partial satisfaction of the requirements for the degree of Master of Science in
Mathematics
by Hedwig Gertrud Knc-:J.cr-
June, 1978 The Thesis of Hedwig Gertrud Knauer is approved:
(William Karush) Advisor
(Joel L. Zeitlin) (Date · Committee Chairman . \.
California State University, Northridge
ii Dedication
To my husband Wolfgang and my sons Tom, Stephen, and Peter
iii Acknowledgement:
I wish to express my gratitude to
Dr. Huriel Wright and to Dr. Joel
Zeitlin for their constant en couragement and help with this project..
iv TABLE OF CONTENTS
Dedication p. iii
Acknowledgment p. iv •ra.ble of Contents p. v
Abstract p. vi
I. Introduction p. 1
II. History p. 3
III. Definitions artd General Notions p. 4 IV. Curves of Constant Width or Rotors
in the Square
A. Circular Arc Constructions
(l) The Reuleaux Triangle and its
Applications p. 12
(2) Reuleaux Polygons and Their
Construction p. 18
B. Non-Circular Arc Constructions:
Involutes of Astroids p. 20
V. General Theorems p. 24
VI. Involutes of Astroids as Curves of p. 36
Const.ant Width
VII. ~-Curves or Rotors in the Equilateral p. 49
Triangle
VIII.Conclusion p. 61
Footnotes and Bibliography p. 63
v ABSTRACT
CURVES OF CONSTANT WIDTH AND L'l-CURVES
by
Hedwig Gertrud Knauer
Master of Science in Mathematics
This thesis deals with curves of constant width (rotors
in a square), a topic which first appeared in 1780 in a tre~
atise by Leonard Euler and reappeared interrnittently in the
late 19·th and early 20th century in widely different mathe matical contexts. After the turn of the century it was dis
covered that L'l-curves (rotors in an equilateral triangle)
are closely related to curves of constant width. About the middle of this century renewed interest in these curves was
evident as part of the theory of convexity~
In this thesis significant examples of curves of con
stant width and L'l-curves are described, constructed, and
parametrized. A variety of theorems are presented which es tablish major characteristics of these curves. The conclu
sion indicates the present state of research in this field.
vi I. Introduction
Just before the invention of the wheel primitive people probably used logs as rollers to transport heavy loads. The logs' circular cross sections kept the under side of the load always the same distance from the ground. "Ob- viously a wheel must be made in the form of a circle with the hub at the center, since any other form will produce an Fig. 1 up-and-dovm motion." [l,p. 163] If rollers are used, how- ever, the movement of the center is not important. The cross section may be any curve that will serve to keep the load the same distance h above the ground. Such a curve which has always the same width h between a supporting line
(in this case the ground) and the "top" - is called a
"curve of constant width". It can be shown that a curve of constant width h revolves inside a square of sidelength h, and it is therefore called a "rotor in a sguare."
Since curves of constant width are rotors in a square, it is natural to ask if there are rotors in an equilateral triangle. Such curves do exist and are called b.-curve~ ("delta-curves"), because the Greek capital letter
~ resembles an equilateral triangle.
In this thesis we will study the properties of both types of curves and some of their applications. Chapter II
1 2
gives a brief history, Chapter III gives definitions and introduces some tools from differential geometry for subse quent development. In Chapter IV several examples of curves of constant width, including the Reuleaux triangle and in volutes of astroids are described and constructed. The next chapter lists and proves several theorems which char acterize curves of constant width. In Chapter-VI for the particular case of involutes of astroids parantetrizations are developed, the theorems are verified and some of the in volutes are graphed. Chapter VII deals with ~-curves and the analogies to the curves of constant width. II. History
Leonhard Euler discovered and studied curves of con- stant width as involutes of tricusped curves (1778). He called them "orbiformes" [2]. Since then, many authors have written about them, yet in 1958 Eggleston was moved to
·1r.rrite 11 it is surprizing, how little is known about them"
[3]. Barbier studied the curves in 1860 in connection with Buffon's needle problem and found the theorem named after him: "All curves of constant width h have the same per1meter• II [4,J. In 1875 Franz Reuleaux published a "Text- book of Kinematics" in which a large section is devoted to curves which revolve in a square (namely curves of constant width) and to rotors in an equilateral triangle (~-curves).
But the emphasis is on geometrical construction and not on analysis. The well known curvilinear triangle consisting of three congruent circular arcs of radius h is named after
Reuleaux [5]. Many papers on curves of constant width were published in the late 19th and the early 20th century, es- pecially in Japanese Mathematics Journals, but written in
French, German, and English. Papers treating rotors in the equilateral triangle analytically appeared around 1915, again mostly in Japanese Journals [6],[7],[8],[9].
3 III. Definitions and General Notions
To begin the discussion of curves of constant width, we have to define the width of a closed convex curve in a , given direction d*): Each point of the curve C is project- ed perpendicularly onto a line n also in the given direction.
These projections fill the closed line segment AB of the line n. The length of the line segment !AB"J is the width h of the curve C in the direction of n. The line nA, perpendicular to n at A has at least one point in common with the curve C, while C lies completely to one side of nA. Similarly this is the case for the line nB, perpendicular to n at B. These two lines are called a pair of supporting lines in the direct- ion of n. A closed curve has exactly two supporting lines
(= one pair of supporting lines) for every direction n. It should be noted that a supporting line does not have to be tangent. We will see later that a Reuleaux triangle has infinitely many supporting lines which are not tangents at each of it's vertices.
The definition of curves of constant width may now be stated more precisely: For a curve of constant width the distance between every pair of supporting lines is always
*)This discussion follows Rademacher and Toeplitz. pl63ff.
4 5
the same distance h. Now, if one considers any two.pairs of supporting lines of a curve of constant width, they form a rhombus (Fig. 3). If the two pairs are in perpendicu lar directions, they form a square of sidelength h. There fore all squares circumscribed around a given curve of con stant width h are congruent
(Fig. 4). This means that a square of sidelength h can be Fig. 3 rotated around a curve of con stant width h in such a way, that it always touches the curve at four points. Since the motion is relative, it follows, that a curve of con stant width can rotate inside a square of sidelength h. Fig. 4
It is important t.o realize, that only for the circle the center of rotation stays the same. For all other curves of constant width the instantaneous center of rotation changes (Fig. 5). At any point. of contact between the sides of the square and the curve, the center of rotation has to lie on the perpendicular to the square's side at that point.
Since this has to be the case for all points of contact, 6
the perpendiculars have to meet in one point M. As the curve rotates, the contact points change, and so does the center of rotation.
Definitions and Notions from
Differential Geometry
In the following we will need some expressions and defi- nitions from Differential Fig. 5 geometry.
The function
{3.1) X ( t) = (x ( t) , y ( t) ) represents a continuous curve if x (t) and y (t) are defined and continuous functions of a parameter t in some closed interval t1 ~ t ~ t 2 • We can change to another parameter r of t is a monotone function of r in the closed interval r 1
< r < Then the interval r ~ r ~ r is mapped onto t 1 2 1 < t < t in a one-to-one way and we can write 2 X (t (r)) -·(X (t(r)), y (t (r))
or X (r) (x (r), y (r)) for r < r ~ r = 1 2 X (t) can also be looked upon as the path of the end point of the vector X (t) from the origin to the point X with the coordinates (x (t), y (t)). The length of this vector is defined as 2 2 (3.2) lx (t) I = /C (t) + y (t) The linear combination of two vectors x , x is defin- 1 2 7
ed as
( 3. 3)
and a, b real numbers.
The scalar product of two vectors x1 , x2 is defined a.s (3.4)
or (3.5)
where x 1 , y 1 , x 21 y 2 are the coordinates of the end points of xll x2, and 8 is the angle between the vectors.
In particular we have for any vector
( 3. 6} --X . X = 1-12X I and if two vectors x1 '· x 2 are perpendicular (I x1 1 I- 0, jx2 1 f o, 8 = 90°)
x1 · x2 = o X(t} is differentiable with respect to t if this is
true for the functions x(t) and y(t) 1 where the derivative is defined as (3.7)
This represents a vector in the direction of the tangent to the curve X(t), positioned at the point (x(t), y(t)). We will also insist lx' (t} I is never zero. Differentiation of a Scalar product of vectors*}[lO] follov1s the "product rule'~:
*) This follows Stoker [10, p. 9. ff]. 8
(3.8)
The arc length s(t) of a curve from a fixed point Xft)
(x(t ), y(t )) to a variable point X(t) = (x(t), y(t)) = 1 1 is defined as ( 3. 9) s (t) -I- . Jx' (-r) . X (T) dT . 1
=I Vlx' (T) I 2 dT tl and
(3.10)
or with (3.2)
{ 3. 11)
If s is used as a parameter - it is a mono·tone function of ds ' t since dt = IX (t) I > 0 -
2 S = )' J(X' (T)) dT sl
= JS /(dX) 2 dT dt sl and from (3.2) and (3.11)
dxl = ds = 1 lds ds Which means that the· tangent vector is a unit vector when the arc length is used as a parameter. A curve is of unit 9
speed 1 X' (t) t = 1 if and only if the parameter gives arc length from some point. We write
( 3. 12) T (s) = with jT(s) I = 1.
The normal vector N (s) .is defined as a unit vector
which is perpendicular to T (s) 1 its direction is 90° in the
counterclockwise sense from T (s).
In the following we will denote with ' differentiation with respect to a parameter t 1 ·differentiation with re- spect to the arc lengths. Since 1¥ (s)l = 1 we have by (3.4)
T (s) • T (s) = 1 and differentiation gives
(3.13) T"· T = 0 which means ~· is perpendicular to T, and in the same di- rection as N. We can introduce a factor K such that dT (3.14) K = ds = K • N. The vector K = T~ expresses the rate of change of the tangent vector when we move along the curve; it is called the curvature vector. The radius of curvature is defined 1 as r - K If G is the angle between the x-axis and the tangent
(Fig. 6) 1 we have
T = (cos G1 sin G) 1 and by definition
N = (-sin G1 cos G). Differentiation of T results in 10
dG T" {(-sin G) • dG =' ds' (cos G) ds) and the length of T" is
-T. I = dG 1 ds"
In (3.14) we defined the length of if· as K and so r dG (3.15) K = ds"
By differentiation of N we find de G) N" = ((-cos ds' X dG {-sin G) ds) Fig. 6 dG (-cos e, sin G) = ds - which is a vector of length K in the direction opposite to
T and we write (3.16) N" = - KT.
{3.14) and (3.16) are called Frenet's Formulas for the special case of a plane curve:
{3.14) if• = KN
{3.16) N" =-KT.
An involute of a curve is obtained by going from each point X(t) = (x(t), y(t)) of the curve in the negative di- rection of the tangent at that point and marking off the arc length between X (t) and a fixed point X (t ). _ In vector form 1 this is written
Y(t) X(t) + (s(-t) - s(t )) · (- T(t)) = 1 11
We will choose s(t1 ) = 0 so that (3.17}. Y = X - (s) · (T} If we visualize a thread stretched along the curve, a fixed point on the thread traces an involute when the thread is unwound while being kept taut.
The evolute of a curve X(t} is the curve
1 - E(t) = X(t) + K{tT N(t) E(t) is the center of the circle best approximating X at the point X(t).
We will also need Minkowski's support function
[lO,p. 353,359]. It is defined as
(3.18} p = X · N This function may be inter-
preted as the projection of X onto N, and therefore it gives the distance of the tangent line at X(t) = (x(t), y(t})
from the origin (Fig. 7).
Fig. 7 IV. Curves of Constant Width or Rotors in the Square
A. Circular Arc Constructions
(1) The Reuleaux Triangle and its Applications.
The simplest noncircular curve of constant width is
the Reuleaux triangle. It is probably the best known one,
since it is the shape of the cross section of the rotary
Wankel engine. This curvi-
linear triangle consists of
three circular arcs of radius h,
each one subtending an angle
of 60°. The points where the
arcs meet are called vertices.
From the geometry of the curve Yl '- it is obvious that it is a T hi curve of constant width h Fig. 8
(Fig. 8). Indeed, for each point X on one of the circular
arcs between the vertices the supporting line is the tangent
n at that point, because it has one point in common with
the circular arc and the rest of the curve lies completely
on one side of it. The parallel supporting line is the line
n 1 parallel ton through the opposite vertex p, which is the center of the circular arc on which X lies. Therefore the
distance between the supporting lines is always h, which
shows that the Reuleaux triangle is a curve of constant 0 width.. The angle at the vertices is 120 because ~ TPR = 0 0 0 9 0 and 4 SPQ = 9 0 . Since ~ RPQ = 6 0 J 4 QPT = ~ RPS = 3 0°
12 13
and4'rPS = 60° + 2 • 30° = 120° Q.E.D. When a Reuleaux triangle rolls on a straight line, the "top" of the curve will always be the same distance above the line. It's vertices, however, describe paths which consist of straight lines, cycloid arcs, and circular arcs (Fig. 9):
··· .. ?z. I :, I .: \Cj(loid • J •• \ I ·.. I .· . \I R ... 1Ih _Jr;
Fig. 9
The vertex P moves from P to P1 on a straight line since it ,--.... is the center of the circular arc QR. From P1 to P2 it moves on a circular arc because the Reuleaux triangle ro- tates around R. From P2 to P3 the point moves on a cycloid arc because it is now a point on the circumference of a circle with radius h, rolling on a straight line. All the other points move on paths which consist of cycloid arcs
(while the Reuleaux triangle rolls on one of the circular arcs) and circular arcs (while the Reuleaux triangle rotates around one of its vertices).
An early application of the Reuleaux triangle was patented in 1914 by an English engineer, H. J. Watts, living in Pennsylvania. He advertised a drill bit for boring 14
"square holes". He used the Reuleaux triangle's ability to rotate in a square. His drill bit was in the shape of a
Reuleaux triangle with a small area left out for cutting edges and the removal of shavings
[12] (Fig. 10). However, since the Reuleaux triangle forms an angle of 120° at each of its ver- tices, it obviously cannot reach the corners of the square and the
"square holes" drilled by this Fig. 10 drill bit have rounded corners.
Consider Figure 11: in order to find the shortest distance bet.1.veen a corner of the square STUV and the Reuleaux triangle PQR revolving inside it, let us assume PQR is
0 T /' / :\ h._ I /2. I \ D \ / h. \ -----\ B ~1;----- 15° ---,I \ ----...~ ~·1-- v A 5 Fig. 11 15
located so that the diagonal SU is an axis of symmetry.
Then the chord RQ (jRQj =h) is perpendicular to US, and
IDOl = juDj = 1/2 · h. PQR forms an equilateral triangle with sidelength h, therefore the length of the altitude is
I DQj = (h/2) · ~ The length of the diagonal is I USI = h · 2. Therefore \Ps\ = (ust- (!unt + !nP~ = hf21 - h/2 - (h/2) \[3' h ({2'- 1/2 - 'JY/2)
= h(.048) Q.E.D.
Also because the center of rotation does not stay the same it is necessary to allow for the eccentric movement of the chuck, and to guide it properly*). There were several pat- ents taken out on this idea and the related idea of a drill which drills triangular holes[l2].
The Reuleaux triangle has also been used as a film transport device[l3, p. 72]. Since continuous motion of a filmstrip would give a blurred picture, one has to find a device that will give a motion interrupted by rests. ·rf a
Reuleaux triangle rotates about one vertex R inside a rigid frame of height h (Fig. 12) the following motion will result for a point P :.
*>F. Reuleaux shows in his book the locus of the center of rotation. 16
Ro-tation of PR Frame with G as in Fig. 12
oo 60° stays in upper position
60° - 180° drops by h
180° 240° stays in lower position
240° - 300° is lifted by h
300° - 360° stays in upper position
This achieves the desired motion. The movement of the frame is then used to transport the filmstrip.
Probably the most wide-
I ly known use of the Reuleaux 1 I L------1 triangle is in the Wankel rotary engine. It is not Fig. 12 clear however, that the con- stant width property is exploited. The motor housing has the shape of an epicycloid or "epitrochoid"[l4]. It is ob- tained by rolling a smaller circle on the circumference of a larger one, while using a point inside the smaller circle as the "generating point". In this case1 with a the radius of the larger circle, b = a/2 is the radius of the smaller circle, and p = b/2 = a/4 is the distance of the generat- ing point from the center of the smaller circle (Fig. 13).
Then the coordinates of the epicycloid are obtained as follows: 17
a a ~ - b X ~d so X = n ~ 0 TI - = 2 ~ X
TI TI - - ~ (1 + ~) = ~(b + a) 2 b 2 b
TI TI = - i3a) - ~ - 3 2 ~ a 2 ~
Since X = (a + b) cos ~ - p sin 0
y = (a + b) sin ~ p cos 0 3a a we have X cos - cos 3 = 2 ~ 4 ~ 3a a y = z- sin ~ - 4 sin 3 ~
The Reuleaux triangle is able
to revolve inside this figure while the vertices always
touch the epicycloid. Seals are placed at the vertices of
the rotor and so the space be-
tween the housing and the rotor is divided into 3 sep- arate spaces, which form the Fig. 13 coniliustion chambers (Fig. 14).
Another application is mentioned in an article by
J. P. Moulton[l5]. The shafts which are used to turn on or off the fire hydrants in Philadelphia have the shape of a
Reuleaux Triangle. When children or other unauthorized persons try to use an ordinary wrench to turn them, the property of constant width makes this impossible (Fig. 15). 18
It takes a special tool, which is supposedly only available to firemen, to turn the shafts.
(2) Reuleaux Polygons
and their con-
struction
A rather general method
U2]of construction of curves Fig. 14 of constant width is1 to start with any number of intersecting lines, shown for n = 4 in Figure 16. The ,--.... first circular arc PsP1 is drawn witt an arbitrary radius jP1Q1 j and the intersection Q1 of the lines n and n as the center. 1 4 The next arc has the intersection Fig. 15
Q of the lines n and n as the center and radius jQ P j, 2 1 2 2 1 and so forth. This construction results in a curve of con- stant width because
IPlPsl -- IPlQlj + IQ1Psl on the line P P line n 1 5 = 1 = jP1Q2j + IQ2Psl
= IP2Q2j + jQ2P6j
= IP2Q31 + jQ3P6j 19
= IP3Q31 + !Q3P7j on the line P3P7 = line n 3 = jP3Q4j + jQ4P7j
= IP4Q41 + I04Psl on the line P4P 8 = line n4 = IP4Qll + I01Psl
= IPsQ1I + jQlPlj on the line n 1 .
~\
Fig. 16
When n is odd, a special case exists, when the dis- tance between all of the consecutive Qi s becomes equal
(h1 = h 2 = h 3 = h 4 = hs = h) and this is used as the radius of the circular arcs (and hence the width h) • This means that the centers of the circular arcs are the vertices of a
"starshaped" polygon (Fig. 17). If the radius of the cir- cular arcs is equal to the distance between the vertices we obtain "Reuleaux polygon~which are generalisations of the 20
Reuleaux triangle.
Furthermore any curve of constant width yields another by addition of a constant length in the direction of the rtormal at each point of the original curve
(Fig. 18). If the original curve has vertices where circular arcs meet at an angle, circular arcs Fig. 17 of radius d with the vertex as center are added. The width of this new curve is h' = h -!- 2d.
B. Non-circular Arc
Constructions:
Involutes of Astroids.
The constructions have so far shown curves which consist of circular arcs. There are however curves of constant width, which Fig. 18 consist entirely of non-circular parts. They are obtained as involutes of astroids with an odd number of cusps*)[l6]. This is actually the oldest type known. Euler[2] discovered that involutes of tricusped astroids are curves of constant width and called them
"orbiformes". Astroids are epicycloids which are obtained as the path of a point on the circumference of a smaller
1:) Hurwitz showed that the number of cusps has to be odd. 21
circle rolling inside a
larger one. If the ratio of the radii is a rational number, a closed curve is obtained. For the ratio 1/3 the result is an astroid with 3 cusps at A1 , A2 , A3 (Fig. 19). Let the arc length for each arc of the
~ astroid (e.g. A1A2 ) be h. The arc of the in- Fig. 19 volute between A2 and A3 is obtained when a thread is held fixed at A1 and unwound ~ from the astroid arc A1A2 , then rewrapped on the astroid .,.--... arc A1A3 .. The next involute arc between A3 and A1 is ob- tained by holding the thread at A2 , the last arc by holding it at A3 . In this way each point P of the astroid A de termines two points Q1 and Q2 of the involute C. Each point Qi lies on the tangent in the direction of ___..,. Ai at a distance equal to the length of the arc PAi, where
A·l is one of the endpoints of the astroid (or one of the cusps) adjacent to P. That means
jPQlj = jPAlj
jPQ2 ! = !PA2 j ,--.. Since the length of the arc PA1 together with the length of 22
...... -. ~ the arc PA2 gives the length h of the full arc A1A2 , .we have IPQll + IPQ21 = IQ1Q21 and IQ1Q2I = h which is constant. A similar argument holds for the points on the other arcs of the astroid. The definition of the in- volute implies that the tangent of the astroid at P is the normal of the involute at Q1 and Q2 , and the tangents of the involute at Q1 and Q2 are perpendicular to Q1Q2 . Since these tangents are a pair of parallel supporting lines for the involute, IQ 1a2 1 = h measures the width of the involute, which is therefore a curve of constant width.
For an odd number of cusps greater than 3, a similar argument applies, hov1ever, the astroids have to be "star- shaped", that is of type a rather than type b in Figure 20.
I I I / / '- /
Type A Type B Fig. 20
Type b would result in a looped involute indicated by the dotted line,which is clearly not a convex curve and not a curve of constant width. 23
Before dealing with the involutes of astroids in de tail in Chapter VI, general properties of curves of constant width and general theorems will be developed in the follow ing chapter. V. General Theorems
The following eight theorems have been selected to highlight the most significant properties of curves of con stant width, concerning width, angle at the vertices, and area:
Theorem I: The distance between any two points of a curve
of constant width h is at most h.
Theorem II: The angle at any vertex of a curve of constant
width cannot be less than 120°.
Theorem III: The chord connecting two points of a curve of
constant width with parallel tangents is per
pendicular to these tangents (it is the normal
to these tangents) .
Theorem IV (Barbier's Theorem): All curves of constant
width h have the same perimeter as the circle
of diameter h, namely ~h.
Theorem V: For two points with parallel tangents on a
curve of constant width h the sum of the radii
of curvature is equal to the width.
Theorem VI: A curve of constant width has only one point
in common with each of its supporting lines.
Theorem VII: The area of a Reuleaux triangle of width h is (1/2). h 2· <~ '{3). Theorem VIII (Blaschke-Lebesgue Theorem) : Of all curves of
constant width h, the circle has the largest
area and the Reuleaux triangle has the small est.
24 25
In the following.we prove the theorems.
Theorem I: The distance between any two points of a I curve of constant width h is at most h. Proof: Let P and Q be t.wo points
on the curve (Fig. 21) and PQ the a' line segment connecting them.
This line segment is part of a
line n. The supporting lines per-
pendicular to n intersect n at P'
and Q' respectively. IP'Q' I = h by
definition. P and Q lie between P'
P and Q. Therefore !POl is at Fig. 21 most equal to h. Q. E. D. r Theorem II: The angle at any vertex of a curve of constant width cannot be less than 120°.
Proof [l,p. 173]: If the angle at the vertex Q is G, then
the bundle of supporting lines
at Q occupies an angle (180° -
8), as can be seen in Figure 22.
The perpendiculars to the two
extreme supporting lines include
bet.ween them the same angle
(180° - 8), and a circular arc
of radius hand length h•(l80°- Fig. 22
8)· rr/180 must lie between them. By Theorem I the length of
the chord of this arc cannot exceed h (the maximum width of 26
the curve) . Since h is also the radius of the, arc PR, the angle "t,.. PQR cannot exceed 6 0°. Therefore
180° - 8 < 60°
- 8 < -120°
8 > 120°
Q. E. D.
Barbier's Theorem.
In 1860 E. Barbier [17] investigated curves of con- stant width and found that all curves of constant width h have the same perimeter as a circle of diameter h, namelyrrh.
There are several proofs of this theoren1. It is possible to use only high school geometry, but the proofs are somewhat a\vbvard - using inscribed and circumscribed polygons.
With the methods of differential geometry the proof is much simpler [11, p. 47]:
Let X represent a closed, twice differentiable*) curve for which the vector of cur- vature always points to the interior (Fig. 23). This may be accomplished by re- versing orientation (and re- placing s hy (l - s) if need Fig. 23
*) This requirement excludes Reuleaux polygons, because at their vertices the first derivative is not defined. However, it is possible to prove the theorem in a diff erent way in that special case. 27
be. Then the tangent vector turns continuously to the left,
if we follow the curve counterclockwise. Every point P =
X(s) has an "opposite" point P X(s (s)) for which the 1 = 1 tangent is parallel and ~ = - T; that means the tangent 1 vectors point in opposite directions. Then we can write
( 5 .1) xl = X + m~T + h·N where (as in Figure 24)
I m(s) I = the distance between the normal lines
I h (5.2) = x· m"T m~· h"N hN" x·1 + + + + Since = X(s {s)) it follows that xl 1 ds ds ds x· = dXl . 1 = . 1 -T 1 1 ds ds Tl ds = ds 1 Using Frenet's formulas we may rewrite (5.2): - ~ dsl = ~ + m"T + mKN + h"N - hKT ds Equating the coefficients of T and N on both sides yields 5 - = 1 + m· - hK as~ 1 or (5.3) 1 + dsl = dm d.s - ds + hK 28 and (5.4) 0 == mK + h" If X is a curve of constant width then h" = ~~ = 0 and cts ( 5. 4) becomes mK = 0. Since K is not 0, we obtain ( 5. 5) m(s) == 0 and formulated this result as r Theorem III: The chord connecting two points of a curve of constant width with parallel tangents is per- pendicular to these tangents (it is the normal to .__ ___these _ t.angents) . From (5.4) we can deduce Barbier's Theorem (Theorem IV): All curves of constan width h have the same perimeter as the circle of diameter h, namely n·h. Proof: The perimeter of the curve is L L L == J I x· (s) I ds == f ds = s=O 0 s s s ds 0 = jo ds + jo 1 ds == ds + jodsl = ] (ds + ds,) ds Ju .!.. 0 0 0 0 0 where s is the arc length from the point with s = 0 to the 0 point with the parallel tangent. From (5.3) we deduce hi<' = 1 + dsl ds d8 ds or h . 1 1 ds = + ds d8 ds ds ( 5. 6) h . 1 ds = ds + ds s TI hence L jo (ds ds ) h·d8 hTI D. == + 1 == J = Q. E. 0 0 29 There is a nice proof of Barbier's theorem for Reuleaux polygons, which consist of circular arcs of radius h (which \-Te considered earlier) . This proof gets the perimeter from the sum of the interior angles and uses only high school geometry [13, p. 259 ff]. Going back to (5.6) we can d lVl. 'd e b y dGds an d h ave t h en ds ds ds ds h - . ds dG + ds 1 dG or ds ds h - de + dG 1 Using the definition of the radius of curvature r (8) = ds/d0 we have (5.7) h = r(G) + r (e) 1 We can express this result as Theorem V: For two points with parallel tangents on curve of constant width h the sum of the radii of curvature is equal to the width h. ~orem VI: A curve of constant width has only one L_____~t in common with each of its supporting lines. Proof [1, p. 168]: Assume that two points P and P of the 1 2 curve C lie on the supporting line s (Figure 25). The other supporting line s' touches C at Q. The distance between s and s' is h. By definition IP1QI =hand P1Q is perpendicular to s at P , and perpendicular to s' at Q 1 (Theorem III). Since P P Q is a triangle, ~ P P Q cannot 1 2 1 2 be a right angle. Therefore IP QI would be greater than 2 30 the side P 1Q = h. By Theor- em I this is not possible and therefore the assumption is false. Q. E. D. Fig. 25 r Theorem VI I : The ar e'a'C!-o_f.:c_,t~h_,e,__R_e_u_l__.e.,..a_u_x __ t_r_i-ang le of I width his (l/2)·h2, (~- {J). l R Proof: In Figure 26 the equi- lateral triangle PQR has the \ I \ 2 " area h ~/4. The circle lh \ I \ sector centered at P has the \ I Q. I \ area h 2· n/6. Therefore the area of each segment (between the chord and the circular arc) is Fig. 26 and therefore the area of the Reuleaux triangle of width h is equal to the area of the equilateral triangle plus 3 times the area of the segment: 31. = (h2 /2} ('IT - '{3-) Q. E. D. At this point we want to recall that the area of the circle of diameter h-and therefore of radius h/2- i~ h 2·TI/4. The following theorem states that the area of this circle and the area of the Reuleaux triangle of v1idth h are extreme cases of the area of all curves of constant width h: The Blaschke-Lebesgue Theorem (Theorem VIII: Of all curves of constant width h, the circle has the larges area, and the Reuleaux Triangle has the smallest. Proof: The first part of the theorem follows from the iso perimetric inequality and Barbier's th8orem: The isoperi metric inequality says that the circle of radius h/2 has the largest area of all curves of perimeter Tih. According to Barbier's theorem all curves of constant width have the same perimeter nh and therefore the circle of radius h/2 has the largest area. There are several proofs of the part of the theorem, which states that the Reuleaux triangle has the smallest area of all curves of constant width h. Eggleston's proof is one of the more recent ones. It requires some lemmas from the theory of convexity, which are now stated: Lemma l (Besicovitch[l8]}. If two points of a curve of constant width h lie on a circle C of radius h then the 32 whole of the smaller arc of C connecting the two points be- longs to the curve or lies inside of it. Lemma 2 (Pal[l9]). Any curve of constant width h can be in- scribed in a regular hexagon of width h. Lemma 3[20]. The area of a convex curve is 27r 1 . 2 ·J r (8) p(8)·d8 0 v1here r(8) the radius of curva·ture p(8) the support function, defined as p = - X . N G the angle of the tangent w.ith the x-axis. Proof of the theorem D (Eggleston) (Fig. 27) : Let R be any curve of constant width I. • • .t .·s h. Let R1 , R2 be Reuleaux • I IR I 2. triangles of width h. Let F I I c . I ABCDEF be a regular hexagon ·., ·,., J circumscribed about R 2 '·· ... R ·. I '• ... (Lemma 2). Let R be so locat 1 A ed that its vertices are at B,D,F (dotted line in Fig. 27), R2 be located with its ver tices at A,C,E (dashed line). Let 8 be the angle of the tangent through a point on R with the initial direction AB. The origin is assumed to lie inside of R, R1 , and R2. R touches the sides of the hexagon BC at I 0 where 8 = ~/3 CD at J where 2~/3 0 G = DE at Ko where 8 = 1T EF at I2 \vhere 8 = 41T/3 FA at J2 where 8 = 51T/3 AB at K2 where 8 = 21T r (8) is the curvature radius and p(8) the support function for R, r 1 (8) is the curvature radius (=h) and p 1 (8) the support function for R1 , r 2 (8), p 2 (8) the same functions for R2 • Since R touches the hexagon's side BC at I and side CD at 0 ~ J , the arc BD of R intersects R ~omewhere in the points 0 1 I 1 and J 1 · r 1 and J 1 lie on a 6ircular arc of radius h (centered at F) and therefore by Len®a 1 the part of this arc between r 1 and J 1 lies inside R. This means that the J tangents (= support lines) for R between I 0 and 0 are fur- ther away from the origin than the tangents on the circular --. arc BD of R1 . We can express this by using the support function (5.8) p (8) 2_ p(8) for 1T/3 < 8 < 21T/3 1 The same argument applies for ...-...... ---.. the circular arc DF of R and arc K I of R for 1r<82_ 41T/3 1 0 2 "'"'- ...---... and the arc FB of R1 and arc J 2K2 of R for Srr/3 <8< 2n. ""- Now we consider for the arc EA of R2 (dashed line) and the ~ arc r 2J 2 of R: for 4n/3 < 8 < Sn/3 34 .....-.... ,.---..._ < and also for AC of R2 and arc K2Io of R2 for 0 e ,< TI/3 or 21T< e '< 71T/3 .,.-...... ----.. for of R and of R for 2TI/3 < e < TI CE 2 JOKO or Sn/3 < e < 31T But p (e) = p (e - TI), which means the same as 2 1 (5.9) pl (e) .::._ p (e + 1T) for TI/3 < e < 2n/3 and 1T < e < 2n/3 and 5TI/3.s_ e < 21T By Lemma 3 the area:zu of R is l/2 ·J r(e)·p(e)·de (5.10) ~ 1/2 15"1... r:+ F) (r(e) p(e) + r(e+TI)• ![: 7r s-it p(e + n ))de .:'> 3 Because of (5.8) we have r(e) p{e) > for these e and because of (5.9) we have r(e + n) p(e + n) > r(e + 1T) p (e) for there 1 quired e. Therefore the integrals in (5.10) are z_F LJ.j' 211 LJ(r(e) +r(e+n)) p (e) de ~-l/2{! + L+ 1 ...,- II "SIJ ~ :5 Using theorem V we can write 2.11 'f.!l. .3 .211 + -t- . = 1/2 { t f J }(h•p1(0) de) T 11 Sll 3 .3 since h is the curvature radius for the Reuleaux triangle we can write :t7i = 1/2 p (8) d8 f rl (8) 1 0 which is the area of the Reulea.ux triangle (again by Lerruna 3). Q.E.D. VI. Involutes of Astroids as Curves of Constant Width. In this chapter we shall construct curves of constant width which arise as involutes of astroids. We shall give. a parametrization of the star-shaped hypocycloids parametri sized by a multiple of arc length, then we will construct an involute of these astroids, using as parameter8 the angle between the tangent line and the x-axis (so that the point opposite 8 is at (8 + n). It is then verified that these involutes are curves of constant width and also that Barbier's theorem holds. In addition a computer assisted graph of some examples is given. We start with hypocycloids which are defined to be the curves generated by a point on a small circle rolling inside a large circle. In Figure 28 let a = radius of the large circle b = radius of the small circle. The center of the large circle is at the origin 0. Let A = '/ (a,O) be the point where the larger circle intersects the x-axis and where the two cir cles touch initially. Let B be the center of the moving X smaller circle. Let P1 be A the point where both circles touch after the smaller one Fig. 28 36 37 has rolled from A to P . Let P be the po.int on the roll 1 2 ing circle which touched the larger circle at A. We will trace the path of this point. ,.....--..,. P P x the angle which subtends the arc 1 2 of the smaller circle w the angle between the negative y-axis and BP 2 . Since the small circle is rolling inside the large one, the arc These relations between the angles hold: w = (TI/2 + ¢) -x = n/2 + ¢- (a/b)¢= n/2 + ¢(1- a/b) = n/2- ¢(a- b)/b sin w = cos ¢ (a - b)/b; cos w = sin ¢ (a- b)/b Therefore the vector from the origin to B is B = ((a - b) cos ¢, (a - b) sin ¢) and the vector from B to P2 is P = (b cos ¢ (a - b)/b, -b sin ¢ (a -b)/b) P2 is the endpoint of the vector which is found by X(¢} = B + P = ((a- b) cos¢+ b cos (¢ (a- b)/b), (a- b) sin¢ -b sin (¢ (a- b)/b)) The coordinates of P2 are x = (a- b) cos ¢ + b cos (¢ (a- b)/b) ( 6. 1) y = (a - b) sin ¢ - b sin (¢ (a - b)/b) This hypocycloid becomes an astroid with an odd number of 38 cusps 2k+l if b/a = m/(2k+l) (m = 1,2,3, ..... ; k = 1,2, ... ) To obtain a curve of constant width as involute of this curve, we have to choose b so that the arc between the cusps after two full revolutions of the small circle is (2~a)/(2k+l). Thi~ assures that there are no other cusps between them, and the astroids are "star-shaped" (Fig. 29). Since the circum- E'ig. 29 ference of the large circle is 2~a and 2 full revolutions of the small circle give 2 · 2~b we must have 2na - 2 · 2~b = 2~a/(2k + 1) or a - 2b = a/(2k + 1) 2b -· a [ 1 - 1/ ( 2k + 1) ] b = (a/2 ) . ( 2k + 1 - 1) 2k + 1 b = (a/2) · [2k/(2k + 1)] = ak/(2k + 1) The coefficients in (6.1) become (a b) = a(l- k/(2k + 1)) = a( 2k + 1 - k) = a(k+l)/(2k+l) 2k + 1 (a - b) /b = a ( k + 1 ) · ( 2 k + 1 ) = ( k + 1) /k (2k + 1) a k Using these in (6.1) leads to ·k + 1 ak x = a 2k + 1 cos ¢ + 2k +l cos (¢ (k + 1)/k) y = a ~k++l sin ¢ - 2 ~~ 1 sin (¢ (k + 1)/k) 39 or X = [a/(2k + 1)] . [(k + 1) coset> + k cos(cp (k+l) /k) J (6.2) y = [a/(2k + 1) J .. [(k + 1) sincp - k sin(cp (k+l)/k)] For 02_cf>2_ 2rrb/a = 2rrk/ (2k+l) we trace one arc of the astroid. To get all (2k + 1) arcs we need therefore O::_cp~ (2k + 1) . • 2rrk/ (2k + 1) = 2kn or 02._ cp ~ 2krr. In Chapter III it was stated that the involute of a curve is obtained by (6.3) Y - X - s · T where X represents the original curve s ·the arc length T the tangent vector of unit length and all of these are functions of cp. In our case we obtain by differentiation of the equations (6.2) dx/dcf> =- a(k + l)/(2k + 1). [sincp + sin(cp(k+l)/k)] dy/dcf> = a(k + l)/(2k + 1) • [coset> cos(¢(k+l)/k)] By using the addition theorems for the trigonometric funct- ions, these equations become dx/dcf> =- a(k+l)/(2k+l) • 2 · sin(¢(k+k+l)/2k) · cos(¢(k-k-1)/2k) =- a(k+l)/(2k+l) · 2 • sin(cp(2k+l)/2k) · cos(cp/2k) dx/d¢ = 2a·(k+l)/(2k+l) · sin(cp(2k+l)/2k) · cos(cp/2k) (6.4) dy/d¢ = 2a· (k+l)/(2k+l) · sin(¢ (2k+l)/2k) • sin(cp/2k) This gives a vector in the direction of the tangent, but not necessarily of unit length. 40 To obtain the angle 8 between the tangef1t and the x-axis, we write tan¢ = dy/dx = dy/d¢ . 1 dxjd¢ = sin(¢/2k) 1 -cos(¢/2k) =- tan(¢/2k) 'Therefore 0 =- ¢/2k and for 0 < cp .2. 2kn 0 < -2k 8 < 2k1T 0 > 8 > - 1T By convention the tangent vector points counterclockwise for increasing (cos 8, sin 8) is pointing in the negative direction of the tangent. To obtain a tangent vector in the counter- clockwise sense we have to .write Fig. 30 ( 6. 5) T = - V = ( - cos 0, - sin 8). At this point it is convenient to change to 8 as a new parameter. This is possible because it is a monotone funct- ion of 8 (and vice versa): d¢/d8 = - 2k t 0 for all ¢. Equations (6.2) change to ( 6. 6) x = [a/(2k+l)] · [(k+l) · cos 2k8 + k·cos 2(2k+l)8] 41 ( 6. 6) y = [a/(2k+l)] · [-(k+l) ·sin 2k8 +k·sin 2(2k+l)8] Also, d¢/d8 = - 2k, so that equations (6.4) become dx/d8 = dx/d¢ · d¢/d8 - = - a• 2 (k+l) I (2k+l) ·[sin (-2k8) (2k+l) /2k] · cose • ( -2k) dx/d8 = -a·4k(k+l)/(2k+l)·[sin8(2k+l)] · case (6.9) dy/d8 = -a·4k(k+l)/(2k+l)·[sin8(2k+l)] · sine This means T = (-cos 8, -sin 8). The derivative of sis (see (3.1)) 2 ds/d0 = V The arc length along one arc of the astroid is then e s = j( a·4k(k+l)/(2k+l) · sin 8(2k+l) • d8 0 = a"4k(k+l)/(2k+l) · l/(2k+l) ·(-cos 8(2k+l))! (6.9a)s = a·4k(k+l)/(2k+l) 2 . (1- cos 8(2k+l)) or 2 (6.9b)s = a 8k(k+l)/(2k+l) .sin2 8(2k+l)/2 where 0 > 8 > - n/(2k+l) One full arc of the astroid has the length s =a 8k(k+l)/(2k+l) 2 ·sin2[(2k+l)/2· (-n)/(2k+l)] 2 (6.10)s =a 8k(k+l)/(2k+l) Using (6.3), (6.5), and (6.9a) the coordinates of the 42 involute are 2 x 1 = x-a 4k(k+l)/(2k+l) · (1 -cos 8(2k+l)) ·(-cos 8) = a/(2k+l) · [(k+l) cos 2k8 + k cos 28· (k+l)] 2 +a 4k(k+l)/(2k+l) · (1 -cos8(2k+l)) · (-cos 8) 2 = a(k+l)/(2k+l) ·[(2k+l) cos 2k8 + k(2k+l)/(k+l) . cos 28•(k+l) + 4k cos8- 4k cos(2k+l)8 · cos 8] = a{k+l)/(2k+l) 2 · [(2k+l) cos 2k8 + k(2k+l)/(k+l) . cos 28 (2k+l) + 4k cos8- 2k(cos (2k+l+l)8 + cos·(2k+l-l)· 8)] = a(k+l)/(2k+l) 2 · [(2k+l) cos 2k8 + k(2k+l)/(k+l) · cos 2·(k+l)G + 4k cos8- 2k cos 2(k+l)8 - 2k cos 2k8] = a(k+l)/(2k+l) 2 · [cos 2k8 + k{2k+l-2·(k+l))/(k+l) · cos 28·(k+l) + 4k cos8] 2 x 1 - a(k+l)/(2k+l) · [cos 2k8- k/(k+l) ·cos 28(k+l) + 4k cos8] 2 y 1 = y- a 4k(k+l)/(2k+l) • (1- cos(2k+l)8) ·(-sinG) -- a/ (2k+l) • [- {k+l) · sin 2k8 + k sin 28 · (k+l)] +a 4k(k+l)/(2k+l) 2 ·[sin8- cos(2k+l)8 · sin 8] = a(k+l)/(2k+l) 2 ·[-(2k+l) ·sin 2k8 + k(2k+l)/(k+l) · sin 28·(k+l) + 4k sine- 2k(sin (2k+l+l)8- sin{2k+l-1)8)] 2 = a(k+l)/(2k+l) · [-(2k+l) · sin 2k9 + k(2k+l)/(k+l) · sin 28· (k+l) + 4k · sine + 2k · sin 2k8- 2k • sin 28· (k+l)] 43 = a(k+l)/(2k+l) 2 · [- sin 2k8 + k{(2k+l) - 2 (k+l) )/ (k+l) · sin 28 · (k+l) + 4k sin 8] 2 = a(k+l)/(2k+l) · [- sin 2k8 - k/(k+l) · sin 28·(k+l) + 4k sin 8] so that the coordinates of the involute of a (2k+l)-cusped astroid are 2 x 1 = a(k+l)/i2k+l) · [cos 2k8- k/(k+l)·cos28•(k+l) + 4k cos 8] (6.11) y a(k+l)/(2k+l) 2 ·[-sin 2k8- k/(k+l) ·sin 28·(k+l) 1 = + 4k sin 8] We can again verify that this is a curve of constant width by using the support function p = - X · N At the point (x1 (8), Yl (8)) the support function p (8) gives the distance from the origin for the tangent line through that point (Fig. 31). Since the tangent at (x1 (G+n), y 1 (8+n))is parallel (both tangents are supporting lines) the sum of the tangent distances to Fig. 31 4 the origin is the distance between the tangent.s and should be constant and equal to the width of the curve, which should be the length of one arc of the astroid. We have to show 2 p(0) + p(0 + rr) =a 8k(k+l)/(2k+l) We start by using the tangent vector of the astroid which is the normal vector of the involute: N(inv.) = T(astr.) = (-cosG, - sinG) and we call the vector from the origin to the points of the involute xl, so that p(G) = - x1 (G) · T(G) We have 2 p(G) = -a(k+l)/(2k+l) · [(cos 2kG -k/(k+l)·cos 2G·(k+l) + 4k cos G) · (-cosG) + (-sin 2kG k/(k+l) sin 2G (k+l) + 4k sin G) · (-sin G)] 2 =- a(k+l)/(2k+l) · [-cos 2kG · cosG + sin 2kG · sinG + k/(k+l) · (cos2G(k+l) ·cos8+sin2G(k+l) ·sinG) - 4k cos2e - 4k sin2e] p(G) =- a(k+l)/(2k+l) 2 · [-cos (2k+l)0 + k/(k+l) cos (2k+2-1)0 - 4k] 2 =- a(k+l}/(2k+l) ·[- cos(2k+l)0 + k/(k+l} cos(2k+l)G- 4k] 2 = a(k+l)/(2k+l) . 45 • [(- (k+l) + k)/(k+l) cos(2~+1)0- 4k] 2 = a(k+l)/(2k+l) • [1/(k+l) cos(2k+l)- 4k] To get p(8 + rr) we need cos (2k+l) (8 + rr) = cos ( (2k+l) 0+ (2k+l) rr) = cos (2k+l)0· cos(2k+l)rr - sin(2k+l)0·sin(2k+l)rr - cos(2k+l)0· (-1) - sin(2k+l)8· (O) = - cos(2k+l)8 Therefore 2 p{G+rr) = a(k+l)/(2k+l) ·[1/(k+l)cos(2k+l)(8+rr) + 4k] = a(k+l)/(2k+l) 2 ·[-l/(k+l)cos(2k+l)8 + 4k] 2 and p(0) + p(8+rr) = a(k+l)/(2k+l) · (4k+4k) =a 8k(k+l)/(2k+l) 2 It is also easily verified that Barbier's theorem holds: since the constant width of the involute is equal to the arc length of the astroid arc, the perimeter of the 2 involute should be q = a(8k(k+l)/(2k+l) · rr We need 2 . dx1/d8 = a(k+l)/(2k+l) [-2k sin2k8 + 2k sin2 (k+l)G - 4k sine] 2 . dy. 1 /d0 = a(k+l)/(2k+l) [-2k cos2k0 - 2k cos2 (k+l)0 + 4k cose] If q is the arc length for the involute, its derivative is dg/d0 = a(k+l)/(2k+l) 2 · 2k · 2 [sin2 2k0 + sin 2(k+l)0 + 4 Sln. 20- +cos2 2k0 + cos2 2(k+l)0 + 4 cos2e 46 -2 sin 2k8 · sin 2(k+l)8 + 4 sin 2k8 · sinG +2 cos 2k8 · cos 2(k+l)8- 4 cos 2k8 · case -4 sin 2(k+l)8 · sinG 1:: -4 cos 2(k+l)8 · cos0] 2 dq/d8 =a 2k(k+l)/(2k+l) 2 · [1+1+4+2 cos(2k+2k+2)8 1 - 4 cos(2k+l)8- 4 cos (2k+2 - 1)8]~ =a 2k(k+l)/(2k+l) 2 · 1 [6 + 2 cos 2(2k+l)8- 8 cos (2k+l)8]~ ') =a 2k(k+l)/(2k+l)~· 2 k [6 + 2 (2 cos (2k+l)8- 1) - 8 cos(2k+l)8] 2 =a 2k(k+l)/(2k+l) 2 · 2 k2 [6 + 4 cos (2k+l)8- 2- 8 cos(2k+l)8] =a 2k(k+l)/(2k+l) 2 4 + 4 cos2 (2k+l) k - 8 cos (2k+l) ) 2 2 dq/d8 =a 4k(k+l)/(2k+l) 2 · ~1 cos(2k+l)8) dq/d8 =a 4k(k+l)/(2k+l) 2 · (1 cos(2k+l)8) The perimeter of the full involute is found by integration: 2.U q = J a 4k(k+l)/(2k+l) 2 · (1- cos(2k+l)8) d8 0 2'JT =a 4k(k+l)/(2k+l) 2 · (8- l/(2k+l) · sin(2k+l)8)] 2 0 =a 4k(k+l)/(2k+l) (2'IT- 0) 2 q =a 8k(k+l)/(2k+l) · 'IT Q. E. D. The following graphs of astroids and their involutes were obtained by using the equations (6.11) in a computer program to print out the coordinates for several successive ------~------~------.------47 n = 2k+l. It is obvious that for the higher Dumbers the astroids approach the \\. \ \ ' \ \ \ l - ·~------+----~~------~~~r--~ ,X k = 1 n = 3 Fig. 32 1----.-- - X -t- \ \ \ \ \ \ k = 2 \\ \\ n = 5 \ Fig. 33 48 /I / /j /I / / I - ~ k = 3 n = 7 Fig. 34 k = 4 n = 9 Fig. 35 VII. ~-Curves or Rotors in the Equilateral Triangle. When Franz Reuleaux wrote his textbook on "The Kinematics of Machinery"[S] he was interested in the motion of pairs of rotating shapes. One element was held station ary while the other rotated a:round or inside of it [ 5, p. 115 ff]. This interest brought him to realize that the Reuleaux triangle, the Reuleaux polygons, and all curves of constant width are rotors in a square. It was only natural for him to investigate rotors in an equilateral triangle (fl.-curves)_. He found one shape with this property, called it the "duangle" (Zweieck) , and gave a nice proof that it indeed revolves inside an equilateral triangle of altitude h (Fig. 37). The duangle consists of two circular arcs of radius h/2, each subtended by an angle of 120°, which means that the center of each circu lar arc lies at the mid- Fig. 37 point of the other. The tangents at the-vertices form an angle of 120°. Later it was found that there exists a second fl.-curve which consists of two circular arcs. This s~ape is called the 6-biangle. It holds a similar position among the fl.-curves as the Reuleaux triangle holds among curves of 49 50 constant width, because it has the least area of all ~-curves. The ~-biangle consists of two circular arcs of radius h, each subtended by an angle of 60°. The angle at the vertices is also 60° (Fig. 38) : by the construct ion 4 CPS = 90° 4 CPS == 60° therefore A CPS - 4 CPQ == 4 QPS == 3 0 ° 4 RPS == 2 ( 4 QPS) == 60° Fig. 38 In order to show that the ~-biangle is a rotor in an equilateral triangle, we have to prove that all circumscrib- ed eq~ilateral triangles are congruent[25]. This is the case, if they all have the same altitude h. Let ABC be any equilateral triangle circumscribed around the ~-biangle RS (Fig. 39) . Let D be the center of the cir- cular arc HPS, and IRDI == Fig. 39 lsDj = h. Let P the contact point between the biangle and the side Bc of ABC. Then we have to show that the altitude IAQj is congruent to jDPj ==h. 51 Proof: BC is tangent at P and we have DPJL BC. Also, RSD is an equilateral triangle by construction, and 4 RDS = ~BAC = 60°. Therefore RADS is a cyclic quadrilateral (points RDS determine a circle; the angle over the chord RS is congruent ~ RAS over the same chord) . In a cyclic quadrilateral opposite angles are supplementary: .4~RAD + 4RSD = 180° 0 4RSD = 60 therefore 4 RAD = 12 0 ° Since 4 ABC + 4RAD = 60° + 120° = 180° AD is parallel to BC, and since4AQB =4DPB = 90° it follows that QPDA is a rectangle in which the sides jAQj = jDPI = h. Q. E. D. 'vhen the biangle rotates inside an equilateral triangle it actually touches the vertices of the triangle. This is possible because the angles at the vertices of the ~-biangle are 60° (Fig. 40). We may recall that the Reuleaux triangle does not quite reach the corners of the square in which it ro- tates (Fig. 11). Fig. 40 Another example of a ~-curve is the curvilinear polygon formed by four circular 52 arcs, each centered at the ---- corner of a square of side length h (Fig. 41). Again it can be shown that all circumscribed equilateral triangles are congruent and have the same altitude h. Other ~-curves can be constructed as a series of circular arcs of equal radius, their number may be any number not divisible by 3. This implies that some rotors in the square are also rotors in the equilateral triangle. An example is the Reuleaux polygon consisting of 5 circular arcs of radius r (Fig. 42), it revolves in a square of side- Fig. 42 length h = r as well as in an equilateral triangle of altitude 3r/2. 53 .Furthermore it can be shown that involu-tes of astroids with n cusps (n not divisible by 3) are ~-curves. For an even number n of cusps of the astroid the involute consists of two separate congruent closed curves, each of which is s .!l-curve. If n/2 is an odd integer, the curve is identical to the involute of the (n/2) cusped astroid, which is also a curve of constant width. The following graph,s were obtained by computer print- out for the coordinates. y ...... ·· I\ ·. . .~ ·. : I \ \ .. I \ . I \ ·. ~ I '- ~. / .. :. / " ,:, ).- :-.._ / : : ...... / ·.: : ,_ .-...... / - . . / ' ,:: =/ \ '- .";-.,. /: :.. ' / f . / : ...... · ...... ·· n = 4 Fig. 43 54 - ,- n = 8 Fig. 44 55 The following theorems are analogies to the theorems for curves of constant width. Theorem IX (Barbier's Theorem for 6-curves): All rotors in an equilateral triangle of altitude h have the same perimeter 2Tih/3; namely the same as the inscribed circle, which has the diameter 2h/3. (The proof will follow). Theorem X: All 6-curves must have angles of more than 60° at points where circular arcs meet at an angle. The 6-biangle is the only 6-curve with two 60° angles at its vertices. Theorem XI (Blaschke-Lebesgue Theorem for 6-curves) : Of all curves which rotate in an equilateral triangle of altitude h the inscribed circle of radius h/3 has the largest area and the 6-biangle has the least. There are three proofs of this last theorem. The first one was given by Fujiwara and Kakeya in 1917[18], the second by Yaglom and Boltiyanski[l3]. A third one, given by B. Weisskopf in 1970 is modeled after Eggleston's proof for the case of curves of constant width[26]. The following proof of Barbier's theorem is modeled after the one for curves of constant width in Chapter V. 56 Theorem IX (Barbier's Theorem for L\-Curves): All rotors in an equilateral triangle of altitude h have the same perimeter 2nh/3, namely the same as the in- scribed circl~e~·------~ Proof: Consider Figure 45. Let X represent a twice differentiable L\-curve. Let X, X1, x2 be the points where the curve touches the equilateral triangle. Let T,T , T be the tangent 1 2 X T vectors (of unit length), Fig. 45 N, N1 , N2 be the normal vectors (also of unit length) at these points. Because of the geometry of the equilateral triangle we have (7 .1) T + Tl + T2 - 0 N + N1 + N2 = 0 Another way to express this fact is ( 7. 2) In order to simplify, we will sometimes use m = V'Ji/2, m2 = 3/4. The fact that the curve revolves inside the equilater- al triangle implies, that the normals at the points of con- 57 tact meet in one point M: the center of rotation ha~ to lie on the normal at each point of ciontact with a sid~ of the triangle; therefore, if there is rotation, the normals meet in one point. Also, for any point inside an equilateral triangle the sum of the perpendiculars from that point to the sides of the triangle is h. So we have for the point M ( 7. 3) where t, t , t are the distances from M to the sides of the 1 2 triangle. We express the stated facts in the equations xl = X + t N + tl· (-Nl) (7. 4) xl = X + t N + t 1 · (m T + N/2) X + t N + x2 = t 2 · <- 'N 2 > (7. 5) x2 = X + t N + t 2 · (- m T + N/2) both for 0 < e < 2n/3 Differentiation gives dXi/ds = dXi/dsi • dsi/ds i = 1,2 = Ti(si) · dsi/ds In particular, differentiation of (7.4) and use of Frenet's formulas gives dX /ds • ds /ds = + N" t + t• N 1 1 1 x· + ti(mT+N/2)+t1 (mT"+N"/2) . ds /ds = T -tKT + t"N" 1 + t• (mT + N/2)+t (mKN-KT/2) 1 1 (-T/2 + mN) . ds /ds T tKT + t":N 1 = 58 Equating the coefficients for T and N on both sides we have for T . (7. 6) 1/2 • ds /ds = 1 - Kt + t m - t K/2 1 1 1 for N = t" + t"/2 + t mK 1 1 or t = 2(m ds /ds -t· - t m~) 1 1 1 Substituting this in (7.6) we obtain - 1/2 · ds1/ds = 1 - K~ + 2m(m ds1/ds - t" - t 1mK} - t K/2 1 2 2 1 - Kt + 2m ds /ds - 2mt"- 2m t K = 1 1 - t K/2 1 = 1- Kt + 3/2 ds /ds - 2mt·- 3/2 t K 1 1 - t K/2 1 1/2 · ds /ds = 1 Kt + 3/2 ds /ds - 2mt·- 2t K 1 1 1 _. 2 · ds /ds = 1 Kt - 2mt" - 2t K 1 1 (7. 7) Differentiation of (7.5} gives dX /ds • ds /ds = + t"N + tN" + t; (-mT + N/2} 2 2 2 x· + t <-mrr· + N"/2) 2 T2 · ds2/ds =.T + t"N - tKT - t;mT + t2N/2 + t (-mKN 2 <- T/2 t;mrr + t;N/2 - t 2mKN - t 2KT/2 Equating the coefficients, \ve have for T ( 7. 8} 59 for N - m · ds /ds t + t /2 - t mK 2 = 2 2 t = 2(- m · ds /ds- t + t mK) 2 2 2 Substituting this in (7.8) gives - 1/2 · ds2/ds = 1 - Kt - 2m(- m ds2/ds - t•+t2mK) - t2K/2 2 = 1-Kt + 2m ds2/ds + 2mt· 2 2m t K-t K/2 2 2 - 1/2 · ds /ds 1 - Kt + 3/2 ds /ds + 2mt· 2 = 2 - 3/2 t 2K - t2K/2 - 2 ds /ds = 1 - Kt + 2mt• - 2t K 2 2 ( 7. 9) ds2/ds = - 1/2 + Kt/2 - mt· + t 2K When (7.7) and (7.9) are added by (7.3) this is ds /ds + ds /ds = - 1 + Kh 1 2 or (7.10) ds /ds + ds /ds + 1 = Kh 1 2 Since K = d8/ds for 0 < 8 < 2rr/3. we can integrate and obtai1;. the perimeter 2/3• TI p = fds 1 + ) ds 2 + rds = s h d8 0 p = h 2TI/3 Q. E. D. 60 If (7.10) is divided by K . . 1/K ds1 /ds + 1/K ds2/ds + 1/K = h ds /dG ds /dG 1 + 2 + 1/K = h or l/K1 + l/K 2 + 1/K = h Since the radius of curvature at each point is defined as r(s) = 1/K(s) we can write (7.11) and express this as Theorem XII: For a ~-curve rotating in an equilaterl al triangle the sum of the curvature radii at the I points of contact with the triangle is equal to the ~l_._t_u__ d_e __ h__ o_f ___ t_h_e __ e_q~u __ i_l_a_t_e_r_a~_tr __ i_a_n_g~l_e_. ______VIII. Conclusion In the early 20th century investigators turned from this point in several directions. Fujiwara and others went on to study rotors in regu lar polygons of more than 4 sides[6],[8]. One of the most prolific writers on this subject isM. Goldberg[9]. A gen eral form of Barbier's theorem holds in all these cases: The length of the perimeter of all rotors in a regular polygon of n sides is equal to the length of the perimeter of the inscribed circle. In addition, there exist general methods to construct rotors in regular polygons, consisting only of circular arcs[9]. However, even the simplest ex amples for n ~ 5 are more complicated than the Reuleaux triangle or the ~-biangle. It is also not known which of the examples for each polygon has the least area. Minkowski[21] made the first intensive study of sur faces of constant width in three dimensions and found gen eral geometric methods of deriving them. In addition he obtained theorems concerning them. A surface of constant width is a rotor in a cube. t-1eissner[22] investigated rotors in all the regular poly hedra and showed that non-spherical rotors exist for the regular tetrahedron, the cuber and the regular octahedron, and that they do not exist for the dodecahedron and the icosahedron. Goldberg's paper on rotors in polygons and polyhedra shows photographs of some of these models. 61 62 Finally Santalo[23] studied non-spherical rotors in the simplex (the analogue of the equilateral triangle and the tetrahedron) and the hypercube for n dimensions. They do exist and can be analytically described. Footnotes and Bibliography 1) H. Rademacher and 0. Toeplitz, The Enjoyment of Mathematics. Princeton University Press, 1966, p. 16~ This book is a translation of Von Zahlen und Figuren, Julius Springer. Berlin, 1933. by the same authors. 2) Leonhard Euler, "De Curbis Triangularibus~. Acta Acad. Petropolitanea. (1778) II p. 3-30. 3} Egglestron, Convexity, Cambridge Univ. Press, 1958, p. 133. 4) E. Barbier, "Note sur le proble~e d'aiguille et le jeu du joint couvert". J. Math. pures apJl1. (2) v. 5 (1860) p. 273-286. 5) Franz Reuleaux, Lehrbuch der Kinematik, Braunschweig, 1875. English translation: The Kinematics of Ma~hinery, Dover Publications, New York. 6) M. Fujiwara, "Ueber die einem Vielecke einbeschriebenen .und umdrehbaren konvexen geschlossenen Kurven". Science Reports, Tohoku Univ., v. 4, 1915, p. 43-65. 7) H. Fujiwara, "Ueber die innen-umdrehbare Kurve eines Vielecks", Science Reports, Tohoku Univ., v. 8, 1919, p. 221-246. 8) M. Fujiwara and S. Kakeya, "On some problems of maxima and minima for the curve of constant breadth and the in-revolvable curve of the equilateral triangle". Tohoku Math. J., v. 11, 1917, p. 92-110. 63 64 9) Ivl. Goldberg, "Rotors in Polygons and Polyhedra". Mathematics of Computation, 14 (1960), p. 229 ff. 10) J. J. Stoker, Differential Geometry. Wiley-Inter science, J. Wiley and Sons, New York, 1969. 11) D. J. Struik, Differential Geometry. Addison Wesley, 1950. 12) Martin Gardner, Scientific American, Feb. 1963, p. 148. 13) Yaglom and Boltiyanski, Ccnvex Figures, Holt, Rinehart and Winston, New York, 1961. 14) J. P. Norbye, The Wankel Engine, Chilton Book Company, 1971. 15) J. P. Moulton, "Some geometry experiences for elementary school children", The Arithmetic Teacher, Feb. 1974, p. 114 ff. 16) A. Hurwitz, "Sur quelques applications geometriques des series de Fourier". Ann. Ecole normal (3) v. 19 (1902), p. 357-408. 17) E. Barbier, "No·te sur le probleme de 1' aiguille et le jeu du joint couvert". J. Math. pures appl. (2) v. 5 (1860}, p. 273-28f,. 18) A. S. Besicovitch, "Measure of asymmetry of convex curves. (II). Curves of constant width". J. of London Math. Soc. (2) 26 (1951), p. 81-93 19) J. Pal. Hath. Ann. 83, (1921), 311. 65 20) . Bonnesen and Fenchel, Theorie der konvexen Koerper. Chelsea Publishing Co., New York, 1948, p. 48: Cauchysche Oberflaechenformel. Also p. 65. Also: Eggleston 3. 21) H. Hinkowski, "Ueber die Koerper konstanter Breite", Rec. Math., Soc. Mat.h. Moscmv, v. 25, 1904, p. 505- 508; Collected Works, v. 2, p. 277-279, 215-230. 22) E. Meissner, "Ueber die durch regulaere Polyeder nicht stuetzbaren Koerper'', Vierteljahresberichte der Naturforschenden Gesselschaft, Zuerich, v. 63, 1918, p. 544-551. 23) L. A. Santa1o, "On the convex curves of constant width in En", Portugeiae Mathematica, v. 5, 1946, p. 195-201 (in Spanish) • 24) G. Tiercy, "Sur les courbes orbiforms", Tohoku Math. ~· 1 V. 18, (1920) f P• 90-101. 25) Ross Honsberger, Mathematical Gems. Dolciani Mathe r~atical Expositions No. 1. The Mathematical Assoc. of Arrteric?, 1973, p. 54 ff. 26) B. Weissbach, "Rotoren im regulaeren Dreieck", Publicationes Mathematicae. Debrecen (Hungary), 1920 {1972)' p. 21-27.