Circular Segment

Circular Segment CIRCULAR SEGMENT

Balmoral Software

Let a circular segment S of radius 1 be oriented with the origin at the midpoint of the chord connecting the endpoints of its arc, and with the center of the circle on the negative y-axis, so that S is symmetric with respect to that axis. If β < π/2 is one-half the angle subtended by the segment arc, then the extreme abscissa points of S are (±sin(β),0), the maximum ordinate point is (0,1 - cos(β)), and the width x height of the bounding rectangle of S is 2sin(β) x 1 - cos(β).

Let t represent the usual parameter, ranging from 0 to 2π through the quadrants in counterclockwise order. In Quadrants I and II, the parametric equations of the segment arc are:

x(φ) = cos(φ)
y(φ) = sin(φ) - cos(β),

where the angle φ is measured from the center of the circle and ranges from π/2 - β to π/2 + β. We can scale t to the angular parameter φ as follows:

Range of t Range of φ

[0,π] [π/2 - β,π/2 + β]

or equivalently,

φ = 2βt/π + π/2 - β

In Quadrants III and IV, the parametric function x(t) of the horizontal chord is linear in t. Combining definitions, we have

Metrics

The perimeter of the segment is the arc length of the circular arc (in radians) plus the chord:

L = 2β + 2sin(β)

The area of S is

A = [2β - sin(2β)]/2

and its centroid ordinate is

Specific Example

In the remainder of this paper, we use

β = 35° = 7π/36
L = 2.368883
A = 0.141019
y_c = 0.072931
Bounding rectangle: 1.147153 x 0.180848

Incircle

A candidate incircle is located on the y-axis between the ordinate extrema of S, and so has radius and center ordinate both equal to

R = d = [1 - cos(β)]/2 = 0.090424

For verification, we have

Inellipse

Using z = 0 in Lemma E,

d/dt x(t)[y(t) - z] = d/dt cos(2βt/π + π/2 - β)[sin(2βt/π + π/2 - β) - cos(β)]

has a zero over [0,π) at t* = 0.682719. The corresponding coordinates are

x* = 0.338539
y* = 0.121800

We then have

For verification,

Circumellipse

The extreme points of the segment create a y-symmetric isosceles triangle with vertices (0,1 - cos(β)) and (±sin(β),0). Using h = 1 - cos(β) and w = 2sin(β) in Lemma TE, a candidate for the circumcircle has parameters

For verification,

Circumcircle

The smallest enclosing circle is one with the chord of S as its diameter, so

R = sin(β) = 0.573576
d = 0

Since the radius of the circumcircle is less than the unit radius of the segment, the circumcircle has greater curvature and thus encloses S.

Summary Table

Figure Parameters Perimeter Area Centroid

Incircle R = [1 - cos(β)]/2 0.568151 0.025687 (0,0.090424)

Inellipse a = 0.390911
b = 0.081200 1.649400 0.099720 (0,0.081200)

Circular segment Width: 2sin(β)
Height: 1 - cos(β) 2.368883 0.141019 (0,0.072931)

Circumellipse a =
b = 2.765929 0.250860 (0,0.060283)

Circumcircle R = sin(β) 3.603887 1.033552

Figure	Parameters	Perimeter	Area	Centroid
Incircle	R = [1 - cos(β)]/2	0.568151	0.025687	(0,0.090424)
Inellipse	a = 0.390911 b = 0.081200	1.649400	0.099720	(0,0.081200)
Circular segment	Width: 2sin(β) Height: 1 - cos(β)	2.368883	0.141019	(0,0.072931)
Circumellipse	a = b =	2.765929	0.250860	(0,0.060283)
Circumcircle	R = sin(β)	3.603887	1.033552

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