Circular Segment

DOWNLOAD Mathematica Notebook CircularSegment

A portion of a disk whose upper boundary is a (circular) arc and whose lower boundary is a chord making a central angle theta<pi radians (180 degrees), illustrated above as the shaded region. The entire wedge-shaped area is known as a circular sector.

Circular segments are implemented in the Wolfram Language as DiskSegment[{x, y}, r, {q1, q2}]. Elliptical segments are similarly implemented as DiskSegment[{x, y}, {r1, r2}, {q1, q2}].

Let R be the radius of the circle, a the chord length, s the arc length, h the height of the arced portion, and r the height of the triangular portion. Then the radius is

 R=h+r,
(1)

the arc length is

 s=Rtheta,
(2)

the height r is

r=Rcos(1/2theta)
(3)
=1/2acot(1/2theta)
(4)
=1/2sqrt(4R^2-a^2),
(5)

and the length of the chord is

a=2Rsin(1/2theta)
(6)
=2rtan(1/2theta)
(7)
=2sqrt(R^2-r^2)
(8)
=2sqrt(h(2R-h)).
(9)

From elementary trigonometry, the angle theta obeys the relationships

theta=s/R
(10)
=2cos^(-1)(r/R)
(11)
=2tan^(-1)(a/(2r))
(12)
=2sin^(-1)(a/(2R)).
(13)

The area A of the (shaded) segment is then simply given by the area of the circular sector (the entire wedge-shaped portion) minus the area of the bottom triangular portion,

 A=A_(sector)-A_(isosceles triangle).
(14)

Plugging in gives

A=1/2R^2(theta-sintheta)
(15)
=1/2(Rs-ar)
(16)
=R^2cos^(-1)(r/R)-rsqrt(R^2-r^2)
(17)
=R^2cos^(-1)((R-h)/R)-(R-h)sqrt(2Rh-h^2),
(18)

where the formula for the isosceles triangle in terms of the polygon vertex angle has been used (Beyer 1987). These formula find application in the common case of determining the volume of fluid in a cylindrical segment (i.e., horizontal cylindrical tank) based on the height of the fluid in the tank.

The area can also be found directly by integration as

 A=int_(-Rsin(theta/2))^(Rsin(theta/2))int_(Rcos(theta/2))^(sqrt(R^2-x^2))dydx.
(19)

It follows that the weighted mean of y is

<y>=int_(-Rsin(theta/2))^(Rsin(theta/2))int_(Rcos(theta/2))^(sqrt(R^2-x^2))ydydx
(20)
=2/3R^3sin^3(1/2theta),
(21)

so the geometric centroid of the circular segment is

 y^_=(<y>)/A=(4Rsin^3(1/2theta))/(3(theta-sintheta)).
(22)

Checking shows that this obeys the proper limits y^_=4R/(3pi) for a semicircle (theta=pi) and y^_=R for a point mass at the top of the segment (theta->0).

CircularSegmentQuarter

Finding the value of h such that the circular segment (left figure) has area equal to 1/4 of the circle (right figure) is sometimes known as the quarter-tank problem.

Approximate formulas for the arc length and area are

 s approx sqrt(c^2+(16)/3h^2),
(23)

accurate to within 0.3% for 0 degrees<=theta<=90 degrees, and

 A approx 2/3ch+(h^3)/(2c),
(24)

accurate to within 0.1% for 0 degrees<=theta<=150 degrees and 0.8% for 150 degrees<=theta<=180 degrees (Harris and Stocker 1998).

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