(Part A) Machinerys Handbook 31st Edition Pages 1-1484

Machinery's Handbook, 31st Edition

SEGMENTS OF CIRCLES 83 Segments of a Circle.— The table that follows gives the principal formulas for dimen- sions of circle segments. The dimensions are illustrated in the figures on pages 79 and 84 . When two of the dimensions found together in the first column are known, the other dimensions are found by using the formulas in the corresponding row. For ex- ample, if radius r and chord c are known, solve for angle a using Equation (13), then use Equations (14) and (15) to solve for h and l , respectively. In these formulas, the value of a is in degrees between 0 and 180 ° . Formulas for Segments of a Circle Given Formulas a , r (1) (2) (3) c 2 r l

α 2 --    

α 2 = sin --    

π r α 180 = -----

h r 1– cos   =

 

c

π c α 360sin = ----------- α 2 --     -----

--     -----

2sin = -------- α 2

r

l

α 4 --    

(4)

(5)

(6)

2 = –-- tan 2 h tan = ------

h c

a , c

π h α α 2 – cos --     = ------------------- 180 1     ----- 180 l 1 α 2 – cos --     πα = --------------------     -----

h 1– cos = ----------- α 2 --     ----

α 4 --     ----

c

l

r

(7)

(8)

(9)

a , h

α 2 --     ---------

360 l sin πα = ----------

180 π -----

l α --

(10)

(11)

(12)

a , l

=

r

c

h

cos –1 1 c 2 2 r 2 – ----     

π 90 --- r sin –1

(13)

(14)

(15)

c 2 r ---    

2

2

r, c

4 r

c

=

=

α

l

2 = – ------------

h r

π 90 --- r cos –1 1

(16)

(17)

(18)

h r – --    

h r – --    

r, h

α

2cos –1 1

2 h 2 r h – ( ) =

=

c

=

l

90 l π R = sin ----    

90 l π r – cos ----        

180 π -----

l r -

(19)

(20)

(21)

h r 1 =

c 2 r

r, l

α

=

4tan –1 2 h c ---    

2

2

(22)

(23)

(24)

c 2 4 h 2 + 360 h ----------

2 h c ---    

=

 

 tan –1

c, h

c + 8 h = ---------- 4 h

α

π

=

l

r

Given

Formula To Find

Given

Formula To Find

(26) Solve Equation (26) for a by iteration a , then α 1– cos = ----------- α 2 --     ---- 180 π ----- l h --

360 π ----- l c -

α = ------

  ----

(25) Solve Equation (25) for a by iteration a , then α 2 sin --  

h, l

c, l

r = Equation (10) c = Equation (11)

r = Equation (10) h = Equation (5)

a Equations (25) and (26) cannot be easily solved by ordinary means. To solve these equations, test various values of α until the left side of the equation equals the right side. For example, if given c = 4 and l = 5, the left side of Equation (25) equals 143.24, and testing various values of α reveals the right side equals 143.24 when α = 129.62°. Angle a is in degrees, 0 < a < 180 Formulas for Circular Segments contributed by Manfred Brueckner.

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