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

Machinery's Handbook, 31st Edition

50

CIRCLE FORMULAS

π L R 180

= ·

Central angle , in degrees φ

π 180

Arc length L R φ = ·

2

2

L φ

K

E F F + 4 8

π 180

2

= ·

=

=

Radius R

L

RL

π

2

=

·

=

Area of sector K R φ

360 2

RL E R F − 2 2 ( ) –

=

Area of segment S

2 φ ( ) ( ( )) sin c os φ 2

E F R F D = − = 2 2 ( )

Chord length

2

2

R E −

4

F R

R = − 1

= −

Depth

2

φ 2 ( ) ( ) φ

/ E R F E R 2 −

Fig. 6b.

=

tan

=

sin

2 2

Annulus R 1 = radius of outer circle R 2 = radius of inner circle Area of annulus = π

(

)

2

2

W R R −

1

2

2 − ( )

φ

π R R 1 2 2

U

=

Area of annulus segment

360

Fig. 6c. Example 2: Find the area of a sector of a circle having a central angle of 30 ° and a radius of 7 cm. Solution: Referring to Fig. 6b, K φ° 360 -----     π R 2 30 360 -----     π 7 2 12.83 cm 2 = = = × × Example 3: Find the chord length E of a circular segment (Fig. 6b) with a depth of 2 cm at the center that is formed in a circle whose radius 12 cm. Solution: The chord length is E 2 F 2 R F – ( ) 2 2 44 = = = = 4 11 = 13.27 cm 2[(2)(12) – 2] Example 4: Find the area S of the segment in Example 3. Solution: First determine angle f , then find arc length L of the segment, and then solve for area S, as follows:

E 2 ⁄ R F – ------- 13.27 2 ⁄ 12– 2 =

φ 2 --     tan

φ 2 --

---------- 0.6635, =

=

=

33.56 °,

67.13 ° =

φ

tan –1 0.6635 =

π

π

L R φ = × = 12 × 67.13° ×

= 14.06 cm

180

180

Area S RL 2 = – ----------- 12 14.06 ( ) 2 = Copyright 2020, Industrial Press, Inc. ---- E R F – ( ) 2

– ------------ 84.36 – 66.35 18.01 cm 2 = = ebooks.industrialpress.com

------------ 13.27 10 ( ) 2

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