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

ELLIPSE CALCULATIONS Machinery's Handbook, 31st Edition

53

Length, Point, and Angle Calculations R 1 = radius of director circle = A 2 B 2 + , R 2 = radius of equivalent circle = AB , P = center to focus distance = A 2 B 2 – ,

A = major radius = B 2 B = minor radius = A 2

2

+

P

2

P

2

2 B

A ----- = distance, origin to latus rectum

A φ = cos , and Y = B

J = any point ( X , Y ) on curve where X = A θ sin

θ cos

B φ = sin

–1 Y B --

–1 X A --

f = angle with major axis = sin

, q = angle with minor axis = 90 ° φ –

   

   

=

cos

L = total perimeter (approximate) = A 1.2 B A --     2

B A --     4 +

+

1.1

L = perimeter (sections) = π 180 -----  

  2 φ AB

Area Calculations

N = total surface area of ellipse = π AB W = area between outer and inner ellipse = π A 1 B 1 A 2 B 2 – ( ) M = area of complement section M = AB π AB 4 – ------ S = area of segment S = AB cos –1 X 1 A ---     X the inverse cosine is in radian measure T+S = combined area of segment S + area T = AB cos –1 X 2 A ---     X that results from the inverse cosine is in radian measure V = area of segment V = R 2 2 sin –1 X A --     XY – ( )

1 Y 1 – , where the angle that results from

2 Y 2 – , where the angle

, where the angle that results from the

inverse sine is in radian measure K = area of sector K = AB cos –1 X A --  

  , where the angle that results from the inverse

cosine is in radian measure Example 2: Find area of sector K and complement area M , given the major radius of ellipse is 10 cm, minor radius of ellipse is 7 cm, dimension X = 8.2266 cm. Solution: Sectional area K: Area K AB cos –1 X 1 A ---     10 7 cos –1 8.2266 10 --------     × × 70 0.6047 rad × 42.33 cm 2 = = = = = Example 3: Find the area of elliptical segments S, T + S , provided that major radius A of ellipse is 10 cm, minor radius B of ellipse is 7 cm, dimension X 1 = 8.2266 cm, dimension Y 1 = 4.4717 cm, and dimension X 2 = 6.0041 cm. Solution: Segment area S is found: S AB cos –1 X 1 A ---     X 1 Y 1 – 10 7 cos –1 8.2266 10 --------     × × 8.2266 4.4717 × – 5.5437 cm 2 = = = Solution: Complement area M: Area M AB π AB 4 = – ------ 10 7 π = 10 7 × × 4 – ------------ × 15.0221 cm 2

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