ELLIPSE CALCULATIONS Machinery's Handbook, 31st Edition
54
Solution: Segment area T + S: φ cos –1 X 2 A --- = =
2 B φ sin 7 1.51072 rad ( ) sin 10.57504 = = =
Y
1.51072 rad,
X 2 A --- X
T S + AB cos –1 = = 64.876 – 33.6097 31.266 cm 2 = = Example 4: Find the area of elliptical segment V if major radius of ellipse is 4 inches, minor radius is 3 inches, dimension X = 2.3688 inches, dimension Y = 2.4231 inches. Solution: Segment area V: R 2 AB = R 2 2 AB 3 4 × 12 = = = V R 2 2 sin –1 X A -- XY – 12sin –1 2.3688 4 -------- 2.3688 2.4231 × ( ) – = = = 7.6054 – 5.7398 1.8656 in 2 = ( ) , ( ) 2 Y 2 – 10 7 × 0.9268 × 6.0041 5.5978 × ( ) – Four-Arc Oval Approximating an Ellipse * .— The method of constructing an approx- imate ellipse by circular arcs, described on page 69, fails when the ratio of the major to minor diameter is 4 or greater. Additionally, it is reported that the method always pro duces a somewhat larger minor axis than intended. The method described below presents an alternative. An oval that approximates an ellipse, illustrated in Fig. 7, can be constructed from the following equations: (1) where A and B are dimensions of the major and minor axis, respectively, and r is the radius of the curve at the long ends. The radius R and its location are found from Equations (2) and (3): (2) (3) r B 2 2 A ---- A B -- 0.38 = A 2 4 --- Ar – Br B 2 4 + – --- R
B 2 -- X = +
B 2 r – = ----------------------
X
A
r
B
R
X
Fig. 7. Four-Arc Oval Ellipse To make an oval thinner or fatter than what is given, select a smaller or larger radius r is chosen than that calculated by Equation (1), and then X and R are found using Equations (2) and (3). * Four-Arc Oval material contributed by Manfred K. Brueckner
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