Machinery's Handbook, 31st Edition
Center of Gravity
229
Area of Trapezoid:
a
A
B
The center of gravity is on the line joining the midpoints of parallel lines AB and DE . c a b h a b d a b h a b e a b a ab b 3 2 3 2 3 2 2 = + + = + + = + + + ^ ^ ^ ^ ^ h h h h h The trapezoid can also be divided into two triangles. The center of gravity is at the intersection of the line joining the centers of gravity of the triangles and the middle line FG . Two cases are possible, as shown in the illustra tion. To find the center of gravity of the four- sided figure ABCD , each of the sides is divided into three equal parts. A line is then drawn through each pair of division points next to the points of intersection A , B , C , and D of the sides of the figure. These lines form a parallelogram EFGH ; the intersection of the diagonals EG and FH locates center of gravity. The center of gravity is on the line that bisects the arc, at a distance a l r c lh c c h 8 4 2 2 # = = + ^ h from the center of the circle. For an arc equal to 1 ⁄ 2 the periphery: . a r r 2 06366 ' = π = For an arc equal to 1 ⁄ 4 of the periphery: . a r r 2 2 09003 ' π = = For an arc equal to 1 ⁄ 6 of the periphery: . a r r 3 09549 ' = π = An approximate formula is very nearly exact for all arcs less than 1 ⁄ 4 the periphery is: a h 3 2 = The error is only about 1 percent for a quarter circle, and it decreases for smaller arcs. The distance of the center of gravity from the center of the circle is: sin b A c A r 12 3 2 3 3 3 # α = = in which A = area of segment.
c
h
e
d
D
E
b
F
C
G
Any Four-sided Figure: C H
G
B
D
F
E
A
Circular Arc:
h
r
c
C
l
a
C
h
a
c /2
c /2
Circle Segment:
r c
b
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