We may locate vm by satisfying Equation 51, which may be written428 Position Of VM 0400302 dxz = 0. But this integral is represented by the shaded area (Fig. 231), which is the equivalent of saying that the segment OCB = the rectangle OK X OB. If OCB were a parabola, OK would exactly equal 2/3 CD. Even with circular arcs, the ratio § is approximately correct if the angle is small. Therefore, for flat circular arcs, draw vm at 2/3 the height of the arc. If necessary, increase the height according to the figures given in the accompanying tabular form:

2a

Ratio

Excess Error

10°

.667

0.04

per cent

15°

.667

0.09

,,

,,

20°

.668

0.15

,,

,,

25°

.668

0.24

,,

,,

30°

.669

0.35

,,

,,

40°

.671

0.62

,,

,,

60°

.676

1.42

,,

,,

90°

.689

3.35

,,

,,

180

.785

17.8

,,

,,

Of course, for full-centered arches in which 2 a = 180°, the error of the § rule is very great, but the tabular values are correct.

Since an elliptic arc may be considered as a circle in which the vertical ordinates have all been shortened by some constant ratio, the same law and same percentage of error will hold true.

For any other curve, particularly multi-centered curves, the position of vm may be found by determining by trial a position such that the summation of equally spaced ordinates is zero.