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PIERHEADS, QUAYS, AND LANDING-STAGES. 2II
Flirther, taking the equation of the parabola in the first phase, viz.,
X2 =
L2
60^
where the two curves intersect, we get
36 •
18Da: - 18L^r2 - LD = 0,
L- L2
or
(6æ - L)8=0.
Therefore the parabola and hyperbola have three point contact at
LD
6’6
The curve may now be traced from the foregoing data.1 This being done
(fig. 183), it remains to find the centre of buoyancy corresponding to any
possible water-line within the limits already specified. Take VS as such a
line and, as in the case of the parabola, bisect that portion of it, fs, which
forms a chord of the hyperbola. Join this middle point C to N the centre
of the curve. Where the line CN intersects the curve is the centre of
buoyancy for the position, and a line drawn therefrom at right angles to
the water-line will cut the original vertical axis in the metacentre. Thus B
and M are the centre of buoyancy and the metacentre respectively, for the
water-line VS.
The last phase of the problem is the same as the first. When the pontoon
is immersed as with its axis JOX (fig. 184) vertical in the initial position,
we have a similar set of conditions to those in which the axis QOK was
initially vertical.
To find the depth (DJ immersed, we have
D^LD or Dj=-^D,
1 The method is shown in fig. 186.