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204
HARBOUR ENGINEERING.
In the initial position, the centre of buoyancy is at Bj in the vertical line
QK passing through O.
Now, suppose the pontoon be acted upon so as to take up an inclination
in which the water line is FOS. The immersed section becomes FHNS, and
the centre of buoyancy which corresponds to this position lies somewhere to
the right of Bp We have to determine its position.
On examination, we see that of the immersed area, a triangular wedge
RFO, representing upward pressure both in amount and intensity, has been
transferred to the other side of the axis KQ, viz., to TSO. Accordingly, by
a principle of mechanics, the centre of gravity of the rectangle RHNT has
been moved along a line parallel to the line joining the centres of gravity of
the equal triangles RFO and TSO, a distance measured by that between the
centres of gravity of the triangles, multiplied by the ratio of the area of one
triangle to the area of the original rectangle. To put this in symbols, let g1
be the centre of gravity of the triangle RFO, and g2 the centre of gravity of
the triangle STO. Then, if B2 be the position of the new centre of buoyancy,
we must have
(1) the line BjB2 parallel to gxg^,
(2) the distance BjB2 equal to g^2 x
area RFO
area RHNT
Since the alteration in the position of B corresponds to the direction of
g^, and therefore is partly an upward movement, we may conveniently find
the locus by co-ordinates, with the initial position of B as origin. A line
through Bj, parallel to RT or JIN, will be the axis of x; the line KQ, the
axis of y.
Draw g-Jix and g2h2 perpendicular to RT.
The abscissa of B2 is proportionate to Ä^j its ordinate to g^+g^.
Call OR, the semi-width of the pontoon, RH, the original depth of
immersion, D; and RF, the extent of emergence, A.