Page 59 - Introduction to Naval Architecture

46 FLOTATION AND STABILITY

Problems in trim and stability
Determination of displacement from observed draughts
Suppose draughts at the perpendiculars are T a and T f as in Figure 4.12.
The mean draught will be T- (T a + T f)/2 and a first approximation to
the displacement could be obtained by reading off the corresponding
displacement, A, from the hydrostatic curves. In general, W 0Lo will not
be parallel to the waterlines for which the hydrostatics were computed. If
waterline W t Lj, intersecting W 0Lo at amidships, is parallel to the design
waterline then the displacement read from the hydrostatics for draught
Tis in fact the displacement to Wj L }. It has been seen that because ships
are not symmetrical fore and aft they trim about F. As shown in Figure
4.12, the displacement to W 0Lo is less than that to W } Lj, the difference

Figure 4,12

being the layer Wj L! L^ W 2 where W 2L2 is the waterline parallel to Wj LI
through F on W 0Lo- If A is the distance of Fforward of amidships then the
thickness of layer=A X t/L where t= T a- T f.
If i is the increase in displacement per unit increase in draught:

Displacement of layer = A X ti/L and the actual displacement

Whether the correction to the displacement read off from the
hydrostatics initially is positive or negative depends upon whether the
ship is trimming by the bow or stern and the position of F relative to
amidships. It can be determined by making a simple sketch.
If the ship is floating in water of a different density to that for which
the hydrostatics were calculated a further correction is needed in
proportion to the two density values, increasing the displacement if the
water in which ship is floating is greater than the standard.
This calculation for displacement has assumed that the keel is
straight It is likely to be curved, even in still water, so that a draught
taken at amidships may not equal (<4 + <^)/2 but have some value 4n
giving a deflection of the hull, <5. If the ship sags the above calculation
would underestimate the volume of displacement. If it hogs it would