508

NICHOLLS'S  SEAMANSHIP  AND NAUTICAL KNOWLEDGE

moved is measured on tne batten, also its perpendicular length PQ
This gives two sides of the right angled triangle P Q R from which /.P,
usually called 0, can be found. Suppose R 0=12 inches and P 0=20
P 0     240 ins
feet then cotangent 0=^ = -——=20 /. 0=2° 52' (from None'*
RQ       12 ms
Nwtiieal Tables).

Fig. 30.
The 10-ton weight may then be shifted over to the other side of the
ship and the experiment repeated as a check on the first trial The
angle of heel is greatly exaggerated in the figure to open out the angle 0.
(iii) The following deduction is then made. It is required to find
the distance between G and M, the positions of which are not yet known.
But G actually moves to G^ in a direction parallel to that of the weight
and is therefore parallel to Q R', G^M is parallel to R P and £.M =^/.P
=/_0 so that triangles P Q R and M G 6^ are similar.
The ship is in equilibrium so her centre of gravity must be in the
same vertical line as her centre of buoyancy and M, by definition, must
be the metacentre.
The shift moment ot G is proportional to the shift moment of w,
that is GGlxW=dxw from which GG^-—   ,   .   .   Equation (1)
where G Gl is the transverse shift of G
W    the ship's displacement in tons
d      the distance through which the weight w has bf en
moved