CABGO AND TRIM 525

total moment , Jrt „ . moment „
If total trim - ITM then 49-8ins.- —^

Moment=49-8x1423=70865 4 foot-tons and—
Distance =moment H-weight=70865 -4^-400=177 ft.

The centre of gravity of the 400 tons weight would need to be
shifted forward 177 feet from its present position to bring the ship on
to an even keel, or 800 tons moved forward half the distance, viz., 88$
feet, would bring about the same result.

STABILITY EQUATIONS.

n r — ^ x w
1 ^ W

Where G Gl is the shift of G

d the distance through which
w the weight has been moved
W the ship's displacement.

•2. G M = £ (^ cot 0.

Where G M is the metacentric height
0 the angle of heel.

3

4. GZ = GMsmQ.

Where G Z is the righting arm.

, L,M. _

Where I.T.M. is the inch-trim-moment

LG M the longitudinal metacentric height
W the ship's displacement
I her length between perpendiculait

30 T2
6. I.T.M. (approx.) =

Where T = tons per inch immersion
B = breadth of ship