696 NICHOLLS'S SEAMANSHIP AND NAUTICAL KNOWLEDGE

(ii) To find height of C.G. before adding 64 tons.

C B at half draught - - - - KB = 4 00 ft

BM = 10 67

KM = 14-67
Given GM = 2 80

C.G. above keel .... KG ** 11'87 ft.

(iii) To find rise of C.G due to adding 64 tons at 16 ft. above keel
16 — 11 87 = 4 13 ft.

V 53760
Displacement in tons at 8 ft. draught ^-^ = —35— "" 1536

New displacement = 1536 + 64 = 1600 tons.
Rise in ship's C.G = 641^'13 = 0'17 It.
.*. KG* — J i-87 + '17 — 12 04 ft.

(iv) To find new BM ~ BXM
» TVT / 573440

1600 X 35

1*0*

(v) New draught.

T.P.I. - LB „ . 210o>< 32 - 16 tons.
35X12^ 420

Increase of draught « 51= 4/7 = *33 ft.
16

.'. New draught 8*33 ft.

(vi) New height of metacentre and new GM .

KBl = 4 17 ft. at half draught.
BiMi » 10-24

« 14-41
— 12-04

***Mi •" 2'37 ft*

8. The solution of this question requires three steps as follows:—
(i) To find longitudinal BM.

Longitudinal moment of inertia « / ~ BL !. » 32 x 2IOt_

**

= 24,696,000
Longitudinal BM = ^ -*4^S?>0- ««» «.

(ii) Moment to change trim 1 inch.
I536 tons

-_ 12 X 210 ft; — " *
Assuming BM «* GAf to be a good approximation*