By proportional methods

CHANGE FORWARD = 69 * 0,4 = 0,21
130

CHANGE AFT = 61 * 0,4 = 0,19
130

FORWARD DRAFT = 5,97 -- 0,21 = 5,76 M

AFTER DRAFT = 6,31 + 0,19 = 6,32 M

MEAN DRAFT = 5,76 + 6,32 = 6,04 M
2

QUESTION 22:

SHIFTING

A PLANK WEIGHT 25,09 KG. WHAT WILL BE THE SHIFT OF IT'S CENTRE OF GRAVITY IF A
WEIGHT OF 17,24 KG IS PLACED A DISTANCE OF 3,66 M FROM IT'S ORIGINAL CG?

ANSWER:

(a) THE CG OF A BODY MOVES PARALLEL TO THE SHIFT OF THE CG OF ANY WEIGHT
SHIFTED WITHIN IT.

(b) THE DISTANCE IT MOVES IS EQUAL TO THE WEIGHT SHIFTED, MULTIPLIED BY THE
SHIFT OF IT'S CG; ALL DIVIDED BY THE TOTAL WEIGHT OF THE BODY.

17,24 * 3,66 = 63,09 = 1,49 M
25,09 + 17,24 42,33

QUESTION 23:

A BEAM CARRIES 2438 KG AT A DISTANCE OF 1,83 M FROM ONE END & THE CENTRE OF
GRAVITY OF THE WHOLE MASS IS 6,09 M FROM THE END. IF THE WEIGHT OF THE BEAM
ALONE IS 4267,47 KG, WHAT IS THE DISTANCE OF THE CG OF THE ORIGINAL BEAM FROM
THE END?

ANSWER:

GG = w * d = 2438,55 * 4,26 = 2,43
W 4267,47

2,434 + 6,09 = 8,524

QUESTION 24:

T P C

FIND THE TONNES PER CENTIMETER IMMERSION OF A BOX SHAPED VESSEL 64 M LONG BY
10,668 M BREADTH.

ANSWER:

T P C = 1,025 * AREA OF THE WATERPLANE (A) = 1,025 * 10,668 * 64,0 = 6,99
100 100

QUESTION 24: (METRIC)

WEIGHT OFF THE CENTRE OF FLOTATION

VESSEL'S LENGHT IS 115,82 M, T P C IS 14,96 TONNES, M T C IS 330,7 TONNES. CENTRE
OF FLOTATION ON THE CENTRE LINE.THERE IS 57,9 TONNES REMOVED 39,62 M AFT OF THE
STERN. ORIGINAL DRAFTS ARE 5,23 M FWD & 5,33 M AFT. FIND NEW DRAFTS.

ANSWER:

LENGTH = 115,82 = 5791 -- 39,62 = 18,29 M
2

MOMENT TO CHANGE TRIM = W * D = 57,9 * 18,29 = 1058,99

CHANGE OF TRIM = 1059,99 = 3,20 CM
330,7

REMOVAL OF WEIGHT THE VESSEL WILL RISE = W = 57,9 = 3,6 CM
T P C 14,96

ORIGINAL DRAFT RISE FWD = 0,23 M AFT = 5,33 M

5,30 5,298
TRIM -0.036 + 0,036
5.162 M 5,334 M

Дата добавления: 2015-07-25