kg calculations

KG CALCULATIONS (Formulae method and moments)

TUTORIAL QUESTIONS

1. Calculate the effect on G of shifting 250 t of cargo downward through 16 m from a position on deck to the lower hold in a ship displacing 22000 t.

(0.182 m down)

2. A ship displaces 24300 t and has a KG of 5.60 m. Calculate the final KG if 180 t is loaded on deck at Kg 9.60 m.

(5.629 m)

3. A ship displaces 12260 t and has KG 6.68 m. A weight of 34 t is discharged from the lower hold, Kg 2.2 m. Calculate the final KG.

(6.692 m)

4. A ship has an initial KG of 6.20 m. If the displacement is 23360 t, what is the maximum amount of cargo that may be loaded on deck in a position Kg 10.40 m to ensure that the final KG does not exceed 6.46 m?

(1541.5 t)

5. A ship displaces 12300 t and has KG 5.84 m. What is the maximum amount of cargo that can be discharged from the lower hold, Kg 1.80 m, to ensure that the final KG does not exceed 6.00 m?

(468.6 t)

6. A ship displaces 2730 tonnes and has a KG of 6.00 m. The ship then loads the following:540 tonnes at 5.0 m above the keel;370 tonnes at 8.5 m above the keel;110 tonnes at 10.4 m above the keel;850 tonnes at 4.6 m above the keel.

Calculate the final KG.(5.930 m)

7. A loaded lighter displaces 856 tonnes and has a KG of 1.5 m. Find the new KG after the following weights have been discharged:

160 tonnes from 2.5 m above the keel;40 tonnes from 3.7 m above the keel;395 tonnes from 1.2 m above the keel.

(1.004 m)

8. A ship leaves port with a displacement of 9060 tonnes and a KG of 5.2 m. During the voyage the following is consumed:

Oil fuel: 200 tonnes from 0.8 m above the keel;320 tonnes from 0.7 m above the keel;

Stores: 98 tonnes from 9.5 m above the keel;FW: 87 tonnes from 10.0 m above the keel.

What will be the KG on arrival at port of destination?(5.377 m)

KG CALCULATIONS (Formulae method and moments) - Tutorial Questions (Rev. 03/09/02) 1

9. The original displacement of a ship was 4285 tonnes, KG 6.0 m. The following is loaded:800 tonnes at 3.6 m above the keel;440 tonnes at 7.0 m above the keel;110 tonnes at 5.8 m above the keel;630 tonnes at 3.0 m above the keel.

Find the new KG.(5.459 m)

10. A ship has a KG of 6.5 m and a displacement of 6020 tonnes. Find the new KG after loading and discharging the following weights:Load: 500 tonnes at 2.5 m above the keel;

850 tonnes at 5.0 m above the keel;220 tonnes at 8.4 m above the keel;

Discharge: 330 tonnes from 5.5 m above the keel;700 tonnes from 2.6 m above the keel;

(6.531 m)

11. A ship displaces 18940 tonnes, KG 6.22 m. Calculate the final KG if the following cargo is worked:Load: 200 tonnes at Kg 8.62 m;

188 tonnes at Kg 2.56 m;46 tonnes at Kg 3.46 m;

Discharge: 562 tonnes from Kg 7.68 m;Shift: 236 tonnes from Kg 4.20 m to a position on deck Kg 12.2 m.

(6.259 m)

12. A ship displaces 16420 tonnes, KG 5.64 m. Cargo is worked as follows:Load: 1500 t at Kg 6.50 m;

1200 t at Kg 5.00 m;900 at Kg 4.20 m;1000 t at Kg 8.20 m;

Discharge: 220 t from Kg 1.50 m.What is the maximum amount of cargo that can be loaded on deck at Kg 12.00 m to ensure that the final KG does not exceed 5.80 m?

(101.8 t)


SOLUTIONS

1. GGv = w d GGv = 250 16 GGv = 0.182 m downward W 22000

2. GGv = w d GGv = 180 (9.60 - 5.60) GGv = 0.029 m upW + w 24300 + 180

Initial KG 5.600GGv (up) 0.029 FINAL KG 5.629

3. GGv = w d GGv = 34 (6.68 - 2.20) GGv = 0.012 m upW - w 12260 - 34

Initial KG 6.680GGv (up) 0.012 FINAL KG 6.692

4. Initial KG 6.200Final KG 6.460 Max. GGv (up) 0.260

GGv = w d 0.260 = w (10.40 - 6.20) Solve the equation for wW + w 23360 + w

0.260 (23360 + w) = w (10.40 - 6.20)

6073.6 + 0.260w = 4.2w 6073.6 = 4.2w - 0.26w

6073.6 = 3.94w w = 1541.5 tonnes

5. Initial KG 5.840Final KG 6.000 Max. GGv (up) 0.160

GGv = w d 0.160 = w (5.84 - 1.80) Solve the equation for wW - w 12300 - w

0.160 (12300 - w) = w (5.84 - 1.80)

1968 - 0.160w = 4.04w 1968 = 4.04w + 0.16w

1968 = 4.20w w = 468.6 tonnes


6. Take moments about the keel.





12. Take moments about the keel, let x = cargo to load on deck at Kg 12.00 m. The final KG is known which is 5.80 m.

Final KG = MomentsFinal Displacement

Therefore: 5.80 = 120008.8 + 12x 20800 + x

5.80 (20800 + x) = 120008.8 +12x 120640 + 5.8x = 120008.8 + 12x

120640 - 120008.8 = 12x - 5.8x 631.2 = 6.2x x = 101.8 t


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