Shifting weight fore and aft
1The lesson in plain words
A shift fore and aft leaves displacement unchanged and makes a trimming moment. Divide by MCTC, then share the change via the LCF.
The trimming moment
Moving a weight w a distance d fore and aft makes a trimming moment w × d, in t·m. Displacement does not change, so the ship does not sink. She rotates about the LCF. Change of trim (cm) = w × d ÷ MCTC, towards the end the weight moved to.
Sharing the change
The end farther from the LCF moves more. AP is the aft perpendicular; FP is the forward perpendicular. Change aft = change of trim × distance from AP to LCF ÷ LBP. Change forward = change of trim × distance from LCF to FP ÷ LBP. One end sinks; the other rises. Split the change equally only when the LCF is amidships.
Signs
Write the old draughts. Add the share at the end the weight moved towards; subtract the other share. Then check: new trim = old trim ± change of trim. Sign errors are the commonest mistake in exams and on board, and this check catches them.
Using it
Officers move ballast or fuel fore and aft to reach a required trim: for a draught limit, propeller immersion or an even-keel draught survey. Turn the formula round: w = required change of trim × MCTC ÷ d.
Side effects
MCTC changes with draught, so use the value for the condition. A ballast transfer also leaves slack tanks while it runs, adding free surface.
2Step by step
Worked example: moving ballast forward
- 1
Data: Δ 15,000 t, MCTC 180 t·m/cm, LBP 140 m, LCF 66.0 m forward of the AP; F 7.577 m, A 8.377 m.
The condition from Trim basics.
- 2
Move 300 t of ballast 60 m forward: moment = 300 × 60 = 18,000 t·m.
Δ is unchanged, so there is no sinkage.
- 3
Change of trim = 18,000 ÷ 180 = 100 cm by the head.
The weight went forward.
- 4
Aft: 100 × 66 ÷ 140 = 47.1 cm. Forward: 100 × 74 ÷ 140 = 52.9 cm.
The LCF is 74 m from the FP.
- 5
New draughts: F 7.577 + 0.529 = 8.106 m; A 8.377 − 0.471 = 7.906 m.
The forward end sinks and the aft end rises.
- 6
Check: 0.80 m by the stern − 1.00 m = 0.20 m by the head. 8.106 − 7.906 = 0.200 m.
The trim check catches sign errors.
3Key data
- Change of trim
- COT (cm) = w × d ÷ MCTC
- Change aft
- COT × LCF from AP ÷ LBP
- Change forward
- COT × LCF from FP ÷ LBP
- Weight to shift
- w = required COT × MCTC ÷ d
- Displacement
- Unchanged by a shift: no sinkage
4Key points
- 1
A shift fore and aft changes trim, not displacement.
- 2
Change of trim = w × d ÷ MCTC, towards the end the weight moved to.
- 3
Share the change about the LCF: the end farther from it moves more.
- 4
Always check the new trim against old trim ± change of trim.
5Frequent mistakes
Wrong: Split the change of trim half forward and half aft.
Right: Share it in proportion to the LCF's distance from each perpendicular. Halves are right only with the LCF amidships.
Wrong: Add sinkage for a weight moved within the ship.
Right: Δ is unchanged. Only the trim changes.
Wrong: Measure d from midships.
Right: For a shift, d is the distance the weight moves. For a weight loaded, it is measured from the LCF.
6Practice case & exam keys
Trim must fall from 0.80 m to 0.20 m by the stern. Ballast can move 50 m forward; MCTC is 180 t·m/cm. LBP is 140 m and LCF is 66 m forward of AP. The bow can deepen by only 0.25 m. Can this transfer meet both limits?
Show the answer
The trim target needs 60 cm by the head: w = 60 × 180 ÷ 50 = 216 t. The bow deepens by 60 × 74 ÷ 140 = 31.7 cm, exceeding the 25 cm allowance. Do not approve this transfer as planned. Replan for both limits, tank capacity and free surface. Confirm final draughts and trim.
Exam keys
- COT = w × d ÷ MCTC; no sinkage for a shift.
- Change aft = COT × l(AP to LCF) ÷ LBP; forward = COT × l(FP to LCF) ÷ LBP.
- Check: new trim = old trim ± COT.
- w = COT × MCTC ÷ d for a required trim.
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