Moving, adding and lifting weights
1The lesson in plain words
Moving a weight moves G parallel to the move. One formula covers it: GG′ = w × d ÷ Δ.
The rule
When a weight w moves a distance d aboard a boat of displacement Δ, G moves parallel to the move by GG′ = w × d ÷ Δ. Use the same units for w and Δ, for example tonnes. Move weight up and G rises; move it sideways and G moves sideways.
Vertical moves
Moving an existing weight upward raises KG by w × d ÷ Δ. At unchanged displacement and draught, GM falls by the same amount. Lowering it does the opposite. High lockers, the coachroof and an outboard boom can put weight above your yacht’s original G.
Sideways moves and list
Moving weight to one side moves G off the centreline. With positive GM, she heels until B is under the new G. For small angles tan θ = GG′ ÷ GM. A small GM turns a small shift into a big angle.
Adding or removing weight
Adding weight moves G towards the new weight; removing it moves G away. With the new total: GG′ = w × d ÷ (Δ ± w), where d is the distance from the old G to the weight. Displacement and draught change too, so check KM before estimating the new GM.
Suspended weights
Once a load hangs freely from a halyard, boom or davit, treat its weight as acting at the point of suspension, not where it hangs. Your yacht’s G moves towards that point by the weighted proportion; it does not move all the way there. An outboard suspension point also shifts G sideways.
2Step by step
Worked example: crew moving to one side
- 1
Data: Δ = 6.0 t, GM = 1.2 m. Four crew, 0.30 t in all, move 2.0 m to port.
Use tonnes for both weights and metres for distances.
- 2
GG′ = w × d ÷ Δ = 0.30 × 2.0 ÷ 6.0 = 0.10 m.
G moves 0.10 m to port, parallel to the crew’s move.
- 3
tan θ = GG′ ÷ GM = 0.10 ÷ 1.2 = 0.083.
The boat heels until B lies under the new G.
- 4
θ ≈ 4.8° to port.
This result lies within the small-angle range used by the formula.
3Key data
- Shift of G
- GG′ = w × d ÷ Δ (parallel to the weight’s move)
- Heel from a shift
- tan θ = GG′ ÷ GM (small angles)
- Add or remove
- GG′ = w × d ÷ (Δ ± w), d from old G
- Suspended weight
- Acts at the point of suspension
4Key points
- 1
GG′ = w × d ÷ Δ, parallel to the weight’s move.
- 2
Moving existing weight up raises G and cuts GM equally at unchanged displacement and draught.
- 3
Sideways shift: tan θ = GG′ ÷ GM; small GM, big list.
- 4
Hanging weights act at the point of suspension.
5Frequent mistakes
Wrong: A dinghy hanging from the boom acts at the dinghy’s own height.
Right: It acts at the boom end where it hangs from. Lift slowly with the boat level, and lower it to the deck soon.
Wrong: A shift of G of a few centimetres cannot matter.
Right: Compare it with GM. On a tender boat a few centimetres give a large heel and less reserve.
6Practice case & exam keys
You lift a dinghy from the deck on a vertical main halyard. Why does stability decrease as soon as the dinghy leaves the deck, even while it is still low?
Show the answer
Once clear of the deck, the dinghy acts at the masthead suspension point. The yacht’s G rises by the dinghy’s weight times the vertical distance from the dinghy’s original centre of gravity to the masthead, divided by total displacement. Displacement is unchanged, so GM falls by the same amount. G does not rise all the way to the masthead. A centreline suspension adds no transverse shift of G.
Exam keys
- Existing weight shift: GG′ = w × d ÷ Δ. At unchanged draught, KG and GM change equally in opposite directions.
- Heel from a transverse shift: tan θ = GG′ ÷ GM.
- Adding/removing weight: GG′ = w × d ÷ (Δ ± w).
- Suspended weight acts at the point of suspension.
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