How much power does trim cost? For a handysize bulk carrier of about 30,000 DWT at 14.5 knots and 9.5 m mean draft, the vessel's own data gives a clear answer: about 4.8% less brake power at 2.0 m by the bow than at even keel, and up to 50% more when trimmed heavily by the stern.
Result
Required brake power at each trim, relative to even keel, computed from the vessel's resistance and propulsion data at 13 trim conditions. Calm water, no wind or current.
Example vessel (main particulars rounded): handysize bulk carrier · DWT approx. 30,000 t · LBP approx. 170 m · loaded displacement approx. 37,000 t at 9.5 m mean draft. Vessel identity is kept confidential.
Handysize bulk carrier (~30,000 DWT) · 14.5 kn · mean draft 9.5 m · hover a point for its value
| Trim (m, + = by the bow) | Brake power vs even keel |
|---|---|
| -4.0 | +49.6% |
| -3.5 | +43.1% |
| -3.0 | +36.4% |
| -2.5 | +29.4% |
| -2.0 | +22.4% |
| -1.5 | +15.6% |
| -1.0 | +9.3% |
| -0.5 | +3.4% |
| 0.0 | 0.0% |
| +0.5 | -2.0% |
| +1.0 | -2.9% |
| +1.5 | -4.2% |
| +2.0 | -4.8% |
Reading the result
Method
For each trim condition, effective power, hull–propeller interaction and the propeller's open-water curve give the propeller operating point and the brake power needed at the chosen speed. Results are interpolated across drafts and trims, and the platform searches for the trim with the lowest power in 0.01 m steps. Our team first delivered trim optimization and energy efficiency studies on 27 commercial vessels in 2013; this case is recomputed with our current web platform. About the platform →
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