25-Nav-A2 Hydrodynamics of Ships (I)_ Resistance and Propulsion · December 2017

Question 3 of 5: Inclining Experiment & Ship's KG

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Question 3: Inclining Experiment & Ship's KG (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Displacement (with inclining weights aboard) $\Delta=3400$ LT, $KM=26.5$ ft, and the five (moment, list-angle) readings above; the 28 LT inclining weight is moved horizontally between readings, with its own centre of gravity at 43 ft above the keel.

Find. The ship's $KG$ at this condition.

Approach. An inclining experiment satisfies $M=\Delta\,GM\tan\theta$ for small angles; zero-correct the readings against the residual list recorded at zero applied moment, fit a straight line through the origin to the corrected data (the "plot" the question asks for), read off the slope $\Delta\,GM$, then take $KG=KM-GM$.

1. Zero-correct the list readings. At zero applied moment the ship already lists 0.2° to port — a small pre-existing asymmetry (off-centre stores, minor list) that must be removed before fitting a line through the origin. Taking starboard positive, subtract this $-0.2^\circ$ baseline from every reading:

   Moment (ft·LT)	Raw angle	Corrected angle
   +880	+2.3°	+2.5°
   +528	+1.2°	+1.4°
   0	−0.2°	0.0°
   −528	−1.5°	−1.3°
   −528	−2.3°	−2.1°

2. Best-fit slope through the origin. With $\tan\theta_i$ computed from the corrected angles, a least-squares fit of $M=k\tan\theta$ forced through the origin gives $$k=\dfrac{\sum M_i\tan\theta_i}{\sum \tan^2\theta_i}=18{,}947\ \text{ft}\cdot\text{LT}$$ (this is the slope plotted in the figure above; the scatter about it is normal experimental noise across five readings).
3. GM and KG as tested. Since $k=\Delta\,GM$, $$GM=\dfrac{k}{\Delta}=\dfrac{18{,}947}{3400}=5.57\ \text{ft}$$ and therefore $$\boxed{KG_{tested}=KM-GM=26.5-5.57=20.93\ \text{ft}}$$
4. Correction for the inclining gear. The stated displacement $\Delta=3400$ LT is the ship with the 28 LT inclining weight aboard. Because that weight starts the experiment centred on the ship's centreline (the zero-moment reading above is exactly the pre-existing list, not a shift caused by the gear), it contributes no net heeling moment before the test and its vertical position (43 ft) is already folded into the 3400 LT, $KG_{tested}=20.93$ ft figure computed above — there is no separate "as-inclined" displacement quoted to correct against. $KG_{tested}=20.93$ ft is therefore reported directly as the ship's $KG$ in this condition.