PE Naval Architecture common mistakes
PE Naval Architecture · Understand the traps and how to avoid them.
Computes GM as KM minus KG and then applies a free-surface correction to a tank that the loading condition shows as pressed up (completely full) or completely empty.
Candidates memorise 'GM = KM − KG − FSC' as a single formula and apply the FSC term mechanically to every tank in the condition, forgetting that the correction only exists when a tank has a free surface. Because the numbers still 'work out', the error is invisible without a physical check.
What to do instead
Before computing any FSC, list every tank in the condition and mark it slack / full / empty. Only slack tanks with a free surface moment contribute. Carry a column in the calculation sheet headed 'slack? (Y/N)' so the decision is visible and auditable.
Takes the righting-lever curve from the hydrostatics book for the design displacement and reads GZ, range of stability and angle of maximum GZ directly off it, without correcting for the actual KG of the loading condition.
Textbook GZ curves are usually plotted for an assumed KG (often the design or light KG). Candidates who have just done a KM−KG calculation assume the curve already reflects 'their' ship, and the curve's shape looks plausible at every displacement.
What to do instead
Rebuild GZ from the KN (or MS) cross-curves: GZ = KN − KG·sinθ, using the KG of the condition being assessed. State the KG used in the curve title whenever you plot it.
Uses the ITTC-57 friction coefficient with speed entered in knots (and sometimes length in metres) so that the Reynolds number is wrong by roughly a factor of two, then propagates the error into the full-scale resistance extrapolation.
Model test reports, speed-power tables and ship particulars are all quoted in knots, so knots is the number on the page. The ITTC line is written in terms of Rn, and candidates plug the 'speed' number straight in without converting.
What to do instead
Write the conversion down before substituting: V [m/s] = V [kn] × 0.5144, and ν [m²/s] ≈ 1.19×10⁻⁶ for fresh water at 15 °C. Compute Rn as a separate labelled line with units attached, then feed Rn into the friction line.
Confuses wake fraction and thrust deduction - uses wT in place of (1 − t) when converting between thrust power and effective power, or uses the average wake for the Taylor wake in a single-screw ship.
Both terms appear in the same powering chain and both are dimensionless numbers close to 0.2–0.4, so they look interchangeable on the page. The definitions are easy to hold separately but easy to swap under time pressure, and the resulting answer is often within a plausible band.
What to do instead
Write the chain with the definitions spelled out in words each time: PE = RT·V; PT = T·VA where VA = V(1 − w). Then PD = PT/ηR and PS = PD/ηD. Check the result against the sanity band PS/PE ≈ 1.6–2.2 for a typical displacement hull; a swap usually pushes it outside.
Computes midship section modulus by dividing the second moment of area by the distance from the baseline (or from the deck) to the extreme fibre, rather than from the neutral axis, and reports a single value rather than deck and keel moduli.
In simple beam problems the neutral axis often lies at mid-depth, so the habit of using the half-depth persists. In a ship section the neutral axis sits well above mid-depth, so the keel-fibre distance is much larger than assumed and the keel modulus is markedly smaller than the deck modulus.
What to do instead
Locate the neutral axis first (sum of first moments / total area from a stated datum), then compute I about that axis. Report both Z_deck = I/y_deck and Z_keel = I/y_keel, with y measured from the neutral axis, and quote which one you used in the bending stress check.
Adds hogging and sagging still-water bending moments to the wave-induced moment without respecting the sign convention, producing a total moment that is the difference of the two components instead of their sum.
Hogging and sagging are described in words in the question, and the candidate converts to ± signs from memory rather than from a defined convention. Definitions of positive BM differ between textbooks and class rules, so the sign that 'feels right' is often the wrong one.
What to do instead
At the start of the calculation, draw a small sketch of the hull with the chosen positive BM direction arrowed, and state it in words ('sagging positive'). Then apply the wave moment with the sign implied by whether the wave crest is amidships or at the ends, not by whether it is called 'wave hogging'.
Uses the roll period formula T = 2πk/√(g·GM) with the radius of gyration in metres and GM in feet (or vice versa), or drops the 2π and reports a period that is about six times too small.
Roll period is one of the few seakeeping formulas candidates recall purely from memory under exam pressure, and the 2π is easy to lose. Mixed-unit data (GM from a stability book in metres, k from an imperial rule-of-thumb) compounds it.
What to do instead
Write the formula with a small unit-check line beside it: T [s] = 2π·k [m] / √(9.81 [m/s²] × GM [m]). Sanity-check the answer against the rough rule T ≈ 0.8·B/√GM for a typical cargo ship before moving on.
Selects a pump, heat exchanger or generating set from the nominal duty point without adding a fouling/service margin, or mixes specific fuel consumption units between g/kWh and kg/h so that the fuel flow is out by a factor related to the density.
Machinery sizing questions present a clean design condition, and the candidate answers the clean condition. Margin conventions and unit conventions live in class rules and maker's data rather than in the calculation itself, so nothing in the arithmetic flags the omission.
What to do instead
State the margin explicitly as a line item (e.g. 'heat exchanger duty × 1.15 for fouling') and carry units through every line: SFC in g/kWh × power in kW gives g/h, then divide by 10⁶ and by fuel density for m³/h. Check the final fuel flow against the ship's daily consumption figure.
Builds a weight estimate by adding a design margin percentage on top of an already-margined lightship weight, or reconciles the deadweight by adjusting the margin rather than the lightship, so that the displacement no longer closes.
Weight estimates are assembled in stages over a long question, and each stage legitimately carries its own allowance. Candidates lose track of which allowances are already embedded and treat the margin as a free variable to make the displacement balance.
What to do instead
Keep a two-column ledger: 'estimated weight' and 'allowance already included'. Apply the design margin once, to the sum of the un-margined items, and state it in the closing line. If displacement does not close, adjust the item that is genuinely uncertain and say so, rather than the margin.
Answers a damage-stability or large-angle question using wall-sided assumptions (GZ = GM·sinθ, or constant KM with heel) beyond the range where they hold, without checking that the deck edge is still immersed or that the waterplane area is unchanged.
The wall-sided formula is the one most often memorised and it gives a clean answer at any angle. The failure is silent because the arithmetic is correct; only the geometric assumption is wrong.
What to do instead
Before using any small-angle or wall-sided formula, state the angle range over which it is valid and check the geometry: is the deck edge immersed? Has the waterplane changed? If yes, use the KN cross-curves or an assumed-forms integration and say why the simple formula was rejected.