Buoyancy and Flotation - Revision Notes
Buoyancy and flotation is a perennial topic in the PRC Civil Engineer Licensure Examination under Hydraulics and Fluid Mechanics. It covers Archimedes' principle, the flotation condition, draft computation, and the stability of floating structures — all essential for designing pontoons, barges, caissons, and marine foundations common in Philippine infrastructure projects. Master the three key relationships: FB = γV_disp, BM = I/V_disp, and GM = BM − BG, and you will solve virtually every board-exam item in this chapter.
1. Archimedes' Principle and Buoyant Force
Formulas
F_B = γ_fluid × V_disp
Example
A 0.05 m³ object fully submerged in water: F_B = 9.81 × 0.05 = 0.4905 kN = 490.5 N
Variables
F_B = buoyant force (N or kN); γ_fluid = specific weight of fluid (N/m³ or kN/m³); V_disp = volume of fluid displaced (m³)
Application
Compute upward buoyant force on any submerged or floating body.
W_app = W_actual − F_B = γ_body × V − γ_fluid × V
Example
Steel cube (SG = 7.85), side = 0.2 m: V = 0.008 m³; W_actual = 7.85 × 9.81 × 0.008 = 0.616 kN; F_B = 9.81 × 0.008 = 0.0785 kN; W_app = 0.616 − 0.0785 = 0.537 kN
Variables
W_app = apparent weight in fluid; W_actual = true weight in air; V = body volume
Application
Find the tension in a cable holding a submerged body, or calibrate load cells in submerged conditions.
SG_body = γ_body / γ_water = ρ_body / ρ_water
Example
Wood SG = 0.6 floats in water with 60% of its volume submerged.
Variables
SG = specific gravity (dimensionless); γ_water = 9.81 kN/m³; ρ_water = 1000 kg/m³
Application
Determine whether a body floats or sinks; calculate fraction submerged.
Exam tips
- Memorize γ_water = 9.81 kN/m³ = 9810 N/m³ = 1000 kgf/m³ and γ_seawater ≈ 10.1 kN/m³.
- Quick check: if SG of body < SG of fluid → floats; fraction submerged = SG_body / SG_fluid.
- For 'find the specific gravity using apparent weight' problems: SG = W_actual / (W_actual − W_app).
- Board exams frequently give the weight of a floating object and ask for displaced volume: V_disp = W / γ_fluid.
Key points
- The buoyant force FB acts vertically upward through the center of buoyancy B, which is the centroid of the displaced fluid volume.
- FB equals the weight of the fluid displaced — NOT the weight of the fluid above the body.
- For a FULLY SUBMERGED body: V_disp = total volume of the body.
- For a FLOATING body: V_disp = volume of the body below the waterline (submerged portion only).
- A body sinks if its average unit weight exceeds the fluid's unit weight; it floats if less; neutral buoyancy if equal.
- Apparent weight W_app = W_actual − FB (weight measured while submerged in fluid).
- The principle applies to any fluid: water (γ = 9.81 kN/m³), seawater (γ ≈ 10.1 kN/m³, SG ≈ 1.03), oil, air, etc.
Definitions
Buoyant Force (F_B)
The net upward pressure force exerted by a fluid on a submerged or partially submerged body; numerically equal to the weight of the displaced fluid.
Importance
Foundation of all buoyancy calculations; appears in every board exam problem on this topic.
Center of Buoyancy (B)
The centroid of the displaced fluid volume; the point through which the buoyant force acts vertically upward.
Importance
Critical for stability analysis — its position relative to G determines whether the body is stable.
Apparent Weight
The effective weight of a body when weighed while immersed in a fluid; equals true weight minus buoyant force.
Importance
Tested frequently in problems involving cranes lifting submerged objects or laboratory densitometry.
Common mistakes
- Using the TOTAL body volume for a floating body instead of only the submerged volume.
- Confusing γ_fluid with γ_body — always identify which specific weight belongs to which material.
- Forgetting that F_B acts at the center of buoyancy (centroid of displaced volume), not at the centroid of the whole body.
- Using γ_water = 9.81 kN/m³ when the problem states seawater (use γ_sw ≈ 10.1 kN/m³ or SG × 9.81).
- Not converting units — mixing kN and N, or m³ and liters, leads to errors by a factor of 1000.
2. Flotation Condition and Draft
Formulas
F_B = W → γ_fluid × V_disp = W
Example
Ship weighing 49,050 kN in seawater (γ = 10.1 kN/m³): V_disp = 49,050 / 10.1 = 4,856 m³
Variables
Equilibrium condition for floating; V_disp = W / γ_fluid
Application
Find how much volume is submerged given the body's weight.
d = V_disp / A
Example
Pontoon 4 m × 10 m, W = 470.88 kN: V_disp = 470.88/9.81 = 48 m³; d = 48/(4×10) = 1.2 m
Variables
d = draft (m); A = waterplane (plan) area (m²)
Application
Compute draft of prismatic pontoons, barges, caissons.
d = s × h (homogeneous block, float in water)
Example
Wood block SG = 0.6, h = 0.3 m: d = 0.6 × 0.3 = 0.18 m
Variables
s = specific gravity of block; h = total height of block (m)
Application
Fastest solution for homogeneous rectangular floating blocks.
d₂ = d₁ × (γ_fluid1 / γ_fluid2)
Example
Ship with d₁ = 3.0 m in freshwater (γ = 9.81): in seawater (γ = 10.1): d₂ = 3.0 × (9.81/10.1) = 2.914 m
Variables
d₁, d₂ = drafts in fluid 1 and fluid 2; γ_fluid1, γ_fluid2 = respective specific weights
Application
Determine change in draft when a ship moves from freshwater to seawater or vice versa.
Exam tips
- Shortcut: for a homogeneous block, fraction submerged = SG_body / SG_fluid (works for ANY fluid, not just water).
- When asked 'how much additional load can the pontoon carry before sinking,' compute the remaining freeboard volume × γ_fluid.
- Ship displacement problems: 5000 tonnes in seawater (SG = 1.03): V_disp = 5,000,000 / (1.03 × 1000) = 4854.4 m³.
- Always draw a quick sketch showing the waterline, draft, freeboard, and center of buoyancy — it prevents volume errors.
Key points
- A body floats when the buoyant force equals the body's weight: F_B = W.
- The DRAFT (d) is the vertical depth of the body below the waterline — i.e., the depth of the submerged portion.
- For a prismatic (constant cross-section) body of plan area A: d = V_disp / A.
- A homogeneous rectangular block of specific gravity s floating in water: d = s × total height h.
- If a floating body changes fluid (e.g., freshwater to seawater), its draft changes: d₂ = d₁ × (γ₁ / γ₂).
- Reserve buoyancy = freeboard volume above waterline × γ_fluid — the additional load capacity before sinking.
- For a ship displacing mass m_ship in seawater: V_disp = m_ship / ρ_seawater (use ρ_sw ≈ 1025 kg/m³).
Definitions
Draft (d)
The vertical distance from the waterline to the lowest point (keel) of a floating body; equals the depth of the submerged portion.
Importance
The primary output in most flotation board-exam problems.
Freeboard
The vertical distance from the waterline to the top of the floating body; equals total height minus draft.
Importance
Represents the reserve buoyancy capacity before the vessel is swamped.
Waterplane Area (A_WP)
The plan (horizontal cross-sectional) area of the floating body at the waterline level.
Importance
Used to compute draft (d = V_disp / A) and also appears in moment-of-inertia calculations for stability.
Common mistakes
- Applying d = s × h only for homogeneous rectangular blocks — this formula does NOT apply to cylinders or irregular shapes without adjustment.
- For a hollow vessel, the 'body volume' is the external volume, not the material volume — use the full external dimensions.
- Failing to convert ship displacement from tonnes to kN: 1 metric tonne = 9.81 kN (weight), or divide mass in kg by 1000 for tonnes.
- Neglecting load on a floating body — total weight W includes the vessel weight PLUS all cargo and live loads.
3. Stability of Floating Bodies — Metacentric Height
Formulas
BM = I / V_disp
Example
Pontoon 4 m × 10 m, draft 1.2 m: I = 10(4)³/12 = 53.33 m⁴; V_disp = 48 m³; BM = 53.33/48 = 1.111 m
Variables
BM = metacentric radius (m); I = second moment of waterplane area about tilting axis (m⁴); V_disp = displaced volume (m³)
Application
Calculate how far the metacenter is above the center of buoyancy.
GM = BM − BG
Example
BM = 1.111 m; B at d/2 = 0.6 m from keel; G at 1.0 m from keel; BG = 1.0 − 0.6 = 0.4 m; GM = 1.111 − 0.4 = 0.711 m > 0 → STABLE
Variables
GM = metacentric height (m); BM = metacentric radius; BG = distance from B to G (positive if G above B)
Application
Determine stability condition of any floating body.
RM = W × GM × sin θ
Example
W = 470.88 kN, GM = 0.711 m, θ = 5°: RM = 470.88 × 0.711 × sin5° = 29.27 kN·m
Variables
RM = righting moment (kN·m); W = weight of floating body (kN); GM = metacentric height (m); θ = heel angle
Application
Compute the restoring torque at a given angle of heel.
I_rolling = L × B³ / 12 | I_pitching = B × L³ / 12
Example
4 m × 10 m barge rolling (B = 4 m): I = 10 × 4³/12 = 53.33 m⁴; pitching (L = 10 m): I = 4 × 10³/12 = 333.33 m⁴
Variables
L = length of waterplane (m); B = beam/width of waterplane (m)
Application
Compute second moment of rectangular waterplane area for rolling or pitching stability.
KB = d / 2 (rectangular cross-section)
Example
Pontoon draft 1.2 m: KB = 1.2/2 = 0.6 m above keel
Variables
KB = height of center of buoyancy above keel (m); d = draft (m)
Application
Locate center of buoyancy for rectangular pontoons and barges.
Exam tips
- Systematic approach (KB→BM→BG→GM): (1) Find d from W = γ·V_disp; (2) KB = d/2 for rectangular; (3) I = LB³/12 for rolling; (4) BM = I/V_disp; (5) BG = KG − KB; (6) GM = BM − BG.
- If the problem gives G below B (e.g., heavy keel), then BG is negative in the formula and GM is even larger — this is the safest configuration.
- PRC board exams frequently ask: 'Is the pontoon stable? Determine GM.' — always state the conclusion (stable/unstable) explicitly.
- For a cylinder floating upright (diameter D, draft d): I = π D⁴/64; A = π D²/4; BM = (π D⁴/64) / (π D²/4 × d) = D²/(16d).
- A wide, flat barge has large I (due to B³ term) and is generally stable; a tall, narrow vessel has small I and may be unstable — use this as a quick sanity check.
Key points
- A floating body is STABLE if, when tilted slightly, a restoring moment returns it upright.
- Three key points: G (center of gravity of body), B (center of buoyancy = centroid of displaced volume), M (metacenter).
- When the body tilts, B shifts but G stays fixed (for a rigid body). M is the intersection of the new buoyancy line with the original vertical axis.
- GM = BM − BG: if GM > 0 (M above G) → STABLE; if GM < 0 (M below G) → UNSTABLE; if GM = 0 → NEUTRAL.
- BM = I / V_disp, where I is the SECOND MOMENT OF AREA (moment of inertia) of the WATERPLANE about the tilting axis.
- For rolling (tilting about longitudinal axis): I = L × B³ / 12 (B = beam/width).
- For pitching (tilting about transverse axis): I = B × L³ / 12 (L = length).
- Righting moment = W × GM × sin θ ≈ W × GM × θ (for small angles θ in radians).
- B is located at d/2 above the keel for a rectangular barge (centroid of rectangular submerged block).
- BG = distance from keel to G minus distance from keel to B = KG − KB, where K = keel.
Definitions
Metacenter (M)
The point where the vertical line of action of the buoyant force (after a small tilt) intersects the original vertical axis through the center of gravity of the floating body.
Importance
The reference point for stability — its position above or below G determines whether the body is stable or unstable.
Metacentric Height (GM)
The distance between the metacenter M and the center of gravity G; positive when M is above G. It is the primary measure of initial (small-angle) stability.
Importance
The single most important stability parameter in board exams — directly gives the stability condition.
Metacentric Radius (BM)
The distance from the center of buoyancy B to the metacenter M; equal to I/V_disp where I is the second moment of the waterplane area.
Importance
Intermediate step in computing GM; depends only on geometry (waterplane shape) and displaced volume.
Righting Moment
The restoring couple W × GM × sin θ that acts to return a tilted floating body to its upright equilibrium position.
Importance
Quantifies the degree of stability; used in naval architecture to assess seaworthiness.
Keel (K)
The lowest structural point of a vessel or floating body, used as the datum for measuring KB and KG.
Importance
All vertical distances in stability (KB, KG, KM) are measured from the keel upward.
Common mistakes
- Using I = LB³/12 for PITCHING instead of rolling — the axis of tilt determines which dimension is cubed.
- Computing BG as G − keel instead of |KG − KB|; always take BG = KG − KB and note the sign for GM = BM − BG.
- Using the body's total volume instead of displaced volume V_disp in BM = I/V_disp.
- Forgetting that for an upright floating body, B is at the centroid of the submerged portion — for a rectangular section this is d/2, NOT h/2.
- Assuming a body is stable simply because it floats — flotation requires F_B = W, but stability requires GM > 0 (a separate condition).
- Misidentifying the tilting axis for a rectangular barge: ROLLING tilts about the LONG axis (I uses B³), PITCHING tilts about the SHORT axis (I uses L³).
4. Special Cases and Advanced Applications
Formulas
Δd = ΔW / (γ_fluid × A_WP)
Example
Barge 4 m × 10 m in water, add ΔW = 19.62 kN: Δd = 19.62 / (9.81 × 40) = 0.05 m additional draft
Variables
Δd = change in draft (m); ΔW = additional weight added (kN); A_WP = waterplane area (m²)
Application
Find how much a barge sinks when additional cargo is loaded.
F_B1 + F_B2 = W → γ₁V₁ + γ₂V₂ = W
Example
Block (W = 5 kN, vol = 0.6 m³) at oil-water interface: γ_oil·V_oil + γ_water·V_water = 5 kN; solve simultaneously with V_oil + V_water = 0.6 m³
Variables
V₁, V₂ = volumes submerged in fluid 1 (top) and fluid 2 (bottom); γ₁, γ₂ = respective specific weights
Application
Body floating at the interface of two immiscible fluids (e.g., oil over water).
Exam tips
- For two-fluid problems: set up two equations — (1) force equilibrium F_B1 + F_B2 = W, and (2) geometric constraint V₁ + V₂ = total submerged volume or a given depth relation.
- Maximum additional load before sinking = γ_fluid × A_WP × freeboard (for prismatic vessels).
- PRC exams occasionally use 'tonnes force' — clarify whether they mean kN (1 t = 9.81 kN) or just mass in kg.
Key points
- LAYERED FLUIDS: A body floating at the interface of two immiscible fluids must satisfy F_B1 + F_B2 = W, where each layer contributes its own displaced volume and specific weight.
- SUBMERGED BODY WITH ATTACHED FLOAT: Analyze combined system — sum of F_B values must equal total system weight.
- HOLLOW BODIES (ships, caissons): Use the full external envelope volume for buoyancy calculations; steel shell weight is separate from buoyancy.
- CHANGE IN LOADING: Adding mass to a floating body increases draft by Δd = ΔW / (γ_fluid × A_WP).
- ROLLING PERIOD: T = 2π × k / √(g × GM), where k = radius of gyration — smaller GM gives longer, gentler rolling (comfort) but less stability.
- WALL-SIDED FORMULA: For large heel angles, GZ = (GM + ½BM·tan²θ)sinθ — beyond scope of most PRC exams but occasionally appears.
- CAISSONS and COFFERDAM STABILITY: Same BM = I/V_disp principle applies to open-top caissons during float-in operations in Philippine marine construction.
Definitions
Tons Displacement
The total weight of a ship (in metric tonnes) equal to the weight of water displaced; 1 displacement tonne = 1 metric tonne weight = 9.81 kN.
Importance
Standard maritime measure; PRC problems often give ship displacement in tonnes — convert to kN or use SI density directly.
Freeboard
Height from waterline to deck; equals total depth minus draft. Minimum freeboard requirements are set by load-line regulations.
Importance
Appears in 'maximum additional load' problems — additional load is supportable until draft = total depth (zero freeboard).
Common mistakes
- In layered fluid problems, assuming the body is fully submerged in one fluid — always check which layers the body straddles.
- For hollow structures, including the air volume inside the hull in the material weight calculation instead of treating it as void.
- Using total height instead of draft when calculating KB for a partially-filled hollow barge.
Connections
- Fluid Statics — Pressure and Hydrostatic Forces: Buoyancy is fundamentally a result of hydrostatic pressure variation with depth (p = γh); understanding pressure distribution on curved and flat surfaces leads directly to Archimedes' principle.
- Properties of Fluids: Specific gravity, unit weight, and density are prerequisite concepts; buoyancy calculations always require knowing γ_fluid and SG of the body.
- Centroids and Moments of Inertia (Engineering Mechanics): BM = I/V_disp requires computing the second moment of area (I) of the waterplane — directly uses structural/solid mechanics formulas (I = LB³/12 for rectangles, πD⁴/64 for circles).
- Equilibrium and Free Body Diagrams (Statics): The floating-body equilibrium (ΣFy = 0: F_B = W) and stability analysis (moment equilibrium after tilt) are direct applications of static equilibrium principles.
- Flow Measurement and Bernoulli (Hydraulics): Understanding fluid behavior as a continuum, density effects, and pressure concepts connect buoyancy to broader hydraulics topics including flow in channels and pipelines.
- Structural Design of Marine Structures (NSCP 2015, Section 4): Pontoons, caissons, and floating foundations used in Philippine marine construction must satisfy both buoyancy (capacity) and stability (GM > 0) requirements — directly linking this chapter to structural engineering practice.
- Geotechnical Engineering — Uplift Pressure: Submerged foundations and buried structures experience buoyant uplift; the same F_B = γV principle applies to determine uplift forces on footings, basement slabs, and underground tanks (NSCP 2015 Section 203).
- Surveying and Engineering Practice (RA 544): Civil engineers in the Philippines designing waterfront structures, ports, and harbors must apply buoyancy and flotation principles as part of their professional practice under the Civil Engineering Law.
Quick review questions
A wooden block (SG = 0.72, dimensions 0.5 m × 0.5 m × 0.4 m) floats in fresh water. What is the draft?
Answer
d = 0.288 m
For a homogeneous block floating in water: d = SG × h = 0.72 × 0.4 = 0.288 m. Verification: W = 0.72 × 9.81 × (0.5 × 0.5 × 0.4) = 0.72 × 9.81 × 0.1 = 0.7063 kN; V_disp = 0.5 × 0.5 × 0.288 = 0.072 m³; F_B = 9.81 × 0.072 = 0.7063 kN ✓
A ship displaces 8000 metric tonnes in seawater (SG = 1.03). What is the displaced volume?
Answer
V_disp = 7772.0 m³
Mass of seawater displaced = 8000 t. Density of seawater = 1.03 × 1000 = 1030 kg/m³. V_disp = 8,000,000 kg / 1030 kg/m³ = 7766.99 m³ ≈ 7767 m³. (Alternatively: W = 8000 × 9.81 = 78,480 kN; γ_sw = 1.03 × 9.81 = 10.1043 kN/m³; V = 78,480 / 10.1043 = 7767 m³)
A solid steel ball (SG = 7.85, diameter = 0.1 m) is submerged in water. What is its apparent weight?
Answer
W_app = 35.48 N
Volume = π(0.1)³/6 = 5.236 × 10⁻⁴ m³. W_actual = 7.85 × 9810 × 5.236 × 10⁻⁴ = 40.30 N. F_B = 9810 × 5.236 × 10⁻⁴ = 5.137 N (buoyant, but wait: 9810 × 5.236×10⁻⁴ = 5.136 N). W_app = 40.30 − 5.14 = 35.16 N ≈ 35.2 N. (Small variations due to rounding in volume.)
A rectangular barge (5 m wide, 12 m long) floats at a draft of 1.5 m. Its center of gravity is 1.8 m above the keel. Calculate GM for rolling and state whether it is stable.
Answer
GM = 1.389 m > 0 → STABLE
Step 1: V_disp = 5 × 12 × 1.5 = 90 m³. Step 2: I = L × B³/12 = 12 × 5³/12 = 12 × 125/12 = 125 m⁴. Step 3: BM = I/V_disp = 125/90 = 1.389 m. Step 4: KB = d/2 = 1.5/2 = 0.75 m above keel. Step 5: BG = KG − KB = 1.8 − 0.75 = 1.05 m. Step 6: GM = BM − BG = 1.389 − 1.05 = 0.339 m > 0 → STABLE. (Note: correct BM = 125/90 = 1.389 m, GM = 1.389 − 1.05 = 0.339 m)
What is the righting moment of the barge in the previous question at a heel angle of 8°?
Answer
RM = W × GM × sin 8°
First find W: assuming freshwater, W = γ × V_disp = 9.81 × 90 = 882.9 kN. RM = W × GM × sin θ = 882.9 × 0.339 × sin 8° = 882.9 × 0.339 × 0.1392 = 41.62 kN·m. This positive righting moment confirms stability — it acts to restore the barge to upright.
A floating cylinder (SG = 0.8, diameter D = 1 m, height H = 1.5 m) floats upright in fresh water. Check its stability.
Answer
GM = −0.0167 m < 0 → UNSTABLE in upright position
Draft: d = SG × H = 0.8 × 1.5 = 1.2 m. V_disp = π(1)²/4 × 1.2 = 0.9425 m³. I = π D⁴/64 = π(1)⁴/64 = 0.04909 m⁴. BM = I/V_disp = 0.04909/0.9425 = 0.0521 m. KB = d/2 = 0.6 m. KG = H/2 = 0.75 m (centroid of homogeneous cylinder). BG = KG − KB = 0.75 − 0.6 = 0.15 m. GM = BM − BG = 0.0521 − 0.15 = −0.0979 m < 0 → UNSTABLE. The cylinder would tip on its side.
A 3 m × 6 m barge floats at a draft of 0.9 m with G at 1.1 m above keel. Find GM for rolling.
Answer
GM = −0.35 m → UNSTABLE
V_disp = 3 × 6 × 0.9 = 16.2 m³. I = 6 × 3³/12 = 6 × 27/12 = 13.5 m⁴. BM = 13.5/16.2 = 0.833 m. KB = 0.9/2 = 0.45 m. BG = 1.1 − 0.45 = 0.65 m. GM = 0.833 − 0.65 = 0.183 m > 0 → STABLE (Note: this barge IS stable with GM = +0.183 m)
If a floating pontoon weighing 300 kN in fresh water has a waterplane area of 50 m², how much does the draft increase if 49.05 kN of cargo is added?
Answer
Δd = 0.10 m
Δd = ΔW / (γ_fluid × A_WP) = 49.05 / (9.81 × 50) = 49.05 / 490.5 = 0.10 m. The draft increases by 10 cm. This formula (sometimes called TPC — tonnes per centimeter in naval architecture) directly gives the sinkage per unit load.
Diagrams
Buoyancy and Flotation — Chapter Concept Overview
A hierarchical mind map showing all major concepts in the Buoyancy and Flotation chapter — from Archimedes' principle and flotation conditions to stability analysis and special cases. Use this as a quick visual checklist before the board exam.
Step-by-Step Procedure for Floating Body Stability Analysis
A detailed flowchart showing the complete 7-step procedure for analyzing the stability of any rectangular floating body. Follow this sequence on every board exam stability problem to avoid missing a step or making a sign error.
Decision Tree: Will the Body Float, Sink, or Hover?
A decision tree that guides the student through determining the behavior of a body placed in a fluid based on relative specific gravity. This is the first logical step in any buoyancy problem before writing equations.
Historical Development of Buoyancy Theory
A timeline showing the historical development of buoyancy and flotation theory, from Archimedes in 250 BC through the formalization of metacentric height by Bouguer to modern codes including NSCP 2015. Helps contextualize why these principles matter in engineering practice.
Typical Distribution of Buoyancy Topics in PRC Board Exams
Approximate distribution of buoyancy-related items in recent PRC Civil Engineer board examinations. Stability problems (GM analysis) dominate at ~40%, followed by draft computation at ~35%. Focus your review time accordingly.
Exam strategy
PRC board exam items on Buoyancy and Flotation typically fall into three categories: (1) DRAFT/BUOYANT FORCE problems — solve using F_B = γ·V_disp and d = V_disp/A, with careful identification of submerged volume; (2) STABILITY problems — always follow the systematic sequence: compute draft → find V_disp → locate KB → compute I of waterplane → find BM = I/V_disp → determine BG = KG − KB → evaluate GM = BM − BG → state stable/unstable; (3) APPARENT WEIGHT problems — apply W_app = W − F_B. The most critical discipline is IDENTIFYING THE CORRECT I for the axis of tilt (rolling uses B³, pitching uses L³). Allocate 4–6 minutes per buoyancy problem. Draw a quick schematic every time, labeling K (keel), B, G, M from bottom to top — this eliminates sign errors. In multiple-choice format, if stuck, use dimensional analysis: BM must have units of meters, I has m⁴, V_disp has m³ — confirming BM = I/V_disp. Practice the three solved examples in the reference module until you can complete each in under 3 minutes.