SHIP STABILITY, THEORY AND PRACTICE · VOLUME TWO · APPLIED STABILITY AND TRIM

Chapter 15 · The Intact Stability Criteria

The rule book meets the curve: six criteria, one worked compliance run, one honest failure, and the energy the area represents.

Every loaded condition a ship sails in must be proved acceptable before she leaves, and the proof is written against a short list of criteria set out in Part A of the International Code on Intact Stability, 2008 (the IS Code), applied through the flag state’s load line and safety regulations. All six general criteria are measurements of one object: the GZ curve. This chapter builds the curve for MV Ninja from her KN tables, measures it with the Simpson rules from Volume One, fills in the compliance table an examiner expects, and then shows what a failing condition looks like, because the most instructive fact about the criteria is that a ship can pass the GM line and still fail the curve.

15.1 The rule book: six lines

The criteria come in three families. Three are areas under the curve, in metre radians: at least 0.055 up to 30 degrees, at least 0.09 up to 40 degrees or the angle of flooding if that is less, and at least 0.03 between 30 and 40 degrees. Two are about the height and position of the curve: the righting lever GZ must be at least 0.20 m at an angle of heel of 30 degrees or more, and the maximum GZ must occur at an angle of not less than 25 degrees. One is about the start: the initial GM, corrected for free surfaces, must be at least 0.15 m. For ships carrying timber deck cargo the Code’s timber criteria (Part A, section 3.3) may be applied instead: a single area of 0.08 metre radians to 40 degrees or the angle of flooding, a maximum GZ of at least 0.25 m, and a GM of not less than 0.10 m at all times during the voyage.

The six lines of the rule bookthe general intact stability criteria of the 2008 IS Code, Part A, section 2.2Area under the GZ curve, 0 to 30 degreesat least 0.055 metre radiansArea under the GZ curve, 0 to 40 degrees (or the angle of flooding if less)at least 0.09 metre radiansArea between 30 and 40 degrees (or the flooding angle)at least 0.03 metre radiansRighting lever GZ at an angle of heel of 30 degrees or moreat least 0.20 mAngle at which the maximum GZ occursnot less than 25 degreesInitial GM, corrected for free surfacesat least 0.15 mthree areas, one height, one position, one starting slope: together they describe a healthy curve
Figure 15.1   Three areas, one height, one position, one starting slope: together they describe a healthy curve.

15.2 From the KN tables to the curve

The booklet cannot print GZ directly because GZ depends on the KG of the day. Instead it prints KN, the lever measured from the keel, at a set of angles for each displacement. One subtraction per angle turns KN into GZ. MV Ninja’s KN table runs at 5, 10, 12, 20, 30, 40, 50, 60, 70 and 80 degrees, for displacements from 8000 t to 30500 t in steps of 1500 t, so every angle the criteria name is printed and the only interpolation is in displacement. At 29751 t the fraction between the 29000 t and 30500 t rows is 751 ÷ 1500 = 0.5007, so KN at 30° = 5.261 − 0.5007 × (5.261 − 5.146) = 5.203 m and KN at 40° = 6.576 − 0.5007 × (6.576 − 6.327) = 6.451 m. When drawing the curve, the GM is erected at 57.3 degrees and joined to the origin as a construction guide: the curve leaves the origin along that line.

GZ = KN − KG × sin of the heelMCA formula sheet, September 2020
Every point on the curve comes from the same subtractionthe booklet gives KN; the loading condition gives KG; the sine does the rest: shown here at 30 degreesKN from the booklet at 30 degrees5.203 mKG × sin 30° = 8.30 × 0.50004.150 mGZ, the righting lever1.053 mGZ = KN − KG × sin of the heel = 5.203 − 4.150 = 1.053 mrun this once per tabulated angle and the whole curve exists
Figure 15.2   One subtraction per angle: the booklet’s KN, the condition’s KG, and the sine between them.
Animation 1 · Building the curve, one subtraction at a time
press play 0° heel 80°
At each tabulated angle a blue KN bar rises, the gold KG × sin bar climbs against it, and the difference, the GZ, is planted as a point. The condition is Worked example 15.1 (KG 8.30 m). By 80 degrees the whole curve exists, drawn from ten subtractions.
10°20°30°40°50°60°70°80°57.3°the GM (2.028 m) erected at 57.3° and joined to the originis the tangent at the origin (the line leaves the top of this plot at 36.7°)MV Ninja’s curve at 9.40 m, KG 8.30 m: what the criteria measureblue shading is the area to 30 degrees; gold is the band from 30 to 40; the peak is 1.116 m at 40 degreesGZ (m)0.40.81.2GZ 1.053 m at 30°maximum 1.116 m at 40°blue area 0.298 mr (needs 0.055); gold band 0.190 mr (needs 0.030); whole area to 40° 0.488 mr (needs 0.090)the whole examination question is: is there enough curve, in the right places
Figure 15.3   The curve for the worked condition, with the two criterion areas shaded and the GM guide at 57.3 degrees.

15.3 Worked example 15.1: the full compliance run

Worked example 15.1

MV Ninja floats at 9.40 m even keel in salt water (displacement 29751 t, KM 10.328 m). Her KG, corrected for free surfaces, is 8.30 m. Using the KN tables, determine her compliance with the intact stability criteria.

GM = 10.328 − 8.30 = 2.028 m (fluid).

KN at 29751 t, interpolated between the 29000 t and 30500 t rows (f = 0.5007): 0.901, 1.807, 2.171, 3.641, 5.203, 6.451, 7.454, 7.975, 8.153 and 8.031 m at 5, 10, 12, 20, 30, 40, 50, 60, 70 and 80°. GZ at each angle = KN − 8.30 × sin of the angle: 10°: 1.807 − 1.441 = 0.366 m; 20°: 3.641 − 2.839 = 0.802 m; 30°: 5.203 − 4.150 = 1.053 m; 40°: 6.451 − 5.335 = 1.116 m; 50°: 7.454 − 6.358 = 1.096 m; 60°: 7.975 − 7.188 = 0.787 m; 70°: 8.153 − 7.799 = 0.353 m; 80°: 8.031 − 8.174 = −0.143 m. The angle of flooding at 9.40 m is 54.0°, beyond 40°, so 40° governs the area criteria.

Area to 30° by the three eighths rule (ordinates 0, 0.366, 0.802, 1.053; multipliers 1, 3, 3, 1): sum of products = 4.557; area = (3 ÷ 8) × (10 ÷ 57.3) × 4.557 = 0.298 mr.

Area to 40° by the first rule (ordinates 0, 0.366, 0.802, 1.053, 1.116; multipliers 1, 4, 2, 4, 1): sum = 8.396; area = (1 ÷ 3) × (10 ÷ 57.3) × 8.396 = 0.488 mr. Area between 30° and 40° = 0.488 − 0.298 = 0.190 mr.

The compliance table: area to 30°, 0.298 against 0.055, complies; area to 40°, 0.488 against 0.090, complies; area 30° to 40°, 0.190 against 0.030, complies; GZ at 30° or more, 1.053 m at 30° against 0.20 m, complies; angle of maximum GZ, 40° (1.116 m) against 25°, complies; GM 2.028 m against 0.15, complies. The ship complies in full, with wide margins throughout: a stiff, healthy condition.

The two Simpson ladders that measure the areasfour ordinates to 30 degrees need the three eighths rule; five ordinates to 40 use the first rule; h converts to radiansArea 0 to 30° (three eighths rule)heelGZSMproduct0°01010°0.36631.09820°0.80232.40630°1.05311.053sum4.557area = 3/8 × (10 ÷ 57.3) × 4.557 = 0.298 mrArea 0 to 40° (first rule)heelGZSMproduct0°01010°0.36641.46420°0.80221.60430°1.05344.21240°1.11611.116sum8.396area = 1/3 × (10 ÷ 57.3) × 8.396 = 0.488 mrarea 30 to 40 = 0.488 − 0.298 = 0.190 mr; degrees become radians through the 57.3
Figure 15.4   The two Simpson ladders: the three eighths rule to 30 degrees, the first rule to 40, and the 57.3 turning degrees into radians.
The compliance table: MV Ninja at 9.40 m, KG 8.30 mrequirement against actual, line by line: this table is the answer the examiner wants to seecriterionminimumactualverdictarea 0 to 30°0.055 mr0.298 mrcompliesarea 0 to 40°0.090 mr0.488 mrcompliesarea 30° to 40°0.030 mr0.190 mrcompliesGZ at 30° or more0.20 m1.053 m at 30°compliesangle of maximum GZ25°40° (1.116 m)compliesGM (fluid)0.15 m2.028 mcompliesthe ship complies in full, with wide margins: a stiff, healthy loaded condition
Figure 15.5   The compliance table, line by line: the answer the examiner wants to see.
Animation 2 · Filling the areas: watching the metre radians accumulate
press play area so far: 0.000 mr
The area sweeps in from zero degrees with a live counter in metre radians. At 30 degrees it checks 0.298 against the required 0.055; at 40 degrees, 0.488 against 0.090; the strip between the two dashed lines is the 0.190 mr band that must reach 0.030. Watch how much of the total arrives after 20 degrees: the criteria guard the middle of the curve for a reason.

15.4 Worked example 15.2: dynamical stability

The areas are not an arbitrary bookkeeping choice. The area under the GZ curve, multiplied by the displacement, is the work the sea must do to heel the ship to that angle: her dynamical stability. A condition with generous areas is a ship that takes real energy to put over, which is why the rule book measures curve area and not just curve height.

Dynamical stability = W × area under the GZ curve to the chosen angleMCA formula sheet, September 2020
Worked example 15.2

Find MV Ninja’s dynamical stability to 40 degrees in the condition of Worked example 15.1.

Dynamical stability = 29751 × 0.488 = 14518 tonne metre radians.

That number is the energy account behind the pass marks: the sea has to deliver all of it to lay her over to 40 degrees.

10°20°30°40°area to 40° = 0.488 mrDynamical stability: the area is energythe shaded area is the work the sea must do to heel her to 40 degrees: multiply by the displacementdynamical stability = W × area under the curve= 29751 × 0.488 = 14518 tonne metre radians to 40°
Figure 15.6   Worked example 15.2: the shaded area is energy, and the displacement converts it to tonne metre radians.
Laboratory 1 · The GZ machine
One subtraction, live: the booklet’s KN at 29751 t (interpolated in displacement between the 29000 t and 30500 t rows of the table; at an angle between two tabulated angles it is read on the straight line between them), minus KG × sin of the heel. Defaults reproduce the 30 degree line of Worked example 15.1.

15.5 Worked example 15.3: when a condition fails

The criteria earn their keep on marginal conditions, and the most dangerous marginal conditions are the ones the GM alone would wave through. Load MV Ninja’s deck high and the whole curve sags, because KG × sin of the heel grows at every angle at once.

Worked example 15.3

The same ship and draught, but with deck cargo and free surfaces the KG is now 10.00 m. Test the condition.

GM = 10.328 − 10.00 = 0.328 m: more than twice the 0.15 minimum, so the GM line passes and the condition looks acceptable from the bridge wing.

The levers say otherwise: 10°: 1.807 − 1.736 = 0.071 m; 20°: 3.641 − 3.420 = 0.221 m; 30°: 5.203 − 5.000 = 0.203 m; 40°: 6.451 − 6.428 = 0.023 m; 50°: −0.207 m. The curve rises only to about a fifth of a metre and falls to zero at about 41°.

Areas: to 30°, sum 3 × 0.071 + 3 × 0.221 + 0.203 = 1.079, area 0.071 mr, complies against 0.055. To 40°, sum 4 × 0.071 + 2 × 0.221 + 4 × 0.203 + 0.023 = 1.561, area 0.091 mr, complies against 0.090 by a thousandth. Between 30° and 40°: 0.091 − 0.071 = 0.020 mr against the required 0.030: FAILS.

The peak: the largest lever, 0.221 m, is at 20° and the lever at 30° is already smaller, so the maximum occurs at 20°, below the required 25°: FAILS. The lever at 30°, 0.203 m, is at least 0.20 m, but only by 3 mm.

Verdict: the condition does not comply, on two criteria, with two more only just passing. Both failures are about the shape of the curve between 20 and 40 degrees, the range of heel a beam sea produces, and neither is visible in the GM. The cure is the next chapter’s subject: find the KG at which every line passes, and load to stay below it.

Animation 3 · The sagging curve: raise the KG and watch the lights change
KG 8.30 m
The KG sweeps from 8.30 m up to 10.10 m and back while the curve redraws live. The six criteria lights along the bottom flip from green to red one by one: the band between 30 and 40 degrees and the peak position go first, long before the GM light even flickers. That order is the whole lesson of this chapter.
10°20°30°40°60°80°0.20 mKG 8.30: compliesKG 10.00: peak 0.221 m at 20°The same ship with deck cargo: KG 10.00 m, GM 0.328 mthe GM still passes; the curve does not: two criteria fail and two more only just passarea 30° to 40°: 0.091 − 0.071 = 0.020 mr against 0.030 requiredFAILSthe largest lever, 0.221 m, is at 20° and the lever at 30° is already smaller: the peak is at 20°, below 25°FAILSGM 0.328 m, area to 30° 0.071 mr, area to 40° 0.091 mr (by 0.001), GZ at 30° 0.203 m (by 3 mm)pass
Figure 15.7   Worked example 15.3: the GM passes, the curve does not. The sagging shape fails the band and the peak position.
Laboratory 2 · The compliance dashboard
The full examination method runs live: GZ table to three decimals, both Simpson ladders, and the six line verdict. Set KG 8.30 for Worked example 15.1 and 10.00 for Worked example 15.3.
Laboratory 3 · Find the limit
The game: push the KG as high as it will go while every criterion still passes. The first line to break tells you which criterion governs this ship at this draught, and the KG where it breaks is her limiting KG, the subject of Chapter 16.

15.6 The angle of flooding

One refinement. If any opening that cannot be closed weathertight would go under at an angle smaller than 40 degrees, the 40 degree criteria stop at that angle instead: the area to the flooding angle must reach 0.09 mr, and the band runs from 30 degrees to it. With a flooding angle of 36 degrees, the convenient ordinates are 0, 9, 18, 27 and 36 degrees with the first rule. The principle is blunt: curve beyond the angle where the sea can get in does not count.

Chapter 15 in five lines

Six criteria, one object: three areas (0.055 mr to 30°, 0.090 to 40° or the flooding angle, 0.030 between), the height and position (GZ at least 0.20 m at 30° or more; the maximum GZ at not less than 25°), and the start (GM 0.15 m fluid).

GZ = KN − KG × sin of the heel, one subtraction per tabulated angle; interpolate KN in displacement between the tabulated rows.

Areas come from the Simpson rules with h in radians: the 10 degree spacing divides by 57.3.

Area is energy: W × area is the work the sea must do, the dynamical stability.

A passing GM can hide a failing curve: the 30° to 40° band and the peak position catch what the GM cannot see.

Test yourself