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CHAPTER 2

STABILITY

2-1 INTRODUCTION

This chapter discusses the stability of intact ships and how basic stability calculations areDefinitions of the state of equilibrium and the quality of stability as they apply to ships are in the following paragraphs.

2-1.1 Equilibrium. A ship floating at rest, with or without list and trim, is in static equilibriumthat is, the forces of gravity and buoyancy are balanced. They are equal and acting in odirections and are in a vertical line with each other.

2-1.2 Stability. Stability is the measure of a ship’s ability to return to its original position wheis disturbed by a force and the force is removed. A ship may have any one of three differenof stability, but only one at a time.

2-1.2.1 Positive Stability. If the ship tends to return to its original position after being disturby an external force, it is stable or has positive stability.

2-1.2.2 Negative Stability. If the ship tends to continue in the direction of the disturbing foafter the force is removed, it is unstable or has negative stability.

2-1.2.3 Neutral Stability. A third state, neutral stability, exists when a ship settles in the orietion it is placed in by the disturbing force. Neutral stability seldom occurs to floating ships, of concern in raising sunken ships because a ship rising through the surface passes througtral condition. While the ship is neutrally stable, even a very small disturbing force may causcapsize.

2-2 TRANSVERSE STABILITY

Transverse stability is the measure of a ship’s ability to return to an upright position after disturbed by a force that rotates it around a longitudinal axis. The following paragraphs defielements of transverse stability and provide a method to calculate the transverse stability teristics of a vessel.

2-2.1 Height of the Center of Gravity (KG). One of the primary concerns in transverse stabiis the height of the center of gravity above the keel. This distance is measured in feet. Iships, the center of gravity lies between a point six-tenths of the depth above the keel and thdeck. The position of the center of gravity depends upon the position of weights in the shchanges whenever weight is added, removed or shifted. To calculate the height of the cegravity, the following steps are necessary:

2-1

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f the

ht of

a. Classify all the weights in the ship.

b. Determine the height of each weight above the keel.

c. Multiply each weight by the height above the keel to determine the moment oweight.

d. Total the weights and the moments of weight.

e. Divide the total of the moments of weight by the total weight to determine the heigthe center of gravity.

EXAMPLE 2-1

CALCULATION OF THE HEIGHT OF THE CENTER OF GRAVITY (KG)

A ship has the following weights on board:

Material Weight (w) Height above the keel (kg)

Ship’s structure 2,000 15

Machinery 500 10

Stores 400 20

Fuel 250 5

Cargo 800 14

What is the height of the center of gravity?

A tabular format is convenient for this type of calculation.

Weight Height above keel Moment of Weight w kg w x kg

2,000 15 30,000

500 10 5,000

400 20 8,000

250 5 1,250

800 14 11,200

Sums 3,950 55,450

2-2

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Height of the center of gravity:

2-2.2 Height of the Center of Buoyancy (KB). The height of the center of buoyancy above tkeel or baseline is another important distance in stability. This distance is measured iBecause the center of buoyancy is the geometric center of the underwater body of the sheight of the center of buoyancy depends upon the shape of the ship. In flat-bottomed fulsuch as carriers and tankers, the center of buoyancy is lower than in finer lined ships, sdestroyers or frigates. Calculation of the location of the center of buoyancy for ship shaplengthy and tedious process. The height of the center of buoyancy is contained in the cuform.

When curves of form are not available, estimates sufficient for salvage work may be madelows. The height of the center of buoyancy is half the draft for a rectangular barge. In a form, the center of buoyancy lies between 0.53 and 0.58 of the draft. A reasonable first apmation of the height of the center of buoyancy that is sufficiently accurate for salvage w0.55 times the mean draft.

2-2.3 Transverse Metacentric Radius (BM). The transverse metacentric radius is the distabetween the center of buoyancy and the metacenter. Transverse metacentric radius is meafeet. It is defined as the moment of inertia around the longitudinal axis of the waterplane atthe ship is floating divided by the displacement volume.

If the shape of the waterplane is known, the moment of inertia of the waterplane can be dexactly. For salvage work, a reasonably accurate approximation may be made by:

I = CIT x L x B3

where:

CIT = The transverse inertia coefficient and is equal to CWP2/11.7

L = Length between perpendiculars

B = Beam

KG = sum of the moments of weight

total weight-----------------------------------------------------------------------------

KG = 55 450 ,3 950,

-------------------

KG = 14.04 feet

BM = I

V-------

2-3

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verti-

The transverse inertia coefficient may also be obtained from Figure 2-1 by entering along thehorizontal scale with the waterplane coefficient, reading up to the curve, then across to thecal scale.

2-4

Figure 2-1. Transverse Inertia Coefficient.

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EXAMPLE 2-2

CALCULATION OF THE TRANSVERSE METACENTRIC RADIUS

An FFG-7 Class ship is 408 feet long with a beam of 44 feet and draws 14.5 feet.Her block coefficient is 0.487 and her waterplane coefficient is 0.754. What is hertransverse metacentric radius (BM)?

a. Determine the transverse inertia coefficient.

b. Calculate the moment of inertia of the waterplane.

l = CIT x L x B3

l = 0.0486 x 408 x (44)3

l = 1,689,096 feet4

c. Calculate the displacement volume.

V = CB x L x B x T

V = 0.487 x 408 x 44 x 14.5

V = 126,768 feet3

d. Divide the moment of inertia by displacement volume to determine the trans-verse metacentric radius.

The value of the metacentric radius derived from the curves of form is 13.4 feet.The calculated value is sufficiently accurate for salvage work.

CIT = C

WP2

11.7------------

CIT = 0.7542

11.7---------------

CIT = 0.0486

BM = lV---

BM = 1 689 096, ,126 768,

---------------------------

BM = 13.32 feet

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cy shiftsthe dis-osition. centerf incli--

NOTE

The expression for transverse inertia coefficient is derived from theanalysis of numerous ships and is a reasonable approximation foruse in salvage. For a vessel or a barge with a rectangular water-plane (CWP = 1.0), an exact calculation is:

2-2.4 Height of the Metacenter (KM). The height of the metacenter is the distance betweenkeel and the metacenter. The height of the metacenter is measured in feet. It is the sumheight of the center of buoyancy and the metacentric radius, that is:

KM = KB + BM

For an upright ship, the metacenter lies on the same vertical line as the center of buoyancycenter of gravity. If the curves of form are available, KM can be determined directly from th

2-2.5 Metacentric Height (GM). The metacentric height, measured in feet, is the distabetween the center of gravity and the metacenter and is the principal indicator of initial staA ship whose metacenter lies above the center of gravity has a positive metacentric heighstable; conversely, a ship with the metacenter below the center of gravity has negative metaheight and is unstable. With the distances KB, BM and KG known, GM can be calculated:

GM = KB + BM - KG

and

GM = KM - KG

2-2.6 Righting Arm (GZ). In an upright ship in equilibrium, the forces of gravity and buoyanact equally in opposite directions along the vertical centerline. As the center of buoyancywhen the ship heels, these two opposing forces act along parallel lines. The forces and tance between them establish the couple which tends to return a stable ship to the upright pThe righting arm is the distance between the lines of action of the weight acting through theof gravity and the force of buoyancy acting through the center of buoyancy at any angle onation. Righting arms are measured in feet. Figure 2-2 shows the righting arm for an inclined stable ship.

I = L x B3( )

12----------------------

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rm toine of

Figure 2-2. Righting Arm.

The length of the righting arm varies with the angle of inclination. The ratio of the righting athe metacentric height, GZ/GM, is equal to the sine of the angle for any small angle. If the sthe angle is represented by Sin θ, the equation can be written as:

NOTE

The sine is one of three natural functions of angles that are used inbasic naval architectural and salvage calculations. The others arethe tangent and the cosine. A table of these natural functions isgiven in Appendix D.

Sin θ GZGM---------=

GZ = GM x Sin θ

or

2-7

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the dis-

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ility.

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This equation provides a convenient means of calculating righting arm for small angles of intion. At angles of heel greater than about ten to fifteen degrees, the metacenter moves awthe centerline and the relationship between the metacentric height and righting arm is noexact. The righting arm at large angles of heel can be determined from the statical stabilitydescribed in Paragraph 2-2.9.

2-2.7 Righting Moment (RM). The righting moment is the couple of the weight and buoyaof an inclined ship. This moment acts to return the ship to an upright position. Righting mommeasured in foot-tons. Because forces creating the couple (the weight and buoyancy of thare equal, opposite and equal to the displacement, the righting moment is the product of placement of the ship and the righting arm or:

RM = W x GZ

The size of the righting moment at any displacement and angle of inclination is a measureship’s ability to return to an upright position. As the righting moment at any displacemedirectly proportional to the righting arm, the righting arm may be used as an indicator of sta

2-2.8 Cross Curves of Stability. The cross curves of stability are a set of curves, each for aferent angle of inclination, that show righting arm changes with displacement. A set of curves of stability for the FFG-7 Class ship are shown in Figure 2-3. Note that a particular heighof the center of gravity has been assumed in computing the curves. The importance assumption is explained in Paragraph 2-2.9. To use the cross curves, enter on the horizontal swith the displacement of the ship, read up to the curve representing the angle of interest aacross to the vertical scale to determine the value of the righting arm.

For example, to obtain the righting arm at 3,200 tons displacement at an angle of 30 degreethe curves of Figure 2-3 along the horizontal scale with 3,200 tons, then:

a. Read up to the intersection with the 30-degree curve.

b. Read across to the vertical scale where it can be seen that the righting arm (GZ)feet.

The principal use of the cross curves of stability is in constructing the curve of statical stab

2-2.9 The Curve of Statical Stability. The curve of statical stability (or simply the stabilitcurve) shows righting arm changes as the ship inclines at a particular displacement. Rightinin feet, is plotted on the vertical scale while the angle of inclination, in degrees, is plotted horizontal scale. The curve of statical stability, when plotted and corrected as described will provide the following information:

2-8

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non-

r that

Figure 2-3. Cross Curves of Stability.

• Range of inclination through which the ship is stable (Range of Stability)

• Righting arm at any inclination

• Righting moment at any inclination

• Angle at which maximum righting arm and maximum righting moment occur

• Metacentric height.

2-2.9.1 Plotting the Curve of Statical Stability. Figure 2-4 is the curve of statical stability takefrom the cross curves shown in Figure 2-3 for a displacement of 3,200 tons. The curve was cstructed by:

a. Entering the cross curves along the 3,200-ton displacement line.

b. Reading up to the angle of inclination and across to determine the righting arm foangle of inclination and displacement.

c. Repeating the last step for each angle plotted in the cross curves.

d. Plotting the values obtained and drawing the curve.

2-9

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ort, the

ve thele; con-entric is theand the

ravity.

orrec-

Figure 2-4. Statical Stability Curve.

2-2.9.2 Height of Center of Gravity Correction. The cross curves of stability are calculated fa particular height of the center of gravity. When the center of gravity has a different heighmetacentric height and the stability curve change. If the actual center of gravity lies aboassumed center of gravity, the metacentric height is decreased and the ship is less stabversely, if the actual center of gravity is below the assumed center of gravity, the metacheight is increased and the ship is more stable. The correction at any angle of inclinationproduct of the difference between the actual and assumed heights of the center of gravity sine of the angle of inclination or:

correction = GG1 x sin θ

where:

GG1 is the difference between the actual and assumed heights of the center of g

Thus, if the center of gravity is two feet above the assumed center of gravity, the ction can be calculated as:

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e twomed belowce

av-cteris-The

gravity

ravity.alcu-

The corrections are plotted to the same scale as the curve of statical stability as shown inFigure2-5. The corrected curve of statical stability is drawn by plotting the difference between thcurves as shown in Figure 2-6. If the actual height of the center of gravity is less than the assuheight, the calculation is done in the same manner, however, the correction curve is plottedthe horizontal axis as shown in Figure 2-7. The new statical stability curve is again the differenbetween the two curves as shown in Figure 2-8.

2-2.9.3 Off-Center Weight Correction. When there is off-center weight and the center of grity is no longer on the centerline, a list results and there is deterioration in the stability charatics, including a reduction in righting arm toward the side to which the ship is listing. reduction in righting arm is equal to the product of the distance between the new center of and the centerline times the cosine of the angle of inclination or:

correction = GG1 x cos θ

where:

GG1 is the distance between the centerline and the new position of the center of gThus, if the center of gravity is 0.5 feet from the centerline, the correction can be clated as:

Angleθ

Sine of the anglesin θ

Height differenceGG1

CorrectionGG1 sin θ

0 0 2 010 0.174 2 0.3520 0.342 2 0.6830 0.500 2 1.0045 0.707 2 1.4160 0.866 2 1.7375 0.965 2 1.9390 1.000 2 2.00

Angleθ

Cosine of the anglecos θ

Horizontal distanceGG1

CorrectionGG1 x cos θ

0 1.000 0.5 0.5010 0.985 0.5 0.4920 0.940 0.5 0.4730 0.866 0.5 0.4345 0.707 0.5 0.3560 0.500 0.5 0.2575 0.259 0.5 0.1390 0 0.5 0

2-11

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Figure 2-5. Correction to Statical Stability Curve, CG is 2 Feet Above Assumed Point.

Figure 2-6. Corrected Statical Stability Curve.

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2-13

Figure 2-7. Correction to Statical Stability Curve, CG is 2 Feet Below Assumed Point.

Figure 2-8. Corrected Statical Stability Curve.

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As is done with the height corrections, the off-center weight corrections are plotted to thescale as the curve of statical stability. The corrected curve of statical stability is drawn by pthe difference between the two curves as shown in Figures 2-9 and 2-10.

The angle at which the corrected curve of statical stability crosses the horizontal axis is theof list caused by the off-center weight.

2-2.9.4 Range of Stability. The range of stability is the number of degrees through whichship is stable or the number of degrees through which the ship can heel without capsizinrange of stability may be measured directly from the statical stability curve and its limit iintersection of the curve and the horizontal axis. For instance:

• In Figure 2-4, the uncorrected stability curve, the range of stability is from 0 degreemore than 90 degrees.

• In Figure 2-6, the stability curve corrected for height of the center of gravity, the ranof stability is from 0 degrees to 77 degrees.

• In Figure 2-10, the stability curve corrected for off-center weight, the range of stabis 20 degrees to 75 degrees.

2-14

Figure 2-9. Correction to Statical Stability Curve for Transverse Shift of G.

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2 is

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Figure 2-10. Corrected Statical Stability Curve for Transverse Shift in G.

2-2.9.5 Righting Arm and Righting Moment. The righting arm at any inclination may be readirectly from the curve. Because each stability curve applies only to a specific displacemerighting moment can be obtained directly for any angle by multiplying the righting arm by theplacement. In Figure 2-6, the maximum righting arm is 1.1 feet, the maximum righting momis 3,520 foot-tons and the angle where the maximums occur is 51 degrees. Similarly, in Figure 2-10 the maximum righting arm is 0.83 feet, the maximum righting moment is 2,656 foot-tonthe angle where the maximums occur is 49 degrees.

2-2.9.6 Metacentric Height. The metacentric height may be obtained directly from the curvstatical stability by:

a. Erecting a perpendicular to the horizontal axis at 57.3 degrees (one radian)

b. Drawing the tangent to the statical stability curve at the origin.

The intersection of the two lines indicates the metacentric height. In Figure 2-11, the metacentricheight of the ship with stability curve 1 is 3.47 feet and that of the ship with stability curve1.47 feet.

2-3 LONGITUDINAL STABILITY

Longitudinal stability is the measure of a ship’s ability to return to its original position after bdisturbed by a force that rotates it around a transverse axis. Longitudinal stability is imporrefloating operations because changes in the longitudinal stability of a stranded or sunkewill not be apparent since the ship does not respond in the same manner as a ship aflochanges must be calculated to ensure salvors have an accurate assessment of the actual nal stability situation.

2-15

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-is toength mid-ng the aboutbtain

e the

r the

Figure 2-11. Metacentric Height from Statical Stability Curves.

2-3.1 Longitudinal Position of Center of Gravity (LCG). The longitudinal position of the center of gravity is as important to longitudinal stability as the height of the center of gravity transverse stability. Its position is determined solely by the distribution of weight along the lof the ship. The longitudinal position of the center of gravity is measured in feet from theships section or the forward perpendicular. It is determined in a manner similar to determiniheight of the center of gravity above the keel in that the sum of the moments of the weightseither the forward perpendicular or the midships section is divided by the total weight to othe desired position. The following steps are necessary:

a. Classify all the weights in the ship.

b. Determine the longitudinal distance of each weight from the reference.

c. Multiply each weight by the longitudinal distance from the reference to determinmoment of the weight.

d. Total the weights and the moments of weight.

e. Divide the total of the moments of weight by the total weight to determine how falongitudinal position of the center of gravity lies from the reference.

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EXAMPLE 2-3

CALCULATION OF THE LONGITUDINAL CENTER OF GRAVITY

A ship has the following weights on board:

Material Weight (w) Distance from the FP (Icg)(Long Tons) (Feet)

Ship’s structure 2,000 225Machinery 500 210Stores 400 201Fuel 250 180Cargo 800 220

What is the longitudinal position of the center of gravity?

A tabular format is convenient for this type of calculation.

Weight (w) Distance from FP (Icg) Moment of Weight (w x Icg) (Long Tons) (Feet) (Foot Tons)

2,000 225 450,000 500 210 105,000 400 201 80,400 250 180 45,000 800 220 176,000

Sums 3,950 856,400

2-3.2 Longitudinal Position of the Center of Buoyancy (LCB). The longitudinal position ofthe center of buoyancy is measured in feet from either the forward perpendicular or the msection. For a ship in equilibrium, the longitudinal position of the center of buoyancy is isame vertical line as the longitudinal position of the center of gravity. At any given time, thonly one point that is the center of buoyancy; the height and longitudinal position are two cnates of that single point. Determination of the longitudinal position of the center of buoyalengthy and tedious, requiring calculation of the underwater volume of the ship and its distion. The longitudinal position of the center of buoyancy may be obtained from the curv

Distance of LCG from FP = sum of the moments of weight

total weight--------------------------------------------------------------------------------

LCG = 856 400,

3 950,-----------------------

LCG = 216.8 feet (or 217 feet) abaft the FP

2-17

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of forcect theravitytion arenter are

s-etacen-rplane

efined

form. In salvage, the longitudinal position of the center of buoyancy is important primarimaking strength calculations when buoyancy must be distributed to place the longitudinations of the centers of gravity and buoyancy in the same vertical line.

2-3.3 Longitudinal Center of Flotation (LCF). The longitudinal center of flotation is the poinabout which the ship trims. It is the geometric center of the waterline plane. The longitudination of the center of flotation is measured in feet from either the midships section or the foperpendicular. In ships of normal form, it may lie either forward or aft of the midships sectiofine-lined ships, the longitudinal center of flotation is usually slightly abaft the midships secThe longitudinal position of the center of flotation is required to calculate final drafts whenchanges. It can be calculated if the exact shape of the waterplane is known or it may be ofrom the curves of form. If the position cannot be obtained, it can be assumed to be amidsh

2-3.4 Longitudinal Metacenter (ML). The longitudinal metacenter is an imaginary point importance in longitudinal stability. Like the transverse metacenter, it is located where theof buoyancy’s lines of action running through the longitudinal center of buoyancy intersevertical line through the center of buoyancy of the untrimmed ship. While the center of gand the center of buoyancy are points whose heights above the keel and longitudinal positwo coordinates of the same point, the transverse metacenter and the longitudinal metaceseparate points, each with its own set of coordinates.

2-3.5 Longitudinal Metacentric Radius (BML). The longitudinal metacentric radius is the ditance between the center of buoyancy and the longitudinal metacenter. The longitudinal mtric radius is measured in feet. It is defined as the moment of inertia of the waterline wateabout a transverse axis divided by the volume of displacement.

If the shape of the waterplane is known, the moment of inertia of the waterplane can be dexactly. For salvage work, a reasonably accurate approximation may be made by:

IL = CIL x B x L3

where:

CIL = The longitudinal inertia coefficient and is equal to (0.143 x CWP - 0.0659)

B = Beam

L = Length between perpendiculars

BML = IL

V----

2-18

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NOTE

The longitudinal inertia coefficient may be obtained directly fromFigure 2-12 by entering along the horizontal scale with the water-plane coefficient, reading up to the curve and then reading across tothe vertical scale.

2-19

Figure 2-12. Longitudinal Inertia Coefficient.

S0300-A6-MAN-010

EXAMPLE 2-4

CALCULATION OF THE LONGITUDINAL METACENTRIC RADIUS

An FFG-7 Class ship is 408 feet long with a beam of 44 feet and draws 14.5 feet.Her block coefficient is 0.487 and her waterplane coefficient is 0.754. What is herlongitudinal metacentric radius (BML)?

a. Determine the longitudinal inertia coefficient:

CIL = (0.143 x CWP - 0.0659)

CIL = (0.143 x 0.754 - 0.0659)

CIL = 0.0419

b. Calculate the moment of inertia of the waterplane:

IL = CIL x B x L3

IL = 0.0419 x 44 x (408)3

IL = 125,212,356 feet4

c. Calculate the displacement volume:

V = CB x L x B x T

V = 0.487 x 408 x 44 x 14.5

V = 126,768 feet3

d. Divide moment of inertia by displacement volume to get longitudinal metacen-tric radius:

BML = l L

V----

BML = 125 212 356, ,

126 768,---------------------------------

BML = 987.72 feet (or 988 feet)

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s- in feet.meta-

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2-3.6 Height of the Longitudinal Metacenter (KML). The height of the longitudinal metacenter is the distance between the metacenter and the keel and is measured in feet. The the longitudinal metacenter is the sum of the height of the center of buoyancy and the longimetacentric radius or:

KML = KB + BML

where:

KML = The longitudinal height of the metacenter

KB = The height of the center of buoyancy

BML = Longitudinal metacentric radius

2-3.7 Longitudinal Metacentric Height (GML). The longitudinal metacentric height is the ditance between the height of the center of gravity and the longitudinal metacenter measuredThe longitudinal metacentric height is the difference between the longitudinal height of center and the height of the center of gravity or:

GML = KML - KG

where:

GML = Longitudinal metacentric height

KML = Height of the longitudinal metacenter

KG = Height of the center of gravity

2-3.8 Trim. The quantities necessary for determining trim and how they are used are explaithe following paragraphs.

2-3.8.1 Trimming Moment. A trimming moment is a moment exerted by a weight anywherthe ship acting about the center of flotation. A trimming moment causes the ship to trim around the center of flotation. There is no special symbol for trimming moment, though M isvenient to use. Trimming moment is measured in foot-tons. For instance, a weight of 2placed 100 feet forward of the center of flotation would have a trimming moment of 2,500tons by the bow. Paragraph 3-2.2.2 provides an example for longitudinal effects of weight adtions and removals.

2-3.8.2 Moment to Change Trim One Inch (MT1). The trimming moment required to causechange in trim of one inch (1") is known as the Moment to Change Trim One Inch. The MT

2-21

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calcu-

the:

found

any ship depends upon the hull form and may be either obtained from the curves of form orlated by:

where:

GML = Longitudinal metacentric heightW = DisplacementL = Length between perpendiculars

As the longitudinal metacentric radius, BML, is easily obtained and is not very different fromlongitudinal metacentric height, GML, it is often used to determine an approximate MT1 by

Additional methods of calculating the approximate moment to change trim one inch can bein Appendix C.

EXAMPLE 2-5

CALCULATION OF MOMENT TO TRIM ONE INCH

What is the approximate moment to change trim one inch for the FFG-7 Classship described in Example 2-4 ?

To calculate the approximate moment to trim one inch:

From the calculation in Example 2-4 :

BML = 988 feetL = 408 feetV = 126,768 cubic feet

MT1 = GML W×( )

12 L×( )----------------------------

MT1 = BML W×( )

12 L×( )---------------------------

MT1 = BML W×( )

12 L×( )---------------------------

2-22

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then has

the

otherw andnd andge at

d by

Displacement can be calculated from displacement volume:

MT1 from the curves of form is 760 foot-tons. The value based on calculation iswithin 10 percent of the actual MT1 and is a reasonable approximation for salvagework.

NOTE

The Greek letter delta (À) is used in conjunction with symbols tomean “the change in” the quantity that the symbol represents.

2-3.8.3 Trim Calculations. In salvage, trim calculations are usually made to determinechanges in draft fore and aft resulting from a change in trimming moment. The calculatiothree parts:

a. Determination of the trimming moment

b. Determination of the total change of trim by dividing the trimming moment bymoment to change trim one inch or:

c. Determination of the new drafts. When a ship trims, one end gains draft while theend loses draft. For instance, a ship that trims by the bow gains draft at the boloses draft at the stern. The total trim is the sum of the amount gained at one elost at the other. As a ship trims about the center of flotation, the amount of chanthe bow is proportional to the ratio of the product of the change of trim multipliethe distance between the forward perpendicular and the center of flotation and thelength of the ship or:

W = 126 768,

35---------------------

W = 3,622 tons

then

MT1 = 988 3 622,×( )

12 408×( )-----------------------------------

MT1 = 730.9 foot-tons

δ trim = trimming moment

MT1------------------------------------------------------

δ T f = δ trim x (FP to LCF)

L--------------------------------------------------

2-23

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appro-

iplied of the

te:

The new draft forward will be the old draft forward with the change added or subtracted as priate:

Likewise, the change in trim aft is equal to the ratio of the product of the change of trim multby the distance between the after perpendicular and the center of flotation—and the lengthship or:

The new draft aft will be the old draft aft with the change added or subtracted as appropria

EXAMPLE 2-6

CALCULATION OF TRIM

The FFG-7 Class ship described in Example 2-5 is floating at a draft both forwardand aft of 14.5 feet. The center of flotation is twenty-five feet abaft the midshipsection. A trimming moment of 25,000 foot-tons by the bow is introduced. Whatare the new drafts?

The first step, the determination of trimming moments, is not necessary becausethat information is already available; i.e., M = 25,000 foot-tons.

To obtain the total change in trim, divide the trimming moment by the moment tochange trim one inch:

Moment to trim one inch is 760 foot-tons from the curves of form; therefore:

New T f = Original T f + δ Tf

δ T a = δ trim x (AP to LCF)

L---------------------------------------------------

New T a = Original T a + δ Ta

δ trim = M

MT1-------------

δ trim = 25 000,

760------------------

δ trim = 32.9""

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To obtain the new draft forward, determine the distance from the FP to the LCF:

then

To obtain the new draft aft, determine the distance from the AP to the LCF:

then

A method to check the accuracy of the draft calculations is to add the change in

FP to LCF = L 2----- + 25

FP to LCF = 408 2

---------- + 25

FP to LCF = 229

δ T f = δ trim x (FP to LCF) L

--------------------------------------------------

δ T f = 32.9 229×

408-------------------------

δ T f = 18.5″ , or 1′ 6.5″

New T f = Old T f + δ T f

New T f = 14′ 6″ + 1′ 6.5″

New T f = 16′ 0.5″

AP to LCF = L 2----- - 25

AP to LCF = 408 2

---------- - 25

AP to LCF = 179

δ T a = δ trim x (AP to LCF) L

---------------------------------------------------

δ T a = 32.9 179×

408-------------------------

δ T a = 14.5″ , or 1′ 2.4″

New T a = Old T a + δ T a

New T a = 14′ 6″ + 1′ 2.4″

New T a = 13′ 3.6″

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S0300-A6-MAN-010

draft forward and the change in draft aft and compare the result with the totalchange in trim. These quantities should be equal. In the previous example thetotal change in trim was 32.9 inches.

Since the sum of the draft changes is equal to the total change in trim, the calcula-tion was performed correctly.

δ T f = 18.5″

δ T a = 14.5″32.9″-------------

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