2-1 A small boat weighing 40 LT has a submerged volume of 875 ft3 when traveling at 20 kts in

seawater. (ρ = 1.99 lb·s2ft4 ; 1 LT = 2240 lb)

(a) Calculate the magnitude of the hydrostatic support being experienced by boat.

FBuoyancy = ρgVsub = 1.99 lb·s2ft4 · 32.2 ft

s2 · 875 ft3 = 56,068 lb · 1 LT2,240 lb = 25 LT

(b) What other type of support is the boat experiencing?

In addition to hydrostatic support, this boat is experiencing hydrodynamic support/lift.

(c) Calculate the magnitude of this other type of support.

FHydrodynamic = Wboat − FBuoyancy = 40 LT − 25 LT = 15 LT

(d) What will happen to the submerged volume of the boat if it slows to 5 knots? Explain your

answer.

Hydrodynamic lift is proportional to V 2boat. As the boat slows, the reduction in hydrodynamic

support will cause the boat weight to exceed the total lift so the boat will sink down (draft, T ,

will increase). As T increases, Vsub increases so Fbuoyancy increases until a new equilibrium is

reached at a new, deeper, draft.

12. hi

2-4 Sketch a profile of a ship and show the following:

(a) Forward Perpendicular

(b) After Perpendicular

(c) Sections, assuming the ship has stations numbered 0 through 10.

(d) Length Between Perpendiculars

(e) Length Overall

(f) Design Waterline

(g) Amidships

13. hi

2-5 Sketch a section of a ship and show the following:

(a) Keel

(b) Depth

(c) Draft

(d) Beam

(e) Freeboard

14. hi

2-6

2-7 For this question, use a full sheet of graph paper for each drawing. Choose a scale that gives the

best representation of the ship’s lines. Use the FFG-7 Table of Offsets given on the following page for

your drawings.

(a) For stations 0-10 draw a Body Plan for the ship up to the main deck. Omit stations 2.5 and 7.5.

(b) Draw a half-breadth plan showing the 4 ft, 12 ft, 24 ft waterlines, and the deck edge.

(c) Draw the sheer profile of the ship.

2-8 A box-shaped barge has the following dimensions: Length = 100 ft, Beam = 40 ft, Depth = 25 ft. The

barge is floating at a draft of 10 ft.

1. Draw a waterplane, profile, and end view of the barge. On each view indicate the following: centerline,

waterline, midships, center of buoyancy (B), and center of flotation (F).

2. On your drawing show the following distances: KB, LCF referenced from the forward perpendicular,

and LCB referenced from amidships.

3. Based on the given dimensions of the barge, determine the following dimensions:

(a) KB KB = T2 = 10 ft

2 = 5 ft

(b) LCF referenced to amidships As ship is symmetric fore to aft, LCF = 0 ft from

(c) LCB referenced to the forward perpendicular LCB = L2 = 100 ft

2 = 50 ft from FP

(d) Height of F above the keel F lies on the waterplane which is T ft above baseline, or 10 ft

2-9 Using Simpson’s Rule calculate the areas of the following objects:

1. Right triangle with base length of a and a height of length b.

Aright triangle =

∫ a

0

y(x) dx =

∫ a

0

b

ax dx =

b

a

x2

2

∣∣∣∣∣a

0

=b

a

a2

2=ab

2

Using three points for Simpson’s First Method x = {(0, a2 , a)

Aright triangle ≈∆x

3

1 · y0 + 4 · y1 + 2 · y2 + 4 · y3 + 2 · y4 + ...

+2 · yn−4 + 4 · yn−3 + 2 · yn−2 + 4 · yn−1 + 1 · yn

Aright triangle ≈

a2

3

[1 · 0 + 4 · b

2+ 1 · b

]=ab

2(Identical to Exact Solution)

2. Semi-circle of radius r.

Asemi−circle =

∫ 2r

0

y(x) dx =

∫ 2r

0

+√r2 − x2 dx =

πx2

8

∣∣∣∣∣2r

0

=πr2

2

Using three points for Simpson’s First Method x = {(0, r, 2r)

Asemi−circle ≈∆x

3

1 · y0 + 4 · y1 + 2 · y2 + 4 · y3 + 2 · y4 + ...

+2 · yn−4 + 4 · yn−3 + 2 · yn−2 + 4 · yn−1 + 1 · yn

Asemi−circle ≈

r

3[1 · 0 + 4 · r + 1 · 0] =

4r2

3(A 15.1% underestimate of the integral)

3. Equilateral triangle with each side having length a.

Aequilateral triangle =

∫ a

0

y(x) dx =

∫ a2

0

√3 x dx+

∫ a

a2

√3 (a− x) dx

=

√3x2

2

∣∣∣∣∣a2

0

+√

3ax−√

3x2

2

∣∣∣∣∣a

a2

=

√3a2

4

Using three points for Simpson’s First Method x = {(0, a2 , a)

Aequilateral triangle ≈∆x

3

1 · y0 + 4 · y1 + 2 · y2 + 4 · y3 + 2 · y4 + ...

+2 · yn−4 + 4 · yn−3 + 2 · yn−2 + 4 · yn−1 + 1 · yn

Aequilateral triangle ≈

a2

3

[1 · 0 + 4 ·

√3a

2+ 1 · 0

]=

√3a2

3(A 33.3% overestimate of the integral)

Prob 10

2-11 Using the FFG-7 table of offsets, calculate the sectional area of station 3 up to the DWL.

Asection = 2 ·A = 2 ∆z3

∑4i=0 S.M.i ·yi = 2 4 ft

3 [1 ·0.68ft+4 ·10.77ft+2 ·14.43ft+4 ·16.31ft+1 ·17.75ft] =

414.96 ft2

2-12 Using the FFG-7 table of offsets, calculate the area of station 6 up to the 24 foot waterline.

As we have unequal spacing between waterlines (up to 24 ft WL), divide the integral into two sub-domains:

Integrate 0− 8 ft with ∆ z = 4 ft and 8− 24 ft with Delta z = 8 ft

Asection = 2[∫ 8 ft

0 fty dz +

∫ 24 ft

8 fty dz

]= 2

[4 ft

3 (1 · 0.68 ft+ 4 · 12.86 ft+ 1 · 19.21 ft) + 8 ft3 (1 · 19.21 ft+ 4 · 21.79 ft+ 1 · 23.33 ft)

]Asection = 902.2 ft2

Prob 13

2-14 Using the FFG-7 table of offsets, and stations 0, 2.5, 5, 7.5, and 10, calculate the location of the longi-

tudinal center of flotation (LCF ) of the DWL referenced to amidships.

∆x =Lpp

n−1 = 416 ft5−1 = 102 ft

Station S.M. yif(Area)

= S.M. · yi

xi

(from FP )

f(Moment)

= S.M. · yi · xi

0 1 0.33 ft 0.33 ft 0 ft 0 ft2

2.5 4 15.52 ft 62.08 ft 102 ft 6,332.16 ft2

5 2 22.61 ft 45.22 ft 204 ft 9,224.88 ft2

7.5 4 20.82 ft 83.28 ft 306 ft 25,483.68 ft2

10 1 12.46 ft 12.46 ft 408 ft 5,083.68 ft2

203.37 ft 46,124.4 ft2

LCF = f(Moment)f(Area) = 46,124.4 ft2

203.37 ft = 226.8 ft from FP

LCF = 226.8 ft− Lpp

2 = 226.8 ft− 204 ft = 22.8 ft aft of midships.

Prob 15, 1615)

16)

2-16 An FFG-7 is floating on an even keel at a draft of 14 feet. Using its Curves of Form, find the following

parameters:

1. Displacement (∆) ∆(Curve 1

)= 107 · 30 LT

1 = 3,210 LT

2. Longitudinal center of flotation (LCF ) LCF(Curve 5

)= 22 ft aft of

3. Vertical center of buoyancy (KB) KB(Curve 3

)= 44 · 0.2 ft

1 = 8.8 ft

4. Tons per inch immersion (TPI) TPI(Curve 4

)= 157 · 0.2 LT

in

1 = 31.4 LTin

5. Moment to trim 1 inch (MT1′′) MT1′′(Curve 6

)= 124.5 · 5.78 ft·LT

in

1 = 719.61 LTin

6. Submerged volume (∇sub) ∇sub = ∆ρg = 3,210 LT

64 lbft3· 2240 lb

LT = 112,350 ft3

2-17 An FFG-7 is floating with a forward draft of 14.9 feet and an aft draft of 15.5 feet. Determine the

following: First, find mean draft, Tm. Tm =Tf+Ta

2 = 14.9 ft+15.5 ft2 = 15.2 ft Use this draft to pull values

from the curves of form.

1. Displacement (∆) ∆(Curve 1

)= 122.5 · 30 LT

1 = 3,675 LT

2. Longitudinal center of flotation (LCF ) LCF(Curve 5

)= 23 ft aft of

3. Moment to trim 1 inch (MT1′′) MT1′′(Curve 6

)= 132 · 5.78 ft·LT

in

1 = 762.96 LTin

2-18 The FFG in problem 15 changes its draft from 14 feet to 15.5 feet. What is the new value of TPI? Why

does this value of TPI change?

TPI(14 ft WL) = 31.4 LTin

TPI(15.5 ft WL) = 164 · 0.2 LTin

1 = 32.8 LTin

As the hull is riding lower in the water, the Awp increases due to the flare in the hull form. As TPI is

directly related to Awp, TPI increases as well.

Prob 19, 2019)

20)

2-20 A DDG-51 is floating on an even keel at a draft of 21.5 ft. A piece of machinery weighing 50 LT is

moved from the center of flotation to a point 150 ft forward of F . What is the change in ship’s trim due to

this weight shift?

Start by finding the value for MT1′′ from the DDG-51 curves of form:

MT1′′(Curve 6

)= 162 · 9.24 ft·LT

in

1 = 1,496.88 ft·LTin

For a longitudinal weight shift: δT = w·lMT1′′ = 50 LT ·150 ft

1,496.88 ft·LTin

= 5.010 in by the bow.