EN400 Ch 4 Prob 13

13. An LCS1 has a displacement of 3000 LT and KG = 14.0 ft. Use Cross Curves of Stability found in the Ship’s Data section.

a. Plot the Curve of Intact Statical Stability for starboard heels for the ship at TCG =

0ft. On the same axes, plot a second curve for the LCS1 in the same condition but with TCG = 1 ft.

b. Compare the two curves. In which condition is the ship more stable? c. What is the permanent angle of list when TCG = 1.0 ft?

Solution:

Step 1: Obtain GoZo for Δ = 3000LT, KG = 0ft from Curve of Intact Static Stability Step 2: Correct GoZo for rise in KG (sine correction):

G1Z1 = GoZo – KGsinφ Step 3: Correct G1Z1 for shift in TCG (cosine correction):

G2Z2 = G1Z1 – TCGcosφ

φ (deg) GoZo (ft) KGsinφ G1Z1 (ft) TCGcosφ G2Z2 (ft) Curve of Int.

Stabil. 14ft *

sin(deg) GoZo –

KG*sin(deg) 1ft *

cos(deg) G1Z1 –

TCG*cos(deg) 0 0 0 0 1.00 -1.00 5 1.4 1.22 0.18 1.00 -0.82 10 3 2.43 0.57 0.98 -0.42 15 4.5 3.62 0.88 0.97 -0.09 20 6.1 4.79 1.31 0.94 0.37 30 9 7.00 2.00 0.87 1.13 40 11.8 9.00 2.80 0.77 2.03 50 14.2 10.72 3.48 0.64 2.83 60 16 12.12 3.88 0.50 3.38 70 16.8 13.16 3.64 0.34 3.30 80 17.2 13.79 3.41 0.17 3.24 90 17.4 14.00 3.40 0 3.40

EN400 Ch 4 Prob 13

a. b. At initial condition (TCG = 0ft), the ship has maximum righting arm of 3.9 feet and a

range of stability 0-90°. After shifting TCG 1ft to starboard, maximum righting arm decreases to 3.4 feet and a range of stability of only 17-90°.

c. With TCG = 0ft to starboard, the ship has a permanent list of 17° (where the curve

crosses the horizontal axis.

EN400 Ch 4 Prob 21

21. A LCS1 class ship suffers a major flood in the forward portion of the ship resulting in the total flooding of 7800 ft3 (i.e no free surface). The flood’s center of gravity is located 142 ft forward of amidships, 8 ft above the keel, and 1 ft starboard of centerline. Prior to the flood, the ship was on an even keel at a draft of 14 ft, KG = 12 ft. Lpp = 324 ft.

Using the curves of form and cross curves of stability determine the following:

a. Weight of flooding water. b. Ship’s KG and TCG of the ship after flooding has occurred. c. Angle at which the ship is listing. d. Ship’s forward, aft, and mean drafts after the flood occurs. e. Compute and plot the ship’s righting arm curve before and after the flood. Use of

a computer is encouraged. f. What affect does the flooding have on the ship’s stability and seaworthiness?

Solution: a. WFLOOD = rgV = 1.99 lb-s2/ft4 * 32.17 ft/s2 * 7800 ft3 * (1 LT / 2240 lb) = 223 LT b. Curves of Form at T = 14ft Δ = 332scale * 10LT/scale = 3320LT

KGNEW = [KGOLD ΔOLD + ∑ (wi kgi)] / ΔNEW = [12ft * 3320LT + 223LT * 8ft] / (3320LT + 223LT) = 11.75 ft

TCGNEW = [TCGOLD ΔOLD + ∑ (wi tcgi)] / ΔNEW = [0ft * 3320LT + 1ft * 223LT] / (3320LT + 223LT) = + 0.063 (to stbd) c. Curves of Form at T = 14ft KMT = 372scale * 0.033 ft/scale = 12.28 ft

GM = KM – KG = 12.28ft – 11.75ft = 0.53 ft

Angle, φ = tan-1 (TCG / GM) = tan-1 (0.063ft / 0.53ft) = + 6.8° (i.e. to starboard)

EN400 Ch 4 Prob 21

d. From Curves of Form for initial draft of 14ft: LCF = - 36 feet (aft of midships) TPI = 275 general scale * (0.12 LT/in / 1 general scale) = 33.0 LT/in MT1 = 195 general scale * (4 ft-LT/in / 1 general scale) = 780 ft-LT/in Change in draft due to Parallel Sinkage: δTPS = w / TPI = 223LT / 33.0 LT/in = 6.76 in * (1ft / 12in) = 0.56 ft (adding weight increases draft) Changes in draft due to Trim: (adding weight fwd of LCF makes bow draft deeper)

δTrim = w*l / (MT1) = w * (LCF-to-midships) / (MT1) = 223LT * 178ft / 780 ft-LT/in = 50.9 in * (1ft / 12in) = 4.24 ft δTrim / Lpp = δT fwd / d fwd = δT aft / d aft d fwd = (FP-to-midships) + (midships-to-LCF) = 162ft + 36ft = 198ft d aft = (AP-to-midships) – (midships-to-LCF) = 162ft – 36ft = 126ft

δT fwd / d fwd = δTrim / Lpp δT fwd / 198ft = 4.24ft / 324ft δT fwd = 2.60 ft (increasing at FP b/c weight is fwd of LCF)

δT aft / d aft = δTrim / Lpp δT aft / 126ft = 4.24ft / 324ft δT aft = - 1.65 ft (decreasing at AP b/c weight is fwd of LCF)

Final changes in draft overall: T fwd = To fwd + δT fwd + δTPS = 14.0ft + 2.60ft + 0.56ft = 17.2 ft T aft = To aft + δT aft + δTPS = 14.0ft + (-1.65ft) + 0.56ft = 12.9 ft

EN400 Ch 4 Prob 21

T mean = T aft + T fwd = (12.9ft + 17.2ft) / 2 = 15.1 ft e. Step 1: Obtain GoZo for Δ = 3000LT, KG = 0ft from Curve of Intact Static Stability

Step 2: Correct GoZo for actual ship in KG (sine correction):

GvZv = GoZo – KGsinφ Plot original ship GZ vs. Angle of Heel

Step 3: Correct GoZo for actual ship KG after flood (sine correction) Step 4; Correct GvZv for shift in TCG due to flood (cosine correction):

G2Z2 = G1Z1 – TCGcosφ

Plot ship after flooding GZ vs. Angle of Heel

CurvesKG = KG old = KG new = TCG new

0 12 11.75 0.063φ (deg) GoZo (ft) KGsinφ GvZv (ft) KGsinφ GvZv (ft) TCGcosφ GtZt (ft)

0 0 0.00 0.00 0.00 0.00 0.06 -0.065 1.4 1.05 0.35 1.02 0.38 0.06 0.3110 3 2.08 0.92 2.04 0.96 0.06 0.9015 4.4 3.11 1.29 3.04 1.36 0.06 1.3020 5.8 4.10 1.70 4.02 1.78 0.06 1.7230 8.8 6.00 2.80 5.88 2.93 0.05 2.8740 11.5 7.71 3.79 7.55 3.95 0.05 3.9050 13.8 9.19 4.61 9.00 4.80 0.04 4.7660 15.4 10.39 5.01 10.18 5.22 0.03 5.1970 16.3 11.28 5.02 11.04 5.26 0.02 5.2480 16.6 11.82 4.78 11.57 5.03 0.01 5.0290 16.8 12.00 4.80 11.75 5.05 0.00 5.05

Pre-Flood Post-Flood

EN400 Ch 4 Prob 21

f. Max righting arm has increased from 5.02ft to 5.24ft. KG lowered from12ft to 11.75ft. TCG of 1ft causes a small list to starboard. Draft increased from 14ft to 15.1 ft. Overall stability increased by adding weight low on the ship. Decreased freeboard and a trim of -4.24 feet (down by the bow) is concerning for seaworthiness.