Ch3...Inclining gear weighs 28 LT and is loaded 43 ft from the keel on the centerline . Inclining moment List Angle 880 ft-LT (stbd) 2.3 deg stbd .

EN400 Ch 3 Prob 9

9. USS FREEDOM (LCS-1) enters a shipyard for an overhaul. As it entered the shipyard, the ship’s displacement was 2860 LT with KG = 19.7 ft, on the centerline. During overhaul the following work was performed.

Removed Items Added Items

Item Weight Kg Item Weight Kg

ASW Fire Control Package 15.0 LT 25.0 ft VBSS Tactical

Control Package 20.0 LT 34.0 ft

Portable Towed Array 6.0 LT 15.0 ft Two 50-cal Small

Arms Stations 8.0 LT 45.0 ft

Engine Upgrades 3.0 LT 8.0 ft Antenna Upgrades 5.0 LT 64.0 ft

a. Determine the ship’s displacement and KG after the overhaul.

b. Determine the ship’s draft before and after overhaul

Solution:

Item w (LT) kg (ft) w*kg (LT-ft) ASW Fire Control Package - 15.0 25.0 - 375 Portable Towed Array - 6.0 15.0 - 90 Engine Upgrades - 3.0 8.0 - 24 VBSS Tactical Control Package

+ 20.0 34.0 + 680

Two 50-cal Small Arms stations

+ 8.0 45.0 + 360

Antenna Upgrades + 5.0 8.0 + 40 ∑ wi = 9 ∑ (wi kgi) = 591

a. ΔNEW = ΔNEW + ∑ wi = 2860LT + 9LT = 2869 LT

KGNEW = [KGOLD ΔOLD + ∑ (wi kgi)] / ΔNEW = [19.7ft * 2860LT + 591 LT-ft] / 2869 LT = 19.84 ft b. From Curves of Form: ΔOLD = 2860LT * (1 general scale / 10 LT) = 286 general scale 12.7 ft ΔNEW = 2869LT * (1 general scale / 10 LT) = 287 general scale 12.8 ft

EN400 Ch 3 Prob 12

12. USS FREEDOM (LCS-1) is floating upright with a displacement of 3100 LT and KG = 18.0 ft. 25 LT of equipment are added to the ship at an average location 30 ft above the keel and 8 ft starboard of the ship's centerline.

a. What is the new KG?

b. What is the new TCG?

c. This new location of G is unsatisfactory. At what transverse and vertical location would you add 20 LT of lead ballast to return to the ship’s original KG and TCG?

Solution: a. KGNEW = [KGOLD ΔOLD + ∑ (wi kgi)] / ΔNEW = [18.0ft * 3100LT + 30ft * 25LT] / (3100LT + 25LT) = 18.10 ft b. “Floating upright” implies the initial TCG = 0ft (i.e. on centerline): TCGNEW = [TCGOLD ΔOLD + ∑ (wi tcgi)] / ΔNEW = [0ft * 3100LT + 8ft * 25LT] / (3100LT + 25LT) = + 0.064 ft (to stbd) c. To restore initial KG, set the revised KG*Δ as ‘old’ and the desired w*KG as ‘new’:

KGNEW = [KGOLD ΔOLD + ∑ (wi kgi)] / ΔNEW 18.0 ft = [18.10ft * 3125LT + kgi * 20LT] / (3125LT + 20LT) kgi = 2.375 ft

TCGNEW = [TCGOLD ΔOLD + ∑ (wi tcgi)] / ΔNEW 0 ft = [0.064ft * 3125LT + tcgi * 20LT] / (3125LT + 20LT) Tcgi = - 9.88 ft (to port)

EN400 Ch 3 Prob 20

20. The following data was taken on an inclining experiment: Ship: LCS-1 Level trim, draft = 14 ft in the light ship condition Inclining gear weighs 28 LT and is loaded 43 ft from the keel on the centerline Inclining moment List Angle 880 ft-LT (stbd) 2.3 deg stbd 528 ft-LT (stbd) 1.2 deg stbd 0 0.2 deg port 528 ft-LT (port) 1.5 deg port 880 ft-LT (port) 2.3 deg port

Determine the location of the ship’s vertical center of gravity in the light ship condition.

Solution:

List Angle (deg) tan φ 2.3 stbd 0.040 1.2 stbd 0.021 0.2 port - 0.003 1.5 port - 0.026 2.3 port - 0.040

From Curves of Form for lightship condition: ΔLIGHT = 335 general scale * (10LT / 1 general scale) = 3350 LT KMT = 372 general scale * (0.033 ft / 1 general scale) = 12.3 ft

ΔINCL = ΔLIGHT + wincl = 3350LT + 28LT = 3378 LT GMINCL = w*t / (ΔINCL * tan φ)

Plot Inclining moment (w*t) vs. tan φ to find slope (w*t / tan φ): 22075 ft-LT/rad

EN400 Ch 3 Prob 20

GMINCL = w*t / (ΔINCL * tan φ) = 22075 ft-LT / 3378LT = 6.53 ft KGINCL = KMT – GMINCL = 12.3ft – 6.53ft = 5.77ft

Use the standard weight add/remove equation to find new (lightship) KG:

KGLIGHT = (KGINCL ΔINCL – wincl kgincl) / ΔLIGHT = (5.77ft * 3378LT – 28LT * 43ft) / 3350 LT = 5.46 LT

EN400 Ch 3 Prob 24

24. USS FREEDOM (LCS-1) is floating at a draft of 13.5 feet with zero trim. Ship length is 324 feet. 50 LT are added at a location 122 feet aft of amidships. Draw a diagram showing the location of the weight added, the parallel sinkage, the final longitudinal center of flotation and the initial and final waterlines to find:

a. Final forward and after drafts

b. Final mean draft Solution: From Curves of Form: (values are essentially the same before and after minor weight addition) Δ = 313 general scale * (10LT / 1 general scale) = 3130 LT LCF = - 27 feet (aft of midships) TPI = 253 general scale * (0.12 LT/in / 1 general scale) = 30.36 LT/in MT1 = 168 general scale * (4 ft-LT/in / 1 general scale) = 672 ft-LT/in Change in draft due to Parallel Sinkage: δTPS = w / TPI = 50LT / 30.36 LT/in = 1.65 in * (1ft / 12in) = 0.14 ft (adding weight increases draft) Changes in drafts due to Trim: (adding weight aft of LCF makes stern draft deeper)

δTrim = w*l / (MT1) = w * (LCF-to-midships) / (MT1) = 50LT * 95ft / 672 ft-LT/in = 7.07 in * (1ft / 12in) = 0.59 ft δTrim / Lpp = δT fwd / d fwd = δT aft / d aft d fwd = (FP-to-midships) + (midships-to-LCF) = 162ft + 27ft = 189ft d aft = (AP-to-midships) – (midships-to-LCF) = 162ft – 27ft = 135ft

δT fwd / d fwd = δTrim / Lpp δT fwd / 189ft = 0.59ft / 324ft δT fwd = - 0.34 ft (decreasing at FP b/c weight is aft of LCF)

EN400 Ch 3 Prob 24

δT aft / d aft = δTrim / Lpp δT aft / 135ft = 0.59ft / 324ft δT aft = 0.25 ft (increasing at AP b/c weight is aft of LCF)

Final changes in draft overall: a. T fwd = To + δT fwd + δTPS = 13.5ft + (-0.34ft) + 0.14ft = 13.3 ft T aft = To + δT aft + δTPS = 13.5ft + 0.25ft + 0.14ft = 13.9 ft b. T m = (T fwd + T aft) / 2 = (13.3ft + 13.9ft) / 2 = 13.6ft

EN400 Ch 3 Prob 27

27. An LCS-1 class ship is pierside with a forward draft of 14.0 ft and an aft draft of 15.0 ft. Lpp = 324 ft. 70 LT of equipment is added 132 ft forward of amidships. Using an appropriate diagram, determine the ship’s final drafts at the FP and AP.

Solution:

T m = (T fwd + T aft) / 2 = (14.0ft + 15.0ft) / 2 = 14.5ft (use this value on Curves of Form) From Curves of Form: Δ = 353 general scale * (10LT / 1 general scale) = 3530 LT LCF = - 36 feet (aft of midships) TPI = 278 general scale * (0.12 LT/in / 1 general scale) = 33.36 LT/in MT1 = 198 general scale * (4 ft-LT/in / 1 general scale) = 792 ft-LT/in Change in draft due to Parallel Sinkage: δTPS = w / TPI = 70LT / 33.36 LT/in = 2.10 in * (1ft / 12in) = 0.175 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) = 70LT * 168ft / 792 ft-LT/in = 14.8 in * (1ft / 12in) = 1.24 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 = 1.24ft / 324ft δT fwd = 0.76 ft (increasing at FP b/c weight is fwd of LCF)

EN400 Ch 3 Prob 27

δT aft / d aft = δTrim / Lpp δT aft / 126ft = 1.24ft / 324ft δT aft = - 0.48 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 + 0.76ft + 0.175ft = 14.9 ft T aft = To aft + δT aft + δTPS = 15.0ft + (-0.48ft) + 0.175ft = 14.7 ft