Questions: 1. Rework example 2, the 17 foot fiberglass runabout, but use the ABYC tables 1a or 1b in Appendix A to determine engine weights and total weight of engine, controls, battery & full portable fuel tank. 17 Foot Fiberglass Outboard Runabout W(h) = Hull Weight = 800 lb (362.9 Kg) Wd = Deck Weight = 200 lb (60.6 Kg) (deck weight includes everything above the gunwale including windshield, deck hardware and accessories) W(hp)= Weight of wood in the hull = 0 PC = Maximum Persons Capacity = 900 lb (362.9 Kg) MC = Maximum Weight Capacity = 1400 lb (635 kg) HP = Maximum Horsepower Rating = 85 HP (63.38 Kw) K(f) = conversion factor for fiberglass = 0.33 K(w) = conversion factor for wood B = Buoyancy of flotation in lb per cubic foot = 60.4 C = Maximum Weight Capacity Dry weight of engine battery and portable fuel tank weight, see col 6 in Table 4, or col 10 in Table 1a and 1b. Step 1: determine flotation to support the boat Fb. Step 2: determine flotation to support the swamped motor Fp. Step 3. determine flotation to support the persons capacity Fc. Step 4. Determine total Flotation Needed Ft 2. What is the dry weight of the 85 hp (63.4 kw) outboard in Problem 1, from table 4, and from table 1a or 1b? a. 352 lb, 447 lb (159.7 Kg, 202.8 Kg) b. 405 lb, 427 lb (183.7 Kg, 193.9 Kg) c. 208 lb, 386 lb (94.3 Kg, 175.1 Kg) d. 550 lb, 602 lb(249.4 Kg, 273.1 Kg) 3. Calculate the amount of foam to support the boat Fb,, the persons Fc and the engine Fp for; A inboard board: Wh = Fiberglass = 612 lb (277.6 Kg) 1
Wd = Plywood = 115 lbs (51.2 Kg) Wc = Equipment hard ware and accessories = 62 lb (28.1 Kg) Engine, drive and battery weight = 1000 lbs (453.6 Kg) Fuel Capacity = 30 gallons = 180 lb (113.6 L, 81.6 Kg) MC = Maximum Weight Capacity = 1400 (635 kg) PC = Persons Capacity = 1400 (635 Kg) 4. If a boat uses air chambers for flotation what value for fresh water would you use to calculate the size of the air chambers. a. 60.4 (27.4 Kg) b. 62.4 (28.3 Kg) c. 64 (29 Kg) d. 58.4 (26.5 Kg) 5. Calculate the amount of flotation needed for a dinghy that uses air chambers. It is rated only for manual propulsion. Wh = 105 pounds (47.6 Kg) Wh Fiberglass = 90 lb (40.8 Kg) WH wood = 15 lb (6.8 Kg) MC = Maximum Weight capacity = 450 lbs (204.1 Kg) PC = Maximum Persons Capacity = 405 lb (183.7 Kg) a. 1.5 cubic feet (42.5 cu cm) b. 1.8 cubic feet (51 cu cm) c. 2.1 cubic feet (59.5 cu cm) d. 2.5 cubic feet (70.8 cu cm) 6. In question 4 what arrangement would be best for the air chambers to achieve level flotation. This boat has two seats, one in the middle of the boat where the rower sits and a seat in the stern along the transom. a. One chamber under the floor. b A chamber in the bow and a chamber under the stern seat. c. One chamber under the middle seat. d. none of the above. 7. If the boat in question 4 had a 2 HP (1.49 KW) (1.49 Kw) outboard motor how much would the flotation requirement increase. Use ABYC table 1a or 1b for engine weights. Use G(s) = swamped weight of the engine. a. 0.8 cubic feet (22.7 cu cm) 2
b. 1.0 cubic feet (28.3 cu cm) c. 0.3 cubic feet (8.5 cu cm) d. 0.4 cubic feet (11.3 cu cm) 8. You have an 18 foot Sterndrive boat. The engine is a 3.0 L Mercruiser Alpha with 135 HP (101 KW). Overall the boat weighs 2235 lb (1014 KG) including engine and sterndrive. The engine/sterndrive package weighs 635 lb (288 kg). The Maximum weight Capacity is 1200 pounds(544 KG), Maximum Persons Capacity is 1100 pounds (499 kg) (8 persons) It has a 28 gallon fuel tank (106 L). The boat uses 2 pound density polyurethane foam for flotation. Wh = 1050 lb (476.3 KG) fiberglass Wd = 250 lb (113.4 KG) fiberglass Whw = 150 lb (68 KG) plywood (in seats, cabinets, flooring) We = 150 lb (68 KG)(windshield, deck fittings, lights etc.) Wm = 635 lb (288 KG) MC = 1200 lb (544 KG) PC = 1100 lb (449 KG) Fuel Tank = 28 gallon, (106 L) = 168 lb (76.2 KG) What is the amount of flotation required to support the boat? 8. What is the amount of flotation to support the engine and sterndrive package? 9. What is the amount of flotation to support the persons? Maximum Weight Capacity = 1200 lb. (504.4 Kg) Fuel tank = 28 gals = 168 lbs. (76.2 Kg) (gasoline weighs about 6 lb per gallon) 10. What is the total amount of flotation required for basic flotation? 3
Appendix A Table 4 from 33CFR183 Weights of Outboard Motors and Related Equipment For Various Boat Horsepower Ratings Boat Horsepower Rating Motor and Control Weight Battery weight Full Portable Fuel Tank Weight column Dry Swamped Dry Submerged 1+3+5 1 2 3 4 5 6.01 to 2 25 20 - - - 25 2.1 to 3.9 40 34 - - - 40 4.0 to 7 60 52 - - 25 85 7.1 to 15 90 82 20 11 50 160 15.1 to 25 125 105 45 25 50 220 25.1 to 45 170 143 45 25 100 315 45.1 to 60 235 195 45 25 100 380 60.1 to 80 280 235 45 25 100 425 80.1 to 145 405 352 45 25 100 550 145.1 to 275 430 380 45 25 100 575 275.1 and up 605 538 45 25 100 750 Transoms Designed for Twin Motors 50.1 to 90 340 286 90 50 100 530 90.1 to 120 470 390 90 50 100 660 120.1 to 160 560 470 90 50 100 750 160.1 to 290 810 704 90 50 100 1000 290.1 to 550 860 760 90 50 100 1050 550.1 and up 1210 1076 90 50 100 1400 Below are tables 1a and 1b published in ABYC Industry Conformity Standard S-30, Outboard Engine and Related Equipment Weights. It is updated on a five year cycle and contains the weights for four stroke outboards.. 4
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Appendix B: Outboard Motorboats Over 2 HP (1.49 KW) Fiberglass Wood Wh = Boat Weight + Wd = Deck Weight + PC = Maximum persons weight Wp = Swamped Motor & Controls + Battery + = From Table 4 33 CFR or ABYC Tables 1a or 1b Dead Weight = Maximum weight Capacity Persons Capacity Motor Controls Battery Wdw = MC PC Wm Ignore if zero or less. Fb = Flotation for Boat Weight (Wh + Wd) X K / 60.4 = Fb Factors: Fiberglass 0.33 (cubic feet of 2 pound density Aluminum 0.63 closed cell foam) Plywood -0.81 ABS Plastic 0.11 x x K /60.4 = Oak, White -0.18 Oak -0.56 Mahogany -0.72 Fc = Flotation For Persons Weight Fc For boats over 550 lb PC = (0.5 X 550) +.125 (PC-550) +.25(Wdw)/60.4 275 + + /60.4 = Fc For boats with less than 550 lb PC = (0.5 x PC) +.25(Wdw) / 60.4 + / 60.4 = Fp = Flotation for swamped motor (Wp ) / 60.4 Wp comes from Table 4 or ABYC / 60.4 = Tables 1a or 1b Amount of Flotation in Cubic feet Ft = Fb + Fc + Fp + + = 7
Inboard Motorboats Fiberglass Wood Wh = Boat Weight + Wd = Deck Weight + PC = Maximum persons weight Wp = dry weight motor + drive + battery (includes motor, drive, propeller, shaft, battery and transmission) = Fb = Flotation for Boat Weight (Wb + Wd) X K / 60.4 = Fb Factors: Fiberglass 0.33 (cubic feet of 2 pound density Aluminum 0.63 closed cell foam) Plywood -0.81 ABS Plastic 0.11 x x K /60.4 = Oak, White -0.18 Oak -0.56 Mahogany -0.72 Fc = Flotation For Persons Weight Fc = 0.25 (Fc1 + Fc2) / 60.4 Fc1 = PC fuel weight 0.25 ( + ) = Fc2 = MC PC ( - ) = Fc = 0.25 (Fc1 + Fc2) / 60.4 0.25 ( + ) /60.4 = Fp = Flotation for swamped motor (0.75 x Wp ) / 60.4 / 60.4 = Amount of Flotation in Cubic feet Ft = Fb + Fc + Fp + + = 8
Outboard Motorboats 2 HP (1.49 KW) or less. Fiberglass Wood Wh = Boat Weight + Wd = Deck Weight + PC = Maximum persons weight Wp = Swamped Motor & Controls From Table 4 33 CFR or ABYC Tables 1a or 1b = Dead Weight = Maximum weight Capacity Persons Capacity Wdw = MC PC Ignore if zero or less. Fb = Flotation for Boat Weight (Wb + Wd) X K / 60.4 = Fb Factors: Fiberglass 0.33 (cubic feet of 2 pound density Aluminum 0.63 closed cell foam) Plywood -0.81 ABS Plastic 0.11 x x K /60.4 = Oak, White -0.18 Oak -0.56 Mahogany -0.72 Fc = Flotation For Persons Weight Fc For boats over 550 lb PC = (0.5 X 550) +.125 (PC-550) +.25(Wdw)/60.4 275 + + /60.4 = Fc For boats with less than 550 lb PC = (0.5 x PC) +.25(Wdw) / 60.4 + / 60.4 = Fp = Flotation for swamped motor (Wp ) / 60.4 Wp comes from Table 4 or ABYC / 60.4 = Tables 1a or 1b Amount of Flotation in Cubic feet Ft = Fb + Fc + Fp + + = 9
Answers: Q1 Step 1: determine flotation to support the boat Fb. K = conversion factor for fiberglass =.33 Fb = (Wh x K) + Wd / 60.4 Where (Wh x K) = ((Whf x Kf) + (Whp x Kp)) There is no wood so Whp X kw = 0 (Wh x k) = (800 x.33) This is a level flotation boat so the deck is not submerged. Therefore use the dry weight of the deck. Wd = 200 lb Fb = (800 x.33) + 200 / 60.4 Fb = 264 + 200 / 60.4 = 7.68 Flotation to support the boat = 7.7 cu ft(218 cu cm) Q1. Step 2: determine flotation to support the swamped motor Fp. Fp = (G/B) S = Swamped weight of engine (Table 1a col 4) = Fp = 380/60.4 = 6.29 Flotation to support the engine = 6.3 cu ft.(178.4 cu cm) Q1. Step 3. determine flotation to support the persons capacity PC C = 900 lb = MC Engine Weight = 1400 602 = 798 lb Fc=.5(first 550 lb of PC)+.125(PC-550) +.25(C-PC Eng) 60.4 Fc=.5(550) +.125(900-550) +.25(1400-900-427) 60.4 10
Fc= (275 + 43.75 + 18.25)/60.4 = 337/60.4 = 5.58 Flotation to support persons capacity= 5.6 cu ft(159 cu cm) Q1. Step 4. Determine total Flotation Needed Ft = Fb + F(m) + F(pc) Ft = 7.7 + 6.3 + 5.6 = 19.6 cubic feet (555 cu cm) 2. b. 405 lb, 427 lb (183.7 Kg, 193.9 Kg) 3. Fb = 612 x 0.33 + 115 (-0.81) + 0.69 X 62 / 60.4 = 2.4695 Fb = 2.5 cubic feet (70.8 cu cm) Fp = 0.75 (1000) / 60.4 = 12.4 Cubic feet (351 Cu Cm) 12.417 Fc =.25 ((1400-180) + (1400-1400)) / 60.4 Fc = 5 cubic feet (141.6 Cu Cm) = 5.0497 Total = 2.51 + 12.417 + 5.0497 = 19.9362 (0.5647 cu m) 4. b. 62.4 (28.3 Kg) 5. Fb = (95 X 0.33) + (15 x -0.81)/ 62.4 Fb = (31.35-12.15) /62.4 Fb = 0.30 cubic feet to support the boat (8.5 Cu cm) Fp = 0.25 x C =.25 x 450/62.4 = 112.5/62.4 = 1.8(51 cu cm) Ft = 0.3 + 1.8 to support the load = 2.1 Cu ft (59.5 cu cm) c. 2.1 cubic feet (59.5 cu cm) 6. b. A chamber in the bow and a chamber under the stern seat. 7. Fc = 0.25(450 30)/62.4 = 105/62.4 = 1.68 Cu ft(47.6 cu cm) 11
Fp = G(s) / 62.4 = 26 /62.4 = 0.42 cu ft (11.9 cu cm) Ft =Fb + Fc + F(m)=0.30 + 1.68 + 0.42= 2.4 cu ft.(68 cu cm) c. 0.3 cubic feet (8.5 cu cm) 8. What is the amount of flotation required to support the boat? Fb = (Wh x K ) + (Whw X K)+ (Wd) +.69(We) 60.4 Fb = (1050 x 0.33) + (150 x -.81) + 250 +.69(150) 60.4 Fb =346.6 121.5 + 250 + 103.5 =578.6 =9.58(9.6)(0.27cu m) 60.4 60.4 9. Fp = G/B G =.75 X Wm =.75 x 635 = 476.25 Fp = 476.25 / 60.4 = 7.88 (7.9) (0.02237 cu m) 10. Fc =.25(Fc1 + Fc2) B Fc1 = Max Persons Capacity Fuel Wt = 1100-168 lb.= 932 Fc2 = Max Weight Capacity Max Persons Capacity = 1200-1100 = 100 Fc = 0.25 (Fc1 + Fc2) =.25((1100-168) + (1200 1100)) B 60.4 Fc = 258 / 60.4 = 4.27 Cubic feet (4.3) (0.218 cu m) 11. Ft = 9.58 + 7.88 + 4.27 = 21.73 (0.615 cu m) 12