Handbook of Ship Calculations, Construction and Operation
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A reference source for shipowners, ship officers, ship and engine draughtsmen, etc....
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HANDBOOKS SHIP CALCULATIONS CONSTRUCTION and OPERATION CHARLES H.HUGHES
m
HANDBOOK OF
SHIP CALCULATIONS,
CONSTRUCTION AND OPERATION
HANDBOOK OF SHIP CALCULATIONS,
CONSTRUCTION AND OPERATION A BOOK OF REFERENCE FOR SHIPOWNERS, SHIP
AND ENGINE DRAUGHTSMEN. MARINE ENGINEERS, AND OTHERS ENGAGED IN THE BUILDING AND OPERATING OF SHIPS
OFFICERS, SHIP
BY
CHARLES
H.
HUGHES
NAVAL ARCHITECT AND ENGINEER
•
»
» «
- >
APPLETON AND COMPANY LONDON NEW YORK D.
1917
Copyright, 1917, bt
D.
APPLETON AND COMPANY
Printed in the United States of America
PREFACE
This
handbook
has
been
compiled with the purpose of assembling in a single publication in convenient form, practical data for everyday reference, for men engaged in the designing, building and operating of ships. Theoretical calculations have been purposely omitted. Shipowners and men in the offices of steamship companies will find particular interest in the sections on Loading and Stowing of Cargoes, Maintenance, Ship Chartering, and To men employed in shipyards the sections
Marine Insurance. on Ship Calcula-
and Hull Construction, Structural Details, Machinery, and Ship Equipment, and the various formulae for making quick In making preliminary designs the calculations will be of use. section on Hull and Machinery Weights, as also the tables giving tions
particulars of all classes of vessels, will be found convenient. Ship officers and marine engineers will find, in the section
on
Machinery, valuable data on the overhauling of boilers, on indicator cards, on the operating of pumps, condensers and motors, and many other practical subjects. They will find useful also the sections on Loading and Stowing of Cargoes, Ship Machinery, and many other subjects.
Marine underwriters, ship brokers and freight brokers convenient data on ship construction and the stowage
will find
sizes of
materials, with a large number of miscellaneous tables. For men engaged in the designing and building of war vessels a section on warships has been included, which describes the .
armor and armament. Although the fundamental calculations for all vessels, merchant and war, are
different classes
and
their
the same, the text contains frequent special references to warships, as on the subject of electric propulsion, electric steering gears, electric winches, etc.
To
the student of naval architecture and marine engineering this handbook offers a broader *collfctipi*, of practical data than
Preface
vi
any other published work. The verj^ latest marine practice is given, and such subjects as electric propulsion, geared turbines, Diesel engines, and oil fuel are fully treated, as are also recent and special types of construction, such as tankers, battle cruisers, submarine chasers, and submarines. The handbook represents many years of collection and classification of material, assembled primarily for the writer's
everyday
use.
The data have been obtained from many
sources (see authorities), not only from textbooks but very largely from technical papers and trade literature. As it is impossible to mention in the text all the
works consulted and used, the writer wishes to make here a general acknowledgment of his indebtedness to many other workers in the He wishes to thank particularly the editors of Intermarine field. national Marine Engineering and Shipping Illustrated. Prof. H. E. Everett kindly revised the section on Freeboard. Mr. J. C. Craven checked Structural Details, while other friends in the trade read over various sections: To Mr. F. G. Wickware, of D. Appleton and Co., he is indebted for the typographical arrangement and many suggestions.
Chas. H. Hughes.
New
York,
June
26, 1917.
CONTENTS (See also Index)
SECTION
I
WEIGHTS, MEASURES AND FORMULA PAGE
— —
Weights and Measures United States and Great Britain Weights and Measures Metric System Decimal Equivalents of Fractions of an Inch Centimeter, Gram, Second System Conversion Table
1
4 5 6 7 8 8 9 10
— Metric Units into United States
Board Measure Water, Weights of Different Units of Feet Board Measure in Timber Inches and Fractions in Decimals of a Foot Water Conversion Table Composition of Salt Water Specific Gravities and Weights of Materials Cubic Feet per Ton or Stowage Sizes of Materials Shipping Weights of Lumber Weights of Miscellaneous Units of Different Products !
.
Bundling Schedule
of
11
12 13 16
18 19
Pipe
Barrels, Sizes of
Horse Powers Equivalent Values of Mechanical and Electrical Units
Comparison of Thermometer Scales Thermometers Circumferences and Areas of Circles Advancing by Eights Involution and Evolution To Extract the Square Root of a Number To Extract the Cube Root of a Number '
Logarithms Powers and Roots of Numbers Circumferences and Areas of Circles Geometrical Propositions Circle and Ellipse, Formulae pertaining to Areas of Plane Figures and Surfaces of Solids
Volumes of Solids Trigonometry—^-Trigonometric Functions Oblique Triangles Trigonometric Formulae Natural Sines, Cosines, etc., Table of Moment of Inertia, Radius of Gyration and Center of Gravity Center of Gravity of a Cross Section of a Ship Center of Gravity of a Water Plane
:
20 20 22 23 24 25 25 26 26 27 28 29 29 31
32 33 36 39 41 42 43 50 50 51
52
Properties of Sections vii
Table of Contents
viii
section n
STRENGTH OF MATERIALS Definitions
—
PAGE Stress, Strain, Tension, etc
70
Strength of Materials, Tables Factors of Safety
71
73
74 75 Beam Formula 76 Beams Under Various Loading Conditions 78 Columns Formulae for 84 and I Columns Safe Loads for 86 Safe Loads for Square and Round Wood Columns 87 Safe Loads for Wrought Iron Pipe, Strong and Extra Strong Columns .... 88 89 Torsional Stresses 90 Springs 90 Tubes, Pipes and Cylinders, Formulae for 91 Bursting and Collapsing Pressures of Wrought Iron Tubes 91 Ship Fittings, Strength of 93 Shearing and Tensile Strength of Bolts ? 94 Tests of Hooks 95 Tests of Shackles 95 Tests of Eye Bolts Tests of Hoist Hooks 96 97 Tests of Turnbuckles 97 Formulae for Davits 98 Stresses in Cranes, Derricks and Shear Legs 101 Rivet Heads and Points 102 Proportions of Rivets 102 Diameters of 102 for Lengths Ordering 103 Signs for Rivets 104 Shearing and Tensile Strength of Steel Rivets Formulae for Riveted Joints 105 107 Shearing Value of Rivets and Bearing Value of Riveted Plates Reduction of Diameters to Inches 108 109 Weight of Cone Head Rivets Number of Cone Head Rivets in 100 lbs 110
Timber, Strength of Beams Neutral Axes, Bending Moment, Shearing Stresses, etc .
—
—
H
SECTION
III
SHIPBUILDING MATERIALS Steel
and Iron
Ill Steel, Methods of Manufacture 112 Carbon, Manganese, Nickel and Alloy Steels Structural Steel Lloyd's, Am. Bureau of Shipping and Am. Soc. of Testing 114 Materials, Requirements 117 Rivet Steel 118 Cast Steel
—
»
Table of Contents
ix PAGE
Iron
Wrought
119 120 120
Iron
Cast Iron Malleable Iron
121
Pickling Steel Plates
121
Galvanizing Fittings and Steel Plates
121
Hundredths of an Inch Weights Standard Gauges United States and Great Britain for Sheets and Plates Diamond Checkered Plates Weights of Sheets and Plates of Steel, Copper and Brass (Birmingham Wire Gauge) Weights of Sheets and Plates of Steel, Copper and Brass (American or Brown and Sharpe Gauge) Sizes of Steel Plates and Heads Sizes and Properties of Structural Shapes Weights and Areas of Square and Round Bars, and Circumferences of Round
122 123 125
of Steel Plates in
—
.
.
.
Bars
Weights
126 127 128 129 143
of Flat Rolled Steel
Bars
149
Non-Ferrous Metals and Alloys 155 155 155 155 155 156 156 156 157 157 157
Copper
Aluminum Zinc
Lead Tin
—
Bronzes Phosphor, Admiralty, Titan, Tobin, Manganese Gun Metal ,
Brasses
Muntz Metal Naval Brass Alloys
Wood Sawing and Seasoning Hardness of Wood, How Measured Table of Relative Hardness of Soft Woods Table of Relative Hardness of Hard Woods
Hard Wood Soft
Wood
157 158 158 159 159 160
Sizes
Sizes
Characteristics,
Weights
and
Specific
Gravities of
Woods used
building
in Ship-
160
,
Miscellaneous Non-Metallic Materials
Oakum
162
Caulking Cotton
Steam Pipe Covering
162 162 162 163
Boiler Covering
163
Portland Cement Insulating Materials
— Magnesia,
Asbestos, Cork, Hair Felt, Mineral
Wool
.
.
.
x
Table of Contents PAGE 164 164 165
Cylinder Covering Tests of Insulating Materials Relative Value of Non-Conducting Materials
SECTION IV SHIP CALCULATIONS Lengths Breadth
,
Depth Draft
Displacement Displacement Curve
Deadweight Registry
Tonnage Cubic Capacity
Tons per Inch Coefficients
of
Immersion
—Prismatic, Block, etc
Wetted Surface Center of Buoyancy ,
•
•
•
•
.
Transverse Metacenter Metacentric Heights Moment of Inertia of a Water Plane about Displacement Sheet Curves of Stability
its
Center Line
'
Notes on Stability
Trim
Moment to Alter Trim To Find the Trim by Trim
Lines Quantity of Water That Will Flow Into a Ship Through a Hole in Her Side. Compartment Flooded, Calculation of Trim, by Trim Lines Compartment Flooded, Calculation of Trim, by Mean Sinkage Center of Gravity of a Vessel, Fore and Aft Position of Center of Gravity of a Vessel, Vertical Position of Heights of the Center of Gravity above the Base Heights of the Metacenter above the Base Effects of
To Find
Moving Weights
the Center of Gravity of a Vessel by
Freeboard Freeboard Freeboard Freeboard Freeboard
Moving Weights
Calculations Calculations for a Shelter
Deck Steamer
Notes
Markings Powering Vessels Approximate I. H. P. to Propel a Vessel Effective Horse Power
Towing Engine Revolutions to Drive a Vessel at a Given Speed Formula for Estimating Speed of a Motor Boat
.
166 166 167 167 168 169 169 169 169 170 170 171 172 172 173 174 176 177 183 189 190 190 192 195 195 196
200 202 202 202 203 204 205 213 218 220 221 222 222 223 223 224 224
Table of Contents
xi
PAGE Resistance
Law
of
225 226 227 227 228 229 231 232 233
Comparison
Surface Friction Constants
Launching Launching Ways Launching Calculations Releasing and Checking Devices Launching Data Declivity of Ways and Launching Velocity
SECTION V
HULL CONSTRUCTION and Organizations Governing Shipping Merchant Vessels
Classification Societies
Types
War
of
Vessels
Armor Armament Types of War Vessels Systems of Construction Frames Reverse Frames
Merchant
of
Vessels.
Shell Plating
Double Bottom Keelsons and Longitudinals Keels
Deck Plating and Coverings Deck Beams Hatchways
«...
Pillars
Stringers
Bulkheads
Stem and Stern Frames Rudders Machinery Foundations Deck Erections Cementing Painting
Wood
Vessels
Carpenter and Joiner Work Interior Painting Tables of Screws, Nails and Spikes Structural Strength Curve of Weights
Curve Curve
of
Buoyancy
of
Loads
Neutral Axis and
Moment
Hogging and Sagging Curve of Shearing Stresses
of Inertia Calculations
-
"
234 238 238 244 245 247 253 255 256 258 261 263 264 264 265 266 268 268 268 272 275 277 277 278 279 282 284 287 290 294 294 296 296 297 298 298
Table of Contents
xii
PAGE
Curve
of
Bending Moments
Specification Headings for Hull Specification Headings for Specification Headings for
Machinery Equipment
Hull Weights, Formulae for Hull Weights of Vessels
Machinery Weights per I. h. p Engine Weights Boiler Weights Weights of Water Tube Boilers Weights of Boilers, Engines and Auxiliaries Weights of Diesel Engines Data on Passenger and Cargo Steamers [reciprocating engines] Data on Passenger and Cargo Steamers [turbines] Data on Excursion Vessels, Tugs, Lighters and Steam Yachts Data on Motor Ships Data on Motor Boats Data on Sailing Vessels with Motors Data on Schooners Data on Schooners with Motors Oil Carriers
Lumber Steamers Trawlers
Dredges Shallow Draft Steamers
Tunnel Vessels
and Horse Steamers and Estimates
Fittings for Cattle Prices, Costs
Prices of Vessels sold in 1916
Estimates for Building a Motor Schooner Estimates for Operating a Motor Schooner Operating Costs of Diesel Engines Repair Costs of Motor Ships Costs of Electric and Refrigerating Systems Prices of
Steam Engines and
,
299 300 301 301 302 303 304 305 306 308 308 309 310, 311 312 314 316 317 318 319 320 320 327 327 327 328 330 330 331 332 334 334 335 336 337 337 338 340 ,
Boilers
Estimates, Preparing Labor Costs
SECTION VI
MACHINERY Steam Definitions— British Thermal Unit, Mechanical Equivalent of Heat, Calorie..
Total and Latent Heat of Steam Saturated Steam Specific,
Superheated Steam
Dry Steam
Wet Steam Steam Table
'
—Properties
of
Saturated Steam
,,,,,,,
341 342 343 343 343 343
344
Table of Contents
xiii
PAGE Fuels Coal Required to Evaporate one Pound of Water Coal Consumption Coal or Oil Consumptioon Evaporation per Pound of Combustible Heat Values of Coal Calorific Value of Coal from its Chemical Analysis Sizes of Coals Heat Values of Wood
349 349 350 350 351 351 351 352 353 353
.
of Fire
Temperature Air Required
for
Combustion
of Fuel
Oil
Crude Petroleum and
Products Fuel Oil, Table of Beaume" Gravity, Specific Gravity, Beaum6 Gravity Specific Gravities and Weights of Oils Definitions Flash Point, Fire Point, Viscosity, etc its
—
Oil for Boilers
Heat Values
of Oil
Fuel Oils for Internal Combustion Engines Lubricating Oil Oil
Burning Systems
etc.,
of
353 354 355 355 356 356 356 357 358 360
Boilers
Types
of Boilers
Scotch Boilers, Proportions of Scotch Boilers, Tables of
Locomotive Boilers
Leg Boilers
Water Tube Boilers Comparison of Fire Tube and Water Tube Boilers Boiler Horse Power Boiler Horse Power Required for an Engine Factor of Evaporation Boiler Efficiency
Gallons of Water Evaporated per Minute in Boilers Boiler Fittings
.
,
Safety Valve
Stop Valve
Feed Water Connections for Scotch Boilers Feed Check Valve Surface and Bottom Blows
Steam Gauges Water Gauges and Cocks Boiler Circulators
Fusible Plugs Injectors and Inspirators Hydrometer
Superheaters Ash Ejectors
:
;
363 363 364 366 366 368 370 371 371 373 374 375 376 376 377 377 378 378 378 379 379 379 380 381 382 384
Table of Contents
xiv
PAGE 385 385 386 386 387
Boiler Operating
Firing
Shutting Off Boilers Overhauling Boilers Boiling
Out
Boilers
'.
Draft
389 389 390 390 391 392 393
Systems
Measurement
Draft Air Pressure Escaping into the Atmosphere of under Velocity Air Required , Blowers Forced Draft Installations Frictional Resistance of Stack of
Marine Steam Engines Types
of
Ratio of Cylinders and Steam Expansion Expansion, Cut-off and Back Pressure Crank Sequences Paddle Wheel Engines Valves
Lap and Lead Valve Travel Valve Mechanism Setting Valves
'
Steam Pressure in a Cylinder at End of Stroke Steam Pressure at Different Cut-offs Cut-off and Coal or Steam Consumption Indicator Cards
Mean
Effective Pressure and I. H. P. Calculations Coal Consumption per I. H. P Engine Formula? Estimated Horse Power
—
Shafting
Cylinders
Connecting Piston
Rod
Rod
i
Pistons
Bearing Surfaces Engine Fittings Thrust and Line Shafting Bearings Engine Room Floors Operating Trials.
394 394 395 396 397 399 400 401 402 404 404 405 405 406 410 411 411 412 413 413 414 414 414 414 418 419 419 421
Propellers
Definitions—Pitch, Driving Face, Projected Area, etc Slip
Table of Propellers Formulae for Slip, Speed, Revolutions and Pitch Rule for Finding Pitch of a Propeller
424 425 426 428 428
Table of Contents
xv PAGE
Finding Helicoidal and Projected Area Propeller Thrust Propellers for Turbine Ships
Rule
for
Formulae for Keys, Nuts, etc Propellers for Motor Boats
-.
Weights of Propellers Speed Table, When Pkch, Slip and R. P. M. Are Given
429 429 429 430 431 432 433
Paddle Wheels Breadth of Floats, etc
Formula
for Slip, Speed,
436 436 437
Revolutions and Pitch
Table of Paddle Wheels
Steam Turbines Turbines Types Geared Turbines Trials of Geared Turbines and Reciprocating Engines Turbo-electric Propulsion Comparative Performance of Different Systems of Propulsion of
Efficiency
Steam Consumption Weights Calculation of Horse Power Steam per Shaft Horse Power Auxiliaries
Steam Plant
Auxiliaries
Definitions— Atmospheric Pressure, Gauge Pressure, etc
Thermodynamics of Condensers Types of Condensers Surface Condensers
Operating
Vacuum and Vacuum Gauge To Find Vacuum under Working Conditions Vacuum and Corresponding Steam Temperature Jet Condensers
Cooling Water Required for Surface or Jet Condenser Keel or Outboard Condensers Air Pump Circulating
Pump
Feed and Filter Tank Steam Traps Feed Water Filter Feed Water Heaters
:
Evaporators
Pumps, Types
of
Reciprocating Pumps Centrifugal Pumps Pumps installed in a Freight Steamer Installing
and Operating Pumps
439 442 443 444 445 446 446 446 447 448 448
:
449 449 450 452 453 454 455 456 456 457 459 459 462 462 463 464 465 467 470 470 474 477 477
Table of Contents
xvi
PAGE Internal Combustion Engines
478 479 480
Fuels
Operation Horse Power Formula Table of Engines (Electric Ignition) Carburetors and Vaporizers
481
Starting
Reverse Gears Lubricating Systems Valves Ignition Systems Timers and Distributors
Magnetos Spark Plugs Motor Trouble Hot Bulb Engines Diesel Engines of Diesel Engines, Steam Engines and Turbines General Features of Diesel Engines
Comparison Operation
Types
of Diesel
Engines
Diesel Engine Installations
Piping, Tubing, Valves
and
482 482 483 484 484 486 487 489 490 491 492 495 495 496 498 501 506
Fittings
Trade Terms
507 Tables of Standard, Extra Strong, and Double Extra Strong Wrought Iron 508, 509 Pipe Boiler Tubes 510, 511, 512 513 Copper Tubes Brass and Copper Tubes 514 518 Brass and Copper Pipe Formula for Working Pressure 518 519 Copper Tubes 519 Bending Pipes and Tubes Flow of Water through Pipes 520 521 Comparative Areas of Pipes 522 Flanges Bolt and Pipe Threads 524 Gaskets 526 526 Nipples and Couplings Unions 526 Materials for Piping Systems 527 528 Valves, Cocks and Fittings
SECTION
VII
ELECTRICITY Definitions— Ohm, Ampere, Volt, Coulomb, etc
Voltage Wires, Calculation of Size of Tables of Sizes of Wires
531
532 533 534, 535, 536, 537
Table of Contents
xvii PAGE
Diameters by Different Wire Gauges Wiring Systems Conduits Switchboards and Equipment Determination of Output '.
. 1 1
Laying Out Electric Installations Wiring of a Steamer Wiring of a Motor Boat Wiring of Gasoline Engines Incandescent
Lamps
Searchlights Primary Batteries
Storage Batteries
Grouping
of Cells
Generating Sets Windings of Generators Engine Horse Power Tables of Direct Connected Sets Operating Notes
J
Electric Motors, Windings of Formula for Horse Power of Motor Weights of Motors Current Taken by Motors Motors for Ship Work. Motor Starting and Controlling Devices
533 539 541 543 543 544 547 559 551 552 553 554 554 557 557 558 558 559 560 561 562 563 564 564 565
SECTION VIII HEATING, VENTILATION, REFRIGERATION, DRAINAGE, PLUMBING AND FIRE EXTINGUISHING SYSTEMS Heat Passing Through a Ship's Side or Bulkhead Heating Systems Heating by Steam Steam Heating Installations Sizes of Radiators Square Feet of Radiation for a Heating Surfaces of Pipes Heating by Thermotanks Heating by Electricity Heating by Special Systems
567 568 568 568 569 570 571 571 573 574
Room
Ventilation
Fresh Air Required Air Pressure
Systems
.
.
;
—Plenum and Exhaust
.
Ventilation of Oil Carriers Ventilation of Engine Ventilators
Rooms
Fans, Types of
Horse Power Required to Drive a Fan
.-...-
575 576 576 578 579 579 580 582
Table of Contents
xviii
PAGE 582 583 584 584 585
Ducts Data on the Escape of Air into the Atmosphere .....' Duct Areas Laying Out Ventilating Systems Loss of Pressure by Friction of Air in Pipes Refrigeration
Heat and Latent Heat of Food Products Horse Power Required for Compressors Operating Notes
587 587 588 591 593 594 594 595 596 596 597 598 599 600 600 600
Drainage Systems Main Drain Auxiliary Drain U. S. Steamboat-Inspection and Lloyd's Requirements
601 602 603 603
Keeping Perishable Products Insulating Materials
Cold Storage Temperatures Compression System Brine Circulating
—
Refrigerants Ammonia and Carbon Dioxide Different Makes of Machines Cooling by Air
and Fittings
Pipe, Valves
Linear Feet of Pipe Required
Capacity of
Ammonia Compressors
Refrigeration Required for Cold Storage Rooms Refrigeration Required for Stored Products Specific
Plumbing 606 609 609
Fixtures
Waste Lines Fresh Water Service Fire Extinguishing
and Alarm Systems 610 611 612 612 613 614 614
General Requirements Fire Main (Water) Sizes of Water Streams
Main (Steam) Sulphur Dioxide System Sprinkler Systems Fire
Fire Alarms
SECTION IX SHIP EQUIPMENT Steering Gears
Steam Steering Gears Electric Steering Gears
Arrangements Installations
•
615 615 616 617 618
Table of Contents
xix PAGE
Transmission
Power Required to Turn a Rudder Pressure on Rudder Steering Chain and Rod Windlasses Steam and Electric Table of Steam Windlasses Winches or Hoisting Engines Steam and Electric Tables of Winches Power Required to Raise a Load Rope Capacity of a Drum Capstans and Gypsy Capstans Steam and Electric
—
—
—
Tables of Capstans
Towing Machines Rope, Trade Terms Hoisting Speeds Knots and Hitches
:
Tension in Hoisting Rope Kinds of Rope, Materials and Strands
Weight and Strength of Manila Rope Weight and Strength of Hemp Clad Wire Rope Weight and Strength of Flattened Strand Hoisting Rope Table Comparing Manila and Hemp Clad Wire Rope How to Measure Wire Rope Weight and Strength of Cast Steel Wire Rope Weight and Strength of Steel Mooring Lines Formulae for Size and Weight of Rope Weight and Strength of Yacht Rigging and Guy Ropes Weight and Strength of Galvanized Steel Hawsers Weight and Strength of Galvanized Ships' Rigging and Guy Ropes Weight and Strength of Galvanized Steel Hawsers Length of Rope Required for Splices Blocks, Types of Wood Blocks for Manila Rope Steel Blocks for Wire Rope Working Loads for Blocks
.
Tackles, Types of
Power Gained with Tackles Chain Pitch, Breaking and Working Strains of Chain Anchors, Types of Anchors for Yachts and Motor Boats Anchors for Steam Vessels Anchors for Sailing Vessels Anchor Cranes U. S. Steamboat-Inspection Requirements Lrfe-Saving Equipment Life-Saving Equipment, Abstracts from Seamen's Bill
—
Capacities of Lifeboats Lundin Lifeboats
Engelhardt Collapsible Lifeboats Life Rafts.
. ,
620 623 623 624 624 625 626 627 628 628 630 630 631 632 632 633 634 634 636 637 638 640 640 641 642 642 643 644 645 646 646 646 648 648 648 650 652 655 656 657 657 658 659 660 663 665 669 670 671 672
xx
Table of Contents PAGE
Life Preservers
672 673 673 673 673 676 676 678 679 680 681 682
,
Buoys Boats Carried by War Vessels Boat Davits Rotating Davits Pivoted Davits Quadrant Davits Heel of a Vessel When Lowering a Boat Rigs of Vessels
•.
,
Wireless Equipment
Storm
•
Oil
Line-carrying
Guns and Rockets
SECTION X SHIP OPERATING Loading and Stowing of Cargoes General Considerations Oil Cargoes
—Bulk
—
Stability,
Winging Out Weights,
etc
Stowage of Oil in Barrels Stowage of Oil in Cases Grain Cargoes Settling of Grain and Angle of Repose Rules of N. Y. Board of Underwriters Board of Trade Requirements Coal Cargoes Effect of Using Bunker Coal Rules of N. Y. Board of Underwriters
Lumber Cargoes
.-.
.
.
.
.
.
Regulations for Carrying Dangerous Articles
MACHINERY OPERATING
683 684 685 686 688 688 688 692 693 693 693 694 695
(see Index)
Maintenance
When Hull
Surveys Are to Be
—
Shell Plating
Made
696 698 698 698 698 699 699 699
Working
Decks Leaking Removing of Rubbish fiom Bilges
'.
Galvanic Action Corrosion in Double Bottom Sea Valves and Outboard Bearings .
Docking.
/
Painting (see Index)
Machinery, Care of Surveys Taking Indicator Cards Care of Boilers Log to Be Kept by Engineer
.
700 700 701 701 702
Table of Contents
xxi PAGE
Ship Chartering 1 Trip Charters Contracts for the Time Charter Charter Forms Preamble Clause
.•
Movement
of Freight
Delivery Redelivery Trading Limits and Insurance Warranties
Speed amd Consumption Berth Terms Cotton Rates Abbreviations
703 704 704 704 707 707 708 708 708 709 709 709
Marine Insurance Insurable Value
711
Ship
711 711 711
Freight.
Goods Policies,
Kinds
Paragraphs in
711 713
.'
of Policies
713 714 715 715 715 715
General Average Particular Average
River Plate, Clause Protection and Indemnity Clause Collision or
Running Down Clause
Inchmaree Clause Export and Shipping Terms Abbreviations and Authorities
Index
Terms
716
Quoted
,
.
.
718 721
ABBREVIATIONS AND SYMBOLS Weights and Measures (U. oz.
S.
and English)
XXIV
ABBREVIATIONS AND SYMBOLS Naval Architecture
F.P.
HANDBOOK OF
SHIP CALCULATIONS,
CONSTRUCTION AND OPERATION
Handbook of Ship Calculations, Construction
and Operation SECTION
I
WEIGHTS, MEASURES AND FORMULAE WEIGHTS AND MEASURES Troy Weight 24 grains
= =
1
12 ounces
pennyweight ounce
1 20 pwts. Used for weighing
and
gold, silver
=
1
pound
1
ounce
jewels.
Apothecaries' Weight 20 grains
= =
1 dram The ounce and pound
3 scruples
8 drams 12 ounces
1 scruple
in this are the
same
= =
as in
1
pound Troy weight.
Avoirdupois Weight 27.344 grains 16 drams 16 ounces
= = =
1
dram
1
ounce
1
pound
2000 pounds 2240 pounds
= =
Shipping Weight 16 ounces
28 pounds 4 quarters or 112 pounds 20 hundredweight or 1 2240 pounds j
= = = =
1
pound
1
quarter (qr).
1
hundredweight (cwt.)
,
t0n (T °
*
/rn
.
1
(lb.)
.
1 short 1
ton
long ton
WEIGHTS AND MEASURES
2
Measure
Shipping United States shipping ton
= =
100 cubic feet 40 cubic feet or o2.14 United States
British shipping ton
=
bushels or 31.16 Imperial bushels 42 cubic feet or 32.72 Imperial
1 register 1
1
ton
bushels or 33.75 United States
bushels
Linear Measure (Land)
= = 3 feet 5£ yards =
12 inches
1 1
yard rod
Other units are:
=
40 rods 8 furlongs or 5280 feet J
1 foot
1
furlong
1
mile (statute)
"1
4 inches
=
hand; 9 inches = 1 span; 1000 link: 100 links or 66 feet or 4 poles 1
= 1 inch; 7.92 inches 1 chain; 10 chains = 1 furlong. =
mils
=
1
Mariner's Measure 6080 feet = 1 nautical mile (knot) = 1 league 1 cable length 3 knots = = cable length 120 fathoms 960 spans = 720 feet = 219.457
6 feet 120 fathoms 1
= =
fathom
1
meters.
^V degree at meridian = .999326 U. S. nautical miles 1852 meters = 6076.10 ft. 1 U. S. nautical mile is the length of one minute of arc of a great 1
international nautical mile
=
=
circle of 1
U.
S.
a sphere whose surface equals that of the earth. Thus nautical mile = 1.15155 statute miles - 6080.20 ft. =
1853.25 meters. 1
1853.19 meters.
The knot .
144 square inches 9 square feet
30
=
British nautical mile
M square yards
= = =
generally adopted
is
Square Measure
1
square foot
1
square yard square rod
1
6080 feet = the one of 6080 feet.
=
1.15152 statute miles
40 square rods = 1 rood 4 roods = 1 acre 640 acres = 1 square mile
Time Measure 60 seconds = 1 minute 60 minutes = 1 hour = 28, 29, 30 or 31 days in computing interest) 365 days m 1 year
24 hours 7 days 1
calendar
month
366 days
= =
1 1
day week
(30 days
=
1 leap
=
1
month
year
MEASURES
3
Circular Measure 60 seconds 60 minutes
m =
1
minute
1
degree
90 degrees
=
1
quadrant
360 degrees
=
1
circumference
Instead of an angle being given in degrees it can be given in radians, one radian being equal to the arc of a circle whose length Thus if R denotes the radius, the circumference is the radius. of the circle
— ;
H X 2 ttR
2 x R, then the circular measure of 90'
similarly the circular
measure of 180°
is
x,
R
60°
-5-,
etc.
An
angle expressed in degrees may be reduced to circular measure by finding its ratio to 180° and multiplying the result by x.
—'——
11
Hence the
An grees
circular
measure of 115°
is
^i
angle expressed in circular measure
may
by multiplying by 180 and dividing by
The
= \^ 10
As
180 for x.
=
.63
lo
£ 10
X
180
=
x
be reduced to de-
x, or
by substituting
84°.
angle whose subtending arc
equal to the radius, or the 180° = 57.2958. unit of circular measure reduced to degrees is is
X
Therefore an angle expressed in circular measure to degrees
X
57.2958
by multiplying by 57.2958.
=
be reduced 2 2 Thus the angle - = o o
may
38.1972°.
Dry Measure 2 pints
=
1
quart
4 pecks
=
1
bushel
8 quarts
=
1
peck
36 bushels
=
1
chaldron
One United
States struck bushel contains 2150.42 cu. ins. or
law
dimensions are those of a cylinder 18}^ ins. diameter by 8 ins. high. A heaped bushel is equal to \% struck bushels the cone being 6 ins. high. A dry gallon contains 268.8 cu. ins. and is J^ of a struck bushel. One U. S. struck bushel x may be taken as approximately \ /i cu. ft., or 1 cu. ft. as f of a 1.244 cu.
ft.
By
its
bushel.
The
British bushel contains 2218.19 cu. ins. or
or 1.032 U. S. bushels.
1.2837 cu.
ft.
WEIGHTS AND MEASURES
4
Measure
Liquid
= = =
4 gills 2 pints 4 quarts
One United
A gallons
31 X
1
pint
1
quart
1
gallon
=
1 barrel
2 barrels or 63 gallons 1 hogshead
States gallon contains 231 cu. ins. or .134 cu.
=
ft.,
o
1 cu. ft. contains 7.481 gallons.
The
British Imperial gallon both liquid and dry contains 277.2' cu. ins. or .160 cu. ft., and is equivalent to the volume of 10 lb of pure water at 62° F. To convert British to U. S. liquid gallon
multiply
by
To
1.2.
convert U. S. into British divide by
1.2.
Metric System
The fundamental
unit of the metric system is the meter, thi unit of length, which is one ten-millionth of the distance from th pole to the equator or 39.3701 ins. From the meter the units o capacity (liter) and of weight (gram) are derived with subdivision of 10 or multiples of 10, the following prefixes being used: milli = 1
—r—
,
centi
for all
=
.
deci -TTji,
1000, myrie so on.
1
=
=
10000.
— 1
,
deca
=
hecto
10,
Thus a millimeter
is
y^™
=
100, kilo
of a meter,
=
am
The
units of meter, liter and gram are simply related, a = 1 liter and 1 liter o practical purposes 1 cubic decimeter
water weighs
1
kilogram at 4° C.
The metric system Belgium,
Bolivia,
is
Brazil,
specified
by law
Bulgaria,
Chile,
in Argentina,
Austria
Colombia, Denmark
Luxemburg Mexico, Montenegro, Norway, Peru, Portugal, Roumania, Servia Siam, Spain, Sweden, Turkey and Uruguay. Finland, France, Germany, Holland, Hungary, Italy, '
Linear Measure 10 millimeters
=
Decimal Equivalents of Fractions of an Inch, and Millimeter-Inch Conversion Table Fract.
WEIGHTS AND MEASURES
6
Volume and Capacity 10 10 10 10 10 10
A
milliliters
1 centiliter
centiliters deciliters
1 deciliter
liters
1
decaliters hectoliters
1
liter is
= = = = = =
1 liter
=
equal to the
decaliter hectoliter
1 kiloliter
.61
6.10 61.02 .353 3.53 35.31
volume occupied by
1
cubic cubic cubic cubic cubic cubic
inches inches inches feet feet feet
cubic decimeter of
water at 4° C.
Weight 10 10 10 10 10 10
milligrams centigrams
decigrams
grams decagrams hectograms
1000 kilograms
One gram
= = = = = = =
1 1
centigram decigram
1
gram
1
decagram hectogram
1 1 1
.154 1.54 15.43 154.3 .220
=
kilogram metric ton
:
= =
grains grains grains grains
pound avoirdupois 2.204 pound avoirdupois 2204.621 pound avoirdupois
the weight of 1 cu. cm. of pure distilled water at a temperature of 39.2° F., or 4° C; a kilogram is the weight of 1 liter (1 cubic decimeter) of water; a metric ton is the weight of 1 is
cubic meter of water.
Centimeter, Gram, Second, or Absolute System of Physical Unit of space or Unit of mass Unit of time Unit of velocity
Measurement distance = 1 centimeter = 1 gram
=
—
-¥ time
Unit of acceleration
= = =
1
second
1
centimeter in
1
second
change of 1 unit of velocity in 1 second Acceleration due to gravity at Paris = 981 centimeters in 1 second. 1 .0022046 m lb. Unit of force = 1 dyne = ^5- gramme = 981 981 .000002247 lb.
A
that force which acting on a mass of one gram during one second will give it a velocity of one centimeter per second. The weight of one gram in latitude 40° to 45° is about 980 dynes,
dyne
is
at the equator 973 dynes and at the poles 984 dynes. Taking due to in British measures the value of g, the acceleration gravity at 32.185 ft. per second at Paris, and the meter as 39.37 ins., then 1
32.185 X 12 - = 981 dynes. gram = .3937
METRIC CONVERSION TABLE Metric Conversion Table
R
WEIGHTS AND MEASURES
8
Unit of work Unit of power
= = =
=
lerg = 1 dyne-centimeter = .00000007373 1 watt = 10,000,000 ergs per second .7373 ft. lb. per second
W
=
h P '
756
*
=
-
00134 h P -
ft. lb.
*
(CGS) unit of magnetism = the quantity which attracts or repels an equal quantity at a distance of Centimeter, Gram, Second
one centimeter with a force of one dyne.
CGS
unit
of
current
electric
=
the
current which, flowing
through a length of one centimeter of wire, acts with a force of one dyne upon a unit of magnetism distant one centimeter from every point of the wire. The ampere, the commercial unit of current,
is
CGS
one-tenth of the
unit.
Board Measure
To
find the
number
multiply the length in
board measure by the breadth in
of feet feet,
in a stick of timber, feet,
by the thickness
in inches.
Example. 2
Find the board measure
of
a piece of timber 20
ft.
long, 2
ft.
wide by
ins. thick.
20
ft.
X
To convert board To convert board
2
ft.
X
2
ins.
=
80 feet board measure.
feet into cubic feet, divide the board feet by 12. feet into tons, divide the board feet by 12,
and multiply the quotient by the weight
timber per cubic Divide the weight in foot, thus giving the weight in pounds. pounds by 2240 to get it into long or shipping tons, or by 2000 to of the
get into short tons. Example. board.
What
A is
schooner has 1,000,000 feet board measure, of yellow pine on the weight of her load in shipping tons?
1,000,000 12
=
83f 333 cu
.
ft .
Yellow pine weighs 38
lb.
per cu.
ft.
38 3,166,654
lb.
=
1415 tons nearly.
Water cubic foot of fresh water weighs 62.42 lb. at its maximum density 39.1° F. One cubic foot of salt water weighs 64 lb. 35.88 cubic feet of fresh water weighs one ton (2240 lb.) 35 cubic feet of salt water weighs one ton One cubic foot of water (fresh or salt) = 7.48 gallons (U. S.) One gallon (U. S.) of fresh water weighs 8.33 lb. One gallon (U. S.) of salt water weighs 8.58 lb. One cubic foot of ice (fresh) weighs 56 lbs., specific gravity .9.
One
BOARD MEASURE
9
Feet Board Measure in Different Sizes of Timber* Length
in Feet
Size in Inches
2x4 2x6 2x8 2 x 10 2 x 12 2 x 14 2 x 16
2*x 12 2*x 14 2*x 16
3x6 3x8 3 x 10
3 x 12 3 x 14
3 x 16. 4 x
4
4 x
e
4 x
8
4 x 10
4 x 12 4 x 14
6x6 6x8 6 x 10 6 x 12 6 x 14 6 x 16
8 x
8
8 x 10 8 x 12
8 x 14 10 x 10 10 x 12
10 x 14 10 x 16 12 x 12 12 x 14 12 x 16 14 x 14
14 x 16
10
12
6? 10
10
12
13*
12
14
16
18
20
13* 16f
16
20
18! 23*
21* 26! 32
24 30
26! 33* 40
29* 36!
44
46* 53* 50
51* 58! 55
20 23J 26| 25 29* 33*
8
24 28 32 30 35 40
15
18
20 25 30 35 40
24 30 36 42 48
13§
16
24 32 26! 33* 40 48 40 46* 56 36 30 48 40 60 50 72 60 84 70 80 96 53* 64 66| 80 96 80 112 93* 83* 100 100 120 116! 140 133* 160 120 144 140 168 160 192 163* 196 186! 224 20
14
9*
28
16
18
36 42 48 45
20
22
24
26
28
30
32
H!
16
24 32 40
17* 26
18!
22
28
21* 32
34! 43* 52
37* 46! 56
20 30 40
60!
65* 74 70 81
48 56 64 60 46! 52* 58* 64* 70 46! 53* 60 66! 73* 80 24 21 30 33 27 36 28 32 36 40 44 48 35 40 45 50 55 60 42 48 60 54 66 72 49 56 63 70 77 84 56 64 80 72 88 96 26 s 29* 32 18! 21* 24 28 32 36 40 44 48 48 37* 42 53* 5S S 64 46! 53* 60 66! 73* 80 56 64 80 88 72 96 84 112 65* 74! 93* 102§ 42 48 54 60 66 72 56 64 80 72 88 96 70 80 90 100 110 120 84 96 108 120 132 144 98 112 126 140 154 168 112 128 144 160 176 192 74! 85* 96 106! 117* 128 93* 106! 120 133* 146! 160 112 128 144 160 176 192 130! 149* 168 186! 205* 224 116! 133* 150 166! 183* 200 140 160 180 200 220 240 163* 186! 210 233* 256! 280 186! 213* 240 266! 293* 320 168 192 216 240 264 288 196 224 252 280 308 336 224 256 288 320 352 384 228! 261* 294 326! 359* 392 261* 298! 336 373* 410! 448 32! 37* 35 40 5
37* 42! 40
69* 65 75§ 86!
50
60 70 80 75
42| 53* 64 74! 85* 80
87* 93* 93* 100 106§ 42 39 45 48 52 56 60 64 65 70 75 80 84 78 90 96 91 98 105 112 104 112 120 128 34 37* 40 42! 52 56 60 64 69* 74! 80 85* 86! 93* 100 106! 104 112 120 128 121* 130! 140 149* 78 84 90 96 104
130 156 182
208
112 140 168 196 224
120 150 180
210 240
138! 149* 160 173* 186! 200
208
224
242! 261* 216! 233* 260 280 303* 34C! 312 364 416
326! 373* 336 392 448
424! 457* 485* 522!
240 280 250 300 350 400 360 420 480 490 560
128 160 192 224 256 170!
213* 256 298f 266! 320
373* 426| 384 448 512 522! 597*
2 ins. X 4 ins. X 12 ft. long contains 8 ft. board measure. Board measure is often abbreviated B. M. * From Mechanical Engineer's Handbook. W. Kent.
Thus a
stick of timber
Inches and Fractions in Decimals of a Foot Parts of
Foot
in
Inches
and Fractions
FRESH WATER Fresh
W
11
WEIGHTS AND MEASURES
12
Weight and
Gallons of Fresh
Size of Different Standard
Water Cubic
Weight in
Gallons
of
a Gallon
Inches in a Gallon
Cubic Weight of a cubic Foot foot of fresh water,
in a
Pounds
English standard, 6.232102 7.480519
10.00 8.33111
277 .274 231.
Imperial or English United States
62 .321
lb.
avoirdupois.
Salt Water
The composition world, but
of salt water varies at different parts of the usually contains the following to every 100 parts:
Pure water
Common
96 2 2.71 54 .
salt
Sulphate of lime Sulphate of magnesium. Calcium bicarbonate Organic matter
08 ..
Magnesium chloride 01 Magnesium bromide About 5 ounces of solid matter are present in one gallon of water, and this density can be expressed as a fraction thus solid
matter
5 oz.
5 oz. 16 X 10
1 gal. water holding it in solution that is, one part in 32 of sea water is solid matter, If an American of 1 gal. lish gallon of 10 lb. is used.
5 oz. 16 X 8.33
,
=
.
12 01
33 salt
1_
if
=
32 an Eng8.33
lb.,
1
26.7
Salt water boils at a higher temperature than fresh owing to its greater density, as the boiling point of water is increased by any
substance that enters into combination with it. The property water has of holding chemical substances, as salts of lime in solufollows that tion, decreases as the temperature increases; from this
steam pressure form more scale than those working at lower temperatures and pressures. boilers carrying a high
density at 39.1° F. or 4° C. The boilsalt water ing point of fresh water at sea level is 212° F. and of Fresh water freezes at 32° F. or 0° C; salt water freezes 213.2. In freezing, water expands. Thus as at a lower temperature.
Water
is
at its
hot water cools
maximum
down from
the boiling point
maximum density,
it
contracts to 39.1°,
while below this temperature it expands again. The British and United States standard temperature for specific has the greatest specific gravity is pure water at 62° F. Water heat of any known substance except hydrogen, and is taken as
its
the standard for
all solids
and
liquids.
Specific Gravities and
Weights of Materials*
Material
Specific
Weight,
Gravity 1
per cu.
100%
Alcohol,
79
Alum
49 107
..
Aluminum, bronze Aluminum, cast Aluminum, sheet
7.7 2.55-2.75
Anthracite coal (broken)
Asphaltum
1.4-1.7 6.7 2.1-2.8 .62-. 65 1.1-1.5
Babbitt metal Barley Barytes Basalt Bauxite
4.5 2.7-3.2 2.55
Antimony Asbestos Ash, white-red
Beech Bell metal Benzine Birch
73-. 75
.
.
Bismuth Bituminous coal (broken)
9.'
Boxwood
74'
.96
8.4-8.7 1.8-2.0 7.4-8.9
Brass, cast-rolled Brick, common (1000 weigh about 3 J tons) Bronze, 7.9 to 14% tin
Camphor 32-. 38
Cedar, white-red Cement, Portland, loose
Chalk
1.8-2.6
Charcoal (piled) Cherry Chestnut Clay, dry Clay, moist Coal see anthracite and bituminous.
70 66
—
Coke Concrete, cement
cast, rolled
Copper Cork Corn
ore, pyrites
478 160 168 47-58 417 153 40 81 456 38 281 184 159 44 503 46 33
608 49 63 534 120 509 62 22 90 137 10-14 42 41 63 110
23-32
—stone—sand
Copper,
2.2-2.4 8.8-9.0 4.1-4.3
".
144 556 262
156
.25
.48
48 93 30
Dolomite Earth, dry loose Earth, packed and moist
2.9
181
Ebony Elm
1.25
1.47-1.50
Cotton, pressed Cypress
1
The
76 96 79 45
.72 specific gravities of solids
and
lb. ft.
liquids refer to water at 4
per cubic foot are derived from average specific gravities. * From Pocket Companion. Carnegie Steel Co.
1Q
(
C.
The weights
Specific Gravities and
Weights of Materials
Material
— Continued
Specific
Weight,
Gravity
per cu.
Emery 2.5-2.6
Felspar Fir,
.51
Douglas (Oregon pine)
Flagging Flax
1.47-1.50
,
Flour, loose Flour, pressed Flint.
.4-. 5 .7-. 8
Gasoline
.66-. 69 2.4-2.6 2.45-2.72 19.25 2.4 2.5 1.9-2.3
Glass, common Glass, plate or crown
Gold, cast, hammered Gneiss, serpentine Granite
Graphite Greenheart straw bales
Hemlock
42-. 52 74-. 84
Hickory Hornblende
3.
88-. 92
Ice
India rubber Iron, cast, pig Iron,
7.2 7.6-7.9
wrought
Ivory Kerosene Lance wood
.66
Lead
11.37 7.3 .86-1.02 1.10
Lead, ore, galena Leather
Lignum
vitae
Lime, quick, loose Limestone Linseed oil Locust
2.5 .73
'.
7.2-8.0 3.7
Manganese Manganese ore Mahogany, Honduras Mahogany, Spanish Maple Marble Mercury Mica .
.
.65
13.6
Muntz metal Nickel. Nitric acid
Oak,
live
.
.
251 159 32 168 93 28 47 164
42 156 161 1205 159 175 131
62 J 2.3
Gypsum Hay and
lb.
ft.
8.9-9.2
91% .
1.5 .95
.
14
159
20 29 49 187 56 58
450 485 114 42 42 710 465 59 83 53-60 165 58 46
475 259 35 53 49 170 849 183 511 565 94 59
Specific Gravities and
Weights of Materials
Material
Specific
Weight,
Gravity
per cu.
Oak, red, black Oats, bulk Oil
65
—see gasoline, petroleum,
Olive
— Continued
32
Oregon pine Paper
.
.51
.70 .87 .79
Petroleum, crude Petroleum, refined Phosphate rock Phosphor bronze Pine long leaf yellow Pine short leaf yellow Pine white
3.2
— — —
.70 .6 .41
1.07 21.1
Pitch.
hammered
Plumbago 48
Poplar Potatoes, piled Quartz, flint
2.5
Rubber, caoutchouc Rubber goods
.92 1.-2.
Rye Salt, granulated, piled
Saltpeter Sand, dry, loose
*
Sand, wet Sandstone
2.2
Shale, slate, piled Silver, cast, hammered Soapstone, talc
10.4 2.6
Spruce, white, black Starch Steel, cast Steel, structural
Sulphur Talc Tallow Tar, bituminous
Teak Tin, cast, Tin ore
.4
1.53 .
.
7.8 1.93 2.6
.82
hammered
Walnut, black Water, fresh Water, salt
7.2 6.4-7.0 .61 1.
1.02
Wheat White metal, Babbitt Wool, pressed Zinc, cast, rolled
Zinc ore, blende
41
etc.
oil
Platinum, cast,
lb. ft.
1.32 6.9 3.9
57 32 58 54 50 200 537 44 38 26 69 1330 140 30 42 165 59 94 48 48 67 90-105 120 147 92 656 169 27 96 493 . 490 125 169 59 75 52 459 418 38 62.5 64 48 456 82 440 253
Cubic Feet per Ton (2240 Le.) op Different Materials* Cu.
Material
Alcohol in casks Almonds in bags Almonds in hogsheads. Aniseed in bags Apples in boxes Arrowroot in bags Arrowroot in boxes .
.
.
.
Arrowroot in cases Asbestos in cases Asphalt
Bacon in Bananas
cases
Barley in Barley in
bags bulk
Beans, haricot, in bags Beans in bulk
.
.
Beef, frozen, packed ....
Beef hung in quarters
.
.
.
Beer, bottled, in cases.
.
.
Beer in hogsheads
Beeswax Bone meal Bones, crushed Bones, loose
:
Books Borate of lime
Borax
in cases
empty, in crates Bran compressed in bales Bran in bags Brandy, bottled, in cases Bottles,
Brandy
in casks
Bread in bulk Bread in cases Bricks
Buckwheat
in bags Butter in kegs or cases
.
.
in cases Candles in boxes Canvas in bales Carpets in rolls Cassia in cases
Camphor
Cellulose
Cement
in barrels
Chalk in barrels Cheese Chicory in sacks Chloride of lime in casks *
From The Naval
ft.
Constructor.
Cu.
Material
per ton
Cider in casks Cigars in cases Cinchona (Peruvian bark) Cloth goods in cases .... Cloves in cases Coal (Admiralty) Coal (American) Coal (Newcastle) Coal (Welsh) Cocoa in bags Cocoanuts in bulk Coffee in bags Coir yarn in bales
80 70 108 120 90 52 70 50 53 17 65 90 59 47 68 47 93 125 80 54 74 45 60 85 50 50 52 85 80 110 55 80 124 155 22 65 70 50 56 43 80 184 240 40 38 70 60 80
ft.
per ton
65 180 140
87 50 48 ,43
45 40 80 140 61 190
Coke
80
Copper, cast
10
10-20 Copper ore 50 Copper sulphate in casks 52 Copperas in casks 85 Copra in cases 270 Cork wood in bales of S. bale U. a Cotton cotton is 54 ins. by 27 by
—
24 to 30
ins.
high de-
pending on the compression, assuming 30 ins.
is
space occupied
25.3
cu.
Average stowage per ton Cotton waste Cowrie shells in bags. Creosote in casks Dates ft.
.
.
.
.
.
Earth, loose
Earthenware
in crates.
Fish in boxes Fish, frozen Flax Flour in bags Flour in barrels Freestone
Fuel oil Furs in cases Ginger Glass bottles
Glassware in crates Granite blocks Gravel, coarse
Grease
G. Simpson.
16
114 170 75 60 43 25 47 95 .60 105 47 60 16
39-40 130 80 85 180 16
23 65
Cubic Feet per Ton (2240 Lb.) of Different Materials Cu.
Material
Guano
Gunny bags Gunpowder
160 70 120 140 100 70 60 85 120 50
Hair, pressed in barrels
Ham
Hay, compressed Hay, uncompressed ....
Hemp Hemp
in bales seed in bags .... Herrings in barrels .... Herrings in boxes Hides in bales Hides in barrels Hops in bales '.
260 39 72 67 36
Ice
India rubber, crude .... Indigo in cases Iron, corrugated sheets Iron, pig
.
Lead, pig
Lead
pipes,
random
Linseed in bags Locust beams in bulk.
.
.
Logwood Manure phosphate ....
—
Maize in bags Maize in bulk Marble in slabs Margarine in tubs Marl Matches Melons
Mineral water in cases Molasses in bulk Molasses in puncheons. .
Mutton
90 220 85 57 84 92 45 51 49 17 69 28 120
Milk, condensed, in cases Millet in bags .
.
.
.
in boxes Pitch in barrels Potatoes in bags Potatoes in barrels Prunes in casks Raisins Rape seed Rice in bags
12
Lemons
,
Oil, lubricating, in bbls. Oil in drums Oil in bottles in cases Oil cake in bags Olives in barrels Onions in boxes Oranges in boxes Oysters in barrels Paint in drums Paper in rolls Peas in bags Phosphate of lime
Pineapples, canned, and
sizes
about Leather in bales Leather in rolls
Oats in bags Oats in bulk
10
40 70 8
'.
Nails, kegs Nitrate of soda Nuts, Brazil, in barrels Nuts, pistachio, in cases Oatmeal in sacks
,
80 45 50 70 25 65 110
Rice meal
Rope
Rum in bottles and cases Rum in hogsheads Rye
in
Salt in bulk Salt in barrels
Saltpeter
61
60 49 75 50 67 77 90 60 16 120 50 42
60 45 55 68 52 52 60 48 62 135 66 70
19
Sand, fine Sand, coarse Sandstone
20 14
83 125 112
Shellac
13
.
Sponge
21
32 90 70 65 78
53 55 37 52 36
bags
Sago
Silk in bales Silk in cases Slate Soap in boxes Soda in bags. Soda in casks
ft.
per ton
28 58 (China clay) in
Lard
Cu.
Material
42 50 50 48
Gum
Ivory Jute Kaolin bags
ft.
per ton
— Cont.
.
.
.'
46 57 54 152
Cubic Feet per Ton (2240 Lb), of Different Materials Cu.
Material
barrels
in
40 54 60 27 40 60 70 34
and
tierces
Tallow in hogsheads. Tamarinds in cases Tamarinds in casks or .
.
.
58 70 45 54 48 57 54 100
in barrels
Tea, China, in chests Tea, Indian, in cases. .
Ties,
.
.
.
.
.
oak
(Rough Lumber in Ash, black Ash, white
Basswood Beech Birch Butternut
Cherry Chestnut
Cottonwood Douglas fir Elm, rock Elm, soft red
lb.
40 150 80 60 110 36 35 52 48 39
39
Beech
51
Fir
Greenheart
Mahogany Wool in sheets Wool in bales,
per 1,000
ft.
pressed.
Lumber
board measure)
3,000 3,000 4,500 3,000 3,500 3,000 3,900 3,900 2,800 4,200 3,000 2,800 4,000
Hemlock Hickory..
Long leaf pine
Mahogany Maple, soft Maple, hard
Oak Poplar, yellow Shortleaf pine
Sycamore Tupelo Walnut
Weight of Green Logs per 1,000
ft.
.
60 65 34 34 260 100
'
Gum, sap
3,200 3,500 2,500 4,000 4,000 2,500 3,800 2,800 2,800 3,300 3,800 3,000 3,300
38 85
Ash
Shipping Weights of American
Gum,
.
Elm
extract
Tapioca
Tar
small bales
Turmeric Turpentine in barrels Vermic. Hi Water, fresh Water, salt Wheat in bags Wheat in bulk Whitening in casks Woods, sawn into planks
.
kegs
Tan
.
15 13
ft.
per ton
Ties, steel Tiles, roofing, in crates. Tobacco, Brazilian, in bales in Tobacco, Turkish,
80
Sulphur in cases Sulphur in kegs Sumac in bags
Cu.
Material
100 50
Starch in cases ......... Stone, paving Stone, limestone Sugar in bags Sugar in hogsheads Sugar in casks Sulphur in bulk
Syrup Tallow
ft.
per ton
— Cont.
board measure
Yellow pine (Southern)
8,000 to 10,000
lb.
Norway pine (Michigan) Hemlock (Pennsylvania), bark
7,000 to
8,000
lb.
6,000 to
7,000
lb.
off
18
WEIGHTS
19
Weights of Miscellaneous Units of Different Products of nails Firkin of butter Chest of tea Barrel of flour, etc. See Sizes of Barrels. Bushel of oysters Bushel of clams Bushel of barley Bushel of beans Bushel of buckwheat Bushel of charcoal Bushel of castor beans Bushel of clover seed Bushel of corn (shelled) Bushel of corn (on cob)
Keg
—
Lb. 100 56 68
80 100 48 60 48 30 50 60 56 70 34 57 32 60 56 45 60
Bushel of malt Bushel of onions Bushel of oats Bushel of potatoes Bushel of rye Bushel of Timothy seed Bushel of wheat 480 Quarter or 8 bushels of wheat Gallon of m3lasses 12 Bale of United States cotton weighs 500 Bale of Peruvian cotton weighs 200 Bale of Brazilian cotton weighs 250 Bale of East Indian cotton weighs 400 Bale of Egyptian cotton weighs 750 Bale of jute weighs 440 One bushel of wheat = 60 lb. = 1.244 cu. ft. Eight bushels of wheat = one quarter = 9.952 cu. ft. = 480 lb. = 4% quarters = 46.43 cu. ft. = 2240 lb. One ton of wheat A case of kerosene oil generally contains two 5-gallon cans or ten 1 -gallon, in the former taking up 2 cu. ft. and in the latter 2.1. Some hold fifteen 1-gallon cans and take up 3.2 -
cu.
ft.
Gallon of honey Gallon of crude oil about 7 bags of sugar (one ton) 1 1 bags of potatoes (one ton)
One bag
83^ 2240 2240 140 4369 2862
of flour
Cord of dry hickory Cord of dry maple Linoleum \i of an inch per sq.
12 :
thick, including cement, weighs 1.5 lb.
ft.
Rubber tiling, fs of an mcn thick, weighs 2 lb. per sq. ft. White tiling, ye of an inch thick, weighs 6 lb. per sq. ft.
WEIGHTS AND MEASURES
20
Bundling Schedule for Buttweld Pipe 1 This schedule applies to buttweld wrought iron pipe only.
Standard Weight Pipe
m.
No.
Size
of Pieces
per Bundle
H H V% A H
App ox/ No /
'
A PP™ X
^BuSdlT
42(Approx.)500 24 450 18 340 12 245
'•
X
1
7 5
140 100
1M
3
1^2..
3
60 58
-
Weight "*
Lb 120 190 190
210 160 168 138 158
Extra Strong Pipe
42 24
Vs
% V% A
18 12 7 5 3 3
l
M1 l l
A A l
l
500 450 330 245 140 100 60 58
157 241 244 266 206 217 180 211
Double Extra Strong Pipe
V K
7
2
5 3 3 3
1
1% 1^2 1
Adopted on June
1st,
126 95
60 60 60
1915, at the suggestion of the National Pipe
215 230 220 310 380 and Supplies
Association.
Barrels There is no standard size of barrel universally adopted either by Great Britain or the United States. In Great Britain an old wine barrel = 26 }/i imperial gallons, an ale barrel = 31J^ imperial = 36^2 imperial gallons. A French gallons and a beer barrel barrique of Bordeaux — 228 liters = 50 imperial gallons. Four barriques
=
1
tonneau.
BARRELS
21
A
barrel for fruit, vegetables and other dry commodities as fixed by a United States statute approved March 4, 1914, specifies ins. long, heads 17% ins. dia., distance between heads staves 28
%
26
circumference 64
outside measurements, representing as nearly as possible 7050 cu. ins. or 4.08 cu. ft., equivalent Besides the above the different states specify to 105 dry quarts. dimensions of barrels for various commodities. the The usual barrel for liquids contains 31% U. S. wine gallons of 231 cu. ins. ins.,
Below
is
a table of
Material Held
wood
ins.,
all
barrels.
Diameter Top and
Diameter at Bilge
Height
Bottom (in3.)
(ins.)
(ins.)
H
21% 19% 25%
28%
Fish....
20
Meat.
Molasses
21% 22%
22% 25% 27%
30 33 35
Salt ....
18% 16 16
21 18
30
Cement Lime
Sugar
.
.
Flour. Oil
.
21
.
.
.
.
Apple
19% 17%
.
.
.
Potato
.
.
.
.
17%
.
.
15
Tar
19%
All dimensions are outside.
The above
18 19
16% 21%
Cubic Feet
30
5.60 4.36 8.37 52 gals. 6.23 8.37 10.04 60 gals. 5.34 3.75 3.75 4.33 3.22 5.60
33
28% 28% 28% 28% 30
barrels are of wood, data from G. A.
Rieley, Cleveland, O.
An
oil
company
(Piatt
& Washburn
Ref'g Co.,
the following figures on the sizes of their
drums
:
Diam. Material
Wood Wood Drum
half barrel (steel)
Half-drum
.
.
.
wood
New
York) gave
barrels
and
steel
WEIGHTS AND MEASURES
22
Horse Powers Horse Power
the unit of power equivalent to raising a one foot in one minute.
(h. p.),
weight of 33,000
lb.
of water evaporated per and at 212° F.
2.64
hour from
watts
746
One
lb.
.746 kw.
horse power
ft. lb.
33,000
550
ft. lb.
per minute per second
heat units per hour heat units per minute
2,545 42.4
Indicated Horse Power (i. h. p.) is the power as measured by an indicator and calculated by the following formula: P = mean effective pressure in pounds per sq. in. on the piston as obtained from the indicator card
L =
A =
N
length of stroke in ft. area of piston in sq. ins.
= number
of single strokes per
minute or two times the num-
ber of revolutions
Then
indicated horse power
(i.
h. p.)
=
QQ nnn oo ,UUU
Brake Horse Power (b. h. p.) is the actual horse power of an engine as measured at the flywheel by a friction brake or dynamometer. It is the indicated horse power minus the friction of the engine.
—See Boilers. —Lloyd's formulae are as follows: Nominal Horse Power
Boiler Horse Power.
(n. h. p.).
D = diameter of p. cylinder in ins. = stroke in ins. s H = heating surface in sq. ft. P = working pressure in pounds per sq. N — number of cylinders 1.
(1)
N
Where the h p n
= ^
~^~
^®
2
/ £> V 100
.
"*"
1000 £* 2 s 590 1500 ( 100 boilers are fitted with forced
P +
If
and heating surface are known s/s H\ where boiler pressure is be-
boiler pressure
'
*
in.
substituted for H/15.
V ^ i
15/
low 160
lb.
zL | where boiler pressure above 160 lb. 15/ or induced draft then H/12
is
is
HORSE POWER Equivalent Values of Mechanical and Electrical Units Unit
23
WEIGHTS AND MEASURES
24 Effective
Horse Power
Thrust Horse Power
(t.
(e.
See Powering Vessels.
h. p.)
h. p.) is the
power delivered by the pro-
peller for the propulsion of the ship. Owing to the friction of the of the and working parts engine shafting, the horse power trans-
mitted to the propeller used by the propeller
Thrust in
lb.
=
=
33000 dist.
is
about J^ of the indicated.
thrust in
lb.
X
dist.
ship travels in ft. in
33000
X
used by the propeller ship travels in ft. in 1 min. h. p.
Comparison of Thermometer Scales Cent.
Horse power 1
min.
THERMOMETERS
25
Thermometers Fahrenheit (F.) thermometer is used in the United States and The freezing point of water is marked 32 and in Great Britain. the boiling at sea level 212, the distance between these points is divided into 180 parts or degrees. 32 parts are marked off from the freezing point downwards, and the last one marked or zero.
used extensively in Europe and in scientific The freezing point of water is marked 0, and the calculations. at sea level 100, and the distance between is divided boiling point into 100 parts or degrees. To convert Fahrenheit readings into Centigrade, subtract 32 To convert Centigrade into Fahrenheit and multiply by Centigrade (C.)
is
multiply by | and add 32. Reaumur (R.) is used in Russia.
taken as
0,
readings into To convert
and the
Reaumur Reaumur
The
boiling point 80.
freezing point of water
To
is
convert Fahrenheit
subtract 32 and multiply by into Fahrenheit multiply
by
f.
f
and add
32.
the temperature be below freezing, "add 32" in the formula becomes " subtract from 32" and "subtract 32" becomes "subIf
tract
from 32."
See table on page 24.
Circumferences and Areas of Circle Advancing by Eighths Diameter
WEIGHTS AND MEASURES
26
MATHEMATICAL TABLES Involution and Evolution
The quantity represented by the letter a multiplied by a quantity represented by the letter b, is expressed a X b or ab. Quantities in brackets thus (a
+
6)
(a
+
b) signify
be multiplied together. To square a number multiply the number by 2 square of 4 (often written 4 ) is 4 X 4 = 16.
To cube a number cube of 4 (written 4
To
3 )
itself.
they are to
Thus
multiply the square by the number. = 4 X 4 X 4 = 16 X 4 = 64.
the
Thus
power of a number, multiply the cube by the number. Fourth power of 4 = 64 X 4 = 256. The nth power of a number as an is obtained by multiplying the logarithm of the number by n and then finding the number 18 corresponding to the logarithm. Thus 5 = log. of 5 X 1.8, and from the table of logarithms find the number corresponding to this find the fourth
logarithm. V~ is the radical sign and either with or without the index figure 2 as indicates that the square root of the quantity under it is
v
Thus the
to be taken. to be taken as
VI
is
2.
\/
indicates the cube root
is
\J~ that the fourth
root as \/256 is 4. root is the square root of the square root, and the sixth the cube root of the square root.
\/8
is
2.
The fourth root
is
Any
logarithm of
number
^/a may be obtained by taking the the number a and dividing it by the index n and from
root of a
as
the table of logarithms finding the corresponding number. To Extract the Square Root of a Number. Point off the given number into periods of two places each beginning with units. If
—
there are decimals, point these off likewise beginning at the decimal point, and supplying as many ciphers as may be requirecj. Find the greatest number whose square is less than the first left-
hand
and place it as the first figure in the quotient. Subtract its square from the left-hand period, and to the remainder annex the two figures of the second period for a dividend. Double the first figure of the quotient for a part ial__di visor. Find period,
how many
times the latter
of the right-hand figure,
is
and
contained in the dividend exclusive
set the figure representing that ber of times as the second figure in the quotient and annex
numit
to
the right of the partial divisor, forming the complete divisor. Mul-
SQUARE AND CUBE ROOT tiply this divisor
by the second
27
figure in the quotient
and subtract
the product from the dividend. To the remainder bring down the next period and proceed as before, in each case doubling the Should figures in the root already found to obtain the trial divisor. the product of the second figure in the root by the completed divisor be greater than the dividend, erase the second figure both from the quotient and from the divisor, and substitute the next smaller
one small enough to make the second figure by the divisor less than or equal to the dividend. Find the square root of 3.141592 figure or
3.141592
1.772
|
+
square root
1
27
214 189
347
2515 2429 3542 8692 7084
To Extract the Cube Root.
—Point
off
the
number hand or
into periods
units' place. of three figures each, beginning at the right Point off decimals in periods of three figures from the decimal point.
Find the greatest cube that does not exceed the left-hand period, write its root as the first figure in the required root. Subtract the cube from the left-hand period, and to the remainder bring down the next period for a dividend. Square the first figure of the root, multiply by 300, and divide the product into the dividend for a trial divisor, write the quotient after the first figure of the root as a trial second figure. Complete the divisor by adding to 300 times the square of the first figure, 30 times the product of the first by the second figure and the square of the second figure. Multiply this divisor by the second figure, and subtract the product from the remainder. Should the product be greater than the remainder the last figure of the root
and the complete divisor are too large; substitute for the last figure the next smaller number and correct the trial divisor accordingly. To the remainder bring down the next period, and proceed as before to find the third figure of the root; that is, square the two figures of the root already found, multiply visor, etc.
If
the trial divisor
another period of three figures, as before.
by 300
for a trial di-
than the dividend bring down and place in the root and proceed
is less
WEIGHTS AND MEASURES
28
The cube
root of a
number
will contain as
many
figures as there
are periods of three in the number. Find the cube root of 1,881,365
1,881,365
|
123.
+
cube root
Squares, Cubes, Square Roots, Cube Roots, Logarithms, Circumferences and Circular Areas of Nos. from 1 to 50
Squares, Cubes, Square Roots, Cube Roots, Logarithms, Circumferences and Circular Areas of Nos. from 51 to 100
GEOMETRICAL PROPOSITIONS number
31
multiplied or divided by any integral power of 10, producing another number with the same sequence of figures, the mantissae of their logarithms will be equal. To find the logarithm of a number take from the table the mantissa corresponding to its If
a
is
sequence of figures, and the characteristic rule given above.
Thus
if
30.53
log. log.
log.
3.053
log. of
305.3 3053.
= =
=
may
be prefixed by the
.484727
1.484727
log. .3053
2.484727
log.
.03053
3.484727
log.
.003053
= =
9.484727 8.484727
-10
7.484727
-10
- 10
only enjoyed by the common or Brigg's logarithms and constitutes their superiority over other systems of
The above property
is
logarithms.
Geometrical Propositions
The sum
of the angles of a triangle
is
equal to 180°.
a triangle is equilateral it is equiangular. In a right-angled triangle the square on the hypothenuse to the sum of the squares of the other two sides. If
A straight
is
equal
from the vertex of an isosceles triangle perpendicular to the base bisects the base and the vertical angle. A circle can be drawn through any three points not in the same line
straight fine.
a triangle is inscribed in a semicircle, it is right-angled. In a quadrilateral the sum of the interior angles is equal to four If
right angles or 360°.
In a parallelogram the opposite sides are equal, as also the opposite
A
angles are equal.
is bisected by its diagonals, which in turn are each other. by If the sides of a polygon are produced, then the sum of the exterior angles is equal to four right angles. The areas of two circles are to each other as the squares of their
parallelogram
bisected
radii.
a radius
perpendicular to a chord, the arc subtended by the chord. If
From
it
bisects the chord
The tangents
drawn are
so
equal. to meets it at one point, a circle tangent perpendicular to the radius drawn to that point.
A
and
a point without a circle only two tangents can be drawn
to the circle.
is
is
straight fine
.
and
it
WEIGHTS AND MEASURES
32
an angle is formed by a tangent and a chord, it is measured one-half of the arc intercepted by the chord. by If an angle at the circumference of a circle between two chords If
subtended by the same arc as an angle at the center between two radii, the angle at the circumference is equal to half the angle is
at the center.
Properties of Circles and Ellipses Circle. is
—The ratio of the circumference of a
circle to its
diameter
3.141592 and is represented by the symbol -k (called Pi) Circumference of a circle = diameter X 3.14159
Diameter of
circle
Circumference of Circumference of
circle
X X X
.88623
.28209 1.1284
=
side ° f eqUal SqUare
/
=
perimeter of equal square Side of square of equal periphery as circle = diameter X .7854 Diameter of circle circumscribed about square = side X 1.4142 Side of square inscribed in circle = diameter X .70711
To
circle
multiply the diameter of the circle by the number of degrees in the arc and this product by .0087266. Or let C represent the length of the chord of the arc and c the length of the chord of half the arc, then the length c
ol
.
find the length of
circle,
—— C
c>c
=
,
the arc
an arc of a
-
o
n Chord .
-.
f
,
u
of the arc
=
o 2
v, X
r radius
ks
X
•
sin
—
an S le
in degrees
r
Rise (the perpendicular di stance from the center of t he chord 4 radius 2 — length of chord 2 to the arc) = radius —
KV
=
o 2
s>
X
aradius
v X
•
sin2
an & —
le in
degrees
^
For areas of segments and sectors see Areas. it
4
=
3.1415926
=
0.497149
log.
=
T.895090
log.
=
T.5028501
log.
=2.2418774
log.
-
log.
1.7581226
AREAS Let
Ellipse.
D = d
major
axis
= minor
axis
Approximate circumference
=
= D X d X = ^r-^ X
Area
33
——
y
3.1416 .78539
3.14159
Areas of Plane Figures and Surfaces of Solids Plane Figures.
= base X 14 altitude = Vs (s — a) (s — b)
Triangle
Parallelogram
Trapezoid
Trapezium
= = =
—
(s
of the three sides a, base altitude
where b and c c)
s
=
J/£
sum
X
altitude
X
\i the
divide into two
sum
of the parallel sides triangles and find area of the
triangles
Circle
Sector of circle
= =
diameter 2
X
.7854
length of arc
_ xX
radius'
=
ic
X
radius 2
X A the radius X angle in degrees = X
radius 2
X
—=-
/3.1416
^^ ^
angle in degrees line forming the segment cuts the Segment of circle. circle, draw lines to the center forming a sector and a center angle A.
—Where the
m ihen area ot
-
segment
radius 2
=
\
XA
in degrees
=
side same area as square: diameter = side diameter Square of same area as circle:
Circle of
~
"180-
.
Sln
A
A
)
X X
1.12838 .88623
=
long diameter X short diameter X .7854 Ellipse Parabola = base X perpendicular height.
%
Regular polygon = sum of its sides X perpendicular from its center to one of its sides divided by 2. Or multiply 3^ the perimeter by the perpendicular from the center to a side. Irregular polygon: draw diagonals dividing it into triangles, find the sum of the areas of the triangles.
Trapezoidal Rule.
ABCD
and
—To
(see Fig. 1),
find the area of a curvilinear figure, as divide the base into any number of con-
venient equal parts, and erect perpendiculars meeting the curve. To the half sum of the first and last perpendiculars add the sum of all the intermediate ones; then the sum multiplied by the common interval will give the area.
WEIGHTS AND MEASURES
34
^
*»
-4^
-1-/9 Figure
=
Let h 2/i,
the
common
y2, etc.,
Then the area
>> /?
*,
\- 6
-\
1
interval
lengths of the perpendiculars to the line
ABC D
=
'V\ 1
h
AD
+ Vi + 2/2+?/3 + 2/4+?/5 +
?/6
)
—
Simpson's First Rule. This rule assumes that the curved line B C forming one side of the curvilinear area A B C D (see Fig. 2) is a portion of a curve known as a parabola of the second order whose equation is y = ax 2 + bx + c.
Figure 2
Divide the base into any convenient even number of parts, and erect perpendiculars to meet the curve. To the sum of the end or ordinates add four times the even numbered perpendiculars ordinates and twice the odd numbered ordinates. Multiply the sum by one-third the common interval and the product will be the area.
Thus the area
ABCD
Or the area
A B C D in Fig.
+ 4^/6+
2/7)
of
in Fig.
2
1
=
-|"
o
=
y
(2/1+
(2/1+
^2/2
+
2/3)
4y2 + 2y s +
4z/ 4
+ 2ij
b
SIMPSON'S SECOND RULE
35
found that areas given by the above approximate rule for curvilinear figures are very accurate, and the rule is extensively It is
used in ship calculations. Example. The ordinates to a curve are 1.5, 3.1, 5.2, 6.0, 6.5, 7.0, 8.1, 8.5 9.0 ft., the common interval is 3 ft. Find the area.
,:iJ
WEIGHTS AND MEASURES
36 Simpson's
first
rule
used more than the second as
is
it is
simpler quite as accurate. Surface of Solids. Lateral surface of a right or oblique prism or cylinder = perimeter of the base To get the lateral length.
and
is
—
X
add the areas
total surface
of the bases to the lateral surface.
Pyramid or cone, right and regular, lateral surface = perimeter of base To get total surface add area of base. Ji slant height.
X
Frustum
pyramid or cone, right and regular parallel ends, = (sum of perimeters of base and top) X Yi slant To get total surface add areas of the bases to the lateral of
lateral surface
height. surface.
Surface of a sphere = 4 ir radius 2 = -k diameter2 Surface of spherical sector = J^ -k r (4 b See Fig. 4. c). 2 = =» Surface of a spherical segment 2 ir r b c2 ) }/i it (4 b .
+
+
.
See
Fig. 5.
Surface of a spherical zone = 2 -k r Surface of a circular ring = 4 x2 R
b. r.
See Fig. 6. See Fig. 7.
Surface of a regular polyhedron (a solid whose sides are equal regular polygons) = area of one of the faces X the number of faces.
Volumes of Solids
—
Prism, right or oblique regular or irregular. Volume = area of section perpendicular to the sides X the lateral length of a side.
—
Volume = Cylinder, right or oblique, circular or elliptic, etc. the lateral length of area of section perpendicular to the sides
X
a side.
—
any prism or cylinder. Volume = area of base X perpendicular distance from base to center of gravity of opposite
Frustum
of
face.
right or oblique, regular or irregular.
Pyramid or cone,
=
area of base
A X
the perpendicular height.
—
any pyramid or cone, parallel ends. Volume = the areas of base and top plus the square root of their
Frustum (sum of
X
—Volume
products)
of
X
A X
the perpendicular height.
Wedge, parallelogram perpendicular height
Sphere.
—Volume
X
face.
—Volume
= sum
of three edges
perpendicular width.
=}ir
(radius)
3
or (diameter) 3
X
.5236.
X
VOLUMES OF SOLIDS
37
Volume =
%
it
r2 b
Volume =
y
it
b 2 (3 r
Figure 4
Spherical Sector
z
-
6)
Figure 5
Spherical Segment
>Q
c
Volume =& x
b (3a 2
*|
Figure 6
Spherical Zone
*
r
—*—
/?
•A>
D
3 Figure 7
Circular Ring 3
Volume = 2
7T
2
22 r*
+ 3^+
46 2 )
WEIGHTS AND MEASURES
38
=
}4
w
r
a b
r2
h
Figure 8
Ellipsoid
Volume = J^ x
Figure 9
Paraboloid
—
Regular polyhedron. Volume = area of its surface X \i the perpendicular from the center to one of the faces. The volume of any irregular prismatic solid may be obtained by dividing it into prisms or other bodies whose contents can be calculated by the above formulae. The sum of the contents of these bodies will give the total volume of the solids. To find the volume of a solid bounded by a curved surface, as the underwater portion of a ship's hull, divide the solid by a series of planes or sections spaced an equal distance apart. The area of each section can be calculated by either the trapezoidal or Simpson's rule, or by means of an instrument called a planimeter. The areas of the sections can be laid off on ordinates which are spaced the same distance apart as the sections which
the body was divided into. A curve is drawn through the points laid off on the ordinates, and the area of the curvilinear figure is the volume of the solid. Example. The areas of cross sections of a ship below the load water apart.
Find the volume in cubic
feet,
line are 1.2,
and 6.6 sq. ft. The sections are 9.5 ft. and the displacement in tons of salt water.
17.6, 41.6, 90.7, 134.3, 115.4. 61.7, 30.4
TRIGONOMETRY Number of Section
39
WEIGHTS AND MEASURES
40
As the sum
of the three angles of a triangle is equal to 180°, any the angle supplement of the other two. Trigonometric Functions. In the right triangle (Fig. 10) if A is
is
—
one of the acute angles, a the opposite c the hypotenuse,
and
sine of angle
side, b
the adjacent side
OBLIQUE TRIANGLES First
Second
Third
Fourth
-f-
— —
—
+ + +
Sine and cosecant
Cosine and secant
41
— —
+
— Tangent and cotangent -f1 The symbol sin" ^ means the angle whose sine is x, and is read inverse sine of x and anti sine of x (also arc sine x). Similarly -1 While the direct functions sine, cos, etc., are cos £, tan~\r, etc. single valued, the indirect are 30° or 150°. If
an acute angle and one
many, thus side or
if
sin 30°
two
=
.5,
but sin
_1
may
A +B =
is
=
sides of a right triangle
are given the other elements can be determined. Let A and acute angles (see Fig. 10), a and b the sides opposite them.
acute angles are complementary, that
.5
B
be
The
Five cases
90°.
be distinguished.
Given
and
c
"
a
"
b
"
a
"
a
A A A
then a b
a
= = =
c sin
A, b a cot A, c b tan A, c
c
A =
sin
b
A =
tan
—
-1
c cos
B
a cosec b sec
A
A
b
=V (c-J-a)
c
= Va2 -f ¥
/
,
-1
—
= = =
y,
(c—a)
Solution of Oblique Triangles. If any three of the six elements (three angles and three sides) of a triangle are known, the remaining three can be found, provided one of the given three is a side.
There are four cases as follows: Case 1. Given one side and two angles. The third angle equals 180° minus the sum of the two given. « sin B a sin C ,i t T.jr If the given side be a then b = —. j- and c = —. r ,
—A
.
,
—A
,
sin
sin
Case
Given two
2.
Then
sides (a
A = B «=
c
V2 V 2
and
(A (A
—A
b)
and the included angle C.
+ B) + Y2 (A - B) +B) -y2 (A-B)
a sin C —. rsin
Given two
Case 3. of them. sin
B =
—
sin
obtuse unless sin ing these
A
sides a
giving
B>1
two values
I>i
and
b,
and the angle
two values
of B,
A
opposite one
one acute and one
which case the data are impossible. and B 2 then
in
,
Call-
WEIGHTS AND MEASURES
42
corresponding to
=
180
—
(A
+
B)i and
Ci
B C =
180
-
(A
+
B) 2
d =
Bh 2,
That
is,
d 2
there are two solutions unless
=
C R = Rai - W (ai - a)
Wi
M max.
+ W =Rot >R Mm&x. PFi + W = Ri or > Ri
M at W
2
IV 9.
Beam
W/
if TFi if
2
= «a 2 -
W (a
2
-
TTi (a 2
w.
M max.
Supported at Ends.
if
TF2
=
Ri or
>
-
a)
—
ai)
i?i
—Uniformly distributed load. R
S770X
(max shear)
M,
=
distance x
M max. at center W max. D max.
R\
w= —
rV-f
STRENGTH OF MATERIALS
82
Beam
10.
Supported at Ends.
—Load increasing uniformly to one
end.
W
R
3
2W
R\ (max. shear)
M,
3
Wx
distance x
3
H)
M max. distance ZV3
Wl
2
9V3" 27
W max. D
Beam
11.
Supported at Ends.
.013044
max.
—Load
fs
2HlT
WP
EI
decreasing uniformly to
center.
R
(max. shear)
M,
distance x
=
W
Ri
2
=
Wx
+
2
3^
z
M max. distance £
12
fs
12
TF max.
I
D
12.
Beam
Supported at Ends.
ZWP
max.
—Load
320
E
I
increasing uniformly to
center.
R
(max. shear)
M,
distance x
= =
W
Ri
Wx
2 |
-
-
2jrA
2 3/y (1_
M max. distance i
TF£ 6
W max.
6/s
D
WEI
max.
/
BEAM SUPPORTED AT ENDS 13.
Beam Supported at Ends.
83
—Uniform load partially distributed. R
(max. shear
<
a
if
=
c)
W
+
(2c
b)
21
W (2a
Ri
M,
=
distance x
>
Mi, distance x
Supported at Ends.
—Uniform
tinuous.
R
a)*
b)
Rb
+ b)]
b) [4al -f b (2c
8/ 2
=
+
(2c
Beam
-
26
+ W
a
8? 2
14.
W (x
Rx
+ b)- = 2a
(a
M max. distance W max.
=
a
W (2x
Rx
W (2c +
<
a or
>a = Rx -
Mi, distance x
4- b)
2c
/
s
+
b) [4a/ 4- 6 (2c
b)]
load partially discon-
W W
(max. shear
if
>
=
Wi)
+
-a)
(21
Wi
c
21
Wi
Ri
-
(21
+ Wa
c)
21
= R
distance x
<
a
Mi, distance x
>
a
M,
=
Rx 2 Wal
Wa >
+
\Va*
2
and
2a
-
W (2x
-
M max. distance x -=
WX*
x
Wi Ca
WI
Wic
Beam
Continuous over
Two
Supports.
RM 2W
''
W max. 15.
2 j
—Two
exterior
(max. shear)
M,
=
W
Ri
2
Wx
distance x
2
Wa
M max. from R to Ri
2 2 f s
TFmax.
= Wa
D, distance a
a
-
(3aZ
12 I
Di, distance
s
sym-
metrical loads.
R
a)
= Wa
E (l
if\
4a 2 )
I
-
2a)2
k
r
STRENGTH OF MATERIALS
84 16.
Beam
Continuous over
Two
Supports.
—Uniformly distributed
load.
R =
Ri
w
*-r, max. shear
Wa
-T
s,
=
ty il/i
at
if
x
R and
a
2
\ )
«0
2Z
=
/2i
/l
- Ix 4W (x T\2-
M, distance x =
^
W
OT
^|/
I (I
- 4a)
= Wa* 21
max.
w
+ center * A/2 at
=
a
if
TF
(*
max. TFi
Wi max. = I
-
max.
—Formula for deflection
is
if
I)
__
a
<
a
>J(V£ —
Z (
V
2
"~
i)
2i-/s
max.
Deflection.
(
o
a*
max. —
> Z V2 — - 4a)
if
_
4a if
a
<
|)
Z (
V£
*-
I)
Beams steel beams
given in section on
under Various Loading Conditions. The depth of rolled should not be less than jfa of the span, and plate girders not than x^.
less
Columns was formerly assumed that the strength of a column depended largely on the condition of its ends. Many engineers now make no difference in their calculations for round-ended, pin-ended and It
square-ended 'columns. Usual factor of safety 5 or 6. Below are formulae for calculating the strength of columns:* (1) Steel
P =
Columns.
on column in pounds, cluding proper allowance for impact
total centrally applied load
A = minimum
in-
area of cross sections in square inches
= =
total length of column in inches r its least radius of gyration Then for steel columns of ordinary length where l/r does not exceed 120 for the principal members, or 150 for the secondary memI
bers,
and where P/A does not exceed 14,000
P = A Formulae from
(l6,000
70
—\
Electrical Engineer's
Handbook.
lb.
\
COLUMNS (2)
85
Cast Iron Columns.
=
d
diameter of circular column or shortest side 01 rectangular column in inches
-r not to exceed 40
a
P = A (3)
(6,100
32
Vi
Timber Columns.
P = A
Long-leaf yellow pine
Short-leaf pine
Or
P = A
and spruce
(l
|1300 I
1100
I
1
-
~\
— xtt,)
taken as the ultimate load in pounds per square inch, then the safe load for a given section may be obtained by multiplying the value of p as found from the formulae given below,* by the if
p
is
area of the section and dividing by the factor of safety. Steel column with both ends fixed or resting on flat supports. .
50000
V = 1
+
Z
2
36000
r2
Steel column with one end fixed and resting on and the other end round or hinged.
supports
50000
V = 1
Steel
flat
+
I
2
24000
r2
column with both ends round or hinged. 50000
V = 1
+
2
I
18000
r2
Cast iron columns solid with both ends fixed or resting on supports, d = diameter of column.
V
=
—
80000
1
*
From Machinery's Handbook
+
I
2
800 d2
flat
STRENGTH OF MATERIALS
86
Columns of
H
and
I
Sections*
(Safe loads in thousands of lbs.)
Allowable fiber stress per square inch, 13,000 50 radii or under; reduced for lengths over 60
lb.
radii.
for lengths of
SQUARE WOODEN COLUMNS
87
Cast iron column, hollow, round, both ends fixed or resting on = outside diameter of column. supports, d 80000 =
—
flat
V
1
+
1
+
/
2
800 d2 Cast iron column, hollow, square, with both ends fixed or resting on flat supports, S = outside dimension of square. 80000 =
V
P
S2
1000
For square wood columns with
flat
supports, the side of the
square being S,
V
5000
= 1
+
P 250 S2
Square Wooden Columns (Safe loads in thousands of pounds)
America Railway Engineering Association Formulae Long-Leaf Pine
—White
Oak
— 1,300
|
\
Length
1
—
ttt-.I
bud/
88
STRENGTH OF MATERIALS Round Wooden Columns (Safe loads in thousands of pounds)
Long Leaf Pine
Length
—White Oak— 1,300
I
1
—
~7g)
TORSIONAL STRESSES
89
Safe Load on Strong and Extra Strong Wrought Iron Pipe
Columns Both Ends Fixed
Factor of Safety Strong
Size
=
6
In Tons of 2000
lb.
STRENGTH OF MATERIALS
90 Springs.
Let
—To determine the
D v—
mean diameter
W
total load in
c
= = =
Then d
=
d
wire for wire springs,
size of steel
in inches of coil.
pounds
diameter of round or side of square steel wire in inches 11,000
3
•
•'
c
To obtain the number of free coils iV when the above data known and the compression C is decided on, use the formula Cd*a
N
W = =
are
£>3
an inch 26 for round (British Admiralty) or 22 (Board of Trade) = 32 for square (British Admiralty) or 30 (Board of Trade) Formula for Calculating Strength of Tubes, Pipes and Thin Cylinders. The one (Barlow's) commonly used assumes that the d a
where
size in sixteenths of
—
elasticity will
of the material at
the different circumferential fibers
have their diameters increased in such a manner that the
length of the tube is unaltered by the internal pressure. = thickness of wall in inches t Let
= S = D = n = p
pounds per square inch allowable tensile strength in pounds per square inch outside diameter in inches internal pressure in
safety factor as based on ultimate strength
P ™ Then ^= ^
2t
~ .
S =
t t
2 St
D n
=
DP —
_
DP
-
s
2t for butt-welded steel pipe
50000
n 60000
n
for lap-welded steel pipe
for seamless steel tubes
28000
n
for
wrought iron pipe
In the above, the thickness of the wall
t
is
assumed to be small
WROUGHT IRON TUBES
91
compared to the diameter. The thicknesses of thin pipes under the same internal pressure should increase directly as their diameters. A cylinder under exterior pressure is theoretically in a similar condition to one under internal pressure as long as it remains a true circle in cross section.
Bursting and Collapsing Pressures of Wrought Iron Tubes [Lukens Iron & Steel Co.]
STRENGTH OF MATERIALS
92
The
strength of a bitt or bollard can be calculated as a beam supported at one end and loaded at the other. Usually a thickness of 13^ ins. is sufficient, but the outside diameter depends on the
hawser that will be used. For steel wire hawsers, should not be less in diameter than four times the circumference
size of the chain or
bitts
of the hawser.
Riding Bitts or Bollards
Dia. in Inches
BOLTS Shearing and Tensile Strength of Bolts
93
STRENGTH OF MATERIALS
94
Tests of Hooks and Shackles Experience has shown that the same brand of iron or steel will not maintain the same tensile strength under various conditions. The following tables give the results of tests of hooks from in. to
%
3 ins. diameter
and
of shackles
from J^
in.
to 3 ins. diameter," the
taken from the catalogue of the Boston & Lockport In the column "Size, Inches," the diameter Co., Boston, Mass. It is suggested that not more of the hook or shackle is meant. than 20% of the tensile strength as given in Column 2 be reckoned as the working load, and on this basis Column 4 is calculated. Ordinarily the hook of a block is the first to give way, and when heavy weights are to be handled, shackles are far superior to hooks. By many tests it has been proven that flattening a hook adds from figures being
12 to
15%
to its ultimate strength.
*
Tests of Hooks [Boston
&
Lockport Co.]
In column Size, Inches, the diameter of the hook or shackle
Size,
Inches
is
meant.
Tests of Shackles
Size,
Inches
X % X X
Tensile Strength, Lb.
15,400 20,500 22,700 40,100 66,380 68,900 78,900 105,900 121,850 126,700 150,600 170,500 230,200 260,500 280,600 498,000
l
m i% IX \%
m XX 2
2M 2X 3
Description of Fracture
Working Load in Lb., Based on 20% of the Tensile Strength
Sheared Shackle Pin
Weldless Eye Bolts (Either plain or shoulder pattern)
Shank
3,080 4,100 4,540 8,020 13,276 13,780 15,780 21,180 24,370 25,340 30,120 34,100 46,040 52,100 56,120 99,600
Drop Forged Hoist Hooks With Eye (Capacity with plain shank the same) Diameter
of
Eye
TURNBUCKLES
97
TURNBUCKLES Drop-forged, with hook and eye, shackle and eye, two eyes, two hooks, two shackles, or hook an'd shackle. Size Turn-
STRENGTH OF MATERIALS
98
The average values given below substituted in the above formula will give a handy equation for calculating the diameter of the davit.
= Hence D =
K
=
\/
W
25 and
a
•
.0812
12000
X R
For davits of structural steel their dimensions must give the same strength as round bar davits as figured with the above formula. Lloyd's rule for boat davits.
L
length of boat
B
beam
of boat
D
depth
of
H
height of davit
S
c
boat
above
its
uppermost point
of support
spread All the above dimensions are in feet, constant = 82 when the davit is of wrought iron and of sufficient strength to safely lower the boat fully
equipped and carrying the d d
maximum number
of
passengers diameter of davit in inches
L
XB
X D
(H +4S)
C
may be calculated as beams, fixed at one end and loaded See also section on Anchor Davits. at the other. Stresses in Cranes, Derricks, and Shear Poles. The stresses Thus in Fig. 13 lay off in any member can be found graphically. Davits
—
^AW^vMT Figure 13
CRANES, DERRICKS, SHEAR POLES
99
distance p to any scale, say 1 inch = 1,000 lb., it representing )he downward force or weight of the load, and draw a parallelogram with the sides b t parallel to B and T so that p is the diagonal.
he
3y scaling
t
the tension in the
compression b in the brace.
tie
T
is
The above
obtained and similarly the also applies to Fig. 14.
z^JJ^W/WAk Figure 14
In a guyed crane or derrick as Fig. 15 the strain in
B
is
B P X— — j^
being that portion of the vertical included between B and T wherever T may be attache 1 to A. If T is attached to B below its I1
4_
L
nL Figure 15
STRENGTH OF MATERIALS
100
extremity, there may be in addition a bending strain B due to a tendency to turn about the point of attachment of T as a fulcrum.
The strain in T may be calculated by the principle of moments. The moment of P is P X c. The moment of the strain on T is
T X
d,
strain
T
on
The is
P X
moment is
If
is
g,
the guy
F X
the strain
tion or
on the guy rope
The moment
c.
is
c
j—
As d decreases the
increases.
strain
moments.
P X
T
therefore the strain on
/,
G X
and
and
is
of
G
by the the load about the bottom is
calculated
horizontal the strain in
P Xc F =
it
If it is inclined,
principle of of the
is
mast
F and
its
moment
the
1
the perpendicular distance of the line of
P Xc G =
The guy rope having
its direc-
the least strain
is
9
the horizontal one F, and the strain in G = strain in F X secant of the angle between F and G. As G is made more nearly vertical g decreases and the strain increases.
where the tie T is not perpendicular to A 1 or The the post A may be omitted and T extended to the ground. forces and the be equations, may applied parallelogram of Another case
(1)
tension in
is
,
—A
P X T —— T = and 1
(2)
compression in
—
P B B = Xj^— ,hold. A 1
Figure 16
Shear poles with guys. See Fig. 16. First assume that the two masts act as one placed at B D and the two guys as one at A 5.
RIVETS AND RIVETING Calculate the strain in tiply half the strain in
101
A B and B D as in the previous case. A B (or B D) by the secant of half the
Mul-
angle the two masts or guys make with each other to find the strain in each mast or guy. (From Mech. Eng'rs Pocket Book. W. Kent)
RIVETS AND RIVETING Different types of rivets are shown in Fig. 17. Pan- and buttonrivets Yi inch in diameter or over have coned or swelled necks
head for
punched
and straight necks
plates,
for drilled.
The advantage
that the diameter of the punched hole on the die side is always slightly larger than on the punched side. In assembling the plates are reversed, and thus with swelled-neck
of swelled-neck rivets
is
rivets the holes are completely filled.
C7 r\
f A
Figure 17.
—Rivet Heads and Points.
A = pan head B = snap or button head, makes a neater appearance than pan = countersunk raised head. E head. C = flush or countersunk flat head. than is required for a plain diameter inch an of rivets. greater tap They are % rivet to the same thickness of plate or shape (Am. Bureau of Shipping Kules^. F = snap point, proportions same as button head. G = hammered point. H countersunk point, proportions same for countersunk head.
D
Y
STRENGTH OF MATERIALS
102
Form
Rivet in outside plating
of
—
Proportions. The proportions of the heads and countersinks vary Tapered neck to with Uit
length in rdation to thick-
ness of plates.
the
societies.
own
different
The U.
Navy
has
its
There are thus no
standard.
.
S.
classification
«
•1NMMMMC0O3O)I^«iOC, 4, etc. From the body plan scale the distances from the center line to the intersection at the second water line and write it down (generally in red ink) under 2 water line as .10; do this for the 1}4 ordinate which is .33, and so on for all the water lines. Multiply the Simpson's multipliers below the water lines by the half-breadths and write the products below. Thus .10 X IjHj
=
.15, .33
X IK =
-50, etc.
each ordinate as for No. write the
sum
in the
1,
.15
Add
+
these products horizontally for
.4
column functions
+
.2
+
of areas.
.4
+
.1
=
1-25
and
Multiply the func-
l by Simpson's multipliers as A, 2, l A, 4, etc., writing the products in the column multiples of areas. Add up this column
tions of areas
l
SHIP CALCULATIONS
178
3256 and multiply it by }4 of each interval and by 3^ of the distance the water lines are apart, thus, X 2 ft. X Vz X 9.5 ft. X 2 (as half-breadths were taken), which will
which
in the present case
is
Y
give the volume of the displacement in cubic feet as 13748, and to convert it into salt water tons divide by 35, as 35 cu. ft. of salt
water weigh one ton.
To
find the fore
and
aft center of
buoyancy, multiply the multiples of areas by their lever arm from the midship ordinate, thus giving forward and after moments as .60 X 5 = 3., 35.22 X 4K = 158.49, etc. Add up the forward moments and the after ones. In the present case the after sum or 3287.92 is the largest, so subtract the forward from it leaving a remainder of 617.95. Multiply 617.95
by 9.5 ft., the distance the ordinates are apart, and the product divided by the sum of the multiples of areas will give the location of the center of buoyancy; in the present vessel it is aft of No. 6, the midship ordinate, a distance of 1.83 ft.
To
find the vertical position of the center of buoyancy, multiply the half-breadth as for 2 W. L. at No. 1, viz.: .10 by Simpson's -
Y
multiplier giving .50; do the same for the next half -breadth, as .33 by 2, giving .66, writing the products in the column to the Continue thus for the other water lines, and add right and so on.
up each as
2.75, 133.33, 170.90, etc., multiplying
son's multipliers as
246.66, 256.35, etc.
with arms
them by Simp-
K, 2, 13^, 4, etc., the products being 1.37, Take moments about the base line which is
J^, 1, 2, 3, etc.,
the
sum
of
which
is
9416.47.
Multiply
9416.47 by the distance the water lines are apart or 2 ft., and divide by the sum of the multiples of areas, the quotient being 5.78 ft., which is the distance the center of buoyancy is above the base line.
The above calculations may be simplified for getting the areas of the cross sections. ment
lay off a table as below:
Station
by using a planimeter Thus for the displace-
CENTER OF BUOYANCY
179
taken) = volume of displacement in cubic feet, which divided by 35 will give the displacement in salt water tons. To find the fore and aft center of buoyancy use a table as follows:
Station
p o
co
00
o
00 00
O
do
O o t* CO lO
03
O ^pqoQHfeOW
184
CROSS CURVES OF STABILITY
of such a density as to bring the steamer to her
homogeneous cargo
summer (4)
D
load
185
line.
same as
(3),
with bunker
coal,
stores
and
fresh water
consumed, approximating to the end of the voyage. (5) E ready for sea, water in boilers, bunker coal, stores and fresh water aboard, and all ballast tanks filled. (6) F same as (5), but with bunker coal, stores and fresh water consumed. (7) G same as (3), but loaded with a coal cargo, part of the 'tween-decks empty. (8)
H
same as
consumed. Cross Curves of
(7),
but with bunker
Stability.*
when
coal, stores
and
fresh water
—These are calculated for two or three
loaded, and loaded with the bunkers empty. Select angles of inclination as 15°, 30°, 50°, 70° and 90°. Prepare body plans for the fore body (see Fig. 23) and
conditions as,
after body,
angles.
the vessel
is light,
and draw on them the load water
Make
the calculation
first for
line
Figure 23
From A
class
and the
inclined
the loaded condition.
book on Naval Architecture, W.
J.
Lovett.
SHIP CALCULATIONS
186
Find the area of each section of immersion and emersion at the assumed inclination, by a planimeter preferably, altho these can be found by Simpson's rules. Mark the center of gravity of each section.
Draw
plane. are taken.
Ordinates
XX
a line perpendicular to the inclined water the line about which the moments of the wedges Prepare a table as follows for the submerged wedge.
This
is
TO FIND THE AREA
187
the difference between the volumes of the submerged and emerged wedges. Make a correction for the difference, laying out a table thus: Si
is
SHIP CALCULATIONS
188
XG
the greater volume is on the emerged side and the center of gravity of it on the emerged side, then the moment
In Si
note
if
obtained has to be deducted from the total moment. If the greater volume is on the submerged side and its center of gravity on the submerged side the moment has to be added to the total moment.
GZ sin
= B
R—B G sin
= B R
BM
Also find
by
BR
this is
setting up inclinations.
BR
same manner
for 30°, 50°, 70° and 90°. for other drafts for all the angles of inclination.
Proceed to find
When
6
in the
of stability may be constructed, found for the different at the different drafts the
done the cross curves
GZ
Run
through each series of spots and these are the cross curves of stability. See Fig. 25. lines
lines
To
construct stability curves (Fig. 26) lay off the inclinations as 15°, 30°, etc., horizontally and vertically the values of found. Draw curves the thus laid off. previously through points
GZ
The
show constant inclination at varying displacements. The stability curves show constant displacement at varying inclinations. The cross curves show the value of the righting arm G Z. A curve of righting moments could also be made showcross curves
W
In [displacement] X G Z). ing the foot-tons (the value of drawn to the the sections are uppermost preparing the body plan continuous deck. If a watertight poop, bridge or forecastle become
immersed at the higher angles of inclination, the value buoyancy should be calculated. As the above curves have been considered with the vessel ary, they are called static curves.
of their
station-
NOTES ON STABILITY
189
—
Notes on Stability. For ordinary vessels the transverse metacenter remains practically unchanged up to 10° inclination. The should not be less than 10 ins. and have a righting value of G
M
arm
of at least 10 ins. at 45°.
30°
/£
SO°
V
£/.3°
70°
&o
Figure 26
An
ordinary seagoing ship should have a range of stability of Stability varies as the square of the breadth and inversely A 300-foot steamer when loaded had a maximum as the draft. 70°.
ins., while a similar one under similar conditions broader had a maximum righting lever of 12 ins. Free-
righting lever of 8
but 2 board
ft.
in stability, as the stability immediately begins to decrease when the edge of the deck gets under water, so that every additional inch of freeboard increases the is
an important factor
vessel's range.
Approximate Formula for Calculating Stability (G Z). Let 6
=
angle the vessel
normal water
GM =
is
inclined, that
line
and the
distance between the
is,
the angle between
inclined
center of gravity
and the
metacenter
BM =
distance
the
metacenter
is
above the center
of
buoyancy.
GZ =
righting
arm
Then
GZ = £
M
sin
.0
+ ^-^- tan
2
sin
to an angle of 30°, provided the ratio of the beam to the draft not abnormally great, the above formulae may be used instead
Up is
of the long stability calculations.
The
values at inclinations of
SHIP CALCULATIONS
190
and 20° are
practically the same as obtained with the usual stability calculations. Trim is the difference between the forward and aft draft of a vessel.
15°
10°,
Thus, suppose a vessel draws 12 ft. forward and 15 ft. aft; then she is said to trim 3 ft. by the stern. Longitudinal Metacenter. Let B (see Fig. 27) be the center of buoyancy when floating on an even keel, L, and suppose the trim of the vessel to change, the displacement being the same, then Bi is the new center of buoyancy. a vertical line meeting Draw B\
—
W
B
M
at
tance
G
M.
M
Then
M
M
the longitudinal metacenter, and the disthe longitudinal metacentric height. is
Figure 27
—
Moment to Alter Trim One Inch. Suppose a weight w is moved from w to w, then the change of trim = W Wi + LLi = (WiS -fS L\) X tan = length of load water line X tan 0. The movement of the weight w causes the center of gravity of = the displacement the vessel to move aft a distance G\ G. Let
W
in tons,
Gi
a
=
G = G
the distance the weight
M
X
tan $
change of trim length of load water
Change
L X w X
WXGM
is
moved, then
^~W and tan $
=
WXGM
line
of trim in feet
a
=
w
=
length of load water line (L)
X
tan 6
=
CALCULATIONS FOR TRIM To
moment
get the
to alter trim one inch substitute in
Therefore the
WXGM L X
12
w X
a
W XGM w X a 12 L W X GM i
change of trim length of load
191
.
,
one inch or
,
line'
moment
t 1 y* ^ z foot,
thus
'
=w X
to alter trim one inch
a=
—
-=
foot-tons.
Example. A 350-ft. steamer, displacement 6700 tons at her designed draft, has a longitudinal metacentric height of 350 ft. If 10 tons of cargo in her forward hold was moved 100 ft. aft, find the change in trim. a/t ++to change u Moment
u trim one inch •
*
=
WXGM — — = Li
Moment
aft
X
1*
from shifting cargo = 10 tons
Hence change
of trim aft
=
^„
=
X
6700 ooO
100
ft.
X X =
350
=
55.8 foot-tons.
1^
1000 foot-tons.
17.9 ins.
o5.8
Approximate Calculations for Trim. to alter trim one inch
—-—-
= Li
X
1^
—In
the formula,
foot-tons,
G
if
M
is
moment assumed
to be equal to L the length of the ship, which is roughly true in the case of ordinary cargo vessels at their load displacements, the
W — foot-tons.
trimming moment per inch becomes
Another approximate formula giving above is the following:
T = A = L = B = V = W= The
results
closer
than the
tons per inch of immersion area of load water plane in square feet
length on the load water line in feet
breadth of ship amidships in feet
volume of displacement in cubic
feet
displacement in tons
height of the longitudinal metacenter above the center of
buoyancy assuming
in ordinary cargo steamers
B^
,, M
^ ** and = G M, i
as
W
ttt
is
B
= v °l-
M
=
A XL 2
.0735
p
°f displacement
^z
= V 05
SHIP CALCULATIONS
192
Then
L 35
x A
moment
the
trim
alter
one
B
*
XV
moment
for the
X
n
X
To Estimate the Displacement Her Designed Trim.
T = y = L = Then the
displacement
A ft.
displacement
steamer 350
3
by the
ins.
draft of 20
is
ft.
is
The displacement ft.
when
Floating
Out
of
length of vessel in feet
extra
displacement curve
for 2
= 190 = 180 =172
of a Vessel
one foot of extra trim
for
=
14
ft. forward and 24 ft. 3 ins. aft, thus loaded she trims 5 ft. by the stern. If aft amidships, and the tons per inch of immersion 35,
ft.
long,
ft.
draws 17
When
stern.
the steamer's displacement?
At a
is
tons per inch of immersion center of flotation aft of amidships in feet
the center of flotation is
,
foot-tons.
y
Example. trimming 7
what
T2
=
^7— 12
X
draft
where n for fine vessels where n for ordinary where n for cargo
T X
f Li
to alter trim one inch
length on water line
9
—
=
A* 30.9 X „ .000175 -77 foot-tons, or tj
=
12
Another formula
,
inch
A2XL
0735 T
to
1)^
ins.
—
*
—
'-
I
her displacement from the
5850 tons. for
3 ins. extra trim
one foot of extra trim
=
=
12
X
35 ;r— o5U
X
14
=
16.8 tons,
and
37.8 tons.
Thus new displacement = 5850
To Find
'-
I
+
37.8
=
5887.8 tons.
the Distance the Longitudinal Metacenter
is
Above the
Center of Buoyancy. Let V = volume of displacement in cubic feet I Q = moment of inertia of water plane about a transverse axis passing through the center of flotation.
Then the
longitudinal metacentric height
B
M = ^.
See Meta-
center s, page 186.
any Mean Draft and Longitudinal Position of the Center of Gravity by Trim Lines or Curves.* L to represent the mean draft for Draw a line See. Fig. 28. * From and Calculations Cons., G. Nicol. Ship
To Find
the
Trim Corresponding
W
to
—
CALCULATIONS FOR TRIM which the trim
On
line is required.
193
B
this line a point
as the longitudinal position of the center of keel,
and a
vessel.
line
NN
Thus the
of
buoyancy
of
it.
is
is
buoyancy drawn representing the midship
distance
taken
is
at a level line of the
BN
represents the distance the center from amidships, which in the present case is forward
Figure 28
The vessel
horizontal distance from
trimming 2
Change
of trim
=
B
by the stern
ft.
of the center of
is
buoyancy
of the
calculated as follows:
length of water line
X
%
tan
(for
see Fig. 27)
change of trim length of water line Gi equals nearly B B h or the distance between the centers of buoyancy before and after the trim has been changed, so
Now G
G G = B B = G M X tan 0. G M is approximately equal l
water
x
line,
then substituting
GGi = B B = G x
M
X
tan
G
to the length of the ship
M
= L and
= L X
tan $
=
change of trim
L
on the
^-=
.«¥
and in the present case this distance is set off from B. Next calculate the position of the center of buoyancy with the vessel trimming 4 ft. by the stern, the same method as just outlined being used, and lay off this distance as B B 2 At Bi and B 2 verticals are erected, and the corresponding trims (2 ft. and 4 ft.) laid off, the same scale being used. Through the points thus found and the point B a line is drawn, which is the trim .
line required.
SHIP CALCULATIONS
194
For forward trims the trim
should be continued below its of the center of buoyancy in that direction. It should be noted that the center of gravity and center of buoyancy are here assumed to travel the same distance level line to indicate the
line
movement
not quite true as B is below G and therefore more remote from M, and moves a greater distance. For very accurate work the distance plotted from B
when a change
towards plus
of trim takes place.
W should be the
BG X
tan
0.
It
is
This
is
calculated travel of the center of gravity not necessary to proceed to this refinement
not worth considering. From the trim line just drawn can be determined any trim up to 4 ft. (other trims, as 6 ft., 8 ft., etc., could be plotted if dein ordinary cases, as the error involved is
due to the movement of weights on board. For if the distance the center of gravity travels aft on account of the movement of the weights be ascertained and plotted from B along the level line C, and a vertical line be erected to intercept the trim fine at D, C D must be the trim by the stern, as the center of buoyancy and center sired),
of gravity are
A of
trim line
always in the same vertical is
displacement
only reliable at is
its
considerable
own
line.
draft,
and when the change is required. For
a new curve
ordinary purposes three conditions are sufficient,
and
light.
Effect of Flooding 'a
Damaged Compartment.
viz., load, ballast,
—To find the
effect
a compartment being thrown open to the sea by collision or other accident, account must be taken not only of the water that would enter if the ship remained in her original position, but also of the additional water which will enter due to the heel, change of trim, of
and sinkage caused by such flooding. When the compartment is wholly under water, and the water is prevented from spreading by a watertight deck or inner bottom the effect is the same as of adding a weight in a known position. To Find the Trim when a Compartment is Flooded. The weight
—
water in the compartment up to the original water line should be found and the parallel sinkage determined assuming the compartment open to the sea and the admitted water placed with its of the
center of gravity in the vertical plane containing the center of gravity of the added layer of displacement. This distance measured in the trim diagram above the height of the original water plane,
from which the level line and corresponding trim line should be drawn. The trim can then be obtained (as described in the paragraph on Trim Lines) by finding the travel aft of the
will give the point
CALCULATIONS FOR TRIM
195
center of gravity, assuming the weight to be translated to
its
true
position. It will next be necessary to calculate the weight of water in the compartment, assuming the surface to rise to the level of the new draft,
and to use
should differ
same way
in the
it
much from
the
first
in
another trim estimate. it
calculation,
may
If this
be necessary
to proceed to a third.
Or instead of the above, which is the trim line method, first determine the amount of mean sinkage due to the loss of buoyancy, and second, determine the change
of trim caused.
Quantity of Water That Will Flow into a Ship Through a Hole in
Her Let
Side.
H
— A = g — V =
distance center of the hole
The volumejn 8
below the water
line in feet
area of hole in square feet acceleration due to gravity (32.16) rate of flow in feet per second
Then ond =
is
V =
\/2 g
H
= S\/H
approximately
cubic feet of water passing through the hole per sec-
VH X A
Example.
A hole having an area of 2 sq.
in the side of a ship. What into her flow per minute?
Cubic feet per second Cubic feet per minute
= 8 \/h X A = 8 \/l X = 32 X 60 = 1920.
1920
Tons per minute =
ft., 4 ft. below the water line was made would be the approximate tons of water that would
..^
=
2
=32.
54 85 tons. .
Calculating the Trim by the Trim Line Method when a CompartAssume a box-shaped vessel 210 ft. long, 30 ft. is Flooded.* beam, and 20 ft. deep, drawing 10 ft. forward and aft. Suppose
—
ment she
is
in collision
Find the
s
draft.
and a compartment at the (See Fig. 29.)
after
end
is
flooded
SHIP CALCULATIONS
196
Using the trim draft.
WL
at 2
and 4
ft.
30 X X—
210
.
10
2y
2.
volume
of the
obtain the trim line at 10
first
W
line,
L and W* L
2
2
=
of displacement L S L 2 is half the vessel's length,
wedge
=
is
—
5
105 1
moves and L
1575 cu.
B
to
tan
then
2)
a
=
w X
=
35 1800
to
to the position
L
ft.,
one
2
and g\
foot, the
in
moving
g2 or
.
140 #ft.
of the vessel's center of
^
a
W XGM 1800
That
2
B B2 = G M X
a B »B r > = 1575 and
vessel
X
-
g
The corresponding movement
B
when
WL
1800 tons, and in passing from the
105X2*
from
those
stern.
aft its center of gravity travels a horizontal distance
is
4
ft.
vessel to be floating in salt water, her displacement
W L the wedge W S W As S L 2
method,
by the
ft.
Assuming the is
line
the level water
is
buoyancy
tan 6
x X
140 1575 140
1800
X
35
X 140 = 35 Q( ft n X 35 ,
the horizontal travel of the center of buoyancy with the trimming 2 ft. by the stern is 3.5 ft. With the vessel 4 ft. by is
the stern, the horizontal travel is double 3.5 ft. or 7 ft. From the above, a trim line can be drawn for the initial draft.
Trim
lines corresponding to other displacements
same manner.
can be obtained
30 is the complete diagram for the vessel and shows cross curves with a range of from 7 ft. 6 ins. to 15 ft. draft. Next begin with the calculation for the bilging. in the
Weight
of
Fig.
water in bilged compartment
—
=
=
85.71
tons/ Parallel sinkage assuming water situated amidships
ment open
to sea
= 85.71X35X12 = 1200 X SO
a
and compart-
.
mS
'
Horizontal travel aft of vessel's center of gravity, assuming the
%
* Center of gravity of a wedge is from the apex. of being 210 instead water the line, t The length of
compartment
flooded
is
10
ft.
long.
ft.,
is
now 200
ft.,
as the
CALCULATIONS FOR TRIM
197
water at the increased draft to move into its true position and the ship's bottom to be intact: new draft of 10' 6" X 10' length X 30' beam = 90 tons w = 35 a
=
W
=
210'
—
—
10' (length of 2 :
compartment) -
original displacement of 1800 tons
Gfla !|a w a «)Xia), 1890
=
,_100
.;
ft.
+ 90 tons =
1890 tons
t7flfl
A.
Figure 30
Referring to Fig. 30 the trim line corresponding to a level line at 10 ft. 6 ins. can be drawn, and by measuring 4.76 ft. along this line from A B, and erecting a perpendicular and scaling it, its length 2 ft. 1034 ins. is the trim by the stern. The drafts will be
Forward = 9'
0^ Aft
11' v
=
1W
0"
+ parallel
sinkage of 6*
-
10' 0"
+ parallel
sinkage of 6"
+
10'
Y2 (2'
10M") =
Vz (2' 10Ji")
=
In the second approximation, start with the vessel in the above 8
SHIP CALCULATIONS
198 trim.
The weight
11.86
X
X
10
30
=
35
D „ , Parallel
of the
water in the bilged compartment
be
101.66 tons.
101.66 tons X 35 smkage = 210 ft •
will
i
cu.
X
X
ft.
30
12
ins.
=
„«„
^
ft
.
ins
'
nearly.
Taking the center of gravity of the water line at the middle of the length of the compartment, then the travel of the vessel's center
of gravity
due to admission of water
101.66
a
W
1800
X~ 2
+ 101.66
laying this off on the trim diagram, on the water scaling up to the trim line, the trim will be found to be ins. by the stern.
5.35
ft. aft.
line,
and
3
2%
ft.
= w X
By
Dividing this equally forward and parallel sinkage, the drafts
Forward
10' 0'
Aft
10' 0"
Calculating the
Trim
and adding
aft,
become
+ 6%" - 1' 7%" = + 6%" + V 734" = Mean
8' 1
12'
6% ins.
as the
W
2Y% W
when a Compartment
by Sinkage — A Flooded.* rectangular lighter 100 long, 40 ft.
ft.
beam, 10
ft.
is
deep,
water at 3 ft. draft, has a collision bulkhead 6 ft. from the forward end. If the compartment forward of this bulkhead is flooded, what would be the trim in the damaged position? floating in salt
(See Fig. 31.) (1)
Determine the amount
of
mean sinkage due
to the loss of
buoyancy.
Determine the change of trim caused. (1) The lighter, due to the damage, loses an amount of buoyancy represented by the shaded part G B, and if it is assumed the lighter sinks down parallel, she will settle down at a water line w I such that volume iv G = volume G B. This will determine the distance x = 94 ft. L. GL = 6 ft., w between w I and (2)
W
w G = w H X 40 volume G B = G L X 40
H
For the volume
ft.
For the
ft.
GL X 40X3 (2) *
Change
X X
18..
of trim.
From Theo. Naval
Architecture, L. T. Attwood.
x 3
ft.
01/
.
Volume
WHEN COMPARTMENT
IS
of displacement in cubic feet
=
Displacement
=
100
X
40 35
X
3
2400 7
=
FLOODED 100
X
342
40
tons,
weight acts through G, the center of gravity, which either end.
X is
199
3
and 50
ft.
this
from
SHIP CALCULATIONS
200
7200
The new water
line
W
1
L
1028
=
Therefore change of trim
66
66
= 15H
ins.
1
will pass through the center of gravity water line wl at K, and the change of trim aft and forward must be in the ratio 47:53,
of the
47
Decrease of draft aft Increase of draft forward
=
100 53 100
3
X 15^ - 7H ins. X 15^ - 8M
ins.
New draft aft = 3 ft. + 2J£ ins. [from (1)] - 7}iins. = 2 ft. 7 ins. New draft forward = 3 ft. + 2% ins. [from (1)] + 8M ins. = ft. 10K ins.
CENTER OF GRAVITY. Coincident with the calculations of the displacement and centers of buoyancy, are made calculations of the fore and aft, and vertical positions of the common center of gravity of the hull, machinery and cargo. The fore and aft position of the center of gravity of all the weights must come over the fore and aft position of the center of buoyancy. If on the first estimate it does not,
then the weights must be shifted until it does. On a profile of the vessel draw a vertical line midway between the forward and aft perpendiculars. Also draw a base line parallel to the water line, for getting the vertical distance of the center of gravity. Except when the keel is given a drag, the base line is
To
taken as the molded find the fore
hull, lay off
and
line of the
frames at the keel.
aft position of the center of gravity of the
a table as follows:
CENTER OF GRAVITY Assuming the moments
moments
aft
— moments
W
201
be greater than those forward then forward aft to
=
distance center of gravity
is
aft
of amidships.
To
find vertical position of center of gravity of the hull, lay off a
table thus
:
Items
SHIP CALCULATIONS
202
Assuming the moments
moments
aft
— moments
be greater than those forward then forward „ ,;•; aft to
=p-
=
distariee ot center of gravity
aft of midships.
To
find vertical position of the center of gravity of a ship, lay off the following table:
Dist. Cent, of
Item
Weight
Grav. Above Base
Moment
Hull Boilers
Engines.
Cargo Cargo
.
in forward hold in aft hold
&c.
M
W The sum
of the
moments
M divided by the sum of the weights
W will give the distance the center of gravity
is
above the base.
Care must be exercised in locating the engines, boilers, cargo, tanks and other weights in a ship. If they are placed too high, the ship will be unstable and if too low she will be very uncomfortable in a seaway, owing to too quick a return to the vertical position.
The
table below gives the heights of the center of gravity of ordinary passenger and freight steamers, and of freight steamers.
EFFECT OF MOVING WEIGHTS
203
moving weights on the center of gravity of a vessel. (1) Suppose the Weight Was Raised.—The distance the center of gravity of the vessel was raised would be found by multiplying the weight moved by the distance it was moved and dividing the result by the total weight or displacement. Effect of
Example. A weight of 30 tons was raised from the hold and placed on the deck steamer at a distance of 20 ft. from its Original position. The steamer had a displacement of 1000 tons. Find the distance the center of gravity was raised. of a
weight X distance displacement
=
The Weight Was Removed.
(2)
30 X 20 1000
—In
=
"
6
ft '
this case multiply the
weight
distance from the center of gravity of the ship, and divide the product by the displacement after the weight was removed.
by
its
Example. A weight of 30 tons 10 1000 tons displacement was removed.
ft.
weight X distance displacement -weight (3)
Adding a Weight.
from the center
below the center of gravity of a ship of was the center of gravity raised?
How much ==
30 X 10 1000 - 30
=
,3
'**
— Multiply the new weight by
of gravity of the vessel
distance
its
and divide by the new
displacement. Example. A weight of 30 tons was placed on board of a steamer with an original displacement of 1000 tons 10 ft. below her center of gravity. Find the distance the center of gravity was lowered.
weight X distance displacement + weight
=
30 X 10 1000 + 30
=
300
=
1030
'
28
ft "
—
Moving a Weight Athwartships. Multiply the weight by the distance moved and divide by the displacement. Example. A weight of 20 tons at the center of the upper deck was moved 10 ft. (4)
to starboard.
The steamer had a displacement
Find the distance
of 1000 tons.
her center of gravity was moved.
weight
X
distance
"
displacement.
20 X 10 1000
=
-2
ft.
to starboard.
—
JVfoving a Weight in Two Directions. The new positions of the center of gravity can be found by using formulae (2) and (4). (5)
Example.
In a vessel of 4000 tons displacement, 100 tons of coal were shifted and 4 ft. 6 ins. vertically. Find
so its center of gravity moved 18 ft. transversely the new position of the center of gravity.
100
By
(2)
the center of gravity will
move
vertically
By
(4)
the center of gravity will
move
horizontally
(Author not known)
X
100
—=
4 5 .
4000
X —
18
-H
=
ft.
45
ft
SHIP CALCULATIONS
204
In this case, however, the angle of heel is usually calculated instead of the distance the center of gravity moves. Thus in the above example assuming the 20 X 10 20 X 10 = steamer had a G of 2 ft. the angle of heel would be 1000 X G 1000 X 2 = .10, consulting the table of natural sines, the angle is found to be 5 degs. 75 mins.
M
M
To Find the Center Even if the position
of Gravity of a Vessel by Moving Weights.* of the transverse metaeenter is known, it
— is
no value
in predicting a vessel's initial stability as the center of gravity of the entire vessel (hull, machinery, and cargo) must be known. The center of gravity can be calculated as outof itself of
lined above, or
it
can be obtained by the inclining experiment as
described below.
A
perfectly calm day should be selected, all the crew ordered off the vessel, all movable weights made fast, and the vessel trimmed
A plumb line is hung down one of the two different hatches), usually as near amidships as possible. At the end of the plumb line a horizontal batten is placed on which can be marked the deviation of the plumb so she
is
perfectly upright. hatches (sometimes two at
line
when the
A weight 3,
vessel
is
inclined.
from port to starboard on the top of weight through a distance of d feet, and the deviation of the plumb line 1 is
shifted
noted.
Weight 2 is shifted from port to starboard on top of weight 4 and the deviation of the plumb line noted. The weights 1 and 2 are then replaced in their original position, the vessel returning to the upright position again. Weight 3 is moved from starboard to port on top of
If
Then
w
W a I
d
GM
= = = = = —
Then (?I = *
and
similarly 4 is the weights are returned to their original position.
deviation of the of 2.
and the moved on top 1
plumb
line noted,
weight moved in tons displacement of vessel in tons deviation of length of
plumb
plumb
distance weight
line
along the batten in
line in ins. is
moved
in
ins. •
ft.
distance between the center of gravity and the transverse metaeenter in ft.
wXdwXdXl Wx *
From Theo. Naval
WXa
Architecture, E. L. Attwood.
FREEBOARD
205
Example. A steamer has a displacement of 5372 tons, and draws 16 ft. 9 ins. forward and 22 ft. 10 ins. aft. Weight used for inclining 50 tons, which was moved 36 ft. Length of plumb line 15 ft. Two plumb lines were used.
SHIP CALCULATIONS
206
roundup
of
deck beams, and for
this vessel the curves read directly.
Deck
erections contribute to safety and are taken account of as a corrective term, as are also other variations from the standard ship.
In practice the freeboard is actually assigned after the ship and usually by one of the classification societies' agents, but
is
its
built
pre-
liminary determination is an important and necessary item in the design of any vessel, as the draft plus the freeboard gives the depth. The complete tables as issued by the British Board of Trade take up a variety of modifications and corrections which are in-
volved by vessels differing from the arbitrarily assumed standard. The following work is based upon the rules directly, although the presentation and the wording are modified. The curves given on Plates 1-4 are a graphical representation of corresponding tables in the rules. Spar deck steamers and sailing vessels are not in-
cluded as these classes are not numerous in present-day designs. The limitations of loading as laid down by the above act (for complete text see publication issued by Marine Department of the
Board
Trade, entitled Instructions to Surveyors, Load Line) are represented by a disk and number of horizontal lines which are cut and painted on the side of the ship amidships as British
shown
of
in Fig. 32.
The upper edge
of each line is the point of
meas-
urement.
The word "freeboard,"
legally,
denotes the height of the side
above the water line, measured at the middle of her length along the load water line. It is measured from the top of the deck at the side. The reserve of buoyancy necessary for flush deck steamers of full scantling and awning deckers are given by the curves on Plate 1 and these curves hold for any and every vessel regardless of proportions. For the standard vessel of these classes and within the dimensions given the freeboards required may be read directly from Plates 2 and 3. For awning deck vessels the freeboards are determined more by considerations of structural strength than by reserve of buoyancy, and indicate the depth of loading beyond which it is probable that first class vessels of this type would be unduly stressed when at sea. Therefore the freeboards and percentages of reserve buoyancy are in excess of what would be required for full scantling vessels. They are measured to the deck below the shelter or awning deck. The freeboards given in the curves are for flush deck vessels in all a ship which has no deck erections, cases, and for the standard ship has a proportion of length to depth of 12, has a roundup of deck of the ship
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