AOITaSV THE aval atichitect's and shipbuilder's pocket-book M ,.08 i& flajYT-OSAWYA *-^niT ,^^,=r^,jr^ A D VKTiTISKMK^^TS MARINE PATTERN Duplex Steam Pumps To Lloyd's Requirements. VERTICAL or HORIZONTAL. HAYWARD-TYLER & CO., Ltd. 99 Queen Victoria Street, LONDON, E.G. Teleerrsirns : " TYLEROX CENT, LONDON." Telephone : 9626 CITY THE NAVAL ARCHITECT'S and SHIPBUILDER'S POCKET-BOOK OF jfonnula^, IRulee, anb ^ablee AND MARINE ENGINEEirS AND SURVEYOR'S HANDY BOOK OF REFERENCE BY jjjf^j^ CLEMENT MACKROW t ( l^ATE MEMBER OF THE INSTITUTION OF NAVAL ARCHITECTS T.ATR LKCTIRER ON NAVAL ARCHITECTURE AT THE BOW AND BROMLEY INSTITUTE AND LLOYD WOOLLARD JtOYAL CORPS OF NAVAL CONSTRUCTORS MEMBER OF THE INSTITUTION OF NAVAL ARCHITECTS INSTRUCTOR IN NAVAL ARCHITECTURE AT THE R.N. COLLEOE, GREENWICH iElefacutlj 1SDiti0n TH0R0U(3i'HLY heviscd WITH A SECTION ON AERONAUTICS NEW YORK THE NORMAN W. HENLEY PUBLISHING CO. 132 NASSAU STREET 1916 ^ \ ^v \ \ PRINTED BY STEPHEN AUSTIN AND SONS, LTD. HERTFORD. PREFACE TO THE ELEVENTH EDITION. The need of a new edition of this Pocket-book haf? arisen through the continual development of the science of Naval Architecture, and the tendency towards standardization and regulation of parts of the structure and equipment of ships. Very many changes have been introduced, and much of the book has been )'e written, but where possible its form has been left unaltered. Its object remains the same as that stated in the Preface to the original edition, viz. to condense into a compact form all data and formulae that are o)-dinarily required by the Shipbuilder or Naval chitect. Amongst the new matter inserted, it is believed that the section on Speed and Horse-power will be useful in enabling ships of ordinary form to be approximately }>owered from the data therein given ; a brief description of modern methods of powering and determining forms suitable from a propulsive standpoint has also been included. The necessity for economizing weight where possible without diminution of strength has led to the sections on Strength of Materials, Kiveted Joints, and Stresses in Ships being considerably extended. In- formation concerning British Standard Sections, Screws, Iveys, etc., has also been added, by permission of the Engineering Standards Committee. Finally, two new sections on Aeronautical matters will be of service, not onlv to those engaged in that modern and rapidly 355437 VI PREFACE. developing branch of engineering, but also to Naval Architects on account of the kindred nature of the subjects, and of the direfet application of many air data to questions relating to the resistance of bodies in water. The remaining subjects treated, which were also included in previous editions, have now been brought completely up-to-date ; the excision of obsolete data has enabled the new matter to be inserted without increase in the size of the book. The new tables of logarithms, etc., it is trusted, "vvill be found of great practical convenience to those using them. The scope and extent of the revision were arranged in the first place with the original author ; although, owing to his death before the comj)letion of the work, the absence of his advice and experience during the later stages has been felt and regretted, the reviser has had the benefit of securing great assistance from many sources during the preparation of the new edition. Among those who kindly contributed, the reviser is greatly indebted to Mr. A. W. Johns, the results of w^hose valued experience have been embodied in various parts of the book ; the new sections * Aerodynamics ' and ' Aeronautics ' are entirely due to him. Considerable aid in the treatment of Speed and Propellers has been rendered by Professor T. B. Abell, while Mr. E. F. Atkinson has supplied useful data concerning small craft and tugs. To these, and to many others to whom reference is made in the course of the book, the reviser tenders his cordial thanks. He also trusts that the numerous correspondents who have offered suggestions and pointed out errors in previous editions may be led to take the same kindly interest in the present revision. L. W. Barnes : January 1, 1916. P K E F ax: E TO THE FIRST EDITION. The object of this work is to supply the great want which has long been experienced by nearly all who are connected professionally with shipbuilding, of a Pocket- Book which should contain all the ordinary Formulae^ Rules, and Tables required when working out necessary calculations, which up to the present time, as far as the Author is aware, have never been collected and put into so convenient a form, but have remained scattered through a number of large w^orks, entailing, even in referring to the most commonly used Formulae, much waste of time and trouble. An effort has here been made to gather all this valuable material, and to con- dense it into as compact a form as possible, so that the Xaval Architect or the Shipbuilder may always have ready to his hand rdiable data from which he can solve the numerous problems which daily come before him. How far this object has been attained may best be judged by those who have felt the need of such a work. Several elementary subjects have been treated more fully than may seem consistent with the character of the book. This, however, has been done for the benefit of those who have received a practical rather than a theo- retical training, and to whom such a book ^s this would be but of small service were they not first enabled to viil PREFACE. gather a few elementary principles, by which means they may learn to use and understand these Formulae. In justice to those authors whose works have been consulted, it must be added that most of the Eules and Formulae here given are not original, although perhaps appearing in a new shape with a view to making them simpler. There are many into whose hands this work will fall who are well able to criticise it, both as to the usefulness and the accuracy of the matter it contains. From sucH' critics the Author invites any corrections or fresh mate- rial which may be useful for future editions. Sl^MMARY OF CONTENTS. -•<>• PAGES. Signs and Symbols .... 1-3 Logarithms ..... 4-7 Trigonometry ..... 7-13 Curves (Conic Sections, Catenary , Cycloid, etc. ) 13-18 Differential and Integral Calcnlns 19-21 Practical Geometry . . 22-36 Mensuration of Areas and Perimeters . . 36-49 Mensuration of Solids . 49-59 Centres and Moments of Figures . . 59-69 Moments of Inertia and Eadii of Gyration . 69-75 Mt'cbanical Principles 76-79 ('(■ntrp of Gravity .... 80 Motioii ...... 81-83 i )ynamics ...... . 84-88 Hydrostatics ..... 88-89 1 >i.s))lacement, etc. .... 90-102 Weight and Centre of Gravity of Shii)s 102-109 Ma])ilit\ 110-143 Waves ...... 143-149 Rolling ...... 150-160 Speed and Horse-power 160-190 Propellers ...... 190-197 Speed Trials and Tables 197-207 Sailing, Force of Wind 208-211 -V r>uiui»ii\.xvx ur v^UiN PAGES. Distances down Rivers . 212-221 Weights and Dimensions of Materials . . 222-237 Wire and Plate Gauges . 238-240 British Standard Sections . . 241-254 Notes on Materials . 255-259 Weight and Strength of Materials . 260-263 Admiralty Tests, etc., for Materials . . 263-284 Lloyds' Tests for Materials . . 284-285 Riveted Joints and Rivets . . 285-294 Braced Structures . 294-300 Shearing Forces and Bending Moments of Beams .... . 300-308 Strength of Materials and Stresse s, etc. : General .... . 309-312 Bending .... . 312-327 Compression . 328-331 Shear .... . 332-337 Miscellaneous . 337-339, 344 Keys and Wheel Gearing . 340-344 Longitudinal Stresses in Ships . 345-352 Mechanical Powers . 353-363 Notes on Steering, Rudders, etc. . 364-371 Launching . 372-378 Armour and Ordnance . 378-388 Notes on Machinery . . 388-390 Notes on Design . 391-394 Fans and Ventilation of Ships . 395-401 Hydraulics . 402-404 Heat . . . 405-406 Aerodynamics (Forces on Plates, etc.) . . 406-431 SUMMARY OF CONTENTS. XI Aeronautics (Notes on Airships, etc.) . 431-448 B. of T. Regulations for Marine Boilers, etc. 448-466 „ ,, for Motor Passenger Vessels 466-469 „ for Ships .... 469-488 Strength of Bulkheads .... 483-487 International Regulations for Preventing Collisions at Sea .... 487-490 Tonnage 490-495 B. of T. Rules, etc., for Life-saving Appliances 495-502 „ ,, for Emigrant Ships . . 502-505 Lloyds' Rules for determining Size of Shafts 506 „ for Ships .... 507-519 „ ,, for Yachts of the International Rating Classes . . . 520-526 Anchors and Cables 527-533 British Standard Pipes and Screws 533-537 Ship Fittings 538-556 Seasoning and Measuring Timber . . 557-561 Miscellaneous Data . . . . . 562-566 Dimensions and Weights of Blocks . . 567-571 Weight and Strength of Hemp and Steel- wire Rope 572-583 Lloyds' Rules for Yards, Masts, Rigging, etc. 584-593 Distances of Foreign Ports from London . 594 Paints, Caulking Varnishes, Galvanizing, etc. 595-605 English and Foreign Weights and Measures 606-627 Decimal Equivalents . . . . . 628-631 Foreign Money 632-633 Discount and Equivalent Price Tables . . 634-635 Useful Numbers and Ready Reckoners . 636-638 xu SUMMARY OF CONTENTS. Tables of Circular Measure Tables of Areas of and Circumferences of Circles ..... Tables of Areas of Segments of Circles Tables of Squares and Cubes and Roots of Numbers ..... Tables of Logarithms and Antilogarithms Tables of Exponential and Hyperbolic Functions . . Tables of Hyperbolic Logarithms Tables of Natural Sines, Tangents, etc. Tables of Logarithmic Sines, Tangents, etc. Index 639-64; 642-65] 652-654 655-69^. 700-70". 708-71( 711-715 716-71S 720-728 725-742 ii,|| MACKROW AND WOOLLAED'S POCKET BOOK FORMULiE, EULIS, AND TABLES FOR NAVAL ARCHITECTS AND SHIP-BUILDERS. SIGNS AND SYMBOLS. ' The following are some of the signs and symbols commonly used in algebraical expressions: — = This is the sign of eqaality. It denotes that the quantities so connected are equal to one another; thus, 3 feet= I yard. + This is the sign of addition, and signifies plus or more ; thus, 4 + 3 = 7. — This is the sign of subtraction, and signifies minus or less ; thus, 4-3 = 1. X This is the sign of multiplication, and signifies multiplied by or into ; thus, 4x3 = 12. -^ or / This is the sign of division, and signifies divided by ; thus, 4-^2 = 2 or 4/2 = 2. {} [] These signs are called brackets, and denote that the quantities between them are to be treated as one quantity; thus, 5{3(4 + 2)-6(3-2)|=5(18-6) = 60. This sign is called the bar or vinculum, and is sometimes used instead of the brackets ; thus, 3(4 + 2) — 6(3 — 2) x 5 = 60. Letters are often used to shorten or simplify a formula. Thus, supposing we wish to express length x breadth x depth, we might put the initial letters only, thus, Ix bx d, or, as is usual when algebraical symbols are employed, leave out the sign x between the factors and write the formula l.b.d. When it is wished to express division in a simple form the divisor is written under the dividend ; thus, (x + y)-rz = ^^-iti^ Z SIGNS AND SYMBOLS. ! , ,! ! , -'^ y THese'-are sigr^s of proportion; the sign : = is to, the sign :: = as ;* thus," 1 ': 3^ ':: 3 : 9, 1 is to 3 as 3 is to 9. < This sign denotes less than ; thus 2 < 4 signifies 2 is less than 4. > This sign denotes more than ; thus 4 > 2 signifies 4 is more than 2. *.* This sign signifies because. .*. This sign signifies therefore. Ux.: ',' 9 is the square of 3 .*, 3 is the root of 9. ~ This sign denotes difference, and is placed between two quantities when it is not known which is the greater ; thus (x ~ y) signifies the difference between .r and ij. , ', These signs are used to express certain angles in degrees, minutes, and seconds ; thus 25 degrees 4 minutes 21 seconds would be expressed 25° 4' 21". JVbte. — The two latter signs are often used to express feet and inches; thus 2 feet 6 inches may be written 2' 6". \/ This sign is called the radical sign, and placed before a quantity indicates that some root oi it is to be taken, and a small figure placed over the sign, called the exponent of the root, shows what root is to be extracted. Thus ^/a or \^a means the square root of a. ^a „ cube „ 4/a „ fourth „ -^^ This denotes that the square root of a has to be taken and divided by h. -— - This denotes that h has to be divided by the square Va root of a. 'a+ ' y: This denotes that the square root oi a+h has to be a + d divided by the square root of a + d. It may also be written y a + b a-\rd thus. Z^^;-^., or h \/a + h oc This is another sign of proportion. Ex. : acch; that is, a varies as or is proportional to b. CO This sign expresses infinity ; that is, it denotes a quantity greater than any finite quantity. Q This sign denotes a quantity infinitely small, nought. L This sign denotes an angle. Ex. : l_ abc would be written, the anftie abc. SIGNS AND SYMBOLS. -3 L This sign denotes a right angle. _L This sign denotes a perpendicular ; as, ab Led, i.e. ah is perpendicular to cd. A This sign denotes a triangle; thus, Aabc,\.e. the triangle abc\ II This sign denotes parallel to. Ex. : ab \\ cd would be written, ab is parallel to cd. f ox F These express a function ; as, a =/(.r) ; that is, a is a function of x or depends on x. / This is the sign of integration ; that is, it indicates that the expression before which it is placed is to be integrated. When the expression has to be integrated twice or three times the sign is repeated (ihus,JX,///) ; but if more than three times an index is placed above it (thus,/"). D ord These are the signs of differentiation ; an index placed above the sign (thus, d') indicates the result of the repetition of the process denoted by that sign. 2 This sign (the Greek letter sigma) is used to denote that the algebraical sum of a quantity is to be taken. It is commonly used to indicate the sum of finite differences, just as the symbol/ is used for indefinitely small differences. g This sign is used to denote the acceleration due to gravity at any given latitude. Its value is about 32-2 in foot-second units and 981 in C.G.S. units. TT The Greek letter pi is invariably used to denote 3-14159 ; that is, the ratio borne by the diameter of a circle to its circumference. e or e This letter is generally used to denote 2-71823, which is the base of hyperbolic or Napierian logarithms. I n or n ! termed ' factorial n ', where n is a positive integer, denotes the product of the series n (n-1) (n-2) . . . 2.1. Thus, [£=3.2. 1 or 6; and [_5_= 5 . 4 . 3 . 2 . 1 = 120. }i denotes the midship section or midship part of a vessel. As the letters of the Greek alphabet are of constant recur- rence in mathematical formulas it has been deemed advisable to append the following table : — Ao Alpha. I I Iota. Pp Rho. B)8 Beta. K K Kappa. 2 = sinh — = — sec * = cosh — = — c c c c The values of the hyperbolic functions (sinli x, cosh a:, etc.) are tabulated on pp. 708-10. Examples showing their application to the catenary are given below. Approximate Equations for flat piece of chain, nearly horizontal. 5 = dip DC = 7/-c; a; = i span. '""2 c '24c''~'»'V2 \ x^ 1 . s* - 82 ^ = 2 2 1 a:^ . 2 82 2-c +2i?=v(2^(^-^); , 1 , s*-82 a/_A_ + 6 '="^^=^^67^::^ ^ = ^ + 6 ^^=^+ 3 X Tension* = ^ total weight x span -f sag at centre. -^oie.— When the points of support are in the same horizontal plane, the catenary is symmetrical about a vertical line ^passing midway between them, and the preceding formulae can be directly employed to determine the particulars of the curve. Ex. 1. — A chain weighing 15 lb. per foot run is suspended between two points at the same level and 100 feet apart. The dip is observed to be 40 feet. Determine the length of chain, the maximum tension, and the inclination at the supports. Dip = c (cosh— - l). Here dip = 40 ; ^=—^ = 50. Hence 40 = c (cosh— - l). By trial, from the tables (p. 709), c = 36 approximately. Length of chain = 2s = 2c sinh— = 135 feet approximately. c * On substituting ' pressure ' for ' weight ', this is applicable to a rope or net under uniform pressure when sag is moderate. EQUATIONS TO THE CATENARY. 17 Maximum tension occurs at supports and is given by — X T = wc cosh 11501b. Angle at supports =

^ X— fo- T .79.. — "1 mZ The radius of curvature {p) at a point is given by — d^/ dx^ dx^ — d^ dy" When tlie curve is tangential to the x axis at the origin, then x^ P is equal to — when x and y are very small. Integral Calculus. Definition. — If ^ is the differential coefficient of a funcfcioD 2 Vr-ith respect to x, then, conversely, z is termed the integral o? y with respect to cc, being denoted Jy dx. A tabl6 of integrals frequently required is appended. dx « or \ !/ . dx dz z ov \ y . dx a* cos X cot X 1 1 x'i 1 -a^ Va;2 + a2 Va2- a;2 Va;2-a2 sec X 1 a2-a;2 tt^/logr^ a sin a; log sin X . _i x sm 1 - a a sinh " ^ a; einx tan a; _J 1 ;Va2: logtan(^+|) 1 , g+a ; 2^ ^°^ ^a: V2aa:-a;2 1 Va;2-a2 Va2 + x2 cosec a; 1 log a; —cos X log sec X a a cosh"' — ■ a xVrt''i + a;2 . a2 2^^"^'' a log tan ^ 1 , x-a n\v.dx-\{f^.]v.dx)dx INTEGRAL CALCULUS. 21 Definite Integrals. — On evaluating the integral \y . dx .\t two constant quantities a and h are separately substituted for a;, and the former result substracted from the latter, the difference is termed a definite integral between limits a and h, being denoted y . dx. Thus {x^.dx = -j, and / x^ dx = — j — = 16i. The following definite integrals are of frequent occurrence: — 2 cose .dd = 1 J B'mO.de = 1 J J J n n-2 ■: cos^ a . cf0 = 7 ^ I - when 00 ^ n is an odd integer. n-1 n-3 3 1 IT , = — ^ • — 2 * • '4*2*2 ^^^° *^ ^^ ^^ ®^'®" integer. 2 sin ^e . cos "0 . de (m and n being integral) _ hn-l){m-3) . ■ . (n-Djn-S) . . ~ {m + n)(in + n — 2){m + n-i) . . . n is odd. when either m or (7;i-l)(m-3) . . . (n-l)(rt-3) . . . tt : (,n^n)(m + n-2){m + n 4) . . . " ¥ ^^"° ^'^ ^"^ 71 are both even. 22 GEOMETRY. PRACTICAL GEOMETRY. 1. From any given 2Joint in a straight line to erect a per pen dicv lar. ( Fig. 1 2 . ) On each side of the point A in the line from which the perpendicular is to be erected set off equal distances xb, Ac ; and from b and c as centres, with any radius greater than Ab or Ac, describe arcs cutting each other at d,d'; a line drawn through dd' will pass through the point A, and Ad will be perpendicular to be. 2. To erect a perjfendicular at or near the end of a line. (Fig. 13.) With any convenient radius, and at any distance from the given line ab, describe an arc, as BAG, cutting the given point in A ; through the centre of the circle N draw the line BNC : a line drawn from the point A, cutting the intersection at c, will be the required perpendicular, 3. To divide a line into any mimber of equal paHs. Let AB be the given straight ^la. 14. line to be divided into a number of equal parts ; through the points A and B draw two parallel lines Ac and DB, forming any convenient angle with ab ; upon AC and DB set off the number of equal parts re- quired, asA-1, 1-2, &c.,B-l, 1-2, &c. ; join A and d, 1 and 3, 2 and 2, 3 J^-"/ S-""/ D and -jk:^^ 1, c and B, cutting AB in a, b, and c, which will thus be divided into four equal parts. 4. To find the length vf any given arc of a circle. (Fig. 15.) With the radius Ac?, equal to one- fourth of the length of the chord of the arc AB, and from A as a centre, cut the arc in c ; also from b as a centre with the same radius cut the chord in b ; draw the line cb, and twice the length of the line cb is the length of the arc nearly. 5. 'To draw from or to the cir- cumference of a circle lines tend- ing towards the centre, 7vhen the centre is inaccessible. (Fig, 16.) Divide the given portion of the circumference into the desired number of parts ; then with any radius greater than the distance of two parts, describe arcs cutting each other as Al, cl GI'OMK'niT. Fig. 17. draw the lines Bl, c2, etc. , which will lead to the centre, as required. To draw the end lines Ar', Yr, from B and E with the same radii as before describe the arc r' , r, and with the radius Bl, from A as centre, cut the former arcs at r', r ; lines then drawn from Ar' and Fr will tend towards the centre, as required. 6. To describe an arc of a circle of large radius. (Fig. 17.) Let A,B, c be the three points through whicli the arc is to be drawn ; join ba and BC ; then construct a flat trian- gular mould, having two of its edges perfectly straight and making with each other an angle equal to ABC. Each of the edges should be a little longer than the chord Ac. In the points A, c fix pins ; and fix a pencil to the mould at b, and move the mould so as to keep its edges touching the pins at A and c, when the pencil will describe the required arc. 7. Another method. (Fig. 18.) Fig. 18. Draw the chord ADC, and draw EBF parallel to it ; bisect the chord in D and draw db per- pendicular to AC; join AB and BC ; draw AE perpendicular to 2 o 2 r~c AB and CF perpendicular to BC ; also draw A;i and c/i perpendicular to AC ; divide AC and EF into the same numbei' of equal parts, and A/?, en into half that number of equal parts ; join 1 and 1, 2 and 2, also B and s, g, and B, and t, t ; through the points where they intersect describe a curve, which will be the arc required. 8. To describe an ellipse, the major and minor axes being given. (Fig. 19.) Let AB bo the major and CD the minor axis, bisecting each other at right angles in tlie centre e ; from c as a centre, with ea as radius, describe arcs cutting AB in f and f', which will be the foci of the ellipse ; between b and F set off any number of points, as 1, 2 (it is advisable that these points should be closer as they approach f). From F and f', with radius Bl, describe the arcs G, G', g", g'". From F and f', with radius Al, describe the arcs H, h', h", h'", intersecting the arcs G, g', g", G"'in the points i, i, I, i, which will be four points in the curve. Then strike arcs from f, f' first with a2, then with b2 ; these radii intersecting will give four more points. Proceed in this way with all the points between e and F; the curve of the ellipse must then be traced through these points by hand. 24 GEOMETRY. Fig. 20. 9. Another method. (Fig. 20.) Let AB and cd be • the axes ; find F, f', the two foci as before ; join CF, cf' ; make an endless thread equal in length to the perimeter of the triangle cff', and passing it round two draw- ing-pins at F and f', draw it taut by means of a pencil-point p, so as to make a triangle pff' equal in peri- meter to cff' ; move the pencil-point P along, keeping the thread taut, and the required curve will be described. 10. Another method. (Fig. 21.) (Ap- proximate.) At 0, the intersection of the two dia- meters, as a centre, with a radius equal — to the difference of the semi-diameters, describe the arc ab, and from J as a centre with half the chord bca describe the arc cd ; from o as centre with the distance od cut the diameters in dr, dt ; draw the lines rs, rs, ts, ts, then from r and t describe the arcs SDS, scs ; also from d and d describe the smaller arcs BAS, SBS, which will complete the ellipse required. This method is applicable when the minor axis is at least § the major. 11. To draw a tangent and a normal to an ellipse at any point. (Fig. 22.) Fig. 22. Let G be the point; from f, f', the two foci of the ellipse, draw straight lines through G and produce them ; bisect the angle made by the pro- duced parts, by on, then gii is normal to the curve ; at G bisect the angle formed by fg and f'g produced, by 15, then IJ will be the tangent to the curve at G. 12. given. Fig. 23. To desorihe an eJUptlo arc, the sp'cn and height betjig (Fig. 23.) (Approximate.) Bisect with a line at right angles the chord or span AB ; erect the perpendicular AQ, and draw the line QD equal and parallel to AC ; bisect AC in c, and aq in n ; make CL equal to CD, and draw the line lcq ; draw also the line nsu, and bisect SD with a line KG at right angles to it, and meeting the line LD in G ; draw the line GKQ, and make cp equal to CK, and draw the line G^2 ; then from G as centre with the radius GEOMKTIIT. 25 Fig. 24. I GD flescribe the arc sd2, and from K and p as centres with the radiu3 ak describe the arcs as and 2b, which complete the arc, as required. This method gives good results for cUipsev^ of all proportions. 13. Another method. (Fig. 24.) Bisect the major axis ab, and fix at right angles to it a straight guide, as bo ; prepare of any mat-crial a rod or staff, def ; at / fix a pencil or tracer so that df is one-half ab, and at E fix a pin so that ef is one-half the minor axis ; move the st-aff, keeping its end d to the guide, and the pin e to ab, and the tracer will describe a half of the arc required. 14. To describe a parabolic aro when its height and base are given. (Fig. 25.) Let CD be the base and ab the height; set them off as shown in the figure, so that cb = CD, and complete the rect- .angle cdfe ; divide EC and fd into any number of equal parts, say three, at a, b, c, and d; join Aa, Ab, ag, Ad ; divide ae, af, bc, and bd into the same number of equal parts at e, g, k, Jn,f, h, I, n; join e/, gh, hi, mn, cutting Ab, Aa, Ac, Ad at q, p, r, and s. A line drawn through c q p A r s y> will i;e the parabola required. 15. Another viethod, . when the directrix and focus are given. (Fig. 26.) Place a straight-edge to the direc- trix ab, and apply to it a square cde ; to the end c of the square fasten a thread, and pin the other end to S the focus, making the length of the thread equal to ce ; slide the square along the straight-edge, holding the thread taut against the edge of the square by a pencil P, by which the curve is de- scribed. Fio 25. A- A k 1 r w '^ 5 \ Fig. 26. IG. To describe a hyperbola, the diameter, abscissa, and double ordinate being given. (Fig. 27.) Let AB be the diameter, BC its abscissa, and DE its double ordinate ; then through b draw gf parallel and equal to de ; draw also DO and ef parallel to the abscissa BC. 26 GEOMETRY. Divide DC and CE into the same number of equal parts, as 1,2, &c., and from the points of division draw lines meeting in A. Divide gd and ef each into the same number of parts as DC or c*E, and from the points of division 1', 2', &c., draw lines meeting in b. Fio 27. /ll'&'uV, _«=• I \ \\\\ . /'// ' ! ! 1 \ \\\ ^JJ // 1 I \ \ \ \ \\>. / ; I I I \ \ \ i3 / / I I \ \ \ / / I I \ V \ \i i._(_ J__L_i _\ \ 6 3 S- I C * ^ a E The points of intersection of the lines 1 and 1', 2 and 2', &c., thus found, will be points in the required curve. 17, Another method, when the fool and a point on the curve are given. (Fig. 28.) A hyperbola is a curve such that the difference of the distances of any point in the curve from the two foci is equal to the transverse axis ; and this pro- perty suggests the following me- chanical construction : — Let P (fig. 28) be any point on the curve, and F and F, its foci; join FF, and produce it, making XX' the axis ; draw pm perpendicu- lar to XX', and produce it to Q, making mq equal to pm ; bisect ff, at c, and produce PC, QC, to CP', CQ', making them equal to CP or CQ. The p, p', Q, q' are four points on the curve. From one of them, say P, stretch two pieces of string pf and PF,, fastening them to the paper at F and F„ and simply knotting them at P ; slip a small bead over them at P, and taking hold of P and keeping the thread taut, slide the bead along the threads, and the bead will describe the curve as far as the axis. Kepeat this process at p', q, and q'. GEOMETRY. 27 18. To describe a rectangular asympfoles and a 'point on the curve. Let ox, OY be the asymptotes, and P the given point. Draw rar parallel to OY, and Ps parallel to ox ; set off any ordinates(generally equi- distant for convenience) 11,22,33, 44,55, 66, and join to the intersections of these ordinatea with PS, cutting PM at 1', 2' hyperbola, given tha (Fig. 29.) Fig. 23. 6 5^ B'', etc.; through 1' draw I'l parallel to ox, cufctinj 11 in through 2', 2'ii cutting 22 in ii, and so on for m, iv, v, and vi ; then P, I, II, III, etc., are points on the required curve. 19. Given five points on any conic to obtain any number of additional points desired. (Fig. 30.) Denote the given points by a, b, c, d, e. Draw any line pap' through A, on which it is required to find a sixth point. Fig. 30. Join AB, DE, cutting at x, and CD cutting pap' at Y. Join BC, cutting XY in z; join EZ, cutting pap' in F. f is the required point on the curve ; by drawing additional linos through A, any number of points on the curve may be obtained., 20. To construct a catenary approximately. (Fig. 30a.) Let E be the lowest point in the curve, oe its parameter, and ox its directrix. Make ae ecjual to OE; then with a as 28 GEOMETRY. centre and ae as radius describe the small arc ef. Join fa and produce it to M and to b, makin* bf equal to FM; then with B as centre and bf as radius describe the small arc fg. Fia. 80a. Join BO and produce it to N and to c, making CG equal to GN ; then with c as centre and CG as radius describe the small arc Gn. Proceed in a similar manner till the curve is of the required length. 21. To obtain hy measurement the length of any direct line, though intercepted by some material object. (Fig. 31.) , Suppose the distance between ^^^ 3^- A and B is required, but the straight line is intercepted by the object G. On the point d, with any convenient radius, describe the arc cc', and make the arc twice the radius dc in length ; through c' draw the line dc'e, and on e describe another arc ff equal in length to the radius dc ; draw the line efr equal to efd ; from r describe the arc g'g, equal in length to twice tho radius rg\ continue the line through rg io m '. then A and B will make a right line, and de or er will equal the distance between dr, by which the distance between ab is obtained, as required. 22. To ascertain the distance geometrically of an in- accessible object on a level plane. (Fig. 32.) Let it be required to find the distance between a and c, A being inaccessible. Produce ab to any point d, and GEOMETKY. 29 bisect «T> m C ; through d draw Fio. 82. Da, making any angle with DA, and take DC and db respectively and set them off on Dor as D 6 and do ; join Be, cb, and a6 ; through e, the inter- section of Be and cb, draw def meet- ing a6 in F ; join bf and produce it till it meets Da in a : then ab will be equal to ab, the distance required. 23. Another method. (Fig. 33.) Fig. Produce ab to any point d ; draw the line T)d at any angle to the line ab ; bisect tlie line Dd in C, through which draw the line B&, and make Cb equal to Bc; join AC and db, and produce them till they meet at a : then ba will equal ba, the distance required. 24. To measure the distance between Uvo objects, both being inaccessible. (Fig. 34.) Let it be required to find the distance ^iq. 34. between the points a and B, both being in- accessible. From any point c draw any line Cc, and bisect it in d ; produce ac and Be, and prolong them to E and F ; take the point E in the prolongation of Ac, and draw the» line EDe, making De equal to de. In like manner take the point F in the prolongation of Be, and make d/ equal to df ; produce ad and ec till they meet in a, and also produce bd and /c till they meet in b: then the distance between the points a and b equals the distance between the inaccessible points a and B. 25. To cut a beam of the strongest section from any round piece of timber. (Fig. 35.) Divide any diameter CB of the circle into three equal parts ; from d or e, the two points of division in CB, erect a perpendicular cutting the circumference of the circle in d or a ; draw CD aad db, also ac equal to db and ab equal to CD : the rectangle abcd will be the section of the beam required. Note. — To get the stiff est beam make cd = ^ cb and proceed as before. FlQ. GEOMETRY. 26. To describe the proper form of a flat plate by which to construct any given frustum of a cone. (Fig. 36.) Let ABCD represent the required frustum of a cone ; continue the lines AC and BD till they meet in E ; from E b-s a centre, with ed as radius, describe the arc dii, and from the same centre, fwith EB as radius, describe the arc Bi; make Bi equal in length to twice agb, equal to the circumference of the base of the cone ; draw the line ei : then BDHI is the form of the plate required. 27. To find the development of the frustum of a right cone tchen cut by an angle inclined to the base. (Fig. 37.) Let ABCD represent the required frustum of the cone ; continue the lines AC and BD till they meet at E ; divide the circumference of the base of the cone into any number of equal parts- say 12 — in the points 1, 2, 3, etc. ; join the projections of these points to E ; next find the development of the Is base of the cone, as shown in the pre- ceding example, and on it set off the same number of points — viz. 12 — and draw lines from them to e ; project the points of intersection of each of the lines El, e2, e3, etc., with the line CD, horizontally on to either of the slant sides (say eb) ; then from e as centre measure the distance down along eb to the foot of each projection and set it off on the corresponding numbers (measuring from e) in the develop- ment : a line drawn through these points will give the curve of the top of the section, as required. 28. To find the development of the frustum of a cylinder when cut by a plane inclined to the base. (Fig. 38.) Let ABCD represent the required frustum of a cylinder ; divide the base into any number of equal parts — say 12 — GEOMETRY. 81 and draw lines through those points on the cylinder parallel to AC and BD ; draw a line efg equal in length to the circum- ftr.rence of tlie cylinder, and divide it into the same number ;>ff j^u'ts ; on each point of division set up perpendiculars to Fig. ns. it, making eh and GK equal in length to bd,, and make Fi equal in length to AC; then take the height at 1 and set it up on the corresponding number on each side of Fi, and so on with each number : a line ti*accd through the points thus obtained will be the curve of the required development. 29. To find the approximate development of any given portion of a segment of a (Figs. 39, 40, and sphere. 41.) Let ABC (fig. 39) be the middle section of the seg- ment, and CFG in the plan (fig. 40) the portion to be developed; bisect ab (fig. 39) in E, and set up the perpen- dicular EC ; divide the arc AC into any given number of equal parts — say, four — and through the points of division draw the lines 1 1, 2 2, etc., parallel to ab ; on the plan (fig. 40) from c Pig. 40. as a centre, with the radius 1 1 taken from fig. 39, draw the arcs 1 1 cutting fc and CG in 1 and 1, and so on with 2 2 and 3 3; draw any line bc (fig. 41), making it equal in length to bc (fig. 39), and on it set off the same number of equal parts; at each point of division draw linos perpendicular to bc, and number them the same as on fig. 39. 82 GEOMETRY. Measure the length of the arc 1 1 in fig. 40, and set off half of it on each side of co on line 1 1, and so on with each arc, including FG ; a line traced through the points thus obtained will give the curve of the sides of the given portion of the segment when it is developed. To describe the curve at the bottom of the figure, take one-fourth of the circumference of the base as a radius, and from F and o as centres describe arcs cutting bo in s ; then from s as centre, with the same radius, describe the arc fbg, which w^U bo the curve of the bottom of the figure, as required. Should the top of the figure be cut off at the lino 1 1 (fig. 39), from s as a centre in fig. 41 describe the arc 1h1, which will be the curve of the top of the figure, as required. 30. To -find the a'p'proxlmate development of any given 'portion of a paraboloid. (Figs. 42, 43, and 44.) Fig. 42. The development is found in the same manner as that of a portion of a segment of a sphere, as described in the last example (No. 29), with but one exception — that is, the length of the radius for describing the bottom curve of the figure, which instead of being equal to one-fourth of the circum- ference, as in example No. 29, is equal to one-half the length of the arc acb (fig. 42) in this example. 31. To find the development of an entablature plale. Let fig. 45 be the side elevation, fig. 46 the front elevation, fig. 47 the plan, and fig. 48 the development of the figure ; divide ado (fig. 46) into eight equal parts, and from the points of intersection draw lines parallel to abc, cutting CD (fig. 45) in the points 1, 2, etc.; on bd (fig. 45) erect a perpen- dicular EC, and from the points on CD draw lines parallel to BED. From fig. 46 take the points 1, 2, etc., on abc and set them off on afc (fig. 47), and erect perpendiculars from afc at these points. From c (fig. 45) along ce measure the points c, 1, c, 2, etc., and set them off on their corresponding lines from afc in fig. 47; draw a line through those points, then measure it with its divisions and set it off in fig. 48 as a straight line GEOMETRY. §3 AEC, and at the points of division erect perpendiculars, con- tinuing them either side of the line akc ; measure tiie distances? 1, 1; 2, 2, etc. (fig. 45), on either side of ce, and set them olf Fig. 45 from AEC (fig. 48) on their corresponding lines, and on their respective sides of aec. These will give the development. 32. To describe a cycloid, the generating circle being given. (Fig. 49.) Let b6 be the generating circle ; draw a line abc, equal to the circumference of the generating circle, by dividing the circle into any number of given parts, as 1, 2, 3, eta., and setting off half that number of parts on each side of B; draw lines from the intersections of the circle 1, 2, 3, etc., 7, 8, 9, 34 GEOMETRY etc., parallel to AC ; set olf one division of the circle outwards on the first lines 5 and 7, two divisions on the next lines 4 and 8, then three on the next, and so on : then the intersection of those points on the lines 1, 2, 3, etc., will be points in the curve. 33. To draw a trochoid or wave-form, the height and length being given. (Fig. 50.) Draw AB equal to the length ; with centre C on ab produced describe a circle whose diameter is equal to the height. Divide the circumference into a convenient number (say 12) of equal parts 0, 1, 2, 3, . . . , CO being vertical. Diride AB into the same number of equal parts f, o, H, . . . From A, F, G, . . . B, draw Aa, f/, 0*7, . . , b6, parallel and equal c 0, c 1, c 2, . . . C 0, respectively. A curve drawn through the points a, f, g, . . . b \% the required trochoid. Fig. 50. liLH *, F 6 i^. To describe an epicycloid, the generating circle and the directing circle being given. (Fig. 51.) Let bd be the generating circle, and ab the directing circle; divide the generating circle into any number of equal parts (say 10) as 1, 2, 3, etc., and set off the same distances round the directing circle; draw radial lines from a through these last points, and produce them to an arc drawn with a as centre and AE as radius, as shown by cccc and c'c'c'c' on the diagram ; draw GEOMETRY. 35 concentric arcs also through all the points on the generating circle, with A as centre ; then taking c, c, c, c and c', c', c', c' as centres, and be as radius, describe ai'cs cutting the concentric circles at 1', 2', etc. : the points thus found will be points in the required curve. 35. To describe a hypo- cyrloid, the generatino circle and the directin(^ circle being given, (Fig. 52.) Proceed as in the epi- <5ycloid, the exception being that the construction lines are drawn within the directing circle instead of outside, as in the epicy- cloid. FiO. 52. ^^ 36. To describe the involute of a circle. (Fig. 53.) Let AD be the given circle, which divide into any equal number of parts (say 12) as 1, 2, 3, etc. ; from the centre draw radii to these points ; then draw lines (tangents) at right angles to these radii. On the tangent to radius No. 1 set off from the circle a distance equal to one part, and on each of the Fig. 53. tangents set off the number of parts corresponding to the number of its radius, so that No. 12 has twelve divisions set off on it (that is, equal to the circumference of the circle) : a line traced through the ends of these lines will be the curve required. 36 MENSURATION OF AREAS AND PERIMETERS. 37. To -find the dip of the horizon. (Fig. o3a.) Let denote the centre of the earth, pb a tang-ent from the eye of an observer looking from a height ap to the earth's surface at B; then B is a point on the horizon: draw PC at right angles to po ; then the angle BPC is called the dip of the horizon. Let OP cut the earth's surface at a, and let the angle bpc be denoted by Q ; with distances in miles, PB = V 2 . AP . AG approximately = V 8,000 x AP ; and Q in degrees = 1-28 Vap. MENSUEATION. I. Mensuration of Areas and Perimeters. 1. To find the area of miy ^parallelogram. (Fig. 54.) EuLE. — Multiply the length by the perpendicular height, and the ~'Tt product will be the area. Thus, if /ij A = the area, a = the length, and ' i^ b = the perpendicular height, tlien A = ah. Fig. 54. * L 2. 2'o find the area of a trapezoid. (Fig. 55.) Fig. 55. ...& Rule. — Multiply the sum of the parallel sides by the perpendicular distance between them; half tha product will be the area. Thus, if A = the area, b and a = the parallel sides, and c = the perpendicular distance between tJiem, then a =. (^±_&)£ ' 2 3. To find the area bf any triangle. (Fig. 56.) Pjq gg EuLE. — Multiply the base by the per- pendicular height ; half the product will be the area. Thus, if a = the area, b = the base, and d = the perpendicular height, O-. — TT^ then A = — - 4. Or, if the lengths of the 3 sides a, h, and c are given, then = \/&{s- a) (5 - 6) (s - c) where 2s = a + 6 + c. 5. To find the area of any regular 'polygon. (Fig. 56a.) Rule. — Multiply the sum of its sides by a perpendicular drawn from the centre of the polygon to one of its sides ; half the product will be the area. Thus if A = the area, c = the number of sides, 6 = the length of one side, and a = the perpendicular, then A = -r- FiG. 56a. MKNSUUATIOX OF ARKAS AND PERIMETERS. 37 Table of Kegular Polygons. A = the angle contained between any two sides. R = the radius of the circumscribed circle. r = the radius of the inscribed circle. s = the side of the polygon. 5s ^2 Name 23M A R = SX r=sx S = RX s = rx Area=8^ .3 Trigon 60° •57735 •28868'l-73205'3-46410' -43301 4 1 Tetragon . 90° •70711 -50000 1-41421 2-00000 1-00000 5 Pentagon . 108° •85065 -68819 1-175.57 1-45309 1-72048 6 Hexagon . 120° 1-00000 •86603 1-000001-15470 2-59808 7 1 Heptagon . 128|° ;i-1.52.38'l-0.S826 -86777, -96.315 3-63.391 8 j Octagon . 13.5° 1-306.56:1-20711 -76537! -82843 4-82843 i) Nonagon . 140° 1-46190 1-37374 -684041 -72794 6-18182 10 Decagon 144° 1-61803 1-.5.3884 -61803 -64984 7-69421 11 Undecagon 147fi° 1-77473 1-702841 -56347 -58725 9-,36,564 12 1 Duodecagon 150^ 1-93185 1-866031-51764 -5359011-19615 6. To find the area of a quadrilateral. Rule. — Multiply the diagonal d by the sum of the two perpendiculars a and h let fail upon it from the opposite angles; half the product will be the area. Thus if A = the area, a and J = the perpendiculars, and ^ s/ ^ Area ^ r2 yjr = d' x - = i j^n 4 Side of equal square =r x \/7r = D x ^v/7r = C x ^^' - Hide of inscribed square =Dx \/^ = cx -\/l= ^/ A x ,^/ - Diameter of equal circle = s x 2^ - Diameter of circumscribing circle = s x \/2 Circumference of circumscribing circle -s x ■jr\/2 Circumference of equal circle = s x 2\/ir 2 Area of inscribed square = A x - 40 MENSURATION OF AREAS AND PERIMETERS. 12. To find the area of a sector of a circle. Rule (I.) — Multiply the length of the arc by the radius of the sector, and half the product will equal the area. A = area of sector, l = length of arc, R = radius. Rule (II). — Multiply the number of degrees in the arc by the area of the circle, and g^^of the product will equal the area. Thus, if a = area, D = number of degrees in the arc a = area of circle, then _ J^ ~ B60 13. To f.nd the area of the segment of a circle. Rule (I). — Find the area of a sector having the same arc as the segment ; then deduct the area of the triangle con- tained betv.-een the chord of the segment and the radii of the- sector. The remainder will be the area of the segment. Thus, if A = the area of the segment, c = the choi-d, and H = the height, then Rule (II). — To two-thirds of the product of the chord and height of the segment, add the cube of the height divided by twice the chord ; the sum will be the area of the segment^ nearly. Thus, /2cH , H^\ Pj(5 gi 14. To find the area of a circular zone. ^,..'>,; (Fig. 61.) ^ "^N Rule. — Find the area of the circle of \ which the zone forms a part, and from it \ subtract the sum of the two segments of the A J circle formed by the zone ; the remainder I will be the area. Thus, if \ — area of the V / zone, a and b = the area of the two seg- ^ ^y ments respectively, and c = area of the circle, then \ = c — {a -\- h). 15. To find the ar&a of a flat circular ring. (Fig. 62.) Rule. — Multiply the sum of the inside and outside diameters by their difference, and the result by "7854 ; MENSURATION OF AREAS AND PERIMETERS. 41 Fra. 62. the product last obtained will be the area. Thus, if A = area of ringf, i) = diameter of large circle, and d = diameter of small circle, then A= •7854{{D + d) {D-d)} 16. To find the area of an ellipse. (Fig. 63.) Rule. — Multiply together the trans- verse and conjugate diameters of the ellipse, and the result by - or -7854 ; the product will be the area. Thus, if A = area of ellipse, a = the conjugate diameter, and 6 = the transverse diameter, then A = -ab X -7854. Fig. 63. 17. To find the area bounded by a rectangular hyperbola, two ordinates and the base. (Fig. 8, p. 15.) Rule. — Multiply the product of either ordinate and the corre- sponding abscissa by the hyperbolic logarithm of the ratio between the two abscissae. Thus the area of ABCD is equal to ab x OB 00 lo; OB 18. To find the area bounded by a cycloid and the line joining the cusps. (Fig. 11, p. 18.) Rule. — Multiply the area of its generating circle by 3 ; or multiply the product of its length and height by f . 19. To find the area bounded by a trochoid and a line joining the crests. (Fig. 50, p. 34.) Rule. — If r be the radius of the rolling circle (or the length divided by 2ir), and r the radius of the tracing circle (or one-half the height from crest to trough), the required area is equal to Trr (2r +r). Note. — The area between the curve and the line joining the troughs is vr {2R-r). 20. To find the area of a segment of a parabola. Rule. — Multiply the base by § of the maximum height. 42 MENSURATION OF AREAS AND PERIMETERS. 21. To find a general expression for the area of any plane curve. Using cartesian co-ordinates, the area intercepted between the curve, the x axis, and two ordinates distant a and 6 from the origin, '6 is equal to the definite integral / ?/ . dx. J a Using polar co-ordinates, the area intercepted between the curve and two radial lines making angles a and fi with ox, is equal to I fr^ . d6. J o. Remark.— A. curve whose equation is given by y = a + bx + cx^ + dx^ + . . . Ka;" is said to be a parabolic curve of the n*^ order. Thus a parabolic of the first order is a straight line ; of the second order a common parabola. Eules for the area of a parabola of any order are applicable also to curves of a lower order, but not in general to those of a higher order. 22. To find the area of a parabola of the third order when three ordinates are given. (Fig. 64.) Rule. — To the sum of the two endmost ordinates add four times the intermediate ordinate ; multiply the final sum by J of the common interval between the ordinates. The ^■^ result will be the area. Thus, if 2/1, Vi, and 1/3 be the ordinates. Ax the common interval, and X the area, then Fig. 64. \ydA \yd 'x=-f{yi + ^y-2+yi)' Note. — This is termed Simpson's first rule, 23. To find the area of a parabola of the third order when four ordinates are given. Rule. — To the sum of the two end- most ordinates add three times the intermediate ordinates ; multiply the final sum by § of the common interval K^ between the ordinates : the result will be the area. Thus, if \ydx = ih.e area, then SAa!/ Fig. 65 y y^ y-? AJB A.r AJC \ydx-- 8 ■(Z/l + 3l/2 + 3?/3-i-2/4). Note. — This is termed Simpson's second rule. MENSURATION OF AREAS AND PERIMETERS. 43 Table showing the Multipliers for the foregoing AND SOME other RuLES, yu yi, ys, etc. = the ordinates, and Ax = the common interval or abscissa between the ordinates. 1. Trapezoidal rule. Area = -—[yi + y^). 2. Rule lor parabola of the third order with three ordinates. Area = -^ (t/i + 4?/2 + l/s) • (Simpson's first rule.) o 3. Rule for parabola of the third order with four ordinates. Area = —— (i/i + 3?/2 + 3?/3 + 2/4). (Simpson's second rule.) o 4. Rule for parabola of the fifth order with five ordinates. Area = -J^i'^Vi + 32?/2 + 12?/3 + 32?/4 + ly^) ■ 5. Rule for parabola of the fifth order with six ordinates. 5 Ax Area = -— {19yi + 75?/2 + 50ys + 50yi + Ibys + IQi/e) . 6. Rule for parabola of the seventh order with seven ordinates . Area = ^ (41yi + 216t/2 + 272/3 + 2721/4 + 27^6 + 2162/6 + 41^7) ■ 7. Approximation for curve with six ordinates. Area = -^^ {O'^yi + y-i + t/a + 2/4 + 2/5 + 0*42/6) . 8. Weddle's approximation for curve with seven ordinates. 3 Aa; Area = — - {yi + 52/2 + 2/3 + 67/4 + 2/5 + 62/6 + 2/7) • 24. To measure any curvilinear area by means of the tra- pezoidal rule. Rule. — To the sum of half the two endmost ordinates add all the other ordinates, and multiply the sum by the common interval ; the result will be the area. Thus, \ydx=Ax(y'-y^ + y2 + ys . . . . + 2/n-i) Remark. — In Bhipbuilding work it is very often convenient to perform the additions in the above rule mechanically, by measuring off the ordinates continuously on a long strip of paper, and measuring the total length on the proper scale. This rule is only approximate, but it is especially useful for getting the areas of the transverse sections in the first rougl? calculations of trim, displacement, etc. 25. To measure any curvilinear area by means of Simpson's first rule. Rule. — To the sum of the first and last ordinates add four, times the intermediate ordinates and twice all the dividing 44 MENSURATION OF CURVILINEAR AREAS. oidinates ; multiply the final sum by ^, the common interval : the result will be the area. Thus /^ ydx = ~{y, 4 43/2 + 22/3 + 4^, + 2^3 + 4^„_i + yn\ Remark. — The number of intervals in this rule must be even. The ordinates which separate the parabolas into which the figure is conceived to be divided, are called dividing ordi- nates, and all the other ordinates except the two endmost ones are called intermediate ordinates. 26. To measure any curvilinear area by means of Simpson' f second rule. KULE. — To the sura of the two endmost ordinates add three times the intermediate ordinates and twice all the dividing ordinates ; multiply the final sum by |, the common interval, and the result will be the area. Thus / ydx = -^-(//, + 3?/2 + 3^3 + 2?/, + 3y, . . . . + 3y„_i + yn). The number of intervals in this case must be a multiple of three. He/nark. — The sequence of the multipliers in the two fore- going rules is obvious. Thus in the first rule the simple multi- pliers are 1.4. 1, but they are combined thus : — 1.4.1 1.4.1 1.4.1 1.4.1 1.4.1 1.4.1 1.4.2.4.2.4 4.2.4.2.4.1 In the second rule the multipliers are 1.3.3.1. 1.3.3.1 1.3.3.1 1.3.3.1 1.3 3.1 1.3.3.1 1.3.3.2.3.3.2.3.3 3.3.2.3.3.1 And iii the same way the multipliers to measure any curvi- linear area may be obtained from the table on p. 43. Simpson's first rule is superior to the second rule in accuracy as well as simplicity. 27. To measure any curvilinear area when subdivided intervals are used. \st. When Si'mpson's first rule is used. Rule. — Diminish the multiplier of each ordinate belonging to a set of subdivided intervals in the same proportion in which MENSURATION OF CURVILINEAR AREAS. 45 the intervals are subdivided. Multiply each ordinate by its respective multiplier as thus found, and treat the sum of their products as if they were whole intervals ; that is, multiply the sum thus found by i of a whole interval, and the product will be the area. 27id. When Simpson's second rule is used. Rule. — Proceed as in the first rule, but multiply by § of a whole interval for the area. Exaviple to Simpson's First Ride. — The series of multipliers for whole intervals being 1 . 4.2.4.2, &c., those for half- intervals will be f, . 2 . 1 . 2 . 1, &c., and for quarter-intervals Remark. — When an ordinate stands between a larger and a smaller interval, its multiplier will be the sum of the two multipliers which it would have had as an end ordinate for each interval. Thus for an ordinate between a whole and a half interval the multiplier is ^ + 1 quarter interval ^- + J = f . Table of Multipliers when Subdivided Intervals ARE used. Simpson's First Rule. Ordinates |0 |l 2 2^ 2i 2 3 3 j3| 3i 4 !4 6 H 2 7 1~ 7h 2 8 1 Multipliers 1 4 1^ n Ordinates i 1 H 2 2^:3 2~|U 4 4 5 1^ 2 6 1 61 I H i 1 7 1 Multipliers 12 1 2 1 Ordinates o|i. 2 2^:3 3}3| 3f 4 H H 4| ^ 4.^ 6 Multipliers 1 u u 2 f li^ 1 1% 1 ^ 2 3 k 1 i Simpson's Second Rule. 1 Ordinates 1 2 3" 3 1 4 4i;5 15^6 Ujl*:2 2i2^ 6| 2f 8 6S 31 6t 3| f 1 7 J 4 J 6 1 Multipliers 1 3 1 Ordinates Multipliers 1 1 1 J 1 1 i! 1 J 2 Jjjijj 2^ 3 31 ^ 4 J if. 1 Ordiftates Multipliers u t 1 1 1 Note. — The ordinates in this table are numbered the same as if they were the number of intervals from the origin. 46 MENSURATION OF AREAS AND PERIMETERS. ThomsoiVs Rule may be used when subdivided intervals are used at each end ; the advantage being that all multipliers except the three end ones are unity ; so also in the common multiplier. Thus the ordicates should be multiplied by ^, J, ^, 1, 1, 1, ... ... , 1, 1, i^, ^, i ; the spacing of the three ordinates at each end being one half that elsewhere. 28. To calculate the area separately of one of the two divisions of a parabolic figure of the second order. (Fig. 66.) Rule. — To eight times the middle ordinate add five times the near end ordinate, and subtract the far end ordinate ; multiply the remainder by 3^2 *^e common interval : the result will be the area. Note. — The near end ordinate is the ordinate at the end of the division of which the area is to found. Ex. : In figure abcd let it be required to find the area of the division acef. Let y^ = the near end ordinate, ^2 = ^^^ middle \y/ ^ S^s ordinate, and ^3 = the far end ordinate ; \ Asc A tx I then \ydx = -j^{oyi + 8^2 - y&). Fig. 66. y 12 29. To measure an area bounded by an arc of a '^plane curve and two radii. (Fig. 67.) Fig. 67. RuLE. — Divide the angle subtended by the arc into any number of equal angular intervals by means of radii. Measure these radii and compute their half-squares. Treat those half -squares as if they were ordinates of a curve by Simpson's first or second rule, as the number of intervals; C may require. Note. — The common interval must be taken in circular measure. (See pp. 8 and 9.) Ex. : In the figure ABC let vi, r^, ra, Vi, rs = the radii, -dd =: the area ; then (ri2 + 4^22 ^ 2rs^ + W + r.5-) A9 n Id 30. To measure any curvilinear area by nteans of Tcheby- cheff's rule. Rule. — Find the middle point of base, and from it set off, along the base, and in both directions, distances equal to- the half length, of base multiplied by the constants given in the Schedule below. Erect ordinates at the points so obtained and measure them. Their sum, divided by the number of ordinates, and multiplied by the length of base is the area required. MENSURATION OF AREAS AND PERIMETERS. 47 Schedule. Number of Ordinates used. Distance of Orainates from Middle o! Base In Fractions of Half the Base Length. 2 3 4 5 6 7 9 •5773 0, ^7071 •1876, -7947 0, -3745, -8325 •2666, ^4225, -8662 0, -3239, ^5297, -8839 0, -1679, -5288, -6010. -9116 Note. — As evident from the Schedule, there is an ordinate at the middle of base, only when an odd number of ordinates is employed. Examples. — With four ordinates. Let ABCD be the figure. Bisect the base AB at E. Calling the half length of base b, set off EF and ef' equal to ^1876 h and eg and eg' equal to ^7947 h. Erect ordinates GL, fk, f'k', g'l' at G, F, f', g' ; and call them 7/1, 3/2, ys, and y*. (Fig. 68.) K if *+« {f- M Then area of figure abcd Vi + y-2 + ys + Vi X 26. With five ordinates. Fig. G'J. E' K (Fig. 69.) -6 -H" ■b" As before, let ABCD be the : figure, e the middle of base, and b its half-length. Set off EF and ef' equal to -3745 b and EG and eg' equal to -8325 6, and erect ordinates B at G, F, e, f', and g', calling them 2/1, 2/2, 2/3. 2/4, 2/«. Then area of figure ABCD = ^^ + ^^^ + V^ + V^ + V^ ^ 26. 5 Note. — This rule can be used for calculating displacements, and fewer ordinates are required for the same degree of accuracy than if Simpson's rule is used. Ten ordinates are usually employed instead of twenty-one, the rule for five ordinates being applied separately to each half of the ship. It is also of great assistance in preparing cross curves of stability. If eight ordinates are used (four repeated), the following " Simpson " sections, assumed numbered from 1 to 21, can be utilized : 2, 5, 7, 10, 12, 15, 17, 20. iS MENSURATION OF AREAS AND PERIMETERS. 31. To measure any curvilinear area by three ordinate^ irregularly spaced. Rule. — Let abcdef (Fig. 70) be the curvilinear area, whose ordinates ab, fc, ed, are yi, 2/2, and ^3. Let af = 7t, and YE = hh, where fe is a ratio. ^^^[7^17.(2-7.) + y, [h + lY + y^[2U-l)] Note.—li AF = 2fe, so that 7c = I, ' Area = -^ (7/1 + 87/2) \ 32. To find the area hetxoeen the first tivo ordinates of a curvilinear area given three ordinates irregularly spaced. EuLE. — The area included between the ordinates ar and cf. " 6A:(/f+i) ^y' ^' ^^^' + ^^ + '^'^ ^^^ + ^^ ^^' + ^^ " y^^ Note. — If AF = 2fe, so that h = l. Fig. 70. AF Area = — (7 7/i + 15?/2-4?/3) If AF = |fe, so that h = 2, Area = -11(167/1 + 217/2-7/3) 33. To obtain a general exjJression for the length of any plane curve. Using cartesian co-ordinates the length intercepted between two points whose ' x ' co-ordinates are a and h is equal to the definite integral / /y 1 + (y ) . dx. This may be obtained by Simpson's rules in a similar way to the area ; t he ' ordinat e ' in this case being replaced by the value of ^ 1 + (-^) or sec. + tan'^ «// at each intersection. Treat this as the ordinate of a solid, and proceed to find its volume by any of the rules given above. The result is the area required. It is desirable that no part of the surface should be approximately perpendicular to the base. Example. — A portion of a ship's side is bounded by two sections 40 feet apart, and two waterlines 8 feet apart. The tangents of the angles (0) made with the middle line at the flections and at one situated midway between them, and the* 58 MEXPURATI-ON OF SOLIDS. tangents of tlie angles (f) made with the middlo lino at the two watorliuGS, and At one midway, are as follows: — W.L. 1 2 3 Section 1 Section 2 Section 3 tan'/> 1 tan'f' tan

tau'/' •30 j -21 •21 i ^22 •]0 1 -22 •35 •31 •14 •16 •17 •18 •48 •44 •25 •06 •07 •09 nd v^l + tan^

"H Vl + tan«(^ + tan2i|/ £ 1 1-06 1-06 107 1-07 111 1-1 4 1^04 4-16 1-06 4-24 1^10 4.4 1 1^03 1-03 7-25 1 1-03 1-03 6-34 4 1-02 1-0 6-5 7-25 25-36 6-5i = 39^] 20 The area of the curved surface is 39-14 x -jt x = 348 square feet. ^ ^ (II) Approximate method. Rule. — Take girths along (say) the vertical sections at equidistant intervals. For each section in the half -breadth plan, note the angle at wjliich the various waterlines cro^s, and estimate the mean slope of the waterlines surroundingi the surface under question. The secant of this mean anglo with the middle lino is termed the modifying factor, and ia multiplied by the girth concerned. These modified girths are then regarded as the ordinates of a curve, whose area ia the surface required. Example. — In the previous example, the girths at Section? 1, 2, and 3 are 8-2, 83, -uv.d 8*7 feet respectively. The respective mean angles of the waterlines with the middle line are '22, '17, and '07. Find the sur face. The first modifying factor is /sjl + ('22)2 or 1*02 ; similarly the others are 1"015 and 1*0 approximately. CBNTHES AND MOMENTS OF FIGURES. 69 No. Girth Modifying Factor Modified Girtli Simpson's Multiplier Product 1 2 3 8-2 83 8-7 102 1015 10 8-4 8-45 8-7 1 4 1 8-4 33-8 8-7 60-9 20 Area of surface = 50-9 x "«- = 340 square feet approximately. CENTRES AND MOMENTS OF FIGUEES. To FIND THE CeNTBES OF GRAVITY OF A FEW SPECIAL FIGURES. 1. Triangle. (Fig. 91.) Rule. — From the middle points of any two Bides draw lines to the opposite angle ; the point of intersection d of these lines is the required centre. Or, trisect the line joining" the middle point of one side with the opposite vertex ; the point of trisection nearer to the base is the required centre. Fig. 92. 2. Trapezoid. (Fig. 92.) Rule. — Bisect ab in e and CD in P and join ef. Produce ab beyond B to H, making Bii = CD, and pro- duce CD beyond c to i, making ci = AB ; then join nr, and where this line intersects ef is the centre of gravity G. Note. — EG is to OF as 2cD -f ab is to 2ab -f cd. If the angles at a and c are right aisles, the distance of a from ac , . AB^ + AB . CD + CD* '' ^'1'^"^ ^ 3(AB + 0D) 3. Quadrilateral. (Fig. 93.) RuLC. — Draw the diagonals AD and cb intersecting in E ; along CB set off of equal to eb, and join fa and fd ; the centre of the triangle afd will be the centre of the quadrilateral. Fia. 98. 60 CICNTIIES AND MOMENTS OF FIGURES. Fig. 93a. Fig. 94. Or, bisect the diagonal bd (fig. 93a) at E ; join ea, eg. Make ef = ^ea and EG =Jec. Join FG, cutting bd at h. Make kg = hf. Then k is the required centre. 4. Circular arc. (Fig. 94.) Rule. — Let adb be the circular arc and c the centre of the circle of wliich it is a part : bisect the arc ab in d, and Join DC and ab ; multiply the radius CD by the chord ab, and divide by the length of tlie arc adb ; lay off the quotient ce upon CD ; then b is the centre required. Fig. 95. 5. Very flat curved lino (approximato). (Fig. 95.) KuLE.— Let ADB be the arc ; draw the chord ab, and bisect it in C; draw cd perpendicular to ab ; make CE equal to § Df CD ; then E will be the centre required. 6. Sector of a niide. (Fig. 96.) Rule. — Let abo be the sector, e its centre; multiply the chord AB by § of the ^ radius CA ; divide the product by the length of the arc: the quotient equals the distance CE set along the line CD, D being at the bisection of the arc ab. 7. Sector of a plane circular ring. (Fig. 97.) Fig. 97. RuLE. — Let CA be the outer and CE the inner radius of the ring ; divide twice the difference of the cubes of the inner ^' and outer radii by three times the difference of their squares; the quotient will be an intermediate radius CF, with which describe the are ff, subtending the same angle with the sector : the centre H of the circular arc FF, found by Rule 4, will be the centre required. CENTRES AND MOMENTS OF FIGURES. 8. Circular segment, (Fig. 98.) Rule. — Let c be the centre of the circle of which it is a part ; bisect the arc AB in D, and join CD ; divide the cube of ^^ half the chord ab by three times the area of half the segment adij ; set ofiE the quotient ce along CD, and E will be the centre required. Fig. 98. 61 9. Parabolic half-segment. (Fig. 99.) Rule. — Let abd be a half-segment of a parabola, bd being part of a diameter parallel to the axis and ad an ordinate con- jugate to that diameter — that is, parallel to a tangent at b. Make be equal to f bd, and draw ef parallel to ad and equal to I aL'. Then f will be the centre of the half -segment. 10. Height of centre 'if semicircle or semi-ellipse from its base. Rule. — Multiply the radius of the semicircle (or that acmi-axis of the ellipse which is perpendicular to the base) by 4, and divide the product by on. 11. Height of centre of parabola from its base. Rule. — Multiply its vertical height by 2, and divide the product by 5. 12. Prism or cylinder with plane parallel ends. Rule. — Find the centres of the ends ; a straight line joining them will be the axis of the prism or cylinder, and the middle point of that line W(ill be the centre required. 13. Cone or pyramid. Rule. — Find the centre of the base, from which draw a line to the summit ; this will be the axis of the cone or pyramid, and the point at J fromi the base along that line will be tlio centre. 14. Hemisphere or hemi- ellipsoid. Rule. — The distance of the centre from the circular or elliptic base is | of the radius of the sphere, or of that semi- axis of the ellipsoid which is parpendicular to the base. 62 CENTRES AND MOMENTS OF FIGURES. 15. Paraboloid. Rule. — The distance of its centre from the base along its axis is ^ of the height from the base. Fia. 100. 16. To find the centre of gravity of any continuous curved line. (Fig. 100.) Ex. : Let ABC be the given curve ; bisect it at B ; join ab and bc, and bisect those chords at the points d and E respectively ; set off fd per- pendicular to AB_, and EG perpen- dicular to BC ; make fh = ^df and GK==^GE, and join hk ; bisect Hk at the point L, vrhich will be a close approximation to the position of the centre of gravity of the curved line ABC. Remark. — If the line is too irregular to permit its two parts to be regarded as flat regular curves, it should be divided into four or eight parts as required. The points corresponding to L in the above figure are found separately for each pair of parts, joined in pairs and bisected ; this process is repeated until only one point remains, this being the required centre of gravity. eules for finding the moments and centres of Figures. The geometrical moment of a figure, whether a line, an area, or a solid, relatively to a given 'plane or axis is the product of the ^nagnittide of that figure, into the perpendictdar distance of its centre from the given plane or axis, and is equal to the sum of the moments of all its parts relatively to the same plane. The centre of an area is determined when its distance from two axes in the plane of the figure is known. The centre o^ a figfure of three dimensions is determined CENTRES AND MOMENTS OF FIGURES. 68 when its distance from tln-ee planes not parallel to one another is known. 1. To find the moinent of an ir^'cgula?' fignre relatively to a given plane or axis. Rule. — Divide the figui*e into parts whose centres are known ; rcultiply the magnitude of each of its parts into the perpendi- cular distance of its centre from the given plane or axis ; dis- tinguish the moments into positive and negative, according as the centres of the parts lie to one side or the other of the plane : the difference of the two sums will be the resultant moment of the figure relatively to the given plane or axis, and is to be regarded as positive or negative, according as the sum of the positive or negative moments is the greater. 2. To find the pe7'pendicular distance of t/ie centre of an irre- gular figure from a given plane or axis. Rule. — Divide the moment of that figure relatively to the given plane or axis by its magnitude ; the quotient will be the perpendicular distance of its centre from the given plane or axis. 3. To find the centre of a figure consisting of two j^a^'ts whose centres are known. (Fig. 101.) Rule.— Multiply the distance between the two known cen- tres by the magnitude of either of the parts, and divide the product by the magnitude of the whole figure ; the quotient will be the distance of the centre of the whole figure from the centre of the other part, the centre of the whole figure being in the straight line joining the centres of the two parts. Ex.: Let abcd be such a figure, M and m the magnitude of its two respective parts, M + m the magnitude of the whole figure, D the dis- tance between the centres M and in of the two parts, and c the centre of the whole figure. Fig. 101. MC = : M X D ■ M + w' Fig. 102. 4. To find the centre of any j^lane area by means of or di nates. (Fig. 102.) Let ABC, the quadrant of a circle, be such an area ; CB the base line, divided into a number of equal parts by ordinates ; AC the transverse axis traversing its origin. 1st. Determine the peipendiciilar distance of the centre of the quadrant from the trans- verse axis in the following manner : — Rule. — Multiply each ordinate by its dis- tance from the transverse axis ; consider the products as ordinates of a new curve of the same length as the given figure : the area of that curve, found by the proper rule, will be the moment of tlie figure relatively to the transverse ^:tTm^^ G4 CENTRES AND MOMENTS Of FIGURES. axis ; this moment, divided by the whole area of the figure, will give the perpendicular distance of its centre from the transverse axis. In algebraical symbols the moment of a plane figure rela- tively to its transverse axis, and found by the above rule, is expressed thus : — fxydx. Note. — In practice it is better to proceed as follows : — Multiply the ordinates first by their multipliers, and then those products by the number of intervals from the origin ; take the sum of those products and multiply it by ^rd of a whole interval squared, if Simpson's first rule is used, by fths of a whole inter- val squared, if Simpson's second rule is used, and so on for the other rules. Example. No. of Intervals Ordinate? 1 2 3 ^ 4 16-0000 1.5-4919 13-8564 12-4900 10-5830 9-3274 7-7460 5-5678 0-0000 Multi- pliers 1 4 2 Products 16-0000 61-9676 20-7846 24-9800 7-93725 9-3274 3-8730 5-5678 0-0000 Products X No. of Intervals from Origin •00000 61-9676 41-5692 62-4500 23-81175 30-31405 13-5555 20-87925 •00000 Interval 150-4376^ Interval - 3 254-54735 Approximate area = 200-58353Approx. moment = 1357-585 M oment 1357-585 ^ c.nna /"approximate perpendicular distance Area 200-5835 \ of centre from the transverse axis. 2tid. Find the perpendicular distance of its centre from tlie base line. KuLE. — Square each ordinate, and take the half -squares as ordinates for a new curve of the same length as the figure ; the area of that curve, found by the proper rule, will be the moment of the figure relatively to the base line : this moment, divided by the w^hole area of the figure, will give the perpendicular distance of its centre from the base line. In algebraical symbols the moment of a plane figure rela- tively to its base line, foimd by the above rule, is expressed thus : — / ^-Ldx. 2 CEiNTRES AND MOMENTS OF FIGURES. E.xamplc. 65 No, of Int^vals Ordiua'.es Half-squares JIultipliers rroducta 1 2 4 16-0000 15-4919 13-8564 10-5830 0-0000 128-0000 119-9995 95-9999 55-9999 00000 1 4 2 4 1 128-0000 479-9980 191-9998 223-9996 00000 Interval 3 Approximate moment 1023-9974 = 1365-3298 Moment 1365-3298 Area20r0624~ = 6-796 J approximate perpendicular dis- \ tance of centre from base. Actual moment = 1365-3 Actual afea ^201-0624 5. To find the centre of a plane area hounded hj a curve and two radii by means of polar co-ordinates. (See lig. 68.) 1st. Determine the perpendicular distance of its centre from a plane ti'aversin/j the pole and at inght angles to one of the hound- ing radii, called the first radius, in the folhiuing manner: — KuLE. — Divide the angle subtended by the arc into a conve- nient number of equiangular intervals by means of radii ; mea- sure the lengths of the radii from the pole to the arc, and multiply the third part of the cube of each of them by the cosine of the angle which they respectively make with the first radius ; treat these products by one of the rules applicable to finding the area of a plane curve (the only difference being that the common interval is taken in circular measure) ; the result will be the moment of the figure relatively to the plane tra- versing the pole : this moment, divided by the area of the figure, will give the perpendicular distance of its centre from the plane traversing the pole. Exam2)le. >-o. of Radii Radii 1 12 2 12 3 12 4 12 5 12 .jiCules of l?!idii 576 576 576 576 576 Anples! with First Radius 0= 5= 10- 15= 20- Ckjsines Products 1-0000 576-0000 -9962 573-8112 -9848 567-2448 -9659 556-3.584 •9397 541-2672 Simpson s Multi- pliers Products Interval in circular measure 3 576-0000 2295-2448 !ll34-4896 2225-4336 j 541-2672 677F4352" -0291 Moment relatively to plane traversing pole = 197-077864 F 66 CENTRES AND MOMENTS OF FIGURES. Moment 197-077864 Area 25-1327 ~ 7-841 /perpendicular distance of centre 1 from plane traversing pole. In algebraical symbols the moment, as here found, is ex- pressed thus : — f cos ede. 2nd. Determine the Tnoinetit of the figure relaiively to tlie first radius precisely in the same ivay as in the foregoing rule, with the exception tluit sines must he used in the itlace of cosines; this moment, divided hy the area of the figure, will give the perpen- dicular distance of its centre from the first radius. Note. — It is usual, in practice, to defer the division of the cubes of the radii by 3 until after the addition of the products. Example. No. of Radii Radii 1 12 2 12 3 12 4 12 5 12 Cubes of Radii 576 576 576 576 Angles with First Radius Sines of Angles Products Simpson's Multi- pliers 0° •0000 -0000 1 5° •0872 50-2272 4 10° •1736 99-9936 2 15° •2588 1149-0688 4 20° -3420 196-9920 1 Products •0000 200-9088 199-9972 596-2752 196-9920 Interval in circular measure 1194-1732 •0291 Moment relatively to first radius = 34*750440 Moment 34-75 044 f perpendicular distance of centre from Area 25-1327 " ^ ^« \ first radius. In algebraical symbols the moment as here found is ex- pressed thus : — / -^ sin QdQ. 6. To find the perpendicular distance of the centre of a solid, hounded on one side hy a curved surface (figs. 87 and 88), from a plane perpendicular to a given axis at a given point. KuLE. — Proceed as in Kule 4, p. 63, to find the moment relatively to the plane, substituting sectional areas for breadths : then divide the moment by the volume (as found by Kule 2, p. 54) ; the quotient will be the required distance. To determine the centre completely, find its distance from three planes no two of which are parallel. CENTRES AND MOMENTS OF FIGURES. 67 Fig. 103. 7. Having the itwment and centre of a ff/urc relativrbi to a given 2>lune^ to find the nciv moment and centre of the fgiue rela- tively to the same plane n-hcn a part of the figure is shifted. (Fig. 103.) In the figure wlk let c be its centre, and zz' a plane with respect to which the moment of the figure is known ; suppose the part avsm to ^^ be transferred to the new position SNL, so as to alter the shape of the figure from wlk to mnk ; let i be the original and H the new cen-" tre of the shifted part : then the momeyit of the p'gure mnk relatively to the plane zz' is found as. follows : — Rule.— Measure the distance, perpendicular to the plane of moments, between the centres of the original and new position of the shifted part, as hd, and multiply it by the magnitude of the shifted part ; the product will be the moment required. The new 2>osition of the entire fgure is then found by the following rule: — Rule. — Multiply the distance between the centres of the original and new position of the shifted part by the magnitude of that part ; that product, divided by the magnitude of the whole figure, will give the distance the centre has traversed in the direction in which the part has been shifted, and in a plane parallel to a line joining the centres of the original and new position of the shifted part, as from c to c' in fig. 103. 8, To find the centre of a wedge-shaped solid (fig. 104) by means of polar co-ordinates. \st. Determine the perpendicular distance of its centre rela" tively to a transverse sectional plane, as pab. Rule. — Divide the y\g. io4, ■ solid by a number of parallel and equidistant planes, as pab, p,a,b,, PjAjB,, tfcc. ; then mul- tiply each sectional area by its distance from the plane pab ; treat the products as though they were the ordinates of a curve of the same length as the figure ; the area of that curve, found by the proper rule, will be the moment of the figure relatively to the plane pab : that moment, divided by the volume of the figure, will be the distance required. 68 CENTRES AND MOMENTS OF FIGURES. 2)id. Determine the i^erpendicular distance of its centre re- latively to a longitudinaL plane j^assing through its edge, as MPM, l)crpendicular to the first radius, pb. Rule. — Divide the figure by a number of longitudinal planes radiating from the ed.^^'e mpm at equiangular intervals (as pp^AA^, PP4CC^, PP4BB4) ; also divide the length of the figure into a number of equal intervals by ordinates, and treat eacli of the longitudinal planes as follows : — Measure its ordinates, take the third part of their cubes, and treat those quanti- ties as if they were ordinates of a new curve ; that is, find its area by one of Simpson's rules : the area of that new curve is termed the moment of inertia of the longitudinal plane in question. Then multiply each moment of inertia of the several planes by the cosine of the angle made by the plane to which it belongs with the plane pb, and treat these products by a proper set of Simpson's multipliers ; add together the products, and multiply the sum by \ of the common angular interval in cir- cular measure if Simpson's first rule is used, and by | if Simp- son's second rule is used. The result will be the moment of the figure relatively to the plane MPM. This moment, divided by the volume of the figure, will be the distance required. The algebraical expression for the moment as found in this rule is ././ ^.3 -^ cos Od.ndB, Srd. Determine the 2)e7'pendicular distance of its centre re- lativehj to a longitudinal plane passing through its edge, and a radius as pp*bb*, by the foregoing rule, with the exception of multiplying by sines instead of cosines. Note. — In practice it is usual to defer the division of the cubes of the radii by 3 until after the addition of the products. 9. To find the centre of gravity of a plane area contained between two consecutive ordi7iates, with respect to the near end ordinate. RuuE. — To the sum of three times the near end ordinate, and ten times the middle ordinate, subtract the far end ordinate, and multiply the remainder by the Fig. 105, square of the common interval. The product, divided by 24, will be the moment about the near end ordinate, p^ On dividing this by the area, the longitudinal position of the centre of gravity is obtained. Ex.\ In fig. 105 let abc be the base, and ad, be, and CF the ordi- nates. Call them y^, 1/2? ^^^ th A" respectively, and let the common interval be denoted by h. Then the moment of tJio area abed about the •h--i MOMENTS OF INERTIA AND RADII OF GYRATION. 69 near end ordinate ad is equal to ^^-^'-t.l^^2":2^«) ^J^'. If this be 24 divided by the area of abed (see p. 46), the quotient will be the distance of the C.G. from ad. For an example, let the ordinates be 62, 8-5, and 94 feet, and the common interval 12 feet. No. of Ordi- nates 1 2 3 Ordinates 6-2 8-5 9-4 Multipliers for Area 5 8 -1 (Interval Products Ordinates Multipliers for Moments Products 310 68 -9-4 89-6 )- 1 6-2 8-5 9-4 Momen 3 10 ~1 (Interval) 18-6 85 -9-4 94-2 .^- 6 Area of portion between 1 and 12 ncluded " 12 . J . =89-6 24 t about 1 = 565-2 Moment 5652 Area 89-6 = 6-308 Perpendicular distance of centre of portion included between Nos. 1 and 2, from No. 1 ordinate. Note.— When the moment of the area is required about the middle ordinate, the above multipliers should be changed to Tyi + 6?/2 -j/3 ^ ^^2^ 1 ; so that moment = 24 Moments of Inertia a.vd Radii op Gyration. 1. To find the moment of inertia of a body about a given axis. Rule. — Conceive the body to be divided into an indefinitely great number of small parts; multiply the mass (or area) of each of these small parts into the square of its perpendicular distance from the given axis : the sum of all these products as obtained will be the moment of the body about the given axis. 2. To find the square of the radius of gyrnlion of a body about a given axis. Rule. — Divide the moment of inertia of the body relatively to the givfM) axis by the mass (or area) of the body. 70 MOMENTS OF INERTIA.. 3. Given the moment of ineHla of a body about an axu traversing its centre of gravity in a given direction, to find its moment (f inertia about another axis parallel to the first. Rule. — Multiply the mass (or area) of the body by the square of the perpendicular distance between the two axes, and to the product add the given moment of inertia. 4. Given the separate moments of ineHia of a set of bodies about pai'allel axes traversing their several centres of gravity, to find the moment of inertia of these bodies about a com,vwn axis parallel to their separate axes. Rule. — Multiply the mass (or area) of each body by the square of the perpendicular distance of its centre of gravity from the common axis ; the sum of all these products, together with all the separate moments of inertia, will be the combined moment of inertia. 5. Given the square of the radius of gyration of a body about an axis traversing its centre in a given direction, to find the square of the radivs of gyration abont another axis parallel to the first. Rule. — Square the perpendicular distance between the two axes, and add the product to the given square of the radius of gyration. 6. To find the moment of ineHia of a plane area, bounded on one side by a curve (see fig. 102), relatively to its base line. Rule. — Divide the base line into a suitable number of equal intervals, and measure ordinates at the points of division; take the third part of the cube of each of these ordinates, and treat those quantities so computed as the ordinates of a new curve : the area of that new curve, foimd by the proper rule, will be the moment of inertia required. In algebraical symbols the above rule is expressed thus : — fta.. Note. — When the moment of inertia is required as a whole, and not in separate parts, it is usual to postpone the division of the cubes till the end of the calculation. 7. To find the moment of inertia of a plane area, bounded on one side by a cnrr:e, relatively to one of its ordinates. Rule. — Multiply each ordinate by its proper multiplier, ac- cording to one of the rules for finding the area of such figures ; then multiply each of the products by the square of the number of whole intervals that the ordinate in question is distant from the MOMENTS OF INERTIA. 71 ordinate taken as the axis of moments : the sum of these pro- ducts, multiplied by | or | the cube of a whole interval, accord- ing as Simpson's first or second rule is used, will be the moment of inertia required. In algebraical symbols this rule is expressed thus :— fx-ij(hv. Example I. Calculation of Moment op Inertia op the Quadrant OP A Circle Relatively to the Base Line. No. of Intervals Ordinates Cubes of Ordinates 3 Multipliers Products 1 2 2i 16-00 15-49 13-86 12-49 1365-33 1238-89 887-50 649-48 1 4 2 1365-33 4955-56 1331-25 1298-96 296-07 270-72 77-58 57-29 0-00 3 n 3f 4 10-58 9-33 7-75 5-57 000 394-76 270-72 155-16 57-29 00 1 r i Interval 3 9652-76 12870-34 Example II. Calculation I OF THE M< 3MENT OF INERI riA OF THE Quadrant No. of Interrals Ordinates Multipliers Products Squares of Nos. of Intervals Products 160000 1 16-0000 000 000 1 15-4919 4 61-9676 1-00 61-9679 2 13-8564 u 20-7846 4-00 83-1384 n 12-4900 2 24-9800 6-25 156-1250 3 10-5830 1 7-93725 9-00 71-4353 3i 9-3274 1 9-3274 10-5625 98-5207 3^ 7-7460 h 3-8730 12-2500 47-4443 3| 5-5678 1 5-5678 14-0625 78-2972 4 S)0000 i 00000 160000 00000 A pproxima te momei Interval 3 It of inert ia = , 596-9288 12734-4810 72 MOMENTS OF INERTIA. Definition. — If a body be conoeivcd divided into an infinite number of parts, and the mass (or area) of each part be multiplied by the square of its distance from a fixed point, the sum of all these products is termed the polar moment of inertia about the point. 8. To find the 'polar moment of inertia of a plane area about a ^oint. i (I) EuLE. — At equal angular intervals sufficient to in- clude the whole area, draw radii from tlie point to the peri^ meter. Treat the fourth power of these radii as the ordinates of a new curve having a common interval equal to the angular interval between consecutive radii expressed in circular measure. One quarter of the area of this curve, found by the proper rule, is the polar moment of inertia required. ! Example. — Find the polar moment of inertia of a semi- circle of 5 feet radius about one end of the diameter, i The polar radii at an angular interval of 15" are 1000, 9-66, 8-66, 7-07, 5-00, 2'59 feet. No. Radius (Radius)^ Multiplier Product 1 10-00 10,000 1 10,000 2 9-06 8,735 4 34,940 3 8-66 5,624 2 11,248 4 7-07 2,498 4 9,992 5 5-00 625 2 1,250 6 2-59 45 4 180 7 — — 1 Common i nterval = -2618 67,610 ■ X t^ X J X -2618 Polar moment of inertia = 1,475 (II) Rule. — If the moments of inertia about two perpen- dicular axes through the point are known, their sum is equal to the polar moment of inertia about the point. definitions. — The product of inertia of an area about two perpendicular axes is the algebraic 6um of each element of area multiplied by the product of its co-ordinates with reference to the two axes. In the first and third quadrants the product of inertia is positive ; in the second and fourth quadrants it is negative. The principal axes of inertia through a point are those axes about which the product of inertia is zero. 9. Given the moments and products of inertia about iivo perpendicular uxes, to find the corresjjonding qtmntities about any two other perpendicular axes. llULE. — It' OX, oy (fig. 106) are the axes, X and Y the moments of inertia about them, and P their product, the moments and MOMENTS OF INERTIA. 73 product of inertia about Ox, Oy' (denoted by x', y', and p'), making a positive angle 6 with the original axes, are given by the following formulae : — x' = X cos" 6 + \ sin" - 2p sin 6 cos 0, y' = X sin- B + Y cos^ + 2p sin 9 cos 0, 80 that x' + y' = X + Y ; and p' = p cos 20- ^Y-X) sin 20. Note. — If ox and oy are principal axes, P = 0, and the formulse become x' = X cos- + Y sin' ; y' = X sin- + Y cos" ; P' = - § (y - x) sin 20. 9 ^ If an ellipse (fig. 106) be drawn having its principal axes ox, oy along the principal axes of inertia, and of magnitude oa, ob equal to radii of gyration about oy and Ox respectively, the radius of gyration about any other axis Oa;' is represented by the perpendicular OM drawn to that tangent of the ellipse which is parallel to the axis Ox' ; the moment of inertia about Ox' is proportional to the square of OM, or equally inversely proportional to the square of the radius op along ox' . The product of inertia about ox', oy' is similarly represented by the product of Oil and MQ, where OQ is con- jugate to OP. 10. Given the moments and product of inertia about two perpendicttlar axes, to find the principal moments and axes of inertia. • Rule. — If x, y, and P are the moments and product of inertia resx>ectively for the axes ox, oy, the angle (reckonecl positively) made by the principal axes ox', Oy' , with the original axes is given by the formula — f o » 2P tan 2t' =-• ^^-^ 74 MOMENTS OF INERTIA, The magnitudes of the principal moments of inertia x', Y' about ox', ot/', are given by — assuming x' to be the least and y' the greatest moment of inertia. 11. Given the moments of inertia about three axes, two perpendicular and one bisecting the angle betiveen them, to find the principal moments and axes of inertia. Rule. — If x, y, are the moments of inertia about the axes ox, oy, and z that about an axis bisecting the angle yox, tlie angle Q (reckoned positively) made by the principal axes Ox't Oy' , with the original axes is given by the formula — Y + X - 2Z tan 20 = — ^— Y~~ The magnitudes of the principal moments of inertia \' , Y' about ox\ oy' , are given by — ^±---Vz^-(Y>X) + ^4^' y' =l±?+\/z'-Z(Y + x) + Y^ + X^ Note.—^inae X + Y = x' + y', the sum of the moments of inertia about any two perpendicular axes is constant. If the area has an axis of symmetry, the principal axes are along and perpendicular to this axis. Ex. — An unequal-sided parallelogram is formed of two right- angled Isosceles triangles of 1 inch side. Find the principal moments and axes of inertia. Take ox parallel to the shorter sides, and oz perpendicular to the longer sides. Then.x = ^\ Y = J ; Z = i^\. By the formulae above tan 20 = - 2 ; = - 32° or 58°, the former corresponding to the least moment of inertia. Greatest i or y' = J + -gj = -218. Least I or x' = J - 2^ = -032. Table of Squares op Eadh op Gyration op a fkw Spectaij Figures. body Rectangle ; sides a and b Square ; side a Triangle ; sides a, h, e ; heights a', b', c' Equilateral triangle ; height d Trapezoid ; height h, parallel sides a and b Trapezoid with two right angles ; parallel sides a and b, perpendicular side h Circle ; diameter a Ellipse : diameters a, b Common parabola; height a, base b perpendicular to axis Sphere ; radius r Spherical shell ; external and internal radii ri and ra Ellipsoid of revolution ; trans- verse semi-axis r Ellipsoid : semi-axes a, b, o Circular cylinder; radius r, length 2a Hollow circular cylinder ; radius — external ri, internal r2 ; length 2a Baiiptic cylinder; semi-axes b, c, length 2a Tone : height 7i, radius of " Else r side a axis through C.G. parallel to side a any axis through C.Q. side a axis through C.G. parallel to side a any axis through O.Q. / . side a axis through C.Q. parallel I axis [ to side a Bide h axis through C.G. parallel i to side h i diameter 1 centre (polar) diameter a f axis of parabola j base b I axis through C.G. parallel I. to base b diameter centre (polar) j diameter !• axis of revolution axis 2a {longitudinal axis transverse diameter through C.G. C longitudinal axis I transverse diameter L through C.G. f longitudinal axis I transverse axis 2b through I. C.G. ' longitudinal axis transverse axis through C.G. transverse axis through base plane of base 8 b» 12 «i 12 «:? 6 18 4! 13 ^ q+Sb 6 • a+b Ti? a'^ + iab + b^ 18* (a-l-b)2 I(a2-l-b2) ai + 2a%+2abS+hi 18 (a+b)^ a2/16 a2/8 ba 16 2Z>2/5 8a2/35 12a2/175 5" 8r5| 6 2(rig-r2g) 6 (ri8-r28) 2ra 5 b2 + C« 6 4 3 ri^-f-ra" 2 4 """S b^ + c'i 4 _c2 oa 4'^8 To'*' j-3 10 20 10 Mcment of Inertia = square of radius of gyration x mass (or area) of the flKure. 76 MECHANICAL PRINCIPLES MECHANICA^L PRINCIPLES. Resultant and Resolution of Forces. 1. To find the resultant of two forces acting through one jMnnt hut not in the same direction. (Fig. 107.) Let AB, AC represent the two forces P and Q acting through the point A; complete the parallelo- gram ABCD : then its diagonal ad will represent in magnitude and direction the resultant of the two forces p and Q. R = resultant. = an a = angle R makes with q. )8 Fig. 107. le P makes with Q. angle R makes with p. R= a/p2 + Q'- + 2.p.q.cos0; ,P sin a = sin 6- sin )3 = sin d R Fig. 108. 2. To find the resultant of any number of forces acting in the same plane and tJi?'ough one point but not in the same direction. (Fig. 108.) Let p, P,, P.^, P3 be the forces acting through the point of application o ; commence at o and construct a chain of lines OP, PA, AB, BC, representing the forces in magnitude and paral- lel to them ; let C be the end of the chain : then a line R joining oc will represent in magnitude and direction the resultant of the forces p, p„ Pg, and Pg. Note. — This geometrical pro- blem is true whether the forces act in the same or in different planes. R = resultant. = angle made by R with a fixed axis ox, a, «„ «.,, See. = angles made by the forces P, P,, P.^, &;c., with ox. 5x = sum of the series of P . cos a + P, . cos «, + Pg . cos a.^, ko. 2y = sum of the series of p . sin a + p, . sin a, + Po . sin a.,, kc. R.cos0 = 2x. R: R . sm d = 2y. (2xy + (2Y)"^ tan e^ 2y 2x cos e = 2x R sin 6=: 5y R* MECHANICAL PRINCIPLES 77 3. To find the resultant of three forces acting through oiie point and making right angles with'one another. (Fig. 109.) Fig. 109. Let OA, OB, OC represent in magnitude and direction the forces x, Y, z acting through one point o ; complete the rectangular solid AEFB : then its diagonal OG will rejtrcsent in magnitude and direction the resultant of the forces x, Y, z. R = resultant, a, 3, 7 = the angles R makes with x, Y, z, respectively. Y = R . cos j8. R = A/'X* + Y'^ + z2. Z = R . cos 7. X = R . COS O. 4. To find the resultant of amj number offoi'ces acting through one point in different directions ayid not in the same plane. Let p, p„ P2, &c., be the forces o, 3,7 ; o„ /8„ 7, ; O2, /3.^, y^, the angles their directions make with three axes jmssing through the point of application and making right angles with one another. R = resultant. 2x = p . cos a + p, . cos a, -f P2 . COS a^ + &c. 2y = p . cos )8 + P, . cos 3, + P2 . cos i8.^ + &c. 2z = p . cos 7 + p, .cos 7, + Pj . cos 72 + &c. R = ^/(2x)2 + (2Y)•^+(2z)^ cos a = 2x R cos j8- 2y R cos 7 = 2z r' N.B. Cosines of obtuse angles are negative. Note. — P cos a, p cos 3, and P cos 7 are termed the components >f the forces in the directions of x, Y, and z respectively. The omponents of the resultant are obtained by adding (allowing for jign) the components of the several forces in their respective Ikections. Parallel Forces. A cou/ple consists of two equal forces, as p and Q (see If. 110), acting in parallel and opposite directions to one another, and is termed a right- or left-handed Fig. 110. couple, according to whether the forces tend to turn in a clockwise direction or the reverse. The moment of a couple is the product of either the forces into the perpendicular distance ab between the lines of direction of the forces. The distance AB is termed the arm or lever of the couple. - — coi hin 78 MECH\NIOAL PRINCIPLES. Fig. 111. Fi3. 112. 5. To find the resultant moment of any number of couples acting upon a body in the same or parallel planes. Edle. — x\d(i together tJie moments of tl:e right- and left- handed couples separately ; the difference between the two sums will be the resultant moment, which will be right- or left-handed, according to which sum is the greater. 6. To find, the resultant of two parallel forces. (Fig. Ill and 112.) The magnitude of the resultant of two parallel forces is their sura of difference, according to whether they act in the same or contrary directions. Let fig. Ill represent a case in which the two forces act in the same direction, and %. 112 a case in which the components act in opposite directions. Let AB and CD represent two forces ; join ad and CB, cutting each other in E; in da (produced in fig. 112) take df =BA ; through F draw a line parallel to the components ; this will be the line of the resultant, and if two lines dg and AH be drawn parallel to BC, cutting the line of action of the resultant in and H, GH will represent the magni- tude of the resultant. Or, numerically, the line of action of the resultant is obtained by adding (allowing for sign) the moments of the two forces about any (point, this being equal to the moment of the resultant ; the perpendicular distance of the line of action from the point is obtained by dividing this moment by the magnitude of the resultant. AF DC. AD DF =- AB. AD GH GH 7. To find the resultant of any number of parallel forces. Rule. — Take the sum of all those forces which act in ont direction, and distinguish them as positive ; tlien take the sum of all the other forces which act in the contrary direction, and distinguish them as negative. The direction of the resultant (positive or negative) will be in that of the greater of these two sums, and its magnitude will be the difference between them. 8. To find the position of the resultant of any number of parallel forces when they act in tivo contrary directions. Rule. — 1st. Multiply each force by its perpendicular dis- tance from an assumed axb in a plane perpendicular to the MECHANICAL PRINCIPLES. 79 linos of action of the forces ; distinguish those momenta into right- and left-handed, and take their resultant, which divide by the resultant force : the quotient v.-ill be the perpendicular distance of that force from tiie assumed axis, 2nd. Find by a similar process the perpendicular distance of the resultant force from another axis perpendicular to the first and in the same plane. 9. To find the resultant of any number of couples not necessarily in a plane. Two couples of equal moments in the same or in parallel planes are equivalent to one another, whatever the magnitudes and positions of the forces composing the couples may be. A couple is therefore conveniently represented by a lino perpendicular to its plane, and of length proportional to its moment ; usually the direction of the lino is taken so that its relation to the direction of the couple is the same as that between the travel and the rotation of a right-handed screw. Note that any two parallel linos of the same magnitude and sense represent the same couple. Rule. — Replace tlie couples by lines as above, giving them their correct magnitudes and direction, and treat these a? forces through a point by Rule 4. The resultant gives the magnitude and direction of the resultant couple. 10. To find the resultant of any number of forces in a plane. Rule. — Treat them as forces through any fixed point by Rule 2, and find their resultant. Calculate also the moment of each force about*^the point, and add them together allowing for the sign of each. The resultant moment divided by the magnitude of the resultant force gives the perpendiculai: distance of its line of action from the point. Definition. — The moment of a force about a line that it does not meet is the product of the component of the force perpendicular to the line with the shortest distance between the line and the line of action of the force. 11. To fifid the resultant of any number of forces, not in one place. Rule. — Resolve the forces parallel to three perpendicular axes as in Rule 4, and find the magnitude and direction of their resultant E. Calculate the moments of each component about the three axes, and treating these as couples find the resultant couple F by Rule 9. Resolve this couple into couples a parallel to the force R, and h perpendicular to B. Resolve the couple H into 2 foroas E^, E2, of which Ej^ is equal and opposite to R, while Rg is equal and parallel to R ; find the position of Ro (not in plane of figure). Then the final resultant is equal to the force R2 combined with the couple a (since R and Rj^ neutralize). The combination of a force in and a couple jibout tlie same line is termed a wrench. 80 CENTRE OF GRAVITY OF BODIES. CENTRE OF GRAVITY. 1. To Jincl the moment of a hody's n-cight relativehj to a given pl/ine. Rule. — Multijjly the w^eight of the body by the perpen- dicular distance of its centre of gravity from the given plane. 2. To find the common centre of gravity of a set of detached bodies relatively to a given plane. KuLE.-— Find their several moments relatively to a fixed plane ; take the algebraical' sum or resultant of those moments and divide it by the total sum of all the weights : the quotient will be the perpendicular distance of the common centre of gravity from the given plane. Note. — When the moments of some of the weights lie on one side of the plane, and some on the other, they must be dis- tinguished into positive and negative moments, according to the side of the plane on which they lie, and the difference between the two sums of the positive and negative moments will be the resultant moment. The sign of the resultant will show on which side the common centre of gravity lies. Let 71', m\ W-, &c. = the several weights. rZ, d\ d\ &c. = the several perpendicular distances of the centres of gravity of w, w', w-, &c., from the plane of moments. D = the perpendicular distance of their common centre of gravity from the plane of moments. _ w< / + w ^d^ + rv'^d^ + &c. ?y + w' +w^ + &c. 3. Ih find the centre of gravity of a body consisting of parts of unequal heaviness. Rule. — Find separately the centre of gravity of these several parts, and then treat them as detached weights by the foregoing rule. 4. To find the distance through which the common centre of gravity of a set of detached weights moves when one of those weights is shifted into a new position. Rule.— multiply the weight moved by the distance through which its centre of gravity is shifted ; divide the product by the sum total of the weights : the quotient will be the distance through which the common centre of gravity has moved in a line parallel to that in which the weight w^as shifted. Let w = weight shifted. \g 6 ^ Angulae Motion. I = mass moment of inertia about axis of revolution, i = angular acceleration. « = angular velocity. = angle turned through. M = angular momentum = l«. G = moments of forces about axis. _ y _ doo d-d dM I^^ote. — If G is expressed in foot-pounds, I must be expressed in the gravitational unit, or is i/gr of the density of the material multiplied by the volume moment of inertia of the body (see pp. 69-75). WOEK AND EnEEGY. The work done by a force on a body is the product of the force by the distance moved resolved along the direction of the force. The work done by a couple is the moment of the couple DYNAMICS. 85 multiplied by the anjjle turned through resolved along the plane of the couple. With the previous notation, if the body runs through distances x, y, z parallel to the axes, and rotates through an angle 0, the work done is vx -{- Q,y -\- ixz + g9, allowing for sign. The energy of a body is its capacity for doing work. Kinetic energy is energy due to motion. With the preceding notation, its amount is ^ m {u^+v'+td^) + 5- i»^ wi and I being expressed in pounds and the result in foot-pounds. Potential energy is energy due to position, and is measured from an arbitrarily fixed datum. A body of height h feet above the sea-level has potential energy of 7iih foot-pounds. A ship has potential energy due both to the height of its centre of gravity and the depth of its centre of buoyancy. Molecular energy, due to heat, electrical state, magnetism, vibration, etc., is frequently waste energy as far as its capacity of doing useful work is concerned. Conservation of energy. The work done on a bodv (other than that involved in a change in the potential energy) in a given interval of time is equal to the increase of its total energy. Power is the rate of doing work. It is equal to Pw + Qy + WW -f Gw, allowing for sign. This is equal to the rate of increase of energy. The practical unit of power is the hojse-poiver, equivalent to 550 foot-pounds per second, or 33,000 foot-pounds per minute. Another unit is the watt, 746 of which are equivalent to one H.P. Uniform Foece in Line of Motion. p = uniform force in pounds weight. ni = mass in pounds. / = uniform acceleration = pgr/m. V = initial velocity in feet per second. s = distance travelled. t = time occupied. V = final velocity. v = Y+ft; v'' = \^ + 2fs', s = \t + W. For retarded motion change /to -/. For motion vertically under gravity f to g or - g, according as the initial motion is downwards or upwards. In that ^ case p = +m. For 'motion down an incline of angle o to the horizontal, replace f hj g tan o. For angular rotation with the notation above, fl being the initial angular velocity, u = n + ^t \ a" = Cl} + 2^9 ; e = m + ii^^ Gravity. g — acceleration due to gravity in feet/second^. A = latitude of the place. h = height above sea-level. 86 DYNAMICS. R = radius of earth in feet = 20,900,000. 27i g = B2 088 (1 + -005302 sin^ (p - -000007 sin'' 2 - — ). R Usually g is taken as 32-2, or 981 in centimetres/second^. Simple Vibration. M = mass in pounds. a = semi-amplitude of vibration. n ■=■ frequency or number of double vibrations per second. E = modulus or force in pounds required to produce unit extension. t — time. X — displacement at time ' t.' / = acceleration at time ' t.* a = a constant. 17 = 32-2. %i=i ~\J -^\ a; = a sin [lirnt + a) ; / M Simple Pendulum. L = length of pendulum in feet. T = time of a single small vibration in seconds. g — acceleration due to gravity = 32-2. T = 7r\/-= -554 v/£: Table giving THE Lengths of Pendulums in Inches 1 THAT Vibrate Seconds IN various Latitudes. Sierra Leone 89-01997 New York 39-10120 Trinidad 39-01888 Bordeaux 3911296 Madras 39-02630 Paris 39-12877 Jamaica 39-03o03 London 3913907 Eio Janeiro 39-04350 Edinburgh 39-15504 Table giving the Times of Vibration for Pendulum swinging through Large Arcs. Angle swung on each side of vertical 80" CO' 90P 120° 150" 180' Actual time of vibration -r Time for infinitely small angle 1017 1073 1-183 1-373 1-762 Infinite DYNAMICS. 87 Compound Pendulum. K = radius of gyration of body about axis of rotation. h = centre of gravity below axis. I = length of equivalent pendulum. I = k-jh. The centre of 'percussion, or point at which a blow struck perpendicularly to the axis will cause no stress at the axis, is situated at a distance I (determined by the above formula) below the axis. Centrifugal Force. F = centrifugal force of body revolving in a circle at a unifcrra speed, or apparent force required to balance that necessary to produce the requisite normal acceleration. w = weight of body. N = number of revolutions per minute. n = number of revolutions per second. V = linear velocity in feet per second. « = angular velocity in circular measure per Kecood. r = radius of circle in feet. g = acceleration due to gravity = 32-2 nearly. _ wu^ _ viraP _ 4wnVV _ v.'nV _ wnV ~ ~gr ~ g ~ 'g ~ ^8154" 2935 Gyroscopic Action. If the axis of a revolving body is made to rotate into a new position, resistance is experienced due to the ' gyroscopic action ' of the revolving mass. Let ab represent in the usual way the angular momentum ia> of a body having a moment of inertia i about the axis of revolution, and an angular velocity u. If this axis is forced to occupy after a short time the position AC, BC represents the chajige of angular momentum. This is e^qual to iw x Lb AC. If this change is effected by turning the axis uniformly with angular velocity w', the rate of change of angular momentum is Icow', which is equal to the moment G of the applied couple. Note that the plajio of G is perpendicular to that of shaft rotation, and of the direction of movement. If i is in weight units (lbs. X feet^), and n and n' are the number of revolutions per minute of shaft rotation, and of bodily rotation, _ I 47r^ nn' _ i.n.n' ^ ~'g^ 3600 " 2935 In the case of a ship going ahead with a right-handed screw, the forces required on the shaft when turning to starboard are downward aft and upward forward ; the re- action on the hull is then such as to cause a slight trim by the bow. 88 HYDROSTATICS. Wl Impact. M2 = the velocities of two bodies before impact (if moving in opposite directions make U2 negative). Vi, 1)2 = the velocities after impact. mi, ni2= the masses. e = coefficient of restitution = ratio of velocity of separa- tion to that of approach. For direct impact, _ ui (mi - g^na) -f ma M2 (1 + e) Vi V2 mi + ma Ui mi (1 + e) + U2 (wia enii) Kinetic energy lost = mi + TTta — ^^1 ^^ (^^1 - ih)" (1 - e^ ) 2g (mi + m^2) Total momentum is unchanged, or 7ni ui + ma«2 — ^'^i Ui + w^2 v^. For oblique impact, resolve the velocities along and perpen- dicular to the line of impact ; treat the components along the line by the above formulae ; the latter are unaltered by the impact. The value of the coefficient e depends to some extent on the shape of the bodies and the velocity of impp,ct, as well as on the material. Approximate values for the impact of like materials are given in the following table : — Material Cast Iron Mild Steel SoffBrasa Lead Elm Glass Ivory e •70 •67 •38 •20 •60 •94 •81 HYDROSTATICS. The density/ of a fluid is the wfeight of a unit volume., Generally it is stated in pounds per cubic foot, or inversely aa the number of cubic feet required to weigh 1 ton. (See tables on p. 262.) The specific gravity of a fluid is the ratio of its density to that of water. Density of a Mixture of Two Liquids. Wi, W2 = densities of the two liquids. w = density of the mixture. mi, m2 = proportion of the two liquids in the volume, ni, na = proportion of the two liquids in the weight. mi w i + 7)12102 _ ni_ mi + r?ta ni Wi ni/wa = mi Wilm2 W2', w = mixture by mixture by + Ha na W2 HYDROSTATICS. 89 Peessure in a Liquid. w = density of liquid in pounds per cubic foot. z = depth below free surface. p = intensity of pressure in pounds per square inch. P = intensity of pressure in pounds per square foot. V = wz ; p = wzjlii. 4 In salt water, w = Gi, v =^ Giz, p = ^z. In fresh water, w = 62-5, p = 62-5^, p = iSSz. If the absolute pressure bo required P and p must be increased by 2120 and 14'7 respectively, in order to allow for the pressure of the atmosphere. Nofe.-^The centre of pressure of an immersed plane surface is that point on the surface through which the resultant pressure acts. Pressure on Immersed Plane Surface. If surface be vertical find the centre of gravity a and take axes Gx horizontal and Gy vertically downwards. Let A = area of plane. h = depth of centre of gravity belowfree surface. w = density of fluid. T = total thrust or pressure on plane. X, y = co-ordinates of the centre of pressure. Then t = wA.h y - Ah 1 ~j / y^ . dx . dy over area. , X moment of inertia of area about Gx. Ah X = — r / a:?/ dx dy over area. = -r X product of inertia of area about Gx, Gy. Ah If the surface and the axis Gy be inclined at an angle 9 to tl;e vertical, T and x are unaltered, but the value found for y should be multiplied by cos 0. Pressure on any Closed Surface. The resultant pressure on the whole immersed surface of a body Is equal to the weight of the water displaced by the body and acts vertically upwards through the centre of gravity of the displaced volume. The upward force is termed the displacement, and the point through which it acts the centre of buoyancy. 90 DISPLACEMENT, DISPLACEMENT, Etc. Computation of a Ship's Displackmext. This consists of computing the volume of the body of the vessel below the water-plane, up to which it is required to know her displacement, by one of the rules used for finding the volume of solids bounded on one side by a curved surface (see pp. 54, 55). Two processe3 are generally made use of in computing a vessel's displacement, as the calculations in each process are required to determine the position of the centre of gravity of displacement, or centre of buoyancy, and also because thp two results (are a check on the correctness of the calculations. One process consists in dividing the length of the ship on the ^ load water-line by a number of equidistant vertical sections, computing their several areas by one of Simpson's rules, and then treating them as if they were tlie ordinates of a new curve, the base of which is the' load water-line. The other process consists in dividing the depth of the vessel below the load water-line by a number of equidistant horizontal planes parallel to the load water-line ; the areas of their several planes are then computed by one of Simpson's rules, knd those areas are treated as if they were the ordinates of a new curve, the base of which is the vertical distance between the load water-line and lowest horizontal plane. As the vessel generally consists of two symmetrical halves, the volume of only half the vessel, below the load water-line, is calculated, the ordinates all being measured from a longi- tudinal vertical plane at the middle of the ship. Usually the portion below the lowest water-line is treated, as are also the stern, rudder, bilgea, keels, oto., as an appendage, its volume being calculated by means of equi- distant vertical sections. The water-lines that are ' snubbed. * or cut short abaft the fore perpendicular or before the after perpendicular are conceived to extend to these perpendiculars, the extra volumes thus introduced being regarded as negative appendages. The displacement of a ship can also be obtained by dividing the length into sections, spaced as required by Tchebycheff's rule ; the integration in a longitudinal direction is effected by simple summation. The water-lines are equidistantly spaced and integrated by Simpson's rules as before. This method is^ generally speaking, more expeditious than is the one pre- viously described, since fewer ordinates can be employed, and half the multiplication is dispensed with. Both methods are illustrated in the displacement Sheet? given on pp. 94 ff. DISPLACEMENT. 91 Determination of a Ship's Centre of Buoyancy for THE Upright Position. The centre of buoyancy is also termed the centre of gravity of displacement, as it occupies the same point as the centre of g^ravity of the volume of water displaced by the vessel, and its position is determined by the rules used for finding the centre of gravity of solids, bounded on one side by a curved surface (see rules, pp. 66 and 67), with the exception that its position need only be determined for its vertical distance from a horizontal plane, and its horizontal distance from a vertical plane ; for the ship consisting of two symmetrical halves, it must necessarily lay in the longitudinal vertical plane in the middle of the ship. Calculation of the centre of buoyancy is generally performed on the displacement sheet (see pp. 94 ff.). Curve of Areas of Sections. This curve (see fig. 115) is of use in designing and in estimating the resistance of a ship, for it fixes the distribution of displacement along the length. Fig. 115. CURVE OF SECTIONALAREAS. Method of Construction. — Compute the area of each transverse section up to the l.w.l.; and set it oif to scale on a base of length. A curve drawn through the tops of the ordinates will form the curve required. Curve of Areas of Midship Section. This curve (see fig. 116) is used to determine the area of the immersed part of the midship section of a vessel at any given draught of water. Method of Construction. — Com\mXe the areas of the midship section from the keel up to the several longitudinal water-planes Fig. 116. KrtHe •/ areat which are used for calculating the displacement ; set these areas off along a base line as ordinates, in their consecutive order, the abscissa? of which re- present to scale the respective distances between the longi- tudinal water-planes: a curve bent through the extremities MLjs" of these ordinates will form the required curve. 92 CURVE OF DISPLACEMENT. Curve of Displacement. This curve is used to determine the displacement a vessel has at any draught of water parallel to the load water-line (see fig 117). Method of Construction. — This curve is constructed in a similar manner to the foregoing curve, with the exception that the ordi- FiG. 117 tcale of Tons nates represent the several volumes of displacement (in tons of 35 cubic feet for salt water, and 36 cubic feet for fresh water) up to their respective longitudinal water-planes. Curve of Tons ter Inch of Immersion. This curve (see fig. 1] 8) is used to determine the number of tons required to immerse a vessel one inch at any draught of water parallel to the load water-plane. To find the displacement per inch in cubic feet at any water- plane, divide the area of that plane by 12 ; and if the displace- FiG. 118. Scale of Tons ment per inch is required in tons, divide by 35 or 36, as the case may be. A = area of longitudinal water-plane in square feet. T = tons per inch of immersion at that water-plane. T = —~ — for salt water ; T = , for fresh water. 115x35 '12x36 COEFFICIENTS OF FINENESS. 93 Bthod of C (instruction. — This curve is also constructed ixi a similar manner to the two foregoing curves, wifch the exception that tlie ordinafces represent to scale the tons per inch of immersion at the respective water-pianos. The coefficients of fineness of a vessel consist of the block coefficients {&), the prismatic coefficient (7), and the midship section coefficient (/t). They are determined from the following equations : — V = volume of displacement in cubic feet. L = length of vessel at load water-line in feet (or length between perpendiculars, according to convention). B = extreme immersed breadth in feet. (Occasionally this is taken as the breadth at l.w.l, in cases where this is less than the extreme breadth.) D = mean draught of water in feet. (Take to top of keel if bar keel.) 2 = Area of midship section up to l.w.l. in square feet. o V V 2 ^ ^ = L":B:^.'"''"i:2'^=i:^.'^ = '>'-'*- Another coefficient sometimes used is that of water-line area (a) given by A = :^— ^ where A is the area of the l.w.l. Usually L.B. tons per inch' this latter being this is expressed as a coefficient equal to -— - A Values of these four coefficients for typical ships are given in the table below. Table 01 ' Coefficients of Fineness. Block Co- Prismatic Mid. Sec. Waterline efficient « V Coefflc't V Coefflc't 2 Coefflc't ^ A Class of Ship Tons ^-I.BD "^=£2 ^-^ ^=Zi per Inch Battleship (modern) . •GO •62 •965 •73 575 Battleship (older) •65 •68 .95 •81 620 First-class Cruiser •56 •62 •90 •68 620 Modern Light Cruiser . •58 •63 •92 •76 550 T -ncdo Boat Destroyer •55 •67 •82 •76 550 a Yacht •52 •565 •92 •69 610 : . • Passenger Steamer •59 •62 •95 •70 600 Large Cargo Vessels . •73 •77 •95 •83 510 Sailing Yacht •2 •5 •4 •75 660 Tug . •58 •61 •95 •76 550 A^o^«.— The ' length ' in warships is the length between perpendiculars. Table showing Method of Computing a Ship Length between perpendiculars, 885 feet ; breadth. 41 feet ; draught at perpendicula 1 Appendage below lowest Water-line Wateb-lines 1 3 ■< li i!i 1-2 it 7W.L. 6W.L. 6W.L. 4W.L. 3W.L 3 Simpson's Multipliebs 3 c i 2 1 2 1 h •9 2 05 3-67 5-24 7-18 9-88 12-48 14-3 15^ 15-7 14-2 11-35 8-1 5-72 •9 4-1 3-67 10-48 7-18 19-36 12^ 28-6 15-65 31-4 14-2 22-7 8-1 11-44 3-9 5-25 •27 10 ~9 8 "7 ~G "5 ~4 "3 "2 1 "o 1 1 3 4 ~5 ~6 "7 1 ~9 1-6 •8 1-6 6-0 9^2 12-8 8-46 •20-6 11-96 26-2 1^76 3^52 3^52 •8 •8 2-6 2-6 4-66 4-66 6-94 6-94 9-4 9-4 11-7 11-7 13-9 13-9 15-86 15-86 17-34 17-34 18-26 18-26 18-7 18-7 18-74 18-74 18-06 18-06 17-1 17-1 15-4 15-4 13-2 13-2 10-3 10-3 6-9 6-9 3-8 3-8 1-5 1-5 •76 •76 •4 52 4-66 13-88 9-4 23-4 13-9 31-72 17-34 33^52 18^7 37-48 18-06 34-2 15-4 26-4 10-3 13-8 3-8 3-0 •38 •8 16 30 6^0 5-3 10^6 79 15^8 10-5 21-0 12-86 25-72 15-04 30-08 16-9 33 • 8 184 36^8 193 38^6 197 39^4 1964 :39-28 •4 60 5-3 15-8 10-5 25-72 15-04 33-8 18-4 38-6 19-7 39-29 19^2 36^92 17-0 •5 •5 31 3-1 5-7 5-7 8-54 8-54 11-2 11-2 13-66 13-66 15-8 15-8 17-6 17-6 190 19-0 199 19-9 20-26 20-26 20-24 20-24 19-9 19^9 193 193 181 181 2 — 6 1 7-2 •3 •27 3-6 7^0 55 11^0 j 3-5 1 11^0 1 5 2 28-7 -45 1^84 3-0 1-5 4-6 2-3 6-4 3-2 8-46 4-23 10-3 5-15 11-96 5-98 13-1 6-55 17 1 22-02 •48 1^76 7-64 7-64 15^28 11 2 52-4 •51 5^34 9-9 19^8 |19^8 27 1 28-72 •54 3-87 12^04 12^04 24^08 141 ,282 28-2 1 15-6 jl5-6 31-2 1 15 2 59-28-57 11^25 35 1 24 -9o •6 7^5 19 2 28-6 -63 18^0 1654 33-08 33-08 17-1 340 16^2 29-6 12-8 20^8 7-3 8-68 2-1 1^6 •38 39 1 251-88 •66 10-33 20-7 9-22 14-8 13-6 6-8 13-46 6-73 12-5 6-25 10-6 5^3 8-3 4-15 5-9 2-95 136 26^92 12^5 •21 • 2 8^3 11^8 3^7 17-1 34-2 17-0 34-0 16-2 32-4 14-8 29-6 12-8 25-6 10-4 20-8 7-3 14-6 4-34 8-68 2-1 4-2 -8 1-6 •76 1^52 20 2 31-4 -66 40 1 28-4 -65 19-2 38-4 18-46 36-92 170 34^0 151 30^2 12-64 25-28 9-5 19-0 6-0 12-0 30 6-0 -76 1-59 19 2 68-1 -67 38 1 32-4 5-58 18 2 57-2 -71 8-14 30-2 164 16^4 32 1 3-9 2-63 •54 23-4 -73 2-88 3-7 1^85 12-6414-34 19-0 11-6 11-6 6-0 1 8-4 i 8-4 60 1 5-0 1 5-0 14 2 36-82 2-16 •75 3-94 •081 22 1^1 4^4 1^0 1^52 •38 23 I 1-0 •5 •76 •38 8 2 — — 10 1 — — 10 — — — •76 -38 •38 15 1^5 1 200^0 279^8S 251^88 2800 125-47 197-54 284-94 337-94 375-88 404 98^77 + 569-83 + 337-94 + 751-76 + 404 6 5 4 3 592^62 -f 2849-40 + 1351-76 + 2255-28 + 808 >}.B.— The dark figures are the ordinates; the light figures under them and also to their right are t] )duct8 of the ordinates by their respective Simpson's multipliers, which are ilaced at the head ai the left of the table ; if each row and column of these products be added together, and the resul Displacement, etc., using Simpson's First Rule. B feet for'd, 14 feet aft, 13 ft. 6 in. mean ; waterlines apart, 2 feet ; ordinates apart, 19 ft. 8 in. 24 46irr 921 _ 14: 20 28!_ 6610' 12 6-32 66 28 26 360 19^ 40^ 20-46 28-92 6-6 6-34 18' 145-71 19-36 40-4 20- 41-0 19-04 2416- 12 12-1 18-0 2-58 Vertical Sectioks 3-3 23-19 44-01 66-29 88-47 109-59 158-6 166-77 170-53 170-24 165-64 157-57 126-83 105-41 80-71 56-07 33-18 14-74 S ft 11 _1 46-38 44-01 132-58 88-47 218-18 128-88 -291-42 158-6 333-54 170-53 340-48 165-64 315-14 144-17 253-66 105-41 161-42 56-07 66-36 7-37 10 16-5 417-42 .352-08 928-06 9-26 11-94 14-46 515-52 16-54 874-26 18-2 530-82 1095-9 317-2 333-54 5381-3 340-48 381-28 20-3 945-42 19-9 576-68 1268-3 632-46 116-24 1129-94 448-56 597-24 73-7 Metacentkes 19-36 20-2 20-5 20-5 19-04 17-76 14-3 120^ -0 •16 1702 3023 4525 6029 7256 8242 8615 8615 7881 6902 5600 4283 2924 1772 729 138 6-6 6-34 1586 1 18-52 1702111 -94 6046128-92 16-54 4525 12058 36-4 8365 20 6902 4283 16 5848 28 Longitudinal 7256 19-36 16484 40-4 8615 20-5 17230 41-0 157621 39-8 19-04 11200|35-52 1772|L2-1 1458118-0 2-5 S2I .2 o(i( 59-4 50-72 129-64 71-64 144-60 66-16 109-2 38-72 40-4 710-48 41-0 40-6 119-4 76-16 177-6 97-44 200-2 96-8 162-0 25-8 2 ©"" n- 10 9 534-6 8 405-7 7 907-4 6 429-8' 5 723-0 J 3 264-64 327-6 2 77-44 1 40-4 1 410 2 81-2 3 358-2 4 304-64 5 888-0 6 584-64 7 1401-4 8 774-4 9 1458-0 258-0 424-48 438-7 6344-06 131488 438-7 103700 9860-24 848-96 + 219-35 = 3231-0 5381-30 710-48 1 _ 962-76 326-52 848-96 - 8706-78 (Continued on next page.) grated by the proper multipliers, and the sums of these products added tocether, the two sums agree if the calculations are correct. In tliis case the sum thus obtained by two methods is 82310. 96 DISPLACEMENT. T^• 1 J. . ,., 3231-0 8 ,„ „„ Displacement - mam solid = gg x - x 19-25 x 2 == 3159 tons. 8706 • 78 8 Moment belo\v L.w.i.. - main solid = g^ x - x 19-25 x4 = 17,026ft.-tons. g Moment abaft 3^ - main solid = 962-76 x - x (19-25)2 x 2 = 18,121 ft.-tons. 200 4 Displacement — lower appendage = xr- x —x 19-25 = 147 tons. do O 125 • 47 C.G. below 7 "W.ii. — lov/er appendage = g^ = -63 feet. Moment abaft ^ - lower appendage = ^ ^ 3 x (19-25)2 = 395 ft.-tons. Below L.W.L. Abaft ^ Item. Tons. Distance. Moment. Distance. Moment. Main solid 3159 17026 18121 Lower appendage 147 12-63 1857 395 Aft 5-7 1-3 7 197-8 1127 Fore „ -6 6-0 4 - 193-6 - 116 Rudder 2-2 8-7 19 194-3 427 Bilge keels •6 9-5 6 Shafting 3-2 6-3 20 153-8 492 Shaft brackets 1-2 6-5 8 168-5 202 Propellers •9 6-7 6 173-1 156 gwell •6 5-8 3 121-5 73 Recess -2-8 5-9 -16 146-6 - 410 Negative appendage a-ft -7-0 11-2 -78 176-0 -1232 Total 3311-2 5-70 18862 5-8 19235 Displacement 3311 tons ; c.B. 5-70' below li.w.li., 5-8' abaft }£. 4 Area of l.w.l. (main portion) = 438-7x-xi9.25 = 11,260 sq. ft. Moment abaft ^ - 32G-52 x |- x (19-25)2 = 161,300 ft.3 Moment of inertia about ^ - 9860-24 x ^ x (19-25)3 = 93,700.000 it* Item. Area. c.F. abaft }^ Moment about ^ Moment of Inertia about )^ Main portion Appendage aft 11,260 100 198 161,300 19,800 93,700,000 3,900,000 Total 11,360 15-95 181,100 97,600,000 11,360 X (15 -95)2 = 2,90q,_p00 Moment of inertia about c.F. = 94,700,000 ft.'* DISPLACEMENT SHEET. 97 11,350 K^ 420~ "" 27 0. c.F, abaft ^ = 15-95'. 1.1 nrm nnf\ BG (with G in li.w.L.) = 6' approx. Tons per inch T • ;,• , 94,700,000 „, , Longitudinal bm = „.,.- ^ -^ = 818 diill X do .*. Longitudinal gm = 818 - 6 = 812. ■J311 X 81'' Moment to change trim 1 inch = - . ^ ^^ = 582 ft. -tons. Transverse bm = |f^'J^ x | x .| x 19.25 = 9-78 ft. 4 Area of midship section = 170-5 X- x2 = 154 sq .ft. No. S.M 'A.'\V.L.(2ft.aboveL.AV.L.) 2W.L. :> W.L. X IJ 1 5 Li V — 1 1 ■|o| 1 1 |J 1 h 1 § t fa 1 " 1 ^ 1 3 .— _ _ •2 •5 2 2 3-45 6-90 41 82 3-16 31 62 31 30 60 3 1 6-65 6-65 294 294 6-04 220 220 5-7 185 185 4 2 9-55 19-10 871 1742 8-94 717 1634 8-54 623 1246 5 1 12-2 12-20 1816 1816 11-66 1581 1581 11-2 1405 1405 6 2 14-65 29-30 3144 G288 14-14 2833 5666 13-66 2549 5098 7 1 16-7 16-70 4657 4657 16-26 4291 4291 15-8 3944 3944 8 2 18-25 36-50 6078 12156 18-0 5832 11664 17-6 5452 10904 9 1 19-3 19-30 7189 7189 19-3 7189 7189 19-0 6^59 6859 10 2 2.J-05 40-10 8060 16120 20-14 8181 16362 19-9 7881 15762 11 1 20-4 ';0-40 8490 8490 20-46 8552 8552 20-26 8316 8316 12 2 20-4 40-80 8490 16980 20-5 8615 17230 20-24 8292 16584 13 1 20-2 20-20 8242 8242 20-24 8300 8300 19-9 7881 7881 14 2 19-85 39-70 7821 15642 19-7 7645 15290 19-3 7189 14378 15 1 19-1 19-10 6968 6968 18-7 6539 6539 18-1 59;i0 5930 16 2 18-0 36-00 5832 11664 17-2 5088 10176 16-4 4411 8822 17 1 16-65 16-65 4616 4616 15-46 3690 3690 14-34 2950 2950 18 2 14-95 29-90 3341 6682 13-14 2274 4548 11-6 1561 3122 19 1 13-05 13-05 22-22 2222 10-56 1174 1174 8-4 593 593 20 2 10-25i 20-50 1(<77 2154 7-1 358 716 5-0 125 250 21 § 6-6 8-30 287 \44 3-24 34 17 1-5 8 2 446-35 1^4148 124701 114291 Function Mult. Product. Mult. Moment, of area. w.r. NT.L. Up yer i., 416-4 4.38-7 424-5 2232 3510 5742 424 5318 3125 2632 5757 _424 5333 Displacement of layer (L.W.L. to A.W.L.) = 5318 x|x 19.25x^^x1 to L.W.L. W.Ii. to A.W.L. Tons. 3311 651 S962 Moment helotv L.w.L, 651 tons. Moment about l.w.l. — 5333x|xl9.25xixl = 652 ft.-tons. Displacement to A.W.L. = 3962 tons. , , 1^210 , ^ , c.b. below L.w.L. = ^ofio'^*"^ 4 19-25 Tonsperin.=446-4x-x-^ =27-3. 1.34148 4 19-25 Transverse BM^g-^^-^ ^ i ^ ~F~ = 8 -28'. 98 DISPLACEMENT SHEET. 2W.L. Function Mult. Product. Mult. Moment, of area. L.W.I. 438-7 5 2194 3 1316 .'W.L. 424-5 8 3396 10 4245 3W.L. 404-4 -1 5590 -1 5561 4 4 404 5186 5157 Tons. Movient about li.w.ii. Up to L.W.Ij. 3311 18862 Toi Layer l.w.l. to 2 W.T,. 634 630 Trai 2677 18232 Displacement of layer (l.w.l. to2w.ry.) = 5186x|xi9-25xf2X^ = 634 tons. Moment about li.w.L. = 5157 x|x 19-25x^^x1 = 680 ft.-tons. Displacement to 2 w.l. = 2677 tons, 18232 c.B. below L.w.L. =^^ =6*82'. 4 1 Q - Q.*} Tons per in. =424 -5 X ->< "T^TT =25-9. 3 420 124701 4 19-25 Transverse BM=2g^-^-^ ^ 3 "^ ""§" = 11-4'. J W.L.. Ci.W.Ii. !W.L. JW.L. Functicn Mult. Product. Mult. Moment, of area. 43S-7 424-5 404-4 438-7 1698-0 25n-i Displacement of layer (li-W.Ii. to 3w.L,.) = 2541 xix 19:^^=1242 tons. 1698-0 808-8 2506 Moment about L.w.li. = „ ^ 4 ^ 19-25x4 2506-8 35 = 2452 tons. Up to L.W.Ii. layer l.wl. to 3w.L. Tons. 3311 1242 Moment about iv.w.ii 18862 Displacement to 3w.L. = 2069 tons. 16410 . c.B. below li.w.ii. = ^..^Q =7-93'. 4 19-25 Tons per in. = 404-4 X -x^^^^ =24-7. Transverse bm= 13-52'. 3 420 114291 4 19-25_ 2069x35 3 3 Explanation of Displacement Sheet (see pp. 94, 95). The length of the ship between perpendiculars is divided into twenty equal intervals, and the immersed depth by seven equally spaced water-planes, the lowest being 2 feet above the keel amidships. Below 7 w.L. is treated as an appendage, it being preferable in all cases not to take the lowest w.l. down to or very near to the keel. The ordinates or half-breadths at the intersections of the vertical cross sections with the horizontal sections are measured off in feet, and set down in dark figures (usually in red) in rows opposite their ordinate number and under their w.l. number. Water-lines that are snubbed or cut away at the ends should be produced to the perpen- diculars by eye for the purpose of these measurements ; the volumes thus added are afterwards deducted as negative appendages. The Simpson's multipliers (halved in order to reduce the labour of multiplication) are placed against the ordinate and water-line numbers ; each 'ordinate is multiplied by the multiplier appropriate to its ordinate number, the result being placed on the right ; it is also multiplied by the multiplier appropriate to its water-line, and the result is placed underneath. Adding the former products in columns gives the functions of the water-planes ; these are multiplied by the appropriate water-line multipliers, and the products then added, giving a number (3231"0 in the text) which is a function of the displacement. The displacement DISPLACEMENT SHEET. 99 ot the main solil is obtained by multiplying this function f x 3^3 x spacing: of w.l.*s x spacing of ordinates ; the factor J is derived from Simpson's first rule applied twice in succession, tha 8 Is 2 for both Bides of ship x 4 for the half multipliers used twice instead of the whole ones ; the 3^5 converts cubic feet into tons (for sea-water). The functions of the water-planes are again multiplied by the number of intervals from the l.av.l. ; the sum of the products (8706'78) being a function of the moment about the l.w.l. The multiplier — | x 5^5 x (spacing: of w.Ti.'s) 2 x spacing of ordinates — gives this moment in foot-tons. Again, the products of the ordinates with the water-line multipliers are abided in rows, the sums being functions of the transverse areas ; these are multiplied by the appropriate ordinate multipliers, and the products added, giving the same function of the displacement (3231*0) as before. These products (headed 'multiples of areas ') are further multiplied by the number of ordinate spacings from amid- ships (station 11) ; the products are added for each end of the ship, and the difference between the suras gives a function (962'76) of the moment of the main solid about amidships. On using the multiplier g X g^5 x spacing of w.l.'s x (spacing of ordinates) 2, the actual moment is obtained in foot-tons. The lower appendage is dealt with by calculating the half-area of each transverse section below the lowest w.l., ani the vertical position of its e.g. They are tabulated on the left, as shown. Each semi-area is multiplied by its Simpson's multiplier, and the result by the number of intervals from amidships ; the functions of areas are also multiplied by the distances of their e.g. below 7 vi^.L. The three results are added in columns, allowance being made for the opposite signs of the two ends of the longitudinal moments, the sums are converted by the correct multipliers as shown. The remaining appendages are calculated by the ordinary rules for volumes and moments of solids, rough approximations being alone required. The 'recess ' is that due to the emergo(nce of the shafts. A table is set forth containing the displacements and moments of each item ; the total displacement and the position of the centre of buoyancy are then found by simple summation and division. To obtain the position of the longitudinal metacentre (see p. 133), each ordinate of the load water-plane is twice multiplied by the number of intervals from amidships. The difference between the sums of the first products for each ends (326"53) is a function of the moment; the multiplier — tx (longitudinal interval)2 — gives the moment of the main portion about amidships. The sum of the second products multiplied by | x (longitudinal interval) 3 gives the moment of inertia of the main portion about amidships. Area of the main portion is obtained from the function of area (438"7) multiplied by the multiplier | x longitudinal interval. In this case the appendage aft, being fairly large, has an appreciable effect ; its area, moment, and moment of inertia are calculated (the last being equal to area x (distance of e.g. from amidship3)2) and inserted in a small table as shown. Thus the total area, the position of the centre of flotation (or e.g. of water-plane), and the moment of inertia about 11 are obtained. The correction necessary to find the inertia about the c.f. (see p. 70) is then introduced, and the longitudinal BM = moment of inertia about c.f. —■ volume of displacement. By assuming an approxi- mate vertical position for the e.g. of the ship, the moment to change trim I inch or "^ ^ gm (long.) .^ obtained. This, together with the 1-2 !> position of the c.f., does not vary greatly with moderate changes of draft ; and they are generally assumed constant. The remaining particulars evaluated are the tons per inch (equal to area of water-plane -r- 420), the area of midship section (equal to function of area for 11 x g x water-line interval), and the transverse BM. To obtain the last, the cubes of the ordinates of the water-plane are multiplied by Simpson's multipliers, and the products added. The DISPLACEMENT SHEET. 101 •am multiplied by ^ x lontjitudinal irterval '.s equal tj thj n^omtrb of inertia of the water-plane about amidships ; on dividing this by the volume of displacement, the transverse bu is obtained (see p. 111). Frequently the displacement, tons per inch, transverse bm, and vertical position of the c.b. are required for other water-lines. Here they are worked out for 'a' w.l. <2 feet above l.w.l.), 2 w.l., and Sw.L. The process consists of finding the volumes and moments of the layers between the respective w.l. and the l.w.l.; and adding to, or subtracting from, the displacement and moment for the l.w.l. The multipliers used at a. w.l. are 6, 8, —1 for volumes, and 7, 6, — i for momenta about middle ordinate (see pp. 46 and 68). For 2w.L>. the multipliers are 5, 8,-1 for volumes, and 3, 10,-1 for moments about end ordinate. For 3 w.li. the ordinary Simpson's multipliers are employed. When the after appendage is large, it is desirable to use the tons p«r inch and C.F., both corrected for after appendage, instead of the " functions of area" taken on p. 97. The transverse BM's are obtained by cubing the ordinates as for the li.w.L. Explanation o? Displacement Sheet using Tciiebycheff's Eule. The horizontal water-lines are spaced equidistantly as with the preceding displacement sheet ; an appendage is left below 7 w.l. The vertical transverse sections are spaced so as to meet the requirements of Tchebycheff's rule (see p. 46), using five ordinates for each half of the length, i.e. ten ordinates in all ; the positions of the sections are indicated at the top of the sheet.* The ordinates are measured from the half-breadth plan ; they are numbered I, II, III, IV, V for the fore end, commencing from amidships, and I a, II a, III a, IV a, Va for the after end, commencing from amidships. The half-breadths are measured off in feet and inserted in the table against the number of the corresponding ordinate, and under the corresponding water- line, in dark figures (usually in red). Under each water-line is set the correct Simpson's multiplier, halved for convenience ; no multi- plier is required opposite the ordinates. The ordinates are first added in columns, the sums being functions of areas of the water-planes. These are mult4plied by the corre- sponding Simpson's multipliers and their sum (1076'82) is a function of the displacement ; the multiplier required to obtain the displace- ment in tons is 2 (for both sides) x § (Simpson's rule for half multipliers) x ^ (salt water) x water-line spacing —^ — . The pro- ducts of the functions of the water-line areas are also multiplied by the number of intervals from the l.w.l.; the results are functions of vertical moments. The sum of these multiplied by | x g^j x (water- line spacing) 2 x ^°^ — is the moment of the main solid about the L.W.L. Each ordinate is afterwards multiplied by its appropriate water- line multiplier, and the products added in rows ; the sums are functions of the areas of the transverse sections. The sum of these gives a function of the displacement, which should be the same as that, previously obtained (1076*82). The differences between the functions of the areas for the fore and after ends, taken in pairs, are written down and multiplied by the levers equivalent to the Tcheby- cheff spacings expressed in terms of half the length ; in the example the functions for the after body are greater in each case than the corresponding ones for the fore body, but, if this is not the case, allowance should be made for sign. The products are functions of moments about amidships ; their sum multiplied by $ x ^ x water- line spacing x ~ is the moment of the main solid about amid- ships expressed in foot-tons. Unless a special body has been constructed with Tchebycheft sections, the calculation of the lower appendage is the same as that in the ordinary displacement sheet, equidistant sections being used. * Alternatively the ordinary * Simpson ' sections numbered 2, 5, 7, 10, 12. 15, 17 and 20, may be taken instead of the exact sections required for Tchebycheff's rule with four ordinates repeated. 102 WEIGHT AND CENTRE OF GRAVITY. The lertiainiag appendages are oalqulated as before; the final table has not been inserted in t&is case. ' The calculation for the transverse metacentre is the same as before except that the cubes of the ordinatea are added direct, and their sum multiplied by — x ■ ^"^* to obtain the transverse moment o ID of inertia. For the cenire of flotation, the differences of the ordinates in pairs are written down, allowing for sign if necGssary, and multiplied by the Tchebycheff levers. The sum of the products is a function of moments ; and the distance of the c.f. abaft amidships is equal to function of moments ^ length. „, ,. ,^1 . , . — z r. 1 X ■ — ,r— The area of the water-plane is equal function of areas 2 f ^ to function of area x 2 x ^"^ ' To obtain the longitudinal moment of inertia, the ordinates are added in pairs, and multiplied by the squares of the Tchebycheff levers ('007, '038, '25, -472, -840). The sum of the products multiplied by a twentieth of the cube of the length is the moment of inertia about amidships. The corrections for the after appendage, and for the position of the c.f., and the calculation of BM are similar to those on the ordinary sheet. The remaining calculations are made as before. WEIGHT AND CENTRE OF GRAVITY OF SHIPS. In the early stages of design, an approximation must be made to the weight of a ship in order to determine whether it is equal to the displacement assumed ; the position of the centre of gravity is also required in order to determine the stability and trim. The weight of a ship is conveniently divided into six items : hull, equipment, machinery, fuel, armour, and arma- inent. In a merchant vessel the two last named are replaced by the load to be carried. The proportions vary greatly in different ships ; those in the table are illustrative of certain types : — Type of Ship. Percentage Weight. i >> g c ^d u Armament or "5 '3 -§ ^ Load. W H ^ ^ - < Battleship . . 34 3 10 3^ 31J 18 First-class Cruiser 38 4 17 7 20 14 Light Cruiser. . 43J 6 20 12 12^ 6 T. B. Destroyer (deep condition) 34 4 34 25 — 3 Steam Yacht . . 61 10 19 10 — — Atlantic Liner . 55 included 27 16 — 2 (Passengers in hull. and Stores) WEIGHT OF HULL. 103 Hlll. First Estimate. The weight of hull is determined to a first approximation * in a variety of ways. In ships of very similar typo it may bo assumed to be the same percentage of the displacement, e.g. 340/0 in battleships, etc. Or it may be compared with the product length X (breadth + depth) amidships, the co- eflBcient being determined from a similar ship, making allow- ance for any great alteration of scantlings. Mr. J. Johnson (in Trans. Inst. Nav. Archs., 1897) published a useful method for approximating to the hull weight of a vessel built to the highest class at Lloyd's or Veritas. If N = a modification of Lloyd's old longitudinal number = Length from after part of stem to fore part of stern post on upper deck beams x {i greatest moulded breadth + depth from top keel to top upper deck beams + ^ midships girth to upper deck stringer}. In spar- and awning-deck vessels the girths and depths are measured to the spar or awning decks ; they are taken to the main deck in one-, two-, and well-decked vessels, w = finished weight in tons of the steel hull. = K (-^) ^ \ioo / Then w or log 10 W = X log 10 (-3^) - ^• Where x and K or A are determined from the table below : — Type of Vessel, Three deck .... Spar deck Awning deck .... One-, two-, and well-deck Sailing 1-40 1-35 1-30 1-30 1-40 0-492 0-576 0-665 0-856 0-410 0-308 0-240 0-177 0-068 0-387 The distance of G abaft the middle of length and above the keel can be estimated from information available for other ships, taking these distances proportional respectively to the length and totsd depth. Detailed Estimate. In the later stages, when scantlings are fixed, the weight and centre of gravity of hull are found as follows: — The hull may be divided into two groups : (1) calculable items forming about 60 0/0 (in a warshij)) of the whole, which include the greater part of the structure, and (2) * judgment ' items, including portions of complicated structure and fittings. The latter can only be assessed by comparison with known weights in a similar ship (recorded weights if possible) ; the centre of gravity of each item can usually be determined ♦ See also under " Design ". 104 WEIGHT OF HULL. with fair accuracy from its position. The former are directly calculated from the scantlings ; the manner of so doing is indicated in a few instances below. To each item 3o/o should be added for fastenings (or such an addition should be made at the end of the calculation). If the stresses on the ship are also required, ifc is con- venient to divide every item into portions wholly before and wholly abaft the midship section. Moments are taken about two fixed planes, one being generally the midship section and the other the l.w.l. or the keel. Outer Bottom Plating. — Assume all of uniform thickness, repeating as necessary for the portions whore the thicknesg is in excess or in defect of that assumed. Divide the length into sections spaced equidistantly, and measure the half-girthp at each section. Apply the method of Rule II, par. 24, p. 58, obtaining the ' modifying factor ' at each section. (This is the ratio of the slant length of the mean water-line intercepted between the sections to their perpendicular spacing.) Find the height of the centre of gravity of each section by dividing it into four equal parts, and proceeding as in par. 16, p. 62. Then arrange the calculations as in the following table for the forward portion of a warship below the armour deck, the percentage for laps, butts, and liners being taken as calculated for an average-sized plate (say 20' x 4'). Transverse Framing. — There are usually several varieties, such as web and ordinary frames, or bracket, lightened plate, and watertight frames. To avoid calculating each one separately, calculate the weight and height of e.g. for a specimen frame at intervals of about ^^ length. Plot these as curves on a base of length of ship, drawing separate curves for each type of frame. The weight and e.g. position can then he read off for each frame, or a mean can be taken for a group of similar frames coming together. Longitudinal Framing. — Usually uniform in section, but the height of e.g. must be taken at equidistant intervals, and the mean taken. Bulkheads. — Main bulkheads are usually thicker towards the bottom, and the e.g. is below the centre of area. Take the minimum thickness of plating and calculate its weight and e.g. without any allowance. Then add the additional thickness, the stiffeners, and allowances for laps, butts, and fastenings ; all but the first item have the e.g. at centre of area. Then find the ratio of the initial to final weights, and of the e.g. below centre of area to the height of bulkhead. If this be done for one or two typical bulkheads, the rest may be determined from the simpler first calculation by allowing the same ratios. The ratio of weight for the main bulkheads of a warship is 1 : 1'9 ; for ordinary below-water bulkhead/* it is 1 : 1 66. WEIGHT OF HULL. 10, ■5 W?I>-Qbl>- C^ GodsiCirHOl 10 CO to ^DO-^OO Oi O500.-l»0 CO iH iH Ol -H C^l C^ G<) Ol CO ft 2, p. !3 -^ g CO ^• 1 1 s § oOtHcqOiOs cooot>oo .—1 3 § o 2 -^"■i^kOrHCiCO C^I'^rHC^lO rH rn H 10 CI 05 -* CO '-T' -T** t^ CO »C rH rH C^ OJ 0^ (M r-( : 1 i-l i-H CO iH ^ 2 •^ -3? 1 oT op o o »0 -^ CO -;C>;D 00 ;0-^C500 ■^ ;?' I' c^ o^rHods CO (^^I^.cocbl^^ CO 1 i-H S >j rH »C CO t^ CO OO Tj< X •»«( X (M tr- !h m l^ fa^ ie 1 I-H O^ fH eo«o 05 «Dt-asoo i II ob 1 rHcibiHlOOS rH GSICOCO""^'^- ?-i 10 iH IM^NCOCOCO -^ rl<-^>«J<'^-<* X X II li 1^^ eo CO CO (M 05 iH i 1 X CI C<1 X II < 09000 X rH :«£) IMI^ 1 a »o . (M X 'lO II B II « ^ II ?^ 5 OC;iQO«OQO up 2-0 9-0 •1-5 1-2 1-5 •5 •4 3-2 •4 11 •9 12-0 •5 2-2 •8 1-0 1-2 •6 60 1-2 1-0 Before 6. C.G. Mmt. 15-6 2-5 64 1-0 55 20 2-5 16 20 1-6 33 55 347-4 45 653-4 54-2 112 120 24-4 29-2 320 Abaft 6. I Above base. C.G. Mmt. C.G. I Mmt, 2 2 •5 11-8 22 50 5-0 250 22 20 1-9 16-5 27-6 650 760 19-2 67-5 IO08O 130 35 18-3 18 48 50 367 48 214 1050-4 200 5-0 1-0 106-2 330 60 2-5 27-5 51-3 900 48-4 15-6 21-0 21-9 10-8 43-2 200 110 f6 870 720 240 1831-3 10 8 746-3 270 18-5 150 18-7 21-5 190 16 5 18 10-0 12 21-0 190 220 190 20-5 165 180 22 22-5 150 17-0 170 17-0 165 1-5 140 20-0 15-3 65 12 12 8-5 9 3 12 12 12 12 12 56 55 7-0 180 1-1 10 110 C.G. abaft 6, 2 14'. Above base, 11'. Take 350 tons. 2 1' abaft 6. 11' above base. 108 WEIGHT OF EQUIPMENT, ETC. Equipment. This is conveniently divided (in warships) as follows : Fresh water (for 10 days allow about 8*7 tons for 100 men) ; provisions and spirits, including tare (for 4 weeks allow about 5*7 tons per 100 men) ; officers' stores and slops ; officers, men, and effects (allow 8 to the ton) ; masts, rigging, sails, etc. ; cables (500 fathoms) ; anchors ; boats ; warrant officers' stores (4 months) ; torpedo net defence. In passenger ships undergoing long voyages allow 1 tonj per 5 persons for passengers' gear, including baggage,- bedding, etc., also '025 ton por day per person (average) for water and provisions. Machinery and Fuel. In the preliminary estimate the weight of machinery is based on the total power. Coefficients for various types of machinery are given on pp. 389, 390. Information obtained from actual ships should also be utilized where possible. The weight of coal assumed is frequently an arbitrary amount less than the full bunker capacity. The full coal storage can be, however, determined from the volume of the bunkers calculated by the rules on p. 54, the areas of the sections being taken to underside of beams only. About 10 or 15 per cent (varying with type and shape of bunker) is then deducted for broken stowage ; the net volume in cubic feet, on being divided by 43 (North Country coal), 40 (Welsh coal), 36 (patent fuel symmetrically stowed) or 45 (patent fuel shot into bunkers), gives the stowage in tons. The weight of liquid fuel is equal to the whole volume in cubic feet divided by 38-5. In all cases the centre of gravity of th3 fuel is the e.g. of the volume (see p. 66). Armour. The weight and position of e.g. of armour in warships are determined by a process similar to that adopted for the hull. If the armour is not specified by its weight par square foot of plate, this can be determined from its thickneS(3> sim^a it weighs 4951b. per cubic foot. Add l^o/o for bolts. Backing is dealt with similarly to the planking of a teak deck. Armament or Load. The weight of guns, mountings, charges, and projectiles are known (see pp. 380-385), and the position of the e.g. of each item can generally be spotted without difficulty. Allow 30 to 40% tare for cartridge cases. SUMMARY OF WEIGHTS 109 The load in a cargo ship is generally determined before- hand. Its e.g. is usually found by assuming the whole space available to be filled with a homogeneous cargo — the assump- tion the most unfavourable to the stability ; the e.g. is then that of the volume of the hold. For passengers without baggage allow 16 to the ton with men, women, and children ; 14 to the ton with men only. In pleasure steamers where the stability can be affected, assume the e.g. of passengers seated to be 6 inches above the seat ; for those standing?, 2 feet above the deck is generally a safe) assumption. The final weight and position of centre of gravity are found by adding together the weights and moments of the several portions as shown in the table below : — Summary OF Battleship Weights (580' x 9C )'x2r6"- -44'deep). Moment Moment Moment Item. Weight. from above below amidships. li.W.Ii. li.W.L. Weight before amid- Tons. Ft.-tons. Ft.-tons. Ft.-tons. ships— Armament .... 2076 236110 43020 1870 Armour 3270 353300 42077 1901 Hull 3770 355340 25473 28796- Machinery .... 820 29500 — 8200 Coal 670 23000 1360 5840 Equipment .... Weight abaft amid- 400 55000 7420 766 1052250 ships — Armament .... 2763 348020 46280 2190 Armour 3585 481250 28848 3563 Hull 4933 697170 24849 38897 Machinery .... 1930 198800 — 19300 Coal 330 21400 640 3060 Equipment. . . . 335 48200 3564 52 1 24882 1694840 1052250 223531 114435 114435 1 642590 109096 Total weight of ship- = 24882 ton s. C.G. abaft y^ - 642590 _ ^ 24882 7 feet. C.G. above L.W.L. - ^^f ^^6 - 4-38 feet 24882 110 STABILITY. Fig. 120 STABILITY. If a ship be slightly disturbed from a position of equi- librium, and if the forces then in operation tend to restore the original position, the equilibrium is termed stable ; if the forces tend to move it further from the original position, the equilibrium is termed unstable ; if it shows no tendency to move away from or return to the original position, the equilibrium is termed neutral. The equilibrium of a ship is always stable as regards vertical deflections causing an alteration of displacement ; the only disturbances that need examination consist of inclina- tions about horizontal axes with the displacement unaltered. Of these the principal are : (a) in'^lination in a transverse plane about a longitudinal axis, and (b) inclination in a longitudinal plane about a transverse axis. The stability in these directions is termed transverse and longitudinal respectively. Transverse Stability. T^ig. 119 is a transverse section of a ship heeled over through a certain angle 6. w'l' is the water-line for the inclined, position, and wl is the water-line for the upright position. These two planes intersect each other in a longitudinal direction, and bound two 2 wedges l'sl and wsw equal in volume to each other, provided the dis- placement remains the same. The wedges are . , called respectively the wedges of immersion and emersion, or the in and out wedges. G is the centre of gravity of the ship and b' her centre of gravity of displace- ment, or centre of buoyancy. The weight of the ship then acts vertically downwards through G, and the resultant pressure of the water acts vertically upwjards through b', these two forces forming a righting couple, the arm of which is GZ — that is, the perpendicular distance between the lines of action of the two forces. The moment of this couple— that is, the weight of the ship, or its displacement, multiplied by the length of the arm GZ— is the moment of statical stability of the ship at the given angle of inclination e. This moment is generally expressed in foot-tons — that is, the weight of the ship in tons multiplied' by the length of the arm Gz in feet. B is tho centre of buoyancy of the ship when upright ; S^ ig the point of intersection of the two water-lines, I the point where the vertical b'm cuts the plane of flotation ; g and g' ^IG. 119 STABILITY. Ill are the centres of gravity of the emerged and immersed wedges respectively, gh and g'h' being perpendiculars dropped to g and g' from the plane of flotation w'l'. The point M. where the vertical line BM, drawn through the centre of buoyancy B when the ship| is in (an upright poaition, cubs the vertical" line B'M, drawn through the centre of buoyancy b' for the inclined position, is termed the tranwerse metacentre when the ship is inclined through an indefinitely small angle, and also when the point of intersection is the same for all angles of heel. If the centre of gravity g is below the metacentre m, the equilibrium is stable ; if a is above M, the vessel is unstable, and will capsize or at least h^ol to a larga angle ; if o coincides with M, the equilibrium is neutral. The intersection of the new vertical through b' is found usually to pass very near the metacentre m for all angles of heel Tip to 10° or 15°. Within these limits the stability lever GZ is equal to GM . sin Q \ or the moment of statical stability ia w . GM sin Q. For moderate angles the stability depends wholly on the value of GM, which is termed the metacentric height. The position of G is calculated by the rules given on pp. 102-9 ; that of M is obtained by the process indicated below. To obtain the height of the transverse metacentre. Assume the angle of heel B to ^ig. 121. be small, let y = half breadth WS or SL at any station. V = volume of either wedge wsw' or lsl'. j7, <7i = c.g.s. of the wedges. h, h\ =^ ^k^ctRcenties. ^«AU6HTp^,'^S-, PER ■p- p-^-— MENT. ~co , ~00 00 oc l6-0>i 4.140 A CO CO CM CM / 15-6 3.964 27 65 Z X Z ^ 14-- 6 3.623 2714 B . 13-6 3.303 26-65 / / 12-7 3 031 C 1 f 'y ^2-3 2.947 D / 11-6 2.677 2615 / 9-6 2,067 24 81 / > c urvet Jf ^ ^ bu oyanc y. A. Deep condition. Coal 725, oil 142, reserye feed 92 tons. B. Normal condition. Coal 450 tons. C. Light condition. No coal ; no consumable stores. D. Condition as inclined. Note. — The curve of buoyancy is generally nearly straight ; 4he tangent of its inclination to the horizontal is equal to 12 X depth of C.B. below W.L. x Tons per inch Tons displacement 114 STABILITY. Type of Ship. Minimum gm in feet. First-class Battleships— Modern . Do. Older types and Cruisers Torpedo Boat Destroyers . . . Torpedo Boats Steamboats Large Mail and Passenger Steamers Cargo-carrying Steamers . . . Tugs 5-0 3-5 2-0 1-5 1-0 1-0 to 2-0* (maintained by water ballast) 2-0 1-5 Very large 1*5 to 6 (depending on sail area) Shallow Draught Vessels .... Sailing Ships In some very large modern liners the gm is greater; e.g. in Aquitania GM is 4 feet. Approximate Formula for Height of Metacentre. Depth of Centre of Buoyancy (Normand's Formula). Depth of C.B. below water-line = J mean draught + volume of displacement . , , ^ , displacement in tons 1 1 7 ^1 ; or i mean draught + -r;; — 7-7 -• — r S X area of water-plane ' *> ° 36 x tons per mch Note. — If a bar keel is fitted, the mean draught should be taken to the top of keel. Alternatively, this depth can be expressed as a percentage of the mean draught, which is about 42 for fine ships, 44 for ordinary battleships, and 46 for many merchant vessels. Distance (bm) between Centre of Buoyancy and Mela- tentre. (greatest beam)^ BM = -^ li -, — K X draught ships, 12 in light cruisers, destroyers, cargo, and passenger steamers, and 11 in steam yachts. In new designs it is advisable to take the value of K found in a similar ship. Stability at Large Angles of Heel. If the heel be so large that the vertical through b' (fig. 119) no longer intersects the middle line at a fixed point, the metacentric method is no more applicable. During the inclination of the ship the centre of buoyancy moves from B to b', and b' lies in a plane parallel to a line joining g and g' . The distance bb' can be found from the where K is approximately 13 in battle- following expression : — bb' = V X gg where v = volume of displacement and v of the wedges ; volume of either STABILITY. 115 V X hh' , . T , . / BR = , where BR is. perpendicular to B M ; , . „ V X hh' . „ and GZ = BR - BG . sin 6 = BG . sin 0, whence Atwood's formula for expressing the moment of statical stahility at any angle Q is fivxhh') , . A M = w-j — - — - - (bg . Sin e)f The moment of statical surface stability at any angle B is BR X w, being the righting moment obtained on the assumption that the e.g. of the ship coincides with B. The angle of heel in fig. 119 is bmb' = lsl', and its sine is equal to BR GZ BM " GM Dynamical stahility is deftnod to be the amount of mechanical work necessary to cause a body to deviate from its upright position, or position of equilibrium. Dynamical stability is expressed as a moment by multi- plying the sum of the vertical distances through which the centre of gravity of the ship ascends and the centre of buoyancy descends (i.e. the vertical separation of G and B), in moving from the upright to the inclined position, by the displacement. In fig. 119 during the inclination of the ship through the angle 6, the centre of gravity has been moved through a vertical height Gii — go, and the centre of buoyancy has been lowered through a vertical distance b'i-ch, and the whole work to do this, or her moment of dynamical stability for the given angle 0, is = w{(gh - go) + (r'i - bh)} = w(b'z - bg) = w(b'r - BG . vers 0) /wv (gh-{-g'h') „, = w(^— ^-- -"^ — ' - BG . vers 0) ; whence Moseley's formula for the moment of dynamical stability at any angle is = ivvigh + g'h') - (w x bg . vers 0), where w is the density of water. The dynamical stability of a ship at any an2ple is the integral of its statical stability at the given angle — that is, if M = the statical stability and U the dynamical stability, then U = ^^1(10, where d0 is a very small angle of heel. The moment of dynamical surface stability is expressed by multiplying the weight of the ship, or displacement, by the depression of the centre of buoyancy during the inclination -that is, for the angle U = w(b'i - bh). 116 STABILITY. The Curve of Statical Stability is a curve used to record the value of the stability lever (Gz) of a vessel at any given angle of heel. Fig. 123. curve of statical stability of an ibonclad with high freeboard Method of Construction. — Calculate the length of the arm of the righting couple, or GZ (see fig. 119), for several succes- sive angles of heel taken between the upright position and that at which the leng'th of the arm becomes zero ; set the lengths of these arms off as ordinates (see fig. 123) from a base line the abscissa) of which represent to scale the respective angles of heel : a curve bent through the extremi- ties of these ordinates will form a curve of statical stability- Note.—The length of the perpendicular at 57'3° (one radian) intercepted between the tangent at the initial portion of the curve aaid the base line is equal to the metacentric height. The Curve of Bynamiccl Stability is constructed in a similar manner to that of the curve of statical stability, with the exception that the various lengths of the arm (b'z — bg) = (b'r-«g vers 6) (see fig. 119) are taken as ordinate? instead of GZ. Or preferably the curve is obtained by in- tegrating the statical curve. The area up to each ordinate of the statical curve expressed in degrees X feet is divided by 67"3° and set up as an ordinate of the dynamical curve. Fig. 124. curve of dynaincaii stability of an ironclad with high freeboard. ^^' '""^ M* tt» «• !«• as* »• 35 " «- 4S» so" SS« «o' 66* lo" 7s' «0' tS »• BCALK tr •'■cncES rta amblb tr tiuk 2sote.—T\\Q angle at which the statical lever vanishes (and at which the dynamical lever is a maximum) is termed the range of stability. STABILITY. 117 Fig. 125. Cross curves of Stability. Cross Curves of Stadility. These curves may be termed 'vertical curves of stability ' ; they consist of curves of righting levers at v.arious draughts or displacements for certain fixed angles of heel. They hold a somewhat similar relation to the ordinary curves of stability as the body plan of a ship does to its water plane. For cross curves (see fig. 125) the righting levers are cal- culated at certain fixed degrees of heel at various displace- ments, and the levers are set up as ordinates from an axis the abscissa} of which represent the displacement at which the levers for the fixed degree of heel are found. A number of such curves are constructed for various inclina- tions, and set off as in fig. 125. Fig. 126. jo> Curve of Stability at soo tns.Displac Mt. T. 15 30 45 60 75 Scale of angle of heel in degrees. For finding such curves at various draughts and angles of Fig. 127. heel, say at 15° (see A / fig. 127), divide the body plan by a number of parallel planes representing various draughls of water or displace- '•^-.Tients. ^ Drop a perpen- ^ dicular through thp w^ point where the highest water-line cuts the middle line of the ship, and then calcu- late (by the methods indicated hereafter) the horizontal distances d^, d^, d^, etc., of Fig. 128. Scale of feet for levers. 113 STABILITY. the centre of buoyancy up to each inclined water-plane from the vertical ab. By assuming the centre of gravity to be at s, and fixed there -for all draughts, the dis- tances f/j, d2, d^, etc., would be the righting levers at the dis- 7oo- placement?, up to the respective water-planes w^, Wg, W3, W4. q, These lengths are then set off 5^00 as ordinates along an axis having ^ the several displacements up to o ^ the water-planes as abscissae. ^^ The actual righting levers can w then be determined, when the r correct positions of the centres o of gravity corresponding to the s various displacements are fixed, ZgQo by multiplying the respective dis- 5 tancej h^, h^, A3, etc., of the h actual centres of gravity g^^ gc,, z 200 .^3, 9i below S by the sine of the ' angle of heel, and adding this length to the arms already found (see fig. 128). The actual righting lever for the displacement up to W4 would be equal to di + /^4 sin 15° = di + ^4,. Up to Ws, W3 it would be equal to dz + Jib sin 15° = ; ^ j i- ic :S| «> ?, » I I I ! I IS u, o LL- Lil o o N e 527 = .-3 I 2 ssaft's i C C ' 4) * •a" * ■ 11 1- ^1 ll Ills I fa's ^U< H I5-, s- 11 li gS§E 9^r5 oJo iliggiiilii-Niil ii^^ SS g IS' ■^ s '-^ 2 I o i « o o « o -^^^ ^^ t II i iii^"g il = 1 .JO p'"2;2gSg £ i3;jlHV I C 5 • c 'si I -3 II i i 124 STABIUTY. sums of the functions of the cubes are multiplied by the sines of the various angles of inclination instead of the cosines; the sum of the products so obtained being divided and multiplied by the same numbers as were used for the statical stability, in order to find the moment of the wedges uncorrected relatively to the respective radial planes. The corrections for the appendages are then made, that for the correcting layer being subtracted in all cases. The moment for the correcting layer is found bj^ multiplying its volume by half its thickness, that being about the vertical height of its centre of gravity from its radial plane. This final result divided by the total volume of displacement will give the length of the arm b' r, from which if BG . vers 6 be deducted, the remainder will equal the length of the arm for the dynamical stability, or the vertical height through which the centre of gravity of the ship has been lifted and the centre of buoyancy depressed. 12. Geometi'ical Mode of Calculating Dynauiieal Stability. — The dynamical stability of a vessel at any given angle of heel is the sum of the moments of the statical stability taken at indefinitely small equiangular intervals up to tlie given angle of heel, and is therefore equal to the area of the curve of sta- tical stability included between the origin of the curve and the angle in question. It must be noticed that the abscissie of a curve of statical stability is given in angles, and therefore the longitudinal interval is taken in circular measure. But, as the lengths of the arms for statical stability are generally used to construct a curve instead of the moments of stability, the area, as above found by the rule from such a curve, will necessarily give the length of the a7'm for dynamical stability and not the moment. Example (see fig. 123). — To find the length of the arm for dynamical stability at an angle of 30° inclination. Angles of Heel de2:rees 10 15 20 25 30 Lengths of Statical Levers GZ •0 •2 •42 •68 •97 1-30 1-66 Simpson's Multipliers Products •0 •8 •81 2-72 M)4 5-20 1-66 l346 ^ of angular interval in circular measure = '0291 1316 11844 2632 Dynamical lever for 30° = -382956 STABILITY. 125 13. Curre of Stability for Light Draught. — The lengths o.f the arms for this curve can readily be approximated from the results obtalijed for the curve in the load condition. In fig. 131 WL 13 the load water-line, and wl the light water-line, for the upright position of the vessel. If the vessel is inclined through an angle 0, and w'l' is the true position of the in- clined water-plane for the load condition, then the true position of the wafer-plane for the light condition will run parallel to it, as w'V. To determine its perpendicular dis- tance from w'l', divide the volume of the layer contained between the light and load water-planes by the area of the assumed inclined water-plane hh', which was found for the inclined load condition. Let B be the centre of buoyancy for the upright load condition, b' for the inclined load condition, and b2 for the inclined light condition, br is perpendicular to the vertical b'm, and br' is perpendicular to the vertical b2m'. Let V equal volume of light displacement. „ V == volume of displacement contained between the light and load water-planes. „ c = distance of centre of gravity of assumed in- clined water-plane from the vertical through A, assumed positive on the emerged side. „ t32 and G'z' = the lengths of the arms of the righting couples for the load and light condition respectively. rrn- , , ■ f . fl , '^(BR - BA sin -^ c) Then g z' = GZ - GG sin e + '- Example.— In. the ship illustrated in the tables (pp. 122, 123) find the lever of statical stability at 30° when light, assuming the displacement diminished by 200 tons, and the e.g. raised by 15 ft, B is 6*5 ft. below original upright w.L. Here v = 200 X 35 = 7000 ; V = 86767 - 7000 = 79800 approx. GZ = l-65; gg' = 1-5; sin 6 = § ; BR = 4-78 ; BA = 6-5 ; C = - 1-16. 7000(4-78 - 3-25 - 1-16 ) GZ' 1-65- -75 + 79800 •93'. Direct Method. Lay a piece of tracing paper over the body plan, and on it draw a trial water-line at the correct inclination. Trace the u^ STABILITY. wedge sections, replacing the curved portions by one or two straight lines approximating ^as closely as possible to the curves. Find, graphically the areas of the triangles and quadrilaterals, and thence determine the volume of each wedge. If these are not nearly equal raise or lower the water-line, and proceed as before until there is practical equality in volume. Find the e.g. of each triangle or quadrilateral (see p. 59) and calculate the moment of its area about anj' line perpen^ dicular to the inclined water-line. Thence find the moment? of the volumes and add. The total moment divided by the volume of displacement is equal to be (fig. 119), whence GZ is at once determined. The direct method is, perhaps, the most convenient one when an integrator is not available Amsler-Laffon's Mechanical Integrator. By means of this instrument, the area, moment, and moment of inertia about any axis, can be obtained for any curvilinear area by tracing its outline with a pointer. Its principal use is that of obtaining the stability of a vessel at various angles of heel and at various drafts. It is usual, when using this instrument, to first calculate the righting lever for a number of displacements atone inclination, say 15°. Then the same for 30°, 45°, and soon ; the cross curves being constructed before the ordinary curves. Let fig. 132 be a body plan drawn for both sides of a ship ; let wl be its upright waterline intersecting the middle line at s. Through S draw inclined waterlines at the required inclinations, and let w'l' be any one of them, say at 15°. The first step is to find the displace- ment at w' l' as it is gene- rally different from that at WL. The pointer is passed (i) round the two end sections, (ii) round the dividing sections, and (iii) round the intermediate sections* ; the pointer in each case passing along the waterline and round the section, as w'l' aw'. Readings are taken at the start and after passing round (i), (ii), and (iii), so that after subtracting, the readings due to each of the three series of sections are known. Reading (ii) is multiplied by 2, and (iii) by 4, and the tw^o products added to reading (i). The total is then multiplied by the common interval and the constant of the instrument and divided by 3 times the square of the scale used. The result is the volume of displacement, w^hich is then reduced to tons. * See Simpson's Rules. STABILITY. 127 If in the same way ST, the line through s, perpendicular to w'l' is made the axis for moments, and the readings for moments are treated in the same way as those for areas, it is evident that tlie final result will be the moment of the under- water portion about ST as axis (obviously, the total must now be divided by 8 times the cube of the scale instead of the square). This divided by the volume of displacement will give the per- pendicular distance of the inclined centre of buoyancy from ST ; that is sz, when b'z is parallel to ST. The righting lever, or GZ, is equal to sz + SG sin Q when G is below s as at g, ; and equal to sz — SG sin d when G is above S. The righting lever gz is set off at its proper displacement on the cross curve for 15°. This is done at different waterlines and the cross curve thus completed. The following is the actual form of the calculation for sz. Sections 10*6 apart. Scale of body ^" to 1 foot. 20 1000' 40 1000' Machine constants Areas Moments Angle of Hekl 15^^ Sections Initial End ordi nates Dividing ordinat. Intermediate „ . Areas 4029 4111 10502 17309 6391 6807 ,"2 fi 82 12782 27228 40092 Moments 982 1398 1819 4 412 421 4 824 1684 2512 Displacement in tons * The 4 multiplier is the reciprocal of the scale of the drawing. 123 STABILITY. Tchebyoheff's rule (see p. 43) can be very usefully cm- ployed instead of Simpson's rule in the above ; the saving of time due to its adoption is, for a complete set of cross curves, more than suflScient to compensate for the time of preparing the special body plan, which need only be drawn in fairly roughly. In this labour may be avoided by using the sections numbered 2, 5, 7, 10,. 12, 15, 17, and 20 from an ordinary body plan whose equidistant sections are numbered 1 to 21. It will be found that these coincide nearly in position with those required with Tchebycheff's rule for 4 ordi nates, repeated. This was pointed out at Inst. N.A., 1915, by Mr. W. J. Luke, Example. — Length of ship, 210 feet ; number of sections, 8 ; scale of body, J" to 1 foot. Machine constants as before. 20 ''lO 1 Displacement in tons = — r- x 16 x ~ x — x area reading. J.UUU o oo 4 = the scale Moment reading ^^ ^ 2 = ratio of Area reading machine co i- stints. Note. — The above or ' all-round ' method is the simplest, since it gives directly the stability lever desired. A more accurate and expeditious method, however, is that known as the * figure-eight '. The pointer is passed around the outline of the wedge sections, only, taking them in the opposite directions on the two sides of the ship ; e.g. commencing at s (fig. 132) the pointer reaches the points l', l, s, w, w', s in the order named. The result is to give the difference of the wedge volumes (by the area reading) and the sum of their moments (by the moment reading). If v be the original volume of displacement, v the increased volume registered by the machine, and M the moment registered (the last two being found from the readings as in the ' all-round ' method), and Bs the distance of the upright C.B. below s, M - V . BSsinS sz = TT^— Formulae for Stability Levers in Special Cases. 1. Ship with concentric circular sections, cylinder. — The metacentric method is here applicable to all angles of heel and statical lever GZ = gm sin d, dynamical lever = gm vers Q. 2. Wholly immersed vessel. — The metacentre and centre of buoyancy are coincident, and the above formulae apply if B be substituted for m. 3. Wall-sided vessel, parabolic cylinder. — Statical lever GZ = sin B (gm + i bm tan^ 0). Dynamical lever = GM (1 - cos 0) + i BM (sec + cos - 2). For BM. its value when ship is upright is intended. STABILITY. 129 Change of Metacentrio Height dub to Small Chang 1:9 IN Dimensions. Let the beam of a sliip bo increased by — of itself, all transverse ordinates being augmented in the same proportion. Similarly let the draught be increased by — of itself. If these changes are moderate, and the height of the e.g. above the keel be assumed to vary as the draught, the increase of metacentric height is given by — where m is the original GM, S/« the increase of gm^ and a is bg. If the beam only be increased, — = 0, and ?"' =-"(!+-"-)... (2) 111 Hi \ 1)U ' If the draught be diminished so as to maintain the same displacement as before, — = — — and -?^=i(3 + i?)...(3) In the preceding case if the total depth be unaltered (the freeboard being increased to compensate for the diminution of draught), and if /i represent the height of the e.g. above the keel, originally, VI n V 111 / ^ If in the preceding case it be assumed alternatively that the freeboard is unaltered (the height of e.g. above keel varying as the total depth as before), and if s represent tha original ratio — freeboard -=- draught, — = - 3 + (4(1- ---) y . . . 5 m n y in ^ 1 + sO In the general case, determine the effect on GM of increasing the beam by one foot, assuming that bm oc lb^/A- The increase of GM roughly varies as that of beam. Example 1. — In making a preliminary estimate of the iimensions of a new design, the beam is assumed 36 feet, the Jistance bo is 8 feet and GM is 1 foot. It is desired to double the metacentric height, while maintaining the draught un* altered. Find the beam required. Using formula (2), Zm = 1, m -^ 1, a — Q, Whence -(1 + 8) = 1, or — = i. Ill ' n 18 180 STABILITY. Therefore the beam required is 36 (1 + ^) or 38 feet. Note that if it is desired not to alter the displacement, the ength must be diminished by ^ of itself. Example 2. — In a battleship having beam 89 feet, mean draught 27 feet, GM 5 feet, a above water-line 6J feet, and BG 18 feet, find the effect on the metacentric height of in- creasing the beam by 1 foot, assuming that owing to a change in the distribution of weights the e.g. is 0*35 feet higher above the water-line in the new design. The displacement and length are assumed unaltered. Using formula (3), m = 6, ni = ■^, a = 18. Whence -y = ^(8-t-~|) = -195 or 5w = -98 feet. But this assumes that the height of G above water-line becomes 6-5 (l + — ) or 6-5 (l - ^7.) or 6-43 feet; it is actually 6 .83 feet. ^ ">^ ^ ^^' Hence the metacentric height is 5 + '98 — (6-85 — 6-43) = 5 -56 feet. Alteration of Stability Curve due to Small Changes in Dimensions. Assume the beam increased by — of itself, and the draught 1 '^1 by — of itself as above. This process is applicable to any two ships of fairly similar form, even if they depart somewhat from exact proportionality. Given the curve of statical stability (Gz) for the first ship, it is required to construct the corresponding curve for the desired vessel without con- structing the body plan or performing the usual calculations. EuLE. — I. Construct the curve of dynamical stability of the first ship by taking areas of the GZ curve (see p. 116). Two or three spots are sufficient, as great accuracy is not required. 2. Corresponding to the angle B at which the stability lever is required in the new ship, determine an angle

therwi8e it is the difference between the changes of draught. To determine the draughts and trim at, the draught marks jiven those at the perpendicular, and the converse. Let L = length of ship between perpendiculars. a, b = distance of forward and after draught marks from amidships. Di, D.2 = draughts at F.P. and A.P. D3, D4 = draughts at forward and after draught marks. + - (Di - Da). D3 = D4 = 2 Di + Da + |(-> Dj. 132 STABILITY. a + b , , D4 - I>8 = — Z (D2 - Di). J-l (a + &) Di = Da ( 2 + & ) — D4 ( I - « ) {a + b)i)2 = D4{^+ a,)-Ds(^~ - b) To determine the displacement of a vessel floating out of her designed trim. Let D bo mean draught amidships, w the corresponding displacement as obtained from the displacement slieet, T the tons per inch, d the number of inches excess of trim by tha stern, L the length in feet between perpendiculars, and c the distance of the centre of flotation abaft amidships in feet. cd" Then virtual mean draught is D -\ — — L Hence the displacement is W H tons. Li Ex. — In a ship where l = 400, c ^ 15, t = 80, the dis- placement deduced from the mean draught is 14,000 tons where the ship has a trim of 2 feet from the bow. If the normal trim bef 1 (Foot by the stern, find the true displacement. Here d = — 36", and increase of displacement is 80X15X36 ,^^, 400 =-108 tons. Hence displacement is 14,000 - 108 -= 13,892 tons. Note. — 1. The distance c expressed as a fraction of the ship's length has the following average values : — Battleship ^, light cruiser ^, t.b. destroyer ^, steam yacht -^, channel steamer, ■^, cargo steamer 1^. 2. The centres of buoyancy and gravity lie abaft the midship section at a distance, which, expressed as a fraction of the ship's length, has the following average values : — Battleship ^, light cruiser -gV. t.b. destroyer ^, steam yacht 1^, channel steamer ^, cargo steamer 0. 3. For a change of trim t without change of displacement, the t tc t tc draught forward is altered by— + — and that aft by 7, — — J L ^ li To -find the changes of draught and trim in passing from salt to fresh water^ and vice versa. Let the symbols w, T, and D above refer to salt water. Let 8 inches be the sinkage in fresh water, and d' the final meap draught. • 9w w Then D' = D + 8/12; i=^^ = ^:^^ It is assumed above that the fresh water occupies 35*9 cubid STABILITY. 133 feet to the ton. If the water is brackish, and occupies 35 + a: cubic feet to the ton, the latter formula becomes 3oT The change of trim is usually very small. If c' be the distance of the centre of flotation abaft the centre of buoyancy, and M the moment to change trim in salt water, the change of trim by the bow on pas.-ing from salt to brackish water is, in X c'w c w inches, ; or ^^^t"^ for fresh water where a; is -9. Ex. — Find the changes of draught and trim in a light cruiser on passing into fresh water if w = 3000, t = 25, M = 650, c' =11. 3000 Increase of mean draught = ■ ^ os^or: "^ 3*1 inches. r^x, *^ . , ., ,. 11X3000 1 o • 1 Change of trim by the bow = ——: — ^^^= 1-3 inches. To determine the positions at which a weight must be added or removed so as to leave the draught at one end constant. EULE. — To maintain constant draught at a distance y abaft (or before) the centre of flotation, place the weight at a distance X before {or abaft) the centre of flotation, where a; = — . If constant draught is desired at either perpendicular, the two points 2 M for the weight are situated at a distance very nearly from the C.F. This distance is about , or about ry: in many ships. To determine the vertical height of the longitudinal meia- "sentre above the centre of buoyancy. Divide the moment of inertia of the water-plane relatively ;o a transverse axis passing through the centre of flotation by ;he volume of displacement (for example, see displucemcni; »heet and explanation on pp. 94, 99). Note. — This height is frequently greater than the ship's ' ■•th, so that i!G is negligible in comparison; then Gil = .i[)proximately. Moment to alter trim of a vessel. — In fig. 133 let the '^ weight P be moved forward through a longitudinal distance I, changing the water-line from wl to w'l'. r_,, , W.GM W X GM X trim in feet , Then PcZ = w.GGi = — T~^ 5 hence trim . ^ 12PfZL ^ n inches = 134 STABILITY. The product vd is the moment causing trim ; if several weights are moved, their moments are added, allowing for sign. Note that the moment to change trim one inch is equal to the expression — zr^r- — . This is fairly constant La Li within moderate changes of draught, and practically unaffected by vertical shifts of weight. Ap])roa.-imate ■formiilce. 2 1. (J. A. Normand). Long, bm = 13,000 t", L = length on L.W.L in feet, B = beam in feet, V = volmne of displacement in cubic feet, T = tons per inch. 2. (Derived from the preceding) ^2 Moment to change trim 1" = 30 — B 3. Moment to change trim 1" = l^b/10,000 in ships ol ordinary form, TL -^ 18-5. Effect of Adding Weights of Moderate Amount. The weights added are supposed insuflBcient to affect appreciably the transverse stability, or to cause relatively large heel, trim, or immersion. EuLE. — Find the distance of each weight from the middle line plane and from amidships. Calculate the moments, positive and negative (weights removed are considered nega- tive), and add Mean sinkage = weight added -r tons per inch. „ , . , „ transverse moment Heel m degrees = 57-3 X tt—^j : r„ ^ displacement x gm Trim in inches = longitudinal moment about centre of flotation moment to change trim 1 inch. __ longitudinal moment about ^Hh yo X c.F. abaft ^j moment to change trim 1 inch using + sign when the net weight [iv] added or subtracted is before amidships. EiTECT OF Adding Weights of Considerable Amount. Eule. — Add the weights and their moments as above, in- cluding in addition the vertical moments required to find the rise or fall of the e.g. The new mean draught is found from the curve of dis- placement, or more accurately from the curve of tons per inch, by estimating the mean tons per inch between the twq water- lines. If necessary make the correction due to the positiot of the centre of flotation as described on p. 132. To obtain the heel find first the vertical position of G ; froir the metacentric diasvam the new GM is obtained. The latera STABILITY. 135 movement (gg') of G is found by dividing the transverse moment by the new displacement. A moderate angle {0) of heel is given by the formula tan 6 = gg'/gm. If 6 is very large, construct a curve of stability for the new condition, using the cross curves, and find by trial the angle 9 at which the relation Gg' = GZ sec 6 holds. For the trim tho method given on the preceding page is usually sufficiently accurate. If, however, the Binkage is very great, construct a curve of moment to change trim 1 inch on a base of draught, also one giving the longitudinal position of the centre of buoyancy. Then at the original displacement if the trim be by the stem, the distance of abaft b is equal to the trim in inches X moment to change trim 1 inch at that draught (found from the curve) -f displacement. Knowing the longitudinal position of B from the curve, that of G 13 obtained. The change in this due to the added weights is then determined ; and the above process, reversed and using the final moment, positions of B and G, and displace- ment, gives the final trim. Examples of the above methods are given in the inclining experiment described below. To Dl'lLRMINE THE VERTICAL POSITION OF A ShIP'S CeNTRE OF Gravity by Experiment. In fig. 134 let mz be the upright axis of a ship ; her centre of gravity then lies somewhere in that axis. M is the metacentre, and GAI its vertical height above the centre of gravity G. If a weight P be moved transversely through a dis- tance PP' = d, it will heel the vessel over through an angle 6, and her centre of gravity will then shift in a direction GO' parallel to that in which the centre of gravity of the weight has been Bhifted. Let mt be parallel to gg' and tg' parallel to gm ; let P = weight shifted in tons, and W = displacement of ship in tons : then MT = GG V y d , and GM = gg' cotan 9 V X d cotan 9. w ' W In practice the ballast is usually in the form of pig iron arranged in two parallel rows on the port and starboard sides ■''©f the upper deck. The method of performing the experiment is illustrated by the calculations below, which correspond to u light cruiser, whoso metacentric diagram i.s given In fig. 121.